Everyday Physics:
Unusual insights into
familiar things
Jo Hermans
To Hanneke, Reinoud and Francine,
whose questions inspired me
to write this book
PREFACE
Some people think that science and technology are only for people with a formal scientific
education, but that’s a misconception. Our society is more and more dependent on technology,
and anyone without a basic knowledge of science is at a disadvantage. Apart from that, it’s often
just nice to understand a bit more about the world around us and to see how things work.
This book gives the inquiring reader scientific insights into everyday things. It contains
a collection of interesting – sometimes surprising – examples, many based on questions
from family, friends and students, often just motivated by curiosity. The choice of topics
is therefore slightly arbitrary.
You don’t have to read this book all in one go, or even from front to back! Chapters are
self-contained and can be read individually, although related topics are loosely grouped
together. Some more complex topics and calculations are shown in boxes in the text,
for advanced readers.
You can do almost all the experiments in the book with simple materials found at home.
Some are just for fun, while others are illustrative and enlightening, but they all show that science
doesn’t have to be boring!
Suggestions and errata: you can suggest new topics to include in future editions, and obtain
the latest list of changes, corrections (and interesting additions) via the book’s webpage at
www.uit.co.uk/everyday-physics
Continue the conversation at #EVPHYS on Twitter.
Published by
UIT Cambridge Ltd
www.uit.co.uk
PO Box 145, Cambridge CB4 1GQ, England
+44 (0) 1223 302 041
Copyright © 2021 UIT Cambridge Ltd
All rights reserved.
Subject to statutory exception and to the provisions of relevant collective
licensing agreements, no part of this book may be reproduced in any
manner without the prior written permission of the publisher.
First published in 2021, in England.
The author has asserted his moral rights under the
Copyright, Designs and Patents Act 1988.
Diagrams prepared by Hayley Wells
hayleywellsillustration@gmail.com
Front cover photographs sourced via Adobe Stock
Photograph credits are on page 301
Design by Mad-i-Creative
biscuits79@icloud.com
The publishers have endeavoured to identify all copyright holders
but will be glad to correct in future editions any
omissions brought to their notice.
ISBN: 9781906860806 (paperback)
ISBN: 9781906860820 (ePub)
ISBN: 9781906860813 (PDF)
Also available as Kindle
Disclaimer: the advice herein is believed to be correct at
the time of printing, but the author(s) and publisher accept
no liability for errors or for actions inspired by this book.
ep-1-1
CONTENTS
PART A I OUTDOOR LIFE
Chapter 01
I
0
Chapter 02
How does GPS navigation work? (With a nod to
Albert Einstein)
Why are some mountain winds so warm?
Chapter 03
What is wind chill?
Chapter 04
Why is ice so slippery?
Chapter 05
Waves at the beach
Chapter 06
How fast do raindrops fall?
Chapter 07
Why don’t fog drops fall?
Chapter 08
Skydiving: how fast can you fall?
Chapter 09
How high will the sun rise today?
Chapter 10
How hot does the sun feel?
Chapter 11
Parallel light beams from the sun
Chapter 12
Summer and winter, why such a big difference?
Chapter 13
Why do you walk the way you do?
PART B I BICYCLE AND CAR
I
Chapter 14
The human engine
Chapter 15
The human energy equivalent of a vacuum cleaner
Chapter 16
How do you keep your temperature constant?
Chapter 17
How efficient is cycling?
Chapter 18
What forces affect a cyclist?
Chapter 19
Can you cycle at 100 km/h?
Chapter 20
How fast can you cycle on the moon?
Chapter 21
Is cycling really harder with a side wind?
Chapter 22
Minimizing your journey time
Chapter 23
The cyclist’s soggy back
1
5
7
11
13
15
19
21
25
29
31
33
37
42
43
47
49
55
59
63
67
69
75
79
Chapter 24
Chapter 25
Chapter 26
Can you get less wet by cycling faster?
Rolling resistance, air resistance and fuel consumption
How many cars per hour can a road take?
PART C I LIGHT AND COLOUR
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
I
Chapter
Chapter
Chapter
Chapter
Chapter
46
47
48
49
50
Chapter 51
Chapter 52
98
Cosy candlelight
Why are incandescent bulbs so inefficient?
Luminous ideas: fluorescent lights and LEDs
Why is the sky blue and the setting sun red?
Two kinds of smoke from the same cigarette?
Swimming pools are deeper than they look
Sunlight filtering through the leaves of trees
How sharply can you see?
Your eye is more sensitive than a camera
Puddles on a dry road
Seeing the sun after sunset
Transparent windowpanes & opaque lace curtains
Seeing clearly underwater
What makes rainbows?
Why are soap bubbles so colourful?
Why are CDs so colourful?
How do holograms work?
Why does the sea look so blue?
What’s special about Polaroid glasses?
PART D I SOUND AND HEARING
83
87
93
I
What do your ears hear?
Why isn’t there more noise pollution?
The energy cost of talking
How can you tell where a sound is coming from?
Discriminating between different voices: the cocktail
party effect
Do you hear better at night?
Can the wind blow sound to you?
99
101
105
107
113
115
119
123
129
133
135
139
143
147
155
159
165
169
173
180
181
185
187
191
195
199
203
Chapter
Chapter
Chapter
Chapter
53
54
55
56
Do noise barriers work?
Can you hear whether the curtains are closed?
Doh-re-mi: the physics of musical scales
Why orchestras go out of tune
PART E I IN AND AROUND THE HOUSE
I
Chapter
Chapter
Chapter
Chapter
57
58
59
60
Why do eggs explode in the microwave?
Can you cool your home with your fridge?
Curve balls, backspin and topspin
How much power can you get from solar energy?
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
61
62
63
64
65
66
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
67
68
69
70
71
72
73
The mystery of the wandering carpets
Common misconceptions about the radiometer
Is thick glass a better insulator than thin glass?
Is there a vacuum inside double glazed windows?
Can you feel energy consumption?
Is a black central heating radiator better than a
white one?
Does black paint get hotter than white?
Does baby really need a hot water bottle?
The lid on the saucepan
Why does the air get so dry in winter?
Why don’t you die of heat in the sauna?
The wine-mixing problem
The wandering tea leaves
I ACKNOWLEDGEMENTS
I HANDY REFERENCE DATA
II
I
I
I
APPENDIX: THE WINE-MIXING
PROBLEM – CALCULATION
TWO MORE EXPERIMENTS
RESOURCES
CREDITS
INDEX
205
207
209
217
220
221
226
229
239
245
247
251
255
261
265
267
269
271
273
277
281
283
286
287
296
298
300
301
302
0
8
Everyday Physics: Unusual insights into familiar things
01
HOW DOES GPS
NAVIGATION
WORK?
(WITH A NOD TO ALBERT EINSTEIN)
I’m riding on the bike paths along the southern California coast and
I’ve lost my way. I do have a detailed map of the region with me,
though. Fortunately there is a crossing with a sign showing that I
am 14 kilometres from Santa Monica and 12 kilometres from Malibu.
I am therefore on a circle around Santa Monica with a radius of
14 kilometres and at the same time on a circle around Malibu with a
radius of 12 kilometres. The two circles intersect in two points. I am
at one of these points. Which one? In this case I’m lucky: one of the
two points lies in the Pacific, and I am not standing in the ocean, so
I must be at the other point. I can now easily find the crossing on
the map. If I also use the distances to the other towns on the sign
in addition to Malibu and Santa Monica, my location becomes even
more reliable, and the location in the ocean is eliminated.
A GPS (Global Positioning System) receiver determines
position in an analogous way but it does it in three dimensions:
not just the location on the map but the height as well, and with
astounding precision. So how does GPS (or its European counterpart,
Galileo) work?
Broadly speaking it works like this: GPS has 24 satellites
orbiting at a distance of 20,000 km above the Earth (Fig. 1) and
making two revolutions about the Earth per day. (The satellites are
not geosynchronous/geostationary, by the way.) Each sends a radio
Everyday Physics: Unusual insights into familiar things
1
signal that can be picked up by the GPS receiver in your car, on your
bike or in your phone.
Fig. 1: GPS uses a
network of 24 satellites.
Credit: After commons.
wikimedia.org/wiki/
User:Yuma
You need two things to determine your position: (1) the
position of each satellite that is visible to your GPS receiver and (2)
the distance to each.
(1) The position of each satellite is accurately monitored by a
number of ground stations spread over the Earth and coordinated
from the “mother station” in the US, where the system was originally
developed for military purposes. These data are transmitted from
Earth to the satellites. In this way each satellite knows its own
position and reports this to your GPS receiver.
(2) Now you need to know the distance to each satellite. Your
receiver finds this by measuring the time it takes for a radio signal
from the satellite to reach the receiver; this is the difference between
the time when the satellite sent the signal and the time your GPS
received it. Multiply the time by the speed of the radio signal (i.e. by
the speed of light) to give the distance. If you know the distance for
(at least) three satellites, then you can calculate your position, just as
in the road-sign story:
2
Everyday Physics: Unusual insights into familiar things
(a) The distance to the first satellite tells you that you are on an
imaginary sphere around that satellite, with a radius equal
to the measured distance.
(b) Similarly, the second satellite yields another imaginary
sphere – around the second satellite (Fig. 2). So you must be
on both spheres simultaneously, that is, on the intersection
of these two spheres. This is a circle, just like two adhering
soap bubbles form a circle.
(c) The third satellite yields yet another sphere, which
intersects the circle in two points. Fortunately one of those
drops out: it lies deep in the Earth or far above it. The
remaining point is your position. Of course there is always
a margin of error in the measurement; if your GPS receiver
can see more satellites and takes them into account, the
error margin will be smaller.
Fig. 2: The intersection
of two spheres is a circle.
It all comes down to measuring time. That has to be very
accurate because the radio signals travel at the speed of light
(300,000 km per second). A timing error of one millionth of a
second would mean an error of 300 metres in your position.
To achieve the required accuracy all satellites and ground
stations have precision clocks (but note – not the GPS receivers) that
are based on an extremely precise and stable transition within a
caesium atom. Clocks of this type attain fabulous accuracies: the best
ones on Earth run so precisely that in 1,000 years from now they will
be out by less than one tenth of one millisecond! The satellite clocks
Everyday Physics: Unusual insights into familiar things
3
need not be that good, but they must be synchronized to within one
billionth of a second.
The clock in your GPS receiver isn’t an atomic clock like the
ones in the satellites and ground stations, and in fact it doesn’t need
to be. By using an extra satellite (i.e. four instead of three), the GPS
receiver can precisely synchronize its (cheap, quartz) clock with the
atomic clocks in the satellites, and so measure the times accurately,
and therefore the distances.
See the Resources appendix for an animation of GPS satellites
orbiting the Earth.
GPS and relativity
We’ve seen that GPS depends on precise time measurement. Now
something unexpected happens: we need Einstein and his theory of
relativity. Einstein himself would never have guessed that his theory
– such an elusive piece of theoretical physics – would ever play such
an important role in something as practical as GPS. It’s a lovely
example of an unexpected and unintended application of cuttingedge fundamental science.
Here’s how it works: according to relativity theory, the clocks
in the satellites don’t run at the rate they would when sitting on
Earth, for two reasons: (1) because they are moving at high speed,
they run somewhat slower, and (2) because they are so high up they
feel less gravity (only a quarter of what they would feel on Earth), so
they run somewhat faster.
The second effect is the greater, and the satellite clocks get
ahead of the clocks in the ground stations by 38 millionths of a
second per day. That does not sound like much, but it means that
after one day the error in calculating a position would be more than
11 km, which would render your GPS useless.
To compensate for this relativistic effect, before launch the
clocks in the satellites are set to run at a slightly slow rate – exactly
slow enough so that after launch and once they are in orbit, they
are precisely synchronous with the clocks in the monitoring stations
on Earth.
Problem solved. Destination reached. With a nod to Einstein.
4
Everyday Physics: Unusual insights into familiar things
02
WHY ARE SOME
ALPINE WINDS
SO WARM?
In the Alps, the Föhn is the warm wind that blows from the
mountains into the valleys, especially in winter. Curiously, even
though it comes down from the mountains it’s warm and dry. Why?
To see what’s happening, let’s first look what happens to dry
air that comes over the mountains (Fig. 1). The air pressure at the
top of the mountain is lower than in the valley below, because there
is less air pressing down, to cause the pressure. So air that streams up
a mountain expands and therefore cools down. (This is the opposite
of what happens in a bicycle pump: in the pump, air is compressed
and therefore warms up.) By rising and expanding, air cools about
1°C per 100 metres, which explains why it gets colder when you
hike up into the mountains. When air rises, say, 2,000 metres,
Everyday Physics: Unusual insights into familiar things
5
it cools by about 20°C. When it comes down the other side of the
mountains it warms at the same rate. The net effect is therefore zero:
no Föhn (Fig. 1).
Fig. 1: In this model
example the dry air
starts out in the valley at
20°C, cools to 0°C at the
top of the mountain, and
warms again to 20°C in
the valley.
With moist air (as in the Föhn) things are different (Fig. 2).
As the air rises up the mountain it cools, just as the dry air did. But
something else happens too: as the air cools, the water vapour in the
air condenses and forms rain, just like morning mist can form above
a meadow as the temperature goes down, or like the water vapour
in your breath condenses if you blow on a cold windowpane. Heat
is released when vapour condenses – the opposite of evaporation
which absorbs heat.
Because of the heat that is released at condensation the rising
moist air doesn’t cool as much as the dry air did. The net result is
that the temperature rises to 27°C as it comes down the other side
of the mountain. And the air is not only warm: it is also dry because
it dropped most of its moisture as rain on the slope while rising.
The relative humidity (the amount of water vapour the air contains
divided by the amount that it can contain) can easily get below 20%
this way. Warm and dry air, that’s what you get.
Fig. 2: The rising
moist air cools down by
only 13°C rather than
20°C. But as it descends
the other side of the
mountain, it warms up
by the full 20°C, just as
the dry air did in Fig. 1.
6
Everyday Physics: Unusual insights into familiar things
03
WHAT IS
WIND CHILL?
The weatherman reports that the temperature is 0°C, but that
the “wind chill” factor is minus 17. What does this mean?
When it is cold outside you feel particularly chilly if there is a
strong wind blowing. The moving wind cools your body more than
stationary air does. This is often called “wind chill”, i.e. subjective
temperature. It is important whenever you skate, ride your bike, sail
or ski; if you ignore it, you could end up with frostbite.
There is a lot of misunderstanding about wind chill. One
newspaper said that if there is enough wind the water in a car
radiator can freeze, even when the temperatures is above 0°C.
Everyday Physics: Unusual insights into familiar things
7
That’s wrong. The radiator temperature will drop until it is the same
temperature as the air and it can’t become colder than that. What
is true, however, is that in a wind the radiator will cool down more
quickly than in stationary air. And what’s true for the radiator is true
for any object suspended in the wind: it will reach the temperature
of the air, period. (The only exception is where something is wet;
then evaporation takes away heat and can lead to excess cooling,
but that’s not what we are talking about here.)
However, for a human being who has to generate heat
constantly to maintain their body temperature of about 37°C the
rate of heat loss is important. The environment is normally cooler
than your body, so your body loses heat to the environment. In
still air, the heat loss is relatively small: even exposed skin is still
enveloped by an insulating air layer a few millimetres thick, which
you carry with you as though it were a sweater (see How do you keep
your temperature constant? p49). That air layer limits the heat losses
and keeps your skin relatively warm.
In contrast, when there’s a wind, it blows away the insulating
air layer, so that the cold gets right up to your skin, which becomes
colder quicker. It becomes just as cold as it would be at a lower air
temperature without wind. That lower temperature is the subjective
temperature or “wind chill”. In other words: it is the temperature
that still air would be to cause the same heat loss as the real
temperature does in the wind.
8
Everyday Physics: Unusual insights into familiar things
Fig. 1 shows the wind chill for various temperatures and
wind speeds in average conditions. The left-hand column shows
the real air temperature; the other columns give the subjective
temperatures in degrees Celsius for various wind speeds. Note that
the wind speed is relative: if you are standing still, and the wind is
blowing at 25 km/h, the effective wind speed is 25 km/h. But if the
air is stationary, and you are cycling through the air at 25 km/h, the
effective wind speed is again 25 km/h, so a moving cyclist will cool
down a lot more than someone standing still.
Fig. 1: Wind chill
values under different
conditions. Wind speeds
of 13 and 20 m/s roughly
correspond to the speed
of a professional cycling
time triallist (47 km/h)
and a skier on a fast
slope (72 km/h).
Wind speed
0 m/s = 0 km/h
Real air
temperature
7 m/s = 25 km/h
13 m/s = 47 km/h
20 m/s = 72 km/h
Subjective temperature (“Wind chill”)
0°C
0°C
−11°C
−17°C
−20°C
−2°C
−2°C
−14°C
−20°C
−23°C
−4°C
−4°C
−17°C
−23°C
−27°C
−6°C
−6°C
−20°C
−26°C
−29°C
−8°C
−8°C
−23°C
−29°C
−32°C
−10°C
−10°C
−25°C
−32°C
−36°C
−12°C
−12°C
−28°C
−36°C
−39°C
We see that it makes a lot of difference whether or not there is
a wind blowing. At 0°C with a 72 km/h wind (or when skiing down
the mountain at 72 km/h), it feels as cold as −20°C would be without
wind. And if the temperature is −12°C at that speed, it feels like −39°C.
It’s like being in Siberia!
If the air temperature is 37°C (i.e. at body temperature), there
is no wind chill. However, you will probably find moving air more
comfortable than still air, because the moving air evaporates some
of your perspiration, and cools you a bit.
Everyday Physics: Unusual insights into familiar things
9
A different way of looking at all this is to realize that while
wind chill doesn’t reduce the temperature, it does increases the rate
of heat loss and this matters to humans and animals. For example,
say you are sitting at rest in still air at 0°C. Depending on body
mass and clothing, you use about 120 joules/second, i.e. 120 watts.
However, when there’s a wind of 20 m/s, you will need 185 watts
to keep your body at constant temperature. If your internal heat
production doesn’t increase fast enough, your temperature will fall
and that’s how you get hypothermia.
10
Everyday Physics: Unusual insights into familiar things
04
WHY IS ICE SO
SLIPPERY?
Why is ice so slippery? It sounds obvious: of course ice is slippery
– that’s why you skate on it. But what exactly makes it so slippery?
It’s not just because it is smooth: glass is very smooth, but it’s not
a good surface to skate on. “As smooth as glass” isn’t the same as
“slippery as ice”.
To slide on a surface you need more than just smoothness
– a layer of water, for example: a surface that is smooth becomes
slippery only when it is wet. You slip on a wet floor, not on a dry one,
so a bit of water is what you need. And indeed, that is precisely what
makes ice slippery: a thin layer of water between skate and solid ice
acts as a lubricant.
The next question is: why would there be water under the
skate when the ice is below freezing? An old misconception is that
this is due to the pressure that the skater puts on the ice: the weight
Everyday Physics: Unusual insights into familiar things
11
of the skater is concentrated on a tiny area – on the few square
centimetres of the skate that touch the ice – and that produces a lot
of pressure. It’s true that, when under pressure, ice melts below the
normal freezing point. That is just plain logic: ice has a larger volume
than liquid water (which is why ice floats). So if you compress ice,
it tends to become liquid, which means that the melting point has
dropped a bit. However, if we calculate that change in melting
temperature, we find it’s very little in the case of the skater – a few
tenths of a degree at most. So the fun of skating would be over as
soon as the temperature is a few degrees below the freezing point.
(Incidentally, this pressure argument would also fail to explain why
a light object like an ice hockey puck slides so well on ice.)
The real answer is that ice has a layer of water at its surface
by nature. Ice is made up of a lattice of water molecules, tightly
bound to their neighbours all around. But at the outermost layer of
molecules things are different. The molecules at the surface have no
neighbours on the outside. Therefore, they are less tightly bound,
have a high mobility and behave like fluid water even at freezing
temperatures. At freezing point, this layer of water on ice has a
thickness of some 200 molecules – about 70 nanometres (roughly
one thousandth of the thickness of a human hair.) That’s not exactly
thick, but it is enough to act as a lubricant. In addition, the friction
between skate and ice releases heat which, in turn, melts some extra
ice, making sliding even easier.
However, as the temperature goes down, the water layer
becomes thinner and the lubrication reduces. If it gets sufficiently
cold, far below −40°C, even the outermost molecules freeze, and
there is no water layer left.
You might now be tempted to conclude that skating is best at
temperatures just below freezing point, but this is not the case. The
ice gets a little soft and loses mechanical strength since it’s closer
to the melting point, and the skate cuts into it, causing some extra
resistance. To skate really fast you therefore must go down a bit in
temperature. The optimum – depending on the type of skate – is
around −7°C or −8°C. That is cold enough to keep the skate from
cutting into the ice too much, but not so cold that the water layer
becomes too thin. So the recipe for beating the world speed skating
record is no wind and −7°C.
12
Everyday Physics: Unusual insights into familiar things
05
WAVES AT
THE BEACH
If you’ve walked along a beach you may have noticed that the waves
always roll in parallel to the beach. For a north–south beach like the
Cape Cod National Seashore this is pretty logical in the case of an
easterly wind. But if the wind is from the north, isn’t it strange that
the waves don’t change direction: they still keep coming in parallel
to the beach? There is a bit of physics behind this. The indirect
cause is that the beach rises gradually. Towards the beach the water
becomes gradually less deep, and that depth influences the speed
of the waves. As the water becomes shallower, the speed goes down
(see box).
Now take those lovely broad waves far out to sea, driven by a
north-easterly wind, so they are travelling at 45° to the coast (Fig. 1a).
As the waves approach the coast, the side of a wavefront that is
further from the shore will be in deeper water than the side nearby
(Fig. 1b). Now remember that the speed at the deep side is higher
than at the shallow side. So, the waves gradually turn in a curve
towards the shore, with the deep part automatically on the outer side
of the curve because it is travelling faster. The wave therefore
turns towards the beach (Fig. 1c) until the speeds of the wave ends
become equal, which only happens when it is parallel to the coast,
where the depths are equal (Fig. 1d). After that there is no more reas
on to depart from the straight course and the waves remain neatly
parallel, provided of course that the gradient of the ocean floor
is more or less equal everywhere. So: wave-surfers should thank
heaven for nicely sloping beaches. And for the laws of physics,
of course.
Everyday Physics: Unusual insights into familiar things
13
Speed of water waves
The speed at which water waves propagate depends on several variables. Two forces
that want to restore a surface to smoothness are surface tension (which makes raindrops
round, for example) and gravity. As soon as a wave crest is formed, both forces want to
suppress it. Therefore, both influence the speed. The surface tension is important only
for small ripples on the water with wavelengths of a few centimetres, so we can neglect it.
For the larger waves, like at the beach, gravity prevails. In this case the ratio of wavelength
to the depth of the water is important. For a shallow beach we can assume that the
wavelength is large compared to the depth, and then the expression for the speed (v) is
simply: v = √ (g D), where g is the acceleration due to gravity and D is the depth.
So the speed of the wave decreases as the depth decreases. For a depth of two metres and
g = 10 m/s2, we find that the speed is about 4.5 m/s. If the depth decreases to one metre,
the speed is about 3 m/s – now only two or three times walking pace.
Fig. 1: Waves
approaching the coast
at an angle turn
gradually as the water
becomes shallower.
14
Everyday Physics: Unusual insights into familiar things
06
HOW FAST
DO RAIN
DROPS FALL?
In a heavy rain shower it is quite striking that the big drops come
first and as the shower wanes the drops are smaller. So it seems
that big drops fall faster than small ones, which is strange, because
in a vacuum everything falls equally rapidly, as we remember from
school. For example, a feather in a vacuum tube falls as fast as a
marble or a coin.
The explanation is that raindrops aren’t falling in a vacuum;
they’re falling in air, so friction (or “air resistance”) plays a role,
acting against gravity. To understand the difference between large
and small drops we have to look at the process of falling. That
process starts as gravity makes the drop accelerate. But as the speed
increases, so does the friction, directed upward, against the force of
gravity. So the net force downward becomes smaller, and therefore
the speed increases less rapidly – according to Newton’s “force is
mass times acceleration”. The speed continues to increase until the
net force becomes zero, at which point the speed remains constant:
there is equilibrium between the downward gravity force and the
upward friction.
The size of the drops is critical. For a falling drop bigger than
about 1 mm, air resistance is proportional to area, i.e. to the square
of the diameter (D2). So a drop 2 times the diameter of another will
feel a resistance 4 times as large at the same speed, and a drop 3
times the diameter will feel 9 times as much resistance. It is the area
Everyday Physics: Unusual insights into familiar things
15
that counts because the air stream becomes turbulent, just as it does
for a cyclist (as we will see on p59) and for a car (p87).
Now suppose a drop falls at a certain speed and suddenly
grows to twice the diameter (because it merges with several other
drops, say). The friction force becomes 4 times as big. But since
mass is proportional to the cube of the diameter (D3 ), the mass has
becomes 8 times as big and thus the gravity force becomes 8 times
as big. So there is now suddenly a greater net downward force, and
immediately the drop starts to accelerate. This continues until the
speed has increased so much that the forces are in balance again.
That’s why large raindrops fall faster than small ones.
16
Everyday Physics: Unusual insights into familiar things
How large do raindrops actually become? Could we have
raindrops that are 10 cm in diameter? To answer that question we
first have to look at how drops behave. Just like a balloon, water
drops like to be spherical, because this gives the smallest possible
surface area for a given volume. It is the attractive forces among
molecules that minimize a drop’s surface area. These forces are the
basis of surface tension, which acts as though an elastic membrane
(like a balloon’s) were wrapped around the drop.
Everyday Physics: Unusual insights into familiar things
17
As a drop falls, it feels the air resistance. For larger and faster
drops the increasing air resistance causes the drops to flatten on the
downward side (Fig. 1). As the drop reaches a diameter of 5 mm
it becomes so unstable that it breaks into smaller drops, which is
why we don’t get raindrops the size of footballs. (This can be nicely
observed in a vertical wind tunnel, in an air stream at just the right
speed to make the drops hang stationary in the air.)
Raindrops stay nice and round up to a diameter of about
3 mm. If they become larger, they flatten on the downward side,
the resistance increases and the drops no longer gain speed (Fig. 1).
Fig. 1: As raindrops
get bigger they tend to
flatten on the underside.
How high are those speeds? For drops 1 mm in diameter the
speed turns out to be 16 km/h and for 3 mm drops it’s 28 km/h,
which is nearly the maximum. For larger drops any further increase
of speed is resisted by the flattening. The fastest drops are those that
approach 5 mm in diameter, which achieve about 29 km/h – very
little faster than 3 mm drops.
So if you cycle fast, you can travel at the same speed as
the fastest raindrops. Of course, they fall vertically and you go
horizontally, so the drops strike you at exactly 45°, from the front,
which is good to know if you want to use an umbrella on your bike.
18
Everyday Physics: Unusual insights into familiar things
07
WHY DON’T FOG
DROPS FALL?
Rain and fog both consist of water drops, but rain falls whereas fog
just hangs in the air. The only difference is the size of the drops.
Large drops fall much faster than small ones (as we just saw in
the preceding chapter, which dealt with drops bigger than about
1 mm diameter). Fog drops fall very slowly if at all. The reason is
that the gravity force on such tiny drops is very small – because it’s
proportional to the mass of the drop, i.e. to the cube of the diameter.
Everyday Physics: Unusual insights into familiar things
19
So far this looks just like the raindrops story. However, the
fog drops are so much smaller and fall so much more slowly that a
different law of friction applies. The air that flows around the little
fog drops is well-behaved, forming nice streamlines that run more
or less parallel; this is called laminar flow, and is less turbulent than
with larger and faster objects. For droplets smaller than 0.1 mm the
friction is directly proportional to the diameter. So when a droplet
becomes half the diameter, it will also feel half the air resistance at
the same speed, but the force of gravity is only one eighth (because
that’s proportional to D3).
Does this mean the fog droplets fall eventually?
Probably, but only very slowly. Take fog droplets of 0.002 mm
diameter. That’s about four times the wavelength of light and
therefore large enough to be seen with your eyes. (You see the
drops as a white cloud.) If you do the fairly complex calculation, you
get a speed of 0.1 mm/s (or 36 cm per hour – a bit more than a
foot per hour). That is not much – less than 10 m in a whole day.
A breath of air will overwhelm the small force of gravity and keep
such droplets afloat.
However, some fog droplets really will continue floating even
if the air is perfectly still, provided they are small enough! This isn’t
so strange – after all, the molecules of air and water don’t fall to the
ground either; instead they form the Earth’s atmosphere. Because of
thermal motion, gas molecules and other small particles continue
floating, so that the atmosphere does not fall on your head, but
stretches above you for many kilometres.
You can calculate how small the droplets must be in order
to continue floating in the Earth’s atmosphere, like gas molecules
do. Because fog drops are much heavier than individual molecules
we cannot expect that they will float just as high and form as thick
a layer as the atmosphere. To form a “droplet atmosphere” of
modest thickness (say one thousandth of the thickness of the real
atmosphere, i.e. a layer of fog having an effective thickness of 8 m
instead of 8 km) the water droplets will contain some 1,600 water
molecules. The diameter of such drops is 4.5 nanometres – about
one ten thousandth of the thickness of a human hair. These are tiny
droplets, so small that individual drops are invisible to the naked
eye. But they don’t fall.
20
Everyday Physics: Unusual insights into familiar things
08
SKYDIVING:
HOW FAST CAN
YOU FALL?
Suppose you bail out of a plane without a parachute. How fast will
you be falling when you reach the ground? Does it matter at what
altitude you bail out?
Everyday Physics: Unusual insights into familiar things
21
As long as the plane doesn’t fly too low (not below about 1,000
metres) the altitude doesn’t matter. Once you start falling, your speed
increases and so does the air resistance. However, the gravitational
force remains constant, and at some point the downward-acting
gravity is balanced by the upward-acting air resistance. Then, just
as with raindrops (see How fast do rain drops fall? p15), the net force
is zero, so your speed remains constant; this speed is called the
terminal velocity, and that’s what we want to determine.
The air resistance of a skydiver – just as for a car or a large
raindrop – is proportional to the frontal area, to the square of the
speed, and to a constant factor Cd, the drag coefficient. The value of
the drag coefficient depends on the aerodynamic characteristics of
the body: the more streamlined the shape, the less drag there is, and
the smaller Cd is.
As every skydiver knows, the frontal area and the drag depend
on the shape you adopt when you fall. The area and the drag are
largest if you spread out your arms and legs, and the terminal speed
will be almost exactly 200 km/h, or 124 mph. That’s bad news: you
have to be really lucky with your landing spot to survive. (Hope for a
hay stack!) If you roll yourself up like a ball, the terminal speed will
be even larger, for two reasons: first, your frontal area decreases, and
second, your aerodynamics improve – you are more streamlined.
The result is that you hit the ground at a frightening 320 km/h.
(Hope for a very big hay stack!)
Can you go any faster? Sure – if you start really high, in
thinner air. That gives a lot less air resistance, since the resistance
is proportional to the air density. Obviously, the nicest place to start
would be at the edge of the atmosphere. But it is freezing cold up
there, and you’d need an oxygen bottle for breathing. Even so, on
16 August, 1960, US Air Force pilot Joseph Kittinger jumped out of
a balloon at 31,300 metres (about 100,000 feet, approximately
three times the cruising altitude of a commercial airliner). The air
pressure at that height is only about 2% of the sea-level value, so
Kittinger started his fall practically in vacuum, with almost zero air
resistance. His “free fall” lasted 4.5 minutes in total. He reached a
speed of 980 km/h. Some people claim that he broke the sound
barrier (about 1,050 km/h at that altitude), but that is most probably
not the case.
22
Everyday Physics: Unusual insights into familiar things
This was an irresistible challenge for Felix Baumgartner
from Austria. On 14 October 2012, dressed in a kind of space suit,
he jumped from a balloon at an altitude of 38,969 metres. During
his free fall of 4 minutes 19 seconds he reached a maximum speed
of 1,357.6 km/h, well above the speed of sound. So he became the
first person who managed to break the sound barrier without using
any sort of engine. But it wasn’t exactly a pleasant trip: sometimes it
takes a bit of discomfort to end up in the Guinness Book of Records.
Now let’s see how a parachute changes things. It increases
your frontal area and thus decreases your speed dramatically. Using
the traditional round parachute (with a surface area of about 60 m2
and Cd of about 0.8) the landing speed is about 18 km/h, which is the
same as jumping off a 1.5 m high wall.
Calculating the terminal velocity
Let’s first look at a skydiver who jumps out of a plane in relatively dense air. After a few
seconds she reaches a constant speed where her weight (mass m × acceleration due to
gravity, g = 9.8 m/s2) is balanced by the air resistance Cd A × (½ ρ v2), where Cd is the
drag coefficient, A the frontal area, ρ the density of the air (= 1.3 kg/m3 at sea level) and v
the speed. This yields v = √ (2 m g / Cd A ρ).
For a fully equipped skydiver we take m = 100 kg, and with arms and legs fully stretched
we estimate A = 1 m2 and Cd = 0.8 for such a poorly streamlined body. Let’s assume that
the skydiver jumps from an altitude of 3,000 metres (10,000 feet). There the air density
ρ is only about 70% of what it is at sea level, or about 0.9 kg/m3. This gives v = 52 m/s or
188 km/h – in fair agreement with the actual value of 200 km/h.
And now for Joseph Kittinger’s and Felix Baumgartner’s record jumps. In free fall in
the absence of air resistance (i.e. very little or no atmosphere) the speed of any object
increases with the acceleration due to gravity, g = 9.8 m/s2. In other words each second
the speed increases by 9.8 m/s. After 30 seconds of free fall this yields a speed of 294 m/s
or 1,058 km/h, practically the speed of sound. In practice the increasing air resistance
will keep the speed somewhat lower. Even so, if the jump is continued some more, the
sound barrier can be broken, as shown by Baumgartner. Will you hear a sonic boom? No,
because there is hardly any air up there and therefore little to propagate sound, so you
don’t hear anything. Anyway, there can’t be a boom because a boom requires a shock
wave, and you can’t have a shock wave without air.
Everyday Physics: Unusual insights into familiar things
23
Home experiment: Magic trick – the slow fall
If you drop a pebble into a vertical pipe one metre long, you can calculate exactly how
long it will take to reach the bottom: 0.45 seconds. Whether it’s a pebble, or a metal
sphere or a coin, and whether the pipe is made of PVC, steel or copper, it makes no
difference: it always takes about half a second, as long as the falling object easily fits the
tube and the air resistance is negligible.
Now you repeat the experiment, but using a small metal sphere instead of a pebble,
and guess what? In the PVC tube it takes half a second to fall as expected, but it takes
nearly ten seconds in the copper pipe. Your audience is flabbergasted.
The explanation is simple: the metal sphere is a strong little magnet. Since the tube
is made of a good electrical conductor (copper), the falling magnet induces electric
currents in the pipe, just as happens in the copper windings of the generator of your
car, or in your bicycle dynamo. The experiment is a simple and dramatic illustration of
Lenz’s law which says that an electrical conductor resists a change of magnetic field by
generating currents that induce an opposing magnetic field. The currents can circulate
nicely in the tube provided this is conductive, and copper is a very good conductor.
(The induced currents in the copper tube actually warm it up, but not enough for your
hand to feel.)
(For this trick, if you use a length of 22 mm copper pipe, your magnet should be about
19 mm diameter.)
We can also view the experiment in the context of the law of conservation of energy.
Part of the kinetic energy of the falling sphere is converted into heat which warms
up the tube by means of electric currents. The reduction in the kinetic energy of the
object emerging from the tube means it comes out of the tube at a lower speed.
24
Everyday Physics: Unusual insights into familiar things
09
HOW HIGH WILL
THE SUN RISE
TODAY?
The Earth’s axis is inclined at 23.5° to the ecliptic – the plane in which
the Earth orbits the sun (Fig. 1).
Fig. 1: The Earth’s axis is inclined
at 23.5°. Cambridge (C) is 52° above
the equator.
Working out the height of the sun at the equator is easy. At
noon on 21 March or 21 September (the two equinoxes) the sun is
directly over your head, i.e. at 90° above the horizon (E in Fig. 2).
But for people in the US, Canada and Western Europe, it is a bit
more complicated. Take Cambridge, England, or Kiska Island, in
Alaska US, or Quebec in Canada, all at about 52° north (C in Fig. 2).
When the sun is directly overhead at the equator, it is 52° lower here.
Measured from the horizon, it is only 90 − 52 = 38° high.
Everyday Physics: Unusual insights into familiar things
25
Fig. 2: At noon on
an equinox, the sun
is directly (90°) above
the equator. For
Cambridge (C) it is
38 degrees above
the horizon.
However, at midsummer (the summer solstice, 21 June), in the
northern hemisphere the Earth’s axis is slanted 23.5° towards the
sun, so we add 23.5° to the height of the sun for places in the northern
hemisphere. Now the sun is directly above the northern tropic, the
Tropic of Cancer (T in Fig. 3). So in Cambridge (C in Fig. 3), the
height is 38 + 23.5 = 61.5°. For the winter solstice, 21 December, we
subtract the same amount: 38 − 23.5 = 14.5°.
Fig. 3: At noon at the
summer solstice, the
sun is directly above the
Tropic of Cancer (T). For
Cambridge (C) it is now
38 + 23.5 = 61.5 degrees
above the horizon.
What about dates in between? Here we can use the lucky
coincidence that there are about as many days in a year (365) as
there are degrees in a circle (360). The Earth proceeds by roughly
26
Everyday Physics: Unusual insights into familiar things
1° each day in its orbit around the sun. It is even simpler in terms of
months: on average the Earth proceeds 360° / 12 = 30° per month.
Now things become straightforward if we remember a bit of
math from high-school. Continuing with the example of Cambridge,
the height that the sun reaches is to a good approximation a sine
curve (Fig. 4) around the average of 38°, with maximum and
minimum amplitude of 23.5°. Now remember that a sine curve
reaches half its maximum at 30° (one month in this case) and needs
another 60° (2 months) to reach its maximum. In everyday terms,
that means that at the equinoxes the day length changes most
rapidly from one day to the next (the slope of the graph is at its
maximum – points E1 and E2), and at the solstices it changes most
slowly (the slope of the graph is zero – points S1 and S2).
Fig. 4: The angular
height of the sun at
different dates. The
slope of the curve is at
its minimum at the two
solstices (S1, S2) and
at its maximum at the
equinoxes (E1, E2).
Everyday Physics: Unusual insights into familiar things
27
So, starting at 21 March, by 21 April the sun has already made
half its gain reaching about 38 + 12 = 50° above the horizon (Fig. 4,
M). It takes two more months for the sun to reach its full maximum
of 38 + 23.5 = 61.5° on 21 June. Two months later, around 21 August, it
passes 50° again, and one month later it is back at 38°.
We have simplified things a little. For example, we assumed
the orbit of the Earth around the sun is a circle, which is not quite
right: the orbit is really an ellipse and the Earth travels a bit faster in
the northern hemisphere’s winter, when the distance to the sun is
a little smaller, and a bit slower in its summer (when the sun is a bit
further away). Therefore, that “21st” date isn’t 100% accurate. And of
course, not all twelve months of the year are equally long. Even the
year is not exactly 365 days, but 365.25 days, so that every four years
a leap year has to be introduced to remain in step. For that reason
these “21st” dates shift a bit more now and then. But in spite of all
these effects, our results are fairly accurate.
28
Everyday Physics: Unusual insights into familiar things
10
HOW HOT DOES
THE SUN FEEL?
Why do you sun-burn so easily while skiing or swimming? There
are two obvious causes. First, you are outdoors all day, which
most people aren’t used to. Second, the sunshine comes not only
from above but it’s also reflected by the snow- or water-surface,
unhindered by trees or buildings as happens in town, nearly
doubling the intensity. In addition, the contribution of the blue sky
under those circumstances is substantial: the great open sky not
only contains indirect blue and violet sunlight, but ultraviolet too.
Two other factors may be involved. The first is geographic
latitude. Suppose you live in northern Europe and for your ski
vacation you travel south to the Alps. For every 100 km southward
the sun rises higher by roughly 1°. (The circumference of the Earth
is 40,000 kilometres, so the distance from pole to equator is
10,000 km. That corresponds to 90°, and 10,000 / 90 gives 111 km
per degree, to be exact.) So if you travel to the southern French Alps
from London, say, then the sun is already 10° higher. Therefore, in
the winter, at midday the sun is about 25° above the horizon instead
of 15°. The distance the sun rays travel through the atmosphere is
then much shorter, and that makes a big difference (Fig. 1). This is
because the sun’s radiation is attenuated by the atmosphere, and
when the sun is near the horizon, it has to radiate through a much
thicker layer of atmosphere than when it is high in the sky. A very
low angle of incidence therefore makes the trajectory through the
atmosphere very much longer, so more of the radiation’s energy is
absorbed. This is especially true for the ultraviolet part of the solar
spectrum that tans or burns your skin.
Everyday Physics: Unusual insights into familiar things
29
The second factor that can make a lot of difference is the
height above sea level. The fact that you get closer to the sun is
irrelevant because the change is at most a 2 km in a distance of
150 million km! What does count, though, is that you leave a lot
of atmosphere below you, so that the amount of atmosphere that
the sun’s rays have to travel through is less. At a height of 1,000 m
the atmospheric pressure is 12% less, at 2,000 m 22% less. And at
3,000 m it’s 31% less (see box). Those are big changes, and they apply
everywhere in the world. On the other hand, the damaging effect
of UV on your skin is not a simple function of height. For example
ozone, an important attenuator of UV radiation, is not evenly mixed
throughout the atmosphere; it is found mostly at heights between
15 and 35 km. So it does not matter greatly if we skip the lowest few
kilometres. Still, using sun screen in the mountains is a must, even
when there’s no snow.
Fig. 1: When the sun is
high in the summer, the
distance the rays travel
through the atmosphere
(AB) is much shorter
than the distance in the
winter when the sun is
lower (CD).
Air pressure and height
The atmospheric pressure initially goes down rapidly with height, about 12% for the first
1,000 metres. After that, the pressure drops more slowly with height. In fact, it goes down
roughly as an exponential according to e−h/8 km, where h is the height in kilometres. At a
height of 1 km, the pressure is then only e−1/8, that is 0.88, of the pressure at sea level – a
reduction of 12% as we saw above. On Mont Blanc, or Mt Vancouver, Alaska (both 4,800 m
high), the air pressure is e−4.8/8 or 55% of that at sea level. At 5,000 m (5 km) the pressure
is halved, which gives a nice rule of thumb: atmospheric pressure halves for every 5 km
you go up. From 5 km to 10 km (the cruising height of aircraft) the air pressure is halved
once more, and is only a quarter of what it was at sea level.
30
Everyday Physics: Unusual insights into familiar things
11
PARALLEL LIGHT
BEAMS FROM
THE SUN
On days with lovely thunder clouds in the sky, you sometimes see
several broad beams of sunshine emerging from holes in the cloud
cover (Fig. 1). The beams don’t look parallel but seem to come from
a point not very far above the clouds, making it seem like the sun
isn’t very far away. However, the idea that the sun is so close is purely
an optical illusion – or rather a perspective distortion. The beams are
actually parallel. The perspective distortion you see here is the same
as you see on a train line: the tracks seem to come from a single point
far in the distance, although in reality they are parallel.
Fig. 1: The sun’s rays seem to come from
a point not very far above the clouds.
Everyday Physics: Unusual insights into familiar things
31
(The comparison isn’t perfect: the sun’s rays do in the end
come from something like a point (the sun) while the tracks do not
– they remain truly parallel for the whole length of the track. But in
practice this makes little difference.)
How parallel are the light beams exactly? The sun is
150,000,000 km from the Earth. The distance between the holes in
the clouds letting the light beams through might be 15 km, which is
1/10,000,000th of the Earth–sun distance. Then the angle between
these two beams is the same as the angle between two strings
that you hold 1 m apart and that are fixed to a nail 10,000,000 m
(10,000 km) away. The curvature of the Earth prevents you from
really holding those strings straight over such a distance, but if you
could, you would see that, for distances of a few kilometres, to all
intents and purposes the strings are parallel.
Conclusion: if we say that sun rays are parallel beams, we are not
far out.
32
Everyday Physics: Unusual insights into familiar things
12
SUMMER AND
WINTER, WHY
SUCH A BIG
DIFFERENCE?
In countries like England or the Netherlands the average
temperature in the summer is about 17°C. In the winter it’s not much
above freezing; 1 or 2°C. That sounds like the summer temperature
Everyday Physics: Unusual insights into familiar things
33
is about 10 times more than winter. Is the sun so much further away
in the northern hemisphere’s winter? No; in fact the sun is closer in
January than in July.
Actually, the question is rather misleading. The difference
between summer and winter is not large at all: it only looks that way
because our Celsius temperature scale has a somewhat arbitrarily
chosen zero point. For a true comparison we should look at the
temperature values with respect to the absolute zero point, in other
words compare the summer and winter temperatures in kelvin
rather than in degrees C. Then the difference is much smaller: about
290 kelvin compared to 275 kelvin, i.e. the summer temperature is
290 / 275 = 1.05 times the winter temperature, not 17°C / 2°C = 8.5
times. When we look at it this way, we should be surprised that the
difference is so small in England or the Netherlands, for two reasons:
1. Winter days are much shorter than the summer’s. In midwinter
the sun is above the horizon for only half the time that it is in
midsummer (Fig. 1).
2. The sun is much lower, which makes a particularly big difference.
In midwinter the sun hardly rises 15° above the horizon. So a
sunbeam of 1 m2 cross-section is distributed over 4 m2 on the
ground (Fig. 2) whereas in the summer, with the sun at its highest
point, the sunbeam is distributed over just 1.1 m2.
Conclusion: the winter sunshine on the ground is many times more
“dilute” than the summer sun.
Fig. 1: The length of
the day in summer (ABC,
left) is much longer than
in winter (ADC, right).
(Point D is Vermont,
USA, at 45° north.)
34
Everyday Physics: Unusual insights into familiar things
Fig. 2: When the sun
is high the same amount
of solar energy is spread
over a relatively small
area and is spread over a
bigger area in winter.
This combination means that, in the south of England or in the
Netherlands for example, during the shortest winter day (around 21
December) only about 1⁄7 of the solar energy is received per unit area
compared to the longest summer day. On top of this, at very low
elevation the sun is also weakened by the much thicker air layer it
has to travel through (see Fig. 1, How hot does the sun feel? p30), so
overall a unit area on 21 December receives hardly 1⁄10 of the solar
energy compared to 21 June.
This prompts the question: why is the temperature difference
between summer and winter not much bigger? Part of the answer
in western Europe is the equalizing effects of the warm Gulf Stream,
and of the atmospheric currents and the entire “weather machine”.
However, the most important reason is that the seasons aren’t long
enough for the land to reach equilibrium with its surroundings
(atmosphere and sea), and this is especially true for maritime
regions. The sea acts as a heat buffer: it has a huge heat capacity, and
this attenuates temperature fluctuations. As a result, the differences
between summer and winter are much smaller than in mid-Siberia
(or Mongolia), for example, which are actually at the same latitude.
The dry Siberian soil heats up and cools down much more quickly
than the sea. In addition, the water in the sea is constantly being
mixed, so the deeper layers contribute to the buffering effect,
something that can’t happen on land.
The slow coming to equilibrium also shows up in the lag of
the seasons. The warmest weeks in the summer on average do not
occur around the summer solstice (approx. 21 June in the northern
hemisphere, 21 December in the southern hemisphere), when the
sun reaches its maximum height, but more than a month later.
Everyday Physics: Unusual insights into familiar things
35
Similarly, in winter the coldest time isn’t the winter solstice, but
quite a bit later.
All in all, we can be glad that the Earth orbits the sun so
rapidly. If it took five times or so longer, the seasons would be five
times longer. The land would have time to cool down fully in winter
before starting to warm up again in spring, and we would have truly
Siberian seasons even in western Europe.
36
Everyday Physics: Unusual insights into familiar things
13
WHY DO WE
WALK THE WAY
WE DO?
The way you walk seems normal because everybody does it. You
swing your arms and legs in opposite directions; when your left leg
goes forward, your left arm goes backward. Same thing on the right.
Why do you do that? Because you unconsciously take account
of a law of physics – that the angular momentum (the amount of
rotation) of a body remains constant in the absence of external
forces. Your arms and legs are not attached to the middle (vertical
axis) of your torso, but to the side. When your right leg moves
forward it must push itself off from the right side of your body, so
the right side of the body is pressed backward (just as there is recoil
if you shoot a gun from your right shoulder). The consequence is
that you tend to turn around your vertical axis. You prevent that by
moving your right arm backward at the same time, in compensation.
Everyday Physics: Unusual insights into familiar things
37
You could move your left arm and left leg forward or backward
at the same time, but then the movement of the arm wouldn’t
compensate for the movement of the leg; instead it would make
the rotation worse. This would happen at every step, and your feet
would have to absorb this rotation in their contact with the ground.
So, walking this way not only looks funny, it’s inefficient too.
The same consideration applies when you walk up the stairs
with a cup of tea, to surprise your partner in the morning. As you
walk up the stairs with the tray in your hands, you inadvertently
move the tray to and fro, because you miss the natural swing of the
arms. Thus, a piece of applied science prevents you from spilling tea
all over the place.
Ten-pin bowlers are especially troubled by this problem
when throwing their heavy bowling ball. However, experienced
bowlers have found a solution: just as they release the ball,
they automatically make an elegant sweep with their free leg
to compensate for the torsion. The law of physics provides an
unintended bit of sporty elegance!
Home experiment: A law with an unexpected twist
With an office chair or a bar stool that swivels smoothly, you can give two nice
demonstrations of the law of conservation of angular momentum. (The chair must
swivel very smoothly – lubricate it to make sure.)
Demonstration 1.
This illustrates the pirouette principle: the more mass we move inwards towards the
axis of rotation, the more the angular speed increases. (You can see this very clearly
with a rotating figure skater. When their arms and one leg are outstretched, they rotate
slowly, but as they draw in their limbs, they rotate faster and faster.)
Raise the seat of your swivel chair to its highest position (to keep your feet off the
floor). Sit on the chair holding a full bottle of water in each hand. Spread your hands
horizontally and get someone to spin you and the chair (Fig. 1a). Now bring the two
bottles towards your chest or – better still – above your head, because that is as close as
possible to the axis of rotation (Fig. 1b): your rotational speed increases. If it goes too
fast you can slow down by extending your arms again.
38
Everyday Physics: Unusual insights into familiar things
Fig. 1a: Spread your hands containing bottles of water, and get someone to spin your chair.
Fig. 1b: Raise the bottles above your head and your rotation speed increases. This is the law
of conservation of angular momentum in action.
During these operations, there is no external force, so the amount of rotation –
the angular momentum – remains constant. However, it depends not only on the
rotational speed, but also on the distribution of mass around the axis of rotation. As
a consequence, if the mass gets closer to the axis, the rotational speed must go up to
compensate. We see the same phenomenon in tornados: as the moving air gets closer
to the “eye” of the tornado, its rotational speed increases.
Demonstration 2.
For this demonstration you need a bicycle wheel that spins freely on its axle. If there’s
a tyre on the wheel, so much the better: what you want is a lot of mass on the outside,
so a tyre on a heavy rim is best.
Everyday Physics: Unusual insights into familiar things
39
Attach a handle to each side of the axle. Make each handle from a metal clothes hanger
from the dry cleaners. Put one side of the axle in one narrow end of a clothes hanger,
and tighten the nut securely, just as if the wheel were in the fork of a bike. Now bend
the clothes hanger at a right angle out from the wheel, and fold the protruding part of
the hanger in two, to make a nice grip. Do the same with the second coat hanger on the
other side of the axle, so you can hold the wheel with both hands (Fig. 2a). Now you’re
ready to go.
We will now use the fact the amount of rotation – the angular momentum – is
characterized not only by its numerical value but also by the direction of the axis of
rotation, and this is included in the law of conservation of angular momentum. (In
other words, angular momentum has magnitude and direction: it is a vector quantity.)
Sit on your swivel chair and using both hands hold the wheel firmly with the axle
horizontal (Fig. 2a). Get someone to spin the bike wheel rapidly. Now tip the rotating
wheel so that the axle is vertical (Fig. 2b). Behold: the whole set-up – you, the chair,
and the bike wheel – starts to rotate! And if you tip the wheel by 180° (i.e. turn it upside
down, Fig. 2c) you start to rotate in the opposite direction!
Here’s why. When the wheel started to rotate with the axle horizontal, there was no
angular momentum in the vertical direction (i.e. no spinning around a vertical axis),
and since there are no external forces the conservation law says that must remain so.
When you tip the wheel axle over vertically, there is now a rotation around a vertical
axis, which must be compensated for by a rotation of the whole configuration (you and
the chair included) in the opposite direction.
a)
b)
c)
Fig. 2: A surprising experiment with a bicycle wheel.
40
Everyday Physics: Unusual insights into familiar things
Home experiment: The reckless wine glass party trick
Take a piece of string about 120 cm long. Tie one end to an ordinary wine glass, and tie
the other end to a wine bottle cork.
Take a wooden stick about 40 cm long; the handle of a wooden spoon or ladle will do
fine. Hold the stick horizontally in you right hand. Take the cork in your left hand and
pull the string to the left, so that the glass dangles just below the stick (Fig. 3a).
Now tell your audience that you’re going to let the cork go. Everyone will think the glass
will crash to the floor!
When you let the cork go (Fig. 3b) the laws of physics come into play: the cork end of
the string wraps itself around the stick, and the glass doesn’t fall to the floor! (Practise
with a big soft cushion under the glass until you are confident – and don’t use your best
glasses!) Hold the stick at a slight angle towards you, to prevent the cork from hitting
the string and instead of wrapping round the stick bouncing back and ruining your
demonstration (and the glass).
What’s happening is a beautiful illustration of the law of conservation of angular
momentum. When you let go (Fig. 3b), the cork is pulled towards the stick, and
simultaneously it falls downward and starts to wind around the stick, getting shorter as
it does so. Now the conservation law kicks in: as the string gets shorter, it revolves more
rapidly (Fig. 3c), just like the pirouetting ice-skater. Before the glass hits the ground, the
cork has safely tied the string around the stick (Fig. 3d) and the glass is saved.
Once again the laws of physics yield another astounding experiment.
a)
b)
c)
d)
Fig. 3: Because of conservation of angular momentum,
the cork wraps the string around the stick and stops the
glass falling to the ground.
Everyday Physics: Unusual insights into familiar things
41
42
Everyday Physics: Unusual insights into familiar things
14
THE HUMAN
ENGINE
We’re going to see how much energy the average person uses when
performing various activities. However, before we do that, we’ll
define a few terms so that we can be precise in our analysis.
Energy, work, heat
When you hit a squash ball you do work, and transfer energy to the
ball. With that energy the ball in turn can do work – break a window,
compress a spring, or move another ball.
So work and energy are interchangeable and therefore
have the same unit: the joule (rhymes with cool), symbol J, in the
international MKS (metre, kilogram, second) system of units (also
known as SI, from the French Système International). One joule of
work is equal to one unit of force (one newton, N) times one unit of
distance (metre, m). In other words: 1 J = 1 Nm (one newton metre).
Work and energy finally convert to heat. (The squash ball
becomes noticeably warmer after playing for a few minutes.) The
unit of heat is also the joule. One calorie (an old unit of heat) equals
about 4.2 joules, so one kilocalorie or kcal (often wrongly called a
calorie when talking about food and diets) is 4.2 kJ.
Power
Power is the rate at which work is done, or energy is used, or heat
is produced. So the unit of power is joules per second. Because this
is such a widely used unit, it is also given the specific name watt,
Everyday Physics: Unusual insights into familiar things
43
abbreviated W: 1 W = 1 J/s. Conversely: one joule is one watt for one
second, or 1 watt second (in the same way that a “person-hour” is
one person working for one hour): 1 J = 1 Ws.
Most people are familiar with watts in the use of electricity.
The speed of rotation of the electricity meter at home is a measure
of the number of watts that are “on” (i.e. are being used) in the house.
If 1,000 W is on for one hour, the energy used is 1,000 watt hours,
that is 1 kWh, which is the unit you see on your electricity bill. You
can convert the kilowatt-hour into “proper” energy units (joules) as
follows: 1 kWh = 1,000 W for 3,600 seconds, that is 3,600,000 watt
seconds (Ws) = 3.6 million joules or 3.6 MJ (megajoules).
The same goes for heat. The wood in a matchstick has a total
energy content of about 2,000 J. If the match burns for 20 seconds,
it’s acting as a heater with a power of 2000 J/20 s = 100 J/s, that is
100 W.
44
Everyday Physics: Unusual insights into familiar things
People at rest
When a lot of people are in an enclosed space like a classroom or
theatre or concert hall, they warm the place up. Each of us produces
heat, because in a sense each of us is a small engine. When we sit
still the engine idles, and all it does is produce heat. The cat knows
this and is glad to crawl into our lap (although we prefer to think it’s
because she loves us).
How much heat do you produce? You can work this out
easily, by considering your food intake. For an adult this is about
2–3,000 kcal per day. (Kilo)calories are dieters’ measures; the physics
units are joules. One kcal = 4.2 kJ, so your food intake is equivalent to
about 10,000 kJ or 10,000,000 J per day. When you don’t do much
physical work, and if your metabolism is functioning properly and
your weight remains constant, all that food is converted into heat.
That happens over the full 24 hours of the day, i.e. over 24 × 60 × 60
= 86,400 seconds. So your average heat production per second is
10,000,000 J/86,400 s = 115 J/s, which is 115 W. So apart from top
athletes or people doing heavy physical labour, a human generates
about as much heat as a 100 W incandescent light bulb. That’s your
resting energy expenditure.
Everyday Physics: Unusual insights into familiar things
45
It’s amusing to see what’s equivalent to an adult’s 10,000 kJ
daily energy consumption. Burning oil or petrol generates 40,000 kJ
per kg, so energy-wise a quarter litre (one cup) of oil supplies an
adult’s energy needs for a whole day. A car travelling at 120 km/h on
the highway uses a quarter litre in about two minutes, compared to
which a human’s energy use is pretty economic.
Here’s another way of looking at it. Of the food constituents
that you eat (carbohydrates, proteins and fats), fat has the highest
combustion value. Because fat is not so different from fuel oil, its
combustion value is about the same as that of oil. This means that
the energy in a quarter kilogram of fat is enough to live on for a
day. Conversely, if you ate nothing, your energy for one day would
come from burning a quarter kilogram of body fat. So be wary of
any slimming advertisements that suggest you can lose more weight
that that every day, especially if you continue to eat a little. (When
you start dieting, you often lose weight rapidly through loss of water,
which is great for morale – “see how many kilos I’ve lost!” – but makes
no long-term difference.)
The polar explorer’s diet
People often expect that solo Antarctic explorers, who have to carry all their food and
fuel themselves, bring only dried “astronaut” foods, but they don’t. They use about 8,000
kilocalories (or 34,000 kJ) a day, so they need a very high calorie diet with minimum
weight that requires minimum cooking – so they take a lot of butter. (“It’s horrible for the
first few days, but then you crave the fat content and eat it like slabs of cheese.”) However,
it’s difficult (and boring!) to live on nothing but butter, so they also take cereals/cereal bars,
nuts, dried fruit, chocolate and salami – and maybe some freeze-dried meals as treats.
46
Everyday Physics: Unusual insights into familiar things
15
THE HUMAN
ENERGYEQUIVALENT
OF A VACUUM
CLEANER
How many watts of mechanical power can a human generate?
Take a guess.
A human is nowhere near as powerful as a car (which is
obvious as soon as you try to push a car). The power of a car engine
is about 40 kW, so a human is far less powerful that that. But how
much less? Is the power of a human 10 watts, or 100 watts – or even
1 kW, which is the power of a vacuum cleaner?
You can estimate the power of a human by looking at the simple
but efficient movement of walking up the stairs. The movement
involved looks a lot like cycling and uses the same muscles. Most of
the effort expended is in raising your body the height of the stairs.
(You can ignore the energy for the horizontal component of your
travel without significantly affecting your calculation.)
It’s easy to calculate the energy needed to gain height. You
just need to know the gravitational force you have to overcome
(your weight), and remember that work done equals force times
(vertical) distance.
Everyday Physics: Unusual insights into familiar things
47
A reasonable value for how fast you walk up a stairway for a
sustained period is one step per second. Compare that with hiking
in the mountains: one 15 cm (6”) high step per second corresponds
to gaining 500 metres height in one hour, which is fairly strenuous
– so one step per second is a reasonable estimate for going up stairs.
The calculation (see box) shows that you generate 100 W
going up the stairs, well short of the power of a car. Note that here
we are talking of mechanical energy only; the total energy used is
a lot higher, because the efficiency of your muscles is well short of
100% and a multiple of that 100 W is released as heat (see How do
you keep your temperature constant? p49).
How much – or how little – 100 W is becomes clear if you
generate electricity with a home trainer. Converting mechanical to
electrical energy can be about 90% efficient, but for simplicity let’s
say it is 100%. So, if you spend 10 hours generating electricity on the
home trainer, cycling all the time, then at the end of the day the total
work done = power × time = 100 W × 10 h = 1,000 Wh or 1 kWh
altogether. You get that out of the wall socket for about 15 pence or
25 cents.
Conclusion: even if electricity is expensive, pedalling on a home
trainer is not a good alternative.
The energy you use going up stairs
The potential energy you gain by rising a height h in the gravitational field with an
acceleration due to gravity g of ≈10 m/s2 is mgh, where m is the mass (and therefore mg is
the weight in newtons). For a step height of 15 cm and a person weighing 70 kg, the work
done for one step is:
mgh = 70 kg × 10 m/s2 × 0.15 m = 105 J
If you do this at a rate of one step per second, then the power is 105 J/s, or roughly 100 W.
48
Everyday Physics: Unusual insights into familiar things
16
HOW DO YOU
KEEP YOUR
TEMPERATURE
CONSTANT?
When you exert yourself physically, e.g. by running to catch a train,
or biking into town, you produce not just mechanical energy, but
also a good deal of heat. Your muscles don’t convert nutrients into
mechanical work at 100% efficiency. (In that respect people are like
car engines: a large fraction of the energy used is wasted in the form
of heat, which has to be removed.) The efficiency of your muscles
depends on the kind of work that you do: for cycling or climbing
stairs the efficiency is about 25%. That means that the rest of the
expended energy (75%) comes off as heat, which is three times as
much as the useful work done. And that is on top of your resting
energy expenditure that just keeps your body systems going.
Everyday Physics: Unusual insights into familiar things
49
To maintain a constant temperature, the heat production
and heat removal have to be equal all the time. If you exert yourself
more, you produce more heat, so the rate of heat removal has to
increase. You could adapt your clothing to your exertion (which
is why athletes wear shorts and singlets) but that’s not suitable for
everyday life. In fact you cool your human engine – remove heat – in
three ways: (1) conduction to the ambient air, (2) radiation and (3)
evaporation. Does the wind not contribute? We tend to think of it as
a separate mechanism, but the only thing the wind does is bring the
cool air closer to your body. It doesn’t penetrate your body to bring
the coolness in, and if you were to look closely enough you would
find that it streams parallel to your skin. Therefore, its only effect is a
steeper temperature gradient close to your skin, which causes more
rapid heat removal by conduction. In short: conduction, radiation
and evaporation are all you’ve got.
The rate of conduction is completely determined by the
temperature difference between the outside of your body and
the environment. (This is also true for radiation, although it’s not
obvious – see the Radiating body box below.) The amount of heat
removed by conduction and radiation remains practically constant,
50
Everyday Physics: Unusual insights into familiar things
as long as the ambient temperature and your body temperature
do not change. (During exercise, the temperature of your skin
increases a little, and gets closer to your core temperature of 37°C
due to increased blood circulation, but it never exceeds your core
temperature.)
Conclusion: heat removal by radiation and conduction hardly
increase even when you exercise hard.
Evaporation is the only heat removal method that you have
more control over, so it’s your main control mechanism for achieving
heat balance. And it’s very effective, because evaporating water
requires a lot of heat. (You know that from your kitchen: if you put a
kettle on the stove and it takes, say, 5 minutes to come to the boil, it
will take another 20-30 minutes for the kettle to boil dry. This shows
that evaporating the water takes about five times as much energy
as heating the water from room temperature, about 15°C, to 100°C.)
The importance of evaporation is illustrated schematically
in Fig. 1, which shows how evaporation affects someone in normal
dress on a home trainer at ambient temperature. The horizontal
Everyday Physics: Unusual insights into familiar things
51
axis shows the muscular work done per second; the vertical axis is
the energy production per second.
►
The green line shows the total energy produced per
second, including both heat and mechanical work.
►
The blue line shows how much energy is produced as heat,
which needs to be removed.
►
The height between the blue and the red lines is the
amount of mechanical/useful work done. This is zero,
(i.e. the green and blue lines coincide) when there is no
muscular work done, representing the resting energy
expenditure.
Fig. 1: Where
the energy goes
when you
exercise on a
home trainer.
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Everyday Physics: Unusual insights into familiar things
For a cyclist to generate 100 W of useful muscular power (AB,
Fig. 1) at an efficiency of 25%, the power required is 400 W. On top
of that there is the 100 W resting energy expenditure, so the cyclist
has to generate a total of 500 W (AC in Fig. 1). Only 100 W is useful,
and the remaining 400 W is heat that has to be removed – BD being
the amount lost through evaporation, and DC the amount radiated
or convected.
When little or no muscular work is done (bottom left of
Fig. 1), the heat removal is largely by radiation and conduction. As
the exertion increases, evaporation becomes progressively more
important, and you sweat more when you work harder. Because
radiation and conduction can’t increase appreciably, evaporation
has to remove practically all the extra heat. (If you can’t increase
the evaporation – as in the simple example of a cyclist on a home
trainer where there’s no rush of passing air to aid evaporation – you
may overheat.)
There’s a lesson here if you swim in water that is too warm. If
the water is cooler than body temperature (37°C), your body loses
some heat to the surrounding water. But if the water is above 37°C,
the only cooling available is from your head that is above water, and
evaporation from such a small area can’t help much. If a swimmer
tries to establish an Olympic record in the 400 metres under those
conditions, they could well come to grief from overheating.
Everyday Physics: Unusual insights into familiar things
53
Radiating body
The amount of heat radiated by a body is proportional to the temperature raised to the
fourth power. It is equal to σ T 4 times the area of radiating surface where σ is the Stefan–
Boltzmann constant and T the absolute temperature in kelvin, K.
However, to find the net radiation transfer to the environment, we must also factor in
the radiation coming back from the environment to you, which in terms of absolute
temperature is not much cooler than you are. Then the net amount of heat you lose is
proportional to the temperature difference. If Tb is your body temperature, and Ts is the
temperature of your surroundings, then the net radiation that you emit is proportional to:
Tb4 − Ts4
Using standard mathematics – remember, a2 − b2 = (a+b)(a−b) – we can rewrite that as
(Tb2 + Ts2) (Tb2 − Ts2)
and by rewriting the second factor again, we get:
(Tb2 + Ts2) (Tb + Ts) (Tb − Ts)
Now the difference in temperature, Tb − Ts, which we’ll call ΔT, is small. So, in the factors with
the plus signs, we can forget the difference between the two, i.e. Tb ≈ Ts ≈ T. That gives us:
(T 2 + T 2 ) (T + T) (ΔT)
which is 4 T 3 ΔT.
We can also view that graphically: if on the T4-curve two points are close together, we can
regard the line joining the two points as a straight line – the tangent to the curve. In the
language of differentials:
Δ(T4) = 4 T 3 ΔT
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Everyday Physics: Unusual insights into familiar things
17
HOW EFFICIENT
IS CYCLING?
While walking is fairly efficient compared to flying and other ways
of locomotion in the animal world, humanity improved transport
performance by a factor of four to five with the invention of the
bicycle. Cycling beats everything in the transport category for the
amount of energy per kilogram moved over a given distance. For a
bike this is about 0.6 kJ/(kg⋅km) (i.e. it uses 0.6 kJ to transport 1 kg a
distance of 1 km) although this depends on the kind of bike and the
speed, as we’ll see in the next chapter.
Everyday Physics: Unusual insights into familiar things
55
Fig. 1 shows how efficient the bike is. In the figure, several
modes of transport are arranged by mass – light ones on the left,
heavy ones on the right. The vertical axis shows the energy used per
displaced kilogram per km. You can see that a human cyclist is about
100 times more efficient than a rat, and about 6 times more efficient
than a human walker.
Fig. 1: Comparison
of the energy different
animals use, per km per
kg transported mass,
arranged from light (left)
to heavy (right) (Data
from SS Wilson, Scientific
American March 1973,
p90).
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Everyday Physics: Unusual insights into familiar things
Why is cycling so much more efficient than walking? This is
partly because the centre of gravity of a cyclist doesn’t move up
or down: when one leg rises the other goes down, so overall
there’s no change. That’s not the case with walking: your centre of
gravity moves up and down about 3 cm with every step, and that
wastes energy.
How much energy does cycling take? The muscular power
it needs is comparable to that of mounting stairs, because cycling
involves much the same movement as climbing stairs. As we saw in
The human energy equivalent of a vacuum cleaner, p47, that requires
about 100 W. From that we can deduce a cyclist’s energy use by
means of a thought experiment. We saw that 100 W means a gross
energy production of 400 W, given a muscle efficiency of 25%,
and that a continuous power output of 400 W is the equivalent of
burning 1 litre of oil per day (see The human engine: People at rest,
p43, where we stated that 100 W = ¼ litre/day). So 1 litre of oil is the
Everyday Physics: Unusual insights into familiar things
57
energy equivalent of a 24 hour bike ride. In those 24 hours you cover
a distance of 24 × 20 km, approximately 500 km.
Conclusion: a cyclist uses roughly 1 litre per 500 km, which is
enormously efficient (especially compared to an SUV at about
50 litres per 500 km). But there’s a catch: you don’t bike on oil or
petrol, but on bread, milk, steak, cheese. Producing all that food in
the first place needed much more energy than the energy (calorie)
value of the food itself. Take milk, for example. The cows have to eat,
they have to be milked, and the milk has to be cooled, transported,
heated to be pasteurized, cooled again, and transported again. If you
add up all these steps, a litre of milk turns out to cost half a litre of
oil before it comes on the table. It is much the same with the rest of
your food.
So while biking is wonderful, and great for staying in shape, to
save energy you might be better off buying a light motorcycle.
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Everyday Physics: Unusual insights into familiar things
18
WHAT FORCES
AFFECT A
CYCLIST?
In places like the Netherlands it is easy to see the forces involved
in cycling: the terrain is flat, so you don’t have to go up hills against
gravity. The bike itself has little resistance if it’s in good condition: the
ball bearings only cause an energy loss of about 1%, a well running
chain about 1.5%, and a derailleur gear at most 5%.
The real forces that resist you, once you are up to speed, are
the rolling resistance and the air resistance. Since these behave quite
differently with increasing speed, we’ll look at them separately.
Everyday Physics: Unusual insights into familiar things
59
The rolling resistance is largely caused by the tyres. They
convert some energy into heat as they are deformed by contact
with the road. The resistance is independent of the speed, and is
proportional to the weight and to the “rolling resistance coefficient”
Cr (see box). The magnitude of Cr relates to the deformation process
of the tyre, and is strongly influenced by the tyre pressure.
It may seem strange that the tyres cause resistance. After
all the elastic rubber returns neatly to its original state after
deformation, so no energy is consumed in permanent deformation.
But the whole dent-and-bulge cycle does consume energy, because
the tyre has to be dented with more force than it delivers when
bouncing back. Therefore, it warms up, something you can clearly
feel easily with car tyres after a long drive.
The air resistance increases with the square of the speed. So at
20 km/h it’s 4 times greater than at 10 km/h. It is the air resistance
that troubles you most when you want to go fast. Also, air resistance
is proportional to the surface area seen from the front (the “frontal
area”). The proportionality coefficient is the “Cd value”, drag
coefficient or air resistance coefficient, which is determined by how
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Everyday Physics: Unusual insights into familiar things
streamlined the shape of the body is. For a cyclist sitting upright, Cd
is about 1, or about 0.88 for a bike racer (and in this case the frontal
area is also smaller, leading to a double gain). For cigar-shaped super
bikes – the HPVs (Human Powered Vehicles) – it is 0.1.
Fig. 1 shows the resistance for an ordinary town bike at
different speeds. At very low speeds, there’s almost no air resistance
so it’s mostly the rolling resistance you have to overcome. At
10 km/h, the air resistance (AB in Fig. 1) is about half the rolling
resistance (AC). At about 15 km/h they are equal (DE = DF).
Thereafter the air resistance dominates. Therefore, for high
speeds streamlining is essential to reduce air resistance. HPVs can
exceed 100 km/h, something an ordinary racing cyclist cannot
even dream of (see Can you cycle at 100 km/hr? p67).
Fig. 1: How resistance
increases with speed,
for a town bike. An
interesting feature
here is that the total
resistance that the cyclist
meets also represents
the energy consumption
per km. This is because
energy = force times
distance, so force =
energy divided by the
distance, or energy per
unit distance. Take the
case where the total
resistance force is 20
newtons; 20 newtons
= 20 joule per metre =
20 kJ per km.
Everyday Physics: Unusual insights into familiar things
61
Since the total resistance increases with increasing speed, the
energy used per kilometre also goes up. Biking fast not only costs
more energy per hour (which is obvious, because you’re working
much harder), but also costs more per kilometre, although it does
save time and keeps you fit. (At a steady 20 km/h you burn about
2 grams of fat per kilometre, see box. If you want to lose weight
by cycling, then you know what it takes.) However, if you have a
particular distance to cover and want to save energy, cycle slowly.
Rolling resistance and air resistance
The rolling resistance can be written as Cr × m g, where Cr is the rolling resistance
coefficient, m the mass and g the acceleration due to gravity. For a town bike Cr is about
0.006, and for a racing bike 0.003. The losses are mainly due to the fact that the force with
which the tyre is dented is greater than the one with which it bounces back; this is a form
of hysteresis (the closed loop integral F·dS is not equal to zero). Another part comes from
“Reynolds-slip” (the friction of interfaces due to the deformation).
The air resistance is directly proportional to the air density, ρ, and proportional to the
square of the speed (unlike that for small raindrops) because the flow pattern is not
laminar but turbulent. From Bernoulli’s law (p + ½ ρ v 2 = constant, where p is the pressure)
we can derive right away that the dynamic pressure in front of a flat surface perpendicular
to the air stream is equal to ½ ρ v 2 . For an object with an arbitrary shape and a frontal area
A, this yields a force Cd × A × ½ ρ v 2 . The air resistance or “drag coefficient” Cd is about 1 for
the ordinary bike and 0.1 for super bikes. The limiting value is about 0.05 for the ideal
streamline, more or less a fish shape. For the relevant speed range Cd can be considered
a constant.
The total resistance for an ordinary town bike (Cr = 0.006; m = 90 kg; Cd = 1.1; A = 0.51 m2) at
20 km/h turns out to be 16 newtons (see Fig. 1). The mechanical energy use is therefore 16
kJ/km. With a muscle efficiency of 25% this means a total energy use of 64 kJ/km, which is
equivalent to burning 1.6 gram of oil per km (assuming a combustion value of 40 MJ/kg),
or 1.6 × 500 g per 500 km, = 800 g/500 km. As the density of oil is about 900 g/litre,
this agrees well with our earlier rough estimate (see How efficient is cycling? p55) for the
oil-equivalent of a cyclist’s energy use of 1 litre per 500 km or 2 grams per km.
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Everyday Physics: Unusual insights into familiar things
19
CAN YOU
CYCLE AT
100 KM/H?
If you want to cycle fast, it’s clear from the figure on p61 (What forces
affect a cyclist?) that you have to minimize the air resistance. Racing
cyclists do this by crouching down over the handlebars and wearing
streamlined helmets. Even better is a reclining “super bike” with a
small frontal area and a perfect streamline.
Everyday Physics: Unusual insights into familiar things
63
Not surprisingly, the top speed you can reach increases with
the power – energy per second – that you can develop for the duration
of the speed test. Official bike-speed records are normally measured
over a 200 m course, with a flying start. We can read off the power
directly from the necessary force in Fig. 1, p61, if we remember that
power is force times distance per second, that is, force times speed.
If we know the force, then we know the power at every speed.
A surprising point is that a cyclist starting from stationary – or a
car or train starting from stationary – is expending zero power while
stationary. This is logical: even when there is a force, as long as there
is no movement, no work is done. However, as every cyclist knows,
getting your bike moving from rest requires a lot of effort, so is there
a contradiction here? No, because how the force is generated differs
between a human and a concrete block, say. Think of a concrete
block placed on the pedals when the bike is stationary: it’s clear that
this expends no energy. However, for a human to press hard against
a wall, or to press down on the pedals, or for a train to start moving,
energy is expended, because of the chemical and/or biological
actions (in your body or in the engine) needed to exert the force.
Fig. 1: Power (in watts)
required to propel a
town bike (red line)
and a human powered
vehicle (HPV) (blue line)
at different speeds.
Fig. 1 shows the power for two different cases – an ordinary
town bike, and an HPV with Cr = 0.0045, m = 90 kg, Cd = 0.10, and
frontal area A = 0.44 m2. The difference is striking. For example,
we see that using the power needed to go 20 km/h on a town bike,
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Everyday Physics: Unusual insights into familiar things
we could reach about 50 km/h on an HPV. The maximum power
that can be developed by a professional cyclist for a short interval is
around 750 watts (one horsepower!). On an ordinary town bike you
soon run up against that limit – at about 40 km/h (X, Fig. 1) – but
an HPV could do nearly 100 km/h (Y, Fig. 1). And many HPVs have
done so. In the 1980s and 1990s a number of HPVs exceeded the
then speed limits in the US of 55 mph (88 km/h). Some speed freaks
got themselves fined on purpose, and became the proud holders of
an “honorary speeding ticket” from the California Highway Patrol.
Everyday Physics: Unusual insights into familiar things
65
The 100 km/h limit was exceeded in 1998, and in October
2002 the Canadian Sam Whittingham rode 130.36 km/h (not at sea
level, but at 407 metres altitude on Battle Mountain, Nevada, where
the air pressure is lower). Meanwhile this record has been broken a
few times. In 2016 it was raised to 144.17 km/h by the Canadian Todd
Reichert, also at elevation in Nevada. (The International Human
Powered Vehicle Association registers these records.)
To cycle faster still, you can almost completely eliminate the
air resistance by riding very closely behind a racing car with a large
flat vertical windshield attached to the back. That’s exactly what
Fred Rompelberg from Maastricht, the Netherlands, did on the
Bonneville Salt Flats, near Salt Lake City, Utah. On 3 October 1995
with a breath-taking 269 km/h he became the fastest cyclist of all
time, even without using a super bike!
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Everyday Physics: Unusual insights into familiar things
20
HOW FAST CAN
YOU CYCLE ON
THE MOON?
Nobody has ever cycled on the moon (yet!). There are no roads and
no oxygen. But let’s just suppose. The moon would be perfect for
a record attempt: there’s no atmosphere (so no air resistance), and
gravity is only 1⁄6 of that on Earth, so the rolling resistance is only
1⁄6 too. Of course you’d need an oxygen tank, which means extra
weight, but that should be manageable.
Everyday Physics: Unusual insights into familiar things
67
So how fast could you go? The calculation is straightforward.
Since the rolling resistance is the only counterforce, we need just a
reasonable assumption for the mass of the “moon racer” and for the
rolling resistance coefficient Cr ; let’s assume m = 100 kg (in view of
the oxygen tank and the space suit) and Cr = 0.0045, the value for a
super bike on Earth and about the average between the value for a
town bike and a racing bike. If you develop 750 watts, as on Earth,
your top speed is 3,700 km/h.
Isn’t that faster than sound? Well, it would be three times the
speed of sound on Earth in air at normal temperatures. But on the
moon you have no problem with the sound barrier because there is
no atmosphere in which sound can propagate. You do have to look
out for lunar hills, though: 3,700 km/h is half the escape velocity, so
if you shoot up a hill, you might launch off into space!
68
Everyday Physics: Unusual insights into familiar things
21
IS CYCLING
REALLY HARDER
WITH A SIDE
WIND?
When you’re cycling, a tail wind is an advantage, and a head wind
definitely makes pedalling harder. But does a side wind coming at
you from right angles give you any trouble? At first thought you
might think that this cannot make any difference, but in fact you’d
be wrong. Strange as it may seem, pedalling is definitely harder in
a side wind.
Everyday Physics: Unusual insights into familiar things
69
The explanation is that the air resistance on a cyclist increases
disproportionally with the speed (see What forces affect a cyclist?
p59). On a completely windless day, if you double your speed the air
resistance is four times greater. That’s easy enough: if you imagine
the air as a sea of small ping-pong balls you have to cross, then
doubling your speed means you hit those balls at twice the speed
(the momentum is doubled). But in addition you meet twice as many
ping-pong balls every second, so that gives a total of four times as
large a force (the momentum transfer per second is quadrupled).
How does the side wind affect things? Let’s take an example:
you cycle at velocity V with no wind at all (Fig. 1a). You feel an effective
wind speed of V in the opposite direction, and the consequent air
resistance, D, say (Fig. 1b).
Fig. 1: How drag
increases with a
side wind.
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Everyday Physics: Unusual insights into familiar things
Now what if there’s a side wind blowing exactly from the right
(Fig. 1c)? Because you are moving forward, effectively the wind is
coming from your front right (Fig. 1d) because of the parallelogram
of forces. Assume that the wind blows so hard that the air speed with
respect to you is doubled, i.e. is 2V. The resulting drag is now four
times as large as when there was no wind, i.e. it’s 4D. What matters is
its component in the direction of motion. That is also much bigger
than you’d have had without wind: twice as much air resistance, i.e.
2D (Fig. 1e). To take a concrete example, if you’re cycling at 20 km/h
and there’s a cross wind of about 35 km/h (force 5 on the Beaufort
scale), you’ll have twice as much wind resistance.
Two other factors make things even worse. With the effective
wind coming from the side, your “frontal” surface area, i.e. the area
facing into the (now sideways) wind, increases quite a lot, which has
a big effect on the air resistance. Second, your shape is no longer the
front profile of a streamlined cyclist, but practically the (completely
unstreamlined!) side surface of cyclist plus bike. And a croucheddown position over the handlebars doesn’t help much, since it
hardly reduces the side-area facing the wind at all.
Everyday Physics: Unusual insights into familiar things
71
If a side wind is to benefit you rather than impeded you, it has
to blow from some angle from behind you. How much of an angle?
That depends on the speed of the wind relative to your speed. In our
scenario of Fig. 1 you feel no net effect if the wind blows at an angle
of 104° (where 0° is straight ahead); if it goes further behind you than
that, you get an overall benefit.
A strong side wind has a big effect on bike road-races like
the Tour de France. Consider a large group of riders. The apparent
wind comes diagonally from the side, so riders benefit by staying
diagonally behind riders in front (much like a group of migratory
birds) in a formation that cyclists call an echelon (Fig. 2).
However, given the limited width of the road, only a small
number riders can do this and get good shelter from the wind, so
a large group breaks into several smaller echelons. Because of the
increased resistance caused by the side wind, groups of less-strong
riders can’t keep up with the faster groups, and the result is often
large time differences between the groups by the end of the race.
Fig. 2: In a side wind,
racing cyclists travel in a
diagonal formation (an
“echelon”).
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Everyday Physics: Unusual insights into familiar things
Home experiment: Healthy blowpipe
Occasionally you come across them at a funfair: a kind of plastic blowpipe with two
exit channels, one above and one below, like a “T” turned on its side. A ring of woolly
fluffy thread is run through it. If you blow into the pipe, the thread starts to rotate: it
comes out the top of the pipe, and goes back in at the bottom – a sight to see (Fig. 3).
Fig. 3: The “rotating
string” blowpipe.
Fig. 4: The airflow
through the blowpipe.
But how does it work? The secret is in the diameter of the exit channels (Fig. 4). The
diameter of the top channel is a bit larger than that of the lower channel. Therefore, the
air speed going up is larger than the air speed going down. The air stream thus pulls the
fluffy thread harder upwards than downwards. The net result is the rotating thread.
With simple tools you can make a pipe like this yourself. Take a 3 × 6 cm rectangular
piece of Plexiglas (Fig. 5), about 1 cm thick. (You can use wood instead, but it’s not as
easy to see that the holes are in the correct place.) Drill a hole about 4.5 mm diameter
in the top, and one about 3.5 mm in the bottom, to take the thread. Drill a bigger third
hole in the side and attach a robust hollow handle to blow into – some plastic tube or
part of an old ball-point pen will do. Make the fluffy thread-ring with a circumference
Everyday Physics: Unusual insights into familiar things
73
of about one metre. To make it easy to see the thread
moving, make the ring from two pieces of different
colours, each about 50 cm long, or use ink to mark
noticeable spots on the thread. Make the knots very
small so the thread can easily pass through the
drilled block.
Fig. 5: Making the
“rotating string”
blowpipe. (Not to scale.)
74
The experiment illustrates a phenomenon that is
important for your health. What goes for air also goes
for liquid, and – within limits – for your bloodstream.
The experiment shows that the stream velocity is
larger in wider tubes than in narrow ones (given the
same pressure difference, of course). To be exact,
for a tube 2 times as wide, the stream velocity is 4
times as large. But that is not all: the streaming takes
place over a larger cross-sectional area, which is
also 4 times as large, so the amount of fluid moved
per second is 16 times as much. This is Poiseuille’s
law, which says that the throughput per second is
proportional to the diameter to the fourth power.
This means also that a 1% smaller diameter implies a
4% smaller throughput (because (1 − 0.01)4 ≈ 0.96). So
it is good for your health to keep your blood vessels
wide and elastic.
Everyday Physics: Unusual insights into familiar things
22
MINIMIZING YOUR
JOURNEY TIME
Some things you automatically do right, because you have an
intuitive understanding of some of the laws of physics without even
realizing it. Take walking: when your left leg goes forward, your left
arm goes backward; it’s the same for the right. That way the net
amount of rotation of your body is zero and you don’t suffer any
unnecessary strain by trying keep your body straight. Cycling is a
bit more difficult, but once you know how to do it, your “automatic
pilot” does exactly what’s needed so you don’t fall off, even when
going round corners. (If you wanted to teach a robot how to do it,
you’d need a huge amount of physics and control software.)
Everyday Physics: Unusual insights into familiar things
75
Here’s an easier problem. Suppose you are an express cycle
courier and every second matters. You have to deliver to someone
at the far corner of a campsite. You have to cross a car park of very
rough gravel, where you can cycle (at 6 km/hour, say), and then an
expanse of grass, where you walk at 4 km/hour. (“No cycling on the
grass.”). What route do you take to minimize your journey time?
Remarkably enough, you intuitively take a nearly optimal route.
What is the fastest route? Of course, you don’t take a straight
line path (X in Fig. 1). It’s the shortest path, but not the quickest,
because you lose the advantage of being faster on the gravel than on
the grass. You cycle a bit further on, but how far? You don’t continue
until the tent is directly in front of you (Y, Fig. 1); you feel somehow
that dismounting a bit earlier is better (Z, Fig. 1), but how do you
work this out exactly?
Fig. 1: Which is the
cycle courier’s fastest
route from the entrance
to the tent? Path X is
the shortest, and path
Y involves the least
walking, but is there
another path Z that’s
quicker than X and Y?
You can see why path Y (A-B-C) isn’t the fastest, by comparing
it with path Z (A-D-C) in Fig. 2. In Z you walk further but cycle less. To
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Everyday Physics: Unusual insights into familiar things
see how much further you walk, draw an arc through B with centre
C, and you see that the red line DE is the extra walking distance. This
increases your journey time by (DE ÷ 4 km/h). To see how much less
you cycle, draw an arc through D with centre A, and you see the
cycling distance is reduced by green line FB, which decreases your
journey time by (FB ÷ 6 km/h). So as long as (FB ÷ 6) > (DE ÷ 4), your
journey is quicker.
Fig. 2: The effect of
walking a little more (red
line DE) may save more
time by reducing the
distance on the bike a
lot (green line FB).
So more generally, it is the relative speed of cycling compared
to walking that matters. Clearly, if walking and cycling were equally
fast, the straight line (X in Fig. 1) would always be best. At the other
extreme, if you walk as slowly as a snail, it’s best to cycle to almost
exactly in front of the tent.
So where exactly is the best place to dismount? Nature points
the way with an analogous case – the path a light beam takes going
from air into water. Fermat’s principle says that light always follows
the path which takes the least time. The light kinks at the transition
to the water because light propagates more slowly in water, whereas
in air it travels almost as fast as in vacuum. (The speed of light in
Everyday Physics: Unusual insights into familiar things
77
water is about three-quarters of that in air.) In glass the speed of light
is about two-thirds its speed in air. Therefore, a light beam entering
glass is just like our cycle courier it travels through air a bit longer to
profit from the higher speed, and enters the glass quite late on, with
a kink, i.e. it is refracted.
Fig. 3: Snell’s law: v1
and v2 are the respective
velocities of light in air
and water.
Credit: After nl.wikipedia.
org/wiki/User:Sawims
78
The best place to get off the bike is given by Snell’s law which
describes the refraction of light at a boundary between two media.
Snell’s law states that the ratio of the sines of the angles of incidence
and refraction is equal to the ratio of the velocities in the two media
(Fig. 3). The cycle courier has to make the same kink as a light beam
when it enters glass where the speed is two-thirds of that in air.
(This is why we chose the slightly unrealistic speeds for cycling and
walking). Doing the calculation shows that your path over the grass
is not perpendicular, but has to make an angle of about 65°. Next
time you do it, notice what you do: in practice, you intuitively get it
almost exactly right.
Long live our intuition for the laws of physics!
Everyday Physics: Unusual insights into familiar things
23
THE CYCLIST’S
SOGGY BACK
Racing bikes don’t have mudguards, so when the road is wet you
get a nice straight muddy stripe down your back. This is caused by
water drops that are thrown in a forward direction from the rear
tyre (Fig. 1). The drops are thrown from the wheel by centrifugal
force, which increases as the speed increases, so fast cyclists have
soggier backs than slowcoaches.
Everyday Physics: Unusual insights into familiar things
79
Fig. 1: Water spraying
from a bicycle wheel.
Credit: After pixabay.com/
vectors/silhouette-wheelcyclist-bike-3067346/
pixabay.com/
users/mohamed_
hassan-5229782
You might think that most of the water leaves the tyre at the
highest point, because this is where it attains its maximum forward
speed (combining the speed of rotation and the forward speed of
the bike). However, it is not the speed with respect to the road that
determines the centrifugal force but the speed of rotation around
the axis, which is the same all around the tyre.
Gravity is involved too: because it always acts downward, the
drops near the lowest part will be easiest pulled off, which promotes
splashing near the spot where the tyre touches the ground. Higher
up, gravity becomes less and less of a help. At the highest point it
even works exactly against drops flicking off the wheel. Even so,
water does splash off, because the centrifugal force overwhelms the
gravitational force.
But which drops wet your back? These are not the drops
leaving the tyre exactly at the highest point. After all, the drops leave
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Everyday Physics: Unusual insights into familiar things
the tyre along an imaginary tangent and then, just like a ball, they
proceed along a ballistic curve or “bullet trajectory” (ignoring any
effect from the wind). When they leave the tyre at the highest point,
they start out horizontally, and they can’t get any higher, so they
never reach your back (Fig. 2).
Fig. 2: Drops thrown
from the top of the
wheel can never reach
your back.
So the bad guys have to let go earlier if they are going to make
you soggy. Just where? That is hard to say. How high above the tyre
is your saddle? How upright is your back? What’s your speed? We’ll
do a detailed calculation in a moment, but a reasonable guess is that
droplets which let go about 45° before the highest point can just
touch the saddle. If they want to make it to your back, they must
follow a steeper trajectory and therefore must let go earlier, at least
about 60° before the highest point.
At what bike speed does the back-splashing begin? To start
with, the centrifugal force has to overcome gravity. At the highest
point that happens above 7 km/h (see box), and then droplets will
let go. At 60° earlier, where the bad guys come from, that happens
already at a lower speed because gravity does not yet fully counter
the centrifugal force. As well as overcoming the gravitational force,
the droplets also need some speed to travel up to your back, and
they manage that only at a higher bike speed than the 7 km/h above;
this turns out to be about 12 km/h. Below this speed the droplets at
best reach the saddle. However, a racing cyclist travels much faster
than 12 km/h, and the price is a wet back!
Everyday Physics: Unusual insights into familiar things
81
Water drops thrown off a wheel
First we’ll work out at what bike speed droplets let go of the wheel at the highest point.
The centrifugal acceleration is ω2 R, where ω is the angular velocity in radians per
second and R is the radius of the wheel. ω = v/R where v is the speed of the outside of
the wheel with respect to the axis (or wheel hub in this case). So we can rewrite the
centrifugal acceleration as v2/R. At what value of v does v2/R = g? In other words, at what
forward speed does the centrifugal force just overcome the force of gravity? That’s a
straightforward calculation:
v2/R = g
so
v2 = R g
so
v = √(R g)
If R = 0.4 m and g = 10 m/s2, this yields v = 2 m/s or just over 7 km/h.
By the way, we see from the equation that for a folding bike with small wheels this speed is
lower, so the droplets are released earlier. For R = 0.2 m, for example, we find v = 1.4 m/s
or 5.1 km/h. And when do you get a wet back with small wheels? We consider droplets
that will let go at 60°. These must be fast enough to gain at least 40 cm in height on their
way to the back. That turns out to be at a bike speed of 3.3 m/s or 12 km/h. Because this is
faster than the 7 km/h that was necessary for the droplets to leave the tyre at the highest
point, this is automatically also enough to release them from a lower point of the tyre.
From a precise calculation with a computer model it turns out that most of the droplets
that hit you are released earlier still. The model also shows that up to 12 km/h you will
remain dry.
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Everyday Physics: Unusual insights into familiar things
24
CAN YOU
GET LESS WET
BY CYCLING
FASTER?
When you cycle in the rain, it’s easy to work out how wet you get
per second. Your topside area is smaller than your front side, so you
catch the least rain when you are stationary. As you start to move,
your frontal area becomes increasingly important, because you
bump into some of the raindrops in front of you, so you get wetter
as you go faster.
However, you’re probably interested in figuring out how wet
you’ll get covering a certain distance – say from home into town. If
you cycle fast you take less time (which keeps you dryer) but you
get the rain slanting from the front (which means wetter). Which of
the two effects wins?
The question is complicated by the wind. If there’s a tail wind,
you can adjust your speed to be the same as the wind’s. That’s surely
optimal because practically only your upper side gets wet.
But what if there’s no wind? To work this out, consider your
top area and your frontal area separately. The top area is easy.
On a horizontal surface a certain amount of rain falls per second,
whether the surface moves or not. That amount is independent of
the speed. The less time you’re in the rain, the less wet your top
area gets.
Everyday Physics: Unusual insights into familiar things
83
It’s different for the frontal area. Think of this as a thin vertical
plate that can move horizontally. When stationary it catches no
rain. When it moves, consider it to be moving through a shower
of stationary droplets. (This is because in a constant shower every
droplet is immediately replaced by another one, so that the number
of droplets in a cubic metre remains constant.) The amount of rain
caught by the moving plate is equal to the volume swept out, which
is the area of the plate times the distance travelled. Therefore,
the wetness of the plate does not depend on the speed but only
on the distance.
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Everyday Physics: Unusual insights into familiar things
Another way of looking at this is to imagine the rain droplets
are suspended motionless in the air and instead give the plate an
equal and opposite vertical upward speed. For the process of wetting
that makes no difference. Because the plate also moves forward
(since you are cycling), it sweeps up a volume of air containing
suspended raindrops. All the droplets in that volume are caught.
Looked at from the side, that volume looks like a skewed rectangle
(Fig. 1) – a parallelogram on its side. The parallel sides are formed
by the plate in the initial and in the final state. The horizontal
distance between them is the distance cycled. How much the
parallelogram is skewed depends on the speed. Now the nice thing
about a parallelogram is that its area depends only on the length
of the parallel sides and their distance apart; the skewness makes
no difference. For the cycling plate this means that the volume
swept up is independent of the speed, dependent only on the start
and end points.
Fig. 1: The volume
swept out by a cyclist’s
frontal area depends
only on the distance
travelled, D, not the
speed V.
Everyday Physics: Unusual insights into familiar things
85
Conclusion: you gain a bit by cycling faster. It makes no difference
to your front side how fast you go, but your upper side stays a bit
dryer, so you gain overall. All of this goes for the ideal case where
raindrops neatly fall vertically. So the answer to the question is yes,
provided there is no wind, cycling faster keeps you dryer.
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Everyday Physics: Unusual insights into familiar things
25
ROLLING
RESISTANCE,
AIR RESISTANCE
AND FUEL
CONSUMPTION
You waste fuel if your car tyres are soft, because soft tyres have a
large rolling resistance. And if you spoil the streamline of your car by
driving with the windows open or by stowing stuff on top, again you
use extra fuel, because the air resistance increases. How exactly do
those resistances influence the fuel consumption?
Everyday Physics: Unusual insights into familiar things
87
For the rolling resistance of a car – the friction to be overcome
to get the car moving – the wheel bearings don’t really play a role,
as we saw for a bike (p59). Rolling resistance is pretty much entirely
caused by the deformation of the tyres on the road.
That may seem odd. After all, isn’t rubber elastic, and doesn’t
it spring back nicely into its original shape after being compressed?
While that is true, some energy is lost in the process of compression
and expansion. The reason is that the force needed to compress
rubber is greater than the push-back force when it expands (see box
Rolling resistance and air resistance, in What forces affect a cyclist?
p62) and the energy difference is converted into heat. ( Just feel how
warm the car tyres get after driving on the highway for a while!) The
resistance that results from all this – the rolling resistance, or rolling
friction – is proportional to the weight of the car, just as for a bike.
It is, in fact, a fixed percentage of that weight, for a given type of
tyre at a certain pressure; this percentage is called Cr , the “rolling
resistance coefficient”.
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Everyday Physics: Unusual insights into familiar things
(The fact that the rolling resistance is proportional to the
weight explains why it’s easier to push a broken-down car if
everybody gets out of it: you’ve reduced the weight so you have
reduced the rolling resistance.)
Let’s try to guess what the value of Cr might be. Would it be
around 1 or 0.1? We can make an educated guess by thinking of it in
a different way based on everyday experience. Suppose your car is
parked on a slope, and you release the brakes. For the car to start
rolling, a slope of, say, 1% would probably do. On a slope of 1%, if you
decompose the gravitational force on the car (i.e. its weight, W) into
a component parallel to the road and one perpendicular, you find
that 1%, i.e. W/100, is parallel to the road (Fig. 1). That force pushes
the car forward. If that is enough to beat the rolling resistance, the
value of Cr must be around 0.01. And sure enough, it is – at least if
the pressure isn’t too low: since all the rolling resistance is down
to deformation of the tyre, it is obvious that Cr will increase as the
pressure goes down.
Fig. 1: On a 1% slope,
1% of the car’s weight W
acts parallel to the road.
(Angles not to scale.)
It is important to note that speed is not involved here. The
rolling resistance is, for all practical purposes, independent of speed.
The air resistance, or drag, is different. It goes up with the
square of the speed; as is the case for a cyclist (see What forces affect
a cyclist? p59), if you double your speed, the air resistance is four
times as high. You might guess that from experience: if you stick your
hand out the window at 100 km/h, the force you feel is much greater
than at 50 km/h. Furthermore, the air resistance is proportional
to the density of the air, so air resistance goes up with barometric
pressure. It also means that in the heat of the summer, the air
resistance is lower than in the winter, given a certain atmospheric
pressure: warm air expands and therefore gets “thinner”. Another
consequence is that the thinner air at higher altitudes offers less air
Everyday Physics: Unusual insights into familiar things
89
resistance. That why it’s easier for cyclists and speed skaters to break
speed records at higher altitudes.
Another factor determining the drag is the size of the car,
or more precisely its frontal area. Low and narrow cars therefore
have an advantage over high and wide ones when it comes to
air resistance.
Finally, the air resistance is proportional to the famous drag
coefficient Cd. While the value of Cd can be as high as 1 for awkwardly
shaped objects that aren’t streamlined at all, it can be around 0.3
for modern cars.
Why are rolling resistance and air resistance so important?
They directly determine the energy consumption as we saw with
cycling (see What forces affect a cyclist? p59). Their sum is the total
resistance that a car experiences when driving on a horizontal road,
and the engine must continuously produce an exactly matching
force to maintain a constant speed. The interesting point is that
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Everyday Physics: Unusual insights into familiar things
this force is equivalent to the work done – or the energy produced –
per unit distance. And that can be directly translated into the
energy consumption of the car (see box Resistance and fuel
consumption overleaf ).
In order to assess the influence of the two resistances on the
fuel consumption, we have to know their relative magnitude. For a
typical midsize car this is shown in Fig. 2.
Fig. 2: Resistance
increases with speed
(midsize car).
When driving through residential areas at low speed (below,
say, 50 km/h), Fig. 2 shows that the rolling resistance dominates.
Clearly, it is advantageous here to have a light car (and a light car also
uses less energy to accelerate up to speed after the frequent stops in
urban driving).
Everyday Physics: Unusual insights into familiar things
91
On the open highway things are different. Here air resistance
dominates by far, so it pays to have a car with a small frontal area (low
and narrow), and streamlining of the car body is of great importance
(driving with open windows or stowing stuff on top can increase
the fuel consumption dramatically). Moreover, since air resistance
increases with the square of the speed, driving fast does use more
fuel. Fortunately, things are not as bad as the steeply rising curve
in Fig. 2 suggests, because the efficiency of the engine improves
at higher speeds, partly compensating for the steep increase in
air resistance. But driving fast remains inefficient. Therefore, a
reasonable speed limit may be a good idea after all.
Resistance and fuel consumption
Fig. 2 shows the behaviour of rolling resistance and air resistance (or drag) with
increasing speed, for a car that is not particularly streamlined. The equations for the two
kinds of resistance are given in the box about rolling resistance and air resistance (p62).
The values used in Fig. 2 are: mass = 1,000 kg; rolling resistance coefficient Cr = 0.01; drag
coefficient Cd = 0.4; frontal area A = 2 m2. Here the figure shows that above 50 km/h the
air resistance dominates (whereas for a cyclist that point was at about 15 km/h.) The unit
for this resistance is the units of force: newton (N).
The resistance is a direct measure of the energy consumption, so as resistance increases
with speed, so does fuel consumption. Fig. 2 shows that at 100 km/h the resistance
is about 500 newtons for this car. Work is defined as force × distance, so force = work/
distance, i.e. force = work per unit distance, i.e. 1 N = 1 J/m, or 1 kJ/km. So our 500 N
resistance is equal to 500 kJ/km. At an estimated efficiency of 20% that comes to a total
energy use of 2,500 kJ/km. Oil and petrol yield about 40,000 kJ per kg, so this works out
at about 6.25 kg per 100 km, or a fuel consumption of about 7 litres per 100 km.
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Everyday Physics: Unusual insights into familiar things
26
HOW MANY
CARS PER
HOUR CAN A
ROAD TAKE?
To avoid traffic jams without building new roads everywhere, we
must use the capacity of existing roads effectively. But how many
cars can a highway actually cope with in an hour – what is its
capacity? Can we increase it?
Everyday Physics: Unusual insights into familiar things
93
At first you might think that driving faster increases the
capacity, because the higher the speed, the greater the throughput.
But this doesn’t work in practice. Imagine a single lane on the
highway, 1 km in length, with all cars travelling at the same speed.
The number of cars the lane can handle per hour depends on the
car density (number of cars per km) as well as the speed. If this
string of cars drives at 100 km/h with 15 cars in each kilometre
stretch, this lane can handle 1,500 cars per hour: the capacity is
speed times car density.
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Everyday Physics: Unusual insights into familiar things
The density must not be too high or it becomes unsafe. What
is a safe density? Traffic rules differ from country to country, but a
typical requirement is to “keep far enough back from the car in front
that you can brake and stop safely”. However, on a motorway this is
too strict, because it is not reasonable to expect that the car in front of
you will stop abruptly, or that a concrete block will suddenly appear
on the road. If you took the above law literally on the highway, then
at 100 km/h you would have to maintain 100 m separation between
cars, reducing the capacity of the lane at high speed. (This distance is
large because the braking distance increases with the square of the
speed: driving twice as fast require four times the braking distance.)
Fig. 1 shows how capacity changes with speed. The red curve
is for the strict separation rule, and shows that we reach maximum
capacity (of 2,000 cars per hour) at 25 km/h. At higher speed
each car makes more progress, but the space between cars is
disproportionally larger, resulting in a capacity reduction.
Everyday Physics: Unusual insights into familiar things
Fig 1: Two scenarios
for the capacity of
a lane: maintaining
distance according to
the book (red curve)
and maintaining
distance according to
the two-second rule
(blue curve).
95
Several countries have a rule of thumb for driving on a dry
straight highway with good visibility, which allows a smaller car-tocar separation. The French, for example, have the two-stripe rule
(“deux traits = sécurité”): keep two stripes visible between you and
the car in front. The British have a similar two-chevrons rule. The
Netherlands applies a two-second rule: keep at least two seconds
back from the car in front. What does that mean for the capacity of
the lane? One car every two seconds means 30 cars in a minute, that
is 1,800 cars per hour. (In reality it is a bit less, because those two
seconds are calculated from the rear bumper of one car to the front
bumper of the next car. We have ignored the length of the car; taking
that into account reduces the capacity. That especially adds up at
lower speeds, because then the in-between space and the car length
are comparable. At higher speeds the in-between space is so large
that the length of the car is negligible.) The blue curve in Fig. 1 shows
how this rule affects capacity.
At speeds of around 50 km/h and above, the two-second
rule yields a greater capacity than the strict braking-distance rule,
because the two-second rule allows cars to drive closer. So the twosecond rule entails more risk than the strict rule, but the risk is
pretty low in good visibility on a dry road, so for the open road it’s a
reasonable compromise.
While the two-second rule is OK for 120 km/h on a clear road,
if there’s congestion (or if it’s likely), 120 km/h is no longer safe so we
have to drop the speed. The red line capacity at 50 km/h is the same
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Everyday Physics: Unusual insights into familiar things
as the blue line at 120 km/h, i.e. 1,700 cars/h. So when the traffic
authorities say “drive at 50 km/h in congestion” it’s reasonable,
because it’s safe, and we haven’t reduced the capacity of the road.
The capacity of a road
Assume that all cars in a lane have the same speed and the same separation from the car
in front. Since the number of cars per hour is equal to the number of cars per kilometre
times the number of kilometres that the row of cars moves per hour, we can write the
capacity C of one lane as C = v n, where v is the speed in m/s and n is the number of cars
per metre. (The result is number of cars per second, which we must later convert to cars
per hour). Now n is equal to the inverse of the amount of space occupied by one car; the
space occupied by one car is equal to the length L of the car itself (say 4.5 m), plus the
space between cars.
For the two-second rule the space between cars is the distance driven in two seconds,
which is τ v where time interval τ is equal to 2 seconds. We then find C = v/(L + τ v). At very
low speeds (bumper to bumper) only L remains. Then the car density is constant, and
C = v/L: the capacity is proportional to the speed. At very high speeds we can neglect L
and that yields C = v/(τ v) = 1/τ. This is ½ per second, or 1,800 per hour, and is independent
of speed (which is why the blue line in Fig. 1 flattens out towards the right).
The strict-rule case is more difficult. The space between cars consists of two parts:
1. The reaction-space t v – the distance you cover in the time t which you take to
react and shift your foot from the accelerator to the brake pedal, where v is the
speed in m/s
2. The actual braking distance v2/2a, where a is the braking deceleration.
(For instance, Dutch law specifies a = 5.2 m/s2).
That gives C = v/(L + t v + v2/2a).
For very low speeds both the braking distance and the reaction distance are negligible,
which makes the in-between space negligible. All that is left is the length L of the car, and
we get C = v/L, exactly as with the two-second rule.
For the other limiting case – very high speeds – only the last term (the braking distance) is
of importance. Neglecting the other terms we see that C is proportional to 1/v. That is why
the right part of the red curve decreases.
In between lies the maximum capacity, at about 25 km/h.
Everyday Physics: Unusual insights into familiar things
97
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Everyday Physics: Unusual insights into familiar things
27
COZY
CANDLELIGHT
Candles are really cosy, adding an air of intimacy and romance to
a relaxed dinner. But originally they were practical light sources,
much handier than oil lamps. In fact, compared to an oil lamp, a
candle does give a lot of light, and some people even say that it is a
super energy-efficient light source. Is that true? How much energy
does a burning candle use?
A candle consumes a surprising amount of energy. This
becomes obvious when you feel the heat produced, which is
considerable. A reasonable guess is that a candle produces as much
heat as a 100 watt incandescent light bulb – both of them can burn
your hand.
However, you don’t have to guess: you can easily measure the
energy used by the candle by finding out how much wax it burns
in a measured time. Combine this with the heat of combustion of
wax (see box Candle heat), and you find that the heat generated per
second is roughly 100 joules per second which is 100 W. However,
the light produced is much less than that of a 100 W incandescent
light bulb, let alone a cluster of compact fluorescent bulbs or LED
(light emitting diode) lamps whose wattage adds up to 100 W.
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99
The wattage of an item is a measure of the energy required to
run it, not the energy it emits in the form of light: an electric radiator
consumes 1,000 W but nobody would call it a good light source.
It radiates primarily in the infrared which your eyes can’t see, but
when we talk about “light”, we mean radiation that can be detected
by your eyes, specifically by the rod and cone photoreceptors in
your retinas. This type of visible radiation is not measured in watts,
but in lumens. The efficiency of a light source is determined by the
amount of light it gives (in lumens) per unit of energy consumed (in
watts), so efficiency is expressed as “lumens per watt”.
From that perspective, a candle performs poorly, producing
about 0.15 lm/W, whereas an incandescent lamp gives about
12 lm/W – 80 times as much. LED lamps do even better: at 80 lm/W
they are more than 500 times as efficient as the candle!
However, that’s not the whole picture, because most electricity
is generated from heat in a power station, which isn’t very efficient.
The electricity that you use in your home represents only about ⅓
of the energy that the power station needs to produce it, so perhaps
the LED lamp is only 500/3 = 167 times better than the candle, but
that’s still pretty good. So while candles may produce a romantic
ambience and even help to heat your home with 100 watts each,
if you want them to produce a lot of light, take them to the power
station, throw them in the furnace, and then rush home and use the
electricity they produce to power an LED lamp.
Candle heat
To find out how much wax a candle uses, weigh it before and after burning it for an hour
or so (making sure you measure the time interval carefully). Not all candles are equal, and
not all flames are the same, but a typical candle burns around 8 grams of wax per hour.
Look up the heat of combustion for candle wax: it’s about 48 kJ per gram, so 8 grams
represents 8 × 48 kJ = about 380 kJ of heat, produced in 1 hour, or 380,000 J in 3,600
seconds = 105 J in one second, which by definition is 105 W.
As we saw, 105 W is about the energy a person uses when not doing heavy exercise. You
can sanity-check this. Using 100 W for one day, i.e. for 24 hours × 3,600 seconds/hour,
gives 100 × 24 × 3,600 J – or 8,640 kJ. That is a perfectly normal diet for an adult (see The
human engine, p43).
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Everyday Physics: Unusual insights into familiar things
28
WHY ARE
INCANDESCENT
BULBS SO
INEFFICIENT?
Moving from candle to incandescent lamp was a big improvement:
no fumbling to find matches, no hot wax spilled on the furniture,
no danger of fire, and a lot more light. Even so, the old-fashioned
incandescent bulb isn’t very efficient: it uses a lot of electricity for
the amount of light it gives.
Everyday Physics: Unusual insights into familiar things
101
Why is that? The efficiency is determined by how much of
the emitted radiation can be sensed by your eye. Comparing the
emission spectrum of the bulb with the eye’s sensitivity will tell us
a lot about the bulb’s efficiency. Fig. 1 shows the emission spectrum.
Fig. 1: The emission
spectrum of an
incandescent bulb
compared to the
eye’s sensitivity.
►
►
The tall thin curve on the left shows that your eye is
sensitive to wavelengths between about 400 and 700
nanometres, which are the colours of the rainbow.
(One nanometre is one millionth of a millimetre, or – for
traditional readers – 10 angstrom units.) The peak lies
about 550 nm, in the yellow/green; that’s where your eye
is most sensitive.
The red curve is the spectrum emitted by the incandescent
wire in the bulb. Its peak lies near 1,000 nm, which is in
the infrared.
The two curves don’t match well: most of the bulb’s emission
falls outside the range of the eye, so its remains invisible – useless as
far as lighting is concerned. An incandescent bulb with an emission
spectrum like this is always inefficient.
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Everyday Physics: Unusual insights into familiar things
Can we get a better match for the two curves? One way is to
make the incandescent filament hotter than the usual temperature
of about 2,700°C, because at a higher temperature not only is more
radiation emitted, but also the colour changes. Think of an oldfashioned poker in the fire: its colour goes from red to orange to
yellow as it heats up. Or think of a dimmed incandescent bulb: as
you turn up the dimmer to make the lamp brighter, it gives more
light but the light is also “whiter”. In other words, as the filament gets
hotter, the emission spectrum contains more shorter wavelengths,
and the peak shifts to the left.
Twice as hot a filament (in terms of its absolute temperature)
would be perfect: the peak would be at half its normal wavelength,
right where the eye is most sensitive. This would require a
temperature of about 5,700°C. That’s approximately the temperature
of the surface of the sun. (So the solar radiation peaks where the eye
is most sensitive. That is not a coincidence, of course. Why might
that be?) Unfortunately, when we heat a tungsten filament beyond
2,700°C we fairly soon run into its melting point, ~ 3,400°C. Even
Everyday Physics: Unusual insights into familiar things
103
approaching the melting point too closely reduces the lifetime of
the incandescent bulb dramatically. Filling the bulb with a halogen
gas helps a bit, but the melting point is a hard limit. And other metals
aren’t the solution: their melting points are even lower. So using
incandescent filaments is just not a good way to generate light.
For photographers, another disadvantage of incandescent
bulbs is that the light is not white. Fig. 1 shows that the emission in
the orange/red is two to three times stronger than in the blue, giving
a warm light, which makes it difficult to see the true colour of blue
objects. You may have noticed this while shopping for clothes in
artificial lighting: sometimes you have to step outside to be sure that
something really is blue, not black.
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Everyday Physics: Unusual insights into familiar things
29
LUMINOUS
IDEAS
FLUORESCENT
LIGHTS AND
LEDS
Lighting based on a glowing filament cannot be efficient, as
explained in the preceding chapter. What we’d like is something that
mimics sunlight more closely. The temperature of the surface of the
sun (where sunlight comes from) is so high that the molecules and
atoms get into a higher energy state – an “excited” state – by colliding
with each other. As the molecules fall back into their unexcited
(“ground”) state, they lose that extra energy largely in the form of
visible light – our sunlight.
Clever light sources emulate this by a combination of
electronics and choice of materials. In fluorescent tubes and
compact fluorescent lamps (CFLs), it’s achieved by a smart choice
of the filler gas. In Light Emitting Diodes (LEDs) it’s achieved by the
choice of the solid material that makes up the diode. The result is
that the input energy is converted directly into radiation and not
– as in the incandescent bulb – indirectly by heating the filament.
They therefore give more light for a given amount of energy – more
lumens per watt (lm/W).
Fig. 1 shows how much better these “new” lamps are. All of
them are more efficient than the traditional incandescent bulb, they
Everyday Physics: Unusual insights into familiar things
105
have a longer life, and there is no glowing filament (which slowly but
surely evaporates, which you notice by the blackening of the inside
surface of the glass).
Candle
Incandescent
bulb
Compact
fluorescent
lamp
LED lamp
Fluorescent
tube
Efficiency (lm/W)
0.15
12
50
80-100
100
Life span (hours)
10
1,000
> 5,000
~ 50,000
10,000
Fig. 1: Efficiency of
various light sources
compared.
These values potentially understate the efficiency of LEDs.
LEDs have the useful property that they can produce light in a single
colour: red, blue or green, which is useful if you want a specific
colour – for example for traffic signals or for a car’s brake lights. In
such cases it’s more efficient to produce the proper colour directly
instead of producing the entire spectrum of colours and filtering out
the unwanted ones with the attendant loss of useful (visible) energy.
If you want to make white light with LEDs, you can do it
by a clever combination of the three colours, or you can use an
intermediate step, where the light of the LED first irradiates a
fluorescent layer, as also happens in fluorescent tubes. And the LED
offers possibilities unthinkable for old-fashioned lamps: a specific
type of LED lamp is already available that can produce any colour
required, by combining a red, green and blue LED and regulating
the current in each LED separately to produce the desired colour.
If the colour of the lighting is not so important, the efficiency
can be extremely high. Sodium street lights, (which give a pure
orange light) deliver about 200 lm/W – about 1,300 times the
efficiency of an old-fashioned candle. What is the highest lighting
efficiency we could ever achieve? That would be when all the light is
radiated at the peak of the eye’s sensitivity curve (around 550 nm,
Fig. 1, Why are incandescent bulbs so inefficient? p101), producing a
yellow/green light, and (from a calculation we haven’t shown) the
efficiency would be 680 lm/W.
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30
WHY IS THE
SKY BLUE AND
THE SETTING
SUN RED?
When there are no clouds, the sky is blue. We take it for granted
because we’re so used to it, but why is it blue? Where does the blue
come from? Indeed, why is the sky bright at all, and not just black,
which is how astronauts in space see the sky? After all, light travels
in straight lines, so unless you’re looking directly at the sun or a star,
you should see no light.
Everyday Physics: Unusual insights into familiar things
107
We must conclude that the atmosphere bends the light rays
from their straight paths. It scatters the light, or part of it. Dust
particles can cause scattering: just think of the glowing dust particles
you sometimes see when you open the curtains a little in the
morning and the sun shines in. But scattering also happens in
completely clean air, which may seem a little strange, because air
is transparent.
How does this scattering come about? Air molecules consist of
charged sub-atomic particles – positive nuclei and negative electrons.
Because light waves consist of an alternating electromagnetic field,
the positive and negative charges are set in motion by the electric
component of the field. (This also happens in the antenna of a TV
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Everyday Physics: Unusual insights into familiar things
or phone.) The charges in an air molecule shift in the electric field:
the positives ones go one way, the negative ones the other way – the
molecules are “polarized” to the rhythm of the alternating field.
Because of the moving charges, the molecules now act as
transmitters themselves: they emit light at precisely the same
frequency (and therefore of the same colour) as the original
incoming light. However, the important difference is that the emitted
light travels in all directions. A small part of the original light is thus
removed from its straight path and scattered by the air molecules,
although that scattered light is very faint.
This explains why the sky is bright, and not black. But why is it
blue? That’s because some colours are scattered more than others.
The scattering is much stronger for violet and blue light than for
orange and red light (Fig. 1). The scattering intensity is inversely
proportional to the 4th power of the wavelength: a wavelength 2
times longer means scattering is 16 times less intense. The wavelength
of blue light is only half that of red, so its scattering is therefore
about 16 times stronger than red’s. This type of “Rayleigh scattering”
(named after the British physicist Lord Rayleigh) occurs as long as
the scattering particles are much smaller than the wavelength of
the light – and they are, because the particles here are air molecules.
(The light’s wavelength is about 500 nm and the diameter of an
oxygen molecule is about 0.33 nm – about 1,500 times smaller.)
Fig. 1: Air molecules
scatter different colours
by different amounts;
blue is scattered much
more than red.
In summary: the sun’s rays contain all the colours of the rainbow.
Most of the rays pass through the atmosphere in their original
direction, but some are scattered, and some of this scattered light
Everyday Physics: Unusual insights into familiar things
109
reaches your eyes. And because blue light is scattered more than
red, overall the sky seems blue. (The surprisingly simple experiment
on p111 illustrates this nicely.)
The same mechanism also explains why the setting sun is so
red. At low solar elevation, i.e. when the sun is low in the sky, the
layer of air that the sun rays have to pass through is very thick (line
CD in Fig. 2) so the scattering is very large. When you look at the
setting sun, the light that reaches your eyes comes in a straight line
from the sun with little or no scattering (because the scattered light
won’t reach you). Not much blue is left because it has been scattered
in all directions. The red part of the spectrum is scattered far less
and simply propagates straight to your eyes, which you perceive as
the beautiful red colour of the setting sun.
Fig. 2: Proportionally,
the Earth’s atmosphere
is thinner than the skin
on an apple. When the
sun is directly overhead,
its rays only have to pass
the short distance AB
through the atmosphere,
whereas at dusk they
have to travel the much
longer distance CD.
We’ve said that at low elevation the sun has to shine through a
much thicker layer of air than at high elevation, but this is true only
because the thickness of the atmosphere is small compared to the
diameter of the Earth (Fig. 2). If the atmosphere were very deep
compared to the diameter of the Earth, the sun’s elevation wouldn’t
make much difference. However, if in a thought experiment we
swept up the rarefied upper layers of the atmosphere into a single
layer that had normal sea-level air pressure throughout, we’d
get a layer about 8 km thick. So the “effective thickness” of the
atmosphere is about 8 km. Comparing this to the diameter of the
Earth of nearly 13,000 km, then proportionally the thickness of the
atmosphere around the Earth is less than the thickness of the skin
on an apple. This is why light has so much further to travel through
the atmosphere as the sun lowers, and the scattering becomes
very large.
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Everyday Physics: Unusual insights into familiar things
Home experiment: The blue sky and the red setting sun
You can demonstrate the selective scattering of the colours in a light beam with a
simple experiment. You need a glass with a flat bottom, a few drops of milk, a piece
of white card and an ordinary flashlight. (An incandescent flashlight is best because it
emits all the colours of the rainbow. How well an LED torch will work here depends on
its emitted colour spectrum; a single-colour LED torch won’t work.) The experiment
works best when the lamp has a beam that is more or less parallel and is narrower than
the base of the glass. (In this particular case the glass was supported by the (relatively
large) torch.)
First, pour water into the glass, and shine the torch through the glass from the bottom
up. The white light of the torch exits the glass at the top and virtually no light is scattered
sideways, and the beam striking the white card above the glass is white. (Fig. 3a). Next
add a few drops of milk, and stir (Fig. 3b). You have now made a medium that scatters
light much more strongly than air. It is no longer transparent but whitish (Fig. 3c). The
scattered light (as viewed from the side of the glass) is a little blue, even though it is lit
by ordinary white light, which is the principle of the blue sky we explained earlier. (In
Fig. 3c, the blue at the edge of the glass is just visible.)
a)
b)
c)
Fig. 3: The plain water lets the white light beam through (a). Add a few drops of milk to the
water (b) and now the transmitted beam only contains orange/red (c).
Everyday Physics: Unusual insights into familiar things
111
If your torch is wide enough, you can support the glass of water on the torch itself
(Fig. 3b). Otherwise, hold the torch in one hand and the glass in the other, and get
someone else to hold the card above the glass.
The convincing proof comes only when you look at the transmitted light. If the
flashlight has a strong, largely parallel beam and the room is dark, you will see
an orange-red spot on the ceiling. It illustrates the setting sun. Alternatively,
you can hold a white sheet of paper just above the glass to view the transmitted
light; it’s clearly redder than the scattered light although it comes from the
same light source. It shows the difference between the light that is scattered
and the light that is transmitted without scattering. (If the transmitted light is
too faint, there is too much milk in the water. A few drops are usually sufficient.)
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Everyday Physics: Unusual insights into familiar things
31
TWO KINDS OF
SMOKE FROM
THE SAME
CIGARETTE?
Curiously, if you look closely at a burning cigarette you see two kinds
of smoke. Blue smoke comes from the burning end, whereas the
smoke that comes out the filter end is white or grey.
The explanation is Rayleigh light scattering, as with the blue
sky. In fact, light scattering is the only reason you see smoke at all
because smoke itself gives no light. Smoke scatters incident light in
all directions. You see smoke best when a bright light, like the sun,
shines on it. The scattered light then contrasts nicely with the dark
background. This is seen most clearly in photos taken into the light.
Everyday Physics: Unusual insights into familiar things
113
What colour will the scattered light be? The reason the smoke
at the burning end of the cigarette is blue is the same as why the sky
is blue. The smoke particles are much smaller than the wavelength of
light – as are the air molecules that produce the blue sky. The degree
of scattering is not the same for all colours (p107): blue dominates in
the scattered light.
Even a small puff of cigarette smoke scatters the light a lot,
because the smoke particles consist of quite a large number of
molecules in a clump, and these scatter disproportionally more light
than a single molecule. (The intensity increases with the square of
the number of particles in the clump.) So even a small quantity of
smoke produces enough scattering to be clearly visible.
Light scattering and photography
You can use your understanding of light scattering to take better photos in smog, or when
there’s a blue haze surrounding mountains in the distance. Use a lens filter that selectively
blocks the short wavelengths, which scatter the most (blue, violet and particularly UV).
An extreme example of this is infrared photography, which only makes use of long
wavelengths, resulting in minimal scattering.
What about the smoke at the filter end of the cigarette? Why
isn’t that blue? The explanation above is valid only for particles
that are small with respect to the wavelength (and at the burning
end of the cigarette, the particles are small). However, as the smoke
travels through the cigarette towards the filter, the particles clump
together and can grow as large as the wavelength of light or even
larger. For such particles all light is scattered more or less equally, or
in other words, the intensity of the scattered light is approximately
independent of the wavelength (whereas for small particles it’s
inversely proportional to the 4th power of the wavelength). So if the
incident light is white, the scattered light is also white – or maybe a
bit grey because the intensity is decreased.
For the same reason clouds are white or grey. They consist of
small water droplets that scatter light just as the big smoke particles
do: white light in, white light out.
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Everyday Physics: Unusual insights into familiar things
32
SWIMMING
POOLS ARE
DEEPER THAN
THEY LOOK
A problem, especially for children, is that swimming pools are
deeper than they seem to be. You stand at the edge of the swimming
pool, look across to the other side, and think: “It’s not deep, I can
jump in safely.” But you are wrong: the water is deep – you’ve been
tricked by the refraction of the light. You’re looking at point D on
Everyday Physics: Unusual insights into familiar things
115
Fig. 1: Why a swimming
pool looks shallow: the
bottom appears to be
located at the extension
(B) of the ray that enters
your eye, and not where
it really is (D).
the bottom of the pool. Light leaves D in all directions, and one
ray reaches your eye E. But this ray is not straight: it bends at the
transition X from water to air, because light rays always follow the
fastest path. By making that bend at X and entering the air sooner
than a straight line from D to E, the ray profits from the greater speed
in air and travels a shorter distance (line DX) in the water where the
speed is lower. (If the light travelled in a straight line from D to E, it
would spend a much longer distance DY in the water.) This is exactly
similar to what the hurrying cycle courier does (see Minimizing your
journey time p75).
Fig. 2: Why a
swimming pool looks
shallower to a child
than to an adult.
Fig. 2 explains why the problem is more serious for a child
than an adult. First, the child has less experience, and may not know
that the apparent depth of water is misleading. But physics also
plays a part. The ray seen by the child (Fig. 2, red ray) comes at a
glancing angle over the water. In contrast, an adult looks at the pool
bottom more nearly perpendicularly, and sees a different ray (Fig. 2,
green ray). For the adult the pool looks somewhat deeper, although
still not as deep as it really is. (You can see this for yourself: look into
a pool when standing; now squat down – the bottom of the pool will
appear to rise!).
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Everyday Physics: Unusual insights into familiar things
If you look down perpendicularly to a point on the bottom of
the pool directly beneath you, will you see no distortion at all? You
might think so because the light ray entering the water precisely
perpendicularly doesn’t bend at all (after all, in which direction
would the bend have to go?).
Everyday Physics: Unusual insights into familiar things
117
But now the fact that you have two eyes comes into play.
Each eye looks at the bottom from a different position. This means
that the rays you use to see the bottom are slightly bent (Fig. 3).
Again the bottom looks closer than it is, although the difference is
small: the real depth is 4⁄3 times the apparent depth. (This 4⁄3 factor
arises because the ratio of the speeds of light in air and in water is
4⁄3.) Granted, the difference is only 30%, but it can make a world of
difference to a child.
Fig. 3: When you
look straight down
into water, the actual
depth is 4/3 times the
apparent depth.
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Everyday Physics: Unusual insights into familiar things
33
SUNLIGHT
FILTERING
THROUGH THE
LEAVES OF
TREES
When the sun is high in the sky and shines through the leaves in a
forest, you get a playful pattern of small and large light spots on the
ground. But there is something special about those little spots: they
are always round. How come? Do the leaves form circular holes?
Everyday Physics: Unusual insights into familiar things
119
That would be unlikely, and indeed isn’t true. The roundness
of the light spots has nothing whatsoever to do with the shape of
the holes that the light passed through. The spots are in fact small
pictures of the sun. Each hole between the leaves acts as a “camera
obscura” (see box). The only thing the holes have in common is that
they are small. Fig. 1 shows how it works. Consider a single point S on
the sun. That point emits light rays in all directions. A single ray just
passes through the hole between the leaves, and hits the ground at a
specific point G, whereas all the other rays are blocked by the foliage.
This happens with each point on the sun: each produces a bright
point on the ground. The result is a picture of the sun, although it’s a
mirror image and upside down.
Fig. 1: An image of the
sun is projected on to
the ground through a
small gap in the leaves
of the trees.
You can use the size of the spots to determine the height of the
trees (or more precisely, the height of the hole between the leaves).
For this you need to know the angular diameter of the sun (Fig. 2)
which is almost exactly half a degree, which is also the angular
diameter of a 1 cm coin at a distance of 100 cm = 1 metre. So, if the
spot on the ground is 15 cm in diameter, the hole the light passed
through is 15 cm × 100 = 15 metres above the ground. It doesn’t
matter much if the hole isn’t perfectly round, as long as it is small
compared to the spot on the ground, and anyway, who can complain
about a camera that cheap?
You can use the small-gap idea as a simple and safe way to
observe a solar eclipse (Fig. 3): when the moon starts to obscure
the sun, you will see many sun crescents on the ground! An easier
way, instead of having to find a convenient forest, is to use a piece of
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Everyday Physics: Unusual insights into familiar things
cardboard with a pinhole in the middle (Fig. 4), and project the sun’s
image on to a piece of white paper on the ground. You can even use
an unmodified kitchen colander to project the sun’s image on to the
ground – try it!
Fig. 2: The angular
diameter of the sun, and
of a 1 cm coin 1 m away
from your eye.
Fig. 3: Images of
partial solar eclipse
projected through the
leaves of a tree.
Credit: commons.
wikimedia.org/wiki/
User:Ellywa Creative
Commons AttributionShare Alike 3.0 Unported.
Unmodified.
Everyday Physics: Unusual insights into familiar things
121
Fig. 4: Using a piece
of cardboard with a
pinhole in the middle to
project the sun’s image.
Camera obscura
“Camera obscura” is Latin for “dark chamber”. The name refers to the phenomenon by
which an image is produced in a dark room from a bright scene or object outside, by
using a small hole in the wall instead of a lens. The “pinhole image”, as it is often called,
is inverted – upside down and right to left (Fig. 5). Leonardo da Vinci wrote about the
camera obscura back in the 16th century. Many tourist destinations have camera obscuras
open to the public (see Resources appendix).
By the way, a normal camera also produces an inverted image. However, the camera lens
is much bigger than a pinhole, and it allows many more light rays to contribute to the
image without the image getting blurred, thus making it much brighter. Nowadays many
camera obscuras use a lens instead of pinhole, to produce a better image.
Fig. 5: How a camera
obscura creates an
image. The pinhole is
at C.
Credit: commons.
wikimedia.org/wiki/
File:001_a01_camera_
obscura_abrazolas.jpg
You can easily make a small camera obscura yourself, with a shoebox. Make a pinhole
in the middle of one of the short sides. In the opposite short side cut a rectangular hole,
and over it stick a piece of tracing paper or kitchen parchment, and put the lid on the
box. That’s it! Indoors at night you can get a reasonable image of a switched-on light bulb.
During the day, if you point your “camera” out the window, you can see the sun’s image
– but don’t look at the sun directly! – and you can probably see (upside-down) images of
buildings too. (If your image is too dim for comfort, try making the pinhole bigger.)
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Everyday Physics: Unusual insights into familiar things
34
HOW SHARPLY
CAN YOU SEE?
Why do you see things? Because the light coming from them reaches
your eyes, of course. Some objects – lamps, candles, the sun – emit
light of their own, but you see most things because they reflect light
from an external source.
How do you see an image of an object? The lens of your
eye casts an image of the object on your retina, where the light is
detected by the photoreceptors (the rods and cones that convert the
light falling on them into nerve impulses that are then processed by
your brain).
The image needs to be sharp, i.e. in focus, independent of the
distance from the object to your eye, but the position where the
image is formed behind the lens depends on how far the object is in
front of the lens (Fig. 1).
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123
Fig. 1: The distance
between the lens and the
image formed depends
on how close the object
being viewed is to the
front of the lens.
Credit: teddy bear:
oNline Web Fonts – www.
onlinewebfonts.com
In a photographic camera, you adjust the focus by changing
the distance between the lens and the film (or digital image sensor).
You can’t do that with your eye, because the diameter of the eyeball
would continually be growing larger or smaller. So instead, your
eye adjusts the power of the lens in your eye, a process called
accommodation. When the muscle is in its relaxed state, the lens is
stretched and becomes less curved (Fig. 2 top), suitable for viewing
distant objects. In contrast, if the object comes closer, your eye
dynamically changes the shape of the lens using the ciliary muscle
that surrounds the lens. When this muscle contracts around the
lens, it exerts less tension on the lens, and the lens, which is elastic,
reverts to its natural convex (more powerful) shape (Fig. 2 bottom).
This is what you need to view something that is close-up. Contracting
the muscle becomes more difficult as you age because the eye
lens becomes less elastic, which is why many older people need
reading glasses.
Fig. 2: The muscle
around the lens of
your eye controls how
powerful the lens is,
so the image is always
focused on your retina.
(Not to scale.)
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Everyday Physics: Unusual insights into familiar things
Who needs reading glasses?
Just as with a camera, you want your eyes to provide a sharp image of an object, whether
it’s close or far away. That doesn’t happen automatically: the power of the lens has to
change according to the distance of the object. For both a camera and your eye, the
formula specifying the lens we need is:
1/o + 1/i = 1/f
where o is the distance from the object to the lens, i the distance from the lens to the
image (or to the retina, in the case of the eye) and f the focal length (Fig. 2). The figure 1/f
is called the strength of the lens, and its unit is dioptres, or inverse metres (m-1).
Fig. 3: How the
object and image
distances (o, i) and
the focal length (f) of
the lens are related.
Because the distance, o, of the object varies, to get a sharp image, either i or f must vary
accordingly. In a camera we change i by moving the lens closer to (Fig. 1 top) or further
from (Fig. 1 bottom) the film or digital image sensor. In contrast, in your eye, i (the diameter
of your eyeball) remains constant, so your eye must accommodate by changing the
focal length f, which it does by contracting or relaxing the muscles controlling the shape
of the lens.
Young people can see objects sharply between infinity and about 0.2 metres. At infinity,
o = ∞, so 1/o = 0, and our formula becomes 1/i = 1/f . At 0.2 m 1/o = 5 m-1, so our formula is
5 + 1/i = 1/f. As i remains constant, only f can change, so that 1/f, the lens strength, must vary
by 5 m-1; in other words, the accommodation range is 5 dioptres.
As you get older, your accommodation range gradually reduces to practically zero, and
adjustments for distance must be made artificially – with glasses. For an eye that sees
sharply at infinity, with reading glasses of 3 dioptres the region in focus will be ⅓ m =
33 cm away, even when the eye does not accommodate at all, and 33 cm is a comfortable
reading distance.
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125
If you can’t see an object sharply enough, it may help to squint.
This makes your pupil narrower, at least in one direction, and blocks
most of the light rays that come through the outer part of the lens.
As a result, the rays entering through the central part of the lens
dominate the image formation, which reduces the effect of any
defects in the shape of the lens near the edges. Moreover, a narrower
light beam sharpens a poorly focused image – which photographers
will recognize is the same effect as increasing the depth of field by
reducing the aperture (reducing the size of the opening that lets the
light into the camera, by narrowing the camera’s diaphragm).
Can you make the eye exceed its performance by making the
pupil artificially smaller than it is by nature? To some extent you can,
by reducing the aperture of your eye to look through very small holes.
Because this reduces the amount of light entering your eye from the
object, the object might need to be more brightly illuminated for
you to see it properly. (Occasionally this can be useful. If you want to
read the wattage printed on an incandescent bulb without turning it
off, you can’t normally do it because the light is too bright. However,
if you clench your fist so there’s only a tiny hole through it, and view
the bulb through this, with one eye, you can read the writing on the
bulb clearly.)
But you can go too far with this approach, and run into
problems with the wave character of light. If you make the pupil
so narrow that it is comparable to the light’s wavelength, the light
waves spread and the image blurs. On the retina each point of the
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Everyday Physics: Unusual insights into familiar things
object is no longer represented as a sharp point but as a broader
spot whose diameter increases as the pupil diameter decreases. (See
the experiment below).
Home experiment: Light waves fanning out
Light is a wave phenomenon, and this limits the sharpness of an image. You can easily
demonstrate this with a very sharp needle and a piece of aluminium foil.
Pierce the aluminium foil to make a tiny hole. (The smaller the better, so use the tip of
a fine darning needle to pierce the foil.) Look through the hole at text in a newspaper.
Even in perfect lighting conditions the letters will appear less sharp than when you
look at them with the naked eye. This is because of the wave character of light.
You can demonstrate the same effect in a different way with a laser pointer. Shine the
laser pointer through the tiny hole in the foil, on to a white piece of paper or a white
wall – the further away the better (Fig. 4).
WARNING! For safety, DON’T look directly into
the laser or even into the reflected laser beam.
Fig. 4: Demonstrating
the wave nature of light
with a laser pointer and
aluminium foil.
You might expect that the narrow laser beam will become
narrower still because it’s passing through a tiny hole
(Fig. 5), but that’s not what happens: in fact the light
spot on the wall is wider than the beam of the laser
pointer, and it’s surrounded by faint concentric rings. The
larger diameter of the central light spot indicates that a
diaphragm that is too small doesn’t give a sharper image,
but in fact the reverse.
Everyday Physics: Unusual insights into familiar things
127
Fig. 5: What
you might expect
to happen with a
laser beam passing
through a pinhole.
The radius of the central spot on the wall turns out to be 1.22 times the ratio between
the distance to the wall and the diameter of the hole, multiplied by the wavelength
of the light. For example, if the distance to the wall is 2 m, the diameter of the hole is
0.1 mm, and the wavelength is 650 nm (red), the central spot is 1.22 × (2,000 ÷ 0.1) ×
(650 × 10-9) m = 1.22 × 13 mm ≈ 16 mm in diameter.
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Everyday Physics: Unusual insights into familiar things
35
YOUR EYE IS
MORE SENSITIVE
THAN A CAMERA
As experienced photographers know, natural daylight is much more
intense than artificial light. If you’re taking photos indoors using a
traditional camera with a film, as evening approaches you need either
a flash to increase the brightness, or a tripod to keep the camera still
while you expose the photograph for longer in order to make the
most of the little light there is. But in the same circumstances, you
can easily read a newspaper! Your eye adapts effortlessly, and you
can still see quite well even when it gets darker still; if there’s a fairly
bright moon, you can find your way comfortably.
Everyday Physics: Unusual insights into familiar things
129
So it’s clear that your eyes have a wonderfully large capacity
to adapt, but exactly how large? You can get a good approximation
of the answer if you know how much fainter moonlight is than
sunlight, and assume that full sunlight is the brightest light your
eye can handle, and dim moonlight the faintest. You can calculate
the brightness of moonlight if you know how big the moon is, how
far it is from the Earth and how well it reflects the sunlight, i.e. how
“white” it is. When you do the sums (see box), it turns out that full
moonlight is about one million times fainter than sunlight.
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Everyday Physics: Unusual insights into familiar things
To see what that means, compare how a traditional photo
camera would handle things. With a given lens, to take two photos
with light intensities that differ by a factor of a million, the exposure
in one case has to be a million times longer than in the other. (You
can reduce that a bit by changing the aperture, but that only gives
you a factor of 128 at most, with a change of 7 f-stops, because every
stop allows twice as much light into the camera as the previous
stop). For ease of calculation, assume changing the f-stop reduces
the exposure by a factor of 100 instead of 128. Accommodating the
remainder – a factor of 1,000,000 / 100 = 10,000 – would have to be
done with exposure time; increasing this by a factor of 10,000 gives
you an exposure of maybe 10-50 seconds. The only way a camera
can manage such a long exposure is with a tripod. In contrast, your
eye doesn’t need a tripod or a 10-second exposure as you walk in the
moonlight: you can see just as quickly as normal. Incidentally, this
calculation also illustrates how amazing our modern digital camera
or smartphone is.
How does your eye achieve the factor of 1,000,000? As with
the camera, a small contribution comes from changing the aperture
– the pupil diameter. It varies at most from 2 to 8 mm in diameter,
a factor of 4, which means a factor of 42 = 16 in area. The remainder
(1,000,000 / 16 = 62,500) is taken care of by the photoreceptors –
the rods and cones – in your retina. These complement each other
beautifully with changing light intensity. When there is a lot of light,
the cones do most of the work (and they are also responsible for
colour vision). When there is little light, the much more sensitive
rods smoothly take over, although they aren’t very sensitive to
colour, which is why things appear grey at night. Clever instrument,
your eye.
Comparing the brightness of the sun and moon
To compare the relative brightness of the full moon and full sun you need to know the
size of the moon, its distance from the Earth and its reflection coefficient (“whiteness” or
“albedo”). Let’s say the solar intensity – quantity of light per square metre – at the moon is
I. (We’ll assume that I is the same at the Earth, because the Earth is approximately as far
from the sun as the moon is.) Then the moon, with radius R, captures πR2 × I. If the moon
is perfectly white (an albedo of 1.0, i.e. 100%) this light is reflected in all directions. Let us
Everyday Physics: Unusual insights into familiar things
131
assume that this is equally strong in all directions, in a “half-sphere” directed towards the
Earth. If we call the Earth–moon distance a, then the moonlight at the Earth is spread over
the surface of a half-sphere with radius a. The surface area of that half-sphere is 2πa2. Per
square metre the amount of light is:
(πR2I)/(2πa2) = ½ (R/a)2 I
Compared to the direct solar intensity I it is therefore attenuated by a factor of ½ (R/a)2.
With the values R = 1,700 km and a = 380,000 km this yields 1.0 × 10-5, i.e. one part in
100,000. But that’s not the whole story. The moon is not perfectly white, its albedo is only
0.1. That makes the intensity of the full moon one millionth of the full sun.
By the way: an albedo of 0.1 means that the moon is a very poor light reflector: you
probably knew that without realizing it. If you see the moon during daytime it is always
much less bright than a white wall, even though both are illuminated by the same sunlight
and both are approximately the same distance from the sun. Furthermore, the albedo
depends on the phase of the moon: at full moon the albedo is largest, while at less than full
moon the albedo rapidly diminishes. That is primarily caused by the shadowing effects of
the rough moonscape and its grainy soil. It is only at full moon that the shadows play no
role. (Strictly, you never see a really full moon, because this happens only when the sun,
the Earth and the moon are exactly lined up, i.e. when there is a lunar eclipse!)
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36
PUDDLES ON
A DRY ROAD
Fantastic as they are, your eyes can sometimes deceive you and
make you see things that aren’t there. One example is the mirage
that people often see in the summer. You are driving on the highway,
the weather is beautiful, the sun is shining and you see what looks
like puddles in front of you on the road. But as you get closer, the
puddles abruptly disappear. Why?
What happens is this: you’re looking at the road, but what
you’re actually seeing is the blue sky, as if a mirror were lying on the
ground. The light rays that reach your eye are not coming from the
road, but from above it. They have been deflected by the air above
the hot road (Fig. 1).
Fig. 1: The light rays above the warm road bend, so while you
think you are seeing the road surface ahead, you’re in fact seeing the
sky above it.
Everyday Physics: Unusual insights into familiar things
133
The reason the light bends is that the speed of light close to
the warm road is just a bit larger than higher up where it’s cooler
(Fig. 2). Why does this temperature gradient bend the light rays?
Pretend that each dotted line in Fig. 2 is a row of people, seen from
above, marching forward with their arms linked. If the people on the
“warm” side take larger steps than those on the “cold” side (because
the speed of the “warm” people is greater), the whole row will bend
towards the cold side. In other words, the faster side is automatically
on the outside of the curve, just as you saw in Waves at the beach
(p13). Light rays behave the same way.
Fig. 2: A light ray
originating from the
sky skims over the
warm road surface
from left to right and
is deflected upwards.
But why is the speed of light greater near the hot road than
higher up? Air expands when heated, and gets thinner as a result.
The thinner the air, the closer it is to a vacuum so the greater the
speed of light, because the speed of light is greatest in a vacuum.
Granted, the differences are minuscule, but over a long distance the
beam is deflected by a large-enough angle to produce this effect.
As you get closer to the “puddle”, you are looking too steeply
downward. The deflection angle of the distant rays is too small to
make it into your eye, so the “puddles” vanish, and you see the actual
surface of the road (Fig. 3).
Fig. 3: When
you look steeply
downwards at the
surface, you see the
surface itself rather
than deflected rays
from further away.
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37
SEEING THE SUN
AFTER SUNSET
You’re standing on the beach. It has been a lovely day and the sun is
about to set. You watch as it just touches the horizon. But that’s not
what’s happening: in fact, the sun set a few minutes ago! How can
that be?
Everyday Physics: Unusual insights into familiar things
135
Here’s how. Light rays can be bent when the light velocity
varies in different parts of the atmosphere, as you already saw in
the previous chapter. There, the velocity differences were caused by
temperature differences which in turn caused density differences.
Density differences also occur in the atmosphere even without
temperature differences, because the air pressure gradually
decreases with height.
Isn’t that a very small effect? Yes, it is, but even though the rays
are bent only slightly (as with the mirage), if you look over a large
enough distance the deflection is noticeable. In Fig. 1, the thinnest
air – and therefore the highest light speed, and therefore the outer
side of the bend – is on the upper side, and the rays bend downward
around the horizon. This is schematically shown in the figure below.
Your eye assumes that the light is travelling in straight lines, and
therefore that it comes along the dotted lines, so you see the sun
higher than it is.
Fig. 1: The differences in air
pressure cause the light rays coming
from the sun below the horizon (A)
to curve, so the sun appears to you
to be above the horizon.
Moreover the sun seems flattened. Its height is only 4⁄5 of
its width. That is because the lowest rays are bent the most since
they travel the longest distance through the atmosphere. You can
understand this by looking at what happens when the sun is directly
overhead. There is no bending of any kind, because the rays from
all points on the sun travel the same distance in the atmosphere. As
the sun lowers, the bending becomes stronger and stronger. The end
result is that your eyes “lift” the sun a bit. You see the sun higher than
it is. And since you see the lower edge raised more than the upper
edge, the sun appears flattened.
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Everyday Physics: Unusual insights into familiar things
ng
)
u
In practice, the behaviour can be more complicated, because
this “air pressure” effect and the mirage effect (p133) can occur
simultaneously. When both effects combine, the setting sun can take
on the most bizarre forms.
A special case of this phenomenon is the “green flash” or
the “green ray”, sometimes visible for a few seconds at the very
last moment of sunset, if the air at the horizon is very clear. What
Everyday Physics: Unusual insights into familiar things
137
happens is that the degree of the “lifting” effect above is different
for different colours. Just as in a prism, the red end of the spectrum
refracts the least (Fig. 2), so the red disappears first. You might
expect that violet remains the longest, but violet light is scattered
so much that there is hardly any left (p107). The middle – the green
part – remains and delivers a flashy sunset.
Fig. 2: When light
passes through a prism,
the red is refracted the
least, and the violet
refracted the most.
Credit: Attribution 2.0
Generic (CC BY 2.0),
Siyavula Education
(www.flickr.com/
photos/121935927@
N06/13580411493)
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Everyday Physics: Unusual insights into familiar things
38
TRANSPARENT
WINDOWPANES
& OPAQUE
LACE CURTAINS
Sometimes you can see through a window from one side and not the
other, but at other times it seems to work the other way round. Lace
curtains on a window are a good example. In the daytime you can
see perfectly well from the inside to the outside, but not vice versa.
Even a bare window shows the same effect to some extent: it is much
more difficult to look in than to look out. Take, for example, a shop
window: from outside it sometimes acts like a nearly perfect mirror,
but you can see out from the shop without any problem..
Everyday Physics: Unusual insights into familiar things
139
Is that how light works? Does it mean the light can’t pass
through the same medium depending on the time of day? Or does
the transmission depend on the direction? Of course not: it is a
simple question of differences in brightness. Your eyes have an
enormous range for brightness (see The eye is more sensitive than
a camera p129). They adapt extremely well, partly by changing the
size of the pupil to regulate the amount of incoming light. But if
there is a bright image close to a faint one, the bright one “wins” –
it takes precedence over the faint one. You can see this if you view
someone’s face with the sun directly behind them: your friend’s face
seems nearly black, although a lot of light from it reaches your eye.
It is simply “outshone” by the much more intense light from the
background. It’s the same for lace curtain and window – a question
of relative brightness.
First consider a bare window. Suppose you are outside looking
in. If the outside light falls perpendicularly on to the windowpane,
about 4% is reflected on the front side and another 4% at the back
side. So a total of 8% is reflected back to you and the remaining 92%
passes through the window (Fig. 1). Now an 8% reflection may not
seem like much, but it’s quite a lot if it is 8% of a bright sky or of a sunlit street scene. Compare this with the background that’s competing
with it. Inside the shop is very much darker than outside (although
that may not be obvious because your eye adapts when you are in
the dimmer indoor light). In reality the difference can easily be a
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Everyday Physics: Unusual insights into familiar things
factor 20 to 100. You notice that when taking pictures, for example:
as soon as you go indoors even in the daytime, you may have to use
the flash.
Fig. 1: For falling light
that is more or less
perpendicular to the
windowpane, about 8%
of the incident light is
reflected and the other
92% is transmitted to the
other side.
So if you look through a window from the outside to the
inside, you see two things simultaneously: (1) the inside scene,
whose light travels – practically unattenuated – to the outside; (2) the
reflection of the street scene with a brightness of about 8% of the
real outside brightness. On a sunny day, the outside brightness can
be 50,000 or 100,000 lux, whereas even a well-lit interior might be
about 500 lux. Because of this large brightness difference, the 8% of
the bright outside scene can be much stronger than the 92% of the
inside scene, so you see the inside only dimly (Fig. 2). If you really
want to see in, you have to bring your nose to the windowpane and
shield the outside reflections with your hands: then the inside light
can easily “win”.
Fig. 2: When you look
in a shop window, the
8% of the bright outside
scene can be much
stronger than the 92%
of the inside scene,
so you see the inside
only dimly. (In this
example, the outside
reflection is nearly ten
times brighter than
the transmission from
inside – 4,000 lux versus
460 lux.)
Everyday Physics: Unusual insights into familiar things
141
This also explains how you can use a shop window as a mirror.
If you are well lit because the street is bright, your reflected image is
nice and bright. Since the light that comes from inside is faint, the
display in the shop window is dark and doesn’t interfere with your
reflected image.
Now let’s go back to the lace curtain. Why can’t you see in
during the day if there is a white lace curtain on the window? The
lace curtain increases the effect of the reflection of outside light.
Suppose that the curtain fabric is 50% holes; the other 50% is white
thread that reflects light. Then about 50% of the outside intensity
returns from the curtain to your eyes – much more than the 8% of
the bare window. The light that passes from the inside to the outside
is negligible compared to it, especially because half of it is blocked
by the lace curtain. You therefore cannot see inside from the outside.
In contrast, seeing out from inside is no problem. Even with the
brightness of the outside scene halved by the curtain, the other half
is more than bright enough for you to see it well. You do however see
the lace curtain as a kind of white haze on the outside scene, vague
because you focus your eyes on the more distant outside image. You
could avoid the haze by using black lace curtains. But that would be
much less effective in preventing the outside world looking in (and
not to many people’s taste).
And what about curtains at night, when the lights are on in the
house and the outside is dark? As before you see well from dark to
light, but now that is from outside to inside, and as before you see
poorly from light to dark (from inside to outside). Clever idea, really,
that white lace curtain.
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39
SEEING CLEARLY
UNDERWATER
If you swim underwater wearing diving goggles, and the water is
clean, everything is nicely in focus; without the goggles, everything
is blurred – you can’t focus properly. So the goggles make the
difference. But how, since the goggles don’t have lenses – just plain
flat glass?
Everyday Physics: Unusual insights into familiar things
143
The secret is in the convex curvature of the cornea (front side)
of your eye, because both it and the eye lens proper (Fig. 1) play a
part in forming the image in your eye. In normal viewing – in air – the
curvature of the cornea accounts for about two-thirds of the total
power of your eye’s powerful lens, i.e. more than the lens itself.
Fig. 1: Both the cornea
(front side) of your eye
and the lens inside
play a part in forming
the image in your eye.
The bending of the ray as it crosses the air/water (i.e. air/eye)
boundary is just the same as you saw with the light ray bending at
the air/water boundary of the swimming pool (see Swimming pools
are deeper than they look p115) as shown in Fig. 2a. The convex
curvature of the front side of your eye is therefore indispensable
to seeing sharply, but when your eye is in water this doesn’t work.
Optically speaking, your eyeball is like water – so optically, it’s just
“water in water” and doesn’t affect the passage of the light ray. In
other words, as the light ray crosses the water/eye boundary, it
passes practically undeviated into the eye (Fig. 2b).
Fig. 2: In air, incoming light rays are
refracted at your cornea, because they
are crossing an air/water boundary (a).
(The angle of refraction is exaggerated for
illustration purposes.) However, when your
eye is in water (b), the incoming light rays
pass into the eye without any deflection.
You have lost the important focusing effect of the eye’s convex
front side, so that the combined effect of cornea plus lens gives a
focused image in air (Fig. 3a) but not in water (Fig. 3b).
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Everyday Physics: Unusual insights into familiar things
Fig. 3: In air (a) the combined power of
cornea and lens focus the image sharply
on your retina. However, in water (b) the
focus is some distance behind the retina
so the image is blurred.
With goggles the eye is in air again, and the situation of
Fig. 3a is restored. However this introduces another transition, at the
water-glass-air boundary of the goggles. What effect does that have?
It doesn’t affect focusing at all because the surfaces aren’t curved but
simply flat (planar) and therefore have no lens effect. However, it
does affect your vision in another way, as we see below.
Why sharks can easily
surprise divers
“Diver attacked by shark” is a common-enough newspaper headline.
Why can a diver be so easily surprised by a shark? The reason is that
when you swim underwater you have a much smaller field of view
than above water. As you look around, you see much less than you’re
used to. If, for example, you want to see what’s happening behind you,
you have to turn your head much further than you think necessary.
The cause is the refraction of light. Under normal
circumstances, i.e. in air, your field of view is extra large, because the
light rays refract towards the centre of the eye, as they enter the eye
(Fig. 4a). Underwater that changes: the light rays transmit practically
straight through the water-eyeball boundary (Fig. 4b) and your field
of view is dramatically restricted. Goggles do not change it that
much – the water/air and air/eye transitions compensate for one
another; while close up the field of view is wider, as shown in Fig. 4c,
at further distances from the eye the field of vision is only slightly
bigger than in water, as a comparison of Fig. 5a and Fig. 5b shows.
Everyday Physics: Unusual insights into familiar things
Fig. 4: In each of the
three parts of the
diagram the size (XY) of
the image on the retina
is the same. In air, you
have a large field of view
(a). In water without
goggles (b), you have a
narrow field of view, so
sharks can easily sneak
up on you. With goggles
(c) your close-up field of
view is slightly wider, but
at further distances it is
almost as narrow as in
water without goggles,
see Fig. 5.
145
Fig. 5: At greater
distances, the field of
view without goggles (a)
is only slightly smaller
than with goggles (b).
The moral of this story is, if you swim underwater – especially
if you have an air tank on your back that restricts your vision even
further – be aware that you are not seeing everything around you –
you have blind spots. To be sure nothing swims up behind you, turn
your head as far as you can to either side, and avoid being surprised!
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Everyday Physics: Unusual insights into familiar things
40
WHAT MAKES
RAINBOWS?
The rainbow has caught people’s imagination throughout the ages. It
is beautifully pure and colourful but also mystical and elusive. And it
literally is elusive – you can neither grab the rainbow nor fly to it. Can
you walk under the rainbow? Walking under your “own” rainbow
(i.e. the rainbow that you are looking at) is no more possible than
jumping over your own shadow. But can you walk under someone
else’s rainbow?
To answer that question we must see what makes a rainbow:
►
You need the sun (or some other strong light source) and
rain (or at least water droplets – from rainfall, or a lawn
sprinkler, or a waterfall).
►
Whenever you see a real rainbow, it is with the sun behind
you and it’s either raining now or it was recently.
►
You never see a rainbow in the middle of a summer’s day
when the sun is high in the sky. The lower the sun, the
more of the bow you see. When the sun is just above the
horizon, you can sometimes see half a circle. And if you are
high enough above ground – in an aeroplane for example
– you can with a bit of luck see a complete circle. (You can
sometimes see a complete circle with a garden sprinkler
rainbow too.)
►
Often there is a second fainter bow with a reversed colour
spectrum outside the “main” bow: while the main bow
has blue below and red above, the secondary bow has the
opposite (Fig. 1).
Everyday Physics: Unusual insights into familiar things
147
►
Fig. 1: A secondary
rainbow above the
“main” rainbow, with
the order of colours
reversed.
Credit: commons.
wikimedia.org/wiki/
User:Alexis_Dworsky,
Creative Commons
Attribution 2.0 Germany
148
A careful observer will note that the sky above the rainbow
is darker than below it. All these features of the rainbow are
explained when we see how a rainbow arises.
The rainbow displays the full spectrum of colours, so it cannot
have been produced by means of reflection only. This is because
for reflection, the angle of incidence is always equal to the angle
of reflection, so it’s the same for all colours; therefore, reflection
can’t separate white light into its individual components. It must
be refraction that’s decomposing the white sunlight, just as with
refraction by a prism. As we will see below, the rainbow arises from
light rays that are refracted in the raindrops and leave the droplets
at a specific angle. We could call that angle the “rainbow angle”: the
angle between the rainbow itself and its imaginary centre, which is
exactly opposite the sun (and hence is under the horizon for a real
rainbow). That angle is about 42° for the main bow.
Everyday Physics: Unusual insights into familiar things
The sun is behind you, so light must be reaching the
raindrops in front of you, and must be refracted (to decompose it
into colours) and reflected (to travel back to you). Let’s see if we
can work out the path of a ray through a raindrop. Fig. 2 shows a
simple suitable path. (By the way, the greater part of the light exits
at the back side, shown by the dotted line (CF in Fig. 2), but that is
not important here.) Because the droplet is in principle spherical,
this sketch is not only for the horizontal or vertical plane but for
all planes through the centre of the droplet. (Different colours
of the rainbow are seen at slightly different angles. This means
that each colour originates from those droplets that produce just
the right refraction for that specific colour. We cover this in more
detail below.)
Tracing or calculating the exact path of the rays is laborious.
It is much easier to measure the angles with an experiment using a
glass of water as a model of a water droplet.
Fig. 2: A ray of sunlight
(A) comes from the sun,
enters the raindrop,
and is refracted (B),
reflected (C), refracted
again (D), and travels
back again (E).
Home experiment: The rainbow angle
With a laser pointer you can easily investigate the path of a “sunbeam” through a “rain
droplet”. Use the laser pointer to imitate the sun rays, and for the raindrop you can
use a cylindrical glass full of water (because optically water and glass are not very
different). From the top this looks like the cross-section of the droplet in Fig. 2. Ideally
choose a glass with thin walls, and at least 8-10 cm diameter. (Don’t use a wine glass,
because its curvature works as a lens and spreads the laser beam.)
Everyday Physics: Unusual insights into familiar things
149
Put a piece of tape over the on/off button of the laser pointer so that it stays on
constantly, and raise it about 2 cm off the table (on a book or something similar). Place
the pointer and the glass so that there is a white wall behind the laser pointer (Fig. 3).
Start with the laser beam just missing the side of the glass (Fig. 4a). Now move the glass
slowly to the left, into the beam (Fig. 4b). You will see a light spot on the wall, which
originates from the reflection of the laser beam from the back of the glass. Slowly move
the glass through the beam and observe the spot on the wall. It keeps moving to one
side until it reaches a maximum deflection; then it comes back towards the centre
again. That maximum corresponds to the rainbow angle, as we discuss below.
Fig. 3: The setup for measuring the rainbow angle using a laser pointer and a glass of
water. On the top right is the orange spot on the back side of the glass where the light beam
reflects against the rear side of the “droplet”. On the bottom left, on the wall, you can see
the same beam that left the glass at the rainbow angle.
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Everyday Physics: Unusual insights into familiar things
Fig. 4: Move the
glass across the
laser beam, so that
the beam “scans”
the whole width of
the glass.
The experiment shows a surprising result. As you slowly scan
the laser ray across the “droplet” (the glass) and look at the light
reflected backward, you see that the deflection angle (relative to the
entering laser beam) never exceeds a maximum value of 42°! Fig. 5
illustrates this: both ray X and ray Z return at a smaller angle than
ray Y.
Fig. 5: The middle ray
(Y) returns at the largest
angle – the rainbow
angle of ≈ 42°.
Everyday Physics: Unusual insights into familiar things
151
This has two consequences when you look at a real rainbow:
►
Standing with your back to the sun and looking at the light
reflected by a rain shower, no light is emitted from droplets in
positions beyond an angle of 42°. So, first of all, this explains
why the sky outside the rainbow appears darker than inside.
►
The rainbow itself arises because at the special deflection
angle of 42° extra light is reflected. Because this angle is a
maximum, direct neighbours on either side of ray Y return
at the same angle and contribute extra intensity. This angle
therefore marks not only the limit of the area where you see
reflected light but also the area that is brightest.
The only thing now needed for a rainbow is the colours.
These arise because different colours refract differently in the
droplets. That angle of 42° is for red; for other colours it is smaller,
for example 40° for blue because it refracts more strongly than
red. Red is therefore on the outside of the bow and blue is on the
inside. In fact, you are looking at different droplets when you see
different colours. Fig. 6 shows this for the case where the sun is right
on the horizon.
Fig. 6: Every colour
of the rainbow comes
from its own droplets.
(These diagrams are
only schematic: the
observer is always at the
centre of the rainbow
circles.)
Normally the sun is higher than in Fig. 6, and you see only
a small part of a circle (Fig. 7). This also shows that you see no
rainbow at all if the sun is too high; our bow is then below the
horizon, so to speak. (In many mid-latitudes, the sun rises above
42° at noon only from April to September, so it’s only then that
this no-rainbow effect can occur.) (See How high will the sun rise
today? p25.)
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Everyday Physics: Unusual insights into familiar things
Fig. 7: The higher
the sun, the less you
see of the rainbow,
because more of it is
“underground” (a). If
the sun is above 42°
you can’t see the
rainbow at all (b).
Everyday Physics: Unusual insights into familiar things
153
SECONDARY RAINBOWS
Sometimes you see a second rainbow (upper bow in Fig. 1) on the
outside of the main bow. It arises from an extra reflection in the
droplets (Fig. 8). For this reason the secondary bow is fainter than
the main bow. Moreover its colours are reversed, because the rays
now pass through the droplet in the opposite direction to Fig. 2 –
they enter at the bottom and come out from the top. The red – which
refracts the least – now comes to you at a smaller angle than the blue:
about 52° for the red and 54.5° for the blue.
With a bit of luck in your laser pointer experiment you can see
at about 52° the (much fainter) maximum due to the extra reflection
seen in Fig. 8. This occurs on the other side of the glass to the
spot you saw earlier, and corresponds to the rainbow angle of the
“secondary beam”, with the reversed spectrum. (You may have to do
the experiment in a darkened room to see the fainter spot.)
A variety of rainbows is possible: no two bows are the same.
This is due to a number of factors that we have ignored up to
now, such as interference and bending of the light, and differences
in size and form of the droplets. This makes rainbows more
interesting still.
However, all these rainbows have one thing in common:
you can’t walk under your own rainbow, which follows from the
explanation above about how they are formed – your rainbow
travels along with you. And while I can walk under your rainbow, I
cannot see that myself.
Fig. 8: The secondary
rainbow arises from
an extra reflection
within the rain droplet.
(Compare this with
Fig. 2, where there
is only one reflection
within the droplet.)
154
154
Everyday Physics: Unusual insights into familiar things
41
WHY ARE SOAP
BUBBLES SO
COLOURFUL?
Clean water is colourless, and adding a bit of soap doesn’t change
that, but as soon as you blow bubbles, the most beautiful colours
appear. You can see colours in a single soap membrane, and even
puddles with oil floating on top show all the colours of the rainbow.
It has to do with the fact that you’re looking at a thin layer.
Everyday Physics: Unusual insights into familiar things
155
These colour effects are a good illustration of the fact that
we can think of light as a wave. Light has a very small wavelength:
about 1/100th the thickness of a human hair. Light shows its wave
character only when it interacts with structures that are of similar
size to the light’s wavelength – i.e. with very small structures. The wall
of a soap bubble is very thin: a few thousandths of a millimetre, or
about 1/10th the thickness of a human hair, so we won’t be surprised
if bubbles display some of the wave properties of light.
Waves can reinforce or weaken one another, in a process that
is called interference. This is clearest for two waves that are the same
height. What happens depends on whether the waves are in phase
(the crest of one wave coincides with the crest of the other), or out
of phase (the crest of one wave coincides with the trough of the
other). If we let these interfere with each other, the result is a wave
with double the height if the two waves are in phase (constructive
interference: Fig. 1a), or no wave at all if they are out of phase
(destructive interference: Fig. 1b).
Fig. 1: Constructive
interference of
waves in phase (a),
and destructive
interference of waves
out of phase (b).
156
You can easily demonstrate interference by dropping two
stones about 1 m apart in a pond or a big puddle of water. The stones
create circular waves, and you can see the two waves interfere.
Science classes demonstrate interference more easily using a ripple
tank, which is a shallow glass tank of water, and vibrating bars or
paddles generating continuous waves on the surface of the water.
Fig. 2 shows the interference of two circular waves; the grey bands
are where destructive interference annihilates the waves.
Everyday Physics: Unusual insights into familiar things
Fig. 2: Interference of
two circular waves in a
ripple tank.
Credit: Andrea
Persephone at English
Wikipedia
Back to the soap bubble. Why do we get interference in
the thin wall of a soap bubble? One part of the incident light is
reflected by the front side of the bubble wall and another part by
the rear side, so the reflected light that you see consists of two
components (Fig. 3).
Fig. 3: The light you
see reflected from the
soap bubble consists of
two components (one
reflected from the front
surface of the film, and
one from the back)
which can interfere
with one another.
Everyday Physics: Unusual insights into familiar things
157
These two light waves can attenuate or amplify each other.
The colours arise because each colour has its own wavelength. At a
given thickness of the bubble wall one colour is reinforced, the other
is weakened or even extinguished completely, so the reinforced
colour predominates. The thickness of the wall varies, and so you
see different colours on different parts of the bubble.
Sometimes something remarkable happens just before the
bubble bursts. All the colours disappear and the bubble darkens
– as though the bubble feels that its end is near and gives up! In
fact the bubble wall has become extremely thin – thinner than the
wavelength of light. No colour can be amplified any more. On the
contrary, the waves reflected from the front are extinguished by the
waves from the back. The reason for this may be a surprise: light
has the peculiar property that its phase reverses when it is reflected
from an optically thicker medium, e.g. at an air/water interface. This
happens only with the light reflected from the front side, and not
with that reflected from the rear side (since that light reflects from an
opposite water/air interface). This is then the only thing that affects
the phase in a layer much thinner than the light’s wavelength, so the
two waves now have opposite phases and extinguish each other. No
light is reflected any more, and the only light you see is whatever is
transmitted from behind the bubble.
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Everyday Physics: Unusual insights into familiar things
42
WHY ARE CDS
SO COLOURFUL?
When ordinary white light passes through a prism it splits into
all the colours of the rainbow. Even when you don’t have a prism
but a thick piece of glass with oblique angles (such as the bevelled
edge of a fancy mirror), you sometimes also see the whole spectrum
from red to violet. Let’s recall how that arises: different colours
refract to different extents – red refracts the least, violet the most.
In other words, the refractive index – the measure of how much a
medium refracts light – depends on the wavelength. While a CD is as
flat as a coin, and light does not even go through it, you nevertheless
see the whole colour spectrum if you hold the CD at an angle to the
light. Why?
Everyday Physics: Unusual insights into familiar things
159
The secret is in the “grooves”. These are so close to each other
that their separation is of the order of the wavelength of light. The
separation is 1.6 μm (1.6 micrometres, or 1.6 millionths of a metre),
and the wavelength of visible light ranges from about 0.4 μm (violet)
to 0.7 μm (red). That’s why unexpected things happen. The digital
information is in the grooves themselves, in the form of pits, and the
pits are relatively dim. Between the pits are “mirroring tracks” (or
“lands”). When light reflects from them they act as sources of light,
and because light is a wave, the reflected light waves can amplify
or extinguish each other. With a whole series of sources at equal
distance in a row (the lands), this happens in a very regular manner.
Fig. 1: An example of
an interference pattern
(caused by two rocks
thrown into a pond
simultaneously).
Because every one of those sources sends out waves at
all angles, you can almost always find two waves extinguishing
one another. But there is an exception: there is a special angle at
which the waves arising from two neighbours differ by exactly one
wavelength (or two or three), so they amplify one another. In fact,
at this angle, waves from all the lands differ successively by exactly
one wavelength, so they interfere constructively and you see a lot of
light, giving the fan of bright peaks in Fig. 1.
And now the colours. Because different colours have different
wavelengths, the direction in which they amplify each other –
and you therefore see light – differs for each colour. The longest
wavelength, the red, is at the greatest angle to the vertical, and violet
the least (Fig. 2). So if you let light fall on a CD from one side, you
will see the whole spectrum reflected. Your CD gives you music and
a low-cost optical instrument into the bargain!
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Everyday Physics: Unusual insights into familiar things
Fig. 2: Waves of
different wavelengths
interfere at different
angles to the vertical.
Everyday Physics: Unusual insights into familiar things
161
Home experiment: The CD as an optical instrument
Take a bright light source that produces a fairly parallel beam, e.g. a halogen spotlight
in the ceiling or the sun shining through a pinhole in a blind, and hold a CD about two
metres away from it. (Tape cardboard over a small window if you don’t want to make
holes in your good blinds! Cardboard is also more opaque than most blinds.) Make the
room as dark as possible apart from your light source.
Use the CD as a mirror (Fig. 3) to reflect the light back where it came from , but offset
by about 15 cm. (Some CDs are better than others for this; a blank CD-R works best.)
Observe the result on a white piece of paper or a white wall or on the ceiling itself.
As with an ordinary mirror, you see a white spot of light in the middle; this is from
rays reflected straight back (i.e. at 180°) and for these, the path difference between rays
originating in neighbouring “grooves” is zero. This is the case for all wavelengths, so
there is no separation of colours, hence the white spot. Beyond the central white spot
you see a colourful circle with the whole spectrum (Fig. 4). This is caused by rays where
the path difference is one wavelength: first comes the violet, because it has the smallest
wavelength (as in Fig. 2); then blue, green, etc., up to red.
Fig. 3: Support a CD on your fingertip under a halogen spotlight in the ceiling.
In fact, you see two colourful circles. This arises because diametrically opposite parts of
the CD both participate. You can easily see this is the case by covering the CD with a piece
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Everyday Physics: Unusual insights into familiar things
Fig. 4: The CD in Fig. 3 gives you a white spot
and two circular colour spectra.
of paper that exposes only a slot
over the middle of the CD (Fig. 5a).
You see two coloured bands instead
of circles, one on each side of the
white spot (Fig. 5b). Next, move the
paper so the slot exposes a strip on
only half of the CD, (Fig. 5c). Now
you see only a single coloured band
at each side (Fig. 5d).
Fig. 5: Cover all the CD except for
a slot right across the diameter (a);
on either side of the central white
spot you a see a double band of
colour (b). Now expose only one
side of the central slot (c): you see
a only a single coloured band on
each side (d).
5a
5b
Everyday Physics: Unusual insights into familiar things
163
5c
If you use a DVD instead of a CD you see
the same patterns, but enlarged because
the “grooves” on the DVD are closer to
each other than for the CD. Provided that
the light source is sufficiently bright, you
can determine how much finer the DVD is
than the CD, using just a ruler. Measure the
distance between the central white spot and
the middle of the coloured band; say this is
C for the CD, and D (which is larger) for the
DVD. Then the ratio D/C is approximately
the ratio between the groove spacing on
the CD and the groove spacing on the DVD.
(We’re using the approximation that for
small angles a we can write sin (a) ≈ tan (a)
– although this isn’t very accurate for the
DVD bands.)
(See the Resources appendix for information
on how to make a spectrometer using an
old CD.)
5d
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43
HOW DO
HOLOGRAMS
WORK?
A good hologram image has an astounding degree of reality and
depth. It is more than a stereo photo, which is two pictures taken
about 6 cm apart (see box) and quite different from a hologram. A
hologram is a single image, but a very special one: it shows an object
from all directions, and you can “walk around” the imaged object,
so to speak. For a hologram on film, which you look through like
a large photographic slide, it’s like looking through a window at a
real object that has a well-defined position behind the window; if
you move your head to one side, you see the object from a different
angle, and parts of it become visible that weren’t visible from the
other viewpoint.
The technique in holography is very different from that in
normal photography. For normal photos a lens forms an image of
the original directly on film or on a light-sensitive chip; the image
contains information exclusively about the projected image,
seen through the lens. For a hologram much more information is
recorded, and instead of a lens, the image is created by the wave
character of light and the associated interference (see Why are CDs
so colourful? p159).
To record a hologram we illuminate the object with a laser.
(The laser has light of only one colour and therefore only one
wavelength, which is important.) We make the narrow laser beam
broader using of a set of lenses and shine the beam through a semiEveryday Physics: Unusual insights into familiar things
165
transparent (half-silvered) mirror. The part of the light transmitted
through the mirror falls on the object and the light scattered by the
object is captured on the film (Fig. 1).
Fig. 1: Making a
hologram 1 – laser light
illuminates the object,
which scatters the light
towards the film. For
a single point on the
object the scattered
light will behave as
spherical waves.
But that’s only half the story. Simultaneously with this
scattered light the direct beam (via the semi-transparent mirror) is
also projected on to the film (Fig. 2).
Fig. 2: Making a
hologram 2 – part of the
laser light is intercepted
by a semi-transparent
mirror and also directed
towards the film.
The rays reflected from the mirror and the rays scattered by
the object combine to create an interference pattern that is captured
on the film (Fig. 3).
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Everyday Physics: Unusual insights into familiar things
Fig. 3: Making a
hologram 3 – the
light from the object
combines with the
light from the mirror to
create an interference
pattern that contains
all the information
about the object. The
interference pattern
due to a single point on
the object results in a
“grating” with gradually
varying spacing.
This pattern contains all information – a complete 3-D image
– of the object because the scattered light (and therefore also the
interference pattern) is completely determined by the object. To
see how this works, consider a single point on the object. This point
scatters the incident light in all directions: it emits spherical waves,
just like a marble that falls into a pond makes circular waves. These
spherical waves interfere on the film with the light that comes
directly from the light source via the semi-transparent mirror.
We can regard the direct light as a plane (flat) wave, just as the
waves that wash from the sea on to the beach are also more or less
plane waves.
The interference pattern that arises from the simultaneous
action on the film of these two waves forms a pattern of light and
dark spots (an optical “grating” as it is called) indicated by the
intersections with the film of the red circular rays (Fig. 3); note that
the distance between the spots increases from left to right.
Now let’s consider how we view the hologram, once the film
has been developed. (We are dealing with a film, which transmits
light through it instead of reflecting it as a CD does. But that doesn’t
matter: just as it makes no difference whether we talk about a slide
or about a photo when discussing the principles of photography.)
We illuminate the developed film (for now the hologram of that
single point) with direct laser light without any mirror. The film with
its grating pattern not only transmits light straight through but also
diffracts it sideways in well-defined directions just as a CD does (as
Everyday Physics: Unusual insights into familiar things
167
we saw in Why are CDs so colourful? p159). Those diffracted rays
come from the direction of the single point on our object (Fig. 4).
Fig. 4: Viewing the
hologram produced
earlier.
Moreover, because the grating has a progressively increasing
distance between the bars, the beams of light will not be parallel but
will spread out, all appearing to come from the single point on the
object as you move your eye back and forth. Fig. 4 explains this: given
a fixed wavelength of the light, the angle at which the outgoing rays
leave the grating depends on the spacing between the neighbouring
“light sources”. Your eye will have the illusion that this single point is
at a very specific spot behind the film.
The same argument applies to each point on the object, so your
eye “sees” the whole object as though it’s really there behind the film.
A surprising feature of the hologram is that each part of it
contains the whole object. The pattern on even a tiny part of the
hologram is formed by interference to which all points of the object
have contributed. Even if you cover part of the hologram, you can
still see the whole object in the remaining part. The only difference
between this image and the entire hologram is that the “window”
through which you observe the object is smaller.
Marvellous that such a small piece of material contains so
much information!
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Everyday Physics: Unusual insights into familiar things
44
WHY DOES
THE SEA LOOK
SO BLUE?
The sky is blue is because of light scattering. If you’re looking at
the sky away from the sun, the rays that reach your eye have been
deviated by the molecules in the air, and the effect of the scattering
is greatest for the blue end of the spectrum (see Why is the sky blue
and the setting sun red? p107).
But why is the sea blue? Is it because it reflects the blue sky?
No, or at least it’s not the main cause: after all, when the sky is
covered with white clouds the Mediterranean doesn’t turn white all
of a sudden. The same goes for mountain lakes full of clear, clean
water: they often seem deep blue, and a cloudy sky doesn’t make
much difference.
Deep-sea diving gives us a clue about what’s happening. If you
dive a few metres underwater wearing goggles, everything still seems
bluish. (This is very noticeable on underwater photos: those beautiful
red fish seem a bit pale now.) Underwater you have to use the flash
to get accurate colours, especially the red – clearly the red part of
the colour spectrum penetrates deep water less well than the blue.
What’s happening is that the light is attenuated selectively
by clear water at a depth of a few metres, even though it is quite
transparent. (Dirty water attenuates all colours equally and does
so much more than clean water.) Water acts like a colour filter that
stops the red a little, and lets blue pass through nearly unhindered.
In other words, it absorbs with “spectral selection”.
Everyday Physics: Unusual insights into familiar things
169
Fig. 1 shows that the absorption is very small for violet and
blue; it increases for green/yellow; and it becomes really dramatic
for red. (Box Absorption explains what the “absorption coefficient”
means.)
Fig. 1: How water’s
absorption of light
varies with wavelength.
Absorption
The attenuation of light through a medium is not linear – the graph of remaining light
intensity doesn’t follow a downward slanted straight line, which would mean that after
a few metres it becomes completely dark. The attenuation is “exponential”, which seems
reasonable when you think about what is happening: every metre of water attenuates a
percentage of the remaining light intensity, not a fixed amount. An exponential decay
curve starts out very steep but levels out more and more as it approaches zero. We’ll see
that the curve looks like the temperature curve of a cooling cup of coffee which also goes
down fastest in the beginning.
The formula for attenuation is:
dI = −α I dx
where dI is the reduction of the intensity I, α the absorption coefficient and dx the thickness
of the layer (of water in this case). The solution is I = I 0 e-αx, where I0 is the intensity at the
water surface (i.e. at x = 0).
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Everyday Physics: Unusual insights into familiar things
Fig. 1 shows the absorption coefficient α, expressed as attenuation per metre. Note that
this does not mean that this proportion of light is removed per metre. If it did, then for
red light, where α is approximately 1 per metre, it would mean complete blackness after
1 metre. Substitute some values into the solution: α = 1 m-1, I0 = 100%, x = 1 m. That gives
I = 1 e-1 = 37%, i.e. the light at 1 m deep is only 37% of what it was at the surface – it has
been reduced by a factor e (just over 2.7) for each metre. After two metres that is already
a factor of about 7.5, so you lose the red quickly as you dive deeper. Fig. 2 shows how the
intensity of red light falls off with increasing depth.
Fig. 2:
Exponential
curve, showing
temperature
against time,
for a cooling
cup of coffee.
Fig. 3:
Exponential
curve, showing
remaining light
intensity against
depth, for red
light attenuated
in clear water. The
slight attenuation
of blue light
is shown for
comparison.
Everyday Physics: Unusual insights into familiar things
171
How does that affect the colour of the sea or a mountain lake?
If you look perpendicularly into the water and the bottom is white
sand or rock, what you’re seeing is actually the sunlight reflected by
the bottom. (Of course you also see the light from the sky reflected by
the water surface, but at perpendicular incidence that is not much
– only about 2% of the incident light is reflected. So if the bottom is
light and the water not too deep, what you’re seeing is mostly light
reflected by the bottom.) That light must go down and up, so x in
our formula is twice the depth. In the light’s journey down and up
it mostly loses the red part of the spectrum, so the water looks
blue! Sounds logical, doesn’t it?
But that can’t be the whole story. After all, even if the water
is very deep and there is no bottom in sight, the water still seems
dark blue (if it’s clean). How does that light come up again out of
the water into your eye? That must be through scattering: light is
scattered by the water molecules themselves and by other small
particles suspended in the water, just as air scatters the sun rays.
Once again the scattering is strongest for the blue, as in the blue
sky, so it is predominantly blue light that emerges from water. That’s
why the deep ocean looks blue – as long as it’s not full of algae and
green seaweed.
Optically speaking, frozen water behaves much the same as
liquid water, and a thick layer of ice preferentially transmits the blue
side of the spectrum. That is why glacier crevices appear blue down
below, and the light in a tunnel through a glacier is blue. Even a layer
of snow can take on a bluish tinge from transmitted light. You can
see this especially clearly when there is a deep hole in the snow.
Snow typically has a grainy structure, which causes light to be
scattered multiple times and therefore take a long path to reach
your eye. That makes for extra absorption, so even a relatively thin
layer of snow can still take on a noticeably blue colour.
172
Everyday Physics: Unusual insights into familiar things
45
WHAT’S
SPECIAL ABOUT
POLAROID
GLASSES?
The inventors of Polaroid glasses combined business sense and
physics: they realized there was money to be made from the fact that
light waves can oscillate in all planes perpendicular to the direction
of propagation. Light waves are like the waves you get (Fig. 1) when
you move the end of a thick rope or a garden hose rapidly back and
forth, or up and down (or in any other direction – the plane of the
oscillation can have any orientation).
Everyday Physics: Unusual insights into familiar things
173
Fig. 1: Waves in
a rope oscillate
perpendicular
to the direction
of propagation.
It’s the same for all electromagnetic waves for that matter. It
turns out that in these waves, the oscillating electric field E can take
on any plane. For simplicity let’s think of two directions: horizontal
and vertical. (The oscillating magnetic field is always perpendicular
to the electric field, therefore we can forget about the magnetic
field.) Oscillations in an arbitrary direction can be considered as
being composed of a horizontal and a vertical component.
With the waving rope, you can filter out parts of the
oscillation through a slot, for example a letterbox in your front door
(Fig. 2, Fig. 3).
What do Polaroid glasses do? The lenses consist of polarization
filters, or Polaroid filters, that transmit only one component and
block the other. In the analogy with the oscillating garden hose we
can compare the Polaroid filters with the letterbox in your front
door. (And you can rotate a filter 90° – just as you can rotate your
door – if you want to get the other component instead.) The light
that gets through the filter is said to be polarized in the particular
plane (Fig. 4).
174
Everyday Physics: Unusual insights into familiar things
Fig. 2: The vertical
waves in the rope are
transmitted by the
vertical mail slot in
the door.
Fig. 3: The vertical
waves in the rope
are transmitted by
the vertical mail
slot but not by the
horizontal slot.
Everyday Physics: Unusual insights into familiar things
175
Fig. 4: In polarized
light, the oscillation
of the electric field is
all in one plane – in
the vertical plane for
vertically polarized
light (a), and in the
horizontal plane for
horizontally polarized
light (b).
A Polaroid filter is really only useful when the light is
significantly polarized and when there is an advantage to stopping
one of the two directions. Ordinary light is barely polarized at all:
all oscillation directions occur. In that case the Polaroid glasses
cannot do more than stop half the light, e.g. the components of the
oscillations in the vertical plane. These glasses then behave like a
normal attenuator, i.e. like ordinary sun glasses.
But in certain situations, Polaroids work especially well:
they can remove troublesome dazzles and help you to see through
a window or into water. Now they are making use of the fact that
reflected light is often partly polarized, especially when it is
reflected by water or glass, and then the Polaroid glasses particularly
attenuate the reflected light, which for example cuts down the dazzle
of oncoming headlights reflected from a wet road surface at night.
If the reflecting surface is nice and smooth, the extent of
polarization depends on the incident angle. There is even one
incident angle – if you look a little under 45°, see box – for which
reflected light is completely polarized. In that case you can completely
176
Everyday Physics: Unusual insights into familiar things
suppress that reflection with Polaroids, for example allowing you
to see clearly something underwater without the image being
overwhelmed by reflection from the bright sky.
What happens at the two extreme angles of incidence:
perpendicular to and grazing along the surface? For perpendicular
incidence it is simple. Because all oscillation directions are
equivalent (since there is symmetry around a line perpendicular
to the surface) there can be no difference between the several
polarizations: they all are reflected equally. From a water surface
about 2% is reflected, and from glass 4%. Because a windowpane has
a front side and a back side, each of which reflect, a total of 8% of
the incident light is reflected.
For grazing incidence there is also no difference between the
two polarizations: both are completely reflected. So if you look over
water at grazing incidence, the surface acts like a perfect mirror and
you see the sky as clearly in the water as above it. The same goes for
glass: if someone stands with their nose against a shop window, you
see the mirror image of that nose as clearly as the nose itself (as long
as your eye is very close to the window).
So Polaroid glasses really work for reflections from water and
glass but only if you view the surface at an angle. Interestingly, pilots
do not wear Polaroid glasses because they do not want to miss the
reflections – the reflected flash from another aircraft may be a useful
warning that it’s too close.
Light reflection
Fig. 5 shows how the percentage of light (the reflection coefficient R) reflected at the
surface of a transparent medium depends on the angle of incidence. (Grazing incidence
is 90° and incidence perpendicular to the surface is 0°.) There are two different curves for
the two polarization directions: one with the electric field parallel to the plane of incidence
(//, blue in Fig. 5, “parallel polarization”) and one perpendicular to that (⊥, red in Fig. 5,
“perpendicular polarization”). The two curves differ substantially. In particular, the blue
curve has no reflection at all at a certain angle (the “Brewster angle”), so at that point the
only light that is reflected is from the red curve – it is therefore completely polarized. For
an air/glass transition (drawn in Fig. 5) that angle is 56°; for an air/water transition it’s 53°.
Why does the reflection at this angle disappear completely?
Everyday Physics: Unusual insights into familiar things
177
Fig. 5: The
percentage of light
reflected from an
air/glass surface
depends on the
angle of incidence,
where grazing
incidence is 90°
and incidence
perpendicular to
the surface is 0°.
The two directions
of polarization are
shown by the
red (perpendicular)
and blue
(parallel) curves.
The reflection disappears because of how reflected light
comes about. When a light ray hits the molecules just under
the surface, the molecules act as small dipole antennas, with
the dipole directed along the electric field. The light ray bends
(Fig. 6), and the molecules responsible for the reflection will be
in that bent ray.
Now look at the parallel polarized light. Its electric field by
definition lies in the plane of incidence (the plane defined by
the incident ray and the perpendicular to the surface). The field
Fig. 6: Snell’s law: v1 and v2 are the respective
velocities of light in air and water.
Credit: After nl.wikipedia.org/wiki/User:Sawims
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Everyday Physics: Unusual insights into familiar things
oscillates in that plane and perpendicular to that (bent) light ray, and so do the dipoles
induced by the field. But then no radiation can be transmitted in that direction because
dipoles cannot transmit radiation along their own axis. Conclusion: at that angle the
reflection must be zero.
As the direction of the bent ray depends on the refractive index, the Brewster angle must
also depend on that. Simple trigonometry shows that the tangent of the Brewster angle is
equal to the refractive index, n, as is quite simply derived using Snell’s law (sin i / sin r = n,
with i and r being the incident and refracted angles).
This also explains the different behaviour between water and glass: for perpendicular
incidence (red curve) water and glass have different values of R (the reflection coefficient).
That is due to the difference in refractive index: n = 4/3 or 1.33 for air/water and about 1.5
for air/glass. The reflection coefficient for perpendicular incidence is given by R = (n – 1)2
/ (n + 1)2. For water this becomes (1/3)2 / (7/3)2 = (1/7)2 = 1/49 = 2%. For glass (1/2)2 / (5/2)2 =
(1/5)2 = 1/25 = 4%. Because of the squaring the reflection for the transition air/glass is the
same as for glass/air. A normal windowpane therefore attenuates a perpendicular beam
in total by 8%.
Everyday Physics: Unusual insights into familiar things
179
180
Everyday Physics: Unusual insights into familiar things
46
WHAT DO YOUR
EARS HEAR?
Sound is basically periodic pressure fluctuations of the air (or more
precisely, periodic compression and expansion of the medium
through which the sound travels). In the days of 78-RPM records you
could demonstrate this easily by sticking the corner of a postcard
in the groove of a rotating record and then you could hear the
music! The postcard just followed the oscillations of the groove that
represent the pitch of the recorded music. The postcard pushed and
pulled the air at the corresponding frequency and that was it. It was
by no means hi-fi quality, and not very loud, but it worked.
Everyday Physics: Unusual insights into familiar things
181
Fig. 1: Sound intensity,
in dB (left axis) and
W/m2 (right axis) at
different frequencies.
Credit: After https://
en.wikipedia.org/
wiki/File:Octave_
Designations_4.svg,
Wikipedia Tagrich1961
182
How loud must sound be in order to be heard by your ears? It
depends on the pitch, or frequency. At a few thousand hertz (hertz
or Hz means “vibrations per second”) your ears have the highest
sensitivity, and can detect quite low intensities. We can express the
intensity of sound in terms of energy flow per unit area, or watts
per m2: the further you are from a loudspeaker, the more “diluted”
the watts that the loudspeaker produces, i.e. the fewer W/m2. In
that respect sound behaves just like light: the further away you are
from the lamp, the dimmer the illumination on your book. At the
frequency where your ears are most sensitive, you can just hear
some 10-12 W/m2 (a millionth of a millionth).
Everyday Physics: Unusual insights into familiar things
Fig. 1 illustrates various intensities of sound at different
frequencies, for the average person. The horizontal axis shows
the frequency, or pitch, on a logarithmic scale. (A piano keyboard
is a logarithmic scale. As you ascend the musical scale in octaves,
you don’t go up in fixed steps of frequency, i.e. not 100 Hz,
200 Hz, 300 Hz… In fact each octave is a doubling of frequency,
so the octaves are more like 110 Hz, 220 Hz, 440 Hz, 880 Hz…
This is why Fig. 1 can accurately show a piano keyboard and its
associated frequencies along the horizontal axis.) On the vertical
axis is the sound intensity. It is proportional to the volume control
of a stereo system.
The vertical axis has a logarithmic scale also: the axis on the
right gives the intensity – in W/m2 as explained above. The axis on
the left gives the intensity in decibels (dB). Each 10 decibels – or 1 Bel
– represents an increase of a factor of 10 in intensity. (By the way,
the unit is named after Alexander Graham Bell, the inventor of the
telephone). The 0 dB reference level is, somewhat arbitrarily, put at
the lowest level normally audible: 10-12 W/m2.
Everyday Physics: Unusual insights into familiar things
183
Each of the solid curves represents a certain subjective
impression of loudness. The lowest curve represents the threshold
of hearing, i.e. the minimum intensity at which you can just hear the
sound. In the most sensitive region, around 3,000 or 4,000 Hz, it is
about 10-12 W/m2, as we said. Notice that the threshold curve rises
steeply at very low and very high frequencies: this means that you
need a higher intensity (more watts per m2) before you can hear
these frequencies. Evidently, your ears are less sensitive to such high
and low frequencies.
The next five solid curves are similar but each one indicates
the intensity needed to create a certain subjective loudness level for
your ears. In musical terms, from the bottom up they correspond to
pianissimo (pp), piano (p), mezzo forte (mf), forte (f) and fortissimo
(ff) respectively. Finally, the uppermost curve corresponds to the
pain threshold, where the intensity is so high that the sound starts
to be painful.
The numbers in Fig. 1 show that your ears are extremely
sensitive – we are dealing with unimaginably small intensities. One
millionth of a millionth of a watt per square metre is terribly little.
Compare it with the lighting in your living room: if the lamps total
200 watts, for example, and they illuminate an area of 20 m2 the
average intensity is 10 W/m2. Even the energy of disco noise is feeble
by comparison!
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Everyday Physics: Unusual insights into familiar things
47
WHY ISN’T THERE
MORE NOISE
POLLUTION?
Excuse me? The last thing you want is more noise. But if you look
carefully at Fig. 1 (repeated from the previous chapter), the above
question makes sense, because Fig. 1 shows the immense range
of intensity your ears can manage. As the vertical axis shows, it’s
approximately a factor of 1012 (a million million) at the mid-range
of around 1,000 hertz. In a way this is not altogether surprising:
while a choir of 100 singers is a lot louder than a single singer, you
don’t perceive an enormous difference with your ears. In fact you
find these 100 singers about 4 times louder than a soloist, not 100
times louder.
The large intensity range your ear can handle is worrisome in
the context of noise pollution. In principle, the intensity of sound is
inversely proportional to the square of the distance to the source.
Your ears can handle an intensity difference of 1 million squared.
So if you were very close to a source of noise that begins to hurt
your ears – say an aircraft at take-off, or a disco loudspeaker – you
would have to move a million times further away from the source in
order to hear it no longer. That ought to make life awfully noisy: you
would, so to speak, be troubled by the din in Durban and the noise
in Nagoya.
In real life, however, there’s much less of a problem, thanks
mostly to the absorption of sound in the air. For every compression
and expansion, a tiny bit of sound energy is lost, with the high tones
Everyday Physics: Unusual insights into familiar things
185
suffering the greatest loss. You especially notice that with thunder:
close-up, you hear a lot of sharp high tones in a thunder, but from
a long distance away the same thunder clap just sounds like a
dull rumble.
Fig. 1: Sound intensity,
in dB (left axis) and
W/m2 (right axis) at
different frequencies.
186
But fortunately even the low tones are attenuated in the
atmosphere. In addition, the sound is weakened by intervening
obstacles. Finally, the Earth is not flat, and over long distances the
curvature of the Earth helps reduce the intensity.
Everyday Physics: Unusual insights into familiar things
48
THE ENERGY
COST OF
TALKING
Fig. 1 (repeated here from earlier) also illustrates something
interesting about sources that produce sound. Take, for example,
sound with an intensity of 60 dB; this is roughly the intensity of
the average conversation, or music in your living room. The righthand scale shows that this corresponds to 10-6 W/m2 (1 microwatt/
m2 ). Suppose the source of this sound – a loudspeaker or somebody
talking – is fairly close, so that the energy is spread out over a surface
of, say, 10 m2. This means that the source is producing about 10-5
watts, or 10 microwatts of sound, which is very little indeed. It
means, for example, that old-fashioned radios, which drew 50 to
100 W of electricity from the outlet, converted only a tiny bit of that
energy into sound – the rest, well over 99% – was emitted as heat. So
we could consider those old radios as electric heaters with a tiny bit
of noise as a by-product.
Everyday Physics: Unusual insights into familiar things
187
Fig. 1: Sound intensity,
in dB (left axis) and
W/m2 (right axis) at
different frequencies.
188
If the sound is produced in the form of speech, then the
average talker in a living room situation also produces some 10
microwatts. Just for fun, let’s calculate how much energy a teacher
delivers during their entire career, or how much energy you produce
talking over your whole life. Take the extreme case and assume that
you talk 24 hours a day, 365 days a year. That’s 24 × 365 hours or, if
we exaggerate a bit for simplicity, 25 × 400 = 10,000 hours a year. If
you live for 100 years, the total is at most 106 or one million hours.
So the total energy you spend talking is 10-5 W × 106 hours = 10 watt
hours or 0.01 kWh. Even with a sky-high electricity cost of 20 pence
per kWh unit, the energy value of talking non-stop for your whole
life is less than a penny!
Everyday Physics: Unusual insights into familiar things
Music reproduction and the
“loudness” control
The shape of the curves in Fig. 1 is important: the lower curves rise
steeply on the left, i.e. at low frequencies. That affects how your
stereo equipment at home reproduces sound. Suppose you are
listening to music and the volume control is up high, giving a sound
intensity of 10-2 W/m2 in your living room; this corresponds to the ff
curve (see What do your ears hear? p182) in Fig. 1. The curve – which
indicates the subjective loudness level – is relatively flat here, so you
perceive all frequencies as about equally loud, from very low to very
high, so the sound has almost the same natural “colour” that it would
have in the concert hall.
What happens if you turn down the volume, so that it
corresponds to the 10-10 W/m2 (pianissimo) curve in Fig. 1? Notice
what happens at the lower frequencies: below about 200 Hz the
intensity is below the threshold, so you don’t hear all the low tones
– unless you selectively adjust the sound intensity for that low
frequency region. On some hi-fi systems there is a special control to
do just that. (It’s sometimes called the “Dolby volume” or “loudness
control” – not to be confused with the volume control.) It selectively
increases the amplification for the low frequency region (and also
somewhat for the very high frequencies) when you turn the volume
down. In this way the richness of the music is preserved over the
entire frequency range.
Everyday Physics: Unusual insights into familiar things
189
Home experiment: Increasing the volume of a tuning fork
To get a loud sound from a tuning fork after striking it you have to press the handle
of the tuning fork against a table or other big hard surface. Otherwise it’s not very
loud, which isn’t surprising for the following reason. For example, consider the tone
produced by a 440 Hz (“A”) tuning fork. Its wavelength is ¾ metres, which is much
bigger than the tuning fork itself, and the pressure fluctuations have time to “shortcircuit” around the prongs of the fork: while one side of a vibrating prong wants to
increase the air pressure, the other side wants to decrease it. The two sides work
against each other, and the result is you don’t get much sound. The same short-circuit
effect causes a loudspeaker without a housing to produce only high tones, because
for the low ones it is just too small. For most people that is not at all obvious: we
often think more in terms of resonances and the sound boxes of musical instruments
and loudspeakers.
Here’s a nice experiment to demonstrate this shortcircuit effect. You need a tuning fork and a large stiff
envelope or a piece of cardboard. Cut out a rectangular
slot in the cardboard, big enough to take one prong
of the tuning comfortably, so that you can easily hold
the resonating fork in the slot without touching the
cardboard (Fig. 2). Strike the tuning fork and gently
move one prong into the slot, as shown.
Sure enough, now the fork sounds much louder. The
short-circuit paths have been adequately damped,
so that the pressure variations on the outside of the
cardboard now effectively originate from one side of
one prong of the fork. This arrangement is comparable
to a loudspeaker mounted in the middle of an old door.
Fig. 2: Placing a tuning
fork in a slot cut in thick
cardboard prevents the
sound waves “shortcircuiting” to the other
side of the fork and
reducing the volume.
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Everyday Physics: Unusual insights into familiar things
49
HOW CAN YOU
TELL WHERE
A SOUND IS
COMING FROM?
How do your ears determine the direction a sound is coming
from? For that you need both ears. Because of their different
positions, each of your ears receives the sound slightly differently.
Consider a sound coming exactly from your left-hand side: not only
does your left ear hear the sound somewhat louder than the right
(because it’s closer), it hears it earlier too. So two mechanisms
are at play: intensity difference and time difference, as shown
schematically in Fig. 1.
Everyday Physics: Unusual insights into familiar things
191
Fig. 1: When a sound
comes from the left,
your left ear hears it
louder (a) and earlier
(b) than the right ear.
(TL is the travel time of
the sound to the left ear,
TR the time to the right.)
Consider the time difference first. If a sound comes from one
side, the nearer ear hears the sound a bit earlier than the other
ear. This mechanism works only for low tones (long wavelengths);
to see why that is, we have to consider the length of the wave. The
length of a single wave is the speed divided by frequency. The speed
of sound is just over 300 m/s; a fairly low tone has a frequency of
about 100 Hz (i.e. 100/s), so the wavelength is 300 m/s divided by
100/s = 3 metres. Such a long wave is much bigger than the distance
192
Everyday Physics: Unusual insights into familiar things
between yours ears, so each of your ears will perceive the same
wave, but in a slightly different phase: one ear hears the beginning of
the wave earlier than the other (Fig. 2).Your brain observes that time
difference with great sensitivity and can work out which direction
the sound is coming from.
For a mid-range tone where the wavelength is equal to the
distance between your ears – about 15 cm – the sound waves from
the side hit the ears in the same phase (although with a whole
wavelength difference). Because there’s no phase difference, the
ears can’t distinguish between this sound and sound that comes
exactly from the front or the back, so your brain can’t tell where
it came from. For this wavelength and even shorter ones, the time
difference is no good for determining direction.
Fig. 2: For sound with
long wavelengths, your
two ears hear different
phases of the wave.
However, there is a second mechanism, which does work for
these short waves – the intensity difference. Short waves, in particular
those smaller than your head, show an intensity difference between
right and left. Take a tone a hundred times as high as before (i.e.
10,000 Hz instead of 100 Hz). Instead of 3 m the wavelength is
only 3 cm, much smaller than your head. When the sound comes
exactly from the left, your head is in the way of your right ear: your
head forms a “sound shadow” and your right ear hears a less intense
sound. Measurements show that the intensity difference can easily
be as much as 25 dB; your ear can detect such big differences without
any problem, and your brain can then work out the direction of the
sound.
It is amazing how precisely you can determine the direction
from which the sound comes. Careful measurements have
Everyday Physics: Unusual insights into familiar things
193
demonstrated that you can do it with a precision of about 2° – the
angle between the ends of a matchbox at a distance of three metres!
By the way, neither of these two mechanisms explains how you
manage to determine whether the sound is coming from the front or
the back, which in fact you are able to do. Probably the shape of your
ear shell plays a part, because it’s not symmetric front-to-back. (In
fact, if you temporarily make the ear shell symmetric by means of an
artificial attachment, it’s much harder to differentiate between front
and back.) Moreover, there is yet another aid: you turn your head to
one side when you’re not sure whether the sound is from front or
back. Turning your head changes the front/back distinction into a
left/right distinction, and your ears can handle that with a good deal
of precision.
Similarly, to distinguish between up and down you rely on the
shape of your ear shell and of your head. It helps to nod your head
when you want to better localize sound in the vertical direction.
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Everyday Physics: Unusual insights into familiar things
50
DISCRIMINATING
BETWEEN
DIFFERENT
VOICES:
THE COCKTAIL
PARTY EFFECT
Background noise can make it difficult to hear a conversation,
especially for older people, even if their hearing is good otherwise. It
becomes hard to isolate one conversation from a sea of voices. This
is sometimes called the cocktail party effect. How does it work and
why do we have more trouble with it as we age?
Everyday Physics: Unusual insights into familiar things
195
First of all, it is a complex phenomenon and it’s not just sound
that’s involved, because lip-reading is also important in following a
conversation – it has been shown experimentally that blind people
have even greater problems following a conversation in a noisy
environment. However, for older people the cause of the problem
is primarily acoustic: the loss of sensitivity to high frequency is key.
Fig. 1 shows how large that loss is for ages 30-85, over the
frequency range from 250 to 8,000 Hz. Age 15 is taken as the
baseline, representing no hearing loss. The curves are averages taken
over large numbers of people. You can see that the loss increases
with age, as expected, but what is striking is that the loss becomes
dramatic for the higher tones. At 8,000 Hz there is typically a loss of
35 dB at the age of 60, and it increases by 10 dB each 5 years.
Fig. 1: Hearing loss
with increasing age.
Credit: Dr JAPM. de Laat,
Audiological Centre,
LUMC, Leiden, The
Netherlands, 2002)
Comparing Fig. 1 with the graph on page 182, it’s clear that by the
time we reach the age of 80 we are practically deaf to high tones
above about 8,000 Hz.
Why does that make it so difficult to follow a specific
conversation in a hubbub?
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Everyday Physics: Unusual insights into familiar things
►
Consonants such as p, t, k, f and s contain highfrequency components, which you can’t hear properly,
so can easily confuse one for another, or even miss
them entirely.
►
Not hearing some of the words reduces your
overall comprehension of the conversation you’re
interested in.
►
It becomes more difficult to isolate that one conversation
from the surrounding babble, because your directional
hearing (see How can you tell where a sound is coming from?
p191) deteriorates.
■ While the time difference mechanism works well for
low tones, as we saw earlier, it’s not so good in enclosed
spaces, because in many function venues the walls, floor
and ceiling reflect low frequencies well (because all
kinds of surfaces absorb low frequencies less than high
ones). The result is that low tones reflect and continue
to reverberate, so the reflected sound dominates,
making it hard to determine the direction of the original
sound. (We know that from our speaker set-up. For
stereo we need at least two speakers, and the effect is
provided primarily by the high tones. The low tones
don’t contribute much to the stereo effect, so a single
subwoofer is sufficient.) Overall, in enclosed spaces low
tones give little directional information.
Everyday Physics: Unusual insights into familiar things
197
■ As we saw, the intensity difference mechanism works
particularly well for the high frequencies. However,
Fig. 1 shows that that is precisely where ears fail
with age.
►
Therefore, neither of the two mechanisms works well
as you age, so you have little directional information to
help isolate one specific conversation from the hubbub of
voices. (The Technical University of Delft has developed a
device that does help. It’s a pair of glasses that selectively
amplifies sound from the direction you’re looking it. It
works surprisingly well and makes it much easier to hear
the person you’re talking to.)
If you don’t feel like wearing glasses, you can still switch to lipreading – hearing with your eyes and seeing what is being said.
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Everyday Physics: Unusual insights into familiar things
51
DO YOU HEAR
BETTER AT
NIGHT?
The night is quieter than the daytime: there’s less traffic, no children
playing, no noise from the neighbours. So it’s not surprising that
overall you notice faint noises more at night than in the daytime.
But not all nights are the same – on some nights you hear the distant
highway much more clearly than on others; one night you clearly
Everyday Physics: Unusual insights into familiar things
199
hear the church bell, but not another night. This happens even
when there is no wind, so wind can’t be the explanation.
Maybe temperature has something to do with it – perhaps it
bends sound as happens with light (which we saw in Puddles on a
dry road, p133). Sound does bend, as with all waves propagating in a
medium. Bending depends on the propagation speed, and the speed
of sound is more strongly temperature-dependent than the speed
of light.
In air, the speed of sound is related to the thermal velocity of
the molecules: they have to pass on the pressure disturbance, and
they can’t do that any faster than the speed at which they themselves
move. Because the thermal speed of the molecules increases with
temperature, the speed of sound increases with temperature. (That’s
true for light too, although the underlying physics is very different.)
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Everyday Physics: Unusual insights into familiar things
On a sunny day, the air just above the ground is warmer than
higher up, so the speed of sound is also greater there. Therefore,
you’d expect sound to bend, in the same way as we saw light
does when it creates mirage puddles. For sound waves moving
horizontally, the warmest side of the wavefront is the fastest, and the
fastest side automatically forms the outer curve, just as we saw with
light waves and waves on the beach (Fig. 1). Therefore, sound from
a source just above ground curves away from the Earth (Fig. 2). As
a result, if you’re standing some distance from the source the sound
tends to go over your head, and you hear the sound less well.
Fig. 1: Light rays
skimming over the
warm road surface are
deflected upwards.
Fig. 2: Sound travels
faster through the
warm low air than
through the colder
air higher up, causing
the sound waves to
curve upwards from
warm ground.
A kind of sound shadow can even form where little sound
penetrates (Fig. 3). The lowest sound wave is the one that just skims
over the ground. This wave then curves up from the ground, and if
you’re far enough away it goes completely over your head, and you
hear it less.
Everyday Physics: Unusual insights into familiar things
201
Fig. 3: If you stand
in the “sound shadow”
caused by the bending
of sound waves on a
warm sunny day, you
will hear less sound.
On a cloudless night things are precisely reversed. The air
just above the ground is colder than higher up, because the ground
easily radiates its heat to the clear sky. The sound rays now curve
down towards the Earth. You hear the sound source better or – if the
noise is troublesome – the sound pollution is worse. So if you hear a
faraway barking dog better at night than in the daytime, you’re not
imagining it – it’s probably just a matter of temperature.
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Everyday Physics: Unusual insights into familiar things
52
CAN THE WIND
BLOW SOUND
TO YOU?
Sometimes in Swiss and Austrian mountain villages, when the
wind blows from a neighbouring valley, you can hear the bells from
the far side of the mountain. Similarly you hear the noise from a
distant motorway better when the wind blows from there. It might
appear the that wind is catching the sound and blowing it to you,
but that isn’t so. The real reason (which isn’t obvious) is twofold.
First, wind direction influences the effective speed of sound – the
speed decreases against the wind (Fig. 1a), and increases in the same
direction as the wind (Fig. 1b).
Fig. 1: When
sound travels
into the wind,
its effective
speed is lower
(a). When
sound travels
with the wind,
its effective
speed is
higher (b).
The second part of the reason is that wind speed increases
with height. Close to the ground the wind speed is relatively low.
Higher up the wind blows harder. (Think of wind turbines: the
higher they are, the better.) So, with the wind, the speed of sound
Everyday Physics: Unusual insights into familiar things
203
increases with height, so the sound waves curve downward (Fig. 2a).
However, for the same reason, against the wind the effective speed of
sound decreases with height. That gives an upward-curving bending
pattern (Fig. 2b). This looks the same as the left-hand side of Fig. 2 in
the previous chapter, because in both the speed of sound decreases
with height. The sound waves therefore curve upward.
Fig. 2: Wind speed
increases with height.
Therefore, with the
wind, the sound gets
faster with height
and therefore bends
downwards (a). Against
the wind, sound gets
slower with height
and therefore bends
upwards (b).
So, a downward-curved ray can pass above obstacles in its
way, which explains the tolling bells in the mountain village (Fig. 3).
It also explains why the motorway is more of a nuisance when the
wind blows from there: the “sound rays” are concentrated as they
bend towards you. In the next chapter we’ll see whether noise
barriers can help.
Fig. 3: A downwardcurving sound wave can
avoid obstacles and be
heard further away than
you’d expect.
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Everyday Physics: Unusual insights into familiar things
53
DO NOISE
BARRIERS
WORK?
Even people living very close to a busy highway like to have quiet
conversations and undisturbed sleep. Noise barriers can help. An
obstacle – such as a noise barrier – gives rise to a sound shadow,
analogous to a light shadow. Inside the sound shadow the sound
is a lot fainter than outside it. (That’s quite noticeable when, for
example, you cycle on a sunken cycle path. When you come back
above ground and can see the cars, the traffic noise is noticeably
louder.) The same applies to a noise barrier: you hear less noise as
long as you are in the “shadow” of the barrier (Fig. 1).
Everyday Physics: Unusual insights into familiar things
205
Fig. 1: When you are
in the sound shadow
cast by a noise barrier,
the sound you hear is a
lot fainter.
But when it’s windy, noise barriers can be particularly
ineffective. The wind speed is always much less than the speed of
sound, so why does it make such a difference?
The reason is not the wind speed itself but the fact that the
speed can vary strongly with height – certainly close to noise
barriers. The wind speed increases with height – there is a “velocity
gradient” in the vertical direction. As we saw in the previous chapter,
this means that the sound waves curve down when travelling with
the wind, and the sound reaches areas behind the barrier that
would normally be in the “sound shadow” (Fig. 2). In fact, just above
the barrier the velocity gradient is extra large and the curvature
therefore more pronounced, making the sound shadow smaller still.
Measurements at the labs of the TNO-TPD research centre in Delft
showed that at moderate wind speeds (4-5 m/s) the noise behind
the shield can be 5-15 dB stronger than when there’s no wind.
Fig. 2: When the
wind is blowing from
the source of the noise,
the sound waves are
bent downwards, and
the sound shadow cast
by the noise barrier is
much smaller, so you
hear more noise.
The only consolation is that when the wind turns and blows
away from the house towards the highway, the effect is reversed.
Then you have less noise than when there is no wind.
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Everyday Physics: Unusual insights into familiar things
54
CAN YOU HEAR
WHETHER THE
CURTAINS ARE
CLOSED?
Unfurnished empty rooms sound hollow; the acoustics are poor
because the walls reflect sound well and don’t attenuate it very
much. You notice the same thing in an enclosed swimming pool, or
in a restaurant with bare walls and a concrete or tile floor. The sound
bounces back and forth many times before it dies out, so in the
restaurant there is so much background noise ricocheting around
that it almost drowns your own conversation.
Everyday Physics: Unusual insights into familiar things
207
Fortunately living rooms usually have better acoustics. The
sound is muffled fairly efficiently by – among other things – the
carpets and curtains. This is because soft surfaces absorb sound
better than hard ones: you’ve probably noticed that thick curtains
absorb sound much better than a bare window. The reason is that
thick curtains have a large microscopic surface, and it is at surfaces
that absorption takes place.
Here’s an interesting question: imagine you’re wearing a
blindfold and enter a room when it is pitch black. You don’t know
whether the curtains are closed or not. But if you make a noise, by
speaking or stamping your foot say, and you listen carefully, then
you should be able to hear whether the curtains are open or closed.
Right? Surprisingly, the answer is no.
In fact open curtains muffle sound as well as closed ones,
because indoor sounds are muffled differently from how the wall
of a house attenuates sound coming from outdoors. For a wall it’s
the mass that matters most: thick heavy walls insulate sound better
than thin light ones. That’s logical: heavy masses are more difficult
to get vibrating than light ones, so a heavy wall transmits less sound
to annoy the neighbours on the other side. But for the muffling of
sound inside a room, materials that are very porous (and thus have a
large surface area on a microscopic scale) are most effective. As we
saw earlier, the reason is that sound waves are absorbed at surfaces –
primarily by friction of the vibrating air – and eventually transformed
into heat. Materials with open structures absorb well because they
have large surface areas. Examples are glass wool, thick textiles and
even plastic foam with an open cell structure (unless it’s painted
– because paint can close the open structure and spoil the soundabsorbing properties).
So thick curtains are good absorbers. It doesn’t matter whether
they are spread over a large area or whether they hang in a bundle
in a corner, because the amount of porous surface is unchanged.
Of course, they must not be in a cupboard (which probably has a
door with a hard surface) but as long as they are easily accessible
to the sound they are good mufflers, and the thicker the curtain the
better. Therefore, the overall conclusion is: you can’t hear whether
the curtains are open or closed, but you can tell if they have been
taken away for cleaning!
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Everyday Physics: Unusual insights into familiar things
55
DO – RE – MI …
THE PHYSICS OF
MUSICAL SCALES
Musical scales are interesting: you may find it pleasant when
certain combinations of pitches are produced, but not others. To
explain why, we need to look at how a musical instrument generates
its notes.
Notes, tones, overtones,
and timbre
A guitar or violin string is stretched tight between two fixed points.
At rest, it’s not vibrating (Fig. 1a) and it produces no sound. When
you pluck the string or play it with a bow, the string vibrates. The
Everyday Physics: Unusual insights into familiar things
209
most common vibration is the one where the middle of the string
vibrates the most (Fig. 1b). There the size of the wave – its amplitude
– is greatest. The tone or note produced by this vibration is called is
the fundamental and is the lowest tone the string produces; let’s say
this frequency is f.
Fig. 1: Fundamental
and overtones of a
vibrating string.
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Everyday Physics: Unusual insights into familiar things
A string can also produce what are called overtones
(Fig. 1b-1e) with frequencies higher than the fundamental tone.
Take the first overtone – the 2f vibration – for example. You can hold
the string in the middle without affecting anything because that
point does not vibrate. Now the length of vibrating string is only half
as long as before (though there are two almost-separate lengths),
so its frequency, or pitch, is doubled. Similarly the second, third,
etc. overtones have frequencies 3f, 4f, etc. (The fundamental tone
is also called the first harmonic (1f ), the first overtone is the second
harmonic (2f ), and so on.)
Which vibration will the string produce? Usually it produces
a mixture of the fundamental and a number of overtones. That
mix determines the colour of the tone, its timbre. For a guitar, for
example, you can influence this by where you pluck the string. If
you pluck it in the middle, the fundamental tone dominates; pluck
it close to an end, and many overtones are added, and the sound is
then coarser. (Flamenco guitarists use this feature a lot to produce
the distinctive flamenco sound.)
Everyday Physics: Unusual insights into familiar things
211
The frequency at which the string vibrates increases if you
tighten the string. The reason is that when you pluck the string now,
the force on it to return to its resting (middle) position is greater – the
string therefore moves faster and will be back in the middle quicker.
In this way, the frequency of the vibration increases, and you get
a higher pitch. For the same reason a short string has a higher
frequency than a long one at the same tension: for a short string, a
deviation of 1 mm from the centre causes a larger force pulling the
string back to the middle than for a long string.
The pitch of the note generated by the string is also affected
by the thickness of the string, or rather its mass. A heavy string
vibrates more slowly – and gives a lower pitch – than a light string
(see box). If you look inside a piano, you can easily see the difference
in thickness between the strings for the low notes and those for the
high notes. You can see the same thing on a violin or guitar, though
it’s not as obvious.
String weight and note frequency
A heavy string gives a lower pitch than a light string. The reason is that the heavier the
string, the more slowly it accelerates for a given force, and the slower the vibration, the
lower the pitch.
The frequency f of a vibrating string is f = (1/2L) × √ (F/μ), where L is the length of the string,
F the tension force, and μ the mass per metre.
Relationships between notes
We said that certain combinations of notes can be pleasant, but not
others – at least to western ears. It turns out that the notes that sound
well together are the overtones for a specific fundamental. Take dohmi-sol (the major triad C–E–G in musical terms). These sounds are all
overtones of the C two octaves lower. Their respective frequencies
are 4, 5, and 6 times that of lower C. They relate as 4:5:6. The whole
musical scale is that way – all simple ratios.
But not quite. Consider the guitar. Metal strips called frets are
embedded in the fingerboard along the neck of the instrument.
When you press a string down between two frets, it shortens the
string to a specific length to give you a specific pitch. If you press the
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Everyday Physics: Unusual insights into familiar things
string at the twelfth fret (just where the sound box begins), the string
length is halved and the pitch is doubled – you get a note exactly one
octave higher.
Look closely at the guitar neck (Fig. 2). Notice that the distance
between frets gets smaller the closer you come to the sound box,
as shown diagrammatically in Fig. 3. Because the frets are at right
angles to the strings, they affect all strings equally, although different
strings have been tuned to very different pitches. It looks as though
every segment between frets reduces the string length by a fixed
percentage and therefore adds a fixed percentage to the pitch. That
is indeed the case; the proof is that by fitting a capo which clamps
all the strings at any desired fret (Fig. 2) and shortens the length of
an “open” string, you can play exactly the same chords as before but
now everything sounds a half tone higher per fret shortened.
Fig. 2: A capo on a
guitar to shorten all the
strings and raise their
open-string pitch.
Everyday Physics: Unusual insights into familiar things
213
Fig. 3: The separation
between the guitar
frets gets successively
smaller as you move
from the head to the
sound box.
Now let’s work out how much the pitch is raised per segment.
Say the “increase factor” is h, i.e. if the note at one segment is f Hz,
then the note at the next segment is f × h, then at the one after that
it’s f × h × h, and so on. We know the pitch is doubled (i.e. is 2f) after 12
segments, so h12 = 2, and using your calculator shows that h = 1.05946.
That looks weird! With such a crazy number how will you get
simple ratios between two notes? For example, how do you get the
interval doh-mi, which is 5:4 or a ratio of 5/4 = 1.25? Four segments
on the guitar gives (1.05946)4 = 1.260 rather than the exact 1.25 you
wanted. For the interval doh-sol you want the ratio 6:4 that is 1.5. On
the guitar the seventh segment gives (1.05946)7 = 1.498, which is very
close but not exactly right.
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Everyday Physics: Unusual insights into familiar things
So the tones are always slightly out of pitch. However, the big
advantage is that you can start at any pitch (i.e. choose any suitable
note to be “doh”) so an accompanist can choose the starting pitch to
suit a singer’s voice.
What about the notes on a piano? If again you want to be able
to start a tune at any pitch, you must make a small concession just as
with the guitar, and fudge the pitch a little. You have to tune the piano
so that the ratio between successive keys (“half tones”) is always
1.05946; this is called “equal temperament” tuning as opposed to the
whole-number ratio scheme (so-called “natural tuning”). A variety
of equal temperament was for the first time applied by JS Bach in his
Das Wohltemperierte Klavier. Equal temperament is now generally
used on keyboard instruments.
How are the other notes affected when you change from one
method of tuning to the other? Fig. 4 shows that the differences are
remarkably small: nowhere are the “errors” that you make by going
from the natural to the even tuning larger than 0.9%. Really musical
people can hear the difference, but it is not huge. Bach didn’t think
so either.
Fig. 4: Comparison of
the relative frequencies
using the two methods
of tuning.
C
D
E
F
G
A
B
C
Natural:
1
9/8
5/4
4/3
3/2
5/3
15/8
2
or
1
1.125
1.250
1.333
1.500
1.667
1.875
2
Even:
1
1.122
1.260
1.335
1.498
1.682
1.888
2
Difference
0
0.003
0.01
0.002
-0.002
0.015
0.013
0
0.0%
0.3%
-0.8%
-0.1%
0.1%
-0.9%
-0.7%
0.0%
Difference %
Everyday Physics: Unusual insights into familiar things
215
Home experiment: The two tones of a mug
Put an empty mug with a handle on the table and gently strike the outside rim with
a spoon. The mug produces a certain tone. Whether you strike exactly opposite (180°
from) the handle or exactly on the side (at 90°), the pitch is the same. But now strike a
spot just between those two (at 45° or 135°): the pitch is clearly higher. Why? Same mug,
same spoon and yet a different pitch.
The explanation must have something to do with where
you strike the mug relative to the handle. What would
happen if the mug had no handle? Fig. 5 shows its
simplest vibration, seen from above: the original circular
shape becomes slightly oval periodically, first in one
direction (red oval) then in the other one (blue oval). The
largest amplitudes are found at four opposing locations:
the maxima of the vibrations, and one of these is where
you struck the spoon.
There are four points that do not vibrate at all; these are
called nodes, which are at +/− at 45° and 135° from the point
you struck (marked “N” in Fig. 5). If you add a handle at
one node, the vibration won’t be affected much, because
the node isn’t vibrating. However, if you stick on a handle
on to a maximum, during a vibration the handle has to
be dragged back and forth. That slows the vibration and
therefore lowers the pitch.
Fig. 5: How a circular
mug vibrates.
It’s worth trying this with a few different mugs. Some
work better than others, probably due to the quality of
the porcelain.
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Everyday Physics: Unusual insights into familiar things
56
WHY
ORCHESTRAS GO
OUT OF TUNE
When several musicians play together, they must ensure they are all
in tune. How they do that depends on the type of instrument.
Everyday Physics: Unusual insights into familiar things
217
As we saw in Doh-re-mi: the physics of musical scales (p209), the
pitch of a string is determined directly by three parameters: (1) the
length of the string, (2) its mass per unit length, and (3) its tension.
The speed of sound in air increases as the temperature increases, but
this makes no difference to how the string vibrates – and therefore
makes no difference to the frequency. Even if the concert hall warms
up, it doesn’t matter.
Wind instruments are different. Consider an organ pipe.
Here the wavelength of the sound is fixed; because of the pipe’s
construction, it generates sound waves of a specific length. Now
pitch = speed of sound ÷ wavelength, so the pitch does depend on the
speed of sound. (A different way of looking at this is that the pitch is
determined by how long a sound wave takes to go back and forth in
the pipe. The more often that happens per second the higher the
pitch, so increasing the speed of sound increases the pitch.)
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Everyday Physics: Unusual insights into familiar things
►
The only one of their three parameters that string players
can adjust quickly is the tension, so they tune their
instruments by loosening or tightening the strings.
►
Wind players tune by adjusting the length of their
instrument, typically by sliding a flexible joint in the
instrument in or out to shorten or lengthen the tube.
►
Organists are out of luck – they can’t adjust all those pipes!
All they can do is ask the rest of the orchestra to keep in tune
with them.
Now suppose the temperature increases. As we saw in Do
you hear better at night? (p199), the speed of sound increases with
temperature. The air molecules move faster; they can pass on
pressure waves more rapidly and so the sound speed increases.
(In fact it is directly proportional to the average speed of the
molecules.) And for a fixed wavelength, as in a wind instrument, a
greater sound speed means a higher pitch. So the higher the
temperature the higher the pitch. To minimize temperature changes
when playing, wind players warm up their instrument beforehand
with their breath. Therefore, some practice before the concert is not
only useful to loosen up the player’s fingers but to warm up the
instrument too.
All in all it’s surprising that so many concerts sound wonderful.
It’s just as well that while they are playing, all those musicians don’t
worry about the laws of physics!
Everyday Physics: Unusual insights into familiar things
219
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Everyday Physics: Unusual insights into familiar things
57
WHY DO EGGS
EXPLODE IN THE
MICROWAVE?
Cooking in a microwave differs from cooking in a traditional oven
in lots of ways:
►
The cooking time in a microwave depends on the amount
of food you put into it: two portions take twice as long as
one. That is actually quite reasonable. For example, if
your oven supplies 700 watts of microwave power and if
that is distributed over twice the quantity of food, it’s not
surprising it takes twice as long.
►
The food is heated from inside rather than from
the outside.
►
There can be large temperature differences between parts
of the food, if the food is stationary in the microwave.
►
Strange things can happen with metal foil and with
decorated porcelain.
►
Thawing frozen food is slow – there’s even a separate
setting for it.
►
Whole eggs explode!
The fundamental difference from an ordinary oven is that
in the microwave the heating occurs not by hot air but by
electromagnetic waves – hence the name microwave. The waves
are in fact radio waves, with a frequency higher than FM radio –
Everyday Physics: Unusual insights into familiar things
221
about 2,450 MHz (2450 million hertz, or 2.45 GHz) compared
to FM’s 100 MHz. The microwave wavelength is around 12 cm.
(Divide the speed of light by the frequency to get the wavelength:
300,000,000 m/s ÷ 2,450,000,000 Hz = 0.12 m.) And just like
ordinary radio waves, microwaves penetrate most materials
including (dry) crockery. (You know radio waves penetrate most
materials because you can listen to a radio almost anywhere – in the
house, in the car, etc.)
Because of their very high frequency, microwaves are strongly
absorbed by water. The reason is that water molecules have one
electrically positive end and one electrically negative end: they are
small electric dipoles. The water molecules therefore want to orient
themselves along the electric field of the waves, like a compass
needle does in Earth’s magnetic field. However, the electric field of
the electromagnetic waves alternates as the wave travels along from
the source, and it alternates at the frequency of the microwaves. So
the molecules flip back and forth at a frequency of 2.5 billion times
per second; they collide with neighbouring molecules, and make
them move even more. So the temperature rises. The net effect is
that all the energy of the electromagnetic waves is absorbed by the
water, and the water heats up. (Other liquids whose molecules are
electric dipoles heat up as well as water.)
Now it becomes clear why thawing is so slow. In ice (and
therefore in frozen food) water molecules do not move freely:
they are locked inside ice crystals. That limits their movement
and therefore the process of heating. Only when a layer of water is
formed that absorbs the waves efficiently does the heating speed up.
(An amusing stunt is to try to boil water in a cup made of ice before
the cup itself melts; put the whole lot in a big enough bowl so you
don’t risk flooding your microwave. Again, the reason this stunt is
possible is the relatively poor heat conduction of ice – something
inhabitants of igloos figured out a long time ago.)
Meals in the microwave can heat unevenly because the
wavelength of the microwaves (12 cm) is smaller than the oven
(typically 30 cm × 30 cm × 20 cm); the waves are reflected by
metal walls, and we get wave interference. Wherever the waves
reinforce each other it becomes extra hot and where they attenuate
each other it remains relatively cool. A rotating turntable reduces
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Everyday Physics: Unusual insights into familiar things
the unevenness: the food goes through “warm” and “cold” spots
alternately. In addition a small rotating metal fan promotes even
heating – not because it stirs the air but because it reflects microwaves
in all directions, so even if the food does not move, the “warm” and
“cold” spots wander about the oven.
Gold edges or similar decorations on the crockery can get
damaged or even cause sparking, whereas a metal spoon hardly
heats up. That seems strange because they are all metals and all
good conductors. When they are exposed to electromagnetic
waves, currents run through them, and the current in turn heats
the metal just as in an incandescent lamp. Why is there such a huge
difference between the gold edge and the spoon? The answer is that
at such high frequencies currents only run in the “skin” of metals,
i.e. in a very thin layer of about one micron (one thousandth of a
millimetre) whereas nothing happens deeper inside the metal. For
the gold layer that is catastrophic: all the heat produced in that thin
layer has nowhere to go; the temperature rises very swiftly; melting,
evaporating, and sparking is then possible. It is different for thick
objects such as spoons: because the spoon is a good heat conductor,
the heat can spread over the whole thickness of the spoon and the
temperature rises only modestly.
However, you still have to be careful with forks and knives
(as opposed to spoons). The sharp extremities can work as
lightning conductors or lightning rods: the electric fields at the
sharp extremities become so strong that an electrical discharge
can easily occur, resulting in a rain of sparks and a damaged fork.
The current only in the skin also explains why the waves are
reflected from the walls of the microwave. Whereas the metal wall is
maybe 2 mm thick, the microwaves penetrate only about 2 microns.
A tiny bit of the microwave energy is absorbed in that thin layer; the
rest is reflected (by the moving electrons that form a kind of screen
for the electric field). The door also reflects, because the holes in the
metal mesh on the door are much smaller than the wavelength of
12 cm. No radiation escapes, provided the door is properly closed.
Why do microwave ovens use the 2.45 GHz frequency
(12 cm wavelength)? Is it an exact resonance frequency of the H2O
molecule? No, that’s not the case. It is just a frequency at which water
Everyday Physics: Unusual insights into familiar things
223
absorbs the radiation well (otherwise the food wouldn’t heat up),
but not so well that the outside of the food heats up the before the
inside has had time to cook. (Furthermore, there are international
agreements that allocate frequencies of radiation for specific uses,
including this one.)
What happens if you turn on the oven while it is empty? Then
there is nothing to absorb the waves. The microwave tube that emits
the waves cannot dissipate its energy and can be easily damaged.
Finally, why do eggs explode in the microwave? If there is
nothing else in the oven, all the energy goes into the egg as heat.
The egg reaches 100°C very quickly but the heating continues, so the
water in the egg boils, and swiftly builds up pressure as in a pressure
cooker. As there is no safety valve on the egg, the pressure keeps
increasing until the shell explodes.
That makes a big mess. And it is somewhat disrespectful to
the chicken.
224
Everyday Physics: Unusual insights into familiar things
58
CAN YOU COOL
YOUR HOME
WITH YOUR
FRIDGE?
It sounds like a good idea: beat the heat using the fridge. This could
be useful in a heat wave, and maybe even give a second useful life to
an unused refrigerator: switch it on, leave its door open and there
you are!
Everyday Physics: Unusual insights into familiar things
225
Unfortunately, it doesn’t work. Even worse, it will act as a heater
instead of a cooler. That sounds strange – a refrigerator as a heater.
But if you think about it carefully, it’s reasonable. All a refrigerator
does is transport heat from its inside to its outside. It is literally a
heat pump. All the heat that it removes from its inside is dumped
neatly outside the case – into that dusty black grid of pipes on the
back of the fridge; this heats up the grid; the grid in turn releases its
heat to the air. So far the net effect seems zero, as far as your home is
concerned. You are simply moving heat, not removing it.
But how does the fridge transport the heat? It uses a (normal)
pump to circulate the cooling fluid (Fig. 1). The fluid is made to
evaporate in pipes (blue in Fig. 1) inside the cold compartment of the
fridge, thereby cooling it. The vaporized fluid condenses outside the
fridge in the black grid (red in Fig. 1) – thus heating it up; the heat is
carried away by the air into the rest of the room. Suppose the pump
is driven by a 300 W electric motor, i.e. it draws 300 watts of power
from the outlet. All this energy is turned into heat: the electric motor
and the pump both get hot. Overall, while the fridge makes some
places cold (the food compartment) and others hot (the motor, and
the dusty grid of pipes, and the room), the net overall effect is that
the fridge is a 300 watt electric heater if it runs continuously – which
it certainly will do if you leave the door open!
Fig. 1: In a fridge, the
pump compresses
(and thereby heats)
and circulates the
cooling fluid. The hot
fluid discards heat
through a grid of pipes
and cooling fans (red),
and the fluid evaporates
in tubes inside the
fridge (blue) and cools
the interior.
226
Everyday Physics: Unusual insights into familiar things
Is there any way to make that fridge cool your home? Sure,
but you ensure that the heat released is dumped outside your home.
You can do that by mounting the fridge facing inwards in an open
window and removing the door (Fig. 2), so that the heat in the grid of
pipes is carried away by the wind outside. In effect what you’ve done
is install your own air conditioning unit (and made the neighbours
wonder about you).
Fig. 2: A fridge in a
window, acting as an
air conditioner.
Everyday Physics: Unusual insights into familiar things
227
So if a refrigerator is a heat pump, could you use it to heat your
home? Sure: reverse the fridge in the window and leave the door
off. The black grid of pipes then heats the room. The cool interior of
the fridge, at a much lower temperature, is warmed by the outside
air (Fig. 3). The net effect is to pump heat from the outside world
into your room. It is heat stolen from the wind. If the outside air is
not too cold and you seal and insulate the gaps around the fridge
in the window, you have, believe it or not, a much more efficient
heating system than a normal electric heater. This is because a
heater produces heat, whereas a heat pump merely moves heat. The
ratio between the heat moved and the energy needed for doing
this depends on the temperature difference and is generally much
greater than 1. It is called the Coefficient Of Performance (COP).
Fig. 3: A fridge in a
window, acting as an
air-source heat pump.
228
Everyday Physics: Unusual insights into familiar things
59
CURVE BALLS,
BACKSPIN AND
TOPSPIN
Why are those volleys or drop shots so successful in tennis or table
tennis? And why does a ping-pong ball stay in place above a blowing
hairdryer instead of blowing away? The answer to these questions
is non-trivial, all the more because these particular phenomena
sometimes contradict our intuition. The explanation lies largely in
Everyday Physics: Unusual insights into familiar things
229
Bernoulli’s law which says that in a gas flow, the pressure is lower
where the flow speed is higher and vice versa. This law holds for
every flowing medium, including liquids. It’s a consequence of the
law of conservation of energy.
Bernoulli’s law has many applications, from flow rates in an
oil pipeline to the pitot tube on the nose or wing of an aircraft to
measure the air speed. But the law also explains many everyday
phenomena like those above, and some can be illustrated by
spectacular experiments.
First we’ll consider a tennis ball travelling to the right (Fig. 1).
With topspin the ball rotates as well as moving forward: the top of
the ball shown in Fig. 1 rotates to the right. A rotating ball flying
through the air with speed V behaves the same as a stationary
ball with the air flowing past it in the other direction with speed V
(Fig. 2), and looking at it from this vantage point makes it easier to
analyze what’s happening.
Fig. 1: A tennis ball
moving to the right with
velocity V.
Fig. 2: A stationary
tennis ball, with air
flowing past it to the
left, at velocity V. The
physics is the same as
in Fig. 1.
As Fig. 2 shows, the bottom of the topspun ball rotates to the
left, i.e. in the direction of the air flow, so the net flow speed is thereby
increased. Bernoulli’s law says higher speed means lower pressure.
On top the reverse is the case: the ball turns against the air flow, the
230
Everyday Physics: Unusual insights into familiar things
air speed is reduced, and so pressure is higher here. High pressure
above and low pressure underneath pushes the ball downwards,
increasing the total downward force (Fig. 3). The end result is that
the topspun ball makes a tighter curve than a ball without the spin
(Fig. 4). Exactly the same applies to a baseball with side spin, except
the side spin makes the ball curve sideways rather than down, which
explains the curve ball.
Fig. 3: The downward
forces on (a) a ball
without spin, and (b)
a ball with topspin.
G = pull of gravity.
P = downward
push because
of aerodynamics.
Fig. 4: Side view of
a tennis ball passing
over a net, struck with
topspin (red line) and
without spin (blue line).
Everyday Physics: Unusual insights into familiar things
231
The description above is not complete because many rotating
objects in a flow of air or water are affected by turbulence, which is
explained by what is called the Magnus effect. An illustration of this
effect is given in the Home experiment below.
Home experiment: The curved path of a rolling cylinder
Objects rotating in air or water are affected by whether the object is spinning, as we saw
above with the baseball and tennis ball. In cases in which turbulence dominates, this is
called the Magnus effect. This describes the sideways force, experienced by a rotating
object as it moves through a gas or fluid.
To demonstrate the effect in water, take a solid cylinder of PVC or acrylic, 2 cm in
diameter and about 5 cm long. (The choice of material is important: high-density
materials such as Teflon or solid metal don’t work well because they sink too fast.) Let the
cylinder roll down a slanted board into a tank filled with water (Fig. 5). An aquarium is
best, so you can view from the side what’s happening. The board should nearly touch the
surface of the water, near the middle of the tank. The cylinder is rotating when it enters
the water, having started to rotate as it rolled down the board. In the water you might
expect the cylinder to continue its forward motion, and hit the bottom some distance
towards the right of the tank. In fact the cylinder curves backward and hits the bottom to
the left of the place it entered the water.
Fig. 5:
Demonstrating
the Magnus effect.
You might expect
the cylinder to hit
the bottom to the
right of (a) which
is directly below
where it entered
the water, but
in fact it hits the
bottom to the left,
at (b).
232
Everyday Physics: Unusual insights into familiar things
Fig. 6:
Demonstrating
the Magnus effect
in air. Wind up a
paper cylinder on
two threads tied
to a stick.
If you don’t have a suitable water tank, here’s a simpler version of the experiment (or
view the video – see Resources appendix). Make a paper cylinder by rolling up a sheet
of paper. Take a wood or plastic rod/stick about 1 m long, and tie two pieces of light
sewing thread near one end, about 15 cm apart (Fig. 6). Roll the threads around the
cylinder about 3 cm from each end, so that the cylinder is gradually wound up the
threads. (Fig. 7a). Hold the stick parallel to and about 1 m above the floor. Now let go
of the cylinder. The unwinding threads spin the cylinder as it falls, and the cylinder
describes an elegant curve in the air (Fig. 7b) – thanks to the Magnus effect.
Everyday Physics: Unusual insights into familiar things
233
Fig. 7: End view of
the paper cylinder
wound up on the
two threads, ready
to be released
(a). The cylinder
unwinding as it falls
(b), describing an
elegant curve due to
the Magnus effect.
Winding in the
opposite direction
(c) results in an
opposite curve.
234
Everyday Physics: Unusual insights into familiar things
If you wind the cylinder up on the other side of the threads
(Fig. 7c), the curve will be to the left of the vertical instead
of to the right as it is in Fig. 7b.
A famous real-world demonstration of this effect was
Anton Flettner’s Baden Baden – the “sailing ship without
sails” that crossed the Atlantic in 1926. Instead of sails it had
two “rotors” – vertical rotating cylinders, driven by electric
motors. (The energy needed to drive rotors reduced
the overall efficiency, so the idea never really caught on,
although there have been a few recent ships powered by
“Flettner rotors”.)
Another illustration of the Magnus effect was the bouncing
bomb used in World War II to destroy dams on German
water reservoirs, in the so-called “Dambuster Raids”.
The bomb was set spinning before being dropped, and
was dropped just short of the dam. Because of the spin,
the bomb as it sank curved in towards the dam, and was
touching the dam when it exploded.
For videos, see Resources appendix, Magnus effect.
Home experiment: Sucking by blowing!
In a nutshell, Bernoulli’s law says that where the air flows fastest, the pressure is lowest.
You can easily create an air flow with a hairdryer, and perform a couple of surprising
demonstrations of objects being drawn to the low-pressure (highest velocity) side of
the flow.
First is a fairly well-known trick with a ping-pong ball. Switch on the dryer (on a “cold”
setting if it has one) and point it vertically upwards. Place a ping-pong ball in the air
stream about 30 cm above the dryer. Most people expect the ball to blow away, but
in fact it remains suspended above the dryer (Fig. 8a). Even more surprising is if you
hold the dryer at an angle to the vertical (Fig. 8b) – the streaming air appears to suck
the ball towards it and the ball still doesn’t fall down. The reason is that the air speed in
the centre of the flow is highest, so the pressure is lowest. If the ball leaves the centre,
it will be pushed back by the higher pressure outside the centre.
Everyday Physics: Unusual insights into familiar things
235
a)
b)
Fig. 8: The ping-
pong ball floats in
the air flow above
the dryer (a), and
remains there even
if you angle the
dryer (b).
Here’s an even more surprising demonstration. Make two disks of foam plastic /
expanded polystyrene about 30-40 cm in diameter, one about 1 cm thick, the other
3 cm thick. In the middle of the thicker disk cut a hole that tightly fits the dryer nozzle
and push the dryer’s nozzle into it (optionally making it more secure with masking
tape or sticky tape). Point the dryer – with attached disk – vertically upwards, but don’t
switch it on yet. Place the second disk on top of the first, and push a match or cocktail
stick through the middle of the second disk to keep it roughly concentric with the
first. Switch on: nothing special happens. Now ask your audience what will happen
when you turn the dryer upside down. Everyone guesses wrong! The disk doesn’t fall
off, but continues to float under the “fixed” disk instead. Thanks to Bernoulli’s law
the air between the disks has a high speed and therefore a relatively low pressure. In
contrast, the outside air has no speed at all, so it has a higher pressure which presses
the disks together.
236
Everyday Physics: Unusual insights into familiar things
Fig. 9: Even though the hairdryer is blowing air, the loose bottom disk is sucked against
the disk fixed to the dryer, and doesn’t fall off.
Home experiment: Oddly bouncing ball
It looks impossible: you bounce a little ball off the floor against the underside of a table
and it returns (Fig. 10) as though you had thrown it against a wall, almost exactly back
into your hand. How can that be, since the underside of the table is completely flat?
Why didn’t the ball continue and come out at the other side of the table?
Clearly something non-obvious is at work here. In fact this is not an ordinary ball but
a superball (often sold in science museum shops). It’s about the size of a squash ball,
but bounces much better. If you drop it from a height, it bounces back to nearly the
same height.
Everyday Physics: Unusual insights into familiar things
237
But there is more: when rotating around its own axis, that rotation (the angular
momentum) is reversed during a collision with a solid surface. This is a special feature
of a superball and it explains the strange bouncing behaviour under the table. What
happens is this. At the first bounce (on the floor, Fig. 10), the ball loses part of its
forward motion and converts it into rotation so that it looks like a tennis ball with lots
of top spin (Fig. 1, p230). At the second bounce (against the underside of the table), that
rotation brakes the forward motion, as is the case with a tennis ball with spin. But here
the effect is even stronger: the ball actually starts to rotate the other way. Therefore,
the effect is twice as large, causing the ball to return the way it came. Then, at the third
bounce (against the floor) this new rotation of the ball repeats that effect once more so
that the ball comes back more or less on its original path.
Fig. 10: The superball bounces back the way it came (solid line), instead of coming out
the other side as expected (dotted line).
What happens if you use an ordinary ball instead? When it collides with the ground or
the table it converts a much smaller part of its forward speed into rotation (spin). The
same laws of physics apply to both types of ball, but the exchange of linear momentum
and angular momentum is different for the two, so that at every bounce the ordinary
ball slows down less. The result is that halfway under the table the ball gets stuck
bouncing up and down under the table and doesn’t come out the other side.
What if you use a perfectly smooth ball? That doesn’t start rotating at all and therefore
doesn’t slow down on collision. It will zigzag between table and floor and reappear on
the other side. (If you don’t have a smooth ball, you can take an ordinary ball and smear
it with soap to make it smooth.)
238
Everyday Physics: Unusual insights into familiar things
60
HOW MUCH
POWER CAN
YOU GET FROM
SOLAR ENERGY?
The sun radiates so much energy that if we express it in common
energy units, we get a number so large it’s hard comprehend.
Einstein’s famous equation E = mc2 says energy and mass are
equivalent, so instead of expressing the sun’s energy output in watts,
we can express it in radiated kilograms per unit time instead. If we
do that, it turns out that the sun loses more than 4 billion kilograms
per second – about the mass of 3 million compact cars – every second.
And it has been doing that for billions of years!
Everyday Physics: Unusual insights into familiar things
239
The sun is an unimaginably gigantic stove. It’s impossible to
imagine a fire 150 kilometres away from where you live that’s so
big it keeps you as warm as the sun does. Let alone at 150 million
kilometres – nearly 4,000 times the circumference of the Earth!
What can we do on Earth with all that solar energy? First,
let’s calculate how large the energy flow is (see box): we find it’s
1,350 watts per square metre just above the atmosphere or – because
the atmosphere absorbs some of the energy as it passes through –
just over 1,000 watts per square metre at the Earth’s surface on a
cloudless day. The solar radiation per square metre is therefore the
equivalent of a 1,000 watt electric heater. The average radiation
per horizontal square metre, averaged over day and night and
over summer and winter, turns out to be about 100 W in the UK;
between 150 and 250 in the US; between 200 and 250 in Australia;
and about 150 in New Zealand.
The description above is not complete because many rotating
objects
in much
a flow ofsolar
air or energy
water arereaches
affected bythe
turbulence,
How
Earth? which
explained by what is called the Magnus effect. An illustration of this
The surface of the sun behaves like a radiator with a temperature, T, of 5,750K (kelvin).
effect is given in the Home Experiment below.
Once you know that, you can calculate the total emitted energy flow. According
to the Stefan–Boltzmann law, that energy flow is proportional to the surface area
<box start>
of the sun and to the fourth power of the temperature. The proportionality constant
Home
experiment: The curved path of a
is σ, the Stefan–Boltzmann constant. So the total energy flow in watts is:
rolling
cylinder
2
(4 π R ) σ T4, where R is the sun’s radius.
Objects rotating in air or water are affected by whether
At
Earth,
at a distance
a from
the sun,
thebaseball
energy isand
distributed over an imaginary
thethe
object
is spinning,
as we
saw above
with
2
π
a
(Fig.
1)
so
the
energy
per
square
sphere
of
surface
area
4
tennis ball. In cases in which turbulence dominates, this is metre is
called
effect. This describes the sideways force,
(4 πthe
R2) Magnus
σ T4
experienced by a rotating object as it moves through a gas
or fluid.
4πa2
or σT4 (R/a)2. Now substituting the radius of the sun (R = 7.0 × 108 m), the sun–
To demonstrate
effect
in9 water,
take
a solid
cylinder and
of the value of the constant
m), the
solar
temperature
Earth
distance (athe
= 150
× 10
PVC
or
acrylic,
2
cm
-8
-2 in
-4 diameter and about 5 cm long. (The
(σ = 5.7 × 10 W·m ·K ), we then find an energy flux density of 1,350 W/m2.
choice of material is important: high-density materials such
as Teflon
metal
don’tthat
workthe
well
because
they sinkover is perpendicular to the
Note
that or
thesolid
“square
metre”
energy
is measured
too fast.) solar
Let the
cylinder
roll down
a slanted
boardBecause
into a of different angles of the sun’s
incident
rays,
not a square
metre
of ground.
incidence at different latitudes and at different times of year, the 1,000 W/m2 is distributed
over a wider area (Fig. 2).
240
Everyday Physics: Unusual insights into familiar things
Fig. 1: At
the Earth, at
a distance a
from the sun,
the energy is
distributed
over an
imaginary
sphere with a
surface area
of 4πa2. (Not
to scale.)
Fig. 2: The calculated energy flow is per square
metre perpendicular to the incident solar rays. On the
ground, the same 1,000 W of energy is spread over a
wider area – e.g. over 1.1 m2 in summer (left), or over
4 m2 winter (right).
Everyday Physics: Unusual insights into familiar things
241
Fig. 3: Nuna7, winner
of the 2013 World Solar
Challenge car race.
Credit: Creative
Commons AttributionShare Alike 3.0
Unported, Wikimedia
user GTHO
242
You can capture this energy in order to heat your home. The
simplest way to do this is to let the sun shine into your windows,
or into a conservatory which warms the air entering your home.
The most efficient way is to use solar-thermal collectors, which are
up to 75% efficient, depending on the installation and the type of
application.
You can also use sunshine to generate electricity, using
solar panels of photovoltaic cells. However, they are less efficient:
common silicon solar cells reach a maximum of about 15%, so of the
1,000 watts incident solar power, these cells give you 150 watts of
electric power at most. More advanced cells (such as used in space
technology, or for the “Nuna” cars (Fig. 3), many-times winners of
the World Solar Challenge car race in Australia) reach 25%, but they
are very expensive; leading-edge research cells in the laboratory
have reached an efficiency of >45%.
Everyday Physics: Unusual insights into familiar things
Another way to “harvest” solar power is biomass – burning
plants or trees to provide heat or, indirectly, electricity. To determine
how useful this is, you need to know how fast things grow, which
in turn is determined by the efficiency of photosynthesis. That of
course depends on the kind of crop, but in temperate areas with
adequate water a reasonable estimate for the annual average for
photosynthesis is 1%, i.e. for each 100 kWh of incident energy, you
grow enough to burn to give 1 kWh of heat. That’s not a lot, but an
advantage of biomass over photovoltaics is that you get “energy
storage” for free: you can easily store wood and burn it when it suits
you, whereas storing large amounts of electricity is very expensive.
In terms of oil “energy equivalents”, the yield of biomass is about one
litre of oil per square metre per year.
Everyday Physics: Unusual insights into familiar things
243
Land area needed for
power generation
For large-scale photovoltaic and for biomass, the area of land
required is important. How much space is needed for a large solar
electric power station? Photovoltaic cells produce electricity directly.
For biomass we have to take into account the transformation from
heat to electricity. Consider a 1,000 MW (1 Gigawatt, 1 GW) power
station generating electricity, averaged over the year.
The space needed with current technologies turns out to
be large:
►
For photovoltaic cells (with 15% efficiency), area needed =
60 km2 (the area of the island of Manhattan, NY).
►
For power generated from biomass (with 1% efficiency),
area needed = 3,000 km2 (a little less than the area
of Long Island, NY, or a little more than the country
of Luxembourg).
If, for example, the Netherlands wanted all its energy to come
from biomass – by continually sowing and harvesting – it would need
an area bigger than the Netherlands and Belgium combined. It’s clear,
therefore, that much research is needed, both in the production of
cheap but high-efficiency solar cells, and artificial photosynthesis
with greater efficiency than nature’s. This is a suitable task for our
politicians to invest in, provided of course they ever look beyond
the next election.
244
Everyday Physics: Unusual insights into familiar things
61
THE MYSTERY OF
THE WANDERING
CARPETS
Wall-to-wall carpets do not wander about, but their smaller relatives
do: loose rugs and runners that are not tacked down can slowly
get out of place, and have to be put back every so often. Curiously,
they always move in the same direction, even if the floor is perfectly
horizontal. Why is there a preferred direction?
Everyday Physics: Unusual insights into familiar things
245
It is because of the laws of mechanics. If you walk over a carpet
at constant speed, nothing happens. Granted, there is a backward
force when you push off with your back foot, and a forward force
when your forward foot lands, but they even out so the net force
is zero (Fig. 1). Even on the most slippery surface the carpet stays
neatly in place. But as soon as you stop, the carpet and the floor have
to deliver enough friction for you to brake. The carpet therefore
tends to shift in the same direction you’re walking in. Even if it can
move only a millimetre, it will.
Fig. 1: (a) When you
step on to the rug, the
force tends to move the
rug forward. (b) When
you step off the rug, the
force tends to move the
rug backward.
But isn’t that shift compensated for as you leave the carpet,
when you walk out of the room for instance? It depends: yes, if
you continue walking in the same direction as before; that’s what
happens to a doormat when you walk in after wiping your feet. But
it’s different if after stopping you go back the way you came: the
“pushing off” as you resume walking repeats the effect caused by
your stopping. Think of a rug in the middle of a room that has only
one door: the rug will wander away from that door, even if it lies in
the middle of the room. In a room with two doors, the rug will drift
away from the door that has the most use.
So show me how a carpet drifts over time, and I will tell you
the typical walking pattern of the people who use that room.
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Everyday Physics: Unusual insights into familiar things
62
COMMON
MISCONCEPTIONS
ABOUT THE
RADIOMETER
Science museums and even shop window displays often display a
radiometer. It consists of a glass bulb with a small horizontal rotor
on a vertical axis (Fig. 1). Each vane is black on one side and white
or silver on the other. When illuminated brightly, the rotor turns like
crazy – see the Resources appendix for a video.
Everyday Physics: Unusual insights into familiar things
247
Fig. 1: The radiometer.
Credit: https://commons.
wikimedia.org/wiki/
User:Haade, CC-BY-SA3.0
<image>crookesradiometer=haade.jpg
A common misconception (repeated even in a popularscience magazine a few years ago) is that the rotation is caused by
the pressure of the light. Granted, light does exert a pressure: to put
it in scientific terms, photons – the quanta or “particles” of light –
have momentum. So the argument goes that the light rays bounce
off the shiny side of the vanes, but are absorbed by the black side
(Fig. 2). The result is a net force that turns the rotor.
But in fact that isn’t correct. The light pressure is tiny. (The
value of the light pressure, given by a detailed calculation, is the
number of watts per square metre divided by the speed of light.)
Even if the vanes are in full sunlight, the force on a vane is a few
tenths of a billionth of a newton (10-7 N). That’s equivalent to the
weight of a tiny piece of hair about 0.01 mm long – so small you can
hardly see it. Such a small force couldn’t conceivably turn the rotor
248
Everyday Physics: Unusual insights into familiar things
so quickly, not even in a vacuum. Moreover, the mill turns the wrong
way to be explained by light pressure: it isn’t the shiny side that is
pushed back, but the black side.
Fig. 2: The wrong
explanation of why
the radiometer
vanes rotate.
The real mechanism is completely different. The only effect
of the incident light is to heat the sides of the vanes. The heating is
uneven: the black side gets a bit hotter than the shiny side. The force
that makes the rotor turn is actually delivered by gas molecules. The
glass bulb is only a partial vacuum and contains some gas at low
pressure. In gas at sufficiently low pressure, funny phenomena (like
this) occur that don’t occur at normal pressures.
Let’s look at exactly what’s happening. The gas pressure is so
low that the molecules hardly ever collide with each other; mostly
they bounce from wall to wall inside the glass bulb, and occasionally
hit the vanes. Because the bulb is symmetrical, there’s no preferred
direction of travel for the molecules, so the number of molecules
striking the shiny side of the vanes is the same as the number
striking the black side. And the molecules have equal speeds, since
they have just bounced off the same glass wall. In other words, when
Everyday Physics: Unusual insights into familiar things
249
molecules strike the vane surface, they deliver equal momentum
whether the surface is black or shiny; so far, there is no net force to
move the vane.
But the above is only for the striking of the vanes; things
are different when a molecule bounces back off a vane. During the
nanosecond or so when a molecule is touching a vane surface,
the molecule takes on the temperature of the surface, and that
temperature determines the take-off speed. Molecules leave the
black (hot) surface at higher speeds than the shiny (cooler) surface
– they push off from the black surface more strongly than the shiny
one (Fig. 3). Therefore, it’s the black surface that is pushed back, not
the shiny one, which is the opposite of what would happen if it was
light pressure that caused the rotation.
Fig. 3: It is the air
molecules that rotate
the vanes. Molecules
striking the (hotter)
black surface rebound
at higher speed than
molecules rebounding
from the (cooler)
white surface.
This phenomenon has been studied in detail, including the
case of higher pressures where the molecules start colliding with
each other, to determine the optimum pressure to make the rotor
turn fastest. (In 1924 Albert Einstein published a paper about this –
see Resources appendix.)
250
Everyday Physics: Unusual insights into familiar things
63
IS THICK GLASS
A BETTER
INSULATOR THAN
THIN GLASS?
Glass is a poor heat conductor. The handle of a glass teacup doesn’t
get nearly as hot as the handle of a tin mug. If glass insulates so well,
could you double the insulation value of a window by making the
glass twice as thick?
Everyday Physics: Unusual insights into familiar things
251
The answer is no. Even though doubling the thickness of
the glass doubles the heat resistance, it makes no difference to the
window. Surprisingly, even a cellophane film of 0.1 mm thick works
just as well as thick glass, which is why this sort of film is used as a
temporary form of double glazing.
The reason is that air insulates far better than glass – it
conducts heat about 40 times less well than glass. In a window, 99%
of the insulating capacity is due to the air layers adjacent to the pane.
Doesn’t the air need to be still to insulate well, whereas air
flows all the time, especially on the outside of a window where the
wind blows? Indeed there is flow – due to convection – even inside.
This is a good thing, because with this air flow the temperature in
the room becomes reasonably even: warm air above the radiator
expands and rises, flows along the ceiling and comes down on the
other side of the room. Without this “natural” convection there
would be even bigger temperature differences in your rooms than
there are now – warm at one chair, cold at the other.
However, this doesn’t affect the insulation properties of the
window, because heat is transported by convection only in the
direction of flow, since it is the travelling air that takes the heat along.
In a layer adjacent to the windowpane, air can only flow parallel to
the pane: it cannot go into it. This layer usually has a thickness of a
few millimetres, and in this layer the heat transport must take place
by conduction, as if the air really were stationary. And a stationary
air layer insulates 40 times better than a glass layer of the same
thickness as we said above.
In fact the air has a double benefit, because there are two
layers, one on the inside of the glass, and one outside. The exterior
layer is slightly less effective – slightly less thick – than the interior
one, due to wind. But together these layers are some 4 to 5 mm
thick, thicker than most panes of glass, so they easily outperform
glass in the insulation contest.
Because air is a much better insulator than glass, in a window
we can regard the glass as a good conductor within which the
temperature is practically homogenous; there is no temperature
difference between the interior and exterior of the pane (Fig. 1,
252
Everyday Physics: Unusual insights into familiar things
inset). The temperature gradient occurs exclusively in the insulating
air layers, so the temperature profile is as shown in Fig. 1. It’s clear
that the temperature of a single pane is quite low. With the external
temperature at −10°C and the interior temperature at +20°C, the
temperature of the glass can go below freezing, and when the winter
is very cold you can see frost on the inside of single-glazed windows!
Double glazing doubles the number of air layers, as long as
the in-between space is at least about 6 mm. (A much wider cavity
doesn’t improve the heat insulation, because convection limits the
insulating action to the layers adjacent to the panes. However, a
wider cavity does improve the noise insulation.) Fig. 2 shows the
temperature profile, with the same inside and outside temperatures
as in Fig. 1. The insulating value is practically doubled: you now have
four air layers instead of two, and an additional advantage is that
the temperature of the inner pane is much higher than in the case
of single glass. That adds comfort, because it reduces the draughts
from the window.
Everyday Physics: Unusual insights into familiar things
Fig. 1: The
temperature profile
across a single-glazed
window. Inset: note
that the temperature
on the inside surface of
the window is almost
exactly the same as on
the outside surface.
253
Fig. 2: Double glazing
insulates a room more
effectively than single
glazing, and the much
higher temperature
on the interior side of
the glazing reduces
draughts from the
window. The curve in
the temperature profile
symbolizes the onset
of convection between
the panes.
Radiation also affects your comfort (see How do you keep your
temperature constant? p49). The side of your body facing the cold
window will be colder than the other side, and this unequal cooling
can feel unpleasant.
What about “secondary double glazing” – a removable pane
of glass, sometimes in a light frame, that’s temporarily fixed over
the window? As is clear from the above, these are just as effective as
real double glazing. In the US, these are called storm windows and
can be installed inside or outside. In the UK they are almost always
installed inside.
Instead of an extra pane of glass, you can get purpose-made
very thin plastic film (like cling film / food wrap) to stretch over
the inside window frame and tape on at the sides. This also works
well. The only real requirement is, after all, that the extra surface
is impermeable to air. Thickly woven curtains that close well are
nearly as good in this respect.
254
Everyday Physics: Unusual insights into familiar things
64
IS THERE A
VACUUM INSIDE
DOUBLE-GLAZED
WINDOWS?
Double glazing is a great invention, substantially improving the
insulation of our homes, as we’ve just seen. It works because you’re
adding an extra layer of air between the two panes of glass, and air
is a good insulator. Even better would be to completely evacuate
the space between the two panes – after all, a vacuum is the best
possible insulator.
That sounds good, but how big is the force on such a pane,
and can the panes stand it? Let’s compute. Atmospheric pressure
corresponds to a mercury column of 76 cm. Mercury is very dense
(13.6 times denser than water), so the weight of such a mercury
column on 1 square centimetre is 76 cm × 13.6 g/cm3 = 1,033 g/cm2
or more than a kilogram! That’s an awful lot and really adds up for
a large surface. Take a window that’s one metre square (which isn’t
particularly big as windows go): its area is 10,000 cm2. So assuming
there’s a vacuum on the other side of the glass, the force on the pane
is a weight of 10,000 kg, the same as about 8 small cars. So a vacuum
between the panes is impossible – there’s no way the glass could
withstand the pressure without breaking. Unless you play a trick. You
can get evacuated double glazing panels that have tiny transparent
pillars inside to keep the panes from collapsing. The manufacturer
says “they are 0.5 mm diameter and 20 mm apart, [and] when
viewed in transmission at a distance of 1-2 metres they become
imperceptible.” (See Resources appendix.)
Everyday Physics: Unusual insights into familiar things
255
The above example illustrates the enormous force of
atmospheric pressure on a relatively large area, which was first
demonstrated by Guericke’s famous 1654 experiment in Regensburg:
he placed two hemispheres together (Fig. 1), evacuated them, and
teams of horses couldn’t pull them apart. Another example is an oldstyle TV with its deep picture tube or “cathode ray tube” (CRT) which
was also a feature of old computer screens. The tube has to maintain
a vacuum to allow the electron beam to move unhindered and form
the picture on the screen; it must therefore be able to withstand the
external pressure without imploding in the middle of your living
room. The atmospheric force on the front of a TV with a screen of
40 by 50 cm (that is 2,000 cm2) is 2,000 kg, so the glass must be
very thick to be strong enough. This explains why for half a century
TVs hardly became lighter in spite of the miniaturization of all other
electronic components: it was simply the weight of the glass.
Back to double glazing. Could you get at least some of the
benefit of a vacuum by lowering the internal pressure a bit, e.g. to
0.9 or 0.8 bar, which might reduce the heat loss by 10 to 20%? Alas,
no: it turns out that the heat conduction of a gas is independent of
the pressure. Here’s why. Say you halve the pressure. Then there are
only half as many molecules left to transport the heat, so you might
256
Everyday Physics: Unusual insights into familiar things
Fig. 1: Guericke’s
original “Magdeburg”
hemispheres (named
after Guericke’s
hometown, where
he later repeated the
demonstration) and
vacuum pump, in the
Deutsches Museum,
Munich, Germany.
Credit: Creative
Commons AttributionShare Alike 3.0
Unported, LepoRello
(Wikipedia)
think the heat conduction would be halved. However, that’s not true:
each molecule now transports heat twice as efficiently because the
molecules hinder one another less – the length of the average path
between two collisions is doubled. The end result is that decreasing
pressure does not affect the heat losses. Insulation is improved only
when the pressure becomes so small – far below a thousandth of an
atmosphere – that the molecules hardly collide at all, but fly from
pane to pane unhindered. Then, the heat conduction becomes
proportional to the number of molecules, and does decrease with
reducing pressure.
Everyday Physics: Unusual insights into familiar things
257
Home experiment: Greedy balloon?
When you’re inflating a balloon, the first few puffs are the hardest, even if you’ve
stretched the balloon a few times, but as soon the balloon reaches a certain size, it’s
easier to inflate. This is surprising, because it’s the opposite of what happens when you
stretch a rubber band: it’s easiest at the beginning. Similarly it’s easiest at the beginning
when you stretch an uninflated tyre, so it can’t be the type of rubber that makes it hard
to inflate the balloon; something else is happening here.
More surprising still is what happens if you connect two identical balloons with a piece
of flexible tubing or garden hose that you can squeeze shut with your fingers (Fig. 2).
Attach the first balloon to the tube, and inflate it to a reasonable size (roughly twothirds of its maximum size); now squeeze the tube shut between two fingers. Inflate
the second balloon somewhat less (about half the size of the first one), and attach it to
the other end of the tube (Fig. 2a). And now for the exciting moment: what happens
when you release your fingers on the tube, so the air can flow freely between the
two balloons? You’d probably expect that the air will distribute itself equally between
the two balloons, and they will end up the same size. After all, that’s what happens
when you pull on two elastic bands that are tied together: they stretch equally
far. However, that’s not what happens here (Fig. 2b): the larger one grows and the
smaller one shrinks!
258
Everyday Physics: Unusual insights into familiar things
The explanation is not that one balloon is greedier than
the other. In fact it relates to something you notice when
inflating an empty balloon: when the balloon is small it
needs a lot of pressure to inflate. To see why that is, we can
view the inflating as pushing two imaginary halves of the
balloon away from each other. To do that you must apply a
force, F say, that’s just big enough to stretch the rubber. You
may remember that force is equal to pressure times area
(because by definition, pressure is force per unit area). Now
the area – the cross-section of the balloon – increases with
the square of the diameter: twice as large a balloon has an
area four times as large an area of cross-section. This bigger
area requires less pressure for the two imaginary halves
to be pushed apart, so the balloon grows more easily. (We
tacitly assumed that the counteracting forces of the rubber
remain constant in the process. This isn’t quite true, but
these forces increase only slightly, as long as we are within
the elastic range of the rubber.)
Everyday Physics: Unusual insights into familiar things
Fig. 2: Two balloons
connected by a tube:
(a) with the connecting
tube closed, and (b)
with the tube open.
259
Thinking about inflating a balloon that has a long sausage-like shape might remind
you of what we’ve just explained. When you blow into a sausage balloon (Fig. 3a),
as it inflates it doesn’t thicken evenly along its length but in fact inflates at one place
(Fig. 3b) until the “elastic limits” are reached, i.e. until the rubber there is stretched so
much that it’s much harder to stretch any more. Only then does it become thick over
the whole length (Fig. 3c).
Fig. 3: Inflating a sausage-shaped balloon (a) causes the balloon to inflate initially at one
place (b) and not evenly along its length; only when fully inflated (c) is the balloon even
along its whole length.
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Everyday Physics: Unusual insights into familiar things
65
CAN YOU
FEEL ENERGY
CONSUMPTION?
Everyone knows you can measure energy consumption, or
look in the instruction manual to find the rating of your electrical
appliances. But in order to make a rough estimate, that isn’t necessary
– most of the time you can feel the energy use via the heat an
appliance produces. After all, according to the law of conservation of
energy, the energy has to go somewhere and that’s nearly always in
the form of heat, so in effect a 1,000 W vacuum cleaner is a 1,000 W
heater, a 60 W lamp is a 60 W heater, and so on. If you want to know
if your DVD player, the hand-mixer or internet router use much
Everyday Physics: Unusual insights into familiar things
261
energy you need only put your hand on it to find out. Appliances
that remain cool cannot be big energy consumers (although even
modest energy consumption used 24/7 does add up over time).
If an appliance has a ventilator for cooling, that’s a bad sign,
because it suggests that the manufacturer expects the appliance to
produce a lot of heat. And things designed to produce heat – central
heating units, boilers, washing machines, dishwashers – are the
really big consumers.
Here’s an example of estimating energy use – for an electric
shaver. It can’t use much energy, because the only place that the
energy can go is conversion into heat – from the electric losses
and mechanical heat of friction in the shaving head. That cannot
be much: during the whole shaving operation of a few minutes the
head barely heats up, and the amount of heat generated is much less
than you’d need to heat a basin of water for a wet shave. So we can
estimate with a rough calculation that an electric shave uses about
as much energy as running the hot water tap for one second. A wet
shave doesn’t save energy – quite the opposite, unless you shave with
cold water.
Of course, when considering electricity, you must also take
into account the efficiency of the power-generating station, and any
further losses that occur before the current gets into your house.
For every joule (one watt for one second) of electricity that comes
into your house, nearly three joules of energy were consumed at the
power station, assuming it runs on natural gas, coal, oil or nuclear,
rather than solar or wind. So we have to multiply the electricity
consumption by three for a true whole-energy comparison with,
e.g., the gas used in the central heating unit that has an efficiency of
nearly 100%, or with the car that runs on petrol or diesel fuel.
Fig. 1 shows a rough comparison of various “appliances”. It
gives round numbers for the power in watts, and the energy use in
terms of litres of oil or petrol per hour, or cubic metres of natural gas
per hour. The figures assume that the appliance is used 24 hours/
day, at the specified power. The conversion losses are included. So
the energy use of a car in terms of power (watts) is taken as five times
the mechanical power of the engine, because the engine efficiency is
taken to be 20%, i.e. the energy delivered by the engine is only 20%
of the energy in the fuel.
262
Everyday Physics: Unusual insights into familiar things
Power in
watts
Equivalent energy
consumption in m3/hour of natural
gas, or litres of oil / hour
Clock (3 W electric)
10
0.001
Shaver (10 W electric)
30
0.003
Candle; small pilot flame;
a human adult
100
0.01
Lamp; laptop (100 W)
300
0.03
PC, TV (200 W)
600
0.06
Large gas ring
2,000
0.2
Vacuum cleaner (1,000W)
3,000
0.3
Kitchen instant water
heater (on natural gas)
10,000
1
Midsize central
heating unit
20,000
2
Generous shower
20,000
2
Car (100 km/h, 16 kW)
80,000
8
Full car, per person
20,000
2
Full airliner, per passenger
400,000
40
The table shows some interesting points. The energy used in
a day by an old-fashioned pilot flame or a candle is enough to watch
TV all evening, or to shower for ten minutes. With the energy used
in a day by a 100 W electric appliance, you can drive ten kilometres.
A full airliner uses 20 times as much fuel per passenger per hour
as a full car on the motorway. Of course, the aircraft goes 10 times
as far as the car in an hour, so it uses only twice as much fuel per
kilometre; therefore driving to your holiday destination in a half-full
car uses about as much fuel as flying there on a full plane.
Everyday Physics: Unusual insights into familiar things
Fig. 1: Comparison
of whole-energy use of
various appliances.
263
Rules of thumb for the running cost of appliances
One watt (1 W) used constantly for a year = 1 W × 3,600 s/hour × 24 hours/day × 365 days/
year = 31,536,000 watt seconds (Ws) = 31,536 kilowatt seconds (kWs) = 31,536/3,600 kWh =
8.7 kWh ≈ 10 kWh. In other words, if you run something for a year that uses 1 W, it uses
about 10 kWh/year. Now for the cost:
►
If a unit (i.e. 1 kWh) costs 10 pence or cents or whatever your currency is, the yearly cost
of your 10 kWh is about £1 – or $1 or, etc.
Using this, you can easily estimate the annual cost of an appliance. If you leave a 100 W
bulb burning in the basement all year, that costs you about 100 × £1 = £100. If you leave
your 8 W internet router switched on 24/7, it will cost you £8 per year.
(If a unit costs 15 pence or cents instead of 10, the yearly cost is 1.5 times the above, so just
multiply up accordingly: e.g. your 8 W internet router will cost 8 × 1.5 = £12/year.)
►
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Everyday Physics: Unusual insights into familiar things
66
IS A BLACK
CENTRALHEATING
RADIATOR
BETTER THAN
A WHITE ONE?
Most central heating radiators are white or another light colour
that suits the interior decor. You might think that’s silly – after all,
don’t radiators yield much of their heat by radiating, and doesn’t a
black surface absorb and radiate heat best? (And that’s why you use
a chromium coffeepot, and pack your baked potatoes in aluminium
foil, because they radiate heat poorly.)
It’s true that a surface that absorbs radiation poorly, such as
a white surface, also emits radiation poorly. (In technical terms, the
absorption coefficient is equal to the emission coefficient.)
Everyday Physics: Unusual insights into familiar things
265
That is actually correct but there is a snag. A surface with a
low absorption will indeed also have a low emission, but at that
wavelength. A surface that you perceive as “white” has little absorption
for visible light, so that also goes for that white radiator. But a heating
radiator doesn’t emit any visible light: it radiates in the far infrared,
with a wavelength about 20 times longer than visible light (about
10 μm compared to 0.5 μm). And the absorption properties at these
very different wavelengths can also be very different.
The surfaces of most radiators are painted and, in the
infrared region we are talking about, most types of paint surfaces
are efficient radiators: they are “black” at those wavelengths even
though are “white” at visible wavelengths. White paint on a radiator
is practically as efficient as black paint.
There are exceptions: metallic paints such as aluminium.
These are fairly “white” in the infrared and therefore poor heat
radiators, just as is the case for most metals themselves. So that shiny
metal coffeepot is certainly clever, and so is the aluminium foil:
they radiate heat poorly and keep the contents warmer for longer.
What if you have an aluminium-painted radiator at home? Well,
it’s not as bad as it seems. The heat that would have been radiated
from a different radiator isn’t lost – it just goes back to the central
heating boiler. However, the temperature of the radiator has to be
raised in order to have the same heat emission. That means a higher
temperature of the water in the system, which reduces the efficiency
of the central heating unit somewhat. Not ideal, but no reason to
rush for your paintbrush right away.
Do radiators radiate?
Many internet articles say that radiators don’t radiate much, and that instead they yield
most of their heat by convection. Is that true?
For radiators consisting of three panels, each behind the other, it is true – about 80% of the
heating is by convection. However, for single-panel radiators, about 50% is by convection
and 50% by radiation. (The radiation percentage in multi-panel radiators is lower because
the front panel blocks radiation from the middle and back panels.)
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Everyday Physics: Unusual insights into familiar things
67
DOES BLACK
PAINT GET
HOTTER THAN
WHITE?
South-facing doors and window-frames have a rough time when the
sun shines. Wood is a poor thermal conductor and can’t conduct
away the solar heat easily; the paint gets hot, expands and shrinks,
and the surface has to be repainted frequently.
Everyday Physics: Unusual insights into familiar things
267
A reasonable question to ask is does the colour of the paint
matter? Does white paint stay cooler than other colours? At first you
might say yes, because white surfaces absorb less solar heat than
dark ones, so they stay cooler. But in fact it’s not that simple. Aren’t
surfaces that are poor absorbers also poor emitters? In other words:
the net effect of colour on temperature may well be zero.
In fact the first, intuitive answer is correct: colour does matter.
You can prove it with a simple measurement. Around noon on
a sunny day in June, my digital thermometer told me that a white
window frame was 43°C and a dark green one was 66°C. That’s
a significant difference. (The real difference may be larger, since
touching the surface with the temperature sensor drains away a bit
of heat from the painted wood, yielding readings which are too low.)
So why is the net effect of colour not zero? The reason is that
our analysis above was incomplete: the crucial point is that whereas
a lot of the absorption by the paint is in the visible range, the emission
(as discussed in the previous chapter) is in the far infrared, where
the paint’s emission properties may well be different. And sure
enough, they are. In the infrared, almost all painted surfaces behave
as if they were black, so while different-coloured paints may have
different absorption for the incoming visible light, they are almost
identical for the infrared emission. The overall effect is that white
paint therefore remains cooler because it absorbs less radiation
than black, while emitting just as efficiently.
This effect is like two buckets, both having the same large hole
in the bottom. One is being continuously filled by a small jet of water
and the other by a larger jet. When the levels reach equilibrium, the
water level (corresponding to “temperature”) in the bucket filled
by the larger jet will be highest. And sure enough, that’s the “black
bucket”. Correspondingly, when warmed up by the sun, white paint
takes less time than black to reach the point where absorption and
emission are balanced. (The black paint has to get hotter before the
two balance.)
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Everyday Physics: Unusual insights into familiar things
68
DOES BABY
REALLY NEED
A HOT WATER
BOTTLE?
People who use a lot of matches notice a problem with big economypack matchboxes: the striking surface often wears out too quickly
so you can’t use the rest of the matches. Why does this happen: on
a bigger box, isn’t the striking surface bigger too? (Don’t worry, this
really does have something to do with Baby’s hot water bottle!)
The secret is in the difference between volume and surface.
The number of matches in the box is proportional to the volume
Everyday Physics: Unusual insights into familiar things
269
(and in all your boxes the matches are the same size). Twice as big
a box means twice the length, twice the width, twice the height, i.e.
2 × 2 × 2 = 8 times the volume. So your big box contains eight times
as many matches as the small one.
Now consider the striking surface, or two striking surfaces if
there’s one on each side. Their area has become twice as long and
twice as wide, therefore 4 times as large. However, you have 8 times
as many matches, so you need the striking surface to last twice as
long, which sometimes it doesn’t.
This is just one example of the importance of the area/volume
ratio when comparing objects that are similar but of different sizes.
For example, this ratio explains why animals don’t become much
bigger than they are. Elephants’ legs already have to be very thick to
support so much weight. Making an elephant 2 times as big would
mean that the cross-sectional area of its legs would be 4 times bigger,
whereas its weight would be 8 times bigger; so mega-elephants’
legs wouldn’t be able to support this double load per unit area. In
contrast, a ladybird has extremely thin legs for a relatively stout body;
as beasties become smaller their weight decreases disproportionally
rapidly, so skinny legs easily support them. Similarly, an ant can carry
eight times its own weight (but don’t expect this from an elephant!).
The same area/volume argument applies to aircraft too.
Simply doubling all dimensions of a Boeing 747 would result in an
aircraft that couldn’t fly: the 4-times-greater-area wings would be too
small to carry the 8-times-greater weight.
Now back to Baby. For people, the area/volume ratio is really
important. We’ll continue to assume that when a child becomes
twice as big, every part of their body becomes twice as big. So
then their surface area is 4 times as big but their weight is 8 times
as big. Adults therefore have a relatively large body volume with
respect to their surface area, whereas Baby has a large surface area
with respect to volume. Baby therefore cools down faster than an
adult, because the amount of cooling is determined by the surface
area, whereas the heat content (or heat production) is determined
by the volume.
Conclusion: when it’s cold, wrap up Baby warmly.
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Everyday Physics: Unusual insights into familiar things
69
THE LID ON THE
SAUCEPAN
The science section of a newspaper posed the following problem:
when you boil water in a saucepan (or cook something in boiling
water), you think that by putting the lid on the pan you’re reducing
the amount of moisture that’s evaporating into your house. However,
when the editor tested this by experiment, they found it made no
difference. A fixed quantity of water evaporated, whether the lid was
on or not.
This seems strange: with the lid on the pan a large part of the
water vapour condenses inside the lid and runs back into the pan.
Nevertheless the net evaporation is the same as without the lid!
Everyday Physics: Unusual insights into familiar things
271
The law of conservation of energy explains this. What exactly
happens in a pan full of boiling water?
►
Your gas burner or electric ring provides a fixed quantity of
heat per second to the pan.
►
Once the water has been heated to boiling point, exactly the
same amount of heat must be removed (i.e. go somewhere
else) because as long as the water is boiling, the temperature
remains constant at 100°C. (This wouldn’t be true if the pan
contained ball bearings or iron filings. In that case, the
pan and contents would continue to absorb heat, and the
temperature would continue to rise.)
►
A small part of that heat is released by the pan to the
environment by means of conduction and radiation. We call
that the heat losses. The rest of the heat – most of it, in fact –
is used to evaporate water.
When you put the lid on the pan, you don’t change the
conduction and the radiation, because they are determined by the
temperature of the pan, which remains constant (at 100°C). So the
quantity of water that evaporates must remain constant, too. (It is
even probable that the shiny lid reduces radiation heat losses, so
that more heat remains and evaporation increases!) The only thing
changed by the lid is that the water vapour escapes with some force
between lid and pan. But that doesn’t change the quantity of vapour.
Now, all this seems to contradict common experience:
everyone “knows” that it’s more economical and reduces the
dampness in the kitchen if you keep the lid on the pan when you
boil water. So what’s the explanation? As you’ve probably guessed,
it’s because the above description isn’t complete: almost intuitively,
you lower the gas burner or cooking ring to the point where the
water just barely boils. Then the cooker is supplying only the energy
to compensate for the conduction and radiation heat losses, and
hardly any for evaporation. Without a lid, there would be a lot of
extra evaporation because the air circulation around the pan blows
the vapour away and more heat is required to replace it. (This is why
people put a saucer on the top of cup of hot coffee that they want to
keep warm.)
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Everyday Physics: Unusual insights into familiar things
70
WHY DOES THE
AIR GET SO DRY
IN WINTER?
The air is dry when it freezes outside, which you might notice because
newspaper rustles, and wood is more inclined to crack. How come?
In winter does dry air flow into the country from somewhere else?
No. Suppose the outside temperature is below freezing, and
it rains. The air is full of water droplets; it cannot become more
humid; it contains as much water vapour as it can hold. The relative
humidity (the amount of water vapour the air contains divided by
the amount that it can contain) is practically 100% (see Why are some
alpine winds so warm? p5).
Now go indoors, where it is 20°C, say. Ultimately, all the inside
air has come from outside – whether through cracks, or deliberate
ventilation. As the air enters and warms up, the quantity of water
vapour in the air remains the same. However, warm air can contain
more water vapour than cold air, so the relative humidity goes
down: the air becomes drier. What’s surprising is how dry the air has
become: it’s much drier than you would expect.
The explanation is the steep shape of the vapour pressure
curve. (The vapour pressure is the maximum pressure that water
vapour can have at a certain temperature; if the pressure were higher
the vapour would condense.) The curve (Fig. 1) gives the value of the
maximum water vapour pressure at different temperatures. Since
we are discussing water vapour in air, we can put it more precisely:
the graph gives the maximum contribution from water molecules
Everyday Physics: Unusual insights into familiar things
273
Fig. 1: Vapour
pressure curve for
water, from 0°C-20°C.
to the total ambient atmospheric pressure (most of which is due
to nitrogen and oxygen). The vertical axis shows the pressure in
millibars (or in hectopascals, which are the same). The growth in
the curve is nearly exponential: it doubles approximately every
10°C (rising from 6 mbar to 12 mbar in 0°C-10°C, and from 12 to
23 mbar for 10°C-20°C). That means that warming air from 0°C to
10°C allows it to hold twice as much water vapour, and at 20°C it
can hold about four times as much as at 0°C. So even if air at 0°C is
saturated with water vapour, when you heat it to 20°C the relative
humidity can’t exceed 25%, which means that the air at 20°C is
quite dry. That’s why the interior air is very dry when the outside
temperature is very low. Dry air in winter is not due to the relative
humidity of the air outside; it’s because it is cold outside. (Of course,
you can accidentally or deliberately make your interior air moister.
In some “passive houses” the ventilation system always leaves the air
very dry, so some owners leave their washing to dry inside the house
to moisten the air.)
Fig. 1 also shows how easily water condenses on cold surfaces
if the interior air is not very dry. Suppose the inside temperature is
20°C and the relative humidity is 50%; that means the water vapour
pressure is 50% of 23 mbar = 11.5 mbar. (Point A on Fig. 1 shows
where the 23 mbar – the maximum possible vapour pressure at 20°C
– comes from.) Assume also that you have a window, or the inside
surface of a poorly insulated outside wall, at 10°C. When the 20°C air
cools down to 10°C, its water vapour pressure is still 11.5 mbar, but
now the maximum possible vapour pressure is 12 mbar (point B on
Fig. 1), so the relative humidity is 11.5⁄12 = 96%. If you have a small
source of extra moisture, the level reaches 100% and you’ll get misty
windows and mouldy walls; to avoid this you need some ventilation
to remove the extra moisture.
Fig. 2 shows the vapour pressure curve for a larger temperature
range. At 100°C the water vapour pressure reaches 1,000 mbar
(actually 1,013 mbar) which is normal atmospheric pressure, which is
why the vapour from water boiling in a pan can lift the lid off.
Just below freezing, the steep trajectory of the vapour pressure
curve continues (Fig. 3). The bottom left of Fig. 3 explains why it
hardly ever snows in Siberia in the middle of winter. It’s not because
it’s too warm but because it is too cold: below −30°C there just isn’t
enough water vapour in the air to make snow.
274
Everyday Physics: Unusual insights into familiar things
Fig. 2: Vapour
pressure curve for
water, from 0°C-100°C.
Fig. 3: Vapour
pressure curve for
water, from −30°C
to +20°C.
Everyday Physics: Unusual insights into familiar things
275
If the air is too dry…
If the air is too dry, you can bring extra moisture into your room. However, it’s not a oneoff operation: because ventilation is continuous, you must add water vapour continuously
too – it’s like fighting a losing battle. For a typical level of ventilation that’s not so easy: a
water container hung on a radiator isn’t very effective unless you find it evaporates many
litres of water per day.
It’s easy to estimate how much water you must evaporate to raise the relative humidity
in a room of volume 100 m3 (3,500 ft3), say. First you need to know how many m3 (or ft3)
of water vapour you get by evaporating one litre (a quarter gallon) of water. When you
evaporate a litre of any liquid, you always get about 1 m3 (35 ft3) of vapour. (The precise
number depends on the liquid; it varies from about 0.7 m3 to 1.3 m3, but for simplicity
we’ll stick with 1 m3.)
Suppose you want to raise very dry air of, say, 10% relative humidity to 60% – an increase
of 50 percentage points. That means you have to add half the maximum amount of vapour
that the air can contain. Point A on Fig. 1 shows that at 20°C that maximum amount is
23 mbar, so you have to add half of that, i.e. 11.5 mbar, or 10 mbar in round numbers.
The atmospheric pressure is about 1,000 mbar. You’re adding 10 mbar of vapour, so
you’re replacing 1% of the air molecules in the room with water molecules. The volume of
your room is 100 m3 so you need 1 m3 of water vapour, and we saw above we get that from
one litre of water. But that only humidifies the air that’s in the room now. A typical level
of ventilation is one “air change per hour”, which means you’ll need one litre of water per
hour – about 2 bucketfuls per 24 hours.
In practice, considerably less humidification will make the room more comfortable, but
don’t expect a miracle from a bowl of water on the radiator.
276
Everyday Physics: Unusual insights into familiar things
71
WHY DON’T YOU
DIE OF HEAT IN
THE SAUNA?
The air in a sauna is extremely dry, due to the heat. If the external air
is at 0°C, there is so little water vapour in the air that when this air
enters the sauna and is raised to 90°C, the relative humidity (RH) is
only about 1% (see box). Even if the air that feeds the sauna comes in
at 20°C, the RH in the sauna will hardly reach 3% – unless somebody
does something deliberately to raise the humidity.
Everyday Physics: Unusual insights into familiar things
277
<box>
Relative
Relative humidity in the
saunahumidity in the sauna
Assume
that the
coming
the sauna
at 0°C and
is
Assume that the air coming into
the sauna
is atair0°C
and is into
saturated
with is
moisture.
Point
saturated
with
moisture.
Point
C
in
Fig.
1
shows
that
the
C in Fig. 1 shows that the water vapour pressure is at most 6 mbar. When that air is then
water
vapour
pressure
is atthe
most
6 mbar.
that the
air
heated to 90°C in the sauna, the
vapour
pressure
remains
same,
6 mbar.When
However,
is
then
heated
to
90°C
in
the
sauna,
the
vapour
pressure
maximum water vapour pressure at 90°C is about 700 mbar (D in Fig. 1), so the relative
humidity = 6 ÷ 700 ≈ 1%. remains the same, 6 mbar. However, the maximum water
vapour pressure at 90°C is about 700 mbar (D in Fig. 1), so
Fig. 1: Point C shows
the maximum water
vapour pressure
(6 mbar) at 0°C, and
point D (700 mbar)
for 90°C.
How warm the sauna really is becomes clear if you swing your
arms back and forth. Normally that would cool them down but it
doesn’t: they get warmer, especially if they are still dry. Even though
the wooden benches are at 90°C, you can sit on them comfortably
because of the poor heat conduction of wood. (This poor heat
conduction makes the heat-flow from the benches to your body very
small; the slow heat-flow means that your body can easily dissipate
the small amount of heat it receives. In contrast, if the benches were
metal, you’d get burnt, nearly as badly as with a pot of boiling water.)
278
Everyday Physics: Unusual insights into familiar things
If you touch the glass of the door, it feels so hot it almost hurts,
although its temperature is much less than 90°C (see Fig. 1a, Is thick
glass a better insulator than thin glass? p251).
In spite of the 90°C heat, you can stand the sauna for quite
a while, for two reasons. First and most importantly, the layer of
air very close to your skin is insulating (see How do you keep your
temperature constant, p49). Second, because it is so dry in the
sauna, your sweat readily evaporates and cools you. Even so, the
evaporation can’t stop you gradually getting hotter: the heat flux
into your body is too big for that. It amounts to about 1.5 kW, as much
as an electric heater produces! To cool that off by evaporation, you’d
have to evaporate two litres per hour (see box), and your body can’t
keep pace with that.
When someone throws water on the coals in the sauna, you
feel a brief surge of heat. This has a number of causes. First, hot
steam at 100°C is produced, and that’s higher than the ordinary
sauna’s normal temperature. Next, that steam moves the air around,
sweeping away the insulating layer and bringing the heat closer to
your skin; this is especially pronounced when the skin is still dry
so that evaporation isn’t cooling you. Moreover, this newly formed
steam moistens the air and reduces the cooling effect of evaporation.
Everyday Physics: Unusual insights into familiar things
279
And water vapour conducts heat better than air, so the now-moist
air layer near the skin transports more heat to your body. Finally,
there is another surprising factor: the latent heat released in the
condensation of water vapour on to the skin is probably the most
important contribution of this heating effect when you throw water
on the coals. The reason is that your skin is the coldest place in the
whole sauna, and the humidity can easily reach 100% near the skin,
so condensation is inevitable, and the latent heat of condensation
is large.
In the sauna
In the sauna your body has to cope with a lot of conducted heat and radiated heat.
The conduction contribution can be written in terms of your surface area A, the heat
conductivity of air λ, the temperature difference between the sauna air and your skin
ΔT, and the insulating air layer d around your skin. The conducted contribution then is:
A λ × ΔT / d. Substituting A = 1.7 m2, λ = 0.025 W⋅m-1⋅K-1, ΔT = 50 K and d ≈ 3 mm yields
700 watts.
The radiated contribution is A σ (T4sauna − T4skin). Substituting A = 1.7 m2, σ = 5.7 ×
10 W⋅m-2⋅K-4, Tsauna = 363 K and Tskin = 310 K, we get 790 watts. So the calculation shows
that in the sauna conduction and radiation are about equally important, as they are
under normal circumstances. However, in the sauna the heat-flow goes the other way,
i.e. you’re taking in heat, not losing it. Moreover, the heat-flows are much larger in the
sauna, because of the large temperature differences and the fact that in the sauna you are
appropriately dressed (well, appropriately undressed) so there are no insulating clothes
to reduce the heat-flow.
-8
Thus in total your body takes in about 1,500 watts, which you have to get rid of
by evaporation. The “latent heat of evaporation”, i.e. the heat needed to change liquid
water to vapour at the same temperature, is 2.4 kJ per gram at body temperature.
To get rid of 1,500 W, the quantity of water to be evaporated per second is 1,500 W /
2.4 kJ per gram = 0.62 gram per second = 0.62 ml/s = 0.62 ml × 3,600 seconds/hour =
2.2 litres (half a gallon) per hour. In practice it will be somewhat less because some heat
goes to raising your body temperature a little bit, especially at the start when you have
just entered the sauna.
280
Everyday Physics: Unusual insights into familiar things
72
THE WINE-MIXING
PROBLEM
This is a typical science quiz problem. You have two identical glasses,
one containing red wine and the other white. Take a spoonful of wine
from the red glass and put it into the white and stir, making it a little
rosé. Now take an exactly equal spoonful of this rosé and put it back
into the red. Which of the two wines is now most “contaminated” by
the other?
To most people the answer isn’t obvious. The spoon of red
that went into the white was, after all, pure red, while the spoon of
white that went into the red was already a bit rosé, so isn’t the spoon
of pure red “more contaminating”? On the other hand, that second
spoon of rosé went into a glass that was less full, so it might have had
more of an effect! Which effect wins?
Everyday Physics: Unusual insights into familiar things
281
The answer is that each glass is just as contaminated as
the other.
You can see why this is so in several different ways. The first
way is to forget the details and just look at the end result instead. You
haven’t destroyed or created any wine, so you have exactly the same
total amount as you started with. Any red wine not in the white glass
must therefore be in the red glass; and as the white glass contains the
same volume now as at the start, any red wine in it has replaced the
same amount of white wine, which must now be in the red glass. So
both glasses have exactly the same level of “contamination”.
A second, more abstract, way of addressing the problem is to
consider the case where you have 100 red balls in one glass and 100
white balls in the other. Move a scoop of red balls to the white glass,
and move a scoop of red/white mix back to the red glass. If there are
now 2 white balls in the red glass, there must be 98 red balls there
(because it’s full) and the only place the other 2 red balls can be is in
the other glass. This also makes it very clear that it doesn’t matter
whether you thoroughly mix the red and white balls (wine) before
transferring between the glasses: it’s always true that any red balls
not in the red glass are in the white, and vice versa, whatever the
concentrations are.
The third and final way is to calculate exactly what is
transferred from one glass to the other. While the calculation isn’t
complicated, it’s long-winded because of all the fractions, so we’ve
shown it only in an appendix (p296).
282
Everyday Physics: Unusual insights into familiar things
73
THE WANDERING
TEA LEAVES
Stir your cup of tea, stop stirring, and then what do you see? When
the liquid has stopped moving, the tea leaves have gathered in the
middle at the bottom, but wouldn’t you have expected the centrifugal
force to push the heavier-than-water leaves or sugar to the outside?
Everyday Physics: Unusual insights into familiar things
283
Centrifugal force does of course play a role; the circulating
liquid does want to go to the outside. However, the centrifugal force
is not equally strong at every height (because the circulation speed is
not the same everywhere), and friction also plays a part:
►
At the top, when the tea is stirred it rotates unhindered and
the centrifugal force can do its work, so the liquid moves
from the centre towards the edge.
►
Near the bottom things are different. The rotation speed is
smaller and therefore the centrifugal force is less. This is
because the rotating tea is hindered by friction against the
cup’s bottom.
The result is that something unexpected occurs. At the top the
tea is pressed towards the outside with a lot of force; at the bottom
the tea is pressed outwards with a smaller force. So the larger force
at the top “wins” and the tea at the top flows towards the outside.
From there it has to go down along the outside, and it comes in again
towards the middle at the bottom where the centrifugal force is
lowest. So overall, in addition to a rotating motion in the horizontal
plane, there is also a circulation pattern in the vertical cross-section
of the cup (Fig. 1).
Fig. 1: In the teacup, the centrifugal force is greater at the top than
at the bottom – due to friction at the bottom – so the tea circulates as
shown, and the tea leaves collect in the middle at the bottom.
284
Everyday Physics: Unusual insights into familiar things
The circulating tea takes the tea leaves with it – towards the
outside at the top, then down the walls, and back into the middle
at the bottom. When the flow speed is high, the leaves keep on
circulating, but when the flow reduces and is no longer strong
enough to lift the leaves back to the top, they are swept into a nice
pile in the middle.
Thus stirring does eventually bring some order into chaos.
Everyday Physics: Unusual insights into familiar things
285
ACKNOWLEDGEMENTS
It is a pleasure to thank all those who reviewed the manuscript for accuracy or comprehensibility
and those who contributed in other ways to the realization of this book: Ellen Backus,
Rinus Boone, Barry Cats, Daniëlle Duijn, Loek Eenens, Huub Eggen, Eric Eliel, Martin van
Exter, Jan, Ineke and Annette Heijn, Hanneke, Reinoud and Francine Hermans, Marjon de
Hond, Gert ’t Hooft, Marieke Huijvenaar, Vincent Icke, Frank Israel, Stuart Johnson, Jan de Laat,
Henriette van Leeuwen, Jan Lub, Tjerk Oosterkamp, Gijs Pellinkhof, Henk and Linda Petersen,
Ellie van Rijsewijk, Edith van Ruitenbeek, Jean Schleipen, Caesar Sterk, Fokke and Free Tuinstra,
and the enthusiastic audience at my “Everyday physics” lectures for the University of the Third
Age (HOVO).
Thanks to Professor Fokke Tuinstra, Delft University of Technology, for material on bike side
winds and on bike-wheel splashing.
Jo Hermans
286
Everyday Physics: Unusual insights into familiar things
HANDY REFERENCE
DATA
The Earth
Circumference: 40,000 km
Atmospheric pressure at sea level: on average 1.013 bar (1 bar =
105 pascal) or about “1 kilogram per cm2”
Density of the air at 20°C: 1.2 kg/m3
Earth’s atmosphere: the effective thickness is approx. 8 km. The
atmospheric pressure falls with height approximately according
to the formula e-h/8 km; it halves roughly each 5.5 km.
Around the Earth
Orbit time of “low” satellites (just outside the Earth’s
atmosphere): approx. 90 minutes
Orbit time of GPS-satellites: 12 hours (at a height of 20,000 km)
Orbit time of a satellite in geostationary orbit, 35,786 km above
sea level and directly above the equator: 24 hours
Orbit time of the moon: 27.3 days (at a height of approx.
380,000 km)
The moon
Average distance to the Earth: approx. 380,000 km
Diameter: approx. 3,500 km
Apparent diameter in degrees: ½ degree
The sun
Average distance to the Earth approx. 150 million km
At the beginning of July the sun is a good 3% further away than
at the beginning of January
Everyday Physics: Unusual insights into familiar things
287
Diameter: 1.4 million km
Apparent diameter in degrees: ½ degree
Mass: 2 × 1030 kg
Temperature at the surface: approx. 5,750 K or approx. 5,500°C
Emitted energy flow: 3.8 × 1026 watts
Emitted energy flow, translated into equivalent loss of mass:
4 billion (4 × 109) kg per second
Solar energy reaching the Earth
Just outside the Earth’s atmosphere: approx. 1,350 W/m2 (measured
perpendicularly to the sun rays).
At the Earth’s surface: approx. 1,000 W/m2 (measured
perpendicularly to the sun rays).
In the northern United States and in central Europe, the average on
a horizontal surface, over the entire year and 24 hours per day, is
typically 150 W/m2.
Energy
(For the definitions of energy, work, heat and power see The human engine p43)
Energy content of:
►
Oil, fat: approx. 40 MJ/kg
►
Petrol/gasoline: 48 MJ/kg or 35 MJ/litre
►
Natural gas: 35 MJ/m3 (higher value) or 32 MJ/m3 (lower value) (The “higher value”
includes the condensation heat of the water produced in the chemical reaction of
combustion. This water is in the vapour phase, so by condensing it out of the exhaust
gases, which is what a “condensing boiler” does, we get extra useful heat.)
288
Everyday Physics: Unusual insights into familiar things
Light
Speed of light in a vacuum: 299,792 km/s, i.e. about 300,000 km/s
or 300,000,000 (3 × 108) m/s
Speed of light in air is only 0.03% less
Speed of light in water: 225,000 km/s
Sound
Speed of sound in air (15°C) : 340 m/s or 1,200 km/h
Speed of sound in water (15°C): 1,470 m/s or 5,300 km/h
Speed of sound in a vacuum: in a vacuum there is no sound!
Prefixes
multipliers
kilo
k
103
thousand
mega
M
106
million
giga
G
109
billion
tera
T
1012
trillion
peta
P
1015
quadrillion
dividers
milli
m
10-3
thousandth
micro
μ
10-6
millionth
nano
n
10-9
billionth
pico
p
10-12
trillionth
femto
d
10-15
quadrillionth
Everyday Physics: Unusual insights into familiar things
289
Rules of thumb
Speeds
To convert miles or kilometres per hour, to metres/second:
100 mph = 44.7 m/s ≈ 50 m/s, i.e.
2 mph ≈ 1 m/s
2 miles ≈ 3 km, so
3 km/h ≈ 1 m/s
►
I cycle at about 5 m/s, which is about 10 mph.
►
The New Horizon satellite is travelling at 36,000 mph, or about 18,000 m/s = 18 km/s.
For comparison, a rifle bullet travels at about 1 km/s and weighs about 10 g, so if New
Horizon bumps into a 10 g piece of space debris, it’s like being shot with a very very
fast bullet.
Rain
A rainfall of 1 mm is the same as 1 litre of rain per square metre.
The advantage of the litres/m2 expression is that it’s easy to see what volume of water a property
will have to handle. For example, if the area of your house and garden is 500 m2, and you have
a rainfall of 8 mm, then the volume of water that has to flow away is 4,000 litres, which is 4
cubic metres.
290
Everyday Physics: Unusual insights into familiar things
Circles and spheres
How many litres of water do you need to fill a spherical goldfishbowl? (Not that it’s advisable to keep goldfish in that sort of bowl.)
How many square metres is a round plot, so you can work out how
much grass seed to buy?
While it’s handy to know the proper formulae for circles and spheres
by heart, it’s not essential. In everyday life when you don’t have to be
very precise, remembering a few simple images is enough.
Take the volume of a sphere. Imagine a cubic box that fits the sphere
exactly. What fraction of the cube does the sphere occupy (Fig. 1)?
Fig. 1: Estimating
the volume of a
sphere by fitting it
within a cube.
At the centre points of the six sides of the cube, the cube and sphere coincide, there is no space
in between. However, in each of the eight corners, there are fairly large parts of the cube that the
sphere doesn’t touch. At a guess, if we add up the volume of these eight pieces, they might be half
the volume of the cube.
Everyday Physics: Unusual insights into familiar things
291
Would that be a good guess? Look at the formula for a moment. The
volume of a sphere of radius r is 4⁄3 π r3. The value of π is 3.14, which
we’ll round down to 3 as we’re not being too precise, so the volume
is about 4⁄3 × 3 × r3 = 4 r3. Now for the cube: each side is 2 r long,
so the volume is (2 r)3, i.e. 8 r3, which is double our approximation
for the sphere. Conclusion: the volume of a sphere is about half the
volume of its surrounding cube, to within a few per cent, because
the inaccuracy of our calculation was small – rounding 3.14 to 3.00,
an error of 14 in 300 or about 5%.
Fig. 2: Estimating
the area of a circle
by fitting it within a
square. The top-left
quarter (marked with
blue dashed lines) is
highlighted in Fig. 3.
Fig. 3: The top-left
quarter of the circle
of Fig. 2.
You can use the same approach for the area of a circle. Draw a square
around it that fits it exactly (Fig. 2).
Clearly the circle doesn’t cover the whole square, but it does cover
much more than half, because in Fig. 3 the triangle marked (A) is
half the area, and area (B) looks about the same size as area (C), so
¾ would be a good guess. Check this using the formula: the area of
a circle is π r2, or approximately 3 r2. The area of the surrounding
square is (2 r)2 or 4 r2. So our guess above of ¾ is good to within a
few per cent.
Easiest of all is the circumference of a circle. Fig. 4 shows that the
circumference of the circle is not much longer than the circumference
of the hexagon that fits inside exactly. The six triangles are all
equilateral (because the vertices are 360 ÷ 6 = 60°), so the sides are
equal to the radius. Therefore, the circumference of the hexagon
is 6 r and the circumference of the circle is about the same – six
times the radius (or three times the diameter), again to within a few
per cent.
If you have to be really precise and need the value of π to many
decimal places, use the mnemonic aid: “How I want a drink,
alcoholic of course, after the heavy lectures involving quantum
mechanics.” Count the numbers of letters in each word, which
gives you 3.14159265358979. Or if you only want the decimal
digits after the initial “3.”:
Fig. 4: Estimating
the circumference of a
circle using a hexagon.
292
I wish I could determine pi.
Eureka, cried the great inventor.
Christmas pudding, Christmas pie,
Is the problem’s very centre.
Everyday Physics: Unusual insights into familiar things
Pi
Here’s π to a thousand decimal digits:
3.14159 26535
89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164
06286 20899 86280 34825 34211 70679 82148 08651 32823 06647 09384 46095
50582 23172 53594 08128 48111 74502 84102 70193 85211 05559 64462 29489
54930 38196 44288 10975 66593 34461 28475 64823 37867 83165 27120 19091
45648 56692 34603 48610 45432 66482 13393 60726 02491 41273 72458 70066
06315 58817 48815 20920 96282 92540 91715 36436 78925 90360 01133 05305
48820 46652 13841 46951 94151 16094 33057 27036 57595 91953 09218 61173
81932 61179 31051 18548 07446 23799 62749 56735 18857 52724 89122 79381
83011 94912 98336 73362 44065 66430 86021 39494 63952 24737 19070 21798
60943 70277 05392 17176 29317 67523 84674 81846 76694 05132 00056 81271
45263 56082 77857 71342 75778 96091 73637 17872 14684 49010 22495 34301
46549 58537 10507 92279 68925 89235 42019 95611 21290 21960 86403 44181
59813 62977 47713 09960 51870 72113 49999 99837 29780 49951 05973 17328
16096 31859 50244 59455 34690 83026 42522 30825 33446 85035 26193 11881
71010 00313 78387 52886 58753 32083 81420 61717 76691 47303 59825 34904
28755 46873 11595 62863 88235 37875 93751 95778 18577 80532 17122 68066
13001 92787 66111 95909 21642 01989
The truth about exponential growth
Just about everyone can misjudge the long-term effect of exponential growth. Exponential
growth is growth where a fixed percentage is added every time period. So compound interest
of 5% per year gives exponential growth. Inflation of 322% per month, as there was in the 1923
hyperinflation in Germany, is also exponential growth.
Test yourself: you have two seconds to answer the following question. Which would you
prefer, $50,000 a day for the next month, or ¢1 today, ¢2 tomorrow, ¢4 the day after, ¢8… for
the next month?
Everyday Physics: Unusual insights into familiar things
293
Now give yourself 20 seconds, to estimate (without a calculator) how much your ¢1 will grow to
after a month. (This is just a variant of the traditional story of the grains on the chessboard. Before
a game of chess, a king asked a sage to name his own reward; the sage asked for 1 grain of rice on
the 1st square, 2 grains on the 2nd, 4 grains on the 3rd, 8 grains on the 4th, etc.) The answer to
what your ¢1 becomes is – assuming 29 days of doubling – $5.4 million. (The royalty-bankrupting
rice amount is 1.8 × 1019 grains, or 18 billion billion!)
The above is about rapid doubling, but even growth with a fixed small percentage can be
treacherous. In 1626 Peter Minuit bought Manhattan for $24. The current estimated value of
Manhattan is about $250 billion, that is about ten billion times as much. Surely such a fantastic
investment was much better than the optimistic 7% per year that the stock market yields over
the long term? To see if that’s correct, calculate how much the stock market would have yielded.
You can use two simple tricks to calculate the answer quickly. The first trick is “the rule
of 70”. This is based on the fact that money invested at 1% compound interest does not
need 100 years to double; because of the effect of compounding, it needs only 70 years.
The rule is:
The doubling time is equal to the number 70 divided by the growth percentage.
So, at a rate of 2% your investment doubles in 70 ÷ 2 = 35 years, and at 7% it doubles in 10
years. The rule gives a very good approximation (although it goes wrong for large percentages).
Accordingly, you can regard every exponential growth as a process of doubling.
The mathematical background here is that the natural logarithm of the number 2 is approximately
0.70, i.e. ln 2 (or loge 2) = 0.693.
The second trick relies on the fact that when you repeatedly double your investment, the amount
increases according to the series
2, 4, 8, 16, 32. 64, 128, 256, 512, 1024…
so after 10 doublings, your investment has increased by a factor of 1,024, giving us our second
rule of thumb:
Ten doublings increase your investment by a factor of 1,000.
These two simple rules make it easy to estimate the consequences of a process of exponential
growth. Look at Peter Minuit’s investment in Manhattan as an example. If he had invested his
money at 7% compound per year:
294
Everyday Physics: Unusual insights into familiar things
1. According to the “rule of 70” his money would have doubled every ten years. In 380 years that
amounts to 38 doublings.
2. According to the second rule, the first 30 doublings yield three factors of 1,000, i.e. 1,000 ×
1,000 × 1,000, that is a factor one billion (1 bn).
That still leaves 8 doublings, which give a further factor of 256, giving a total factor of 256 bn,
about 25 times better than the 10 bn factor actually achieved, i.e. $24 × 256 bn = $6,144 bn
instead of the actual $250 bn. So poor old Peter Minuit didn’t make the best investment decision:
he ought to have shopped around for a fund with a yield of 7% compounded per annum, and
invested his $24 there instead.
Here’s an even more surprising example – world population. Around 1970 the population was
growing at a little less than 2% per year. While that doesn’t sound very much, you can see what
its long-term effect will be, using a thought experiment. Suppose that the world started very
modestly 2,000 years ago, with only two people – maybe Adam and Eve at the time of Christ, to
jumble our biblical history a bit. And suppose the growth since then has been 2% per annum.
Would the population be much lower than the current 8 bn, or about the same, or even higher?
Let’s use our rules to make a quick estimate. The “rule of 70” says 2% per annum yields a doubling
every 35 years, or about three doublings per century, which is 60 doublings in 20 centuries. Our
second rule says 60 doublings give us 60/10 = 6 factors of 1,000, one after the other, or 1018, i.e.
one billion billion! Even though a precise calculation gives something slightly less – 3 × 1017 people,
starting with two people in the year zero – that’s still 50,000,000 times as much as the actual
population. Clearly the world population actually grew very much less than 2% per annum over
the centuries. Until the year 1600 the growth was probably lower than 0.1% per annum. However,
the most important conclusion from this example is that exponential growth, even with a modest
percentage, in the long term leads to impossible situations. For example, if the world population
really did reach 3 × 1017 we’d have 2,000 people per square metre (or only 600/m2 if we spread
them over the oceans too).
The “rule of 70” gives a nice insight into what happens within a human lifetime which, for
simplicity, we’ll say is 70 years. Each 1% growth of anything then means a doubling in “a
lifetime”, and the growth percentage gives the number of doublings. E.g. if the volume of rubbish
incinerated goes up by 5% per year, then our grandchildren will need 25 = 32 times as many
incinerators as we have today.
In brief: while we regard growth as indispensable, long-term growth at a constant compound
rate is impossible.
Everyday Physics: Unusual insights into familiar things
295
APPENDIX: THE
WINE-MIXING PROBLEM –
CALCULATION
This appendix shows the third – detailed calculation – way to see that the red and white wine
glasses in The wine-mixing problem (p281) are equally contaminated.
For simplicity assume that each glass contains 100 ml of wine and that your spoon contains 7 ml.
(In fact it doesn’t matter what size the spoon is, as you’ll see.) The table below shows the number
of millilitres in the white (W) and red (R) glasses respectively, at each stage of the process.
WHITE GLASS
RED GLASS
W
R
R
W
Initial situation
100
0
100
0
After one spoon of red
added to white:
100
7
93
0
At this point the white glass contains 107 ml, of which 100 are white and 7 are red. So in the
white glass:
►
The W concentration is 100:107, i.e. 1 ml of this rosé contains 100/107 ml of white, and a
7 ml spoon contains 7 × (100/107) = 700/107 ml of white.
►
The R concentration is 7:107 i.e. 1 ml of this rosé contains 7/107 ml of white, and a 7 ml
spoon contains 7 × (7/107) = 49/107 ml of red.
These white-glass concentrations remain unchanged because you don’t dilute the white glass any
further. Now use these concentrations to see how much red and white are transferred to the
red glass. To keep it simple, just look at the contamination in both glasses: in either glass there is
700/107 ml of the “wrong” kind. QED.
296
Everyday Physics: Unusual insights into familiar things
WHITE GLASS
W
Remove one spoon of
“white” (actually rosé)
from the white glass
100 − 7 ×
(100/107)
= 100 –
(700/107)
And add it into the
red glass:
(unchanged)
100 –
(700/107)
R
RED GLASS
R
W
7 – (49/107)
= 700/107
93
(unchanged)
0
(unchanged)
(unchanged)
= 700/107
93 + (49/107)
= 100 –
(700/107)
0 + (700/107)
= 700/107
Everyday Physics: Unusual insights into familiar things
297
TWO MORE EXPERIMENTS
Home experiment: Twin divers
Here’s a trick that was popular at children’s parties in the past. The “magician” holds
in her hand a sealed, flat-sided glass bottle filled with water, containing a primitive
model diver that can be sent to the bottom of the bottle on command, and back again
on command. What the children see is no magic trick but the stealthy use of the laws
of physics.
The trick is easier to demonstrate with a plastic water bottle. The “diver” consists of a
vertical glass tube open at the bottom only; an eye- or nose-dropper from the drugstore
complete with squeezer works very well. Fill the bottle nearly to the brim with water,
and suck enough water into the diver so that it just barely floats in the bottle. Cap
the bottle, and you’re in business. Squeeze the bottle: the pressure increases, more
water enters the diver and it sinks to the bottom! Squeeze less hard and it rises to the
surface again. (To do this with a cylindrical glass bottle is impossible, but you can do it
with a flat-sided glass bottle, and it’s much more impressive than with a plastic bottle,
although it requires a fairly strong hand! The alternative is to seal the top of the bottle
with a piece of balloon and discreetly put your thumb on it to increase the pressure.)
You can make a game out of this by
using two divers in the same bottle,
each with a hook of bent copper
wire, one facing up and one facing
down (Fig. 1). Fill the down-hook
diver as before, but fill the up-hook
diver with enough water so that it just
hovers on the bottom of the bottle.
The challenge is to rescue the lower
diver with the upper one. It’s not so
easy, so the child who manages it is
rewarded with the biggest piece of
birthday cake, and later goes on to
study physics.
298
Fig. 1: Rescuing a “diver” in a bottle by
manipulating the pressure.
Everyday Physics: Unusual insights into familiar things
S
Home experiment: Super simple electric motor
You can build an electric motor in five minutes! The only part you need that you
wouldn’t find around the house is a small super-magnet in the form of a flat disk or
– easier still – of a sphere of about 2 cm diameter. You can buy these cheaply on the
internet – search for “super magnet”. Any handyman probably has the others parts: a
short (< 25 mm) flat-headed steel screw, an AA or AAA battery, and a piece of copper
wire with bare ends.
Fig. 1 shows how to assemble your motor. (It also shows the field lines of the magnet.)
The magnet sticks on to the screw, and if the screw isn’t too long, the point of the screw
sticks to the bottom (the negative terminal) of the battery.
Fig. 1: How to connect the components of
your motor, showing the magnetic field from
the spherical magnet.
With the battery in one hand and the wire
in the other, close the circuit by holding
one bare end of the wire against the top of
the battery (the positive terminal) and the
other bare end loosely against the side of the
magnet as a kind of sliding contact. Now the
screw and magnet start to rotate like mad.
Friction is minimal, thanks to the pointed
screw and the smooth bottom of the battery.
But why does the magnet rotate? It is
because of the Lorentz force, a force that
works perpendicular to a current in a
magnetic field. The magnetic field is vertical
in this case and the sliding contact ensures
the horizontal current (shown by the white
arrow in Fig. 2).
Fig. 2: The current flows horizontally from
the lower end of the wire into the spherical
magnet. With the magnetic field directed
vertically, the Lorentz force acts perpendicular
to both and makes the sphere turn.
Everyday Physics: Unusual insights into familiar things
299
RESOURCES
►
Animated image of GPS satellites orbiting the Earth:
https://commons.wikimedia.org/wiki/File:GPS24goldenSML.gif
►
Camera obscuras that you can visit:
Public access: Camera obscuras open to the public
https://en.wikipedia.org/wiki/Camera_obscura#Public_access
►
How to make a spectrometer using an old CD:
Papercraft Spectrometer Intro Kit
https://publiclab.org/wiki/papercraft-spectrometer
►
The Magnus effect:
■ Video of rolled paper cylinder:
https://upload.wikimedia.org/wikipedia/commons/e/e1/08._Цилиндар_што_паѓа_и_
ротира.ogv (or search Wikipedia for “Magnus effect”, and click on the relevant video on
the right-hand side).
■ Video of a home-made Flettner rotor-powered toy car:
Flettner Rotor Sail Cart // Homemade Science with Bruce Yeany
https://www.youtube.com/watch?v=XsGin7CFaF8
■ Video: What Happens When a Spinning Basketball is Thrown Off a Dam!
https://www.youtube.com/watch?v=QtP_bh2lMXc
■ The bouncing bomb:
Article: https://www.smithsonianmag.com/history/how-british-engineer-made-bombthat-could-bounce-on-water-180969095/
Video: Dambusters | Building the Bouncing Bomb | 2 of 2
https://www.youtube.com/watch?v=8IeGYkwVIWw
►
Crookes radiometer video:
https://en.wikipedia.org/wiki/Crookes_radiometer and click on the video in the
Thermodynamic explanation section.
►
Albert Einstein’s autograph manuscript of his paper on how a radiometer works:
Zur Theorie der Radiometerkräfte
https://alberteinstein.info/vufind1/Record/EAR000034044
(Click on the image of the document, to read it.)
►
Evacuated double glazing description:
https://www.pilkington.com/en-gb/uk/products/product-categories/thermal-insulation/
pilkington-spacia
300
Everyday Physics: Unusual insights into familiar things
CREDITS
►
Jo Hermans:
p299.
►
Mad-i-Creative:
p67*, 69*, 96, 126, 155, 201*, 251, 255, 265* (x2), 281* (x2).
►
Pexels: https:/www.pexels.com/
p17, 19, 37, 47, 49 (x2), 59, 83, 89, 112, 113, 119, 161, 169, 180, 181, 189, 191, 195 (x2), 197, 199, 207
(x2), 211, 225, 244, 245, 267.
►
Pixabay: https:/www.pixabay.com/
p7, 8, 10, 11, 15, 21, 31, 33, 44, 51, 57, 67, 67*, 69*, 90, 94, 96*, 98, 99, 101 (x2), 103, 115, 117, 119,
126*, 129, 133, 147, 155, 158, 159 (x2), 173 (x2), 185, 191, 205, 209 (x2), 214, 215, 231, 239, 245, 261,
265*, 267, 269 (x2), 287, 288 (x3).
►
Pxfuel: https:/www.pxfuel.com/
p21, 33, 99, 105, 107, 199, 203, 220, 221, 265*, 271, 273, 277 (x2), 281* (x2).
►
Unsplash: https:/www.unsplash.com/
p0, 1, 5, 13, 19, 25, 27, 29, 32, 50, 53, 55, 59, 60, 63 (x2), 65, 71, 75 (x2), 79, 84, 86, 87 (x2), 93 (x2),
107, 108, 113, 115, 123 (x2), 129, 130, 135 (x2), 137, 139 (x2), 140, 143 (x2), 146, 183, 187 (x2), 201*,
217 (x2), 218, 225, 227, 229 (x2), 239, 251, 261, 271, 283 (x2), 287.
►
Wikimedia Commons: https://commons.wikimedia.org/
p37, A man walking. Photogravure after Eadweard Muybridge, 1887. Credit: Wellcome
Library, London, CC BY 4.0.
p165, Hologram on a 100-Euros bill, Heike Löchel / CC BY-SA 3.0.
p205, Circular road, in Třebíč, Czech Republic, Credit: Frettie, CC BY 3.0.
p235, Bouncing bomb tranining - IWM FLM. Credit: Official film maker / IWM staff
photographers / Public domain.
p247, A collection of Crookes radiometers at the Royal Society, London. Credit: The wub, CC
BY-SA 4.0.
p247, Radiometer Crookes - Oberweißbach-Thüringen. Credit Nieuw~commonswiki, CC BYSA 3.0.
p256, 2016-03-05 Ablenkeinheit by DCB. Credit: DCB, Wikimedia Commons, CC BY-SA 3.0.
p290, National Weather Service Standard Rain Gauge measuring tube and measuring dip
stick in use. Credit: Famartin, CC BY-SA 4.0.
* Edited by Mad-i-Creative
Everyday Physics: Unusual insights into familiar things
301
INDEX
Page numbers in bold refer to boxed text and home experiments; page numbers in italic refer
to Figures and captions.
A
B
absorption coefficient 170–1,
265, 268
acceleration 15, 16, 23, 48, 62,
82, 91
aerodynamics 22, 231
air
dry see dry air
insulating property 252, 255, 279, 280
air density 22, 23, 62, 89, 136,
287
air molecules 108–9, 109, 114,
200, 219, 249–50, 250
air pressure 5, 22, 30, 30, 136,
255, 256, 276, 287
air resistance 15, 17, 20, 22, 23,
59, 60–1, 61, 62, 63, 66, 70, 87, 89–92, 91, 92
air resistance coefficient 60–1
aircraft
area/volume ratio 270
energy consumption 263,
263
aluminium foil 265, 266
angular momentum, conservation of 37–8,
38–41,
39, 40, 41, 238
ants 270
area/volume ratio 269–70
atomic clocks 3–4
baby, keeping warm 270
Baden Baden sailing ship 235
balloons 17
inflation 258–60
balls
bouncing balls 237–8
bowling balls 38
curve balls 231
superballs 237–8
suspension in hairdryer airflow 235–7
topspin 230–1, 230, 231
Baumgartner, Felix 23
Bell, Alexander Graham 183
Bernoulli’s law 62, 230, 235,
236
bicycle pumps 5
bikes/biking see cycling
biomass 243, 244
black paint 265–6, 268
bloodstream 74
blue sea 169–72
blue sky 29, 107–10, 111–12, 133
body temperature 8, 10, 50–4,
280
bouncing bomb 235, 300
Brewster angle 177, 179
302
C
caesium atoms 3
calorie 43
Everyday Physics: Unusual insights into familiar things
camera obscuras 120, 122,
300
candles
energy efficiency 99–100, 106, 263
wax burn per hour 99, 100
carpets, wandering 245–6,
246
cars and driving
air resistance (drag) 87,
89–92, 91, 92
braking distance 95, 97
energy consumption 46, 87,
91, 92, 262, 263, 263
engine power 47
“Nuna” cars 242, 242
“puddles” on a dry road
133–4, 133, 134
radiator, frozen 7–8
rolling resistance 87–9, 89,
90–2, 91, 92
traffic density 93–7, 95, 97
tyres 87, 88
cathode ray tubes (CRTs) 256
CDs 159–64, 167–8
centrifugal force 79, 80, 81, 82,
283–5, 284
cigarette smoke, colour of
113–14
circles and spheres 3
area estimation 292, 292
circumference estimation
292, 292
volume estimation 291–2,
291
clocks, atomic 3–4
cloud colours 114
Coefficient of Performance (COP) 228
coffeepot, chromium 265, 266
colours
blue sea 169–72
blue sky 29, 107–10, 111–12,
133
CDs 159–64
cloud colours 114
LED colour spectrum emission 106, 111
rainbows 147, 148, 149, 152,
154
setting sun 110, 112, 137–8
soap bubbles 155–8
condensation 6, 271, 274, 280
conduction 24, 50–1, 52, 222,
223, 252, 257, 267, 272, 280
convection 53, 252, 253, 266
curtains
insulating property 254
lace curtains 139, 140, 142
sound absorption 208
curve balls 231
cycling
air resistance 59, 60–1, 61,
62, 63, 66, 70, 90
back-splashing 79–82, 80,
81, 82
echelon 72, 72
efficiency 55–8, 56, 62, 75
fast 63–6
home trainers 48, 51–2, 53,
200
HPVs 61, 64, 65, 66
minimising journey time
76–8, 76, 77
on the moon 67–8
power 53, 64–5, 64
in the rain 18, 83–6, 85
rolling resistance 59, 60, 61,
62, 67, 68, 90
side winds 69–72, 70, 72
speed records 64, 90
super bikes 61, 62, 63, 68
tail wind 69, 83
Everyday Physics: Unusual insights into familiar things
303
tyres 60, 62
wind chill 7, 9
D
daylength, seasonal 34–5, 34
decibel (dB) 183
dipoles 178, 179, 222
Dolby volume 189
double glazing 252, 253, 254,
255–7
evacuated 255–7, 300
secondary glazing 254
drag coefficient 22, 23, 60–1,
62, 90
dry air
humidification 276, 279–80
mountain air 5–6, 6
saunas 277–80
winter air 273–4
E
ears and hearing
age-related hearing loss
196, 196, 198
cocktail party effect 195–8
directional hearing 191–4,
192, 193, 197
hearing glasses 198
hearing threshold 182, 184
intensity difference mechanism 191, 192,
193,
198
lip-reading 196, 198
night-time 199–200, 202
sound sensitivity 182, 182, 184, 185
time difference mechanism 191, 192–3, 197
Earth
atmosphere 20, 29, 30, 110, 110, 287
304
axis 25, 25, 26
data 287
orbit 26–7, 28, 36
eggs in the microwave 224
Einstein, Albert 4, 239, 250
electric motors 226, 299
electric shavers 262, 263
electricity 24, 44, 48, 100, 223,
243, 244, 262
electromagnetic fields 108,
109, 222
electromagnetic waves 174,
221–2, 223, 224
elephants 270
emission coefficient 265, 268
energy 288
consumption, feeling 261–2
definition 43
energy consumption comparisons 262–3,
263
human engine 45–8
law of conservation of energy 24, 230, 261,
272
resting energy expenditure
45, 49, 52, 53
work and 43
equator 25, 26, 29
equinoxes 27, 27
evaporation 6, 50, 51–2, 52, 53,
271, 272, 279, 280
exponential growth 293–5
eye
cornea 144, 144
lens 123, 124, 125, 126, 144,
144
photoreceptors 100, 123, 131
sensitivity 102, 102, 103, 106,
129–31, 140–1
see also vision
Everyday Physics: Unusual insights into familiar things
F
falling
process 15–22
slow fall 24
fats, dietary 46
Fermat’s principle 77–8
figure skating 38
fog drops 19–20
Föhn 5, 6
food, conversion into heat 45
force
unit of 43
work and 43, 47
friction 12, 15, 16, 20, 62, 88,
246, 262, 284
fridges 225–8
cooling your home with
227, 227
heat pump action 226, 226,
228, 228
heating your home with
228, 228
G
glaciers 172
glass
heat conduction 251
insulation properties 251–2
see also windows
glasses
hearing glasses 198
Polaroid glasses 173–7
reading glasses 124, 125
GPS (Global Positioning System) 1–4
gravity 14, 15, 19, 20, 22, 23, 48,
57, 62, 80
lunar 67
grazing incidence 177, 178
Guericke, Otto von 256
Gulf Stream 35
H
hairdryer airflow 229, 235–7
heat
definition 43
latent heat 280
heat pump 226, 226, 228, 228
hertz (Hz) 182
hiking 5, 48
holograms 165–8
home experiments
angular momentum, conservation of 38–41
balloon inflation 258–60
blue sky and red setting sun 111–12
bouncing ball 237–8
CD as an optical instrument
162–4
electric motor 299
hairdryer airflow 235–7
Magnus effect 232–5
mug, vibration and pitch of a 216
rainbow angle 149–50
reckless wine glass party trick 41
“rotating string” blowpipe
73–4
the slow fall 24
tuning fork, increasing the volume of a 190
twin divers 298
wave nature of light 127–8
household appliances
energy consumption 263
running costs 264
see also specific appliances
human engine 45–8
heat loss/removal 8, 9, 10,
50–1, 52, 53, 54
Everyday Physics: Unusual insights into familiar things
305
humidity 6, 273, 274, 276, 277,
280
humidification 276, 279–80
relative humidity (RH) 6,
273, 274, 276, 277, 278
hypothermia 10
I
ice 172
slipperiness 11
thawing 12, 222
ice skating 11–12
infrared 100, 114, 266, 268
insulation 8, 251–2, 253, 254,
255–7, 279, 280
interference 156–8, 156, 157,
160, 160, 161, 165, 166, 167, 167, 168, 222
J
joule ( J) 43, 44, 45
K
kilocalorie (kcal) 43, 45
kilowatt-hour (kWh) 44
kinetic energy 24
Kittinger, Joseph 22
L
ladybirds 270
laminar flow 20
Lenz’s law 24
Leonardo da Vinci 122
light
absorption coefficient
170–1, 265
attenuation 169–70, 171, 171
306
brightness, relative 140,
141, 141
Fermat’s principle 77–8
polarization 174, 176–7, 176,
178–9, 178
Snell’s law 78, 78
speed of light 3, 78, 134, 289
white light 106, 114, 148, 159
light bulbs/lamps
fluorescent lamps 99,
105–6, 106
incandescent bulbs 99, 100,
101–4, 102, 106, 126
LED lamps 100, 105–6, 106
light waves 108–9, 126, 127–8,
156, 248
interference 156–8, 156, 157,
160, 160, 161, 165, 166, 167, 167, 168
oscillation 173–4, 174, 175,
176, 176, 177
prism, passage through a
138, 138, 159
reflection 140, 141, 142, 148,
149, 154, 158, 162, 172, 176, 177–9
refraction 77–8, 108, 115–18,
116, 118, 133–4, 133, 134, 136, 138, 138, 144,
145, 148, 149, 149, 159
scattering 108, 109–10, 109,
111–12, 113–14, 166, 166, 167, 169, 172
lip-reading 196, 198
lumens 100
M
Magdeburg hemispheres 256,
257
magnets 24, 299
Magnus effect 232–5, 300
matchboxes 269–70
mechanical energy 48, 49, 62
Everyday Physics: Unusual insights into familiar things
megajoule (MJ) 44
microwave cooking 221–4
mirage effect 133–4, 136, 137
moon
albedo 131, 132
brightness 130, 131–2
cycling on the 67–8
gravity 67
moon data 287
reflection coefficient 131
multipliers and dividers 289
muscles 47, 48, 49, 62, 124, 125
muscular work 49, 52, 53, 57
music and musical instruments
musical scales 183, 209
stringed instruments 182,
182, 209–15, 210, 213, 214
wind instruments 218, 219
N
Newton, Isaac 15
newton (N) 43
noise barriers 205–6, 206
noise pollution 185
O
oscillation 173–4, 174, 175, 176,
176, 177
ozone 30
P
painted surfaces 265–6, 268
parachutes 23
“passive house” ventilation
274
photography 104, 114, 122,
124, 125, 126, 129, 131, 141, 165, 167
photons 248
photoreceptors 100, 123, 131
photosynthesis 243, 244
photovoltaic cells 242, 243,
244
pi (ϖ) 293
ping-pong ball, and hairdryer airflow 235–7
pirouette principle 38
pitch (frequency) 181, 182, 182,
183, 189, 197, 198, 210, 211, 212, 212–15, 215,
216, 218, 219
Poiseuille’s law 74
Polaroid glasses 173–7
power, definition 43–4
power stations 100, 244, 262
prism 138, 138, 148, 159
R
radiation 50–1, 53, 54, 100,
102, 105, 179, 254, 265, 266, 272, 280
see also specific forms of
radiators
black vs white 265–6
electric 100
radiation/convection 266
radio waves 3, 221–2
radiometer 247–50, 300
radios 187
rainbows 147–54
colours 147, 148, 149, 152, 154
darker sky above 148, 152
features of 147–8
no-rainbow effect 152, 153
rainbow angle 148, 149–52,
150, 151, 154
secondary rainbows 147,
148, 154, 154
raindrops 14, 15–18, 18, 83–6,
149, 149
Everyday Physics: Unusual insights into familiar things
307
rainfall 19, 290
Rayleigh scattering 109–10,
111–12, 113–14
reading glasses 124, 125
reckless wine glass party trick 41, 41
records, 78-RPM 181
reflection 29, 140, 141, 142,
148, 149, 154, 158, 162, 172, 176, 177–9, 197, 207
reflection coefficient 131, 177,
179
refraction
light 77–8, 108, 115–18, 116,
118, 133–4, 133, 134, 136, 138, 138, 144, 145,
148, 149, 149, 159
sound 200, 201, 201, 202,
202, 204, 204, 206, 206
refractive index 159, 179
Reichert, Todd 66
relativity theory 4
resources 300
resting energy expenditure
45, 49, 52, 53
Reynolds-slip 62
ripple tank 156, 157
rolling resistance
cars 87–9, 89, 90–2, 91, 92
coefficient 60, 68, 88, 92
cycling 59, 60, 61, 62, 67,
68, 90
Rompelberg, Fred 66
“rotating string” blowpipe 73–4
S
satellites
GPS satellites 1–4, 2, 287,
300
New Horizon 290
orbit times 287
saunas 277–80
308
shock waves 23
skiing 7, 9, 9, 29
skydiving 21–4
Snell’s law 78, 78, 178, 179
snow 172, 274
soap bubbles 3, 155–8
sodium street lights 106
solar eclipse, observing 120–1,
121, 122
solar panels 242
solar power, harvesting 242–4
solar radiation 29–30, 30, 35,
35, 103, 239–40
energy flow 240–1, 288
solar-thermal collectors 242
solstices 26, 27, 27, 35–6
sonic boom 23
sound
absorption 185–6, 197, 208
intensity 182–4, 182, 185, 186,
187, 188, 189, 191, 193, 198
noise barriers 205–6, 206
noise pollution 185
pitch (frequency) 181, 182,
182, 183, 189, 197, 198, 210, 211, 212, 212–15,
215, 216, 218, 219
reflected sound 197, 207
sound barrier 22, 23, 68
“sound shadow” 193, 201,
202, 205, 206, 206
wind-blown 203–4, 203
see also ears and hearing; music and
musical instruments
sound, speed of 23, 200–1,
203–4
in air 289
temperature-dependent
200–1, 218, 219
in water 289
sound waves 190, 192–3, 218
Everyday Physics: Unusual insights into familiar things
amplitude 210, 210, 216
refraction 200, 201, 201,
202, 202, 204, 204, 206, 206
spectacles see glasses
spectrometer 164, 300
speed
acceleration 15, 16, 23, 48,
62, 82, 91
conversions 290
terminal velocity 22, 23
speed skating 90
squash (sport) 43
squinting 126
stair climbing 38, 47–8, 48,
49, 57
Stefan-Boltzmann law 54, 240
stereo equipment 189, 197
stringed instruments 182, 182,
209–15, 210, 213, 214
fundamental and overtones
210, 210, 211, 212
harmonics 211
pitch 210, 211, 212, 212–15,
215, 218, 219
timbre 211
tuning 215, 215, 217–19
summer and winter temperatures 33–6
sun
brightness 130, 132
elevation 25–8, 25, 26, 27,
29, 30, 35, 110, 136, 152
“green flash” phenomenon
137–8
parallel light beams 31–2, 31
radiation see solar radiation
setting sun 110, 112, 135–8,
136
solar eclipse 120–1, 121, 122
sun data 287–8
sunlight, filtered 119–20, 120
surface temperature 103,
105, 288
winter sunshine 34–5, 34, 35
sunburn 29, 30
surface tension 14, 17
swimming 29, 53
goggles 143, 145, 145, 146
shark attacks 145–6
underwater vision 143–6,
146
swimming pools 115–18, 116,
118, 144
T
talking, energy cost of 187–8
tea leaves in a cup 283–5
temperature
body temperature 8, 10,
50–4, 280
seasonal temperatures 33–6
wind chill 7–10
ten-pin bowling 38
tennis ball, topspun 230–1,
230, 231
terminal velocity 22, 23
thunder 186
tree height, estimation 120
tuning fork 190
TVs 256, 263
twin divers 298
U
ultraviolet (UV) radiation 29,
30
V
vacuum 15, 134, 249, 255–7
Everyday Physics: Unusual insights into familiar things
309
vacuum cleaners 47, 261, 263
vision
accommodation 124, 125
mirage effect 133–4, 136, 137
sharp 123–6, 124, 125
squinting 126
underwater 143–6, 144, 145,
146
see also eyes
W
walking
angular momentum 37–8,
75
efficiency 55, 56, 57
up stairs 38, 47–8, 48, 49, 57
water
boiling 51, 271–2, 274
fog drops 19–20
molecules 222, 273–4
raindrops 15–18
waves 13–14
see also condensation; evaporation; ice;
water vapour
water vapour 6, 273, 276
heat conduction 280
vapour pressure 278, 278
vapour pressure curve 273,
274, 274, 275
watt (W) 43–4
waves, water
interference 156, 156, 157
parallel to the beach 13, 14
speed 14
weight-loss diets 46
white light 106, 114, 148, 159
white paint 265–6, 268
Whittingham, Sam 66
wind 50, 200
310
mountain winds 5–6
side winds, cycling and
69–72, 70, 72
wind chill 7–10, 9
wind speed 9, 9, 203, 203,
204, 206
wind instruments 218, 219
wind turbines 203
windows
insulation 251–2, 253, 254,
255–7
as mirrors 142, 177
reflection 140, 141, 177
temperature profile 253,
253, 254
transparency 139, 140, 141,
141, 142
see also curtains
wine-mixing 281–2, 296–7
wood, and heat conduction
267, 278
work, definition 43
world population growth 295
World Solar Challenge car race 242
Everyday Physics: Unusual insights into familiar things
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