Dimensions
500.101
Dimensional Reasoning
Dimensions
500.101
Dimensions and Measurements
•
“Dimension” is characteristic of the object, condition, or event and is
described quantitatively in terms of defined “units”.
•
A physical quantity is equal to the product of two elements:
– A quality or dimension
– A quantity expressed in terms of “units”
•
Dimensions
– Physical things are measurable in terms of three primitive qualities
(Maxwell 1871)
• Mass (M)
• Length (L)
• Time (T)
Note: (Temperature, electrical charge, chemical quantity, and luminosity were
added as “primitives” some years later.)
Dimensions
500.101
Dimensions and Measurements (cont.)
– Examples
• Length (L)
• Velocity (L/T)
• Force (ML/T2)
• Units
– Measurements systems--cgs, MKS, SI--define units
– SI units are now the international standard (although many
engineers continue to use Imperial or U.S.)
Dimensions
500.101
SI Primitives
DIMENSION
UNIT
SYMBOL for UNIT
Length
meter
m
Mass
kilogram
kg
Time
second
s
Elec. Current
ampere
A
luminous intensity
candela
cd
amount of substance
mole
mol
Dimensions
500.101
SI Derived units
DESCRIPTION
DERIVED UNIT
SYMBOL
DIMENSION
Force
newton
N
mkg/s2
Energy
joule
J
m2kg/s2
Pressure
pascal
Pa
kg/(ms2)
Power
watt
W
m2kg/s3
Dimensions
500.101
•
Dimensional analysis
Fundamental rules:
– All terms in an equation must reduce to identical primitive dimensions
– Dimensions can be algebraically manipulated, e.g.
– Example:
•
s
1 2
at
2
L
T 2  L
2
T
Uses:
– Check consistency of equations
– Deduce expression for physical phenomenon
L
T
T
L
Dimensions
500.101
Dimensional analysis
distance s = s0 +vt2 + 0.5at3
constant = p + ρgh +ρv2/2
volume of a torus = 2π2(Rr)2
sin( ) sin(  ) sin( )


2
2
a
b
c2
Dimensions
500.101
Deduce expressions
• Example: What is the period of oscillation for a pendulum?
 L
Possible variables: length l [L], mass m [M], gravity g  2 
T 
i.e. period P = f(l , m, g)
Period = [ T ], so combinations of variables must be
equivalent to [ T ].
g
l
m
Dimensions
500.101
Dimensional analysis (cont.)
• Example: What is the period of oscillation for a pendulum?
 L
Possible variables: length l [L], mass m [M], gravity g  2 
T 
i.e. period P = f(l , m, g)
Period = [ T ], so combinations of variables must be
equivalent to [ T ].
Only possible combination is
~
l
g
g
l
Note: mass is not involved
m
Dimensions
500.101
Quantitative considerations
• Each measurement carries a unit of measurement
– Example: it is meaningless to say that a board is “3” long. “3”
what? Perhaps “3 meters” long.
• Units can be algebraically manipulated (like dimensions)
• Conversions between measurement systems can be
accommodated, e.g., 1 m = 100 cm,
m
1
cm
100 

cm or
100 m
or
3m  3m 100
Example:
cm
cm
 3 100 m 
 300 cm
m
m
Dimensions
500.101
Quantitative considerations (cont.)
• Arithmetic manipulations between terms can take place only with
identical units.
Example:
but,
3m  2cm  ?
3m 100
cm
 2cm  302 cm
m
Dimensions
500.101
Buckingham Pi Theorem (1915)
•
Pi theorem tells how many dimensionless groups define a
problem.
•
Theorem: If n variables are involved in a problem and these
are expressed using k primitive dimensions, then (n-k)
dimensionless groups are required to characterize the
problem.
Example: in the pendulum, the variables were time [T],
gravity [L/T2], length [L], mass [M] . So, n = 4 k = 3. So, only
1 / 2 describes the system.
one dimensionless group
t g / l 
Dimensions
500.101
Buckingham Pi Theorem (cont.)
• How to find the dimensional groups:
– Pendulum example:
where a,b,c,d are coefficients to be determined.
1  f ( 2 , 3 ,, nk )
In terms of dimensions:
 1  t a l b g c md
T a Lb ( LT 2 )c M d  T ( a2c ) L(bc ) M d  M 0 L0T 0
Dimensions
500.101
Buckingham Pi Theorem (cont.)
Therefore:
a - 2c = 0
b+ c=0
d=0
Arbitrarily choose a = 1. Then b = -1/2, c = 1/2, d = 0.
This yields
g
1  t
 const.
l
Dimensions
500.101
Buckingham Pi Theorem (cont.) Oscillations of a star
A star undergoes some mode of oscillation. How does the frequency of
oscillation ω depend upon the properties of the star? Certainly the density ρ
and the radius R are important; we'll also need the gravitational constant G
which appears in Newton's law of universal gravitation. We could add the mass
m to the list, but if we assume that the density is constant, then m = ρ(4πR3/3)
and the mass is redundant. Therefore, ω is the governed parameter, with
dimensions [ω] = T-1, and (ρ; R; G) are the governing parameters, with
dimensions [ρ] = ML-3, [R] = L, and [G] = M-1L3T-2 (check the last one). You can
easily check that (ρ; R;G) have independent dimensions; therefore, n = 3; k = 3,
so the function Φ is simply a constant in this case. Next, determine the
exponents:
[ω] = T-1 = [ρ]a[R]b[G]c = Ma-cL-3a+b+3cT-2c
Equating exponents on both sides, we have a - c = 0; -3a + b + 3c = 0; -2c = -1
Solving, we find a = c = 1/2, b = 0, so that ω = C(Gσ)1/2, with C a constant. We
see that the frequency of oscillation is proportional to the square root of the
density, and independent of the radius.
Dimensions
500.101
•
“Dimensionless” Quantities
Dimensional quantities can be made “dimensionless” by “normalizing” them with
respect to another dimensional quantity of the same dimensionality.
Example: speed V (m/s) can be made "dimensionless“ by dividing by the velocity of
sound c (m/s) to obtain M = V/c, a dimensionless speed known as the Mach number.
M>1 is faster than the speed of sound; M<1 is slower than the speed of sound.
Other examples: percent, relative humidity, efficiency
•
Equations and variables can be made dimensionless, e.g., Cd = 2D/(ρv2A)
•
Useful properties:
– Dimensionless equations and variable are independent of units.
– Relative importance of terms can be easily estimated.
– Scale (battleship or model ship) is automatically built into the dimensionless
expression.
Dimensions
500.101
Dimensionless quantities (cont.)
– Reduces many problems to a single problem through normalization.
Example: Convert a dimensional stochastic variable x to a
non-dimensional variable
x 
x

to represent its position with respect to a Gaussian curve--N(0,1),
e.g., grades on an exam
Dimensions
500.101
Proof of the Pythagorean Theorem
The area of any triangle depends on its size and shape, which can be
unambiguously identified by the length of one of its edges (for example,
the largest) and by any two of its angles (the third being determined by
the fact that the sum of all three is π). Thus, recalling that an area has
the dimensions of a length squared, we can write:
area = largest edge2 • f (angle1, angle2),
where f is an nondimensional function of the angles.
Now, referring to the figure at right, if we divide a right triangle in two
smaller ones by tracing the segment perpendicular to its hypotenuse and
passing by the opposite vertex, and express the obvious fact that the
total area is the sum of the two smaller ones, by applying the previous
equation we have:
c2 • f (α, π/2) = a2 • f (α, π/2) + b2 • f (α, π/2).
And, eliminating f:
c2 = a2 + b2, Q.E.D.
Dimensions
500.101
Scaling, modeling, similarity
• Types of “similarity” between two
objects/processes.
– Geometric similarity – linear
dimensions are proportional; angles
are the same.
– Kinematic similarity – includes
proportional time scales, i.e.,
velocity, which are similar.
– Dynamic similarity – includes force
scale similarity, i.e., equality of
Reynolds number (inertial/viscous),
Froud number (inertial/buoyancy),
Rossby number (inertial/Coriolis),
Euler number (inertial/surface
tension).
Dimensions
500.101
Scaling, modeling, similarity
• Distorted models
– Sometimes it’s necessary to
violate geometric similarity:
A 1/1000 scale model of the
Chesapeake Bay is ten
times as deep as it should
be, because the real Bay is
so shallow that, with
proportional depths, the
average model depth would
be 6mm, too shallow to
exhibit stratified flow.
Dimensions
500.101
Dimensions
500.101
Scaling, modeling, similarity
• Scaling
– What’s the biggest elephant? If one tries to keep similar
geometric proportions, weight  L3, where L is a characteristic
length, say height.
– However, an elephant’s ability to support his weight is
proportional to the cross-sectional area of his bones, say R2.
– Therefore, if his height doubles, his bones would have to
increase in radius as 22 R, not 2R.
– [Note: A cross-section of 8 R2 = (22 R)2]. So, with increasing
size, an elephant will eventually have legs whose cross-sectional
area will extend beyond its body
Dimensions
500.101
Biological scaling