Contextualized Learning Activity: Motion
Introduction:
In the following exercises and discussion, we will assume ideal situations. That is, we do not
take into account any losses due to friction in the gear systems, sprockets, and chains. The
efficiency of any gear train can be calculated by measuring output and comparing it to the
expected output by calculation.
Motion
Beginning with the basics of motion – speed = distance x time (s = d x t) we can look at a motor
with angular velocity of x and translate that in to a forward velocity of a vehicle.
We can then “change” the velocity by way of gears. A transmission converts the velocity of the
motor to a velocity that is desired. Example: Your 18 speed mountain bike. As you change the
gears, you increase torque (force) when going up a hill or you increase speed when going down a
hill or on a straight.
Exercise 1.1: Cut out a circle of radius 2.00 cm and another of 4.00 cm. Mark a point on the
edge of each circle. Now start rolling the smaller circle on a ruler lining up the mark on the
beginning mark of the ruler. See diagram below.
Record where the mark again meets the ruler. This can also be verified by calculation. The
distance the circle travelled was of course the circumference. Calculate the circumference of
each circle and record.
1
Circumference = 2 * Pi * R
Small circle —> 12.56 cm
Larger circle —> 25.12 cm
Notice that the distance the smaller circle travels is half the distance the larger circle travels. This
explains why two gears, one half as big as the other, have a gear ratio of 2:1. The smaller gear has
to spin twice to cover the same distance covered when the larger gear spins once.
Most gears that you see in real life have teeth. The teeth have three advantages:
1. They prevent slippage between the gears. Therefore, axles connected by gears are always
synchronized exactly with one another.
2. They make it possible to determine exact gear ratios. You just count the number of teeth in
the two gears and divide. So if one gear has 60 teeth and another has 20, the gear ratio when
these two gears are connected together is 3:1.
3. They make it so that slight imperfections in the actual diameter and circumference of two
gears don't matter. The gear ratio is controlled by the number of teeth even if the diameters
are a bit off.
In the example below, the DRIVER has 60 teeth and because it is the largest we say that it
revolves once. The DRIVEN gear has 30 teeth. Simply divide 60 teeth by 30 teeth to work out
the number of revolutions of the driven gear.
Gear Ratio
= (Distance moved by Driver) / (distance moved by load)
= Driver / Driven
= 60 T / 30T
=2
Therefore the gear ratio is 2:1 That is, the driven gear goes around twice for each revolution of
the driver. The gear ratio is proportional to the ratio of the gear diameters and inversely
proportional to the ratio of gear speeds.
2
Exercise 1.2:
Calculate the gear ratio for each of the following questions.
2a)
1b)
See the table of common gear sizes (number of teeth) and
corresponding ratios in Appendix A.
Ryan, V, “Working out Gear Ratios”, 2001,
TechnologyStudent.com,
http://www.technologystudent.com/gears1/gears5.htm, April 26,
2008
Exercise 1.3:
a)
If the driver (pedal) gear has 80 teeth and the driven
(sprocket) gear has 20 teeth, what is the gear ratio?
b) If the driver gear has 100 teeth and the driven gear has 50
teeth, what is the gear ratio?
c) If the driver gear has 60 teeth and the driven gear has 20 teeth,
what is the gear ratio?
d) In the three above questions, what has the gearing done? Increased torque or speed?
2. Gear Trains
To create large gear ratios, gears are often connected together in gear trains, as shown here:
3
The gear on the right (purple) in the train is made of two gears. A small gear and a larger gear
are connected together, one on top of the other. Gear trains often consist of multiple gears as
shown in the next two figures.
In the case to the left, there are four gears in the train with the purple (2nd gear from left) turning
twice the rate of the blue (left most) gear. The green gear turns at a rate twice that of the purple
and so on, with the red gear turning twice that of the green. Therefore, the red gear will travel at
2 x 2 x 2 = 8 times the speed of the blue gear. The gear train shown below has an even higher
gear ratio.
In this train, the smaller gears on top of the large gears are one-fifth the size of those they are
connected to. That means that if you connect the purple gear to a motor spinning at 100
revolutions per minute (rpm), the green gear will turn at a rate of 500 rpm and the red gear will
turn at a rate of 2,500 rpm. You could also turn this around and attach a 2,500-rpm motor to the
red gear to get 100 rpm on the purple gear.
Here is how it would look mathematically: You have two gear ratios of 5:1. For the first one you
will have:
Driver RPM Output

Driven
RPM Input
5
x

1 100 RPM
x  500 RPM
If you can see inside your power metre on the side of
your house, and it's of the older style with five
mechanical dials, you will see that the five dials are
connected to one another through a gear train like this,
with the gears having a ratio of 10:1. Because the dials
are directly connected to one another, they spin in opposite
directions (you will see that the numbers are reversed on dials
next to one another). Check your metre when you get home
and report on whether it is of this style or not. You can also
check your gas metre in the same fashion.
Exercise 2.1:
Given the following gear train, what would the overall ratio
be as well as the output RPM.
a)
4
1:3, 1:4, 1:5 with a driver at 10500RPM
b)
1:2.2, 1:3.3, 1:4.4 with a driver at 3990 RPM
c)
5:1, 6:1 with a driver at 100 RPM
d)
4:1, 3.3:1, 2.5:1 with a driver at 10 RPM
3. Planetary Gears
There are many other ways to use gears. One specialized
gear train is called a planetary gear train. Planetary gears
allow the output shaft to be in the same axis as the input
shaft.
Let's assume you need a gear ratio of 6:1 with the input
turning in the same direction as the output. One way to
create that ratio is with a three-gear train as follows:
In this train, the blue gear has
six times the diameter of the
yellow gear (giving a 6:1 ratio). The red gear is used to change the
direction of the rotation so that both the input gear and output gear are
travelling in the same direction. The size of the red gear does not
matter. However, if you want the axis of the output gear to be the
same as that of the input gear you can use a planetary gear system as
shown below. A common place where this same-axis capability is
needed is in an electric screwdriver.
In this gear system, the yellow gear (the sun) engages all three red gears (the planets)
simultaneously. All three are attached to a plate (the planet carrier), and they engage the inside
of the blue gear (the ring) instead of the outside. The use of three red gears (planets) makes this
gear train extremely rugged. The output shaft is attached to the blue ring gear, and the planet
carrier is held stationary -- this gives the same 6:1 gear ratio as above but the input and output
shafts will be in the same axis. Not shifted as in the one above.
Planetary gear systems can be found in sprinklers and inside automatic transmissions.
Brain, Marshall. "How Gear Ratios Work." 20 November 2000. HowStuffWorks.com.
<http://science.howstuffworks.com/gear-ratio.htm> 26 April 2008.
5
Brain, Marshall. "How Oscillating
Sprinklers Work." 23 May 2000.
HowStuffWorks.com.
<http://home.howstuffworks.com/s
prinkler.htm> 01 May 2008.
Nice, Karim. "How Automatic Transmissions Work." 29 November 2000. HowStuffWorks.com.
<http://auto.howstuffworks.com/automatic-transmission.htm> 01 May 2008.
Exercise 3.1:
Given the following gear ratios in a planetary gear, calculate the output speed based on the input
speed.
a)
b)
c)
d)
6:1, 5 RPM
10:1, 2 RPM
12.3:1, 4.5 RPM
10:1, 4.6:1, 10 RPM
Application 3.1
Example: Given the example drive train in section 2 of 5:1, and 5:1 and attach the final gear to a
10 cm wheel with a ratio of 1:2, what will the resulting speed be with an input rpm of 100?
The overall gear train will
12.5 or 12.5:1
6
Driver RPM Output

Driven
RPM Input
12.5
x

1
100 RPM
x  1250 RPM
speed  1250 RPM 10cm   / RPM
 3.9 10 4 cm / min
 6.5m / s
give a ratio of 5 x 5 / 2 =
1.
A motor with an output shaft that has an RPM of 5000 is
connected to the following gear train: 1:2, 1:3.3, 1:2.3 and
the wheel size is 20 cm diameter, what would the speed of
the machine be?
2.
If the same machine has three speeds with an overall ratio of 1:3, 1:4, & 1:12 connected to
the 20 cm wheel, what would the “ground” speed be for each setting?
3.
2 CIM motors at 4000 rpm under load are connected to the following gear train: 1:3.3,
1:3.3, 1:1.6 that drives a wheel of 15 cm diameter. What is the RPM of the wheel?
4.
How fast will the robot in number 3 travel?
5.
How far will the robot move in 15 seconds?
6.
How quickly will the robot travel 16 m?
4. Torque
Torque is a force that causes a rotation or turning of things. It can be thought of as a “rotational
force”. You generate a torque any time you apply a force using a wrench. Tightening a nut is a
good example. When you apply a force at the end of the wrench, the force creates a torque on the
nut, which tends to turn the nut.
The torque and speed of a gear train are inversely proportional. That is, as you increase the speed
with a gear, you decrease the torque. Remember the mountain bike in section 1.
Example: Using the same example from 3.1 we see that the speed of the output gear is 12.5 times
faster than the input. The torque will be the opposite. If the torque under normal load of the
input motor is 0.45 Nm then the output torque will be:
7
InputTorque
Driver

Driven OutputTorque
12.5 0.45 Nm

1
x
x  0.036 Nm
x  3.6  10  2 Nm
The SI units for torque are Nm (Newton - metre). Force times distance (distance from fulcrum to
applied force).
Exercise 4.1
Given the following overall gear ratios, calculate the output torque based on the input torque.
a)
6:1, 1.0 Nm
b)
10.5:1, 2.0 Nm
c)
1:2.3, 25 Nm
d)
1:104, 0.50 Nm
8
Exercise 4.2
Given the following, calculate the missing values for each situation.
Driver
Teeth
a)
12
b)
12
c)
24
d)
9
Driven
Teeth
Gear
Ratio
1:5
Input
RPM
Output
RPM
100
4
300
1000
Output
Torque
Nm
6.0
55
100
Input
Torque
Nm
2.0
2.4
80
0.43
1.2
5. Encoders
Have you ever wondered how you can determine how far a robot or motorized machine has
moved? Well that is the job of an "encoder". An encoder is a device that encodes information so
that your computer processor and program can interpret them. If you want to count rotations of a
wheel with a tab that pushes a switch every rotation you have just encoded information. This is a
form of a mechanical encoder.
There are also magnetic and optical encoders in common use. Most automotive distributors now
use a solid state form of the old mechanical “points” which uses the "Hall effect", where the
magnetic properties of a metal-toothed wheel rotate past this encoder.
Optical encoders use light to switch on and off. There are also rotary and linear encoders.
The type of sensor you use depends on your application. However, in industry and robotics,
optical encoders are used due to their speed and reliability, as opposed to a mechanical encoder
whose mechanical components may eventually deteriorate.
Quadrature encoders
A quadrature encoder is simply an encoder that has four states or positions. The word
“quadrature” has the root of “quad” standing for four. A regular encoder has two states which are
“off” and “on”, or the popular binary values 0 and 1. A quadrature encoder has four. If you think
in binary you immediately will realize that to represent four states you need at least two bits.
Example: 01 or 11.
Quadrature encoders use two switches, called the “A” phase and the “B” phase. Essentially
quadrature encoders supply counts to you. You will receive the signals of phases A and B in
some pattern like 00, 01, 10, 11 (or better known as 0, 1, 2, 3) The most interesting aspect of a
quadrature encoder is that it can tell you direction. Let’s say that the pattern 0, 1, 2, 3, 0, 1, 2,…
is forward. Then the pattern 0, 3, 2, 1, 0, 3, 2,… is backward! If you keep track of these state
changes, you can track distance moved.
By the way if you think that this technology is not in common use… then take apart your
mechanical mouse (at your own risk, of course). Go ahead, open it up by twisting the cover
around the ball, and then take out the ball. You may also be able to find an old (not in use) ball
mouse and take the screws out of the bottom. Look inside and you will see two shafts that the
ball rolls against at right angles to each other. Then look to the ends of each shaft and you may
see the encoder, and, depending on the type of mouse, you may also see an encoding wheel.
Mechanically, the shift in encoder phases can be accomplished in optical encoders by having two
"mask wheels" through which light shines. Or there can be two light sensors mounted around one
mask wheel at an angle that creates the phase shift.
Robot Application
In any application, one must first select a location of the encoder. Once selected, you must do
some math to select the appropriate encoder for the given application.
Start by finding the motor’s maximum RPM (revolutions per minute) without load from the
specification sheet or manufacturer. Also if possible, measure how fast the robot travels a given
distance, then using known gear ratios and drive wheel circumference compute the motor shaft’s
10
RPMs under load for the robot’s maximum speed. The difference in speed is the effect of friction
within your drive train. You should have both numbers for a check. For sake of example we will
use an RPM of 5.1 under no load and 4.3 under load. For examples of the above calculations, see
the example below
Next, identify the resolution that you want the robot to have. Its OK to be unreasonable. I
suggest that you try a number of different resolutions when all is said and done.
For example, choose the resolution to be 1/8”. Then compute how many increments of the
selected resolution are required for the drive wheel to rotate once. Given a 28” circumference
divided by 0.125” resolution we got 224 increments or "state changes", “ticks”, or interrupts per
drive wheel revolution.
At the robot’s maximum theoretical and observed speed how many times did the wheel rotate per
second? Our computed speed was 5.1 and 4.3 drive wheel revolutions per second, receptively.
Multiply the number of interrupts per drive wheel revolution by the drive wheel rotations per
second to get the number of interrupts per second. Believe it or not, numbers like 1,000 to 2,000
interrupts per second are practical. From our observed wheel speed of 4.3 rev./sec. * 224
interrupts/rev. = 936 interrupts/sec.
Now you begin selecting which encoder to use. Encoders are described by the number of
positions, cycles or phase changes per revolution of the encoder. They are basically all the same
thing. As a rule, the fewer phase changes per revolution, the less you will pay.
Compute the resulting number of encoder revolutions per wheel revolution.
Next take the theoretical and observed target number of interrupts per second and divide it by the
number of encoder revolutions per second to determine the number of encoder interrupts per
revolution needed, (interrupts / sec.) / (rot./sec.) = (interrupts/rot.)
Now comes the judgement part of the design. Look at the catalogues to find an encoder that
comes close to the desired state changes per revolution. Some encoder makers seem to like using
powers of 2 so numbers like 8, 16, 32, 64, 128 and 256 are popular. But numbers like 24, 25, 50,
75 or 100 are common as well. It all depends on the manufacturer. Now compute your "real"
resolution. Given the number of state changes per rotation of the encoder selected times the
number of revolutions of the encoder per drive wheel rotation, compute the number of state
changes per drive wheel rotation, and finally your real resolution. Also compute the number of
interrupts per second given the selected encoder.
In this design process you have to iterate your design multiple times to find the best fit. It is most
important to be certain that the number of interrupts per second is reasonable with the motors
under load. The unloaded state just makes certain that you don’t damage the encoder by spinning
it too fast should your drive wheel lose grip. But realize that if the drive wheel ever loses its grip,
your positional information will be unreliable.
Quadrature Encoders
February 18, 2004 Version 0.1
By Daniel Katanski, Programming Mentor, Team 240,
http://www.chiefdelphi.com/media/papers/1490
11
Driver sprocket
Encoder on shaft of driver
Wheel sprocket (driven)
Wheel
Andy Mark Super Shifter
Example 5.1:
CIM motor with 4300 RPM normal load with a 12 tooth gear coupled to a 40 tooth gear. On the
same shaft a 15 tooth gear drives a 48 tooth gear which is on the same shaft as a 15 tooth
sprocket connected via chain to a 24 tooth sprocket which is bolted to a 10 cm diameter wheel.
The speed of this drive train is 1.32 m/sec. The encoder on the shaft of the sprocket shaft is a 32
pulse Grayhill encoder. That is, the encoder will “tick” 32 times every revolution. A tick
translates into 6.1 mm of robot movement.
Speed = 4300 RPM*12/40 * 15/48 * 15/24
=250 RPM
Speed = 250 RPM/60 sec/min
= 4.2 RPS
d = 4.2 RPS *10 cm * PI
= 132 cm/sec
= 1.32 m/sec
encoder speed = 4300 RPM*12/40 * 15/48 / 60 sec/min
= 6.72 RPS
= 6.72 RPS * 32 ticks /R
= 215 ticks / sec
encoder distance /tick = d/encoder speed
= 132 cm/sec /215 ticks/sec
= 6.1 mm / tick
Exercise 5.1:
a) Based on the a robot with CIM motors with 4000 RPM under load and a drive train with a
gear ratio of 1:20 and wheels that are 10 cm in diameter, what encoder should be chosen?
b) With an encoder of 32 ticks, how far will it travel each tick?
How many ticks will it take to travel 8.0 m, 4 m , 2 m ?
12
Other applications of encoders:
Motor speed
Motor speed can be determined by an encoder attached to the drive shaft of almost any
application.
Web Speed/Tension control
On a conveyor system, the speed of the conveyor and the tension on the belts or web of the
conveyor can be determined by encoders.
Linear Measurement/Cut-to-length
Steel can be moved along a conveyor type system a specific distance so that it can be cut-tolength. An encoder can determine the exact length or distance the motor has turned.
Position Measurement - Conveying, arm motion etc
Robotic arms can be moved to specific positions or turned specific amounts by the use of
encoders for feedback.
6. Gear Tooth Sensors
Computerized Odometers
If you make a trip to the bike shop, you most likely won't find any cable-driven odometers or
speedometers. Instead, you will find bicycle computers. Bicycles with computers like these have
a magnet attached to one of the wheels and a pickup attached to the frame. Once per revolution
of the wheel, the magnet passes by the pickup, generating a voltage in the pickup. The computer
counts these voltage spikes, or pulses, and uses them to calculate the distance travelled.
If you have ever installed one of these bike computers, you know that you have to program them
with the circumference of the wheel. The circumference is the distance travelled when the wheel
makes one full revolution. Each time the computer senses a pulse, it adds another wheel
circumference to the total distance and updates the digital display. Many modern cars use a
system like this, too. Instead of a magnetic pickup on a wheel, they use a toothed wheel mounted
to the output of the transmission and a magnetic sensor that counts the pulses as each tooth of
the wheel goes by. Some cars use a slotted wheel and an optical pickup, like a computer mouse
does. Just like on the bicycle, the computer in the car knows how much distance the car travels
with each pulse, and uses this to update the odometer reading.
One of the most interesting things about car odometers is how the information is transmitted to
the dashboard. Instead of a spinning cable transmitting the distance signal, the distance (along
with a lot of other data) is transmitted over a single wire communications bus from the engine
control unit (ECU) to the dashboard. The car is like a local area network with many different
devices connected to it.
13
7. Application
1)
Given the following motor specs, design a drive that will produce a wheel speed of 4.0
m/s. Give the gear sizes and ratios as well as the wheel size. Justify and explain each
selection. Include a quadrature encoder and justify why it was chosen. Grayhill’s optical
encoders from Digikey.com would be a good place to look. CIM Motor specifications:
Under no load the output shaft has a speed of 5300 RPM and under normal load it has an
RPM of 4300.
2)
A conveyor system, on a 15 cm diameter pulley, is required to move 1.0 m in 1 second. If
a motor with normal load spins at 1000 RPM’s, design a gear train that will produce the
desired speed. How often would a 128 pulse encoder pulse if connected to the motor
shaft? How often would it pulse if connected to the driven shaft? How far does the
conveyor travel for every pulse of the encoder connected to each of these locations?
3)
A cut-to-length system is required to move a piece of steel exactly 3.45 m each time.
Based on a motor speed under load of 500 RPM and a desired conveyor speed of 0.85
m/s, design a gear train that will produce the desired speed and suggest an encoder that
will provide adequate feedback to control the system. The output pulley is 18 cm in
diameter. Justify each decision.
4)
The picture on the right is an automated vacuum seal
machine. The belt moves the product under the vacuum
unit and then seals it. The system needs to move a
precise number of increments based on the product being
processed. A redundancy mechanism is built in with
electronic eyes to see that the product is within the
bounds. If the motor on the conveyor has an RPM with
load of 2500 RPM, and the conveyor is to move between
1.0 m in 3 seconds and 3.0 m in 10 seconds, create a gear
train that will produce the desired output and suggest an encoder size that will mount on
the output gear.
5)
The foam cut off saw as pictured here has an indexing increment of 25.4 mm. That is, it
can move 25.4 mm each time or any multiple of that. The speed of the conveyor is 24.5
m/min and it has a motor that rotates at
1000 RPM under load. Design a gear
train and size the encoder required to
receive the output specified.
14
Appendix
A. Common Gear Sizes and Ratios
32 pitch
Teeth
16
18
20
24
40
16
1.000
18.000
1.250
1.500
2.500
18
0.889
1.000
1.111
1.333
2.222
20
0.800
0.900
1.000
1.200
2.000
24
0.667
0.750
0.833
1.000
1.667
40
0.400
0.450
0.500
0.600
1.000
24 Pitch
Teeth
12
15
16
20
21
24
36
48
72
12
1.000
15.000
1.333
1.667
1.750
2.000
3.000
4.000
6.000
15
12.000
1.000
1.067
1.333
1.400
1.600
2.400
3.200
4.800
16
0.750
0.938
1.000
1.250
1.312
1.500
2.250
3.000
4.500
20
0.600
0.750
0.800
1.000
1.050
1.200
1.800
2.400
3.600
21
0.571
0.714
0.762
0.952
1.000
1.143
1.714
2.286
3.429
24
0.500
0.625
0.667
0.833
0.875
1.000
1.500
2.000
3.000
36
0.333
0.417
0.444
0.556
0.583
0.667
1.000
1.333
2.000
48
0.250
0.312
0.333
0.417
0.438
0.500
0.750
1.000
1.500
72
0.167
0.208
0.222
0.278
0.292
0.333
0.500
0.667
1.000
20 pitch
Teeth
12
15
16
20
23
24
25
32
36
48
60
12
1.000
15.000
1.333
1.667
1.917
2.000
2.083
2.667
3.000
4.000
5.000
15
12.000
1.000
1.067
1.333
1.533
1.600
1.667
2.133
2.400
3.200
4.000
16
0.750
0.938
1.000
1.250
1.438
1.500
1.562
2.000
2.250
3.000
3.750
20
0.600
0.750
0.800
1.000
1.150
1.200
1.250
1.600
1.800
2.400
3.000
23
0.522
0.652
0.696
0.870
1.000
1.043
1.087
1.391
1.565
2.087
2.609
24
0.500
0.625
0.667
0.833
0.958
1.000
1.042
1.333
1.500
2.000
2.500
25
0.480
0.600
0.640
0.800
0.920
0.960
1.000
1.280
1.440
1.920
2.400
32
0.375
0.469
0.500
0.625
0.719
0.750
0.781
1.000
1.125
1.500
1.875
36
0.333
0.417
0.444
0.556
0.639
0.667
0.694
0.889
1.000
1.333
1.667
15
48
0.250
0.312
0.333
0.417
0.479
0.500
0.521
0.667
0.750
1.000
1.250
60
0.200
0.250
0.267
0.333
0.383
0.400
0.417
0.533
0.600
0.800
1.000