Teacher Answer Key: Wheels and Distance
Introduction to Mobile Robotics > Wheels and Distance Investigation
Condition 1: Standard Wheels - Calculate Distances
Record your measurements in the table on the last page of this worksheet.
1. Diameter of the wheel
For the standard wheel, the diameter is 5.5cm. Students should measure the diameter of the wheel at
the thickest point (where it actually touches the table while rolling) using a ruler or similar measuring
device.
2. Circumference of the wheel
Students should calculate circumference using the formula C = pi * diameter. For the standard wheel,
this is 17cm (note that there are only 2 significant figures here because our initial measurement only had
2; students should be sure that their recordings preserve significant figures).
3. Number of motor rotations
This is found in the program. The given program for this lesson specifies rotational distance in degrees,
so students will have to convert to full rotations to find this value. The default program goes for 720
degrees, which is 2.0 rotations (2 significant figures in 720 degrees).
4. Distance traveled (hypothetical)
Distance traveled is equal to the length of wheel circumference as it “unrolls” along the ground as the
wheel turns. One full turn of the wheel will move the robot one circumference’s distance. Therefore,
students will find the distance traveled by multiplying circumference by the number of rotations.
17 cm * 2.0rotations = 35 cm
(34 cm if rounded early; all calculations should be done before rounding to significant figures)
Condition 1: Standard Wheels - Run and Measure
Record your calculations:
5. Actual distance traveled in each trial
Students will measure this. Based on the measurements taken in the steps on this page, students
should fill in data for the three trials for the Standard Wheel in the data table on their worksheet. For
convenience, these boxes are numbered in the table to indicate that there are 3 trials within this condition.
Answers should be in cm, as indicated in the column header.
Condition 1: Standard Wheels – Evaluate the Hypothesis (1)
6. Look at the data in your table.
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i. Did the robot go the exact same distance in all three trials? Why or why not?
Probably not. Generally, the robot will not have gone the exact same distance in all three trials, due to
the many factors collectively called “random experimental error”. This includes such factors as
unpredictable wheel slippage, motor power variation, and tiny bits of human error in aiming the robot
directly down the measured path.
ii. Calculate the average distance that the robot went with these wheels and this program.
Average Distance
(for 3 trials)
=
distance 1 + distance 2 + distance 3
3
This is just the average (arithmetic mean) of the values of the three trials for this condition. Averaging the
data helps to reduce the effects of random experimental error.
For the sample data provided in the table on the last page:
Average = (34.5cm + 37cm + 33cm)/3 = 34.8cm
iii. What is the purpose of averaging the three distances?
To reduce error. Averaging the results of the three trials is intended to reduce the effects of the random
error, and give a better sense of what the “true” distance is. Using the arithmetic mean (average) works
well, because for example, a slightly-too-high value will cancel out a slightly-too-low value when they are
averaged, leaving a more “correct” value in general.
7. How far off was the experimentally measured average from the predicted value? Find the
Percent Error (% error) of the measured value compared to the predicted value using the
following formula:
Students will calculate this.
% Error is a quick way to estimate how well a prediction matches the actual data. A large % error
indicates that the experimentally gathered data did not match the prediction well, possibly due to large
amounts of experimental error (usually systematic error, which consistently throws off the data in a
specific way), or possibly because the prediction was simply wrong.
Students should find a fairly low % error value, because the hypothesis is theoretically correct in this
case. Since students are working with LEGO and not multi-million dollar lab equipment, reasonable %
error values can range up to about 20%.
In the case of the sample data, we found:
(|35cm – 34.8cm|)/35cm = .006 or 0.6%
8. Look back at your table.
i. Was the average of the distances you measured close to what Dr. Turner’s hypothesis
predicted it would be?
Students should draw on their % error calculations and general sense of data trends to draw conclusions
based on their calculations, but generally, the average measured distance and the hypothetical
distance will be very similar.
ii. Does this support the hypothesis? Why or why not?
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This result supports the hypothesis if the % error calculations are low, meaning that the measured
distances are similar to those predicted by the hypothesis.
iii. Is this set of trials alone enough to prove or disprove how valid the hypothesis is, in general?
It is not enough to prove the hypothesis, by any means.
In fact, hypotheses can never be totally proven, they can only be supported by evidence. A hypothesis
that is supported by enough evidence is generally accepted as a theory, but is always subject to revision
or replacement if future experimentation shows that it is not adequate.
Some students, particularly those who had bad data due to incorrect measurement technique, may
conclude based on their bad data that the hypothesis is in fact incorrect. This would be a well-reasoned
conclusion based on poor data, and the two processes (data gathering and data analysis) should be
evaluated independently. However, you will want to encourage students with bad data to go back and
resample their data at this point (perhaps by making a class wide comparison of results) and rethink their
conclusions, or the rest of the investigation will not make much sense.
9. The instructions tell you to measure from the front of the robot to the back of the line. Why
shouldn’t you measure the space “between” the robot and the line (from the back of the
robot to the front of the line) instead?
You want to measure the actual displacement (distance traveled) of the robot. The correct way to do
this is by picking one spot on the object and measuring how far that part of the robot moved from start to
finish. We chose the front tip of the robot as our “spot”. The “back of the line” represents the starting
location of the robot’s tip. You should then measure the distance to the final location of the tip of the
robot.
Students will often intuitively measure the distance from the back of the robot to the front of the line,
because it’s the distance “between” the two.
A quick illustration of why this is wrong: imagine a robot design so long that its back end doesn’t even
cross the line, even if it has moved the correct distance. What is your distance then? Clearly you can’t
measure the closest point between the robot and the line anymore, because the robot isn’t even off the
line. And you can’t say that the robot has failed to move, because it certainly did move. The only way to
fairly measure how far the robot has gone is by comparing how far the same point on the robot went; in
the investigation, it is the front of the robot to the front of the robot.
If measurements hadn’t been taken from the front of the robot before it started to the front of the robot
after it finished every time, then data points may be skewed. For instance, if for two of the three trials a
group measured from the front of the robot after it had finished to the back of the starting line (where the
front of the robot was before it started), but the other trial was measured from the back of the robot after it
had stopped to the back of the starting line, this one piece of data would be misleading and not actually
represent the distance traveled, not to mention skewing the entire data set.
Condition 2 – Small Wheels: Calculate Distances and Run
Record your measurements in the table on the last page of this worksheet.
10. Diameter of small wheel
The diameter of the small wheel is 3.0cm. This is the same procedure as for standard wheels, just
measuring the small wheels instead.
11. Circumference of small wheel
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The circumference of the small wheel is its diameter times pi, or 9.4cm.
12. Number of motor rotations
The program hasn’t changed at all, so the number of rotations is still 2.0.
13. Distance traveled (hypothetically)
The program still runs for 720 degrees = 2.0 rotations.
Its theoretical distance is therefore 9.4cm * 2.0 rotations = 19 cm.
14. Distance traveled in each trial (actual)
Measured as before, following the instructions on the page.
Condition 2 – Small Wheels: Evaluate the Hypothesis (2)
15. What is the average distance the robot ran with these wheels? Is this average a good
representation of the data you gathered in this Condition, or does the data look nothing like
the average?
Average Distance
(for 3 trials)
=
distance 1 + distance 2 + distance 3
3
For the sample data:
Average = (21cm + 20cm + 18cm)/3 = 19.7cm
If the students gathered the data using good measurement techniques, the data should all be fairly
similar, and the average should be a good representative value.
16. Find the Percent Error (% error) of the measured values compared to the predicted values
using the following formula:
% Error calculations should be done again for the small wheels. Trials should be similar to the predicted
values, resulting in low % errors, below 20%.
In the case of the sample data, we found:
(|19cm – 19.7cm|)/19cm = .039 or 3.9%
17. Look at the data in your table.
i. Were the average measured distances about what Dr. Turner’s hypothesis predicted they
would be?
If students measured correctly, the experimental values should come out quite similar to the
theoretical predictions. Again, a reasonable % error cutoff for “similar” might be around 20% with the
measuring techniques and technologies used in this lesson.
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ii. Do you think you have enough evidence to reasonably accept or reject how valid the
hypothesis is now?
With additional evidence that the hypothesis was accurately able to predict for both different wheel sizes,
the body of supporting evidence is growing. It is up to students to decide whether this is really enough. In
the purely scientific sense, this is far too small a sample size, but in the classroom sense, this is pretty
good.
iii. If so, do you accept or reject it? If you are not sure, what additional testing could you do to
help you decide?
For purposes of the rest of this investigation, students will be operating under the assumption that
this hypothesis is acceptable for general use. In general, this is the basic accepted theory in realworld use.
If students are interested in a larger sample size, they can take more trials or combine results with the
rest of the groups in the class. If the classroom has other wheels of different sizes handy, students can
measure and run tests with these wheels to increase the number of experimental conditions.
Analysis and Conclusions: Conclusion
18. In her fax, Dr. Turner proposes a relationship for calculating distances traveled by a robot,
based on the robot’s wheel size. You have been testing this hypothesis to see how well it
predicts the distance your robot travels.
i. Does the data you have collected support the hypothesis, or are the two very dissimilar?
Yes, the hypothesis is supported by the evidence.
The hypothesis in equation form is Distance traveled = rotations x circumference. Students found the
difference between the predictions of this hypothesis and the actual values they measured themselves.
The %error calculations should have shown that the two were very similar.
ii. What is your conclusion regarding the hypothesis? If you had to decide, would you say it is
correct or not?
This is something students have already somewhat decided in the Evaluate the Hypothesis phase for
Condition 2. Students should generally conclude that the hypothesis is, in fact, correct, based on the
fact that the predictions matched the data with a reasonable degree of certainty in two out of two cases.
iii. Summarize the steps you took and data you gathered to investigate, and explain how they
led to your conclusion.
Students should summarize the main steps in the experiment here.
•
Calculated the theoretical distance the robot should go for a certain number of rotations in a
program, using the hypothetical relationship
•
Run the program and measure the actual distance traveled by the robot under those conditions
•
Average the experimental data to get a representative value
•
Compare the hypothetical and measured average values using a goodness-of-fit test like percent
error
•
Repeat predict and measure steps with a different sized wheel
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•
Compare results again, and continue to evaluate the hypothesis’ ability to predict the data
correctly
19. Configure your robot using what you feel are the best all-purpose wheels for your
classroom. Measure the wheels on your robot.
i.
How many centimeters does your wheel travel per 360 degree rotation?
Let us assume that a student has chosen the default wheels as the best “all-purpose” wheels. There are
no set criteria for making this decision, though students should be able to justify their choice if asked.
Default wheel:
The wheel turns a distance equal to its circumference every time it rotates. The circumference of the
default wheel was already found earlier during the experiment. The wheel travels 17 cm every time it
rotates.
ii. In order to travel 10cm, how many degrees does the wheel need to turn? Show your work.
There are several different methods of solving this problem. Here are a few examples.
Ratio method:
The robot travels 17cm in one full rotation. If you can find what portion of 17cm the 10cm distance is, then
you can find what portion of a turn the wheel needs to make.
10 cm x fraction of a rotation
=
17 cm
1 rotation
x = 10 /17 = 0.59 of a turn
.59 rotations ⎛ 360 degrees ⎞ 210 degrees
⎜
⎟=
1
rotation
⎝
⎠
Algebra method:
distance traveled = circumference of wheels × number of rotations
10 cm = 17 cm × number of rotations
number of rotations = 10 /17 = .59 rotations
.59 rotations ⎛ 360 degrees ⎞ 210 degrees
⎜
⎟=
⎝ 1 rotation ⎠
iii. In order to travel 20cm, how many degrees does the wheel need to turn? Show your work.
Over the next few questions, students should begin to notice a pattern in the procedure needed to find the
number of degrees. The number of cm changes, but the steps taken to find the equivalent number of
degrees of rotation does not.
Ratio:
The robot still travels 17 cm in one full rotation. If you can find what portion of 17 cm the 20 cm distance
is, you know the portion of the turn.
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20 cm x fraction of a rotation
=
17 cm
1 rotation
x = 20 /17 = 1.2 turns
1.2 rotations ⎛ 360 degrees ⎞ 420 degrees
⎜
⎟=
⎝ 1 rotation ⎠
Algebra:
distance traveled = circumference of wheels × number of rotations
20 cm = 17 cm × number of rotations
number of rotations = 20 /17 = 1.2 rotations
1.2 rotations ⎛ 360 degrees ⎞ 420 degrees
⎜
⎟=
1
rotation
⎝
⎠
iv. How many degrees to travel 30cm?
Hopefully by now it is becoming apparent to students that the 17 cm and the 360 degrees conversion are
the same no matter how many cm you wish to go.
Ratio:
30 cm x fraction of a rotation
=
17 cm
1 rotation
x = 30 /17 = 1.8 rotations
1.8 rotations ⎛ 360 degrees ⎞ 640 degrees
⎜
⎟=
⎝ 1 rotation ⎠
Algebra:
distance traveled = circumference of wheels × number of rotations
30 cm = 17 cm × number of rotations
number of rotations = 30 /17 = 1.8 rotations
1.8 rotations ⎛ 360 degrees ⎞ 640 degrees
⎜
⎟=
⎝ 1 rotation ⎠
v. Look back at the procedure you used to find the number of degrees in the last 3 questions. If
you want the robot to travel X cm, describe what mathematical operations (e.g. add 2 then
divide by 5) you must perform on X in order to find the number of degrees.
In every case, regardless of which method students used, they ended up taking the number of cm,
dividing by 17 (cm in the circumference), and multiplying by 360 (degrees in a rotation). If desired,
this can even be simplified into a single step of multiplying by 21.2. Therefore, multiplying X cm by 21.2
will give you the number of degrees.
This value represents the conversion ratio between centimeters and degrees for this wheel. Like (10 mm
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= 1 cm) or (5280 ft = 1 mile), (21.2 degrees = 1 cm) is just a factor that can be used to convert cm to
degrees. The operations involved in the conversion are what students should focus on, because those
steps will be necessary each time students wish to convert a distance into the appropriate number of
wheel rotations.
vi .Does the procedure you found in part (v) work for any robot? If not, what robots will it work
for?
The procedure that students found only works for robots using the wheel they chose. Using a
different-sized wheel would change the factors used in the calculation. Similarly, although students
probably will not know this yet, there can be no change in the gear ratio of the robot – the rotations
controlled by the program are actually motor rotations, so if gears were introduced that caused the motor
to turn at a different rate than the wheel, the conversion equation would change again.
20. What is the advantage of being able to control your robot’s movement distance in
centimeters as opposed to rotations?
Major advantages include the ability to navigate by driving a known distance (cm is a unit of distance,
while rotations are not), and the ability to use common tools such as rulers or metersticks to plan
how far the robot needs to move.
A central theme in the history of measurement has always been the issue of standards. In order for a
measurement to mean something, it must be compared to a standard. When there are different systems
of standards, confusion can result. This remains an issue in the U.S. today, where cultural and economic
factors have resulted in a mixture of measurement systems in common use. Programming the robot to
operate in terms of a major system of standards makes good sense for communication purposes.
21. Technically, the Wait For block in your program waits for motor rotations, and not wheel
rotations. In this Investigation, we have been considering them to be interchangeable. What
physical characteristics of the robot allow us to make this simplification?
The fact that the motor and the wheel always turn the exact same number of times together lets us make
this simplification.
We can say that motor rotations and wheel rotations are equal because they are, thanks to the robot’s
gear ratio. The robot’s wheels are not actually on axles that are turned by the motor. Instead, on the
axles that go through the motors, there are gears, and these gears mesh up with the gears that are on the
axles that go through the wheels. All of the gears have the same number of teeth and are the same
diameter. The gear ratio is 1:1. That means that for each revolution of the motor, the wheel will also make
one full revolution. If the gear ratio were not 1:1 (i.e. the gears on the two axles had different numbers of
teeth), then one full motor revolution would not necessarily produce one full wheel revolution.
Analysis and Conclusions: Exercises
22. Tracy measures the wheels on her robot and finds that they are 2.3cm in diameter. Her
program is shown in the picture below.
i.
How far should her robot go?
Her robot should go 14.5 cm.
2.3cm diameter * pi = 7.2cm circumference
720 degrees rotation (read from the program in the picture) = 2 rotations
7.2cm * 2 = 14.5 cm traveled (14.4 cm if students rounded early)
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This is a review question, asking students to follow the same sequence of steps they used during the
guided portion of the investigation.
ii. Is her robot likely to go exactly that distance when she runs it? Why or why not?
The robot probably won’t go exactly that distance due to random experimental error.
23. Rodney likes monster trucks. He replaces the wheels on his robot with wheels that are four
times as large (in diameter) as the wheels on his old one, but leaves the program the same.
How many times farther (or shorter) will his robot run than the old one?
Four times as far.
This question can be solved using either ratios or simple algebra, whichever is more appropriate to the
math level of the class. Because the program is staying the same, you can leave it out as a consideration.
Theoretical values should be used, since this is a predictive problem.
Ratios method:
Compare two of the existing conditions to see how their relative wheel diameters compare to their relative
distances traveled. For example, look at the ratio relationships of the standard wheel and the small wheel
(conditions 1 and 2).
diameter of standard wheel 5.5cm
=
= 1.8
diameter of small wheel
3.0cm
distance traveled with standard wheel
35cm
=
= 1.8
distance traveled with small wheel
18.8cm
As you can see, the ratio of the diameters is the same as the ratio of the distances traveled. The problem
asks for a prediction about wheels that are four times as large. The ratio of the diameters is 4:1, therefore
the ratio of the distances traveled should also be 4:1. The new robot should travel 4 times as far as the
old one.
Algebra method:
Begin with equations expressing the distance traveled in each condition.
distance traveled with old wheels = circumference of old wheels × number of rotations
distance traveled with old wheels = diameter of old wheels × π × number of rotations
distance traveled with new wheels = circumference of new wheels × number of rotations
2.
distance traveled with new wheels = diameter of new wheels × π × number of rotations
1.
The problem stipulates that the diameter of the new wheels is 4 times the diameter of the old wheels.
diameter of new wheels = 4 (diameter of old wheels )
distance traveled with new wheels = diameter of new wheels × π × number of rotations
4.
distance traveled with new wheels = 4 (diameter of old wheels) × π × number of rotations
3.
Find a substitute for (diameter of old wheels) in terms of (distance traveled with old wheels):
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distance traveled with old wheels = diameter of old wheels × π × number of rotations
5. distance traveled with old wheels
= diameter of old wheels
π × number of rotations
Substitute and simplify:
6.
distance traveled with new wheels = 4 (diameter of old wheels ) × π × number of rotations
distance traveled with old wheels
distance traveled with new wheels = 4 (
) × π × number of rotations
π × number of rotations
distance traveled with new wheels = 4 (distance traveled with old wheels)
The robot with the new wheels should go 4 times as far as the old one.
Reality check:
A robot with 4x larger wheels would likely encounter problems starting and stopping due to lack of torque
in the motors. Real-world measurements are therefore likely to be skewed relative to the predictions,
based on error caused by the inability of the robot to stop quickly. Also, wheels that large may interfere
with other parts or features on the robot, as they may not have the clearance to spin freely.
24. Many wheels are constructed from a tire (black) and a hub (grey). Mark’s program runs 720
degrees using the assembled (hub+tire) wheel. However, he wants to run the robot using
just the grey hub as a wheel, without the black tire.
i.
How many degrees must the new hub-only wheel turn in order to go the same distance?
The hub-only robot must go 1008 degrees to move the same distance as the original program.
Students must interpret the diagram in order to find the necessary wheel dimensions to do this problem.
The hub has a total diameter of 3.0cm. The wheel is a little trickier, because students will need to realize
that the diagram only shows the width of the tire on one side (0.6cm), but the diameter across the whole
wheel includes tire on both sides, so the tire thickness actually needs to be counted twice (1.2cm total).
Note that the dimensions on the wheel in this problem are modeled after the actual standard robot wheel,
but students should go by the diagram and not their own measurements for this problem.
The original wheel with both the hub and tire had a diameter of 4.2 cm. The diameter of the hub alone is
3.0 cm.
The original (hub+tire) robot goes 4.2 cm * pi * 720 degrees = 26.4 cm.
To go 26.4 cm using the 3.0 cm hub instead, the robot must go:
rotations =
26.4 cm
= 2.8 rotations
3.0 cm × π
2.8 rotations ⎛ 360 degrees ⎞ 1008 degrees
⎜
⎟=
⎝ 1 rotation ⎠
ii. What other issues might he encounter in using the hub as a wheel by itself?
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The surface of the hub is very slick and will not give very good traction. As a
consequence, the robot may not move as much as the wheel spins, especially on hard and/or
slippery surfaces.
iii. How would these other issues affect the actual distance the robot travels?
Slipping will make the robot go a shorter distance than the wheel rotations indicate for a few
reasons. First, it will waste motion spinning the wheels without actually moving the robot the whole time
(like running a car on ice). Second, it will most likely not go straight, and the wasted sideways motion will
result in less distance traveled. There are other possible consequences; any well-reasoned answer
should be acceptable.
25. You are working as the mechanical engineer on your team, building a brand new wheeled
robot to explore the Atacama Desert in Chile. You have just found a new wheel that seems
to perform better on the rugged terrain of the salars there. You decide that this wheel will be
used on the next prototype of the robot that will be deployed for testing. What information, if
any, must you communicate to the other members on your team when you make this
change to the design? What might happen to the robot’s behavior if you do not make these
points clear?
If the diameter of the wheel changed, the team needs to know about it, so they can reprogram to
go the same distance.
This is an open-ended question asking students to summarize the importance of the effects and
relationships they have been exploring in this lesson in a teamwork context.
Primarily, students should note that if the new wheel has a different diameter from the old wheel, they will
need to notify the programmer on the team to make the appropriate adjustments to the distance-moving
code on the robot. Failure to do so could cause the robot to become very confused about its actual
location, because it will think it has gone a different distance than it actually traveled.
Students may also note that changing the wheel (especially with the motive of improving performance on
rugged terrain) may result in different traction and slippage characteristics for the robot, which are also
factors that the programmers must be kept aware of in order to be able to properly estimate odometry
(distance-traveled) data for the robot. Again, failure to inform other members of the team will result in the
robot ending up in the wrong place (potentially stuck on a rock or off a cliff!).
Students who think a little harder about the question may also realize that they need to inform the project
management and potentially the other project staff. Project managers need to keep track of the cost of
the project, and purchasing new wheels would constitute a potentially significant expense, and may
impact the selling price of the final robot. The tires themselves may also have significantly more mass, or
take up more space, both factors that affect the logistics of transporting the robot. The effects of simply
changing the wheels on the robot can be quite far-reaching!
As an additional activity, this question provides an opportunity to talk about terrain, traction, and the
reliability of encoder-based navigation. It can also open the door to student research into what exactly the
Atacama Desert is, allowing cross-curricular linkages to geography, Earth Science, and a great many
other disciplines.
26. Diana’s robot travels 7.85cm using the program shown at left.
i.
What size are the wheels on her robot?
Wheel Diameter = 2.50cm
The program shows 360 degrees of rotation. A simple unit conversion or recall of the conversion ratio
shows that this is exactly 1.00 rotation of the wheel. The robot goes 7.85cm in 1.00 wheel rotation.
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Students must then either reason backwards or use the numeric relationships they have established to
find the circumference and diameter of the wheel.
Ratios and Proportions
(Distance/Circumference) = (Rotations/1 Rotation)
7.85cm/Circumference = 1
Circumference = 7.85cm
Diameter = Circumference/pi = 7.85cm/3.14 = 2.50cm
Algebra
7.85cm = 1.00rotation
7.85cm = 1.00 x wheel circumference
Wheel circumference = 7.85cm
Wheel circumference = pi x wheel diameter = 7.85cm
Wheel diameter = 7.85cm / pi = 2.50cm
ii. How far will her robot go with the following program? Show your work.
Distance = 15.7cm
This question is essentially checking whether students recognize that the distance the robot travels is
also controlled proportionally by the number of degrees set in the program. The code shown makes the
robot run 720 degrees of motor rotation, which is 2.00 turns of the wheel.
Ratios and proportions
2.00 turns is double the 1.00 turn that it took for the robot to go 7.85cm, therefore the robot will travel
double the distance, or 15.7cm.
Algebra
Rotations = Degrees turned/360
Rotations = 720/360 = 2.00
Distance = Rotations * Circumference = Rotations * Diameter * pi
From previous question, Diameter = 2.50cm
Distance = 2.00 * 2.50cm * pi = 15.7cm
27. Jeanne’s robot travels 65cm using the program shown below.
i.
What size are the wheels on her robot?
Jeanne’s Wheel Diameter = 3.7cm
The program runs 2040 degrees of rotation and travels 65 cm. The first step would be to find out how
many wheel rotations this is (factor-label unit conversion shown, see Helper Links for more information on
this method).
2040 degrees ⎛ 1 rotation
⎜
⎜ 360 degrees
⎝
⎞ 5 2 rotations
⎟= 3
⎟
⎠
Now, you need to figure out what sized wheel would go 65 cm in 5-2/3 rotations.
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distance traveled = diameter of wheels × π × number of rotations
2
65 cm = diameter of wheels × π × 5 rotations
3
65 cm
= 3.7 cm
diameter of wheels =
2
π ×5
3
Jeanne’s robot has wheels with a 3.7 cm diameter.
ii. Jack’s robot goes the same distance with the program below:
What size are his wheels?
Jack’s Wheel Diameter = 7.3cm
There are two ways to solve this problem. The robot goes the same distance in 1020 degrees of rotation
rather than 2040. Students may immediately recognize that this is half the number of rotations. They can
then reason that going the same distance in half the number of rotations means the circumference must
be twice as big. If that’s the case, then the diameter would be twice what it was on Jeanne’s robot,
making Jack’s wheels 7.3 cm in diameter (7.4 cm if students rounded the last answer).
A more systematic way of finding the answer would be to simply solve the equations again.
1020 degrees ⎛ 1 rotation
⎜
⎜ 360 degrees
⎝
⎞ 2 5 rotations
⎟= 6
⎟
⎠
distance traveled = diameter of wheels × π × number of rotations
5
65 cm = diameter of wheels × π × 2 rotations
6
65 cm
= 7.3 cm
diameter of wheels =
5
π ×2
6
28. You are your team’s lead programmer hard at work refining your robot’s movement code the
day before it needs to compete in a hallway navigation challenge. Your friend, the team’s
mechanic, stops by and informs you that he seems to have misplaced the wheels for your
robot during cleanup last night. He also says the local Wheels R Us will not have any
replacements in stock for at least two weeks. There are several sets of 3in diameter wheels
lying around the workshop, but they are not the same as the 2.7cm diameter ones that you
had been planning to use. Your old code is shown at right. What must you do in order to get
your robot in working order to perform the challenge tomorrow?
As a strictly mathematical exercise, the team would need to change the Wait For block to wait for
3420 degrees instead of 9600 degrees.
From a theoretical perspective, the only change that needs to be made is to the Wait For block to account
for the different sizes of the wheels. Note that the new wheel size is specified in inches and not
centimeters, so a conversion must be made there first.
3 inches ⎛ 2.54 cm ⎞ 7.62 cm
⎜
⎟=
⎝ 1 inch ⎠
© Copyright CMU Robotics Academy
13/15
New Wheel Circumference = pi * diameter = pi * 7.62cm = 23.9cm
Next, students need to figure out what the appropriate cm-to-rotations conversion ratio is for this new
wheel size. This proceeds as before. Students should know from previous problems that they can easily
recalculate this number by simply dividing by the circumference and multiplying by 360.
360 degrees/ 23.9 cm = 15.0 degrees-per-cm. Multiply by 15.0 to make the new conversion.
But first students will have to find out how far the robot was intended to go with the original wheels, which
were 2.7cm in diameter.
2.7cm * pi = Old Wheel Circumference = 8.5cm
(9600 degrees/360 degrees-per-rotation) = 26.67 rotations
26.67 rotations * 8.5 cm-per-rotation = 226.7cm
So they originally wanted the robot to travel forward 226.7cm. Now with the new wheels, to travel the
same distance, students need to divide that distance by the circumference of the new wheel to find the
appropriate number of rotations.
226.7cm/23.9cm-per-rotation = 9.5 rotations
9.5 rotations = 3420 degrees
So, theoretically speaking, the team would need to change the number of degrees that the motors turn for
to 3420 degrees.
On the practical engineering side of matters, there are also matters of documentation and most
importantly, testing. The physical characteristics of the wheel are likely different in more ways than
simply diameter. Different friction characteristics can make a big difference in real-world performance.
Similarly, depending on motor power, changing the size of tires will affect the robot’s ability to accelerate
or decelerate, which may require programming adjustments to compensate. You may wish to discuss in
general the impact and consequent undesirability of last-minute changes in engineering design.
© Copyright CMU Robotics Academy
14/15
Table: Wheels and Distance
Introduction to Mobile Robotics > Wheels and Distance Investigation
This unit includes a worksheet where students capture data and write conclusions; below you
will find the answer key for the worksheet.
Red = exact answers
Purple = sample answers
Condition
Wheel
Wheel
Number
Diameter Circumference of wheel
(cm)
(cm)
rotations
in
program
Standard
Wheel
5.5
17
2.00
Theoretical
(predicted)
distance
traveled in
program
(cm)
35
Small
Wheel
3.0
9.4
2.00
18.8
Actual
distance
traveled
(cm) in
each
trial
1) 34.5
2) 37
3) 33
1) 21
2) 20
3) 18
Average
actual
distance
traveled
(cm)
34.8
19.7
Useful Equations
pi = approx. 3.14
Circumference = Diameter * pi
Diameter = Radius * 2
1in = approx. 2.54cm
1 rotation = 360 degrees
© Copyright CMU Robotics Academy
15/15