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Servomotors are widely used electro-mechanical
components. For example, they are used in
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steering systems for scale model helicopters
and other radio controlled cars, robots, and
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aircraft, including military "drones".
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A servomotor includes an electric motor, circuitry
to control its speed and direction, and gearing
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to attain the high torques needed to apply
moderately large forces over relatively short
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linear displacements. The gearing inside this
servo is a compound gear train with four stages.
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In this video, we will understand how linear
algebra can help us to predict the losses
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and understand the design trade-offs when
converting high speeds and low torques into
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low speeds and high torques using a gear train.
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This video is part of the Linearity Video
Series. Many complex systems are modeled or
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approximated linearly because of the mathematical
advantages.
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Hi, my name is Dan Frey, and I am a professor
of Mechanical Engineering and Engineering
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Systems at MIT.
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Before watching this video, you should be
able to identify and describe the forces and
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torques that act on a system.
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After watching this video and some practice,
you will be able to:
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Model the forces and torques in a gear train
as a system of linear equations.
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Combine these linear equations into a matrix
equation.
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First, let's get a quick overview of the design.
Here's a servomotor with the case opened.
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Inside the case is a DC motor with a small
pinion gear on it. The pinion has 10 teeth
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and is mated to a gear with 72 teeth. So the
first pair provides a speed reduction of 7.2:1.
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This large, 72 tooth gear is part of a single
molding with a 10 tooth gear.
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The 10 tooth gear on top of the 72 tooth gear
is mated to a plastic gear with 48 teeth as
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you see here. So that stage provides a speed
reduction of 4.8:1. The 48 tooth gear is molded
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to another 10 tooth gear.
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The 11-tooth gear is mated to a black 36-tooth
gear. This 36-tooth gear is molded to a 16-tooth
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gear. The 16-tooth gear is mated to this black
42 tooth gear, which is splined onto the output
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shaft.
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The overall sequence of gear pairs is: 10
teeth mated to 72 teeth, 10 mated to 48, 11
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mated to 36 and finally 16 mated to 42 teeth
on the splined output shaft.
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The overall gear ratio is the product of all
the gear ratios of the gear pairs in the train.
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Therefore about 326 to 1 as shown by this
formula.
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Based on the motor's output and some measurement
and calculation, we would expect an output
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torque of 6.8 Newton meters in an idealized
gear train.
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But there are frictional losses at every stage
of the power transmission process, so we guess
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the output shaft will provide substantially
less torque than that. So let's start by measuring
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the maximum force at the output shaft, and
convert that to an output torque.
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We attach weights to a servo horn that has
been connected with the output shaft on the
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servomotor. An electrical power source (in
this case, a battery) is connected through
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a radio controller.
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Here you see the servomotor lifting 1 kg.
Here it is lifting about 2kg. And here it
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lifts even 3kg! This lightweight (100 gram)
servomotor produces large amounts of torque,
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which can apparently lift more than 30 times
its own weight.
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In this set-up, the servomotor will lift a
4.42kg weight starting from a 90 degree angle.
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As the motor lifts the weight, the readout
on the scale should change.
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Here you see the initial weight, with slack
in the string, is 9.75 lbs. This is about
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4.42kg. With the motor pulling at maximum
capacity, the scale reads 3.70 lbs, which
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is 1.68 kg.
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A calculation tells us that the maximum torque
available at the output shaft according to
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our tests is 1.47 Newton meters.
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We can use linear algebra to work out a better
estimate. The force transmission and frictional
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losses in each step of the train can be modeled
by a set of linear equations, (summing forces
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and summing torques on the rigid body). For
the three rigid bodies that are comprised
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of two gears molded together, we will need
three equations (sums of separate x and y
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forces and sums of torques).
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The pinion gear and last gear in the train
are simpler, because there is only one gear
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in the body, so we can combine each of those
into a single equation.
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The equations modeling the entire train can
be assembled by linking together the five
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sets of equations with four additional equations
to link each mating pair, so the overall system
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will have 15 equations--- 3 sets of 3 each
(9 total) plus 1 each for the input and output
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gears plus 4 equations that model the connections
between the mating gears. The solution to
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that set of equations helps us to estimate
performance of the machine and also gives
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us insight into its design.
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Let's make the model of just one gear body
now, say the 10 tooth and 48 tooth molded
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gear seen here. One way to model the system
is to posit that there are three unknown forces
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acting on this single body of two gears molded
together.
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We can name these forces F-TL2, F-TS2 , F-ShaftX2,
and F-ShaftY2. The force F-TL2 is a force
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tangent to the gear pitch circle on the larger
of the two gears. The T is for Tangent, the
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L is for Large, and the 2 indicates that this
is the second compound gear in the train.
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The force F-TS2 is a force tangent to the
gear pitch circle on the smaller of the two
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gears. Again the T means tangent, and the
S here means small.
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Recall that these tangential forces really
come from forces normal to the gear teeth.
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These gears are designed so that all of the
gears in this servomotor have a pressure angle
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of 20 degrees. So there is a X component to
this force, the separation force between the
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mating gears, which has magnitude given by
FTL2 and FTS2 times tan 0.35, where 0.35 is
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20 degrees expressed in radians.
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The forces F-ShaftX2 and F-ShaftY2 are the
X and Y components of the normal force to
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the shaft, which is applied to the gear it
supports.
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There are also frictional forces associated
with the normal force supporting the shaft,
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but they are not separate unknowns, they link
the X and Y components normal forces by the
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friction coefficient, mu2.
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This completes the description of the unknown
forces in our model. Those four forces appear
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in three different linear equations as represented
by this matrix:
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Take a moment to verify that these equations
balance the forces and torques.
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The
first row in the matrix represents a sum of
the forces on the gear in the X direction.
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The second row represents a sum of the forces
on the gear in the Y direction. The third
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row represents a sum of the torques or moments
on the gear in the direction parallel to the
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gear shaft.
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Now let's work on the model of the next gear
in the train -- the one with 16 teeth and
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36 teeth molded together. Again, we model
the system by positing that there are four
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unknown forces acting on this single body
having two gears molded together and we can
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name these forces FTL3, FTS3 , FShaftX3, and
FShaftY3. And we obtain this matrix.
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How can we link these two matrices together
to get a model of the second and third gears
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combined? We see that the smaller gear on
body 2 is in contact with the larger gear
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on body 3. According to Newton's third law,
the reaction force on body 2 should be equal
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and opposite to that on body 3.
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In fact, the situation is more complex. As
the gears enter mating they slide into engagement.
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As the gears depart contact, they slide back
out. There are losses at the interface that
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are complex to model, and we will represent
them simply using the efficiency in the average
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transmitted force.
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Now we can join the model of the second gear
and the third gear into a single system of
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linear equations represented by this matrix:
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The row in the middle is like "glue" holding
the two models together. The force on the
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large gear on body 3 is equal to the force
on the large gear on body 2 except a penalty
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is applied for inefficiencies. The principal
mechanism for loss of power and torque is
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sliding friction -- in this case, shearing
the lubricant on the gear faces.
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Before going on to assemble more of the model,
it is worth inspecting our work so far for
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patterns. Note that the top right and bottom
left of our matrices are filled with 3x4 matrices
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of zeros. That is a clue that our matrix is
shaping up to be banded in structure. Although
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every variable affects every other variable,
the influences propagate locally (in some
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sense). For example, the frictional losses
on shaft 2 do influence the frictional losses
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on shaft 3, but only indirectly through their
effect on the force between the two bodies.
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Because we arranged the variables in our vectors
in a way that respects this structure, our
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matrix has a band in the middle from the top
left corner to the bottom right. We will try
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to keep this up as we continue building the
model. It will clarify interpretation of the
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model and also aid in efficiency and stability
of the computations needed to solve it.
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Now it's your turn, take a moment to write
a matrix that models the force and torque
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balance on the first rigid body gear in the
gear train.
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This matrix is completely analogous to the
matrices we obtained in for the 2nd and 3rd
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gear bodies.
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When we add the pinion gear and fourth gear
into the model, we need to take additional
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care. These will have slightly different structure.
The gear connected to the pinion on the motor
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has an applied torque due to the motor. The
reaction at the output shaft is an externally
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applied torque, due entirely to the large
fourth gear. It is possible to combine these
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into a single equation for each gear.
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Pause the video and try to write out the full
system of linear equations for the overall
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4 gear train system.
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Placing these into the model as the first
and last rows, the overall system of linear
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equations is:
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Note that the matrix has 15 rows and 15 columns.
This should be expected as the system should
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be neither overdetermined nor underdetermined.
Given a particular torque input at the motor,
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there should be a single value of output torque
consistent with our gear train model and its
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deterministically defined parameters.
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By solving this system, we find unique values
for all the unknowns. When we assembled the
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equations, we guessed the direction of each
force in the free body diagram. So, it's useful
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to inspect the solution for negative values.
For the parameter values we chose, the reaction
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force at the shaft of the third body in the
x direction turns out to be negative.
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We should ask ourselves, did we guess wrong
about the direction of the force, or is there
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something wrong with our matrix? Looking at
the forces computed it seems that the friction
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in the x direction due to support of the large
y reaction on this gear was more dominant
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that we expected.
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It's sensible to run some sanity checks in
a case like this. If the friction on that
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gear were very low, would the sign be positive
as expected? We ran that scenario, and the
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sign did become positive. So our guess about
the direction of the reaction force was wrong.
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Our assumptions for the magnitude of the friction
were too low. But it seems the model is behaving
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in reasonable ways.
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Now that we have a reasonable level of confidence
in the model, we can begin to use it to explore
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the design decisions that the engineers made.
If we put in reasonable values for the gear
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mating efficiency, such as 96%, and reasonable
values for the friction coefficients at the
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bushings, such as 0.3, we find an overall
efficiency of the gear train is about 52%.
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This is in reasonable agreement with our simple
measurements. We found that the maximum torque
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available at the output shaft is 1.47 Newton
meters, which would imply an efficiency closer
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to 25%.
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But our measurement used an overhanging load,
which caused the shaft to bend. Our simple
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model makes a large and optimistic assumption
of loading the servomotor with torques only.
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It's not surprising that the answers differ
substantially. It suggests installing the
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servo so that bending loads are supported
elsewhere, not in the servo itself.
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The gears in automotive transmissions with
similar ratios are much more efficient, perhaps
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90% to 95% efficient. But for a compact and
inexpensive gear train, this design performs
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well, especially since the plastic the gears
are molded from adds a great deal of rolling
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friction.
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To summarize, we built an engineering model
of a servomotor gear train using systems of
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linear equations. The matrix representations
helped us to explore the interactions among
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variables in the system like separation forces
between gears and friction at the shafts.
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Since the 15 by 15 system of equations is
so fast and easy to solve on a modern computer,
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we could run a large number of "what if" scenarios.
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We hope this video helped you see how linear
algebra can be used to make and understand
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engineering design decisions.