# How do I calculate DC motor speed for a given load?

Suppose I have a robot of a given mass, and I'm choosing between 2 different wheels and 2 different motors to put on it. For each wheel I have the diameter, and for each motor I know the stall torque and free speed. How would I figure out which motor and wheel combination will make the robot move the fastest?

My calculations show to use the big motor with the big wheels, but the small motor with the small wheels goes faster than the big motor with the small wheels. I am not sure my calculations are correct, I need to know the correct way to work this problem.

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Hi cyberike! While I am no expert on DC motors your problem should not be too hard to solve. Could you clarify a bit what you mean by "move the fastest"? Are you asking for maximum acceleration, top speed or fastest travel time for a given distance (incl. breaking at the end)? Also if you have done calculations already, why not include them here. From the DC motor perspective there are different torque/speed/power relations possible, so for a realistic estimate you need a diagram like this (amci.com/stepper-motor-control/images/…) for your motor. –  Alexander Apr 23 '12 at 16:42
Following is the information Alexander asked for. unfortunately, I don't know how to do equations the way ja72 has in his reply, so I have to describe. I am looking for the top speed (after acceleration) of the robot itself. –  Cyberike Apr 23 '12 at 19:50
I started with the formula from the MIT website (given below by pygmalian): ω = ω0 (1 - τ/τ_0 ) which comes basically from the motor characteristic curve and shows that a shunt wound DC motor speed decreases with an increased load on the motor. I used the same formula to calculate torque load on the motor that ja72 gives in another comment below, τ = rF (F is the weight of the robot, a constant, and r is the radius of the wheel), so the torque required from the motor increases as wheel radius increases. This seems logical, but this is the part I was trying to confirm. –  Cyberike Apr 23 '12 at 20:17
Basically, it says that the motor has to work harder for each rotation. As long as I am using the formulas correctly, I am fine. However, if I am not using them correctly I need to know it before I present it to my class. Since the load on the motor goes up with larger wheels, the motor slows down for the larger wheels according to the first formula. My students would have to plug the numbers into each formula to see if the motor slows down more (or less) than the increased movement you get per rotation from the larger wheels. The point is, big wheels on a robot are not always better. –  Cyberike Apr 23 '12 at 20:18
If you neglect the acceleration and are interested in top speed only the answer is easy: $\text{speed} = 2\pi \cdot \text{radius} \cdot \text{revolutions}/\text{second}$, so big wheels with motor with high rpm will always be faster. This totally neglects inertia and torque though and is not a very educative treatment of the problem. –  Alexander Apr 23 '12 at 22:28

Regardless of the size of the wheels, and ignoring air resistance, if the motor is making $P = T(\omega)\;\omega$ power then the acceleration is

$$a= \frac{T(\omega)\; \omega}{m v}$$

The motor speed is $\omega = \frac{v}{r}$ where $r$ is the wheel radius. If the torque at $\omega=0$ is $T_0$ and the motor speed at $T=0$ is $\omega_0$ then the torque function is

$$T(\omega) = T_0 \left( 1- \frac{\omega}{\omega_0}\right)$$

The time it takes to reach a certain speed $v$ is

$$t = \int_0^v \frac{1}{a}\;{\rm d}v$$ $$t = \int_0^v \frac { m v } { T_0 \left( 1- \frac{v}{\omega_0\,r}\right) \frac{v}{r} } \; {\rm d} v$$ $$t = \frac{m \omega_0 r^2}{T_0} \ln\left(\frac{\omega_0 r}{\omega_0 r - u} \right)$$

or

$$v(t) = v_0 \left( 1 - \boldsymbol{e}^{-\frac{T_0}{r} \frac{t}{m v_0} } \right)$$ where $v_0 = \omega_0\,r$ is the theoretical top speed.

So to get to $99$% of top speed you need

$$t_{99} = \frac{m\, r^2\, \omega_0} {T_0}\; 2\ln(10)$$

From here you plug in your values and see which one has the highest top speed and which one has the highest acceleration (least time).

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Thank you for your response. I have one more question: How do you convert weight load to torque load on the motor given that motor force is applied at radius r of the wheel? By the way, I am working from a speed vs. torque relationship similar to me.mtu.edu/~wjendres/ProductRealization1Course/… –  Cyberike Apr 19 '12 at 19:49
The tractive force is $F=T/r$ and the linear speed is $v=\omega \, r$. –  ja72 Apr 19 '12 at 19:54
@Cyberike I definitely agree with ja72 about independence of tractive force on mass and on the fact that you need to have some kind of motor characteristics (torque vs frequency)... –  Pygmalion Apr 20 '12 at 14:51
@Cyberlike See also, physics.stackexchange.com/questions/23020. Maybe you should invite Alexander to provide his insight. –  Pygmalion Apr 20 '12 at 14:53
@Cyberike I took liberty to invite Alexander myself. –  Pygmalion Apr 20 '12 at 15:00

This look suspiciously like a homework question, so we're allowed to discuss concepts but not just answer your question (or your professor won't be too pleased :-).

Your question doesn't say what the rolling resistance is e.g. what is the wind resistance the robot feels as it starts moving? If there is no rolling resistance both motors will simply accelerate until they reach their free speed, and in that case you want the highest free speed with the largest wheels.

If there is rolling resistance then the motor will accelerate until the rolling resitance is the same as the force generated at the wheels i.e. the torque times the wheel radius, so it's the torque rather than the free speed that matters.

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This is not a homework question, this is for a classroom tutorial on robotics that I am making (as a teacher). I want to make sure I am using the right formulas. By the way, I am not trying to make it too complicated, so moment of inertia answers are out. –  Cyberike Apr 19 '12 at 15:05
I guess that there is no rolling and wind resistance, that would make the task quite difficult, but torque might be important to get the robot that accelerates faster. –  Pygmalion Apr 19 '12 at 15:07
@Cyberike This thingy has kinematic part too. Let's say one robot has larger acceleration and smaller top velocity, and the other the opposite. Then results depends on the competition track length! –  Pygmalion Apr 19 '12 at 15:08
Oh you're teacher. Luckily, I work at technical universities and bump into civil and mechanical engineers (with PhDs) every day. If I'd had your problem, I would simply ask the most competent of them. –  Pygmalion Apr 19 '12 at 16:32

The solution $might$ be dependent on the context. Mechanical engineers would do power calculation first. What is prevalent about these motors, torque or power or frequency?

Concerning torque $\tau = r F$, so $F = \tau/r$ and motor with larger torque and smaller wheels give larger force (push), while the car is accelerating.

Concerning power $P = F v = M \omega$ the size of wheels is irrelevant (well at least for the time car is accelerating) and motor with larger product of torque and frequency will do the best, while accelerating.

Concerning frequency $v = \omega r$, larger frequency and bigger wheels give larger speed.

I thing the first two can give you which shall accelerate faster, while the last one which robot shall have higher final speed.

Let's suppose that power cannot be calculated using free speed. Then you have four possible robots with four possible accelerations and four possible top speeds. Then results depends on the track length!

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