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I'm trying to study the free body diagram of this differential-drive mobile robot in case of straight trajectory (same torque applied on both of rear wheels) and pure rolling motion

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I started studying the whole system, in which i have the inertial force, vertical force (weight), two friction force Ft for the rear wheels and a friction force Fta for the castor wheel. Then i have three normal forces, two on rear wheels, one on castor wheel.

So far i have three equations (one along x and one along y, one about torques with respect the center of mass) and 7 unknowns.

In order to solve the system, should i disconnect the system (wheels and platform)? How can i solve it?

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Wheel variables:

FTrr = longitudinal force right wheel

FTrl = longidutinal force left wheel

FTca = longitudinal force on castor wheel

Nca = normal force on the castor wheel

Nrr = normal force right wheel

Nrl = normal force left wheel

M = robot mass

d = distance between wheel center - center of mass

e = distance between castor wheel center - center of mass

h = distance between center of mass - ground

a = acceleration


Equations about x, y and torque with respect the center of mass (positive if clockwise):

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Active right/left wheel (after i broke up the system):

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tau = actuator torque

r = radius

FTr = longitudinal force

Nr = normal force

Rx = reaction force between platform and wheel about x axis

Ry = reaction force between platform and wheel about y axis

P/n = inertial load on the right/left wheel (N = amount of load on each wheel)

J = moment of inertia

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  • $\begingroup$ You're right. Just edited. $\endgroup$ – xWalle Jun 2 at 22:24
  • $\begingroup$ You may have to add the equations of rolling without friction (a type of non-holonomic constriant). $\endgroup$ – Futurologist Jun 20 at 19:49
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The first thing I can see is in your $M\ddot{x}=\Sigma{F}$ equation as well as your free body diagram. You are confusing the force being exerted by the wheel with the force exerted on the robot. Free body diagrams only show the forces exerted on the body, not forces the body is exerting. The drive wheels on the robot exerts a force in the negative direction (left) on the ground. The ground then exerts an opposite and opposing force on the wheel via in the positive direction (right). That's what moves the robot forward.

You also seem to be treating your caster wheel as if it were a propulsion wheel. But it is not. Caster wheels provide no propulsion and exert a drag force that slows the robot down in the direction of motion. Which in this case would be a force exerted on the robot in the negative direction. So the direction of the caster wheel was correct but only incidentally in that you were treating it just like all your caster wheels (which were incorrect).


Also, your ma is pointing to the left which is unusual. Not necessarily wrong, but unusual if you are trying to solve for a. Presumably you expect your robot to move forward, which is to the right (positive direction). So you would presumably "guess" that is the direction of the acceleration. If your calculated acceleration is positive, it means you guessed the direction correctly. If the calculated acceleration is negative, it means you guessed incorrectly and the direction is oppposite of what you guessed.

Of course, you could have guessed that acceleration is to the left in which case you would draw the vector as you have. Your calculated acceleration would be positive (which we just know intuitively) which means that you guessed the wrong direction and the acceleration is to the right. We won't always know ahead of time what the directions are so what you drew is not necessarily wrong, but the problem in this case is simple enough that we do know so I can only assume you are misunderstanding something.

Alternatively, you could list all unknowns as being positive regardless of which direction you think is correct. In that case, the polarity of the result will give you the direction. You could also do the opposite and define the unknowns as all in the negative direction. In that case, it would be like guessing all knowns are in the negative direction so a positive result would mean you guessed correctly and a negative result means you guessed incorrectly.


For your wheel FBD, you are again mistaking the force that the wheel exerts on the ground with the force that the ground exerts on the wheel. The things I have to say about the direction you defined $j\theta$ definition follows the same logic as what I said for $ma$ above. Typically in the x-y coordinates CCW is defined as positive due to cross product. The best case scenario for how you defined $j\theta$ would be if you followed the cross product convention and just assigned all unknowns to be positive so that they polarity would give you the direction then that would be correct.

But if that's not what you intended then you took the unusual convention of guessing in a strange direction when your intuition knows better similar what I described above for ma. In this case, you guessed CCW then your calculated $\theta$ would be negative indicating the rotation is actually CW, opposite of what you guessed.

You need to check that your Rx direction is what you intended. Because when rolling to the right, the axle exerts a force on the wheel hole to the left.

You're double counting the weight of the robot on the wheels by having both inertial load and $R_y$.

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  • $\begingroup$ Got it, thank you! What about the single active wheel? I have doubts about the inertial load $\endgroup$ – xWalle Jun 2 at 22:38
  • $\begingroup$ @xWalle Was editing. Still had more to say when you commented. $\endgroup$ – DKNguyen Jun 2 at 22:40
  • $\begingroup$ @xWalle That's why I couldn't figure out what it was. You're double-counting I think. If that is inertial load then what is Ry supposed to be? $\endgroup$ – DKNguyen Jun 2 at 22:58
  • $\begingroup$ @xWalle Yes, that is wrong because the platform is sitting on the axles which are in turn sitting on the wheel. It's the same force exerting a downward direction on the wheel hole. Unless you had the weight of the axle separate from the weight of the platform, but I doubt you intended that. $\endgroup$ – DKNguyen Jun 2 at 23:01
  • $\begingroup$ Got it, thank you... again! So i have to consider Ry and Rx (opposite with respect Rx and Ry onto the wheel hole) also on the platform without wheels, right? Yes, i just considered platform (with axle) and wheels (without axle) $\endgroup$ – xWalle Jun 2 at 23:04

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