about the torque calculation of a mecanum wheel
 

hi everyone,
I'm new here and interested in the mecanum wheeled robot. I have read the torque calculation document from Ether, in which says the actual force (along the roller axis) is Fc=1.414T/r, T is the torque of motor.
But in the article "kinematics and dynamics modeling of a mecanum wheeled mobile platform", can be found at "http://ieeexplore.ieee.org/xpl/login.jsp?tp=&arnumber=4749608&url=http%3A%2F%2Fie eexplore.ieee.org%2Fxpls%2Fabs_all.jsp%3Farnumber% 3D4749608" , it is saied that Fc is one component of T/r, means Fc=0.707T/r. Some other articles also say so.
The above is what confused me. Could anyone try to explain it or do I just understant it in a wrong way?
PS, I do want to konw how does the force Fs (vertical to the T/r) come out.
Thank you a lot.

Re: about the torque calculation of a mecanum wheel
 

I can't access the document you referenced on the ieee site, even after trying a number of edits to your "quoted url", so this is a bit of a guess:

Ether's value for force Fc along the roller axis is correct, and is calculated from the torque applied to the wheel shaft and the radius of the wheel. If it were anything else, the torques around the shaft of the wheel applied by and to the wheel wouldn't be equal and opposite.wouldn't add up.

I suspect that the ieee document has a different definition for Fc, and possibly T and r. For example, T could be torque around the vertical axis through the center of the robot, and r could be the horizontal distance from the center of the robot r to the center of the wheel. Can you provide a cleaner link, or at least more solid definitions of Fc, Fs, T, and r as used in the ieee document?

Re: about the torque calculation of a mecanum wheel
 

Thank you for your reply.
here's another article with open access, in page 3:
http://arxiv.org/pdf/1211.2323.pdf

or search the google schoolar " Modelling of the motion of a Mecanum-wheeled vehicle" by E Matsinos.

Re: about the torque calculation of a mecanum wheel
 

I'm at the office today, so I can't look at it in detail until this evening, but this document has several serious issues, in particular the directions of the rollers as they make contact with the ground - all of the force diagrams (most particularly figure 3) show forces in an "X" configuration, but the rollers should contact the carpet in a "<>" diamond configuration if they have an "X" configuration as viewed from above (and specified in figure 2).
In Fig 3 wheel 2, Fp should point "northwest" and Ft should point "northeast". This also causes them to swap strafe directions in the table. I'm surprised they were able to get any rotation at all.

Re: about the torque calculation of a mecanum wheel
 

Thank you very much.

The first article has been cited by the second one.
Here is the first one with an open access.
http://researchspace.csir.co.za/dspa...dle/10204/2771
or
http://researchspace.csir.co.za/dspa...lale2_2008.pdf

Re: about the torque calculation of a mecanum wheel
 

While Matsinos does decompose the "driving force" F_i into F_i,p and F_i,t improperly because he assumes that it is transverse to the wheel axis (it is not), this error is inconsequential, as when he performs experiments, he does not control torque or force, but angular speed.

The problem I related above is consistent throughout the work. Though the photograph of figure 1 and the overhead diagrams of figures 2 through 6 all depict the proper roller orientation, When he decomposes the force parallel to the roller axis into x and y, he depicts the axis of a roller at the top of the wheel, not one in contact with the floor.

The consequence of this error is recognized within the paper in the paragraph below equation 14 on page 8 (Matsino's emphasis):
I have not yet had a chance to review the experimental data to see if the wheels were installed as pictured or as analyzed (or perhaps some other way).

Re: about the torque calculation of a mecanum wheel
 

His error is fundamental.

He takes a component of force (F_i), breaks it into 2 new components (F_i,p and F_i,t), throws away one of those components (F_i,t), and then breaks the remaining component (F_i,p) into yet two new components (F_i,y and F_i,x).

This is a common high-school physics error.

The proper way to analyze the wheel is to start with the net force (of the floor acting on the bottom of the wheel, in the plane of the floor). This net force F_i,net has a known direction: parallel to the axis of the roller touching the floor.

Then break that net force F_i,net into forward F_i,y and sideways F_i,x components.

The forward force F_i,y will be equal to tau/r, where tau is the driving torque on the wheel and r is the wheel radius. This known value of F_i,y can then be used to compute the magnitude of F_i,x and F_i,net.

Re: about the torque calculation of a mecanum wheel
 

While I agree that it is a fundamental error, it is inconsequential (I should have specified that I mean this within the scope of the paper) because neither the original F_i, nor any other variables dependent on it, are ever measured or controlled. Had the controlled parameter been current to the motor (which is directly related to the output torque), the experiment would have revealed it through the calculation of an unrealistic gearbox efficiency (or other similar parameter) to offset the error.

The difference would also have been of consequence if the vehicle were run at speeds/torques which cause significant slip of the rollers against the floor, but for which the erroneous calculations would have predicted good traction.

Re: about the torque calculation of a mecanum wheel
 

OK.

But arguably, it also has consequences within the scope of the paper for its readers, who may assume the paper has been properly vetted and who may be misled/confused.

Re: about the torque calculation of a mecanum wheel
 

The likelihood of confusion deepens. With the exception of the last two rows, the results in table 1 are what is predicted with the improper wheel orientation shown in the force diagrams of figures 2 through 6. As shown, the wheel rotations indicated in the last two rows would not result in any rotation of the vehicle unless the center of gravity and geographic center were offset in the X direction. This is depicted in figure 6, but it is not clear from a casual reading of the text that this was a necessary condition to achieve the stated rotation.
At first glance (and for me, second and third), the translations reported in table 2 seem to be at odds with the results of table 1. In all 32 cases, wheels 1 (right rear) and 3 (left front) are rotated equally in the positive direction, and wheels 2 (right front) and 4 (left rear) are rotated by the same amount as each other, but less than the magnitude of rotation of wheels 1 and 3. Using the directions depicted in figure 5 (and not contradicted in figure 6), this would lead to a westerly (negative x) translation. All of the results in table 2 report a positive X offset, that is toward the east. The resolution of this inconsistency is in the final, parenthesized sentence at the end of the second paragraph of section 3.1:
That is, that the wheels are installed in the standard mecanum configuration depicted in the graphics of figures 1-6, not the configuration shown in the force diagrams of figures 2-6.

At this point, I'll back away from this paper, rather than trying to figure out if the mistakes and inconsistencies are maintained or repaired in the analysis of table 3 and statement of conclusions.

Re: about the torque calculation of a mecanum wheel
 

The original article begins with the same fallacy, and presents additional fallacies as it goes along.
In the introduction, Tlale defines "effective" variables in equations 1 and 3, but does not describe the derivation, or sufficiently define the meaning to be entirely certain that he is following the same "lost force" fallacy. He never again uses these definitions, though he does use the word effective several times with similar meaning.

Figure 4 and the discussion between equations 4 and 5 clearly depict the fallacy that the "applied torque" must be decomposed into an effective andi an ineffective part and that the "ineffective" part simply disappears. To what object was this force applied? What object received the reaction (Newton's third law) force? If the answer to these is "nothing", then did that force ever exist? If not, how can it have been part of the "applied torque"? The answer, of course, is that it never did exist and the force countering the "applied torque" is not in the direction perpendicular to the wheel axle.

As shown here and in Ether's paper, the frictional force at the roller/carpet interface is essentially parallel to the roller axis. However, ignoring roller friction, the only three forces acting on the wheel are this contact force, the applied torque of the motor, and the forces necessary to constrain the wheel to stay on its axle. In the steady-state case (that is, all wheel speeds, the motion of the CoM and the rate of rotation about the CoM are all constant), The forces and torques on the wheel relative to its axis must add up to zero (otherwise the wheel speed would change). Presuming essentially frictionless wheel bearings, the torque in the plane of the wheel applied by the floor to the wheel must be equal and opposite to the torque applied by the motor/gearbox to the wheel. Since the force is not applied in a purely rotational direction, it must be greater than rF, and in particular rF/sinα, not rFsinα. As Tlale does not present any quantitative results, there are again no numerical opportunities for this inconsistency to be realized within the scope of Tlales paper.

Section IV (turning) is IMO an even worse job of physics. While there is a clear reference to a "pure turning motion" in the second sentence, only one of the cases presented (4 wheels) is at all likely to be about the center of mass, and therefore "pure" rotation.

FWIW, there are two cases which may produce pure rotation with two driven wheels: When wheels on the opposite corners are spun in opposite directions. Presuming that the other two wheels are at least somewhat free to rotate, the robot will rotate as well. These would be expressed as "+ 0 - 0", "- 0 + 0", "0 - 0 +", and "0 + 0 -" in the tables of either paper.

Finally, do not think that because the force at the floor is greater than rF we are getting something for nothing. Since this force is still a frictional force limited to be μW_i, a small value for sinα reduces the amount of torque which may be applied at the motor/gearbox without losing traction.