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Moment(torque) help (bycicle crankset)

 
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KansaiRobot



Joined: 13 Apr 2011
Posts: 1
Location: Japan

PostPosted: Thu Apr 14, 2011 2:08 am    Post subject: Moment(torque) help (bycicle crankset) Reply with quote

Hello everybody.
I would appreciate your help in something I have been thinking all day (and now I am doing an allnighter, without result )

Please take a look at the pic at the end. It is a bycicle crankset. Now


1) There is a force sensor in the center. It can sense Fx, Fy and Fz and also Mx, My and Mz. (where F are forces and M moments (or torques))

2) The rider applies a Total force (Ftot) in the pedal. This Total Force can be decomposed as (F"x, F"y, F"z)-> just a notation.

My problem is that I have to find Ftot and the only data I have is the force and moments measured at the sensor.
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Now, I have done this so far:

1) I can see that Mz and Mx are generated ONLY by F"y. So I guess I can find F"y quite easily:

||F"y||= ||Mz||/(radius of crank)


2) I can see that My is generated by BOTH F"x and F"z.

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Now how can I get F"x and F"z from My (torque data)???


On related notes what is the relationship between the forces at the pedal(F"x, F"y, F"z) and the forces at the sensor? (Fx, Fy, Fz)
THIS is KEY I think!

So far I am thinking that Fx and Fz are proportional (if not equal!??) to F"x and F"z. If that is the case I can find both by geomety from My only.


Any help would be greatly appreciated. I am running late on a deadline, so I will be infintely grateful for your comments and help

Thanks
Kansai
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hepcj
Site Admin


Joined: 23 Jun 2007
Posts: 125

PostPosted: Thu Apr 14, 2011 10:35 am    Post subject: Reply with quote

If have understood your question, then the forces applied at the pedal will be the same because the elements of crankshaft are part of a rigid structure.
The torque generated at centre of rotation will be the force x the perpendicular distance to the centre of rotation.
tau =F x r where the times symbol is the vector cross product. tau, F and r are vectors. F is the force, r is the distance from the centre of rotation. This will decompose into three equations one for each direction, although since it is a crankset of a bicycle you might consider only the torque generated in the plane of rotation of the crank as the other torques will be resisted by the crankset bearing.
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