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    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 attached pic. It is a bycicle crankset. Now

    1) There is a 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 in the sensor.
    ----------------------------------

    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 M? is generated by BOTH F"x and F"z.

    ----------------------

    Now how can I get F"x and F"z from My (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!
    Is Fx=F"x? or Fz=F"z? I dont remember any axiom or theorem on this...

    Any help would be greatly appreciated

    Thanks
    Kansai
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    You are correct regarding the values of the moments.


    My is the sum of the effects of both Fz and Fx
    The moments are, of course equal to the forces times the perpendicular distances (x and z) from the pivot as shown.

    So long as you ignore the mass and weight of the crank and pedals themselves, then as far as I can see, the forces Fx, Fy and Fz will be equal to F"x, F"y and F"z

    The resultant force is the vector sum of the three components.

    As you have not written the whole question I can't be quite 100% certain.
    I assume you have included all the relevant information.
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    Thank you very much for your reply. I would like to ask two more questions:

    1) if Fx,Fy ,Fz and F"x, F"y and F"z are the same, then what is the point in having or measuing the moments?

    2)I know this will sounds stupid but what is the physical principle, law or rule that justifies that those forces are equal? The thing is my boss is asking me not to tell him they are just equal but explain him WHY they are equal. It has been already three days in the company without a shower,almost no sleep and I just want to go home. So a help on this will be GREATLY appreciated.

    Thanks a lot

    Kansai
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    (Original post by KansaiRobot)
    Thank you very much for your reply. I would like to ask two more questions:

    1) if Fx,Fy ,Fz and F"x, F"y and F"z are the same, then what is the point in having or measuing the moments?

    2)I know this will sounds stupid but what is the physical principle, law or rule that justifies that those forces are equal? The thing is my boss is asking me not to tell him they are just equal but explain him WHY they are equal. It has been already three days in the company without a shower,almost no sleep and I just want to go home. So a help on this will be GREATLY appreciated.

    Thanks a lot

    Kansai
    The physical principles are Newton's Laws of Motion as explained here.



    You can apply the same physical principle to all 3 diagrams.

    -Fx is equal and opposite to Fx by Newton 3 and is also equal and opposite to F"x by Newton 1. Therefore, F"x must be equal to Fx
    This is only true if you ignore the weight of the crank etc. If you include the weight of the crank then Fx and -Fx will be greater than F"x by an amount equal to this weight when the pedal is vertical as in the diagram.
    I imagine that in a real cycle you would need to include this weight.
    As torque depends on the force then you need to measure both.
    It's the torque that is transmitted through the chain to the wheels and then becomes a force at the wheel rims. This force drives the bike forwards. It's value depends on the gear system and diameter of the wheels etc.
    It's a force that the cyclist actually applies to the pedal.
    So both force and torque are important.
    It starts and ends with a force, but it's torque that is the intermediary and transmits this force from the cyclists foot to the road.
 
 
 
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