Moment(torque) help (bycicle crankset)

Replies

• 

  yadavundertaker mohit

  i think that your F"z and F"y should be zero. because when you apply force at paddle it is only in one direction and a moment generate which rotate the crankset.
• 

  KansaiRobot

  Thanks for the reply. Actually F'y can not be 0,or the bycicle wont advance at some point since it is a tangencial force. Anyway, what i am intersted to know is in the general case what is the relation between the forces in the pedal and the measurements in the sensor and why?
  
  Please anybody help. I have been three days without sleeping in the company and i just want to go home. A help would be greatly appreciated.
• 

  ISHAN TOPRE

  hello Kansai robot,
  
  What is the problem in finding resultant force(Ftot)? you know Fx, Fy, Fz find the resultant i.e; Ftot.
  What say? I think it is easy.
  
  @Yadav: What you say? Am I right. Can you explain the problem in your own words?
• 

  yadavundertaker mohit

  if you take centre of crankset as origin(let you are sitting on it and viewing the force applying on pedal) in a relative form then what we see is only a downward force acting at a particular fixed position causing a moment which rotate the crankset.
  its my thinking i donot know it is right or wrong...waiting for your suggestion.
• 

  KansaiRobot

  ThanKs for the amswer guys!
  
  Ok, the problem is that Ftot is the sum of F'x F'y and F'z. If i had them, the problem would be super simple.
  
  But i dont have them. All i have are measurements of Force and Torque in another point ( the center of the system).
  
  Now my question is:
  
  Are those "forces"(measurements) the same as the ones i am applying at the pedal? And if they are why? What is the physics principle?
  Can forces on a rigid body be applied at any point ( without taking into account rotational and moment genearation-obviusly forces at thencenter wont generate torque)
  
  Help😐
• 

  ISHAN TOPRE

  Hi Kansairobot,
  
  Your problem I think is simple. Firstly let me make my point clear,the forces acting at center and forces acting at the pedal will be different
  
  But you can easily find the forces acting at the pedal by taking moment of forces( which are acting at center) with respect to pedal.
  For that you will need to know the distance between 'center' and 'center of gravity of pedal'.
  
  If you have that distance well and good but if you don't have that the still you can find it out. Because you also know the torque (i.e; moment at the center).
  
  Got it? 😁
• 

  KansaiRobot

  ishutopre

  Hi Kansairobot,
  
  Your problem I think is simple. Firstly let me make my point clear,the forces acting at center and forces acting at the pedal will be different
  
  

  
  
  Well, that puts me in a trouble😕
  

  But you can easily find the forces acting at the pedal by taking moment of forces( which are acting at center) with respect to pedal.
  For that you will need to know the distance between 'center' and 'center of gravity of pedal'.
  
  If you have that distance well and good but if you don't have that the still you can find it out. Because you also know the torque (i.e; moment at the center).
  
  Got it? 😁

  Actually that is not so easy. It was in fact my original idea and it works for Mz and Mx. In that case I can easily find F'y because both Mz and Mx depends on F'y (a tangencial force that generates both moments)
  
  for example:
  
  Mz= (radius of the crank) * F'y.
  Mx=((distance to the center)*F'y.
  
  Now the REAL Problem comes when I want to find F'x and F'z. A combination of BOTH of them generates the remaining moment My. so we have
  
  My=f(F'x,F'z)...
  
  how to find the other relation then?
  
  -------------------------------------------------------------------------------------
  
  or are you saying that I should do something like:
  My= (distance)*F'x=(radius)*F'z??????😐
  
  can we do that? decompose the momentum like that? I am not really sure and if it is possible what is the name of the physics law for that (I have to provide my boss with the exact reason to do everything here)