## Friday, October 2, 2009

### MetriGear Vector Pt 3: Forces and Torques

A single point mass can aquire three independent momentum coordinates, for example along principal axes x, y, and z. The macroscopic rigid body (i.e. not compressable or deformable) can additionally aquire momentum associated with rotation about these axes. With each momentum component, there are associated force (for translation) or torque (for rotation) components.

For modeling purposes, we can consider each pedal a rigid body. There are then two times six equals twelve total degrees of freedom, constituting six force degrees of freedom and six torque degrees of freedom.

For one pedal, the forces and torques associated with these degrees of freedom might be listed:
1. Fx
2. Fy
3. Fz
4. Tx
5. Ty
6. Tz
where Fx is the force through the center-of-mass along the x-axis, Tx is the torque about the x-axis passing through the center of mass, and similarly for the axes y and z.

For two pedals, one can then define separate forces and torques for the left and right pedals:
1. Flx
2. Fly
3. Flz
4. Tlx
5. Tly
6. Tlz
7. Frx
8. Fry
9. Frz
10. Trx
11. Try
12. Trz
where Flx is the force through the center-of-mass along the x-axis of the left pedal, where Frx is the force through the center-of-mass along the x-axis of the right pedal (note this is a different x-axis; it points in the opposite direction), and similar for other axes and for the torque values.

A side comment: it's not trivial that two pedals would have twelve degrees of freedom. Treating the single pedal as a simple rigid body is an approximation. But the pedals aren't independent of each other; they are constrained in their motion by the crankset. Additionally, while the rigid body may have six degrees of freedom of momentum, it does not imply only six force components are possible. Higher order force components may also be applied. For example, you could apply a positive-z force to the center of a pedal body and a negative-z force along the body perimeter, net force zero. But these higher-order force moments are perhaps of less interest than the moments listed here, and in any case may be unmeasurable by piezoelectric sensors in the spindle. On the other hand, I believe it's plausible (but I don't know) that the Vector would be able to report the force components described here.

So these may be a reasonable set of "measurables". However, rather than use these values, the degrees of freedom can also be quantified using any orthogonal linear combination of these values. For example:
1. Frx + Flx
2. FrxFlx
3. Fry + Fly
4. FryFly
5. Frz + Flz
6. FrzFlz
7. Trx + Tlx
8. TrxTlx
9. Try + Tly
10. TryTly
11. Trz + Tlz
12. TrzTlz

The advantage of this partitioning of the twelve degrees of freedom of the two pedals is that it makes clearer that one and only one degree of freedom is contributing to propulsion. Recall that it's being assumed the right pedal is at 3 o'clock, the left pedal at 9 o'clock. So in this instance, a positive-z force on the right pedal with a negative-z force on the left pedal rotates the pedals and drives the bike forward. This mode is described by the component Frz-Flz.

Since the bike moves in the positive-y direction, you might think a positive-y force on both pedals (Fry+Fly) would also drive the bike forward. However, I already mentioned it's being assumed here that Newton's Third Law implies a counter-force is being applied to the saddle and/or handlebars, and thus in the net the bike does not experience propulsion from such a force.

Okay, so there's the propulsive force. More on the other forces and torques later.