Metrigear Vector Pt 6: Forces and Bending (2)
Back to measuring forces on the pedal: a reminder of the force diagram:
Force diagram of pedal
Assume the sensors measure the torque (the bending moment) on the pedal spindle at each of the associated points, x1 and x1. Forces F1 and F2 are applied at the points x3 and x4. The goal is to determine F3 and F4 from the measurements τ1 and τ2.
Simply enough: two equations, two unknowns (F1, F2):
τ1 = F1 ( x3 ‒ x1 ) + F2 ( x4 ‒ x1 ),
τ2 = F1 ( x3 ‒ x2 ) + F2 ( x4 ‒ x2 ).
Note the second equation is the same as the first with the x1's replaced with x2's, and the τ1 replaced with τ2. Multiply the top equation by ( x3 ‒ x2 ), and the bottom by ( x3 ‒ x1 ):
( x3 ‒ x2 ) τ1 = F1 ( x3 ‒ x1 ) ( x3 ‒ x2 ) + F2 ( x4 ‒ x1 ) ( x3 ‒ x2 ),
( x3 ‒ x1 ) τ2 = F1 ( x3 ‒ x2 ) ( x3 ‒ x1 ) + F2 ( x4 ‒ x2 ) ( x3 ‒ x1 ),
and substract the two and combine terms:
( x3 ‒ x2 ) τ1 ‒ ( x3 ‒ x1 ) τ2 = F2 [ ( x2 ‒ x1 ) ( x4 ‒ x3 ) ],
yielding the solution
F2 = [ ( x3 ‒ x2 ) τ1 ‒ ( x3 ‒ x1 ) τ2 ] / [ ( x2 ‒ x1 ) ( x4 ‒ x3 ) ].
Now for the other solution, swap F1 and F2, and swap x3 and x4, then move terms around to get rid of excess negatives:
F1 = [ ( x4 ‒ x1 ) τ2 ‒ ( x4 ‒ x2 ) τ1 ] / [ ( x2 ‒ x1 ) ( x4 ‒ x3 ) ].
Of primary interest is the total force, trivially adding the previous two results, and simplifying:
F1 + F2 = (τ1 ‒ τ2) / ( x2 ‒ x1 ).
Wow. That's embarrassingly simple.
Interestingly, the only position data you need to know here is the separation of the two torque sensors. This is known, of course, assuming the electronics are inserted straight into the hole, and the hole is itself straight within the spindle. Both of these are likely good approximations within the 1.5% net error budget of the Vector, since errors in orientation only contribute second order to errors in separation along the spindle axis. You don't need to know how far the electronics are from the crank arm, nor how far the pedal body is from the sensors.
Also of interest is the torque on the pedal body:
( F2 ‒ F1 ) ( x4 ‒ x3 ) / 2 =
[ ( [ x4 + x3 ] / 2 ‒ x2 ) τ1 ‒ ( [ x4 + x3 ] / 2 ‒ x1 ) τ2 ] / ( x2 ‒ x1 ).
Wow -- I hope I got all of that correct (I rarely do). Anyway, here you need an additional bit of information, which is the separation of the center of the pedal body from the sensor. This is a parameter which would need to be determined by a calibration step. Fortunately, accuracy on this torque number probably isn't important. It's power we really want to nail down, and that depends on propulsive force, not how hard the pedal is being twisted.
Luckily you do not need to know how far anything is from the crank arm. If you did, for example inserting a spacer under the pedal to change the Q-factor would mess up the result. Perhaps more significantly, bending of the crank arm may yield an effective pivot point beyond the point where the pedal is screwed into the crank, and the position of this virtual pivot may depend on pedal force. So since the result doesn't depend on the location of the pivot point, it may be relatively insensitive to crank flex.
So what do I get out of all of this analysis? Basically that with good bending data taken at two points along the spindle, you can get the average force on the pedal body without knowing the distance to the pedal body. That's very nice. Additionally, you can derive the torque exerted on the pedal body about its center if you know the distance from the bending sensors to the center of the pedal body. This could be calibrated for maximum accuracy, or it could be more crudely estimated given the lack of relative importance of this value.
Force diagram of pedal
Assume the sensors measure the torque (the bending moment) on the pedal spindle at each of the associated points, x1 and x1. Forces F1 and F2 are applied at the points x3 and x4. The goal is to determine F3 and F4 from the measurements τ1 and τ2.
Simply enough: two equations, two unknowns (F1, F2):
τ1 = F1 ( x3 ‒ x1 ) + F2 ( x4 ‒ x1 ),
τ2 = F1 ( x3 ‒ x2 ) + F2 ( x4 ‒ x2 ).
Note the second equation is the same as the first with the x1's replaced with x2's, and the τ1 replaced with τ2. Multiply the top equation by ( x3 ‒ x2 ), and the bottom by ( x3 ‒ x1 ):
( x3 ‒ x2 ) τ1 = F1 ( x3 ‒ x1 ) ( x3 ‒ x2 ) + F2 ( x4 ‒ x1 ) ( x3 ‒ x2 ),
( x3 ‒ x1 ) τ2 = F1 ( x3 ‒ x2 ) ( x3 ‒ x1 ) + F2 ( x4 ‒ x2 ) ( x3 ‒ x1 ),
and substract the two and combine terms:
( x3 ‒ x2 ) τ1 ‒ ( x3 ‒ x1 ) τ2 = F2 [ ( x2 ‒ x1 ) ( x4 ‒ x3 ) ],
yielding the solution
F2 = [ ( x3 ‒ x2 ) τ1 ‒ ( x3 ‒ x1 ) τ2 ] / [ ( x2 ‒ x1 ) ( x4 ‒ x3 ) ].
Now for the other solution, swap F1 and F2, and swap x3 and x4, then move terms around to get rid of excess negatives:
F1 = [ ( x4 ‒ x1 ) τ2 ‒ ( x4 ‒ x2 ) τ1 ] / [ ( x2 ‒ x1 ) ( x4 ‒ x3 ) ].
Of primary interest is the total force, trivially adding the previous two results, and simplifying:
F1 + F2 = (τ1 ‒ τ2) / ( x2 ‒ x1 ).
Wow. That's embarrassingly simple.
Interestingly, the only position data you need to know here is the separation of the two torque sensors. This is known, of course, assuming the electronics are inserted straight into the hole, and the hole is itself straight within the spindle. Both of these are likely good approximations within the 1.5% net error budget of the Vector, since errors in orientation only contribute second order to errors in separation along the spindle axis. You don't need to know how far the electronics are from the crank arm, nor how far the pedal body is from the sensors.
Also of interest is the torque on the pedal body:
( F2 ‒ F1 ) ( x4 ‒ x3 ) / 2 =
[ ( [ x4 + x3 ] / 2 ‒ x2 ) τ1 ‒ ( [ x4 + x3 ] / 2 ‒ x1 ) τ2 ] / ( x2 ‒ x1 ).
Wow -- I hope I got all of that correct (I rarely do). Anyway, here you need an additional bit of information, which is the separation of the center of the pedal body from the sensor. This is a parameter which would need to be determined by a calibration step. Fortunately, accuracy on this torque number probably isn't important. It's power we really want to nail down, and that depends on propulsive force, not how hard the pedal is being twisted.
Luckily you do not need to know how far anything is from the crank arm. If you did, for example inserting a spacer under the pedal to change the Q-factor would mess up the result. Perhaps more significantly, bending of the crank arm may yield an effective pivot point beyond the point where the pedal is screwed into the crank, and the position of this virtual pivot may depend on pedal force. So since the result doesn't depend on the location of the pivot point, it may be relatively insensitive to crank flex.
So what do I get out of all of this analysis? Basically that with good bending data taken at two points along the spindle, you can get the average force on the pedal body without knowing the distance to the pedal body. That's very nice. Additionally, you can derive the torque exerted on the pedal body about its center if you know the distance from the bending sensors to the center of the pedal body. This could be calibrated for maximum accuracy, or it could be more crudely estimated given the lack of relative importance of this value.
Comments
What if the sensors don't measure bending moment?
I'm trying to wrap my head around Jim's reference, but I think what he's saying is Bernoulli's idealized cantilever fails to adequately describe a relatively short hollow tapered cylinder like a pedal spindle, at least within the 1.5% error budget, and that off-diagonal elements of the stiffness matrix are needed to really solve the problem (not just the scalar term used in the simple bending analysis). So maybe a second order calibration is needed.
But if they were instead configured to measure shear strain, which is a change in vertical position along the axis of the pedal, then you're right that the driving force would be force, not a force times distance.
So I guess the question is then whether it's relatively easier and more reliable to measure bending or to measure shear distortion. It's a hollow tapered cyclinder, so I think one would need to run some real numbers (preferably FEA) to see if there's a strong enough shear signal without second-order contamination from bending stress, and vice-versa.
My money's would be on the bending moment. I suppose I could ask, but it's more fun to guess.
It's funny I'm so interested in this topic. It sort of gives me an excuse to exercise some basic physics on a topic (bicycles) which I really enjoy.
Why are two strain gauges used to measure overall force? Could the same be accomplished with only one for each axis?