I needed a cam-lever like this:

But I couldn't find one with enough "throw" - what McM-C calls clamping distance.

Taking the opportunity to have some fun, I decided to make it myself. The cam would be just be an eccentric piece of round - i.e., the center of rotation offset from the stock center:

The clamping distance is the difference between d1 (unclamped) and d2 (clamped) when rotated CCW 90°. The tricky part is determining the offset (o) required. Which is where the algebra comes in. We know:

o = r (radius) - d1

d1 = d2 - clamping distance,

d2 = SQRT(r^2 - o^2) (thanks, Pythagoras)

The radius and clamping distance are given, leaving 3 unknowns and 3 equations. Documenting the algebra involves a lot of squares & square roots, which is ugly in text-only, so I'll skip that & get to it:

o = r - d1

d1 = (-b + SQRT(b^2 - 4*a*c) ) / 2 [quadratic formula] where a = 1, b = r - clamping distance, c = 0.5*(clamping distance^2)

for 1.25 round & a clamping distance of 0.125, the required offset is 0.141.

Note - this is for 90° rotation, which gives the right-triangle & its Pythagorean relationship. The offset can be figured for other than 90°, but it would involve the law of cosines & much messier algebra.

But I couldn't find one with enough "throw" - what McM-C calls clamping distance.

Taking the opportunity to have some fun, I decided to make it myself. The cam would be just be an eccentric piece of round - i.e., the center of rotation offset from the stock center:

The clamping distance is the difference between d1 (unclamped) and d2 (clamped) when rotated CCW 90°. The tricky part is determining the offset (o) required. Which is where the algebra comes in. We know:

o = r (radius) - d1

d1 = d2 - clamping distance,

d2 = SQRT(r^2 - o^2) (thanks, Pythagoras)

The radius and clamping distance are given, leaving 3 unknowns and 3 equations. Documenting the algebra involves a lot of squares & square roots, which is ugly in text-only, so I'll skip that & get to it:

o = r - d1

d1 = (-b + SQRT(b^2 - 4*a*c) ) / 2 [quadratic formula] where a = 1, b = r - clamping distance, c = 0.5*(clamping distance^2)

for 1.25 round & a clamping distance of 0.125, the required offset is 0.141.

Note - this is for 90° rotation, which gives the right-triangle & its Pythagorean relationship. The offset can be figured for other than 90°, but it would involve the law of cosines & much messier algebra.

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