rear rim seems to rub

[Joe Riel]

One final thought. Assuming that k is selected such that
the left and right pads move equally, then the slope of
the line of action of the reaction force at the auxiliary
pivot is given by

Fy/Fx = 1/Gf - 1

where
Gf is the force gain = 1/2*(L1/L2 + L3/L4)

Assuming L1/L2 = 1 and L3/L4 = 2, then

Fy/Fx = -1/3

This is a line that moves upward and to the left. The main pivot
should lie along this line so that no torque is applied to the
half-arm that supports the auxiliary pivot. A crude measurement
gives -20 degrees, which is close enough (tan 20 = 0.36 ).

--
Joe Riel

 

[[email protected]]

Joe Riel writes:

> One final thought. Assuming that k is selected such that the left
> and right pads move equally, then the slope of the line of action of
> the reaction force at the auxiliary pivot is given by

> Fy/Fx = 1/Gf - 1

> where
> Gf is the force gain = 1/2*(L1/L2 + L3/L4)

> Assuming L1/L2 = 1 and L3/L4 = 2, then

> Fy/Fx = -1/3

> This is a line that moves upward and to the left. The main pivot
> should lie along this line so that no torque is applied to the
> half-arm that supports the auxiliary pivot. A crude measurement
> gives -20 degrees, which is close enough (tan 20 = 0.36 ).

So what is the conclusion? I don't understand what this has to do
with the mechanical advantage of the caliper to input force, which as
I said is the ratio of cable movement to pad movement in small angles
about the operating point. Is the above result agree or disagree with
Luns's measurement of that ratio.

As I suggested, since the centering of the caliper is guaranteed by
its design, and if the center bolt isn't loose in the fork crown, the
pads close equally and with equal force onto the rim of the wheel. I
don't understand what the disagreement is about other than the exact
mechanical advantage which varies because this isn't a pure "bell
crank" but is affected by cosine effect.

Jobst Brandt

 

[Joe Riel]

Joe Riel <[email protected]> writes:

> One final thought. Assuming that k is selected such that
> the left and right pads move equally, then the slope of
> the line of action of the reaction force at the auxiliary
> pivot is given by
>
> Fy/Fx = 1/Gf - 1
>
> where
> Gf is the force gain = 1/2*(L1/L2 + L3/L4)
>
> Assuming L1/L2 = 1 and L3/L4 = 2, then
>
> Fy/Fx = -1/3
>
> This is a line that moves upward and to the left. The main pivot
> should lie along this line so that no torque is applied to the
> half-arm that supports the auxiliary pivot. A crude measurement
> gives -20 degrees, which is close enough (tan 20 = 0.36 ).

Dang, made a silly error; substituted the F instead of f into
the computation for Fy. The corrected expression is considerably
more complicated:


Fy/Fx = (2*L7*L2*L4+L1*L4^2+L1*L4*L2-L2*L3*L4-L2^2*L3)/L7/(L2*L3+L1*L4)

assuming k = kopt.

With L1/L2 = 1 and L3/L4 = 2 this reduces to

Fy/Fx = 2/3 * ( 1 - 1/2*(L2 + L4)/L7 )

Fortunately, the computed value of Fy/Fx is about the same; I
need to make some accurate measurements to see how will it agrees.

--
Joe Riel

 

[datakoll]

"From birth, man carries the weight of gravity on his shoulders. He is
bolted to earth. But man has only to sink beneath the surface and he
is free."
- J. Cousteau

I had wondered why the dual pivots replacing center pulls were
difficult. And why the bike shop's english explaination of setup was
incorrect.

good solvent lubrication after cleaning, then checking for equal
movement - no binding one side - and the too great bearing clearance
cures the problem.

I reach back and bash it with my hand once or twice from the opposite
side.

 

[Luns Tee]

In article <[email protected]>, Joe Riel <[email protected]> wrote:
>Joe Riel <[email protected]> writes:

> Fy/Fx = (2*L7*L2*L4+L1*L4^2+L1*L4*L2-L2*L3*L4-L2^2*L3)/L7/(L2*L3+L1*L4)
>
> assuming k = kopt.

Pull L7 to the side and divide numerator and denominator of
what's left by L2*L4:

Fy/Fx = (2*L7 + L1/L2*L4 + L1 - L3 - L2*L3/L4) /(L1/L2+L3/L4) / L7

Now note that L7 = L3-L1 : substitute this in and things
collapse quickly from there:

= (L3-L1 + L1/L2*L4 - L2*L3/L4) /(L1/L2+L3/L4) /(L3-L1)
= (L1/L2+L3/L4)*(L4-L2) /(L1/L2+L3/L4) /(L3-L1)
= (L4-L2)/(L3-L1)

This is the slope of the line between pivots, regardless of what
angle that happens to be at. There is no magic angle for the pivots to
be at.

This is as should be expected: with the rim exerting balanced
forces on the two pads, and the cable's force on the two arms also
being balanced, the caliper as a whole has no force on it for a
mounting bolt/basepiece torque to react against.

-Luns

 

[Joe Riel]

[email protected] (Luns Tee) writes:

> In article <[email protected]>, Joe Riel <[email protected]> wrote:
>>Joe Riel <[email protected]> writes:
>
>> Fy/Fx = (2*L7*L2*L4+L1*L4^2+L1*L4*L2-L2*L3*L4-L2^2*L3)/L7/(L2*L3+L1*L4)
>>
>> assuming k = kopt.
>
> Pull L7 to the side and divide numerator and denominator of
> what's left by L2*L4:
>
> Fy/Fx = (2*L7 + L1/L2*L4 + L1 - L3 - L2*L3/L4) /(L1/L2+L3/L4) / L7
>
> Now note that L7 = L3-L1 : substitute this in and things
> collapse quickly from there:
>
> = (L3-L1 + L1/L2*L4 - L2*L3/L4) /(L1/L2+L3/L4) /(L3-L1)
> = (L1/L2+L3/L4)*(L4-L2) /(L1/L2+L3/L4) /(L3-L1)
> = (L4-L2)/(L3-L1)
>
> This is the slope of the line between pivots, regardless of what
> angle that happens to be at. There is no magic angle for the pivots to
> be at.
>
> This is as should be expected: with the rim exerting balanced
> forces on the two pads, and the cable's force on the two arms also
> being balanced, the caliper as a whole has no force on it for a
> mounting bolt/basepiece torque to react against.

Thanks. After posting this last night (actually, the original, post, with
the math error) I thought about it and suspected that, as you say, there
shouldn't be any special position. However, I didn't prove it. Thanks
for working out the simplification.

--
Joe Riel