Analysis of Dual Pivot Brake

jim beam <spamvortex@xxxxxxxxxxxxxxx> writes:

Luns Tee wrote:
Yes, it is getting boring, however I repeat my question because
despite having been asked it over half a dozen times, you have yet
to give any sort of answer that is consistent with the fundamentals
of mechanics. A freely pivoting 1:1 lever has forces on its ends
in a 1:1 ratio, but you seem to believe that it can support a 2:1
ratio instead.

Your question is reasonable, and I don't understand jim's response
(the lever is not "locked by the rim"). However, your question seems
flawed. Near as I can tell, there is no "freely pivoting 1:1 lever"
in the assembly.

The following diagram is a simplified sketch of a dual-pivot brake.
The capital O's are pivot points, the small o is where the two
arms contact. The pivot on the left is actually the center pivot;
I've shown it like this to avoid dealing with non 90 degree angles.

-------------O-----+
| |
| o
-------------|-----------O
| |
| |
| |
| +-------+ |
+-| rim |-+
+-------+

Here are the relevant dimensions and applied forces on each arm
(F is the force against the rim, T is the cable tension, f is the
force between the two arms).

<----L1-----><--L5->
-------------O-----+
| ^ | |
v | | o
T | | ^
L2 | |
| | f
| |
| |
v +<--F

T f
^ |<-L6->
| v
-------------------------O ^
<----------L3----------->| |
| |
| L4
| |
F-->+ v

At equilibrium, sum of torques around each pivot must be zero.

(1) +T*L1 - F*L2 + f*L5 = 0
(2) -T*L3 + F*L4 + f*L6 = 0

Solving for F and f we get

(3) F/T = (L1/L5 + L3/L6)/(L2/L5 + L4/L6);
(4) f/T = (L3/L4 - L1/L2)/(L5/L2 + L6/L4);

For the brake to operate properly, f > 0, which implies
that L3/L4 > L1/L2 (the nominal gain of the L3/L4 arm
must be greater than that of the L1/L2 arm).


To simplify (3), let k be a design parameter that varies from 0 to 1,
and set

(5) L7 = L5 + L6 (horizontal distance between pivots)
(6) L5 = k*L7
(7) L6 = (1-k)*L7

Then we get

(8) F/T = ((1-k)*L1 + k*L3)/((1-k)*L2 + k*L4)

With k=0 [contact at left (center) pivot]

(8a) F/T = L1/L2

with k=1 [contact at right (aux) pivot]

(8b) F/T = L3/L4

So the effective gain varies from L1/L2 to L3/L4 as the design
parameter k varies from 0 to 1.

--
Joe Riel
.

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