Re: Analysis of Dual Pivot Brake
- From: Joe Riel <joer@xxxxxxxxxx>
- Date: Mon, 19 Nov 2007 20:45:07 -0800
jim beam <spamvortex@xxxxxxxxxxxxxxx> writes:
Joe Riel wrote:
As a followup, we can look at the relative pad movement. Let x1 be
the movement of the left pad, x2 the movement of the right pad. For
small movements, the vertical movement of the contact point of the
two arms is
dy = dx1*L5/L2 = -dx2*L6/L4.
The (negative) ratio is
-dx1/dx2 = L4*L5/L2/L6.
Expressed in terms of k we get
-dx1/dx2 = L4/L2*k/(1-k)
Ideally we'd like this ratio to be 1, so that both pads open equally.
that is ideal, but presumptive. the y-arm is actually free to compress
independently.
Solving for k we get
k = L2/(L2+L4)
Plugging that into the gain ratio and simplifying we get
F/T = 1/2*(L1/L2 + L3/L4)
So, with k set so the pads move equally, the gain of
the brake is the average of the gain of the two
lever arms without the connection (the connection is required
for operation.
the connection is required to help swing the two arms together when
free, but under load, that connection is small beer compared to the
force exertion from compressing the rim. the connection never sees more
than caliper spring loading.
The force is considerable and directly computed from my previous formulae.
If the "optimal" ratio I suggest is used, then it is
f/T = (L3/L4 - L1/L2)*1/2*(L2+L4)/L7
Assuming L3/L4 = 2 and L1/L2 = 1, this gives
f/T = 1/2*(L2+L4)/L7.
Note that the force is proportional to the applied cable tension. The
spring has nothing to do with it, its only function it to open the
arms. If what you said were true, then with the spring disconnected
and the brake applied hard there would be zero force between the little
screw and the pad that it hits. Tightening it would be easy.
Try it. I've done so...
--
Joe Riel
.
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