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Force on a disk brake caliper

Force on a disk brake caliper

Force on a disk brake caliper

(OP)
Here is a puzzle that I've found two different solutions to.  The problem is they vary by a factor of two!  Any help would be appreciated.

If I calculate the required torque on a wheel to stop my car at a specified decelleraton, I can then calculate the required force that must be applied by the caliper. I'll call this tangential force P.

The caliper piston applies a force, F, to each of the pads.

The basic equation for friction force is that the friction force is equal to (friction coef.) x the normal force.  In this case F is the normal force, and it is applied to both pads.

Now comes my question:

Is P = U x F, or

is P = U x 2F           where U = friction coef.

Fred Puhn (Brake Handbook)uses the second equation.

My Mark's Handbook (8th ed) shows a funky disk brake for a lifting device, and they seem to use the first equation.

I'm with Fred Puhn, as I see F acting on two brake pads, and therefore the caliper must resist 2 x UF.

There are also a couple of spreadsheets floating around on Yahoo groups, and these both use the first equation.

I'm hoping someone with a machine design textbook might look this up.

Thanks

David H
Calgary
 

RE: Force on a disk brake caliper

It is 2F. Imagine two separate brake pads applied by independent screw jacks.

Cheers

Greg Locock

RE: Force on a disk brake caliper

It's eaiser to see if you write it:

P = 2 x (uF)

There are two pads, each of which are pressed against the disk with the same force F.

RE: Force on a disk brake caliper

Greg,

I am flying off on a small tangent now, but what bearing do you feel brake cailper location has on long. weight transfer of pro/anti dive suspension systems??

This should be an additive effect (pos. or neg. depending on the set-up) but is its magnitude large enough to warrant its consideration in the suspension design process??

RE: Force on a disk brake caliper

Surely the calliper force is internally resolved in the strut/bearing, and so is invisible to the rest of the suspension, if the strut is stiff enough.

Cheers

Greg Locock

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