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
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
Cheers
Greg Locock
RE: Force on a disk brake caliper
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
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
Cheers
Greg Locock