Force on a disk brake caliper 1

[slopesoar4fun]

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

 

[GregLocock]

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

Cheers

Greg Locock

 

[MintJulep]

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.

 

[Erebus]

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??

 

[GregLocock]

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