On Sat, 16 Mar 2019 15:56:04 +0000 (UTC), Ralph Barone
wrote:

James wrote:
On 16/3/19 10:32 am, John B. Slocomb wrote:


It comes to mind that as the braking resistance is applied at the
contact of the tire and the road and that the resistance is applied to
the wheel hub that a much stronger disc brake wheel would be required
than when using a rim brake as the ratio between the disc brake disc
and the contact with the road is approximately 26.5"/7.5" (average 2
sizes of disc rotor) = about 3.5 ratio while a rim brake is only about
an inch and a half difference so say 26.5/23.5=1.1 ratio. Based on
braking forces it would appear that a disc brake wheel would have to
be roughly 3 times stronger than a rim brake wheel.

36 spoke, cross three, wheels anyone?

But of course an ATB is already so heavy that the addition of strong
wheels is rather a matter of bringing coals to Newcastle.


Well, you don't build a disc brake wheel with radial spokes.



I missed commenting to John earlier. Your visualization of the problem
isn't quite right. With rim brakes, the spokes do not see any braking
forces, since the brakes sit between the spokes and the tire.

I see. True the spokes see no breaking force but they do see a force
simply because they connect the rim to the bicycle and the bicycle
(and rider) represent a certain amount of inertia which has to be
overcome.

No matter
what size of disc brake you have, the spoke forces are the same, since the
braking force is applied to the hub, which then transfers it through the
spokes to the rim/tire. In the limit, the hub is stationary and the wheel
is sliding. In this case, the size of the disc determines how much
tangential force is applied to the disc, but the force on the spokes is
determined by the number of spokes, the lacing pattern and the hub
diameter.

Exactly. As I said, 36 spokes laced "cross three" :-)

--
Cheers,
John B.