Hillhater said:
if you look at the data the highest ADDITIONAL (tensile) load any spoke is subjected to is 40 N (10lbf) !
That's right, and it's why I said "change in tension" instead of "increase in tension". Normal stainless steel spokes are usually good to withstand tensile forces close to four times as high as the static tension we aim for. So they don't fail from gross overload; generally they fail from fatigue. Force cycle depth is relevant to the way they fail, as is the number of loadings and unloadings.
Chalo said:
Where my braking load figures (estimates) come from...
Using your (way underestimated) 200lbf force at the tyre, that is a torque of 225 ftlbf ( 13.5" radius on a 26" rim+tyre)
That torque is transmitted through the (18) spokes to the hub at a diameter of 2.3" ( my disc hub), resulting in a tangential force of 2350 lbf at that diameter.
share that 2350 lbf between 18 spokes, and you get approx 130 lbf per spoke extra load due to braking.
And as i said before... i believe your initial 200lbf braking force is totally underestimated to the extent that these spoke loads could almost be doubled !.
First of all, the 200 lbf is based on how much braking force it takes to tip the bike, which is a hard limit on both acceleration and deceleration. The figure I have seen most often for that force is 0.7g maximum, though there is some variation according to wheelbase and weight distribution. If it were possible to brake harder than that, forks would simply not work under forceful braking. I bend steel forks to straighten them relatively often by sitting on the ground, bracing my feet against the crank arms, and pulling the fork legs. I don't have to exert myself very hard to bend a fork straight, and yet the only person I know who has bent forks from braking forces is me.
You ignore the fact that you're not changing the tension in 18 spokes, but in 36 of them. The torsional force is divided by 36.
Now ,..please explain how you are going to get similar torque figures from any pedal effort ??
It should be obvious that the force required to tip the bike backwards under pedaling force alone is similar to that required to tip the bike forwards under braking alone. But let's do the numbers for a normal non-outlying case.
A 200 pound rider is riding in 22/34 low gear with 7 inch cranks. To push away from a dead stop or to surmount a step, he can stand on the forward pedal and even pull up against the bars to increase his apparent weight on the pedal. So in terms of hub torque, that's say 300 pounds momentary pedal force applied at 7/12 of a foot radius, multiplied by a gear ratio of 34/22. That's about 270 lbs-ft at the hub. A typical low-flange rear hub has a 44mm spoke circle, so the force transmitted at a 22mm radius is about 3750 pounds, divided among 36 spokes, for a net change in tension of 104 pounds force per spoke. But in practice, the forces never go nearly that high on a two-wheeler.
If you've ever tried to pedal that hard in that low a gear, you know that the result is just to lift the bike's front wheel off the ground. (Also, if you've ever tried to pedal that hard in that low a gear against a fully locked rear brake, you'd know that no brake can resist the force.) So in practice, the limit of force at the contact patch is less than what I've calculated. That's the same principle as the maximum usable braking force being limited to no more than the force required to tip the bike in the other direction.
I frequently have to take freewheels off of jackshaft hubs from pedicab trikes, where loads can be heavy and there is no chance of wheelies. I have to use a long box wrench, because two sides of the freewheel remover tool are not strong enough to withstand the force required. The edges of the tool have been twisted into a helix from the force of the six-sided wrench. I weigh at least 325 pounds, and I have to surge my whole weight against the other end of the wrench to get them to break free. That is similar to the peak torque these hubs have seen in their service lives to tighten them on.
