Originally posted by TJ:
......
But, roughly...to roll at a constant velocity (not slip), if
v = velocity
R = radius
w = angular velocity
f = frictional force
N = normal force
v = Rw
......
Much better.
But still not true.
Slippage is determined by weight on the tire vs coefficient of friction. That means that a "revolved" axle half (-w- tire and wheel) has about the pushing force of a 200# man, at best (that's being generous). I've never seen a 200# man push anybody out of a real jam on the trails. Neither will a "deployed" revolver.
I think you mean this:
Originally posted by TJ:
......
But, roughly...IF you are rolling at a constant velocity (not slipping)...
....v = Rw
......
Big
if.....
I still have to post that graph. Haven't had time yet.
......
Ok, the dashed line is standard ride height. The weight is weight exerted against the spring by the axle. The "unload" point is labeled. The graphs wer done without regard to spring rate and are for illustration purposes. The "unload" part is what I don't like.....
Originally posted by TJ:
.......What is happening at the pictured "unloading point" that you don't like is unclear......
No, it's quite clear to me. It's not something I plan on modeling further, though. The non-linear characteristic should be enough of an eye-opener for most science-saavy folks. Most people will undestand that that point occurs at the worst possible time for a rig that is on off-camber terrain. This unloading has been observed and documented numerous times over the course of several years.