Joe Riel said:... Do you corner anywhere near that lean angle? Few do.
Ok, so you first establish that on flat pavement with no surface gunk friction is very high. And
you argue that few folks ride anywhere near this limit on corners. So for most people fc is not
that high.
Quoted message said:Second, because the total force on the contact patch is the vector sum of the cornering force and
the braking force, which are applied perpendicular to each other, it is possible to apply
considerable braking force while barely changing the total force. For example, let the braking
force (fb) be 20% of the cornering force (fc). The total force is then
You then argue that 20% of fc (which we established is not that high) is a "significant braking
force". For some reason, I disagree. :-)
Quoted message said:ftot = sqrt(fc^2 + fb^2)
= fc*sqrt(1+(fb/fc)^2)
~ fc*(1+(fb/fc^2)/2) for fb << fc
= fc*(1+(2/10)^2/2) = 1.01*fc
Here is a much easier way of solving (and understanding) this, with the bonus of getting a more
accurate answer:
ftot = sqrt(fc^2 + fb^2)
Assume that fc = 1unit. Assume that fb = .2*fc = .2units
ftot = sqrt(1^2 + .2^2) = 1.02
Now, if you assume that the rider is cornering somewhat conservatively, because they fear that there
might be a patch of sand or tar on the corner somewhere, then you can assume that fc is not too
high. What happens if they are also going down a steep hill, and want to brake to maintain their
speed? If fc=fb, then this changes to:
ftot = sqrt(1^2 + 1^2) = 1.41
You have just lost 40% of the margin of safety you planned to have in the corner.
I do agree that the best way to learn this is not through math, but through practice. I found that a
great way to learn how to deal with low traction conditions in corners is to go out and ride just
after a fresh snowfall (before the plows come by). Very low traction, loads of fun, and if you
happen to fall then the snow offers some padding...
Chris
--
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