First of all sorry for my bad english.
My problem is that im making car game where car moves only forward and
backward and im trying to understand how to calculate traction force
when tires are slipping. So cars tires are always slipping expect if
there is no accelration, right? I am using slip ratio curve found here
and if i understood right it is always same amount of force delivered at
same slip ratio assuming that load to tires is constant. If it is like
this I need slip ratio to calculate traction force. From the same
website I have used this formula to calculate slip ratio: slip
ratio=(w*r-v)/abs(v), where w is tires angular velocity, r is tires
radius and v is cars velocity. But the thing is how i calculate angular
velocity if there is slip? I havent tried any solution yet cause im
quite confused now. But I have thought simplifying this by assuming that
cars slip ratio is 0 until traction force exceeds static friction and
then calculating angular velocity using this equation a=M/J where a is
angular accelration, M=moment on wheel and J=moment of inertia. From
that I could calculate angular velocity like I can calculate velocity
from accelration but I am not very familiar about moment of inertia so
can I use it in situations like this? Im really confused about this now
and I hope you understood my poor explanations so please help me and
correct if i was wrong about something. I have tried to find solution to
this from many websites but maybe im just dumb.. or lack skills of
finding right things.
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I think ‘slip ratio’ approach is flawed. Slip ratio is derived quantity
and highly unrelated to problem of wheel dynamic.
There are two main regime of wheel rolling. First is static friction
without slipping of tire over road. Nonzero ‘slip ratio’ in this regime
is due to tire deformation rather than actually slipping. So you can
safely assume slip = 0. In this regime friction force is arbitrary with
maximal magnitude proportional to normal force (all applied to point of
contact of tire with road).
Second regime is kinetic friction where tire actually slips. This regime
is energy (and tire material) consuming as opposed to first. In this
regime amount of slip determined by wheel dynamic and friction force
proportional to normal force and directed opposite to surface slip
velocity. Coefficient of kinetic friction is smaller than of static
friction and depend on speed of surface slipping.
Wheel angular velocity in regime of real slipping determined mostly by
engine. You have wheel rotating with angular velocity w under of load of
kinetic friction angular momentum M and engine (with transmission) gives
you angular acceleration.
Thanks for your response!
I got that static friction part. About kinetic friction you said that
“Coefficient of kinetic friction is smaller than of static friction and
depend on speed of surface slipping.” In school we always calculated
exercises where static and kinetic friction coefficients were constants
but now when you say it would make sense.
Tires spin easier when there is much slip. But for locked tires or if
pushed sideways kinetic friction coefficient is still consant, am I
And about that angular accelration didn’t I still need to use moment of
inertia to calculate angular accelration?
School exercises usually assume that coefficient of kinetic friction
equals to the coefficient of static friction and doesn’t depend on
speed. Sideways slipping coefficient is not constant either. Friction
force doesn’t distinguish between lengthways and sideways, it knowns
only about relative surface speed. But I think for game you can neglect
dependence on speed but not difference of magnitude of coefficients.
As for moment of inertia… well, if you want physically correct
description you must calculate moments of inertia. But not only moment
of wheel, you must account moments, friction forces, forces of
interaction for every gear in transmission and engine. It’s possible to
replace all this machinery with effective moment of inertia Jeff but
exact value depends on current transmission gear and clutch.
Ok thanks again. I think am making progress now.