> > Richard, I don't get it. Could you send me a replay in which you show
> > me how to do this? When I turn the wheel in the direction of the
> > slide, I simply turn around.
> > Thanks,
> > Andre
> No, really, it works. I've shown with some physics arguments on this
> newsgroup why it should work a while ago (basically it can be shown that
> the torque of the outside front wheel will produce a smaller torque at
> high yaw angles when it is turned into the turn further, thus
> stabilising the spin). I'll check deja to respost it again.
> Just try it out, though, the next time you will enter what is otherwise
> an irrecoverable spin. Opposite lock works to a certain point, once that
> becomes innefective the turning of the wheel into the turn helps. It's
> really amazing and counterintuitive, but can save you lots of time in
> races.
> -Gregor
Reposting from deja.com why opposite lock doesn't work at high car yaw
angles:
The second effect to consider is more complicated and may actually
contribute more. Let us assume that most of the cornering force comes
from the outside wheels (this is
true for the cars in GPL which have a relatively high ceter of
gravity). In order to stop a spin we need to produce a torque on the car
that opposes the direction of the
spin. Let us also assume that the lateral force on the tyre does not
depend too much upon its orientation, which is also true at high slip
(yaw) angles for the tyres of the
time. These are all plausible simplifications and may be argued, but it
is easiest to show the idea with them. Anyone who has ever worked in
science will know this kind of
approach.
Let us assume a right hand turn. With no lock the situation looking
from above on the front wheels looks like this:
Picture 1:
_ _
| | F | |
| |----> | |
|_| |_| ^
\ / \
\ / \
\ / \
\ / \ velocity vector
O c.g.
Let us consider this a balanced situation, so that the torques produced
by all wheels (including rears) are 0 and the car maintains a slip
(towards the up-left direction) if no
steering corrections are applied. The force on the outside wheel is for
the sake of simplicity considered negligible. Consider now the case when
the driver tries to recover by
dialing some opposite lock:
Picture 2 (okay, so the lines are not perpendicular, imagine them to
be)
:
^
F /
/\ /
\ \/
\ \
\/
\
\
\
O c.g.
As the force becomes closer to perpendicular to the vector between the
center of gravity (c.g.) while the magnitude of the force stays roughly
the same, the torque actually
increases so the net torque of all tyres is not 0 anymore but actually
contributes to the direction of the spin. Only if the magnitude of the
force diminishes enough to cancel
this effect (the tyre gets back into the normal slip angle regime) does
the opposite lock help, otherwise it can actually cause harm and you
would be better of dialing in some
further lock INTO the spin.
-Gregor
