I'll need to poll the audience or call a friend... :-)
Kent Rafferty
> Anyway, here it is:
>
> You have 2 cars at a ProSolo.
>
> Car A weighs m, and puts out f ft/lbs in a perfectly flat torque curve.
> Car B weighs 2m, and puts out 2f ft/lbs in a perfectly flat torque curve.
>
> Both cars have identical gearing/tire diameter such that Car B is putting
out
> twice the thrust as Car A. For purposes of this problem, we assume that
the
> thrust is the same as the engine torque (yes, I'm mixing units - I don't
care.
> Car A has a thrust of f, Car B has a thrust of 2f)
>
> Given that acceleration = force/mass, and given that Car B has 2f and 2m,
we
> determine that f/m = 2f/2m So in other words, both cars accelerate at the
same
> rate.
>
> At the end of the straight, both cars are travelling at an identical
velocity V.
> Immediately following the straight is a turn with radius R.
>
> Now both cars may be travelling at the same speed, but Car A has a
momentum of
> mV, whereas Car B has a momentum of 2mV. In order for Car B to maintain
the same
> speed through the turn as Car A, it must either be capable of maintaining
a
> higher lateral acceleration. (Or is the lateral acceleration value
uneffected,
> but Car B needs more lateral grip to maintain that acceleration?)
>
> How much higher?
>
> If both cars are capable of maintaining 1G laterally, and both cars have
the
> same amount of grip, then Car B must slow down. By how much?
>
> Anybody have the solution?
>
> DG
>
>
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