I've been tossing this problem around in my head for the last couple of days -
I've got the concept, but I'm having trouble putting an equation on it.
It's pissing me off - I know this stuff, dammit!
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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