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This has got me thinking...check this out:
Let's say we're cruising along in 1st gear (in our 5spd) somewhere in the middle of the max torque range, throttle to the floor. Theoretically we should be commanding the maximum allowed torque, which per the link is 230 N-m crank torque. 1st gear ratio being 3.38:1 and final drive ratio 4.05:1. Thus with a perfectly efficient drivetrain, we should see:
230*3.38*4.05 = 3148 N-m axle torque
Of course the drivetrain isn't perfectly efficient. Let's say there's 15% loss in there (should be conservative, I've read claims of 12% from Saab). So now we've got:
3148*0.85 = 2676 N-m axle torque
Given stock tires 205/55-16, this means a wheel/tire radius of:
16*.0254/2 + .205*.55 = 0.316 m
Therefore we've got a potential linear motive force of:
2676/0.316 = 8468 N at the front wheels
Let's assume quite optimistically that somehow 60% of the car's weight stays planted on the front wheels. Furthermore we'll assume two loaded front seats to bring the mass up to a nice round 1500 kg. That means weight on the front tires is:
1500*9.81*0.60 = 8829 N
Thus to deliver my motive force of 8468 N, I need an effective tire friction coefficient of:
8468/8829 = 0.96
Assuming I don't have racing slicks on the Saab, no way. My wheels should be spinning if my crankshaft is putting out 230 N-m
In reality, I'd estimate that we're losing something like 1000 N of weight to the back wheels during this kind of acceleration, meaning the weight on the front is more realistically something like 7500 N. Even if I assumed a big time drivetrain loss of say 25% (knocking down motive force to about 7500 N), I'd still need a friction coefficient on the order of 1.0 to maintain traction.
Seeing as my front wheels don't slip under hard acceleration in 1st gear (assuming I'm not trying to turn :-), what gives?
Cheers,
'Roo
posted by 128.42.176...
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