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Re: Top speed?
Alright, I'm doing the computations for the fastest theoretical lap.
I'm saying the straight is 27ft = 8.22m long.
A friction-limited single lap of overdrive includes:
A) a half-circle of radius r and length pi*r
B) a straight with [(8.22- 2r)/2]m of acceleration followed by (8.22 - 2r)/2]m of deceleration for a total of (8.22-2r) distance
A) another half-circle of radius r and length pi*r
B) another straight
Total distance: 2*pi*r + 2(8.22-2r) = 2*pi*r + 16.44 - 4r
Time tc for the circles will be determined by the centripetal equations.
tc = pi*r / vc
Fc = Ff = uFn = umg = -9.8mu
-9.8mu = -mvcvc/r
9.8u = vcvc/r
9.8ur = vcvc
vc = sqrt(9.8ur)
tc = pi*r / sqrt(9.8ur)
Time ts for the straights will be determined by straight-line acceleration and deceleration, which means we need the kinematic equations. Starting and ending speed for an ideal lap will be vc (our exit speed from the corners). Our acceleration will be friction-limited. Our distance is half the straight, which is (8.22-2r)/2 or 4.11-r
F = ma
Ff = ufn = 9.8mu
9.8mu = ma
9.8u = a // this is our acceleration due to friction
d = vc*t + 0.5*a*ts*ts
4.11-r = sqrt(9.8ur)*0 + 0.5*(9.8u)*ts*ts
4.11-r = 4.8*u*ts*ts
sqrt( [4.11-r]/[4.8u] ) = ts
So now that we know our time for the half-circles (tc) and straights (2ts), we can add it all together. There are 2 half-circles, 2 acceleration phases and 2 deceleration phases in a single friction-limited lap.
ttotal = 2tc + 4ts
ttotal = 2*pi*r / sqrt(9.8ur) + 4*sqrt( [4.11-r]/[4.8u] )
If you like calculus, sub in a known friction coefficient for u and then solve for the optimum turning radius. Since I have long since forgotten how to do a complicated derivative like I'd need to do here, I'm just going to sub in r = 13ft = 3.96m and u = 1.3.
ttotal = 4.12 sec (laps per minute: 14)
Cornering speed = vc = sqrt(9.8ur) = 7.1m/s = 23fps
Edit: After graphing the result for u=1.3, I see that a 3.96m radius turn is very close to the WORST you can do for a friction-limited lap. You could be doing 2 second laps if you had a 50cm turn radius. However, the power output required to accelerate a 50kg robot at 1.3gs and slow it back down twice per lap is ludicrously high. Specifically, if you accelerated a 50kg robot at 1.3g for the 4.11m and decelerated just as fast for 4.11m, then you're looking at this much work:
W = F*d
W = 1.3*9.8*50*8.22 = 5236J of energy expended
Since we know our time for the 8.22m straights is 2*ts = 2*sqrt( [4.11]/[4.8u] ) = 1.62s for the whole straight. This means your power output will be 3.2kW, which at 12 volts means you're drawing 268amps on the straights, assuming a perfectly efficient drivetrain and no resistance.
Last edited by Bongle : 07-01-2008 at 14:03.
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