Top speed?

[Bongle]

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.
t_c = pi*r / v_c
F_c = Ff = uFn = umg = -9.8mu
-9.8mu = -mv_cv_c/r
9.8u = v_cv_c/r
9.8ur = v_cv_c
v_c = sqrt(9.8ur)
t_c = 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 = v_c*t + 0.5*a*t_s*t_s
4.11-r = sqrt(9.8ur)*0 + 0.5*(9.8u)*t_s*t_s
4.11-r = 4.8*u*t_s*t_s
sqrt( [4.11-r]/[4.8u] ) = t_s

So now that we know our time for the half-circles (t_c) and straights (2t_s), we can add it all together. There are 2 half-circles, 2 acceleration phases and 2 deceleration phases in a single friction-limited lap.
t_total = 2t_c + 4t_s
t_total = 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.

t_total = 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*t_s = 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.

[WaterFreak]

Originally Posted by Bongle

a perfectly efficient drivetrain and no resistance.

this is where is all goes awry, since that does not exist

[Bongle]

Originally Posted by WaterFreak

this is where is all goes awry, since that does not exist

Along with the need to draw more amps than our breakers provide, the ability to accelerate hard enough that your tires rather than your motors are the limiting factor, the ability to turn controllably without losing velocity, the assumption that nobody would get in your way while you lapped, etc. It was all just a thinking exercise to figure out the absolute fastest laptime possible. The next step would be to constrain it so that your power output is less than 120amps and see what happens then.