I did some calculations that may help in estimating the battery pack size.
Given that we have the 0-60 time (1.9 seconds), and the total wheel torque (10,000 Nm) provided by Tesla, and we have the tire sizes provided in the pictures above, we should be able to make some reasonable assumptions and calculate the weight of the vehicle.
Given information:
Tires are Michelin Pilot Sport Cup 2, sizes are 265/35R20 front (diameter 27.3 in), 325/35R21 rear (diameter 28.7 in).
Total Torque at the wheel hubs: 10,000 Nm
0-60 MPH time: 1.9 sec
Tire diameters: Front 0.69342 m, Rear 0.72898 m
Let's assume power distribution is 33% front axle, 67% rear axle during max acceleration.
Let's also assume that the acceleration is linear from 0-60, which is probably close to correct given that the torque is held constant at maximum by the inverter.
Let's assume that the system is powerful enough to maintain maximum torque all the way to 60 MPH. (It may not. If it doesn't, then the vehicle will have to be lighter than this calculation will show, thus this calculation is an upper bound on the vehicle weight).
Let's calculate acceleration in m/sec^2:
Vf = at + V0
Vf = 60 MPH = 26.8224 m/sec
V0 = 0 MPH = 0 m/sec
t = 1.9 sec
a = 14.117 m/sec^2 (1.44 G !)
Let's calculate the force on the vehicle given the torque, tire diameters and power distribution:
F = Torque / distance
F = Torque / (% Power Dist front * (Tire diameter front / 2) + % Power Dist rear * (Tire diameter rear / 2) )
F = (10000 Nm) / (0.3333 * (0.69342 / 2) + 0.6667 * (0.72898 / 2) )
F = 27,889.07 N
Compute mass of the vehicle:
F = ma
m = (F/a)
m = (27889.07 / 14.117)
m = 1975.56 kg
m = 4355.4 lbs
I believe that's enough mass to hold a 200 kWh battery and a light vehicle structure.