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Let me try something here:
Drag force = drag coef x speed^2
So horsepower used against drag = drag coef x speed^3
If both cars are using same power at top speed then
Drag coef zl1 x top speed zl1 ^ 3 = drag coef zle x top speed zle ^3
If zl1 top speed is 202mph and zle is 178 then drag coef ratio is (178/202)^3= 0.684
Meaning the zl1 has 68.4% of the drag of the zle or the zl1 gets to use 31.6% more torque accelerating rather than fighting aero drag.
If at 101mph, the zl1 is at half top speed, meaning it's using 1/8th (drag power is cubic) of its power to fight drag which 650hp/8=81hp. This speed would require the zle to use 81hp/.684=118hp. So at 101mph, it's as if the zl1 has 118hp-81hp=37hp more than the zle.
Super oversimplified, but maybe a good ballpark estimate?
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