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Based on the calculations I have done, upside-down racing seems to be possible. But in my mind, and certainly others too, it seems outlandish and impossible. Here is what I have come up with, please check it over and see if what I am assuming is correct. And answer my question with calculations of your own as well!

While racing a car you'll experience three vertical forces (the normal force of the track, gravitational force, and negative lift) summarized as follows:

$F_N$ - $mg$ - $F_L$ = 0

Let's assume some fictional values for a car going around a circular track:

$F_L$ = $m\Bigl({v^2\over \mu_sR} - g\Bigr)$ where $m = 600kg$, $v = 29m/s$, $R = 100m$, and $\mu_s = 0.75$

This will give us an $F_L$ value of $663.7N$. And since we now that $F_L$ is propotional to $v^2$ we can calculate the negative lift for, let's say, $90 m/s$.

Therefore, ${F_L90\over F_L}$ = ${(90m/s)^2\over (29m/s)^2}$

This yields a negative lift of approximately $6600N$.

Finally, since we know $F_g = mg = (600kg)(9.8m/s^2)$ we get $F_g = 5880N$.

With these values, we find out that upside-down racing is possible. The car would stay driving on the "ceiling" theoretically if, and only if, it maintains a speed of $90m/s$ $(= 324km/h = 201mi/h)$. Or possibly a bit of a lower speed since $6600N$ is quite a bit over $5880N$.

If you see anything wrong with my process or have anything additional to add, please answer the question for your own!

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Not really an answer, but there's a few things wrong. First, when you're upside-down, the normal force is in the same direction as gravity, so you should have $F_N + mg - F_L = 0$. But analyzing the normal force isn't necessary here. Your expression for normal force is incorrect. If the cars are not accelerating vertically, the normal force will exactly cancel the amount of lift that exceeds the pull of gravity. This must be so by $\Sigma F = ma$. So all you need to analyze is how much lift you can get. This will be dependent on how the cars are designed. Currently, racecars are designed to not have lift because they would fly off the track. You'd need a new design.

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  • $\begingroup$ Thanks for the response! One quick question I have though; when you say race cars are designed to not have lift, are you accounting for the fact that they would have negative lift too? In my calculations, the $F_L$ represents the negative lift a car produces, somewhat like downward thrust. $\endgroup$ – ThatOneNerdyBoy Oct 13 '19 at 17:29
  • $\begingroup$ @ThatOneNerdyBoy Oh wait yeah I was mixed up on that. Cars normally produce a downwards force which is like a lift into the track above $\endgroup$ – HiddenBabel Oct 13 '19 at 17:43
  • $\begingroup$ So then my calculations make sense, or is there still somewhere I am going wrong? $\endgroup$ – ThatOneNerdyBoy Oct 13 '19 at 18:25
  • $\begingroup$ If your track is flat, all you need to do is consider lift vs. $mg$ $\endgroup$ – HiddenBabel Oct 13 '19 at 19:22

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