Aerobatic pilot here.
I can almost see how to do this, although it’d be a WAAY tricky.
As folks have said, once the cars are up to speed there’s enough aerodynamic force available to more than offset gravity and make the cars stick to the underside of an inverted track. And making the engine & transmission operate upside down is just more engineering, bog-standard stuff from the aerospace world.
And as to crashes, the rest of the pack has no need to slow down. The moment anything interferes with the crashing cars’ aerodynamics, gravity takes over and they fall away from the track, leaving it completely free of obstacles and debris. Chain reaction collisions would be limited to one or two cars in immediate proximity. In this regard (alone) upside-down racing would be safer than rightside up racing.
The problem is how to get started. Obviously the downforce is related to speed, so there’s some minimum speed below which gravity is stronger and the cars fall from the track.
So we’ve got to get up to speed rightside up and then transfer somehow to upside down. And that’s the rub.
I’m going to put “downforce” in quotes hereafter to emphasize the fact we’re dealing with two different coordinate systems, one aligned with Earth gravity & the other aligned with the car’s present orientation. One of the hard things for non-flyers to learn to to keep straight is the difference between the two systems. While flying, particularly aerobatics, there are always two “downs”: little-d down from your head to your feet, and big-D Down from the sky to the Earth’s center. And they’re usually not aligned.
The critical point to always bear in mind is that gravity is always pulling towards the center of the Earth, while the “downforce” of the car is always pulling from the driver’s head toward the wheels. On a normal track those forces are in the same direction & hence additive. On an inverted track they’re in opposite directions and hence subtractive.
But what about during the transition? There are two ways to get from rightside up to inverted: a 1/2 loop & a 1/2 roll. The roll is easy in an airplane, but hard in a car. At the 90 degree point (called “knife edge” in flying) where the track would be “banked” to 90 degrees, gravity is still pulling down, but the “downforce” is straight sideways & the cars will instantly slide to the bottom of the track, net of their tires’ ability to generate cornering forces. But they need 1G of cornering force to stick to the track, and that’s gonna be real hard to pull off.
A 1/2 loop would be easier. Somebody mentioned the old “motorcycles inside a sphere” act. That’s an example of centrifugal (I know, I know, but it makes for an easier explanation) force. Those bikes & riders never feel a net negative G; they’re always being pulled to the outside of the sphere since the cetrifugal force generated by their speed is greater than gravity.
Now in those acts the sphere is very small, say 20 feet diameter, and the bikes aren’t going all that fast, say 20 mph.
Doing the same thing at 150 mph would be problematic. That’s a speed comparable to the small airplanes that do competitive areobatics and that you see at airshows. Notice the loops they do are several hundred feet high. The turn radius is driven by the speed times the centrifugal forces the pilot/aircraft can stand. The cars would actually have a lower maximum G force they could stand, implying an even larger radius 1/2 loop would be required.
So we could have an ordinary oval track the cars use to gain speed, and then once at 150-200 mph they’d enter a chute that is a giant on-ramp in the shape of a 1/2 loop . The cars zoom up this half loop chute and are fully inverted at say 800 feet above the ground. Now what?
Getting them back down to ground level so the spectators can watch and so the falls from the track aren’t guaranteed fatal is gonna be tough.
Certainly the 1/2 loop could be instead a 5/8ths loop and now they’re headed back downhill at 45 degrees on the underside of the track. Their “downforce” is offset from gravity by 45 degrees, so their “grip” on the track is pretty tenuous right now. Only about 70% (1/SQRT(2)) of their “downforce” is working directly against gravity. So the net force holding them to the track is (.70 * “downforce” - weight).
Now we get back to near ground level. The track (& drivers) needs to pull out of their dive. But now centrifugal force is working to pull them away from the track. So the turn radius in the vertical dimension must be pretty huge to avoid reducing their already tenuous grip on the track too much.
My gut feel is it can’t quite be done, at leat not with cars which are driven by their wheels. It takes a horrendous amount of energy to push a car through the air at 150+ mph. That energy is transferred to the track via the traction of the wheels. That traction is proportional to the net “downforce.” If the net is small, say a couple of hundred pounds, the cars will not be able to transfer enough power to the track to maintain speed.
Sorta like trying to accelerate hard on ice. The lack of traction precludes putting very much power through the wheels into the pavement.
In all, I’ve written waay to many words to come to a half-baked almost -comnclusion, but what the heck, I’m gonna hit submit anyhow; electrons are cheap
F1 cars accelerate very smartly from a dead stop with no aerodynamic downforce and the one G of gravity alone to provide tractive force. Power transmission will be the least of my worries when I’m racing up on the lid…