# Indy cars and anti-gravity **URL:** \ **Category:** Factual Questions\ **Created:** [April 10, 2004, 1:58am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263 "2004-04-10T01:58:32Z")\ **Posts on this page:** 16\ **Page:** 2
**Author:** ![Princhester](https://avatars.discourse-cdn.com/v4/letter/p/3e96dc/32.png) [@Princhester](https://boards.straightdope.com/u/Princhester)\ **Post date:** [January 16, 2005, 11:01am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/21 "2005-01-16T11:01:24Z")
Someone else started another thread on this topic, and I’ve directed them to hear. On re-reading this thread (after finding it) I had another thought for what I think would be a very practical solution to the problem of getting the cars into the inverted position, which avoids both the difficulties of the straight roll and the half loop. What got me thinking of the idea was a matchbox car track I had when I was a kid that had a corkscrew rolling section. Because a corkscrew roll has an element of centrifugal force (yes, I know, I know) “down” toward the track at all times, that should resolve the difficulties of getting sufficient force to hold the car to the track as you go through the 100 degree point. Also, one of the reasons the matchbox track worked so well was because the track had a lip and the cars were guided by that. My suggestion (if sideways forces are a problem) is that our upsidedown racers have guide wheels on one side (the outer side for the corkscrew) and there is a rail for the guide wheels to run on, on the outer side of the track. So that if the cars start to slide off the corkscrew at some point, they just get guided along the rail till they are properly upside down and stable. ---
**Author:** ![Princhester](https://avatars.discourse-cdn.com/v4/letter/p/3e96dc/32.png) [@Princhester](https://boards.straightdope.com/u/Princhester)\ **Post date:** [January 16, 2005, 11:03am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/22 "2005-01-16T11:03:01Z")
That’d be “here”. ---
**Author:** ![Ale](https://avatars.discourse-cdn.com/v4/letter/a/f19dbf/32.png) [@Ale](https://boards.straightdope.com/u/Ale)\ **Post date:** [January 16, 2005, 3:39pm UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/23 "2005-01-16T15:39:57Z")
Well, it could be tested comparatively cheaper with an RC car. One of those little buggers running up the walls and on the ceiling would be quite a show. ---
**Author:** ![Thin\_Ice](https://avatars.discourse-cdn.com/v4/letter/t/aeb1de/32.png) [@Thin\_Ice](https://boards.straightdope.com/u/Thin_Ice)\ **Post date:** [January 16, 2005, 4:05pm UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/24 "2005-01-16T16:05:27Z")
You just need to build a NASCAR oval on its side so one straight is on the ground and the other is in the air with the two 180 degree semicircles connecting the two straights. Another thought: magnetic rubber tires and metallic road surface. ---
**Author:** ![Rick](https://avatars.discourse-cdn.com/v4/letter/r/9f8e36/32.png) [@Rick](https://boards.straightdope.com/u/Rick)\ **Post date:** [January 16, 2005, 5:13pm UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/25 "2005-01-16T17:13:54Z")
Don’t any of you guys remember the Chaparral 2J AKA the [vacuum cleaner?](http://www.mulsannescorner.com/chaparral2j.html) More downforce that you can shake a stick at. I miss the old Can-Am series. ☹ ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [January 17, 2005, 2:25am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/26 "2005-01-17T02:25:31Z")
Quoth **LSLGuy** : > [@](#): > > 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. Don’t apologize for using centrifugal force. You used it correctly, and I agree that in this context, it’s probably the simplest explanation. So there’s absolutely nothing wrong with what you have there. One might as well apologize for treating gravity as a force. This goes for **Princhester** , too. ---
**Author:** ![GawnFishin](https://avatars.discourse-cdn.com/v4/letter/g/e9bcb4/32.png) [@GawnFishin](https://boards.straightdope.com/u/GawnFishin)\ **Post date:** [January 17, 2005, 4:05am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/27 "2005-01-17T04:05:54Z")
Of course if **Princhester** would tell us whether there is motor racing in Australia or not, then we’d know that cars can race upside down. 😃 Someone had to say it. > [@Rick](#): > > Don’t any of you guys remember the Chaparral 2J AKA the [vacuum cleaner?](http://www.mulsannescorner.com/chaparral2j.html) More downforce that you can shake a stick at. > I miss the old Can-Am series. ☹ I believe the [Brabham BT46B](http://www.rasf1m.net/teams.html#Q6) was based on the Chaparral. > [@](#): > > the car actually squatted down on to the track if the throttle was blipped while it was stationary. ---
**Author:** ![Princhester](https://avatars.discourse-cdn.com/v4/letter/p/3e96dc/32.png) [@Princhester](https://boards.straightdope.com/u/Princhester)\ **Post date:** [January 17, 2005, 5:39am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/28 "2005-01-17T05:39:41Z")
> [@GawnFishin'](#): > > Of course if **Princhester** would tell us whether there is motor racing in Australia or not, then we’d know that cars can race upside down. Do you mean upside Down or upside down? 😉 ---
**Author:** ![No\_Disguise](https://avatars.discourse-cdn.com/v4/letter/n/9f8e36/32.png) [@No\_Disguise](https://boards.straightdope.com/u/No_Disguise)\ **Post date:** [January 17, 2005, 8:37am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/29 "2005-01-17T08:37:19Z")
This is fun. Maybe you could put wings, landing gear, and small jet engine on the cars. They’d be able to take off like normal airplanes, get up to speed, ‘land’ on the upside-down track, and then switch to wheel drive only. They could make it so the turbine engine only delivers enough thrust for let’s say 2/3 the acceleration forces and top speed they can reach employing the conventional motor-driven wheels. This way, they could leave the track whenever they wanted – like for tight corners which they could navigate like airplanes – but those who could carry the most speed through the corners would be able to get back to the track surface most quickly and get under wheel-driven power again. The cars that spend the most time on the track would be the fastest. They could even use those wings to glide down for pit stops, and at the end of a race as well. This sounds completely feasible even with today’s technology to me. Can I patent this? ---
**Author:** ![spinky](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@spinky](https://boards.straightdope.com/u/spinky)\ **Post date:** [January 17, 2005, 10:15am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/30 "2005-01-17T10:15:20Z")
> [@LSLGuy](#): > > The tough part happens just after you pass knife edge and are at say 100 degrees of roll. At that point your “downforce” vector is only 10 degrees above the horizontal and only sin(10 deg) = 17% of your “downforce” is directed against gravity. I think your logic is flawed. Do the same calculations for 0.01 degree past the knife edge, and a mere 0.02% of your downforce is directed against gravity. Does that mean you fall off the wall? In fact, your rationale leads to the conclusion that you’d fall off the wall at 90 degrees too, since exactly 0% of the downforce is directed against gravity in that case. ---
**Author:** ![spinky](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@spinky](https://boards.straightdope.com/u/spinky)\ **Post date:** [January 17, 2005, 10:17am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/31 "2005-01-17T10:17:17Z")
Oh, one big practical problem with the International Upside Down Racing Association would be that racers _do_ occasionally have to slow down (e.g. to avoid wrecks)… ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [January 17, 2005, 11:07am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/32 "2005-01-17T11:07:47Z")
> [@ntucker](#): > > Oh, one big practical problem with the International Upside Down Racing Association would be that racers _do_ occasionally have to slow down (e.g. to avoid wrecks)… Yeah, but in IUDRA races, crashes immediately drop off the track, so no problem there. IUDRA drivers, by a rapid process of natural selection, would soon unlearn the instinct to stand on the brakes in an emergency… ---
**Author:** ![Princhester](https://avatars.discourse-cdn.com/v4/letter/p/3e96dc/32.png) [@Princhester](https://boards.straightdope.com/u/Princhester)\ **Post date:** [January 17, 2005, 1:18pm UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/33 "2005-01-17T13:18:58Z")
> [@ntucker](#): > > I think your logic is flawed. Do the same calculations for 0.01 degree past the knife edge, and a mere 0.02% of your downforce is directed against gravity. Does that mean you fall off the wall? In fact, your rationale leads to the conclusion that you’d fall off the wall at 90 degrees too, since exactly 0% of the downforce is directed against gravity in that case. You miss the point. Up to 90 degrees the car has to _slide_ laterally to fall down. The track (and hence the tyres) are between you and where gravity is trying to take you. Therefore,as long as your tyres are generating sufficient cornering (ie lateral) resistance that is going to keep you on the track. But once you are past 90 degrees, there ain’t nothing below you, and the tyres no longer help. There is no component of cornering resistance opposing gravity. The car can fall straight down, without the tyres so much as squealing. You now have to rely entirely on “downforce” (which has now become upforce) but that isn’t going to start having a large up component till you roll around well past the 90 degree point. I don’t know that the problem LSLGuy raises is fatal, but it’s correct in principle. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [January 18, 2005, 1:24am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/34 "2005-01-18T01:24:39Z")
Actually, I don’t think that **LSLGuy** ’s problem is correct, even in principle. As long as you have a normal force between the two surfaces, you’ll also have a frictional force parallel to the surfaces. Let’s say that the car is on the left, and the wall is on the right, and draw a free body diagram for the car. You’ll have four forces (ignoring air resistance and thrust from the wheels): Gravity, aerodynamic force, normal force, and friction. Gravity, of course, acts straight down. Aerodynamic force will act “down” in the car’s reference frame, which means almost horizontal to the right, and a little bit up. Normal force will act in the opposite direction (to the left and a little bit down), and will be almost as large as the aerodynamic force (as the track approaches vertical, the normal force will approach the value of the aerodynamic force). Finally, friction will be acting upward along the surface of the track, or upward and a little to the left, and its value can be up to mu times the normal force (mu is called the coefficient of friction, and will typically have a value close to 1, or maybe a little higher for racecar tires. Frictional force can be smaller, if that’s all that’s needed to balance the forces). Let’s look at the case where the track is just barely past vertical, 90.00…01 degrees. Gravity is, of course, equal to the weight of the car, and we’ve already established that the aerodynamic force could be, say, three times the weight of the car. In this case, the normal force would then also be three times the weight of the car, and assuming mu = 1 for simplicity, we find that the frictional force can be as much as three times gravity, easily enough to keep the car from sliding off. If you’d like to perform an experiment, an analogous situation can be had with a magnet stuck to a piece of metal (the magnetic force here substitutes for the aerodynamic force; all other forces are equivalent). If you stick a magnet on the side of your refridgerator, the magnet stays on even if you tilt the fridge forward. ---
**Author:** ![spinky](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@spinky](https://boards.straightdope.com/u/spinky)\ **Post date:** [January 19, 2005, 11:03am UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/35 "2005-01-19T11:03:32Z")
> [@Princhester](#): > > You miss the point. Up to 90 degrees the car has to _slide_ laterally to fall down. > … > once you are past 90 degrees, there ain’t nothing below you, and the tyres no longer help. The car still has to slide down the wall past 90 degrees, too. Imagine that just after you pass 90 degrees, the wall completely disappears. Now you’ve got gravity pulling you down, but three times as much force directing you sideways and slightly upwards. Which direction do you think the car will travel? Not straight down. Change your frame of reference to the driver’s POV. You’ve got 3x gravity pushing straight down, and 1x gravity pushing off to one side and slightly up. Do you lift off the road? Not by a longshot. In order for the car to leave the track, you need to demonstrate a net force in the direction normal to the track (“up” according to the driver’s frame of reference). You’ve got three times the weight of the car in one direction (towards the track), and you have much less than 1G in the other (gravity, pulling straight down, is directed far off the normal when the car is at 100 degrees of bank). ---
**Author:** ![Servo](https://avatars.discourse-cdn.com/v4/letter/s/49beb7/32.png) [@Servo](https://boards.straightdope.com/u/Servo)\ **Post date:** [January 19, 2005, 2:23pm UTC](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263/36 "2005-01-19T14:23:44Z")
How about a reverse Mag-Lev? You could eliminate the transition, use cars that are designed to only run one way up, and have a nice little safety measure as a bonus. There’s plenty of metal in the cars, just put some big electromagnets under the track surface. The cars could transmit speed information [or even better force values from sensors in the suspension] to a Track Control System. Once everyone’s generating enough track-side force the TCS turns off the magnets and the race is on. Accident? Engine failure? Turn the magnets back on and everyone can slow down. Or not, just let them fall off the track [I guess there has to be some element of excitment]. If we can’t get the IUDRA off the ground [Ha] in real life, at the very least I think we’ve got a great idea for one of those movies in the “futuristic sport” genre. [Previous page](https://boards.straightdope.com/t/indy-cars-and-anti-gravity/239263.md?page=1)