Does a tyre produce downforce?

[strad]

Like all pictures made by aerodynamicists, it uses a colour palette designed by old hippies.

I think I feel insulted.
HAHAHAHAHA
.
@ hardingfv32: ref the second paper, Is that the paper with the wheel with 15 holes in it?
I'm not sure I'm finding that exact paper

[xpensive]

@Ciro,
I have no problems at all with your approach to this issue, au contaire, I'm mighty impressed.

Knowing me, it takes a lttle effort to do that.

The only problem is with the resulting numbers, 6.3 kN indicates an average pressure differential of almost 30 kPa (0.3 Bar),
diameter 660 and width 335, which is an insane number at any speed, let alone at 20 m/s (72 km/h or 45 mph).

The A380 aircraft has a wingload of 6.6 kPa, but that's at 800+ km/h.

[Twaddle]

@Ciro

Bearing in mind I'm much less of an aerodynamicist than you, I do have a couple of thoughts on why your results don't look right.

1. There is air under the tyre.
There is air under the contact patch of the tyre, but not 100% of it. I can't tell you exactly how much, but since both surfaces aren't perfectly smooth there's space for some air under there.

Air pressure exerts such large forces on things that if there was no air under things that we think of as being in contact with the floor, you wouldn't be able to lift anything. I'm sure you know this, but because air is so good at getting everywhere we have little day to day experience of large differences in air pressure and it's easy to forget the implications of it. 1 atmosphere is roughly equivalent to the pressure resulting from gravity acting on a 4m depth of concrete.

2. Tyres are quite narrow.
You're going to get greatly exaggerated pressure differentials due to the limitations of 2D analysis. The air isn't actually forced into such large velocity changes overthe whole tyre width since as you get towards the edges it can much more easily move around the sides. Again I can't tell you how much to allow for this, but for something as narrow as a tyre, I expect a significant drop in efficiency in producing aerodynamic forces.


Basically I can't help you calculate it more accurately, but I expect that the tyre will produce some lift. Just much less than you're calculating in your second case at the moment.

[hardingfv32]

No, the wrong item is coming up. I guess that is why I save PDF's.

PM me with Email and I will forward. The best paper I have ever found on this narrow subject.

Brian

[marekk]

With what ever little knowledge that I have, I have the following hypothesis. Kindly correct me if I am wrong.

When a cylinder is rotated in a clockwise direction with a streamline flow over it, it produces lift ( any aerodynamic text book has it's derivation) . So, as a converse, when the cylinder rotates anti clockwise ( in this case, the tyre of the car), it produces downforce.

Thats right. Called Magnus effect.
What's not so obvious is amount of generated lift/downforce. For F1 tyre at 100 m/s free external flow speed and 50 rotations per second it's more than 4000 kgf. (calculated as for a 335 mm wide part of infinite cylinder, source http://www.grc.nasa.gov/WWW/K-12/airplane/cyl.html).

Strong enough to drive quite big objects:

[xpensive]

Look marekk, 4000 kgf over four 660/335 mm tires would indicate a pressure differential of 45 kPa or 0.45 Bar!

Totally unrealistic, when the dynamic pressure at 100 m/s is no more than 6 kPa.

[marekk]

Look marekk, 4000 kgf over four 660/335 mm tires would indicate a pressure differential of 45 kPa or 0.45 Bar!

Totally unrealistic, when the dynamic pressure at 100 m/s is no more than 6 kPa.

Well, the eqaution (a bit oversimplified, but good for 1st order) is simply
L = p*V*2*Pi*r*Vr (p is density, r radius and Vr ratational speed) and gives lift per unit of length. 4000 kgf is for 1 tyre

I didn't said this apply to a tyre rolling on the ground - it won't, of course, mainly due to lacking space (and time) to develop full scale vortices around rotating cylinder.

[xpensive]

I'm afraid that you misunderstand things here, circumferance has nothing to do with it, just take the load and divide it with the wheel's cross-section area and you will get the average pressure differential between over and under the wheel.

Really not that difficult at all.

[marekk]

I think i understand this quite well, my first airplane model (long, long time ago) looked like this one:



It's not taht much about circumference - it's about enough space around rotating cylinder (and time) to build big, strong vortices.

Openwheelers tend to have lot of space over and no space at all under the tyres for most of the time, so i don't think downforce due to Magnus effect will be that significant, and air pressure on the tyre/tarmac interface influences nett forces more.

But there are exceptions: