LCP

I.e., on a 30-degree banked corner, how many more lateral G's does the 30 degree banking allow you to endure (assuming you could corner at 1g without the banking)?

ryoung

g = g₀ / cos(angle)

If a 30 degree bank and one g without banking,

g = 1.0 / cos(30)

g = 1.0 / 0.866

g = 1.16

LCP

OK, that makes sense, but it just "seems" too low IMO.

NASA racer

which adds adhesion by adding "downforce" to your vehicle.

LCP

And for instance, if there was only a 5 degree banking (which on a road course is huge and can let you corner a lot faster than you otherwise would on a neutral camber corner), by using only the cosine angle formula Randy posted, you could only increase lateral cornering G's 0.4% (the cosine of 5 degrees is 0.996; 1/0.996=1.004).

ryoung

Until I thought about a winged Indy car on slicks exiting an oval corner along the wall at 200 mph. The car could probably corner at 3+g on the flat at that speed, and with 30 deg banking it becomes 3.5+g. With a wall only a few inches away, it probably doesn't feel low at all.

Mark Dalen

You are missing a major factor - the g's created by cornering also add to the normal force of the car. Let's use a 45 degree bank. There is .707 g normal force and -.707 g cornering (weight of the car trying to fall down the banking. Now for whatever cornering load you can carry (we will call that G) you have that force (G) acting in the + direction laterally, and .707 G adding to the normal force.

So, corneing force is G - .707 (times the weight which is equivalent to 1 g) and normal force is .707 G + .707 w (which again is equvalent to 1 g. Sloving for G yeilds ~4.8g's (through the perpendicular to the track. I did this quickly so I may have missed something and this is based on tires that can do 1.0g so the coefficient of friction is 1.0.

But you can't really do this - mostly because when you load a tire that much, the coefficient of friction falls way off.

Bottom line is I don't think you can use a formula to estimate this, unless you are at very low banink angles. If you are taling 5 degrees, then the loading is not that high so the coefficient of friction would be relatively constant and this formula would work. Right now, I am too lazy to derive the formula for any banking angle.

Mark Dalen

My net connection went down for a while, so I had nothing better to do, so here is the equation. I think I have it right - this is based on tires that do 1.0g level on a banked track of angle a. G is the mazx g's you can pull, measured parallel with the car. This is the G associated with the velocity so measured g would be lower by sin(a) due to the tilt of the car (In other words, 0 g's would be measured when there is some speed, not at 0 speed.) Again, this does not factor the drop off in coefficient of friction which is VERY significant as loads go up.

G = (cos(a) - sin(a))/(1-cos(a))

LCP

This is what I was imaging would be the case. I just couldn't formulize it.

LCP

I just tried it with 45 degree banking, and it came out as zero, and many other angles seemed out of whack too.