Beyonce Knowles

Does anybody know if the relation between skidpad g's and the maximum speed around a circle of a given radius is linear or exponential? I can't remember the formula offhand, but I think I remember seeing a radical in there.

Injuneer

Lateral acceleration in "G" = V^2 / (14.97 x R)

V is velocity in MPH
R is the radius in FEET

Beyonce Knowles

Well I guess that makes it linear then.

Steve0

The velocity is squared. That would make it a exponential relationship between force and velocity. (ie, if you double the velocity, the force actually goes up by a factor of 4)

The only linear relationship in that equationis between the between radius and force.

Injuneer

Well said.

z71camaro

Also g= 1.225xR / TxT

R= Radius of skidpad
T= Time in seconds to complete full circle

With this anybody can figure their own g's.

Beyonce Knowles

Well, thats what I figured, but when I plotted it and it had a perfect linear regression, but the exponential regressian had an r value of .98 Maybe somebody here who is a little better at math than me could answer what I am trying to get at: Will the benefit from increasing skidpad from say, .88 to .90 be the same increase as going from .90 to .92? For instance, with a .88 g skid pad I can take a corner at at 50 mph and with a .90 g skidpad I can take that same corner at 52 mph, at .92 g's will I be able to take it at 54 mph or will it be like 55 or 56 mph due to the exponential nature of it (or 53 or 52.5 if the exponential goes the other way?) There now, that's pretty clear

EDIT: after reviewing Steve0's reply, it looks like in order to double speed around a corner, you would have to quadruple the skidpad. Right?

EDIT AGAIN: So this would mean that the increase from .90 to .92 would be less than the increase from .88 to .90?

Steve0

Yes, to double your speed around any given corner, you would in theory be pulling four the amount of lateral G's. Real world factors may throw this off a bit, but the physics is pretty solid at atleast predicting what to expect.

What exactly did you plot?

Force = (mass * velocity^2)/radius

If you plot that in a graphic calculator, letting y be the force and X be velocity you will get a porabola of some sort. The slope will be increasing as your velocity goes up. That said, yes, the increase in force from .90 to .92G's will be less of a velocity incerase than going from .88 to .90Gs.

This difference in velocity may be very small, but its still there.