hmm. I thought that in practice traction "circles" were really ellipses. So in most cars you're acceleration will be bigger fore/aft. Especially while braking.

and your right, keeping your tyres along the outside edge of the traction "area" requires a very smooth arc.

 

I'm speaking specifically about the tires capabilities, not the cars. A very front heavy car like ours will not be able to balance the weight as smoothly or easy as say a lighter, more balanced miata. but in a perfect world, assuming perfect weight balance F/R as well as corner to corner, you should be able to produce the same G forces on all four corners.

 

Suspension geometry will also affect the real world circle. Tyres have one optimal camber per air pressure (at least from what i've read). A turn that requires a 10 degree movement of the tire will have a different camber than one that requires 20 degrees. Hence, the traction circle is almost always different for each turn.

I am sure that real race chassis compensate for this, but i cant imagine anyone getting it perfect.

one last nitpick ... i think the shape of the contact patch affects the shape of the traction circle. So even in your perfect little world I'm still right (ok, i had to say one arrogant thing)

 

I should have known better than to try and have a discussion with someone from Joyzie!

I hear what you're saying about the traction being different for each turn, thats not really what I'm talking about though. Think about it like this... placing a tire/wheel on a flat surface, camber and slip angle aside, if you somehow put 2,000 lbs of force on the tire vertically, you should, in a perfect world, be able to apply the same amount of force in any direction before the tires loses grip. does that make sense?

 

yup, that makes sense. So what are we gonna discuss about this? lol

here is a question ... i think i know the answer, it seems pretty straight forward ... but i think I'm missing something, anyway:

An Evo and a Miata both reach 1g on on an identical skid pad driving on a circle of the same radius ... blah, blah, blah ... which had the faster speed?

 

wasn't sure about the definition of g-force. I think they'd be the same.

 

is that like which is more a pound of feathers or a pound of lead?

if you already know the answer to the questions, then I'm going to GUESS the one traveling faster would be the miata since its lighter. I have an explaination for this but I don't want to type it all out incase I'm way off, but all things being equal it basically comes down to the weight of the car. Am I right or wrong?

my understand is that G-force is just the force of gravity at work. SO 1g is the force of gravity... 1.5g is .5x the amount of weight being applied to something, 2g is twice the amount of weight being applied and so on. Thats my understanding anyway, I could be off though.

 

This is what i was wondering. But it seems that the proper g-force units are m/s^2. Meters over seconds squared, which is simply acceleration.

since, force = mass X acceleration, the miata simply requires less force to maintain the same speed. This would translate to thinner tires, ect. Basically you could calculate the g-force if you knew the speed and radius of the turn.

Your examples above are correct, you can interchange weight and force. For lateral acceleration there is no gravity, so the measurements are relative to gravity. 2g = 2 * (9.8xxxxx m/s^2)

 

another cool thing about the friction circle is how its directly related to the steering angle, which is why in all those driving classes/schools they talk about unwinding the wheels AS you roll into the power. <--- that right there is easy to do on a daily basis in any car and IMO really helps you with throttle control and maximizing traction coming out of a turn with is sooo important to fast lap times.

 

Kenny, what's your address?

 

it is cool, that's what i was talking about earlier. This is one thing i need to improve this year.

sorry, i don't need any dirty pictures.

 

An object moving in a circle has an acceleration toward the center of the circle given by:
a = v^2/r
where “v” is the velocity of the object in m/s
and “r” is the radius of the circle in meters.

 

word.

so replace speed with velocity ... close enough!

 

Whats the difference between speed and velocity?

 

velocity has a direction, so it is actually two numbers ( a vector)