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### #ActualPochyPoch

Posted 10 September 2012 - 10:59 AM

Hi, K_J_M !

Thank you for your answer ! I already downloaded your top down racing game, but haven't had many time to play it yet ! It seemed a little hard to me at first sight

If i understand you, then a static but spinning car would generate velocities for tires that would be with 0 longitude, and some value in x depending on the direction of rotation. Something like that : http://en.wikipedia....circular_motion

I found it to be true if the tire is aligned with the radius of the circle. As the wheels are not centered, but rather on the side of the car, when they are not steering, then the radius from wheel attach point to car CG will not be perpendicular to the wheel heading.

If you draw a path following the front right wheel positions, it is indeed a perfect circle, and the velocity of that point (blue vector) is indeed tangent to circle. But as my wheel is not aligned with the radius, the projection of the velocity according to steering (wich is 0), will give me both lateral and frontal velocities, wich in turn will give me a slip angle not equal to +/- 90°. The only case i have such results is if :
a) There is only 1 front wheel, centered on the chassis (like a bike, or tricycle)
b) The wheel is aligned with the radius coming from CG
c) The car is sliding lateraly

That is a little confusing to me, so i am wondering if :
• I should have 0 longitidunal velocities as long as the tire is describing a circular path, regardless of wheel steering. The steering would then be "used" in slip angle calculation like the formula from marco suggest it :
			slip = Atan(latVel / |frontVel|) - signOf(frontVel) * steerAngle

• I should have a longitudinal & lateral velocities (excepted for previous a,b,c cases) and my calculations are OK until there (the wheel front/lat velocities are the green and red vectors on the picture).
I am suspecting it is not the only thing that goes wrong with my simulation but it would really helped me to know which way is the right one

And again, thanks for your help, it is much appreciated !

### #2PochyPoch

Posted 10 September 2012 - 10:55 AM

Hi, K_J_M !

Thank you for your answer ! I already downloaded your top down racing game, but haven't had many time to play it yet ! It seemed a little hard to me at first sight

If i understand you, then a static but spinning car would generate velocities for tires that would be with 0 longitude, and some value in x depending on the direction of rotation. Something like that : http://en.wikipedia....circular_motion

I found it to be true if the tire is aligned with the radius of the circle. As the wheels are not centered, but rather on the side of the car, when they are not steering, then the radius from wheel attach point to car CG will not be perpendicular to the wheel heading.

If you draw a path following the front right wheel positions, it is indeed a perfect circle, and the velocity of that point (blue vector) is indeed tangent to circle. But as my wheel is not aligned with the radius, the projection of the velocity according to steering (wich is 0), will give me both lateral and frontal velocities, wich in turn will give me a slip angle not equal to +/- 90°. The only case i have such results is if :
a) There is only 1 front wheel, centered on the chassis (like a bike, or tricycle)
b) The wheel is aligned with the radius coming from CG
c) The car is sliding lateraly

That is a little confusing to me, so i am wondering if :
• I should have 0 longitidunal velocities as long as the tire is describing a circular path, regardless of wheel steering. The steering would then be "used" in slip angle calculation like the formula from marco suggest it :
		slip = Atan(latVel / |frontVel|) - signOf(frontVel) * steerAngle

• I should have a longitudinal & lateral velocities (expect in previous a,b,c cases) and my calculations are OK until there.
I am suspecting it is not the only thing that goes wrong with my simulation but it would really helped me to know which way is the right one

And again, thanks for your help, it is much appreciated !

### #1PochyPoch

Posted 10 September 2012 - 10:48 AM

Hi, K_J_M !

Thank you for your answer ! I already downloaded your top down racing game, but haven't had many time to play it yet ! It seemed a little hard to me at first sight

If i understand you, then a static but spinning car would generate velocities for tires that would be with 0 longitude, and some value in x depending on the direction of rotation. Something like that : http://en.wikipedia.org/wiki/Uniform_circular_motion

I found it to be true if the tire is aligned with the radius of the circle. As the wheels are not centered, but rather on the side of the car, when they are not steering, then the radius from wheel attach point to car CG will not be perpendicular to the wheel heading.

If you draw a path following the front right wheel positions, it is indeed a perfect circle, and the velocity of that point (blue vector) is indeed tangent to circle. But as my wheel is not aligned with the radius, the projection of the velocity according to steering (wich is 0), will give me both lateral and frontal velocities, wich in turn will give me a slip angle not equal to +/- 90°. The only case i have such results is if :
a) There is only 1 front wheel, centered on the chassis (like a bike, or tricycle)
b) The wheel is aligned with the radius coming from CG
c) The car is sliding lateraly

That is a little confusing to me, so i am wondering if :
• I should have 0 longitidunal velocities as long as the tire is describing a circular path, regardless of wheel steering. The steering would then be "used" in slip angle calculation like the formula from marco suggest it :
	slip = Atan(latVel / |frontVel|) - signOf(frontVel) * steerAngle

• I should have a longitudinal & lateral velocities (expect in previous a,b,c cases) and my calculations are OK until there.
I am suspecting it is not the only thing that goes wrong with my simulation but it would really helped me to know which way is the right one

And again, thanks for your help, it is much appreciated !

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