Physics Help Forum Roll angle

 Dec 2nd 2014, 09:15 AM #1 Member   Join Date: Jul 2014 Location: KY Posts: 41 Roll angle In a race car I am trying to determine the roll angle of the front and rear suspensions individually as they go through the corners. I am using the formula of RA=(0.975*Right Front Shock Travel-0.975*Left Front Shock Travel)/Distance between front shocks. Where 0.975 is compensating for the angle of the shock.(I read this in a racing publication) Rf shock travel is 2.75" Lf shock travel is 2.50" Distance between shocks is 33" If needed, the LF spring is 200 RF spring 250 LR spring is 185 RR spring 250 Am I on the right track??? Last edited by bei77; Dec 2nd 2014 at 09:18 AM.
 Dec 2nd 2014, 11:05 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,320 That formula will give you the roll angle in radians, but be careful to use the correct sign for the shock travel. If turning left I would expect the left front shock to compress while the right front extends, so the difference between left and right shock travel is 2.5-(-2.75) = 5.25 inches. The roll angle would then be: Angle = 0.975*5.25/33 = 0.162 radians Now multiply by 180/pi to get the angle into degrees: 0.162 radians x 180/pi = 9.2 degrees. I can't vouch for the accuracy of the 0.975 factor, but that would be appropriate if the shocks are at an angle of 13 degrees from vertical.
 Dec 2nd 2014, 11:20 AM #3 Member   Join Date: Jul 2014 Location: KY Posts: 41 Ok, thanks. In these cars, because of the way the shocks are valved, when we go into a corner the front end goes down and is held there by the shocks till acceleration. Would spring rate come into play?
Dec 2nd 2014, 12:11 PM   #4
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,320
 Originally Posted by bei77 Would spring rate come into play?
No. If you know the shock compression/extension numbers, you don't need spring stiffness to try to guess at the roll angle.

 Dec 2nd 2014, 01:08 PM #5 Member   Join Date: Jul 2014 Location: KY Posts: 41 By guessing you mean not using the exact angle of the shock?
 Dec 2nd 2014, 01:25 PM #6 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,320 What I mean is: the only reason to think shock or spring stiffness has anything to do with determining roll angle is if you want to try to calculate the roll angle based on the car's mass, center of gravity, and the magnitude of centripetal acceleration in a corner. But you don't need to do it that way, since you have the actual data for shock extension/compression in the turn.
 Dec 2nd 2014, 01:39 PM #7 Member   Join Date: Jul 2014 Location: KY Posts: 41 Are you saying I can predict roll angle based on that data. Here's what I'm trying to get to. Front and rear suspension have different roll angles. The goal for a well balanced set up is to have both the front and rear work together at the same, or as close to the same, roll angle. If I could predict that by using a formula. Then adjust it by changing the spring rate or wheel weight in the formula, I could have the car set up better before I get to the track. Car weight is 2800 Lateral G force is 1.2 Center of Gravity 16"
 Dec 2nd 2014, 02:22 PM #8 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,320 This is very similar to the problem we worked on a few months ago regarding corner balancing. When the car is in a turn the centripetal acceleration of the CG causes forces to be applied to the outside springs and removed from the inside. It wouldn't be too difficult for you to take the spreadsheet from the work of a few months ago and modify the corner weights to reflect these forces. From that you can determine the spring extension/compression for a given centripetal force. One correction to what you wrote - the roll angle for front and rear must be the same - otherwise the car body would have to twist.
 Dec 2nd 2014, 04:12 PM #9 Member   Join Date: Jul 2014 Location: KY Posts: 41 What would I have to do to make that work? Thanks for the correction.
 Dec 3rd 2014, 01:24 PM #10 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,320 You can get a very good approximation using this approach: Given: h: the height of the CG above the attachment point of the strut to the suspension. Do you have this value? I'm guessing it's probably about 6 inches... L = the distance between strut attachment points (33 inches from previous posts) m = car mass = 2755 lb from previous posts K_R = combined spring constant for right side of car = K2+ K4 = 250 + 250 = 500 lb/in K_L = combined spring constant for left side of car = K1 + K3 = 200 + 185 = 380 lb/in a = centripetal acceleration of the car as it goes around a corner. You can calculate this from v^2/R if you have numbers for velocity and the radius of the corner. Or we can simply use values such as 1 g = 32 ft/s^2 to see what the roll angle is under 1 g of cornering load. The additional compression force to be carried by the front and rear outside strut when in the turn is: Delta_F = mah/L And the reduced force on the two inside struts is -Delta_F. The increased deflection of the outside of the car is then Delta_F/K_r, and the reduced deflection on the inside strut is Delta_F/K_L. Hence the total change in height due to the roll across the span of the strut attachment points is Delta_y = Delta_F/K_R+Delta_F/K_L And the angle of roll is Delta_y/L For the numbers supplied I calculate that the roll angle in a 1g turn is approximately 2 degrees. Of course that's based on my guess of the height of the CG above the strut attachment points. Hope this helps!

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