Calculating Roll Bar Resistance

1. Jan 19, 2012

Jas1159

Can somebody help to validate my formula? I believe the following can be used to calculate anti roll bar resistance;

(Roll Bar Rate / (Wheel movement / Relative roll bar pickup movement)^2) * Track Width^2 / 2*Pi / 180 = Roll Bar Resistance

My knowledge is limited so other than pondering torque equations etc i do not know how this formula was derived and have been told my bar resistance could be incorrect....

My calculation is

(1618.035m.kg / (0.025m / 0.024m)^2)) * 1.39^2 / 2Pi / 180) = 24.563m.kg/degree

I would greatly appreciate it if somebody could explain where this equation has been derived from or if it is correct?

2. Jan 19, 2012

Ranger Mike

Question - you want to know the resistance applied to the Anti Roll Bar (ARB) as the wheel goes thru Bump or Droop?

3. Jan 19, 2012

Jas1159

Hi Mike, thanks for the reply.

I am trying to calculate the impact a roll bar has on chassis roll / weight transfer and simply need to understand whether the calculation i did gives me a good representation in reality. otherwise as i am sure you know i could design a bar that will useless or too efficient/rigid.

This is one of many calculations that when accompanied by corner weights lead to the amount of weight transfer and therefore chassis roll in degrees while cornering at 1g.

I have also calculated the spring resistance in a similar way (shown below) to obtain Total resistance.

(pring rate / (wheel movement / Relative spring movement)^2) x Track Width^2 / 2Pi / 180 = Spring Resistance

4. Jan 19, 2012

Jas1159

I dont want to confuse the issue further as i really need to finish these calculations in order to complete my disertation :0 however, I can explain why i have this issue...

Following the Allan Stanliforth book weight transfer calculations i obtain this calculation for spring resistance;

(SF / (WmF / SmF)^2) x TF^2 / 2pi / 180 = ArF = 2890 in.lbs/degree

and this for roll bar resistance;

(BF x ((WmF x BmF )^2) x (TF^2) x (Pi)) / 180 = BrF = 4675 in.lbs/degree

Now during the next step i need to add these two figures together to give me the total front resistance and the same for the rear.. once i have these two figures i can calculate a percentage transfer from front to rear and inner to outer wheel.

Anyway.. the book is in imperial, when i convert to metric i notice a huge error where my roll bar resistance = x10^-6 while my spring rate maintains a reasonable unit. also when converting the spring rate answer from imperial to metric and I get the same answer as when running through the whole equation in metric... now this is different for the roll bar calculation where i get a rediculus answer and therefore making further calculations incorrect.

you will notice in the above equation for spring resistance is largely devide and times, now the bar resistance is all times... must be an error?! also the 2pi becomes pi..... so if i use the spring rate equation for bar rate I get comparable answer! i just need to validate that the equation is correct otherwise my roll bar and weight transfer could be way off.

The difference when using the spring resistance formula for roll bar resistance is an answer from 4675 to 236 approximately /10 /2

Sorry for the essay i really appreciate your help.

Last edited: Jan 19, 2012
5. Jan 19, 2012

Ranger Mike

without more information on the ARB formula I can not offer valid comments..BUT..I do know this..you can not add the spring rate to the ARB rate to calculate requirement to resist load transfer..it will not work.

6. Jan 24, 2012

Kozy

IIRC the ARB formula contains the elasticity modulus of steel where the springs do not, if you change all other units but do not change the modulus you will get the huge error you found.

So far as I can see it will? The formulas contain the respective motion ratios of both components so those forces are at the wheels and so adding them should work.

Last edited: Jan 24, 2012
7. Jan 24, 2012

Ranger Mike

kozy.. it will not work.
What you need to do ultimately is to figure out the resistive force necessary to counter weight transfer during cornering, braking etc..your assumption is not going to work..
the answer is simple and readily apparent..and a prime teaching point when dealing with handling problems..anyone want to take a stab?

8. Jan 24, 2012

Kozy

I'm sorry, that reply has baffled me somewhat.

The load transfer is determined by the track, weight, CGH and acceleration. You are correct in saying that adding the spring and bar rates does not alter that. What I meant was adding them does give you the roll resistance of that axle.

The total roll resistance from the front springs and bar and rear springs and bar will give a total roll resistance. The front/rear distribution then determines how much of the elastic load transfer is apportioned to each axle.

I'm not sure what you are saying will not work? Is this a semantics issue?

9. Jan 24, 2012

Ranger Mike

you are correct in the language barrier is in effect..Kozy, my friend..and you have almost reached the correct solution..

if you read post # 19 on Race Car suspension class on Mech Eng forum here..
example as noted..

This particular car weighs 2800 lbs. of 35% of weight will transfer under 1.3 G
and 75% will be on front end due to engine weight and corner loading

75% of 980 lbs. = 720 front end weight
divided by three to determine wheel rate ( two front springs and sway bar )
so we need wheel rate of 240

Wheel rate = (Length of A-arm divided into inside frame mount point to center of spring mounting point) squared times spring rate.

The key here is that you can not just add ARB weight and wheel rate...you need to add ARB rate and Wheel rate and Wheel Rate

10. Jan 24, 2012

Kozy

I see, you are going at it from the opposite direction to me, calculating the rates required from wheel loads. I start with all the rates and logged G data and work out the wheel loadings to determine handling balance.

I can see where you are coming from with the two springs thing however to my recollection, this is not how it is written in the book in question. If a 350lb spring is specified I do not recall reading anywhere that this is to be doubled for an axle on accord of their being two springs, however there is a similar argument regarding halving the bar rate as each end only moves half of the total displacement. A 350lb/in spring rate is not the same as a 350lb/in bar as each end would only move 1/2" and so would actually be 175lb/in per side. I think we are talking about the same thing expressed differently?

This aside, I don't get where your 75% comes from in above workings. Surely this is dependant on the rates?

Last edited: Jan 24, 2012
11. Jan 25, 2012

Ranger Mike

The point i am coming from is to estimate the proper springs and ARB required to handle weight transfer for a particular race track. Most race cars are set up pretty close to 50% front 50% rear weight. When entering a turn 25% of the sprung weight will shift to the front so we use a figure of 75% in the formula. Not perfect but a good starting point

12. Jan 25, 2012

Kozy

Ah I get you, mine are done on pure steady state so no longitudinal transfers.

Been putting work on the combined long./lat. spreadsheet off for a while now!