Oscar Gonzalez

11-16-2002, 01:29 PM

Does anyone know if there are any swaybar programs. Different configurations for swaybars? What do judges look for in swaybars? Any help would be really appreciated.

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Oscar Gonzalez

11-16-2002, 01:29 PM

Does anyone know if there are any swaybar programs. Different configurations for swaybars? What do judges look for in swaybars? Any help would be really appreciated.

Marc Jaxa-Rozen

11-16-2002, 01:58 PM

I'm not too sure if this is what you're looking for, but solid steel sway bar stiffness can be found by the following equation:

K (lb/in) = 500,000 D^4/((0.4244*A^2*B) + (0.2264*C^3))

Where D is the bar diameter, B is the length of the center element, A is the effective lever arm length (i.e perpendicular to the center bar) and C is the total length of the side levers.

K (lb/in) = 500,000 D^4/((0.4244*A^2*B) + (0.2264*C^3))

Where D is the bar diameter, B is the length of the center element, A is the effective lever arm length (i.e perpendicular to the center bar) and C is the total length of the side levers.

Mark Ortiz

11-17-2002, 01:18 PM

When using this formula, it is important to be aware that it gives you pounds per inch of motion at ONE END of the bar, assuming the other end to be stationary -- not per inch per end, or per inch per wheel. Or stated another way, the formula gives you pounds per inch of position DIFFERENCE between the two lever ends -- an inch per end being two inches of difference.

It is useful to know what the bar equates to in terms of wheel rate in roll, so that we can calculate how much we can reduce wheel rate due to springs, while maintaining wheel rate in roll, or what spring-derived wheel rate increase would be required to equal the effect of the bar.

To do this, we need to multiply the value from the equation by two, and then multiply it again by the square of the lever-end-to-wheel motion ratio.

The inclusion of lever bending deflection in the equation is good, but it does assume identical material for the torsional (center) element and the lever arms.

It is useful to know what the bar equates to in terms of wheel rate in roll, so that we can calculate how much we can reduce wheel rate due to springs, while maintaining wheel rate in roll, or what spring-derived wheel rate increase would be required to equal the effect of the bar.

To do this, we need to multiply the value from the equation by two, and then multiply it again by the square of the lever-end-to-wheel motion ratio.

The inclusion of lever bending deflection in the equation is good, but it does assume identical material for the torsional (center) element and the lever arms.

Nigel Lavers

11-18-2002, 08:13 PM

While we're on the topic... I've encountered a bit of a dilemma.

I am unsure of what assumptions Milliken made in RCVD when he derived the formula for anti-roll bar calculations (chap 21). For example, to determine the safety factor I was going to assume that the worst case scenario is when the bar is under full twist. This would be when the left wheel is under full bounce and the right wheel is under full droop. Would this state of roll ever arise? After some review I don't think it will and I seem to want to trust Milliken's formula.

But the question still remains, what assumptions does he make about the loading condition on the bar? He sets a maximum stress value of 100,000 psi. How does this come into the equation?

Nigel

I am unsure of what assumptions Milliken made in RCVD when he derived the formula for anti-roll bar calculations (chap 21). For example, to determine the safety factor I was going to assume that the worst case scenario is when the bar is under full twist. This would be when the left wheel is under full bounce and the right wheel is under full droop. Would this state of roll ever arise? After some review I don't think it will and I seem to want to trust Milliken's formula.

But the question still remains, what assumptions does he make about the loading condition on the bar? He sets a maximum stress value of 100,000 psi. How does this come into the equation?

Nigel

awhittle

11-19-2002, 07:34 PM

The worst case is one tire at full jounce and the other at equiliberium sway bar pulling up and the spring and gravity pushing down. The unladen tire will never be at full droop when the laden is at full compression.

Andy

Andy

Oscar Gonzalez

11-21-2002, 11:39 PM

Thank you Marc, Mark, Nigel, and awhittle for your info. I really appreciate it.

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