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Mark Ortiz Automotive is a chassis consulting service primarily serving oval track and road racers. This newsletter is a free service intended to benefit racers and enthusiasts by offering useful insights into chassis engineering and answers to questions. Readers may mail questions to: 155 Wankel Dr., Kannapolis, NC 28083-8200; submit questions by phone at 704-933-8876; or submit questions by e-mail to: email@example.com. Readers are invited to subscribe to this newsletter by e-mail. Just e-mail me and request to be added to the list.
ANGULAR ROLL STIFFNESS IN SPECIAL CASES
I have some questions related to the chassis newsletter.
I understand the formula [in last month’s issue, relating angular roll stiffness Kφ to linear wheel rate in the roll mode Kroll] and I agree but I have some questions that I think will be interesting for a lot of potential readers.
1. This formula works when spring stiffness is the same for both wheels. I mean, if on the outer side there is bump stop contact, then this formula does not work unless you can split inner and outer wheels. Am I right?
2. In the case that inner wheel is not in contact with the ground, typical in saloon cars or even in some formula vehicles that have very low suspension travel, then I think the formula has to take into account only one wheel stiffness. Again am I right?
3. If suspension has some preload, then it has to be included in this formula.
4. If monotube or even double tube shock absorbers are used with high gas pressure, it is not clear how to take that into account. It is supposed as the vehicle is in equilibrium then there is no need to take that into account.
From last month, here is the equation in question:
Kφ = ½ * Kroll * t2 * π/180 (1a)
Kroll = (2 * Kφ) / (t2 * π/180) = 360Kφ / πt2 (1b)
Or, approximating 180/π to three significant figures:
Kφ = ½ * Kroll * t2 / 57.3 (1c)
Kroll = 2 * 57.3 * Kφ / t2 (1d)
Kφ = angular roll resistance, lb-in/deg
Kroll = linear wheel rate in the roll mode, lb/in
t = track width, inches
Taking the last question first, I have addressed the question of gas spring forces in shocks in the past. It is vital not to confuse force with rate. Typically, shock builders or dyno service providers will tell you the gas spring force at one point in the displacement interval used in the dyno test. This will typically be one endpoint of the dyno stroke or the midpoint.
That is not the rate of the gas spring. The rate of the gas spring is the rate of change of that force with respect to displacement, not the value of the force at a single point. To get a measure of that, we need to know the force at a minimum of two points. Gas springs are non-linear, so the measured rate will vary depending on the points chosen. However, the rate in pounds per inch will generally be much smaller than the instantaneous force in pounds.
That is, the gas spring in a shock is a soft spring with a lot of preload. In general, its rate is much less than that of the tires. Therefore, if we are seeking to improve accuracy by including rates other than that of ride springs and interconnective springs (e.g. anti-roll bars), the first thing we need to take into account is tire compliance. If we are treating the tires as rigid, we can safely ignore the rate of the gas springs in the shocks; we have bigger modeling inaccuracies to worry about.
Suspension having preload can mean various things:
1. It could merely mean having static wedge in the car. That is common in oval track cars, but fairly uncommon in road racing.
2. It could mean having some preload on the ride spring(s) at full droop. That is common in road cars. If a spring compressor has to be used to assemble the suspension, the system has preload at full droop.
3. It could mean having a “zero droop” setup, in which the suspension is at the droop stop in static condition and thus cannot extend, but will compress further with any addition of load. In other words, the suspension is exactly topped out at static.
4. Or, it could mean that at static condition the suspension is compressed so much that it will not move with a slight additional load, but will if the load gets beyond some threshold. At static condition, the suspension is topped out, and then some.
There can be various reasons the inside wheel may be off the ground. The suspension at that corner may be topped out, but it doesn’t necessarily have to be in all cases. With independent suspension, it is possible for a stiff anti-roll bar to hold the inside wheel up.
With a beam axle, it is possible for geometric anti-roll to hold the inside wheel up. In some cases, the suspension may even have a negative or inward roll displacement, even with the car rolled outward: the body is tilted out of the turn, and the axle is tilted out of the turn more, so the displacement at the suspension amounts to inward roll: inside shock compressed more than outside.
The inside wheel may be off the ground because there is vertical acceleration in combination with lateral acceleration, due to the car negotiating a bump, a curb, or a crest.
Whatever the reason may be, we know that if the inside wheel is airborne, all the available tire loading at that end of the car is on the outside wheel.
Let’s consider a relatively simple case, where the road surface is smooth and level, there is no aero loading, the c.g. is laterally centered, the only acceleration is lateral (y axis), and the car has no static wedge. If the inside front tire is off the ground, that means the front end is exerting all the roll resisting moment on the car that it can. The front suspension may still have the ability to displace in the roll mode, and have a calculable angular rate in that mode of displacement, but it can’t exert any additional roll resisting moment on the sprung mass, and it can’t increase front load transfer or reduce rear load transfer.
In this condition, if we add further lateral acceleration, the resulting added roll moment must be resisted entirely by the rear suspension. Beyond the point of inside front wheel lift, the body’s rate of roll angle change with respect to lateral force change is greater than it is short of the point of wheel lift; the rate of rear lateral load transfer change with respect to lateral force change is greater; the rate of front lateral load transfer change with respect to lateral force change is zero.
Front load transfer is then 50% of the total front wheel pair load. At static, 50% was on each front tire, and now 100% is on the outside one. Rear load transfer is the total for the car, minus the front. Front anti-roll moment is 50% of the front sprung weight times the track width. The rear anti-roll moment is whatever additional amount is needed to react the total sprung mass overturning moment.
Now let’s consider a condition similar to the one above, except the inside front wheel is on the ground and the setup is zero-droop: just barely at the droop stop at static. Let’s also suppose, for simplicity, that the suspension is independent and has no geometric anti-roll, so it doesn’t jack up at all in response to lateral force.
Assuming that everything but the springing devices is rigid, the suspension cannot displace in pure roll. The wheels cannot move equal amounts in opposite directions, because the inside suspension cannot extend. The suspension can displace in roll, but only by displacing an equal amount in ride at the same time. To have an inch per wheel of roll displacement, we have to have two inches of compression at the outside wheel, since we cannot have any extension at the inside wheel. The ride and roll components are additive on the outside wheel. On the inside wheel, they are subtractive and exactly cancel each other.
The load increase required at the outside wheel to produce this condition is then the force required to create an inch of roll displacement and an inch of ride displacement. The equation for Kφ is then
Kφ = ½ * (Kroll + 2Kride) * t2 * π/180 (2a)
Or, approximating 180/π to three significant figures:
Kφ = ½ * (Kroll + 2Kride) * t2 / 57.3 (2b)
It will be apparent that the system has much greater angular roll stiffness than it would if the inside wheel’s suspension could extend. This squares with common sense, or intuition: if half of the system can’t give, the system as a whole is stiffer.
What if instead of being topped out at static, the suspension is bottomed out? What if we create a “zero bump” suspension, instead of “zero droop”?
The situation is very similar to the previous case, except that the suspension needs to extend in ride in order to displace in roll, rather than compress. We have one side that is up against a stop and can’t displace, while the other side can move. Kφ is calculated the same way. The sprung structure has to rise in order to roll rather than drop, but otherwise the dynamics are similar.
Very often, we have a situation where the suspension comes up against a bump or droop stop at some point in its travel that we are likely to encounter, but it is not at that point at static. In the case of a suspension that has preload against its droop stops at static, we can have a situation where the suspension is immobilized until forces reach a threshold, and then it can move. All of these are cases of what we might call two-stage behavior. The system behaves one way up to some force or acceleration threshold, and at that point its characteristics change.
Assuming roughly linear behavior both below threshold and beyond, we approach analysis by calculating what lateral acceleration gets us to threshold, in the particular situation we’re examining. If we are modeling an acceleration that is beyond threshold, we then know by what increment it is beyond threshold. We calculate the displacements and load changes at threshold, based on the system’s properties short of threshold, and then the additional displacements and load changes for the increment beyond threshold, based on the properties in that range.