Car Physics - Lateral Front Roll Center Location

I've been reading up on some formulas that calculate the effect of roll center heights and Neutral Roll Axis for steady state weight transfer but I haven't seen any formulas for calculating how the laterel location of the roll center effects weight transfer or ride/roll rates. From what I understand the lateral weight transfer would remain the same but ride/roll rates would be differant. In effect the lateral location of the roll center controls how much of the weight transfer is effected by the springs or directly into the suspension linkages. Kind of like the way anti-dive works but for roll. Any help or insight would be appreciated.

Not much time now, so I'll come back later and try to attack this from a different angle. For now this might get you thinking though enough to apply it to what you have on roll centers already. Usually lateral roll center movement only happens during cornering, or more specifically, when the body rolls. I don't think it's deliberately designed into many cars in the rest state. Not sure if that's what you meant though. When standing behind the car and looking at towards the front, if you took the normal "center" position of the roll center and moved it towards the right, you could get an idea of what happens by seeing that you've got two roll centers in a sense (not literally, but as far as jacking forces go you could say this is the same thing.) The line going from the left tire through the roll center now crosses the center of the car at a slightly lower height (the line's slope is less), while the line pointing from the right tire through the roll center now crosses the center of the vehicle at a higher position. In effect, at that end of the car, when you move the roll center to the right you basically get the effect that you have raised the roll center as the right side tires see it and have lowered it for the left side tires. Sorry if that leaves you hanging. Somebody else will probably cover this by the time I get back, but until then maybe it'll be something to chew on. Todd Wasson Racing Software http://PerformanceSimulations.com My car sim http://performancesimulations.com/scnshot4.htm

Heheheee, Joel, (its me Sting32) Your thinking too hard old buddy, just get in the car and stomp the gas!!! ;) "Joel Brown" <[email hidden]> wrote in message news:Jkt7a.144883$[email hidden]... > I've been reading up on some formulas that calculate the effect of roll > center heights and Neutral Roll Axis for steady state weight transfer butI > haven't seen any formulas for calculating how the laterel location of the > roll center effects weight transfer or ride/roll rates. > > From what I understand the lateral weight transfer would remain the samebut

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> ride/roll rates would be differant. In effect the lateral location of the > roll center controls how much of the weight transfer is effected by the > springs or directly into the suspension linkages. Kind of like the way > anti-dive works but for roll. > > Any help or insight would be appreciated. > >

Thanks for the response - looks like you were up kind'a late. From what I've read in Gillespie's book he talks about it as 2 roll centers and where it crosses the center of the car is the height. But I work with stock car people and they think of it as where the 2 roll centers intersect each other and they will deliberately change it from center to as much as 20"+ to the right. Although they typically stay somewhat close to the center - usually within 6" static. I'm more concerned with ride/roll rate prediction as I think the steady state lateral weight transfer should remain fairly consistent as long as I put in the correct roll center height. "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > Not much time now, so I'll come back later and try to attack this from a > different angle. For now this might get you thinking though enough toapply it > to what you have on roll centers already. > > Usually lateral roll center movement only happens during cornering, ormore > specifically, when the body rolls. I don't think it's deliberatelydesigned > into many cars in the rest state. Not sure if that's what you meantthough. > > When standing behind the car and looking at towards the front, if you tookthe > normal "center" position of the roll center and moved it towards theright, you > could get an idea of what happens by seeing that you've got two rollcenters in > a sense (not literally, but as far as jacking forces go you could say thisis > the same thing.) > > The line going from the left tire through the roll center now crosses the > center of the car at a slightly lower height (the line's slope is less),while > the line pointing from the right tire through the roll center now crossesthe > center of the vehicle at a higher position. > > In effect, at that end of the car, when you move the roll center to theright > you basically get the effect that you have raised the roll center as theright > side tires see it and have lowered it for the left side tires. > > Sorry if that leaves you hanging. Somebody else will probably cover thisby

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> the time I get back, but until then maybe it'll be something to chew on. > > > > > > > Todd Wasson > Racing Software > http://PerformanceSimulations.com > My car sim > http://performancesimulations.com/scnshot4.htm >

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>Thanks for the response - looks like you were up kind'a late. From what >I've read in Gillespie's book he talks about it as 2 roll centers and where >it crosses the center of the car is the height. But I work with stock car >people and they think of it as where the 2 roll centers intersect each other >and they will deliberately change it from center to as much as 20"+ to the >right. Although they typically stay somewhat close to the center - usually >within 6" static. > >I'm more concerned with ride/roll rate prediction as I think the steady >state lateral weight transfer should remain fairly consistent as long as I >put in the correct roll center height. > >

Hi Joel. First of all, I am not an expert in this. I'm going stricly on theory as I use in my own simulator work and have observed from that, which I do not is right enough to guide real car setup work such as you're doing. It's very possible that I am missing something that could completely make the rest of this post bunk. (I.e., if Mario Andretti tells you I don't know what I'm talking about, listen.) With my disclaimer out of the way, I'll continue as though what I "currently know" actually reflects reality (which is quite improbable). Second, I just got back from the bar. :-D Third, I'll make this up as I go and throw around calculations helter-skelter that may be meaningless tomorrow. Either way, this will be a very long, long, drawn out post that may server more for my own education than anything else. If it turns out to be useful, cool. It would be helpful to concretely define some car that we're working with in order to analyze this, as I do not know the rules of thumb and have no experience with oval track Nascar type cars to be able to make generalizations that you might get from crew members. Another reason this might be bunk :-D You described a Nascar type race car, so I'll just pull some numbers from my behind that I think would roughly describe a vehicle you might be dealing with. We're interested in the lateral roll center location's effects on handling. More specifically, the effect of this on roll/ride rate at one end of the car May I quote something here?

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>I've read in Gillespie's book he talks about it as 2 roll centers and where >it crosses the center of the car is the height. But I work with stock car >people and they think of it as where the 2 roll centers intersect each other >and they will deliberately change it from center to as much as 20"+ to the >right. Although they typically stay somewhat close to the center - usually >within 6" static. >

The "2 roll centers" are really called "instant centers." The point of intersection you are talking about is the real "roll center." In other words, from the diagrams you are looking at there are lines drawn extending from the upper and lower arms that intersect at a point. This point is the "instant center," as seen in the front/rear view of the car. The "roll center" is more indicative of the point where the lines you saw cross the vehicle centerline. Either way, I think you know perfectly well what you're referring to, but it's good to make a distiction between the roll center and the two instant centers. I'll expand on the roll center just to make sure we're talking about the same thing. Look at one side of the suspension in front view. Two lines could be traced along the upper and lower A-arms (one line each) that would intersect somewhere. The could intersect outside of the wheel (I've seen this in an RC car) or inside the wheel. This intersection point is the "instant center." A line drawn from the center of the contact patch to that instant center will intersect a line drawn the same way from the opposite suspension at the "roll center". On a car that has identical suspension geometry on each side (left and right) of the car, the lines drawn from the center of each contact patch to its respective "instant center" will cross the center of the car at the same height. What's important here is where those two lines intersect. That point is the roll center, but in a symetrical case, this "point" will be in the center of the car. When body roll is present, the instant centers move, so that point can move up, down, left, or right of course. Ok, I just explained everything you already knew, but am just making a distinction between "instant center" and "roll center." Don't forget I've had a couple and may ramble :-D What happens if we move this point to the right as seen from the rear of the car? Take a look at your diagrams and see where the instant centers are. Forget about everything except the lines going from the contact patches to he instant centers. (Also think of roll center location, the intersection of those two lines.) All this hoopla about roll centers and instant centers comes down to one thing: Jacking forces. Imagine for a moment what happens when the roll center is at ground level. There could be a net lateral force of 1000 lb acting along the ground plane at the right tire in the left direction (we're in a left turn and watching this from the rear). All of this force acts on the car in the lateral direction (along the ground; the force doesn't point up or down at all). What if the roll center is somewhat above ground level? What matters is not the location of the roll center, but the angle between the contact patch and a line drawn through the roll center when trying to find what's happening at one wheel. This angle is the same if you extended the line through the instant center. The angle between the line that extends from the contact patch to the instant center of that wheel's suspension is what dictates the jacking force. That's so important I'll say it again: The angle between the line that extends from the contact patch to the instant center of that wheel's suspension is what dictates the jacking force. This jacking force is the portion of the vertical force that acts directly through the suspension links and bypasses the springs, dampers, and anti-roll bars, and is what causes the change in roll stiffness you get from the roll center being above or below ground level. How in the world can we see what will happen if we move the roll center left or right? Basically we should be able to figure this out by forgetting about the roll center location entirely. Instead, we can use the propose roll center location to give us our "vertical angles" (described in the two repeated statements above) that are required to calculate jacking forces. I need a real example now. Let's invent a car. Its front track width (as defined by distance between center of contact patch locations across each "axle", or more specifically, the distance between the force centroid locations across each "axle") is 60 inches, or 5 feet. Our car's center of gravity is located half way between each wheel and raised 24 inches (2 feet) above the ground. For simplicity, let's assume our car can accelerate laterally at 1 g. In addition, we have 50/50 front rear weight distribution. The vehicle's weight is 3217 lb, or as I commonly say (despite what physicists tell me is proper) it has a mass of 100 slug. I hope you're not European :-) If this car is in a steady state left turn in a corner with no banking (not speeding up or slowing down, and it has "set" into the corner where the dampers/shocks will not be influencing anything), then the total lateral force acting on the car is: force = mass * acceleration force = 100 * 32.17 (1g) force = 3217 lb The same as the weight of the car: 3217 lb. Since the weight distribution is 50/50 front/rear, the front will provide half the cornering force and the rear will provide half of it. I know you know this already, I'm just doing it this way for my own mental exercise. Bear with me please :-D What's all this stuff about the roll center height/location? It's suppose to effect something, right? It does, but we won't see exactly where yet. Let's imagine that the roll center is located precisely in the center of he car WHILE the car has assumed whatever roll angle it assumes while pulling us around the corner at 180 mph (usually not the case; the roll center usually moves left or right). What effect does the height have? What is this? The roll center height effects how much of the weight is transferred directly through the suspension links. I'll cheat and look up the formula for weight transfer from Fred Puhn's "How to Make Your Car Handle": Weight tranfer = force * cg height / track width I'm only concerned with the weight transfer at the front end right now, where the lateral force is 1/2 of the total force (3217 * 0.5 = 1608.5 lb) Weight transfer = 1608.5 * 2 feet / 5 feet (60 inch or 5 foot track) Weight transfer = 643.4 lb This "weight transfer" is the total vertical force change in both the left and right rear wheels. We're in a left turn (most of the weight is on the right wheel), so we can calculate the left/right wheel vertical forces like this: Front vertical force when stationary : 3217 * 0.5 (50% front weight distribution) = 1608.5 lb When stationary, both left and right front wheels would have a vertical force equal to half this: 1608.5 / 2 = 804.25 lb Left vertical force when not cornering = 804.25 Right vertical force when not cornering = 804.25 We can now add the weight transfer to find the current vertical force at each end: Left vertical force while cornering = 804.25 - 643.4 Right vertical force while cornering = 804.25 + 643.4 Left vertical force while cornering = 160.85 lb Right vertical force while cornering = 1447.65 lb For simplicity, let's assume the body only rolls 1 degree in this situation. We could then say that the roll stiffness is 1608.5 lb*ft/deg for the front suspension. This would assume the rear is identical. ------------------------- Let's raise the front roll center now: ------------------------- There isn't a very clear cut mathematical answer like I wish there was, but the basic effect should be visible. But, we should be able to see why the roll stiffness changes when the front roll center is raised. If we raise the roll center to a height of 6 inches, or 0.5 feet, we will now introduce jacking forces into the picture. What matters here is the angle between the line from the contact patch through the roll center or instant center of each wheel. The *angle* matters here, and NOT the actual roll center or instant center location. If our roll center is located 1/2 way between the front wheels at a height of 0.5 feet, the angle between the contact patch and the roll center will be: angle = arc_tangent (0.5/2.5) angle = arc_tangent (.2) angle = 11.3 degrees And some of you thought trigonometry would never be useful ;-) This angle is important. The lateral force acts entirely along the same direction it did before, but now it's also causing an upwards force on wheel and a DOWNWARDS force on the other because of the orientation of the suspension arms. For simplicity's sake, we'll go ahead and look at the situation as though things were nice enough to simply add this jacking force where we are now to the whole picture. This isn't quite right in this situation, but it should still be useful for visualizing what happens: vertical force = lateral force * tangent of (11.3 degrees) If we assumed that our two tires each were pulling "1g", or in other words producing the same lateral force as vertical force, we could approximately calculate the change due to this jacking force and arrive at a reasonable approximation of the change in weight transfer due to the elevated roll center. This is NOT the exact answer really, but it should give a sense of scale anyhow and illustrate what is going on when you move the roll center up: From before: Left vertical force while cornering = 160.85 lb Right vertical force while cornering = 1447.65 lb If these are all causing 1g acceleraion, then the lateral forces at each tire are identical: Left lateral force = 160.85 lb Right lateral force = 1447.65 lb vertical force (jacking force) = lateral force * tangent of (11.3 degrees) For each wheel (the jacking force acting in addition to the regular weight transfer): Left vertical force (jacking force) = 160.85 * -0.2 Right vertical force (jacking force) = 1447.65 * 0.2 Left vertical force (jacking force) = -32.17 Right vertical force (jacking force) = 289.53 Note that the left wheel, the inside wheel, is actually trying to pull DOWN on the chassis here. Regardless, we can now roughly calculate the new total vertical forec at each wheel so we can estimate how much this would effect the "normal" weight transfer (with no roll center height). Left total vertical force = 160.85 -32.17 Right total vertical force = 1447.65 + 289.53 Left total vertical force = 128.68 lb Right total vertical force = 1737.18 lb Well, weight a minute! The total weight on the front of the car is only 1608 lb! What the heck happened? What this calculation means is that there are two forces pushing up on the front of the car. These are acting in addition to the springs and anti-rollbar forces. At this moment in time, they will push the front of the car upwards, causing the springs to extend (so they don't produce as much force) and eventually equalize so we have a new weight transfer situation. With the front roll center raised, the front of the car will RISE while rolling. I.e., the entire car lifts up a little while cornering. Hence the term "roll center"; it doesn't roll right along the ground plane. Ok, this is getting complicated and off the top of my head I can't think of a way to get the "right" weight transfer, but I think it would still be useful to see the ratios of right/left weight at this moment with and without the roll center height: With ground level or no roll center: Left vertical force while cornering = 160.85 lb Right vertical force while cornering = 1447.65 lb Right divided by left: 1447.65 / 160.85 Right divided by left: 0.9 We have 9 times as much weight on the right wheel as the left if the roll center is at ground level. What if it was raised to our 6 inch (0.5 foot) height? Left total vertical force = 128.68 lb Right total vertical force = 1737.18 lb Right divided by left: 1737.18 / 128.68 Right divided by left: 13.5 We have 13.5 times as much weight on the right wheel as the left if the roll center is raised to 0.5 times at this moment in time. In reality, the front of the car will rise, and the body will assume a smaller roll angle because the left tire is not pushing upwards as hard (twisting the body to the right), and the right tire is pushing up much harder ((twisting the body to the left). Once things even out, we'll have more weight transfer at the front than before. ------------------------------------------------ What happens if we now move the roll center to the right by 12 inches (1 foot)? Remember our initial "no roll center" weight distribution for our 3217 lb car cornering at 1g: Left vertical force while cornering = 160.85 lb Right vertical force while cornering = 1447.65 lb We'll start from that and do the same calculations as we did with the raised roll center, only this time we will move it towards the right tire 1 foot. First, we need to know the angles from the contact patches to the roll center. They will be different now because the roll center is no longer at the center of the chassis. Honestly, I don't know what to expect, and these calculations do not really give the new steady state vertical forces at each wheel, but I think the trends should be interesting to see if we proceed anyway: We now have two different angles as described. The roll center is located 1.5 feet inboard of and 0.5 feet above the right tire's force centroid (center of contact patch): right tire angle = arc_tangent (0.5/1.5) right tire angle = 18.33 degrees The roll center is located 3.5 feet inboard of and 0.5 feet up: left tire angle = arc_tangent (0.5/3.5) left tire angle = 8.13 degrees Here are our "no roll center" lateral forces from before: Left lateral force = 160.85 lb Right lateral force = 1447.65 lb We have to calculate two different jacking forces as before, but this time we have two different angles. We don't actually need to know the angle, we just need to know the 0.5/1.5 = 0.3333 and 0.5/3.5 = 0.1429 quantities. Left vertical force (jacking force) = 160.85 * -0.1429 Right vertical force (jacking force) = 1447.65 * 0.3333 Left vertical force (jacking force) = -22.99 lb Right vertical force (jacking force) = 482.5 lb Let's stop a second. These are the portions of the force acting vertically directly through the suspension links because the roll center is located 1 foot to the right and 0.5 feet above ground level at the chassis centerline. Let's compare these to what we had with the roll center raised 0.5 feet, but still at the chassis' centerline: Roll center 0.5 feet up and 1 foot right: Left vertical force (jacking force) = -22.99 lb Right vertical force (jacking force) = 482.5 lb Roll center 0.5 up and center: Left vertical force (jacking force) = -32.17 lb Right vertical force (jacking force) = 289.53 lb From this, even though the numbers do NOT reflect the true steady state values that this car will observe, it appears that moving the roll center towards the outside of the corner increases front weight transfer. That would generally cause understeer in a vehicle with an open or limited slip open to a degree, but could cause oversteer with a tighter rear, I suppose, in the same manner increased caster could. Note that the inside, lightly loaded tire's jacking force pulled down on the chassis a bit less with the roll center off to the right, but the much more influential outside tire had a much steeper angle (18.3 degrees vs. 8.3 degrees) between the contact patch force centroid and the roll center. In conclusion, it would probably be best to focus mostly on the height at which the line drawn from the contact patch to the outside tire's instant center. Moving the roll center to the outside of the turn increases the "roll center height" in the context you were speaking of at the outside, but decreases it at the inside. However, there should be a double-whammy effect, where the outside tire has more cornering force to cause jacking and at the same time the angle increases more quickly. Sorry if that post was way too long. This was an interesting problem that I've wanted to work through in a general way like this for some time, and you've provided the opportunity and motivation to do so :-) I'm involved in a radio controlled car simulator project now where we are using manufacturer specified suspension geometry in real time. It's facinating to watch the instant centers and roll centers move in real time, and puts a whole new spin on things. Most noteworthy with one design was how the front roll center rises and falls much more quickly than the rear, giving more oversteer off throttle or under braking than you'd otherwise get, and more understeer on throttle during corner exit than you'd otherwise see. Either way, the roll centers moved towards the inside corner substantially. I found myself watching the angle the lines from the outside tire made on their way to the roll center more than anything. That should be the most important thing, but keep in mind I'm not an expert in this field :-) Hope that helps. Todd Wasson Racing Software http://PerformanceSimulations.com My car sim http://performancesimulations.com/scnshot4.htm

>The angle between the line that extends from the contact patch to the instant >center of that wheel's suspension is what dictates the jacking force. > >That's so important I'll say it again: > > And this time I'll say it correctly :-D : The angle between the line that extends from the contact patch to the instant center of that wheel's suspension **and the ground** along with the lateral force is what dictates the jacking force. It's a bit hard to imagine an angle unless there are two lines :-D Sorry 'bout that. Todd Wasson Racing Software http://PerformanceSimulations.com My car sim http://performancesimulations.com/scnshot4.htm

Thanks for the info Todd. I'm going to work through some numbers you gave me along with the work I've done so far and see what I come up with. If you're interested, I'll let you know how it goes. I have quite a bit of data to compare against and I'm hoping to get some telemetry data very soon too. What I've developed at this point is working for NASCAR Late Model car setups. I had a chance to show it to a NASCAR Busch team and they want to try it also. But they do some interesting things to the front roll center and I'm not sure I model it correctly enough for them. You've definetley given me some things to think about and try. Thanks again... you can go back to the bar now <g>. "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > >The angle between the line that extends from the contact patch to theinstant

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> >center of that wheel's suspension is what dictates the jacking force. > > > >That's so important I'll say it again: > > > > > > And this time I'll say it correctly :-D : > > The angle between the line that extends from the contact patch to the

instant > center of that wheel's suspension **and the ground** along with thelateral > force is what dictates the jacking force. > > It's a bit hard to imagine an angle unless there are two lines :-D Sorry'bout

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> that. > > > > > > > > Todd Wasson > Racing Software > http://PerformanceSimulations.com > My car sim > http://performancesimulations.com/scnshot4.htm >

On 02 Mar 2003 11:44:09 GMT, [email hidden] (J. Todd Wasson) wrote: ... >Hi Joel. >Second, I just got back from the bar. :-D Did you know you talk too much when you're drunk? ;-) Now I have to read this small book you've written, lol. Ruud van Gaal Free car sim: http://www.racer.nl/ Pencil art : http://www.marketgraph.nl/gallery/

>>Hi Joel. >>Second, I just got back from the bar. :-D > >Did you know you talk too much when you're drunk? ;-) > >Now I have to read this small book you've written, lol. :-D Todd Wasson Racing Software http://PerformanceSimulations.com My car sim http://performancesimulations.com/scnshot4.htm

>Thanks for the info Todd. I'm going to work through some numbers you gave >me along with the work I've done so far and see what I come up with. If >you're interested, I'll let you know how it goes. I have quite a bit of >data to compare against and I'm hoping to get some telemetry data very soon >too. > Please do share! >What I've developed at this point is working for NASCAR Late Model car >setups. I had a chance to show it to a NASCAR Busch team and they want to >try it also. But they do some interesting things to the front roll center >and I'm not sure I model it correctly enough for them. > I'd love to hear more if you get time. If you can send some geometry info on the cars, I'll try it in my simulator and see how it works. Something might be learned just by watching the things move around. Maybe I could send screen shots of all that in action. >You've definetley given me some things to think about and try. Thanks >again... you can go back to the bar now <g>. Thanks ;-) Todd Wasson Racing Software http://PerformanceSimulations.com My car sim http://performancesimulations.com/scnshot4.htm