Great GRM Math Gods, I Throw Myself Upon Your Tender Mercies

But they should differ between the spring setup and the swaybar setup.

When a pure spring car undergoes torque, the inside wheel is not attempting to press down on the outside wheel. In a swaybar car, the inside is affecting the outside in that manner.

Tires do not torque a swaybar, the roll of the body torques the swaybar. The tires are simply the contact point with the ground. It's all shown there in the drawings.

Follow through the drawings in order. Think about each one until you understand what you are seeing, then move on to the next one.

My head hurts now.

Thanks a lot, stoopid math.

foxtrapper wrote: Tires do not torque a swaybar, the roll of the body torques the swaybar. The tires are simply the contact point with the ground. It's all shown there in the drawings. Follow through the drawings in order. Think about each one until you understand what you are seeing, then move on to the next one.

The outside swaybar link sees a downward load from the inside link, right? Doesn't the tire, by extension, see the same thing? Disconnect the swaybar and the swaybar links (by extension tires) should see no such load.

I understand the conventional wisdom, I really do. But I think it's either wrong or there's something happening with the swaybars I don't understand. AngryCorvair seems to accept my hypothesis, and so does Steve Hoelscher (or rather, I stole his hypotheis), the thing I'm looking for now is either a reason they are wrong or a formula showing how they are right.

The hypothesis being this, for two identical cars under identical side loading, a purely spring car will transfer less weight to the side than a spring and bar car.

Deleted, let me try something else later.

DaewooOfDeath wrote: AngryCorvair seems to accept my hypothesis, and so does Steve Hoelscher (or rather, I stole his hypotheis), the thing I'm looking for now is either a reason they are wrong or a formula showing how they are right. The hypothesis being this, for two identical cars under identical side loading, a purely spring car will transfer less weight to the side than a spring and bar car.

no, i do not accept that hypothesis.

weight transfer is a function of lateral force, CG height, and track width. period.

I can't help but think that there is some confusion arising from attempting to use simplified models which in some ways are perpendicular to reality.

For instance, in this example, there are forces being applied by the antiroll bar, but there is no differential displacement between the wheels.

If each wheel is compressed the same 5", the sway bar is having no effect.

There's nothing mathematically wrong with it, but you will never have forces from the antiroll bar without the car actually rolling.

Begin optional and possibly rambling part of post:
I also wonder whether some of DoD's math/visualization issues stem from thinking of the bar as having its own wheel rate, when its all to do with the difference between wheels. In his first example with the 500 lb/in springs and 500 lb/in bar with one wheel up 1" and the other down 1", I would have thought the example impossible because you'd sum to 1500 lb for a 1000-lb model because the bar was deflected 2". Well, in reality you couldn't deflect the bar 2" because the outside spring would've stopped pushing it when it got to full extension. In short, the compression and extension values weren't compatible with the rates involved. It wouldn't happen in reality. Sometimes taking a concept to its logical conclusion can make it easier to visualize, but you have to be careful not to make bad assumptions.

OK. I'm on track. For starters, Steve Hoelscher's work is cool. But please note that he's just working ball park #s for the springs, and trial-and-error tweaking that with sway bars. He's not coming anywhere close to calculating what you seem to want. Here's where I would start to get what you want. (btw, both your spring forces and your sway bar forces are based on suspension travel, which is a factor both of straight-forward weight transfer AND vehicle roll tendencies. That is, it's based on linear forces and the torque from Center of Mass around Instant Roll Center. Are you sure you want to continue this? ) Really, go back to Adam's book, work your way through chapters 2 and 3. You're gonna have to know your Center of Mass height and your Instant Roll Center path. From that I'm pretty sure I can get you a differential equation that will allow you to plug in values for spring rate and bar rate, but if you also want to adjust ride height, we're gonna need to build a computer program. Which leads me to believe someone else already has. (pause as I go to another desktop to play with Speed Wiz) I'm not sure I couldn't do what you want by inputting guess-and-try # into Speed Wiz, but that would take a full line-up of chassis dimensions. My thought is that your model you are looking at is way too simple, so you really can't "see" what's going on. What you think should be a simple question seems to me to be pretty complicated.

AngryCorvair wrote:

DaewooOfDeath wrote: AngryCorvair seems to accept my hypothesis, and so does Steve Hoelscher (or rather, I stole his hypotheis), the thing I'm looking for now is either a reason they are wrong or a formula showing how they are right. The hypothesis being this, for two identical cars under identical side loading, a purely spring car will transfer less weight to the side than a spring and bar car.

no, i do not accept that hypothesis. weight transfer is a function of lateral force, CG height, and track width. period.

Then I misunderstood this.

so to answer your last question, YES the tires most definitely do see the forces from the bar as you have described them. vertical force is increased on the outside tire and decreased on the inside tire.

Please elucidate.

If the tires are seeing vertical load from the swaybars' resisting roll, how is that not weight transfer?

pstrbrc wrote: OK. I'm on track. For starters, Steve Hoelscher's work is cool. But please note that he's just working ball park #s for the springs, and trial-and-error tweaking that with sway bars. He's not coming anywhere close to calculating what you seem to want.

He's actually getting rid of the bars completely, on the theory that they increase side to side weight transfer and therefore reduce the total mechanical grip available to the system.

Supposedly he has calculated this stuff. Supposedly there is a formula for determining how much weight transfer a swaybar adds as a result of resisting roll, I just don't have access. The drawbacks to living in Asia and so forth...

Here's where I would start to get what you want. (btw, both your spring forces and your sway bar forces are based on suspension travel, which is a factor both of straight-forward weight transfer AND vehicle roll tendencies. That is, it's based on linear forces and the torque from Center of Mass around Instant Roll Center. Are you sure you want to continue this? )

Absolutely, I'd like to continue. I have roll center heights mapped front and rear. I have a rough maximum cornering G and I have a precise amount of body roll (2.7 degrees) reached at maximum cornering G.

Really, go back to Adam's book, work your way through chapters 2 and 3. You're gonna have to know your Center of Mass height and your Instant Roll Center path.

Center of mass is roughly 22 inches. Roll center is 3.5 inches above ground front and slightly more than 7 inches above ground in the rear.

From that I'm pretty sure I can get you a differential equation that will allow you to plug in values for spring rate and bar rate, but if you also want to adjust ride height, we're gonna need to build a computer program. Which leads me to believe someone else already has. (pause as I go to another desktop to play with Speed Wiz) I'm not sure I couldn't do what you want by inputting guess-and-try # into Speed Wiz, but that would take a full line-up of chassis dimensions. My thought is that your model you are looking at is way too simple, so you really can't "see" what's going on. What you think should be a simple question seems to me to be pretty complicated.

I would definitely get lost at the differential equations part of this.

The model is attempting to show a single thing, the vectors the swaybars put on the struts (and therefore the tires). Ie, I'm assuming Hoelscher is correct and trying to conceptualize it. The frustrating thing is that I know my math is wrong but I don't know how.

AHA! Got it. Yeah, I think it's do-able. Let me get some work done, and I'll be back. Actually, I think you're asking an interesting question. I'd like to see if the math supports his contention about a spring-only being more consistent from surface to surface. I think we'll need a different model than the one that's been used above, just to make the math more intuitive. To actually crunch #'s I'll also need track width. Weight distribution? Really need height of Center of Mass axis where it crosses above the axle centerline, but we can get some ballpark # without it. Is this on the Daewoo? That is, fwd sedan? If so, I can get a fudge value for this. If we're just trying to prove a concept, the precise values won't matter too much.

BTW, do you know the roll center location at 2.7* of body roll?

Daewooofdeath,

Lets try it this way, no words, just pictures. Don't add anything to it, just follow what I'm showing. When you think you understand it, try describing it. We'll go from there and build up your understanding.

Now lets build a car.

pstrbrc wrote: AHA! Got it. Yeah, I think it's do-able. Let me get some work done, and I'll be back. Actually, I think you're asking an interesting question. I'd like to see if the math supports his contention about a spring-only being more consistent from surface to surface. I think we'll need a different model than the one that's been used above, just to make the math more intuitive. To actually crunch #'s I'll also need track width. Weight distribution? Really need height of Center of Mass axis where it crosses above the axle centerline, but we can get some ballpark # without it. Is this on the Daewoo? That is, fwd sedan? If so, I can get a fudge value for this. If we're just trying to prove a concept, the precise values won't matter too much. BTW, do you know the roll center location at 2.7* of body roll?

Yep, this is all from the Daewoo. Track width is 68 inches front and 66.5 rear. Weight distribution is, depending on fuel load/what I have in the trunk etc, approximately 62/38. In race situations it might actually be a little more nose heavy because I prefer to run with low fuel load and I gutted the trunk.

Center of Mass axis ... I'm not familiar with this concept.

Fudge away, in other words. Or, if you want more information, just let me know and I'll get it. I have virtual arms, wheelbase, motion ratios, spring rates - all kinds of stuff.

As for the roll center location at body roll, I have no idea. I got that figure by taking a picture of the car at max cornering g, steady throttle, high mu surface and then blowing that picture up and using a compass.

foxtrapper wrote: Daewooofdeath, Lets try it this way, no words, just pictures. Don't add anything to it, just follow what I'm showing. When you think you understand it, try describing it. We'll go from there and build up your understanding.

These are all vector paths. If I'm not mistaken, they show the effects of torque (the red arrows that always point different directions on different sides), the static loads, the loads of joints and the effect of the ground pushing up against the load, which is pushing down.

I realize this is all out of order. Sorry, Chrome doesn't let me look at pictures and reply at the same time.

Now lets build a car.

Now we add the complicating factors of springs, which are the green arrows. The springs, in the first couple pictures, are simply resisting vertical load. In the next pictures, they resist torque. In the final pictures, we get a second set of "springs" - swaybars - that resist the torque by creating a downward torque vector on the compressing side and an equal/opposite upward vector on the drooping side.

In this, we agree perfectly, I think. It is in the ramifications of those added swaybar vectors that I don't think we are understanding each other.

Let me attempt to illustrate my point in a way you can easily replicate at home.

Look at these two football players. For the purposes of this exercise, the white guy in the dark green jersey is the side loading force, which causes the torque vector which causes the black guy in the white jersey to tilt sideways. The black guy represents our suspension across a single axle. His left leg is the outside suspension and his right leg is the inside suspension.

Now, try replicating this position with a friend/spouse/child where you are the black guy and the helper is the white guy. Keep the "spring rates" (your legs) at a constant stiffness. Feel the amount of load that is transmitted to your "outside tire." You will be leaning over, just like your torque vectors above predict.

Now, attempt to replicate the action of a swaybar. Increase the "spring rate" of your outside leg and soften your inside leg. Take note about how this both straightens you up (reduces body roll) AND increases the load on your outside foot.

This is why I think sway bars increase lateral weight transfer.

BTW, if doing this exercise involves too much public humiliation, I totally understand.

Two football players, four feet. That's a full car with four wheels. That adds another dimension.

My drawings are of one football player. A Segway if you will.

With a complete car, you can take torque from the rear and transfer it to the front. Resulting in more torque in the front. You didn't create new torque, you simply relocated it from one end of the car to the other. The total torque for the car stays the same. Same the two football players.

foxtrapper wrote: Two football players, four feet. That's a full car with four wheels. That adds another dimension. My drawings are of one football player. A Segway if you will. With a complete car, you can take torque from the rear and transfer it to the front. Resulting in more torque in the front. You didn't create new torque, you simply relocated it from one end of the car to the other. The total torque for the car stays the same. Same the two football players.

I think you are misunderstanding me. We are not changing torque. The white guy's force is a constant in this example. We're ignoring the white guy's "suspension" and treating him as a torque vector. The black guy is the only one we're paying attention to.

So, if he wants to reduce the amount of roll the torque vector is causing him, he will increase the stiffness of his outside leg and soften the inside leg. This is exactly what a swaybar does when it transfers spring rate from the inside spring to the outside spring. If you act it out, you will see that the process of transferring spring from the inside to the outside increases the load on the outside while reducing the load on the inside.

The swaybar, in other words, resists roll by causing weight transfer.

DaewooOfDeath wrote: The swaybar, in other words, resists roll by causing weight transfer.

You know, I think you're right. The prelim sketches and doodle math seem to support that. Give me a little more time, and I should have worked this out. Life has been hectic!

DaewooOfDeath wrote: So, if he wants to reduce the amount of roll the torque vector is causing him, he will increase the stiffness of his outside leg and soften the inside leg. This is exactly what a swaybar does when it transfers spring rate from the inside spring to the outside spring. If you act it out, you will see that the process of transferring spring from the inside to the outside increases the load on the outside while reducing the load on the inside. The swaybar, in other words, resists roll by causing weight transfer.

Yes, but compressing the outside spring and extending the inside spring has exactly the same effect (more force on spring = more compression and more force on that tire). And both torque via the swaybar and spring compression/extension are caused by body roll which in turn is caused by the roll torque (neither springs nor bar are proactive, both require displacement of the wheels to provide any change in force resisting roll; i.e. you have to have some roll in order to develop any roll-resisting force).

The sway bar doesn't cause weight transfer any more than the springs do, just by a different mechanism.

For a given cornering force (a given roll torque) the body will roll further until that force reaches equilibrium with the roll resistance of the springing mediums (both springs and sway bars combined).

Any given amount of force to counter roll can be achieved with only springs, with springs and sway bar, or even with a monoshock and sway bar in which the springs contribute nothing to roll resistance.

In the simplifications we've made, we are ignoring that the CG moves outward during roll. Fine, it's a small thing and we can ignore that to get the big picture.

In steady state cornering, if we are not going to look at the full car and front vs rear rolls stiffness but instead look simply at weight transfer, and as long as we are going to continue to ignore the movement of the CG due to roll, the simplest diagram showing weight transfer looks just about like this:

This is from here, and they're talking about braking, but the vector math applies. Bf and Br in the image are front and rear braking force. For our purposes, view the diagram as Bf + Br = cornering force.

If we consider the "weight of vehicle" vector to be a 1G unit, and draw the "cornering" (Bf + Bg in this gif) to the same scale, when the cornering force increases to the point that the "Weight Transfer Vector" is pointed from the CG directly to the outside tire's contact patch, at that point the outside tire will carry 100% of the load, and the inside will have 0.

I apologize if what this thread didn't need was another voice, but it seemed possible to me that we were still hoping to dislodge just the right piece of perspective to make things fall into place, and you never know what will do that...

DaewooOfDeath wrote: What I am trying to understand is if using a roll bar to counteract body roll contributes significantly to weight transfer across a single axle, not from front to rear, or rear to front.

You really can't break it up this way.

There is in many cases less influence across a single axle then there is from rear to front, or vise versa.

When a cornering load transfers to force on the right front tire, it reduces load on the left rear. Therefore the REAR anti-roll bar contributes to the front body roll.

Obviously there is a lot more to it, but...

The sway bar will generate understeer by increasing the load on the outside front tire. Once the tire hits its circle of friction limit, it slides. Bingo: understeer.

At rest, with both sides of the suspension at equal compression, the sway bar does nothing. It's just along for the ride. The same holds true if a bump affects both wheels equally.

Once the car begins to turn, the weight begins to shift to the outside due to the body's weight rolling to the outside. Once this happens, the sway bar is deflected upward at the control arm end on the outside. The distance it deflects under a given load is entirely dependent on the bar's physical properties and the length of the 'arm' from the chassis mount point to the link to the control arm.

The force required to cause that deflection is the amount of additional weight (or force, if you prefer) transferred to the outside tire by the bar. More force required to deflect = stiffer, less force = softer.

Example: Let's say the bar takes 250 pounds of force to deflect 2 inches at the control arm end. Once the outside suspension deflects the bar 2" upward the bar is transferring that 250 pounds to the outside tire.

But the bar does NOT change the total weight on both front tires, only when and how much weight is transferred to the outside tire by the bar.

Does this help?

It's possible to get real close to swaybar actions by going stupid stiff with the springs. Basically, the stupid stiff springs take over the job of transferring weight (or force, or body roll) to the outside tire that the sway bar was doing.

But as Hoeschler points out, that's also very dependent on roll centers and suspension geometry. Rough (very rough) oversimplification of his ideas: go stupid stiff on the springs and drop the roll center to minimize body roll, then the control arm angle at rest needs to be such that as the car does roll the limited amount it's allowed it can't jack itself up. That would generally mean the outer ball joint would be higher than the inner pivot.

Something else... by resisting the roll of the body, the sway bar also minimizes the movement of the roll center. But that's another whole discussion.

OK, I've just wadded up my 20somethingth page off the legal pad. This opens up a whole can of worms. HOWEVER...

I'm gonna just start off with the first question: Does a sway bar actually add more weight to the outside tire, when only considering a single axle? Answer: No, if you could isolate an axle. The springs and sway bar merely attach the chassis to the axle, and the distance between the chassis and the axle are irrelevant to the force at the tire's contact patch. However, this is only relevant to a car that has a pivot at the center of mass. Yeah. A John Deere 4wd tractor. Otherwise, if one end of the chassis wants to use more suspension travel to come to stasis than the other end, there will be a torque around an axis that runs through the roll centers, loading one outside wheel with the other end's weight. And the more I work on the math, the more I think that this is the real value of roll centers: The moment-arms of the torque that each end exerts on the other really make the difference. A high rear roll center combined with a low front roll center, and a high front Center of mass combined with a low rear center of mass, make a fwd compact sedan w/struts a pig to really make handle. (If, however, one could lower the inner suspension pickups in the rear, just bringing them down 2" lowers the roll center from 7" to 2.75", which is lower than the front. Escorts are the FWD compact sedan I'm familiar with, and you should see how easy it's gonna be to test this!)

(whew) And I'm still working on the calculations! However, I can tell you that I'm beginning to see the justification for Hoelscher's argument that springs-only are more consistent than springs+bars. The calculation of this torque battle of the roll centers at 1g come out much different than at .5g, when using springs+bars. When using springs-only, it looks remarkably close. Now, for the complicated part. The car: 2000#, 60/40 58" track 100" wheelbase Center of Mass is at 40" behind and 22" above the front tire contact axis. Spring rate: front-180#/in, rear-120#/in Roll Center Height, front-3.5", rear-7" (just in case anybody's wondering, DaeWoo Of Death's questions got me crawling around a couple of cars in my garage. These #'s are a combination of what he gave above and specs of a chump-ready '95 Escort Wagon. ) Now, I do wonder where the Axis of Mass passes over the front and rear tire axis. I have the math to do this, now I need chassis scales. (shrug) But let's work with what I have.

I'm gonna have to clean up the math and the graphics, but I have real work to do. So, in my absence, discuss.

I think what pstrbc is saying about the roll center is that this is the point which determines how much weight goes where. Rough example: in a turn if the roll center stays centered between the wheels then (roughly, assuming symmetrical weight) a percentage more than half of the weight goes to the outside wheel. That number would be due to the distance from the roll center to the tire being equal on both sides.

Now if the roll center moves toward the outside it now has a shorter 'lever' to act on the tire, so there is still weight transfer only a lower percentage than if the roll center moved AWAY from the outside tire.

Or I could be 100% full of E36 M3, it wouldn't be the first time.

By the way, the sway bar can also be said to be pushing UP on the car body. Either way, the force generated has to go somewhere ('for every action,' etc) and it's probably most convenient for our purposes to think of it as adding to the outside front tire's load.

DaewooOfDeath wrote: AngryCorvair wrote: so to answer your last question, YES the tires most definitely do see the forces from the bar as you have described them. vertical force is increased on the outside tire and decreased on the inside tire. Please elucidate. If the tires are seeing vertical load from the swaybars' resisting roll, how is that not weight transfer?

dude, that is weight transfer. the bar increases the vertical load on the outside tire, and it decreases vertical load on the inside tire.

AngryCorvair wrote:

DaewooOfDeath wrote: AngryCorvair wrote: so to answer your last question, YES the tires most definitely do see the forces from the bar as you have described them. vertical force is increased on the outside tire and decreased on the inside tire. Please elucidate. If the tires are seeing vertical load from the swaybars' resisting roll, how is that not weight transfer?

dude, that *is* weight transfer. the bar increases the vertical load on the outside tire, and it decreases vertical load on the inside tire.

OK, physics here. The springs and bar do not increase weight on an outside tire from an inside tire on the same axle. The springs and bar just allow the relationship of the body to the wheels to change, based on the same forces that would be there if the body was mounted solidly on the axle. However, the force that the spring pushes down an the axle is equal to the force that the body presses down on the top of the spring, and the force that the bar pushes down on the axle is... get this... equal to the force that the body pushes down on the bar at the bushing, and the force that the bar exerts upward on the other side is... get this...equal to the force that the body pushes upward on the bushing on that side. Think about this, guys. unfasten the bar at the bushings and tell me what happens. Now just duct-tape the bushings back down. Waddya think is gonna happen? Now. By using springs and bars appropriately you can shift the weight a tire is carrying to the opposite corner. But, if you're talking same axle, springs bars actually keep the weight on the inside corner, at least comparing "stiff" springs and bars to "soft" springs and no bars. How? because a soft suspension allows the center of mass to move outward as the body rolls. stiff springs and bars keeps the center of mass directly over the centerline. But it's not really much.