Weight transfer and roll resistance.

[Ben Ogle]

First off I should explain why this is in this forum. Its in no way an article or a teg tip and I am not asking assistance in working on my suspension (but I do have questions). It relates to racing/race cars (well all cars, but we only care because of racing/competition, right?) and fits in no other forums so here it is.

I don’t know if everyone read the tire stagger thread where I just argued with mythos EF/DA for 3 pages, but it was that thread that once again sparked my interest in the suspension tuning books I bought a while ago. In looking for quotes to counter mythos with I found that I missed a lot of stuff through the first read. That and, at the time, I didn’t understand some of the things the authors were saying. I am not saying that I now understand everything because if I did I wouldn’t have made this thread. I do understand more but there are some parts that are still a little greek. I am also making this thread because typing this stuff out makes me understand better. Gnome sayin? Yeah, so on with it, eh?

I’m gonna start with the terms.

Center of Gravity – Also called the center of mass and I’m gonna abbreviate it CG. The CG is the most center concentration of mass on the car. If you picked up the car in this spot on a worthy string all 4 tires would leave the ground at the same time. If you decided to balance the car on a pole, the car would rest on the top of pole at the CG. The CG also has a height and this is the most important axis of the CG. (man I need a pic, lol)

Mass Centroid Axis – If you sliced the car into little sections (like a loaf of bread) each section would have its own CG height. The mass centorid axis is an axis created if all these little CG’s were connected together.

Roll center – The front and rear suspension each have their own roll center locations. These roll centers are in the same plane as the respective suspension, and are determined by the suspension geometry. To find the roll center you get a front or rear view (an elevation, if you will, of whichever end you want to find the RC for) of the suspension, extend the upper and lower arms into the car until they intersect, then draw a line from the center of the tire’s contact patch to the intersection of the extended arms, and where that last line drawn intersects with the center of the car is the roll center (technically its where this line from the contact patch intersects the line from the other tire’s contact patch, but when the car is sitting the roll center is in the center of the suspension).

picture stolen from Fred Puhn's "How to Make Your Car Handle"


Roll axis – Think of connecting the roll centers to each other by an imaginary line. This line is the roll axis.

Moment arm – This is the distance from the roll center to the CG’s height.

Roll couple – This is the torque generated by the moment arm about the roll center when weight is transferred laterally.

Now, when we go around a corner weight is obviously transferred laterally. It goes from the inside tires to the outside ones. This we can see and feel in the form of body roll. But I think it should be noted that body roll does not cause this weight transfer. The centrifugal force of the car trying to go off on a tangent causes weight transfer and weight transfer causes body roll.

We can calculate the total, overall weight transfer of the car by this equation:

Lateral load transfer (lb) = (Lateral acceleration (g) x car weight (lb) x CG Height (in))/track width(in)

As you can see, the equation doesn’t take the type or stiffness of the suspension into account. That’s because it doesn’t matter for the overall lateral weight transfer.

Now, in the words of Carroll Smith “Lateral load transfer is a bad thing.” This is because “any transfer of load from one tire of a pair to the other reduces the total tractive capacity of the pair.” So we obviously want to reduce lateral load transfer.

That overall load transfer can be broken down into 3 different kinds of load transfer, unsprung weight transfer (wheels, knuckles, brakes, etc…), weight transfer through the roll centers, and weight transfer of the sprung mass. If you add them all up you get the overall lateral load transfer. Each is important in its own respect but the weight transfer of the sprung mass is the most important to us because we can tune it.

Unsprung weight transfer – This is the least important of the three. Some weight is transferred because of it, and there isn’t much you can do about it.

Weight transfer through the roll centers – When we go around a turn, there is weight transfer that goes directly through the roll centers to the other, outside tires. There is more of this kind of weight transfer as the roll centers go up. Also, the heavier end (ie the front) gets more of this kind of weight transfer. Our cars’ roll centers are fairly low, especially when lowered. So this isn’t as bad as it could be.

Weight transfer of the sprung mass – Typically this type of weight transfer takes up most of the overall weight transfer, which is a good thing cause we can tune it. The portion of sprung weight transfer that each end gets can be changed by roll stiffness. The higher the roll stiffness is, the more sprung weight is transferred at that end. Some quotes on this are:

“The ratio of front to rear sprung weight transfer is directly proportional to the ratio of front to rear roll resistance.” (“Competition Car Suspension” by Allan Staniforth, P204)

“The suspension with the highest roll stiffness will receive the largest portion of weight transfer caused by body roll.” (I think what he means by “weight transfer caused by body roll” is “weight transfer of the sprung mass” because that’s what he is talking about when he says that)(“How to Make Your Car Handle” by Fred Puhn P41)

Now is when the stuff gets interesting (and where I am a little and maybe even a lot confused). What happens is the remaining centrifugal force (the portion of the force that isn’t creating the other 2 types of weight transfer) pulls the mass centroid axis outward and, because the tires stick to the ground, the mass centroid axis creates a torque around the roll axis. This torque is called the roll couple. The front and rear suspensions each have their own roll couple and the front and rear roll stiffness resist their respective roll couples. Roll couple equations from Fred Puhn’s book:

Front roll couple = (Front roll stiffness/Total roll stiffness) x Total roll couple

Rear roll couple = (Rear roll stiffness/Total roll stiffness) x Total roll couple

Total roll couple = Front roll couple + Rear roll couple

According to these, as the stiffness goes up at either end so does the total roll couple and the roll couple at that end.

Carol Smith says, “The greater the resistance of the springs, the less roll will result – but there will be no significant effect on the amount of lateral load transfer because the roll couple has not been changed and there is no physical connection between the springs on opposite sides of the car. The same cannot be said of the resistance of the anti-roll bars. In this case, because the bar is a direct physical connection between the outside wheel and the inside wheel, increasing stiffness of the anti-roll bar will both decrease roll angle and increase lateral load transfer.” How is the roll couple not changed? If the roll couple is not changed, how do springs then change the nature of the car? Could someone please explain this?

I guess what I am getting at is the “rates” people talk about. Smith says, “If the roll axis at one end of the car is further below the mass centroid axis than it is at the other end, then that end of the car will have a greater roll moment and therefore lateral load transfer will take place more quickly at that end, and traction will suffer.” I assuming that when he says “more quickly” he means “sooner.” Example: lets say the outside tires have infinite grip, if you increase the lateral acceleration to infinity both the inside tires are going to lift off the ground (lets say both at the same time) and eventually the car is going to roll over. If we add rear roll stiffness to our car, the rear tires will transfer their sprung weight more quickly and will come off the ground sooner than the fronts. Is this right? So if it is the rates of weight transfer that actually make the car more understeer or oversteer prone then adding rear stiffness just raises the rear roll couple. But the only way it could do that is by raising the centrifugal force. I dunno.

If it is just about rates then my guess is that as the rear stiffness goes up so does the roll couple (regardless of type of roll resistance). The only way the roll couple can go up is if the force on the moment arm goes up (cause the moment arm doesn’t really change). Force = mass x acceleration so when the force goes up the acceleration has to go up because the mass says the same. Acceleration == rate. So the more rear roll resistance there is the faster (sooner) the weight transfer at rear happens.

Also, if the roll stiffness tuning is just about changing the rates of weight transfer then does that mean that the other theory of transferring more sprung weight on the stiff end is wrong? Or does it just mean that it transfers the same amount it would’ve eventually anyway, just sooner, so it thus transfers more at some specific instant (see example 2 paragraphs up).

This took waaaaaaaaaaaaaaay too long to type. I hope someone can reply with some help (marc!?!, GspeedR?, Johnny?, Robbie?).

Ben

[cspeed]

ben, great writeup/discussion. i read through it all, particularly the last few paragraphs of your discussion that includes the example you offered. i'm not as technically fluent as you or many others on this board about suspension theory, but i like to think common sense and experience is another means to understanding. taking this into account, i agree with what you were implicating in your discussion.

you said: "Also, if the roll stiffness tuning is just about changing the rates of weight transfer then does that mean that the other theory of transferring more sprung weight on the stiff end is wrong? Or does it just mean that it transfers the same amount it would’ve eventually anyway, just sooner, so it thus transfers more at some specific instant"

i don't think 'wrong' is a correct term for it. 'wrong' would be where you are trying to overcome the laws of physics/geometries of the suspension--which doesn't work. but i think this part of the discussion would be definitely driver theory/belief/preference because i'm sure transferring sprung weight vs. increasing roll resistance (via thicker anti-roll bars), for example, 'feel' different when you're out on a course (also, the type of course will will have its influences, too). any given change over another may be progressive or more subtle, for example, and that would depend on the driver's preferences.

sorry i can't add anything particularly scientific, hopefully others will help out constructively. thanks for taking the time ben.

[Integ5150]

fakkk... that ish hurt my eyes....

*bookmarks* archivethis

[GSpeedR]

Originally posted by Ben Ogle
Now is when the stuff gets interesting (and where I am a little and maybe even a lot confused). What happens is the remaining centrifugal force (the portion of the force that isn’t creating the other 2 types of weight transfer) pulls the mass centroid axis outward and, because the tires stick to the ground, the mass centroid axis creates a torque around the roll axis. This torque is called the roll couple. The front and rear suspensions each have their own roll couple and the front and rear roll stiffness resist their respective roll couples. Roll couple equations from Fred Puhn’s book:

Front roll couple = (Front roll stiffness/Total roll stiffness) x Total roll couple

Rear roll couple = (Rear roll stiffness/Total roll stiffness) x Total roll couple

Total roll couple = Front roll couple + Rear roll couple

According to these, as the stiffness goes up at either end so does the total roll couple and the roll couple at that end.

Carol Smith says, “The greater the resistance of the springs, the less roll will result – but there will be no significant effect on the amount of lateral load transfer because the roll couple has not been changed and there is no physical connection between the springs on opposite sides of the car. The same cannot be said of the resistance of the anti-roll bars. In this case, because the bar is a direct physical connection between the outside wheel and the inside wheel, increasing stiffness of the anti-roll bar will both decrease roll angle and increase lateral load transfer.” How is the roll couple not changed? If the roll couple is not changed, how do springs then change the nature of the car? Could someone please explain this?

I was also a bit confused at this. I believe that it has to do with the connection between the two wheels. Since automotive springs can't apply an extending force to the chassis (in droop), springs only apply forces to the outer tire in bump. The roll bar does apply a "jacking-type" force to the inside tire. The change in roll couple (I think) comes from the fact that the roll moment lowers since the anti-roll bar forces the inside tire into bump (control arms bend up and instant center lowers). The outside tire is obvioulsy in bump in both cases. I don't have a model to check this, but I'm guessing that (in steady-state cornering, which Smith is addressing), the change in roll center height from the outer tire going into bump is negated by the inner tire going into droop. The difficulty I have is why the force component of the roll couple hasn't decreased in both cases.

I guess what I am getting at is the “rates” people talk about. Smith says, “If the roll axis at one end of the car is further below the mass centroid axis than it is at the other end, then that end of the car will have a greater roll moment and therefore lateral load transfer will take place more quickly at that end, and traction will suffer.” I assuming that when he says “more quickly” he means “sooner.” Example: lets say the outside tires have infinite grip, if you increase the lateral acceleration to infinity both the inside tires are going to lift off the ground (lets say both at the same time) and eventually the car is going to roll over. If we add rear roll stiffness to our car, the rear tires will transfer their sprung weight more quickly and will come off the ground sooner than the fronts. Is this right?

I think that when he talks about transfer happening faster, he actually means "more rapidly". This causes a reduction in traction at the tire, because of how a tire functions. It takes a finite amount of time for the tire to "readjust" to a different force and the less time it has, the less grip it can produce (molecular adhesion). So the tires don't like rapid transitions. That's just the problem that has to be dealt with slip angles: essentially if you turned the wheel incredibly slowly, the tire would never operate at a slip angle. You can't do this in real life, since you'd fly right off the track. It is true that the tractive capacity of the end with high roll stiffness is dimished faster and thus the other end has more work to do.

So if it is the rates of weight transfer that actually make the car more understeer or oversteer prone then adding rear stiffness just raises the rear roll couple. But the only way it could do that is by raising the centrifugal force. I dunno.

Or change the roll moment at the rear, which would happen under squat. So technically the car should have the least total grip while turning and accelerating. the roll moment lowers as the rear goes under bump, all the grip from the fronts are at the rear, and all the grip at the inside rear is over at the outiside rear (don't know why I added that).

If it is just about rates then my guess is that as the rear stiffness goes up so does the roll couple (regardless of type of roll resistance). The only way the roll couple can go up is if the force on the moment arm goes up (cause the moment arm doesn’t really change). Force = mass x acceleration so when the force goes up the acceleration has to go up because the mass says the same. Acceleration == rate. So the more rear roll resistance there is the faster (sooner) the weight transfer at rear happens.

I'm getting a bit lost here. The rate of load transfer is a change in forcewith respect to time, which isn't an acceleration. The change in rate would be the "acceleration", though it is usually described as the second derivative of position.

Also, if the roll stiffness tuning is just about changing the rates of weight transfer then does that mean that the other theory of transferring more sprung weight on the stiff end is wrong? Or does it just mean that it transfers the same amount it would’ve eventually anyway, just sooner, so it thus transfers more at some specific instant (see example 2 paragraphs up).[/QUOTE]

You'll have to elaborate on this a bit. Are you asking about the distribution of load transfer front and roll based on roll stiffness ratio, or the f/r distribution of load tranfer rate. If it's the latter, I can't help you.

You may want to PM "descartesfool" over at H-T.com. He's answered a few of my questions and could better answer yours than I could. Possibly post this in the similar thread I made at H-T.com: here

[Ben Ogle]

johnny, thanks for the input. I want to test and compare the feel of a rear spring increase to a rear sway increase. unfortunately when I autox I am too concerned with driving to really "feel" the car. poop.

GSpeed, Are you sure its the moment arm that changes enough to worry about? I have this picture sitting here (if you have Fred Puhn's book look on page 37, otherwise I can post it if you want) that shows the roll centers of an a-arm suspension at 0* and 5* roll (aour cars probably dont roll much more). The roll center moved laterally only. And it looks like the moment arm got a tad larger because of it. but it really looks like it is independent of the type of roll stiffness applied. Plus, since the sway bar provides the "jacking" force to the inside wheel shouldnt it lower the roll center faster than the CG thus creating a larger moment arm? I dunno.

The difficulty I have is why the force component of the roll couple hasn't decreased in both cases.

Are you saying the force component stays the same? Or does it increase? Do you think it should decrease because of the resistance of the springs?

Onto the disection of my confusing original post.

Originally I thought that the idea of the rates of weight transfer and the idea of the stiff end getting more of the weight transfer were 2 different, conflicting ideas. But since 3 of the 4 suspension books say the former is true (Smith only hints at it, he never really comes out and says it) then I figured they had to be related. thats where this came from:

Or does it just mean that it transfers the same amount it would’ve eventually anyway, just sooner, so it thus transfers more at some specific instant (see example 2 paragraphs up).

I cant really think of another way to explain it, except maybe looking at the example again. I am still thinking the sooner thing applies.

lets say the outside tires have infinite grip, if you increase the lateral acceleration to infinity both the inside tires are going to lift off the ground (lets say both at the same time) and eventually the car is going to roll over. If we add rear roll stiffness to our car, the rear tires will transfer their sprung weight more quickly and will come off the ground sooner than the fronts.

Ok. So lets say the car with the untouched suspension lifts both the inside tires at 1.5g so at 1.4g all 4 tires will be on the ground. Now lets say the car with the increased rear roll resistance was to hit the 1.4 g mark and it lifted the rear tire. The 2nd car would have an increase in the "rate" of weight transfer because all the rear sprung weight transferred sooner than in the first car. Also at the point that both cars hit 1.4g the 2nd car wouldve transferred more weight at the stiff end thus validating the "stiff end gets more wieght transfer" theory. Hopefully that cleared up that part of my post. note: this is just a guess, but it may be true. look at these quotes from RR98ITR on this thread: http://www.honda-tech.com/zerothread?id=285747

"Increasing rear roll stiffness will do nothing to change this fact, and will only unload the inside rear earlier in the cornering sequence"

"Under lateral acceleration the end of the car with the highest roll stiffness will transfer the highest percentage of its inside cornerweight to the outside wheel. "

The problem I have, if my rate guess is right, is that I dont understand how the rear transferring its weight "sooner" helps the front end. I suppose it could limit the roll angle for the whole car (well, it does) which would keep the front pair "flatter" (or more squarely(sp)) on the ground and would let the inside front to do more work.

GSpeed, please dont think I am arguing with you anywhere. I am just trying to understand. I have some more reading i'm gonna do before I reply more (found a couple of threads, if you have more post them up. h-t searching is down). Maybe it will answer some of my questions so I dont have to post the ones that are easily found.

Ben

[SE-37K]

This is all BS, just put 255/40-15's on the front and 215/40-17's on the back and you're set.

Oh don't forget to remove your sway bars and attack the touge with the courage of a hunting lion.

[Ben Ogle]

haha. lets not start a war.

Ben

[GSpeedR]

Ben,
I don't mind if you're arguing with me. This is stuff that is difficult for mw to grasp, and discussing it probably would help everyone. I want to re-read a few chapters before I respond though.

[GSpeedR]

Originally posted by Ben Ogle
GSpeed, Are you sure its the moment arm that changes enough to worry about? I have this picture sitting here (if you have Fred Puhn's book look on page 37, otherwise I can post it if you want) that shows the roll centers of an a-arm suspension at 0* and 5* roll (aour cars probably dont roll much more). The roll center moved laterally only. And it looks like the moment arm got a tad larger because of it. but it really looks like it is independent of the type of roll stiffness applied. Plus, since the sway bar provides the "jacking" force to the inside wheel shouldnt it lower the roll center faster than the CG thus creating a larger moment arm? I dunno.

I haven't read Puhn, but I'm assuming that he's not accounting for a swaybar in his diagrams. In roll without a sway, the inner wheel goes into droop while the outer wheel goes into bump, which is why there isn't really any vertical displacement of the roll center. With the swaybar, the innner wheel is not able to extend (springs should be still enough to prevent it from going into bump tho) and so the roll center will move both laterally and vertically. The overall displacement of the roll center (both laterally and vertically) should be a significant change to the roll couple. This is what I believe happens in steady-state cornering, but I've never read a book that "told" me that: I've had to deduce it (which means it could be wrong).

Are you saying the force component stays the same? Or does it increase? Do you think it should decrease because of the resistance of the springs?

So the above paragraph is my response to why the r in r x F changes. As for the F part, my intuition says that both the springs and the swaybar resist the inertial force of the car in roll...so I would think it should be lower in both cases. But the books all say otherwise: The force component of "just springs" does not change since the roll couple doesn't change. So, I don't know why, which sucks.

I'll have to respond to the other stuff later.

[GSpeedR]

Originally posted by Ben Ogle
I cant really think of another way to explain it, except maybe looking at the example again. I am still thinking the sooner thing applies.

I still have to dissagree with the definition of "rate" here. I'll explain why below.

Ok. So lets say the car with the untouched suspension lifts both the inside tires at 1.5g so at 1.4g all 4 tires will be on the ground. Now lets say the car with the increased rear roll resistance was to hit the 1.4 g mark and it lifted the rear tire. The 2nd car would have an increase in the "rate" of weight transfer because all the rear sprung weight transferred sooner than in the first car. Also at the point that both cars hit 1.4g the 2nd car wouldve transferred more weight at the stiff end thus validating the "stiff end gets more wieght transfer" theory. Hopefully that cleared up that part of my post. note: this is just a guess, but it may be true. look at these quotes from RR98ITR on this thread: http://www.honda-tech.com/zerothread?id=285747

Well, the rate of weight transfer explains itself: a rate is a time derivative. It determines how something changes with time (not trying to give a lesson in calc, but explaining my argument). Increasing the "rate" of anything, means that it occurs faster for the given period of time. Both cars will be transferring weight well before they hit 1.4g's: a car begins to transfer weight as soon as it accelerates. However, the 2nd is transferring faster, so when it approaches 1.4g's it has already transferred more than the 1st car. So both begin transferring at the same time (as soon as the wheel is turned), but car #2 (increased rear roll stiffness) is doing it faster and thus lifts the rear inside wheel before car #1.

"Increasing rear roll stiffness will do nothing to change this fact, and will only unload the inside rear earlier in the cornering sequence"

I can completely see why this appears to mean "sooner". The rear inside is unloaded because it has a higher rate of transfer = it happens faster. This transfer began at the same time in both cars, yet is completed (no more weight to transfer = tire is unloaded) "earlier" in the cornering sequence of the second car.

"Under lateral acceleration the end of the car with the highest roll stiffness will transfer the highest percentage of its inside cornerweight to the outside wheel. "

This is because the higher the roll stiffness, the higher the WT rate. The higher rate in the rear means that it transfers the same weight faster OR it means it transferred more weight in the same amount of time. So the distribtion of WT changes since the rear is transferring faster than the front and there is a finite amount of weight.

The problem I have, if my rate guess is right, is that I dont understand how the rear transferring its weight "sooner" helps the front end. I suppose it could limit the roll angle for the whole car (well, it does) which would keep the front pair "flatter" (or more squarely(sp)) on the ground and would let the inside front to do more work.

This involves the assumption that we have to make since we aren't using specialized computer programs: The roll angle of the front is equal to the roll angle of the rear. So the car is torionally rigid, which is not completely true, but pretty damn close. So by limiting the angle at one end, we limit the entire car's angle.

How this helps the front end, is really complicated and I don't quite know how it works. The rear wheels are basically towed (as RR98itr explains) by the car. The goal is to allow rotation in a car that is setup to understeer. If the front transferred weight faster then it would reduce front grip faster than rear and we are no longer able to rotate as well.

GSpeed, please dont think I am arguing with you anywhere. I am just trying to understand. I have some more reading i'm gonna do before I reply more (found a couple of threads, if you have more post them up. h-t searching is down). Maybe it will answer some of my questions so I dont have to post the ones that are easily found.

[Ben Ogle]

we are on the same page with the rate thing. I have been trying to explain the same things you just said, you just did a better job.

sooner == less time b/t no WT and full WT at same lateral accel, so therefore the "rate" has to increase.

so now the main questions i have are:

How do the springs and sway bars really relate to the roll couple? (I am assuming the above equation is wrong b/c the springs supposedly dot change roll couple)

How come the sway bars change the roll couple but springs do not?

If the roll couple is not changed by the springs then what is? (the rate, or %age of WT at a given time)

Does the rate of weight transfer relate to the roll couple or are they independent? (like can you tune the car without changing one or the other)

If they are not independent then how do springs change the way the car handles? (if they relate somehow then changing the springs would change the rate and the roll couple. But it has been said that they do not change the roll couple so if they are related then the springs would not change the rates.)

What I am getting at is that if the rates and roll couple are independent of each other then there is a fundamental difference between the anti-roll bar and the springs. That means that there would be a largish difference in the way the car behaved depending on the type of roll resistance chosen and the saying "sway bars act jst like springs in lateral acceleration" (which I have read in a few books) would be very untrue.

Ben

[GSpeedR]

Originally posted by Ben Ogle
How do the springs and sway bars really relate to the roll couple? (I am assuming the above equation is wrong b/c the springs supposedly dot change roll couple)

My thinking on this is that the swaybar increases lateral load transfer because it physically lessens the weight on the outside wheel by pulling it off the ground. This weight must go to the other side, which it does. In terms of

How come the sway bars change the roll couple but springs do not?

I am thinking that the roll couple is only dependent on the mass and acceleration of the sprung mass and the moment arm. So the springs and bars resist roll, but don't have an effect on the roll couple. We must change the mass, the lateral acceleration or the roll center height to do that. The swaybar does change the roll center height while the springs by themselves don't.

If the roll couple is not changed by the springs then what is? (the rate, or %age of WT at a given time)

Here is a response when I asked basically the same question:

Quote, originally posted by GSpeedR »

I would also guess that the roll stiffness has an effect on how quickly the lateral load tranfer happens, but did not find enough info. in Prepare to Win or Tune to Win.

Posted by descartesfool:
"The speed or rate at which load transfer occurs is a much more complex subject. It depends of course on the roll moment of inertia, which is a little difficult to measure. The Yaw moment of inertia can me measured by placing a car on a suspended platform and seeing how fast it oscillates, but this is a little more difficult to do for the roll moment about the roll axis. Another factor is of course the dampers. The outside front wheel is going into bump (low speed bump damping) while the inside rear wheel is going into rebound (low speed rebound damping) and the other two corners are compressing/extending as well. Jacking has to be considered. The damping force depends on how fast the rear and front are rolling which depends on roll axis inclination and roll resistance balance. Another factor is that for an inclined roll axis, with the front roll centre lower than the rear, the body moves to the outside at the rear as the car corners causing a yaw moment. The real process of cornering needs a far more complex model than those used for steady state cornering. Plus you need a good tire model, since that is what counts in the end.

"

Does the rate of weight transfer relate to the roll couple or are they independent? (like can you tune the car without changing one or the other)

If they are not independent then how do springs change the way the car handles? (if they relate somehow then changing the springs would change the rate and the roll couple. But it has been said that they do not change the roll couple so if they are related then the springs would not change the rates.)

I would say that regardless of whether springs change load transfer, they do change roll resistance: and camber change. So they do change the handling.

Here's a site that I found helpful and will save me from copying it all WT site

What I am getting at is that if the rates and roll couple are independent of each other then there is a fundamental difference between the anti-roll bar and the springs. That means that there would be a largish difference in the way the car behaved depending on the type of roll resistance chosen and the saying "sway bars act jst like springs in lateral acceleration" (which I have read in a few books) would be very untrue.

In terms of roll resistance, that comment is true. The big picture is more complex, it seems. The sprung mass causes most of the WT...