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DaewooOfDeath
DaewooOfDeath Dork
1/24/12 3:08 a.m.

So I'm trying to figure out force vectors in a suspension and I have nowhere near enough math skill to make my own formulas. Here's the problem I'm trying to solve.

http://www.cardomain.com/ride/3870883/1998-daewoo-nubira/page-17#15545644

I can't get the bigger version of the picture to post, so you might have to follow this link.^

In this scenario, am I right to assume the normal force on the compressing wheel is 1,000 lbs and the normal force on the drooping wheel is 0?

On the other hand, if there were no swaybars and the springs were 1,000 lb/inch the normal force would remain 1,000/1,000 lbs on each tire, right?

Further, is there a way I can calculate how much weight transfer a swaybar causes in cornering?

Edit, fixed the picture.

foxtrapper
foxtrapper SuperDork
1/24/12 7:35 a.m.

Your picture is actually indeterminant, but we can make reasonable assumptions and compensate.

Fundamentally, your answer is yes. Which is why eventually the inside tire can sometimes lift off the ground. At which point it's exerting zero pounds force downwards. The swaybar does transfer force back and forth between the sides. The exact function having to do with the lengths of the lever arms, rigidity of the tube, etc.

A swaybar doesn't transfer weight, it transfers force. Be careful here, the terms get used interchangeably, when in fact they aren't. This can frequently lead to confusion. The calculations involve the deflection of the suspension. Trying to calculate it all up front on a spreadsheet requires a lot more information than you've got showing in that drawing. But, you can do a pretty good job using a picture and measuring deflection, then calculating from that deflection.

DaewooOfDeath
DaewooOfDeath Dork
1/24/12 12:30 p.m.

Perhaps I don't understand the difference between the forces being transferred and the loads being transferred.

The reason I am asking this is because I've heard two diametrically opposed explanations of how swaybars work.

Explanation one says they don't transfer any weight from the drooping, inside tire to the compressing, outside tire. Given the force vectors, given the fact that the sway bar is pushing down on the outside strut and lifting up on the inside strut, I can't imagine how this could be true. However, this seems to be conventional wisdom none the less.

Explanation two says that swaybars transfer load from the inside to the outside tire and reduce mechanical grip (as compared to a spring only setup) as a result. This explanation says that you can calculate the amount of extra weight transfer a sway bar causes, but nobody has been able to come up with said formula.

So, I'd either like to know what happens to that force that gets transferred to the strut but somehow doesn't end up loading the outside tire (conventional wisdom explanation) or I'd like to get the formula for figuring out how much load transfer the swaybars cause compared to a spring only setup.

foxtrapper
foxtrapper SuperDork
1/24/12 2:30 p.m.

Stand with your feet shoulder width. Your weight is equally spread between your feet. You feel the weight of your body equally on both feet. You are now a car.

To simulate a turn, have someone start to push on your shoulder, gently. You feel the force. It causes a torque and a change in the force at your feet. You are cornering gently.

Be flexible at the waist. Like you have no anti-rollbar. You bend over as they push you. You lean a lot. Notice how different you feet feel. Notice how easily you roll over and fall down. You don't corner well this way, you are a 52 Buick, and are embarassed about yourself.

Stand up again, and be stiff at the waist, like you have a nice stout anti-rollbar. Let the person push your shoulder, but don't bend. Don't lean into the push, just stand still. Notice that you don't shift as much now when pushed. You are a cornering machine! You are a Corvette! None the less, eventually you are pushed over. Even a Corvette has limits.

Do it a few more times, pay attention to your feet. Soon you'll grasp the answer.

Remember, a lifted inside tire still has a suspension spring attached to it. That spring can put a whole lot of force across that anti-swaybar to the outside wheel.

youngfg
youngfg New Reader
1/24/12 3:07 p.m.

Weight transfer is a function of CG height, track width, and cornering force. Notice spring rate, sway bar rate is not in there. Stiffer springs and sway bars will only affect the amount of body roll. The amount of weight transfer from the body roll is trivial.

DaewooOfDeath
DaewooOfDeath Dork
1/24/12 3:44 p.m.
youngfg wrote: Weight transfer is a function of CG height, track width, and cornering force. Notice spring rate, sway bar rate is not in there. Stiffer springs and sway bars will only affect the amount of body roll. The amount of weight transfer from the body roll is trivial.

How is this possible?

Think of it this way, you are grabbing the struts on a car. You push down on one strut and lift up on the other. This force is transferred to the wheels, no? You have transferred weight from the drooping wheel to the compressing wheel, no?

How does this differ from the actions of a swaybar?

Springs don't do this by themselves, but swaybars seem to. That's why, in my original drawing, I assumed that the normal force on a spring only suspension with one inch of compression on one side and one inch of compression on the other would be 1000lbs/1000lbs.

But look at the swaybar vectors. With 500lb springs and 500 lb swaybars, the normal force on the drooping wheel would be 0, unless I'm getting something wrong conceptually.

If I am getting something wrong conceptually, I would like to figure out where the force vector goes when the swaybar resists roll.

youngfg
youngfg New Reader
1/24/12 7:14 p.m.

When I get time tomorrow, I will type up an example that will explain it better.

youngfg
youngfg New Reader
1/25/12 7:42 a.m.

Think of the front wheel pair, also imagine that there is no suspension, like a go kart, and you are cornering at such a speed with the combination of your track width and CG height that the inside tire is just barely touching the ground, all of the weight has transferred to the outside wheel. There is no body roll because the suspension can't move. So in this case if both tires had statically 500lb on them, now the inside tire has 0lbs., and the outside tire has 1000lbs.

Now lets add a suspension. Say the wheel rate is 500lbs/in. Statically each side of the suspension would be compressed 1 inch. If you took a crane an slowly lifted the car each wheel would droop 1 inch before there was 0lbs. on it. The wheel may fall father if the suspension had more travel but the spring would be loose and rattle around at 1 inch of droop.

Now lets go around the same corner that transfers all the weight to the outside tire. Now the outside tire has 1000lbs on it so the suspension on that side will compress one more inch, for a total 2 inches of compression. The inside tire will have 0lbs. on it, so it will extend 1 inch. So the car has rolled up one inch on one side, and down one inch on the other.

This is all I have time for right now, maybe it will help.

AngryCorvair
AngryCorvair SuperDork
1/25/12 11:54 a.m.

In reply to youngfg:

i will give you half credit for your first answer, and an additional 20% for the second. what's missing is the spring and bar at the other end of the car and how they influence the vehicle behavior.

the suspension at each end of the car has some roll stiffness due to the springs and bars. increasing the roll stiffness at one end of the car increases the amount of load transfer at that end of the car, while simultaneously decreasing the relative jounce/rebound travel of the suspension at that end of the car.

to the OP: someone is doing a poor job of explaining "conventional wisdom" to you. or you're doing a poor job of grasping it. i mean, if they didn't transfer load, couldn't we just throw them away and make our cars lighter?

re. the formula for how much load a bar transfers, you'd need to know some material characteristics and some dimesions from the vehicle. example: for a given lateral acceleration (say, a 0.5g turn) there will be some amount of suspension displacement, which will result in some vertical displacement of the end link, which will result in some angular displacement (ie twist) of the bar. you need to know the angle that the bar twists (remember to consider the displacement of both ends), and the torsional stiffness of the bar. and depending on the lateral distance between the end link attachment of the bar and the chassis bushing on the bar, you may find that the bending of the cantilevered portion of the bar is non-negligible and you'll have to include that as well.

youngfg
youngfg New Reader
1/25/12 2:05 p.m.

In reply to AngryCorvair:

I will also give you half credit.

"Think of the front wheel pair"

Also explain geometric anti-roll, you should be very familiar it.

foxtrapper
foxtrapper SuperDork
1/25/12 2:57 p.m.

DaewooOfDeath,

Let me try a different approach, since the last one didn't help apparently. Lets work with your original drawing and set some points for clarification.

There are no springs shown, and the vehicle is sitting on two perfectly rigid tires, riding on the ground.

The blue arrows represent vehicle weight, and it's uniformly distributed between the two tires.

The red arrows represent torque forces, which is why they are unequal.

If we let the red arrows equal the blue arrows, the car will be right on the edge of tipping. No motion will happen, but the forces on the tires will be different side to side. The ground tire on the left will have no load (blue - red), the ground under the tire on the right will have double load (blue + red). Again, no motion will happen.

This is a great example of the totally inflexible swaybar. Zero deflection takes place. Only when the red arrow exceeds the blue arrow will any motion happen, and that motion will be to lift the left side tire off the ground.

Now lets add some springs under the tires for deflection.

Lets first put the vehicle onto those springs under its own weight only, the blue arrows.

The springs will deflect from that weight. The deflection will be equal side to side. The car sits level.

Now lets add a little torque, the red arrows. We'll start small. The vehicle will twist to the right on those springs. The downward force on the left is reduced (blue - red), and the downward force on the right is increased (blue + red).

Now lets increase the torque and let the red arrows equal the blue arrows. Just like before, the car will be right on the edge of tipping. The car will also twist more on the springs. The ground under the tire on the left will have no load (blue - red), the ground under the tire on the right will have double load (blue + red).

Sound familiar? It's exactly the same as the first, with no roll. The only thing that changed is some body roll on the springs was introduced. The load on the contact patch under the tires never changed from the introduction of the springs, or roll.

Did that help any?

pstrbrc
pstrbrc New Reader
1/25/12 6:52 p.m.

Here's my two cents worth. Buy Herb Adam's "Chassis Engineering". He explains exactly this in the first couple of chapters. For $16.29 + shipping to wherever you are, (Korea, Indonesia?) it's money you have to spend to understand this. And getting it from him means you don't have to wonder if we're right.

DaewooOfDeath
DaewooOfDeath Dork
1/25/12 11:42 p.m.

I don't think you guys are understanding me. pstrbrc, I have read that book, btw.

I understand that COG and lateral G cause weight transfer. I understand how a swaybar counteracts body roll and I understand that body roll does not contribute significantly to weight transfer.

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.

Let me try to rephrase.

Assume we are dealing with a Segway that has been equipped with McPherson struts. This eliminates the front/rear axle complicating factors. Let's also do away with cornering forces. For our purposes here, let's just assume the hand of God is compressing the suspension on one side and causing it to droop on the other.

Now, if our suspension has no swaybars there will be no weight transfer in this situation, correct? The drooping spring will push downward with 1000 lbs of force and the compressing spring will also push down with 1000 lbs of force.

Correct?

Now, let's change the springs to 500/lb springs and install 500/lb swaybars for the same net roll resistance. In this case, when the hand of God compresses the first side, it sees 500lbs of resistance from the spring and 500lbs of resistance from the swaybar (which is pressing down on the strut) for a total of 1000 pounds. However, on the drooping side, the wheel load should drop to zero because the 500lb spring is cancelled out by the 500lb swaybar, which is pushing UP on the strut with 500lbs of force.

Correct?

I know there is a problem with my math here somewhere, but cause according to this, I've made 1000lbs magically disappear with the swaybar, but conceptually, I don't see how a swaybar could do anything but transfer weight across an axle from the unloaded side to the loaded side.

Let me fix my drawing and try again. Those 45 degree lines coming off the wheels are supposed to represent struts.

Remember, all the lb/inch values I'm using are at the wheel rates so we can avoid calculating wheel rates from spring rates.

foxtrapper
foxtrapper SuperDork
1/26/12 7:15 a.m.

DaewooOfDeath
DaewooOfDeath Dork
1/26/12 8:06 a.m.

Foxtrapper, thanks for the illustrations. They are certainly better than mine.

But my question remains, what happens to the force vector, from the swaybar, pushing down on the outside wheel and up on the inside wheel?

Think of it from the strut's point of view. If the strut is on the outside, the bar is pushing down on it. If the strut is on the inside, the bar is trying to lift it up. Do the tires not see this force?

DaewooOfDeath
DaewooOfDeath Dork
1/26/12 9:12 a.m.

BTW, I'm not the one who came up with this idea of swaybars causing weight transfer across a single axle. I got the idea from a 7 time autocross champ. It also seems like the Ariel Atom is built this way. If you want a longish read, I would be happy to post it.

foxtrapper
foxtrapper SuperDork
1/26/12 9:26 a.m.

A swaybar doesn't change the values the tires see, it changes how those values get to them.

Look at my drawings, the last two in particular. A swaybar was added, but the loads at the tires never changed. Weight and torque stayed the same, how they were transmitted to the tires changed.

Without a swaybar, both weight and torque go through the springs alone. 100% weight and 100% torque, all through the springs.

With a swaybar, 100% weight still goes through the springs, but now torque is shared between the springs and swaybar. Lets say 50% through the spring, and 50% through the swaybar.

In both cases, 100% of the weight and 100% of the torque end up at the tires. So as far as the tires are concerned, nothing changed.

AngryCorvair
AngryCorvair SuperDork
1/26/12 10:01 a.m.

DoD, it might help to think of the antiroll bar as resisting the opposing suspension motion rather than resisting the body motion. of course it does both; i'm just proposing that you think of what it does to the suspension rather than what it does to the body.

essentially, the bar is just another spring, and it has a spring rate. the deflection is proportional to the force applied. the bar is just an additional path through which some amount of the force is reacted.

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.

foxtrapper
foxtrapper SuperDork
1/26/12 10:16 a.m.

Lets see if this helps any.

DaewooOfDeath
DaewooOfDeath Dork
1/26/12 10:18 a.m.

It turns the torque into weight transfer ... right? If not, what happens to the force the swaybar resists by pushing down on the outside strut?

Ie, it turns torque into weight transfer.

Let me post the information where I got this idea.

Steve Hoelscher's setup advice/swaybar manifesto.

Ok, so its a pure competition car. No/little compromise. So the starting point is a target "total roll". As a general rule a strut car doesn't react well to body roll due to camber issues and lateral roll center movement. A target of 1.5~1.75 degrees total roll is typical. Total roll is a function of grip, wheel rate, moment arm length and roll stiffness. Keeping the car flat also minimized the severe bump steer issue the Mk1 chassis has.

We'll start with grip. Current generation STS tires are going to generate something in the 1.1g of grip, vs 1.3+ of say a Hoosier A6. So the STS car will require slightly less roll stiffness than a CSP car to achieve the same total roll. We need some roll to give the driver feedback and prevent the outside tires from being overloaded on turn-in and transition. Shocks will play a key roll here.

Next is ride height because that determines the length of the moment arm (the distance between the roll axis and CG). The CG acts through the moment arm (like a lever) to roll the car about the roll axis. The roll axis is the line drawn through the front and rear roll centers. The CG height is basically fixed in relation to the body but the roll axis is a function of the control arm and strut angle and therefore ride height. The roll axis height drops at a significantly higher rate than ride height. So we have to be careful about how low the car goes.

Before we can choose spring/bar rates we need to set the ride height to determine roll axis height and therefore moment arm length. Ideally, we want the roll axis to be 1 to 2 inches above ground with the rear at about 1" and the front at about 2". Since we won't be modeling the car in software, our target is to simply have the roll centers above ground level and that the front be above the rear so the roll axis is reclined toward the rear of the car.

So our starting point is to deal with the things that we can’t really change easily. Spring/bar rates are free so we can choose rates that work with the rest of the setup. Ride height is basically free so we can raise/lower the car to our advantage. What we can’t change as per the STS rules is the control arm geometry. So we need to set the control arm angle to optimize roll axis location and camber curve. In reality, its not control arm angle but instead the angle of the line between the inner control arm pivot bolt and the center of the outer ball joint. You can clearly see the front ball joint pivot is well above the centerline of the control arm. Same for the rear but somewhat less offset.

To achieve our previously stated geometry, we want the front virtual control arm (the line between the inner pivot and the ball joint center) to be roughly parallel to the ground. In the rear, the ball joint should be slightly above (maybe ¼”) the inner pivot. Unfortunately, this is likely to set the nose of the car noticeably higher than the rear. This may be adjusted later when we know the severity of the body’s rake. Actual ride height and roll center location will be effected by tire diameter. This virtual control arm angle is about the best compromise location to optimize the camber curve assuming less than 2 degrees of body roll. Its doesn’t address the bump steer problem but that will be addressed later.

-Steve Hoelscher

Part Two

Now that we have control arm angle (roll axis) set, we will address wheel rates. Realize that spring rates and wheel rates are related but not equal. The wheel rate is the result of the wheel’s mechanical advantage over the spring. This is expressed as a linkage ratio. As an example, the linkage ratio of a typical strut suspension is about 1.1:1. That is, the wheel moves 1.1” for every 1” of strut movement. Now this changes as the suspension moves through its range of travel and typically the mechanical advantage is at its maximum at full droop and goes down as the suspension compresses. While this is a generalization, it is basically the nature of the Mk1’s suspension. Its also important to remember that front and rear suspensions seldom have the same linkage ratio.

Its also important that we understand that a wheel rate includes the action of a swaybar if fitted. Swaybar rates are figured much the same. The bar’s wheel rate is the bar’s spring rate multiplied by the linkage ratio. Most Mk1 swaybars connect to the strut, so the linkage ratio is the same as that of the spring. And finally, the wheel rate is the wheel rate of the spring plus the wheel rate of the bar.

Now we return to the first installment and consider total roll rate. We targeted ~1.5 degrees of total roll. Now we need to figure a total roll rate that yields that amount of roll with the given roll axis height. Total roll rate is the total of the front and rear wheel rates. Front wheel rate + rear wheel rate = total roll rate. So how do we determine a total roll rate? I have developed my own basic starting point that is simple and effective. I have proven this method to work well for me with many different cars in both autocross and road racing. As noted previously, a favorite assumption of many people is to choose spring rates that equal corner weights. The problem with this method is it ignores actual vehicle dynamics. Most such setups would include a front (and sometimes a rear) swaybar. The front swaybar, if sized accordingly could provide enough additional front roll stiffness to give adequate handling balance. For our purposes here, we don’t care if its swaybar, spring rate or a combination of both, we are only concerned with the resulting wheel rate. Later we will decide on the split between spring and bar rates.

Back to determining total roll rates. I noted I had a simple formula for a starting point. That formula is: ½ the total vehicle weight divided by the inverse of the weight distribution. An example for a 2200 lbs car with 44/56 front to rear weight distribution (a rough estimate of your car's weight and distribution):

2200 / 2 = 1100 lbs

1100 * .56 = 616 lbs/in front wheel rate

1100 * .44 = 484 lbs/in rear wheel rate

The reason for inversing the weight ratio is to offset the car’s rear weight bias. If each end of the vehicle had the exact same level of grip, the rear would break away first because the higher weight would overcome the available traction earlier. Therefore a rear weight bias car will naturally oversteer therefore more front roll stiffness is necessary to counteract that natural tendency. How much front roll bias varies but all things being equal (and they never are) basing the offset in roll ratio on the inverse of the weight bias gets you pretty darned close.

Now you are probably stunned at the high wheel rates, but we aren’t through yet. First, remember the second paragraph regarding grip? STS tires don’t generate the same level of grip that A6 Hoosier do. As a result, lower spring rates are necessary to produce the desired amount of body roll. The difference in G loading is about 20~25%, so lets reduce our wheel rates by that amount

616 front * .8 = 492.8 lbs/in

484 rear * .8 = 387.2 lbs/in

To round off the rates: 500 lbs/in front and 400 lbs/in rear.

Now lets apply these to the MR2 chassis and its dynamics. Experience and testing are valuable here and I can apply my experience and make a couple of assumptions. First, that the low roll centers resulting from the current ride height and the high CG result in a long moment arm and a lot of body roll for a given wheel rate. Second, that the front has a longer moment arm than the rear. So we need a little more front roll stiffness and a little more total. Therefore we will bump the rates up slightly, especially in the front. I would think that using the wheel rates as spring rates would be enough of a bump and then add say 50 lbs/in to the front. The resulting spring rates:

550 lbs/in front

400 lbs/in rear.

Next we will address swaybars.

-Steve

Think about it this way. A mid/rear engined car is a fwd car turned around backwards. On a fwd car, they run softer front springs and bars and REALLY stiff rear springs and bars to get something like a normal handling balance. AND, to reduce inside front wheelspin on corner exit.

So if a mid/rear engined car is a fwd car backwards, why not just turn the fwd setup round backwards and put it on the mid/rear engined car?

Or think about it this way. Control body roll with the end opposite the weight/drivewheels. This creates mechanical grip on the drive axle. Now it doesn't matter which end has the drivewheels/weight.

Funny how sometimes the most obvious things aren't that obvious because its so far outside our existing pool of knowledge.

What I have posted above is basically the process I sat down and put together back in 1994 when I realized I wasn't making any progress using the "conventional" tuning methods (softer springs, big swaybars and stiff shocks). I derived and refinded this method over the following couple of seasons. I even put together a spreadsheet that would let me look at the entire setup on one page. Then I could change a spring rate or swaybar rate and see how that change rippled through the entire setup.

It was an eye opening experience for me. I could try lots of different combinations on paper, pick one to test and note the change in feel and performance. Then compare that to the spread sheet. It didn't take long to correlate results with changes and then make changes and predict results. Accurately. A very powerful too.

I now have a much more sophisticated method in use on the DP car. But then Prepared rules give you so much freedom you need a more sophisticated process. But my simple Lotus 123 spreadsheet was a revelation at the time.

The car was transformed in the space of a single season. After floundering around in the middle of the trophies for my first half dozen trips to Nationals, this process put me on the path that lead directly to my first Championship in '98. Now I don't credit this for my first Championship but it certainly put me in a position to win. Car setup isn't a substitute for learning to win. But you can't win if the car isn't there.

-Steve

Now that we have a starting point for individual wheel rates we can now decided how we want to divide the wheel rate between spring rate and swaybar rate. To make an informed decision we must understand the function of the swaybar and how it interacts with the springs. A swaybar is nothing more than a torsion bar (spring) that has either end attached to each wheel of a single axle. It’s the twisting action of the bar that is the torsion spring. Because it only applies when a one wheel of the axle moves independently of the other, it has only moderate effect on ride quality.

To understand the swaybar’s effect on the springs and chassis we must understand how the springs interact with the chassis as well. We already understand how they affect body roll but they also effect ride quality and that is what we will now address. There are three basic principles that apply. First is ride frequency. Ride frequency is the rate at which the chassis reacts to input, a bump in the road. Ride frequency is expressed as Hz. A soft ride frequency would be about 1 Hz or one cycle per second. A cycle being the car passing over a bump, the chassis reacting and then returning to its original state. A stiff ride would be a frequency of about 2 Hz and a very stiff ride would be 2.5 to 3 Hz. We could easily calculate ride frequency based on the sprung weight of the car (total weight carried by each axle – unsprung weight of each axle = sprung weight) and the wheel rate of the springs on each axle. However, for our exercise here its only necessary that we understand the concept. The final part of the concept is that each axle has its own ride frequency based on the sprung weight and wheel rate of each axle.

Ride frequency is used to determine the second principle favored speed. Favored speed is the road speed at which the two ends of the car return to their original state after the car passes over a bump. To achieve a positive favored speed (a speed greater than zero, and yes you can have negative favored speeds) the front ride frequency must be lower than the rear so that the front returns at the same time the rear does for a given speed. Achieving this effect will produce the best ride quality for that speed and desired stiffness. The favored speed can be set at any target speed by tuning the front and rear spring rates to achieve the ride frequencies that produce the effect at the target speed. This effect works for both stiff and soft rides at most any speed and is why your stiffly sprung car feels smoother as speed increases. Most manufacturers set the favored speed using spring rate to achieve a smooth ride at their desired target speed (usually between 50 and 70 mph), then use swaybars to tune the ultimate handling balance and body roll.

The final principle we need to understand is center of suspension. Center of Suspension (CoS) is the point on the chassis at which, if weight was applied, both ends of the car would compress exactly the same amount. As an example, if the front and rear springs had exactly the same wheel rate then you could push down exactly half way between the two axles and the front and rear would compress the exact same amount. If the front springs were softer than the rear, the center of suspension point would be moved rearward to accommodate the softer front spring rate. The weight of the car then acts through the center of suspension via the Center of Gravity (CG). The difference in the car’s CG and Center of Suspension is a moment arm (which is basically a lever) just as the difference in CG and roll center. So the CG acts through the moment arm to compress the suspension in reaction to a bump. If the CG is behind the CoS, then the rear suspension is compressed more for a given load. The inverse is also true, if the CG is ahead of the CoS, the front suspension is compressed more. And like body roll, the longer the moment arm (the greater the distance between the CoS and the CG), the more the suspension on that end is compressed for a given load.

To minimize excessive body movement in response to bumps, we want the center of suspension as close to the CG as possible. This minimizes the length of the moment arm and therefore the CG's leverage over the CoS. For the Mk1, the CG is near the rear axle so we would have to have front springs that were much softer than the rear to achieve a center of suspension anywhere near the CG. Then to achieve good ride quality we set a desired favored speed and then tune the spring rates to suite. For the Mk1, setting a favored speed of say 50 mph, the front springs would have to be slightly softer, or the rear springs slightly stiffer, that that which would locate the center of suspension exactly at the CG.

This is the method the manufacturers use to tune the ride quality and handling balance of their cars. When we consider what we learned about roll ratios in the earlier installments its easy to see why the manufacturers use a hefty swaybar and soft springs on the front of the Mk1 (or most cars for that matter). Now that we understand how and why Toyota setup the stock Mk1 we can determine where we need to go to improve the car. Clearly we are not as concerned with ride quality because we are setting up a competition car. However, it is necessary that the car be able to track smoothly over bumps so as to not upset the car enough to loose traction.

-Steve

Part Five

Now that we understand the function of the springs and how they interact with the body to determine ride quality and body roll we can now directly address the swaybars. To achieve our target body roll and roll ratio we have chosen specific wheel rates. However, the rates we have chosen would result in a very stiff ride if we used only springs to achieve that wheel rate. To soften the ride we need to lower the spring rates, especially in the front (to achieve a positive favored speed) however, that will upset our chosen roll ratio so we add back wheel rate by supplementing spring rate with swaybar rate.

From installment #2 we had chosen 550 lbs/in front and 400 lbs/in rear spring rates. Its easy to now choose a front spring rate that will produce a more comfortable ride. Say 300 lbs/in front springs would give us a positive favored speed and move the center of suspension closer to the CG of the car, just aft of the center point between the two axles. We can then add back the spring rate by installing a swaybar with a rate of 250 lbs/in.

250 lbs/in front bar + 300 lbs/in spring = total rate of 550 lbs/in.

That looks pretty easy. However, its hard to find an off the shelf swaybar that has exactly the rate you want. In reality, unless you want to fabricate a custom bar every time you want to test a different setup, you need to calculate the rates of the available bars and then determine how much spring rate you need to achieve the desired total rate. So lets say you measure your existing swaybar and find it has a rate of 200 lbs/in. The spring rate you would want is determined by subtracting the existing bar rate from the target spring rate.

550 lbs/in target rate – 200 lbs/in bar = 350 lbs/in springs.

Easy enough. But we have a problem here. Swaybars are not the dynamic equivalent of springs. A swaybar transfers load from the inside tire to the outside tire and thus reduce mechanical grip as they add spring rate. And the effect is not linear. The stiffer the bar in comparison to the springs, the greater the effect (loss of mechanical grip). To give an example of the effect, if we set our proposed STS2 car up using the target data we have assumed above using the target spring rates without any swaybar, the car should have good balance. However, if we achieved that same target spring rate using a front swaybar, the car would tend to understeer more than if we used only springs and no swaybar. And the greater percentage of the total front spring rate the bar accounted for, the more the car would understeer. I noted as much early in the thread.

We now must choose how much bar we want to use for our final setup. As the reader may know, I don’t use swaybars on my DP car. Nor did I use swaybars on my previous racecar, a DSP X1/9. For me it is far easier to manage the setup of the car without swaybars. I also prefer the feel of the car without swaybars. It has long been my thought that; because a swaybar reduces mechanical grip, why would you want to put anything on the car that reduces mechanical grip?

With this simple method, it is easy to compare the effect of the front bar by comparing the same total spring rate using just springs and no front bar to the same total spring rate incorporating a front bar. I have done extensive testing and have proven to my own satisfaction that the theory is in fact accurate. The same total front spring rate achieved using a front swaybar will produce more understeer than the same total front spring rate achieved using springs only. In addition, the effect of the front bar changes based on the level of grip the surface offers. As a result, the car does not have consistent balance from surface to surface or even from run to run as the tires heat up from and the surface cleans and heat up throughout the day. I have found that my no-swaybar setup is very consistent on different surfaces and conditions seldom if ever requiring any changes to setup. At most, a pound or two of air pressure is all that is needed to tune the balance even in the most extreme of conditions. In fact, I don’t even change the setup for rain. All I have to do is bolt on the rain tires and the car is fine.

If one chooses to use a front swaybar, the effect resulting from the loss of mechanical grip will have to be accounted for by softening the front springs enough to bring the balance back to neutral. Choosing the amount of swaybar to use is now easy and dependant on driver taste. If the driver prefers a smoother/softer ride, use a very stiff front swaybar and subtract the front bar rate from the total spring rate to determine the required front spring rate. Testing can then determine how much less spring rate is necessary to bring the handling balance back to neutral. One could also compromise and use a very soft front bar, thus minimizing the loss of mechanical grip.

-Steve

Part Six

Remember this number: 150%

So now is easy to see why one might choose, as I have, not to use any swaybars at all. I have spent more than 10 years working on this setup theory and testing the various permutations. Somebody else may not like the setup but I know its effective. The above method is exactly how I got to the point I am now. I came to this method by first using a spread sheet to aid in computing wheel rates and roll ratios accurately. I have since transitioned to a more scientific application of this theory using Susprog3d. I can now figure total roll based on lateral grip, ride height, CG height, roll axis and spring rates very accurately. I can also figure camber curves and roll center movement dynamically. A very powerful tool. But there is no substitute for testing and it is through testing that I have tailored my setup to suite my personal tastes. Now I will throw out much of what we have covered to this point.

I prefer a car that has slight understeer and is VERY stable in transition. I DO NOT want the car to feel loose in a slalom or fast transition. I know that a car with a significant rear weight bias, that is loose in transition, is slow. So my car has a lot of front roll stiffness to achieve my preference. So lets throw out all of the theory above and look at some simple data gathered from testing. Pulling off the swaybars and just sticking springs under the car, what works?

I decided I needed an objective comparison. I wondered if I could I derive a better, faster setup by pure testing alone, ignoring my theory. A number of test days at my favorite venue (Hunt Stage Field in Ozark, Alabama)

http://maps.google.com/maps?f=q&hl=...052872&t=k&z=14

where Wire Grass Region events are run and you can make all of the runs you want, when you want, enabled a lot of valuable data to be gathered. The basics can’t be ignored so the ride height must be set so the roll centers are close, as noted previously, and the camber curves have to be in a reasonable range. So those two elements determine ride height. What about springs? We need the car flat and we determined a total roll based on the weight of the car and potential grip. From installment #2:

2200 lbs car / 2 = 1100 lbs of total roll rate

So pulling data from my test notes. We need 1100 lbs of total roll stiffness less 20% for STS legal tires.

1100 x 80% = 880 lbs

So out came the box of springs and I started making runs. I have a spring inventory that covers a range from 300 lbs/in to 700 lbs/in in 50 lbs increments. I quickly start narrowing down the spring rate combination that produces the best balance. Once the spring rates are close, I start on roll center locations, ride height and camber curves. Then back to spring changes to further refine the balance and confirm the data. Final balance and dynamics are tuned with toe settings and tire pressure.

The testing results tell me that, for most mid-engined cars, making the front roll rate 150% of the rear is a good target. Therefore, given the above data:

350 lbs/in rear springs x 150% = 525 lbs front springs.

and

350 lbs/in rear springs + 525 lbs/in front springs = 875 lbs of total roll rate.

Close enough.

The next installment will cover shocks and dynamics.

-Steve

Tangent 1

Feel? Feel is subjective, its the clock is not.

One of the things you will have to come to terms with is how the car will feel. if you are used to stock class cars, or cars with big swaybars and soft springs you are in for something of a shock the first time you drive the car this way. While I can't speak to the feel of the Bridgestones I can for the Hoosier radials.

I don't know what you have been driving but the first time out with this kind of setup might be a shock. The amount of feedback is likely to overwhelm you. You are likely to be able to feel every tiny imperfection in the pavement. Every seam, crack, dip, gap, etc... I used to joke that I could run over a dime and tell if it was heads or tails.

It may take a few runs but you should adapt as you get acustomed to the sensation. Also, the car should feel more lively and precise. Turn-in should be improved along with the car's response to steering wheel inputs.

If you have any questions before then ask them now as I will be in Daytona for the Grand Am race Wednesday and Thursday. And if you are watching (Thursday night at 8:00pm eastern on Speed Channel) pull for the #26 GT Porsche of Gotham Competition.

I will eagerly await your report.

-Steve

Yes, swaybars should be used as a trim device only. Never the primary means of roll control. Most modern, purpose built, racecars use very light swaybars. Most are adjustable and are either on or off.

They are used as a method of adjustment to correct for changes in fuel load, tire wear and surface changes.

-Steve

I spent a lot of time working on this methodology. Figuring out all of the interactions and putting the elements together into a functioning package took a lot of work on the computer and testing.

When I first put a setup like this on my DSP car (back in '94) I got lots of funny looks and lots of people telling me it wouldn't work. That it couldn't work. Often by people I was beating.

Back then, the conventional wisdom was: Springs should be as soft as possible, that they only held the car up. It was the shocks and bars that were used to control body roll. That just didn't make sense to me. But, because I assumed I didn't know as much as the experienced guys (and I certainly didn't at the time) I put a big f'ing front swaybar on my car. It was awful. Then I put an even bigger front bar at the advise of a respected engineer. It was worse.

On the way home from Nationals that September, I decided I needed to better understand chassis tuning. A month later, I tested my first "all spring" setup just as a baseline to try and understand what "just springs" did and then add swaybars back into the equasion. That event was a revelation and I never put the bars back on. And never looked back.

-Steve

I loved that car (Fiat X19). It was a truely great racecar. My theories on the barless setup were developed, tested, refined and proven on that car.

That car won more than 30 National Tours, 4 National Championships and every single event it entered from the season opener of 1998 to Nationals of 2001 where I finished 2nd to Mark Daddio on that fatefull day of 9/11/01 when the event was stopped after day 1 of DSP (we ran 1st heat Tues). It then won every event entered though Nationals '03 where it finished 5th and was retired. So, from the beginning of the '98 season to Nationals of '03 the car won every event entered except the '01 Nationals where it finished 2nd. That is one of the best records in the history of the sport. Note that during that period of time DSP was one of the largest classes with more than 40 cars turning out at Nationals each year.

It was the ground work done on that car that is now being successfully applied to my DP Mk1 and verno-dub and micaceli's Mk1s.

I should also note that there are a few clubracers successfully running with my barless setups. -Steve

Part Seven

The more dependant the setup is on swaybars for tuning balance, the more changing conditions effect the car's balance and response.

A well balanced, all spring setup, with properly tuned shocks, will be very consistant from surface to surface and as conditions change.

One of the things I see regularly is SP and Prepared guys at big events diving under the car after each run changing the swaybar adjustments, shock settings, tire pressures, etc... trying to dial the car in. You will likely NEVER see me do that to my cars unless its a test session and I am working on testing setup changes. That's because my car is so consistant from surface to surface, hot, cold, sunny, rain, etc...

This works equally well on full road courses. The one thing to keep in mind is that on a full road course, the shocks play a much bigger role in controlling body motion. You might find that because the shocks are at the limits of their capabilities that they may have trouble controlling body motion on the higher speed bumps.

I have run my autox setup on full road courses many times and found it to work just as well there at in autox. I have also used this setup on a number of track and road racing cars. In fact, the 2nd place finisher in H Prod at last fall's "Runoffs" was running my no bar setup.

I would recommend doing the 600f/350r version.

Trust me guys, it won't push. It really won't. You might find slightly more understeer in very high speed sweepers taken at full throttle. But that can be tuned out with toe settings.

-Steve

DaewooOfDeath
DaewooOfDeath Dork
1/26/12 10:21 a.m.
AngryCorvair wrote: DoD, it might help to think of the antiroll bar as resisting the opposing suspension motion rather than resisting the body motion. of course it does both; i'm just proposing that you think of what it does to the suspension rather than what it does to the body. essentially, the bar is just another spring, and it has a spring rate. the deflection is proportional to the force applied. the bar is just an additional path through which some amount of the force is reacted. 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.

Sorry, I didn't see this before. Is there a formula for determining how much vertical force changes based on spring rates, rollbar rates and inches of suspension deflection?

DaewooOfDeath
DaewooOfDeath Dork
1/26/12 10:25 a.m.
foxtrapper wrote: Lets see if this helps any.

Ah, this is very helpful. I need to split the torque and vertical loads into different math problems. But I think you're missing one thing from photo three.

The 100 lbs of torque being resisted on the outside wheel would have a corresponding 100 lbs of torque lifting up on the outside wheel, no? Or at least, we need that -100 lbs to balance the equation.

foxtrapper
foxtrapper SuperDork
1/26/12 10:41 a.m.
DaewooOfDeath wrote: Ah, this is very helpful. I need to split the torque and vertical loads into different math problems.

Sokath, his eyes uncovered!

Remember, we are playing with very simple models here. In a full car, it becomes more complex.

But I think you're missing one thing from photo three. The 100 lbs of torque being resisted on the outside wheel would have a corresponding 100 lbs of torque lifting up on the outside wheel, no? Or at least, we need that -100 lbs to balance the equation.

300=500-100-100

500 vehicle mass

100 50% torque up through the spring

100 50% toruq up through the sway bar

DaewooOfDeath
DaewooOfDeath Dork
1/26/12 10:54 a.m.
foxtrapper wrote:
DaewooOfDeath wrote: Ah, this is very helpful. I need to split the torque and vertical loads into different math problems.

Sokath, his eyes uncovered!

Remember, we are playing with very simple models here. In a full car, it becomes more complex.

But I think you're missing one thing from photo three. The 100 lbs of torque being resisted on the outside wheel would have a corresponding 100 lbs of torque lifting up on the outside wheel, no? Or at least, we need that -100 lbs to balance the equation.

300=500-100-100

500 vehicle mass

100 50% torque up through the spring

100 50% toruq up through the sway bar

I misspoke.

No torque (1000) = 500 right + 500 left

Torque plus 100 lb sway bars (1000) = 300 (inside sprung weight) - 50 torque inside swaybar + 500 (outside sprung weight) + 100 torque in spring + 50 torque outside swaybar

I think this is what it should say, since the swaybar is acting equally on both struts.

E36 M3, that doesn't work either. How do we account for the lifting that the swaybar puts on inside spring? Seems to me lift must equal roll resistance at the other side.

foxtrapper
foxtrapper SuperDork
1/26/12 12:44 p.m.
DaewooOfDeath wrote: How do we account for the lifting that the swaybar puts on inside spring? Seems to me lift must equal roll resistance at the other side.

It's all there, equal and balanced.

The weight of the vehicle is shown as positive, and acts downward.

The torque values are downward on the right, and are added to the weight.

The reciprocal torque values are upwards on the left, and are subtracted from the weight.

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