Lets Talk About Sway Bars

My car has gone through some changes since Ihave owned it, and i have been driving it more in the last year or so. It had some damage to the left front control arm when i got it and in the process of repairing it,I found the sway bar was bent. So the front sway bar was removed and thrown away when the I replaced the control arm. It is almost impossible to find a used one here locally in here in OKC. Since the engine was swapped out and is a tad heavier in the front end, i need a sway bar. My plans were to use sprint car torsion bars, since they come in several different sizes ( lengths and thicknesses). I have attempted research the science behind sway bay spring rates vs. bar thickness vs. solid vs. hollow with no prevail. Can someone explain how sway bars are rated as in coil springs are rated in lbs. per inch of compression. Are sway bars rated they same way? And if so, what is the standard rating. My car weighs about 2780 with me in it and I am a solid 250. It also weighs about 200 lbs. heavier in the front. Any help will definately help. Thank you.

http://www.auto-ware.com/calcs/swaybar.htm

Solid vs hollow is purely weight related. Both act the same.

Ranger50 wrote: Solid vs hollow is purely weight related. Both act the same.

Wall thickness does count into the equation. It isn't as significant as you would think though. A slight bump in overall diameter is well worth the weight savings to get the same spring rate.

To get the stiffness of a hollow bar, figure the stiffness of a solid tube with the same OD and subtract the stiffness of a solid tube with the same ID.

Sway bars are rarely a simple torsion bar with a straight arm, so the calcs can get interesting. It's not typical to see a bar rating.

Here is a chart on the relative stiffness (%) as bar size changes.

What I could really use is a good explanation of how to make use of this sort of rate info for sway bars.

I wind up with brain stew pretty quickly when I try to cocktail-napkin the interrelationship of the inner wheel's spring, the outer wheel's spring, and the sway bar's springy connection between the two as the wheels experience different displacements...

It seems like it should be pretty simple. It probably wouldn't be bad to make up a table of wheel loads at different displacements. What I've been trying to work out is an equation for the spring rate at each wheel; It seems like it ought to just be a springs-in-series thing; something like the outer wheels' rate would be its own spring rate plus (1/insideRate + 1/swayRate)^-1, but among other issues, that doesn't take into account stuff happening to the inside wheel...

Or maybe it doesn't make sense to try to do that at all.

Race Tach did an excellent article on swaybars that I have laying around somewhere.

In reply to 93EXCivic:

If you find it, do let us know what issue; it's probably one I'm missing, but I do hang onto those...

The simple way is to make it adjustable.

Other than that, it's not simple to calculate. We understand dia. matters, but so does the length of the arms, length of the bar, # bends, pre-load, type of attachment, relation to spring rate, roll center of car, weight of the car, live load, cornering speed, geometry at opposite end of car, metalurgy, etc.

They are pretty easy to make adjustable, so racers and people who are trying to dial it in close typically guess, then adjust later.

Deathtrap, I'm gonna go out on a limb and say that that SBC is a bit much for your poor little Corolla's unibody (but I LIKE it! ). I'd first look at some chassis stiffening (maybe use the cage to help?), then do an adjustable bar.

DEATHTRAP wrote: My car has gone through some changes since Ihave owned it, and i have been driving it more in the last year or so. It had some damage to the left front control arm when i got it and in the process of repairing it,I found the sway bar was bent. So the front sway bar was removed and thrown away when the I replaced the control arm. It is almost impossible to find a used one here locally in here in OKC. Since the engine was swapped out and is a tad heavier in the front end, i need a sway bar. My plans were to use sprint car torsion bars, since they come in several different sizes ( lengths and thicknesses). I have attempted research the science behind sway bay spring rates vs. bar thickness vs. solid vs. hollow with no prevail. Can someone explain how sway bars are rated as in coil springs are rated in lbs. per inch of compression. Are sway bars rated they same way? And if so, what is the standard rating. My car weighs about 2780 with me in it and I am a solid 250. It also weighs about 200 lbs. heavier in the front. Any help will definately help. Thank you.

I am not sure that a swaybar is the appropriate way to counteract the front end being heavier. I would think you would want to do that with springrate.

I might be way off on this, so take that for what it is worth.

Rob R.

Well the sbc isn't going to be in the car much longer due to poor oil pressure conditions. An aluminum block 6.0l is being built so the 200 extra lbs. aren't gonna be an issue for long. The weight should be close to 50/50. I have no idea what size the original sway bar was, i didn't pay attention. I am not looking the my car to handle like a slot car just want a little more stability. These sprint car torsion bars are about 75-100 each and i have no idea where to start in the stiffness. Most of the info that was posted was for already knowing sway bar deminsions. And i don't have that. It seems like the torsion bar idea may cost me a little aggrivation and losts of money. I do appreciate the input.

Bars can be computed the same as spring rates. The lever arms also have an effect on the rate.

Would there be a formula for this? Obviously the longer the arm the more leverage. But would there be a standard formula? Like using for a coil spring?

While it is very easy to calculate the torsional rigidity of a straight bar, it's completely useless for what you want to achieve. Unless you figure out the entire system, that information is useless. Like knowing the spring rate of your coil springs, when you need to know the wheel rate.

What car is it for, I'm sure someone here could find the dimensions of the stock swaybar, and you could start from there.

ransom wrote: In reply to 93EXCivic: If you find it, do let us know what issue; it's probably one I'm missing, but I do hang onto those...

November 2008

In reply to 93EXCivic:

Thanks!

Too bad I've got October 2008 then a giant gap to July 2009... D'oh.

Okay, I will start there. What is wheel rate? I have never heard of this 'til today. I am assuming the rate of wheel travel or the distance of wheel travel during suspension travel during during flex?

You might say the wheel rate is the effective spring rate. If your spring is not at a 90 degree angle to the forces, like your coilovers are at an angle, the wheel rate isn't equal to the spring rate. IIRC at a 45 degree angle the wheel rate is half (?) of the spring rate.

And then there's the frequency rate. . .

Okay. The angle of the spring to true vertical will equal the wheel rate in a percentage? Right? Now, how does this relate to finding what I'm looking for? Or am I wasting my time and just need to throw down cash for an aftermarket sway bar? Oh fanfoy, my car is an 81 four door corolla with a SBC. Not the topical project car you can just go find at a salvage yard here in OK. Most of them have already swong from the crane and have been put into a nice cube shape by now.

Oh what is IIRC?

In reply to DEATHTRAP:

"If I Recall Correctly"

Wheel rate is the important result you get from spring rate, since it's what the tire actually sees. Spring rate tells you exactly what you think it does about how much force it takes to compress the spring a given distance (e.g. it takes 250 pounds to compress a 250 lb/in spring one inch).

If that spring is mounted square to the arm but only halfway along it's length, the wheel has a significant mechanical advantage when it compresses that spring. Here we hit one really important part: Whatever that ratio is between arm length and how far out the spring is mounted, that is squared to find the relationship between wheel rate and spring rate.

To give an example, sticking with our 1/2-way spring mounting, the wheel rate is half the spring rate (EDIT: That was an incredibly poor choice of words on my part. I should have said the wheel force is half the spring force at a given displacement, because the whole thing we're establishing is that this is only part of the rate calc), BUT you also multiply that by the fact that to move the wheel one inch you only compress the spring 1/2 inch (because the arm only moves half as much where the spring mounts). So, a 400 lb/in spring would provide a wheel rate of 100 lb/in, and a 1000 lb/in spring would provide a wheel rate of 250 lb/in. If the spring were mounted 1/3 the distance from the inner pivot to the end of the arm the wheel rate would be 1/9 the spring rate.

Angling the shocks is a separate thing, and of course... blargh, I was about to start in with concentric circles and some trig, but I suspect this PDF (pages 5 and 6) probably does it better than I would.

Hopefully someone will cover frequencies. I don't have time to look up a refresher right now, but IIRC it's more or less the natural frequency of the suspension's springs and the weight it's carrying, in the same way that if you taped pennies to a dashboard bobblehead it would move back and forth a bit more slowly (would have a lower frequency)... I know Alan Staniforth covers this and has some guidelines about frequencies to shoot for for road and race cars...

Also hopefully, someone with more practical experience will give you a more informed and useful version of what I just wrote. I believe it's all accurate, but I'm just regurgitating readings...

Take a solid rod and twist it. Now take a piece of pipe or even conduit of the same diameter and twist it. Notice how much more rigid the pipe or conduit is.

The calculation is not one of simply subtracting the material from the calculation. Solid rods are not stiffer than pipe of the same diameter.

This is the same basic reason an I beam is more rigid than a comparable sized solid timber.

In reply to foxtrapper:

What he (she?) said. IIRC, it's about the surface area, right?