How do I determine the effect the sway bar has on the wheel rate? Are there formulas or tests I can use to approximate the wheel rate generated from My after market sway bar versus a stock bar versus no rear bar? Can I translate this into equivalent spring rates? Obviously this will become more difficult because not all bars of the same diameter have the same torsional rigidity so actual tests would be useful.

I hope I'm asking the right questions

 

I don't know how this would conceptually play out. Roll rate and wheel rate have to be considered separately, in my way of thinking (which may not be very complex). Look at the extreme case, where they ARB is completely out of the picture - the zero roll situation, going in a straight line. In simple jounce and rebound, the ARB has NO rate - no influence on wheel motion control. As cornering loads (and therefore body roll, or differential wheel motion, become an issue), the ARBs come into play, but this varies dramatically depending on state (transitional vs. steady), the geometry of the corner, and a bunch of other variables. I don't think that there is an easy answer to your question. Not only would actual tests be useful, they are the only way to determine what actually works.

Kirk

 

Yeah, I was afraid of that. I was imagining that there would some sort of relation between the the lateral load imposed on the vehicle based on the turn. This relation would be based on the rigidity(sp?) of the bar. My instincts want to tell that this is a linear relationship (like a spring)... I'm not sure if that's true. So when both wheels (that are linked with the bar) are compressed at the same time you have just the linear rate of the springs affecting the wheel rates. As the body starts to roll in a turn (we'll imagine a steady state sweeper for now). The sway bar begins to rotate and in affect begins to increase the wheel rate for the wheel right?

I'm trying to figure this out... but my brain hurts!

 

Another thing I just thought of...And again I'm not sure what the effect is (or if there is an effect).

When you turn the suspension compresses on the outside and the swaybar begins to twist The bar in effect is going to try and lift the inside wheel right? Well, Since the inside wheel is sprung, wouldn't some of the force of the swaybar be directed at trying to compress the inside wheels spring? So a softly sprung car with a HEFTY rear sway bar will more easily lift its inside rear wheel? Does that sound right?

EDIT : another thought... Since more traction can be achieved when all 4 wheels are on the ground (versus three wheeling with lots of force on the outside tires) would it make moresense to run a heavy spring setup with smaller a smaller swaybar?


[Modified by Watkinsm3, 3:31 PM 10/15/2002]

 

I can answer one of your questions....right now I do have soft springs in the rear with a beefy bar, and I don't lift the inside wheel. also, as you go into a turn, the sway does act on the inside wheel but more so on the outside. Imagine the sway bar as a spring going from wheel to wheel (which is really all a sway bar is). another thing to think of is that in our FWD applications the rear wheels are useless, they are just along for the ride...so it doesn't matter if you're three wheelin' it or both wheels have contact.

 

You generally want to lift the inside rear wheel actually. Say you have a 2000 lb car so it's static r/l weight distribution is 1000 lbs per side. When you turn you transfer more weight (say 400 bls) to the outside wheels. Now you have 1400 lbs on the outside wheels and 600 on the inside, so there is less weight on your inside drive wheel.

By lifting the rear inside wheel, you put all the inside weight on the front wheel so it still carries 600 lbs in our example as opposed to say 400 lbs if we insisted on making the rear inside wheel do some work. By lifting the inside rear you still transfer the same amount of weight but less is lost from the inside front, so you get more traction to accelerate out of the turn. I hope that makes sense.


edited to minimize stupidness


[Modified by robbin, 3:59 AM 10/16/2002]

 

over a certain limited range of displacement, I would imagine the ARB to have linear behavior, but due to the nature of the bar being a torsion spring in which the lever arm distance can change with changes in suspension deflection, I imagine it's tough to get a single meaningful number.

If I had to devise a 'simple' test to try to get a single number that might have some meaning, I guess you could do a back-to-back test of "bar disconnected", and "bar connected" (or vice-versa) in which you jack up one wheel and measure forces and deflections. There are a bunch of ways you could do it, but heck, you really don't need to even measure forces, I guess.. just jack one wheel off the ground and measure the amount of 'body roll' the car has. This data is all kinds of polluted, though with effects of structural rigidity and front roll rates, etc.

still rambling, if you could put the car up on a lift and push up on one wheel and measure the deflection of the other, if you have linear springs, you can get an torsional spring rate number for that suspension deflection domain.

just rambling.. sorry..

 

I'm not sure not sure I agree with either of you (robbin or carl_aka_carlos) ... Do you think these rear tires are just along for the ride?


The amount of cornering force a tire can generate is related to the amount downward force (weight) on it. If I remeber this is not a linear relationship. There for as the weight shifts to the outside, overall grip will increase up until a point. I've gotta do some research on this... see if I can find the exact equations for it...

EDIT : Shameless excuse to post a pic of my car!!


[Modified by Watkinsm3, 6:45 AM 10/16/2002]


[Modified by Watkinsm3, 6:52 AM 10/16/2002]

 

MechE00,
This is what I was looking for... kinda...

So I could use this to approximate what spring rates I would need to run with the stock bar or no bar to generate a similar wheel rate under lateral loads... (with a certain degree of error).

 

You're right, the relationship between weight and grip is not linear and by lifting the inside rear you make the outside rear do all the work to laterally accelerate the rear of the car. In this state, it is likely that the outside rear will have the most weight on it of the 3 tires still touching the ground, and therefore its coefficient of friction will be lowest. This is not generally a problem in FWD cars however because that outside rear has relatively little mass to accelerate through the turn. Because of this the ouside rear is not working any harder than either of the fronts in a well tuned car.

PS- This theory kinda relies on my suspicion that lateral acceleration increases exponentially with mass, which may very well be wrong. It's been a while since I've taken a physics class.


[Modified by robbin, 4:17 PM 10/16/2002]

 

Here's an excellent white paper that may help you with some of your questions:

http://www.ozebiz.com.au/racetech/theory/wttrans.html

 

ok....you guys are starting to get into a discussion of suspension theory of which I don't know much....however, I do know what set-up works, but I've realised that I don't know what about that set-up works....i need to get some good suspension theory/tuning books...any suggestions?

sorry to hijack your thread
you're car looks good by the way...I'm assuming thats the first right hander on the north course...were you full throttle through there? I was but I was in a EP RX-7

 

Carrol smith's books? I think he knows something about cars...

That is the first right hander on the north course (its the only spot that makes sense to me at least). I was full throttle to the next left hander where I had small lift so I could setup for the next right hander. That section was a blur!!!

 

Hi Matt,

Funny that you mention this -- I have been on exactly the same quest (for a non-Honda DSP car ). I have been very happy with my suspension under most circumstances, but decided I wanted to add an aftermarket adjustable bar. The question I have been asking is, "what is the effect of these bars of different dimensions in 'equivalent' spring rate?". While I haven't totally answered my own question yet, I have gotten to some order-of-magnitude calculations that are helping me to see the light. Allow me to share some of these rough calculations, maybe it will help you some:

I started out with the equations for a torsion bar (a swaybar is basically 2x a torsion bar in terms of a FBD):


////wall////
------------
| |
|r|
| | L
| |
| |___
|_____| <-- F = force (down into the screen) causes disp. x
...D

Starting with the equation for twist of a bar, theta = T * L / J * G,
and plugging in T = F * D ; J = (Pi/32)*r^4 (polar moment of inertia for
a round bar) ; and sin theta = x / D, solve for F:

F = Pi * G * ASIN (x/D) * r^4
....---------------------------
32 * D * L

G = Bulk modulus (about 140 GPa for steels)
r = bar radius (0.010 m for my application)
x = displacement of the end of the bar (solve at 0)
D = bar end length (appx 0.2 m)
L = bar length (appx 0.6 m to the midpoint)

Next, I did some estimates for my own car. These calculations were to determine the absolute maximum displacement of the endlinks based on 100% total weight transfer (car up on 2 wheels)



So I knew I was dealing with bar end displacements of about 2.5-3.0 inches. Time to plug some numbers in for different bars at different displacements;



The numbers in the last 3 columns are forces at the endlinks in lbs. So, what have I learned? Take a look at the front bar numbers at 2.5" disp: Stock 203#, RD 211#, 298#, 455#. Deltas are +8#, +95#, +252#. Now, consider that these forces are adding to one wheel and subtracting from the other, so the actual are probably 2x this amount (here is where my understanding admittedly gets fuzzy).

In any case, I concluded that: For my car, position #1 on the bar is like stock, #2 is like adding about 50# of spring (98#/2), and position #3 is like adding 125# to the front springs.

This is not for a Honda, and it's not a clean set of calculations, and I am sure I made a lot of inaccurate assumptions. But in my mind it gave me an approximate and order-of-magnitude answer. Maybe it can help you out.

I also have all these calculations in an excel spreadsheet. You may be able to adapt it to suit your needs.

Any comments?

(edit - tables don't post so good, do they? Added screenshots)


[Modified by whitney, 10:59 PM 10/16/2002]

 

Question : For the displacement. This would seem to me to be the net displacement of the two sides. So If the right side suspenion did not move at all and the left side was displaced 2" with the front stock bar then you generate 160lbs... So if the left compresses 2" and the right decompresses .5" the net result is 2.5" of displacement?

I'm no engineer and no expert so I can neither confirm or deny the validity of your statements. Here is what I got out of your numbers.. I liked your assumption that the deltas should be divided in half because the bar is working against both wheels. So for the stiffest bar up front the rates are equivalent to 100lbs at 2", 125lbs at 2.5", 150lbs at 3". This shows a linear progression in stiffness and because I'm a geek a I made a makshift graph with all three bars graphed out....

Keep in mind I used the unadjusted data for the graph.Now I make guesses at some data points based on the equation of the line and one last assumption. We'll use the deltas so we'll be comparing it to stock right?

With the stiffest setting on the front bar.
Assumption : with 0 net displacement the bar will generate 0lbs of force.
Guess : data points suggest 0lbs at 0", 25lbs at .5", 50lbs at 1", 75lbs at 1.5", 100lbs at 2", 125lbs at 2.5", 150lbs at 3"
Irrational conclusion : Bar at the stiffest setting behaves like a linear 50lbs per inch spring.

With the middle setting on the front bar.
Assumption : with 0 net displacement the bar will generate 0lbs of force.
Guess : data points suggest 0lbs at 0", 10lbs at .5", 19lbs at 1", 29lbs at 1.5", 38lbs at 2", 48lbs at 2.5", 57lbs at 3"
Irrational conclusion : Bar at the middle setting behaves like a linear 19lbs per inch spring.

These numbers are in addition to the stock numbers... of course. These dissagree with your numbers because you where looking at it over 2.5" of displacement... does that make sense? you declared a 125lbs for the front bar but this was at 2.5" of displacement. so 125/2.5 = 50lbs per inch.

The premise seems good. I'll have to see how this works out for my car... so you have a plug-n-chug version of that spread sheet?

EDITS : to make sense out of some of my gibberish


[Modified by Watkinsm3, 12:40 PM 10/18/2002]

 

I should have listed the two springs rates as such...

Stiffest setting on front bar : 50lbs per inch of net displacement between the two sides.
Middle setting on front bar : 19lbs per inch of net displacement between the two sides.

Just thought I'd put this in there because it might be the most important part to remember

 

unfotunately, when you pick up the rear tire, you loose all roll resistance because the car is pivoting in a plane uncontrolled by the asb's. you may actually get all 600lbs on the inside front, but the outside rear won't be carrying 700lbs anymore. it will be carrying less, and the outside front will be carrying more. thus defeating the whole purpose of the rear bar of reducing loading on the outside front. having just driven a vw setup to "pick-up" the inside rear to get rotation, i have come to the conclusion that this is the most assinine setup around. the car slalomed like **** and had very poor at the limit responsivness. unlike my car which keeps all 4 on the road.

nate

 

matt, talk to chris s. on this too. last we talked, he is planning on going to the "spring based" setup as well, so he may have actually calculated how much more spring he'll need to maintain his current balance. i'm thinking of going to the spring based setup as well. at a bare minimum, i think i'm going back to the stock front bar, since the pic i have from topeka on the south course doesn't even look like i've compressed the suspension at all. how does 500f/700r, stock front bar and possibly no rear bar. if 700 ain't enough to get it to rotate, say hello to some 900's!

nate <-who apparently dislikes any semblence of a comfortable street car

 

You're on the right track with trying to figure out a number that would represent the stiffness of the swaybar. However, keep in mind that the stiffness of a torsion bar (ARB) cannot be represented with a lbs/in figure. What you are really looking for is the wheel rate of the swaybar. And the only number that will have any meaning is the TOTAL wheel rate:

Total wheel rate = spring wheel rate + antiroll bar wheel rate + wheel rate of any suspension binding

This will give you a much better picture of relative roll stiffness between front and rear