# Effects of sway bar stiffness on rough, uneven roads

I just read a really nice write-up on sway bar physics. I also watched a video discussing suspension behavior on rough roads.

Let's assume the following:

• A front-wheel drive vehicle being driven around a corner at a speed which, on a dry, smooth, level road, would be just about the maximum speed it could take before beginning to understeer.
• For simplicity, any sway bar changes would be done to the front and rear in such a way that the TLLTD would not be affected.
• Shocks, struts, springs would not be changed.
• By "rough" roads I mean typical non-perfect driving conditions that you may encounter on a daily basis: Think patches, seams, and potholes on highways, think ripple, ruts, and depressions (e.g. typical road wear near stop signs, on roads frequented by trucks, etc.) on urban roads, post-construction patches, stripped roads being prepped for resurfacing, raised manholes, drain depressions, that kind of thing. It's a wide definition, but I don't mean off-road or post-apocalyptic conditions.

In this case, how would a stiffer set of sway bars affect vehicle handling on rough, uneven pavement? Every discussion of suspension theory and physics I see usually seems to assume good road conditions.

For example, consider the scenario above, cornering left at speed, then in the turn I hit a fairly large, say 2-3cm deep pot hole with the front left wheel.

From my limited understanding the effect of a sway bar that was too stiff would be one of the following, either:

1. The left strut would expand into the pothole, exerting downward force on the wheel.
2. Via the sway bar, some of this would also be transferred to the right side, exerting an upward force on the right side of the body.
3. On exiting the pothole, then, something... complicated would happen that I can't figure out.

Or:

1. The left strut would want to expand into the pothole.
2. The left side expansion would be limited via the swaybar by the downward force present on the right side due to the turn.
3. The left wheel would then take longer to regain contact with the ground, causing the right wheel to experience more lateral force (that was no longer being absorbed by the left wheel), and the car would more readily understeer. And maybe some other complicated thing would happen.

Am I on the right track with one of those assessments? What would the effect be?

Also as a (perhaps too broad) corollary question: What impact should rough road conditions have when deciding on an ideal sway bar configuration?

• Note that a stiffer sway bar is akin to a stiffer spring. Think of it like this: What would happen if you put stiffer springs? Now apply that to a spring that affects both corners at the same time. I can't give you a definite idea, but this will hopefully shed some light. – race fever Mar 20 '16 at 21:27
• This is a great question, btw. I wish I had more knowledge about suspension dimensions to give you a great answer. – Pᴀᴜʟsᴛᴇʀ2 Mar 30 '16 at 1:49
• I'd really like to see the great answer to this one. – DucatiKiller Apr 1 '16 at 1:38

tl;dr: stiffening one of the sway bars on a car will cause that end to be more likely break loose in response to transients.

At a high level, the sway bar acts as a spring just like any other. You can disassemble the sway bar problem by considering a piece at a time. For example, imagine that one end of the sway bar is attached to the wheel assembly at one end but is fixed to an immovable point on the other. If you try to move the wheel assembly up or down suddenly (as would happen with transient bumps and dips in your example), the bar would try to rotate on its pivot points. If the other end weren't affixed to anything, the bar would obviously just freely rotate. However, since it is bolted down in this example, the bar acts as a torsion spring, resisting the twisting action. The further the bar tried to twist, the greater the resulting toque that the bar would exert in the opposite direction. This is translated into a greater spring force on the wheel assembly itself.

Of course, we don't bolt the ends of sway bars to the frame. We connect them to suspension points at either end. As such, they're now coupled to the whole damped spring system that was already there. Again, if we add a force to one wheel, the sway bar will try to rotate on those pivot points. This will result in an equivalent force being exerted on the other wheel assembly (if you try to raise the right wheel, the sway bar will try to raise the left wheel).

Here's where we starting getting into the key points of your question: remember that springs only exert forces when they are moved from their rest state. For the sake of this discussion, let's stick to linear springs:

F = k * d

where F = Force, k = the spring constant and d = distance or deflection. The equivalent for torsion springs is:

T = k * theta

where T = torque, k = a different spring constant and theta = the angle of twist. In both of these cases, you can see that the more you compress, extend or twist the spring, the greater the resulting force or torque. What's more important: if you don't move the spring, there's no force at all. So, for the sway bar to exert any force on the wheel that you're considering, it has to have cause the spring on the other wheel to be deflected (compressed or extended). This is critical: the sway bar doesn't do anything until it has caused something to happen on the other side of the car.

Another way of saying this is that sway bars make your four wheel independent suspension significantly less independent.

Let's restate your original problem in a way that we can break it down. Imagine a single pair of wheels with their springs and an attached sway bar. This is a magic sway bar on which we can dial in a variety of torsion constants (ranging from limp spaghetti to rigid steel I beam). Now we exert a sideways force on this whole contraption that is just slightly less than the limit of a single tire (i.e., if there were only one tire contact patch on the ground, it would almost slide, but with two it does not).

Now turn the magic sway bar down to its near-zero rigidity setting and bump one wheel (e.g., lift its contact patch off the ground suddenly) while the sideways force continues. The opposing wheel is almost completely unaffected by this bump and so its tire contact patch is undisturbed. Since we carefully selected the sideways force to be just less than that required to push the tire sideways, the system is unaffected.

Now set the magic sway bar to effectively infinite rigidity. Now, when we lift one wheel, the other wheel is likewise raised. Since both tires lose contact, the whole system starts sliding sideways.

Reality is, of course, somewhere in between but this sort of thought experiment makes the point: if you lift one wheel, the sway bar is going to try to raise the other as well. This results in that whole end of the car feeling like it's breaking loose.

Practical real life example: when I had a FWD Integra, I tried this exact experiment. My rear sway bar had three settings that allowed me to control the stiffness (really they affected the leverage that the rest of the suspension had on the sway bar but the result was effectively the same). This gave me four possible stiffness settings to experiment: no bar + three increasingly stiff bar choices. There is one particular off ramp nearby that I could use to try tight legal turns. What I found was that increasing the stiffness would decrease the quality of the ride over bumps and increase the feeling that the back end would hop out (try to oversteer).

## Adding a sway bar will result in a harsher ride

I've tried to explain the mechanics with this side-by-side graphic:

## Explanation

• When a wheel encounters a ditch, the weight of the vehicle on that wheel will cause it to deflect downwards. This causes the suspension spring to extend, causing a resistive force to act in the opposite direction.

• Adding a sway bar introduces an additional resistive force in the mix, which reduces the amount of spring-resistive force. This results in a smaller spring deflection compared to when no sway bar was present.

• Less spring deflection means that the vehicle body will want to follow the wheel into the pothole more than when the sway bar was not present.

## Since a picture's worth a thousand words

This is the effect of the sway bar

## It's little surprise that sway bars and off-roading rarely go together

You want the chassis to be a little twistable and flexible.

Else parts would break under duress quite readily.

Actually sway bars are bolted to the chassis at a point somewhere near the lower control arms (in the front, on the subframe). Therefore it essentially makes each corner stiffer, individually. So with that being said, there is a "zero" position (where the suspension is when the car is parked), so the further the suspension travels away from that zero position, either positive or negative, the more resistance there is (I guess you could think of it like a pry bar?). So if you have a small diameter sway bar, it bends easily allowing more suspension travel, where when you get into some of the bigger diameter sway bars, the wheel will just stay in the (essentially in the zero position) air if it is in a hole. The goal here really is to add or take away grip, usually when the suspension is loaded up. The softer the setup, the more grip there is.