Swaybar Attach Points, Symmetry

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With a swaybar, the distance from the pivot (attached to the car frame) to the end link attachment is the moment arm or leverage it has when twisting the bar. Attaching further away from the pivot gives more leverage, making the bar easier to twist, which is a softer setting.

Many swaybars have multiple attachment points, giving them adjustable stiffness. For example, if they have 2 attachment points, they claim to be "2-way" adjustable. I think such a swaybar is actually "3-way" adjustable because there is no need to attach symmetrically. For example, attach one side (doesn't matter which) to the far point (soft setting), the other to the close point (stiff setting), and you'll get a stiffness in between the two settings.

Question: if you attached asymmetrically like this, would the bar still respond symmetrically to L and R body roll? I think it would.

The stiffness is measured in how hard it is to twist the bar from the end links. The end links can't tell how long their moment arm is. They can only tell how hard the bar is to twist. That torque is the NET result of the length of BOTH end link moment arms or attachment points.

For example, suppose the length of the short & long attachment points is 6" and 12" respectively. If both end links are attached to the long arm, that's 12" each, so each end link (L or R) moves the other through 24" of moment arm. If one of them is moved to the short attachment point, then the total is 6" + 12" = 18" of moment arm. Each end link moves the other through 18" of moment arm. Neither can tell whether its own side is connected to the 6" or the 12" point. If you're connected to the short side, then the swaybar twists more on the opposite side because it has a longer lever arm, and vice versa. Yet since the entire swaybar freely moves up and down (lubricated bushings), it is the total amount of TWIST that matters, which is the same either way.

I'm curious what you think... agree? disagree?
 
Seems like to whatever extent the attachment length varies, to that extent you are unevenly loading that side of the suspension before even considering sway. That would also suggest that since the suspension isn't symmetrically loaded, that it would sway asymmetrically too, since the goal of a swap bar is to balance the suspension load, or at least more so than without it.

Why even bother? Surely one or the other adjustment is going to be acceptable or else why did you buy (this aftermarket?) sway bar.
 
With a swaybar, the distance from the pivot (attached to the car frame) to the end link attachment is the moment arm or leverage it has when twisting the bar. Attaching further away from the pivot gives more leverage, making the bar easier to twist, which is a softer setting.

Many swaybars have multiple attachment points, giving them adjustable stiffness. For example, if they have 2 attachment points, they claim to be "2-way" adjustable. I think such a swaybar is actually "3-way" adjustable because there is no need to attach symmetrically. For example, attach one side (doesn't matter which) to the far point (soft setting), the other to the close point (stiff setting), and you'll get a stiffness in between the two settings.

Question: if you attached asymmetrically like this, would the bar still respond symmetrically to L and R body roll? I think it would.

The stiffness is measured in how hard it is to twist the bar from the end links. The end links can't tell how long their moment arm is. They can only tell how hard the bar is to twist. That torque is the NET result of the length of BOTH end link moment arms or attachment points.

For example, suppose the length of the short & long attachment points is 6" and 12" respectively. If both end links are attached to the long arm, that's 12" each, so each end link (L or R) moves the other through 24" of moment arm. If one of them is moved to the short attachment point, then the total is 6" + 12" = 18" of moment arm. Each end link moves the other through 18" of moment arm. Neither can tell whether its own side is connected to the 6" or the 12" point. If you're connected to the short side, then the swaybar twists more on the opposite side because it has a longer lever arm, and vice versa. Yet since the entire swaybar freely moves up and down (lubricated bushings), it is the total amount of TWIST that matters, which is the same either way.

I'm curious what you think... agree? disagree?

The angle of the end links will change so you will continually have some pull on the sway bar. it's important to keep the end links angling symmetrically, especiall for very stiff bars
 
... Seems like to whatever extent the attachment length varies, to that extent you are unevenly loading that side of the suspension before even considering sway. That would also suggest that since the suspension isn't symmetrically loaded, that it would sway asymmetrically too, since the goal of a swap bar is to balance the suspension load, or at least more so than without it. ...
It does seem so at first thought. But thinking about it further, after drawing a free body diagram with Newton's 3rd law in mind, I came to the opposite conclusion. Maybe I'm wrong, which is why I'm asking here.

The swaybar freely pivots up and down, so it never presents any suspension load until some body roll starts to happen (car is turning L or R), or more generally, the L and R sides apply different forces. At that point, the body has to twist the swaybar in order to roll. The resistance is the total twist or torque, which is based on the attachment points on BOTH sides.

The angle of the end links will change so you will continually have some pull on the sway bar. it's important to keep the end links angling symmetrically, especiall for very stiff bars
Ah, yes. So even if asymmetric arm lengths still provide symmetrically balanced forces, the angles between the end link and the swaybar arm should be the same, else the force exerted by the swaybar will act differently on the suspension.
 
The distance between the different holes on the sway bar and the pick up on the suspension isn't equal, so even in a steady state you will always be tranferring load
 
The distance between the different holes on the sway bar and the pick up on the suspension isn't equal, so even in a steady state you will always be tranferring load
They might (or might not) be equal. If the swaybar is positioned so that the attachment holes are all equidistant from the end link, like different points along a circle all being equidistant from its center (the endlink here being the radius of the circle). This depends on the geometry how the bar and endlinks are positioned relative to each other.

Pragmatically, if you can connect the end links on L and R sides, each to different length points on the swaybar, without twisting the bar, then how could it exert any steady state load? If L and R sides move up and down together, they simply move the swaybar up and down freely without trying to twist it.

PS: I can see another problem with asymmetric attachment. An endlink attached closer to the bar's pivot (the short arm / stiff setting) will try to move the bar MORE for a given displacement. Suppose you hit a bump that shifts L and R sides equally up 3 inches. If the L side end link connects to the bar's stiffer hole, it's closer to the pivot, and tries to shift the bar up MORE than the R side, which is connected further from the pivot (soft setting). So, a bump causing equal displacement of both L and R sides, will try to twist the bar. Which you don't want.
 
As the suspension moves, the far and near hole will be different distrance from the pick up, no question about it. theres 2 circles with different centers. one for the sway bar, one for the end links and the one for the sway bar has 2 radii depending on the side. The bar will always be twisted except for one sweet spot.

Connect one end link on the swaybar, then connect the other. unless the hole and end link line up perfectly, you are transferring load. This means if yuou turn one way yourr sway bar will transfer more load the faster you go, but turn the other way and it will first reduce transfer, go to zero and then start transferring again to the other side with increasing turning speed

If the different distance isnt an issue, yes, you could put one side far and one side short.
 
The holes you choose do not have to be symetric. Also the links themselves can be adjusted different lengths to preload one side or the other to suit the expected roll load that will be applied across the car. Example of why you may do this is racing with weight mainly on drivers side of the car. the bar can be made to be stiffer in one rolling direction than another. I have a SXS car with big mushy shocks i do this to dial in the directional handling.
 
As the car sways, a certain torque develops in the bar which will always be equal and opposite at both ends. This will only become an equal and opposite up and down force on the wheels if the lever arms are the same length. If one arm is longer then that wheel will move farther up (down) than the other one moves down (up). You probably don't want that.

This is not the same as setting the link lengths unequal to pre-load the bar when the car is sitting still. That only transfers static force from one side to the other, which affects the relative height. The same thing could be done with the springs if they are adjustable.
 
As the suspension moves, the far and near hole will be different distrance from the pick up, no question about it. theres 2 circles with different centers. one for the sway bar, one for the end links and the one for the sway bar has 2 radii depending on the side. The bar will always be twisted except for one sweet spot. ...
Exactly. So, when the endlinks are asymmetrically attached, any time the suspension tries to move away from that sweet spot (up or down, even if equal on both sides), it will twist the bar which will resist that motion. In this case (when asymmetrically attached) the sway bar adds to the overall spring rate even when both sides move together.

The big question: is this force applied by the swaybar equal on both sides, even when the endlinks are attached asymmetrically? Others have debated this question. Here's one long yet detailed discussion that I found worth reading:

The net is that they actually tested it and found that the forces are asymmetric. Yet the differences weren't as big as one might expect. I find the most plausible explanation for the asymmetry is what @mk378 mentions above. When a swaybar is connected asymmetrically, the torques are still equal and opposite (Newton's 3rd law) but the moment arms are not, so the forces applied are not.
 
Part of the force doesn't go into twisting the swaybar or bending the arm, but pushes up against the sway bar rubbers, here also the assymetry plays a role. so the efficiency of the transfer isn't 100%
 
Part of the force doesn't go into twisting the swaybar or bending the arm, but pushes up against the sway bar rubbers, here also the assymetry plays a role. so the efficiency of the transfer isn't 100%
Nascar setup would eliminate a lot of the play and give close to 100% efficiency.
product1_246-2.jpg
 
The big question: is this force applied by the swaybar equal on both sides, even when the endlinks are attached asymmetrically?

It's not a big question. It's an obvious one except that you are rejecting reality.

No. You were wrong and still are. You cannot pretend a sway bar is free floating once it is attached at different lengths, unless it is only free because of excessive slop and even then, it's detrimental to even handling.

It is impossible for it to distribute sway evenly when you have different leverage on each side.

You are ignoring physics.
 
It's not a big question. It's an obvious one except that you are rejecting reality. ...
If it's obvious, there must be a lot of dense people because I find quite a bit of debate on this topic. I linked to one of the better discussions above.

... No. You were wrong and still are. ...
Wrong about what exactly? In asking the question? In researching other opinions and tests? In concluding that the forces are asymmetric?

... You are ignoring physics.
By making a free body diagram and applying Newton's laws in order to find the answer, I'm ignoring physics? Interesting.
 
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