Variable damping affect squat/roll?

[KSaiyu]

I was watching a re-run of Top Gear the other day, and JC was driving the Enzo, and he mentioned that it has a variable damper system that helps prevent the car from squatting under acceleration. This made me think - I thought dampers could only control the speed of the chassis movement/weight transfer, not actually control how much the chassis moves in relation to the weight transfer, and that it was only the springs that could adjust the squatting? I know a measurement (at least in Richard Burn's Rally) for the damper settings is kN/mm/s - indicating how fast it can convert the energy (and transfer the weight), so how can it possibly affect how much the car squats or rolls in a corner.

I looked up on webpages (heres one http://www.sae.org/automag/techbriefs_12-99/04.htm) and they all say that dampers can, in fact control how "flat" the car remains during cornering.

So, does anyone know how this is possible and where I'm getting it wrong?

 

[///M-Spec]

Sure. Shocks resist motion just like springs do. If the shocks are set firm enough, it can resist motion greater than what the springs will allow. This is because the load must first overcome shock resistance before the springs will compress. If the springs don't comrpess, then there is no dive, squat or roll.

For example, say you have a 3000 lb. car with soft 250 lbs./inch springs on the front. But say the front shocks are valved aggressively enough to effectively cancel compression at say.. 300/80 (you can do the math on this, but I honestly don't know how). When a portion of the car's weight shifts to the front, the weight must first overcome the shock's resistance to motion before it can compress the springs.

I don't know that it's a good idea to substitute shocks for springs, and wager most suspension designs do NOT. But in the case of computer controlled variable rate shocks like the Enzo's, PASM on the 911/Boxster or MSRC on the Corvette, it all depends on how the computers are programed and what the dynamic range of the shock system is. If the computer instructs the shock to be stiffer than thes springs (provided it is within the shock's dynamic range), then it should reduce body motions.


M

 

[Duke]

Thanks for answering that, ///M. I was struggling to put it that clearly.

Ask any clapped-out slammer Civic driver you see bobbing along on cut springs what happens when you substitute dampers for springs - you blow the struts out and have no damping at all.

But in a car that's designed for it, as ///M mentions, it's a viable (if expensive and complicated) method of managing the car's attitude.

 

[///M-Spec]

Thanks for answering that, ///M. I was struggling to put it that clearly.

Ask any clapped-out slammer Civic driver you see bobbing along on cut springs what happens when you substitute dampers for springs - you blow the struts out and have no damping at all.

But in a car that's designed for it, as ///M mentions, it's a viable (if expensive and complicated) method of managing the car's attitude.

No problem. It reminded me I need to read up on the math behind how valving is measured, though. I've always wanted to know.

Incidently, most sports/performance cars have anti-squat and anti-dive characteristics built into the suspension geometry nowadays, so it doesn't seem to make much sense to me to have the shocks do it.

Not that I'd second-guess the designers of the Enzo, of course. Though I would happily second-guess a complete wanker like JC.


M

 

[KSaiyu]

Sure. Shocks resist motion just like springs do. If the shocks are set firm enough, it can resist motion greater than what the springs will allow. This is because the load must first overcome shock resistance before the springs will compress. If the springs don't comrpess, then there is no dive, squat or roll.

For example, say you have a 3000 lb. car with soft 250 lbs./inch springs on the front. But say the front shocks are valved aggressively enough to effectively cancel compression at say.. 300/80 (you can do the math on this, but I honestly don't know how). When a portion of the car's weight shifts to the front, the weight must first overcome the shock's resistance to motion before it can compress the springs.

I don't know that it's a good idea to substitute shocks for springs, and wager most suspension designs do NOT. But in the case of computer controlled variable rate shocks like the Enzo's, PASM on the 911/Boxster or MSRC on the Corvette, it all depends on how the computers are programed and what the dynamic range of the shock system is. If the computer instructs the shock to be stiffer than thes springs (provided it is within the shock's dynamic range), then it should reduce body motions.


M

I see, that does clear it up, thanks.

Btw, what are units are the 300/80 you gave in, I've tried looking to see what dampers are measured in and see how they might compare to GT4's 1-10 (doubtful I think....)

 

[///M-Spec]

I see, that does clear it up, thanks.

Btw, what are units are the 300/80 you gave in, I've tried looking to see what dampers are measured in and see how they might compare to GT4's 1-10 (doubtful I think....)

It should be newtons over distance (thought the exact units escape me at the moment, probably tenths of meters)... basically how much force will compress or extend the shock x distance. Newton-meters would be close, but not quite. First number is rebound, second number is compression. There are other ways of expressing shock rates, I just picked the one I remembered. Shock technology is complex stuff and I'll quickly admit a yawning gap of knowledge on the technical side.


M

 

[retsmah]

The SI units for a damper are N*s/m , for a spring its N/m.

Spring force is F = -kx, where x is how much you've displaced the spring from it's natural length, k is the spring constant.

Damper force is F=-cv, where v is velocity and c is damping constant.

So a damper won't effect how much the car squats under acceleration, it will change how fast it happens Although if the squatting takes, say, a minute, then it might appear that it has stopped it, just because the car can't accelerate very hard for a full minute.

 

[///M-Spec]

^ good stuff to know, retsmah.

N*s/m

s=? seconds?


M

 

[retsmah]

Yeah, seconds. It just has to cancel out with the velocity units to produce units of force.

edit: The units could also be written as kg/s.

 

[KSaiyu]

The SI units for a damper are N*s/m , for a spring its N/m.

Spring force is F = -kx, where x is how much you've displaced the spring from it's natural length, k is the spring constant.

Damper force is F=-cv, where v is velocity and c is damping constant.

So a damper won't effect how much the car squats under acceleration, it will change how fast it happens Although if the squatting takes, say, a minute, then it might appear that it has stopped it, just because the car can't accelerate very hard for a full minute.

So then....the variable in the damper force equation is only velocity - is this the speed at which the energy is transferred to and from the suspension system?

Also, how does the measurement for the damping force fit into the SI unit - is the damping constant the value of N*s/m, the damper rate?

Thanks for all the help

 

[retsmah]

in F=-cv, v is the speed at which the damper is being compressed / expanded. Think of it like a little hand pump you would use to pump up a basketball, except without a ball attatched to it. If you push the pump in slowly, it will not resist much at all. However, if you try to push it fast it pushes back a lot harder.

A variable damping system would change the damping constant c, if somebody says damping rate I would assume they mean the same thing. If you apply a force to a damper with a relativley small damping constant it will compress quickly. If you increase the damping constant, and apply the same force, it will compress slower.

I'm not quite clear on what you are asking with the units. N*s/m will be the units for the damping constant of a damper. So if you have a damper with the 10 N*s/m damping constant, and you want to compress it at a rate of 10m/s, the damping force would be - (10 N*s/m) * (10 m/s) = - 100 N (or 22lbf). The negative sign indicating that it is the opposite direction of motion.

 

[KSaiyu]

Ahhhh, that makes sense, especially as I found this http://forums.nasioc.com/forums/archive/index.php/t-60195.html too, I think I get it now, thanks a lot.

 

[KSaiyu]

So a damper won't effect how much the car squats under acceleration, it will change how fast it happens Although if the squatting takes, say, a minute, then it might appear that it has stopped it, just because the car can't accelerate very hard for a full minute.

****, last question - could you explain that last sentence, do you mean that if the V is reduced in F=-cv this will effect acceleration....?

 

[neanderthal]

the squatting when you accelerate takes only few seconds, depending on many factors. if the squatting lasts for a full minute, then effectively, since the "squat" is so long, you would think that it has been eliminated.

think of how a knife cuts; force/ area. increase the size of the cutting surface enough and you lose the ability to cut. you are still applying the same force.

 

[skip0110]

****, last question - could you explain that last sentence, do you mean that if the V is reduced in F=-cv this will effect acceleration....?

Not really, based on what ///M-Spec said, if c is great enough F will also be large enough at small velocities to oppose the squat almost entirely. The "highly overdamped" situation.

In normal driving, I believe c should be set for "critical damping", so the car does not bob up and down, but it is the springs that are resisiting the motion.

And here I was thinking that weight transfer was a good thing. Especially when trying to launch on street tires.

 

[KSaiyu]

Wow this is older that I remember....

Anyway, I was looking at a certain youtube video

and it got me looking into all kinds of webpages until I came back to this thread (what brought it back was remembering where I had seen force-time before and realising this was something called "impulse"....apparently), and a question that I was to thick to put into words first time around.

I came to this conclusion while watching the video, correct me if I'm wrong. From the airbag example I saw that:

Force (F) X time (s) = Mass (M) X Change in Velocity (∆m/s)

So using this damper I figured I could apply it, say in a car accelerating and squatting

F(damper force)Xs(seconds)=MX∆m/s(change in momentum caused by the resistance to the squatting, the result of an impulse)

So with a normal damper system it could be:

FXs = MX∆m/s
10X1 = 10 since it HAS to be equal.

Now if we were to increase the damping force, but not change the time, in effect the "sports" setting on adaptive damping or a stiffer damper:

FXs = MX∆m/s
20X1 = 20 now because the damping force has gone up, so has the impulse and consequently the change in momentum.

My question is, would this momentum change caused by the impulse as a result of the force from F=-cv help resist motion in the same way as the force from F=-kx, since I assume both of these forces are measured in Newtons and both must be pushing the same way?