27 replies to this topic

[Ray Bell]

Is there some means of determining what unsprung weight would come from a torsion bar?

The Lotus 72, for instance, used these, with both inner and outer bars so they could mount at the point where the load was applied.

Mercedes' W196 used long bars that fed the loads through the chassis.

I don't think it's reasonable to say that a torsion bar doesn't have any unsprung weight, but how would you go about calculating what it has?

[Engineguy]

Figure the inertial effect of a bar that size and divide by two... 'cause one end of the bar is rotating the full number of degrees the attached arm does, and the other end moves not.

[Ray Bell]

The issue I have with accepting this is that it's a twisting motion with no actual movement otherwise...

So how do we work that out? The weight of half the bar divided by the length of the lever through which it operates?

[red300zx99]

If anything I would only suspect a portion of the lever arm weight. Think of a car in bump and rebound(both wheels), your turning the bar and moving part of the weight of the arms, plus some friction. Ignoring the friction all you have is the weight of the arms. Sort of like the a-arm unsprung weight, remember not the full weight of the a-arm is unsprung, or the majority of a push-rod assembly for that matter. Now if only one wheel is in bump, the torsion bar is adding resistance, but not weight.

[Greg Locock]

This is quite odd. I agree the answer is half the weight of the lever, so that represents a substantial reduction in unsprung mass compared with a coil spring.

So if unsprung mass is so important (one of the most repeated mottoes of the ride and handling world, one that I don't necessarily agree with) how come so few people make the effort?

Would someone with a circuit car like to bolt a 1 kg mass to each hub and tell us what the real difference was, compared with carrying 4 kg on the body? I've asked our ride and handling guys to do it, but they haven't so far, as I remind them every time they start wittering on about unsprung mass.

[McGuire]

Having been involved in a number of racing series where the rules required rear live axles, I have seen efforts to reduce the unsprung weight back there. Some have been extreme and expensive (exact replica titanium rear end housing, for example) but the results are usually disappointing and the $$$ surely would have been better spent elsewhere.

It's fun to have one of those housings around to keep under the bench, though. Here, catch.

[Engineguy]

Originally posted by Greg Locock
This is quite odd. I agree the answer is half the weight of the lever, so that represents a substantial reduction in unsprung mass compared with a coil spring.

So if unsprung mass is so important (one of the most repeated mottoes of the ride and handling world, one that I don't necessarily agree with) how come so few people make the effort?

Would someone with a circuit car like to bolt a 1 kg mass to each hub and tell us what the real difference was, compared with carrying 4 kg on the body? I've asked our ride and handling guys to do it, but they haven't so far, as I remind them every time they start wittering on about unsprung mass.

It’s just a matter of the effective inertia of all the moving suspension parts, as applied at (and including) the moving end of the damper. That includes the rotational inertia of the torsion bar itself (about half of what it would be if the bar was not anchored) divided by the lever length, as well as the calculated effective inertia of the lever. All the “half the weight of” assumpions assume a very simple case of constant cross-sections of components, common pivot points, etc. Not to mention the complications of components that constantly change their movement ratio (relative to wheel travel) throughout their travel.

The inertia of unsprung mass is only a problem to the extent that you cannot control (damp) it, or the extent you compromise to damp it. The stiffer you have to make the damper to control excess unsprung mass, the harder it is for “good” movement to happen. If you’re cornering and the tire encounters a bump (or even an undulation) the suspension needs to move. If your stiffer damper (and the extra unsprung mass’s inertia itself) doesn’t allow the travel to take place as quick as needed, the tire carcass is forced to distort instead… that’s not good for the cornering traction. I would note that professional race teams have shock dynos in their transporters, and the difference between winning teams and losing teams is often attributed to a better understanding of their dampers.

How important is the reduction of unsprung mass? Clearly less unsprung mass is better than more. I don’t know quantitatively though. Apparently not as important as massive (=heavy) brake discs and the widest (=heavy) wheels and tires you can fit. Obviously not as important as making suspension components that are beefy (=heavy) enough to absolutely prevent the possibility of catastrophic failure.

By the way, I think that the inertia advantage seen by using a torsion bar rather than a coil spring is offset by the added inertia of its lever and link, or the beefier wishbone if the wishbone acts as the lever. For example, racing 911s had very heavy wishbones when they carried the bending load imposed by the coaxial torsion bar. When they eliminated the torsion bar in favor of a coil over the strut (Carrera RSR Turbo, 935) it freed them to use scary thin wishbones that looked like they were made from McDonalds drinking straws. Also, from a total vehicle mass standpoint, a coil-over-shock or coil-over-strut feeds both the spring load and the damper force to the same chassis point. Torsion bars may require extra or increased structure at least at the anchored end if not both ends of the bar.

[Todd]

Originally posted by Greg Locock
This is quite odd. I agree the answer is half the weight of the lever, so that represents a substantial reduction in unsprung mass compared with a coil spring.

Wouldn't the unsprung mass calculation for a constant rate coil spring be similar? One end of it is fixed to the chassis of the car as well.

[red300zx99]

Originally posted by Todd


Wouldn't the unsprung mass calculation for a constant rate coil spring be similar? One end of it is fixed to the chassis of the car as well.

Yep

[Greg Locock]

If you assume a 1:1 motion ratio then for a constant rate coil spring the contribution to unsprung mass is just half the mass of the /spring/. What interests me is that the torsion bar seems to get round that.

The best way I know of determining the unsprung mass is to check the wheelhop frequency, which is proportional to (effective rate at contact patch /unsprung mass) ^.5

That rate is dominated by the tire radial stiffness for production cars, maybe not on race cars.

Actually I think this 50% rule may be a crock, I'll whack a model of this together later on today. I have a suspicion that there may be a v^2 relationship in there.

[D-Type]

Let's go back to basics:

The term 'unsprung weight' is a misnomer, it should be 'unsprung mass' since this is what we are talking about - the inertia of what's on the end of the spring.

A torsion bar is simply a section of a coil spring of infinite radius. Or conversely a coil spring is a torsion bar curled up. So all they are is different forms of spring.

As to whether a spring (or torsion bar) is part of the sprung or unsprung component goes back to the definition. We have a two-body system separated by a spring. One side of the spring is 'unsprumg', the other side is 'sprung'. So we have to ask ourselves on which side of itself the mass of the torsion bar or spring is. There can only be one logical answer, namely half must be considered 'sprung' and half 'unsprung'

The idea that a spring has an additional 'spring inertia' is flawed. Although it takes a force to compress/stretch a spring, the moment the force is released it shortens or expands again, there is no delay due to any inertia that has to be accelerated other than that due to its mass.

The basic reason that low unsprung mass improves road holding is that the lower the mass, the lower the inertia to be accelerated or ******** for a wheel to follow an undulation exactly.


(I think)

[Greg Locock]

Wrong.

Or at least, you have ignored a very very significant factor.

[Ray Bell]

Originally posted by D-Type
Let's go back to basics:


.....The basic reason that low unsprung mass improves road holding is that the lower the mass, the lower the inertia to be accelerated or ******** for a wheel to follow an undulation exactly.


(I think)

You think quite correctly...

[D-Type]

Originally posted by Greg Locock
Wrong.

Or at least, you have ignored a very very significant factor.

What? Please clarify.

[McGuire]

The lighter the unsprung mass, the greater its frequency...which can be rather more difficult to damp.

We actually have two layers, if you will, of sprung vs. unsprung masses between the vehicle and the road: between the vehicle and the suspension, and between the suspension and the tire....and the tire has own springing and damping properties as well.

It is also necessary to regard unsprung mass in proportion to sprung mass. Watch a small, unloaded trailer being pulled down the road. Dinky tires and wheels, no brakes, so unsprung mass is very low. However, with the trailer unladen it is enormous in proportion to sprung mass...so at highway speeds the trailer bounces around as if it has no suspension at all. here, lightening the unsprung mass is not likely to improve things.

[Todd]

Originally posted by McGuire
However, with the trailer unladen it is enormous in proportion to sprung mass...so at highway speeds the trailer bounces around as if it has no suspension at all. here, lightening the unsprung mass is not likely to improve things.

Isn't this mainly the result of spring rates chosen for the trailer in it's laden state?

[Greg Locock]

It turns out I was thinking far too hard about this. Just consider the CG of the spring as the wheel moves. The CG of a linear coil spring moves by half its deflection. The CG of a torsion spring does not move. Therefore the coil spring gains kinetic energy which the torsion spring does not.

Each spring also gains some rotational kinetic energy, but in the scheme of things that is a relatively trivial number.

This ignores the additional KE of the lever for the torsional spring, of course.

[wegmann]

I won't add any technical info to this conversation, but Greg has at least the support of Gordon Murray, who believes there's an optimum ratio of sprung to unsprung mass. That being said, he also believes most road cars have more than the optimum amount of unsprung mass. The source is that book on the McLaren F1, but I don't have the quote and page number with me.

[McGuire]

Originally posted by Todd


Isn't this mainly the result of spring rates chosen for the trailer in it's laden state?

Sure. You have a sprung mass, an unsprung mass, and a spring medium and damper(s) negotiating the forces between them...with the whole point of the exercise being to keep the unsprung mass planted on the ground, as the roll, pitch, rebound etc forces of the sprung mass are trying to jerk it off the ground.

[mat1]

Originally posted by D-Type


A torsion bar is simply a section of a coil spring of infinite radius. Or conversely a coil spring is a torsion bar curled up. So all they are is different forms of spring.

(I think)

That is not completely true, as the coil spring is not just a curled up torsion bar, but also deflects in its length. The spring effect of this deflection is small. But there is a movement of mass, which is not existing in a torsion bar.

mat1

[Ray Bell]

So the torsion bar doesn't twist at all?

Something has to be accounted for...

[Greg Locock]

They both twist along the axis of the wire. The coil spring equations are all about torsional behaviour.

[mat1]

Originally posted by Ray Bell
So the torsion bar doesn't twist at all?

yes, it does, but it does not bend, as a coil spring does (if seen as a curled torsion bar).

mat1

[Ray Bell]

Naturally enough I know that...

I was wondering about the logic of the previous statement. There must be a movement of mass in a torsion bar.

[mat1]

Originally posted by Ray Bell
Naturally enough I know that...

I was wondering about the logic of the previous statement. There must be a movement of mass in a torsion bar.

Sorry Ray, i misunderstood you.

Of course there is a movement o mass in a torsion bar, but it is just a circular movement (the twisting). That is all, and in terms of "moved" mass it is very small, negligible I think.

Maybe I am again misunderstanding you, because probably this is obvious to you?

mat1

[Ray Bell]

Sure, it's small, but it's there... that was what I was getting at...

There must be a way of calculating or estimating just what contribution this makes to what we'd normally call 'unsprung weight'.

But when you look at it reasonably, it does seem that Champman's setup on the Lotus 72 (copied from early postwar Vauxhalls, I believe) was a very good one.

[Greg Locock]

OK, the polar moment of inertia of a tube is its mass*r^2. Halve that, as only one end is moving. SO, for a lever arm length L, moving by a (the wheel accelertion), you'll get an angular acceleration a/L, and apply a force F giving a torque FL, so the referred unsprung mass m=F/a

and FL=mass*r^2/2*a/L

so m=mass*r^2/2/L^2

And r <
God knows how many mistakes I've made in that!

[Ray Bell]

I won't be checking, that's for sure!