Lower unsprung weight, what is the advantage?

Lower unsprung weight, what is the advantage?

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Sport Coupe

415 posts

153 months

Saturday 4th October 2008
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An enjoyable read, thank you.

Rocket Pepper

1,281 posts

171 months

Saturday 4th October 2008
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Jeez Rich, I was just going to say contact patch and leave the rest for others to argue.

Interestingly, we beat this up a bit a few years ago as my Buell has carbon fibre Dymags. We were dynoing a Land Speed Buell in Colorado readying for Bonneville and the conversation got around to even better unsprung weight brought about by a wheel such as mine giving a higher horse power. I believe we concluded that there may well be a small increase seen on the dyno due to a lighter rear wheel but essentially there would be no relevance to power gain unless the saving in weight was quite significant. As far as I'm aware there is no formula in race math to show such increase or not in power, though I'm sure a formula exists that would lend itself to such an equation. In other words though, as you so well described, unsprung weight is all about contact patch behaviour and nothing really to do with power gain as so many people imagine.

Mave

6,371 posts

170 months

Saturday 4th October 2008
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Rocket Pepper said:
Interestingly, we beat this up a bit a few years ago as my Buell has carbon fibre Dymags. We were dynoing a Land Speed Buell in Colorado readying for Bonneville and the conversation got around to even better unsprung weight brought about by a wheel such as mine giving a higher horse power. I believe we concluded that there may well be a small increase seen on the dyno due to a lighter rear wheel but essentially there would be no relevance to power gain unless the saving in weight was quite significant. As far as I'm aware there is no formula in race math to show such increase or not in power, though I'm sure a formula exists that would lend itself to such an equation.
???? There is no way using lighter wheels gives you more power. If you see more power indicated on a dyno, then it's to do with how the dyno calculates drivetrain losses.

catso

13,698 posts

222 months

Saturday 4th October 2008
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Callan.T89 said:
Reducing the mass of the wheels also reduces there "giro" effect, which is basically always trying to keep the car going in a straight line so improves turn in.
On a Motorcycle lighter wheels make a massive difference as the rotating wheel acts as a gyroscope keeping it going in a straight line so any weight reduction improves turn in and handling no end.

Since a car doesn't lean into corners I imagine the benefit is not so great but some turn-in benefit should still be noticable.

beer

catso

13,698 posts

222 months

Saturday 4th October 2008
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Mave said:
Rocket Pepper said:
Interestingly, we beat this up a bit a few years ago as my Buell has carbon fibre Dymags. We were dynoing a Land Speed Buell in Colorado readying for Bonneville and the conversation got around to even better unsprung weight brought about by a wheel such as mine giving a higher horse power. I believe we concluded that there may well be a small increase seen on the dyno due to a lighter rear wheel but essentially there would be no relevance to power gain unless the saving in weight was quite significant. As far as I'm aware there is no formula in race math to show such increase or not in power, though I'm sure a formula exists that would lend itself to such an equation.
???? There is no way using lighter wheels gives you more power. If you see more power indicated on a dyno, then it's to do with how the dyno calculates drivetrain losses.
But anything that gives less weight improves power to weight ratio so would feel more powerful on the road - where it matters. But I have heard that lighter components (wheels, flywheel etc) on bikes can show a small improvement on dyno figures probably due to being able to pick up quicker?

SlipStream77

2,153 posts

146 months

Saturday 4th October 2008
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[quote=HiRich]


Interesting Post

[quote]

HiRich, is that antirollbar actually attached to the centre of the leading edge of the lower front wishbone? It must be putting quite a lot of stress on there.

Great post by the way, so essentially f = ma smile

El Guapo

2,787 posts

145 months

Saturday 4th October 2008
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snotrag said:
All this extra weight increases the forces going through your shock absorber.

By reducing this, you can reduce the damping forces needed, making the shocks more sensitive, and reduce the need for such tight seals, which reduced friction/stiction.
I'm afraid that none of this is true. Forces going through the damper are down to the mass of the vehicle, suspension geometry, speed & the size/nature of the imperfection in the road. And the stuff about 'tight seals' and friction/stiction is gibberish. Sorry.

Rocket Pepper

1,281 posts

171 months

Saturday 4th October 2008
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Mave said:
???? There is no way using lighter wheels gives you more power. If you see more power indicated on a dyno, then it's to do with how the dyno calculates drivetrain losses.
talking brake dyno's here......

Assuming you take out the alloy wheel then put in the carbon wheel (same tyre etc[weight]) and repeat a number of pulls consecutively, swapping back and forth between wheels until you're convinced the dyno is as accurate as can be, how would you explain a power increase, no matter how slight, shown with the lighter wheel?

And just to stir the Hornets nest. Why can the dyno show different readings if you strap a vehicle down tighter and/or play with tyre pressures?

Like I said, we talked this one to death.

singlecoil

28,742 posts

201 months

Saturday 4th October 2008
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Tyre Smoke said:
Psimpson7 said:
ride quality is one thing it can have a big affect on.

Edited by Psimpson7 on Friday 3rd October 11:31
I was thinking more of track cars, particularly the beautiful big Healy that I had the pleasure of fitting a clutch and gearbox into at Spa last weekend.
Cars with live rear axles have significant unsprung weight at the rear. Not such a problem on smooth racetracks though. Nuisance on normal road when you hit a bump going around a curve.

sminky

41 posts

163 months

Saturday 4th October 2008
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How come you don't see inboard brakes to save unsprung weight? i.e. discs on the driveshafts before the joints

Rocket Pepper

1,281 posts

171 months

Saturday 4th October 2008
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sminky said:
How come you don't see inboard brakes to save unsprung weight? i.e. discs on the driveshafts before the joints
It's still unsprung weight unless the driveshaft passes through a bearing mounting or carrier support after an inboard brake set-up.

flemke

22,470 posts

192 months

Saturday 4th October 2008
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Rocket Pepper said:
sminky said:
How come you don't see inboard brakes to save unsprung weight? i.e. discs on the driveshafts before the joints
It's still unsprung weight unless the driveshaft passes through a bearing mounting or carrier support after an inboard brake set-up.
Yes, but the % that is unsprung goes down as the weight goes farther inboard.

Different effects of inboard brakes depending on F v R, but they include:

Inboard brakes are out of the airflow and thus more difficult to cool
Increased total weight
At rear, inboard brakes will add heat to diff, gearbox, engine
Brakes exposed to leaking oil from diff, gearbox...
Broken half-shaft will result in total loss of brakes at that corner
Brake apparatus gets in way of under-floor aerodynamics

Rocket Pepper

1,281 posts

171 months

Saturday 4th October 2008
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flemke said:
Yes, but the % that is unsprung goes down as the weight goes farther inboard.
I don't see how it can. If the weight is transferred to the contact patch, then the contact patch is seeing that weight or what happens to that weight as the suspension loads and unloads. Placing some of the weight towards the inner most pivot and further away from the wheel is still the weight. All you're doing is moving weight along a lever. The weight doesn't change.

Maybe the speed at which the suspension works will change, but I can't see it at gone 10 on a Saturday evening.

mat205125

15,847 posts

168 months

Saturday 4th October 2008
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Your suspension is there to control the body movement, not to control the wheel movement - The wheels should ideally just follow the contours of the roads surface. The unsprung components do have a mass, and a mass in motion has momentum, therefore the spring and damper must control that momentum as well as the body ... Less unsprung mass means that the suspension can keep the wheels on the ground easier and concentrate on controlling the body.

Spinning masses also have a gyroscopic force / energy (remember holding a spinning bike wheel by the axles as a kid?). The larger the mass of the spinning components, the greater the forces required to move and turn these pieces.

Basically, anything that you want to turn, move, accelerate, brake or control, is best off being as light as possible

flemke

22,470 posts

192 months

Saturday 4th October 2008
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Rocket Pepper said:
flemke said:
Yes, but the % that is unsprung goes down as the weight goes farther inboard.
I don't see how it can. If the weight is transferred to the contact patch, then the contact patch is seeing that weight or what happens to that weight as the suspension loads and unloads. Placing some of the weight towards the inner most pivot and further away from the wheel is still the weight. All you're doing is moving weight along a lever. The weight doesn't change.

Maybe the speed at which the suspension works will change, but I can't see it at gone 10 on a Saturday evening.
At the extremes, if you had a 20kg weight that you could magically slide to any point along the distance between the CV-joint and the outside of the wheel, don't you think that more work would be required to move the wheel if the weight were at the outside of the wheel than if it were at the CV-joint?
As you say, you're moving the weight along the lever; when it's at the inside, more of it is borne by the CV-joint.
Don't worry about the semantics of "unsprung weight". Think of it intuitively. You want the weight at the wheel to be as little as possible, for obvious reasons. With an inboard disc, the weight at the wheel is easier to lift.
Another intuitive example: take two matchboxes and a 1"x1" timber that's 5 feet long. put a matchbox under each end of the timber. Put a brick on the timber near one end of the timber, an inch from the end. The matchbox at that end is going to collapse, right? At the same time, the box at the other end will be unscathed, even though it was bearing as much "unsprung weight" as the end that collapsed. Replace the crushed box with a fresh one, shift the brick to the opposite side, and the remaining original box will collapse, whilst the replacement will remain stable.

Rocket Pepper

1,281 posts

171 months

Sunday 5th October 2008
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I understand that totally, but as such the analogy you present isn't representative of a wheel placed at the outside of the hub because it is in reality only supported on the end of the plank with the furthest away matchbox offering a fixed pivot to swing around whilst the nearest matchbox is suspended. In other words, taking your analogy as a generalisation to a CV joint on a front wheel drive vehicle or a drive shaft cup from a differential on a rear wheel drive vehicle, neither of these components is meant solely as a load bearing point for the suspension components and/or steering/drive components to be carried on. This instead is the purpose of for example the inner ends of a wishbone(s) or the hanging design of a McPherson strut. So whilst your analogy works, it isn't representative of the components we are referring to or their respective locations.

Because of such, if we take your analogy and instead mount the timber to a fixed point at one end and a spring balance at the other, I'm not clear as to how transferring the weight of the brick along the timber acts in relation to the spring balance as if this were a vehicle it presents two problems for me. Where do you place these components to make them sit further away from the wheel if they have to by design locate along the length of the suspension components somewhere to make a difference in how their weight displacement affects the wheels behaviour to the road surface (and therefore the contact patch behaviour), and what is the opposite force at work by moving the weight load further inboard of the wheel. That is to say, if the weight is beneficial to the contact patch through suspension control by moving some of the weight further inboard and away from the wheel, is there not an opposite force that behaves in keeping the contact patch from the road surface. Now this presents another question, given such is all about suspension control through damping and rebound damping, is there any benefit at all through moving the weight away from the wheel if the weight is best placed nearer the wheel to gain control over rebound damping. In other words, if you move the weight away from the wheel where is the ideal point between damping and rebound damping speeds that presents itself best to contact patch behaviour.

I believe the better goal is to try what all performance tuners probably do, which is to save overall unsprung weight primarily rather than try to displace it, or some of it. But yes, one would imagine more of the weight placed further away from the wheel would be beneficial, but by how much in the available space of the vehicles componentry I'm not convinced is of much benefit unless like in a Jag IRS set-up where by the braking components are placed entirely differently from what is conventional, and therefore their weight bearing is different altogether from what we are discussing here.

Phew I talk some rubbish, lol!!!!

flemke

22,470 posts

192 months

Sunday 5th October 2008
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Rocket Pepper said:
Phew I talk some rubbish, lol!!!!
I wouldn't say that, but you do seem to like a long sentence. wink In places, it's hard for me to follow what you're saying.
I cannot think of another way of describing what I was trying to say. In the many racing cars over the years that have had inboard brakes (usually rear only), the primary reason for locating them there has been to reduce unsprung weight (sometimes also space requirements), despite the compromises that brought to cooling. It doesn't happen anymore because they interfere with the diffuser, and also because carbon/carbon brakes don't have the weight penalty of iron.

Cheers.

Rocket Pepper

1,281 posts

171 months

Sunday 5th October 2008
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Read it slower tongue out

Zad

12,093 posts

191 months

Sunday 5th October 2008
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"Dear Mother, I am writing this slowly, as I know you can't read very fast..."

When an wheel etc meets a bump and is deflected upwards, it takes kinetic energy to raise that object. If it were a smooth ramp then theoretically it would all become potential energy, and thence back to kinetic energy as it dropped again. Unfortunately in reality most of it gets lost in the tyre, dampers etc as heat. Because it is losing energy, the bump effectively tries to slow down the unsprung mass, but it is of course connected to the car, which in turn looses a small amount of inertia. Displacing less mass means less inertia is lost.

Steel wheels often respond better than alloys to accidental abuse (and traffic calming) too, as they have a small amount of elasticity that alloys just don't have.


Scuffers

20,887 posts

229 months

Sunday 5th October 2008
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Rocket Pepper said:
Mave said:
???? There is no way using lighter wheels gives you more power. If you see more power indicated on a dyno, then it's to do with how the dyno calculates drivetrain losses.
talking brake dyno's here......

Assuming you take out the alloy wheel then put in the carbon wheel (same tyre etc[weight]) and repeat a number of pulls consecutively, swapping back and forth between wheels until you're convinced the dyno is as accurate as can be, how would you explain a power increase, no matter how slight, shown with the lighter wheel?

And just to stir the Hornets nest. Why can the dyno show different readings if you strap a vehicle down tighter and/or play with tyre pressures?

Like I said, we talked this one to death.
this is a problem of rolling roads in general...

unless you do a power run at a relatively slow acceleration rate, you are always goign to have the rotational inertia of the drive-train to account for, and this is not insignificant.

even at a relativley slow rate (say 250Rpm/Min) it's still going to be a factor, then think about the inertia of the rollers themselves, yes, you can try and fudge it in software, but it's still there.

also, whilst you on the subject of wheel weights, remeber it's not just the total mass of the wheel/tyre (and have you ever put a tyre on the scales?) it's where the weight is, ie. if it's in the hub of the wheel, it will have less rotational interta than if it's round the rim.