
Unsprung Weight
#1
Posted 19 July 2001 - 17:47
Last night I was reading about porsche´s new PCCB (ceramic composite brakes) and, amongst other multiple benefits, the article stated that such materials reduced unsprung weight in 50%.
The question is, can someone explain to me how does unsprung weight benefit/pejudicates a car performance?
Do F1 cars have unsprung weight?? if so, which parts incorporate such weight (wheels, tires, brake rotors and calipers.. etc.)
Thanx in adbvance
cj
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#2
Posted 19 July 2001 - 18:35
Unsprung weight has a considerable effect on ride & handling performance. The unsprung weight is proportional to the forces and inertia which the springs/dampers must deal with. If a heavy car 'corner' hits a bump, it will generate large forces which have to be damped and absorbed and this takes a certain amount of time which is, of course, too long in motoring terms. Therefore, by minimising unsprung weight, suspension & steering tend to react faster.
Finally, the greater forces caused by heavier components demand beefier brackets, bolts, etc.
p.s, How would racecar designers fare in 'Scrapheap Challenge' ('Junkyard Wars' to our US cousins)?
#3
Posted 20 July 2001 - 09:27

To explain we have to go back about twenty years to the era before the FIA decided to remove skirts and other such dynamic devices that kept the underside of the car in close proximity to the road. As a result suspensions were designed that limited large amplitude suspension movements to keep the underside of the car at a uniform distance from the road. In effect we are watching high powered F1 Go-Karts with the bottom surface of the tire acting as the unsprung weght for low amplitude, high frequency bumps, and the rest of the vehicle acting as unsprung weight over larger amplitude variations. If look at many high speed photos of a modern F1 car riding over a curb you will see the wheels leaving the ground and this is due to lack of suspension compliance, not unsprung weight.
But some compliance is built into F1 suspensions to accomodate driver handling characteristes, curb riding and offroad excursion stresses. And the small suspension displacements allowed result in an upward displacemnt wheel/hub/suspension components which due to their significant masses, tend to continue away from the downward side of the bump profile resulting in a loss of traction during the return phase of the bump. The greater the effective mass of the wheel/brake/hub/control arms, the greater the loss of traction in the down-side mode of the bump.
But we still have to consider that the upside of the bump where a higher unsprung weight will increase traction during this phase! If your not asleep by now you will have realised that unsprung weight is not as simple as most text-books descriptions and Colin Chapman realised.
Colin Chapman being the guru of unsprung weight in the sixties resorting to such deadly artifices as spidery suspension components and inboard brakes connected to the wheel with delicate drive shafts that scared the hell out of such brave men as Jochiem Rindt, a WDC who some say left this world due to a deficiency in this area.
The previous poster mentiond the effect of rotation as increasing effective mass??? Actually this increase would be to gyrational forces only that effect the wheel/tire etc when their rotational plane is changed This can have significant effects if the suspension is designed so that the plane of wheel rotaion is changed with vertical bump displacment but this shouldn't be a problem with the large radii control arms used in a modern suspension system.
#4
Posted 21 July 2001 - 23:49
#5
Posted 22 July 2001 - 01:46
Originally posted by cjpani
Hi fellas:
...
The question is, can someone explain to me how does unsprung weight benefit/pejudicates a car performance?
...
Thanx in adbvance
cj
Unsprung weight affects the car in two big ways (the third is the overall weight of the car): 1) rotational inertia, and 2) vertical inertia (or suspension inertia).
On a side note, about 10 years ago I bought a set of aftermarket wheels and tires (+1) for my car. The Tire Rack wouldn't give me a quote on the different weights of the wheels, so I just went with the best looking pair that fit my budget.
While putting on the new wheels, I took a bathroom scale and weighed the old and new tire/wheel combinations (old = 195/60-14 Miichelin XGT-H4 on 14x6 Nissan wheels, New= 205/50-15 Dunlop SP8000 on 15x7 BBS). The weight difference was 10 lbs, or 50% (20 vs. 30 lbs, IIRC).
Anyway, the effects of unsprung mass became pretty apparent whenever I changed wheels and tires in the summer and fall.
Suspension Inertia: basically, it's harder for the springs to move the suspension up and down as unsprung mass increases. This means that the suspension has a tough time keeping the tires in contact with the ground on rough roads (the suspension doesn't react fast enough). In turn, this leads to poorer handling (through less grip) on rough roads and a rougher ride. Lowering unsprung weight increases grip and ride quality on and road that's not perfectly smooth.
Rotational inertia: it takes more energy to spin those larger wheels and tires. Essentially, what you have is a larger, heavier flywheel. That was easy for me to feel under acceleration: it took more clutch slippage to get the car going smoothly from a stop. In addition, the car seemed to accelerate just a hair slower.
Likewise, it's harder to slow down a [fly]wheel when it weighs more. So, increasing unsprung weight (at the wheel/tire -- the uprights would have no effect) increases braking distances.
For a race car (on a smooth circuit), unsprung weight really only affects performance if we're talking about the rotational mass. [unless I missed something obvious]
For a street car, in my experience, I'd be reluctant to add aftermarket wheels and tires, and would only do it if the car was vastly undertired. I wouldnt' go above +1. R&T or C&D did an article a few years ago where they tested OEM, +1, +2, and +3 sizes. The +1 size was a nice improvement. Performance was a smidgen better for +2 than for +1, but they were starting to get some tradeoffs with unsprung weight. The +3 size was a complete waste. IMHO, your biggest bang would be to go from OEM to a more high performance tire, or even +1/2 (say, from 195/60-15 to 205/55-15).
I'm not sure how my answer turned from F1 to street cars...
#6
Posted 22 July 2001 - 01:50
Originally posted by muppet
Yelnats, why exactly do F1 teams not use inboard front brakes now?
I bet that the considerable heat that the brakes produce would be a good reason to keep them outboard. Could you imagine tucking both of those red-hot carbon fibre disks in the nose? Talking about a hot foot!
#7
Posted 23 July 2001 - 07:20
BTW, how did the Benetton brake work?
mat
#8
Posted 23 July 2001 - 16:23
Thanks a lot guys, It´s now much more clear for me:up:
Thx.
cj
#9
Posted 24 July 2001 - 15:48

Can anyone update us on the above point?
The Benneton brake used a lateral drive between the two front wheels to limit the locking of the inside front on corner entry but it seemed to offer little benifits. Perhaps because braking was a small issue to a team the was suffering more from a lack of GO than a shortages of STOP!
#10
Posted 24 July 2001 - 19:24
#11
Posted 25 July 2001 - 16:49
Another point, we know we can consider the tires on an F1 car to be a large percentage of the overall spring rate, now what would be the advantage of having the tires flex more then the suspension? Knowing that they are undamped springs.
#12
Posted 27 July 2001 - 14:49
Multi-post rigs do more than tune suspension in areas that are effected by unsprung weight. They are also concerned with dynamic weight transfer under cornering and braking, ride height under areodynamic loading, reaction to bumps and curbs, all issues that have huge effects on car handeling.
Tires are not completely undamped as rubber has significant hysterisis effects and sidewalls and belts act to damp the tires to a certain extent.
#13
Posted 27 July 2001 - 17:10
mat
#14
Posted 27 July 2001 - 17:53
Originally posted by mat1
Yelnats and desmo; thanks, I didn't know that about the Benetton system. I understand the front wheels were linked together by an axle. Does that mean (apart from slip of course) both front wheel always turned with the same speed?: And if so, wasn't that a disadvantage in cornering?
mat
Ave !!!
The link wasn´t stiff, it was slip limited with what I believe was like WV-Syncro viscose fluid clutch. But sure one would think it induced some understeer.
- Oho -
#15
Posted 27 July 2001 - 20:04
Originally posted by Oho
The link wasn´t stiff, it was slip limited with what I believe was like WV-Syncro viscose fluid clutch. But sure one would think it induced some understeer.
I see. Thanks for the explanation.
mat
#16
Posted 27 July 2001 - 22:28
The other reason is of course the rotational inertia issue.
#17
Posted 28 July 2001 - 20:25
Why fabricate uprights out of bloody hard to weld titanium? Why continue with CFRC brakes when metal discs and carbon metallic pads give similar stopping power with more feel?
Ben
#18
Posted 30 July 2001 - 19:16
All I'm saying is that when and were it is possible teams will aim to reduce unsprung mass and unsprung centre of mass height.
As for wind tunnel operation; I think I read in the winter that McLaren had 3000 hours of time for their car while Minardi had 130!!!
Ben
#19
Posted 30 July 2001 - 20:08
As previously noted, absolute weight is always important, while mounting the brake caliper horizontaly lowers the CG. It probably allows the designer to place all the upright attachment features on the top & bottom of the upright as well.
As for CF brake disks vs. Steel, an article in Racecar a few years ago reckoned that, whilst more expensive, CF brakes cost the same in the long run. They are less liable to shatter than steel and are a hell of a lot lighter.
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#20
Posted 30 July 2001 - 20:16
As for McLaren's horizontal caliper I reckon this makes mounting the lower wishbone much harder but this is a sacrifice worth making. Having seen some good close ups of the arrangement I'm sure a suspension designer would not prefer it all things being equal.
Ben