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sprung vs unsprung

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Old Aug 20, 2003 | 10:24 AM
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sprung vs unsprung

anyone knows how much 45 lbs of unsprung(sp?) weight is in sprung weight?? thanx
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Old Aug 20, 2003 | 10:34 AM
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1951 Caterpiller D6
none?
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Old Aug 20, 2003 | 11:18 AM
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Re: sprung vs unsprung

Originally posted by thuged_out
anyone knows how much 45 lbs of unsprung(sp?) weight is in sprung weight?? thanx
if you are referring to your 19" CL wheels, i think they weigh more than 45 lbs with tires.... i think the rear wheel alone is about 30lbs, and the tire is going to be at least 25...
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Old Aug 20, 2003 | 12:40 PM
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actually the difference between the combined weights of my original wheels with tires and 19" CL's with tires is 45 lbs. I heard that every pound of unsprung weight is equal to somthing like 4-5 lbs of sprung weight. just wanted to verify
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Old Aug 20, 2003 | 12:59 PM
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Originally posted by thuged_out
actually the difference between the combined weights of my original wheels with tires and 19" CL's with tires is 45 lbs. I heard that every pound of unsprung weight is equal to somthing like 4-5 lbs of sprung weight. just wanted to verify
so... for all 4 corners, the total added weight is 45lbs?

hmm... it be interesting to find out about the sprung weight issue. What kind of drain is that on performance? Is there some formula to calculate it?
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Old Aug 20, 2003 | 01:10 PM
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Originally posted by mmgrad
so... for all 4 corners, the total added weight is 45lbs?

hmm... it be interesting to find out about the sprung weight issue. What kind of drain is that on performance? Is there some formula to calculate it?
CL's are about 11-12 lbs heavier each for the rear wheels with tires and 10lbs heavier for the fronts. i cant give you any numbers as far as drain on perf goes, but after i put the orig 17's on my car this weekend, it feels much quicker of the line and in general. the car feels much lighter. i've forgotten how much faster my car used to be.
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Old Aug 20, 2003 | 01:45 PM
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depends on the weather
not sure about the sprung vs. unsprung equation but, for every 4-5lbs. of unsprung weight is requires 1 hp more to achieve the same performance so, with that in mind you have lost approx. 10 - 11 usable HP
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Old Aug 20, 2003 | 01:52 PM
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Originally posted by Luke@tirerack
not sure about the sprung vs. unsprung equation but, for every 4-5lbs. of unsprung weight is requires 1 hp more to achieve the same performance so, with that in mind you have lost approx. 10 - 11 usable HP
thanx for clearing it up. it sure feels that i lost more than 11 hp.
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Old Aug 20, 2003 | 02:05 PM
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Originally posted by Luke@tirerack
not sure about the sprung vs. unsprung equation but, for every 4-5lbs. of unsprung weight is requires 1 hp more to achieve the same performance so, with that in mind you have lost approx. 10 - 11 usable HP
in that case, better get a pulley kit to get back what i use to have in terms of performance...

i guess its hard to have your cake and eat it too... unless u wanna shell out big bucks for ultra light forged wheels... oh well... i sit in traffic more than i drive anyways... so, as long as it looks good...
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Old Aug 20, 2003 | 08:38 PM
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Now I see what you are asking. I have seen the ratio listed at 1 unsprung to 6 sprung and HP equivilence between 7-11.
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Old Aug 24, 2003 | 10:13 AM
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Also remember when dealing with unsprung weight that in a rear wheel drive car only the rear wheel weights will add to the equation during acceleration. The front wheel weight will hurt during braking.
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Old Aug 24, 2003 | 10:44 PM
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Unsprung weight matters whenever the tire is busy on the pavement...what you're saying only applies if the front or the rear tires are off the ground.
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Old Aug 24, 2003 | 11:25 PM
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i kinda agree with what Zeppelin is saying about the unsprung weight... It should only be directly effect by the mass surrounding the rear wheels since that is where the power is generated.

The added weight on the front wheels should be negliable. it would be treated essentially like weight inside the car, like the driver/passenger, or something in the trunk. though added weight in general will effect performance, the unsprung weight directly attached to the power generating wheels should be the most power robbing weights...

I think heavy 19" wheels would hurt an AWD car more than a FWD or RWD car...
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Old Aug 25, 2003 | 04:20 AM
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WRONG!!!

In the case of acceleration,the mass of the front wheels/tires doesn't just go away,in spite of your perceptions.The front wheels still have to be spun up and moved by the car's engine.The same holds true when braking...the power required to do so is transmitted through the chassis...
Negligible?
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Old Aug 25, 2003 | 11:00 AM
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Re: WRONG!!!

Originally posted by Steve Clark
In the case of acceleration,the mass of the front wheels/tires doesn't just go away,in spite of your perceptions.The front wheels still have to be spun up and moved by the car's engine.The same holds true when braking...the power required to do so is transmitted through the chassis...
Negligible?
i'm not saying that it IS negligible, but wouldn't it be treated like excess weight, like from a passenger in the car or something?

I'm not trying to be argumentative, I'm trying to learn here myself.
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Old Aug 25, 2003 | 02:11 PM
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It's really simple...Newton's 2nd Law of motion

Every part of the car is accelerated by the engine's output alone. Whether sprung or unsprung,it's going along for the ride. If the particular components (ie wheels/tires) are rotating masses,the energy needed for acceleration is greater....you're changing its' rotational speed (accelerating) and its' linear speed (more accelerating).No free lunches!...Even the air in the cabin takes energy to be moved.
The concept of the front wheels' masses not affecting acceleration is a new one.
This is another good reason to avoid heavy wheel/tire combos.
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Old Aug 26, 2003 | 02:07 AM
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Re: Re: WRONG!!!

Originally posted by mmgrad
i'm not saying that it IS negligible, but wouldn't it be treated like excess weight, like from a passenger in the car or something?

I'm not trying to be argumentative, I'm trying to learn here myself.
A spinning wheel accumulates energy and requires certain work done to get spun. Rear wheels are spun by the engine via transmission, whereas the front wheels are spun by pushing their axes which is complemented by the friction forces occuring between the ground and the tires. We've just dealt with the "unsprung weight" (also referred to as "rotational inertia"). Now, the whole car, *including all 4 wheels*, needs to be set in linear motion - which also requires some work to be done which is equal to the kinetic energy of the moving car. This is our "sprung weight".

To illustrate all this even better, let's make this moving car stop. Let's suppose that we lift it of the ground for clarity: the car is flying with all 4 wheels spinning. Now, to stop it, we need to do the opposite of what the engine had done to accelerate it. First let's stop its forward motion - put a wall in front of it: the heavier the car, the greater the impact. The wheels also figure - as simple weights. We've just dealt with the "sprung weight". Now, the wheels are still spinning - we need to stop them, in other words, convert their accumulated energy into heat (that's what the breaks and the tires help achieve). Done! We've just dealt with the "unsprung weight". Obviously, the heavier the wheels, the more energy will be converted into heat. In this hypothetical model which uses nothing but energy conservation principle it absolutely doesn't matter whether the wheels are on the driving axle or not.
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Old Aug 26, 2003 | 02:58 AM
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Re: Re: Re: WRONG!!!

Originally posted by vadim
A spinning wheel accumulates energy and requires certain work done to get spun. Rear wheels are spun by the engine via transmission, whereas the front wheels are spun by pushing their axes which is complemented by the friction forces occuring between the ground and the tires. We've just dealt with the "unsprung weight" (also referred to as "rotational inertia"). Now, the whole car, *including all 4 wheels*, needs to be set in linear motion - which also requires some work to be done which is equal to the kinetic energy of the moving car. This is our "sprung weight".

To illustrate all this even better, let's make this moving car stop. Let's suppose that we lift it of the ground for clarity: the car is flying with all 4 wheels spinning. Now, to stop it, we need to do the opposite of what the engine had done to accelerate it. First let's stop its forward motion - put a wall in front of it: the heavier the car, the greater the impact. The wheels also figure - as simple weights. We've just dealt with the "sprung weight". Now, the wheels are still spinning - we need to stop them, in other words, convert their accumulated energy into heat (that's what the breaks and the tires help achieve). Done! We've just dealt with the "unsprung weight". Obviously, the heavier the wheels, the more energy will be converted into heat. In this hypothetical model which uses nothing but energy conservation principle it absolutely doesn't matter whether the wheels are on the driving axle or not.
thats a good explanation too... thanks for the technical description.

Like i said, i'm just trying to learn a few things here and there. Thanks for the lesson, that includes you Steve. Thanks for the insight guys... this forum comes equipped full of knowledge...
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Old Aug 26, 2003 | 02:18 PM
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Well said,Vadim...Sprung versus unsprung.

The sprung weight is whatever is sitting on top of the suspension springs.The unsprung parts are whatever stays planted on the pavement.
Decreasing unsprung weight has several benefits-
1) the car simply weighs less,so it will stop,start and turn with less effort.
2) if the weight is a rotating mass,it takes less power to change said mass's speed.This applies to anything from the piston crown to the tire contact patches.
3) any reduction in unsprung weight (tires/wheels/brakes/etc) can benefit handling and ride quality.Shocks have an easier time damping the wheel motion (less mass to control),the tire follows the road surface better,and less NVH is transmitted to the chassis.

This is a case of "less is more".Perhaps now some owners will reconsider mounting larger wheels and tires.I think it's dumb to spend money for looks that will make my car slower.
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Old Aug 30, 2003 | 06:55 AM
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One of the car mags did a test to demonstrate precisely this topic. The optimal wheel / tire size was 17 inch for lateral G's, accel., and braking. They compared with tires/ wheels constant width, rev.'s mile, and brand/model.

Conclusion:

17" is the best performing combination of sizes available. The only reason to go larger is to fit rotor/caliper needs. These too should be sized appropriately for the energy you are trying to dissipate. Cooling the brakes works better than bigger. Unless you can't lock your brakes, or can't finish the race on one set of brake pads, you should look to cooling first.
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Old Aug 30, 2003 | 11:40 AM
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Originally posted by Lou Nielsen
One of the car mags did a test to demonstrate precisely this topic. The optimal wheel / tire size was 17 inch for lateral G's, accel., and braking. They compared with tires/ wheels constant width, rev.'s mile, and brand/model.

Conclusion:

17" is the best performing combination of sizes available. The only reason to go larger is to fit rotor/caliper needs. These too should be sized appropriately for the energy you are trying to dissipate. Cooling the brakes works better than bigger. Unless you can't lock your brakes, or can't finish the race on one set of brake pads, you should look to cooling first.
Does this conclusion apply to ALL automobiles? I find it hard to believe t it does.
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Old Aug 30, 2003 | 11:54 AM
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Originally posted by Lou Nielsen
One of the car mags did a test to demonstrate precisely this topic. The optimal wheel / tire size was 17 inch for lateral G's, accel., and braking. They compared with tires/ wheels constant width, rev.'s mile, and brand/model.

Conclusion:

17" is the best performing combination of sizes available. The only reason to go larger is to fit rotor/caliper needs. These too should be sized appropriately for the energy you are trying to dissipate. Cooling the brakes works better than bigger. Unless you can't lock your brakes, or can't finish the race on one set of brake pads, you should look to cooling first.
do you have a link or a copy of this article you could post online for us?
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Old Sep 3, 2003 | 05:03 PM
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The best wheel diameter and tire size depend a lot on the chassis of your car. So where a 17" wheel/tire combo might work best on one chassis a 18" wheel/ combo may be better on another. If the camber curves for the wheels are bad then the taller side wall tire afforded by a 17" wheel will deflect more and provide better grip.

But if the same car were to have a better camber curve as the suspension moves and enough spring and shock to control everything then maybe a 19" wheel/tire combo would work better given the smaller sidewall ht. of the 19" tire.

Jeff
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Old Sep 4, 2003 | 10:20 AM
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The article appeared 1-2 yrs ago. I did not keep the copy.

The performance considerations were not only for cornering force, or turn in response. Granted turn in is improved with a short sidewall tire. There is less effect on ultimate grip.

The conclusion was based on total performance as related to lap times. Also measured were skid pad results, steering transition, accel., and braking.

The larger the wheel / tire combination, generally the heavier and as the mass distribution is moved farther from the center, the effects of that mass are more pronounced.
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