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I read once a ratio between removing say wheel/tire weight compared to sprung weight like spare and jack. Say I remove 10lbs. from my wheel/tire combo what is that compared to body weight. I believe there is a ratio, like 3:1 so removing 10lbs. from wheels is like removing 30 from the body. Anybody have that info? Thanks, Mark
 

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Removing one pound of ROTATING weight is generally considered to be worth 3-4 lbs of static weight. Wheels, clutch, and flywheel are the usual places to reduce rotating weight.

Sprung vs. unsprung has to do with handling and ride quality. I'm not aware of any ratios around it. Generally, you want unsprung weight as low as possible, though.
 

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Sprung vs. unsprung has to do with handling and ride quality. I'm not aware of any ratios around it. Generally, you want unsprung weight as low as possible, though.
soo....the lightest rims and suspension parts help the most??.....


I barely understood this thread...so im tryng to....
:confused:
 

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Believe it or not, I understood it. Just can't answer it. When I was racing in Triathlons it was common knowledge that the first place you took weight off of your bike was the rotating parts. (wheels, gears, tires, chain, pedals). Once you had those down as low as possible you started on the rest.

I'd assume the same rule applies to cars.
 

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I think the general rule of thumb is: unsprung weight slows you down about 3 times as much as normal weight on the car. so if 100# of weight in the trunk slows you down .1 second in the 1/4 mile, then 100#'s of unsprung weight will slow you down .3 seconds in the 1/4 mile.

10 pounds per wheel X 4 wheels = 40 pounds unsprung weight
40 pounds unsprung weight X 3 = 120 pounds normal weight
 

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There are two types of weight that are most desirable to decrease.

Unsprung weight
Rotational weight
Some things fit into both, like wheels and tires.

corner-carvers forum has a lot of good handling info.
 

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driller said:
I think the general rule of thumb is: unsprung weight slows you down about 3 times as much as normal weight on the car. so if 100# of weight in the trunk slows you down .1 second in the 1/4 mile, then 100#'s of unsprung weight will slow you down .3 seconds in the 1/4 mile.

10 pounds per wheel X 4 wheels = 40 pounds unsprung weight
40 pounds unsprung weight X 3 = 120 pounds normal weight
No no no no no no. ;)

It's rotating vs. non-rotating weight that makes the difference. Let's say your wheels weigh 40 lbs each. You spend the dough and get some that weigh 30 lbs each. That's 40 lbs of rotating weight you've removed. That's the equivalent of removing 120 lbs of static weight when it comes to acceleration.

Detailed explanation: Let's look at a 40 pound wheel in the trunk. You have to expend a certain amount of power to move that weight down the track.

Now, let's look at that same 40 lb wheel, but this time it's installed instead of in the trunk. You have to expend that same amount of energy as before to get it down the track. But you also have to expend energy to rotate it. The energy expended to rotate the tire is a function of its polar moment of inertia, which is a function of the square of the radius. Larger wheel equals larger polar moment of inertia equals more energy to rotate it.

Now, let's move to sprung versus unsprung. First, the definition: sprung weight is supported by the springs; unsprung isn't. Generally, you count the weight of the spring and shock as half and half.

For unsprung weight, let's keep it easy and use the brake caliper, since it doesn't rotate. Let's say the caliper and bracket assembly weighs 20 pounds. This is completely unacceptable, so you replace it with an assembly that weighs 5 lbs. You removed 15 lbs of unsprung weight. What's the effect on acceleration? None at all. But the car will probably ride and handle better, since there is now a larger percentage of sprung weight. Sprung weight can be controlled and manipulated by the suspension; unsprung weight cannot.

Clearer now?
 

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Wonderful explanation! I feel like I'm back in Physics class again!

Rotational inertia is proportional to the square of the radius and is directly proportional to the mass. However, it is also important where the mass is concentrated within the radius. For example, a solid cylinder with the SAME MASS as a hollow hoop of the same radius would have less inertia (only one half as much, to be exact). Therefore, not only do the mass and radius of the wheel matter, but also where that mass is concentrated. A fifteen inch rim with a thirty inch tall tire would have more mass concentrated at the center than a 20 inch rim with a thirty inch tire, and therefore would accelerate faster.

When it comes to unsprung weight, it is easier to visualize the suspension as what pushes the wheels down, as opposed to what holds the car up. When there is less unsprung mass, springs that provide the same force are able to accelerate that mass faster (force=mass x acceleration). Therefore, the tires make contact with the ground sooner after impacts with potholes and bumps. This is the point of any suspension, as the wheels cannot accelerate the vehicle without touching the ground. This means the car will handle better with the same springs, or that softer springs can be used without impacting handling, thus achieving a softer ride. Needless to say, I'd take the better handling!
 

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The energy expended to rotate the tire is a function of its polar moment of inertia, which is a function of the square of the radius. Larger wheel equals larger polar moment of inertia equals more energy to rotate it.
-stare....blink blink-

eh?

sssooooo....yeah...i'll just lighten my car with whatever i can afford.....
 

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Knice4.697TB said:
-stare....blink blink-

eh?

sssooooo....yeah...i'll just lighten my car with whatever i can afford.....
big rims suck up power.
 

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Tobey said:
No no no no no no. ;)

It's rotating vs. non-rotating weight that makes the difference. Let's say your wheels weigh 40 lbs each. You spend the dough and get some that weigh 30 lbs each. That's 40 lbs of rotating weight you've removed. That's the equivalent of removing 120 lbs of static weight when it comes to acceleration.

Detailed explanation: Let's look at a 40 pound wheel in the trunk. You have to expend a certain amount of power to move that weight down the track.

Now, let's look at that same 40 lb wheel, but this time it's installed instead of in the trunk. You have to expend that same amount of energy as before to get it down the track. But you also have to expend energy to rotate it. The energy expended to rotate the tire is a function of its polar moment of inertia, which is a function of the square of the radius. Larger wheel equals larger polar moment of inertia equals more energy to rotate it.

Now, let's move to sprung versus unsprung. First, the definition: sprung weight is supported by the springs; unsprung isn't. Generally, you count the weight of the spring and shock as half and half.

For unsprung weight, let's keep it easy and use the brake caliper, since it doesn't rotate. Let's say the caliper and bracket assembly weighs 20 pounds. This is completely unacceptable, so you replace it with an assembly that weighs 5 lbs. You removed 15 lbs of unsprung weight. What's the effect on acceleration? None at all. But the car will probably ride and handle better, since there is now a larger percentage of sprung weight. Sprung weight can be controlled and manipulated by the suspension; unsprung weight cannot.

Clearer now?
Thanks for the clarification of the definition. When most people refer to this topic they are thinking wheels/tires. I was just dumbing it down for public consumption. :tongue:

Math and physics. That's what it's all about. Everything else is simply voodoo.
 

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Don't forget weight transfer :D Most important part in my opinion.
 

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driller said:
Thanks for the clarification of the definition. When most people refer to this topic they are thinking wheels/tires. I was just dumbing it down for public consumption. :tongue:

Math and physics. That's what it's all about. Everything else is simply voodoo.
I'm all for simplifying, but in this case, sprung/unsprung has a meaning also. Don't forget that for manual transmission guys, the flywheel and clutch are really good places to get rid of rotating weight. If you've never heard how fast an engine revs with a light weight clutch flywheel combo, it's something else. The four cylinder race car I used to have had such a setup, with a small diameter clutch. A normal little blip of the throttle would shoot the RPMs up to 3500.

Silenced said:
Don't forget weight transfer Most important part in my opinion.
If you are traction limited, it's an important factor. Whether it does more for your ET than reducing rotating weight, etc., only testing can tell. Probably depends on your power output.
 

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Tobey said:
I'm all for simplifying, but in this case, sprung/unsprung has a meaning also. Don't forget that for manual transmission guys, the flywheel and clutch are really good places to get rid of rotating weight. If you've never heard how fast an engine revs with a light weight clutch flywheel combo, it's something else. The four cylinder race car I used to have had such a setup, with a small diameter clutch. A normal little blip of the throttle would shoot the RPMs up to 3500.
Agreed. Other examples include light weight pulleys, internal reciprocating engine components, hollow camshafts...
 

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For that matter, having lighter reciprocating mass makes possible higher RPM engines, which allows a greater mechanical advantage to be maintained through gearing, which means faster acceleration, assuming that the power band continues to increase at the higher RPM's. In the case of most stock motors, replacing these components would be of little value, as the power and torque have already begun to drop by the stock redline. However, heavily modified engines can benefit greatly from a stronger, LIGHTER bottom end.
 
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