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Discussion Starter #1
haven\'t seen this in awhile

For every pound of rotating mass you reduce, about how much hp should you free up or see on the dyno.

First off I did a search and didn't find what I was looking for. I know I seen it on this site and I'm sure someone knows.
 

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Discussion Starter #2
Re: haven\'t seen this in awhile

someone has to know
 

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Re: haven\'t seen this in awhile

Hey chris or ducman could you move this topic to some place you think it will get answered. Thanks!
 

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Re: haven\'t seen this in awhile

Dude, Why don't you give people more than 35 minutes on a Friday night!
 

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Discussion Starter #6
Re: haven\'t seen this in awhile

What else do you want to know welding rod?
 

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Re: haven\'t seen this in awhile

his question is fine, the wording might be bad though.
he wants to know how much horsepower will show up with every 1 pound reduction in rotating mass.
 

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Re: haven\'t seen this in awhile

You dont get any additional horsepower from reducing rotating wieght. You get decreased inertia. It takes less power to rotate the assembly faster and it takes less power to make it rotate slower but the engine still makes the same power at a specific RPM. Thats what iv been yelling at people for using these lightwieght flywheels... inertia is good for drag racing...
 

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Re: haven\'t seen this in awhile

Originally posted by noresull:
Dude, Why don't you give people more than 35 minutes on a Friday night!
<font size="2" face="Verdana, Arial">What? U mean everyone isn't sitting at their PC logged on to FJ on a Friday night....oh yeah, that's only me (being married with children, and the wife works nights).
 

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Re: haven\'t seen this in awhile

Josh - I assume you want to do the figuring for a flywheel or crank pulley? If so I have the ballpark formula at work. If I remember I will bring it home and post on Monday night if no one else does before then. It will require item diameter, gear ratio, and differential ratio.

The formula gives approximate fixed weight savings equivalency. From that you should be able to roughly equate HP gain equivalency.
 

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Re: haven\'t seen this in awhile

From memory - no guarantees here - I remember for tires and wheels 1 pound should have the rough equivalence of 2.5 to 3.5 pounds of non-rotating weight (the exact number I assume would depend on tire/wheel diameters and weight distribution/distance from center of rotation).

If this is true, a 5 pound per tire/wheel combo savings would figure something like this: 5 lbs x 4 corners x 3 (approximation) = 60 pounds equivalent

If car weighed 2800 pounds and had 110 WHP the PWR would be about 25:1. So that would equate to about 60/25 or about 2.4 HP during accelertation.
 

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Re: haven\'t seen this in awhile

Josh - Here is the formula for approximate weight saving equivalence for flywheels/crank pulleys as taken from SCC magazine a year or so ago (I reconfigured it slightly as the text version in the magazine seemed to be correct while the written formula appeared to have an error in it):

.5 x difference in FW weight x (FW radius x gear ratio x final drive ratio / tire radius)^2

If you want to check your math a 9 pound FS Focus FW should be the rough equivalent of 200 some odd pounds in 1st gear.

[ 11-18-2002, 10:09 PM: Message edited by: Welding Rod ]
 

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Re: haven\'t seen this in awhile

OK, found the article with specifics. Heres an example for first gear:

.5 x 12.5 pounds less x (5.5" x 3.667 x 3.82 / 12.1")^2 = ~ 253 pounds

Their formula uses "flywheel weight" instead of "flywheel weight difference" but their example answer correspondes with using "flywheel weight difference" (?).

Another thing, according to SCC a pound on the tire = 2 pounds on the car and a pound on the wheel = 1.5 pounds on the car. This is more conservative than what I have read in other magazines before, but this mag deals with little cars with short tires so that may account for the difference.
 
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