Rotational weight = how much dead weight?

[Hawgman]

Been investigating this question and there doesn't seem to be any clear cut formula anywhere that I have found so far to answer the question.

1lb of rotational weight would equal to how much dead weight at the track? I know it has to do with rotating inertia. Just haven't really found a consistant formula to try to calculate it with. Best I could find was every pound of rotating mass equals 4 to 8 lbs of dead weight. That is way too vague to be accurate.

 

[Freaknazty]

i dont think it could be hawg due to where the weight it located in reference to the hub . in other words a 50lb 17" wheel will still take less force to spin than a 50lb 20" wheel due to where the extent of the weight is . could be wrong but that's my oppinion anyway lol

 

[Hawgman]

i dont think it could be hawg due to where the weight it located in reference to the hub . in other words a 50lb 17" wheel will still take less force to spin than a 50lb 20" wheel due to where the extent of the weight is . could be wrong but that's my oppinion anyway lol

No, what you are presenting is correct. Thus my reference to rotating inertia. But what if we are discussing something other than wheels? i.e. drive shaft for instance. Or a flywheel/flex plate. Or pulleys. I am talking any rotational weight. Not just tires and wheels.

 

[don_w]

And to complicate it even further, there is "sprung" and unsprung" weight too. The wheels are unsprung, while the driveshaft, etc. is sprung.

Bottom line, it's likely too complicated to come up with a "clear cut" formula that works in every situation.

 

[Hawgman]

Yup! Would be damn cool though if we could get some sort of idea for the different types of rotational weight. We all generally agree that 100lbs of dead weight should equate to about a tenth in the quarter mile. If we could find different examples for the different types of rotating mass (sprung, unsprung, diameter to center, etc.) that might help some people to estimate how much weight savings of a particular type is going to equate to at the track. i.e. take a hundred pounds of dead weight out, gain a tenth. Take a hundred pounds off of your wheels, gain three tenths, and so on.

 

[US-1]

Even the "100 lbs = 1/10th" rule is misleading. Depends on the individual car. On my car in N/A trim (dead stock) 100 lbs was eighteen hundredths (.18) in the quarter. With the twin turbo system 100 lbs is six hundredths (.06).

 

[EagleStroker]

You also begin to look at different factors especially with the flywheel, at a dragstrip lightweight isn't always your friend, depends on the driver/car. There's a reason there's no cut and dry answer!

 

[Dread53]

The equations you are talking about get almost as complex as those about vibration. Every different shape has a 'moment of inertia' coefficient taken into account, and all of those assume consistent mass throughout the shape. Take the driveshaft, for instance. If you wanted a 'rough' idea you would have to assume the weight distrbution was constant, but you would still have to guess what diameter that average weight occured at. Then it becomes a simple ring calculation. Otherwise, buy a super computer.

 

[alloy6ix]

Even the "100 lbs = 1/10th" rule is misleading. Depends on the individual car. On my car in N/A trim (dead stock) 100 lbs was eighteen hundredths (.18) in the quarter. With the twin turbo system 100 lbs is six hundredths (.06).

yeah I've heard that one too. and also I read somewhere the ratio is 1:7lbs of sprung to unsprung weight. then again, that might have been posted by one of the ppl you were correcting at MM.