DEFINITIVE ANSWER ON OEM WHEEL WEIGHT AND OFFSET FOR BREMBO 2011 GT? - Page 3 - MustangForums.com



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Old 10-19-2010, 01:25 AM   #21  
lawman1
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But 0.2 or 0.3 is a wildly optomistic guess. I'd put the difference closer to 0.050, maybe 0.075 seconds.
In retrospect, I agree with Norm. .2 -.3 seconds is a bit much. I was drinking.
It has been said, although I have not seen the empirical evidence, that every reduction of 100 pounds is worth .1 in the quarter. It has also been said that every pound of unsprung mass eliminated is worth 10 pounds of sprung weight. I am not sure either of those statements are true. In fact the more I think about it the latter especially has to be false. When it comes to forward momentum mass is mass, whether sprung or unsprung. I would tend to think also that as for rotational weight, wheels that are heavier at the hub would be easier to spin than wheels that carried their weight on the rim. In short. there are way too many variables to be making blanket statements not based on empirical evidence. I generally don't believe anything until I see the proof. Must discipline myself for talking above my pay grade. So sorry for the misinformation.
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Old 10-19-2010, 06:47 AM   #22  
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I guess you have to know which ones can be trusted . . .
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Originally Posted by lawman1 View Post
It has been said, although I have not seen the empirical evidence, that every reduction of 100 pounds is worth .1 in the quarter.
That one is actually pretty good. Maybe not 0.100 vs 0.095 good, but still good enough for conversational and "order of magnitude" purposes. The Hot Rod Magazine articles where they cut away most everything not necessary to drive, hold the axles together, and keep what was left from collapsing under its own weight - more or less support that correlation (IIRC, ~3000# chopped off the Caddy in "Caddyhack" bought about 3 seconds).

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It has also been said that every pound of unsprung mass eliminated is worth 10 pounds of sprung weight. I am not sure either of those statements are true. In fact the more I think about it the latter especially has to be false.
You're right. This is the one you can't trust. The effect of rotating weight also depends on how fast it is being accelerated rotationally. A lb on the flywheel really could be worth 10 lbs of static weight, although only in the lower gears where the engine rotational acceleration is typically between 8 and 14 times as much as the wheel rotational acceleration.


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When it comes to forward momentum mass is mass, whether sprung or unsprung. I would tend to think also that as for rotational weight, wheels that are heavier at the hub would be easier to spin than wheels that carried their weight on the rim.
Absolutely.

Though in most cases there won't be huge differences in the proportion of mass in the wheel center vs in the 'barrel' among wheels that will stand up under road use. IOW, while this is also a real effect, such variation is kind of a "second order" thing (read: much smaller than the difference between considering and not considering the rotational effects at all).


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Last edited by Norm Peterson; 10-19-2010 at 06:54 AM. Reason: fix quote
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Old 10-19-2010, 10:20 AM   #23  
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.... It has also been said that every pound of unsprung mass eliminated is worth 10 pounds of sprung weight. ....
I suspect that this may have originally been based on the effects on handling when the ratio of sprung to unsprung weight is changed. A car with a lower ratio of unsprung weight to sprung weight will do a better job of keeping the tires in proper contact with the ground.
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Old 07-02-2011, 06:57 AM   #24  
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Default Caliper measurements

Theres lots of talk about offsets and all but I think one thing that is missing is the pad height of the brembo package. Two wheels with the same offset but not enough brake caliper clearance one might fit and the other might not. So does anyone know exactly what the caliper diameter is from the center of the hub and how much it sticks out from the face of the hub?
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