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SVT Shelby GT500
shelby gt500 12.48@110
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<blockquote data-quote="Black2003Cobra" data-source="post: 4021603" data-attributes="member: 3159"><p>As far as the question about the relationship between the 60 foot time and weight, the short time goes as the square root of race weight and is inversely proportional to the square root of rear-wheel torque.</p><p></p><p>“It can be shown that” a decent estimate of the 60-foot time is given by the expression,</p><p></p><p>60’ time = 0.397*[(w*Dtire)/(TQ*GR)]^0.5</p><p></p><p>where w = race weight in pounds, Dtire = tire diameter in inches, TQ = engine torque in lb-ft as meas'd on a chassis dyno (use peak for best-case estimate), and GR = total gear ratio. For the Shelby, GR = 3.31*2.97 = 9.83. In deriving the formula, I assumed a roll out of 12 inches. A little more or less won’t change the answer much. The derivation also assumes PERFECT traction! TQ can be corrected for ambient conditions.</p></blockquote><p></p>
[QUOTE="Black2003Cobra, post: 4021603, member: 3159"] As far as the question about the relationship between the 60 foot time and weight, the short time goes as the square root of race weight and is inversely proportional to the square root of rear-wheel torque. “It can be shown that” a decent estimate of the 60-foot time is given by the expression, 60’ time = 0.397*[(w*Dtire)/(TQ*GR)]^0.5 where w = race weight in pounds, Dtire = tire diameter in inches, TQ = engine torque in lb-ft as meas'd on a chassis dyno (use peak for best-case estimate), and GR = total gear ratio. For the Shelby, GR = 3.31*2.97 = 9.83. In deriving the formula, I assumed a roll out of 12 inches. A little more or less won’t change the answer much. The derivation also assumes PERFECT traction! TQ can be corrected for ambient conditions. [/QUOTE]
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SVT Shelby GT500
shelby gt500 12.48@110
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