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02-28-2011, 10:44 PM | #309 | |
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'08 Carrera S 6MT Guards Red/Black ext leather, Carbon fiber pkg, sport exh, sport chrono +, PASM, Nav, Bose, 19" forged turbos, red tranny tunnel
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03-01-2011, 03:39 AM | #310 | |
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Just because something accelerates faster does not imply it must have a larger force applied to it. I don't go often either but they are something like every other week. I can't seem to locate the thread on it with any recent activity... I would like to go to one again sometime soon.
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03-01-2011, 01:26 PM | #311 | |
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In a vehicle with lighter wheels vs. a heavier set, when it crosses the "line" at some distance ahead of the initial take-off position, it will have built more RPM's due to the same force (Tq) being applied to a lighter rotating object. If the power to the ground is measured exactly when the two vehicles cross this "line" then we will see that the vehicle with the lighter wheels has built up more power than the one with the heavier set. I understand that the engine is not making more power, but the relative power to the ground has increased.
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03-02-2011, 02:54 AM | #312 | |||
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Now onto yours...Your first sentence above is completely incorrect but the second sentence is correct. Any force on any mass makes it move unless overcome by friction or other non conservative forces. For example an any can push a frictionless locomotive, it is just that its acceleration will be incredibly small. I think what you were trying to say is something along these lines: for a given level of acceleration a change in the accelerated mass requires a corresponding change in the force. Quote:
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03-02-2011, 03:28 AM | #313 | |
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For a review the losses on a car are truly losses you can not get that energy back. Masses and rotating components only store energy (during acceleration). Yes their energy is lost say when braking but that is not relevant during our hypothetical drag races. Losses are: -Tire losses -Drivetrain losses (transmission, rear end, axles). These are frictional losses that convert power into heat and them it is "lost". -Aerodynamics. Really aerodynamics enters the equations differently. Aerodynamics simply produce a counteracting force to the drive force that acts more or less across the entire body of the car. But you can also describe this as a loss - the air is actually heated. Just multiply the aero force by the vehicle speed and you get the aero "power loss". -Auxiliaries: AC, steering pump, water pump, etc. Some might be part of the cars power spec some might not be. -Clutch: In a MT these only are during the shifts. All of the losses above except tires and aero go into the overall drive train efficiency and the figure is multiplied by the engine crank torque and divided by wheel radius and vehicle mass to get the primary net drive force. Here the "loss" it typically around 15% and hence the efficiency is 85%. Again these values vary by both engine speed and gear but the typically reported percentages are the peak values at redline. Hope this helps. It is not trivial at all.
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03-02-2011, 10:15 AM | #314 | |
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03-03-2011, 01:37 AM | #315 |
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Cheers, it was a good an valuable discussion . I always learn and find new ways to think about the "right" or "best" way to relate automotive performance back to the relevant physics.
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