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02-28-2008, 04:47 AM | #23 | |
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02-28-2008, 07:05 AM | #24 | |
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02-28-2008, 08:10 AM | #25 |
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F1 cars aren't really that powerful. I would bet a Nascar makes more total HP. F1 cars win the lb/hp battle. 600-800 hp in a 1400 lb car with amazing traction control is crazy. Imagine how fast the turbo F1 cars were when they were pushing 1000-1400 hp in a similarly sized car with no traction control. Nate
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02-28-2008, 09:19 AM | #26 | |
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This is purely an agreement amongst some select manufacturers and not always the case. There isn't a law either in Germany or Europe prescribing the electronic speed limiter. The Mercedes SLR is not unrestricted, the Audi R8 is unrestricted too. Porsche generally do not limit the top speeds of there cars, but they do restrict the Cayenne Turbo S purely due to the tyres.
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02-28-2008, 09:35 AM | #27 | |
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That is quite something to behold :biggrin: |
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02-28-2008, 11:45 AM | #29 | |
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02-28-2008, 11:48 AM | #30 | |
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02-28-2008, 01:30 PM | #32 |
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Correct. but, at the speeds we are talking about over 90% of the power is used to overcome drag.
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Alan
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02-28-2008, 01:31 PM | #33 |
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02-28-2008, 03:34 PM | #34 |
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I can't verify the statement is true, but I can tell you how to figure it out:biggrin:
Wow, I'm really not in a helpful mood right now
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02-28-2008, 06:50 PM | #36 |
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i did the calculations before for a chevy equinox (it was for a hybrid electric vehicle design competition so 0-60 time and vehicle weight were the design variables). i dont have all the values needed to do it for a 1er but really if you change the mass, Cd and frontal area its close enough. these are the values of the forces acting on that vehicle at 140mph (where drag is at peak)..
mass: 1536kg inertial mass factor: 1.05 (this keeps us from having to do sloppy dynamics) Cd: 0.33 Af: 1.86 m^2 Tire rolling resistant coefficient (this should be more, this value is for low rr tires found on hybrid project cars etc) = 0.01 rolling resistance force (mass*gravity*Crr) = 150.7 N aerodynamic drag force (what weve been talking about) = 1442.7 N inertial force (m*Imf*acceleration) is a bit trickier because we have to account for rate of acceleration. i used the times from the C&D test fitted to the simplified nonlinear relationship v=Vm(t/tm)^z where Vm=60mph and tm=4.7s(as tested) approximated the z value to be around 0.5 (gives a 25.6 0-140 time). the average acceleration during the last second using that model is about 2.76 m/s^2. thus.. inertial force = 4451.3 N this sounds unreasonably huge inertial force but assuming a wheel radius of 0.34m this comes out to a required torque of 361lb/ft. which is a bit much if we can theoretically get to 150mph, but i dont think i need to tell you my model isnt perfect. so, what have we learned? if the car is going at constant velocity, yes, 90% of the forces acting on the car are from aerodynamic drag (fuckin spot on, actually). but if we are trying to accelerate in any way, that takes a hell of a lot more effort than anything else involved. i really should have spent that 30 minutes on my thesis. |
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02-29-2008, 03:23 AM | #37 | |
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02-29-2008, 05:38 AM | #38 | |
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It also helps to explain why I run into an aero wall before later model 911s. Up to 120 I can gap them; but, over 130 they have closed the gap, and by 140 they are gone. On the other hand, maybe it doesn't; but, another three day at VIR in two weeks I should be able to check out real world application.
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02-29-2008, 02:30 PM | #39 | |
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03-01-2008, 12:40 AM | #42 |
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You need almost no power for acceleration provided it is non-zero. As a practical limit, you'd need acceleration to reach top speed within some time limit, like 2-3 minutes or so, but that number above seemed crazy high.
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03-01-2008, 01:42 PM | #43 |
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im not really followin you although that sounded too high to me too. double checked when i did it and all but really thats the simplest calculation in there, F=ma...
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03-01-2008, 08:40 PM | #44 |
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