996 Turbo / GT2 Turbo discussion on previous model 2000-2005 Porsche 911 Twin Turbo and 911 GT2.

996 X50 vs. Gallardo..Which is quicker??

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Old Nov 2, 2005 | 10:37 PM
  #61  
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Originally posted by allanlambo
Honestly, im terrible at math, but do yourself a favor, imagine yourself standing at the end of a 150mph run way. Now imagine one car going buy at 150mph and you instantly click your stop watch and wait to 2 seconds until the next car passes. If you think only 4 car lengths will seperate the 2, you are delerious.

Also factor into your math, that when when the Gallardo its 150mph marker, the 360 is going no where near that same speed.

Talking with some drag racers, they said the general rule was figure just in the 1/4 every tenth of a second= 2 car lengths.
Allan, the math is much better than trying to imagine something that doesn't make sense (i.e., a car reaching a SPEED of 150 MPH 2 seconds before another does not mean it's 2 seconds DISTANT from the other). Doesn't it sound strange to you that two cars which are perfectly even at 118/115 MPH are now 16 cars apart only 10 seconds later? Use the math. It will save yourself a headache.
 
Old Nov 2, 2005 | 11:25 PM
  #62  
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Originally posted by Bill S
Allan, the math is much better than trying to imagine something that doesn't make sense (i.e., a car reaching a SPEED of 150 MPH 2 seconds before another does not mean it's 2 seconds DISTANT from the other). Doesn't it sound strange to you that two cars which are perfectly even at 118/115 MPH are now 16 cars apart only 10 seconds later? Use the math. It will save yourself a headache.
10 seconds is an eternity. No ifs ands or butss, the Gallardo will be substantially farther ahead than 4 car lengths.
 
Old Nov 3, 2005 | 02:56 AM
  #63  
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Actually, general rule is ~1 car lengths per .1 sec...

Assuming a car is 15 ft long, at one second, that is 10 car lengths, or 160ft. 160ft/sec is ~102mph.

So, if you're standing at the end of the drag strip when one car crosses at 102mph, keeping velocity constant for both cars, in 1 second, TEN car lengths will have gone by before the 2nd car passes.
 
Old Nov 3, 2005 | 03:08 AM
  #64  
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Originally posted by Bill S
Allan, the math is much better than trying to imagine something that doesn't make sense (i.e., a car reaching a SPEED of 150 MPH 2 seconds before another does not mean it's 2 seconds DISTANT from the other). Doesn't it sound strange to you that two cars which are perfectly even at 118/115 MPH are now 16 cars apart only 10 seconds later? Use the math. It will save yourself a headache.
You are correct... reaching a velocity 2 sec before another does not mean it's 2 sec of distance. You're constant is mph, and Allanlambo is arguing something completely unrelated; that is the 2 second gap in arriving at distance x.
 
Old Nov 3, 2005 | 10:16 AM
  #65  
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Whatever it may be, absolutley no way is the CS only for 4 cars behind at 150 mph with a 2 second gap.
 
Old Nov 3, 2005 | 11:42 AM
  #66  
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Allan... the 2sec gap he is referring to is the time dif between reaching a certain mph... so, if G is at 150mph, CS may be at 149mph, and while both are still speeding along, it takes another 2 sec for CS to reach 150mph... the mph dif between two won't result in a lot of car lengths.

You are talking about 2 second DELAY in reaching distant X... in this case, 2 sec is a life time... at 150mph, or 220 ft/sec, over the period of 2 sec, you are talking ~30 car lengths. You guys are "arguing" different things... apples and oranges.
 
Old Nov 5, 2005 | 12:37 PM
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Originally posted by allanlambo
Honestly, im terrible at math
Allen, that may be the first thing you've said I've ever agreed with!

Gary
 
Old Nov 5, 2005 | 12:39 PM
  #68  
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Originally posted by tdf360
Allen, that may be the first thing you've said I've ever agreed with!

Gary
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