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car completing a loop?


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#1 SecretDestroyer

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Posted 31 December 2003 - 07:20

Would it be possible for a car of any kind to go through a loop like a skateboarder?
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#2 SalutGilles

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Posted 31 December 2003 - 07:27

just look at the pounds of downforce produced by an F1, and you'll see that it could drive around upside down all day as long as it was going fast enough.

as for the loop, if skateboarders (who produce very little downforce) can do it, a car certainly could. Now, you'd probably break the front wing off, but you'll do that starting next to Schuey more often than not anyway.

the only question would be the speed necessary, downforce or no.

#3 Wuzak

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Posted 31 December 2003 - 10:21

I remember Christian Fittipaldi didn't require a platform to do a loop in his Minardi (Italian GP 1993)!

Obviously the loop would have to be of sufficient radius that the front and rear ends of the car didn't scrape the ground. The car would not need its aero to keep going, if the car was going fast enough.

#4 desmo

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Posted 31 December 2003 - 18:38

My nascient empirical vehicle dynamics research conducted near the prestigious University of California at Berkeley under strict scientific rigors using my Hot Wheels TM set during the "turbulent" late '60s suggests that such a stunt is indeed feasible.

#5 McGuire

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Posted 31 December 2003 - 18:58

I say let's not give Humpy Wheeler or the France family any ideas.

#6 diego

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Posted 31 December 2003 - 20:32

Yes

#7 naes

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Posted 31 December 2003 - 20:33

I remember from my A-levels that at a set speed, the same for all masses, any object will go round the loop irrespective of downforce generated or other factors.

Sean

#8 SalutGilles

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Posted 31 December 2003 - 20:34

not to pry, but you wouldn't mind sharing that speed with us, would you?

#9 McGuire

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Posted 01 January 2004 - 01:20

Originally posted by SalutGilles
not to pry, but you wouldn't mind sharing that speed with us, would you?


Enough speed to produce something greater than 1 g of centrifugal force, which is dependent upon both speed and loop radius. The larger the loop's radius, the greater the speed required, V^2/r.

#10 desmo

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Posted 01 January 2004 - 01:58

Originally posted by McGuire


Enough speed to produce something greater than 1 g of centrifugal force, which is dependent upon both speed and loop radius. The larger the loop's radius, the greater the speed required, V^2/r.


Sometimes physics is poignantly elegant. :)

#11 Bladrian

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Posted 01 January 2004 - 05:01

Originally posted by SalutGilles
, but you'll do that starting next to Schuey more often than not anyway.



Oh, FFS. Grow up.

#12 SalutGilles

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Posted 01 January 2004 - 08:32

Originally posted by Bladrian



Oh, FFS. Grow up.


Wha???

what is FFS?

I'm a Schumacher fan, plain and simple, but I can still take the piss...

#13 McGuire

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Posted 01 January 2004 - 12:35

Originally posted by desmo


Sometimes physics is poignantly elegant. :)


V^2/r always chokes me up a little too. sniffle. Give me a moment, I'll be fine.

#14 Greg Locock

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Posted 01 January 2004 - 21:43

FFS= for franklin's sake

#15 Ben

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Posted 02 January 2004 - 00:26

:lol: :rotfl:

Ben

#16 alexbiker

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Posted 02 January 2004 - 13:39

FFS= for franklin's sake




LANGUAGE!

Alex