TL_Type_S

I got into an argument w/ a buddy this weekend about how you CAN brake-torque an auto, FWD car. He's of the belief that the only true way to brake-torque is in a RWD car, and to let the back wheels slip a little while the front wheels are still locked, right before your launch. What is the true meaning of brake-torque, and can it apply to automatic FWD cars?

Second: At 1/4 mile speeds, say 90mph to 96mph .. what does 0.1 second really amount to in car lengths --> he argued that a WRX pulling 14.4 in the 1/4 will only be about four/five carlengths behind a Vette running 13.5 (w/ the .9 seconds difference). I argued that it'd be about eight or nine car lengths .. Tom?

Oh and Tom -- does the trans axle have anything to do w/ a FWD auto not being able to "brake-torque"? Thanks in advance!

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Tom2

Okay, since you asked, here are my answers:

1- Brake torque- I'm not sure if you're getting this confused with actually spinning the tires, or holding the car still to bring the revs up. Serious racers (at the track) will install line locks, which will lock the front brakes only, allowing the car to do a nice long burnout while remaining stationary.

But if you're already at the line in your Acura, you can easily brake torque it to bring the revs up against the torque converter, to leave the line harder/at a higher rpm. If this is what you meant, then I agree with you.

2- Any car moving at about 100 MPH will travel 15 feet in .1 seconds. Since a car length is approx 15 feet, then it is safe to say that .1 seconds is about equal to a victory of 1 car length, all else being equal. Of course this is not exact, only an estimate. I'd have to say that a car running 13.5 vs another car running 14.4 would have a difference of about 9 car lenghts.

In my book, that's getting SMOKED!

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99 M3
dead CL-S

Tom2: The Voice Of Reason and Captain of the Bullshit Police.

TL_Type_S

Tom, thanks for the reply. He continues to insist that brake-torquing doesn't exist for auto, FWD cars .. that whatever we do to launch the Acura faster is just "applying the brake and bringing up the revs...not brake torque."

Peace

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2002 Acura TL-S
1999 MB E320 4-matic
M3_Killer .. in the snow
1998 Isuzu Rodeo
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soopa

Well.. wouldnt that only be true if the cars were traveling 100mph for the entire 1/4 mile?

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gavriil

Hey sorry to intrude here but Tom your math is simplistic.

A car at the track is doing 100 mph ONLY at one moment in time and in this case AT THE EXIT. To reach 100mph it has to go from 0 all the way to 99.99mph, etc. I am sure you get my point.

From my experience, cars that will do the 1/4 mile from 13.5 to 15 seconds will have the difference you described which is 0.1 second for a car's length. So I agree with you Type S and Tom.

Brake torquing is all correct and I agree with you both.

By the way, if you are talking about a C5 Vette, I think the time is slightly pessimistic. They will do 13.4 in a slow day. There has been numerous 13.1 for C5 coupes and even high 12s for some.

I dont get the last sentence of your post though Type S.

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Gabriel
CL Type S
Mag Racer #1

[This message has been edited by gavriil (edited 05-27-2001).]

TL_Type_S

Gavril,

Thanks for the reponse man. This guy, my buddy, has no idea of what he's talkin about, i emailed him w/ this info -- he "agreed unwillingly" about the WRX issue. However, he still disagrees about the brake-torqueing.

About the last line: He mentioned that the transaxle in a FWD automobile prevents the driver from brake-torquing the car -- meaning to keep the back brakes locked while spinning/burning the front tires. I told him that it was not the "transaxle," rather, it is the brake power split .. 80% up front, 20% in the rear.

I did the math in front of him too for 100mph:

( [ 100m/h X 5280ft/m ] / 3600 s/h )
= 146.6 ft/s
(146.6 ft/s) / (15 ft/carlength) = 9.78 carlengths/second

9.78 CL/sec * .9 seconds = 8.8 carlengths
8.8 ~= 9 ... 9 CL/.9 sec = 1 carlength/.1 sec

But then I thought about it, and say the 1/4-mile speed for the Corvette is:

106 mph w/ a 1/4-mile time of 13.6 seconds

And the WRX say.. is 14.4 seconds, 93 mph
(according to the guy who ran on the WRX forum)

By the time the C5 hits the 1/4 in 13.6 seconds, it's going 106 mph (please tell me where my logic is wrong, Tom, Gavrill). Now, .8 seconds, the WRX is gunna cross that line too --> since their is no way for us to know the acceleration of the C5 or the WRX at those moments, let's say for the next .8 seconds (14.4 - 13.6), that Corvette is locked at 106 mph. So it's travelling at roughly 1 carlength per .1 seconds as Tom mentioned.

But, w/ the WRX close to 90 mph at that point, let's lock the WRX at 90 mph for instance w/ about .8 seconds left to go in the 1/4-mile. The difference b/w the C5 and the WRX is 106 - 90 = 16 mph .. if you do the above math .. it amounts to 1.25 carlengths per .1 second. But then again, the C5 was always gaining distance on the WRX, and we cannot assume that both cars are fixed in speed at any time around the 1/4-mile mark. So, maybe it's just a drag-racer's rule of thumb that each .1 second is about 1 carlength.

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2002 Acura TL-S
1999 MB E320 4-matic
M3_Killer .. in the snow
1998 Isuzu Rodeo
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" TARGET=_blank>Click Here for Pics!
</A>

[This message has been edited by TL_Type_S (edited 05-27-2001).]

Tom2

No, not really. The reason is that your 1/4 mile speed is taken at the end of the track. Two cars racing will both end up with a 1/4 mile top speed. I realize that the math is far from exact science or anything, but the fact of the matter is that to estimate the number of car lengths between two racing cars that run around the 100 MPH mark, each 1/10 of a second is about equal to 15 ft, or 1 car length.

I'm not saying that this is an exact measurement, only an estimate.

Trust me, I've seen it at the track XXXXX amount of times.

By the way, if you watch top fuel dragsters, you can throw this theory out the window, because of the fact that they're going over 300 MPH at the end of the 1/4 mile.



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99 M3
dead CL-S

Tom2: The Voice Of Reason and Captain of the Bullshit Police.