Roadrunner

For you math wizards,

If a car can run 15 second quarter mile with trap speed speed of 93 mph, and same car was moded and improvement resulted in a 14 second quarter mile time. Assume same launch.

What would that moded car's trap be and how many feet would the moded car win by?

StealthTL

iTrader: (1)

Here is a little calculator for your equation......

courtesy of Mazdamaniac.com!

Quarter mile time calc....

S

crazy4h20

Trap speed = 99.6 mph
Gap = 122.8 feet

Feel free to point out errors... (1 correction already made)

Assume linear acceleration (doing otherwise makes thing REALLY complex)...

a1, a2 = acceleration rates for each pass
v1, v2 = trap speeds (velocity) for each pass
t1, t2 = time for each pass
d = distance traveled

Assuming zero velocity at start:
d = 1/2 * a * t^2

Assuming constant acceleration, speed at a point in time is:
v = a * t
a = v / t (this will be substituted below)

Since the distance traveled is the same for each pass:
1/2 * a1 * t1^2 = 1/2 * a2 * t2^2
a1 * t1^2 = a2 * t2^2 (multiply both sides by 2)
(v1 / t1) * t1^2 = (v2 / t2) * t2^2 (substitute for acceleration)
v1 * t1 = v2 * t2 (time cancels from each side)
v2 = v1 * t1 / t2 (solve for v2)

v2 = 93 mph * 15s / 14s
v2 = 99.6 mph

--------------------------------------------------------------

For distance between passes, we can ask where was car 1 when car 2 passed the finish.

Use the same variables as above plus,
x = starting point

Key values are:
t = -1s
v1 = 93 mph
a1 = 93 mph / 15s

for conversion, 60 mph = 88 fps (feet per second)

Velocity at t = -1s is
v = a1 * t + v1

Again, assuming linear acceleration, position in space is:
d = 1/2 * a1 * t^2 + v * t + x

We'll use the finish as our point of reference, so
d = 0

0 = 1/2 * a1 * t^2 + (a1 * t + v1) * t + x (substitute velocity at t = -1s)
x = - 3/2 * a1 * t^2 - v1* t (combine a1 terms and solve for x)
x = -3/2 * (93 mph/15s * 88 fps/60 mph) * (-1 s)^2 - (93 mph * 88 fps/60 mph) * (-1 s)
x = -3/2 * (93/15 * 88/60) + (93 * 88/60)
x = -13.6 + 136.4
x = 122.8 ft

epitrochoid

brings back memories of AP physics...lol

Seenitall

Ya this is one great board-we are not only
car enthusiasts, but there are a lot of VERY bright folks here.

Roadrunner

Thanks....I had little doubt the answer(s) would be forth coming.

RX-8 friend

Unfortunatly the quicker 1/4 time often is the result of a better launch, in which case all the above very good math is out the window! Also, the shift time usually swamps the accelleration - ie if you took 1 sec. to shift each of three gears on one run and then took 1/2 sec. on the next run your 1/4 time would improve by 1.5 sec. but there would actually be no difference in accelleration.

Charles R. Hill

iTrader: (7)

Since two cars can run identical e.t.'s and have different trap speeds, much of the conversation seems moot to me. Trap speed and e.t. are measuring different aspects of performance. Example; my '70 Nova ran 14.6@106. Then, I switched to a Powerglide and 10" converter. I picked up 1.3 seconds with no change in trap speed.

Charles