Okay,
It's been a LOOOOOOOOOONG time since I was in High school physics, so I'm relearning this stuff as I go, so pardon any silly-stupid mistakes I'm making here. I'm trying to make a Time to distance graph for a Flash animation based on QTR Mile Time Slips. Now in order to properly animate the results, I have to know time to distance every 10 feet or so.

So let's assume I want to figure out how many seconds it takes a car to hit 50' assuming the Car's 60' time is 2.134. Now I know that I will have to fudge things a little and run on average acceleration rather than actual. Reason being that We only have eight Variables captured from the run (R/T, 60', 330', 8th ET, 8th MPH, 1000, ¼ ET, ¼ MPH). However, from 2.134 Seconds @60' we can calculate that the Initial Velocity (Vi) is "0" and the final Velocity (Vf) is 28.116 Ft/Sec. From that the average acceleration is 13.175 Ft/Sec. I assume this is right? (I'll Use "^" for Delta)

Average Velocity= ^Displacement/^Time or V=^D/^T
Average Acceleration = ^V/^T or (Vf-Vi)/^T

So:
Vi= 0Ft/0 Sec = 0 Ft/Sec
Vf = 60Ft/2.134 Sec = 28.116 Ft/Sec

A= (28.116-0)/2.134

A= 13.175

So First Off, Hopefully I'm correct so far, as my numbers fall to hell from here. Now I will assume that Acceleration stays constant (Even though I know it does not in a drag race) just to keep the math easy. I want to figure out how much Time (T) it takes for the car in question to hit 50 Ft assuming it's accelerating at 13.175 Ft/Sec. Or to make it look like a high School Word Problem.

Solve: If a Corvette is stationary (Vi=0) and begins to accelerate at a rate of 13.175 ft/sec, how many seconds will it take to reach 50 Feet?

So I have some formulas:

(1) Vf=Vi+AT
(2) D=ViT + ½ AT²
(3) D=VfT - ½AT²
(4) Vf²=Vi²+2AD
(5) D=.5(Vf+Vi)T

So I want to solve for time "T", but I don't know Final Velocity "Vf" So I can do:

Vf=Vi+AT or T=(Vf-Vi)/A
Now I don't have Vf so I'll use Formula #4 Vf²=Vi²+2AD or Vf=sqrt(Vi²+2AD)

So:.
T=(Vf-Vi)/A is the same as...

T=( sqrt(Vi²+2AD)-Vi)/A

Plugging in the numbers...

T=( sqrt(0²+2(13.175)(50))-0)/13.175
Or Time @ 50' = 2.755 Seconds.

Now Obviously this is WRONG Because it can't take 2.755 Seconds to get to the 50' mark when everything was calculated for hitting the 60' mark in 2.134 Seconds, so What did I do wrong?

-Adam

 

Okay, let's make our variables

D1 = 60ft
D2 = 50ft
T1 = 60ft time
T2 = 50ft time
A = avg acceleration over 60 feet
Vi = Initial Velocity = 0
Vf = Final Velocity at 50ft


Okay - first formula

D1 = Vi*T1 + 1/2*A*(T1)^2

Since Vi = 0 the term = 0 also

so we get

D1=1/2*A*(T1)^2

or

formula 1 A=2(D1)/(T1)^2

That will give us the average acceleration over the 60ft mark

Now our next formula - let's find the final velocity at the end of the
50 foot mark

(Vf)^2 = (Vi)^2 + 2A(D2)

Vi = 0 so the term = 0

(Vf)^2 = 2A(D2)

or

formula 2 Vf = sqrt(2A(D2))


Now we need to find the time it takes to reach that final velocity at a given
acceleration - so use

(D2) = 1/2(Vf + Vi)(T2)

Vi = 0

(D2) = 1/2(Vf)(T2)

formula 3 T2=2(D2)/(Vf)

okay, T2 is what we want - so now let's substitute back in for what we don't know - Vf - using formula 2

we get

T2 = 2 (D2)/(sqrt(2A(D2)))

Okay, we got rid of the Vf - but we still don't know the term A - so let's substitute in for that using formula 1

T2 = 2(D2)/(sqrt(2*(2(D1)/(T1)^2)*(D2)))

Which is our final formula - we can plug in the
constants D1 and D2 to get

T2 = 2 (50)/(sqrt(2*(2(60)/(T1)^2)*50)))

or

T2 = 100 / sqrt(12000/((T1)^2))


let's try it out using your
T1 = 2.134 value

T2 = 100 / (sqrt(12000/((2.134)^2)))

T2 = 100 / sqrt(2635.071573)

T2 = 100/51.333 = 1.9481


Whew!




[Modified by ChrisB, 4:41 PM 4/11/2002]

 

basically he's quoting AVERAGE acceleration numbers and forgot to also put distance per time SQUARED. since cars don't accelerate constant rate, he really needs some data acquisition (maybe there's a feed off some g-tech like device). measured "g" force as a function of time-->acceleration as a function of time --> integrate (calculus operation on functions) acceleration twice --> get position as function of time --> let's say the position curve is plotted time on he horizontal axis and distance on the vertical axis --> draw a vertical line to the time axis from the point on the position curve that intersects with 50 foot mark and that's your exact answer!!!

Greg
:cool:

 

Chris,
Thank you for straightening this out for me. It looks like those numbers work pretty well. Thanks for taking the time to type all that up to, I’m sure it took a while. Hopefully in a month or so everyone can see the fruits of my labor on this. Should be a pretty cool animation, we’ll have to see :)

-Adam