depends on how fast you're trap speed is.

 

really, interesting.
How about this my trap speed at the track was around 89-91( I think ).

 

i'm sure there's a mathematical way to figure that out but i suck at physics, so my information wouldn't be all to accurate.

 

Have you guys heard the same thing?
Reason I ask this is because of some runs my buddies and I have done on some empty roads. I have been to the track and was trying to make some comparisons. My 2 friends lined up(both ls's both with I/H, one had 16"s) and the one with 16's lost by about a car or so. I ran with him(gsr-I/H/C/E) and I beat him by about 1 car or so. He has a good driver and ran my car faster than I could(used a gtech to determine this). Well we are going to the track and I was wondering if his driving skill would have that much inpact on his times). I am believing that it should only because he could run my car faster than myself. My goal this year is to improve on my times, especially the lauches as those seem to most important. I am excited to seem him run his ls at the track...My buddies and I going up together for a killer day! Could be like 6 tegs!

 

V = Velocity...although since direction is constant velocity just means speed. It would have to be the speed when you crossed the finish line.
D = Distance...or length of the car
T = Time or how long it takes for the car to travle one car length
Here's the very basic derivation of the formula:
V = D/T
TV = D
D/V = T
So if you want the time one car lenght takes take the length of the car and divide it by the speed at the end of the race.

 

Here is an example of above:
Say you finished the race at 40 m/s (about 90 mph) and the car was 5.5 meters long (about 18 feet long) I just made these numbers up...I have no idea of the actual length. So you take 5.5 m divided by 40 m/s and you get .1375 seconds.

So...in conclusion...a car lenth ain't much

 

With this reasoning when the difference in time is .17s my buddy should have been a little over a car length ahead(7m). When I beat a car by 1.36s I should have been ahead by 54m or 12 car lengths? That sounds kind of believable for sure. It didn't seem like he was that far away but my buddy did beat my by around that, maybe even less...it was almost a year ago and so hard to remember but I think it was a car length. My guestimate would be that one car length at 88-90mph would equal .12-.14s at the finish line, for a teg that is.
Sound right?

 

does anyone know the length of an integra?

 

With this reasoning when the difference in time is .17s my buddy should have been a little over a car length ahead(7m). When I beat a car by 1.36s I should have been ahead by 54m or 12 car lengths? That sounds kind of believable for sure. It didn't seem like he was that far away but my buddy did beat my by around that, maybe even less...it was almost a year ago and so hard to remember but I think it was a car length. My guestimate would be that one car length at 88-90mph would equal .12-.14s at the finish line, for a teg that is.
Sound right?

Don't forget about bracket racing. You don't always leave the starting line at the same time. It all depends on your dial in. As a general rule, if both drivers were perfect, they should both cross the finish line at the very same time, when it comes to bracket racing.

 

V = Velocity...although since direction is constant velocity just means speed. It would have to be the speed when you crossed the finish line.
D = Distance...or length of the car
T = Time or how long it takes for the car to travle one car length
Here's the very basic derivation of the formula:
V = D/T
TV = D
D/V = T
So if you want the time one car lenght takes take the length of the car and divide it by the speed at the end of the race.

you're not accounting for acceleration. that would work if it were a constant velocity problem but its not.

you could use x=xo+vot+.5at^2, where x is the distance, xo is 0, vo is 0, t is time, and assuming you knew the average acceleration of each car. you could then get a rough estimate of the difference in distance.

and to find average acceleration you could use
v^2-vo^2=2a(delta_x), where v is the final velocity, vo is 0, and delta_x is the distance (.25 miles).

but keep in mind it isnt going to be exact because acceleration isnt constant.

 

you can try to figure it out, you need the measurements of your car, how long and how many ft are in a 1/4 mile, your speed and his speed and now to figure out the equation (is that how it's spelled???). lets say you were doing 60 mph to make it easy on the math. That means you would do a 1/4 mile in appx. 15 seconds (this is all just figuring)and a 1/4 mile is 1320 ft. So if you divide the feet by the seconds you get 88ft per second. OK, here is where it gets tricky, now how long is your car. Lets say 10ft for easiness of math.... now if you are both traveling at the same speed and he is 1 car length behind you (on your rear bumper) now if I did this right he will be approximately 1/9th of a second behind you. 10 is about 1/9th of 88. So about .09 would be a round about number!!! Now the tricky think is if he was doing a different speed!!!

 

Sean, i dont think that the incredible acceleration rates of our cars in 4th gear are enough to matter over such a short time.

while you do make a good point for an exact number, the accleration factor wont really come into play until you have a very fast car.

spaceboy, you ask how much of a difference driver skill will make? it wins and loses races. a good driver can make a .2-.3 second improvement over an average driver. ive seen larger gaps than that at my local 1/8 mile, so figure it to be almost .5 in the 1/4.

 

alright guys...I got a litte bored...so Click Here

Go look at it as big as you can. In Windows 2000 (I think)...if you let the mouse hang over the pic it should expand. Here is the basic idea though:

my bad... the second column should read "1 car length".

 

These are excellent thoughts. I always wonder because you will hear someone say I beat them by a car length. Well what is a car length...It is like someone having a car that is .5seconds faster(stock cars). At the track what would this equate to? Should it be 1 car, 2, or even 3 cars?
Just an interesting question to me that is all.