1 Car Length = 37rwkw (50rwhp)

[BOSE]

I have been there and seen it, it is about correct,

my car XR6T ute auto 265.4rwkws 12.7@109mph

other car XR6T sedan auto 308rwkws 12.2 115mph

distance was 1 car length across the line

it is also worth noting that a 30rwkw gain is worth approx 1/2 a second down the 1/4 mile

my car stock 14.0sec, 195rwkws

2nd cai and edit 13.3sec, 235rwkws

injectors, high flow cat and custom tune 12.7sec, 265.4rwkws

as mentioned above other XR6T 12.2sec, 308rwkws

<{POST_SNAPBACK}>

Surely that cant be right .... so in otherwords it took your sedan .5 a second to travel one car length at 115mph .... 184 km/h .... or 51 m/s ??

in half a second it would have travelled approx. 25m ... not 2.5-3m or whatever the length of the car is.

[BAturb]

I am mearly commenting on what I have personally seen and achieved and I have the time slips to back up the claims

[shockF6]

you may have time slips but BOSE has mathmatics on his side. 25M in .5 of a second is not 1 car length for your 30kws

[C.S.I]

I'm with Blueboost, as far as 2 similar T's with dissimilar power. Gotta be much more than 1 car length in it.

You traval a long way in 0.5 or 0.6 secs at 100mph+

Alan

[Blonk]

Here goes -

Because the time travelled by both vehicles is relative as we are only trying to calculate the difference in 1 car length (I have used 5m) we need to calculate the acceleration..

Mass equals (from memory) - 1770kgs

Time equals (stock 1/4 mile) - 14 seconds

Distance travelled - 400m (car 1) 395m (car 2)

Car 1

To assess acceleration

s=ut+1/2(at)at

400=0*0.5*a*14*14

a = 4.1

driving force

f=ma

1770*4.1=7257 N

power

P=W/t

P=7257*400/14

P=207.5kw

Car 2

(changing the distance by 5m less)

acc.

s=ut+1/2(at)at

395=0*0.5*a*14*14

a = 4.02

driving force

f=ma

1770*4.02=7115.4 N

power

P= W/t

7115.4*400/14

P= 203kw

So 1 car length with stock cars equals 4 kw's over 400m - Mathematically.

As I said .. There are too many variables and I would say from this traction is a major variable to start with.

Now my brain hurts

[C.S.I]

Faaaaaaaaaaaaaaaaaaaaark, are you for real

[Blonk]

Simple laws of motion... I can calculate the termianl velocity of both vehicles if I gave a shiit

[MYWPN]

BAturb you said that he passed you on the line, If he was behind you with that power it sounds like your reaction time was alot quicker which would explain the 1 car length at the end.

[blueboost]

Here goes -

Because the time travelled by both vehicles is relative as we are only trying to calculate the difference in 1 car length (I have used 5m) we need to calculate the acceleration..

Mass equals (from memory) - 1770kgs

Time equals (stock 1/4 mile) - 14 seconds

Distance travelled - 400m (car 1) 395m (car 2)

Car 1

To assess acceleration

s=ut+1/2(at)at

400=0*0.5*a*14*14

a = 4.1

driving force

f=ma

1770*4.1=7257 N

power

P=W/t

P=7257*400/14

P=207.5kw

Car 2

(changing the distance by 5m less)

acc.

s=ut+1/2(at)at

395=0*0.5*a*14*14

a = 4.02

driving force

f=ma

1770*4.02=7115.4 N

power

P= W/t

7115.4*400/14

P= 203kw

So 1 car length with stock cars equals 4 kw's over 400m - Mathematically.

As I said .. There are too many variables and I would say from this traction  is a major variable to start with.

Now my brain hurts

<{POST_SNAPBACK}>

cannot argue with that, sounds about right, but that is an average over the whole 400 m, and that was the question asked. but in reality the improvment in power is normaly in the last half of the rev range. so in laymans terms, if you improve your power 4klws all the way through your power band. you should improve 1 car length.

but in reality you do not normaly improve it down low right through to red line.

so I have found that about 7-8 kws most times gives you 1mph and about a tenth,

but if you lose 4kws down low to get your 8 kws up high, you don't get much of an improvement.

[Lawsy]

I haven't done the math yet - I think I'll leave it up to Lawsy. However, as a general rule of thumb, it sounds about right for two cars of the same weight and which are operating within their traction limits. But if you started talking in extreme cases then of course it wouldn't stand up scientifically due to so many variables being involved.

Rather than try and explain it with maths, I'll just give you a graph with relevant information and you can figure it out for yourself.

I think the data points can be easily verifiable as real cars without too much hassle...

I would just like to point out though that an easy way to verify this is to take the 1000 foot mph, then the final mph, take the average (because acceleration is pretty linear between these 2 points for most cars, even if you have a gear change) and then you can easily work out the distance by multiplying the time by the average speed accross those 2 points.... Or something like that, I know what I'm on about, its up to you to figure out if what I've just typed is actually what I'm thinking

Anyways.

My quality graph to follow

As you can clearly see, its no where near correct as 1 car length is about 4 meteres.... Well roughly 45rwkw made a difference of nearly 30-40 meters... And another few rwkW made another 40-50 odd meter difference....

Infact looking at it again, having roughly 130rwkW more than standard made a difference of about 90 meters.... Which is about right because if you work it out mathematically, you have to be going about 150km/h to travel 100 meters in just over 2 seconds... About right.