sti vs evo rolling start
Originally Posted by silvery_eagle
i'm going to make another idiotic post here.
unless you're running on high boost at 60mph on 3rd gear (maybe 6000rpm)
and then "punch" it until 100mph the time i guess would be very similar as 0-100 minus 0-60.
Inertia force or momentum i dont know the term will drag you down when you're starting at 60mph with no boost... and let's not get into the turbo lag ok? whenever you punch it at 60mph there will be turbo lag ...
so, time( 0-100 ) - time(0-60) != time(60-100)
unless you're running on high boost at 60mph on 3rd gear (maybe 6000rpm)
and then "punch" it until 100mph the time i guess would be very similar as 0-100 minus 0-60.
Inertia force or momentum i dont know the term will drag you down when you're starting at 60mph with no boost... and let's not get into the turbo lag ok? whenever you punch it at 60mph there will be turbo lag ...
so, time( 0-100 ) - time(0-60) != time(60-100)
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From: Minnesoter
Originally Posted by silvery_eagle
i'm going to make another idiotic post here.
unless you're running on high boost at 60mph on 3rd gear (maybe 6000rpm)
and then "punch" it until 100mph the time i guess would be very similar as 0-100 minus 0-60.
Inertia force or momentum i dont know the term will drag you down when you're starting at 60mph with no boost... and let's not get into the turbo lag ok? whenever you punch it at 60mph there will be turbo lag ...
so, time( 0-100 ) - time(0-60) != time(60-100)
unless you're running on high boost at 60mph on 3rd gear (maybe 6000rpm)
and then "punch" it until 100mph the time i guess would be very similar as 0-100 minus 0-60.
Inertia force or momentum i dont know the term will drag you down when you're starting at 60mph with no boost... and let's not get into the turbo lag ok? whenever you punch it at 60mph there will be turbo lag ...
so, time( 0-100 ) - time(0-60) != time(60-100)
Edit: i asked this question on a physics forum, check it out:
http://www.physicsforums.com/showthr...866#post488866
Last edited by twinevo; Mar 11, 2005 at 01:07 PM.
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From: Minnesoter
Originally Posted by grendel
go watch that lambo vs evo vid and pay close attention to the race between the evo and the rental car.
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From: ft. worth, Texas
but that doesnt help this guys question. What about the torque each have, the gear ratios, all that plays a big part of a rolling start. just because you can beat a car 0-60, or 0-100 doesnt mean you will automatically win from a roll.
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From: Minnesoter
Originally Posted by grendel
but that doesnt help this guys question. What about the torque each have, the gear ratios, all that plays a big part of a rolling start. just because you can beat a car 0-60, or 0-100 doesnt mean you will automatically win from a roll.
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From: Minnesoter
Originally Posted by SterlingEvo
I guess here's the answer to the original question... They're about equal 
http://www.physicsforums.com/showth...8866#post488866
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From: Long Beach
This is a big quote from Krab at physicsforumscom
No. The times would not be different. Why should they be? Different would mean that the 0-60mph time would depend upon what you did after you hit 60mph. That contradicts causality.
The numbers are assuredly NOT linear. Linear would mean that if 0-60mph is 4.8 sec, 0-130mph would be 4.8*130/60=10.4 sec. It would not be linear even if your wheels exerted a constant force to the road. This is because of aerodynamic drag. That's also the reason you top out at some speed. So if 135 mph is top speed, the time needed to reach it (or just higher than it) is infinite.
In fact, your tires do not exert a constant force, because vehicles are power-limited. So the force to drive the car is power divided by speed. Power varies a bit because you cannot stay at peak power except with a continuously-variable transmission, but this is not a large effect. So a good description of an accelerating vehicle is given by
m(dv/dt) = P/v - bv^2
where b is an aero drag coefficient, m is mass, P is power (to the road). You will notice acceleration is zero when . The v that satisfies this formula is of course top speed. So you can rewrite the equation of motion as
m(dv/dt) = P(1/v - v^2/ v^3 inf)
Which is nicer because you don't need to worry about how to find b. Solve this equation and you have v as a function of time. No road test needed. I've done this and it agrees with car magazine test reports. (Well actually it's a little more complicated than this because the force does not go to infinity at zero velocity as P/v would indicate, so you also need the torque in first gear.)
Originally Posted by evolved8
Hi all. Newbie here who has a simple question.
In car magazines, you see these speed tests they do. For example, they'll list a 0-60 time, 0-100 time, and a 0-130 time. My question is, would the times be different if you just made one run from 0-130 and mark down the times when you hit 60, 100, and 130 as opposed to doing three separate runs of 0-60, 0-100, and 0-130.
Hi all. Newbie here who has a simple question.
In car magazines, you see these speed tests they do. For example, they'll list a 0-60 time, 0-100 time, and a 0-130 time. My question is, would the times be different if you just made one run from 0-130 and mark down the times when you hit 60, 100, and 130 as opposed to doing three separate runs of 0-60, 0-100, and 0-130.
Thank you for your help. The bottom numbers I used were just random. The actual numbers WOULD be linear if you graphed them out. I'm a car enthusiast who is comparing the performance of two cars. This helps.
Here are the actual numbers
0-60 mph 4.8
0-100 mph 12.6
0-130 mph 25.0
Here are the actual numbers
0-60 mph 4.8
0-100 mph 12.6
0-130 mph 25.0
In fact, your tires do not exert a constant force, because vehicles are power-limited. So the force to drive the car is power divided by speed. Power varies a bit because you cannot stay at peak power except with a continuously-variable transmission, but this is not a large effect. So a good description of an accelerating vehicle is given by
m(dv/dt) = P/v - bv^2
where b is an aero drag coefficient, m is mass, P is power (to the road). You will notice acceleration is zero when . The v that satisfies this formula is of course top speed. So you can rewrite the equation of motion as
m(dv/dt) = P(1/v - v^2/ v^3 inf)
Which is nicer because you don't need to worry about how to find b. Solve this equation and you have v as a function of time. No road test needed. I've done this and it agrees with car magazine test reports. (Well actually it's a little more complicated than this because the force does not go to infinity at zero velocity as P/v would indicate, so you also need the torque in first gear.)
no magazine racers allowed on this scenerio.
Between those two, it's all about the driver.
If the driver were the same person in mirror? depends on the car, some cars are stong from the factory, some aren't.
if they had the same situation, i got my money on evo, until you hit 4th gear.
Their gear box sucks a ss.
Between those two, it's all about the driver.
If the driver were the same person in mirror? depends on the car, some cars are stong from the factory, some aren't.
if they had the same situation, i got my money on evo, until you hit 4th gear.
Their gear box sucks a ss.
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