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05092002, 11:15 PM  #1 
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mathematics. this may be of some challenge...
Hey guys I was using this gtech thing. I wanted to setup a graph on my acceleration (avg) over 60mph and the final 1/4 mile feet..
here it is. I am looking for an equation that would find out how many feet my car covered trying to get to 60 (WOT). Info: Time: 7.78 seconds Speed: 60mph Distance Covered: x feet Other Info: 1/4mile: 15.94 Speed: 90mph Distance Covered: 1320 feet I am looking for feet covered in the 060 test. I would appreciate any help! Thanks Eric 
05092002, 11:32 PM  #2 
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I can;t think of any possible way to do the math there. I don;t think it is possible without knowing average acceleration (g's)and even then it's near impossible.
But. I;m not a math major However, I'm guessing it is around 650 feet or so. Eric 
05092002, 11:44 PM  #3 
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I would say that the average g's pulled to get to 60 would be about .29 (.35  first gear) (.28 second) (.25 third)

05092002, 11:46 PM  #4 
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First you need to estimate acceleration, you can do this by using the average acceleration given as:
a=Final Velocity / Time , must convert MPH into m/s then distance = 0.5 * a * time^2 example a=26.82 m/s / 7.78s= 3.44 m/s^2 distance= 0.5 * 3.44 * 7.78^2 = 104 meters = 341 feet of course this assumes no wheel spin and a standing start! Unfortunately this is the best you can do with the information you have and it maybe way off depending on the actual circumstances. Last edited by romanom; 05102002 at 08:56 AM. 
05092002, 11:47 PM  #5 
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There is no answer to your question because your rate of acceleration is not consistent. If you look at a chart of G's under WOT, you'll see what I'm talking about. It makes an arc, going up around torque peak and then back down up to redline. Then when you shift to the next gear, it goes down a little and makes a similar arc but at a lower level on the graph.
But anyway, it's not possible to figure it out. 
05092002, 11:52 PM  #6 
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best you can do is with average acceleration
Oops, Romanom beat me to it.
Larry 
05092002, 11:54 PM  #7 
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I'm feeling a little lazy, but I do believe there is an answer to your question. If I remember right, it is the integral of something or other, using 0mph [initial] to 60mph [final] over the 7.78s [time interval] that you ran.

05092002, 11:58 PM  #8  
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Re: best you can do is with average acceleration
Quote:


05102002, 02:24 AM  #9 
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The distance will be the integral of the velocity vs time. Since we dont have much info we have to assume the velocity increases linearly from 0 to 60 (constant acceleration). Therefore, Velocity will be a function of t, whose slope is 60/7.78, so we take the integral of this line from t=0 to t=7.78 and we arrive at the equation romanom provided. That's where the calculus comes in .

05102002, 03:10 AM  #10 
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Has any magazine done acceleration specs on an RS? If so we could get a better model for the acceleration from 0 to 60. Just a thought.

05102002, 10:48 AM  #11 
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No, none of that is correct. My first post is correct in that there is no answer to your question. The math in the other posts assumes a constant rate of acceleration, which absolutely does not apply here. You can have two cars with identical 060 times, but they have traveled a different distance over that time. The Impreza RS was always a good test for this because of its lowend torque. When you compare it to other cars with similar 060 times, say a new Honda Civic SI, the Impreza would cover more distance over that time. Just take a look at their 030 times compared. The Impreza is .5.8 s faster to 30 than the Civic, which gives it a huge leap on it. But the Civic has a good topend and a good hp/weight ratio, so in mph it does catch up right around 60 mph, but it's still trailing the Subaru.
Another good way to think about this is the 1/4 mile times. In this example, the distance is guaranteed to be the same. So if the math in the previous post was to hold true, two cars with identical 1/4 mile times would also have identical trap speeds. Anybody who's ever looked at 1/4 mile times knows this to not be true. We'll use the Impreza RS and Civic Si as good examples again. These cars have similar 1/4 mile times, but the Civic has a trap speed of around 34 mph faster than the Subaru. The explanation for this is the same as the first paragraph. And by the way, yes, I am a math expert. 
05102002, 02:08 PM  #12  
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05102002, 03:53 PM  #13 
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I have something similar to what you're looking for on my site:
http://www.carvideos.com/performanc...?id1=117&id2=0 I start at 30 mph because anything under that is very unreliable. I know the site's a little slow lately. It's overloaded with traffic. I'm going to have to add another DSL soon. 
05122002, 02:51 PM  #14 
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You can't find how far the car travelled mathematically. You can't get it by integrating anything because you have no function to integrate. The acceleration isn't constant in one gear never mind two gears.
Two solutions, log instantaneous acceleration vs time. (gtech does this but only to output time and speed, it could output distance but doesn't) OR Measure the damn distance covered to get to 60. Erice1984, you used a gtech to get your numbers, go back and do them again. Then measure the distance travelled. That's really the only simple way to get the answer you want. wj 
05122002, 06:42 PM  #15 
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this sounds like mathematician vs engineer.
you guys want the exact answer, we provide the best answer with the given information. yes the constant acceleration is an (false) approximation, but with a few more sampling points as suggested by djvapor you can construct a piecewise linear curve to integrate, or if you really wanted, use some higher order interpolation to construct a curve. Then you'll really start getting close without fancy shmancy equipement. sorry this post was unecessary i just dont like to get called wrong especially when i say what i did is based on assumption. and yes, i am a math expert 
05132002, 10:03 AM  #16  
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05132002, 10:40 AM  #17 
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Yes, you're right. It does give you a good approximation. I'd be willing to bet the true distance is within 7 or 8 ft of the calculated number. The gear shifting is usually not that significant on short distances like this, but the launch will play a major role in this distance. If you get a true AWD launch, with a little wheelspin and the car leaps forward without the engine bogging down, then the distance will be slightly longer (even though 60 comes up quicker) than your estimate. But if you do a street launch with a cluch slip and the engine bogs just a little, the distance will probably be slightly shorter. This can get very complicated.
What was the purpose of figuring out the distance in the first place? I don't see the point of a graph with only 2 sets of data. 
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