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  #1  
Old 02-11-2004, 07:54 PM
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0 - 60 MPH Calculator

It actually works! Well for about 80% of the cars that I tested.

here is the formula: 0-60 time = (weight of car in kilogrammes)/(maximum bhp of car * 0.9)

reference: http://www.cars-cars-cars.org/0-60-Times-Calculator.htm

That is just what i found recently surfing the internet. I put some numbers for the SVX, and it turned out to be very close to actual times. That makes me wonder, does having AT tranny really matters?

3580 Lb = 1623.86 kg / (230 bhp * 0.9) = 7.84 s

that is only 7% error. from real time of 7.3 s.

Ok, this formula looks kind of rigged, so I opened the Car And Driver, and started calculating theoretical 0-60s and practical. It turned out the out of about 10 cars - 8 of them were very close to the specs. The only car that was way out of specs is Forester XT. With calculated 0 - 60 at 7.2 and actual 5.4.

So.... my conclusion is... SVX is way too heavy.
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  #2  
Old 02-11-2004, 08:01 PM
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Re: 0 - 60 MPH Calculator

Quote:
Originally posted by zamorush


So.... my conclusion is... SVX is way too heavy.

Way too heavy for what? Remember what the car was designed for.
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  #3  
Old 02-11-2004, 09:32 PM
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Quote:
Way too heavy for what? Remember what the car was designed for.
No, i know what it is designed for. I am just trying to see if 0-60 might change with Manual tranny. Well...... it might a bit, but not that much.

Plus, a sports car have to be a sports car. Even if it is a luxury car. I say put another 100 hp. to SVX to 330. Or make it just a bit lighter, but still luxury. Hey, it would be more then perfect. I am not saying it is not perfect now. I actually like it the way it is, but as a teenager.... well grown up one already, I want more..... as always.

"No matter how fast you are, there always will be someone who is faster then you." (Me.)
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Old 02-12-2004, 03:05 AM
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Anybody study Physics or Applied Maths at School?

This is Newtons Second Law: Acceleration is proportional to Force. More commonly written as Force = Mass x Acceleration.

This can be rewritten as a=F/m.

acceleration is change in velocity over time (v-u)/t

(v-u)/t = F/m

t=m(v-u)/F

v-u is constant because we are talking 0-60 time.

Then you have to play with some constants to convert from SI units to the everyday american ones and you end up with.

time = mass / force * some constant.

Which is exactly the formula you gave.

Phil.
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Old 02-12-2004, 07:43 AM
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I would adjust that 0.9 factor depending on wich type of tranny and engine I have.

Suppose you have a ECVT tranny, then the acceleration is fairly constant over the whole run, but if you have "gears", the acceleration "changes" depending on what rpm you are.

The same goes with an AT, i.e longer "shifttimes" (less acceleration) and so on.

The power curve of the engine is also a factor, a NA engine increases it's HP when the rpm increases, but a turbo engine gives "full" power earlier and less depending on the rpm, so a turbo engine should have a diffrent factor than a NA engine.

If we try to find a suitable factor for MT versus AT and Turbo versus NA, this "tool" can be useful.


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  #6  
Old 02-12-2004, 08:19 AM
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elninoalex took 1 second off his 0-60 times when he installed a 5MT. Probably a combination of less drivetrain loss (no TC) and the 4.11 gearing that his car has.
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Old 02-12-2004, 05:33 PM
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Re: Anybody study Physics or Applied Maths at School?

Quote:
Originally posted by b3lha
This is Newtons Second Law: Acceleration is proportional to Force. More commonly written as Force = Mass x Acceleration.

This can be rewritten as a=F/m.

acceleration is change in velocity over time (v-u)/t

(v-u)/t = F/m

t=m(v-u)/F

v-u is constant because we are talking 0-60 time.

Then you have to play with some constants to convert from SI units to the everyday american ones and you end up with.

time = mass / force * some constant.

Which is exactly the formula you gave.

Phil.

Thanks Phil.

You must be an engineer.

Larry III (Mech. Eng'r.)
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  #8  
Old 02-12-2004, 06:00 PM
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Quote:
Originally posted by mbtoloczko
elninoalex took 1 second off his 0-60 times when he installed a 5MT. Probably a combination of less drivetrain loss (no TC) and the 4.11 gearing that his car has.
My car seems to produce abuot the same. The luanch and then a small gain, with the AT having suprior top end at a certain point.
phil
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  #9  
Old 02-13-2004, 12:46 AM
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Quote:

This is Newtons Second Law: Acceleration is proportional to Force. More commonly written as Force = Mass x Acceleration.

This can be rewritten as a=F/m.

acceleration is change in velocity over time (v-u)/t

(v-u)/t = F/m

t=m(v-u)/F

v-u is constant because we are talking 0-60 time.

Then you have to play with some constants to convert from SI units to the everyday american ones and you end up with.

time = mass / force * some constant.

Which is exactly the formula you gave.

Phil.
You are absolutely right Phil, it is pure physics. For some reason I never thought about it myself. Now that I did, I am very skeptical about it being Newton¡s Second Law, here is why.

From your formula t = mass / Force * constant. However you never specified what force is applied to the car? I am assuming it is Mass * Gravity? Plus, the (v - u) should not be constant, because it is change in velocity according to your derivation. Thus, it is delta V. Then the formula follows ==»

Time = (mass * V)/(mass * acceleration), or if we are taking units as follows:
Time = seconds = s
Mass = kilograms = kg
Distance = meters = m

Then s = (kg * m/s) / (kg * m/s^2) = seconds. That is what suppose to happen I believe. However, when you substitute everything, as in mass of the car, gravitational acceleration, and change in velocity, which in this case, 0-60mph or (0-26.6 m/s)/2, then the answer that we get is totally unrelated to anything, because then it won¡¦t depend on the mass of the car at all.

So here is my solution to this rather complicated problem:

Ok, taking the original formula: time = mass / bhp * constant.

Now, using units:

s = kg / (((kg * m)/s^2) * m/s) * C,

where bhp or hp = (((kg * m)/s^2) * m/s) = N *m/s = J/s = Watt

Therefore when simplified: s = s^3/(m^2 * C), which is apparently the formula that has to work, however for it to work, constant C has to have units of (s^2/m^2). And again, if you look at it, mass of the car is being canceled out, which is not suppose to happen. So my guess is, that the constant C has also be some sort of Force of gravity or frictional force for this formula to work properly, because otherwise it just does not make any sense.

How else would you divide kilograms/ horsepower * constant to get a time in seconds?
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  #10  
Old 02-13-2004, 03:51 AM
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Zamorush,

I am convinced this is Newton's Second Law. I'm afraid I don't really understand your working.

The Force I am referring to is the force produced by the engine to accelerate the car. It's 230bhp multiplied by a constant which determines how much of that bhp actually gets transmitted to the road multiplied by some constant to convert that to Newtons. Gravity neither helps nor hinders the acceleration on a flat road (and in any case is constant).

v-u is constant. Acceleration is final speed (v) - initial speed (u) over time (t). Since we are talking 0-60 time, (v-u) is (60 - 0) * a constant to convert that to m/s.

If you think about it, it can only be N2L. We are calculating the acceleration of a mass, the only tricky bit is finding the right constant (fudge factor) to convert the bhp produced by the engine to the resultant force in newtons that is actually being applied to accelerate the mass.

This constant would have many components, including the frictional forces you mention. But it would be too complex to calculate it. I'll bet they came up with the 0.9 by feeding in the figures from hundreds of different cars and taking an average.

Have I convinced you?

Phil.
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  #11  
Old 02-14-2004, 01:46 AM
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dont forget to add the weight of the driver. that puts a stock svx above 8 seconds, and has me at 7.6
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