[QUOTE]Originally posted by Elistan
[B]Zoran, you're almost there but you're not quite correct.

 

I'm sorry, but you obviously refuse to understand what the hell I am saying.

Did you ever look at the dyno graph? Your horsepower changes with RPM . It's usually a near-linear increasing curve.
So when you upshift your RPMs drop, and so your horsepower drops as well. I am not talking about peak numbers, I am talking about ACTUAL wheel HP numbers at any given speed/RPM.

That is true, but that is not a valid argument for what you are trying to say.

When you turn 4000 RPM in both gears you are going at two vastly different speeds, and so you cannot compare the acceleration in those two instances. One car going 4000rpm in 4th gear (about 60mph) and the other going 4000rpm in 5th gear (about 80mph). What kind of race is that?

[QUOTE][B]
I'm actually flabbergastd that someone who thinks he's so intelligent can come in here, insult those who are correct and then have the audacity to spew this, people

 

Here Zoran, this will help:

 

[QUOTE]Originally posted by FYRHWK1
[B]

I'm actually flabbergastd that someone who thinks he's so intelligent can come in here, insult those who are correct and then have the audacity to spew this, people

 

OK, while I completely agree with Destiny2002's original post, I can understand why some people just glaze over it. He's using Work and Power definitions in his formulas, which might not be clear to not-so-profficient-in-physics people in this thread.

I'll get to the exact same formula as he did, but I'll use the factors that we all know: wheel torque, wheel diameter, gearing ratio, engine torque, engine HP, engine RPM.

Here it goes:

Let's say your car is going at a speed V.

How do you determine the force by which your car can accelerate at that speed? You need to find out the RPM's, gearing and the engine torque, right?

So you're going V mph, and let's say that translates into WRPM wheel rpms (when you calculate the wheel/tire diameter and cirumference):

WRPM = V/(2*R*Pi) ....... (1) (where R is your wheel radius)

Your wheels are rotating at WRPM.

You know what gear you are in , and the total gear ratio for that gear is GR.

Your engine rpms will be WRPM*GR, right?

RPM = WRPM * GR, ......... (2)

So now you know your engine rpm at speed V.

Let's say you have a dyno of your car, and you know the torque at all rpms. Let's say your engine torque at the calculated rpms is TQ.

What is your wheel torque? Engine torque * gearing, right?

So your wheel torque is

RWTQ = TQ * GR ..............(3)

From (2), we can express gearing: GR = RPM/WRPM, and when we substitute that into (3), we get:

RWTQ = TQ*RPM/WRPM

Since TQ*RPM = HP, and replacing WRPM with (1), we get:

RWTQ = HP/V * 2 * R * Pi ....... (4)

What is the FORCE that pushes your car forward?

That's wheel torque, divided by the wheel radius, right?

F = RWTQ/R = HP/V * 2 * Pi

F = 2Pi * HP/V

What is the acceleration? The force that pushes the car forward, divided by the car's mass:

A = F/m = 2Pi * HP/(V*m)

Acceleration = Constant * HP/(mass*speed)

Q.E.D.

 

 

Destiny and Zoran - great posts. Even with a background in physics and two engineering degrees, I still have to think hard about some of this. (Well, not the basics, but the conversions going on in the transmission and the conversions from rotational forces to linear...) But then, my degrees are electical and computer. Now, if you convert the problem into electrical form, it becomes intuitively obvious!

I think the problem here is that most people have limited understanding of physics and react more to the kick-in-the-pants they either feel or don't. Maybe a discussion of the dynamics at very low rpm - starting off the line - between large displacement high low-end torque vs small displacement high revving engines would help.

Tony

 

[QUOTE]Originally posted by Destiny2002
[B]Since the maximum power accelerating the object is equal to the peak horsepower from the engine (minus frictional losses which are similar between vehicles),

 

 

That has nothing to do with the fact that you accelerate harder in 3rd gear due to greater torque mulitplication, the aerodynamic drag of 80 MPH over 60 is not that great. The reason for greater accleration in 3rd is due to higher torque multiplications, period, i've nothing more to say because you're both obviously more interested in reading equations from textbooks instead of looking at basic automotive facts proven every time you drive your car, when you're done trying to show off how much of a bookworm you are then maybe this can be a real discussion.