I have a couple of issues with or questions about Skant's post. I'm not trying to degenerate into arguments, or "I know more than you," because I probably don't. Maybe I do. Who knows? Discussion is healthy, and how we learn. I feel, though, that Skant's post is inconsistent in places.
First, an agreement:
Skant
Torque at the drive wheels is what really matters.
Exactly what I said when I talked about being able to dyno the torque at the wheels in each gear. OK.
Skant
B) They know an oversimplified (and fairly erroneous) saying and think that covers it. The one most commonly quoted seems to be "Horsepower gives you top speed, torque gives you acceleration". These folks are generally
bewildered when the real world doesn't behave anything like how they thought it would.
I'm not bewildered by this concept, it's exactly what was tought to me, drilled into my skull with heavy tools.
Skant
Greater torque means both faster acceleration and higher top speed.
Don't know. I've been taught by them what's s'posed to know, faster acceleration, but nothing to do with top speed.
Skant
Greater horsepower at the engine means you can have greater torque at the wheels
This is where inconsistency begins for me. You can't just throw away RPM. Horsepower includes RPM, torque doesn't. We'll get to that momentarily.
Skant
Peak engine horsepower is king here. Additional top speed can not be attained by trading off rpm for more torque multiplication. So it will just have to be brute force and lots of it. Horsepower would have to be quadrupled in order to double the top speed.
How is that different from the "Horsepower gives you top speed" part of my statement? And technically, force is torque, horsepower is power. So by brute force I assume you mean brute power. Picky, yes, but it seems words for force and words for power are being used interchangably, which doesn't help an explanation of the difference.
Skant
Horsepower is simply the torque multiplied by the rpm.
Absolutely. Let's actually prove the mathematical relationship between horsepower and torque, which we can do without actually bothering with units (I'm told I like to over-simplify)
:
In physics, power is the amount of work done per unit of time. Work is force times displacement. (Displacement means how far something is moved, not how big an engine is. If the force and the displacement are not the same direction, there is a factor to account for the difference in directions of the force vector and the displacement vector.) Of course we all know that velocity is displacement per unit of time. Hmm...
Power = work / time
Work = force * displacement
So Power = force * displacement / time
Since velocity = displacement / time, then reduce to:
power = force * velocity.
Our force and velocity are rotational, so power = torque * RPM.
By now the reader may be asking so what? What has this to do with acceleration? Well, absolutely nothing, which is my point.
In his section "POWER CURVES" Skant has these statements, describing a "torque" car and a "horsepower" car:
Skant
But overall, the car accelerates faster because there's more total area under the entire torque curve. So you have a car that accelerates faster
despite reduced peak horsepower.
then
Skant
But again the total area under the power curve is greater overall and so again, the car accelerates faster.
Why did we change curves? You can't prove a possibly incorrect point by using incorrect data. Where this section is misleading is when Skant talks about increasing power at low RPMs. If you have increased the power at low RPM, you've increased the torque, because HP = torque * RPM. If HP at an RPM is greater, torque has to be greater. It's the torque that improves acceleration.
As I think back on that magazine article, there is another curve that can be plotted, speed vs. force required to maintain that speed. That would be a curve beginning very low at the left rising very slowly, but curving upwards as speed builds, as a function of the drag increasing as the velocity squared. Where that curve meets the cascades, you have terminal velocity. It would look like this:
The force available from the powertrain is equal to the force of wind resistance. Now, since that happens on a torque graph, it would seem that if I had more torque, I could go faster. Well, yes I could, but as soon as you stop accelerating, you're not talking about just force (torque) any more. You're talking about work, because you're using that force to displace something (air). To maintain a velocity (as opposed to reaching it in the first place) you're talking about power. That's how it's been shown to me, and that's how I understand it. I will admit that it's been a while since I've used any of this math, since my work building servers and PC networks doesn't need any of this kind of engineering. It was made clear to me at the time, though, that once velocity becomes constant, you're looking at a different set of variables, and you're looking at power (horsepower, watts, whatever), not force (as in weight or torque).