I think I got all that, but I still don't understand/agree with the max acceleration at peak torque. I mean, it makes sense from a science point of view, but it doesn't make sense to me in an experimental point of view.
For example, I just got back in from driving my granddad's 04 Mach 1. It is bone stock, making 320 ft-lbs of torque at 4250 rpm and 305 horsepower at 6000 rpm, according to Ford. I put it in 2nd gear, and go WOT from a very low rpm/speed. Just to begin with, it is pretty slow getting off the line, as the 4 valve motor isn't know for producing much low end power. At about 3000 rpm, it is getting going a little better. By 4000 rpm, it is really starting to hit its power band. At 5000, it is pulling HARD. 5000 to 6000 rpm goes by really quick, and at 6000-6500, it is pulling like a freight train (no joke). Bottom line, it pulls MUCH, MUCH harder at 6000 rpm (peak horsepower) than at 4250 (peak torque). This is not an opinion; it accelerates harder at 6000 rpm than at 4000 rpm.
Bottom line is that your experiment doesn't have an accurate enough measurement tool. The ass-o-meter isn't that reliable, unfortunately.
If the max acceleration occurs at peak torque, why is this? I understand the science (I think), but my theoretical prediction doesn't meet my practical experience.
Let's revisit the science. The math for this becomes much more complicated and involves derivatives/integrals to explain thoroughly. I can do this math, and you might be able to interpret it if I am articulate/eloquent enough in my explanation of the math, but I don't think math is ever as good as a conceptually correct answer to a question. So here's my best attempt:
Conceptually imagine what is happening at peak torque: the engine is at maximum volumetric efficiency. That is to say that the cylinders are filling themselves with as much fuel and air with each stroke as the engine is able to. The engine does not get as much air through the system in a single stroke at either lower or higher RPM. When the spark lights the air/fuel, the explosion is the cylinder hits harder than at any other RPM. The peak cylinder pressure is the highest it will ever be. The cylinder traps this air and
forces the cylinder down. This is a linear force. This linear force is applied through the rod to the crank journal that is some distance from the centerline of the crankshaft. A force applied at a distance about a "fixed axis," which in this case is the centerline of the crank, is the definition of torque - or the definition I prefer, "torque = rotational force."
Because cylinder pressures are highest, it makes sense that this is the peak torque. The distance of the force that the rod applies from centerline of the crankshaft is determined only by the rod journal. It never changes. Thus the only thing that can increase or decrease the torque is the linear force that is applied at that journal. That linear force is entirely due to cylinder pressure. Thus, peak cylinder pressure = peak torque. Now, imagine stopping your motor for a moment in time and imagine the forces being applied by all of those rods. This is a 4 stroke engine; so at any given point you'll have 2 rods on the power stroke, and 2 on each other stroke. It's the force applied by those 2 pistons that you feel at that moment. They are the only things creating force in the motor at that instant. The harder they push, the more the whole system, vehicle, driver, and all, will accelerate.
I think you understand all of that already, and so it makes sense to you now that peak torque is when the car accelerates the hardest. What confuses you, and most people is how you can do any better than this, or if more hp is better, then how come it doesn't cause the car to accelerate even more in the same gear...
The difference at peak horsepower is that the cylinder does not fill as completely as it does at peak torque with each stroke, but there are a lot more strokes in the same time. So it doesn't have as high a cylinder pressure. It doesn't place as large a force on the piston, rod, or crank during the power stroke. However, the cylinder is actually moving a lot more air and fuel in the same amount of time, because it completes more revolutions in that given time period. Thus, your engine is actually producing more energy.
So why aren't you accelerating faster at the top-end of a gear? To understand all of this, it's important to remember what power really is. Power is not directly what accelerates a vehicle. It is only the ability to do work in a set amount of time. What is work? Work is the application of force to an object
over a distance that the object moves, not to be confused with torque, which is a force applied
at a distance from a rotating point. Normal work involves a linear distance - the car moves a mile while a force is applied (F*d). Rotational work uses, well.... a rotation - how many rotations that a force is applied over(TQ*Rotations).
Now, what's happening in a gear at high RPM where you're making more power, believe it or not, actually is more impressive than what's happening at peak torque, in an important way. While acceleration in a gear, say 4th, is greater at peak torque, the work being performed actually is greater at the peak horsepower, which I'll attempt to conceptually demonstrate without going entirely through some very confusing math.
As an example, let's talk through what's happening in 4th gear. Let's say peak torque is completely flat from 3300-4000 RPM. Let's also say peak power is flat from 6000-6500 rpm. Further, let's also say that since more force is being applied at peak torque, it takes an identical amount of time, in seconds, to accelerate over either of those RPM bands, even though the lower one covers 200 RPM more. Let's just say it takes 10 seconds to cover those RPM bands because you're in 4th. Even though you've accelerated more in the low power band in 10 seconds, the distance covered in the high power band during those 10 seconds is considerably higher. Let's make it easy and define distance as the distance that the car travels during that time. How fast is your car going to be moving at 6000 RPM? By comparison to 3300, pretty damned fast, right? In fact, the car's velocity is almost twice as fast (6000/3300=1.82). How much less torque will it be making at 6000 RPM? It should be making considerably more than half, right? So since we know work is a force over a distance (W=F*d). We can deduce conceptually that more work is done in the high power band in the same 10 second time frame because well more than half the force is being applied at almost double the distance.
From an engine perspective, the distance will be the number of rotations that are completed in those 10 seconds. Obviously, the high powerband is going to complete somewhere around nearly twice as many rotations in the 10 second period as the low powerband (6000rpm*10s/60s > 4000 *10/60). The force applied over that distance is the torque the engine makes. Again, nearly twice as many rotations, with much more than half as much torque = more power (tq * rotations = work)
A little less force applied over a lot more distance = more power. Your motor did indeed produce more work in the same amount of time at the top of the powerband. More "stuff" was accomplished, so to speak.
That's about all I can think of to explain it. Hopefully, I did an adequate understandable job.
I've got some other ideas that I can use to explain this, but it's almost 4am here in Afghanistan, and I'm getting to tired to think straight.
I could do the math and explain the numbers to prove the matter, but that's not basic stuff, anymore, and I'm probably a bit rusty. I actually made a "dragstrip" calculator when I was in college that calculates this stuff very quickly and attempts to estimate 1/4 mile time given vehicle weight, its gearing, average rear-wheel torque in each gear (from a dyno graph), wind resistance, and an approximate 60' time. It's pretty cool and I got it to predict just about every car within a few tenths of a second. Some were very very close. This model was great because it let's the user play around with gearing, weight, 60' times etc... to see how they would impact the car's ET and trap speed. If any of you have any interest in checking it out, I'll email it. I'd like to further develop it one day, but there are so many other awesome calculators out there now with engine programs that I just didn't see the point. I got an A on the project, though.
Chris
