A coworker of mine sent me a link to this over in the "Audi World"

http://s4wiki.com/wiki/Torque_and_HP

Good examples, pretty much relates to everything I've heard but I wanted to know what you guys thought regarding the L98 description in there....

 

Cool article. It goes to show that higher RPM torque rules with proper gearing.

Aaron

 

Excellent article. It shows the L98-Tuned port strength and weakness: low end grunt but running out of suds just about mid-range. I have read any number of places the TPI was originally designed to boost the ordinary drivability of the 305" engine in a heavy vehicle, such as a station wagon/full sedan from, say, idle to 2000.

The constant conflict we C4-owners are up against is picking up the upper RPM range and power without losing the low end power. Have our cake and eat it too. The article points out the only way to do that is to raise the torque curve thoughout the entire range. Simply shifting the torque curve to the right kills low end.

But if we are willing to put in steeper gears along with a shift of the power curve to the right, then we are in and out of the low rpm range so quickly we get a net gain even if we have only moved the torque curve to the right. But in this case we also need more rpms or we end up at max power before we get to the finish line. I thought the example in the article comparing the L98 to the LT1 was an excellent illustration of this.

 

<------------


Not a problem here

 

I well written article that tells it how it is. He should have added that your cars best average excelleration is where you maximize the HP under the operating curve when your racing.

 

Great article - thanks for sharing.

 

that article has been around on the 'net for a long time and pasted from website to website.

its a good intro into understanding the basics.

 

Yes its been around for a long time. Was more support for me when i decided to make a 8000 rpm stroker and the results are proving worthwhile !

 

That article doesn't clarify anything. It's way too long winded and brings up more questions than answers.


It's hard to keep it short and sweet, but here's my attempt to explain myself in another thread. The last part sums it up better:


"Torque tells you everything about how fast the car is! A car will accelerate at the EXACT rate as the torque curve of the engine. "

 

So basically gear your car for whatever you plan on using it for, wether it be a flour mill or the quarter mile.

 

right on. One reason why I tossed the 4.09s in favour of 3.73s.

 

that article mostly gets the bottom line correct, but isn't wholly correct in how it gets there. i have tried to explain this plenty of times, and usually no one believes me, but i'm compelled to keep saying it!

here's my writeup i keep handy so i just have to cut and paste my argument:


*** Talk about Torque ***

Torque is an *instantaneous* twisting force. Look at the units. It is not energy, and there is no time component. It is in terms of force at the end of a lever.

We can't really measure an instantaneous force. To do so requires removing the time component. This means we would be attempting to represent the Universe as having one less dimension than it really does.

That's an abstract thought, but reality bears it out. Consider, for instance, using a strain gauge to measure the torque/force. We could cut out a section of the driveshaft and replace it with a strain gauge. The only way we could get meaningful data is to average it over time, or the gauge would show wildly varying spikes and valleys of force.

"But an engine dyno measures torque!" I've yet to see anyone correctly state how an engine dyno "measures" torque. The almost universal thought is that the dyno measures torque, and hp is calculated using rpm. (I'm talking purely about engine dynos here, quite a few people correctly understand that inertia wheel dynos are measuring power.)

It might be true that the dyno reports torque, but think about this very carefully. The engine dyno is in fact inherently measuring power. The engine's crank spins a "brake" at 100% slippage (or the engine block would be spinning wildly!). They're not like brakes we have on the wheels of our cars, but they're exactly equivalent functionally.

So picture the engine directly coupled to a brake rotor, as if the rotor were the flywheel. Then picture a fixed caliper on that brake rotor. Since there is 100% slippage between the rotor and pads, there's no direct transmission of force. What is actually happening here is that engine power is dissipated via friction in the brake, and the resultant force is measured.

In other words, What the dyno does to is measure the amount of axial force that must be applied to the caliper in order to limit the engine's speed or acceleration. We call that the torque output since, by virtue of equal and opposite reaction, the axial caliper force should equal the force the engine puts to the rotor. (Keep in mind, what's being measured in this case is the force the caliper applies to the rotor, NOT the other way around!) But the only way that force is occurring in the first place is through kinetic friction, and that means all you're really measuring is some component of energy!

Let me digress a bit and show another case where abstract thought follows real physical example:

This force peaks when the heat output and engine power output are in equilibrium, of course. If the engine is accelerating, then "torque output" measured at the caliper is lower. This is a crucial bit. Torque output is lower because you have to release the braking force on the caliper in order to allow the engine to accelerate. Said another way, if you release the caliper braking force, the axial force required to keep the caliper from spinning with the rotor drops, right? (The correct answer is "right.") So measured torque output varies depending on whether the engine is holding a constant rpm, or accelerating.

The really interesting part about the previous paragraph is that it appears to perfectly answer why measured power varies depending on the engine's rate of acceleration. This just goes to show that while torque must absolutely follow all laws of physics, it does not tell the whole story!


*** Talk about Power ***

Power: Whether Joules per second, or horsepower, or kilowatts, or whatever, Power is still just a measure of how much work is being done for a given unit of time. "Work" can also be defined in many different units. I'm going to ignore the energy lost to heat and focus on work done by the engine on the car or dyno. In this case Work is equal to the change in Energy, or deltaE. Power, then, is deltaE divided by time. Note that power is basically a measurement comprised of time and energy, when the most elemental concepts in the Universe are time, energy, and mass.

Let's examine where energy goes in a car. (Again, ignore heat from the
engine, but not from other effects.) There's kinetic energy in the rotation of the engine and driveline parts. There's energy in friction to both moving parts, and to the atmosphere (aerodynamic drag). And there's kinetic energy of the moving car itself, equal to half of the mass times the square of the velocity. Or, commonly depicted as:

KE = 1/2 * m * v^2

Ok, so lets go back to the rotational kinetic energy, and give a formula for that:

KE = 1/2 * I * w^2

This is where I (for Inertia) is a measure of how the mass is distributed in a rotating system, and w is equal to the angular velocity, or how fast the system is turning.

*** Compare Them ***

Comparing Torque vs HP: As a car's, and an engine's, speed changes, the energy changes. This energy has to come from somewhere. If speed goes down, the energy is lost to friction (heat), and work, if it speeds up the energy comes from converting the chemical energy of gasoline (or whatever fuel).

Power is a measure of how much energy can be put by the engine to the car in a given amount of time. So more power means more energy can be converted in the same time interval, or, the same amount of energy can be converted in less time. (If you understand that, then it's already clear why power is the only truly meaningful measure of an engine.)

Theoretical is all well and good, but here are some examples with hard numbers to illustrate the difference between torque and horsepower.

For the first example, consider a car with 500ft-lbs of flywheel torque when running at 3400rpm. Consider, in comparison, a second car which only makes 300ft-lbs of flywheel torque, but spinning at 6000rpm. The two cars are side by side, going equal speeds, the first car is at 3400rpm, the second car at 6000rpm. What happens? Which car wins?

There are a few things we can do in order to ease the calculations. First, since torque is ft-lbs, we can make the tire's radius one foot, so there's no gearing ratio multiplier. This means the force at the tire, in pounds, is equal to flywheel torque times the transmission ratio times the final drive ratio. We need to calculate the gearing in order to determine the force at the tire. (Take careful note that this step is mandatory.) Let's assume the car is traveling at 40mph.

If the radius of the tire is 1ft, then the circumference is:

2 * pi * 1 = 6.2832ft

So, to go 40 miles, the tire turns:

(40miles) * (5280ft/mile) * (1revolution/6.2832ft) = 33613.4 revolutions

If we go 40 miles in one hour:

(33613.4revolutions) * (1 hour / 60 minutes) = 560.2 rpm

Now we can calculate the overall gearing, since we know both the tire rpm, and the engine rpm:

Car1: 3400 / 560.2 = 6.07:1

Car2: 6000 / 560.2 = 10.71:1

(Do you see where this is going?)

Ok, so now let's see what the force at the tire's contact patch will be for each car:

Car1: 500 * 6.07 = 3035lbs of force

Car2: 300 * 10.71 = 3213lbs of force

The acceleration is equal to the force divided by the mass. Let's assume a mass that makes the numbers pretty, so 3035lbs. So, acceleration:

Car1: 3035 / 3035 = 1g

Car2: 3213 / 3035 = 1.06g

Now, let's take another approach entirely. I'm going to do it all from the perspective of horsepower, and see what happens. I may have to make some of the same assumptions, but I'll see how much calculating I can drop and still get a number. We need the horsepower numbers, of course, but my aim is to show the difference between knowing torque, and knowing horsepower. So we can't count this next calculation as part of what I need, since I'm assuming we'd have the horsepower numbers to begin with.

Car1: 500 * 3400 / 5252 = 323.7hp

Car2: 300 * 6000 / 5252 = 342.7hp

Ok, here we go:

Power = Work / time

Work = Force * distance

But, time divided by distance is equal to speed, so:

Power = Force * speed

Or:

Force = Power / speed

Wow, that's nice. All I need is the power, which I know, and the speed, which I also know! But, the problem is units. Again, we assume we know the conversion factors, so it's not something extra we have to calculate. For instance, in the Power equation, it's in Joules per second (and 1 J/s = 1 Watt). The speed is in meters per second. The force is in Newtons.

40mph = 17.878 meters / second

327.7hp * 745.7watts/hp = 241383 joules / second

342.7hp * 745.7watts/hp = 255551 joules / second

1lb force = 4.448 Newtons

Car1: 241383 / 17.878 = 13502 Newtons = 3035lbs

Car2: 255551 / 17.878 = 14294 Netwons = 3213lbs

There you have it. I won't even bother calculating the acceleration, since the forces came out identically to the first set of calculations.

I didn't have to know the rpm, or the gearing. I knew the hp, and the vehicle speed, and I told you what the accelerative force is. Horsepower turns out to be a very real, very defining term to explain and predict vehicle performance. Knowing torque, I also had to know the rpm, and the overall gearing, before I could calculate a meaningful performance figure.

Note that rpm has time in it, so I had to add the time dimension before torque was meaningful. (Remember what I said earlier?) Also, note that the flywheel torque did not translate to vehicle acceleration. All by itself, flywheel torque ends up being meaningless, otherwise the car with more torque would have accelerated harder.

From this illustration, it should be clear this holds up regardless of the car's gearing. So whether you compare cars of different or equal gearing, the one with more hp at a given speed will accelerate harder at that given speed. So the trick is to have a car with more horsepower at every point being used. Or, more realistically, optimize such that the rpm vs time maximizes the area under the horsepower curve. (Either a little more time at much higher hp, or a lot more time with a little more hp, however you can visualize what I mean by maximizing area under the power curve.)

Now, there is one phenomenon which is built into everything we just calculated, but which might not be immediately obvious. This is an important distinction, in case you glossed over the nitty-gritty above: The horsepower determines the acceleration only at a given speed. If you change the speed, it's a different rate of acceleration. Keeping power constant at 328hp means the car will actually accelerate less and less the faster you go, even ignoring friction and aerodynamics.

This is how both F = m*a and Power = Work / time are both true at the same time.

Lastly, let me point out that all industrial engines are rated only in hp. This is because industry and engineers know that only power has any meaning in determining the work the engine is capable of.

-michael

 

In other words, you want the engine in the rpm range that produces the most power in order to get the most acceleration?
Brilliant.

Now, that's a pretty bold statement....


Larry
code5coupe
(feeling particularly antagonistic today...)

 

Let's just race!

 

If this really is true, a vehicle that produces 300 ft-lb at 2,000 rpm (114 hp) and a continuously variable transmission should be able to pull harder and turn better 1/4-mile times than an engine that produced 250 ft-lb at 6,000 (286 hp) also with a CVT because either engine should alway be able to run at max torque.

If this is truly the case, don't you think that the automakers would be taking advantage of this fact and CVTs would be a way of life by now?

 

i'm not sure to whom that's directed, but my argument is exactly the opposite. the combination with the most power at a given speed will accelerate harder. the overall gearing is irrelevant (power inherently accounts for that). that's what the whole exercise shows, so i hope you're responding to something someone else said!

CVTs have other issues, but that's a different topic. i don't think that technology is "ripe" yet.

-michael

 

Below is an excerpt from the link.

"Now, what does all this mean in carland?

First of all, from a driver's perspective, torque, to use the vernacular, RULES :-). Any given car, in any given gear, will accelerate at a rate that *exactly* matches its torque curve (allowing for increased air and rolling resistance as speeds climb). Another way of saying this is that a car will accelerate hardest at its torque peak in any given gear, and will not accelerate as hard below that peak, or above it. Torque is the only thing that a driver feels, and horsepower is just sort of an esoteric measurement in that context. 300 foot pounds of torque will accelerate you just as hard at 2000 rpm as it would if you were making that torque at 4000 rpm in the same gear, yet, per the formula, the horsepower would be *double* at 4000 rpm."


So, in my example, a 300 ft-lb @ 2,000 (114 hp) rpm engine with proper gearing should out pull the 250 ft-lb @ 6,000 rpm (286 hp) engine because the transmission could always keep the engine at peak torque. As such, it would make no difference at what speed the torque is produced, only that there is more avaliable. This being the case, the persuit of high rpm torque (and horsepower) is unwarranted..., assuming that you buy into the premise that it is the torque and not the power that really accelerates the car.

 

Absolutely a misconcept........

For maximum performance you would design the CVT to put the engine at maximum horsepower not maximum torque!!!!!!!

The 286 horsepower engine would blow away the 114 horsepower engine...

A CVT that allows the engine to stay at the maximum horsepower rpm would in fact be the ideal transmission....

 

The key to his statement is "Any given car, in any given gear".....

This is why a broad flat torque curve is good, because the acceleration will be constant from minimum to maximum rpm in each gear.......the missing key here is the importance of gearing......

With proper gearing maximum horsepower gives you maximum acceleration.....

Ultimate transmission would be CVT that keeps the motor at maximum horsepower..

 

So then, you belive that it is power and not torque that accellerates the car. Not saying that you (or the article) are right or wrong, but this is contrary to the article.


So, who is correct?