DuraMax really 300hp?? and 520lb torque?

Wow! This has been extremely interesting, and just a little bit over my head!

ZF, if I understand you correctly, about the only other thing that could affect performance between various 300hp vehicles with different torques, after the rear axle that is, would be tire size and vehicle weight, correct? That would be a couple of variables that could change things.

If your three 300hp with different torque engines puts the same torque to the wheels thru the transmission and rear axle, but one has larger tires which decreases the torque, and that vehicle also weighs more, meaning more mass to move, then it would be at a disadvantage to an otherwise equal truck, right?

Just trying to see if I'm following this correctly.

Excellent and very perceptive comment!

Torque is defined as ft/lbs ... which means the number of pounds of force at a 1 foot radius from the center of the hub.

If you have say, 2000ft/lbs of torque at the rear wheels, and you had tires that were exactly 2 feet high (i.e. a 1 foot radius), then the tire will exert 2000lbs of forward force onto the pavement.

But, our tires are closer to two and a half feet high. This is functionally equivalent to applying less gear reduction. In other words, because the radius is more than 1 foot, we go a little farther every revolution but we have a little less torque.

How much does that decrease the force applied to the ground? The simplest way to figure this out is to just scale it with the ratio of the difference in size. So just take 30" / 24" = 1.25, which means a 30" tire is 25% taller geared than a 24" tire. Therefore 2000 ft/lbs of torque will be 2000 / 1.25 = 1600lbs of force where the rubber meets the road on a 30" tire. Likewise, if a given rpm gives us 70mph with a 24" tire, it'll give us 70 * 1.25 = 87.5mph with a 30" tire. So it did the exact same thing as applying less gear reduction.

Rule 1 in physics is force = mass * acceleration, or more commonly, f=ma. Acceleration then is a=f/m. If you know the force on the rear tires, and the mass of the vehicle (which has a constant relationship to weight on earth), you can literally figure out the acceleration. So yes, weight is a major factor in acceleration, no question.

There are a whole bunch of other factors at work, too, a truly accurate calculation isn't this simple, this is just close. The biggest one has got to be "area under the curve". Essentially all these examples have been at specific data points, i.e. "X" hp, "Y" torque, "Z" rpm gives you this. But in the real world, most of us don't have continuously variable transmissions. So you have to "integrate" and find the area under the curve for the range of rpm that we operate over. That integrated horsepower is more meaningful than peak horsepower.

And then there are even more factors. For example, rotating inertia (i.e. a big flywheel) goes up with the square of the radius, so it has a bigger effect than static inertia. Essentially, it stores some of our power, which means that it'll rob some away while we're accelerating, which makes us accelerate slower. This whole thing is the subject of debate with dyno guys, whether a brake dyno or an inertia dyno gives a more true representation of torque and power. Other forces skew the results, too, like wind resistance.

I love your question. It shows you've thought it through.

I took 4 semesters of physics. Hard courses, and I never really liked it that much. But it's handy for understanding things like this.


[This message has been edited by ZFMax (edited 01-17-2002).]

Let me get this straight, my old 6.0 Liter Vortec had 300 HP too. If I had only changed the gearing from in it from 4.10 to 3.73 I could have improved my performance to that of the Duramax AND saved $40,000.

Wow I am dumb after all.......



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MA,

I'm still not sure that I'm following this correctly, but, if what ZF has said is true, then you may be partially correct. If in fact the 6.0L engine was 300hp, you could in fact change the rear axle to make it pull equal to the Duramax. However, since the 6.0L creates its power in a higher RPM band, you would have to LOWER the ratio to do so. In other words, you would have to go from a 4.10 to a 4.56, or something even lower, I'm not a math guy, so somebody else figure it out. Geared properly, the 6.0L could pull the same load at the same speed. However, it would be way up there in RPM's where the HP is in order to accomplish this, thus, wearing the engine out quickly, sucking the fuel, being very inefficient, etc.

That's why the Duramax is still a better choice. Not necessarily because it simply has more torque, but because it can produce more power at a lower RPM = HP. That way you can tow the same load up the same hill at the same speed but at a lower rpm, using less fuel, being much more effecient, even though both motors have the same HP. I think I'm starting to understand this!

Basically, you're both right. Torque without HP is meaningless, but a higher torque motor can move a load better because it can do so more effeciently....am I close on this?

ZF,

Your remark on the heavy flywheel reminded me of an old fashioned gas engine that my Grandfather and I own.....it's a 1905 vintage International 2HP "hit and miss" engine. It has two, huge flywheels on it. It only fires every one in a while, when it slows down to the point that the governor allows the exhaust valve to close and allow the combustion process to take place. It just sits there and spins for the longest time purely on inertia. It was used to run a water pump back in the old days. This thing probably cranks out a lot of torque, but very little HP, simply because it can't accelerate worth a darn.

Very interesting stuff.

MABurns,

I believe that your old 6.0l gasser would perform as well as a Duramax, as long as you can keep it up around 5000RPMs http://www.62-65-dieselpage.com/ubb/wink.gif

Kevin

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Mark, you're getting there, you went the wrong way on that gearing, but you're getting the idea.

If you accept that hp = (torque * rpm) / 5252, and if you accept that gear reduction divides rpm and multiplies torque, and you know how to multiply and divide, the answer to your question is within your grasp:

1) Pick a target speed. Doesn't matter what it is. It's easier if you just use rear wheel rpm as the units, but you can convert that to mph if you want.

2) Pick an engine hp, torque, and rpm. It can be anything, just make sure it's legal. It MUST follow the formula hp = (torque * rpm) / 5252.

3) Divide the engine rpm by the desired rear wheel rpm to get the needed gear reduction to go that speed. Gear reduction does divide rpm, right?

4) Multiply the engine torque by that amount of gear reduction to get the torque you'll have at the rear wheels. Gear reduction does multiply torque, right?

Now just repeat steps 2 through 4 for ANY other engine hp/torque/rpm combination. Or even another hp/torque/rpm point on the same motor, doesn't matter. As many times as you want. Just make sure your target speed stays the same and your combination of hp/torque/rpm are legal. Otherwise it's an invalid comparison.

Look at the results.

Show me one example where a motor with less horsepower than another motor ended up with more torque at the rear wheels, at the same rear wheel speed. Just one, and you win the debate.

Do you not believe that rear wheel torque determines how fast you can go, or accelerate, with a given load?

Or do you believe that somehow this process is flawed?

There is nothing even remotely fuzzy about this math. It's just working with the basic relationship of torque, rpm, and horsepower, and the effect of gearing. It couldn't be simpler. And it's dead nuts right.

I fully realize this is a non-intuitive concept to you, the idea that extra rpm can be turned into more torque. But it's true.

I certainly didn't buy my 300hp Duramax because I think somehow my 300hp is stronger than another 300hp, just because mine is made with higher torque and lower rpm. That would be foolish, 300hp is 300hp, all you have to do is gear the other motor deeper and you can get just as much torque out of it at the same truck speed. If that's why you bought your truck then I'd have to say yes, you did waste your money.

I bought a 300hp Duramax because it offers better fuel efficiency and more longevity than a 300hp gasoline truck. That's worth the premium I paid for the diesel motor, to me. I don't feel like it's wasted money at all.

Arveetek, at least someone is finally getting it! 300 HP is 300 HP, for acceleration the Average HP across the RPM band that is used IE 2000-3300 RPM's is what provides the differences in acceleration, a low torque motor that must make very high RPM's to produce its peak HP, looses too much steam with every downshift, and spends many RPM's at lower power.

A 6.0L that is geared properly, to run at peak power at a given speed should be able to maintain the same as a duramax at peak power with the same load. But who wants to have to Run around towing at 5000 RPM's with the 6.0L, imagine the longevity, fuel milage, and noise of that motor. Also, if the Duramax produces 520 ft lbs at 1200 RPM's, and the 6.0L produces 300 ft lbs at 1200 RPM's, the Diesel has 73% more HP available at that RPM to get the load started.

Enough said

Hunter

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Hunter, you're right, but keep in mind that the gas motor is geared deeper, which mitigates the problem of getting the load moving and takes away much of our advantage there.

ZF,
Why don

I think the difference here is where you want your rpm to be.

Sure the gasser will out-pull the diesel, but geared down to 6000 rpm isn't the most desirable place to be. I think that's the difference. Higher torque brings your rpms down to reasonable levels.

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Hoot: Exactly!

Mark: My posts weren't intentionally condescending, I genuinely thought you still weren't getting it because of some grossly inaccurate things you said. So I was trying to simplify as much as I could and present more ways of looking at it. I'm thrilled to see that you've got it and if your anger at me is the price, well, okay. But you've come a long way on this subject, from where you were to where you are. I think that's great.

I don't think you're a moron at all. Not even really sure what the word means. You strike me as someone who overcame his preconceived notion about a tricky subject. To me that means you're smart.

I think if you examine my posts a little more carefully you'll see that I wasn't trying to say all 300hp engines are the same, only that they're all capable of putting the same torque to the rear wheels for a given ground speed. Nothing else. There are probably a few places where I neglected to specifically qualify "horsepower is what matters" by saying "for performance", but it was within the context of the discussion.

Of course the motors aren't the same. If they were the same, who in the world would pay an extra $5000 for a diesel? Not me!

They run at different rpms, they use different fuels at different rates, they have different power curves, etc.

Anyhoo, it's been fun, and despite your anger, I'm gratified that you've got it.

I think this horse is dead. Hopefully bbugg has got it by now, too. If not, let's just agree to disagree, this is getting ridiculous.

[This message has been edited by ZFMax (edited 01-18-2002).]

WAIT A MINUTE!!!

I hate to beat a dead horse, but, last night I thought I was beginning to understand all of this. But now I think I'm back to my original conclusion.

Okay, so let's say we have two different motors, both are 300HP, however, one makes 300HP at a low rpm, and the other makes 300HP at a high rpm. You could say the first engine is the Duramax, and the other is the 6.0L gasser.

Now, granted, we could gear the 6.0L truck to pull the same as the Duramax. However, we would have to be running high in the rpm band all the time where the HP is, correct?

The Duramax can pull the same load at a much lower rpm where the power is. Both trucks putting the same torque to the rear wheels.

Now, if the 6.0L was put in the Duramax truck, it wouldn't pull worth a darn, it would be geared wrong. So, using math, we could apply some sort of gear reduction to the 6.0L to make it pull like it had the Duramax.

But, who wants to drive a truck with low gears and a screaming engine? That's why the Duramax is a superior engine to the 6.0L, because it is more effecient, because it makes its power at a lower rpm. Unless I missed something, isn't this a direct result of the high torque of the Duramax? Isn't the high-torque, low rpm motor going to do a better job than the low-torque, high rmp 6.0L?

When this discussion started, it basically was "Torque is what is important" vs. "HP is what is important." Well, one without the other is meaningless.

Maybe to somebody who understands physics, they can deduce what they need from just the HP figure. But, what most of us want to know, is, at what rpm is that HP made?

If we're looking at several trucks to buy, and all of them are 300HP, we want to buy the truck that makes that 300HP at the lowest rpm, to make life easier. Wouldn't the easiest way to figure out which did so would be to look at the torque figures? Wouldn't the highest-torque 300HP be able to produce the power at a lower rpm, thus being able to give us the best effeciency and drive-ability?

The example of the V-8 gasser in the semi truck is a little unfair, I think. Sure, we could apply enough gear reduction to make it work, but it would do an awfully lousy job, the engine wouldn't last long, etc. etc., so the high-torque engine would be a better choice.

Basically, if I thought this thru right, both arguments are correct.

If one truck has a lower HP than another, it's not gonna make it to the top of the hill first.

If both trucks have the same HP and are geared properly, they'll both get to the top of the hill together. However, if one has more torque than the other, it can do a better job in the long run, and be much more pleasant to operate, though still not beating the other.

These are just the thoughts floating around in my head! http://www.62-65-dieselpage.com/ubb/biggrin.gif

I posted my last remarks before ZF's last post, so now I see that I'm leaning towards being right!!

You're right, clearly you've got it!

There's only one little line I have a tiny bit of an issue with, but the rest of your text clearly shows you understand.

When this discussion started, it basically was "Torque is what is important" vs. "HP is what is important."

You gotta hang "for performance" onto that! I'm pretty sure I did that in every place I said HP is what's important.

Well, one without the other is meaningless.

For understanding how much torque can be put to the rear wheels for a given ground speed (i.e. performance), only the horsepower is meaningful. It can be made of ANY combination of torque and rpm at the engine, and we use gear reduction to convert it to the combination we want at the rear wheels.

It would be more accurate to say "torque without rpm is meaningless", at least with respect to performance. And the two of them together is called horsepower.

But of course, as you clearly point out, for considerations other than the performance, torque is VERY meaningful, because it tells us the engine makes it's power at low rpm. And that's a WONDERFUL thing!!! It makes the truck worth at least $5000 more in my opinion!

Posts like yours are very gratifying to me, thanks.

[This message has been edited by ZFMax (edited 01-18-2002).]

There is no right or wrong. What ZFMax has proven is that given one constant (5252) you can manipulate the other vaiables to get the desired ouput (Not trying to imply deception here).
He simply used the formula to say that given a constant HP rating (300) and the constant of 5252 you can prove that the same horsepower can get different amounts of torque to the ground through gear reduction. What I tried to prove (and didn't do a good job)was that "Horsepower is not what is important"
It is the combination of the two at a given RPM and how their curves relate to each other throughout the RPM range (they always cross @ 5252).

I feel that what is important is HP and torque (both) down low in the RPM curve. Diesel engines can due this. A gas engine can do it (through gear reduction), but at a higher RPM which normally comes at a cost(efficiency and longevity).

There is not right answer, because it isn't one over the other. There is a relationship between the two, and they can't be seperated and you can't take one over the other.



[This message has been edited by MABurns (edited 01-18-2002).]

"A steam engine turn fairly slowly. 600 rpm is not an unusual or "slow" turning speed. Don

Doggone it, Mark, you're saying inaccurate thigs again. I really thought you had it. I'm NOT trying to belittle you here, but the statements are wrong.

What ZFMax has proven is that given one constant (5252) you can manipulate the other vaiables to get the desired ouput (Not trying to imply deception here).

Keep in mind that I didn't just make up the 5252. It's the definition of a horsepower. 1 foot/lb of torque at 5252rpm is 1 horsepower. Always. For all motors.

He simply used the formula to say that given a constant HP rating (300) and the constant of 5252 you can prove that the
same horsepower can get different amounts of torque to the ground through gear reduction.

No, actually I've been pretty focused on showing that at a constant HP rating, you get the same torque to the ground for a given ground speed, regardless of the torque rating of the motor.

What I tried to prove (and didn't do a good job)was that "Horsepower is not what is important" It is the combination of the two at a given RPM and how their curves relate to each other throughout the RPM range (they
always cross @ 5252).

How the horsepower and torque curves relate to each other throughout the rpm range does not vary from one motor to the next. It's always exactly the same. There's a formula that describes the relationship. hp = (torque * rpm) / 5252. Always. For all motors.

I feel that what is important is HP and torque (both) down low in the RPM curve.

Saying both the torque and the horsepower matter at some certain rpm makes no sense. There's a fixed relationship between those two numbers at any given rpm. If two motors have the same torque at the same rpm, they have the same horsepower at that rpm. Always.

You've made a point of saying that torque and hp always cross at 5252rpm, which is correct. But what about 1500rpm? Did you know that at 1500rpm, torque is always 3.5 times the horsepower? And at 2000rpm, torque is always 2.6 times the horsepower? And at 10,000rpm, torque is always 52.5% of the horsepower?

There's a very fixed relationship between torque, rpm, and horsepower. This relationship does not vary with different kinds of motors.

[This message has been edited by ZFMax (edited 01-18-2002).]

ZFMAX, Your #48 post says it all as far as I am concerned. My DMAX will pull the wheels of my 5 th. wheel from 1500 to 3000 RPM. At 3000 no one but another DMAX is going to keep up on the hills. Later! Lone Eagle

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2002 2500 HD EC,SB,Back Country Bars, B&W Hitch, Pace-Edwards Roll Top Cover.

OK,
I have to enter into this in my uneducated redneck mechanic way.
My question is this, Back in the mid 70's I had a D***e W200, a 3/4 ton 4wd truck. It ran fine, accelerated well, but wouldn't pull the hat off of your head. I replaced the cam with an "RV" cam and took the 4bbl and manifold off and on with a 2 bbl. I lost horsepower, the truck wasn't nearly as quick in acceleration, but it would pull better. I had always assumed that I had decreased horsepower and increased torque. At least that is what I had been told by all the other rednecks.
What you are telling me I believe is that I lost peak power period, just simply moved the peak power point down to a lower RPM. and the fact that the truck would pull better was actually an illusion?
Lastly if TQ. is irrelevant, why do all of the manufactures quote it?

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Q: What would've happened if you had geared it down instead of moving the engine's powerband down?

A: For a given ground speed, you would've had MORE torque to the rear wheels.


FACT 1: for understanding how much torque you can put to the rear wheels at any given speed, how much you can multiply the engine's torque by (with gearing) is just as important as how much torque it produces.

FACT 2: For any given target vehicle speed, the more rpm the engine has, the more you can gear it down and the more you'll multiply it's torque.

FACT 3: For evaluating performance, engine torque alone only tells half the story. What good is it to know how much torque you have if you don't know how much you can multiply it by, for a given target speed?

FACT 4: For evaluating performance, engine rpm alone only tells half the story. For a given target speed, what good is it to know how much you can multiply the torque by if you don't know the torque?

FACT 5: Because it's the product of BOTH the torque produced AND the amount we can multiply it by for a given target speed (aka rpm), horsepower alone fully describes the performance capability of a motor. Torque and rpm are both built into the number.

FACT 6: For any given ground speed, a truck will always pull hardest if it's geared to put it's motor at it's horsepower peak, because that's the point where the combination of torque and rpm are the highest.


Anyone disagree with any of those? Well, they're all easily provable facts.

If someone's genuinely interested in understanding these facts, I'll continue trying to help. But I have to ask you to let go of preconceived notions, and approach it logically, not emotionally. Otherwise we'll just be wasting our time.

[This message has been edited by ZFMax (edited 01-19-2002).]