BigBake

[quote]Originally posted by 383LQ4SS:
<strong>Only enough horsepower is applied to the shaft to maintain 100% RPM. If you watched the ITT or T5 or whatever your temp gauge was you would see it rise as Tq increased. Also watch fuel flow...it would also rise as temp increased. This is HP increase. You cannot apply 5000 hp to a prop that has no load and not expect it to overspeed. Ive been working on various aircraft and helos for 16 years and have lots of experiance in this dept.

As far as the Tq wrench...I didnt say no energy was expended...I said no work was done, at least in a practical matter. Two different things.</strong><hr></blockquote>

I do not doubt your expertise, but in the same manner then, you would also know that there is 5 forces acting on prop at all times. This keeps things in check. You are correct that you do see rise in T5 and fuel flow, when you apply torque, but you also know that engine is allowed 7% operational variance due to load. So the engine can actually see higher than 100% effiency.
Well you did say that you could apply 300 ft lbs of torque to a torque wrench all day long and no works was done. I just said do it for 5 hours and tell me how you feel!
<img src="gr_grin.gif" border="0">
While ratcheting down on a bolt the torque wrench shows absolutley no torque being applied until the bolt starts meeting resistance. Once you meet resistance you start applying torque, this is measured work. A bolt torqued to 300 ft lbs already, and you want to increase it to 305 ft lbs, still requires the energy to get to that 300 and then requires more to exceed that, to get the 305 ft lbs desired. Just like a car sitting still, it has weight and drag resisting it's movement. It requires force applied to get movement.
This thread has been very educational so do not think I am taking offense to any of it.

BigBake

Well ChrisB I am talking in theory here, but you are correct 5000 horsepower is applied, but I forgot to mention you are also applying 800 ft lbs of torque also, because of air resistance, and the 4 other forces acting on a prop. So I should have not said nothing was being applied. My mistake!
As far as your theory on suspended weight. It is not the same thing. Now take that weight Lets say 500 pds, and put it on the end of a stick and hold it there. Now place a saw horse in the middle, now you hold on to the stick for an hour or 2. Eventually you will give out, and during this process you will give less and less resistance to the weight and eventually you will lose out to the force of torque. The weight is horsepower, the potential, the torque is applied force to you.

383LQ4SS

iTrader: (33)

Bake that doesnt make sense. Dont put a person in place to hold the weight. Thats completely different. Use you 500lb weight in a stick with a sawhorce fulcrum...but instead of a person holding it....tie it off to an anchor on the floor. No work will be done....no HP is present. Unless you untied it and it beagan to fall. X amount of work went to get the weight into position. Once tied off...there is no HP or energy or whatever being used. Just potentail energy. Once untied and allowed to drop the energy that is exerted will be proportional to the energy it took to get the weight into place minus friction in the pulleys and other small factors.

Colonel

No movement = no HP. You can apply that TQ to a wrench for 15 million years if you want and if nothing moves, you've still done NO WORK. No different than a rock sitting on the edge of a cliff. Until it moves it is doing no work.

Same goes for holding a bowling ball perfectly still. You may get tired but unless you actually move the ball no HP is taking place no matter how tired you get. What is makin' your arms tired? Is it HP from the ball? No, not at all. It's the TQ that gravity is applying to your arms. What happens when you finally give out and drop the ball? HP! (But you didn't make it. Gravity did.)

Scalpel

iTrader: (4)

I think some people need just a simple list of definitions to set them in the right frame of mind. I'm not trying to insult anyone's intelligence but I feel that by doing this we can get most people on the same page.

Force is the "push" required to have an object of mass, m, accelerate, thus:
Force = mass/acceleration kg*m/s^2
Torque equals force multiplied by distance in the form of ft*lb
Power is a measure of: Work/time, and in our case of car discussion Horsepower = 550 lb*ft/sec

Thus we have the equation for HP
Horsepower = (Torque * RPM)/5252

In this equation, you can't have one variable without the other. Thus without HP there is no TQ, without TQ there is no HP, (assuming we are at some given RPM with all these variables <img src="gr_grin.gif" border="0"> ) etc, all these variables are needed for this equation to hold true. If you eliminated, say TQ, so substitute a ZERO (0) in for TQ, then you have ZERO HP. It's that simple <img src="gr_stretch.gif" border="0"> TQ and HP aren't contradictory terms yet they are both necessary evils that go hand in hand <img src="gr_stretch.gif" border="0">

I think a basic understanding of these definitions and this equation is needed before others make educated extrapolations of any kind. Many have hit it all square on the head, i.e. Colonel and ChrisB to name a few.

Ok, now on with the battle (which I am siding with ChrisB and Colonel) <img src="gr_grin.gif" border="0">

[ February 03, 2002: Message edited by: NastyC5 ]</p>

Colonel

Let's go back and review this link I posted earlier. It does a pretty good job of explaining a few things... <img src="gr_stretch.gif" border="0">

http://www.vettenet.org/torquehp.html

383LQ4SS

iTrader: (33)

Hey Colonel....I finally read that link. Its great, but one thing. It sais the greatest acceleration occurs at Tq peak. I would think it would be Hp peak. Or did I misread what he said?

Scalpel

iTrader: (4)

[quote]Originally posted by 383LQ4SS:
<strong>Hey Colonel....I finally read that link. Its great, but one thing. It sais the greatest acceleration occurs at Tq peak. I would think it would be Hp peak. Or did I misread what he said?</strong><hr></blockquote>

What you read is correct. Acceleration is equal to Force/Mass, a = F/M so in our case, Torque is our force. Horsepower is simply the amount of work (Force X distance) per unit time. Therefore the maximum torque would make tne numerator (Force) very large while mass stayed the same, therfore you get a great acceleration as your force (torque) increases. Please note, this was a very general application <img src="gr_stretch.gif" border="0">

[ February 03, 2002: Message edited by: NastyC5 ]

[ February 03, 2002: Message edited by: NastyC5 ]</p>

ChrisB

[quote]Originally posted by 383LQ4SS:
<strong>Hey Colonel....I finally read that link. Its great, but one thing. It sais the greatest acceleration occurs at Tq peak. I would think it would be Hp peak. Or did I misread what he said?</strong><hr></blockquote>


Nope, that is correct.

http://www.speeddemonmotorsports.com...torquevsHP.htm

is something I typed out which I thought summed it up very well also <img src="gr_stretch.gif" border="0">

Anyway, look at everything from the perspective of "delivered torque" or torque at the rear wheels.


This is going to be a product of flywheel torque and drivetrain multiplication.

If we fix ourselves in one gear than torque at the rear wheels follows the curve of flywheel torque. So max acceleration occurs at peak torque.

but we are not fixed in one gear, we have a few to choose form. So if you have the choice of max acceleration in second gear @ the torque peak, or downshifting to first and being at the hp peak - well, rear wheel torque will be greater in first gear at the hp peak than second gear at the torque peak.

The is because torque delivered is the product we are maximizing, and that is flywheel torque * gear multiplication

even if we reduce flywheel torque we are increasing gear multiplication to such an extent that the overall acceleration is greater.

Now obvioulsy there is a balance here - you can't just keep increasing the gears. So how do we know when we have hit the perfect balance between gear multiplication and flywheel torque?

HP is proportional to this property. gear multiplication is determined at any point by what rpm youc an turn - so it is analgous to velocity.

Power = force * velocity

or HP = torque * rpm * (K)

(K = constant, 1/5250 here).

So as HP goes up so does the potential product of torque * gear multiplication.


Chris Bennight

ramairJP

iTrader: (3)

[quote] Torque doesn't get you down the track because torque does no work. Horsepower does the work. <hr></blockquote>

Actually Patrick, that is a fundamentally incorrect statement. Torque by definition is work... it's a force time a distance (foot-pounds). Horsepower is just that, power, which is work times time. In equation form:

work = torque = force x distance
horsepower = power = work x time

The best analogy I can think of to compare torque and horsepower is a shovel full of dirt. The size of the shovel would represent how much torque you have, and how fast you shoveled would represent your horsepower. So, you can have a small shovel and work your *** off with a lot of scoops to produce a certain ammount of work completed, or you have have a big shovel and go nice and slow to get the same ammount of work done. Of course, the 10 billion foot-pounds with no hp that colonel talks about would be the Caterpillar trackhoe with the operator on break... lol.

When talking about torque concerning acceleration, all you can do with torque is find the instantaneous acceleration of a given mass. For example, if you know that torque rating of a motor at a given rpm (say 300 lb-ft), the gear ratios in the tranny and rear end, and the diameter of the tires, then you can calculate the acceleration at a given instant. In this case, the best instantaneous acceleration would be at peak torque, but that's only for an instant. When you add the time element, however, you get horsepower. This is one of the best reasons that when you're starting a race from a roll you want to be as close to peak torque as possible so you can get the best initial push. After that... it's allll power. IMO, in drag racing the guy that comes with the combo of the best tracting and best power to weight ration won't ever get beat.

John

[ February 03, 2002: Message edited by: ramairJP ]</p>

BigBake

[quote]Originally posted by ramairJP:
<strong>

Actually Patrick, that is a fundamentally incorrect statement. Torque by definition is work... it's a force time a distance (foot-pounds). Horsepower is just that, power, which is work times time. In equation form:

work = torque = force x distance
horsepower = power = work x time


John

[ February 03, 2002: Message edited by: ramairJP ]</strong><hr></blockquote>

Thank you, you have worded it well. This is exactly what I have been trying to say. You can tell me all you want about the torque wrench theory, you still have to exert 300 ft lbs of pressure to the bolt to ensure that it is torqued to 300 ft lbs. It does not have to move, because resistance is fighting you.
Just like the weight and the sawhorse. Move the sawhorse closer to the wieght and guess what you have increased the effective torque that you can apply to the weight. Just like gear ratios.

BigBake

By the way, have any of you actually used a dial or digital torque wrench before? I mean you do know that when you hit 300 ft lbs on the dail you have to hold it there to keep maintaining 300 ft pds of torque because once you remove force it goes back to 0. Maybe the only reason I know this is because when we torque props on at 3000 ft lbs of force you have to make the person torqueing it hold it for verification to the quality assurance rep. Believe me this guy does not like holding it for very long. Yes we have to use a gear reducer to get to that high of a torque. It would be an awfully long breaker bar if we did not.

Scalpel

iTrader: (4)

[quote] Torque by definition is work... it's a force time a distance (foot-pounds). Horsepower is just that, power, which is work times time. In equation form:

work = torque = force x distance
<hr></blockquote>

Torque is NOT WORK. Torque is a rotational Force.
Work is defined as Force x distance.
Torque is the cross product of Force and your Lever arm (d, or distance from point of application of force).
So we have Tq = R*F sine (theta)
R = lenght of lever arm from point of application of force
F = Force applied (newtons, etc)
Sine (theta) is the angle of the application of the force, and we can assume the angle to be 90 degrees so sine of 90 degrees is equal to 1.
Thus, Tq = R*F
There is no distance moved in this equation. You only have the radius of the lever arm and the force applied. Therefore, no work is done although you may be sweating like a hog, you did NO work.

You have to move something a certain distance for ANY work to be done. If try to lift a chair bolted to the floor you do no physical work, except burn some calories. Now if you lift a 10lb chair 1 foot off the ground, you have done 10 ft*lb's of work.

This is a great site to clear a lot of this up, torque wise:
http://www.howstuffworks.com/fpte.htm
http://www.howstuffworks.com/framed....que/index.html

[ February 03, 2002: Message edited by: NastyC5 ]

[ February 03, 2002: Message edited by: NastyC5 ]</p>

ChrisB

[quote]Originally posted by BigBake:
[QB]Thank you, you have worded it well. This is exactly what I have been trying to say. You can tell me all you want about the torque wrench theory, you still have to exert 300 ft lbs of pressure to the bolt to ensure that it is torqued to 300 ft lbs.QB]<hr></blockquote>


But that 300 ft-lbs of torque doesn't have to come all from flywheel torque. Horsepower allows us to use gearing to take advantage of the greater rate at which torque can be produced at a higher hp level.

The engine may only have 150ft-lbs of torque - not enough for that 300ft-lb breakpoint you give us above. But if I take that 150ft-lbs of torque and multiply it by a 2:1 total drive ratio then I already have that 300ft-lbs of torque I need, yet at the flywheel only 150ft-lbs were made.

If you want to look at torque you have to look at the product of flywheel torque * gear multiplication. This is the connection you are missing.


If torque is what matters tell me what the potential acceleration would be for a car that made an average of 300ft-lbs of torque on it's pass down the 1/4. Kinda difficult.

Now try it with an average of 300hp down the 1/4. Much easier?


Chris

Scalpel

iTrader: (4)

[quote]you still have to exert 300 ft lbs of pressure to the bolt to ensure that it is torqued to 300 ft lbs. It does not have to move, because resistance is fighting you.
Just like the weight and the sawhorse. Move the sawhorse closer to the wieght and guess what you have increased the effective torque that you can apply to the weight. Just like gear ratios.[/QB]<hr></blockquote>

When you say you have to apply 300 lb-ft of pressure, you mean to say force. Pressure and force are two entirely different concepts. Pressure is Force/Area and Force is Mass*Acceleration or in our case of the force being Torque, it is Tq = R*Fsine(theta)
I know this is nit-picky but if terms are thrown around and become synonymous with the wrong topics then everyone starts getting confues. I.E. Referencing torque being synonymous with work when they are two fundamentally different concepts.
Anyway, even though you may be applying 300 lb-ft of torque on the torque wrench to hold the position your are still doing NO work in the physics sense. Yes, you're muscles are doing work via their muschle fibers' actin and myosin heads attaching and reattaching and causing you to sweat, you're still not doing physical work in the Newtonian world.
I'm not trying to flame anyone or insult anyone, I just feel that people need to be accurate with their terminology or else it all spirals out of control.

<img src="graemlins/gr_cheers.gif" border="0" alt="[cheers]" /> <img src="graemlins/gr_cheers.gif" border="0" alt="[cheers]" />

[ February 03, 2002: Message edited by: NastyC5 ]</p>

BigBake

[quote]Originally posted by ChrisB:
<strong> The engine may only have 150ft-lbs of torque - not enough for that 300ft-lb breakpoint you give us above. But if I take that 150ft-lbs of torque and multiply it by a 2:1 total drive ratio then I already have that 300ft-lbs of torque I need, yet at the flywheel only 150ft-lbs were made. Chris</strong><hr></blockquote>


Unfortunately when we throw around horsepower and torque figures, we are talking about dyno numbers which are achieved at 1:1 drive ratio. So the gearing goes out the door. In order for you to figure out horsepower you have to know the torque, which is what the dyno reads.
Maybe this is getting you guys confused. I have seen it posted a few times in this thread that you can have 1,000,000 pds of torque but the car is not going to move. Well this is a very false statement. To have torque you have rotational force applied so you are going somewhere unless the vehicle weighs 1 trillion tons. If it weighs 3,400 pds you are going somewhere very fast, if you could ever get traction.

ChrisB

[quote]Originally posted by BigBake:
<strong>Unfortunately when we throw around horsepower and torque figures, we are talking about dyno numbers which are achieved at 1:1 drive ratio.<hr></blockquote></strong>

Actually no, a 1:1 tranny ratio is normally used, but the final drive ratio still depends on the rear end ratio.

But that is irrelevant - that actual valves may change due to greater/lesser inertial and frictional losses, but you are still going to have peak hp, etc. at the same point.


[quote] So the gearing goes out the door. In order for you to figure out horsepower you have to know the torque, which is what the dyno reads. <hr></blockquote>

Hmm, okay, try this for me. Get on a dynojet and leave off the spark plug lead. Tell me if you get only a torque or only a horsepower reading. To cut to the chase you only get a horsepower reading. Sure you could determine torque if you wanted to also, but you don't have to - and a dynojet measures horsepower directly.

[quote]
Maybe this is getting you guys confused. I have seen it posted a few times in this thread that you can have 1,000,000 pds of torque but the car is not going to move.<hr></blockquote>

Please reference where that was posted. Because of the way an engine works you can not make torque but no horsepower - simply because it has to rotate to work.
This doesn't mean torque implies horsepower - I believe the situation you are intending to reference is putting 1,000,000 ft-lbs on a wrench, yet if the wrench doesn't rotate then you have no work done, thus no power. If you disagree with this please show me the mathematics on how you can have work or power without applying a force over any distance.


[quote][b] Well this is a very false statement. To have torque you have rotational force applied so you are going somewhere unless the vehicle weighs 1 trillion tons. If it weighs 3,400 pds you are going somewhere very fast, if you could ever get traction.[/QB]<hr></blockquote>

No, a torque (rotationl force) does NOT mean work has to be done. Look at the example of a wrench on a stuck nut. You are confusing yourself with this analogy because of the way a car works (you have to perform work to generate torque because of the way an engine is setup.


I repeat my challenge, if torque is what matters please tell me what the estimate ET/MPH would be for a car, say 3000lbs, making an average of 300 ft-lbs of torque down the track.

Chris Bennight

BigBake

[quote]Originally posted by ChrisB:
<strong> I repeat my challenge, if torque is what matters please tell me what the estimate ET/MPH would be for a car, say 3000lbs, making an average of 300 ft-lbs of torque down the track.

Chris Bennight</strong><hr></blockquote>


It would be estimated at 13.20@105.12 mph.
Also I went back and checked against my own vehicle and this worked out prety damn close, within 4hp.

[ February 03, 2002: Message edited by: BigBake ]</p>

ChrisB

[quote]Originally posted by BigBake:
<strong>It would be estimated at 13.20@105.12 mph.
Also I went back and checked against my own vehicle and this worked out prety damn close, within 4hp.
[ February 03, 2002: Message edited by: BigBake ]</strong><hr></blockquote>


Please show me the math behind that.

I will give you a hint, it's impossible.

Chris

ChrisB

Okay, look at it like this


Power = Work/Time

work = force * distance

Power = force * distance / time

force = mass * acceleration

Power = mass * acceleration * distance / time

acceleration = distance / time^2

Power = mass * distance^2 / time^3


Time = (mass * distance^2 / power)^1/3


That formula will give you an estimated ET for a given average HP value (make sure all your units cancel). Distance would of course be 1/4 mile.


That's true, correct?

Okay, let's go back to my original example - 300ft-lbs of torque average. Let's say that was at an average of 10500rpm. So that was an average of 600hp.

Now let's say it was at an average of 5250 rpm. That's an average of 300hp.

So you are saying that it doesn't matter if one has 600hp, the other 300hp, since the both put down an average of 300ft-lbs of torque they will both run about a 13.2@105.

Plug the values (600 and 300) in my formula above - you will get radically different ET's - as you would in real life. Yet you are claiming they would both be the same.

Again, please provide me with your mathematical proof.

Chris