What most people seems to lose sight of is the origin of these rating numbers. When someone say their engine is 400 hp and 700 ft-lbs of torque, what that REALLY means is it's making 400 PEAK hp at XXXX rpm and 700 PEAK ft-lbs of torque at YYYY rpm. XXXX rarely equals YYYY (in other words, peak hp is almost always obtained at a different rpm than peak torque).

 

The important difference is that in Physics, it's only work if 1) it moves, and 2) it moves in the direction of the force applied to it. For example, if you climb stair to a height of 100ft, you only did 100ft times your weight of work-- no matter if you walked 20yds or 20 miles in the process.



Hohn,

It sounds like your example of work would only apply if you've added potential energy to an object by moving it to a higher elevation. Whenever something moves it is responding to a force acting on it, or many forces acting on it that average out to move the object in a given direction. How could something move in a direction that was different than a multiple of the forces acting on it? In your example, if a car was driven around the block and parked in the exact same place no work would have been done. But that does not take into account all the frictional losses required to move it around the block. So work was done. If an object is gliding along out in space no work is being done. If you climb a stair, as in your example, you have added some potential energy to yourself because you have moved to a higher elevation. You also made some heat, displaced some air, flexed some cloth and leather, etc, etc,.
I'm trying to factor in the energy used to overcome all the losses along the way. All the losses required to keep, say, a truck rolling along a 60MPH on level ground. It may not fit neatly into the definition of work but it does take horsepower. What would be a better way to describe that scenario in terms of work? Just horsepower over time?

Wetspirit

 

That is why it's so important to do all of your runs in a 1:1 gear ratio. Most people will use 6th gear because it loads the engine which raises the boost which is what gives us HP on these trucks on an inertia dyno. Using 5th though is the best gear and then tricking the dyno into raising boost levels is the best way to make a run on an intertia chassis dyno.

 

You are correct.HP is for acceleration. assuming you maintain the same torque and increase Hp all you get is faster acceleration,but no more pulling power.,otherwise why would the mid-duty 5.9 be rated at 235 Hp with a much higher GVWR rating than the 325 HP 5.9,simply because the mid duty trucks are not interested in how fast they can get from stoplight to stoplight.

 

.....did you want the explanation of General Relativity or Special Relativity? Completely different derivations are required...
Pretty cool stuff, too........

 

arghhhh what is this thread? my eyes hurt

 

By definition, horse power and torque will always be equal at 5250 rpm.

 

Sorry!! make that 5252 rpm ... and RC51 already mentioned that in the post which started this thread.

 

Great discussion. I agree with the chicken or the egg statement.

But what I have figured out after years of exhaustive research is that if you can accelerate the drums on my dyno from ~75 mph to ~140 mph in ~2.8 seconds, you generate enough torgue and hp that everyone is behind you on the drag strip.

 

that's why an 18 wheeler can pull HUGE loads with only 400-500 horsepower... because they have unreal torque and a zillion gears...

personally, I'd rather not have a 500rpm powerband and have to shift 18 times in my little 3/4 or 1 ton...

and if you make 500-600rwhp with one of our 5.9's, the torque numbers are going to be insane anyway, so I don't see why people freak out over them... 600-700ftlbs is PLENTY enough to pull a good load w/ these trucks... they'll do 800-1000 without batting an eye.

all I care about is horsepower, the torque will be there with these engines.

 

"all I care about is horsepower, the torque will be there with these engines."

But that's not the point of the discussion, as your example is like saying "all I want is the faster strongman, so strength doesn't matter when it comes to moving loads." If all you care about is horsepower then get a toy truck with a 3L V6 with a turbo, put a bigger axle on it, and run it at high rpm, as it'll be cheaper than truck with a 1000 lb turbodiesel :^)

The big diesels have a narrow rpm range and lots of gears, but a lower torque engine with more hp would need even more gears to move larger loads.

 

I think you're misunderstanding me. You don't have to add potential energy to an object to do work on it. But most times this will be the case.

If I exert a Force of 2000lbs on a car for 100 ft, then it will end up with kinetic energy IF the 2000lbs was sufficient to move the car.

If you drove a car around the block and parked it in the same place, you WILL have done work upon it. But you have to look at it from the perspective of FOUR DIFFERENT EVENTS: travelling down each respective leg of the block.

We have to keep in mind the definition of work-- force X distance.


Say you lift a 50# box off the floor 3 feet, then carry it 100 feet across the room. How much work is done?

Work requires both a force AND a distance. So, first we have the 50# force (actually more, because you can't lift 50# with a 50# force; is must be >50#) that you exerted for 3 ft to lift the box. So even though it had to be more than 50#, let's just call it 50.

You did 150 lb-ft of work to lift the box. Notice that it makes no difference if you took 3 seconds or 30 minutes to lift the box-- the same amount of work was done.

How much work was performed carrying the box across the room? NONE!! ZERO ZIP NADA! Why?

Because in the direction of travel, the force acting on the box was ZERO. Essentially, the box is floating in the air and has no forces of friction acting upon it. Since there are no forces to restrain the movement of the box across the room, then any force > zero will move the box laterally across the room. Since the force required to move the box approaches zero, we call it zero. Therefore, W=FxD also becomes ZERO as well.

Now, obviously it's more tiring to carry a box 100ft across a large room than to carry that same box 10ft. But in physics, this isn't WORK, this is ENERGY. ENERGY is the ability to sustain workload.


Anyway, if you were to DRAG the box across the floor, the would be different. You would have a force a friction to overcome (defined as the coefficient of friction times the normal force (in this case the box's weight)). In this case, you would be doing WORK on the box without adding potential or kinetic energy to it.

In all scenarios with the box, you're only doing work if the box actually moves.

You can sort of see the same phenomenon in the rotational world. If you look at the end of an engine's spinning output shaft (crank flange), you can see how the exact centerpoint of it never actually moves. In other words, imagine progressively smaller orbits around the center until you are actually AT the center and aren't moving.

At the exact centerpoint of the output shaft, there is NO TORQUE, and there's NO MOTION as well.

So, from a certain point of view, there's no TORQUE without movement, either-- JUST like the linear example where there's no work performed if the object doesn't move.

JH

 

Actually, a higher HP engine with lower tq would need FEWER gears, but the RATIOS would overall have to be lower.

Let's say we have two opposite extremes, a 700hp mining truck (big Cat diesel), and a 700hp NASCAR car. The mining truck makes peak hp @ 1500rpm. The NASCAR needs 8000rpm for peak HP.

Since they have identical HP, they are capable of moving the SAME LOADS, at least on paper-- given the fact that you can create TQ from HP by sacrificing time.

Regardless, though-- the number of gears required in the gearbox is a function of the RPM range where the engine makes useable power. The mining truck only makes useable power from say 800- 1500 rpm. That's a powerband of 700rpm.

The NASCAR would make useful power from 6000 to 8000 rpm, and require FEWER gears.

The NUMBER of gears and the RATIO of gears are two entirely different concepts. A high-hp but low TQ engine needs FEWER numbers of gears, but they must be at a much lower RATIO to get the execute the time-for-torque trade.

So, to recap Hohn's Fundamentals:
1) Engines make tq, not HP
2) The number of gears required is a function of the width of the powerband in terms of RPM.

 

"Regardless, though-- the number of gears required in the gearbox is a function of the RPM range where the engine makes useable power. The mining truck only makes useable power from say 800- 1500 rpm. That's a powerband of 700rpm.

The NASCAR would make useful power from 6000 to 8000 rpm, and require FEWER gears."

Which is my point, are you going to rev the nascar engine to 7 grand and pop or slip the the clutch ? Where low torque gas engines are used in place of higher torque diesels one sees either the same number of gears but a lower top speed due to the gearing, or the same top speed but more gears to get it there. Torque is handy when it comes to moving large loads, as even though the peak hp for the high torque engine may be lower than a high hp low torque enhine, the hp at the lower rpm off of idle is typically higher.

 

What gives us "power"? Now, I'm not talking about the formula that we all know of hp= tq*rpm/5252

Instead, I'm referring to the PERCEPTION of power. What makes something FEEL fast?

Short and sweet, it's RATE OF FORCE APPLICATION. Yes, that's the textbook defintion as well, but let me explain.

The textbook defines the rate of force application is *frequency*, or in our case, engine RPM.
But this doesn't explain vehicle acceleration. For example, if a rev an engine to 5252rpm, I always have 1lb-ft of tq for every HP at the point. But does this correlate to vehicle acceleration? NOT AT ALL!!
After all, you can steadily cruise at 5252 rpm and experience no vehicle accleration at all.

So we need a better way to think of this "rate" of force application. ENGINE RPM IS NOT IT, but it *is* related as we'll see.

The problem is that we are dealing with DYNAMIC circumstances that the definition of HP (the equation) does NOT address. One way to illustrate this is with a flywheel. If you bolted a very heavy flywheel (say, 100#), to your engine, you would note a DEFINITE change in acceleration. The engine would be a lot slower to accelerate, and so would the vehicle as a result.

But has the engine's output changed at all? NOT ONE BIT. It still makes the same TQ at the same rpm at every point in its rev range. It just takes more time to transition from one rpm to another, but tq output at every rpm is the same, given constant rpm

AHA! Behold the fundamental flaw of "the equation". It doesn't address a CHANGE in rpm-- just rpm.

Now, the above mentioned flywheel example WILL show up on different dynos. On a dyno where the engine has to accelerate the f'wheel as well as the tires, you will see less hp. This makes sense because some of the power went to accelerate the f'wheel and therefore not to the rear wheels.

Conversely, a load dyno will read higher than normal HP because the energy stored in the flywheel tricks the dyno into "seeing" more power TQ than the engine makes. The key difference between the two dynos being whether it measures the engine BEFORE or AFTER it's already spun up the flywheel and its rotating mass (rotational inertia).

So the PERCEPTION of power is different than power. The formula gives you power. What gives you the PERCEPTION?? Physics also gives us the answer for the perception, if we know where to look.

First is position. The rate of change in position is called velocity. It's simply the change in position over time. Units are commonly miles per hour.

Take another derivative and you get accleration. In other words, how fast can you change your velocity? So we add another time element-- miles per hour PER HOUR. Or, we just call it miles per hour squared (mi/hr^2).
Or, you'll often see ft/sec^2.

The last one is what gives us our perception of power. It does this by creating a FORCE that we experience. Since F=MA and our bodies have mass, we experience a Force when we undergo Acceleration. The stronger the force we feel, the more "powerful" we say a vehicle is, because it is accelerating us at a faster rate.

So, we know that the speed that the vehicle will accelerate with (and thus, the force we will feel) is DIRECTLY related to the rate at which the wheels will accelerate (albeit under load). The rate at which the wheels will accelerate is a fuction of the FORCE applied to them (TORQUE!!), F=MA and the wheels have mass, so accelerating them requires force. So it is with each component working our way up to the engine. You can increase accleration with NO CHANGE in engine output by lightening the wheels, tires, gears, axles, driveshaft, transmission components, etc.

What we come to is that acceleration (and hence, perceived power) is a function how quickly the engine can gain RPM when under load. That's it. Nothing else matters. Given two engines, the one that accelerates a known load from pt A to pt B the quickest will be the better "performing" engine EVERY SINGLE TIME.

That's why DynoJets are such great tools. Because even if they "power" they measure isn't accurate, they are superb at measuring "rev gain under load", which is ALL THAT MATTERS. Of course, the trick is using the proper load. Almost all DJs are setup with rollers that are too light.


So you see, there's a lot more to it than just "power" or "tq" because we're dealing with dynamic situations. On paper, my truck should accelerate the fastest in 1st and 2nd gears, because it's putting the most tq to the ground.

But there's so much rotational inertia in those lower gears that my truck actually accelerates harder in THIRD gear than in 1st or 2nd. The "formula" wouldn't predict this, even though it's true.

We have to shift our thinking away from the formula and back to the derivation-- accleration is change in velocity over time. The greater the change for a given time OR the less the time for a given change, then the more "power" you will feel!

H