mark kibort

thats is why when the inevitable torque vs HP discussion comes up, i bring up the two cars with the same HP and one with half the peak torque as the other.

Then, if that doesnt sink in , i usually bring up the infinitely variable gear box. Now, since you found a different application that sets rpms for economy, and not power all the time, maybe you should look at even closer ratio gear boxes. think if you had instantaneous shifting, and you had a 10 speed. where would the engine RPMs end upafter launch in a drag to say, 130mph. max torque, or max HP. If engine torque was so important, wouldnt you need at least look at it. based on this example, the car would accelerate to a speed matching max hp (or a little past it) this way, the engine would stay near max hp for most of the time of the run through all the gears. NOT MAX TORQUE. REASON, Max hp is the max capability of the engine to do work over time. (work=force times displacement, distance , etc and HP is the rate of doing work.)
Without doing work or doing it very fast nothing happens or it happens slowly. Isnt the point we want to do work very quickly, this way we get to the finish line first. we can do work slowly, and your car will not accelerate very fast, agreed? 500000000ftlbs, over 1/4mile but done over one week, is not much HP and is a very slow ET, RIGHT?? " give me a long enough lever, and i can lift the world " famous quote. meaning torque is meaningless without a time/rate attached to it!
HP takes into account all that you need to see how fast you will be. why do you think most pro racing orgs, use HP to weight as the major qualifier?
torque has to be taken into consideration, as the area under the usable hp curve will be greater if the curve starts higher due to a higher peak torque. But ill tell you first hand, equal hp and weights make for a pretty good race regardless of their shapes.

MK

m21sniper

"no, the 500/500 has a max hp at 5000rpm, this means that max torque is going to be in the 3500 range
the torque peak of the 10,000rpm max hp engine would be in the 7000rpm range"

Let's look at that formula used to determine HP a bit more closely.

Horsepower = Torque*RPM/5252.

So we're taking our engines observed torque peak(aka "full load torque"), multiplying it by the RPM at which it occurs, and dividing that by the constant of 5252 to determine horspower.

For the 500/500, through a little detective work, we discover that the engine's torque peak is 5,250rpm.

500lb-ft x 5250rpm = 2,625,000 divided by 5252= 499.8 HP

Now for the 500/250:

250lb-ft x10250rpm =2,625,000 divided by 5252=499.8 HP

So our 500/500 engine will deliver it's maximum torque at 5250rpm, and the 500/250 will deliver it's maximum torque at 10,250rpm.

Let's do one better and 'invent' a 500hp/750lb-ft engine.
We need to know what RPM will produce the 750lb-ft and still reconcile with 500hp.

It turns out that the RPM speed required to reconcile the problem is 3500.

750lb-ft x 3500rpm = 2,625,000 divided by 5252 = 499.8 HP

For a 500/1000 engine:

1000lb-ft x 2625rpm = 2,625,000 divided by 5252 = 499.8 HP

Only 1 specific rpm speed for any given torque rating will reconcile with the 500hp.

For the 500/250 it's 10,250rpm
For the 500/500 it's 5250rpm
For the 500/750 it's 3,500rpm
For the 500/1000 it's 2,625rpm

HP is constant in all four motors, 499.8....all that we're changing is the torque peak, and the RPM at which that figure MUST be recorded to reconcile with 500hp.

If we change the engine RPM at which the torque peak is delivered, by definition we're changing the HP rating.

Note how all four engines end up at 2,625,000 before the constant of 5252 is calculated.

They have to....or they won't be 499.8hp.

Notice also how the drop in RPMs is completely proportional as the Torque is increased at a constant HP. Likewise, it is completely proportional as Torque is decreased based on a 500hp constant.

When a dyno does it's thing it's measuring Torque at any given speed, and using the above formula to convert the delivered torque into HP.

The dyno doesn't measure HP...it's not an actual force, ie, there's nothing to measure. It measures torque, which is an actual force of nature....and converts it to HP via mathematical equation.

m21sniper

"we all agree that its the torque that is the driving force (no pun intended) However, im looking at the system, you are looking at the engine.
max torque in 3rd gear is not the fastest rate of acceleration, it would be in the prevoius gear at max HP!"

The fastest momentary rate of acceleration in G's will be delivered at peak torque in the lowest available gear(ie, when torque multiplication is at it's greatest), from a standing start.(remember acceleration is a comparative force. The moment of greatest momentary acceleration will almost always be from a complete stop.)

For the two engines we've been discussing, the theoretical moment of greatest acceleration will occur from launch in 1st gear. For the 500/500 that's at 5250rpm in 1st, and for the 500/250 it's 10,250rpm in 1st.

At those speeds, using our earlier discussed gearing, the 500/250 and the 500/500 will both be delivering a theoretical 2000lb-ft of torque x the mechanical torque multiplication of 1st gear(lets call it 2:1).....or 8000lb-ft.

From a rest both vehicles if launched at their optimal RPMs will produce the exact same amount- and the maximum possible amount given their gearing- of acceleration.

Getting back to HP vs Torque again....we've both agreed that it is torque that is the actual physical force at play...not horsepower.
For all practical intents and purposes, there could be no such thing as HP, and nothing would change.

Had that catchy marketing term never been invented, our vehicles would still accelerate at the exact same rate because what's making them move is F(force) created by the fuel air charge explosion that's transferred from the pistons to the con rods, and from the con-rods to the crankshaft where it's converted to torque via the reciprocating crankshaft.
The torque is then applied to the flywheel, then to the driveshaft, then through the pinion, into the ring gear, through the axles, where finally, the torque is applied to the wheels as a twisting force(which is the definition of torque).
Note that at all times post the F to torque conversion this is a reciprical force, just as torque must always be...or it wouldn't be torque- it would be F.

I agree that HP is useful in that it is a convenient way to measure the work that can be done via the delivered torque in a set amount of time, but it's not a real force. It's simply a measurement of a real force....that force being torque.

Most simply stated, when you are using a HP curve to set up gearing, you are in all actuality using torque.

mark kibort

WHERE IN THE HECK DID YOU COME UP WITH THIS??????
AGAIN,, Max torque does not happened in the cross over point using units of ftlbs. if you used NM, the cross over point could be 4k. oz/in, and they would never cross. ITS AN ARBATRARY NUMBER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
want proof, just look at your peak torqu at 5250, its 5000hp! remember i said, max hp was 500hp at 5000rpm?
see my inserts below!>>>>>>>>>>>>>>>>>>>>>>>>>>

Mk

[QUOTE=m21sniper]"no, the 500/500 has a max hp at 5000rpm, this means that max torque is going to be in the 3500 range
the torque peak of the 10,000rpm max hp engine would be in the 7000rpm range"

Let's look at that formula used to determine HP a bit more closely.

Horsepower = Torque*RPM/5252.

So we're taking our engines observed torque peak(aka "full load torque"), multiplying it by the RPM at which it occurs, and dividing that by the constant of 5252 to determine horspower.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>peak torque has NOTHING do do with the cross over point. S4s for example have peak torque at 3-4k and crossover at 5250 when using ftlbs for units of torque.

For the 500/500, through a little detective work, we discover that the engine's torque peak is 5,250rpm.
>>>>>>>>>>>>>>>BZZZZT wrong again

500lb-ft x 5250rpm = 2,625,000 divided by 5252= 499.8 HP
>>>>>>>>>>>>>>>>also wrong. dangerous assumptions

Now for the 500/250:

250lb-ft x10250rpm =2,625,000 divided by 5252=499.8 HP
>>>>>>>>>>>>>>same as above , just doubled for the double max rpm proportions

So our 500/500 engine will deliver it's maximum torque at 5250rpm, and the 500/250 will deliver it's maximum torque at 10,250rpm.

Let's do one better and 'invent' a 500hp/750lb-ft engine.
We need to know what RPM will produce the 750lb-ft and still reconcile with 500hp.

It turns out that the RPM speed required to reconcile the problem is 3500.

750lb-ft x 3500rpm = 2,625,000 divided by 5252 = 499.8 HP

For a 500/1000 engine:

1000lb-ft x 2625rpm = 2,625,000 divided by 5252 = 499.8 HP

Only 1 specific rpm speed for any given torque rating will reconcile with the 500hp.

For the 500/250 it's 10,250rpm
For the 500/500 it's 5250rpm
For the 500/750 it's 3,500rpm
For the 500/1000 it's 2,625rpm

HP is constant in all four motors, 499.8....all that we're changing is the torque peak, and the RPM at which that figure MUST be recorded to reconcile with 500hp.

If we change the engine RPM at which the torque peak is delivered, by definition we're changing the HP rating.
>>>>>>>>>>>>>>>>>>this is true. 500 500 was just an arbatrary set of numbers , again for the point. to be more accurate, we should have used 6000rpm/12,000 as peak HP of 500 and peak torque could be anywhere, more than likely in the 4,000rpm range again. LOOSE THE 5250 rpm cross over, its confusing you!

Note how all four engines end up at 2,625,000 before the constant of 5252 is calculated.

They have to....or they won't be 499.8hp.

Notice also how the drop in RPMs is completely proportional as the Torque is increased at a constant HP. Likewise, it is completely proportional as Torque is decreased based on a 500hp constant.
>>>>>>>>>>>>>>>no , as torque remains constant, HP goes up at a 45degree angle , as torque goes down at a 45degree angle, HP is constant.
( I think this is a good response to the above, but maybe im not reading you corrrectly)

When a dyno does it's thing it's measuring Torque at any given speed, and using the above formula to convert the delivered torque into HP.
>>>>>>>>>>>>>>>no, its not measuring torque at a given speed, if we are talkng about a dynojet 248e, we are talking about rate of change of speed. that gives us the instantaneous acceleration (over a small period of time, so it is still change in velocity) that then gives a torque value and THEN based on the over all speed, we (the computer) multiplies out HP.

The dyno doesn't measure HP...it's not an actual force, ie, there's nothing to measure. It measures torque, which is an actual force of nature....and converts it to HP via mathematical equation
>>>>>>>>>>>>>>>>>>>>>>>yes, but HP gives you the force over a distance (work) over time, =HP. HP gives you the relative torque numbers that will be acting on the wheels at ANY GIVEN SPEED!!!!! this is KEY to the discussion and point!!!!!!!!

mark kibort

Read my inserts very carefully! please.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

>>>>>>>>>>>>>>>yes in the end, its torque that accelerates the vehicle. more accurately put, its the net force! but, agian,im looking at what is put to the wheels and you are stillhung up on the flywheel torque. since folks dont say, my car has 2200ftlbs of torque at 700rpms on the tires, or 60mph and then 500ftlbs at 1500rpm at the tires at 120mph, you can just say, " i have 500hp. you see, torque can be grossly misleading. its just ONE factor of what it takes to accelerate a mass or (car) HP encompasses much more and is more relavant.. again, look no farther than race cars. as a racer, i keep the engine at as close to max HP as possible ALL the time. if you dont, you leave time on the table. max torque is meaningless. you dont need to know about it, and its the thing that is truely marketing. max hp, then tells you you will have more torque to the wheels at any speed over a operational range and time.

think about it!

mk

Randy V

The torque wars continue, and as usual, it's Kibort leading the charge...

bcdavis

I feel sorry for the poor bastard who searches through this thread looking for NOS info...

Lance J

i about to call randy to shut it down lol

Chazz

bc, that would be me. I sure was looking forward to some answers to the original question. Mark, your opinions, though entertaining, are not, by my reckoning, firmly rooted in classical kinematic theory.

Joe F

Chazz said: " Mark, your opinions, though entertaining, are not, by my reckoning, firmly rooted in classical kinematic theory."

That would be kibortian theory to you.

m21sniper

Mark, i didn't write the mathematical formula for determining a HP rating from a known torque rating bro....it is a universally accepted scientific formula. So if you don't like the formula, it's not even me you need to convince....it's the scientific community that codified it.

I've shown you the math for four separate engines, all with 500hp, the only difference being the level of torque in the four engines.

You seem hung up by the fact that i'm using the HP/Torque Cross-over point(5252rpm). It's a known scientific constant, and part of the formula, it must be used if one wishes to determine the correct figures.

If you want to know why it's part of the formula, here is a detailed answer:

If you are interested in knowing why the 5252 must be used, and what it means, read the following:

"Horsepower = Torque X (RPM/5,252)

This formula is the result of combining several formulas into one. First, 1 horsepower is defined as 550 foot-pounds per second. The units of torque are pound-feet. So to get from torque to horsepower, you need the "per second" term. You get that by multiplying the torque by the engine speed. But engine speed is normally referred to in revolutions per minute (RPM). Since we want a "per second," we need to convert RPMs to "something per second." The seconds are easy -- just divide by 60 convert minutes to seconds. Now what we need is a dimensionless unit for revolutions: a radian. A radian is actually a ratio of the length of an arc divided by the length of a radius, so the units of length cancel out and you're left with a dimensionless measure. You can think of a revolution as a measurement of an angle. One revolution is 360 degrees of a circle. Since the circumference of a circle is (2 x pi x radius), there are 2-pi radians in a revolution. To convert revolutions per minute to radians per second, you multiply RPM by (2-pi/60), which equals 0.10472 radians per second. This gives us the "per second" we need to calculate horsepower. We need to get to horsepower, which is 550 foot-pounds per second, using torque (pound-feet) and engine speed (RPM). If we divide the 550 foot-pounds by the 0.10472 radians per second (engine speed), we get 550/0.10472, which equals 5,252. So if you multiply torque (in pound-feet) by engine speed (in RPM) and divide the product by 5,252, RPM is converted to "radians per second" and you can get from torque to horsepower -- from "pound-feet" to "foot-pounds per second."
http://www.musclecarclub.com/library...ne-terms.shtml

Again, horsepower is determined by Torque*RPM/5252.

if Torque is Y and peak torque occurs at X rpm, then HP must be Z.
Likewise, if peak torque is Y and peak horsepower is X, peak torque rpm must be Z.

IOW, you can use this formula as a way to determine either A) absolute HP based on Torque*rpm/5252, or we can use it in the form of a mathematical question to determine the RPM at which the peak torque must occur in order to conform with a known peak HP and torque rating.

We do that by writing the formula as a question plugging in our known variables.

Ie, 400lb-ft*X rpm/5252 = 400hp.

It's using precisely that problem that allowed me to determine when our hypothetical engines produced their torque peak. There is only one correct answer for any given HP/Torque combination, or any given Torque@RPM combination.

We can alter the input numbers anyway you want, they will always reconcile with one another using the formula HP= torque*rpm/5252.

Some more examples.

For example:

Engine 1: 200hp, 800lb-ft

written in the form of a problem: 800lb-ft * X rpm/5252 = 200hp

ANSWER: 800lb-ft x 1310rpm/5252 = 199.54hp
(look at the HP/Torque ratings and where the peak torque of this particular engine is- looks an AWFUL lot like a high-displacement diesel engine, doesn't it?)

Engine 2: 800hp, 200lb-ft

200lb-ft x 21,000rpm/5252 = 799.7hp
(In this example we have a small displacement, ultra-high output, ultra-high revving motor... like a radical rotary or alky drag bike engine)

The next three examples are 'balanced designs', ie, the HP and torque peaks are identical. I can boldy predict before i ever do the math that the torque peak RPM for all three must be identical...5252rpm. Let's see if my assertion holds up:

Engine 3: 300hp/300lb-ft

300lb-ft * X rpm/5252= 300hp

300lb-ft x 5252rpm/5252= 300hp

---------------------------------------------

Engine 4: 400hp/400lb-ft

400lb-ft* X rpm/5252= 400hp

400lb-ft x 5252rpm/5252 = 400hp

---------------------------------------------

Engine 5: 600hp/600lb-ft

600lb-ft * X rpm/5252 = 600hp

600lb-ft * 5252rpm/5252= 600hp

IF WE CHANGE THE RPM THAT THE PEAK TORQUE OCCURS AT- EVEN IF THE ABSOLUTE PEAK TORQUE RATING IS CONSTANT, WE ARE CHANGING THE CHARACTERISTICS OF THE ENGINE, IE, IT'S HP RATING.

By using HP=torque*rpm/5252 we can not only determine the HP of the engine....but it's nature as well.

If it has loads of torque at a low RPM we know it's a high displacement low-revving engine, and it will have a cooresepondingly low HP rating.

If it has very little torque, but it comes at a high RPM range, we know the engine is a low displacement high revving engine, and it will have a cooresepondingly high HP rating.

You used 500/500 and 500/250 as examples.

We can see by doing the math that the 500/500 engine is a relatively large displacement engine with relatively good revving potential(probably about a 7000+ rpm redline).

We can also see that the 500/250 engine is much smaller in displacment, and that it makes all it's power very high in the RPM band, with a redline probably well in excess of 13,000rpm.

Arguing against established mathematical formulas without producing conflicting verifiable mathematically obtained data of your own is not exactly the best way to convince someone you're right...

m21sniper

"yes in the end, its torque that accelerates the vehicle."

It most certainly is.

"more accurately put, its the net force!"

Again, agreed...it is net torque that accelerates the vehicle.
Force(F) is actually an entirely different scientific property. Force is what accelerates the pistons. That force is converted to torque via the crankshaft. From the point the F is converted to torque at the crank, it is torque that is driving the vehicle, but yes, it is F that generates the power to drive the pistons so ultimately, it is F that really drives the vehicle.

"but, agian,im looking at what is put to the wheels and you are stillhung up on the flywheel torque."

I'm not hung up on flywheel torque, i've only been using flywheel numbers as a way to simplify things.

If you want we can break it down as a net figure in each gear based on the previously discussed 500/500motor.

Peak torque*.6* selected gear multiplier* rear end multiplier= net torque.
(NOTE: .6 indicates a parasitic loss of 40% of flywheel power through the drivetrain, which is just an arbitrary number i picked out of the blue. For this exercise i will randomly assign a 1st gear ratio of 2:1, a 2d gear ratio of 1.7:1, a third gear ratio of 1.4:1, and a top gear ratio of 1:1. Rear end gearing ratio will be 4:1)

1st gear:500*.6= 300*2= 600*4= 2400lb-ft peak net torque

2d gear:500*.6=300*1.7= 510*4= 2040lb-ft peak net torque

3d gear:500*.6=300*1.4= 420*4= 1680lb-ft peak net torque

4th gear:500*.6=300*1= 300*4= 1200lb-ft peak net torque

Those torque peaks will all occur at 5250rpm, in each gear.

"since folks dont say, my car has 2200ftlbs of torque at 700rpms on the tires, or 60mph and then 500ftlbs at 1500rpm at the tires at 120mph, you can just say, " i have 500hp."

Agreed, and i'm not arguing that. I've already agreed that HP is a useful tool for determining the characteristics of an engine as well as it's ability to do work. I have not dismissed HP as useless, i am merely pointing out that from a scientific standpoint, it's really quite meaningless so long as you know the engines torque curve across the RPM band, and you also know the proper formulas.

"its just ONE factor of what it takes to accelerate a mass or (car)"

The other factors being the RPMs at which peak torque is achieved, the weight of the vehicle, frictional drag, and gearing.

"HP encompasses much more and is more relavant."

HP allows us to take an engine of known power and estimate precisely how much work it can do in a set amount of time, so from that standpoint, yes, it is very useful. In all actuality what is important is Watts(the scientifically accepted PI)....but that just doesn't sound as catchy, does it?

Hence the very cool sounding phrase... "Horsepower".

"max torque is meaningless. you dont need to know about it, and its the thing that is truely marketing. max hp, then tells you you will have more torque to the wheels at any speed over a operational range and time."

If you insist.

Let me ask you this then. If we don't know the torque rating of our engine, or the RPM at which it occurs...how are we going to determine our 'all important' HP rating?

Answer: We can't.

Torque is not meaningless, not from a scientific standpoint, not from a physics standpoint, and not from a practical standpoint.

So long as we know the torque of an engine at all points of the RPM band, and the proper formulae, we don't need HP to tell us anything.

HP or Watts just make it all much simpler.

m21sniper

M21:"So we're taking our engines observed torque peak(aka "full load torque"), multiplying it by the RPM at which it occurs, and dividing that by the constant of 5252 to determine horspower."

Mark: "peak torque has NOTHING do do with the cross over point. S4s for example have peak torque at 3-4k and crossover at 5250 when using ftlbs for units of torque."

My response: I never said it does.

M21:"For the 500/500, through a little detective work, we discover that the engine's torque peak is 5,250rpm."

Mark: "BZZZZT wrong again"

My response: Sorry man, not only is it right, but i proved it was right with mathematical computation.

If you want to convince anyone otherwise, i suggest you do i t with a recognized mathematical formula, because "Bzzzzzt wrong again" after someone just showed you the math is a pretty silly reply.

You say i'm wrong? Prove it. With math.

M21: "500lb-ft x 5250rpm = 2,625,000 divided by 5252= 499.8 HP"

Mark: "also wrong. dangerous assumptions"

Sorry mark. Not only is the above 100% factually correct(and demonstrably so), but there is NO ASSUMPTION. Unless you think that mathematics is subject to interpretion. I hate to point this out........but it's not.

500lb-ft x 5250rpm = 2,625,000 divided by 5252 IS 499.8HP.

END OF STORY.

M21:"250lb-ft x10250rpm =2,625,000 divided by 5252=499.8 HP"

Mark: "same as above , just doubled for the double max rpm proportions"

Indeed it is the same as above, doubled for proportions. A good indicator that our 500/250 motor is precisely 1/2 the displacement as our 500/500 motor with an identical VE.

Note however that 10,250 is NOT, repeat NOT, the engines Redline, it is the RPM at which the Torque peak is achieved. The redline will be much higher.

Remember the formula determines peak HP based on peak torque at a known RPM. It does not determine peak RPM.

M21:"HP is constant in all four motors, 499.8....all that we're changing is the torque peak, and the RPM at which that figure MUST be recorded to reconcile with 500hp.
If we change the engine RPM at which the torque peak is delivered, by definition we're changing the HP rating."

Mark: "this is true. 500 500 was just an arbatrary set of numbers , again for the point. to be more accurate, we should have used 6000rpm/12,000 as peak HP of 500 and peak torque could be anywhere, more than likely in the 4,000rpm range again."

The formula will work on any engine of any output as long as we have either A) Peak HP and Peak torque, or B) Peak torque and the RPM it occurs

Mark: "LOOSE THE 5250 rpm cross over, its confusing you!"

Sorry Mark, math doesn't work that way. I'm not in the slightest bit confused(which is probably pretty evident to anyone reading this thread). In order to determine HP from Peak Torque*RPM the scientific constant of 5252 must be included. If we leave out 5252 we're not determining peak HP at all. In fact, we're not determining anything if we leave the 5252 constant out.

The constant of 5252 must be included in the formula, or it's no formula at all.

It's not that i'm confused, it's that one can't argue with mathematics....so you're looking for a way around the formula by alledging that i'm 'confused'.

I'm not confused Mark, just right.

Every rating i've given you is 100% scientifically accurate and was derived by a known and universally accepted scientific formula.

m21sniper

Chazz: "Mark, your opinions, though entertaining, are not, by my reckoning, firmly rooted in classical kinematic theory."

Ignoring the possibility of continuing mis-communication, i must agree.

I posted the formula and mathematical equations i used to support my side of the argument. What more can i do?

mark kibort

really? they are not opinions, they are plain and simple facts and truths.

find one of the "opinions" that you do not agree with. do some research and let me know.

This is quite simple stuff, especially, when most of us that have studied it in past lives, dont even remember most of the "real" physics. how hard is area under the HP curve to understand. How hard is it to accept an Obvious example that flyweel torque, if not matched with RPM, tells very little about the system performance. and that HP, being a rate of doing work, and work being torque or force x a displacement , is what we are talking about when we want to know what combination will accelerate a load (mass at the fastest rate)

there is always this confusion and myopic view of torque, yes at the end of the drive, its the average torque, or area under all the torque curve profiles dicated by the gear ratios that will determine the rate of acceleration of a car.
However, just looking at engine torque is as meaningless as looking at only its max capable RPMS. agree??? hope so.

that is the ONLY point here and it has been taken so far out of context in this thread, i dont think anyone knows what to think (except for the folks that do really understand the system i speak of)

MK