Horsepower vs. Torque?
 

I've been trying to find that really long thread that compared HP vs. torque - does anyone remember it? If I recall there were some really intelligent posts from one of our members that talked about how the 2 are inter-related, there was an RPM where they both met in the hp/torque band, it was something like 4545 or 5454 or something like that...

Anybody remember it? I have been searching but for the life of me can't find it...

This post has a formula like you mentioned.

http://www.peachparts.com/shopforum/showthread.php?t=46515&highlight=horsepower+torque+formula

The rpm is 5252.

The only thing that matters in an engine is power. Power has 2 components, torque and rpm. These combined make up the power number known as horse power.

For example, if you have an engine making 200 ft/lbs of torque at 2000 rpm, the power is 200x(2000/5252) = 76hp

If you have an engine making 100 ft/lbs of torque at 4,000 rpm, the power is 100x(4000/5252) = 76hp

Thus both engines in this scenario are at those moments in rpm are delivering exactly the same power and would deliever exactly the same acceleration results.

Think of riding a bicycle and torque being the tension that you are exerting on the chain. RPM (cadence) is how fast you are turning the pedals. If you are exerting 50 lbs of tension on the chain and suddenly change that to 100 lbs without changing your rpm, you have doubled the power you are delivering. Likewise, if you keep exerting 50 lbs of tension on the chain but double the rpm you turning the pedals, you have also doubled you power.

To see the power of an engine, you must look at the hp numbers, which account for both torque and rpm. If an engine delivers 1,000 ft lbs of torque at 100 rpm, it is making 19 hp. If an engine delivers 50 ft lbs of torque at 4000 rpm, it would be delivering 38 hp and be making twice as much power.

Thanks for the excellent explanation - that 5252 was the magic number and helped me find it. The thread mentioned above was great, but this is the one I recall being really impressed by:

http://www.peachparts.com/shopforum/showthread.php?t=70074&highlight=torque+5252

The important issue to understand is that "power" is energy per unit time. Mechanical energy at the crankshaft is converted into vehicle kinetic energy via drive thrust at the wheels, so the more energy you apply per unit of time (which is power) the faster the car will gain kinetic energy(which means greater acceleration), and 150 HP is the same whether it is from a gasoline engine that makes 150 lb-ft torque at 5252 RPM or a turbo diesel that makes 300 lb-ft torque at 2626 RPM.

Duke

Probably an unnecessary reply I make here given the quality responses above, but my two cents anyway: As said, if you graph torque & horsepower simultaneously you will see a certain rpm where the curves cross. Torque, I believe, can be called "engine twisting power". For example, a 1967 Mopar 440 wedge motor with 375 hp, had very good torque at lower rpms (in most, if not all engines, maximum torque is seen to rise, then drop, sooner than horsepower. This is why the 440 could, if 440 is stock hi-performance, in the same chassis/transmission/gear ratios, etc, would beat a 426 hemi in the quarter mile. The hemi developed it's torque later, but above 3000 rpm the hemi is gaining on that 440. Couldn't beat a 440 in the quarter mile, generally, but if the race were a half mile, the hemi has come into it's own and devastates the 440. This difference in the quarter mile vs the half is because the hemi has more horsepower. Many may differ, perhaps with good reason, but it is a fair intuitive statement to make that "torque determines acceleration, horsepower determines top speed".

i bet with appropriate gearing the hemi would beat the 440 in the quarter.

tom w

No, as I explained previously, it's horsepower that accelerates the car. The shape of the power curve is important as is gearing. The area under the power curve in the rev range determined by gear spacing and shift points is the total energy input to the vehicle. The higher it is, the faster the car will achieve speed and distance.

An engine that produces peak torque and power high in the rev range needs shorter overall gearing and closer spaced ratios to have the same average power input through the gears compared to a high torque low revving engine of the same peak power such as the examples I listed previously.

Duke

HHMMM, I am guessing...
 

...by that answer your research does not include Wankel, turbine or any propeller driven airplane? I ask to be sure we are all on the same page IE: ONLY piston/direct (no electric) drive, as there are VERY few constants accross the board. But, as always, F=MA, E=IR, PV=nRT etc. the rest are only 'favorite flavors' :silly: BTW here's a little aside to muddy up the mix. In Gemrnay, and most EU, car engine max output is listed in kW (you mean they went back to the original derivation) - power defined in watts?

You're right Tom. I was just making the point with the same gearing/tranny, etc. Now, if both the wedge & the hemi are set up tougher, the hemi wins even in the quarter. Part of the problem, a large part, is that if the owner fooled with the linkage between the two Carter 4 barrels on the Hemi, they generally messed it up by thinking they need to be put exactly in sync as regards when the primarie's opened. That is not the case. The rear carb primarie's, if I remember correctly, provide more of the fuel at idle due to their location relative to the cylinder banks. Just comparing the 440 wedge with the 426 Hemi as regards potential, well, the 440 was a tough engine to beat compared to any stock Detroit Iron, but the Hemi with a more appropriate camshaft, properly configured tube headers, is a Monster.

...doesn't make any difference. If you have power at a shaft, HP= TxN/5252. You can get the same horsepower with high torque and low revs or low torque and high revs. Aircraft engines are rated by "shaft horsepower", which is the power at the output shaft that drives the propellor, so it includes the reduction gearbox, which is required for turbines and big recips.

To drive a ship or aircraft the power is applied to a propellor that delivers a propulsive force. In a car the propulsive force is generated at the tire/pavement interface.

At steady vehicle speed on level ground (or level flight) the required horsepower is equal to force times velocity in feet per second divided by 550. If you know the total drag force, that times velocity times propellor efficiency (in the case of aircraft and ships) times the proper conversions for dimensional homogeneity is delivered shaft horsepower. If more power is delivered the vehicle accelerates, if less it deccelerates.

Duke

Back in the "BC era" - before computers - engine lab dynos were usually electric - essentially a big DC generator with a large resistance grid that you used to control the load. Water brake dynos were also in use, but the old DC generator type was easier to control for lab work, especially if you wanted to control speed at various constant values over a wide range of loads.

The dynamometer frames were mounted on bearings that allowed them to rotate freely, but they were constrained by a lever arm that rested on a scale.

To do a WOT power run, you set the load high then opened the throttle fully and adjusted the load to achieve a steady speed. Then the operator read the scale and RPM. He then reduced load to allow the engine to accelerate and then added more load to stabilize at the next test speed, read the scale and RPM, and so on and so forth at more test speeds up to the maximum speed planned for the test.

So what you ended up with was a data set of RPM and scale load reading in pounds, and knowing the lever arm length allowed you to compute the torque at each RPM point. Then, using torque and the famous HP = T X N/5252 formula you computed horsepower and could then plot both the torque and power curves versus RPM.

So where did this formula come from?

Answer: Basic physics.

"Work" (which is the same as energy) is a force acting through a distance or a torque acting through an angle. If I use one "unit" of force, say pounds, to push a brick across a table one unit of length, say feet, I have expended one unit of work or one foot-pound, which was dissipated as one foot-pound of friction energy.

Similarly, if I use a one foot lever and one pound of force to rotate a shaft that has some friction resistance through one unit of arc length (which is that same distance as the radius) I have expended one foot-pound of energy that was also dissipated as one foot-pound of friction energy. Recall that pi is the ratio of the circumference of a circle to its diameter, which can also be expressed as circumference is equal to 2Pi times radius. One radian is defined as the length of arc along the circle equal to the radius, so there are 2pi radians in a complete circle.

I other words, if I apply a force on a one foot lever arm through a complete revolution, the distance is two pi feet so the work is 2pi time the force. If I apply the force on the lever arm for ten revolutions the total work is ten times the work for one revolution.

Since torque and work/energy have the same fundamental units the current convention is to use "force-length" or pound-feet for torque, and "length-force" or foot-pounts for work/energy to avoid confusion in what we are dealing with.

Recall that power is work or energy over a period of time, so a torque, T, in pound-feet, from a shaft that rotates a complete revolution, which is 2pi radians, over a time period, t, produces power as follows

Power = torque(2pi)/t

If our old fashioned dyno shows 50 lbs at a scale with a two foot lever arm, which is 100 lb-ft at 6000 RPM ( time period of .01 second for one revolution) the power produced is

100(6.28)/.01 = 62,800 ft-lb/sec

Or we can convert to ft-lb/minute by multiplying by 60 and the result is 3,768,000 lb-ft/minute.

Back in the early eighteeth century, James Watt, inventor of the steam engine, spent a considerable amount of time studying horse drawn pumps that pumped water out of coal mines. The horses were harnessed to a capstan and walked around in a circle.

Watt determined that a typical draught horse could deliver 33,000 ft-lb/minute of power continuously over a long period of time and DEFINED this amount of power as ONE HORSEPOWER.

So the above relationship can be rearranged as follows where N is engine speed in revolutions per minute.

HP = T(2pi)N/33000 = TxN/5252

There are many units for power. We are still stuck with Watt's definition from the archane English system of units. Nations that have adopted the SI (Metric) standard use kilowatts and 1KW = 1.34 HP, or 1HP = 0.746 KW.

In the above example our 100 lb-ft of torque at 6000 RPM is delivering 3,768,000/33,000 = 114 horsepower, which is the same as 100 x 6000/5252. In other words 100 pounds of force is being applied with a one foot lever arm over 6000 revolutions in one minute - force over a distance, in this case using a one foot lever arm a total of 6000 times in a circle in one minute.

The power would be the same if it was 400 lb-ft at 1500 from a truck engine.

The common inertia chassis dyno of today continuously records instantaneous angular acceleration and rotational speed of the drum, and this data combined with the known drum rotational inertia is input into a simple software formula that computes instantaneous power. The famous formula, rearranged to HP(5252)/RPM is then used to compute torque from the power data. The basic output data is HP and Torque versus MPH, but most shops have an inductive pickup that they can attach to a plug wire so HP and torque versus engine RPM can be displayed and printed.

So the old fashioned electric dyno "output" is load on the scale that we use to compute torque and then with the RPM data compute power. An inertia dyno continuously records instantaeous roller acceleration and RPM, and uses this data to compute instantaneous power and then torque and plots out a continuous curve of both. Same end, different means.

Modern water brake lab dynos can be programmed to allow the engine to accelerate and measure power in the same way. The usual acceleration value is 300 RPM/sec. It's a lot easier to get a torque/power chart using the inertia technique than the way I did BC - a lot easier on both the operator and the engine!

Duke

Duke,
I very much appreciate the valid information you've given & it is not lost on me that you have a deep academic background in the fundamentals of what we are discussing. I raise an argument with you, not so much to say you are wrong, but for you to show me I am wrong. This is my honest thinking & if I'm incorrect I would like to know that; especially since this topic is of fundamental academic & practical importance.
So, as you say, we start with Watt's equation, replacing N with rpm:
HP=horsepower
T=torque
rmp=revs per minute
HP=(T*rpm)/5252
So, correct me if I'm wrong, though this isn't the point I wish to eventually make: So, when we graph both HP & T as a function of rpm, the curves intersect at 5252 rpm. Now, I realize that if one knows the value of HP, then the value of T can be directly derived, which seems to me important that the two phenomena (even being metrics, it still seems phenomenology to a large extent). And vice versa. Yet, we want to distinguish the two, and we can, though they are equal at 5252rpm, the two curves can have very different trajectorie's.
Now, this is where I believe we differ. I made the statement that it is Torque that determines Acceleration. In the graph of T&HP as a function of rpm, if we add Acceleration as a function of rpm (speaking of a car here where we ignore air density, road friction, etc) to our graph, my understanding is that the Acceleration curve exactly falls atop the Torque curve. Gearing has no effect on this.
For example, 100 foot pounds of torque will yield precisely the same acceleration (ignoring air density & all the rest) whether that 100ft/lb torque is produced at 2000rpm or 4000rpm; yet, per the formula, HP has doubled at 4000rpm as we arbitrarily hold Torque at 100. The doubling in HP, holding Torque the same, yields no further gain in Acceleration. So, this is just a restatement of my original "it is torque that determines acceleration".
HP, as you related in using shaft HP in an engine driving a propellor; in our car, this would relate, to my thinking, to the top speed of our car.
To beat the dead horse a bit further, though I certainly allow for my being incorrect, graphing acceleration, horsepower & torque as a function of rpm, the torque curve and acceleration curve fall upon one another across the measured rpm range. (edit) If I am incorrect that the acceleration curve of an object in free space (which is our hypothetical car that is not affected by air density, frictional power losses, etc) precisely overlaps the torque curve, but not the hp curve, I still argue this view is correct to at least a first approximation. I think the better way to state it, as you used Areas Under the Curve, one sees that the torque AUC more closely predicts acceleration than the HP AUC. As a last thought, when we go back to HP= (T*rpm)/5252 it is obvious that if we know either one of HP or T we can directly derive the other. I'm not oblivious to the fact of forum members describing relationships to speed, gear ratios, etc, that become of prime importance if one were to build a car with certain desired characteristics. But, it is not without some truth, be it great or small, that if one builds an Indy car, one wants a car with high HP capable of running at high rpm, maximizing the HP AUC; if one builds a drag car one wants to maximize the Torque AUC, though this is not to say that rpm is meaningless. It seems to me that both T&HP are brother's or cousins of Force, and I do apologize for not refreshing myself on Newtonian Mechanics, other than the particular case argued here. So, if at any given rpm, and knowing either of the other two degrees of freedom, we know the last, this, I believe, tells us that we shouldn't go hog-wild in our belief that T & HP are two metrics which should enjoy some special status, that is, one over the other.
So, where is my error in fact or logic to say, "Torque determines acceleration."?
(edit) In re-reading all the posts I see it said that Torque in combination with the rest of Watts equation determines HP, or even vice versa. How can we say that either T or HP is dominant as neither has any independent status; one metric yields the other metric. So, to give one or the other, again (I'm sorry), special status as regards the contribution to acceleration, one need to measure both over time as well as (rpm), then calculate the AUC's of both HP & T. It is building an engine where T has the greater AUC that yields maximal acceleration. This is always the case in the purest sense, leaving gearing & the rest out of it. Both HP & T AUC's need be considered when building our car optimized for certain tasks.

I think you missed the fundamental relationship that Power = force x velocity. Torque does not take time into account. Power does, and any calculation involving a dynamic variable, such as vehicle acceleration or velocity, must include time. Force and torque are not a dynamic quantities unless they act over a distance in some time period, which is power.

It's a difficult concept to comprehend, and not at all easy to explain.

Sitting down with a basic physics tests and reviewing the concepts of work, energy, and power might help you out.


Duke

Neither torque nor horsepower explicitly takes time into account in the discussion. The torque or horsepower is calculated instantaneously for rpm X, then for rpm x+1, etc, until we generate the discussed curvilinears. Yet, in practice, such as in the acceleration of an automobile from idle rpm to it's maximum rpm, we now have taken time into account and can calculate the torque across that entire time period; this is no different than Power(horsepower) as regards the temporal element.
Yes, it is true to say that one can calculate velocity by measuring the distance travelled over a defined time. But, it is not a necessary condition that time be considered. Any variable can be described as to it's value as a function of time, or it can be described in an instantaneous fashion.
If one is to say that velocity is unequivocallyknown by measuring the distance a body travels over a known non-zero period of time, that is false, in principle. A Body A, at t=0 resides at Point A. One hour later that same body is at Point B, which is 100 miles from Point A. Does one then say the velocity of Body A is 100mph? I can envision Body A travelling that 100 miles over a time period of one hour, but sometimes having an instantaneous velocity of 200mph, sometimes 0mph. So, 100mph is only assured to be an average velocity. In principle, the true velocity of that body, if it be 100mph over the distance of 100 miles, can only be stated if all (which is impossible, in practice & principle) instantaneous velocitie's are known (instantaneous velocities have no time nor distance components, just a relationship with some arbitrary reference frame, even if it be a reference frame fixed nicely for our little measurement exercises). So, what is an instantaneous velocity? It is a velocity that has no finite time component. It has a time component, but since the time component is always zero, no time vector exists for each and all velocity measurements. Power is no different. Any fondness to name this or that variable as "dynamic" and another not to be so is not good thinking and is rife with arbitrariness and assumptiveness. It is the instantaneous values we have, and that is all we have. We can graph the instantaneous values as a function of time surely, but you are not correct to say that you have taken time into account when measuring power or velocity or torque or acceleration, or anything else for that matter. As soon as we take non-zero time into account we only have the average value of whatever we are measuring as expressed over an arbitrary interval of time and have buried the instantaneous values, which alone have any physical meaning. As soon as we throw time in the mix we lose the true velocity (or power or torque, etc) and are left with an average.

I think if you spent as much time with a basic physics test as you spent typing your posts, it would be beneficial.

Duke

This is where you are confused.

Torque at the rear wheels determines acceleration.

Torque from the engine passes through a gearbox and a rear axle and is increased based upon those gear trains.

If I have an engine that produces 100 lb.-ft. torque at 6000 rpm, I can use a transmission with 2:1 gears, as an example so that my output speed is 3000 rpm and the torque to the pavement is 200 lb.-ft.

If I have an engine that produces 100 lb.-ft. torque at 3000 rpm, this same engine will put only 100 lb.-ft. torque to the pavement with 1:1 gears. The output speed from the transmission, 3000 rpm, is the same for both engines.

But, the engine that produces the torque at 6000 rpm will accelerate the vehicle at twice the speed of the engine that produces the torque at 3000 rpm.

The engine that produces 100 lb.-ft. torque at 6000 rpm has twice the horsepower as the engine that produces 100 lb.-ft. torque at 3000 rpm.

I spent a thousand hours at Argonne Nat'l Lab during grad school, though my PhD is in neurobiology. I spent about 100 unofficial hours at Fermi Lab during this same period. Your response gave me no indication of what understanding or thoughts you have or don't have regarding instantaneous velocitie's, or instantaneous value's of the other metrics and how they relate in relevant ways to your remarks on xxx as a function of time. Time spent typing posts, good point. I could have shortened it considerably. Leaving out any condescending(sp?) remarks kept it a bit shorter.

Brian,
I agree with you as regards gearing and "torque at the rear wheels determining acceleration". I'm sorry I didn't make it clear that gearing, frictional losses, etc were left out so as to speak of torque in a stripped down sense. I should not have said the torque curve falls atop the acceleration curve exactly, but it does, I believe, follow the torque curve in general form and has a much different relationship, in form, to the hp curve. And, yes, I agreed with Watt's equation so agree with your last sentence. Since seeding the Watt's equation with a value assigned to one or the other, but not both, of torque and horsepower, we automatically can generate the same graphs, regardless of which variable we provide with a seed value, this tells us that hp & torque are so intrinsically linked so as to be cautious in making too bold of statement regarding either. I was guilty of that.
....separate thought: I think, without re-reading my post, I said it makes no difference at what rpm the engine produces maximum torque, it is that rpm which produces maximum acceleration; this is not so with horsepower. The gearing variable, with no good reason not to do so, only to simplify the discussion, I left out.

This discussion is now well beyond hopeless. If one does not understand basic 400 year old physics, which includes the concepts of work, energy, and power in addition to Newton's Laws you'll just never "get it".

You can pretend you're Aristotle and engage in a soliloquy of "natural philosophy", but the bottom line is that all this was codified nearly 400 years ago and has been taught in basic physics courses at universities and high schools ever since.


Duke

It makes a significant difference at what rpm produces maximum torque. The higher the rpm at which the maximum torque is produced.........the faster the acceleration of the vehicle.

The higher the rpm at which the maximum torque is produced......the greater the horsepower.

You're still hung up on engine torque........it's the torque to the pavement that pushes the vehicle.

05 e320 cdi made more torque (@lower rpm) and less hp (lower rpm) than the gasser e320.
The cdi was faster to 60 but slower in the 1/4.
Will someone explain the physics as to why the cdi was faster?:D

Acceleration is a function of the time-average thrust force at the tire/pavement interface. And since you know from this thread that thrust force = HP/V you can determine the thrust force at any speed/gear if you have RWHP and gearing data. The higher the average thrust force, the more kinetic energy the vehicle will gain.

Unless you have a CVT that allows the engine to rev to the power peak and stay threre, the thrust force will vary with applied power, which is a function of the shape of the power curve and gearing including the intergear ratios.

Turbo diesels tend to have flat torque curves, so they make more power at low revs than a naturally aspirated SI engine of equal peak power that typically has narrower torque bandwidth, so the TD would probably be quicker off the line (which could be offset by turbo lag).

Once you reach peak revs in first gear and the transmission shifts power falls off, so average power through several gears is a function of both gear spacing and the shape of the power curve, and the vehicle that can deliver the highest AVERAGE power, not the one with highest peak power, will accelerate faster.

Duke

A twin engine airliner is cruising at 35,000 feet at a true airspeed of 500 MPH. From available technical data the pilot knows that the power setting he selected yields exactly 10,000 pounds of thrust from each engine.

How much horsepower is each engine producing?

With the information in this thread and a calculator, it's a fairly simple calculation.

Anyone feel free to respond, and please go through your logic and calculations for the benefit of others.

My intent is education.

Duke

Ramblin Wreck to the Rescue
 

F= 2 x 10000 lbs. = 20000 lbs.
Velocity = 500 miles/hr x 1 hr/3600 sec x 5280 ft/mile =733 feet/second

:. HP = 20000 lbs x 733 ft/sec / 550 (ftlb/sec/hp) = 26655 hp :beer: Go Jackets!

And by the way, real Germans still measure their car's output in PS, not Kw.

26655 ÷ 2 engines = 13327.5 hp / engine.;)

Excellent! And I'd like to offer some further discussion, as jet engines offer an opportunity to show a simpler perspective.

At steady speed and altitude the thrust of the engines is exactly equal to total drag force, so the "thrust horsepower" is simply F x V, with suitable attention to dimensional homogeneity.

Ultimately what accelerates a car is thrust at the tire pavement interface, but since an automotive engine's available power is dependent on shaft speed, and gearing establishes the relationship between shaft speed and vehicle velocity, how does this relate to drive thrust?

Very simply, Horsepower = F x V.

If you do chassis dyno pulls and plot HP versus MPH you can use the above relationship to determine the drive thrust curve, and it could be easily programmed into dyno software. Available drive thrust is much more useful for vehicle performance analysis. Most guys focus on "horsepower", even though they really don't understand what it is, and they obsess over peak horsepower even though it's seldom achieved in driving.

Do the test in all gears and you have a complete characterization of maximum available drive thrust at any speed in any gear.

If you have drag force data available, which is determinable from various road and lab tests, you can determine how much excess thrust force is available, over and above total drag at any speed and in any gear, to accelerate the vehicle, and acceleration is given by Newton's Second Law, a=F/m.

This is how drag racing simulation programs work, and they are fairly accurate for a wide variety of cars.

Jets are simple because their output is measured directly in pounds of thrust, so we don't have to go though all the convoluted calculations with power at a shaft and gearing to determine available drive thrust.

The Toyota Prius has a CVT, and if you floor the throttle from a dead stop the engine revs up to a constant speed, and it accelerates more like a light airplane than a car because thrust force is maximum at initial acceleration. The initial high available force to accelerate the car is reduced steadily as speed increases and a greater proportion is required to overcome drag, so acceleration moderates with greater speed. It's somewhat complicated by the electric assist, but the electric motor delivers near constant power at WOT regardless of engine speed as long as the batteries are at full charge. There is probably also some variation of power with speed due to variation of CVT efficiency.

The "ideal car" would have a high efficiency (at all speeds) CVT. At any speed, if you want maximum acceleration you just floor the throttle. Revs would increase nearly instantly to peak power and peak power would delivered to the rear wheels continuously - no need to downshift and wait for engine revs to get into the peak power range to achieve maximum drive thrust and acceleration.

The Prius CVT does not allow the engine to rev to the peak power revs, initially. It stablizes somewhat above peak torque. This is probably to reduce, noise, wear, and keep BSFC at a low level. At peak power speed most automotive SI engines have higher BSFC than near the peak torque speed because friction horsepower increases with the square of speed, so most automotive engines achieve best full load BSFC in the vicinity of peak torque speed. So the Prius will not allow access to peak power at normal speeds in order to lower fuel consumption.

At high speed the Prius engine revs may close in on peak power revolutions, but I only did 0-60 tests. With the Prius modest power it will probably only achieve around 100 MPH, and it would probably take nearly two miles to achieve, which is typical of low drag, low power cars.

High powered high drag cars like F1 achieve their considerably higher top speeds in much shorter distance - probably no more than about a mile with a high downforce setup.

Duke

I might add that a gas turbine powered airliner falls under this scenario. The engine achieves maximum power within 10 seconds and the acceleration of the airplane is at its maximum within 10 seconds of brake release.

It's not possible to use the thrust formula to compute horsepower in this instance because the airplane is not at constant speed.

As the airplane builds speed, the thrust from it's engines continually decreases while, simultaneously, the force due to friction across the fuselage and wings increases. At the end of the runway, the thrust available for acceleration of the airplane is less than half of the thrust available when the airplane is sitting with locked brakes.

I find airliners fascinating in this regard........and calculation of their available horsepower at any given point in the acceleration of the airliner to be quite a challenge.

Good job guys!
 

I am really impressed! That calculator must be red hot from all this punching!

Now, seriously. Max horsepower and max torque are intrinsic and independent characteristic of an engine. Because of the relationships stated here many times already, you cannot have one without the other. A car with very low torque will take a long time to accelerate to a high speed no matter how much power is available. It will stop accelerating when the rpms (and velocity, P = drag x V) are such that max power has been reached. Conversely a car with high torque and low power will accelerate very quickly to a low speed, the one at it will reach max power (as before). So, in mi humble opinion, Torque will determine the max possible acceleration, power will determine the maximun speed.
Mi $0.0002

JL

Unfortunately, not.

A motorcycle is a classic example of a machine with relatively low torque but very high horsepower. Small engine that revs very high. With proper gearing, it puts massive torque to the pavement and accelerates very quickly.

I'll take a vehicle with high horsepower and low torque and beat the pants off your vehicle with high torque and low horsepower, provided that I get a decent set of gears to work with. Provided that the gears are available, higher horsepower will win the race every time.

Duke explained that the gears are not always perfect, so, it's possible for a vehicle with lower horspower and slightly higher torque to be beaten by the converse, but, it's not the norm.

The converse is also true.

Brian, you cannot have it both ways. It either has low torque or massive torque. Understanding that torque is torque at the wheels of course. That is the ony one that matters.

JL

I was speaking of torque from the engine.

Torque at the wheels is precisely what drives the vehicle. The point of high horsepower and a good set of gears is what allows you to have maximum torque at the wheels.

BTW, nobody quotes torque "at the wheels". It's always the engine output torque and horsepower that are discussed. The torque at the wheels is subject to all the powertrain losses and would need to be measured on a chassis dyno as Duke explained.

Of course you are right. Nobody will disagree with that. I thought the point of disagreement was wether torque or power determines acceleration. In that sense, as you very well pointed out, the only torque that matters is the one at the wheels, that is why I used it. And I do not believe you need high power to get high torque (at the wheels). Just for high speed.

JL

My understanding is also we were discussing whether torque or horsepower determine's acceleration. Leaving out gearing, frictional losses, etc in the drivetrain seemed to me sensible as long as one knows there is no acceleration without driveshafts and wheels. Torque is a measure of twisting power at the crank that is transferred to the flywheel. We can limit the discusiion to end at the flywheel with no loss in discussing the basic questions. Given Watt's equation, we are told that if we know either torque or horsepower we can determine the other. Equations are equalities Watt's equation, in my opinion, is a simple transformation. Given that, I believe we see that torque and horsepower are not so much separate metrics, but are shifting one's gaze to either horsepower or torque, one totally dependent on the other. One of my points is that we are basicaly looking at one spinning crank or flywheel and measuring it's torque or horsepower. But, whichever we measure, we automatically know the other. So, they're not such separate entitie's even though we can design and build engines which generate differet hp & torque trajectorie's as a function of rpm.

It's drive thrust at the wheels that determines acceleration.

If the tire rolling radius is 12" 500 pounds of thrust at the wheels is 500 lb-ft of torque at the axle. Let's establish the final drive gear ratio at 2.5:1 and assume no loss in the axle, so axle input at the drive shaft is 500/2.5 = 200 lb-ft.

The relationship P=FV still holds, but we can rewrite it as:

F=P/V

or

F = TxN/5252V (we're still talking about torque and power at the rear wheels)

Now substitute an engine that makes half the torque, 250 lb-ft, but at twice the speed, and we shorten the axle ratio from 2.5:1 to 5:1, so that peak torque occurs at the same road speed.

At this road speed which setup provides the greatest drive thrust? Which one provides the greatest RWHP?

Duke

Sure thing, in fact engines are taylored to meet certain hp and torque requirements. It is a matter of changing rotating masses, crank lengths, pistons etc. (mostly for torque) and displacement, rpms, engine breathing, etc. (mostly for hp). And now we have those hybrids with those electric motors that give you gobs of torque regardles of rpms. In my opinion though, for a particular engine, horsepower depends on torque, but torque does not depend on horsepower.

JL

If I understand your post correctly the first one , as you state in the post, provides 500 lbf of thrust. The second one, by my calculations 250 lb-ft x 1ft = 250 lbf, unless I misunderstood.

JL

PS I just reread your post and now I am really confused. I am going to bed and let's continue tomorrow. Maybe then I can think straight.

JL

Well, the fact that you consider this an opinion and not "fact" means there is hope.

HP = TxN/5252, or using algebra to rearrange

T = 5252xHP/N

My 2.6L M103 is rated at 162 lb-ft of torque at 4800 RPM and 158 HP at 5800. How much HP does it produce at 4800? How much torque does it produce at 5800?

At 5800 is torque a unique function of power? If not what determines torque at this engine speed? How about at any other engine speed?

You're answer to the previous question was not correct. Give it another try with a freshly rested mind.

Here's another brain exercise. You're checking wheel bolt torque. You apply 80 pounds of force on your torque wrench, which has a lever arm of one foot, but the nut does not move. How much work have you done? How much power did you apply?

Hint: Review the definitions of work and power in previous posts.

Duke

This is mucho mucho fun!
 

Well let's see. I went to my Physics 101 book, I cleaned the dust off and got out my big manifying glasses and read the answer:
0 (Zero)
0 (Zero)

Am I right prof?

Now, here is another brain exercise:

Contestant A climbs in his car with tremendous 1000 HP engine and a puny 200 lb-ft of torque.

Contestant B climbs in his car with puny 200 HP engine and a tremendous 1000 lb-ft of torque.

Both cars have the same weight, gearing etc. and similar torque, HP vs rpm curves.

They both release the clutch at, lets say, 2000 rpm.

Which one will have a higher initial acceleration?

Hint: Review the all the definitions you can find on your Physics 101 book. (Then, burn the book).

Is that 1000 hp at 0 mph or 100mph?

P.S. please post pics of the cars...

Those are good guesses. I'll have to figure out the mph for the torque too. I will borrow a digital camera and send pics later.

JL

It's a fruitless exercise because the two vehicles would have vastly different gearing. Furthermore, they cannot have similar torque, HP vs. rpm curves.

With proper gearing, the 1000 HP engine would blow the doors off the 200 HP engine. To think otherwise is ridiculous.

Yes it is a fruitless and ridiculous exercise, but the physics are correct -ridiculous does not prevent it from being so. The fact is that we are talking about this engine, that engine, my engine; they may all have different characterstics and seem to have different behaviors, but the underlying physics are the same (obviosuly the interpretation of the physics is not). Duke's questions about power and work done were very astute and reveal a clear understanding of the issues. You cannot have power or work without motion, but you can have torque. Similarly you can have acceleration despite the fact that your velocity is zero (otherwise we would never get going).
It was not my intent to further confuse the issue, but I seem to have succeded at it. Sorry.

JL

I had no idea I was essentially creating another oil thread...:D

Amazing, isn't it? This is one of the things that makes this forum so great!

JL

There is one fundamental difference in an oil thread when compared to this thread. An oil thread is founded on opinions. This thread is founded on engineering facts.

There can be no dispute with the facts as Duke has provided them in such an eloquent fashion. Basic physics is indisputable.

Correct! Unless a force acts through a distance - either linearly or an arc of a circle - no work is done, nor is any power expended.

Brian pointed out the inconsistencies in your example. It's just not a valid exercise, even for academic purposes.

You still haven't anwered the question in post #35, yet. Do you want to give it another try? Anyone else?

Duke

Joselu, where are you?
 

Obviously the car making the lower torque will win, because the 1000 ft-lb machine will have blown itself up at 2000 rpm, since you stated it can only deliver 200 hp. At 2000 rpm, it would be producing 380 hp, which of course it can't do. :smoking:

Hey, I am back!
 

Big cigar for Tom! You caught it, nobody else did!
As far as post 35, I am not sure I got it right. First half I get. We get 500 lbf at each wheel. Second part I am not sure. If the 250 lb-ft is the torque at the wheel (as it appears) then the force on each wheel will be 250 lbf. As I see it, the first car (assuming the same weight) will accelerate faster because of a larger driving force, regardless of the rpms. If the 250 lb-ft are measured somewhere else let me know and I will try to understand.

JL

From the description of the engines, the first is a low revving engine, say 500 CID. The second is a high revving engine, say 250 CID. The key is the following statement.

"Now substitute an engine that makes half the torque, 250 lb-ft, but at twice the speed, and we shorten the axle ratio from 2.5:1 to 5:1, so that peak torque occurs at the same road speed."

Both engines are making the same RWHP at road speed. Though the "small" engine makes only half the torque, with suitable gearing it delivers the same RWHP, so both will accelerate at the same rate if the throttle is floored at this road speed.

Again, I stated this example to illustrate that POWER is what counts to accelerate a vehicle.

If a one-liter motorcycle engine rated at 120 HP @ 12000 was substituted in a small sedan for the original 2-liter engine rated at 120 HP @ 6000, acceleration performance and top speed would be the same provided gearing was shortened accordingly. If the two-liter was geared to yield 3000 revs at 60, the one-liter would have to be geared to yield 6000 @ 60.

The same argument can be made for substitution of a low revving four-liter industrial engine that is rated at 120 HP @ 3000 in which case you would gear it to yield 1500 @ 60.

Assuming the car required 120 HP to achieve 120 MPH, all three combinations would achieve 120 MPH at engine revs corresponding to their peak power.

Duke