TQ vs. HP

[Mad_Matt]

Torque VS HP

Well it funny this comes up in every Diesel Fourm. The answer is in the formula below. I did nto bother to read all the posts so here it goes.

HP = Torque * RPM / 5,252

 

[Bob_Fout]

Well it funny this comes up in every Diesel Fourm. The answer is in the formula below. I did nto bother to read all the posts so here it goes.

HP = Torque * RPM / 5,252

Unfortunetly that function doesn't tell you what the real-world implications are, like...What does torque actually do? What does HP do? What are HP and torque? How do they work together?

 

[ShinyOwen]

Well, thanks to TDIMeister, I just learned more about the functioning of the tq vs. hp arguement than I ever thought that I would tonight.

Best of luck with the exam, but it doesn't sound like you've got much to worry about!!!

 

[Mad_Matt]

Torque

IT does becasue the higher in the reves your peak Torque is produces is the key. It all about the multiplication effect. More Available TQ higher in the revs = more HP wich is what realy does the work. TQ is nothing more than the measurement of distance. 1HP= 33000ft/lbs per minute HP is the busniess of getting it done.

Practicle application. A dodge Diesel will beat a Ford Diesle in a seld pull 99 time out of 100 becasue even though the PSD can produce more raw tq it can't produce or maintain TQ out side of a vey narrow RPM band. The Dogde motor makes it most HP and TQ at 5K revs where the PDS mokes its TQ at 2.5K. The dodge can pull harder, longer and faster even though the ford may have more TQ available. The only way to counter the multipliers is to add more gears to keep you RPMs and TQ in the optimal range and make up for the lower multiplication factor. You can't argue with the numbers. remember all other things being considered equal in the origional post.

.

 

[jackbombay]

HP = work done.

Torque without RPM = nothing.

Torque + RPM = HP = work done.

In this case the work is accelerating the car.

The only reason torque numbers are relevant to acceleration in any way is because you can use them in conjunction with an associated RPM to determine the HP being produced, the amount of HP produced is the determining factor for acceleration.

Example: the torque peak for VE TDIs (pre 03') is at 1900 RPM, the HP peak is at 3500 RPM, we all know that our TDIs pull harder at 3500 RPM than at 1900 RPM even though the engine produces less torque at 3500 than 1900.

 

[nicklockard]

Excellent discussion here. Jackbombay, the 'area under the curve' explanation is very well put.

Question of my own:

torque has units like work: N-m or ft-lbs
Work has units of Joules (N-m or ft-lbs)

work is force applied over a distance (N= force, m= distance)
torque is *vector cross product* of force vector and length vector (N= resultant force vector, m= resultant length vector

But torque is NOT work (thanks to david594 for straightening my rusty self out in another thread last week )

Torque is often described as "the ability to do work", or sometimes: "a work-like quantity."

How would you explain torque and work? Can you think of a similar visual to the integral example of yours above? I like the mention of context to RPM. Can you include that in the visual?

 

[jackbombay]

How would you explain torque and work?

Hmmm, how's this, work involves torque, but torque does not necessarily produce work.

An illustration of torque not doing work would be the image below, the dumbell on the end of the steel bar is creating 1000 lb/ft of torque, but no work is getting done because nothing is moving, or more specifically, rotating. There is certainly force/torque there, but over zero distance which equals zero work.

In the next image the torque produced by the 1000 pound dumbell will produce work, specifically it will pull the hair out of the fellows head that is positioned right under the clamping mechanism that the 1000 pounds will raise up once the weight is released.

Can you think of a similar visual to the integral example of yours above?

How was that? I just hope my work doesn't get plagerised and end up in every engineering textbook on the planet.

 

[david_594]

That drawing is amazing.

 

[TDIMeister]

That drawing is amazing.

It also shows that if you imaginarily move the fulcrum point far enough in the direction toward the dumbbell end, there is a point beyond which even the 1000# dumbbell won't be able to prick the hair off Jackbombay's head!!!

 

[jnecr]

Example: the torque peak for VE TDIs (pre 03') is at 1900 RPM, the HP peak is at 3500 RPM, we all know that our TDIs pull harder at 3500 RPM than at 1900 RPM even though the engine produces less torque at 3500 than 1900.

I'm sorry Jack, but I believe this is incorrect. Your maximum acceleration will always occur at your torque peak because that is where the maximum ability to do work is. In this example we would have to consider that the gear ratios are the same for the 1900RPM and the 3500RPM.

Edit: Yeah, I'm being more sure of this now, with a CVT you would be able to accelerate fastest at the point of peak HP because at that point your gear ratios can be reduced the greatest to take advantage of the engine's RPMs. But with a normal tranny your point of greatest acceleration occurs at torque peak because the entire range of the engine has the same gear ratio...

Now.. if we took a CVT and put it on two engine that both produced a max torque figure of 200lbft but Car A at 2000RPM and Car B at 5000 RPM. If we have the CVT hold the engines steady at their torque peaks, Car B would be faster because of lower gear reduction making use of the engine's RPM...


I heard HP defined as the "rate at which you can do work." But maybe I'm thinking of some other unit...

 

[Fix_Until_Broke]

Now.. if we took a CVT and put it on two engine that both produced a max torque figure of 200lbft but Car A at 2000RPM and Car B at 5000 RPM. If we have the CVT hold the engines steady at their torque peaks, Car B would be faster because of lower gear reduction making use of the engine's RPM...

I hope CarB is faster - it has 2.5 times as much power.

F=m*A or F=m*(L/t/t) where F=force, m=mass, A=acceleration, L=length and t=time
with mass being constant
Force times length is torque
Torque for a time is power
Power for a time is energy

A 1Hp weedeater engine can make 200 ft-lbf of torque as can a 5000 hp ship engine - it's all in the gearing.

 

[jackbombay]

Your maximum acceleration will always occur at your torque peak because that is where the maximum ability to do work is.

Torque peak is irrelevant with regards to work output though, you MUST know the associated RPM of the torque output to determine the amount of work being done, once you have a graph that plots torque vs. RPM you can calculate the "work curve" otherwise know as the HP curve.

Besides, with regards to the originall question TDIMeister wrote,

the answer is simple and it is FACT: They will accelerate at the SAME rate. Argument closed.

The question was very clear that car A had twice as much torque as car B yet TDIMeister, who is teaching automotive engineering classes in Germany, right? states very confidently that the car with half as much torque at twice the RPM will be exaclty as fast as the car with twice as much torque at half the RPM, why? Because the work output (HP) is the same.

In this example we would have to consider that the gear ratios are the same for the 1900RPM and the 3500RPM.

Let me get this straight, if there were 2 identical 2002 TDIs that were both stock and one was driving in 3th gear at 1900 RPM and the other at 3500 RPM in 3th and both of them "floored it" you believe that the car that started at 1900 RPM would have a higher rate of acceleration?

 

[jackbombay]

A 1Hp weedeater engine can make 200 ft-lbf of torque as can a 5000 hp ship engine - it's all in the gearing.

You guys are waaayyyy to hung up on torque numbers, sure a weedwacker engine can make 1,000 lb/ft of torque after gearing, AT WHAT RPM??? Lets assume that your weedwacker has a 2 HP engine, if you ran it at it's peak power output through the appropriate gearbox you could get 1000 lb/ft of torque on the output end of the gearbox at 10.5 rpm.

But the above was completely irrelevant as gearboxes dont make any power they just convert it from low RPM high torque power to some other combination of RPM and torque or vice versa. What the question that started this post is dealing with is torque and power output of an engine before it goes through a gear box, the gearbox, however it is geared will never make more power, sure it can make more torque, but at alower RPM so there is no more power present.

If however you can make a gearbox that increases the torque output of an engine while maintanining or increasing the input RPM you better watch your back because I'm sure big oil has a contract on your life

 

[TDIMeister]

The question was very clear that car A had twice as much torque as car B yet TDIMeister, who is teaching automotive engineering classes in Germany, right?

Sorry to disappoint, but I'm studying it, not teaching.

Let me get this straight, if there were 2 identical 2002 TDIs that were both stock and one was driving in 3th gear at 1900 RPM and the other at 3500 RPM in 3th and both of them "floored it" you believe that the car that started at 1900 RPM would have a higher rate of acceleration?

Well this starts to become a more difficult question with more variables at play. I don't know exactly how fast a 2002 TDI is operating in 3rd gear @ 1900 RPM (too lazy to calculate it right now), but for the sake of the discussion I'll estimate it to be 30 MPH. At 3500 RPM, the second 2002 TDI, also in 3rd gear, would be running 84 percent faster, or 55 MPH from my initial guesstimate. The drag component at 55 MPH is about 3.4 times greater (1.84^2) than at 30 MPH. Now it is not clear which car will accelerate at the greater rate, because the demand to overcome drag is that much higher.

The most germane analogy to investigate is to have 2 TDIs travelling exactly side by side and at exactly the same speed at the initial point of interest. One car might be in 3rd gear riding near it's torque peak RPM. Another might be in second and hypothetically at the power peak RPM. IF AND ONLY IF at the very instant they both simultaneously gun the go-pedal, neglecting turbo lag, rotational inertia, etc., the car operating at peak power will have the acceleration edge.

BUT BUT BUT if both hold the same gear, moments after that the car already operating at the power peak loses it's lead and then at a predictable crossover point falls behind. Why? Because the power curve is already falling off from it's peak while the other car's power curve is still building up.

Moral of the illustration is that there are other factors to be considered (torque/power curves, etc.), but the power available at the driving wheels for a given starting point of wheel rotational speed will govern the torque at the wheel and therefore the tractive force at the contact patch and therefore how quickly you accelerate.

 

[jackbombay]

Sorry to disappoint, but I'm studying it, not teaching.

Ooops, its still safe to say that you know more than all (most?) of us about the subject, IMO at least.

EDIT- I took the following quote to mean you were creating the exam, not answering it,

on Thursday I am writing an exam in Automotive Engineering 1: Longitudinal Dynamics that will cover the exact same subject.

Well this starts to become a more difficult question with more variables at play. I don't know exactly how fast a 2002 TDI is operating in 3rd gear @ 1900 RPM (too lazy to calculate it right now), but for the sake of the discussion I'll estimate it to be 30 MPH. At 3500 RPM, the second 2002 TDI, also in 3rd gear, would be running 84 percent faster, or 55 MPH from my initial guesstimate. The drag component at 55 MPH is about 3.4 times greater (1.84^2) than at 30 MPH.

Yea, I realised that but left it out, the acceleration rate is probably quite close between 3rd at 1900 RPM and 3rd at 3500 RPM, but when one considers the difference in drag there is clearly more power to the wheels at 3500 RPM.

 

[bhtooefr]

Umm...

I know that this is likely a 1/4 mile, so this would be insignificant, but...

Would the fuel consumption of car B be higher? If so, then my vote's for it - sure, at the beginning of the race, they weigh the same, but at the end, car B would have a few ounces on car A.

 

[jnecr]

Let me get this straight, if there were 2 identical 2002 TDIs that were both stock and one was driving in 3th gear at 1900 RPM and the other at 3500 RPM in 3th and both of them "floored it" you believe that the car that started at 1900 RPM would have a higher rate of acceleration?

Instantaneous rate of acceleration would be higher on the car that started at 1900 RPM, yes. Get a G-meter and test it out, I guarantee that's what you'll find.

 

[jackbombay]

Instantaneous rate of acceleration would be higher on the car that started at 1900 RPM, yes. Get a G-meter and test it out, I guarantee that's what you'll find.

Cute, you really think your car puts more power to the ground at 1900 than 3500 RPM? The fact that there is 3.4 times as much of an aerodynamic penalty at 3500 RPM than 1900 RPM in 3rd gear does not factor into the acceleration you feel in any way?

 

[jackbombay]

Instantaneous rate of acceleration would be higher on the car that started at 1900 RPM, yes. Get a G-meter and test it out, I guarantee that's what you'll find.

Do you think Atomic Sushi or any of the other drag racers here wind it up to 4500-5000 RPM when drag racing or do they shift at 2500 RPM to get the revs back down to 1900 as soon as possible as thats where you believe the rate of acceleration is greatest?

 

[Fix_Until_Broke]

You guys are waaayyyy to hung up on torque numbers, sure a weedwacker engine can make 1,000 lb/ft of torque after gearing, AT WHAT RPM??? Lets assume that your weedwacker has a 2 HP engine, if you ran it at it's peak power output through the appropriate gearbox you could get 1000 lb/ft of torque on the output end of the gearbox at 10.5 rpm.

But the above was completely irrelevant as gearboxes dont make any power they just convert it from low RPM high torque power to some other combination of RPM and torque or vice versa. What the question that started this post is dealing with is torque and power output of an engine before it goes through a gear box, the gearbox, however it is geared will never make more power, sure it can make more torque, but at alower RPM so there is no more power present.

We're all saying the same thing - torque is not revelant, horsepower (force per time) is what accelerates a mass

Jackbombay - Ii agree with your statement above, I was just presenting it another way. Sorry if I caused any confusion

 

[jnecr]

Well this starts to become a more difficult question with more variables at play. I don't know exactly how fast a 2002 TDI is operating in 3rd gear @ 1900 RPM (too lazy to calculate it right now), but for the sake of the discussion I'll estimate it to be 30 MPH. At 3500 RPM, the second 2002 TDI, also in 3rd gear, would be running 84 percent faster, or 55 MPH from my initial guesstimate. The drag component at 55 MPH is about 3.4 times greater (1.84^2) than at 30 MPH. Now it is not clear which car will accelerate at the greater rate, because the demand to overcome drag is that much higher.

You can leave drag out of it, assume that we are in a vacuum (well.. except for the engine itself ).

The most germane analogy to investigate is to have 2 TDIs travelling exactly side by side and at exactly the same speed at the initial point of interest. One car might be in 3rd gear riding near it's torque peak RPM. Another might be in second and hypothetically at the power peak RPM. IF AND ONLY IF at the very instant they both simultaneously gun the go-pedal, neglecting turbo lag, rotational inertia, etc., the car operating at peak power will have the acceleration edge.

Without a doubt, the car in second gear will accelerate quicker.

But that is a different scenario than the one of instantaneous acceleration of cars in the same gear travelling at different speeds in a vacuum.

The real question: If you were to plot a logitudinal acceleration curve would it mirror the dyno plot of the torque curve or the horsepower curve?

 

[bhtooefr]

Still, I'd like an answer to my question before I vote.

Would Car B use more fuel, assuming that the engines use the same cycle, and use the same type of fuel?

 

[jnecr]

Still, I'd like an answer to my question before I vote.

Would Car B use more fuel, assuming that the engines use the same cycle, and use the same type of fuel?

Everything being theoretical I would have to say that the fuel consumption would be the same as I believe you can relate fuel consumption to HP not torque. But then it would all depend on the engines... Is one a V8 and then other an I-4? Who knows?

 

[jackbombay]

If you were to plot a logitudinal acceleration curve would it mirror the dyno plot of the torque curve or the horsepower curve?

It would mirror the HP curve, really.

 

[Fix_Until_Broke]

Still, I'd like an answer to my question before I vote.

Would Car B use more fuel, assuming that the engines use the same cycle, and use the same type of fuel?

More power will use more fuel over the same time period - assuming equal thermodynamic efficiency

 

[jnecr]

Do you think Atomic Sushi or any of the other drag racers here wind it up to 4500-5000 RPM when drag racing or do they shift at 2500 RPM to get the revs back down to 1900 as soon as possible as thats where you believe the rate of acceleration is greatest?

Because of gear redution, you can accelerate faster in nearly all of second gear than you can in third gear... Man I wish somebody with a g-meter could just do some basic measurements...

 

[jackbombay]

Everything being theoretical I would have to say that the fuel consumption would be the same...

An engine spinning at 5000 RPM will have about twice the reciprocating losses as the one spinning at 2500 though. The higher reving car will likely use more fuel, but it would depend a lot on the engines your question is one about effeciencies, X tq at X RPM can have varying efficiency rates depending on type of engine, type/grade of fuel, etc...

 

[nicklockard]

Jackbombay, great picture

But I was hoping for some visual like a graph that explains the RPM dependence context of the usefulness of torque:

i.e. 200 ft-lbs of torque at 3000 rpm is a hell of a lot more useful for accelerating the car than 200 ft-lbs at 2000 rpm (in graph form, diagram, chart, or plot of some kind.)

PS: I hear good things about Propecia

 

[jnecr]

http://wahiduddin.net/race/dynotest.htm

about half way down the page you will find the acceleration curves of a vehicle in each gear. Please notice the units on the side of the graph

Using the engine torque data, along with the transmission ratios and the tire diameter, it is possible to calculate the rear wheel thrust in each gear. For the following plot, I have ignored the variations in efficiency of the different transmission gears and have also ignored the effects of windage and oil whipping in the transmission at high speed and have also ignored the rotational inertia of the the engine, tranny, wheels, etc..​

The five curves in the plot below depict the rear wheel static thrust that is available in each of the five gears. If you divide the thrust in pounds by the weight of the car, you will find the acceleration in g's. The dashed line on the plot depicts typical forces due to rolling friction and aerodynamic drag.​

​

 

[jackbombay]

Because of gear redution, you can accelerate faster in nearly all of second gear than you can in third gear... Man I wish somebody with a g-meter could just do some basic measurements...

You seem to be avoiding the question, but not intentionally I think.

Gearing is irrelevant, we are talking about power out the crank of the engine so the gearbox does not matter.

Here is the Million dollar question for you.

You are drag racing, if you loose the drag race you will loose your job and your house, obviously there is a huge incentive for you to win, at what RPM will you shift your car? Assuming of course you are driving a pre 2003 5 speed VW TDI.