TQ vs. HP

Which car would win in a drag race?

  • Car A will win because it has twice as much torque.

    Votes: 211 51.6%
  • Car B will win because the engine spins faster.

    Votes: 80 19.6%
  • The cars will tie.

    Votes: 118 28.9%

  • Total voters
    409

Fix_Until_Broke

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jnecr said:
Originally Posted by jackbombay
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...
It is true that you can accelerate faster in 2nd than 3rd as you say, but it has nothing to do with gear reduction - it's moastly due to aerodynamic drag (assuming same vehicle). You are delivering the same ammount of power to the wheels, just a larger % of it is going to overcome drag than accelerate the mass of the vehicle.
 

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Fix_Until_Broke said:
It is true that you can accelerate faster in 2nd than 3rd as you say, but it has nothing to do with gear reduction - it's moastly due to aerodynamic drag (assuming same vehicle). You are delivering the same ammount of power to the wheels, just a larger % of it is going to overcome drag than accelerate the mass of the vehicle.

Hah, that's real funny, you honestly believe that? In which case would you like to answer the question of why we do dynos in third gear rather than 1st or 2nd??
 

jackbombay

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jnecr said:
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

[/left]
That is not a good example because the TQ peak is very close to the HP peak, but it will work to show where your misunderstanding is.

the link that jnecr posted said:
For example, in the Spec Racer Ford, keeping the engine turning above 4000 rpm but below 5600 will get maximum acceleration
Now letslook at the HP curve,



and now the TQ curve,



From 4000 to 5600 RPM the torque does nthing but decrease, while if you look at the HP curve the peak HP output is centered between 4000 to 5600 RPM. In short you link proves my point, that max acceleration is acheived when the area under the HP curve maximised not area under the TQ curve.
 
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jackbombay

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jnecr said:
Hah, that's real funny, you honestly believe that? In which case would you like to answer the question of why we do dynos in third gear rather than 1st or 2nd??
"We" run dynos in 4th because it is a one to one ratio and eliminates trading HP for TQ.

Running a DYNO in 2nd would give you higher TQ numbers at the expense of RPM, so you end up with the same power output.

I'm going to headbutt some nails now.
 

jnecr

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jackbombay said:
"We" run dynos in 4th because it is a one to one ratio and eliminates trading HP for TQ.

Running a DYNO in 2nd would give you higher TQ numbers at the expense of RPM, so you end up with the same power output.

I'm going to headbutt some nails now.
No.. you won't "trade" torque for HP, both figures will increase at a ratio that will be the same as the gear reduction ratio. If, of course, you plot the curve in usual format with posting engine RPM rather than wheel RPM.
 

Fix_Until_Broke

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Quote:
Originally Posted by Fix_Until_Broke
It is true that you can accelerate faster in 2nd than 3rd as you say, but it has nothing to do with gear reduction - it's moastly due to aerodynamic drag (assuming same vehicle). You are delivering the same ammount of power to the wheels, just a larger % of it is going to overcome drag than accelerate the mass of the vehicle.

jnecr said:
Hah, that's real funny, you honestly believe that? In which case would you like to answer the question of why we do dynos in third gear rather than 1st or 2nd??
It probably depends on the dyno - if it is on rollers, my guess is the limit of traction available between the tires and rollers. If it is on a hub mounted dyno, I don't know the answer. I would think that one would do a dyno run in the gear that is 1:1 with the engine, that would minimize transmission losses due to not driving the counter-shaft. The above are my thoughts - I have not spent any time on a dyno, although I would like to think I am familiar enough with the principles to discuss them intellegently. I'm open to correction.
 

bhtooefr

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Let's say that they're both 4-cylinder diesels, running fuel from the same source, the same batch, and the same pump.

Tentatively, I'll say Car B by a few millimeters.
 

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Well the answer is you run a car on a dyno in a gear that is closet to a 1:1 ratio is because if you run it in 1st gear (let's say it's a 4:1 reduction) you will get a torque rating of something like 600lbft of torque...
 

jackbombay

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jnecr said:
No.. you won't "trade" torque for HP, both figures will increase at a ratio that will be the same as the gear reduction ratio. If, of course, you plot the curve in usual format with posting engine RPM rather than wheel RPM.
You need to re-read this thread, you are not understanding the concepts being discussed.

Gear ratios greater than 1 (1st 2nd and 3rd gears in a VW TDI) DO exchange HP for TORQUE, not HP and TQ output of the engine, but RPM and TQ output of the transmission.

Did you read my explanation of how a 2 HP weedwacker engine canmake 1000 lb/ft of torque? It can because it trades HP for TQ at the expense of RPM that is why the weedwacker can produce 1000 lb/ft at 10.5 RPM, the weedwacker engine cannot possibly run at 10..5 RPM that is why the a gearbox is necessary to "trade" Hp for Tq at the expense of RPM
 
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TDIMeister

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jnecr said:
Without a doubt, the car in second gear will accelerate quicker.
Yes, and why is that? Is the gearbox some sort of magical device that creates energy from nothing? No. All it's doing is just converting torque-in to torque-out with a proportional change in the input and output shaft speeds. The common denominator is that the power going into the gearbox is still the same as the power (magnitude) going out of it (well, minus losses, of course).

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.
Vacuum or no vacuum :) , I believe I covered what you are saying in my first post in the thread.:

IF the question were reworded and you set the gear ratios to be the same between the two cars and you stay in the same gear during the acceleration run, then you change everything, and then YES, engine TORQUE alone will determine the acceleration because now you've set the relationship of the rotational speeds of the wheels relative to the engine between the two cars to be constant, and then, the engine with the greater torque wins the race.
 

jackbombay

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jnecr said:
Well the answer is you run a car on a dyno in a gear that is closet to a 1:1 ratio is because if you run it in 1st gear (let's say it's a 4:1 reduction) you will get a torque rating of something like 600lbft of torque...

BINGO, you are very close to getting this now. Why do you suppose you would end up with 600 lb/ft on a dyno in 1st gear? Because the tranny tradded away a lot of RPM for torque, which has nothing to do with the output of the engine. Sure you had more torque at the wheels but with less RPM at the wheels so you don't have any more power than you would have if you ran the dyno in 4th gear

Are you familiar with the following equation?

HP = TQ X RPM/5252
 

Fix_Until_Broke

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jnecr said:
Well the answer is you run a car on a dyno in a gear that is closet to a 1:1 ratio is because if you run it in 1st gear (let's say it's a 4:1 reduction) you will get a torque rating of something like 600lbft of torque...
Yes - and your horsepower will be the same (assuming transmission losses are equal for each gear ratio)

HP = (Torque * RPM)/constant of your choice

The dynomometer measures torque and speed and then calculates horsepower. It does not care what drives it, what gear it's in, etc as long as the mechanical limits of the dyno are not exceed.
 

jnecr

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Fix_Until_Broke said:
HP = (Torque * RPM)/constant of your choice
Constant of your choice?!?! Do you think they just plucked a number out of space? The number must be 5252 if you want to convert torque to HP. There is a long equation as to how this number was reached and you can find it on the internet, but I don't really think we need to go into that...


Originally Posted by jackbombay, a question that he REALLY wants JNECR to answer, pretty please :)
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.
Well?:)
Right around 4700RPMs...


So with that mainly unrelated question out of the way.. :)

The original question that we are now discussing is:
Is a TDIs instantaneous acceleration higher at 1900RPM or at 3500RPM?
We can reword this in mulitple ways:
Will a car's instantaneous acceleration be higher at it's HP or torque peak?
Will a car's longitudinal acceleration plot mirror it's torque dyno curve or it's HP dyno curve?


I finally found a good example to show my side of the discussion (not argument, mind you.. :) ):
So we'll go to number 3 on the wording, "will a car's longitudinal acceleration plot mirror it's torque dyno curve or it's HP dyno curve?"
Well.. there's a very simple conclusion if we examine how a dyno works in the first place.
What does a dyno really measure?
The fact is there is no way of directly measuring power - all types of dynamometer measure torque and then power is calculated from the formula BHP = Torque (ft/lbs) x rpm/5252.
A Dyno measures the force put on the rollers. F=ma (Force equal mass * acceleration)
Force on the rollers is therefore the roller mass multiplied by the acceleration. This force is multiplied by the radius of the roller itself to give torque at the wheels using the following equation:
T = Fr
Where T is torque; F is force; r is radius of application

Therefore we can easily show that acceleration is directly related to torque because, after all, that is how we arrive at the torque curve to begin with.

If it were the other way around (HP related to acceleration) we would directly measure HP on a dyno and have to have an equation similar HP = Tq*RPM/5252.

So have I made my point? If I haven't I will be more than pleased to explain this again in a different example, just gimme time to think of one... But in reality I think this is probably the best one I'll ever come up with.
 

jackbombay

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nicklockard said:
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.)
It may seem crazy, but the graph you desire is nothing more than a graph of the HP output vs. RPM of an engine.

nicklockard said:
PS: I hear good things about Propecia ;)
No no, that was not me depicted in my drawing, I'm far skinnier than that guy.
 

TDIMeister

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It seems to me each and everyone contributing in this thread, by training or by intuition, knows a piece of the puzzle to settle the TQ vs. HP question, but are getting hung up on semantics (and I can consider myself guilty for it as well, for which I should get back to preparing for my exams! :eek: ) because we're not starting from the same boundary conditions that would help us start on the same plane of understanding.

Ultimately, yes, jnecr is correct: the thrust at the wheels is purely dependent on the torque at the wheels. At the exact same engine RPM and exact same gear, the engine that produces more torque WILL always result in greater acceleration. I believe we are and have always been unanimous in agreement of that concept.

Through this magic of gear reduction, however (for this illustration I will use a nice round number like 5), we are partially correct in saying that the torque at the wheels equals the the torque the engine develops times 5. But what we fail to remember is that the RPM at the wheels is now equal to the RPM at the engine divided by 5. In the same way, though some other gear ratio, it is entirely possible that a car that is otherwise entirely identical, but whose engine produces lower torque but at a higher RPM, can match or exceed the acceleration of the car with much higher torque at a low RPM. What will determine which car accelerates quicker is POWER:

HP = Torque * RPM / 5252

So easy and yet so hard to understand, isn't it?

Going back to the very first post by Jackbombay, the question was posed: the driving tires of Car A sees a constant 200 WHP (high torque, low RPM); the driving tires of Car B sees a constant 200 WHP (low torque, high RPM). Which one accelerates quicker? All the information about the engine, CVT, and any further thoughts of torque curve, driveline losses, yadda yadda yadda are just, at the end of the day, smokescreens for the purposes of the question.

Now, pay attention:
For car A, 200 HP at the wheels rotating at, say 1000 RPM (the wheels, not the engine), represents 1050 lb.ft. If the tires are 24 inches in diameter (1 foot in radius -- God, I hate Imperial units), then this represents a thrust or accelerative force of 1050 lbf.

For car B, 200 HP at the wheels rotating at, say 1000 RPM (the wheels, not the engine), represents 1050 lb.ft. If the tires are 24 inches in diameter (1 foot in radius), then this represents a thrust or accelerative force of 1050 lbf.

Golly gee, they're the same!
 

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Note that my argument was that the cars started out the same weight, but if the 5K RPM engine uses more fuel, then it would win, as it loses weight at a faster rate.
 

jackbombay

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jnecr said:
Right around 4700RPMs...
Why would you shift at 4700 RPM if your car accelerates faster at 1900, shifting at 4700 would put your next gear at about 3000 RPM, well above the point at which you cliam max accel occurs, 1900 RPM.

jnecr said:
A Dyno measures the force put on the rollers. F=ma (Force equal mass * acceleration)
Force on the rollers is therefore the roller mass multiplied by the acceleration. This force is multiplied by the radius of the roller itself to give torque at the wheels using the following equation:
T = Fr
OK!!! We are very close to being peaches and cream.

Max power at the wheels for any gear is acheived when the engine is producing max HP. You, as far as I can tell, have been interchanging wheel TQ and engine TQ without noting that you have done so.

Of course 600 lb/ft of wheel torque will accelerate the car faster than 300 lb/ft at the wheels for any given wheel RPM.

In any given gear, lets use first as an example, acceleration will be greatest when wheel power is at its peak, peak wheel power IS produced when the engine is at its HP peak.

3... 2... 1 ... Epiphany?

Edited for accidental interchange of "power" and "torque" and in effect posting a post that took the oposite position that I have taken this whole thread... Ooops.
 
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Fix_Until_Broke

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TDIMeister summed it up nicely

I won't pick on the details that I believe are in error in jencr's post because it will just cloud the ultimate realization we have all been trying to convince each other of in our own way. Power = (Torque x Rotational Velocity)/Constant. (Pick any units you want that satisify the equation)

On otherwise similar vehicles, the vehicle with the most power at the wheels will accelerate the fastest starting from the same initial velocity.
 

TDIMeister

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jnecr said:
Fix_Until_Broke said:
HP = (Torque * RPM)/constant of your choice
Constant of your choice?!?! Do you think they just plucked a number out of space? The number must be 5252 if you want to convert torque to HP. There is a long equation as to how this number was reached and you can find it on the internet, but I don't really think we need to go into that...
I think the point F_U_B was trying to make -- and one which is appreciated because SAE horsepower and pounds feet are not the only measurements for power and torque, respectively, and not everyone in the globe deals with power and torque in those units -- was that the "constant of your choice" depends on what units of power and torque you choose to deal with. If you deal with kilowatts and newton metres, you would get a different constant. If you deal with PS and kilogram metres, you get yet another different constant.
 

jackbombay

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TDIMeister said:
Ultimately, yes, jnecr is correct: the thrust at the wheels is purely dependent on the torque at the wheels.
not the torque at the engine.

TDIMeister said:
At the exact same engine RPM and exact same gear, the engine that produces more torque WILL always result in greater acceleration.
I want to clarify this for Jnecr.

Meister is saying that if 2 engines are both spinning at 3000 RPM and one is making 200 lb/ft and the other is making 250 lb/ft the one making 250 lb/ft is making more power.

Can any of you write me a prescription for propecia?
 

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jackbombay said:
Why would you shift at 4700 RPM if your car accelerates faster at 1900, shifting at 4700 would put your next gear at about 3000 RPM, well above the point at which you cliam max accel occurs, 1900 RPM.
Because the car will accelerate faster in 1st gear at 4700RPM than it will at any RPM in 2nd gear! :)


Max torque at the wheels for any gear is acheived when the engine is producing max HP. You, as far as I can tell, have been interchanging wheel TQ and engine TQ without noting that you have done so.
Wheel torque and engine torque are just a correlation factor away. There's nothing to do with HP here. Here's is where you are getting confused: Max torque at the wheels for any gear is acheived when the engine is producing MAX TORQUE. End of discussion on that.
 

Fix_Until_Broke

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jackbombay said:
Max torque at the wheels for any gear is acheived when the engine is producing max HP. You, as far as I can tell, have been interchanging wheel TQ and engine TQ without noting that you have done so.

Of course 600 lb/ft of wheel torque will accelerate the car faster than 300 lb/ft at the wheels.

In any given gear, lets use first as an example, acceleration will be greatest when wheel torque is at its peak, peak wheel torque IS produced when the engine is at its HP peak.
Max torque at the wheels for any gear is achieved when the engine is producing max TORQUE, otherwise HP in does not equal HP out

Tin * nin = HPin = HPout = Tout * nout (assuming no losses in gears)
100*2000=20000=200*1000
 
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jnecr

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TDIMeister said:
I think the point F_U_B was trying to make -- and one which is appreciated because SAE horsepower and pounds feet are not the only measurements for power and torque, respectively, and not everyone in the globe deals with power and torque in those units -- was that the "constant of your choice" depends on what units of power and torque you choose to deal with. If you deal with kilowatts and newton metres, you would get a different constant. If you deal with PS and kilogram metres, you get yet another different constant.
Gotcha, and I think this would be a much easier discussion if we used SI units, because they are so damn easy to work with...
 

Fix_Until_Broke

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TDIMeister said:
I think the point F_U_B was trying to make -- and one which is appreciated because SAE horsepower and pounds feet are not the only measurements for power and torque, respectively, and not everyone in the globe deals with power and torque in those units -- was that the "constant of your choice" depends on what units of power and torque you choose to deal with. If you deal with kilowatts and newton metres, you would get a different constant. If you deal with PS and kilogram metres, you get yet another different constant.
Exactly - Thanks
 

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jnecr said:
Because the car will accelerate faster in 1st gear at 4700RPM than it will at any RPM in 2nd gear! :)
We'll make it easy and give the benefit of the doubt and say also 4700 RPM in 2nd gear for the other car also. :) Neglecting drag (driving in a vacuum :) ) the power at 4700 RPM is the same at 1st gear or second. But because of gear reduction, the rotational speed of the drive wheels is less in first than in second. Same power over slower rotation equals more torque equals more force equals more instantaneous acceleration. But while you're enjoying the high instantaneous acceleration of sitting in first gear, the guy in the car beside you upshifts to a gear that gives less acceleration but will still leave you in the dust while you run out of revs. But I digress. :D

Wheel torque and engine torque are just a correlation factor away. There's nothing to do with HP here. Here's is where you are getting confused: Max torque at the wheels for any gear is acheived when the engine is producing MAX TORQUE. End of discussion on that.
I'm not going there again!
 
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jackbombay

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jnecr said:
Because the car will accelerate faster in 1st gear at 4700RPM than it will at any RPM in 2nd gear! :)
Why would you intentionally handicap youself (according to what you believe) in a race where you have so much to loose by winding the engine up to 4700 RPM when you believe acceleration is maximised at 1900 RPM?

jnecr said:
Max torque at the wheels for any gear is acheived when the engine is producing MAX TORQUE. End of discussion on that.
Fix_Until_Broke said:
Max torque at the wheels for any gear is achieved when the engine is producing max TORQUE, otherwise HP in does not equal HP out
Yes, I had interchanged power and torque in that last post accidentally:eek:.
 

TDIMeister

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`Tooef and others, you have asked for a revision for the TDI quiz. Guess what questions will be asked?!?! :D

There will be a WHOLE section of questions dealing strictly with torque versus power. For one of them I think I'll just copy Jackbombay's question at the top of this thread verbatim :p Study now and study hard! :D

http://www.tdiclub.com/Quiz

Answers here.
 

jnecr

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jackbombay said:
Why would you intentionally handicap youself (according to what you believe) in a race where you have so much to loose by winding the engine up to 4700 RPM when you believe acceleration is maximised at 1900 RPM?
Jack, this is why our TDIs don't do the quarter mile well.



You would want to shift wherever the thrust lines intersect. On our cars the lines would look very different because our torque peaks would never be used... I'll see if I can't get a graph of our cars made, shouldn't be all that hard...
 
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