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TDI Power Enhancements Archives on TDI Power Enhancement related items.

View Poll Results: Which car would win in a drag race?
Car A will win because it has twice as much torque. 218 51.66%
Car B will win because the engine spins faster. 81 19.19%
The cars will tie. 123 29.15%
Voters: 422. You may not vote on this poll

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Old February 11th, 2006, 21:10   #31
Mad_Matt
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Arrow 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
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Old February 11th, 2006, 21:15   #32
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Quote:
Originally Posted by Mad_Matt
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?
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Old February 11th, 2006, 21:21   #33
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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!!!
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Old February 11th, 2006, 21:27   #34
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Smile 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.

.
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Old February 11th, 2006, 22:12   #35
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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.
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Old February 11th, 2006, 22:34   #36
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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?

Last edited by nicklockard; February 11th, 2006 at 22:36.
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Old February 12th, 2006, 00:35   #37
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Quote:
Originally Posted by nicklockard
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.



Quote:
Originally Posted by nicklockard
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.
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Old February 12th, 2006, 02:27   #38
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That drawing is amazing.
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Old February 12th, 2006, 06:31   #39
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Quote:
Originally Posted by david_594
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!!!
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Old February 12th, 2006, 08:01   #40
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Quote:
Originally Posted by jackbombay
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...
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Old February 12th, 2006, 08:52   #41
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Quote:
Originally Posted by jnecr
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.
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Old February 12th, 2006, 09:59   #42
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Quote:
Originally Posted by jnecr
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,

Quote:
Originally Posted by TDIMeister
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.

Quote:
Originally Posted by jnecr
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?
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Old February 12th, 2006, 10:00   #43
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Quote:
Originally Posted by Fix_Until_Broke
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
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Old February 12th, 2006, 10:32   #44
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Quote:
Originally Posted by jackbombay
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.

Quote:
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.
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Old February 12th, 2006, 10:43   #45
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Quote:
Originally Posted by TDIMeister
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,

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

Quote:
Originally Posted by TDIMeister
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.
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