yes I know, maybe not a question you'd like to hear. But I'm new, and you gotta learn stuff sometime. So can people help me understand what torque is?

 

Clickme

Type in your question and learn away!


[Modified by B18C-EJ1, 2:48 AM 10/13/2002]

 

Google is your friend.

http://www.physics.uoguelph.ca/tutor...que.intro.html
http://www.slcc.edu/schools/hum_sci/...or/2210/torque/

Start there, and then get into engine dynamics.

 

its what my car lacks

 

Whatever it is, in the end it comed down to this: a car with "torque" makes power at low rpms, a car with no "torque" does not make power at low rpms.

 

You know there's such a thing as high end torque without low end, right?

 

Yea thats usually created on our cars with a turbocharger or nitrous :D

 

Yes, that's why I put "torque" in quotes. My point is that if a car makes 1000lb-ft of torque at 10,000rpm and 2lb-ft at 4000rpm, a V8 domestic guy is still going to say "this car has no torque".

I simply stated the meaning that the word has taken with the common man, as it relates to cars.

 

Torque is "twisting force"

Your axles are twisting, therefore they are under a load of torque.

Don't think of tension, which is "push-pull force"

 

What makes you think anyone thought of it as a "push-pull force"?
Are you answering the original question or commenting on something I wrote?


[Modified by Lsos, 8:10 AM 10/14/2002]

 

Torquie = Force x Distance.

So if you (say 180 lbs) stand on the end of a 1 foot breaking bar to remove your axle nut you are putting 180 ft-lbs of torque on that nut. Now if you put a cheeter bar over the breaker bar that was two feet long, T=180lbs x 2 ft and you would get 360 foot pounds. Or if I stood on the 1 foot breaker bar F= 300lbs x 1 or 300 ft lbs.

Now you have heard that torque is what accelerates a car. Here is where it comes from. Take the torque at the flywheel (lets assume 100 ft lbs), now we meet the transmission and differential. When a gearset slows down the shaft, the torque is increased by the inverse ratio of speeds. In other words, if in first gear we slow the engine down by a 3/1 ratio, that 100 ft lbs is increased to 300 ft lbs. Now if the differential and final drive are another 3/1 ratio that 300 ft lbs becomes 900 ft lbs.

Now the tire gets that torque to the road and turns it into a pure force again. Just use our little equation above and say the tire is 24 inches in diameter, than the radius is 12 inches or 1 foot. That becomes 900 ft lbs/1 ft = 900 lbs accelerating your car forward. (With no gear and friction losses considered.)

For acceleration F=mass x acceleration. And that is how it goes. The guy or gal with the most torque to the ground for the most time wins. (assuming both cars are the same mass) You can also see why drag racers try so hard to make their cars light. Reduce the mass 10% or increase the force at the road 10% gets you about the same benefit.

Have fun learning, we all started only knowing who Mom was.........

Hope this helps you some.
Regards,
BigMoose

 

Couple of links
http://ubermensch.org/Cars/Technical/hp-tq
http://www.rs-productions.com/RSP_Mo.../hptq/hptq.htm

 

I have seen those links and variations thereof all over the place. Unfortunately they are full of misinformation. The major one is the claim that a car's acceleration exactly matches its torque curve. This statement is a direct violation of the The First Law of Thermodynamics, which states that energy must be conserved.

Following the logic of the statement which I claim is wrong, an engine making 300lb-ft of torque and 100 hp will accelerate a car three times faster than an engine making 100lb-ft of torque and 300hp.

How can a car that's receiving 223800 joules per second of mechanical energy be accelerating three times slower than a car that's receiving 74600 j/s of mechanical energy? Where is 149200 joules of mechanical energy per second going?

I have found one link that backs up what I say, and it saddens me that there aren't any more:

http://www.leecao.com/honda/vtec/commonmiss.html

I won't claim to be an expert on thermodynamics, and so I invite anyone who has some understanding of it to refute what I wrote.


[Modified by Lsos, 9:11 AM 10/16/2002]

 

That link doesn't say anything that challenges the fact that an engine accelerates at a rate that matches it's torque curve.

And there is no energy "missing" in your comparison. The 100 lb/ft. of torque engine will be faster. You're just ignoring gearing. With 300 hp, it's an indication that that smaller amount of torque is made at a very high rpm (15,756 rpms, to be exact). That means that the 100tq/300hp engine is going to be geared way more aggressively than the 300tq/100hp engine (which makes it's peak hp at 1,750 rpms). That gearing gives a huge mechanical advantage in the transmission, which is invisible at the crankshaft.

Horsepower as an indication of a car's acceleration works under the assumption that the car is geared appropriately.

 

Tq vs Hp Primer

 

That's actually already been linked to in this thread. You'll find that exact same description in about 50 different places on the net - and rightly so, it's one of the best out there.


[Modified by Daemione, 1:32 PM 10/16/2002]

 

torque: The instantaneous measure of rotational energy, measured about a virtual lever arm 1ft long.

horsepower: Torque over time, or the amount of work capable of being done with a given rate of energy output. Stated another way, horsepower is 550 lb/ft of torque per second.

Think of torque as Energy (rotational energy). Energy is defined as force x distance here, such as 50 lbs of force applied to a 2ft lever arm would equal 100 lb/ft of torque. Further, think of horsepower as Power, which for this discussion is energy x time. That same 50 lbs applied to a 2ft lever arm, rotating a crankshaft at 8000 RPM, would equal 152hp.

The most important thing in this discussion is examining why horsepower is the single most important factor to a vehicle's accelerative ability (or more truly, the vehicle weight to horsepower output ratio), yet has no direct application to the physics of acceleration. To understand this, you must first understand what makes a car accelerate.

Acceleration is a time based event, but the force (energy) behind it is NOT. The energy of acceleration is not just torque either, it's effective torque to the wheels, which is found by multiplying engine torque at a specific RPM by the gear ratio you are in and the final drive ratio (shown here in this example)....
Example A:
first gear transmission ratio: 3.42:1
final drive ratio (differential reduction): 4.27:1
engine torque at 5000 RPM: 150 lb/ft
150 x 3.42 x 4.27 = 2190.5 lb/ft effective torque

So in that example, our meager torque output is greatly magnified to about 14.6 times it's original amount of energy. Just as the 2ft lever arm in the earlier example multiplied the force input by 2, the gear reduction here multiplied the torque output by 14.6. Now if you know gearing, you also know that the final drive ratio is fixed for a given differential, but the higher the gear you select, the lower (numercially speaking) the transmission gearing becomes. This is why acceleration always feels weaker in the next higher gear; torque multiplication through gearing is reduced, thus reducing the energy available to accelerate the car.

So why is any of this important? Because, if you understand how important gearing is to the equation, you understand why torque output, taken by itself, is meaningless in estimating accelerative ability. A typical deisel engine in a car outputs near to (or over) 300 lb/ft of torque, yet is almost never as fast as gasoline engines making 100 lb/ft less torque at peak. This is because deisel engines cannot run at high RPM, and therefore cannot take full advantage of torque multiplication through gearing. Since most cars are geared to redline at similar speeds in a given gear (usually around 35-40 mph in first, 60-70 in second, etc...), the engine's redline and power output at high RPM are actually indicative of it's gearing. So let's say we have two engines, Engine A makes 300 lb/ft of torque at all RPM and 300hp at redline, while Engine B makes 150 lb/ft of torque at all RPM and 300hp at peak. Given the earlier math that hp = torque x 5252/ RPM, one can see that engine B must be turning twice the RPM as Engine A to make the same amount of horsepower with only half as much torque to work with. And this is extremely important to realize because, if Engine B's drivetrain is geared exactly twice as short (numerically higher), the effective torque to the wheels will be precisely the same at any speed relative to the Engine A drivetrain. Which means that, given the vehicles weigh the same and we throw out variables like drag and traction, they will both have EXACTLY the same accelerative energy available to move the car forward. Which means they will both accelerate at exactly the same rate, even though the first motor makes twice the energy of the second.

And this is why cars like the S2000, what with it's meager torque output, can run side by side a car of around the same horsepower but with much more torque. To state this simply, horsepower ratings are like a clean summed up guesstimate of a car's accelerative ability, because they nicely package in torque and gearing into one number. And while this is sligthly oversimplified (given I only have so much room to type), that really is how it works. Look at any two cars with about the same horsepower output and you will see they both have about the same acceleration curves... regardless of torque output. Although the S2000 only has about 150 lb/ft of torque to work with at peak, it's gearing makes full use of this torque to motivate the car. Which is why it still feels quick at 4000 RPM, when it's a full 3500 RPM away from it's torque peak.

So to sum up, torque is the only thing an engine produces. It is the direct measure of instantaneous energy the engine makes to provide acceleration, and is the only thing directly important to the physics of acceleration. And yet it is also the most useless single number to look at when judging accelerative ability of a car, since 1000 lb/ft of torque means nothing unless you've got gear multiplication (read: RPM ability) behind it.

Conversely, horsepower output is a direct derivative of torque, and has no direct bearing or application to the physics behind acceleration. Yet it is the single most important number in judging the accelerative ability of a vehicle, because it so nicely packages torque output, RPM ability and gearing into one roundabout number.

And this strange dichotomy is the reason why manufacturers always give both figures, and why people will continue to post excellently informative threads like this on every atuomotive message board across the world. So after reading this, all I hope is that you come away with a newfound respect for both horsepower and torque numbers, and do not immediately dismiss any engine for focusing upon healthy production of one over the other. After all, everything has it's application, the S2000 motor is a gem because it packages 240hp into a 320 lb engine/transmission combo, while the BMW 3.2L I6 weighs much more but has a wonderfully broad and muscular torque curve. Hope this helps explain things, peace.


Ps-As one last thing, I want to say that increasing torque output over a large RPM range continues to be the most effective way to increase a vehicle's accelerative ability. You can increase horsepower output without chaning maximum or even average torque levels (by moving the powerband to a higher RPM range), but this will never be as effective in cost or extra speed as simply making more torque. And that's because you have to adjust gearing to take advantage of this newfound RPM potential, which isn't cheap. This is why forced induction setups are so effective at improving acceleration with minimal cash outlays, and why making an NA car go as fast will take MUCH more time, effort and money to acheive.


[Modified by texan, 10:45 AM 10/16/2002]

 

I'm challenging this!

If you've ever been in a car with a b16, which has a pretty much flat torque curve, you'll realize this is not true. It starts to feel quick at around 5000rpm, even though the torque is very close to what it was at around 2000rpm, where it feels like a dog. The difference in torque is probably no more than 10ft-lbs!

The links state that the car with more torque will feel faster, and yet the car with more horsepower will come out first. I'm saying this is not true. I'm saying that the car with more horsepower will feel faster and will come out first.

It is precisely the statement "a car accelerates at a rate witch exactly matches its torque curve" that violates the First Law of Thermodynamics. If a b16 was accelerating a car at the same rate at 2000rpm as at 8000rpm, which all these links imply, there's a hell lot of unaccounted for energy at 8000rpm.

 

Thank you, Texan, as always that was a very clear explanation.
A stock b16 pulls roughly 70 lb/ft. of torque at 2,000 rpms, and about 93 lb/ft. at it's peak just after it hits VTEC. This is about a 25% difference. When that torque difference is magnified through a low gear, it will feel very drastic. Take 1st gear in a B16 (FD of 4.4, gear ratio of 3.23). At 2,000 rpms, you're putting down 995 lb/ft. of torque. At 5,000 rpms, that figure will leap to 1,322 lb/ft.

So the rate of acceleration follows the torque curve, but the amount that torque curve is exaggerated is determined by the gear ratio.

And of course if you're chugging along in 4th gear at 2,000 rpms, & drop it into 3rd to go up to 5,000 rpms, not only are you at a meatier part of the torque curve, but you've improved your gear ratio substantially by going down a gear. The result is a lot more torque to the ground.

Quote:

The links state that the car with more torque will feel faster, and yet the car with more horsepower will come out first.

I honestly don't know where you're reading that. It's not in the link you provided, nor on any of the others linked to in this thread.

Quote:

It is precisely the statement "a car accelerates at a rate witch exactly matches its torque curve" that violates the First Law of Thermodynamics. If a b16 was accelerating a car at the same rate at 2000rpm as at 8000rpm, which all these links imply, there's a hell lot of unaccounted for energy at 8000rpm.

Again, I don't understand why you think there's energy being unaccounted for. Assuming the torque rating & gear ratio are exactly the same, yes, the car will pull just as hard at 2,000 as it does at 8,000 rpms. What other factor would influence this? (okay, maybe wind resistance, since you'll be going faster at 8,000 rpms.)

 

anyone with a honda (including myself), couldn't tell ya !

good explinations tho guys

 

"First, each car will push you back in the seat (the fun factor) with the
same authority - at least at or near peak torque in each gear. One will
tend to *feel* about as fast as the other to the driver, but the LT1 will
actually be significantly faster than the L98, even though it won't pull
any harder."

That is taken directly out of the links. Not my link, the other links.

My problem with this is that at 8000rpms there's more than 4x the mechanical energy being dumped into the car, and yet it's not accelerating any faster. The speed of a car can be expressed in terms of how much energy it has:

ME=1/2mv^2

This means that if a car has 4x more energy than another, it will be traveling faster. This also means that if more energy is being put into the car, it will be accelerating faster. And yet here they are saying it won't.

It's the same with everything involving energy. You put heat energy into water faster, it will change temperature faster. You put electrical energy into a capacitor faster, it will change charge faster. You put mechanical energy into a car faster, it will change speed faster. If it does not, then it leaves open the possibility of water having a 1000000000j of heat in it, yet having the same temperature as the same amount of water having 5j in it.

The links state that between time A and time B, despite more energy being put out by one engine, both equal weight cars will accelerate just as fast. Therefore, both cars will be traveling just as fast at time B, despite one car carrying more energy than another. This is impossible.


[Modified by Lsos, 10:38 PM 10/16/2002]

 

You're taking that quote out of context. If you read the next 3 or 4 paragraphs, it explains why. The two cars will pull just as hard as each other at their respective torque peaks, and in fact be just as fast as eachother at their torque peaks. But the LT1's curve is broader & has a higher horsepower peak, which means it can stay in the lower gears longer before it benefits itself to upshift. That makes it faster than the L98.
Here's where you're going wrong. RPM's do not equal mechanical energy (in the context that you're using it). The energy being outputted by an engine is represented by it's torque figure, that's it's sole purpose. An engine putting out 100 lb/ft. of torque at 2,000 rpms is putting out the exact same amount of energy as an engine putting out 100 lb/ft. of torque at 8,000 rpms.

Like Texan and all the articles have indicated, the vastly different horsepower rating (which is affected by RPM) is only useful when it's combined with gearing.


[Modified by Daemione, 6:15 PM 10/16/2002]

 

Texan has a great explanation of it all. That link everyone passes about the Torque vs HP using the LS1 and other car isnt the best way of explaining torque and HP. Texan (or anyone else who understands this) should writeup an article explaining it more clearly.

You need to understand that energy has no relation to the acceleration of an object. It's only the measure of how much energy the object has at any given moment. Also I believe HP is a measure of Work, not Energy (anyone 2nd/correct?). Read up on some basic physics concepts.

You use car speed and mass to determine the amount of Mechanical or Linear kenetic energy in the system (car). And if even more, you dont use 1/2 mv² to determine the kenetic energy the engine produces, since it is rotational energy, you use 1/2 I(omega)².

--------
if you need to see an example..

4000rpm engine speed vs 8000rpm engine speed. flat 100ft-lb of torque throughout the entire rev band, same gearing, assuming all frictional/air resistance losses are constant for both systems (thou it really isnt).

I know the terms ME/etc probably arent correct to use, but i'm using the ones you used to help u understand what it all means.

At 8000rpm of engine speed, there is twice the amount of HP being produced, BUT the speed of the car is TWICE as fast (since gearing was the same). And ill just say that there is 4x the amount of ME produced as you say... well if the speed of the car is doubled vs the 4000rpm car, w/ 1/2 mv² for kenetic linear energy, then you would need 4x the amount of ME to maintain that speed. Thus no extra energy or watever is being put to accelearting the car any faster than the 4000rpm system.

 

That's where our thinking takes two different paths.
RPMs * torque / 5252 = One horsepower = 746 watts = 746 joules / second
Horsepower is in fact a direct measure of how fast mechanical energy is being put out.

Pump 100 horsepower into a car for a minute and it will be carrying 4476000 joules of mechanical energy. Pump half the horsepower into it, it will be carrying half the energy.

By the formula ME = 1/2mv^2, the 100hp car will be traveling 1.4 times as fast.
Use any physics formulas to arrive at the same answers.

By the way, how can you agree with texan when you're saying "So the rate of acceleration follows the torque curve" and he's saying "Look at any two cars with about the same horsepower output and you will see they both have about the same acceleration curves... regardless of torque output."?

I don't know if he's agreeing with me, but I don't think he's agreeing with you either.

Type SH: Energy does in fact play a huge role in the acceleration of an object, as long as the equatiuon ME=1/2mv^2 holds true. Just take the derivative of that equation and you've got acceleration.

By definition, energy is the ability to do work. Horsepower is a measure of how fast energy is being put out, or how fast work can be done.

8000rpms is twice the speed as 4000rpms only in the same car. Imagine two different cars going 100mph, one with 10000hp and 1lb-ft of torque, the other with 1hp and 10000lb-ft of torque. Your argument does not explain what would happen in this situation.


[Modified by Lsos, 6:20 AM 10/17/2002]

 

yea, Lsos is right in pointing out that 100ft-lb of torque at 2000rpm is NOT the same <U>energy</U> as it is at 8000rpm, but it is the same amount of Force.

Problem w/ debating any torque/HP/theoretical topic is that people use the wrong terms. =\

-edit-
hmm.. i dont recall learning that the derivative of KE = Acceleration. Any chance you can break down the proof in a nutshell?

f = 1/2 mv²
which gives
f' = mv

which is momentum, not acceleration.

-------
for the situation w/ 10000HP and 1ft-lb of torque, and 10000ft-lb of torque and 1hp,
the setup w/ 10000HP will be putting down 10000 times more acceleration force than the other one, because its doing 10,000 times more work than the other in the same time frame.

You are using diff car setups there though, no way for me to prove that they will behave the same.

----------
for your example here, the thing is that engine speed is directly related to horsepower. Double the engine speed, double the HP. Double the engine speed, double the car speed (in any given gear).

If you want to pump a constant 100HP amount of work into a car for 1 minute to get the total of 4476000 joules, the torque curve will need to be downsloping for as the engine/car speed goes up (to maintain the steady 100HP input of energy for the minute). That is why it will be accelerating slower as the speed increases.

Cut the HP you pump in half, cut the torque you need to get that HP in half, thus cutting the acceleration in half.

For that example, if you were to apply a steady 100ft-lb of torque over the whole minute, acceleration should be constant and HP would rise linearly w/ the engine/car speed (assuming no losses in the system anywhere).

Think of gravity, the earth's gravity force is for the most part, pretty constant (the variation in altitude is like negligable). w/o air resistance, if u drop any object, it will fall at a constant rate and have an increasing speed because the force of gravity is constant. Same w/ the car, if you apply a constant force, w/ no wind resistance or frictional losses, it will accelerate at a constant rate and have an increasing speed.

Even thou HP increases as the engine speed increases, a constant force still means a constant acceleration.


[Modified by TypeSH, 8:19 AM 10/17/2002]