Horsepower Vs. Torque ???
Rotary Enthusiast
Joined: Jan 2002
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From: 15143
HP vs. Torque - the debate that will never end...
I like playing with numbers, so let's try a pair of different engines in the same car. Nothing else changes. Because the diffential ratio and tires are the same, whichever engine can spin the diff harder should accelerate harder. But do we look at power or torque at the diff? At any given road speed, the diff's rpm is fixed, so hp = torque * rpm / 5252 tells us that power and torque are directly linked. Increase one and the other goes up by the same percentage. So it doesn't matter which we examine at the diff. OTOH, that doesn't necessarily mean that the same thing applies at the engine.
One engine makes more torque, but has a lower redline so power is the same. To keep things brief, I'll just see which accelerates harder at 40 mph. At 40, the differential input must be spun at 2000 rpm.
Engine 1) 200 ft-lb at 6000 rpm = 228 hp
Engine 2) 300 ft-lb at 4000 rpm = 228 hp
To get 2000 rpm at the pinion, Engine 1 uses 1st gear (a 3:1 ratio) while Engine 2 uses 2nd (2:1). So both put 600 ft-lb to the pinion and must accelerate the same! And, check THIS out: 600 ft-lb at 2000 rpm = 228 hp! Looks like acceleration matches power, not torque.
If you continue playing around like this, you can find that the linear force (in pounds) pushing your car forward = hp * 375 / mph. In my example, both cars have 1710 lbf and identical acceleration (in G) = 1710 / weight.
Don't argue with me that torque is more important than power, until you find an equation to predict 1/4-mile trap speed from only weight and torque.
The really good clue is that every performance number we care about includes time in some way: miles per hour, elapsed time in seconds, etc. But torque doesn't include any time unit, so it can't predict these things for us, while power can.
And a couple notes from above.
Ummm, yeah... Seems pretty unfair to blame power for that one. More top speed must come from more power or less drag. Torque won't do it unless it also gives more power.
But NOT for the amount of air they move. Compare an N/A 13B to any 2.6L piston engine.
The above math refutes this, and I promise that I can find the shift points faster from a power curve than you can run the torque and gear ratios through a calculator.
I like playing with numbers, so let's try a pair of different engines in the same car. Nothing else changes. Because the diffential ratio and tires are the same, whichever engine can spin the diff harder should accelerate harder. But do we look at power or torque at the diff? At any given road speed, the diff's rpm is fixed, so hp = torque * rpm / 5252 tells us that power and torque are directly linked. Increase one and the other goes up by the same percentage. So it doesn't matter which we examine at the diff. OTOH, that doesn't necessarily mean that the same thing applies at the engine.
One engine makes more torque, but has a lower redline so power is the same. To keep things brief, I'll just see which accelerates harder at 40 mph. At 40, the differential input must be spun at 2000 rpm.
Engine 1) 200 ft-lb at 6000 rpm = 228 hp
Engine 2) 300 ft-lb at 4000 rpm = 228 hp
To get 2000 rpm at the pinion, Engine 1 uses 1st gear (a 3:1 ratio) while Engine 2 uses 2nd (2:1). So both put 600 ft-lb to the pinion and must accelerate the same! And, check THIS out: 600 ft-lb at 2000 rpm = 228 hp! Looks like acceleration matches power, not torque.
If you continue playing around like this, you can find that the linear force (in pounds) pushing your car forward = hp * 375 / mph. In my example, both cars have 1710 lbf and identical acceleration (in G) = 1710 / weight.
Don't argue with me that torque is more important than power, until you find an equation to predict 1/4-mile trap speed from only weight and torque.
The really good clue is that every performance number we care about includes time in some way: miles per hour, elapsed time in seconds, etc. But torque doesn't include any time unit, so it can't predict these things for us, while power can.
And a couple notes from above.
At top speed you are using alot of power but feel no acceleration.
(snip)
You accelerate when you have more power available than you need to maintain your current speed.
(snip)
You accelerate when you have more power available than you need to maintain your current speed.
Rotories have lot's of torque ( even if its not important ) espessially for their size
your theoretical optimum shift point is where torque multiplied through the gear box and rear end drops below what is if you were to shift into the next gear at that speed. which can not be found by reading a power graph because, in a power graph there is no torque line number to calculate this from
Full Member
Joined: Mar 2002
Posts: 168
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From: Monterey Bay Area (Capitola) California
Hallelujah and praise the car gods that we've got some engineers in here!....or atleast people that understand that the numbers game means everything and that quoting race car drivers or "uhh I feel the torque more" doesn't mean **** when you can prove it with mathematics and physics. (No personal flames intended)
Someone said they buy the truck with the most "torque" because thats what gets things moved. Well basically what you're buying is the truck with most horsepower in the low end. The only difference is that they know that consumers are stupid and thats why they don't advertise their trucks as having "200 horsepower at 2000rpm" they advertise it as having "x amount of torque at 2000rpm." Which is really the same thing.
Tell me honestly, if you were looking at 2 different trucks and their two ads ran like this.
Truck 1: "It's loaded for heavy duty pulling with its 400 ft-lbs of torque at 2000rpm!"
Truck 2: "Perfect for pulling the heaviest of equipment, it's engine puts out an astounding 228 horsepower at 2000rpm"
Tell me which truck you would buy for towing. Which one sounds like it's got the ***** you're looking for.
Someone said they buy the truck with the most "torque" because thats what gets things moved. Well basically what you're buying is the truck with most horsepower in the low end. The only difference is that they know that consumers are stupid and thats why they don't advertise their trucks as having "200 horsepower at 2000rpm" they advertise it as having "x amount of torque at 2000rpm." Which is really the same thing.
Tell me honestly, if you were looking at 2 different trucks and their two ads ran like this.
Truck 1: "It's loaded for heavy duty pulling with its 400 ft-lbs of torque at 2000rpm!"
Truck 2: "Perfect for pulling the heaviest of equipment, it's engine puts out an astounding 228 horsepower at 2000rpm"
Tell me which truck you would buy for towing. Which one sounds like it's got the ***** you're looking for.
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Joined: Sep 2001
Posts: 163
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From: Portland, OR
[QUOTE]Originally posted by sbertolone your theoretical optimum shift point is where torque multiplied through the gear box and rear end drops below what is if you were to shift into the next gear at that speed. which can not be found by reading a power graph because, in a power graph there is no torque line number to calculate
this from
[B]
WRONG. Torque should not be confused with force (an extremely common misconception). Torque is force times distance (or in the case of rotation, a "twist") the same units as WORK. When I said torque is "force with a twist", I did not mean "twisting force". Torque = the tendancy of a force to ROTATE a rigid body about a fixed axis. No rotation, no net torque! When you tighten your lug nut to 110 ft pounds, once the nut stops rotating there is no net torque, even though force is still being applied. Neither work nor torque contain imformation about rate, thus torque tells you nothing about acceleration unless you know rpm (for a given car's mass and gearing). If torque was force, then you'd be right, you want to maximize force for maximal acceleration (remember force = mass * acceleration), but alas, torque is not force. Force is mass times acceleration (pounds * feet/second^2). If you're interested in the relationship between torque and angular acceleration, net torque = moment of inertia * instantaneous angular acceleration. As far as work and energy in rotational motion go, Power = torque * instantaneous angular VELOCITY. Ahh, velocity contains rate information (phew!). Angular velocity is directly related to RPM.
The rear wheel power curve already contains all the information you need. Besides, torque is usually measured on the dyno through the drive train. Just maximize the area under the power curve considering the rpm drop for each shift point and you will have your best ¼ mile time (assuming you can maintain traction)
.
[QUOTE]
ok just because a force has been applied to your car doesnt mean its going to move you have to apply that force over a distance and a kick is a very short distance will very unlikely to overcome
the static friction of your car sitting. therefore no work will have been done because your car has
not moved allthough some force has been expelled on your carbut it was over 0 distance so if this
kid kicks your car a 100 times a second but cant move your car 0 work has still be done.
[B]
If you believe that then try standing behind a jet turbine. Sure the impulse of a single molecule hitting you is next to nothing but when the rate of collisions gets high enough you will feel the work being done.
[QUOTE]
Mechanical energy must not be expressed in newton metres (N * m) but only in joules (J). The
former unit is used only for torque or moment of force, question is do you know the difference.
this was taken from my 9th grade physics book.
[B]
A Joule comes from "the mechanical equivalent of heat" which was discovered by James Joule in the mid 1800's. A joule has the energy equivalent of 0.2389 calories. One BTU is equivalent to 1055 Joules. By definition though, a Joule (J) IS a Newton meter (N * m). Ninth grade physics books simplify things for ninth graders to understand. You might want to trade it in for a more advanced and accurate book. Energy is the capacity to do work, and has the same units as work (force over distance, or in SI units kilogram * meters^2/seconds^2= Nm=J). There are two types of energy in the classical sense, potential and kinetic. If you lift a one pound object 10 feet high, you have done 10 foot pounds of work. This is because the mass of the object requires force (1 lb = 4.448 Newtons) acting on it to push it against gravity. Now the potential energy of the object (relative to the ground) is 10 foot pounds. You can convert that into Joules if you like, but it is all the same, distance and force (force is mass times acceleration, acceleration is rate of change in velocity, velocity is distance/time). You can now convert that into kinetic energy by dropping the object (potential energy becomes kinetic energy). It will then accelerate at 9.8 meters/ sec^2 due to gravity. Just before it hits the ground, it will have kinetic energy equal to the potential energy it had before you dropped it, ½ mass * velocity^2. Guess what, the units again are mass, distance^2/time^2, which are Jouls. FYI one Joule = 0.738 ft lbs. One hourse power = 550 ft lbs / sec. (Pounds in these equations refer to force, not mass. A one pound object strapped to your chest pushes against your chest with one pound of force when you accelerate at 9.8 meters/ sec^2 in your car, or rather it would if your car could pull 1 "g" of linear acceleration (6.5 second 1/4 mile at constant acceleration).
The reason you may feel slightly less acceleration at the power peak is because you are moving faster and it takes more energy to accelerate you (kinetic energy = ½ mv ^2). Since energy goes up with the square of velocity, you tend to accelerate more slowly the faster you go. Ever drive a car with a cvt, or a snowmobile for that matter? Maximum acceleration is definitely at peak power, not peak torque when you take gearing out of the picture.
You can't believe everything you read on the web. So if you don't have a solid physics background (sbertolone), you should check out your story before you start misinforming the readers on this forum. But, I will say that this has been a good debate, torque is difficult to understand and a lot of people are confused about it (Probably because of the torque wrench which people equate with force).
I've said too much already. Bye.
this from
[B]
WRONG. Torque should not be confused with force (an extremely common misconception). Torque is force times distance (or in the case of rotation, a "twist") the same units as WORK. When I said torque is "force with a twist", I did not mean "twisting force". Torque = the tendancy of a force to ROTATE a rigid body about a fixed axis. No rotation, no net torque! When you tighten your lug nut to 110 ft pounds, once the nut stops rotating there is no net torque, even though force is still being applied. Neither work nor torque contain imformation about rate, thus torque tells you nothing about acceleration unless you know rpm (for a given car's mass and gearing). If torque was force, then you'd be right, you want to maximize force for maximal acceleration (remember force = mass * acceleration), but alas, torque is not force. Force is mass times acceleration (pounds * feet/second^2). If you're interested in the relationship between torque and angular acceleration, net torque = moment of inertia * instantaneous angular acceleration. As far as work and energy in rotational motion go, Power = torque * instantaneous angular VELOCITY. Ahh, velocity contains rate information (phew!). Angular velocity is directly related to RPM.The rear wheel power curve already contains all the information you need. Besides, torque is usually measured on the dyno through the drive train. Just maximize the area under the power curve considering the rpm drop for each shift point and you will have your best ¼ mile time (assuming you can maintain traction)
. [QUOTE]
ok just because a force has been applied to your car doesnt mean its going to move you have to apply that force over a distance and a kick is a very short distance will very unlikely to overcome
the static friction of your car sitting. therefore no work will have been done because your car has
not moved allthough some force has been expelled on your carbut it was over 0 distance so if this
kid kicks your car a 100 times a second but cant move your car 0 work has still be done.
[B]
If you believe that then try standing behind a jet turbine. Sure the impulse of a single molecule hitting you is next to nothing but when the rate of collisions gets high enough you will feel the work being done.
[QUOTE]
Mechanical energy must not be expressed in newton metres (N * m) but only in joules (J). The
former unit is used only for torque or moment of force, question is do you know the difference.
this was taken from my 9th grade physics book.
[B]
A Joule comes from "the mechanical equivalent of heat" which was discovered by James Joule in the mid 1800's. A joule has the energy equivalent of 0.2389 calories. One BTU is equivalent to 1055 Joules. By definition though, a Joule (J) IS a Newton meter (N * m). Ninth grade physics books simplify things for ninth graders to understand. You might want to trade it in for a more advanced and accurate book. Energy is the capacity to do work, and has the same units as work (force over distance, or in SI units kilogram * meters^2/seconds^2= Nm=J). There are two types of energy in the classical sense, potential and kinetic. If you lift a one pound object 10 feet high, you have done 10 foot pounds of work. This is because the mass of the object requires force (1 lb = 4.448 Newtons) acting on it to push it against gravity. Now the potential energy of the object (relative to the ground) is 10 foot pounds. You can convert that into Joules if you like, but it is all the same, distance and force (force is mass times acceleration, acceleration is rate of change in velocity, velocity is distance/time). You can now convert that into kinetic energy by dropping the object (potential energy becomes kinetic energy). It will then accelerate at 9.8 meters/ sec^2 due to gravity. Just before it hits the ground, it will have kinetic energy equal to the potential energy it had before you dropped it, ½ mass * velocity^2. Guess what, the units again are mass, distance^2/time^2, which are Jouls. FYI one Joule = 0.738 ft lbs. One hourse power = 550 ft lbs / sec. (Pounds in these equations refer to force, not mass. A one pound object strapped to your chest pushes against your chest with one pound of force when you accelerate at 9.8 meters/ sec^2 in your car, or rather it would if your car could pull 1 "g" of linear acceleration (6.5 second 1/4 mile at constant acceleration).
The reason you may feel slightly less acceleration at the power peak is because you are moving faster and it takes more energy to accelerate you (kinetic energy = ½ mv ^2). Since energy goes up with the square of velocity, you tend to accelerate more slowly the faster you go. Ever drive a car with a cvt, or a snowmobile for that matter? Maximum acceleration is definitely at peak power, not peak torque when you take gearing out of the picture.
You can't believe everything you read on the web. So if you don't have a solid physics background (sbertolone), you should check out your story before you start misinforming the readers on this forum. But, I will say that this has been a good debate, torque is difficult to understand and a lot of people are confused about it (Probably because of the torque wrench which people equate with force).
I've said too much already. Bye.
Rotary Enthusiast
Joined: Jan 2002
Posts: 859
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From: 15143
Torque should not be confused with force...
Force is mass times acceleration (pounds * feet/ second^2). If you're interested in the relationship between torque and angular acceleration, net torque = moment of inertia * instantaneous angular acceleration.
Maximum acceleration is definitely at peak power, not peak torque when you take gearing out of the picture.
Full Member
Joined: Sep 2001
Posts: 163
Likes: 0
From: Portland, OR
Contradiction? I do contradict myself sometimes but I don't think I did here. That's the problem with "pounds". Pounds can mean either mass, or force, which are two different things. Sorry for the confusion. I should have stuck to SI units. I meant f=ma=kg*m/s^2, not f=ma=N*m/s^2!
blah blah yada. my horse is bigger then yours and could eat the 30 mile lever arm. obviously neither matter. its all about rpm. if you can rev higher then you can go faster. look at f1 cars. 
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