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03-16-2005, 11:26 PM   #31
ChrisV
The Big Meaney

Join Date: Mar 2004
Location: People's Republic of Maryland
Posts: 3,717
Quote:
 Originally Posted by Wally Forget all the tripe. Fuel is rated in joules and the when burnt at a rate of one joule per second releases a watt of power. How that power is applied whether it be for cooking steaks or to make a wheel turn is a product of that power. Without power torque cannot exist, but power can exist without torque. You cannot have more or less torque than power simply because they are different measurement, like oranges and lemons although they may share common elements. Just because someone decided to scale a graph to conveniently correlate with a 5252 derived constant does not mean one is greater than the other. The Torque (ft-lbs) = HP x 5252/rpm equation can be rearranged whichever way you like. If you use the same nonsense internet mechanics do by rearranging the the equation to prove hp is dependent on torque, you could equally argue power is dependent on rpm, which we all know is not true.

Just like in any math, you can rearrange the equation many ways. But that isnt' what we are talking about here and you should know it.

Torque is tangential force * the distance from the fulcrum. Power is defined as work per unit of time.

Applying 1 lb of force 1 ft from the fulcrum for a complete revolution will lead to;

W = F*2*pi*r = 1 lb * 2*pi * 1 ft = 2*pi lb-ft = 6.283 lb-ft

If it takes one minute to complete this revolution, then the power is;

P = W / time = 6.283 lb-ft / min

1 hp is defined as 550 lb-ft / s = 33,000 lb-ft / min

Therefore, applying 1 lb-ft of torque in one minute (1 rpm) = [6.283 lb-ft / min] / [33,000 lb-ft / min] = 1 / 5252 of 1 hp.

From this you can then calculate the number of hp from any given torque and rpm:

hp = torque (lb-ft) * rpm / 5252

No one "scaled a graph conveniently" to make that happen.

Where does the equation HP=TORQUE * RPM / 5252 come from? We will use Watts observation of one horsepower as 150 pounds, 220 feet in one minute. First we need express 150 pounds of force as foot pounds torque.

Pretend the force of 150 pounds is "applied" tangentially to a one foot radius circle. This would be 150 foot pounds torque.

Next we need to express 220 feet in one minute as RPM.

The circumference of a one foot radius circle is 6.283186 feet. ft. (Pi x diameter 3.141593 * 2 feet)

The distance of 220 feet, divided by 6.283185 feet, gives us a RPM of 35.014.
We are then talking about 150 pounds of force (150 foot pounds torque), 35 RPM, and one horsepower.

Constant (X) = 150 ft.lbs. * 35.014 RPM / 1hp

35.014 * 150 / 1 = 5252.1

5252 is the constant.

So then hp = torque * RPM / 5252

"The word horsepower was introduced by James Watt, the inventor of the steam engine in about 1775. Watt learned that "a strong horse could lift 150 pounds a height of 220 feet in 1 minute." One horsepower is also commonly expressed as 550 pounds one foot in one second or 33,000 pounds one foot in one minute. These are just different ways of saying the same thing. Notice these definitions includes force (pounds), distance (feet), and time, (minute, second). A horse could hold weight in a static position but this would not be considered horsepower, it would be similar to what we call torque. Adding time and distance to a static force (or to torque) results in horsepower. RPM, revolutions (distance) per minute (time), is today's equivalent of time and distance. Back to horses, imagine a horse raising coal out of a coal mine. A horse exerting one horsepower could raise 550 pounds of coal one foot every second."

Of course, the problem, as someone else pointed out in another thread, is that "Watt was simply observing that the horse lifted the 150 lb. object by 220 feet in one minute. We can't change that. Had Watt simply observed a bigger (say, a Clydesdale) or smaller horse (like Ford used to measure the '99 Cobra!) the definition of HP would be different."

The reason we dont measure internal combustion engine horsepower directly is that engies make a rotary motion, not linear motion. And unless the engine is geared down, the speed at which they do work (time and distance or RPM) is too great for practical direct measurement of horsepower. It seems logical then that the solution was to directly measure torque (rotational force eventually expressed in pounds at one foot radius) and RPM (time and distance, i.e. distance in circumference at the one foot radius) and from these calculate horsepower. Torque and RPM are easily measured directly.

Therefore when discussing internal combustion engines, torque IS the measured force, and hp is ALWAYS the calculated amount.
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 03-17-2005, 12:00 AM #32 Wally I Know More Than You     Join Date: May 2004 Location: Oz Posts: 2,564 I agree that measuring, for instance, tractive effort is convenient, but newtons is not newton metres nor is it watts. There is no need to even think about torque when deriving power from newtons over time. I'll demonstrate the graphing in a diffrent way. Here is a nissan sss, from Newcastle Oz, dyno readout conveniently in british units. Where is the crossover at 5252?: I guess you were trying to convey is the relationship of torque/hp by this: hp = torque * RPM / 5252 @ 5252 rpm hp = torque as rpm increases and power is static then torque decreases or if torque is to remain static the power input must increase? __________________ "She gave me a look only a mother could give a child." Last edited by Wally : 03-17-2005 at 12:48 AM.
03-17-2005, 01:09 AM   #33
DodgeRida67
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Join Date: Feb 2004
Location: Jackson, Georgia
Posts: 2,056
Quote:
 Originally Posted by ChrisV Torque is work, hp is work over time. Without time, torque doesn't do anything. Without force, hp doesn't exist.

One small detail I'd like to point out about that. Work and Torque aren't exactly the same thing. Torque is the lb-ft. Work is the ft-lb. Movement must occur for work to be done, torque doesn't have to have movement.

Hobo seems to have this confused in the "torquing my valve cover" (or whatever the name is) thread. Torque a bolt and you are doing work that's ft-lb. Ah, this is for another thread.

I'm just pointing out that torque isn't exactly work. It doesn't fit when used mathematically.

03-17-2005, 04:55 AM   #34
vwhobo
CF's Anal Orifice

Join Date: Feb 2003
Location: Redneck Hell
Posts: 8,630
Quote:
 Originally Posted by DodgeRida67 One small detail I'd like to point out about that. Work and Torque aren't exactly the same thing. Torque is the lb-ft. Work is the ft-lb. Movement must occur for work to be done, torque doesn't have to have movement. Hobo seems to have this confused in the "torquing my valve cover" (or whatever the name is) thread. Torque a bolt and you are doing work that's ft-lb. Ah, this is for another thread. I'm just pointing out that torque isn't exactly work. It doesn't fit when used mathematically.
No, the only thing I'm confused about is why those of us with knowledge make any attempt at helping individuals like you learn. It's pointless because you're convinced that at age 17 you already know everything but we still try. However, here's something else for you to try to comprehend.

http://www.physics.uoguelph.ca/tutor...que.intro.html
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03-17-2005, 05:28 AM   #35
windsonian
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Join Date: Jan 2005
Location: Down Below
Posts: 2,081
Quote:
 I'm just pointing out that torque isn't exactly work. It doesn't fit when used mathematically.

Of course torque isn't work!!! But when you apply a torque to do some work, then at any instant in that process, you are using power.

That's why the speed is in the equation ... the distance in the torque units is different to the distance in the work units. For example, if we forget that we're turning something with torque, and instead pretend we're moving something in a straight line:
Rather than torque we substitute force in Newtons [N] (sorry, we work in metric here in oz)
For work (energy) we have Nm (or joules [J])
For power we have Watts [W] (or J/sec)
For speed we have m/sec (instead of rpm (or radians/sec) in turning case)
For distance we have m

Force x distance = work [N*m = Nm = J]
Force x distance / time = Force x speed = Power [N*m/s = J/s = W]

Therefore, the power used at any instant = the force used to move an object at a certain speed x that speed.
Work done = the force used to move an object x the distance you've moved the object
Force = the force you are applying to the object to move it at the speed corresponding to the power being used.

These can all be converted to rotational motion using torque instead of force, and radians/sec (rpm*2pi/60) instead of m/sec and angle turned in radians (revolutions * 2pi) instead of distance in m.

They quite clearly are linked in the theory.

 03-17-2005, 03:10 PM #36 ChrisV The Big Meaney     Join Date: Mar 2004 Location: People's Republic of Maryland Posts: 3,717 Wally Your graph shows two different measurements of energy, and no measurement of torque. From Hobo's link: "Note that the SI units of torque is a Newton-metre, which is also a way of expressing a Joule (the unit for energy). However, torque is not energy." Your graph thus doesn't display torque, merely two different units of energy. That's why the curves are very similar in shape and don't cross over. __________________ I'm not mean. You're just a wuss. www.midatlantic7s.com
 03-17-2005, 10:48 PM #37 Wally I Know More Than You     Join Date: May 2004 Location: Oz Posts: 2,564 OK let me put it yet another way: The simplified imperial formula is hp = ftlbs-force * RPM / 5252 when the rpms hit 5252 the result is power = torque *1 Now lets take the metric equivalent kW = N-m *RPM/ 9551 when the rpms hit 9551 the result is power = torque *1 In both instances we are dealing with the same effects, but with different units of measurements. As you can easily conclude if we used the same convention for graphing that is used for imperial power/torque curves the metric torque curve would not intersect the power curve until 9551 rpm. As most daily grind engines run out of puff well before 9551 rpm we would not expect to see any intersection on our graph if it's x axis ceased @ 6500 rpm. In so far as the shape of the tractive effort curve, do you really think it's shape will change by multiplying it by a constant rolling tyre radius and dividing by a constant drivetrain ratio? I don't think so somehow, so the curves are indeed valid and the bottom one is in fact the torque profile with a different scale. __________________ "She gave me a look only a mother could give a child." Last edited by Wally : 03-17-2005 at 10:59 PM.

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