what is lb-ft?
- Thread starter [R]viper
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Means foot-pounds. Measures tourqe. Tourque is "twisting force".
I still haven't bothered to figure out how what the tourque equivilants in GT2 to US tourque numbers.
For example, I've got cars with 650hp, but 100ft-lbs of whatever TQ measurement they use......how much is 100ft-lbs in US ft-lbs measurments????
I still haven't bothered to figure out how what the tourque equivilants in GT2 to US tourque numbers.
For example, I've got cars with 650hp, but 100ft-lbs of whatever TQ measurement they use......how much is 100ft-lbs in US ft-lbs measurments????
- 10,840
These are not my words...
Torque vs Horsepower
The word horsepower as it relates to machines, was defined 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." So imagine a horse raising coal out of a coal mine. A horse exerting one horsepower could raise 220 pounds of coal one foot every minute.
One horsepower is also expressed as 550 pounds one foot in one second ot 33,000 pounds one foot in one minute. These are just different ways of saying the same thing.
Here is another way horsepower could be directly measured. Say you have a horse hitched to a plow. In the hitch is a spring scale (like a fish scale). The horse pulls the plow one foot every second and you see 550 pounds on the scale. That would be one horsepower.
So horsepower can be directly measured. However there is a problem directly measuring horsepower of modern day internal combustion engines because they produce rotary motion not linear motion. And unless the engine is geared down, the speed at which they do work (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. Early dynamometers used a brake device to load the engine. A torque arm was attached to this brake's stator. The brake's rotor was coupled to the engine's crankshaft. A spring scale or other measuring device connected the torque arm to the stationary fixture holding the engine and brake. During a test the brake's application loaded the engine. Torque and engine rpm were observed and recorded. Click here for a description of how this happens on our dyno.
So on modern day dynamometers horsepower is a calculated value. It's important to remember the dyno measures torque and rpm and then from these calculates horsepower. On the dyno it takes more water flow to the water brake to increase the load on the engine being tested. As the test engine's torque rises more water flow is needed. As the test engine's torque drops less water flow is needed. The dyno's water brake does not respond to horsepower. Major adjustments to water flow are needed as an engine crosses its torque peak but none are needed as it crosses its horsepower peak. In other words the water flow to the brake during a dyno test follows the engines torque curve and not its horsepower curve. Torque is what twists the tire, prop, or pump. Horsepower helps us understand an amount or quantity of torque. (Torque + time)
Now if we are measuring torque and RPM how can we calculate horsepower? 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
--------------------------------------------------------------------------------
Here is another way; Remember we know 150 foot pounds and 35.014 RPM = one horsepower
1hp is to 150 ft.lbs. * 35.014 RPM as X hp is to observed ft.lbs.torque * observed RPM
Example; We dyno test and observe 400 ft.lbs. torque at 5000 RPM
1 hp is to 150 ft.lbs. * 35.014 RPM as X hp is to 400 ft.lbs. * 5000 RPM
When we cross multiply X hp * (150 ft.lbs. * 35.014 RPM) = 1hp * (400 ft.lbs. * 5000 RPM)
X hp * (5252 ft.lbs. RPM) = 1 hp * (2,000,000 ft.lbs. RPM)
Divide both sides by 5252 ft.lbs. RPM
X hp = 1 hp * 380.80
X hp = 380.80 hp
Horsepower = torque x rpm / 5252
Here's an interesting bit of trivia; below 5252 rpm any engine's torque will always be higher than its horsepower, and above 5252 rpm any engine's horsepower will always be higher than its torque.
Torque vs Horsepower
The word horsepower as it relates to machines, was defined 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." So imagine a horse raising coal out of a coal mine. A horse exerting one horsepower could raise 220 pounds of coal one foot every minute.
One horsepower is also expressed as 550 pounds one foot in one second ot 33,000 pounds one foot in one minute. These are just different ways of saying the same thing.
Here is another way horsepower could be directly measured. Say you have a horse hitched to a plow. In the hitch is a spring scale (like a fish scale). The horse pulls the plow one foot every second and you see 550 pounds on the scale. That would be one horsepower.
So horsepower can be directly measured. However there is a problem directly measuring horsepower of modern day internal combustion engines because they produce rotary motion not linear motion. And unless the engine is geared down, the speed at which they do work (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. Early dynamometers used a brake device to load the engine. A torque arm was attached to this brake's stator. The brake's rotor was coupled to the engine's crankshaft. A spring scale or other measuring device connected the torque arm to the stationary fixture holding the engine and brake. During a test the brake's application loaded the engine. Torque and engine rpm were observed and recorded. Click here for a description of how this happens on our dyno.
So on modern day dynamometers horsepower is a calculated value. It's important to remember the dyno measures torque and rpm and then from these calculates horsepower. On the dyno it takes more water flow to the water brake to increase the load on the engine being tested. As the test engine's torque rises more water flow is needed. As the test engine's torque drops less water flow is needed. The dyno's water brake does not respond to horsepower. Major adjustments to water flow are needed as an engine crosses its torque peak but none are needed as it crosses its horsepower peak. In other words the water flow to the brake during a dyno test follows the engines torque curve and not its horsepower curve. Torque is what twists the tire, prop, or pump. Horsepower helps us understand an amount or quantity of torque. (Torque + time)
Now if we are measuring torque and RPM how can we calculate horsepower? 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
--------------------------------------------------------------------------------
Here is another way; Remember we know 150 foot pounds and 35.014 RPM = one horsepower
1hp is to 150 ft.lbs. * 35.014 RPM as X hp is to observed ft.lbs.torque * observed RPM
Example; We dyno test and observe 400 ft.lbs. torque at 5000 RPM
1 hp is to 150 ft.lbs. * 35.014 RPM as X hp is to 400 ft.lbs. * 5000 RPM
When we cross multiply X hp * (150 ft.lbs. * 35.014 RPM) = 1hp * (400 ft.lbs. * 5000 RPM)
X hp * (5252 ft.lbs. RPM) = 1 hp * (2,000,000 ft.lbs. RPM)
Divide both sides by 5252 ft.lbs. RPM
X hp = 1 hp * 380.80
X hp = 380.80 hp
Horsepower = torque x rpm / 5252
Here's an interesting bit of trivia; below 5252 rpm any engine's torque will always be higher than its horsepower, and above 5252 rpm any engine's horsepower will always be higher than its torque.
- 10,840
Originally posted by askia47
Wow...where did you get the info?
You can find anything you want on the internet. Unfortunately, GTPlanet may not be the best place to find everything, tho.
Just search for "horse power" or "torque lb ft" and see what you get back.
Lb ft torque is basically the same as HP. The higher the more powerfull. Lb ft applies to combusion engines because it's a more intuitive and accurate way to measure it.
I played GT1 & saw kg-m (not lb-ft) there. It gave me an idea.
Viper has 664 N-m. It's 489,6 lb-ft (at 3700 rpm both).
N-m = F.l = m.a.l
In Earth conditions: N-m = m.g.l, so then:
664 = m. 9,8 . l
664/9,8 = m.l
67,8 = m.l (in kg-m)
1lb = 0,4536kg
1ft = 0,3048m, so:
67,8/0,4536 = 149,47
149,47/0,3048 = 490,38
490,38 is almost 489,6
That's the answer for people who use metric system or N-m.
Viper has 664 N-m. It's 489,6 lb-ft (at 3700 rpm both).
N-m = F.l = m.a.l
In Earth conditions: N-m = m.g.l, so then:
664 = m. 9,8 . l
664/9,8 = m.l
67,8 = m.l (in kg-m)
1lb = 0,4536kg
1ft = 0,3048m, so:
67,8/0,4536 = 149,47
149,47/0,3048 = 490,38
490,38 is almost 489,6
That's the answer for people who use metric system or N-m.
- 10,840
One among many stupid things about America is that fact that we can't get on board with the metric system. It's so much more logical and consistent with the rest of the world.
When I think about the Mars lander that failed because of a discrepancy between metric and American measurements I wanna puke
- 254
Why are there different measurement units?? WHY CAN'T WE ALL JUST GET ALONG PEOPLE?!?! Apparently since I'm canadian, (eh?) i'm suppose to measure myself in meters and weigh myself in kilograms.......well I'm 5'8" and 125 lbs. I have NO IDEA what i am in meters and kg, and frankly I don't care.
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