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How is foot-pounds of energy defined?
How are base temperature units found on other planets?Oberth is confusing me. Apparent violation of conservation of energyLow Energy Transfer within Earth-Moon systemIs specific orbital energy a constant? How does this affect the semi-major axis?Save energy in plane changing orbital maneuverWhat precisely is downrange distance - how is it defined mathematically?What is a “synodic frame”? Can one be defined for an elliptical orbit?How is the altitude of a satellite defined, given that the Earth is not spherical?How to interpret and use characteristic energy C3 in this case?is it possible to use earths magnetic field as a energy source while in space?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Tom Lodgson's classic textbook Orbital Mechanics: Theory and Applications frequently measures energy in foot-pounds, which I'd always thought to measure torque. The book doesn't define it. Is it the archaic foot pound-force [sic], or the gravitational potential energy of one pound hoisted one foot in a constant gravitational field, or something else? (Is this unit still in use in aerospace or in other fields?)
orbital-mechanics history measurement
edited 18 mins ago
Manu H
1,55414 silver badges31 bronze badges
asked 14 hours ago
Camille GoudeseuneCamille Goudeseune
7805 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Tom Lodgson's classic textbook Orbital Mechanics: Theory and Applications frequently measures energy in foot-pounds, which I'd always thought to measure torque. The book doesn't define it. Is it the archaic foot pound-force [sic], or the gravitational potential energy of one pound hoisted one foot in a constant gravitational field, or something else? (Is this unit still in use in aerospace or in other fields?)
orbital-mechanics history measurement
edited 18 mins ago
Manu H
1,55414 silver badges31 bronze badges
asked 14 hours ago
Camille GoudeseuneCamille Goudeseune
7805 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Tom Lodgson's classic textbook Orbital Mechanics: Theory and Applications frequently measures energy in foot-pounds, which I'd always thought to measure torque. The book doesn't define it. Is it the archaic foot pound-force [sic], or the gravitational potential energy of one pound hoisted one foot in a constant gravitational field, or something else? (Is this unit still in use in aerospace or in other fields?)
orbital-mechanics history measurement
edited 18 mins ago
Manu H
1,55414 silver badges31 bronze badges
asked 14 hours ago
Camille GoudeseuneCamille Goudeseune
7805 silver badges18 bronze badges
$endgroup$
Tom Lodgson's classic textbook Orbital Mechanics: Theory and Applications frequently measures energy in foot-pounds, which I'd always thought to measure torque. The book doesn't define it. Is it the archaic foot pound-force [sic], or the gravitational potential energy of one pound hoisted one foot in a constant gravitational field, or something else? (Is this unit still in use in aerospace or in other fields?)
orbital-mechanics history measurement
orbital-mechanics history measurement
edited 18 mins ago
Manu H
1,55414 silver badges31 bronze badges
asked 14 hours ago
Camille GoudeseuneCamille Goudeseune
7805 silver badges18 bronze badges
edited 18 mins ago
Manu H
1,55414 silver badges31 bronze badges
asked 14 hours ago
Camille GoudeseuneCamille Goudeseune
7805 silver badges18 bronze badges
edited 18 mins ago
Manu H
1,55414 silver badges31 bronze badges
edited 18 mins ago
Manu H
1,55414 silver badges31 bronze badges
edited 18 mins ago
Manu H
1,55414 silver badges31 bronze badges
1,55414 silver badges31 bronze badges
asked 14 hours ago
Camille GoudeseuneCamille Goudeseune
7805 silver badges18 bronze badges
asked 14 hours ago
Camille GoudeseuneCamille Goudeseune
7805 silver badges18 bronze badges
asked 14 hours ago
Camille GoudeseuneCamille Goudeseune
7805 silver badges18 bronze badges
7805 silver badges18 bronze badges
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Is it... the gravitational potential energy of one pound hoisted one foot in a constant gravitational field...?
Yes indeed it is!
To be energy, the pound has to be parallel to the foot.
$$E = int mathbfF cdot d mathbfs$$
To be torque, the pound has to be perpendicular to the foot
$$tau = mathbfr times mathbfF$$
1 foot-pound (or pound foot) of energy is 9.81 (m/s^2) / 2.2 (kgf/lb) / 3.3 (ft/m) = 1.35 Joules.
answered 14 hours ago
uhohuhoh
48.6k22 gold badges195 silver badges631 bronze badges
$endgroup$
1
$begingroup$
@RussellBorogove in physics they are, and have to be, the same thing. The order can not matter, so I'm forced to pound my foot and stick to my principles.
$endgroup$
– uhoh
14 hours ago
1
$begingroup$
The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
$endgroup$
– Camille Goudeseune
14 hours ago
$begingroup$
@CamilleGoudeseune good guess! I've added a reference to your question. For the value, I used 9.81 m/s^2 but I assume if you use a standard gravity $g_0$ of 9.80665 m/s^2 and whatever the exact conversion from feet to meters is, you'll get Wikipedia's 1.355818 Joules
$endgroup$
– uhoh
13 hours ago
2
$begingroup$
Delightful pun... and the unwashed rabble (of which I must thus be a member) also use lb-ft and ft-lb interchangeably.
$endgroup$
– Camille Goudeseune
13 hours ago
1
$begingroup$
Re, "parallel,"..."perpendicular",... That description somewhat masks the truth. To be energy, the object to which the pound of force is applied must move one foot in the direction of the force. To be torque, nothing needs to move, but the pound of force must be applied to a point that is one foot away from the axis of rotation, and the radius and the axis and the force all must be mutually perpendicular.
$endgroup$
– Solomon Slow
12 hours ago
|
show 2 more comments
$begingroup$
According to Wikipedia foot-pound and foot-pound-force are synonymous:
The foot pound-force (symbol: ft⋅lbf or ft⋅lb) is a unit of work or energy in the Engineering and Gravitational Systems in United States customary and imperial units of measure. It is the energy transferred upon applying a force of one pound-force (lbf) through a linear displacement of one foot.
It’s equal to 1.356 Joules.
The torque unit is the pound-foot, not foot-pound.
answered 14 hours ago
Russell BorogoveRussell Borogove
101k3 gold badges353 silver badges436 bronze badges
$endgroup$
$begingroup$
But pound-foot is used for torques too.
$endgroup$
– Uwe
14 hours ago
$begingroup$
Yes, I was mistaken about lb-ft vs ft-lb. I'll also admit that wikipedia's claim ft-lbf = ft-lb was true thirty years ago when Tom started lecturing. Let's see what others report...
$endgroup$
– Camille Goudeseune
14 hours ago
2
$begingroup$
Per @uhoh's answer, for both energy and torque, when you decompose the mathematical shorthand in the integral and the cross product, you wind up with the units being a force times a distance. Multiplication is commutative, so force times distance is equivalent to distance times force. Distinction between pound-foot and foot-pound isn't physics, it's semantics.
$endgroup$
– Tom Spilker
10 hours ago
add a comment |
$begingroup$
Foot-pound or pound-foot are synonymous, and represent the arithmetic product of pound (force) and foot (length).
The pound (force) is the weight of one pound (mass) at the Earth's surface (somewhat imprecise because Earth's gravity field varies depending on your location, and the effective weight of an object will be influenced by the centrifugal force due to Earth's rotation, again dependent on location).
As a unit of energy, it is the energy of applying a one pound force over a distance of one foot. It is equivalent to raising a one-pound mass one foot in height, well... because.
As a unit of torque, it is the torque resulting from a one pound tangent force applied at a distance of one foot from the axis of rotation.
The same could be said of the Newton-meter (or meter-Newton, but it's never expressed that way); as a unit of energy, it is a one Newton force applied over a distance of one meter; as a unit of torque, it is a one Newton tangent force applied at a distance of one meter from the axis of rotation, except that the Newton is specifically a unit of force with a precise definition where pound may be either force or mass and pound (force) lacks a precise definition.
The measurement system which includes pounds and feet has a long history. When it developed, the variability of Earth's gravitational field and its impact on the weights and measures which depended on it wasn't understood, wasn't measurable, and/or wasn't significant for the engineering problems of the time. Culture, history, and familiarity keep these weights and measures in use despite the awkwardness and the advantages of metric.
answered 7 hours ago
Anthony XAnthony X
10.3k1 gold badge42 silver badges85 bronze badges
$endgroup$
$begingroup$
+1for thorough answer/explanation
$endgroup$
– uhoh
1 hour ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Is it... the gravitational potential energy of one pound hoisted one foot in a constant gravitational field...?
Yes indeed it is!
To be energy, the pound has to be parallel to the foot.
$$E = int mathbfF cdot d mathbfs$$
To be torque, the pound has to be perpendicular to the foot
$$tau = mathbfr times mathbfF$$
1 foot-pound (or pound foot) of energy is 9.81 (m/s^2) / 2.2 (kgf/lb) / 3.3 (ft/m) = 1.35 Joules.
answered 14 hours ago
uhohuhoh
48.6k22 gold badges195 silver badges631 bronze badges
$endgroup$
1
$begingroup$
@RussellBorogove in physics they are, and have to be, the same thing. The order can not matter, so I'm forced to pound my foot and stick to my principles.
$endgroup$
– uhoh
14 hours ago
1
$begingroup$
The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
$endgroup$
– Camille Goudeseune
14 hours ago
$begingroup$
@CamilleGoudeseune good guess! I've added a reference to your question. For the value, I used 9.81 m/s^2 but I assume if you use a standard gravity $g_0$ of 9.80665 m/s^2 and whatever the exact conversion from feet to meters is, you'll get Wikipedia's 1.355818 Joules
$endgroup$
– uhoh
13 hours ago
2
$begingroup$
Delightful pun... and the unwashed rabble (of which I must thus be a member) also use lb-ft and ft-lb interchangeably.
$endgroup$
– Camille Goudeseune
13 hours ago
1
$begingroup$
Re, "parallel,"..."perpendicular",... That description somewhat masks the truth. To be energy, the object to which the pound of force is applied must move one foot in the direction of the force. To be torque, nothing needs to move, but the pound of force must be applied to a point that is one foot away from the axis of rotation, and the radius and the axis and the force all must be mutually perpendicular.
$endgroup$
– Solomon Slow
12 hours ago
|
show 2 more comments
$begingroup$
Is it... the gravitational potential energy of one pound hoisted one foot in a constant gravitational field...?
Yes indeed it is!
To be energy, the pound has to be parallel to the foot.
$$E = int mathbfF cdot d mathbfs$$
To be torque, the pound has to be perpendicular to the foot
$$tau = mathbfr times mathbfF$$
1 foot-pound (or pound foot) of energy is 9.81 (m/s^2) / 2.2 (kgf/lb) / 3.3 (ft/m) = 1.35 Joules.
answered 14 hours ago
uhohuhoh
48.6k22 gold badges195 silver badges631 bronze badges
$endgroup$
1
$begingroup$
@RussellBorogove in physics they are, and have to be, the same thing. The order can not matter, so I'm forced to pound my foot and stick to my principles.
$endgroup$
– uhoh
14 hours ago
1
$begingroup$
The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
$endgroup$
– Camille Goudeseune
14 hours ago
$begingroup$
@CamilleGoudeseune good guess! I've added a reference to your question. For the value, I used 9.81 m/s^2 but I assume if you use a standard gravity $g_0$ of 9.80665 m/s^2 and whatever the exact conversion from feet to meters is, you'll get Wikipedia's 1.355818 Joules
$endgroup$
– uhoh
13 hours ago
2
$begingroup$
Delightful pun... and the unwashed rabble (of which I must thus be a member) also use lb-ft and ft-lb interchangeably.
$endgroup$
– Camille Goudeseune
13 hours ago
1
$begingroup$
Re, "parallel,"..."perpendicular",... That description somewhat masks the truth. To be energy, the object to which the pound of force is applied must move one foot in the direction of the force. To be torque, nothing needs to move, but the pound of force must be applied to a point that is one foot away from the axis of rotation, and the radius and the axis and the force all must be mutually perpendicular.
$endgroup$
– Solomon Slow
12 hours ago
|
show 2 more comments
$begingroup$
Is it... the gravitational potential energy of one pound hoisted one foot in a constant gravitational field...?
Yes indeed it is!
To be energy, the pound has to be parallel to the foot.
$$E = int mathbfF cdot d mathbfs$$
To be torque, the pound has to be perpendicular to the foot
$$tau = mathbfr times mathbfF$$
1 foot-pound (or pound foot) of energy is 9.81 (m/s^2) / 2.2 (kgf/lb) / 3.3 (ft/m) = 1.35 Joules.
answered 14 hours ago
uhohuhoh
48.6k22 gold badges195 silver badges631 bronze badges
$endgroup$
Is it... the gravitational potential energy of one pound hoisted one foot in a constant gravitational field...?
Yes indeed it is!
To be energy, the pound has to be parallel to the foot.
$$E = int mathbfF cdot d mathbfs$$
To be torque, the pound has to be perpendicular to the foot
$$tau = mathbfr times mathbfF$$
1 foot-pound (or pound foot) of energy is 9.81 (m/s^2) / 2.2 (kgf/lb) / 3.3 (ft/m) = 1.35 Joules.
answered 14 hours ago
uhohuhoh
48.6k22 gold badges195 silver badges631 bronze badges
edited 1 hour ago
answered 14 hours ago
uhohuhoh
48.6k22 gold badges195 silver badges631 bronze badges
answered 14 hours ago
uhohuhoh
48.6k22 gold badges195 silver badges631 bronze badges
answered 14 hours ago
uhohuhoh
48.6k22 gold badges195 silver badges631 bronze badges
48.6k22 gold badges195 silver badges631 bronze badges
1
$begingroup$
@RussellBorogove in physics they are, and have to be, the same thing. The order can not matter, so I'm forced to pound my foot and stick to my principles.
$endgroup$
– uhoh
14 hours ago
1
$begingroup$
The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
$endgroup$
– Camille Goudeseune
14 hours ago
$begingroup$
@CamilleGoudeseune good guess! I've added a reference to your question. For the value, I used 9.81 m/s^2 but I assume if you use a standard gravity $g_0$ of 9.80665 m/s^2 and whatever the exact conversion from feet to meters is, you'll get Wikipedia's 1.355818 Joules
$endgroup$
– uhoh
13 hours ago
2
$begingroup$
Delightful pun... and the unwashed rabble (of which I must thus be a member) also use lb-ft and ft-lb interchangeably.
$endgroup$
– Camille Goudeseune
13 hours ago
1
$begingroup$
Re, "parallel,"..."perpendicular",... That description somewhat masks the truth. To be energy, the object to which the pound of force is applied must move one foot in the direction of the force. To be torque, nothing needs to move, but the pound of force must be applied to a point that is one foot away from the axis of rotation, and the radius and the axis and the force all must be mutually perpendicular.
$endgroup$
– Solomon Slow
12 hours ago
|
show 2 more comments
1
$begingroup$
@RussellBorogove in physics they are, and have to be, the same thing. The order can not matter, so I'm forced to pound my foot and stick to my principles.
$endgroup$
– uhoh
14 hours ago
1
$begingroup$
The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
$endgroup$
– Camille Goudeseune
14 hours ago
$begingroup$
@CamilleGoudeseune good guess! I've added a reference to your question. For the value, I used 9.81 m/s^2 but I assume if you use a standard gravity $g_0$ of 9.80665 m/s^2 and whatever the exact conversion from feet to meters is, you'll get Wikipedia's 1.355818 Joules
$endgroup$
– uhoh
13 hours ago
2
$begingroup$
Delightful pun... and the unwashed rabble (of which I must thus be a member) also use lb-ft and ft-lb interchangeably.
$endgroup$
– Camille Goudeseune
13 hours ago
1
$begingroup$
Re, "parallel,"..."perpendicular",... That description somewhat masks the truth. To be energy, the object to which the pound of force is applied must move one foot in the direction of the force. To be torque, nothing needs to move, but the pound of force must be applied to a point that is one foot away from the axis of rotation, and the radius and the axis and the force all must be mutually perpendicular.
$endgroup$
– Solomon Slow
12 hours ago
1
1
$begingroup$
@RussellBorogove in physics they are, and have to be, the same thing. The order can not matter, so I'm forced to pound my foot and stick to my principles.
$endgroup$
– uhoh
14 hours ago
$begingroup$
@RussellBorogove in physics they are, and have to be, the same thing. The order can not matter, so I'm forced to pound my foot and stick to my principles.
$endgroup$
– uhoh
14 hours ago
1
1
$begingroup$
The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
$endgroup$
– Camille Goudeseune
14 hours ago
$begingroup$
The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
$endgroup$
– Camille Goudeseune
14 hours ago
$begingroup$
@CamilleGoudeseune good guess! I've added a reference to your question. For the value, I used 9.81 m/s^2 but I assume if you use a standard gravity $g_0$ of 9.80665 m/s^2 and whatever the exact conversion from feet to meters is, you'll get Wikipedia's 1.355818 Joules
$endgroup$
– uhoh
13 hours ago
$begingroup$
@CamilleGoudeseune good guess! I've added a reference to your question. For the value, I used 9.81 m/s^2 but I assume if you use a standard gravity $g_0$ of 9.80665 m/s^2 and whatever the exact conversion from feet to meters is, you'll get Wikipedia's 1.355818 Joules
$endgroup$
– uhoh
13 hours ago
2
2
$begingroup$
Delightful pun... and the unwashed rabble (of which I must thus be a member) also use lb-ft and ft-lb interchangeably.
$endgroup$
– Camille Goudeseune
13 hours ago
$begingroup$
Delightful pun... and the unwashed rabble (of which I must thus be a member) also use lb-ft and ft-lb interchangeably.
$endgroup$
– Camille Goudeseune
13 hours ago
1
1
$begingroup$
Re, "parallel,"..."perpendicular",... That description somewhat masks the truth. To be energy, the object to which the pound of force is applied must move one foot in the direction of the force. To be torque, nothing needs to move, but the pound of force must be applied to a point that is one foot away from the axis of rotation, and the radius and the axis and the force all must be mutually perpendicular.
$endgroup$
– Solomon Slow
12 hours ago
$begingroup$
Re, "parallel,"..."perpendicular",... That description somewhat masks the truth. To be energy, the object to which the pound of force is applied must move one foot in the direction of the force. To be torque, nothing needs to move, but the pound of force must be applied to a point that is one foot away from the axis of rotation, and the radius and the axis and the force all must be mutually perpendicular.
$endgroup$
– Solomon Slow
12 hours ago
|
show 2 more comments
$begingroup$
According to Wikipedia foot-pound and foot-pound-force are synonymous:
The foot pound-force (symbol: ft⋅lbf or ft⋅lb) is a unit of work or energy in the Engineering and Gravitational Systems in United States customary and imperial units of measure. It is the energy transferred upon applying a force of one pound-force (lbf) through a linear displacement of one foot.
It’s equal to 1.356 Joules.
The torque unit is the pound-foot, not foot-pound.
answered 14 hours ago
Russell BorogoveRussell Borogove
101k3 gold badges353 silver badges436 bronze badges
$endgroup$
$begingroup$
But pound-foot is used for torques too.
$endgroup$
– Uwe
14 hours ago
$begingroup$
Yes, I was mistaken about lb-ft vs ft-lb. I'll also admit that wikipedia's claim ft-lbf = ft-lb was true thirty years ago when Tom started lecturing. Let's see what others report...
$endgroup$
– Camille Goudeseune
14 hours ago
2
$begingroup$
Per @uhoh's answer, for both energy and torque, when you decompose the mathematical shorthand in the integral and the cross product, you wind up with the units being a force times a distance. Multiplication is commutative, so force times distance is equivalent to distance times force. Distinction between pound-foot and foot-pound isn't physics, it's semantics.
$endgroup$
– Tom Spilker
10 hours ago
add a comment |
$begingroup$
According to Wikipedia foot-pound and foot-pound-force are synonymous:
The foot pound-force (symbol: ft⋅lbf or ft⋅lb) is a unit of work or energy in the Engineering and Gravitational Systems in United States customary and imperial units of measure. It is the energy transferred upon applying a force of one pound-force (lbf) through a linear displacement of one foot.
It’s equal to 1.356 Joules.
The torque unit is the pound-foot, not foot-pound.
answered 14 hours ago
Russell BorogoveRussell Borogove
101k3 gold badges353 silver badges436 bronze badges
$endgroup$
$begingroup$
But pound-foot is used for torques too.
$endgroup$
– Uwe
14 hours ago
$begingroup$
Yes, I was mistaken about lb-ft vs ft-lb. I'll also admit that wikipedia's claim ft-lbf = ft-lb was true thirty years ago when Tom started lecturing. Let's see what others report...
$endgroup$
– Camille Goudeseune
14 hours ago
2
$begingroup$
Per @uhoh's answer, for both energy and torque, when you decompose the mathematical shorthand in the integral and the cross product, you wind up with the units being a force times a distance. Multiplication is commutative, so force times distance is equivalent to distance times force. Distinction between pound-foot and foot-pound isn't physics, it's semantics.
$endgroup$
– Tom Spilker
10 hours ago
add a comment |
$begingroup$
According to Wikipedia foot-pound and foot-pound-force are synonymous:
The foot pound-force (symbol: ft⋅lbf or ft⋅lb) is a unit of work or energy in the Engineering and Gravitational Systems in United States customary and imperial units of measure. It is the energy transferred upon applying a force of one pound-force (lbf) through a linear displacement of one foot.
It’s equal to 1.356 Joules.
The torque unit is the pound-foot, not foot-pound.
answered 14 hours ago
Russell BorogoveRussell Borogove
101k3 gold badges353 silver badges436 bronze badges
$endgroup$
According to Wikipedia foot-pound and foot-pound-force are synonymous:
The foot pound-force (symbol: ft⋅lbf or ft⋅lb) is a unit of work or energy in the Engineering and Gravitational Systems in United States customary and imperial units of measure. It is the energy transferred upon applying a force of one pound-force (lbf) through a linear displacement of one foot.
It’s equal to 1.356 Joules.
The torque unit is the pound-foot, not foot-pound.
answered 14 hours ago
Russell BorogoveRussell Borogove
101k3 gold badges353 silver badges436 bronze badges
edited 14 hours ago
answered 14 hours ago
Russell BorogoveRussell Borogove
101k3 gold badges353 silver badges436 bronze badges
answered 14 hours ago
Russell BorogoveRussell Borogove
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answered 14 hours ago
Russell BorogoveRussell Borogove
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$begingroup$
But pound-foot is used for torques too.
$endgroup$
– Uwe
14 hours ago
$begingroup$
Yes, I was mistaken about lb-ft vs ft-lb. I'll also admit that wikipedia's claim ft-lbf = ft-lb was true thirty years ago when Tom started lecturing. Let's see what others report...
$endgroup$
– Camille Goudeseune
14 hours ago
2
$begingroup$
Per @uhoh's answer, for both energy and torque, when you decompose the mathematical shorthand in the integral and the cross product, you wind up with the units being a force times a distance. Multiplication is commutative, so force times distance is equivalent to distance times force. Distinction between pound-foot and foot-pound isn't physics, it's semantics.
$endgroup$
– Tom Spilker
10 hours ago
add a comment |
$begingroup$
But pound-foot is used for torques too.
$endgroup$
– Uwe
14 hours ago
$begingroup$
Yes, I was mistaken about lb-ft vs ft-lb. I'll also admit that wikipedia's claim ft-lbf = ft-lb was true thirty years ago when Tom started lecturing. Let's see what others report...
$endgroup$
– Camille Goudeseune
14 hours ago
2
$begingroup$
Per @uhoh's answer, for both energy and torque, when you decompose the mathematical shorthand in the integral and the cross product, you wind up with the units being a force times a distance. Multiplication is commutative, so force times distance is equivalent to distance times force. Distinction between pound-foot and foot-pound isn't physics, it's semantics.
$endgroup$
– Tom Spilker
10 hours ago
$begingroup$
But pound-foot is used for torques too.
$endgroup$
– Uwe
14 hours ago
$begingroup$
But pound-foot is used for torques too.
$endgroup$
– Uwe
14 hours ago
$begingroup$
Yes, I was mistaken about lb-ft vs ft-lb. I'll also admit that wikipedia's claim ft-lbf = ft-lb was true thirty years ago when Tom started lecturing. Let's see what others report...
$endgroup$
– Camille Goudeseune
14 hours ago
$begingroup$
Yes, I was mistaken about lb-ft vs ft-lb. I'll also admit that wikipedia's claim ft-lbf = ft-lb was true thirty years ago when Tom started lecturing. Let's see what others report...
$endgroup$
– Camille Goudeseune
14 hours ago
2
2
$begingroup$
Per @uhoh's answer, for both energy and torque, when you decompose the mathematical shorthand in the integral and the cross product, you wind up with the units being a force times a distance. Multiplication is commutative, so force times distance is equivalent to distance times force. Distinction between pound-foot and foot-pound isn't physics, it's semantics.
$endgroup$
– Tom Spilker
10 hours ago
$begingroup$
Per @uhoh's answer, for both energy and torque, when you decompose the mathematical shorthand in the integral and the cross product, you wind up with the units being a force times a distance. Multiplication is commutative, so force times distance is equivalent to distance times force. Distinction between pound-foot and foot-pound isn't physics, it's semantics.
$endgroup$
– Tom Spilker
10 hours ago
add a comment |
$begingroup$
Foot-pound or pound-foot are synonymous, and represent the arithmetic product of pound (force) and foot (length).
The pound (force) is the weight of one pound (mass) at the Earth's surface (somewhat imprecise because Earth's gravity field varies depending on your location, and the effective weight of an object will be influenced by the centrifugal force due to Earth's rotation, again dependent on location).
As a unit of energy, it is the energy of applying a one pound force over a distance of one foot. It is equivalent to raising a one-pound mass one foot in height, well... because.
As a unit of torque, it is the torque resulting from a one pound tangent force applied at a distance of one foot from the axis of rotation.
The same could be said of the Newton-meter (or meter-Newton, but it's never expressed that way); as a unit of energy, it is a one Newton force applied over a distance of one meter; as a unit of torque, it is a one Newton tangent force applied at a distance of one meter from the axis of rotation, except that the Newton is specifically a unit of force with a precise definition where pound may be either force or mass and pound (force) lacks a precise definition.
The measurement system which includes pounds and feet has a long history. When it developed, the variability of Earth's gravitational field and its impact on the weights and measures which depended on it wasn't understood, wasn't measurable, and/or wasn't significant for the engineering problems of the time. Culture, history, and familiarity keep these weights and measures in use despite the awkwardness and the advantages of metric.
answered 7 hours ago
Anthony XAnthony X
10.3k1 gold badge42 silver badges85 bronze badges
$endgroup$
$begingroup$
+1for thorough answer/explanation
$endgroup$
– uhoh
1 hour ago
add a comment |
$begingroup$
Foot-pound or pound-foot are synonymous, and represent the arithmetic product of pound (force) and foot (length).
The pound (force) is the weight of one pound (mass) at the Earth's surface (somewhat imprecise because Earth's gravity field varies depending on your location, and the effective weight of an object will be influenced by the centrifugal force due to Earth's rotation, again dependent on location).
As a unit of energy, it is the energy of applying a one pound force over a distance of one foot. It is equivalent to raising a one-pound mass one foot in height, well... because.
As a unit of torque, it is the torque resulting from a one pound tangent force applied at a distance of one foot from the axis of rotation.
The same could be said of the Newton-meter (or meter-Newton, but it's never expressed that way); as a unit of energy, it is a one Newton force applied over a distance of one meter; as a unit of torque, it is a one Newton tangent force applied at a distance of one meter from the axis of rotation, except that the Newton is specifically a unit of force with a precise definition where pound may be either force or mass and pound (force) lacks a precise definition.
The measurement system which includes pounds and feet has a long history. When it developed, the variability of Earth's gravitational field and its impact on the weights and measures which depended on it wasn't understood, wasn't measurable, and/or wasn't significant for the engineering problems of the time. Culture, history, and familiarity keep these weights and measures in use despite the awkwardness and the advantages of metric.
answered 7 hours ago
Anthony XAnthony X
10.3k1 gold badge42 silver badges85 bronze badges
$endgroup$
$begingroup$
+1for thorough answer/explanation
$endgroup$
– uhoh
1 hour ago
add a comment |
$begingroup$
Foot-pound or pound-foot are synonymous, and represent the arithmetic product of pound (force) and foot (length).
The pound (force) is the weight of one pound (mass) at the Earth's surface (somewhat imprecise because Earth's gravity field varies depending on your location, and the effective weight of an object will be influenced by the centrifugal force due to Earth's rotation, again dependent on location).
As a unit of energy, it is the energy of applying a one pound force over a distance of one foot. It is equivalent to raising a one-pound mass one foot in height, well... because.
As a unit of torque, it is the torque resulting from a one pound tangent force applied at a distance of one foot from the axis of rotation.
The same could be said of the Newton-meter (or meter-Newton, but it's never expressed that way); as a unit of energy, it is a one Newton force applied over a distance of one meter; as a unit of torque, it is a one Newton tangent force applied at a distance of one meter from the axis of rotation, except that the Newton is specifically a unit of force with a precise definition where pound may be either force or mass and pound (force) lacks a precise definition.
The measurement system which includes pounds and feet has a long history. When it developed, the variability of Earth's gravitational field and its impact on the weights and measures which depended on it wasn't understood, wasn't measurable, and/or wasn't significant for the engineering problems of the time. Culture, history, and familiarity keep these weights and measures in use despite the awkwardness and the advantages of metric.
answered 7 hours ago
Anthony XAnthony X
10.3k1 gold badge42 silver badges85 bronze badges
$endgroup$
Foot-pound or pound-foot are synonymous, and represent the arithmetic product of pound (force) and foot (length).
The pound (force) is the weight of one pound (mass) at the Earth's surface (somewhat imprecise because Earth's gravity field varies depending on your location, and the effective weight of an object will be influenced by the centrifugal force due to Earth's rotation, again dependent on location).
As a unit of energy, it is the energy of applying a one pound force over a distance of one foot. It is equivalent to raising a one-pound mass one foot in height, well... because.
As a unit of torque, it is the torque resulting from a one pound tangent force applied at a distance of one foot from the axis of rotation.
The same could be said of the Newton-meter (or meter-Newton, but it's never expressed that way); as a unit of energy, it is a one Newton force applied over a distance of one meter; as a unit of torque, it is a one Newton tangent force applied at a distance of one meter from the axis of rotation, except that the Newton is specifically a unit of force with a precise definition where pound may be either force or mass and pound (force) lacks a precise definition.
The measurement system which includes pounds and feet has a long history. When it developed, the variability of Earth's gravitational field and its impact on the weights and measures which depended on it wasn't understood, wasn't measurable, and/or wasn't significant for the engineering problems of the time. Culture, history, and familiarity keep these weights and measures in use despite the awkwardness and the advantages of metric.
answered 7 hours ago
Anthony XAnthony X
10.3k1 gold badge42 silver badges85 bronze badges
edited 6 hours ago
answered 7 hours ago
Anthony XAnthony X
10.3k1 gold badge42 silver badges85 bronze badges
answered 7 hours ago
Anthony XAnthony X
10.3k1 gold badge42 silver badges85 bronze badges
answered 7 hours ago
Anthony XAnthony X
10.3k1 gold badge42 silver badges85 bronze badges
10.3k1 gold badge42 silver badges85 bronze badges
$begingroup$
+1for thorough answer/explanation
$endgroup$
– uhoh
1 hour ago
add a comment |
$begingroup$
+1for thorough answer/explanation
$endgroup$
– uhoh
1 hour ago
$begingroup$
+1 for thorough answer/explanation$endgroup$
– uhoh
1 hour ago
$begingroup$
+1 for thorough answer/explanation$endgroup$
– uhoh
1 hour ago
add a comment |
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