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foot-pounds of energy?


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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}







3












$begingroup$


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?)










share|improve this question











$endgroup$





















    3












    $begingroup$


    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?)










    share|improve this question











    $endgroup$

















      3












      3








      3





      $begingroup$


      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?)










      share|improve this question











      $endgroup$




      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 12 hours ago







      Camille Goudeseune

















      asked 13 hours ago









      Camille GoudeseuneCamille Goudeseune

      7805 silver badges18 bronze badges




      7805 silver badges18 bronze badges

























          3 Answers
          3






          active

          oldest

          votes


















          6












          $begingroup$


          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.



          $$E = int mathbf{F} cdot d mathbf{s}$$



          To be torque, the pound has to be perpendicular to the foot



          $$tau = mathbf{r} times mathbf{F}$$



          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.






          share|improve this answer











          $endgroup$











          • 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
            12 hours ago






          • 1




            $begingroup$
            The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
            $endgroup$
            – Camille Goudeseune
            12 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
            12 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
            11 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
            10 hours ago



















          1












          $begingroup$

          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.






          share|improve this answer











          $endgroup$















          • $begingroup$
            But pound-foot is used for torques too.
            $endgroup$
            – Uwe
            13 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
            12 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
            8 hours ago



















          1












          $begingroup$

          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.






          share|improve this answer











          $endgroup$


















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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$


            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.



            $$E = int mathbf{F} cdot d mathbf{s}$$



            To be torque, the pound has to be perpendicular to the foot



            $$tau = mathbf{r} times mathbf{F}$$



            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.






            share|improve this answer











            $endgroup$











            • 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
              12 hours ago






            • 1




              $begingroup$
              The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
              $endgroup$
              – Camille Goudeseune
              12 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
              12 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
              11 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
              10 hours ago
















            6












            $begingroup$


            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.



            $$E = int mathbf{F} cdot d mathbf{s}$$



            To be torque, the pound has to be perpendicular to the foot



            $$tau = mathbf{r} times mathbf{F}$$



            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.






            share|improve this answer











            $endgroup$











            • 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
              12 hours ago






            • 1




              $begingroup$
              The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
              $endgroup$
              – Camille Goudeseune
              12 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
              12 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
              11 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
              10 hours ago














            6












            6








            6





            $begingroup$


            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.



            $$E = int mathbf{F} cdot d mathbf{s}$$



            To be torque, the pound has to be perpendicular to the foot



            $$tau = mathbf{r} times mathbf{F}$$



            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.






            share|improve this answer











            $endgroup$




            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.



            $$E = int mathbf{F} cdot d mathbf{s}$$



            To be torque, the pound has to be perpendicular to the foot



            $$tau = mathbf{r} times mathbf{F}$$



            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.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 6 mins ago

























            answered 12 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
              12 hours ago






            • 1




              $begingroup$
              The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
              $endgroup$
              – Camille Goudeseune
              12 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
              12 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
              11 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
              10 hours ago














            • 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
              12 hours ago






            • 1




              $begingroup$
              The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
              $endgroup$
              – Camille Goudeseune
              12 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
              12 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
              11 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
              10 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
            12 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
            12 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
            12 hours ago




            $begingroup$
            The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo!
            $endgroup$
            – Camille Goudeseune
            12 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
            12 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
            12 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
            11 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
            11 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
            10 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
            10 hours ago













            1












            $begingroup$

            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.






            share|improve this answer











            $endgroup$















            • $begingroup$
              But pound-foot is used for torques too.
              $endgroup$
              – Uwe
              13 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
              12 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
              8 hours ago
















            1












            $begingroup$

            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.






            share|improve this answer











            $endgroup$















            • $begingroup$
              But pound-foot is used for torques too.
              $endgroup$
              – Uwe
              13 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
              12 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
              8 hours ago














            1












            1








            1





            $begingroup$

            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.






            share|improve this answer











            $endgroup$



            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.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 13 hours ago

























            answered 13 hours ago









            Russell BorogoveRussell Borogove

            101k3 gold badges353 silver badges436 bronze badges




            101k3 gold badges353 silver badges436 bronze badges















            • $begingroup$
              But pound-foot is used for torques too.
              $endgroup$
              – Uwe
              13 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
              12 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
              8 hours ago


















            • $begingroup$
              But pound-foot is used for torques too.
              $endgroup$
              – Uwe
              13 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
              12 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
              8 hours ago
















            $begingroup$
            But pound-foot is used for torques too.
            $endgroup$
            – Uwe
            13 hours ago




            $begingroup$
            But pound-foot is used for torques too.
            $endgroup$
            – Uwe
            13 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
            12 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
            12 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
            8 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
            8 hours ago











            1












            $begingroup$

            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.






            share|improve this answer











            $endgroup$




















              1












              $begingroup$

              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.






              share|improve this answer











              $endgroup$


















                1












                1








                1





                $begingroup$

                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.






                share|improve this answer











                $endgroup$



                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.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 5 hours ago

























                answered 6 hours ago









                Anthony XAnthony X

                10.3k1 gold badge42 silver badges85 bronze badges




                10.3k1 gold badge42 silver badges85 bronze badges

































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