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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?)

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  • $\begingroup$ It's certainly true that the same set of units can describe multiple situations. Take a look at, e.g., some of the tables of constants in the NIST standard SP811 $\endgroup$ – Carl Witthoft Aug 6 '19 at 13:27
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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.

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    $\begingroup$ @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 Aug 5 '19 at 15:38
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    $\begingroup$ The last line here emphasizes that my two guesses for the unit's meaning are equivalent. Bingo! $\endgroup$ – Camille Goudeseune Aug 5 '19 at 15:39
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    $\begingroup$ 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 Aug 5 '19 at 16:17
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    $\begingroup$ 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 Aug 5 '19 at 17:38
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    $\begingroup$ I guess that would be obvious to a physicist. Maybe not so obvious to somebody who is struggling to understand how torque and energy can appear to be specified with the same units (i.e., to the guy who asked the original question.) $\endgroup$ – Solomon Slow Aug 6 '19 at 15:09
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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.

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  • $\begingroup$ +1 for thorough answer/explanation $\endgroup$ – uhoh Aug 6 '19 at 4:16
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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.

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  • $\begingroup$ But pound-foot is used for torques too. $\endgroup$ – Uwe Aug 5 '19 at 15:10
  • $\begingroup$ 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 Aug 5 '19 at 15:28
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    $\begingroup$ 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 Aug 5 '19 at 19:14

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