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The poundal (symbol: pdl) is a unit of force that is part of the foot-pound-second system of units, in Imperial units introduced in 1877, and is from the specialized subsystem of English absolute (a coherent system).
- $ 1\,\text{pdl} = 1\,\tfrac{\text{lb}_m \cdot \text{ft}}{\text{s}^2} $
The poundal is defined as the force necessary to accelerate 1 pound-mass to 1 foot per second per second. 1 pdl = 0.138254954376 N exactly.
Background Edit
English units require re-scaling of either force or mass to eliminate a numerical proportionality constant in the equation F = ma. The poundal represents one choice, which is to rescale units of force. Since a pound of force (pound force) accelerates a pound of mass (pound mass) at 32.174 049 ft/s^{2} (9.80665 m/s^{2}; the acceleration of gravity, g), we can scale down the unit of force to compensate, giving us one that accelerates 1 pound mass at 1 ft/s^{2} rather than at 32.174 049 ft/s^{2}; and that is the poundal, which is approximately ^{1}∕_{32} of a pound force.
For example, a force of 1200 poundals is required to accelerate a person of 150 pounds mass at 8 feet per second squared:
- $ 150\,\text{lb}_m \cdot 8\,\tfrac{\text{ft}}{\text{s}^2} = 1200\,\text{pdl} $
Base |
force, length, time | weight, length, time | mass, length, time | |||||
---|---|---|---|---|---|---|---|---|
Force (F) | F = m·a = w·ag | F = m·ag_{c} = w·ag | F = m·a = w·ag | |||||
Weight (w) | w = m·g | w = m·gg_{c} ≈ m | w = m·g | |||||
System | BG | GM | EE | M | AE | CGS | MTS | SI |
Acceleration (a) | ft/s^{2} | m/s^{2} | ft/s^{2} | m/s^{2} | ft/s^{2} | gal | m/s^{2} | m/s^{2} |
Mass (m) | slug | hyl, also called “metric slug” or “TME” | lb_{m} | kg | lb | g | t | kg |
Force (F) | lb | kp | lb_{F} | kp | pdl | dyn | sn | N |
Pressure (p) | lb/in^{2} | at | PSI | atm | pdl/ft^{2} | Ba | pz | Pa |
The poundal-as-force, pound-as-mass system (Absolute foot-pound-second system) is contrasted with an alternative system in which pounds are used as force (pounds-force), and instead, the mass unit is rescaled by a factor of roughly 32 (Gravitational foot-pound-second system). That is, one pound-force will accelerate one pound-mass at 32 feet per second squared; we can scale up the unit of mass to compensate, which will be accelerated by 1 ft/s^{2} (rather than 32 ft/s^{2}) given the application of one pound force; this gives us a unit of mass called the slug, which is about 32 pounds mass. Using this system (slugs and pounds-force), the above expression could be expressed as:
- $ 4.66\,\text{slug} \cdot 8\,\tfrac{\text{ft}}{\text{s}^2} = 37.3\,\text{lb}_F $
Note: Slugs (32.174 049 lb_{m}) and poundals (1/32.174 049 lb_{F}) are never used in the same system, since each exists to solve the same problem and will cancel each other out; both should not be used together.
Rather than changing either force or mass units, one may choose to express acceleration in units of the acceleration due to Earth's gravity (called g). In this case, we can keep both pounds-mass and pounds-force, such that applying one pound force to one pound mass accelerates it at one unit of acceleration (g):
- $ 150\,\text{lb}_m \cdot 0.249\,\text{g} = 37.3\,\text{lb}_F $
(See Unit force-mass foot-pound-second system.)
Expressions derived using poundals for force and lb_{m} for mass (or lb_{F} for force and slugs for mass) have the advantage of not being tied to conditions on the surface of the earth. Specifically, computing F = ma on the moon or in deep space as poundals, lb_{m}·ft/s^{2} or lb_{F} = slug·ft/s^{2}, avoids the constant tied to acceleration of gravity on earth.
Conversion Edit
newton (SI unit) | dyne | kilogram-force, kilopond | pound-force | poundal | |
1 N | ≡ 1 kg·m/s² | = 10^{5} dyn | ≈ 0.10197 kp | ≈ 0.22481 lb_{F} | ≈ 7.2330 pdl |
1 dyn | = 10^{−5} N | ≡ 1 g·cm/s² | ≈ 1.0197×10^{−6} kp | ≈ 2.2481×10^{−6} lb_{F} | ≈ 7.2330×10^{−5} pdl |
1 kp | = 9.80665 N | = 980665 dyn | ≡ g_{n}·(1 kg) | ≈ 2.2046 lb_{F} | ≈ 70.932 pdl |
1 lb_{F} | ≈ 4.448222 N | ≈ 444822 dyn | ≈ 0.45359 kp | ≡ g_{n}·(1 lb) | ≈ 32.174 pdl |
1 pdl | ≈ 0.138255 N | ≈ 13825 dyn | ≈ 0.014098 kp | ≈ 0.031081 lb_{F} | ≡ 1 lb·ft/s² |
The value of g_{n} as used in the official definition of the kilogram-force is used here for all gravitational units. |
See also Edit
ReferencesEdit
- Obert, Edward F., “Thermodynamics”, McGraw-Hill Book Company Inc., New York 1948; Chapter I, Survey of Dimensions and Units, pages 1-24.
- ↑ Lindeburg, Michael, Civil Engineering Reference Manual for the PE Exam
- ↑ Wurbs, Ralph A, Fort Hood Review Sessions for Professional Engineering Exam, http://engineeringregistration.tamu.edu/tapedreviews/Fluids-PE/PDF/Fluids-PE.pdf, retrieved October 26, 2011