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While acceleration is a vector quantity, g-force accelerations ("g-forces" for short) are often expressed as a scalar, based on the vector magnitude, with positive g-forces pointing downward (indicating upward acceleration), and negative g-forces pointing upward. Thus, a g-force is a vector of acceleration.
The standard acceleration of gravity or standard acceleration of free fall, often called simply standard gravity and denoted by ɡ 0 or ɡ n, is the nominal gravitational acceleration of an object in a vacuum near the surface of the Earth. It is a constant defined by standard as 9.806 65 m/s 2 (about 32.174 05 ft/s 2).
The conversion for the poundal is given by 1 pdl = 1 lb·ft/s 2 = 0.138 254 954 376 N (precisely). [1] To convert between the absolute and gravitational FPS systems one needs to fix the standard acceleration g which relates the pound to the pound-force. [citation needed] =
In engineering and physics, g c is a unit conversion factor used to convert mass to force or vice versa. [1] It is defined as = In unit systems where force is a derived unit, like in SI units, g c is equal to 1.
Peak ground acceleration can be expressed in fractions of g (the standard acceleration due to Earth's gravity, equivalent to g-force) as either a decimal or percentage; in m/s 2 (1 g = 9.81 m/s 2); [7] or in multiples of Gal, where 1 Gal is equal to 0.01 m/s 2 (1 g = 981 Gal).
The gravitational acceleration vector depends only on how massive the field source is and on the distance 'r' to the sample mass . It does not depend on the magnitude of the small sample mass. This model represents the "far-field" gravitational acceleration associated with a massive body.
The pound-force is the product of one avoirdupois pound (exactly 0.45359237 kg) and the standard acceleration due to gravity, approximately 32.174049 ft/s 2 (9.80665 m/s 2). [ 5 ] [ 6 ] [ 7 ] The standard values of acceleration of the standard gravitational field ( g n ) and the international avoirdupois pound (lb) result in a pound-force equal ...
The gravity g′ at depth d is given by g′ = g(1 − d/R) where g is acceleration due to gravity on the surface of the Earth, d is depth and R is the radius of the Earth. If the density decreased linearly with increasing radius from a density ρ 0 at the center to ρ 1 at the surface, then ρ(r) = ρ 0 − (ρ 0 − ρ 1) r / R, and the ...