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(48G) Earth's Acceleration Estimation by Latitude +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Software Libraries (https://www.hpmuseum.org/forum/forum-10.html) +--- Forum: General Software Library (https://www.hpmuseum.org/forum/forum-13.html) +--- Thread: (48G) Earth's Acceleration Estimation by Latitude (/thread-25019.html) |
(48G) Earth's Acceleration Estimation by Latitude - Eddie W. Shore - 2026-05-01 09:54 The true earth gravity force depends on several factors, such as latitude and altitude. This estimation takes the latitude (north/south) into account. The constant 9.80665 m/s^2 is an accepted average, the true force varies. g_Earth = g_45-(g_poles + g_equ)/2*cos(lat*π/90 radians) Simplified: g_Earth = 9.806 - 0.026*cos(lat*π/90) g_45 ≈ 9.806 m/s^2 g_poles ≈ 9.832 m/s^2 g_equ ≈ 9.78 m/s^2 Take the cosine of (latitude * π / 90) radians gLAT: Code: << RAD HMS→ 90 / π * →NUM COSInput: 1: latitude in D.MMSS (degrees, minutes, seconds) No need to enter units Output: 1: Earth's gravity at latitude_m/s^2 with unit object attached Example: Input: 1: 20.2214 (20°22'14") Output: 1: ≈ 9.7863_m/s^2 Source: Grainger Engineering Office of Marketing and Communications. (answer written by Rebecca H.) (2016, November 21). “How gravitational force varies at different locations on Earth.” Illinois. https://van.physics.illinois.edu/ask/listing/64061. Retrieved March 10, 2026. RE: (48G) Earth's Acceleration Estimation by Latitude - Thomas Klemm - 2026-05-01 12:58 You could use DEG instead of RAD: Code: « DEG HMS→ 2. * COS .026 * 9.806 SWAP - '1_m/s^2' * » |