Duration of eclipse calculation

[WafflesToo]

Just something I figured out for myself the other day and thought it might be of some interest to someone else out here. Kind of nice to know how many batteries you need to keep your kethane scanning or coms relay satellites operational as they pass through the dark-side of the planet.

All you need to know is the Gravitational Parameter of the body you're orbiting, the radius of the body you're orbiting, and the semi-major axis of your orbit (works best with circular- or nearly circular orbits.)

First you need to find your orbital period (either by calculating it as below... or by looking it up in the KER window, you cheater)

T = 2pi * SQRT(a^3/u)

Where

T = Orbital Period (in seconds)

a = Semi-Major Axis (in m)

u = Gravitational Parameter (in m^3/s^2)

Next, find the arc of your orbit that will be in the shadow of the planet

w = 180 - 2 * SIN-1 (SQRT(a^2 + 2ar)/(a+r))

Where

w = arc width (in degrees)

a = ave altitude above surface of orbit (in meters)

r = radius of orbited body

Which we plug into a percentage formula to find the amount of time we'll spend there (in seconds)

s = Tw/360

Where

s = time spent occluded from the sun (in seconds)

T = Orbital period (in seconds)

w = arc width (in degrees)

So, if we're orbiting Kerbin in a nearly circular orbit with an average altitude of 75km...

T = 2pi * SQRT(675000^3/3.35316 * 10^12) = 1900 seconds (about-ish)

w = 180 - 2 * SIN-1 (SQRT(75000^2 + 2 * 75000 * 600000)/(75000 + 600000) = 125 (ish) degrees

Which means we'll be spending

1900 * 125 /360 = 660 seconds (11 minutes) in the dark.

Hope someone finds this helpful.

Edited November 16, 2013 by WafflesToo

[politas]

Once I discovered that LibreOffice Calc was returning the inverse Sine in Radians with ASIN(), I got a spreadsheet to make these calculations quick and easy.