# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Time Sight and Ex-Meridian by Bygrave**

**From:**Lars Bergman

**Date:**2021 Apr 7, 14:27 -0700

Tony,

with latitude φ, declination δ, hour angle t, altitude h and reduction to the meridian x, you have

sin(h+x) = sin φ · sin δ + cos φ · cos δ

sin h = sin φ · sin δ + cos φ · cos δ · cos t

Subtraction gives

sin(h+x) - sin h = cos φ · cos δ · (1 - cos t)

sin h · cos x + cos h · sin x - sin h = cos φ · cos δ · 2·sin^{2}(t/2)

With x and t small quantities you can approximate to

x·cos h = cos φ · cos δ · 2·(t/2)^{2} with x and t expressed in radians. Then

x = (cos φ · cos δ / cos h) · t^{2 }/ 2

For x expressed in minutes of arc and t in degrees, you must multiply the right hand side by 60·180/π · (π/180)^{2} which equals π/3. Thus

x = (cos φ · cos δ / cos h) · π/6 · t^{2 } with π/6 = 0.52, Greg's formula.

Lars