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

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Re: Contrail and Aircraft distance
From: Lars Bergman
Date: 2021 Oct 12, 10:29 -0700

Kermit,

A triangle with corners at the centre of the earth (assumed spherical), the observer and the aircraft, has the sides
r, d, r+h with the angle at the observer 90°+α

This gives (r+h)2 = d2 + r2 - 2·d·r·cos(90°+α)

As h << 2·r, and cos(90°+α) = -cos(90°-α) = -sin α ≈ α when α is a small angle in radians, this gives

d ≈ -r·α + √( (r·α)2 + 2·h·r )

We have h ≈ 42000 / 6080 = 6.9 nautical miles, and r = 180·60 / π = 3438 nautical miles, and α=0.76°= 0.013

(1) without refraction

d ≈ -45+ √( 1998 + 47444 ) = 177 nautical miles

(2) with 19' refraction

In this case α → α - 19' = 0.76° - 19' = 0.44° = 0.0077 and d ≈ -26+ √( 701 + 47444 ) = 193 nautical miles

Lars

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