## Introduction

In a previous post (Ref.1) I described how I developed an interest in Celestial Navigation and in my last post (Ref.2) I started explaining what I believe to be the important points such as why CelNav is still relevant and the GP Geographic Position. In this post I will cover the Navigational Triangle and the Trigonometric Solution. The On Line Course starts Feb 3rd.

## Navigational Triangle

Let’s consider a practical example. An observer is located at Point C and the SunGP at that date/time is located at Point A. Point B is the North Pole. Figure 1 shows this situation. BC, BA and CA are great circle arcs on the spherical Earth. BC and BA by definition are Meridian or Longitudinal lines.

In order to solve a spherical triangle, we need to have several pieces of information (Ref.3, Ref.4, Ref.5). For a plane triangle, there is the Sine Law and Cosine Law:

sinA/a = sinB/b = sinC/c

a^2 = b^2 + c^2 – 2*b*c*cos(A)

Similarly for a spherical triangle there is a Sine Law and Cosine Law:

sin(A)/sin(a) = sin(B)/sin(b) = sin(C)/sin(c)

cos(a) = cos(b) + cos(c) + 2*sin(b)*sin(c)*cos(A)

Additionally there is the Napier Analogies:

tan[(b+c)/2] = [cos((B-C)/2)/cos((B+C)/2)]*tan(a/2)

So CelNav basically boils down to solving this triangle depending on the information given. Let’s say that in Figure 1 we took a sextant reading of the Sun so we have Ho and we also took a bearing all at a given time. We are located at C, but don’t know where that is. Given the time, we can determine the GP located at A. Since we know LatA, we can determine the co-latitude = (90deg – LatA).

c = co-latA in rad * Re, where Re = 6371Km

From the previous post we know that the arc distance AC = Zd the Zenith Distance. We also know the Azimuth of the GP so we know angle C:

C = 360deg – Az.

So this is a case of a spherical triangle with 3 independent sources of information = two sides and a non-contained angle (b,c,C). So using a combination of the Sine/Cosine/Napier laws we can solve for a, A, B.

a=2*atan[tan((b+c)/2)*cos((B+C)/2)/cos((B-C)/2)]

A = asin[sin(C)/sin(c) * sin(a)]

B = asin[sin(C)/sin(c) * sin(b)]

Depending on the type of measurement taken, the spherical triangle will change for example you might have two sides and a contained angle.

## Trigonometric Solution

There are many ways to solve the spherical triangle. In my EBook/Course I use ScicosLab which is an Open Source Math program. You can also use a programmable calculator. The Nautical Almanac and AP3270 provide convenient tables to allow you to solve various aspects of the Nautical Triangle using just a paper and pencil, which is very important to know for a practical backup. You need to be able to determine where you are if all electronics fail.

Celestial Navigation Basics Land Sea & Air Course next April 7th 2021.

## References

#1. – “Celestial Navigation Basics Land Sea & Air – Course_a”

#2. – “Celestial Navigation Basics Land Sea & Air – Course_b

#3. – “Scilab &ScicosLab – Open Source Math for CelNav”

#4. – “Scilab & ScicosLab Open Source Math for Plane Triangle”

#5. – “Scilab & ScicosLab Open Source Math for Spherical Triangle”

Please send your comments, questions and suggestions to:

jclark@clarktelecommunications.com