Thought I might put some info here and other things I've been joting down in my notebook over time. If there's a webpage out there that you think might be of interest to me please let me know. Comments, mistakes etc are always welcome, Paul.

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• Selected Problems in Interstellar Navigation
  
• Interstellar Navigation
  
• Navigation and guidance in interstellar space
  
• Relativistic Navigation Needed for Solar Sails
  
  
• Star Tracker 1 | Star Tracker 2
• Inertial Stellar Compass
  
• HEASARC: Coordinate Converter
  
• XNAV: Navigating By X-Ray Pulsar | Spacecraft Navigation Using X-ray Pulsars | Paper | USNO | IEEE/ION PLANS
  
• The Orientation of the Local Interstellar Magnetic Field
  
• Kalman filter
  

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Interstellar fix using position tori A position torus using two stars, we could be anywhere on the exterior surface of this torus if we measured the angle between the stars Three position tori using six stars, where all three tori intersect is our position
Ground work

It is first useful to look into the following known position fixing method used in standard navigation which uses Horizontal Sextant Angles (HSA) to obtain a fix at sea provided we have some landmarks we can identify and are marked on the chart. We measure the angular separation between lighthouses for eg by using the sextant horizontally to give us the angle. Let A, B and C shown below be lighthouses on land, we are somewhere off the coast and wish to fix our position on the chart.
As an example we measured the HSA between A and B to be 30° and the HSA measured between B and C to be 50°. Before going through the mathematical proof of the method let us describe first how we go about using it. We draw a line on the chart between A and B. Then draw a line from A which is 90° - HSA° from this line so in this case we draw a line 90° - 30° = 60° from the line AB to seaward with the aid of a protractor. Similarly draw another line from B 60° from BA. These lines intersect at D shown below forming the triangle ABD. If we did this correctly the triangle ABD is an isosceles triangle as the angle DAB = DBA = 60° in this example and hence AD = BD. We now need our drawing compass, place the center of the compass at D then draw a full circle with a radius AD or DB which are equal. The circle will intersect A and B. This is known as our position circle. We could be anywhere to seaward on this blue circle and we will measure a HSA between lighthouses A and B to be 30° (we'll see later why) however we don't know where on this circle we are (hence the naming position circle). What if we repeated this method but this time let us measure the HSA bewteen B and C. Ideally the vessel is stopped or moving very slowly between these measurements. We measure a HSA of 50° between B and C and plot our second position circle using the above method. As before we must be anywhere to seaward on this green position circle for a HSA between B and C to be 50^o. Since we must also be on the blue position circle as well the only position where this qualifies is where both position circles intersect: we have fixed our position on the chart. Notice there is another location where both position circles also intersect however this is at lighthouse B and is on land so we can rule this out. Any slight error in measurement in the first two HSAs (vessel rolling in heavy seas, instrument errors etc) using the sextant will affect our fix and we need to plot a third position circle to confirm our position. Ideally all three position circles should intersect at one spot however in practice sometimes they don't. We aim to get the smallest "cocked hat". We mark our position in the middle of the "hat" or closest to any dangers such as nearby reefs, rocks etc. A very large cocked hat indicates that one or more of our HSAs has a large error or we are using the wrong charted landmark! However very accurate position fixes can be obtained with this method see: Analysis of Random Errors in Horizontal Sextant Angles.

In the example given above we have measured the HSAs to be less than 90° and in this case the center of the position circles lie on our side. What if our measured HSA is over 90°? Let's say we measured a HSA between two lighthouses A and B to be 120°. In this case the center of the position circle lies on the opposite side of the line joining the lighthouses from the vessel and we take 120° - 90° = 30° for the base of the triangle as shown below: We could be anywhere on this position circle to seaward. Notice the arc where our position could be is smaller now than the landward arc as opposed to our previous HSAs when they were less than 90° the arcs that were part of the position circle to seaward were greater then the landward arcs (we'll talk more about these these major and minor arcs later). Let's now look at the situation where our measured HSA = 90°.
We are somewhere on the ocean at night and the only landmarks closeby are two small islands with lighthouses. With the measured HSA between these lighthouses of 90°, the center of the position circle is now at the bisecting point of the line joining the lighthouses (we'll see later why), there is no isosceles triangle as with the previous examples because 90° - 90° = 0° and the line AB is the diameter of the circle. Note that we could be anywhere on this position circle (except the islands) because in this case we could also have measured a HSA of 90° if we were on the other side of the islands since there is water there as well.

We have covered the three situations where the measured HSA < 90°, HSA = 90° and HSA > 90° and how to construct a position circle on the chart in each case.

Let us now look at the mathematical proof of the above method and see why it works. A number of theorems concerning circles apply here:

Background:
An inscribed angle is an angle formed by two chords in a circle which have a common endpoint. This common endpoint forms the vertex of the inscribed angle. The other two endpoints define what we call an intercepted arc on the circle. The intercepted arc might be thought of as the part of the circle which is "inside" the inscribed angle. (See the pink part of the circle in the picture above.)

A central angle is any angle whose vertex is located at the center of a circle. A central angle necessarily passes through two points on the circle, which in turn divide the circle into two arcs: a major arc and a minor arc. The minor arc is the smaller of the two arcs, while the major arc is the bigger. We define the arc angle to be the measure of the central angle which intercepts it.


Proposition: The measure of the intercepted arc (equal to its central angle) is exactly twice the measure of the inscribed angle.
In Excursions in Geometry, C. Ogilvy gives a good practical problem: Let's say that AB above is a screen for a drive in theater and the professionals informed the owner on the optimum angle θ (theta) that the screen AB should present to the viewer. But only one customer can sit in the preferred spot V directly in front of the screen. The owner is interested in locating other points, U, from which the screen subtends the same angle θ. The answer is the circle passing through the three points A, B and V. AUB is measured by the same arc as AVB, the angle at U is the same as the angle at V. This is also our Horizontal Sextant Angle that we measured between the lighthouses.

Proof:
Let O be the center of a circle. Choose two points on the circle, and call them V and A. Draw line VO and extended past O so that it intersects the circle at point B which is diametrically opposite the point V. Draw an angle whose vertex is point V and whose sides pass through points A and B.

Angle BOA is a central angle; call it θ. Draw line OA. Lines OV and OA are both radii of the circle, so they have equal lengths. Therefore triangle VOA is isosceles, so angle BVA (the inscribed angle) and angle VAO are equal; let each of them be denoted as ψ.

Angles BOA and AOV are supplementary. They add up to 180°, since line VB passing through O is a straight line. Therefore angle AOV measures 180° − θ.

It is known that the three angles of a triangle add up to 180°, and the three angles of triangle VOA are:

180° − θ
ψ
ψ

Therefore

Subtract 180° from both sides,

where θ is the central angle subtending arc AB and ψ is the inscribed angle subtending arc AB. The inscribed angle is only defined for points on the major arc (the longest path around the circle between the two given points) hence the inscribed angle is undefined in the shorter (minor) arc, see this applet to see what happens (you'll need java enabled in your browser). We'll go through another theorem later to find the angle in the minor arc.

Following from above let's look at inscribed angles with the center of the circle in their interior:
Given a circle whose center is point O, choose three points V, C, and D on the circle. Draw lines VC and VD: angle DVC is an inscribed angle. Now draw line VO and extend it past point O so that it intersects the circle at point E. Angle DVC subtends arc DC on the circle.

Suppose this arc includes point E within it. Point E is diametrically opposite to point V. Angles DVE and EVC are also inscribed angles, but both of these angles have one side which passes through the center of the circle, therefore the theorem from the above theorem can be applied to them.

Therefore

then let



so that

Draw lines OC and OD. Angle DOC is a central angle, but so are angles DOE and EOC, and

Let



so that

From our above theorem we know that θ1 = 2ψ1 and that θ2 = 2ψ2. Combining these results with equation (2) yields

therefore, by equation (1),

This is known as the Central Angle Theorem: the measure of the inscribed angle is always half the measure of the central angle. The inscribed angle is our HSA that we measured with our sextant.


Proposition: If A, B and C are points on a circle where the line AC is a diameter of the circle, then the angle ABC is a right angle.
Proof:

We use the following facts:

    - the sum of the angles in a triangle is equal to two right angles (180°),
    - the base angles of an isosceles triangle are equal.
Let O be the center of the circle. Since OA = OB = OC, OAB and OBC are isosceles triangles, and by the equality of the base angles of an isosceles triangle, OBC = OCB and BAO = ABO. Let α = BAO and β = OBC. The 3 internal angles of the ABC triangle are α, α + β and β. Since the sum of the angles of a triangle is equal to two right angles, we have

then

or simply

This is known as the Thales' Theorem: The diameter of a circle always subtends a right angle to any point on the circle. This applies to the situation where we measured a HSA of 90°.


Proposition: Opposite angles in any quadrilateral inscribed in a circle are supplements of each other. (Their measures add up to 180°).

Proof:
From above, the main result we need is that an inscribed angle has half the measure of the intercepted arc. Here, the intercepted arc for Angle(A) is the Arc(BCD) and for Angle(C) is the Arc(DAB).

Arc(BCD) = 2 x Angle(A) and Arc(DAB) = 2 x Angle(C)

Arc(BCD) + Arc(DAB) = 360°

2 x Angle(A) + 2 x Angle(C) = 2 x [Angle(A) + Angle(C)] = 360°

Angle(A) + Angle(C) = 180°

Hence opposite angles in any quadrilateral inscribed in a circle add up to 180°.

Let us now look back at the situation where the measured HSA was 120° and see why we need to draw the triangle opposite to us. Here's the situation again:
Recall we started off by drawing the base of the triangle with angles of 120° - 90° = 30°. Following from the inscribed angles theorem we went through previously, if we were anywhere on the major arc (landward) we would measure a HSA of 60°. However we can't be on this side as our measured HSA = 120°. From the above theorem we just found that opposite angles in any quadrilateral inscribed in a circle add up to 180°. This is only true if we are on the minor arc since 60° + 120° = 180°.

From the above theorems we can now see why the method is correct and why we need to be on the position circle for a given measured HSA between two landmarks.