# DRAWING ASTRONOMICAL FIGURES

## 14 PLANES AND PROJECTIONS

This is a convenient time to introduce a few remarks about the drawing of figures suitable for illustrating problems in Astro-navigation. These figures are used not only to illustrate texts such as this for the purpose of learning the theory of nautical Astronomy, but also to assist in the solution of navigational problems. Students must therefore not only be able to understand the various figures and diagrams used in Astro-navigation, but also be able to draw figures illustrating specific navigational circumstances for themselves. This is not particularly difficult, as this section will show, and the ability to draw these figures will prove of inestimable value both towards a greater understanding of the subject and for the practical solution of navigational problems at sea.

It is impossible to project a sphere on to a plane (i.e. flat) surface without considerable distortion. This may be readily demonstrated by cutting an orange in half, scooping out the fruit from the skin and placing the hemisphere on a table. The skin cannot be flattened out without first snipping it inwards at intervals round the edges, thus causing wedge-Shaped spaces which produce an increasing distortion toward, the circumference, only the central part of the skin remaining, normal.

A plane is a flat or level surface. Most of the figures drawn to illustrate Astro-navigation represent spheres - either the celestial sphere or the Earth - and of necessity must be drawn on a plane surface, i.e. a flat sheet of paper.  When a sphere is drawn in this way, at most only half the sphere (a hemisphere) can be represented at any one time. Now there are an infinite number of ways in which a sphere can be cut exactly in half, and fig. 15-1 shows only four of these. It can be sliced vertically through the centre, it can be sliced vertically at right angles (or any other angle) to the first cut, it can be sliced horizontally through the centre, or diagonally at any angle through the centre. Since each of these cuts passes through the centre of the sphere, the circle representing the circumference of the flat surface of the hemisphere, thus produced is a great circles and the figure is said to be drawn on the plane of this great circle.

Thus fig. 14-2 (a) is said to be drawn on the plane of the observer’s meridian, the observer being in N. Lat., and it shows the Earth and the celestial sphere as they would appear as it seen from the west. Fig. 14-2 (b) is also drawn on the plane of the observer’s meridian, but shows the appearance as if seen from the E., in which, for convenience, the Earth and the observer are indicated by a single point 0.

Fig. 14-1 of this Chapter is drawn on the plane of the great circle at right angles to the celestial equator, while fig. 14-2 is drawn on the plane of the Greenwich meridian.

When it is convenient to show the whole visible sky, the figure must be drawn on the plane of the celestial horizon, as if the celestial sphere were seen from a position directly above the observer’s zenith (fig. 14-3), Z therefore appears at the centre of the circle which represents the celestial horizon; the N.-S. and E.-W. lines divide this circle into four equal parts, and the celestial equator appears as a curve through W, Q and E.

Fig. 14-4 is a diagram attempting to illustrate the several planes involved in the Astro-navigational triangle PZX.  The diagram itself is drawn on the plane of the observer’s meridian NPZQS, indicated by the vertical line shading but SEX (Shaded grey) is the plane of the celestial or rational horizon, of which N is the N. point; S is the S. point and E the E. point.

Z = the observer's zenith

P  = the elevated pole

X  =  a celestial body

C  =  the centre of the Earth

QEQ1  =  the celestial equator or equinoctial

Referring to the body X, we have: -

AX  =  its altitude   = ACX

ZX  =  its zenith distance = ZCX

DX  =  its Dec.

PX =  its polar distance  = PCX

PZ =  the observer’ co-Lat.  = ZCP

ZPX =  L.H.A. of X  = arc of equator QD

PZX = Azimuth of X  = arc of horizon NA

When looking at a sphere or globe only the surface area in the line of vision between the eye and the centre of the globe is accurately seen, and the distortion becomes ludicrous at the circumference. Various methods, called projections are adopted to try to reduce the errors which result from representing the curved surface of a sphere on a flat sheet, each system possessing some special property of delineation for particular purposes, but all of them having disadvantages and limitations. The Mercator projection, for example, possesses more navigational advantages than any other and is therefore adopted for sea charts.

In drawing figures representing the Earth and celestial spheres for Astro-navigation the same problems arise.  It is impossible to project the whole of a sphere on to a plane surface and at most only half a sphere can be shown at one time and of that hemisphere only a small part can be drawn in its true proportion, the remainder being more or less distorted according to the principle of the projection adopted.

A natural projection assumes the observer’s eye to be situated at a fixed position usually external to the sphere, and the points and circles projected on to a plane surface in the direction of the visual rays leading towards them. Such a projection is the orthographic projection and the stereographic projection. Both are difficult to draw. Both involve distortion.

## 60. THE EQUIDISTANT PROJECTION

A more convenient and quicker system of figure drawing and one from which it is possible to obtain, by measurement, nearer approximations to the desired results than is possible by the stereographic projection is the Equidistant Projection or method of least error.

In this projection the eye is supposed to be situated on a diameter produced beyond the sphere to a distance of 1/N2 times the radius. The further semicircle is then divided into arcs of 10°, and the visual rays to these points, when projected to the plane at right angles to the diametrical line of sight, divide its diameter into nine parts that are so nearly equal to each other that they may be considered so for the purpose of illustrating problems in Astro-navigation.

As with the stereographic projection, the plane chosen for the equidistant projection is the plane of the celestial horizon because this enables the whole of the observer’s visible sky to be represented. The scale of degrees on the meridian (NZS) is practically equal throughout, as indeed it is on all the great circles passing through Z, which are vertical circles represented by straight lines on the projection, so that any convenient scale of equal parts may be used when measuring angular distances from the observer’s zenith.

Referring to fig. 14-6, and measuring with a pair of compasses or dividers, it will be noticed that the elevation of the pole (NP) on the plane of the observer’s celestial horizon is equal to the Lat. (ZQ), which is Lat. 40°N, and that the Dec. of the body X1, which is 20°N, when measured as an arc of the meridian QX on the plane of the celestial horizon, is exactly the same measure of equal parts as QY, which represents the Dec. of body Y1 in 20°S. The scale on the observer’s meridian NZS is uniform throughout.

Fig. 14-5 shows the construction of the equidistant projection while fig. 14-6 shows the resultant figure drawn on the plane of the observer’s celestial horizon. The straight line NZS represents the observer’s meridian, and WZE is his prime vertical the point Z being the observer’s zenith. The bounding circle NESW is the observer’s celestial horizon of which N is the N. point, E is E., S is S. and W is W.. The arc WQE is the celestial equator so that QZ represents the observer’s Lat..

The point P is the celestial pole (90° from the celestial equator at Q), the elevation of which (NP) is equal to the observer’s Lat. QZ. The points X and Y represent two celestial bodies, X with Dec. 20° N and Y with Dec. 20° S, situated on the observer’s celestial meridian.

Before going on to describe the practical construction of figures on the equidistant projection, fig. 14-6 enables us to refer to some fundamental notions relative to finding the Lat. from a body when it is crossing the observer’s meridian, a subject that is discussed more fully later in this chapter. The question of finding the Lat. by measuring the altitude of a body on the meridian resolves itself into finding the arc QZ.

In fig. 14-6, the arc SX is the True. Alt. of the body X above the S. point of the horizon, and by subtracting this arc from 90° (SZ) we get the zenith distance of the body X (ZX). The arc QX is the angular distance of the body X N. of the celestial equator, i.e., its Dec., so that the Lat. of the observer QZ = arc ZX + arc QX, in other words: - Lat. = Zenith Distance + Dec.

With regard to the celestial body Y, however, which is also on the observer’s meridian the zenith distance is the arc ZY and the Dec., which is south of the celestial equator in this case, is the arc QY, so that the Lat. of the observer QZ = arc ZY - arc QY, in other words: -

Lat. = Zenith Distance - Dec.

But suppose the body X in fig. 14-6 had a larger Dec. and was bearing N. from the observer at Z, as shown in fig. 14-7 In this case the altitude of X on the meridian (NX) would be observed by sextant and subtracted from 90° to give the zenith distance ZX, but the Dec. of the body X is the arc QX N. of the celestial equator, so that the Lat. of the observer QZ = arc QQX – arc ZX, in other words: - Lat. = Dec. - Zenith Distance.

Summing up these three circumstances, it can be said that the Lat. of an observer obtained by measuring the altitude of a body on the meridian is either the sum of the Zenith Distance and the body’s Dec., or the difference between the Zenith Distance and the body’s Dec. This could be written: - Lat. = Zenith Distance + ̃ Dec.

In practice, the decision whether to add the zenith distance to the Dec., whether to subtract the Dec. from the zenith distance or whether to subtract the zenith distance from the Dec. is best resolved by means of a diagram or figure, hence the value of possessing the ability to draw figures illustrating problems in Astro-navigation.  How to draw figures on the plane of the celestial equator using the simplest and most convenient method, the Equidistant Projection, will now be explained. The instruments required for drawing figures are a pair of compasses, dividers, a protractor, a ruler and a set of nautical tables.

## 61. TO CONSTRUCT A FIGURE ON THE EQUIDISTANT PROJECTION GIVEN LATITUDE AND DECLINATION.

Ex.:  Construct a figure on the Equidistant Projection for Lat. 40° S, Dec. of body 20° N, and from it find by measurement the zenith distance and altitude of the body when on the meridian. (refer to fig 14-8)

Method:

1. Describe a circle with nine equal parts as radius. Nine-tenths of an inch is a convenient scale to use, as each tenth will represent 10° as there are 90° from the zenith to the horizon.
2. Draw in the observer’s meridian NZS as a vertical line bisecting the circle, and the prime vertical WZE as a horizontal line bisecting the circle.
3. Take 40°, which is four equal parts from the scale and lay off with the same opening of the compasses or dividers the Lat. QZ (Z below Q because we are in South Latitude) and the elevation of the pole SP (which is always equal to the Lat.).
4.  Draw in the arc WQE, the celestial equator, the centre of which is always on the meridian NZS (produced outside the circle if necessary) and can be found by trial and error.
5.  Take 20°, that is two equal parts from the scale, and lay off QX the Declination, north of the equator. The spot X is the position of the body on crossing the meridian and its meridian zenith distance (ZX) measures 60°. (see fig 14-8).
6. The altitude is the complement of the Zenith Distance (i.e. 90º – Zenith Distance) so that the respective altitude of X is represented on the figure by SX = 30º. The maximum when the body is on the meridian.