#### 17 THE NAVIGATIONAL PZX TRIANGLE

In § 1-4 of this study we introduced the concept of the celestial sphere, on which the entire theory of Nautical Astronomy is based, and named the various points, circles and arcs upon this sphere. In § 6-8 we discussed the measurement of time, which is vitally important to the Astro-navigator in determining the position of the celestial body he is observing. And in § 9-13 we described how a navigator measures the altitude of a celestial body above the sea horizon with his sextant, and by applying certain corrections, obtain the True. Alt. of that body above the celestial or rational horizon. In § 14-17 we shall attempt to draw these three important threads together and to show how a navigator uses them to determine his position on the Earth’s surface.

Also in § 14-17 we shall show how a navigator can draw figures or diagrams to assist him to solve navigational problems, and we Shall describe the two simplest types of sights a navigator can use to find his position, involving the barest minimum of calculation: a Lat. by meridian altitude and solar noon Long..

Since many of the terms that are used in § 14-17 were introduced and defined in § 1-5 of this study, and are vital to an understanding of the subject, it is strongly recommended that students refresh their memory by going through § 1-5 again very carefully making sure they understand all these terms and definitions thoroughly, before proceeding further with § 14-17.

##### 68. THE GEOGRAPHICAL POSITION OF A CELESTIAL BODY The conception of nautical Astronomy the student has to develop is that an Observer on the Earth’s surface looks outward into space and imagines all the celestial bodies to be projected on to the interior concave surface of a celestial sphere having the same centre as the Earth. Further, at the time of observation the bodies are supposed to be momentarily stationary and are viewed outward from the centre of the Earth so that they appear to be projected on to the Earth’s surface as well as on to the celestial sphere.

If a line is drawn from a celestial body to the Earth’s centre (XC in fig. 17-1) the point where this line cuts the Earth’s surface is called the Geographical Position G.P. of the body. In fig. 17-1, x is the G.P. of the celestial body X. To an observer at x, the body would thus appear to be exactly overhead or to have a True. Alt. of 90°. That is, the body would be at the observer’s zenith.

If an observer were to observe a celestial body at his zenith, his terrestrial position would coincide with the observed body’s G.P. at the time of the observation. If, therefore, the observer knows the G.P. of the body at the time of a zenith observation (from data in the N.A.) his own position would thus also be known, this coinciding with the body’s G.P. This interesting but somewhat rare navigational sight provides the only case whereby a vessel’s position may be found from a solitary observation, and formed the basis of primitive Polynesian navigation by means of which the intrepid voyagers of the South Seas were able to make long sea journeys with navigational precision.

##### 69. THE NAVIGATIONAL (PZX) TRIANGLE

Fig. 17-2 represents the fundamental conception of nautical Astronomy and should be closely studied in conjunction with the following description. This figure must be thoroughly understood, together with the various terms before proceeding further with the subject.

The figure shows two spheres, the inner one representing the Earth and the outer one the celestial sphere, CP being the common axis and C the common centre from which points can be projected on to the surfaces of both spheres. In the description, which follows, CAPITAL letters represent the apparent positions of celestial bodies, circles and arcs on the celestial sphere and the corresponding lower case letters represent their respective geographical positions on the Earth’s surface.

P and PI represent the celestial poles and p and p1 the Earth’s poles. PGQP1, PXAP1, and PZBP1 are celestial meridians, and on the Earth pgqp1, pxap1, and pzbp1 are the corresponding terrestrial meridians.

The celestial equator (Q1Q) is the plane of the Earth’s equator extended to the celestial sphere, so that CqQ, CaA and CbB all lie on the same plane.

The arc QA corresponds to arc qa and both subtend the angle QCA⎫

The arc QB corresponds to arc qb and both subtend the angle QCB⎬

The arc QB corresponds to arc qb and both subtend the angle QCB ⎭

With one important difference! However….

– that the celestial arcs can only be expressed in degrees, minutes and seconds, whereas the arcs on the Earth’s surface may also be expressed in nautical miles. Suppose, for example, the arc AB = 30° on the celestial sphere, then the arc ab on the Earth’s surface is also 30°, but an arc of 30° on the Earth’s surface can also be expressed in linear measure: 30° x 60 = 1800 nautical miles

On the celestial sphere. X is the position of a celestial body and x its corresponding G.P. that is to say, the body X is vertically over positions on the Earth’s surface, both being projected along the line.  CxX.

The Declination of a body is the arc of the celestial meridian between a celestial body and the elevated pole, or 90º – Declination. PX represents the polar distance of the body X, and this arc corresponds to px on the Earth’s surface since both arcs subtend the same angle PCX, so that the polar distance of X = arc PX = arc px.

The elevated pole is the one above the horizon. The North Pole is the elevated one for an observer in N. Lat., and the South Pole, being below the horizon, would be his depressed pole. The Zenith is the point on the celestial sphere vertically over the observer. i.e., 90° from the observer’s celestial horizon. In the figure, z is the position of an observer on the Earth’s surface and a line from the centre of the Earth through this position (Cz) produced on to the celestial sphere gives point Z, the observer’s zenith. But bz represents the Lat. of z on the Earth and this arc corresponds to the arc BZ on the celestial sphere as both subtend the same angle BCZ at the centre of the Earth, so that Lat. z = arc bz = arc BZ.

The Zenith Distance is the angular distance of a celestial body from the observer’s zenith measured on the. vertical circle through the body, in other words 90° – True. Alt. The zenith distance of the body X in fig. 17-2 is ZX. but this figure does not show the relationship between zenith distance and altitude, which can be seen better on figures 17-3 (a) and (b). Fig. 17-3 (a) shows the terrestrial and celestial spheres similar to fig. 17-2. but whereas fig 17-2 in drawn on the plane of the Greenwich Meridian (pgqp1) with the elevated poles (p and P) is drawn on the plane of the observer’s meridian (NPZQS in fig. 17-3 (a) equivalent to PZBP1 on fig. 17-2) with the observer’s zenith (Z) uppermost. In fig. 17-3 (a) NAWS is the observer’s celestial horizon, and in fig. 17-3 (b) the same figure is shown on the plane of the celestial horizon as it viewed from a point directly above Z. In both figures 17-3 (a) and (b), the arc AX is the True. Alt. of the body X above the celestial horizon, and ZX is the zenith distance or 90° – True. Alt. Reverting to fig. 17-2. it will be seen that the arc zx on the Earth’s surface is a terrestrial great circle joining the G.P. of Z and X on the celestial sphere. The arcs ZX and zx are both equal to angle ZCX. the arc ZX measures the angular distance of the body X from the zenith, and the arc zx measures not only the angular distance between x and the observer. but also the great circle distance in miles between them. The Co-Latitude is the arc of a celestial meridian between the observer’s zenith and the elevated pole. On the Earth’s surface. arc pz is equal to the celestial arc PZ since both subtend the same angle PCZ and PZ is therefore the co-Lat. of the observer’s position and is equal to 90° – Lat. of z. We can therefore say that Co-Lat. = arc PZ = arc pz = 90° – Lat. of z (or 90° – bz).

The student will now begin to see how the three sides of the celestial spherical triangle PZX are projected downwards on to the Earth’s surface to form the three sides of the terrestrial spherical triangle pzx. This triangle is known as the pole-zenith-star triangle or the Astro-navigational triangle, because as will be shown, it forms the basics of determining a position line at sea. A navigator can determine the three sides of this triangle as follows: –

Side PX = px = Polar Distance    = 90°  – Dec. obtained from the N.A.

Side ZX = zx = Zenith Distance  = 90°   – Altitude obtained from sextant observation

Side PZ = pz = Co-Lat.                   = 90° – Lat. obtained from the vessel’s D.R. position

Fig. 17-4 shows two spherical triangles and fig. 17-5 two plane triangles, similar to each other in which the angles P, Z and X of the outer triangles are respectively equal to p, z and x of the inner triangles, but in the plane triangles the corresponding sides are not equal to each other.

In the spherical triangles, however, the accrual measures of the respective sides are equal and these values may be transferred from one triangle to the other, thus arc PZ = arc pz, arc ZX = arc zx, arc PX = arc px, not in length but they are equal to the same number of degrees, minutes and seconds of arc.

Having examined the sides of the PZX triangle, let us now take a look at the angles, P, Z and X. Reverting back to fig. 17-2, it will be seen that angle ZPX is the angle at the pole contained between the meridian of the observer (PZ) and the meridian of the body (PX), and this has already been defined as the L.H.A. of the body X. The angle at P of the Astro-navigational triangle is therefore simply the L.H.A. of the observed body. In fig. 17-2, g is the G.P. of Greenwich and G is the celestial counterpart of this position, pgqp1 being the Greenwich meridian and PGQP1 the Greenwich celestial meridian. The arc QA or the angle APQ is therefore the Greenwich Hour Angle (G.H.A.) of the body X, while its terrestrial counterpart qa or the angle qpa is the Longitude of the G.P. of the body X.

The second angle of the PZX triangle, the angle at Z is the azimuth of the observed body. It will be remembered that the azimuth of a celestial body is a measure of the angle at the observer’s zenith (or the arc of the horizon) between the vertical circle through the elevated pole and the vertical circle through the body, named according to whether the body lies to the E. or W. of the observer’s celestial meridian and measured from N. or S. according to whether the observer is in the northern or southern hemisphere respectively, In fig. 17-3, NPZQS is the vertical circle through the elevated pole (P). and AXZ is the vertical circle through the body X, so the azimuth is the angle PZX. In this particular figure the azimuth would be named N. so-many-degrees W. because the observer lies in N. Lat. and the body X lies to the W. of the observer’s meridian. In fig. 17-2, the azimuth (angle PZX) would be named N. so-many-degrees E., because in this case the body X lies to the E. of the observer’s meridian.

##### 70. THE FUNCTION OF THE ASTRO-NAVIGATIONAL TRIANGLE (PZX)

We stated In § 01-04 that the fundamental features of modern Astro-navigation is the relating of the celestial position of a heavenly body at a given instant of time using the horizon system, with its position at the same instant of time using the equinoctial system. The co-ordinates employed in defining a celestial position using the equinoctial system are Dec. and hour angle. The Astro-navigational triangle (PZX), which we have described above, is the connection between the two systems.

Fig. 17-6 illustrates a typical PZX triangle related to the observer o and the celestial body X. It shows the celestial sphere with the Earth at its centre, o being an observer in the northern hemisphere and Z his zenith. The point p is the Earth’s North Pole and p the elevated pole. HH1, is the observer’s celestial horizon and QQ1, is the celestial equator. Arc AX  = Dec. of X and        Angle ZPX = L.H.A. of X

These two co-ordinates define the body’s position using the equinoctial system, the Dec. (AX) being readily convertible into the polar distance PX.

Arc BX  = altitude of X         Angle PZX = azimuth of X

These two co-ordinates define the body’s position using the horizon system, the altitude BX being readily convertible into the zenith distance ZX.

The zenith distance of the celestial body X is, in minutes of arc, equal to the terrestrial great-circle distance between the observer and the G.P. of X. This distance (x 0 in fig 17-6) is the radius of a circle of equal altitude of the body X, shown as a pecked circle in fig. 17-5. A small arc of this circle, when drawn on a navigational chart, is an Astronomical position line, somewhere on which a navigator may fix his vessel’s position.

The Astro-navigational triangle (PZX) is, therefore solved by the navigator when he wishes to obtain an Astronomical position line. Note carefully that the result of a PZX triangle computation is a line of position and not a point of position. An Astronomical position line is the projection on to a navigational chart of part of a circle of equal altitude, the centre of which is located at the G.P. of the observed body and the radius of which is, in minutes of arc, equal to the zenith distance of the observed body.

An essential process in Astro-navigation is timing an altitude observation of a celestial body using a chronometer the error on G.M.T. of which is known. If the G.M.T. of the observation of X is known, the G.P. of X may be found, since the G.H.A. and Dec. of all the celestial bodies used in navigation are tabulated against G.M.T. in the N.A. For example, if the elements for the body X in fig. 14-6 at the G.M.T. of observation were G.H.A. 29° 50.4 and Dec. 53° 16.1 N. then the G.P. of X (x in fig. 17-6) would be Lat. 53° 16.1 N., Long 29° 50.4 W.

Similarly, since the arc ZX has been obtained by means of a sextant (90° – True. Alt.) and this corresponds to the terrestrial great circle distance ox in fig. 17-6, accordingly this distance is also known.

If, therefore, the bearing of o from x can be found, the position of o, the observer’s position, can also be found. Nowhere is the meaning of the small word if more important than it is in the foregoing sentence. There is no way of finding the bearing of o from x unless at least three parts of the PZX triangle are known.

The only known parts are the sides PX (the polar distance – which can be found from the Dec. in the N.A.), and ZX (the zenith distance – found by sextant measurement). The third side of the triangle (PZ) is not known unless the observer knows his Lat., and at sea he is unlikely to know this at the time of the observation. The angle at o, which is equal to the angle PZX (the azimuth), cannot be found from a compass observation to the degree of accuracy necessary for ascertaining the vessel’s position. The angle at p, which is equal to the angle ZPX (the L.H.A. of the observed body), cannot be found unless the observer’s Longitude is known, and again the observer at sea does not generally know his Longitude at the time of observation.

It the navigator does know if his vessel’s Lat. precisely it is an easy matter for him to find his vessel’s Longitude from an altitude observation. Since the Lat. is known, the three sides of the PZX triangle are also known from a single observation, so that the angle ZPX may be computed. This angle, being the L.H.A. of the observed body, when compared with the G.H.A. of the body for the G.M.T. of observation, will give the vessel’s Longitude: –

G.H.A. of body ~ L.H.A. of body = Longitude of Observer.

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