DETERMINING THE CULMINATION OF A CELESTIAL BODY/Prilog problemu odredivanja kulminacije nebeskoga tijela

Headnote

Summary

The meridian passage time does not always coincide with the culmination time of the celestial body. The difference is in the local hour angle. At the culminate the celestial body is crossing the observer's celestial meridian and is said to be in the upper transit. Often, culmination is used to mean upper culmination. A few celestial body altitudes measured around local apparent noon can give a more accurate position fix than using single measurement. In the following article the problem solving equation based on the interpolation polynomial in the Lagrange's form will be shown. This method can easily be solved using a programmable hand-held calculator.

Key words: Nautical astronomy

Sazetak

Meridijanska visina nece uvijek odgovarati kulminacijskoj visini nebeskoga tijela. Razliku cini promjena satnog kuta. Nebesko tijelo kulminira prolazeci nebeskim meridijanom opazaca, za sto se cesto koristi izrazom "gornji prolaz". Mjerenje nekoliko visina nebeskog tijela u blizini nebeskog meridijana podloga je za tocniji racun opazaceve pozicije (geografske sirine). U rjesavanju problema u clanku se koristi interpolacijskim polinomom u Lagrangeovu formatu. Pretpostavlja se uporaba racunala.

Kljucna rijeci: Nauticka astronomija

(ProQuest: ... denotes formulae omitted.)

INTRODUCTION / Uvod

The procedure of obtaining ship's position at sea through observation of the highest altitude in the meridian is based on the assumption that the observed altitude is a symmetrical function of time. Nothing the time of maximum sextant altitude is not sufficient for determining the time of meridian passage because the celestial body appears to 'hang' for a finite time at its local maximum altitude. A few celestial body altitudes measured around local apparent noon give a more accurate position fix than using single measurement. Since 1 min of time corresponds to 15 min of arc, the longitude estimate will be less accurate than the latitude one. Time measurements should be exact and the sextant arc is to be returned to zero between sights. A high-quality micrometer sextant would improve accuracy. The problem solving equation based on interpolation polynomial in the Lagrange form will be evaluated.

CULMINATION OF A CELESTIAL BODY / Kulminacija nebeskoga tijela

The meridian passage time does not always coincide with the culmination time of the celestial body. The equation of a circle of equal altitude is:

H^sub o^ = f (δ, LHA) = arcsin(sinφsinδ + cosφcosδcosLHA) (1)

Ho - observed height or altitude of a celestial body; LHA - Local Hour Angle, δ - declination;

φ and λ are the coordinates of a dead reckoning (assumed) position.

The differentiation of equation (1) with respect to time (t) gives:

dH^sub 0^ = dφcosZ^sub n^ + dδcos π - dLHAsinZ^sub n^cosφ (2)

where Zn stands for the azimuth and p is the parallactic angle of a celestial body.

In the culmination a celestial body attains its highest or lowest observed altitude i.e. the function H0=H0(t) has its extremum. The mathematical condition for it is the vanishing of the derivative dH0 /dt. For that reason the values dφ, dδ and dLHA have to be calculated accurately. As the several observed heights include all these rates of changes, this forms the basic concept of the following method.

APPLICATION OF INTERPOLATION POLYNOMIAL IN THE LAGRANGE FORM / Primjena interpolacijskog polinoma u Lagrangeovu formatu

Interpolation polynomial in the Lagrange form is particularly applicable for a non-equidistant nodes i.e. nonequal time differences between consecutive sights. For a three observed off culmination heights the polynomial reads:

... (3)

where xi - time of observation; f(xi) - observed height.

The expression (3) is the equation of parabola:

L^sub 2^(x) ≡ f(x) = ax^sup 2^ + bx + c (4)

with vertex at ... (5)

Solving for latitude requires the declination and the zenith distance. As the vertex time and height are deduced from (5), well known nautical astronomy rules are simply applied.

AN EXAMPLE / Primjer

The following example illustrates this process. On 8th April 2006 a vessel at the assumed position:

Lat = 30°06'N; Long = 039°W, proceeds at course 41° with speed 15 kts. A fixed star Betelgeuse was observed as per table below:

Applying expressions (3), (4) and (5) we get culmination time and height (vertex):

Hculm = 67°12.8';UT = 19h22m. Consequently, latitude equals Lat = 30°11.7'N.

CONCLUSION / Zakljucak

The shown parabola fits the rate of change of altitude and its peak shows culmination height and culmination time respectively. Albeit, in a given example, a fixed star and slow steaming vessel have been used, resulting in consecutive heights' differences not been substantial, the shown mathematical model gives good result in latitude. Obtaining a vessel's longitude is less accurate as the difficult part lies in determining the precise moment of culmination. Since 1 min of time corresponds to 15 min of arc, the longitude estimate will be less accurate than the latitude one. This method can easily be solved using a programmable hand-held calculator.

References

REFERENCES / Literatura

1. Bowditch, N.: American Practical Navigator-Vol I, DMAHTC, Washington, 1984

Rukopis primljen: 22. 1. 2008.