*A simple and uniform method for obtaining the latitude from bodies on, or near, the meridian*

WHEN using the formula, L = Z+d, for meridian altitude and reduction sights, the different precepts are not easy to remember, so much so, that often a projection on the plane of the meridian is drawn to show how the various elements should be combined. In addition, an approximate altitude must be calculated in order to find the height of eye correction in preparing a constant.

The following method, using the haversine-cosine formula, gets away from the usual precepts; all sights being worked alike in the case of the upper transit of any body. There is no separate operation required to calculate an approximate altitude. The work is extremely short and simple. The haversine-cosine formula for the calculation of the line of position is as follows:

hav z = (hav t cos L cos d) + hav(L~d)- When the body is on the meridian, t, the angle of the body from the meridian, becomes zero, the body bears north or south and the value, (hav t cos L cos d), drops out leaving:

hav z= hav (L~d) z= (L~d)

also, z = 90—He

substituting, we get:

90— Hc= (L~d)

Hc = 90— (L~d)

This Hmc is the calculated altitude of the body when on the meridian, with the height of eye correction and I.C. applied to it. The difference between the sextant altitude of the body on the meridian (H) and the meridian constant (Hmc) gives an altitude intercept (a) in minutes of latitude towards or away from the body according as Hs is greater or less than Hmc. This altitude intercept is measured from the nearest even degree of latitude which was assumed at the beginning of the problem, and gives the latitude.

The following problem will illustrate the method:

(1) The Navigator of a ship predicts his D.R. position at local apparent noon on 2 March 1927 to be Lat. 38°-02'-45" N, Long. 61°-02'-25" W, I.C.+ l'-00", Ht. of eye 65 feet. He prepares a constant for the meridian altitude of the sun. He takes a sextant altitude of the sun at LAN. 44^{o}-23^{,}-00^{,,} and finds the latitude. Sun bearing south.

h m s

LAT of LAN 12-00-00

Long West 4-04-09

GAT of LAN 16-04-09 (2 Mar)

Eq T (sign reversed) (+)12-24.0

GCT of LAN 16-16-33.0 (2 Mar)

Dec. of sun 7°-24:5 S

Latitude assumed 38 -00 N

In order then to prepare a constant for a meridian sight, it is only necessary to combine our nearest even degree of latitude and the body’s declination (if like signs, subtract; if unlike, add), and subtract that amount from 90 degrees, to obtain He. Look up the height of eye correction using He and apply it and add the I.C. to He with signs reversed, to get Hmc (meridian constant.)

L+d 45 -24.5

90_______

He 44 -35.5

Corr. for Ht. of eye & IC (~) 8.2

Hmc (meridian ens’t) 44 -27.3

Hs 44 -23.0

1285

(Sun bearing S) a= 4.3 Away LAN latitude 38 -04.3 N#

When the body is near the meridian, the value, “hav t cos L cos d,” is equal to a t^{1} and Hc = 90— (L~d) — a t^{J}. To obtain a reduction constant, find a t^{2}, and subtract it from Hmc (meridian constant) to obtain Hrc (reduction constant). Apply to Hrc the sextant altitude Hs at the time of the reduction sight to obtain the altitude intercept. This altitude intercept gives the latitude at the time of sight. The following problem will illustrate the method: (2) During evening twilight on July 8 1927 the navigator of a ship in D. R. Lat. IP-OS' N, Long. 53°-30' W, observes the star Arcturus as follows: W 6^{h}-45^{m}-00" P.M., C-W 4^{h}-01^{m} -05^{8}, chronometer error l^{m}-20^{s} fast, I.C.+ l'-OO", Ht. of eye 65 feet, and sextant altitude 81^{o}-36'-30''. The star bears almost north and the navigator works the sight as a reduction to the meridian as follows:

| h m s |

w | 6-45-00 |

C-W | 4-01-05 |

c | 10-46-05 |

cc | (-) 1-20 |

| 10-44-45 |

GCT | 22-44-45 (8 July) |

RAM O | 19-59-47.9 |

Tab. Ill | 3-44.2 |

GST | 17-48-17.1 |

Long. W | 3-34-00 |

LST | 14-14-17.1 |

RA * | 14-12-20.9 |

HA * | 0-01-56.2 |

a =12 | a t |

Aquino has said, “the only true interpretation of any sight is a line of position.” This is just as true of meridian altitudes and reductions to the meridian as of other sights. When the latitude is obtained from a meridian altitude the line of position, being at right angles to the bearing of the body, runs east and west and defines our latitude. When we use the reduction to the meridian method, we obtain the latitude at which our line of position crosses the longitude we have used in working our reduction sight, and the line runs at right angles to the bearing of the body.

Dec. * Latitude L~d | (+) (+) | 19°-33.'8 N 11 -00 N 8 -33.8 90 |

He IC & Ht. Eye c. | 81 -26.2 (+) 7.0 | |

Hmc a t |
| 81 -33.2 (-) 0.8 |

Hrc Hs |
| 81 -32.4 81 -36.5 |

(Star bearing N) Latitude | a = 4.’1 Towards. 11°-04:l N# |

When using the reduction method with high altitudes it is desirable to find the bearing of the body and to lay off our line of position perpendicular to the bearing through our D.R. longitude and the latitude obtained from the reduction sight.