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Sights: Taking and Reducing

Sights: Taking and Reducing. Taking Sights. Day time sights offer the advantage of a clearly visible horizon Early morning sights have been successful because as long as there are bright planets or the moon, the horizon becomes visible as it becomes more light

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Sights: Taking and Reducing

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  1. Sights: Taking and Reducing

  2. Taking Sights • Day time sights offer the advantage of a clearly visible horizon • Early morning sights have been successful because as long as there are bright planets or the moon, the horizon becomes visible as it becomes more light • Evening sights of the moon and planets work but the visibility of the horizon is a problem • For sights at night, work an expected sight reduction to determine ZN , the bearing, to make sure one of the channel islands is not in the background, making bringing down the sight to the horizon bringing down the sight to the top of the land

  3. Solving the Navigational Triangle • The 3-D view of the celestial horizon and celestial equator coordinate systems • Nautical Almanac Sight Reduction, NASR • The divided triangle • Navigational Mathematics • Napier’s rules • Examples • 0<LHA<900 : Arcturus 26 July 2007 • 2700 <LHA< 3600 : Jupiter 26 July 2007 • 900 <LHA< : Venus 26 July 2007

  4. USPS Navigation Text: NASR

  5. Bowditch Online Ch 21: Navigational Mathematics

  6. Napier’s Rules

  7. Napier’s Rules: Upper Triangle • Take the two sides on either side of the right angle, i.e. A and B in Diagram 2-16, the divided triangle. Continue around the triangle, using the complements of the other three parts: Co-LHA, Co-Co-L, and Co-Z1 to get Napier’s Diagram. Co-LHA B Co-Co-L A Co-Z1

  8. Complements for a right triangle Sin q = a/c = cos (900 – q) c q b 900 a

  9. Napier’s Rules • Considering any part as the middle part, the two parts nearest it in Napier’s Diagram are considered the adjacent parts, and the two farthest from it the opposite parts. • The sine of a middle part equals the product of the cosines of the opposite parts

  10. Napier’s Rules • Sin A = cos Co-LHA*cos Co-Co-L • Sin L = cos A*cos B, cos B =sin L/cos A • Sin Co-Z1 = cos Co-LHA*cos B

  11. B Co-LHA Co-Co-L A Co-Z1

  12. Example: Arcturus 26 July 2007, sight # 13 LHA = 290 01.7’ C0-Dec Co-L L = 340 24.4’ N Dec = 190 08.7’ N Eq. Co-H For NASR method use Assumed L = 340 Assumed LHA = 290 G

  13. Example Arcturus NASR • Sin A = cos Co-LHA*cos Co-Co-L = • =sin LHA*cos L = sin LHA*sin Co-L • =sin 290 *sin 560 • A = 23.69860 ~ 230 41.9’ • Cos B = sin L/cos A = sin 340 /cos 23.69860 • B = 52.36052 = 520 21.6’ • Sin Co-Z1 = cos Z1 = cos Co-LHA*cos B • Cos Z1 = sin LHA*cos B = sin 290 *cos 52.36052 • Z1 = 72.80

  14. Arcturus NASR

  15. Lower Triangle Co-F A Co-Co-p Co-Z2 Co-Co-H 900 = B + Co-F + Dec, where Dec = 190 09’ Co-F= 900 – B – Dec = 900 – 52.360520 – 19.150 Co-F = 18.489480 F = 900 – 18.494480 = 71.510520 = 710 30.6’

  16. Arcturus NASR

  17. Lower Triangle Sin H = cos Co-F*cos A = cos 18.489480 *cos 23.698600 HC = 60.274060 = 600 16.4’ And since Ho = 160 02.6’ after correcting the height of the sextant for instrument correction, dip, and the main correction, HC – HO = 600 16’ – 600 03’ = 13’ away Sin Co-F = cos Co-Z2 * cos Co-Co-H sin Z2 = sin Co-F/cos H = sin 18.489480/cos 60.274060 Z2 = 39.7599540 , Z = Z1 + Z2 = 72.80 + 39.80 Z = 112.60 , ZN = 3600 – 112.60 = 247.40

  18. Bearing • Note that ZN = 2470 agrees with reduction by the law of cosines, while reduction by NASR Tables yields ZN = 2490

  19. Arcturus NASR

  20. Jupiter 26 July 2007 LHA=3570 06.1’, tE =3600 – LHA = 020 53.9’ Co-L G L = 340 24.4’ N Co-F Eq Dec 210 25.5’ S 900 = B + Co-F -Dec Co-H For NASR use assumed L = 340 assumed LHA = 3570

  21. Jupiter: Napier’s Circle, Upper Triangle Sin A = cos Co-tE *cos L = sin tE * cos L = sin tE * sin Co-L = sin 30 * sin 560 A = 2.48676 = 20 29.2’ Sin L = cos A*cos B Sin 340 = cos 2.48676*cos B B = 55.96356 = 550 57.8’ Sin B = cos L*cos Co-Z1 sin B = cos L* sin Z1 Sin 55.963560* =cos 340 sin Z1 Z1 = 88.30 Co-tE B Co-Co-L A Co-Z1

  22. Jupiter NASR

  23. Lower Triangle Co-F = 900 – B + Dec = 900 – 550 58’+ 210 26’ = 550 28’ F = 900 – 550 28’= 340 32’ Sin H = cos A* cos Co-F =cos 2.48676*cos 55.46667 HC = 34.496210 = 340 29.8’ HC – H0 = 340 30’ – 340 08’ = 22’ A sin Co-F =cos Co-Z2 * cos H =sin Z2 *cos H Sin 55.46667 = sin Z2 *cos34.49621 Z2 = 88.30 ZN = Z = Z1 + Z2 =88.3+88.3=176.6 Co-F Co-q Co-Co-H A Co-Z2 Z2 Co-F Co-H q

  24. Bearing • Once again Z2 for the lower triangle is different for Napier’s method versus Tabular NASR • Arcturus Z2 = 39.80 (Napier) Vs. 38.60 (NASR) • Jupiter Z2 = 88.30 (Napier) Vs. 88.60 (NASR)

  25. LHA = 920 92.1’ C =LHA Venus Oblique a = Co-L Co-Dec b = Co-Dec G Co-L L Z A Co-H Dec Eq B c = Co-H Assumed L = 340 Assumed LHA = 930 Dec = 60 50.5’ N Co-Dec = 830 09’ Cos c = cos a*cos b + sin a*sin b* cos C = cos 560 *cos 83.150 + sin 560 *sin83.150 *cos 930 c = Co-H = 88.646750 = 880 38.8’ HC = 900 – Co-H = 10 21.2’ Ho – HC = 10 26’ – 10 21’ = 5’ Toward

  26. Venus NASR

  27. Venus Oblique • Sin C/sin c = sin B/sin b • Sin LHA/sin Co-H = sin Z/sin Co-Dec • Sin 930 /sin 88.646750 = sin Z/sin 83.150 • Z = 82.60 • ZN = 3600 – Z = 2770

  28. http://titulosnauticos.net/astro/chapter10.pdf

  29. Cos c = cos a*cos b + sin a* sin b*cos C

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