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1 Sights: Taking and Reducing. 2 Taking Sights Day time sights offer the advantage of a clearly visible horizon Day time sights offer the advantage of.

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Presentation on theme: "1 Sights: Taking and Reducing. 2 Taking Sights Day time sights offer the advantage of a clearly visible horizon Day time sights offer the advantage of."— Presentation transcript:

1 1 Sights: Taking and Reducing

2 2 Taking Sights Day time sights offer the advantage of a clearly visible horizon 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 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 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 Z N, 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 For sights at night, work an expected sight reduction to determine Z N, 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 3 Solving the Navigational Triangle The 3-D view of the celestial horizon and celestial equator coordinate systems The 3-D view of the celestial horizon and celestial equator coordinate systems Nautical Almanac Sight Reduction, NASR Nautical Almanac Sight Reduction, NASR The divided triangle The divided triangle Navigational Mathematics Navigational Mathematics Napier’s rules Napier’s rules Examples Examples 0<LHA<90 0 : Arcturus 26 July 2007 0<LHA<90 0 : Arcturus 26 July 2007 270 0 <LHA< 360 0 : Jupiter 26 July 2007 270 0 <LHA< 360 0 : Jupiter 26 July 2007 90 0 <LHA< : Venus 26 July 2007 90 0 <LHA< : Venus 26 July 2007

4 4

5 5

6 6 USPS Navigation Text: NASR

7 7 Bowditch Online Ch 21: Navigational Mathematics

8 8 Napier’s Rules

9 9 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-Z 1 to get Napier’s Diagram. 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-Z 1 to get Napier’s Diagram. A B Co-LHA Co-Co-L Co-Z 1

10 10 Complements for a right triangle a b c  Sin  = a/c = cos (90 0 –  90 0

11 11 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. 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 The sine of a middle part equals the product of the cosines of the opposite parts

12 12 Napier’s Rules Sin A = cos Co-LHA*cos Co-Co-L Sin A = cos Co-LHA*cos Co-Co-L Sin L = cos A*cos B, cos B =sin L/cos A Sin L = cos A*cos B, cos B =sin L/cos A Sin Co-Z 1 = cos Co-LHA*cos B Sin Co-Z 1 = cos Co-LHA*cos B

13 13

14 14 A B Co-LHA Co-Co-L Co-Z 1

15 15 Example: Arcturus 26 July 2007, sight # 13 G Eq. L = 34 0 24.4’ N Dec = 19 0 08.7’ N LHA = 29 0 01.7’ C0-Dec Co-H Co-L For NASR method use Assumed L = 34 0 Assumed LHA = 29 0

16 16 Example Arcturus NASR Sin A = cos Co-LHA*cos Co-Co-L = Sin A = cos Co-LHA*cos Co-Co-L = =sin LHA*cos L = sin LHA*sin Co-L =sin LHA*cos L = sin LHA*sin Co-L =sin 29 0 *sin 56 0 =sin 29 0 *sin 56 0 A = 23.69860 ~ 23 0 41.9’ A = 23.69860 ~ 23 0 41.9’ Cos B = sin L/cos A = sin 34 0 /cos 23.69860 Cos B = sin L/cos A = sin 34 0 /cos 23.69860 B = 52.36052 = 52 0 21.6’ B = 52.36052 = 52 0 21.6’ Sin Co-Z 1 = cos Z 1 = cos Co-LHA*cos B Sin Co-Z 1 = cos Z 1 = cos Co-LHA*cos B Cos Z 1 = sin LHA*cos B = sin 29 0 *cos 52.36052 Cos Z 1 = sin LHA*cos B = sin 29 0 *cos 52.36052 Z 1 = 72.8 0 Z 1 = 72.8 0

17 17 Arcturus NASR

18 18 Lower Triangle A Co-F Co-Co-p Co-Co-H Co-Z 2 90 0 = B + Co-F + Dec, where Dec = 19 0 09’ Co-F= 90 0 – B – Dec = 90 0 – 52.36052 0 – 19.15 0 Co-F = 18.48948 0 F = 90 0 – 18.49448 0 = 71.51052 0 = 71 0 30.6’

19 19 Arcturus NASR

20 20 Lower Triangle Sin H = cos Co-F*cos A = cos 18.48948 0 *cos 23.69860 0 H C = 60.27406 0 = 60 0 16.4’ And since H o = 16 0 02.6’ after correcting the height of the sextant for instrument correction, dip, and the main correction, H C – H O = 60 0 16’ – 60 0 03’ = 13’ away Sin Co-F = cos Co-Z 2 * cos Co-Co-H sin Z 2 = sin Co-F/cos H = sin 18.48948 0 /cos 60.27406 0 Z 2 = 39.759954 0, Z = Z 1 + Z 2 = 72.8 0 + 39.8 0 Z = 112.6 0, Z N = 360 0 – 112.6 0 = 247.4 0

21 21 Bearing Note that Z N = 247 0 agrees with reduction by the law of cosines, while reduction by NASR Tables yields Z N = 249 0 Note that Z N = 247 0 agrees with reduction by the law of cosines, while reduction by NASR Tables yields Z N = 249 0

22 22 Arcturus NASR

23 23 Jupiter 26 July 2007 G LHA=357 0 06.1’, t E =360 0 – LHA = 02 0 53.9’ Eq L = 34 0 24.4’ N Dec 21 0 25.5’ S Co-L Co-H Co-F 90 0 = B + Co-F -Dec For NASR use assumed L = 34 0 assumed LHA = 357 0

24 24 Jupiter: Napier’s Circle, Upper Triangle A B Co-t E Co-Z 1 Co-Co-L Sin A = cos Co-t E *cos L = sin t E * cos L = sin t E * sin Co-L = sin 3 0 * sin 56 0 A = 2.48676 = 2 0 29.2’ Sin L = cos A*cos B Sin 34 0 = cos 2.48676*cos B B = 55.96356 = 55 0 57.8’ Sin B = cos L*cos Co-Z 1 sin B = cos L* sin Z 1 Sin 55.96356 0 * =cos 34 0 sin Z 1 Z 1 = 88.3 0

25 25 Jupiter NASR

26 26 Lower Triangle Co-F Z2Z2 Co-H A Co-F  Co-  Co-Co-H Co-Z 2 Co-F = 90 0 – B + Dec = 90 0 – 55 0 58’+ 21 0 26’ = 55 0 28’ F = 90 0 – 55 0 28’= 34 0 32’ Sin H = cos A* cos Co-F =cos 2.48676*cos 55.46667 H C = 34.49621 0 = 34 0 29.8’ H C – H 0 = 34 0 30’ – 34 0 08’ = 22’ A sin Co-F =cos Co-Z 2 * cos H =sin Z 2 *cos H Sin 55.46667 = sin Z 2 *cos34.49621 Z 2 = 88.3 0 Z N = Z = Z 1 + Z 2 =88.3+88.3=176.6

27 27 Bearing Once again Z 2 for the lower triangle is different for Napier’s method versus Tabular NASR Once again Z 2 for the lower triangle is different for Napier’s method versus Tabular NASR Arcturus Z 2 = 39.8 0 (Napier) Vs. 38.6 0 (NASR) Arcturus Z 2 = 39.8 0 (Napier) Vs. 38.6 0 (NASR) Jupiter Z 2 = 88.3 0 (Napier) Vs. 88.6 0 (NASR) Jupiter Z 2 = 88.3 0 (Napier) Vs. 88.6 0 (NASR)

28 28 Venus Oblique G LHA = 92 0 92.1’ Eq L Dec a = Co-L C =LHA A c = Co-H B Cos c = cos a*cos b + sin a*sin b* cos C = cos 56 0 *cos 83.15 0 + sin 56 0 *sin83.15 0 *cos 93 0 c = Co-H = 88.64675 0 = 88 0 38.8’ H C = 90 0 – Co-H = 1 0 21.2’ H o – H C = 1 0 26’ – 1 0 21’ = 5’ Toward Co-H Co-L b = Co-Dec Assumed L = 34 0 Assumed LHA = 93 0 Dec = 6 0 50.5’ N Co-Dec = 83 0 09’ Co-Dec Z

29 29 Venus NASR

30 30 Venus Oblique Sin C/sin c = sin B/sin b Sin C/sin c = sin B/sin b Sin LHA/sin Co-H = sin Z/sin Co-Dec Sin LHA/sin Co-H = sin Z/sin Co-Dec Sin 93 0 /sin 88.64675 0 = sin Z/sin 83.15 0 Sin 93 0 /sin 88.64675 0 = sin Z/sin 83.15 0 Z = 82.6 0 Z = 82.6 0 Z N = 360 0 – Z = 277 0 Z N = 360 0 – Z = 277 0

31 31

32 32http://titulosnauticos.net/astro/chapter10.pdf

33 33 Cos c = cos a*cos b + sin a* sin b*cos C


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