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Sights: Taking and ReducingSights: Taking and Reducing

22

Taking SightsTaking Sights Day time sights offer the advantage of a clearly Day time sights offer the advantage of a clearly

visible horizonvisible horizon Early morning sights have been successful Early morning sights have been successful

because as long as there are bright planets or because as long as there are bright planets or the moon, the horizon becomes visible as it the moon, the horizon becomes visible as it becomes more lightbecomes more light

Evening sights of the moon and planets work but Evening sights of the moon and planets work but the visibility of the horizon is a problemthe visibility of the horizon is a problem For sights at night, work an expected sight reduction For sights at night, work an expected sight reduction

to determine Zto determine ZN N , the bearing, to make sure one of the , the bearing, to make sure one of the channel islands is not in the background, making channel islands is not in the background, making bringing down the sight to the horizon bringing down bringing down the sight to the horizon bringing down the sight to the top of the landthe sight to the top of the land

33

Solving the Navigational TriangleSolving the Navigational TriangleThe 3-D view of the celestial horizon and The 3-D view of the celestial horizon and

celestial equator coordinate systemscelestial equator coordinate systemsNautical Almanac Sight Reduction, NASRNautical Almanac Sight Reduction, NASR

The divided triangleThe divided triangleNavigational MathematicsNavigational Mathematics

Napier’s rulesNapier’s rulesExamplesExamples

0<LHA<900<LHA<900 0 : Arcturus 26 July 2007: Arcturus 26 July 20072702700 0 <LHA< 360<LHA< 3600 0 : Jupiter 26 July 2007: Jupiter 26 July 200790900 0 <LHA< : Venus 26 July 2007<LHA< : Venus 26 July 2007

44

55

66

USPS Navigation Text: NASRUSPS Navigation Text: NASR

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Bowditch OnlineCh 21: NavigationalMathematics

88

Napier’s RulesNapier’s Rules

99

Napier’s Rules: Upper TriangleNapier’s Rules: Upper TriangleTake the two sides on either side of the Take the two sides on either side of the

right angle, i.e. A and B in Diagram 2-16, right angle, i.e. A and B in Diagram 2-16, the divided triangle. Continue around the the divided triangle. Continue around the triangle, using the complements of the triangle, using the complements of the other three parts: Co-LHA, Co-Co-L, and other three parts: Co-LHA, Co-Co-L, and Co-ZCo-Z1 1 to get Napier’s Diagram.to get Napier’s Diagram.

A

BCo-LHA

Co-Co-L

Co-Z1

1010

Complements for a right triangleComplements for a right triangle

a

b

c

Sin = a/c = cos (900 –

900

1111

Napier’s RulesNapier’s RulesConsidering any part as the middle part, Considering any part as the middle part,

the two parts nearest it in Napier’s the two parts nearest it in Napier’s Diagram are considered the adjacent Diagram are considered the adjacent parts, and the two farthest from it the parts, and the two farthest from it the opposite parts.opposite parts.

The sine of a middle part equals the The sine of a middle part equals the product of the cosines of the opposite product of the cosines of the opposite partsparts

1212

Napier’s RulesNapier’s Rules

Sin A = cos Co-LHA*cos Co-Co-LSin A = cos Co-LHA*cos Co-Co-LSin L = cos A*cos B, cos B =sin L/cos ASin L = cos A*cos B, cos B =sin L/cos ASin Co-ZSin Co-Z1 1 = cos Co-LHA*cos B= cos Co-LHA*cos B

1313

1414

A

BCo-LHA

Co-Co-L

Co-Z1

1515

Example: Arcturus 26 July 2007, sight # 13Example: Arcturus 26 July 2007, sight # 13

G

Eq.

L = 340 24.4’ NDec = 190 08.7’ N

LHA = 290 01.7’

C0-Dec

Co-H

Co-L

For NASR method useAssumed L = 340

Assumed LHA = 290

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Example Arcturus NASRExample 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=sin 290 0 *sin 56*sin 5600 A = 23.69860 ~ 23A = 23.69860 ~ 230 0 41.9’41.9’ Cos B = sin L/cos A = sin 34Cos B = sin L/cos A = sin 340 0 /cos 23.69860/cos 23.69860 B = 52.36052 = 52B = 52.36052 = 520 0 21.6’ 21.6’ Sin Co-ZSin Co-Z1 1 = cos Z= cos Z1 1 = cos Co-LHA*cos B = cos Co-LHA*cos B

Cos ZCos Z1 1 = sin LHA*cos B = sin 29= sin LHA*cos B = sin 290 0 *cos 52.36052*cos 52.36052

ZZ1 1 = 72.8= 72.800

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Arcturus NASRArcturus NASR

1818

Lower TriangleLower Triangle

ACo-F

Co-Co-p

Co-Co-H

Co-Z2

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’

1919

Arcturus NASRArcturus NASR

2020

Lower TriangleLower TriangleSin 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’ awaySin 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

2121

BearingBearingNote that ZNote that ZN N = 247= 2470 0 agrees with reduction agrees with reduction

by the law of cosines, while reduction by by the law of cosines, while reduction by NASR Tables yields ZNASR Tables yields ZN N = 249= 24900

2222

Arcturus NASRArcturus NASR

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Jupiter 26 July 2007Jupiter 26 July 2007

G

LHA=3570 06.1’, tE =3600 – LHA = 020 53.9’

Eq

L = 340 24.4’ N

Dec 210 25.5’ S

Co-L

Co-H

Co-F

900 = B + Co-F -Dec

For NASR use assumed L = 340

assumed LHA = 3570

2424

Jupiter: Napier’s Circle, Upper TriangleJupiter: Napier’s Circle, Upper Triangle

A

BCo-tE

Co-Z1

Co-Co-L

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 BSin 340 = cos 2.48676*cos BB = 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

2525

Jupiter NASRJupiter NASR

2626

Lower TriangleLower Triangle

Co-FZ2

Co-H

A

Co-F

Co-

Co-Co-H

Co-Z2

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’ Asin Co-F =cos Co-Z2 * cos H

=sin Z2 *cos HSin 55.46667 = sin Z2 *cos34.49621

Z2 = 88.30

ZN = Z = Z1 + Z2 =88.3+88.3=176.6

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BearingBearingOnce again ZOnce again Z2 2 for the lower triangle is for the lower triangle is

different for Napier’s method versus different for Napier’s method versus Tabular NASRTabular NASRArcturus ZArcturus Z2 2 = 39.8= 39.80 0 (Napier) Vs. 38.6(Napier) Vs. 38.60 0 (NASR)(NASR)

Jupiter ZJupiter Z2 2 = 88.3= 88.30 0 (Napier) Vs. 88.6(Napier) Vs. 88.60 0 (NASR)(NASR)

2828

Venus ObliqueVenus Oblique

G

LHA = 920 92.1’

Eq

L

Dec

a = Co-L

C =LHA

Ac = Co-H B

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

Co-H

Co-Lb = Co-Dec

Assumed L = 340

Assumed LHA = 930

Dec = 60 50.5’ NCo-Dec = 830 09’

Co-Dec

Z

2929

Venus NASRVenus NASR

3030

Venus ObliqueVenus Oblique

Sin C/sin c = sin B/sin bSin C/sin c = sin B/sin bSin LHA/sin Co-H = sin Z/sin Co-DecSin LHA/sin Co-H = sin Z/sin Co-DecSin 93Sin 930 0 /sin 88.64675/sin 88.6467500 = sin Z/sin 83.15 = sin Z/sin 83.1500

Z = 82.6Z = 82.600

ZZN N = 360= 3600 0 – Z = 277– Z = 27700

3131

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

3333

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