Sights: Taking and Reducing
description
Transcript of Sights: Taking and Reducing
11
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
77
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
1616
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
1717
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
2323
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
2727
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
3232
http://titulosnauticos.net/astro/chapter10.pdfhttp://titulosnauticos.net/astro/chapter10.pdf
3333
Cos c = cos a*cos b + sin a* sin b*cos C