hp-19c & 29c solutions navigation 1977 b&w

7/29/2019 HP-19C & 29C Solutions Navigation 1977 B&W

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INTRODUCTION

This HP-19C/HP-29C Solutions book was written to help you get the most from your calculator.The programs were chosen to provide useful calculations for many of the common problemsencountered.They will provide you with immediate capabilities in your everyday calculations and you willfind them useful as guides to programming techniques for writing your own customized software.The comments on each program listing describe the approach used to reach the solution andhelp you follow the programmer's logic as you become an expert on your HP calculator.You will find general information on how to key in and run programs under "A Word about ProgramUsage" in the Applications book you received with your calculator.We hope that this Solutions book will be a valuable tool in your work and would appreciate yourcomments about it.

The program material contained herein is supplied without representation or warranty of any kind.Hewlett-Packard Company therefore assumes no responsibility and shall have no liability,consequential or otherwise, of any kind arising from the use of this program material or any partthereof.

HEWLETTPACKARD COMPANY 1977


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TABLE OF CONTENTS

DISTANCE TO OR BEYOND HORIZON , , , , , , , , , , , , , , , , 1This program computes the distance to an object of known heightwhose base is obscured by the horizon. It also computes distanceto horizon for a given height of eye and the distance of visibilityof an object of known height.

DISTANCE BY HORIZON ANGLE AND DISTANCE SHORT OF HORIZON , , , 5This program calculates the distance between an observer and anobject given (1) the vertical angle between its waterline and thehorizon observed from a known height of eye or (2) the object'sheight and its subtended angle.DISTANCE OFF AN OBJECT BY TWO BEARINGS I I This program computes the distance off an object given two bearingsand the distance run between the bearings.

9

VELOCITY TRIANGLE , , , , , , , , , , , , , , , , , , , , , , 12This program provides interchangeable solutions for vector additionproblems such as heading, current, and course problems.COURSE TO STEER , , , , , , , , , , , , , , , , , , , , , , , 15This program computes course to steer given initial position, finalposition, speed, current, and magnetic deviation and variation.ESTIMATED TIME OF ARRIVAL , , , , , , , , I I I I I , , 18This program computes ETA's given distance, speed, departure dateand time, and time zones. It also computes speed required to make

an ETA given distance, date and time of arrival and departure, andtime zones.GREAT CIRCLE NAVIGATION*, , , , , 21This program calculates the great circle course and distance be-tween two points.GREAT CIRCLE COMPUTATION , , , , ,24This program computes the latitude corresponding to a given longi-tude on a great circle passing through two given points.COMPOSITE SAILING , , , ,27

This program computes, for two points, the longitude at which agreat circle is tangent to a given limiting parallel.RHUMB LINE NAVIGATION* , , , , , , , , , ,30This program calculates the rhumb line course and the distance fromorigin to the destination.


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RHUMB LINE DEAD RECKONING This program calculates DR position given course, speed, and elapsedtime from the last fix or DR position. 34

CELESTIAL NAVIGATION AND DEAD RECKONING 38This program keeps track of the positions of a ship and also a celes-tial object. Given time and sextant altitude, an updated DR andaltitude intercept and azimuth are output.SIGHT REDUCTION T A B L E ~ 42This program is a replacement for various published sight reductiontables (e.g.: H.O. 214, H.O. 229, etc.).

* THIS PROGRAM ALSO APPEARS IN THE HP-19C/29C APPLICATIONS BOOK.IT HAS BEEN INCLUDED H E R E ~ IN SLIGHTLY MODIFIED F O R M ~ FOR THESAKE OF COMPLETENESS.


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DISTANCE TO OR BEYOND HORIZON

This program computes the distance toan object of known height whose baseis obscured by the horizon and whosetop subtends a sextant altitude hs withthe horizon. The sextant altitude iscorrected for index error and height ofeye. Additional features are the calculation of the distance to the horizonfor a given height of eye and the distance of visibility of an object ofheight H above sea level.EQUATIONS:

~ tan ha ,\ H-HE tan haD=V\2.46 x 10-;+ 0.14736 - 2.46 x 10- 4Dhor = 1.144 I HEDvis = 1. 144{ /"HE + y'ff)

whereD = distance to object, nautical milesDhor= distance to horizon, nautical milesDVis= distance of visibility, naut. milesH = height of object beyond horizon, feetHE= height of eye, feetha= hs + IC - 0.97 I ~hs= sextant altitude, D.MSIC= index correction, M.m


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2EXAMPLE 1:The height of eye of of an observer is9 feet above sea level, how far away ishis horizon?EXAMPLE 2:An observer "bobs" Farallon Light onthe horizon and finds his height ofeye to be 16 feet. The light is 358feet above sea level. How far is theobserver from the light? (Accuracy isaffected by abnormal refraction)EXAMPLE 3:The top of a lighthouse, whose base isobscured by the horizon, is known to be300 feet above sea level. It is foundto have a sextant altitude of 2 5 ~ 6 abovethe horizon. The height of eye is 20feet and the sextant requires an indexcorrection of + 1 ~ 3 .What is the distance to the lighthouse?What is the distance to the horizon?It has been determined that the luminousrange of the light is "strong", nowcompute its visibility for the givenheigttt of eye.

SOLUTIONS:(1)

(2)

(3)

9.08 ST02(;S823.43 ***

16.88 ST02358.88 ST03r;S82R/S26.22 ***1.38 ST0128.88 ST02388.88 ST038.2536 (;S816.28 u*(;S825.12 ***/S24.93 u*

n.m.

n.m.

n.m.n.m.n.m.


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3User InstructionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

l . Key in the program DD2. Store pertinent data: DDIndex correction (minutes) IC ~ o = J

Height of eye (feet) HE ~ I T JHeight of object (feet) H LriQJ I 3 I

3a. Enter sextant height (D.MS) and compute hs ~ o = J p. (naut. mi.)distance to objectand/nr I II I

3b. ComputE distance to horizon and distance I SB II 2 I Phor (naut mi.of visibility ~ D pvis(nautmi.I II I

DDDDDDI II IDDDI IDDDI IDDI II ID [ ~ t-------L = ~ [ ~ t-----

[ = = J [ = _ ~ t- --------[ = ~ C J 1---------------- C - J [ = ~ J ---------[--] [----]

----- ---"- ----------_ .r---l,- ---1


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4 ProgralD Listings81 'LBLl hs(D.MS) 48 x 1.14482 hso 49 L5TX83 RCLl 58 RCL284 RCL2 51 IX85 IX 52 x86 53 R/5 *** D87 9 54 + hor.8 7 55 R/S ** D .V1S89 x18 -11 612 813 14 + hao15 TffH16 217 8 419 628 EEX21 CHS22 423 24 510525 RCLJ26 RCL227 -28 9 738 431 7 ** "Printx" may be inserted before "R 5".32 3 *** "Printx" may be used to replaceJJ 6 "RIS".J4 J5 IX36 +P I x2 + y2J7 RCL5J8 -39 R/S ** 048 ,LBL241 RCLJ42 IX43 144 45 146 447 4

REGISTERS0 1 Ie 2 HE 3 H 4 5 Used6 7 8 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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5

DISTANCE BY HORIZON ANGLE ANDDISTANCE SHORT OF HORIZON

This program calculates the distancebetween an observer and an object when(1) the vertical angle between itswaterline and the horizon has been ob-served from a known height of eye or(2) the object's height is known, together with its subtended angle.This program also calculates the heightof an object if its subtended angle anddistance from the observer are known.EQUATIONS:

o = HEtan(hs + Ie + .97 ~o = Htan (hs + Ie)

whereo = distance to object, feetHE= height of eye, feetIe= index correction, M.mH = height of object, feeths= sextant altitude, D.MS

NOTE:hs < 10' may make 0 unreliable due toatmospheric conditions when verticalsextant altitude between object andhorizon is taken.EXAMPLE 1:The sextant altitude between the water-line of a buoy and the horizon is foundto be 21:4. The observer has a heightof eye of 22 feet and the sextant requires a +1:7 index correction. Howfar is the observer from the buoy?EXAMPLE 2:The sextant altitude 'subtended by thebase and the top of a 41 foot lighttower is 56!2. The sextant requires a-1:9 index correction. How fa r is theobserver from the light tower?EXAMPLE 3:A vessel is anchored 2015 feet from anobserver. The sextant altitude betweenthe vessel's waterline and truck of mastis 1015!2. There is no index error. Howhigh is the truck of the mast above thewaterline?


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6SOLUTIONS:

(1)1.78 STOt

22.88 ST028.2124 GSB12735.25 *** f t .R/S8.45

***n.m.

(2)-1. 98 STD141.88 EHTt8.5612 GSB2

2595.58***

f t .R/S8.43***

n.m.

(3) 8.88 ST012815.88 EHTt1.1512 GSB344.88 U* f t .


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7User InstructionsINPUT KEYS OUTPUTSTEP INSTRUCTIONS DATA/UNITS DATA/UNITS

l . Key in the program DD2. Store pertinent data: DDIndex correction (minutes) Ie GiQ]w

Height of eye (feet) HE Gru]u=J -3a. Enter sextant height (O.MS) and convert to hs I (iSB II ] I n(fppt\

distance DD3b. (Optional) convert to nautical miles ~ L J n(n


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8 ProgralD Listings81 tLBU hs82 -+If83 RtU84 RCL29S IX16 .87 988 789 )t:18 +11 '5B812 RtL213 x:y14 . D15 'T0916 tLBL2 hs H17 -+If18 RtU19 bSBt2f21 tLBL9 D22 R/S ** D(ft.)23 624 BZ5 726 627 ** D(n.m.)28 R/529 tLBL3 hs D(ft.)38 ~31 Rt1.l32 GSBB33 x34 R/S ** H ( f t . )35 tLBLI36 637 838 .39 +48 TAH41 RTH

** "Printx" may bE inserted before " R / ~ "

REGISTERS0 1 Ie 2 HE 3 4 56 7 8 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 \ 2324 25 26 27 28 29


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9

DISTANCE OFF AN OBJECT BY TWO BEARINGS

To determine the distance off an objectas a vessel passes i t , observe twobearings on the bow and note the dis-tance run between bearings. The program calculates the distance off theobject when i t is abeam and at the timeof the first and second bearings.

EQUATIONS:

o - 10 .abeam - 2 slno01 = I abeam Isin RBIwhereRBI = First relative bearingRB2 = Second relative bearingDrun = St = Distance runS = speed of vesselt = time in minutes01 ,0 2 = Distance at time of firstor second bearingDa= Distance when abeam

EXAMPLE 1:A lighthouse bears -026 (26 counterclockwise) at 1130 and -051 at 1140.Our speed is 15 knots. How far will webe of f the light when i t is abeam? Howfar off were we at 1130 and 1140?EXAMPLE 2:A buoy is sighted bearing 015 on thebow, after a 3 mile run i t bears 105.What was its distance when abeam?SOLUTIONS:

( 1 )

(2)

-26.88 EHTt-51. 88 SS8115.88 EHTt8.18 ~ S B 34.68

***J.2.59***RJ2.82***

15.88 EHTt1 ~ S . 8 8 ~ S B l3.88 GS82

RJ.RJ

8. ***

01O20abeam

oabeam


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10 User InstructionsSTEP INSTRUCTIONS INPUT KEYSDATA/UNITS

l . Key in the program DD2. Store bearings (D.d): DD

at tl RBI IE N T t l Dlat t" RB., ~ I 1 II FntF.r ciic:tanc-p run (naut mi .) RY'un I GSB II 2 Iand compute distances DD

OR I ILJ3b. Enter speed (knots) and time (H.MS) and S I ENTt II I

compute distances t=t2-tl ~ [ I J4. Display remaining distances: I -l- II I~ DDDDDI II IDDDDf- - - DDDDDDI II IDD

I II IDDL ~ DDDL ~ I II II I

OUTPUTDATA/UNITS

Di (naut.mi.)

Dl (naut.mi.)D2(naut.mi .)Da(naut.mi .)


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12

VELOCITY TRIANGLE

This program is an interchangeablesolution for the vector addition prob-lem. Given any two of the vectors shown,the program computes the third.SOLUTION:

5.813 ST0128.58 5T0268.88 EHTt2.88 (;5B198.88 ENTt3.88 (;5B3

(;SBS1.82 *** knots

Compass course is corrected on inputfor magnetic variation and deviation.True course is decorrected on outputto yield compass course. Remember toupdate the values used for variation(changes with location) and deviation(changes with heading).EXAMPLE:A vessel is making 2 knots throughthe water, steering 060 by the compass.The magnetic variation is 20.5E andthe deviation is 5E. Calculate theset and drift of the current if thevessel is making good 3 knots on acourse of 090 0 T.


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13User InstructionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

l . Key in the program DD2. Store compass corrections: DDDeviation (negative if west) +Devo GIQ][L]

Vari ati on (neaati ve if west) +Varo GNJ[LJ3 Enter anv two of the three vectors' L ~ I I3a Headina - DD

rnmni'l


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14

81 fLBLlB2 STD583 x:vB4 STD385 RCLl56 +87 RCL288 +89 SSB918 ST0411 R/S12 fLBL213 STa914 X:Y15 STa816 R/S17 ftBl318 STa719 x:y28 510621 R/S22 fLBL423 S58724 RCLe25 RCL926 CHS27 SSBe28 510429 RCLl38 -31 RCL232 -33 SS8934 510335 x:y36 S10537 R/538 fLBLS39 GS8748 RCL441 RCLS42 CH543 SSBt44 STa845 X:Y46 S10947 R/S48 fLBL649 RCL4

0 16 CMG 7.2 .318 1924 25

ProgralD ListingsEnter heading 58 RCL5

51 XI,YI52 S1.853 x:v54 RCL855 RCL956 'SB8 CMG57 S106Nonnalize angle 58 x:y59 5107 SMG6B R/SEnter current 61 fLBL762 RCL663 RCL764 XI,YI65 S1.8Enter course 66 X:Y67 R1H68 fL8U X2,y2,YI6978 S+.8Compute heading 71 R.72 +73 RC.8 XI+X2,YI+Y274 +P75 x:y76 GSB9 Nonnalize angleCt 77 R1H78 fLBU79 388 681 "onnalize angle 82

Cc 83 +P84 x:r L,360Speed 85 X(8?86 STa8Compute current 87 X:Y -88 R. 19 R1H98 fLBL8

91 + 360 + 192 RTHSetDriftCompute course

REGISTERSDev 2 Var 3 Cc 4 Ct 5 SpeedSMG 8 Set 9 Drift .0 Used .1.4 .5 16 17

20 21 22 2326 27 28 29


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COURSE TO STEERThis program calculates a course tosteer given your location, the locationwhere you want to go, your boat's speedthrough the water, and the set and driftof the current.

DESTINATION

SOURCE

EXAMPLE:A vessel making 6 knots through thewater is at (45N. 12440'W) and shewishes to steer a course toward (4440'N,12410'W). The magnetic variation is

2 0 ~ 5 E and there is a 2 knot currentsetting 090. What course should shesteer.SOLUTION:

e.88 ST0128.S8 ST0298.88 ST082.88 ST094S.e8 ENIt124.48 EHTt44.48 EHH

124.18 ~ S B l29.213 *** Dist., n.m.6.88 ~ S B 2

1 2 ~ . 9 j u* Course to steer, degreesrus7.38 tUp. . S Speed made good,knots4.88 u* Transit time, H.MS

15


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16 User InstructionsSTEP INSTRUCTIONS INPUT KEYSDATA/UNITSl . Key in the program D I I2. Store compass corrections: DD

Deviation (negative if west) +Devo WQ J [ L JVariation (negative if west) ~ V a r o WQJI 2 I

3. Store current vector: I II ISet SetO WQ J u = JDrift Drift/kno siSTO II 9 I

4. Enter positions and compute distance between I II ISource latitude Ll(D.MS) IENT t lDSource lonqitude \, (O,MS) I EtUt II IDestination latitude L?(O,MS) ~ DDestination lonqitude \.,(O,MS) ~ W

5. Enter speed through the water and comoute S knots ~ WI ~ o m n a c ; c ; eourC;f> to c;tf>f>r I II I

6. Compute speed made good ~ D7. Compute time to reach destination ~ I IDD

D I IDDI II IDDI 1/ IDDDDC ~ DL ~ I II II I

OUTPUTDATA/UNITS

nic:t (nm 'Ceo

SMG(knots)t(H.MS)


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ProgralD Listings 17A2L2AILI 58 ST0781 *LBU 51 GSB7 SMG,CMB82 L2 52 CHS83 x:r 53 'SB884 54 RCUes ST.1 55 -86 RJ. Al 56 RCL287 x:r 57 -88 ~ 58 '589 Cc89 - 59 /?/518 CHS All - A21 68 RC.! Dist

11 ST.2 61 RCL712 RJ 62 R/S SMG (knots)13 Lt 63 .14 S-.l U + 64 ~ H " S Time (H.MS)IS + 65 R/S Normalize:16 " 66 tLBL917 Ang., latitude 67 3 o < angle < 36018 COS 68 619 RC.2 69 828 x 7821 RC.1 .; X2 + y2 7122 72 x:y23 6 73 X{8?24 8 Convert to n.m. 74 GT0825 x 75 x:r26 ST. 1 76 RJ.27 x:y 77 RTN28 GSB9 78 tLBL829 STD6 CMG 79 +38 x:y Dist. (n.m.) Be RTH Vector add31 R/S Enter speed and 81 tLBLS X2,Y232 tLBL2 82 -+RT7 ST05 Compute Cc 83 5+.8'v34 RCL6 84 RJ35 RCL8 8S + YI + Y236 RCLE 86 RC.8 Xl + X237 - 87 +P38 SIH 88 x:y r39 RCL9 89 GSS9 e48 x ,,-98 RTH e41 RCL5 91 tLBL7 Vector add42 92 -+R XI,Yl43 SIH-I 93 ST.844 - 94 x:y45 GSB9 Ct 9S RCL846 RCL5 96 RCt!!47 GSB7 97 RTH48 GSSe49 x:y SMG

REGISTERS0 1 Dev 2 Var 3 4 5 Soeed6 CMG 7 SMG 8 Set 9 Drift .0 Used .1 Used.2 Used .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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18

ESTIMATED TIME OF ARRIVAL

This program computes the time of ar-rival at the next port in local zonetime and GMT, when distance, speed,departure date and time in local zonetime are input.It also computes speed required tomake a given ETA, when distance, dateand time of departure in local zonetime and date and time of desiredarrival in local zone time are input.All computations can be made in GMT bystoring zeros in registers 4 and 5 andentering GM T times.EXAMPLES:1. A vessel departs San Francisco at0030 on the 2nd of January localtime, bound for Guam,distance 5,146miles. What will be the date andtime of arrival Guam local time,and GM T at 15.5 knots.2. If the same Vessel makes the samedeparture time and wishes to arrivein Guam at 0700 on the 16th ofJanuary local Guam time, what speedis required.NOTE:Zone time of San Francisco is +8 andGuam is -10.REFERENCE:This program is adapted from HP-65Users' Library program #02185A byCapt. Kenneth R. Orcutt.

SOLUTIONS:

(1) '5146.88 STa1

(2 )

1'5.5& ST028.&8 ST04-1&.8& ST052.88:3& t;SBl16.143& u* Local time

R/S16. 8438 *** GMT

2.883& ENTt16.8788 t;SB215.8582 *** Knots


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User Instructions 19STEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

1. Key in the program DD2. Store pertinent data: DDDistance (nautical miles) Dist ~ C l J

Speed (knots) Speed ~ I 2 IDeparture time zone (negative if east) DTZ [iiQJ I 4 IArrival time zone (negative if east) ATZ ~ Q

3. To compute arrival time and date: I IL ] DD.HHMM3a. Enter departure time and date in DD.HHMM DD.HHMM I GSB II 1 I (1 oca 1 t i m ~format and run. e.g., 2:30 p.m. Jan 3 DDwould be written 3.1430 [ ~ I I

3b. For GM T time and date of arrival ~ D D D . H H M M ( G ~Note: If arrival is in the next month, DD

subtract the number of days in the DDmonth. L _ ~ I I

4. To compute soeed reauired to make a aiven DDETA: D [ ~

4a Enter deoarture time and date DD HHMM G ~ L J ~ - - - - - -4b Enter arrival time and date and comoute DD JfHMt1 [ ~ L ~ ] ~ e d J m o t s .

soeed [---] [----]- - - - --- - - . - - - - - - -- -NOTE- If arrival is in the next month - - [----] [----1- - - - ----- ----- -

enter date olus the number of davs - --- [-----1 r- ----]-- - -_. - -- --- -in the month r - - - - - - - - ----

[ ~ - ~ - J 1- -- -l----------- i-------- -- - [- - ]1- ---- - - _ J -

- - ------ ._------ 1- - - - - - - --[---- --- J [ 1 --

- - - - - - - - - - - - - - - - -- - --- -- - -- -- - - - - - [ --_II 1- - - - - - f - - - - - - - - - - - - - - - - - - - - - - - - - - - - ----- - - - - - L I [ II I 1


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21

GREAT CIRCLE NAVIGATION

This program calculates the great circledistance between two points and theinitial course from the first point.Coordinates are input in degrees-minutesseconds format. The distance is displayed in nautical miles and the initialcourse in decimal degrees.

NOlSTANCE

(I.;"A,)

EQUATIONS:o = 60 COS-I [sin LI sin L2 + cos LI cos ~ cos (A2 - AI)]

C = COS-I [Sin L2 - sin LI cos (0/60)]sin (0/60) cos LI

{ C; sin(A2 - AI) < 0C1 = 360 - C; sin (A2 - Ad ;;;. 0

where:

REMARKS:

LJ, AI = coordinates of initial pointL2 A2 = coordinates of final pointo = distance from initial to final pointC j = initial course from initial to final point

- Southern latitudes and eastern longitudes must be entered as negativenumbers.- Truncation and round off errors occurwhen the source and destination arevery close together (1 mile or less).

- Do not use coordinates located atdiametrically opposite sides of theearth.- Do not use latitudes of +90 or -90.- Do not try to compute initial headingalong a line of longitude (L 1=L 2 ).

This program assumes the calculator isse t in DEG mode.EXAMPLE 1:Find the distance and initial course forthe great circle from Tokyo (L3540'N,A13945'E) to San Francisco (L3749'N,A12225'W).EXAMPLE 2:What is the distance and initial greatcircle course from L3353'30"S,A1823'lO"E to L40027'lO"N, A7349'40"W?SOLUTIONS:(1)

(2)

:'5.4e P1'!"-f-1:9.45 ENTof-:7.49 ENiT

122.25 GSBl446e.e4 H t (D,n.m)

p .s54.:? H.t. (Ci,dec.deg.)

-::.5:5:5& EHTt-18.2318 ENTt4e.2?18 EHTt7:3.4948 (;581676?e9 ,:h: (D,n.m.)p .-s:84.48 .U t (C dec deg )i ' . .


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/'

22 User InstructionsSTEP INSTRUCTIONS INPUT KEYSDATA/UNITS

l . Key in the program. 0 02. Kev in latitude and lonoitude of orio'ln L1(O.MS) IENTt 10A1(O.MS) IENTt 103. Kev in latitude and lonoitude of L2(O.MS) IENTt 10

destination A2(O.MS) I II I4 Calculate distance and initial r.ourC:E> ~ [ DIRjS II I

I II I0 0I II I0 00 00 0I II I0 00 00 00 00 00 1 I0 0I II I0 00 0L J O 0 1 II II I

OUTPUTDATA/UNITS

LlA1L2A2

O(n.m. )Ci(dec.deg)


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PrograDl Listings 2381 tLPL1 4B SIN82 4983 STD3 58 CDS-I C84 PJ. 51 RCLB85 52 sm86 STOt 53 X(8?87' RJ 54 t;T0988 55 N89 STD4 56 318 P-l 57 611 58 812 STD2 59 x:y13 RCL2 68 - ** Ci4 SIN 61 PTN15 PCL! 62 tLPL916 SIN 6J 1U17' 64 RTN ** C.\" 118 PCL219 cas28 PCL121 COS22 x23 RCL:!24 PCL425 -26 STa827' C[lS2B \:029 +38 STD5:?1 C[lS-I32 STOrS3J 634 e35 ** 036 p.'c37 PCL!38 SIN ** "Printx" may be inserted39 PCL2 before "RIS" and "RTN".48 SIN41 PCL542 x4J -44 RCL245 CDS4471 PCL6

REGISTERS0 1..2 - Al 1 L2 2 L, 3 A? 4 Al 5 COS 0/606 7 8 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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24

GREAT CIRCLE COMPUTATION

This program computes the latitudecorresponding to a specified longitudeon a great circle passing through twogiven points.

EQUATIONS:

INTERMEDIATE POINTS(Li,A i)

GlEAT CIRCLEROUTE

coordinates of initial point= coordinates of final pointL2,A2)

(Li,Ai) =NOTES:

coordinates of intermediatepoint

The program does not compute alonglines of longitude (AI = 1.2).

EXAMPLE:A ship is proceeding from Manila toLos Angeles. The captain wishes touse great-circle sailing from L1245!2N,A12420!lE, off the entrance to SanBernardino Strait, to L3348!8N,A120 0 07!lW, five miles south of SantaRosa Island. Find the latitudescorresponding to 1) A = 160 0 34'W; and2)' A = 180.SOLUTION:

12.4512 EHTt-124.2886 ENTt

'J'J. 4848 ENTt128.8786 r;SBl168.34BB (;SB241.2188 U t ON188.8888 r;SB239.4133 Ut ON


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STEP

1.2.

3.

4.

-c--------

25User InstruetionsINSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

Key in the program DDEnter positions: DDInitial latitude (negative for south) L (D.MS) I NTt IDInitial longitude (negative for east) Al (D.MS) I NTt II IFinal 1atitude (negative for south) L2(D.MS) LririJ I IFinal longitude (negative for east) A2(D.MS) ~ ~Enter intermediate longitude and compute Ai(D.MS) I SB I ~ J Licorresponding latitude I II IRepeat step 3 as desired. DDI II I

DDDDDD

L ~ I IDDDDDDL ~ C J - - ~ - - - - - -D [_ J - - ~ - - ~ - - -

L = ~ [===J ---------- [--] [- ---- - - - ~ - --_._------------ - ~ ~ -

[ ~ - ~ - l r - I- - ~ - ~ - - - - ,- II I----------- - - - - - - - - - - - - - - - - - - - I - - ~ - - - ~ ~ - - - L II I---- ---- - - - - - -- - - - -- -- - - - - -

, II Ir - - - - - - - - ~ - - - - - - - - - -- - - - - -- - - - - ---- - - I II II II I


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26 ProgralD Listings81 *LBLl 1..2 L2 Al Ll82 ,*H83 5m384 Rl85 +H86 510187 Rl88 '*H89 510418 RJ.11 ,*H12 5102 ** A.13 R--S 114 *LBL215 +H16 510817 RCL418 -19 SIH28 RCLl21 TAH22 x23 ReL8Z4 RCL?25 -25 SIN27 RCL228 TAH29 x38 -?J ReL?32 RCL4n -34 SIH35 .36 TAH-I37 ,*H1f5 ** L. ** "Printx" my be inserted38 R/S 1 before "R/< II .

REGISTERS0 1 L2 2 Ll 3 1..2 4 Al 56 7 8 A; 9 .0 .1.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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27

COMPOSITE SAILING

When the great circle would carry avessel to a higher latitude than desired, a modification of great-circlesailing, called composite sailing, maybe used to good advantage. The com-posite track consists of a great circlefrom the point of departure and tangentto the limiting parallel, a course linealong the parallel, and a great circletangent to the limiting parallel andthrough the destination. This programcomputes, for each of two points, thelongitude at which a great circlethrough the point is tangent to somelimiting parallel.

EQUATIONS:, , -1(tan LlAV I = Al + cos tan Lmax

where(L l ,Al) = initial position(L2,A2) = final position(Lm ,AVl) = point at which limitingax parallel is met

(Lmax ,Av2) = point at which limitingparallel is leftSI = sgn (1.. 2 - AI)S2 = sgn (11..2 - Al I - 180)

sgn(x) = {+l-1EXAMPLE:

x ~ 0x < 0

A ship leaves Baltimore bound forBordeaux (Royan), France. The captaindesires to use composite sailing fromL 3 6 5 7 ~ 7 N , A 7 5 4 2 ~ 2 W one mile south ofChesapeake Light to L 4 5 3 9 ~ l N , A 1 0 2 9 ! 8 W ,near the entrance to Grande Passe del'Quest, limiting the maximum latitudeto 47N.Required:(1) The longitude at which the limit

ing parallel is reached.(2) The longitude at which the limiting parallel should be left.SOLUTION:

36.5742 ENTt75.4212 ENTt45.3986 ENit1. 2948 GS8147.8888 I;S8238. 168? *HR. S18.5653 H*

Av I (D.MS)Av2 (D.MS)


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28 User InstructionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

l . Key in the program DD2. Enter positions: DD

Initial lat itude (negative for south) Ll(O.MS) IENTt IDInitial longitude (negative for east) IAI (O.MS) IENTt IDFinal latitude (negative for south) 2(O.MS) IENTt II IFinal longitude (negative for east) A2(O.MS) ~ [ L J

3. Enter latitude of limiting parallel Lmax(O.MS) IGSB II 2 I AvdO.MS)and compute AVl (limiting parallel is met) I II I

4. Compute A .,(limitinCl parallel is left) [ B LL JD Av.,(O.MS)I II IDDDDDDI II IDDDDDDDDDDD I IDDI II IDDDDDDI II II II I


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PrograUl Listings 2981 fLBU 5882 +H 51 cos-r83 5T03 A2 L2 A1 L1 52 RCL6 sgn84 RJ. 53 )(85 -+H 54 RCL!86 ST01 55 GTOe87 RJ.8e -+H89 5T0418 RJ.11 -+H12 5T0213 R/514 tLBL2 Enter Lmax1516 5T0517 RCL218 TAN19 X:V2821 COS"22 RCB23 RCL424 - A2 - A125 fHTt26 fIBS IA2 - A11 A2 - A127 128 829 838 -31 x32 fHTt33 f I B ~34 . + 1-35 ST0636 CHS

~ ' l ' )(38 RCL439 fLSLS48 +41 142 -+R Nonnalize43 -+P Angle44 X:Y45 -+HffS ** AV1,AV26 RlS

** "PRINTX" may be inserted before "RI I47 RCLl4849 RCL5REGISTERS

0 1 L2 2 L1 3 A2 4 A1 5 tanLmax6 7 8 9 .0 .1son.2 .3 .4 .5 16 1718 19 20 21 22 2324 25 26 27 28 29


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30

RHUMB LINE NAVIGATION

This program is designed to assist inthe activity of course planning. Yousupply the latitude and longitude ofthe point or origin and the destination.The program calculates the rhumb linecourse and the distance from origin tothe destination.Since the rhumb line is the constantcourse path between points on the globe,i t forms the basis of short distancenavigation. In low and midlatitudesthe rhumb line is sufficient for vir-tually all course and distance calculations which navigators encounter.However, as distance increases or athigh latitudes the rhumb line ceasesto be an efficient track since i t is notthe shortest distance between points.The shortest distance between pointson a sphere is the great circle. How-ever, in order to steam great circles,an infinite number of course changes arenecessary. Since i t is impossible tocalculate an infinite number of coursesat an infinite number of points, severalrhumb lines may be used to approximatea great circle. The more rhumb linesused the closer to the great circle dis-tance the sum of the rhumb line distanceswill be. The Great Circle Computationprogram may be used to calculate inter-mediate course change points which canbe linked by rhumb lines.Latitudes and longitudes are input indegrees-minutes-seconds. Course isdisplayed in decimal degrees. Southernlatitudes and eastern longitudes areinput as negative numbers.

~ Intermedlalepoints (Lt. AI)(L2' A2) ~ found b y Rhumb Line sequence

Great Circle --.. t L3. A3)/"--- RhumbUnes----"'"

EQUATIONS:I 7T (>\1 - A2)C = tan - - , - - : : - : : : - : : - - ~ - = - - - - : - : - - ; - - : - - ; - - - - - - : : - : - : : - - : - - : - - ; - : - - : - : -180 (In tan (45 + 'h Lz) - In tan (45 + 'h Ld)

{60 (A2 - AI) cos L; cos C = 0

D= (L. - L I ) .60 - ; otherwisecos Cwhere:

(L I AI) = position of initial point(Lt. Az) = position of final pointD = rhumb line distanceC = rhumb line course

REMARKS:No course should pass through eitherthe south or north pole.Errors in distance calculations maybe encountered as cos C approacheszero.Accuracy deteriorates for very short1egs.This program assumes the calculatoris set in DEG mode.


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EXAMPLE 1:What is the distance and course fromL3524'12IN,A12502'36"W to L41009'12"N,A14722'36"E?EXAMPLE 2:What course should be sailed to travela rhumb line from L213'42"S,A17907'54"E to L527'24"N,A17924'36"W? What is the distance?SOLUTIONS:

(1)

(2)

35.2412 ENTt125. 8236 ENTt41.8912 ENTt

-147.2236 G5814135.68 **Jj: (DIST.,n.m.)274.79 U1/:

+-2.1342 ENTt

-179.8754 ENTt5. 2724 ft.ITt

179.2436 1;5B1

(C,dec,deg.)

469.31 *"** (DIST.,n.m.)18.73 *** (C,dec,deg.)

31


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32 User InstructionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

1. Key in the program DD2. Key in latitude and longitude of origin L1(O.MS) IENTt IDAdo.MS) IENTt 1 03. Key in latitude and longitude of L2(O.MS) IENTt ID

destination A2(O.MS) I II I4. Calculate distance and course ~ D J O(n.m.)[RIDL] C(dec.deg.)

NOTE: I II ISouthern latitudes and eastern longitudes DDmust be input as negative numbers. I II I

DDDDDDI II IDDDI IDD1 IDDDI IDDDI II IDDDDDDI II II II I


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ProgralD Listings 3381 tLPLl 48 LN82 49 RTN83 ~ T O J A2 se tLBL884 IN S1 Je5 L2 S2 6 E to W 360 - C86 ST01 53' 887 RJ Al 5-i R C L ~88 55 fiBS89 ST04 S6 -113 R.J. S7 :+:LPL'11 S8 fiBS12 Sro2 Ll 59 STU613' RCL4 AI- A2 68 114 ReL? tI_ 815 - 62 816 STae 63' RCL817 ., Make -180 < AI- A2 64 PBS-18 < 180 6S X ~ Y ' ? is [AI-A2] > 18019 SIN 66 (;SB6 If so subtract from28 SIN-' 67 ReL! 36021 9 613 COS22 e 69 x2 j 78 STD724 Pi 71 RCL!25 x 7" RCL22f. RCL! 73 -27 (;S89 74 RCLS28 RCL2 7529 t;SB9 76 g ~ e " is C = 90?38 - 7771"'. 78 ENTt32 ~ 1 C 79 >-:=81)33' sros 88 RCL?3'4 ReLe 81 f

SIN 82 836 SIN-I 83 )i37 X(8? 84 fiBS ** Distance38 GTOS x


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30

34

RHUMB LINE DEAD RECKONING

121 W 30' 120" wnr 'I J I 'I 'I 'I 'I r', bR'lI-12011 FIX~ ~ ,m18 1330 DR L' 08'/D ~ \ \ \ \ \ \ \ 1 \ \ I \ I \ I I I ~ 1 1 I 1 1 / 1 1 /f////!jt

ifo' ~ \ \ ~ \ ,.. -+- .d... \ ~ \ \ \ ~ ~ FOR NORTH LATITUO '"~ ~ ~ ~ ~.,x-... , - $ ; . ~ C31i6 S20.. , ~ i f > COlO 30'1510 It A'''-'. S 17 1823 DR

ii'"I, ~ #. ' I~ ~ jThis program calculates a ship's DRposition given the ship's course, speed,and elapsed time from the last fix orDR position. The DR position is storedso that on subsequent legs just course,speed, and elapsed time need be enteredto obtain the updated DR position. Theprogram may be used for both small andlarge area DR problems.EQUATIONS:The updated position (L,A) is given byfollowing a loxodrome (rhumb line) fromthe initial position (Li,Ai) for a dis-tance determined by the speed and time.L = L,. + S cos C60

1 8 0 t a n C ( l n t a n ( 4 5 + ~ ) - l n t a n ( 4 5 ~ A.+-----------------------------

7T

A = C r 90 or 270 (LirL)A' - SsinC . C=90 or 270 (L,'=L)1 60cosLi '

where:Li =L =Ai =A =S =C =

=NOTES:

initial latitude (N, positive;S, negative)updated latitudeinitial longitude (W, positive;S, negative)updated longitudeship's speed, knotsship's course, degreesthe time (H.MS) between initialand final positions .

1. The program cannot follow ameridian over a pole.2. The program loses accuracy andgets incorrect answers whenwithin 0.5 of a pole.EXAMPLE (Fig. 1):A vessel's position is L3349!lN,A120052!OW at 1200. If she steams asshown,what is her position at eachtime?

TIME C S DR1200 L3349 106"N.A120052 ' OO"W1330 120 15 knots (L3337 ' 51"N.A120028 '34"W)1510 240 15 knots (L3325 ' 21"N.A120054 ' 32"W)1823 90 17 knots (L3325 ' 21 "N.A11949 ' 01 "W)1955 355 20 knots (L3355 ' 54"N.A11952 ' 14"W)


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35

SOLUTION:

n.4986 ENTt128.5288 1;58113. 3888 ENTt12.8888 I;SB3

128.8888 ENTt1 ~ . 8 8 e 8 J;SB23!-.3751 u*R/S 1330 DR128.2834 ***15.1888 ENT'/'13.3888 I;SB3

248.8888 ENTt15.8888 J;SB233.2521 u*

R/S 1510 DR128.5433 ***18.2388 ENT'/'15.1888 J;SB398.8888 ENT'/'17.8888 J;SB233.2521 u*R,"S 1823 DR119.4982 u*19.5588 ENTt18.2388 J;5B3

355.8888 ENT'/'28.8888 (;S8233.5554

***."S 1955 DR119.5214 u*


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36 User InstructionsINPUT KEYS OUTPUTSTEP INSTRUCTIONS DATA/UNITS DATA/UNITS

l . Kev in the program 002. Enter initial position: 00Latitude (negative for south) Li,O.MS I ENT t IO

Longitude (negative for east) Ai,O.MS [iJ[LJ3 Enter shiols way and compute DR: I II I

Time on course* tot,H.MS I ENT t lOCourse Co I ENT t lOSpeed S,knots I GSB II 2 I L.O.MS

GLSJO A.O.MSI II I00

* To find the difference between two times T1,H.MS IENT t lOTl may be less than or greater than T" T2 ,H.MS [iJ[LJ H.MSbut if the times are in different days, I II Iadd 24 hours to the smaller time. 000 1 I

000 1 I00I II I00I II I000000I II II II 1


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38

CELESTIAL NAVIGATION AND DEAD RECKONING

Thi s program allows you to update avessel's position and correct i t usingsights on a celestial object. Theprogram is started with your latitudeand longitude and the object's GHA anddeclination all determined for the sametime. Then when any other time is keyedin , the corresponding DR is computed.If a sight is taken at that time, theresulting altitude may be entered intothe calculator to yield an interceptand azimuth. The DR may be moved accordingly if desired.The dead reckoning technique used is midlatitude sailing which, while not asaccurate as rhumb line dead reckoning,is sufficiently good for most purposes.Altitude intercepts "toward" are considered to be positive, even though carefulreading of Bowditch would indicate theopposite. By using this convention, i tis easy to compute the intercept terminus (most probable position or MPP).The program contains a useful subroutine,GSB 7, which can be used for translatingalmanac entries in degrees, minutes andtenths (DM.M) to decimal degrees (D.d).REFERENCE:This program is based on private communications with Paul E. Shaad of Sacramento,California.EXAMPLE:On February 19, 1975, a ship is steamingon course 240 at 17 knots. At 1800 GMTher dead reckoning position is 42N,135W.Compute her position at 2115.Her navigator shoots the Sun from a heightof 65' (dip = 7!8). At 2340 he obtainsa sextant altitude of 2825'36". Computethe altitude intercept and azimuth andcorrect the ship's DR.

SOLUTIONFrom The Nautical Almanac we take theSun'slGHA and declination at 1800 and1900 GM T and also the Sun's semidiamter.

G.M.T. SUNG.H.A. Dec.d h , 0 ,

19 18 86 31.5 S11 18.519 101 31.5 17 .6S.D. 16.2

8631. 588P GSB7 GHA at 180086.5258 *HSTat

-1118.5e8e GSB7 DEC at 1800-11.3883 *uSTD216.2888 GSB7 Semi diameter8.2788 u*ST0342.8888 ST04135.8888 ST(l57.8888 GSB78.1388 **']f:ST06248.8888 STD717.8888 fNIt68.8888 -

Lat. \Long. PositionDip of the at 1800horizonCourseSpeed

8,2833 f** Speed converted toSTD8 degrees per hour

18. 8888 ST09 Ti me18131. 5888 GSB?18!.5258 **f:

8631.5888 GSB786.5258 f**15.8888 f**ST. 1-1117.61388 GSB?-lL29J3 *** ;-1118.5888 GSB7-11. 3883 u*

8.8158 u*5:.2

Calculation ofRate of Changeof GHA

Calculation ofRate of Changeof DEC


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Now that the setup is compelte, you can dead reckon and reduce sights all daylong.

21. 15ee ~ S B 141.3223 *H:g:y

1 3 5 . 8 4 ~ 9 tU

23.4888 ~ S B 141.1158 ***B6.513428.2536 ~ S B 2-4."3689 ***

\-... -," ' \ ~ r

New timeLatitude lNewLongitude)

Position at 2115

New timeLatitude fNew Position at 2340LongitudeSextant altitude

Altitude intercept219.4574 u* Azimuth

~ S B 7-8.8728 *tt

bSB341.1512 ***x:y136.4752 t. u

Intercept converted to degreesLatitude fIntercept terminus on Line of PositionLongitude

39


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40 User InstruetionsINPUT KEYS OUTPUTSTEP INSTRUCTIONS DATA/UNITS DATA/UNITS

l . Key in the program DD2. Store the following values DDGreenwich Hour Angle of object GHA,D.d GIQJ[LJ

~ n e g a t l V e i f east) DDDeclination of object (negative if south) DEC,D.d LriQJ I 2 ISemi-diameter of object (negative if U.L.) SD,D.d GIQJ[LJLatitude (negative if south) L,D.d I TO II 4 ILongitude (negative if east) A,D.d I TO II 5 IDip of the horizon Dip,D.d GIQ]wCourse C D.d I TQ II 7 ISpeed (in knots divided bv 60) S D.d GruJ[LJTir.Je t.H.h. GroJ[LJRate of change of GHA* Qha.D.dlhl GIQ][LJRatp of chanap of dpc* dec. D.dA'lr I TO II 2 I

"). Enter a new time and compute new DR t new H.MS GSaJ[LJ L .D.MSL X ~ y " I A ,D.MS4. Enter sextant altitude and compute inter- h D.MSs, DDcept and azimuth GSaJI 2 I f---a .m;~ D Zn.D.d

5. Update DR to MPP after GSB 2 in Step 4 I SB II 7 I f- a.D.d( i . e.: An in y; a in x) ~ [ L J L.D.MS~ I I A.D.MS* Enter GHA or DEC for some time Value,.D MS c ; ~ D Valuel,D.dEnter GHA or DEC for one hour earlier V ~ l u e 2 , D . M S ~ D Value2,D.d

OR DD Rate D.d/hrEnter GHA or DEC for some time Valuel,DM.M ~ ~ I 7 I Valuel,D.dEnter GHA or DEC for one hour earlier Value2,DM.M I SB II 7 I Value2,D.dIRate D.d/hr!


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81 tLBU132 otHB3 PCL9134 x : ~ 'l5 STC9Bf X:r'13?BS ST. 489 P':.1II? xnl. ST+!12 PC.413 PC.214 x10::: ST+2'-'16 PI" . . 't17 RCLS18 x19 RCL?28 v" vr.- ;21 *LBLJ22 otP2J S1'+424 22526 RCL427 V"\ '. :+- .2829 COfJI?31 ST-532 peL!'!3 PC!.53435 ST08~ , t , RCL5., 7 otH!'!S. I., 0.. \.' PCL4J9 o t H " ' ~4e PTN41 tLBL242 otHdJ peLf4445 fNH46 -:r{jN47.. ...,xdS 649 "'1

0 LHA6 DiD 7.2 .3dec.18 1924 25

PrograUl ListingsUpdateTime

UpdateGHAAndDEC

UpdateDR

New ANew LComputea and Zn

GHA 2C 8

Ho .42026

DECS

He

51?5152 Pf"'17, \. .......53 +54 ST.J55 P':U?56 PCL2S7 rllC"_ L ' ~ Sf' otP59 PCL468 C"T... . .561 g:y62 otT?-6J C"T ': Of64 RJ.65 ,PC. 5f- Rr:!..2f7 SIN6S otR69 C"T'-' ; .578 RJ71 PC.47?.. +7? SIN-I74 C"T 4'-' .75 RJit PC.5., 7i ;7S ot?79 P!81381 n082 e8J +84 Re.?85 R':.48687 600,..t.,. B

}{ge PTN91 t.LBL?92 ~ H r " S9J [[Xo. ..,. 29596 -tH97 p-:rJ{

REGISTERS3 SD9 t.5 Used2127

ConvertDM .mtoD d

4 L 5.0 .116 1722 2328 29

41

Aqha


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42

SIGHT REDUCTION TABLE

This program calculates the computedaltitude, He, and azimuth, Zn, of acelestial body given the observer'slatitude, L, and the local hour angle,LHA, and declination, d, of the body.It thus becomes a replacement for thenine volumes of HO 214. Moreover, theuser need not bother with the distinc-tions of same name and contrary name;the program itself resolves all ambiguities of this type.EQUATIONS:

EXAMPLE 1:Calculate the altitude and azimuth ofthe moon if its LHA is 239 154"W and itsdeclination 1351 ' 06"S. The assumedlatitude is 3320'N.EXAMPLE 2:Calculate the altitude and azimuth ofREGULUS if its LHA is 3639 118"W and itsdeclination is l212142"N. The assumedlatitude is 3330 ' N.EXAMPLE 3:- 1Hc=sin [sin d sin L + cos d cos

Z; sinLHA


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43

(3) 44.348e ENH26. 8888 ENTt158.8888 {;SBl-16.3725***

(d,D.MS)IUS339.4 H* (LHA,dec.deg.)

1")7l ... ,,': .i. i -fH+462.7*** (GHA,dec.deg.)283.4

-fHPlS2S9.2

***{SHA,D.MS)


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44 User InstruetionsSTEP INSTRUCTIONS INPUT KEYS OUTPUTDATA/UNITS DATA/UNITS

l . Key in the program 0 02. Input the following: 0 0Observer's latitude .L(D.MS) IENTt10

Declination d(D.MS) IENTt10Local hour angle LHA (D.W ) I II I

3. Calculate: 0 0Altitude IGSB II 1 I Hc(D.MS)Azimuth IR/S II I Zn(dec.deg.l)OR: 0 02. Input: I II IObserver's latitude L(D.MS) IENTt10Altitude Hc(D.MS) IENT1'10Azimuth Zn(D.MS) 0 0

3. Calculate: I II IDeclination [ r u [L ] d(D.MS)Local hour angle l l iUO LHA (dec.dl g

0 0I II I0 00 1 I0 0I II I0 00 00 0I II Ir II I


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ProgralD Listings 45e1 tLBU 48 PJ132 L 49 3133 5T02 S13 684 PJ St 885 52 X"V. ,86 STOt d 53 - ** Zn87 RI 54 RTN88 ~ f - I 55 tLBLB89 ST08 LHA 56 RJ.18 SIN ** ZnS7 RTN11 PCL! 58 R/S12 SIN17- x14 PCLt!15 COS16 PCL!17 COS18 x19 PCL228 COS21 x22 +23 ST0324 SIN-I Hc,dec.deg.25 ST0426 ~ H " S Hc,D.MS27 FIX4p/S *** ** "PRINTX" may be inserted before "RT28 II29 rrX1 *** "PRINTX" may be used to replace "RI38 PCL! II31 SIN32 RCL333 PCLf!34 SIN35 x36 -37 PCLe38 COS3948 PCL441 COS4243 cos-r44 PCL24S SIN Z46 X


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hp-19c & 29c solutions navigation 1977 b&w

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