《航海学》课程参考文献（地文资料）CHAPTER 17 AZIMUTHS AND AMPLITUDES

CHAPTER17AZIMUTHS ANDAMPLITUDESINTRODUCTION1700.CompassChecksmeasured and calculated azimuthsand amplitudes of celestialbodies.ThedifferencebetweenthecalculatedvalueandAt sea, the mariner is constantlyconcerned aboutthe ac-the value determined by gyro measurement is gyro error.This chapterdiscusses these procedures.curacyof thegyro compass.Thereare several ways to checkTheoretically,theseprocedures work with any celestialthe accuracy of the gyro.He can, for example, compare itwith an accurate electronic navigator suchasan inertial nav-body.However, the sun and Polaris are used most oftenigatonsystem.Lacking a sophisticated electronic navigationwhen measuring azimuths, and the sun when measuringsuite, he can use the celestial techniques of comparing theamplitudes.AZIMUTHS1701.CompassErrorByAzimuthOfTheSunpending upon whether the actual LHA is greater orlessthanthebaseargument.RecordtheZDiff.forMariners usePub229,SightReductionTables forMa-the incrementofLHA5.Correct the base azimuth angle for each increment.rine Navigation to compute the sun's azimuth.Theycomparethe computed azimuthto the azimuthmeasuredExample:withthecompasstodeterminecompass error.IncomputingIn DRlatitude 33°24.0N,the azimuth of the sun isan azimuth, interpolatethetabular azimuth angle for thedifferencebetween the table arguments and the actual val-096.5°pgc.Atthe timeofthe observation,the declinationues of declination,latitude,and local hour angle.Do thisofthesumis20013.8N:thelocal hourangleofthesunistriple interpolation of theazimuth angleasfollows:316°41.2:Determinecompasserror.1.EntertheSight ReductionTableswiththe nearestSolution:integral values of declination,latitude,andlocalSee Figure 1701 Enter the actual value of declination,hourangle.Foreachof thesearguments,extractaDRlatitude,and LHA.Round each argument to the nearestbaseazimuthangle.wholedegree.Inthis case,roundthedeclinationandthelat2.Reenterthetableswith thesamelatitude and LHAitude down to the nearest whole degree. Round the LHA uparguments but with the declination argument 1oto the nearest whole degree.Enter the Sight Reduction Ta-greateror less than thebasedeclination argument.bles withthese wholedegree arguments andextract thebasedepending upon whether the actual declination isazimuth value for these rounded off arguments. Record thegreater or less than the base argument. Record thebase azimuth value in the table.difference between the respondent azimuth angleAsthefirststepinthetripleinterpolationprocess,inand the baseazimuthangleand label itas the azi-crease the value of declination byjoto 2jo because themuthangledifference(ZDiff)actual declinationvalue wasgreaterthanthebasedeclina-3.Reenter thetables withthe basedeclination andtion.Enter the Sight ReductionTables with the followingLHA arguments, but with the latitude argument 1oarguments:(l)Declination=21°;(2)DRLatitude-330greater or less than thebase latitude argument, de-(3)LHA =317°.Record the tabulated azimuthforthesepending upon whether the actual (usually DR)arguments.latitude isgreater or less than the base argument.Asthesecond stepinthetripleinterpolationprocess,Record theZDiff.for the incrementof latitudeincrease the value of latitude by to 34 because the actu-4.Reenterthetables with the basedeclinationand lat-alDRlatitudewasgreaterthanthebase latitude.Entertheitude arguments, but with the LHA argument 10SightReduction Tableswiththefollowing arguments:(l)Declination=20°;(2)DRLatitude=34°:(3)LHA=3170greater or less than the base LHA argument, de-283

283 CHAPTER 17 AZIMUTHS AND AMPLITUDES INTRODUCTION 1700. Compass Checks At sea, the mariner is constantly concerned about the ac￾curacy of the gyro compass. There are several ways to check the accuracy of the gyro. He can, for example, compare it with an accurate electronic navigator such as an inertial nav￾igaton system. Lacking a sophisticated electronic navigation suite, he can use the celestial techniques of comparing the measured and calculated azimuths and amplitudes of celes￾tial bodies. The difference between the calculated value and the value determined by gyro measurement is gyro error. This chapter discusses these procedures. Theoretically, these procedures work with any celestial body. However, the sun and Polaris are used most often when measuring azimuths, and the sun when measuring amplitudes. AZIMUTHS 1701. Compass Error By Azimuth Of The Sun Mariners use Pub 229, Sight Reduction Tables for Ma￾rine Navigation to compute the sun’s azimuth. They compare the computed azimuth to the azimuth measured with the compass to determine compass error. In computing an azimuth, interpolate the tabular azimuth angle for the difference between the table arguments and the actual val￾ues of declination, latitude, and local hour angle. Do this triple interpolation of the azimuth angle as follows: 1. Enter the Sight Reduction Tables with the nearest integral values of declination, latitude, and local hour angle. For each of these arguments, extract a base azimuth angle. 2. Reenter the tables with the same latitude and LHA arguments but with the declination argument 1° greater or less than the base declination argument, depending upon whether the actual declination is greater or less than the base argument. Record the difference between the respondent azimuth angle and the base azimuth angle and label it as the azi￾muth angle difference (Z Diff.). 3. Reenter the tables with the base declination and LHA arguments, but with the latitude argument 1° greater or less than the base latitude argument, de￾pending upon whether the actual (usually DR) latitude is greater or less than the base argument. Record the Z Diff. for the increment of latitude. 4. Reenter the tables with the base declination and lat￾itude arguments, but with the LHA argument 1° greater or less than the base LHA argument, de￾pending upon whether the actual LHA is greater or less than the base argument. Record the Z Diff. for the increment of LHA. 5. Correct the base azimuth angle for each increment. Example: In DR latitude 33° 24.0’N, the azimuth of the sun is 096.5° pgc. At the time of the observation, the declination of the sun is 20° 13.8’N; the local hour angle of the sun is 316° 41.2’. Determine compass error. Solution: See Figure 1701 Enter the actual value of declination, DR latitude, and LHA. Round each argument to the nearest whole degree. In this case, round the declination and the lat￾itude down to the nearest whole degree. Round the LHA up to the nearest whole degree. Enter the Sight Reduction Ta￾bles with these whole degree arguments and extract the base azimuth value for these rounded off arguments. Record the base azimuth value in the table. As the first step in the triple interpolation process, in￾crease the value of declination by 1° to 21° because the actual declination value was greater than the base declina￾tion. Enter the Sight Reduction Tables with the following arguments: (1) Declination = 21°; (2) DR Latitude = 33°; (3) LHA = 317°. Record the tabulated azimuth for these arguments. As the second step in the triple interpolation process, increase the value of latitude by 1° to 34° because the actu￾al DR latitude was greater than the base latitude. Enter the Sight Reduction Tables with the following arguments: (1) Declination = 20°; (2) DR Latitude = 34°; (3) LHA = 317°

284AZIMUTHSANDAMPLITUDESTab*BaseBaseCorrectionActualZZ Diff.ArgumentsZIncrements(Z Diff x Inc.+ 60)0.39Dec.20°20°97.8096.49-1.4°13.813.8° N33°33° (Same)97.8°98.9+1.1°24.0+0.49DR Lar24.0'N316°41.2317°97.8°97.1°0.7°18.80.2°LHABase Z97.8°Total Corr.0.1°Corr.0.1°(L)z*Respondentforthetwobase arguments andN97.7° E1°change from third base argument, in verticalZn097.7°order of Dec.,DR Lat., and LHA096.5°Zn pgc1.2°EGyro ErrorFigure 1701.Azimuth by Pub. No. 229.Record the tabulated azimuth for these arguments.Next, determine the increment for each argumentAs the third and final step inthetriple interpolationby taking the difference between the actual values ofprocess, decrease the value of LHA to 316because the ac-each argument and the base argument. Calculate thetual LHAvalue was smaller than the base LHA.Enter thecorrectionfor each of the three argument interpola-Sight Reduction Tables with the following arguments: (l)tions by multiplying the increment by theZ differenceDeclination = 20°; (2) DRLatitude =33°: (3) LHA =316°and dividing the resultingproductby 60.Record the tabulated azimuth for these arguments.The signofeachcorrection isthe sameas the signCalculate the Z Difference by subtracting the base az-ofthe corresponding Zdifference used to calculate it.Inimuth from thetabulated azimuth.Be careful to carrythethe above example, thetotal correction sums to-0.:Apply this value to thebase azimuthof97.8°to obtaincorrect sign.the true azimuth 97.70. Compare this to the compassZDifference=TabZ-Base Zreadingof096.5°pgc.Thecompass error is1.2°EAZIMUTHOFPOLARIS1702.Compass Error By Azimuth Of Polaris045°00.0WLongitudeLHA Aries161°25.4The Polaris tables in theNautical Almanac list the azi-muthof Polarisforlatitudes between theequator and65°N.Figure2011 inChapter 20 shows thistable.CompareaSolution:compassbearingof Polaristothetabularvalueof Polaris toEnter the azimuth section of the Polaris table with thedetermine compass error.The entering arguments for thecalculated LHA of Aries. In this case, go to the column fortable are LHA of Aries and observer latitude.LHA Aries between 1600 and 1690Follow that columndown and extract the valuefor the given latitude. Since theExample:incrementbetweentabulated values isso small, visual in-OnMarch17,1994,atL33°15.0'Nand045°00.0Wterpolation is sufficient.Inthis case,theazimuthforPolarisat 02-00-00 GMT, Polaris bears 358.6°T by compass. Cal-forthegivenLHAof Ariesandthegiven latitude is359.3°culate the compass error.Date17 March 1994359.3°TTabulatedAzimuth02-00-00358.6°TTime (GMT)Compass Bearing204°25.4Error0.7°EGHA AriesAMPLITUDES1703.Amplitudesobserver's horizon intersects the celestial equator. SeeFigure 1703.Calculate an amplitude after observing a body on eitherA celestial body's amplitudeis the arcbetween thethe celestial or visual horizon.Comparea body'smeasuredobservedbodyonthehorizonandthepointwherethe

284 AZIMUTHS AND AMPLITUDES Record the tabulated azimuth for these arguments. As the third and final step in the triple interpolation process, decrease the value of LHA to 316° because the ac￾tual LHA value was smaller than the base LHA. Enter the Sight Reduction Tables with the following arguments: (1) Declination = 20°; (2) DR Latitude = 33°; (3) LHA = 316°. Record the tabulated azimuth for these arguments. Calculate the Z Difference by subtracting the base az￾imuth from the tabulated azimuth. Be careful to carry the correct sign. Z Difference = Tab Z - Base Z Next, determine the increment for each argument by taking the difference between the actual values of each argument and the base argument. Calculate the correction for each of the three argument interpola￾tions by multiplying the increment by the Z difference and dividing the resulting product by 60. The sign of each correction is the same as the sign of the corresponding Z difference used to calculate it. In the above example, the total correction sums to -0.1’. Apply this value to the base azimuth of 97.8° to obtain the true azimuth 97.7°. Compare this to the compass reading of 096.5° pgc. The compass error is 1.2°E. AZIMUTH OF POLARIS 1702. Compass Error By Azimuth Of Polaris The Polaris tables in the Nautical Almanac list the azi￾muth of Polaris for latitudes between the equator and 65° N. Figure 2011 in Chapter 20 shows this table. Compare a compass bearing of Polaris to the tabular value of Polaris to determine compass error. The entering arguments for the table are LHA of Aries and observer latitude. Example: On March 17, 1994, at L 33° 15.0’ N and 045° 00.0’W, at 02-00-00 GMT, Polaris bears 358.6°T by compass. Cal￾culate the compass error. Solution: Enter the azimuth section of the Polaris table with the calculated LHA of Aries. In this case, go to the column for LHA Aries between 160° and 169°. Follow that column down and extract the value for the given latitude. Since the increment between tabulated values is so small, visual in￾terpolation is sufficient. In this case, the azimuth for Polaris for the given LHA of Aries and the given latitude is 359.3°. AMPLITUDES 1703. Amplitudes A celestial body’s amplitude is the arc between the observed body on the horizon and the point where the observer’s horizon intersects the celestial equator. See Fig￾ure 1703. Calculate an amplitude after observing a body on either the celestial or visual horizon. Compare a body’s measured Actual Base Arguments Base Z Tab* Z Z Diff. Increments Correction (Z Diff x Inc.÷ 60) Dec. 20° 13.8' N 20° 97.8° 96.4° –1.4° 13.8' –0.3° DR Lar. 33° 24.0' N 33° (Same) 97.8° 98.9° +1.1° 24.0' +0.4° LHA 316° 41.2' 317° 97.8° 97.1° – 0.7° 18.8' –0.2° Base Z 97.8° Total Corr. –0.1° Corr. (–) 0.1° Z N 97.7° E *Respondent for the two base arguments and 1° change from third base argument, in vertical order of Dec., DR Lat., and LHA. Zn 097.7° Zn pgc 096.5° Gyro Error 1.2° E Figure 1701. Azimuth by Pub. No. 229. Date 17 March 1994 Time (GMT) 02-00-00 GHA Aries 204° 25.4’ Longitude 045° 00.0’W LHA Aries 161° 25.4’ Tabulated Azimuth 359.3°T Compass Bearing 358.6°T Error 0.7°E

285AZIMUTHSANDAMPLITUDESsouth and the sun is rising,then subtract the Table 23 cor-rectionfromtheobservedamplitudeThe following two sections demonstrate the procedurePNfor obtainingthe amplitude of the sun on both the celestialandvisiblehorizons1704.AmplitudeOfThe SunOnTheCelestial Horizon2Example:ZenithAlThe DRlatitude ofa ship is 51°24.6'N.The navigatorBodyobserves the setting sun on the celestial horizon.Its decli-nation is N190 40.4:Its observed amplitude is W32.90N.BEqutor(32.go"northofwest."or302.90)HorieonRequired:Compasserror.Solution:Interpolate in Table 22 for the sun's calculated ampli-tude as follows.See Figure 1704.The actual values forlatitudeanddeclinationareL=51.4oNanddec.=N19.670Find the tabulated values of latitudeand declination closestFigure 1703.The amplitude is the arc (A)between thetotheseactualvalues.Inthiscase,thesetabulatedvaluesareobservedbodyonthehorizonandthepointwheretheL = 51°and dec.=19.5°. Record the amplitude correspond-observer's horizon intersects thecelestial equator.ingto thesebase values,32.0°,as thebase amplitude.Next, holdingthe basedeclination value constant at19.5,increase the value of latitudeto the next tabulatedamplitudewithanamplitudeextractedfromtheAmplitudevalue:N520Notethatthis value of latitude was increasedtable.The difference between the two values representsbecausethe actual latitude valuewasgreaterthanthebasecompasserror.value of latitude.Record the tabulated amplitudeforLGive amplitudes the suffix N if the body from which it52°anddec.=19.5°:32.80.Then,holdingthebaselatitudewasdeterminedhasanortherndeclinationandSifithasavalueconstantat5j,increasethedeclinationvaluetothesoutherndeclination.Givethe amplitudes theprefixEifthenexttabulatedvalue:20o.Record thetabulated amplitudebody is rising and W if the body is settingfor L = 5/°and dec.=20°: 32.90.The values in theAmplitude table assume that the bodyThelatitude'sactual value(51.4)is0.4of the waybe-is on the celestial horizon. The sun is on the celestial hori-tween thebase value (51°)and the value usedto determinezon when its lower limb is about two-thirds of a diameterthetabulated amplitude(52°).Thedeclination'sactual val-above the visible horizon.The moon is on the celestial ho-ue(19.67)is0.3of thewaybetweenthebasevalue(19.50)rizon when its upper limb is on thevisible horizon.Planetsand the value used to determine the tabulated amplitudeand stars are on the celestial horizon when they areapprox-(20.0°).Todeterminethetotal correctiontobaseamplitudeimately one sundiameterabove thevisible horizonmultiplytheseincrements (0.4and0.3)bytherespectivedifference between the base and tabulated values (+0.8 andWhen using a body on the visible, not celestial, hori-zon, correct the observed amplitude from Table 23 Apply+0.9,respectively)andsumtheproducts.Thetotal correc-tion is +0.6°.Add the total correction (+0.6)to the basethis table's correction to the observed amplitude and nottoamplitude (32.00) to determine the final amplitude (32.60).the amplitude extracted from the Amplitude table.For thesun,a planet or a star,apply this correction to the observedCalculate the gyro error as follows:amplitude in thedirection away fromthe elevated pole.IfW32.9°NAmplitude (observed)pgc-usingthemoon, applyone-half of theTable23correctionAmplitude (fromTable22)W32.6°N-inthedirectiontowards theelevatedpole.0.3°WCompassErrorNavigators most oftenusethe sun when determiningamplitudes.Therulefor applyingtheTable23corrections1705.Amplitude Of The Sun On The Visible Horizontoasun'sobservedamplitudeissummarizedasfollows.Ifthe DRlatitude is north and the sun is rising,or if theDRExample:latitude is south and the sun is setting,add theTable 23 cor-rection totheobserved amplitude.Conversely,if theDRThesameproblemassection1704,exceptthatthesunlatitudeisnorthandthesun is setting,ortheDRlatitudeisissettingonthevisiblehorizon

AZIMUTHS AND AMPLITUDES 285 amplitude with an amplitude extracted from the Amplitude table. The difference between the two values represents compass error. Give amplitudes the suffix N if the body from which it was determined has a northern declination and S if it has a southern declination. Give the amplitudes the prefix E if the body is rising and W if the body is setting. The values in the Amplitude table assume that the body is on the celestial horizon. The sun is on the celestial hori￾zon when its lower limb is about two-thirds of a diameter above the visible horizon. The moon is on the celestial ho￾rizon when its upper limb is on the visible horizon. Planets and stars are on the celestial horizon when they are approx￾imately one sun diameter above the visible horizon. When using a body on the visible, not celestial, hori￾zon, correct the observed amplitude from Table 23 Apply this table’s correction to the observed amplitude and not to the amplitude extracted from the Amplitude table. For the sun, a planet, or a star, apply this correction to the observed amplitude in the direction away from the elevated pole. If using the moon, apply one-half of the Table 23 correction in the direction towards the elevated pole. Navigators most often use the sun when determining amplitudes. The rule for applying the Table 23 corrections to a sun’s observed amplitude is summarized as follows. If the DR latitude is north and the sun is rising, or if the DR latitude is south and the sun is setting, add the Table 23 cor￾rection to the observed amplitude. Conversely, if the DR latitude is north and the sun is setting, or the DR latitude is south and the sun is rising, then subtract the Table 23 cor￾rection from the observed amplitude. The following two sections demonstrate the procedure for obtaining the amplitude of the sun on both the celestial and visible horizons. 1704. Amplitude Of The Sun On The Celestial Horizon Example: The DR latitude of a ship is 51° 24.6’ N. The navigator observes the setting sun on the celestial horizon. Its decli￾nation is N 19° 40.4’. Its observed amplitude is W 32.9° N. (32.9° “north of west,” or 302.9°). Required: Compass error. Solution: Interpolate in Table 22 for the sun’s calculated ampli￾tude as follows. See Figure 1704. The actual values for latitude and declination are L = 51.4° N and dec. = N 19.67°. Find the tabulated values of latitude and declination closest to these actual values. In this case, these tabulated values are L = 51° and dec. = 19.5°. Record the amplitude correspond￾ing to these base values, 32.0°, as the base amplitude. Next, holding the base declination value constant at 19.5°, increase the value of latitude to the next tabulated value: N 52°. Note that this value of latitude was increased because the actual latitude value was greater than the base value of latitude. Record the tabulated amplitude for L = 52° and dec. = 19.5°: 32.8°. Then, holding the base latitude value constant at 51°, increase the declination value to the next tabulated value: 20°. Record the tabulated amplitude for L = 51° and dec. = 20°: 32.9°. The latitude’s actual value (51.4°) is 0.4 of the way be￾tween the base value (51°) and the value used to determine the tabulated amplitude (52°). The declination’s actual val￾ue (19.67°) is 0.3 of the way between the base value (19.5°) and the value used to determine the tabulated amplitude (20.0°). To determine the total correction to base amplitude, multiply these increments (0.4 and 0.3) by the respective dif￾ference between the base and tabulated values (+0.8 and +0.9, respectively) and sum the products. The total correc￾tion is +0.6°. Add the total correction (+0.6°) to the base amplitude (32.0°) to determine the final amplitude (32.6°). Calculate the gyro error as follows: 1705. Amplitude Of The Sun On The Visible Horizon Example: The same problem as section 1704, except that the sun is setting on the visible horizon. Figure 1703. The amplitude is the arc (A) between the observed body on the horizon and the point where the observer’s horizon intersects the celestial equator. Amplitude (observed) pgc = W 32.9° N Amplitude (from Table 22) = W 32.6° N Compass Error 0.3°W

286AZIMUTHSANDAMPLITUDESRequired:a)Bodyonthecelestial horizon:Compass error.sin d-Amplitude= sinLcos LSolution:Interpolate in Table 23to determinethe correction forwhere d = celestial body's declination and L = observ-the sun on thevisiblehorizon as follows.See Figure1705..er's latitude.Choose asbase values of latitude and declinationthetabu-larvaluesoflatitudeanddeclinationclosesttotheactualb) Body on the visible horizon:values. In this case, these tabulated values are L=51°Nand dec.=20°Record thecorrection correspondingtosind-sinL sin h-thesebasevalues,I.jo,as thebase correction.Amplitude= sincos L coshCompletingtheinterpolationprocedureindicates thatthebase correction(l.1°)is the actual correction.where d = celestial body's declination, L=observer'sApplythis correctioninaccordance withtherulesdis-latitude,andh=-0.7cussedinsection1703.Sincethevessel'slatitudewasnorthandthesunwassetting,subtractthecorrectionfromtheUsing the same example as in section 1704, d =observedamplitude.TheobservedamplitudewasW32.9N.19.67°NandL=N51.4°.Ifthesunisonthecelestial ho-SubtractingtheI.jcorrectionyieldsacorrectedobservedrizon, itsamplitude is:amplitudeofW31.8°N.Fromsection1704,thetabularamplitudewasW32.6°N.Isin19.67°Calculatethegyroerrorasfollows:W32.6°NAmplitude= sin[cos51.4°Amplitude(fromTable22)=W32.6°NAmplitude (observed)=W31.8°NIf the sun is on the visiblehorizon, its amplitude is:CompassError0.8°E-1sin19.67°sin51.4°sin0.7°1706.AmplitudeByCalculationAmplitude=ssincos51.4°coS-0.70As an alternativetousingTable22and Table23,use=W 33.7° Nthefollowingformulastocalculate amplitudes:ActualBaseDiff.Inc.Base Amp.Tab. Amp.CorrectionL=51.4°N51°32.0°32.8°+0.8°0.4+0.3°19.5032.90dec=19.67°N32.00+0.900.3±0.3°+0.6°TotalFigure1704.Interpolation inTable22forAmplitudeDiff.ActualBaseBase Amp.Inc.Tab. Amp.Correction51°1.1°1.1°0.0°0.40.0°L=51.4°N20°1.1°1.0°-0.100.20.0°dec=19.67°NFigure1705.InterpolationinTable23forAmplitudeCorrection

286 AZIMUTHS AND AMPLITUDES Required: Compass error. Solution: Interpolate in Table 23 to determine the correction for the sun on the visible horizon as follows. See Figure 1705. Choose as base values of latitude and declination the tabu￾lar values of latitude and declination closest to the actual values. In this case, these tabulated values are L = 51° N and dec. = 20°. Record the correction corresponding to these base values, 1.1°, as the base correction. Completing the interpolation procedure indicates that the base correction (1.1°) is the actual correction. Apply this correction in accordance with the rules dis￾cussed in section 1703. Since the vessel’s latitude was north and the sun was setting, subtract the correction from the observed amplitude. The observed amplitude was W 32.9 N. Subtracting the 1.1° correction yields a corrected observed amplitude of W 31.8° N. From section 1704, the tabular amplitude was W 32.6° N. Calculate the gyro error as follows: 1706. Amplitude By Calculation As an alternative to using Table 22 and Table 23, use the following formulas to calculate amplitudes: a) Body on the celestial horizon: where d = celestial body’s declination and L = observ￾er’s latitude. b) Body on the visible horizon: where d = celestial body’s declination, L = observer’s latitude, and h = – 0.7°. Using the same example as in section 1704, d = 19.67° N and L = N 51.4°. If the sun is on the celestial ho￾rizon, its amplitude is: If the sun is on the visible horizon, its amplitude is: =W 33.7° N Amplitude (from Table 22) = W 32.6° N Amplitude (observed) = W 31.8° N Compass Error 0.8° E Amplitude sin d cos L - –1 = sin Amplitude sind L – sin sin h cos L cos h - –1 = sin Amplitude sin19.67° cos51.4° - –1 = sin = W 32.6° N. Amplitude sin 51.4 19.67° – sin 0.7 °sin– ° cos 0.7 51.4° cos– ° - –1 = sin Actual Base Base Amp. Tab. Amp. Diff. Inc. Correction L=51.4°N 51° 32.0° 32.8° +0.8° 0.4 +0.3° dec=19.67°N 19.5° 32.0° 32.9° +0.9° 0.3 +0.3° Total +0.6° Figure 1704. Interpolation in Table 22 for Amplitude. Actual Base Base Amp. Tab. Amp. Diff. Inc. Correction L=51.4°N 51° 1.1° 1.1° 0.0° 0.4 0.0° dec=19.67°N 20° 1.1° 1.0° -0.1° 0.2 0.0° Figure 1705. Interpolation in Table 23 for Amplitude Correction