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Per Nautical Almanac :
Per Nautical Almanac :

© Declin., May 5th, 10° 19'27' N + 16'55" Declin., Vurch 4th 6° 19' 18"S.–23' 9" 360° : 16' 55" :: 23° :/

1 4 300° : 23' 9" : : 10° : 2' 31" /

1' 4", cor, of declin. s

3 31 the correction

Sun's declin. at time

16 30 31 0 15 41 S. Ans.

of observation Observed altitude

50° 12' 40" S. The correction here is subtractive, as the longitude is west and the declination decreasing. Dip

4 4 See article ASTRONOMY for a table of the sun's

50 8 36 declination with a table to adapt it to subsequent

Semidiameter

15 52

}

vears.

50 24 28

43

90

39

56

To find the latitude at sea by meridian altitude.

Refraction 48" – paral. 5“ cor. Take the true altitude froin 90°, and the remainder will be the zenith distance, to be called True altitude

50 24 45 S. north when the object is south, and south when the object is north of the zenith. Reduce the declination to the time of observation. Then, if Zenith distance

35 15 N. the declination and zenith distance are both north Declination

15 30 31.V. or both south, their sum is the latitude ; but, if one is north and the other south, their difference Latitude

5 46 N. is the latitude; and it is always of the same de

In observations for the latitude at sea it is nomination as the greater.

customary to coosider the sun's semi-diameter To explain this rule, let A C (plate II. fig. 8) be the line of intersection of the planes of the the nearest minute.

as always 16', and to take all the corrections to

When it is not stated that meridian and rational horizon; and F D the the altitude is taken by a back observation, a intersection of the planes of the meridian and fore observation is always understood. © inequator; let Z be the zenith, P the north, and E

dicates the sun's lower, and the south pole ; and let S be the true place of a

his upper limb; celestial object on the meridian ; then SA is its

) the moon's lower, and ) her upper limb); meridian altitude, S Z its meridian zenith dis- -+ indicates that the number to which it is aftance, and S F its north declination. If S be fixed is increasing, and — that it is decreasing. the place of the ohject, then SA is its meridian Erample 2.-If on February 5th, 1828, at 2h. alutude, S Z its zenith distance, and S' F its 49m. apparent time, as deduced from a chronosouth declination. If S" he the place of the meter, regulated for Greenwich time, the meobject, then S” C is its meridian altitude, S" Z ridian altiiude of ) be observed to be 49° 35' S., its zenith distance, and S” I' its north declination. height of the eye twelve feet; required the latiNow Z F, the latrude ZS + SF=2 $ - tude! SF= F S“ – ZS"; which is the rule.

In this example the Greenwich time is 2h. 49m. Erumple 1.--If the meridian altitude of the past midnight of February 1th. Now by the Nautisun's lower limb be 50° 12' 40" S. on May 5th, cal Almanac, the moon's declination, semi-diame1828, in long. 23° W., height of the eye seven- ter, and parallax, at midnight of that day, with their teen feet; required the latitude ?

variations in the succeeding twelve hours, stand as under:

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90

35 30 Cor, of altitude.

Zenith distance
Declination

40 8 5 N.
2 32 37 S.

Latitude

37 35 28 N.

The moon's semidiameter as given in the Nau- therefore appear under a greater angle. This tical Almanac, is that which she would appear to augmentation of her semidiameter may be taken have if seen from the centre of the earth; if she with sufficient exactness for practical purposes be seen upon the horizon she will of course be at sea from the following table :-nearer to the zenith of the observer, and will

Moon's apparent

altitude.

0° 3° 6° 12° 15° 18° 21° 27° 30° 36° 39° 45° 48° 54° 60° 72° 30°

9" 10" 11" 12" 13" 14" 15" 16"

Augmentation of
)'s semidiameter.

0" 1" 2" 3" 4" 5" 6" 7" 8"

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To altitude 49° we find from this table that the augmentation of the moon's semidiameter is 11"; hence, in the above example, the semidiameter ought to have been 15' 21" ; but it is only in finding the longitude by lunar observation, that such nicety respecting the moon's semidiameter becomes of any practical importance in the business of navigation.

Erample 3.— If the meridian altitude of Aldebaran be 28° 5' 10" N. on February 18th, 1828, height of the eye fourteen feet; required the latitude?

Observed alt. 28° 5' 10" N.
Dip

3 41

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27 59 42 N.
30

4 10 37

11 29

62
16

45

51

.

Refraction
Zenith dist.

0 18 S.
Declination
9 15 N.

3 59 8

Co-declination 45 48 32
Latitude

3 S.
To find the latitude by the meridian altitude of

Latitude 49 47 40 N. er object below the pole.

To find the latitude from the observed altitude To the altitude add the complement of the de- of a celestial object, near the meridiun, the time of . clination, and the sum will be the latitude, of the observution being known. same name with the declination.

If the object is the sun, the apparent time is For (fig. 8, plate II.) F P and Z C are equal, the meridian distance ; for any other object, to being quadrants, and, if the common part Z P be the apparent time add the sun's right ascension, omitied from each, there remains 2 F the latitude and from the sum subtract the right ascension of =PC. Now if s'"' be the object below the the object, and the remainder is its meridian dispole, then S" C is its altitude, S" D its declina

tance. If the meridian distance is more than tion, and S"" P the complement of its declination; twelve hours, subtract it from twenty-four hours. and S'"C +S"P= PC' the latitude.

Then add together the cosine of the meridian In taking the sun's declination, it must be no- distance, and the tangent of the polar distance and ticed that he is on the meridian twelve hours the sum, rejecting the tens from the index, will after noon.

be the tangent of arc first, which will be obtuse Erample 1.-If the altitude of 0 on the meri

or acute according as the polar distance is obdian below the pole be 10° 50' on May 20th, tuse or acute. Again, add together the secant of 1828, in long. 14° E., the height of the eye six- the polar distance, the sine of the altitude, and teen feet; required the latitude ?

the cosine of arc first, and the sum, rejecting the Time of observation

12h. Om. tens from the index, will be the cosine of arc se

cond. The sum or the difference of arcs first Long. in time E Greenwich time of observ.

and second will be the co-latitude; and that

latitude must be taken which agrees most nearly 24h.:12' 16" :: 11h. 4' : 5' 39"

with the latitude by account, which will always O's declin., May 2006; } 20° 2° to +12 16%

be known nearly enough to determine which of Correction

the imputed latitudes is the true one.

Let P A (fig. 9, plate II.) be the meridian on Declination

which the first point of Aries is, PS that on which the sun is, P x that on which the star is,

at the moment of observation. Then if P M be Co-declination

the meridian of the place of observation, SPM 69 52 17

is the apparent time, A PS the same right as

56

11

4

5 39

20 7 43
90

App. time

cension, AP, the stars, and x P M the star's me

b. n. s. ridian distance. Now APS+SPM-A P r: Time of observation 7 3 10

PM, which is the rule for finding the meridian Long. by acc. W. 2 20
distance of a star; and the apparent line SPM
is evidently the meridian distance of the sun. Greenw. time by acc. 9 23 10
Next, let. A B(fig. 10 plate II.) be the object's

h. m. $. altitude, A Z its zenith distance, AP its polar

O R. A. March 9th, 23 19 20 + 3 42 distance, P Z the colatitude, and on P Z, or PZ

1 26 produced, let fall from A the perpendicular AC. Then rad. cos. P = cot. A P. tan. PC; or

O's R. A.

23 20 46 rad. cos. P tan. A Picos. P tan. PC=

whence, 24h. : 3m. 40s. : : 9h. 23m. : 1m. 26s, com. : cot. AP

R P is the object's meridian distance, and PC is O's R. A.; *'s R. A. 4h. 26m. 5s. *'s pot arc I in the rule. Again cos AP: cos A Z(=sin.

dist. 73° 50' 46'. cos. PC:sin. A B

h. m. s. AB):: cos. PC:cos. CZ = cos. AP

7 3 10 cos. PC:sin. A B. sect. AP,

O's R. A.

23 20 46
where C Z is arc
rad. 2.

6 23 56 II in the rule. And when Z and P are on the

*'s R. A. 4 26 5 same side of C the difference is Z P, otherwise their sine is Z P the colatitude.

*'s mer. dist. 1 57 51 = 29° 27° 45' Erample 1.-If the altitude of O be 35° 2', at Oh. 48m, 12s. P. M. October 7th, 1828, in lat.

Obsd. alt. 53° 12' 0" by account 47° 50' N., long. 20° W., height of

Dip.

3 33 the eye fifteen feet, required the latitude ?

53 8 27 Time of observation 0 48 12

Refr.

43 Long. in time W. 20

True alt. 53 7 44 Greenwich time by acc. 2 8 12

Mer. dist. 29° 27' 45" cos. 9.939857 24h. : 22' 58" : : 2h. 8m.: 2' 3" corr. of declin. Pol. dist. 73 50 46 tan. 10.538120 48° 12' = 12° 3', meridian dist.

sect. 10-55601 O's declin. Oct. 7th. 5° 35' 39'S X 22' 58" 71° 35' 56". Arc I tan. 9.477977 cos. 9-19923. Correction. 2 3

Alt. 53° 7' 44" sin. 9.90303

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h. m.

S.

1

5

16

5 37 42

24° 49' 1

Arc II. cos. 9-9379** 90

Co-lai. 46 46 55 Diff. arcs I and II.
95 37 42 Polar dist.

90
Obsd. alt. 35° 2' 0'
Dip.
3 49

43 13 Latitude N.

Gwen the time of observation and the altitude 34 58 11

the pole star to find the latitude. Semid.

3

Find the meridian distance of the star as in the

preceding method of finding the latitude. If the 35 14 14 Refr. 1' 21"— par. 7"

meridian distance is less than 90° consider it 1 14

a course, if it is between 90° and 270° conside

the difference between it and 180° as a course; True alt. 35 13

if it is more than 270o consider the difference Mer. dist. 12° 3' cos. 9.990324

between it and 360° as a course. Then will Pol. dist. 95 38 tan. 11:005955 sect. 11.008057 this course, and the polar distance of the star ia

minutes, enter a traverse table, and the different? 95° 46' arc. I. tan. 10-996279 cos I. 9.002069 of latitude will be the correction to be applied

to the true altitude to obtain the latitude; addiAlt. 35° 13' sin. 9-760927 tive when the ineridian distance is more than 9

and less than 270o, otherwise subtractive. 5349 Arc II. cos. 9.771053 For, let P (fig. 11 plate II.) represent the pole,

MNA the horizon, at BCD, the circle in which Co-lat. 4157 diff. arcs I and II

the pole star moves round the pole, a circle which 90

from its smallness may without any important

error in navigation be considered as a plane ott, 483 latitude N.

Then P N is the latitude, and, when the star in Erample 2.- If the altitude of Aldebaran be at O, its polar distance O P taken from its alt 53° 12° ‘at 7h. 6m. 10s. apparent time, March tude o n leaves P N the latitude. When it is a 9th, 1824, in long. by account 35° W., lat. 44° 180, it is directly below, as at 0 it was about N., height of the eye thirteen feet

, required the the pole, and the polar distance 180 P added to latitude?

its altitude 180 N gives P N the latitude. When

or

tance.

h. m.

3

the star is at 90 or 270 its altitude is the same as thus. Take the angle intended between the ship's that of the pole. When it is at A or D, EP course and the sun's bearing at the first observaor HP taken from the altitudes A Mor DQ tion as a course, and with it and the distance run leaves P N the latitude; when at B or C, PFO on the interval as a distance the correction of the PG added to B M or C Q gives P N the lati- first altitude for the ship's change of place will tude. Now A () is the star's meridian distance be found in the column of latitude. at A, O A B, () BC, and O BCD, its meridian Having now got the polar distance, the true distance at B, C, and D, respectively. The dif- altitudes, both for the place at which the second ference between () Band 180° is B 180, the mea- was observed, take half the interval between the sure of the angle B PG and similarly A PE, two times of observation, and, reducing it into G PC, and D P H, are the angles which taken as degrees, call it the half elapsed time, and proa course with A P, BP,C P, or DP, as a distance, ceed to compute the latitude at the place of the give the corrections P E, P F, PG, and PH; the second observation by the following rule:corrections to be applied to "he altitudes AM, 1. Add the sine of the half elapsed time to the BM, CQ, and DQ, respectively to obtain the sine of the polar distance, and the sum, rejecting latitude PN.

ten from the index, will be the sine of arc first. Erample. The altitude of Polaris was 52° 2. Add the secant of arc first to the cosine of 27' 35"; at 7h. 24m. apparent time February 2d, the polar distance, and the sum, rejecting ten 1829, the sun's right ascension at the time being from the index, will be the cosine of arc second, 21h. 2m., height of the eye fifteen feet; required which will be acute or obtuse, like the polar disthe latitude ?

3. Add together the cosecant of arc first, the App. time 7 24

cosine of half the sum of the true altitudes, and O's R. A. 21 2

the sine of half their difference, and the sum, re

jecting the tens from the index, will be the sine
Sum
4 26

of arc third.
*'s R. A.
0 59

4. Add together the secant of arc first, the

sine of half the sum of the true altitudes, the co*'s mer. dist. 3 27

52° nearly

sine of half their difference, and the secant of arc

third, and the sum, rejecting the tens from the Obs. alt. 52° 27' 35"

index, will be cosine of arc fourth. Dip

49

5. The differences between arc second and arc

fourth is arc fifth when the zenith and the elevated 52 23 46

pole are on the same side of the great circle pasRefr.

41

sing through the apparent places of the sun at

the two times of observation, otherwise their

52 23 5 With 52° as a course and

sum is arc fifth; but arc fifth can never exceed

90°. 99', *'s polar dist. the corr.

6. Add the cosine of arc third to the cosine is 61'

1 1 0

of arc fifth, and the sum rejecting ten from the

index will be the sine of the latitude. Latitude 51 22 5

Note 1.- When there is any doubt whether To find the latitude from two altitudes of the the zenith and elevated pole are on the same or sun observed on the same day with the time elapsed different sides of the great circle passing through between the observations.

the places of the sun, the latitude may be comWe take the following direct rule for solving puted on both suppositions, that being considered this important problem from Mr. Riddle's Trea- as the true latitude which agrees most nearly tise on Navigation and Nautical Astronomy. with the latitude by account, which will always

Take the sun's declination for the Greenwich be known nearly enough for that purpose. This time which corresponds to the middle time be- additional computation will give very little tween the two observations, and if it is of the same trouble, as it is only arc fifth and its cosine that name with the latitude subtract it from 90°, if of will require alteration. a different name add it to 90°, and the sum or the Note 2.- By this method the latitude may be remainder will be the sun's polar distance. De- found from two altitudes of the same fixed duct the true altitudes from the observed ones, star; but, if the interval be in solar, it must be and, if the ship has been sailing in the interval reduced into sidereal time, which may he done between the observations, the first altitude must with sufficient exactness by adding to the elapsed be reduced to what it would have been if taken time ten seconds for every hour or one second at the same place with the second. This correc- for every six minutes. tion may be obtained by the following proportion, Notc 3.- It will expedite the calculation if all As radius is to the distance sailed, so is the co- the logarithms that occur at the same opening of sine of the angle included between the course the book are taken out at the same time, and in the interval and the bearing of the sun at the any little mistake in the observations will produce first observation to the correction of the first al- the less error in the result, the nearer the greater titude; to be added to it if the ship sails within altitude is to the meridian. less than eight points of the sun's bearing, but Note 4.- It is a curious circumstance in the subtracted from it if the course is more than eight history of this problem that almost all who have points from the sun's bearing. Or the correction written upon it have introduced a correction in may be more readily made from a traverse table, the wrong angle for the change of longitude be

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tween the observatione ; a correction which can dillerence of those angles; and rad. :1B :: 003. nave no existence if ihe times are marked as they BIC: AC, the correction of the first altitude. are presumed to be by the same watch, and the If the ship run from A to B, from the sun, then altitudes are reduced to what they would have AC' computed in the same way is the correction been if taken at the same place.

This error is of the first altitude, subtractive, to reduce it to found in Mackay's Navigation, and in bis Treatise what it would have been if taken at the same on the Longitude, and it was in the earlier editions place as the second. of Norie's Navigation, but has been omitted since llaving now obtained the altitudes, both at the 1821. It is found in the article Navi ation in place of the second observation, lei Z(fig. 1.3 the Encyclopædia Britannica; and even in the plate II.) be the zenith; P the pole ; A and B Encyclopædia Metropolitana, a work now in the the places of the sun at the two times of obsercourse of publication. It has probably been vation; AC, BD, the altitudes, AZ, BZ, the owing to this remarkable circumstance wat sea- ze with distances; AP, BP, the equal polar dismen have been so little disposed to rely on the tanies; Z PA, and 2 P B, the times of observisresults of double altitudes.

tion, or the meridian distances of the sun at the Demonstration of the rule.

times of observation. Then APB is the elapse Let Q, represent the sun at the place of the time, and, if P E be a perpendicular on AB, ship at the first observation, NAS the meridian, then Ab and AP B will both be bisected by it; and B the place of the ship at the second obser- whence APE or E P B will be the half elapsed valiou. Ou A Q drop the perpendicular BC;

Join E Z, and from Z on E P draw the then the ship in runuing from A to B bas ad- perpendicular ZF. Then A Eis arc first, E Pare vanced AC directly towards the point on which second, ZF arc third, E Farc fourth, and E Pare the sun bore when the ship was at A ; therefore fifth ; which as the figure is drawn is the differAC added to the altitude observed at A will ence of arcs second and fourth ; but, if Z and P give what that altitude would have been if ob- had been on different sides of A B, F P woul.] served at B. Now NAQ is the bearing of the have been the sum of E P and EF. sun and N AB the ship's course, and B A C the

time.

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sin. Ap.sin. APE Now rad. sin. AE sin. AP •sin APE; or sin. B E

;

rad. cos. AP=

R R. cos. AP sect. A E ·cos. AP cos. A E.cos. E P, or cos. EP=

; cos. A EZ = siu. ZEF= COS. A E

R (os. A2-COS. A E. cos. EZ sin. AC - cos. A Ecos. EZ sin. AE.sin. EZ

sin. A E sin. EZ'

cos. 2 EB

cos. A EZ cos. BZ - cos. B E cos. E Z sin. BD - COS AL·cos. E Z sin. Z EF

llence sin, BE-sin. EZ

we have

sin. AE sin. EZ these two equations: viz. sin. AE·sin. E Z.sin. FEZ

= sin. AC

cos. A E.cos. EZ And sin. A E sin. E Z·sin. FEZ

= sin. BD.

cos. A Ecos. EZ Or sin. ACE =cos. A Ecos. E 2 + sin. A E sin. E Z • sin. FEZ sin. BD = cos. A Ecos. EZ sin. A E sin. E2.sin. F EZ

AĆ + BD A C . BD Whence sin. AC sin. BD-Zcos.

sin.

= sin. AE: EZ: sin. FEZ 2

2

AC + B D AC-BD = 2 sin. AE sin. 2 F and sin. AC + sin. B D = 2 sin.

-- 2 cos. 2

2 AE· cos. C 2 2 cos. A E.cos. EF.cos. FZ.

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COS.

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cos. A E.cos. FZ

rad. 3.

cos. Z F, cos. FP Lastly, rad. cos. Z P, or rad. sin. .at. = cos. ZF.cos. F P, whence

= sin. lai.

rad. Erample 1.-On April 4th, 1828, in lat. 35° N., long. 28° W., by account, at 8h. 10m. 40s. A. V. the alt. of was 38° 12' +, bearing by compass E. S. E; and at 11h. 22m. 145. A.M., the alt. of

was 60° 18' +, the ship having run in the interval S. W. by S. four miles per hour, height of the eye sixteen feet, required the latitude ?

Here the ship is running from the sun, within seven points of that opposite to his bearing; and the distance run between the observations is about thirteen miles; whence the correction of the first altitude is about 3'.

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