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EXAMPLE III. 1870, March 21st, A. M. at ship, in latitude 55° N., the observed altitude of the Sun's 1.1. was 10° 55', I. E. — 4 50”, height of the eye 20 feet; when a chronometer shewed 20d. 20h. 19m. 42s., which was slow on Green. mean time 37m. 278. on Dec. 16, 1869, and gaining 68.8 daily. Find the longitude.

d. h. m. s. From Dec. 16 ) Time by chron. 20 20 19 42 to Mar. 20)

= 94 days. 8}

6.8 daily rate. Error, slow + 37 27

752 20 20 57 09 Gain in interval

10 45 True Gr. mean time 20 20 46 24

12 | 4 | 639:2
8

3:4
1 2.3

•3 60 964,5.2

10 45

564

hours.

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Obs. alt. 10° 55'00" True alt. 10° 57' 10"
I. E.
4 50

Lat. 55 00 00 Secant 0.241409 10 50 10

P. dist. 89 46 54 Cosecant 0.000003 Cor. + 7 00

2) 155 44 04 True alt. 10 57 10 1 Sum 77 52 02 Cosine 9.322607

Remainder 66 54 52 Sine 9.963757 Hour angle 4 43 57 (XXXI.) 9.527776 d. h.

S. Apparent noon at ship 21 00

00 Hour angle

4 43 57 (A. M. subtract.) App. time at ship

20 19 16 03 Equation of time +

7 20 Mean time at ship

20

19 23 23 Mean time at Greenwich 20 20 46 24 Longitude in time

1 23 01

60

4) 83 01 Longitude 20° 454 West.

S.

EXAMPLE IV. 1870, December 1st, A. M. at ship, in latitude 30° N., the observed alt. of the Sun's 1. l. was 30° 27', I. E. + 1'20”, height of the eye 15 feet; when the chronometer shewed 1d. 6h. 15m., which was fast for Green. mean time 7m. 41s. on July 30th, and losing 582 daily. Find the longitude.

d. h. m. From July 30 Time by chron.

124 days. 1 6 15 00 to Dec, 1

5•2 daily rate. Error, fast

7 41 1 6 07 19

215 Loss in interval +

620 10 46

Hours. True Gr. mean time 1 6 18 05

644:8

1:3 6,0 164,0:1

10 46

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Obs. alt. 30° 27'00" True alt. 30° 39' 17"
I. E. + 1 20

Lat. 30 00 00 Secant 0.062469 30 28 20

P. dist. 111 52 38 Cosecant 0.032478 Cor. + 10 57

2) 172 31 55 True alt. su 39 17 1 Sum 86 15 57 Cosine 8.813667

Remainder 55 36 40 Sine 9.916600
Hour angle 1 59 53 (XXXI.) 8.825214

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VARIATION BY AZIMUTH.

1. If P. M. at Ship put down the day, hour, and minute, as given in the question; but if A. M. at ship, put down a day less and add 12 to the hours, for the astronomical time at ship; apply the Longitude in time, if W. add; if E. subtract, for the Greenwich mean time.

2. Take out from page 2 and correct the Declination as in previous questions, adding it to 90° if Lat. and Dec. are different names, and subtracting it from 90° if they are of the same names, for the Polar Distance.

3. Correct the obs. alt. for index error and for Table IX.

4. Add together the true alt., lat., and polar dist., divide their sum by 2, and take the difference between this } sum and the polar distance.

5. Take out the secant of the true alt., secant of the lat., cosine of the į sum, and cosine of the remainder; add these 4 logs together and divide their sum by 2. The result is the sine of half the true azimuth. Multiply the } azimuth by 2 to get the true azimuth. The following form will show this :: T. alt.

Secant 0
Lat.

Secant 0
P. dist.
2) Sum.

Cosine
Remainder

Cosine

2) * True azimuth

Sine

2 True azimuth 6. The true azimuth is always named from N. or S. towards E. or W., N. or S. opposite to the latitude; that is, lat. N., azimuth is S.; and lat. S., azimuth is N.; E. if A. M. at ship, W. if P. M.

7. Put the magnetic azimuth under the true azimuth ; then if they are both reckoned from the N., or both from the S., the difference is the variation; but if one is N. and the other S., take the true azimuth from 180° 00' and the remainder will be the true azimuth, reckoned from the same point as the magnetic. Then the difference is the variation as before.

8. The variation is east when the true is the right of the magnetic azimuth, and west when it is to the left.

Add.

1 Sum

EXAMPLE. I. 1870, October 21st, at 5h. 42m. P. M. mean time at ship, in latitude 40° 10' S., longitude 30° W.; the sun's bearing by compass was S. 76° 20'' W. The obs. alt. Sun's l. 1. was 13° 10' 20", height of the eye 16 feet. Find the variation.

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Variation

M

10 36 East

EXAMPLE II. 1870, November 17, at 9h. 49m. A. M., mean time at ship in latitude 45° 15' N., longitude 30° 15' E.; the sun's bearing by compass was S. by E. E. The obs. alt. of the sun's l. l. was 20° 19', I.E. + 5' 20". Height of the eye 20 feet. Find the variation.

d. h. m.

Long. 30° 15' E.

4 6,0 121 00

Mean time at ship 16 21 49
Long. in time

2 1 E. Mean time at Green. 16 19 48

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