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This problem is useful when the moon is the body observed, and when the ship is steaming at a rapid rate.

It has already been stated on p. 365 that the maximum altitude of the sun, moon, and planets is not always the meridian altitude. If the change in declination is in the direction which would raise the altitude, and the rate of change in declination is greater than the rate of fall in the altitude after the object has passed the meridian, the altitude must increase, and this increase will continue until the rate of change in declination is exactly equal to the rate of fall in the altitude. The maximum altitude will occur at this instant. In a similar manner, if a vessel is approaching the celestial body with a high speed, the vessel will be continually changing its horizon, and the altitude of the body will increase until the rate of speed of the vessel in latitude is the same as the rate of fall in the altitude. Should these—the rate of change in the declination, and the rate of speed in latitude—combine together so that they both tend to raise the altitude, it will be easily understood that the maximum altitude will take place many minutes after the body has passed the meridian, and that it may differ considerably from the meridian altitude. Should the combination produce a diminution of the altitude, and if this rate of diminution is greater than the rate of rise in the altitude, then the maximum altitude may occur several minutes before the body comes to the meridian of the observer. As vessels at the present time reach a speed of more than 30 knots an hour, it becomes necessary, for safe navigation in all vessels of high speed, to be able to calculate the correct latitude by using the maximum for the meridian altitude ; or to calculate at what time before or after the time of maximum altitude the body was on the meridian of the observer. This problem, therefore, becomes a special case of THE REDUCTION TO THE MERIDIAN.

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(Lat. + dec.) is the zenith distances computed from latitude by D.R. and declination.

c" represents the combined change in declination and rate of speed in latitude in seconds, in one minute of time.

The speed in knots per hour in a meridional direction is equal to the speed in seconds (") per minute. Thus, if a vessel is going due south 24 knots an hour, this is equivalent to 24” per minute.

RULE.-Enter the Traverse Table with the true course, and speed per hour, in the distance column, and take out the number in the latitude column; this will be the change of latitude in seconds !") per minute.

Take from the Nautical Almanac in the case of the sun or planet the “ Var. in ih.” and divide it by 60 for the change in declination in seconds ( per minute. In the case of the moon, take the Var. in iom.” and move the decimal point one figure to the left for the change in declination in seconds (“) per minute. For a star there is no change in declination.

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If the vessel and the celestial body are moving in opposite directions, that is, the vessel going S. and the celestial body N., or vice versa, add together the change in latitude and change in declination in seconds (TM) per minute. If the vessel and the celestial body are moving in the same direc! tion, that is, both going N., or both going S., take the difference between the change in latitude and change in declination in seconds () per minute. Call this sum or difference (c).

Find the Greenwich time corresponding to the time of meridian passage, and take from the Nautical Almanac the declination of the body and correct if for the Greenwich time. Add the latitude and declination if of opposite name, but subtract them if of the same name.

To find the time from the Meridian. From the logarithm of (c) subtract the logarithm of goo, call this (x). Add together L cos. latitude, L cos. declination, and L co-sec. (latitude + declination), reject all tens, and call the sum of the logarithms (). Then subtract (y) from (x), and the remainder is the L sin. of the time from the meridian when the maximum altitude occurs.

To find the correction of the maximum altitude.—To the L sin. found in the previous part, add (x), the sum is the L sin. of twice the correction : 01, the correction can be found as in reduction to the meridian.

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To find if the maximum altitude is before or after the time of meridian passage.—When the vessel is going towards the celestial body the time will be after the time of meridian passage, except when the body is moving in the same direction as the vessel, and the rate of change in declination is greater than the rate of speed in latitude. When the vessel is going away from the celestial body, the time will be before the time of meridian passage, except when the body is moving in the same direction as the vessel, and the rate of change in declination is greater than the rate of speed in latitude.

Always add the correction to the maximum true altitude; the sum is! the meridian altitude at the place where the altitude was observed. Then find the latitude in the usual way, using the declination corrected for the time from the meridian.

Example.A vessel is steaming due south at the rate of 30 knots an hour; what will be the meridian distance of the moon at the time of her maximum altitude, and the correction to be applied to the maximum altitude to reduce it to the meridian altitude ? Lat. by D.R. 50° N., decl. 1° N. Var. in Iom.” 175" N. Declination increasing.

If the maximum true altitude is 41° 5' 39" S., what is the latitude at the time of observation, and at the time of meridian passage ?

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As the vessel is going towards the moon, the moon attains her maximum altitude 14m. 135. after the time of meridian passage, and the correction of the altitude is 5' 37".

The moon's declination will have increased in the time 17".5 X 14:22 = 248":85 = 4' 9", thus making it 1° 4' 9' N.

The vessel is altering her latitude 30" per min. ; in 14m. 135. aer latitude will have altered 14:22 x 30" 426":6 = 7'7".

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Examples for Practice Example 1.—December 19th, 1890 ; p.m. at ship; in lat. by acct. 57° 5' N., long. 3° W. ; steering S. 20° W. (true) 20 knots an hour; the moon's maximum altitude of the lower limb was observed to be 31° 34' 50" S., i.e. —2' 5" ; height of the eye 20 feet ; required the latitude when the moon was on the meridian of the ship, and also at the time of taking the altitude.

Ans. Latitude when on the meridian 57° 8' 25" N.; latitude at the time of taking the altitude 57° 4' 15" N.

Example 2.-April 1st, 1890, in lat. by acct. 51° 20' N.; long. 9° 30' W., steering S. 10° W. (true) 25 knots an hour; the sun's maximum altitude of the lower limb was observed to be 43° 7'0" S., i.e. +1' 10"; height of the eye 18 feet; required the latitude at the time of observation.

Ans. Latitude 51° 18' 13" N.

Example 3.-October ist, 1890 ; in lat. by acct. 48° 30' N., long. 5° 15' W., steering north (true) 30 knots an hour; the sun's maximum altitude of lower limb was observed to be 38° 5' 0" S., i.e. —2' 20"; height of the eye 21 feet; required the latitude at the time of observation.

Ans. Latitude 48° 28' 4" N.

LATITUDE BY AN ALTITUDE OF THE POLE STAR OUT OF THE

MERIDIAN

Finding the latitude by Polaris, or the Polar star, is a form of the reduction to the meridian, for which, however, special tables which simplify the computation are provided in the Nautical Almanac. When

When you have no Nautical Almanac you can use Tables “ Latitude by Altitude of Pole Star.”

RULE.-I. You must know the sidereal time of observation, i.e. the right ascension of the meridian at the place; for which purpose you may have the apparent time at ship, the mean time at ship, or the mean time at Greenwich, and the longitude ; with which you can find the required sidereal time of observation.

2. To the observed altitude of the star apply the index error, dip, and refraction in the usual way; the result will be the true altitude, from which subtract the constant I', for the reduced altitude.

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3. Turn to Tables I., II., and III. (pp. 483-485 in the Nautical Almanac for 1890) made expressly for the Pole star. The first correction will be additive or subtractive according to the sidereal time; the second aud

third corrections are always to be added. The application of the three corrections to the reduced altitude gives the latitude.

The nearer Polaris is to the meridian, either above or below the pole, when the observation is taken, the less will be the error in latitude arising from an error in the time. The time of the star's passing the meridian can be obtained from Table " Apparent Time of Principal Stars Passing Meridian of Greenwich," and the “ Explanation.”

By Brief Rule and Tables “ Latitude by Altitude of Pole Star."-For the sidereal time of observation at place, see paragraph 1 above.

To solve the problem without the aid of the special tables in the Nautical Almanac, or Tables "Latitude by Altitude of Pole Star" proceed as follows:

1. Find the Greenwich date. Correct the altitude as usual,

2. Take out the right ascension and declination of Polaris for the Greenwich date. Find the polar distance and reduce it to seconds, which call (P).

3. Correct the R. A. M. S., to which add Mean time at ship. This will be the sidereal time at place. From the sum subtract the star's right ascension, the result is the hour angle, which call (h).

4. Add log. of p to the log of co-sine h, the result will be the first correction in seconds.

If the sidereal time at ship be between 6 and 18 hours add the correction to the altitude, if otherwise subtract the correction.

5. Find sine 1" (p sine h)? tangent a. To log. p add log. sine h, square this and add log. tangent of a and the constant log. of (i sine 1") 4:384545 to the product.

The result is the second correction, always additive.
The formula is l p cos. h + 1 sin. I" (p sin. h)? tan. a

where l=latitude.

p=polar distance in seconds.
h=the hour angle.
a=the altitude.

The usual method of finding the latitude is shown in the following examples.

From the observed get the true altitude, applying index error, and using Table “ Star's Total Correction.”

Then to the true altitude apply the correction from Table " Latitude by Altitude of Pole Star,” adding or subtracting, as directed by the sign plus or minus. A small further correction from Table “Correction of Latitude by Altitude of Pole Star," always additive, will give the latitude.

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