290. To convert arc (or longitude) into time.

RULE LXXVI.

Multiply the degrees, minutes, &c., by 4; this turns the degrees (°) into minutes (") of time, minutes (') into seconds (*) of time, and the seconds (") into thirds (t) of time; or in other words, mark the resulting figures thus:-Those under seconds (") thirds (†), those under minutes (') seconds ($), those under degrees (°) minutes (TM), and those to the left of the latter, hours (1).

NOTE.-Instead of thirds it is customary to use tenths of seconds, in which case the thirds must be reduced to tenths by dividing by 60 (see Rule XVI, page 54).

Four times 15" are 60", which contains 60 once and o over; write this remainder down under the seconds (") and mark it thirds (t) as directed in the Rule, carrying the r. Again, 4 times 18' are 72, and the 1' carried makes 73; 60 goes in 73 once and 13 over; write this remainder (13) under the minutes (') and call them seconds (•) and carry the 1. Again, 4 times 12 are 48, and 1 carried makes 49; write this under degrees () and mark it minutes (m): whence the time corresponding to arc 12° 18′ 15′′ is 49m 13′′ot.

Ex. 3. Turn 77° 2' 10" into time.

Convert 25° 15′ 16′′ into time.
25° 15′ 16′′
4

1h 41m 18 4t

Four times 16" are 64", which contains 60 once and 4 over, and according to Rule this remainder placed under seconds (") becomes thirds (t), and the 1 is to be carried. Again, four times 15' are 60 and I carried makes 61, which contains 60 once and 1 over; write the remainder 1 under minutes ('), and carry 1: four times 25 are 100 and 1 carried gives 101, and 60 into 101 goes once and 41 remainder, which remainder being placed under degrees (2) gives minutes (m) and the 1 carried on being placed to the left of the latter is marked hours (h); whence 1h 41m 14t is the time corresponding to the arc 25° 15′ 16′′.

What time corresponds to

77° 2' 10"
4

Ex. 4.
127° 32′ 40"?

6.

84° 42′ 30′′; and o° 34'.

0° 13′′5; 51° 10′ 12′′; 156° 52′: 178° 49′ 45′′; and o° 41'7.

TO CONVERT TIME INTO LONGITUDE.

m

291. It has been shown (No. 289, page 224) that 4" of time are equivalent to 1° of arc; hence it is evident that if we bring any given time into minutes, and divide by 4, we shall have the corresponding arc in degrees, minutes, and seconds. This is the reverse of the last process.

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RULE LXXVII.

Reduce the hours and minutes into minutes, after which place the seconds, &c., then divide all by 4, and the quotient will be the degrees, minutes, &c., of the corresponding are; or, in other words, after dividing by 4, mark the resulting figures thus:-Those under minutes (TM) degrees (°), those under seconds (*) minutes (), those under thirds (t) seconds (").

Multiply the hours (8h) by 60, and adding the minutes (17) to the product gives 497; divide the result by 4, the quotient is 124°, with remainder 1. Again, multiply the remainder just obtained (10) by 60, and to the product add the seconds of time, viz., 35, the sum is 95; then divided by 4, the quotient is 23' (minutes of arc) with remainder 3. Next multiply this last remainder by 60, the product is 180, to which add the 3ot; and the sum 210, divided by 4, gives 52" of arc, and remainder 2, to which annex a cypher and divide by 4, the quotient is 5 of seconds

of aro.

Multiplying 6h by 60, and adding the 24 to the product, gives 384 as the sum; the quotient of this, divided by 4, is 96°, with no remainder. 43 divided by 4 gives quotient 10' with remainder 3: remainder 3 multiplied by 60 gives 180, which divided by 4 gives quotient 45": therefore, 96° 10' 45" is the are which corresponds to 6h 24m 43.

Ex. 4. What is the equivalent arc to 9h 25m 37°?

9h 25m 37°
60

4)565 37° ot

141° 24′ 15′′

Ex. 6. Convert 11h 39m 50.7 into arc.
11h 39m 508*7
60

4)699m 50o 42t

174° 57' 40"*5

ཅི|o;

Multiply 11 by 60, and to product 660 add 39", dividing the sum, viz., 699 by 4 gives 1740, with remainder 3; this remainder (3) multiplied by 60, and 50 added to product gives 230; this sum divided by 4 gives 57, with remainder 2; remainder 2 multiplied by 60, and 42 added, gives sum 162, which divided by 4 gives 40", and remainder 2; remainder 2, with a cypher annexed and divided by 4, gives quotient 5; whence the aro corresponding to 11h 39m 50*7 is 174° 57' 40" 5.

GREENWICH DATE.

REDUCTION OF GREENWICH DATE.

292. Def.-The Greenwich Date is the day and time (reckoned astronomically) at Greenwich corresponding to a given day and time elsewhere. It is necessary to find the Greenwich date before the information contained in the Nautical Almanac can be made available, because all the elements there tabulated are given for time at the meridian of Greenwich.

As in almost every computation of nautical astronomy we are dependent for some data upon the Nautical Almanac—and these are commonly given for Greenwich-it is generally the first step in such a computation to deduce an exact or, at least, an approximate value of the Greenwich astronomical time. It need hardly be added that the Greenwich time should never be otherwise expressed than astronomically.

The Greenwich Date is found at once from a chronometer, the error and the rate of which is known; but it can also be found by means of the approximate time at place and the approximate longitude.

293. To find the Greenwich Date, the time at any other place and the longitude being given.

RULE LXXVIII.

1o. Express the ship time astronomically (Rule LXXIV, page 223). 2°. Convert the longitude into time (Rule LXXVI, page 225).

3°. In West longitude.—ADD longitude in time to ship time; the sum, if less than 24 hours, is the corresponding Greenwich date on the same day with the ship date; if greater than 24 hours, reject the 24 hours, and put the day one forward.

4°. In East longitude.-From ship astronomical time SUBTRACT longitude in time, if less than the hours, minutes, &c., of the ship date; the remainder is the corresponding Greenwich date of the same day as the ship date; if the longitude in time be greater than the hours, minutes, &c., of ship astronomical, ADD 24 hours to the latter, and put the day one back before the subtraction is made.

5o. When it is noon at the place. The longitude in time, if West, is the Greenwich date (apparent time); but if East, SUBTRACT the longitude in time from 24 hours; the remainder is the Greenwich date (apparent time) after noon of the preceding day.

(a) From this last it is evident that when the sun is on a meridian in West longitude, the Greenwich apparent time is precisely equal to the longitude, that is, the Greenwich apparent time is after the noon of the same date with the ship date, by a number of hours, &c., equal to longitude. When the sun is on a meridian in East longitude, the Greenwich apparent time is before the noon of the same date as the ship date, by a number of hours equal to the longitude in time.

NOTE.-A bad habit prevails in writing dates, of separating the month and day from the hours, minutes, and seconds. The day of the month should always precede the minor divisions of time which give the precise instant of the day intended.

EXAMPLES.

Ex. 1. November 9th, at 4h 10m P.M., apparent time at ship, longitude 32° 45′ W. : required the corresponding time at Greenwich, or the Greenwich date.

In Ex. 4, the added longitude advances the day of the month. (This illustrates latter part of 3° of the Rule). In Ex. 5, a day (or 24 hours) is borrowed before the subtraction is made, since the longitude in time exceeds the astronomical ship date, thus making the days of the month at Greenwich one less than at the place. (This illustrates the latter part of 4° of the Rule).

In Ex. 10, the hours, &c., of longitude to be subtracted are to be taken from a borrowed day, or 24 hours, thus making the day of the month at Greenwich one less than at the place. (See 5° of Rule).

Required the Greenwich date in each of the following examples:

294. The Nautical Almanac* or Astronomical Ephemeris contains the right ascension, declination, &c., of the principal heavenly bodies for given equidistant instants of Greenwich time; the right ascension and declination of the sun and planets, for example, being given for every day at noon (o1 oTM o3) at Greenwich, while for the moon these elements are given for every hour. Before we can find from the Almanac the value of any of these quantities for a given local or ship time, we must find the corresponding Greenwich date (Rule LXXVIII, page 227). Where this time is exactly one of the instants for which the required quantity is put down in the Ephemeris, nothing more is necessary than to transcribe the quantity as there put down. But when, as is mostly the case, the time falls between two of the times in the Ephemeris, we must obtain the required quantity by interpolation, it being requisite to apply a correction to that taken from the Almanac, in order to reduce it to its value at the given instant. To facilitate this interpolation the Almanac

The French Ephemeris, La Connaissance des Temps, is computed for the meridian of Paris, the German Beliner Astronomisches Jahrbuen for the meridian of Berlin. All these works are published annually several years in advance.