Lectures in Navigation Part 19

When the s.h.i.+p is moving, the problem is more difficult. At first thought you might imagine that all you would have to do would be to take the difference between the L.A.T. of the morning sight and 24 hours, calculate the distance the s.h.i.+p would run in this time and from that determine the longitude you would be in at noon. Then proceed as in the case of the s.h.i.+p being stationary. But such a calculation does not take into consideration the easting or westing of the s.h.i.+p itself. Suppose that at the morning sight the L.A.T. is found to be 20h-10m-30s. If the s.h.i.+p does not move, it will be 3h-49m-30s to noon. But suppose the s.h.i.+p is moving eastward. Then, in addition to the speed at which the sun is approaching the s.h.i.+p, there must be added the speed at which the s.h.i.+p is moving toward the sun--i.e. the change in longitude per hour which the s.h.i.+p is making, expressed in minutes and seconds of time. Likewise, if the s.h.i.+p is moving westward, an allowance must be made for the westing of the s.h.i.+p. And this change of longitude in minutes and seconds of time must be subtracted from the speed of the sun's approach since the s.h.i.+p, in going west, is traveling away from the sun.

There are various ways to calculate this allowance for the s.h.i.+p's speed, among the best of which is given in Bowditch, Art. 403, p. 179. Another, and even easier way, is the following, which was explained to the writer by Lieutenant Commander R.P. Strough, formerly head of the Seamans.h.i.+p Department of this School:--

1. Take the morning sight for longitude when the sun is on or as near as possible to the prime vertical.

2. Subtract the L.A.T. of the morning sight from 24 hours. This will give the total time from the morning sight to noon if the s.h.i.+p were stationary.

3. From the course to noon and speed of the s.h.i.+p, figure the change in longitude per hour in terms of seconds of time. For instance, suppose a s.h.i.+p were steaming a course of 275 at the rate of 11 knots per hour in approximately 38 North lat.i.tude. The change of longitude per hour for this speed would be 14' of arc or 56s of time.

4. Now the sun travels at the rate of 60 minutes or 3600 seconds per hour. To this hourly speed of the sun must be added or subtracted the hourly speed of the s.h.i.+p according as to whether the s.h.i.+p is going in an easterly or westerly direction. If, as mentioned above, the s.h.i.+p is steaming a course of 275 (W 1/2 N) and hence changing its longitude at the rate of 56s per hour, then the net rate of approach of the sun per hour would be 3600s - 56s, or 3544s per hour.

5. Divide the total time to noon from the L.A.T. of the morning sight (expressed in seconds of time) by the net rate of approach of the sun per hour. The result will be the corrected time to noon--i.e. the time at which the sun will be on the s.h.i.+p's meridian when the s.h.i.+p is changing its longitude to the westward at the rate of 56s per hour.

6. One more step is necessary. To the watch time of the morning sight, add the corrected time to noon. The result will be the watch time of Local Apparent Noon. Thirty minutes before will be the watch time of 11:30 A.M. and at 11:30 A.M. all deck clocks should be set to the local apparent time of the place the s.h.i.+p will be at local apparent noon.

The following example ill.u.s.trates the explanation just given and should be put in your Note Book:--

Example:--At sea, August 7th, 1919. About 7:30 A.M. by s.h.i.+p's time, position by observation just found to be Lat.i.tude 30 05' N, Longitude 58 08' W. WT of morning sight 6h-53m-13s A.M. C-W 4h-37m-21s. CC + 3m-38s. Course 275. Speed 11 knots. TZ N 90 E. What will be the Watch Time of Local Apparent Noon?

WT 6h -- 53m -- 13s A.M.

+ 12 ------------------- 18 -- 53 -- 13 C-W 4 -- 37 -- 21 ------------------- CT 23 -- 30 -- 34 CC + 3 -- 38 ------------------- G.M.T. 23 -- 34 -- 12 Eq. T. -- 5 -- 42 ------------------- G.A.T. 23 -- 28 -- 30 Lo. in T. 3 -- 52 -- 32 ------------------- L.A.T. 19 -- 35 -- 58 24 -- 00 -- 00 ------------------- Total time to Noon 4h 24m 02s

Course -- 275 Change in Lo. per hr.-- 14', 56s.

3600s -56 ----- 3544s, Net rate of approach of sun

4h 60 ---- 240m + 24 ----- 264m x 60 = 15840s

15840s + 02 ----- 15842s, Total time to noon.

3544) 15842 (4.47 hours 14176 ----- 16660 14176 ----- 24840 24808 -----

Corrected time to Noon 4h -- 28m -- 12s WT of A.M. sight 6h -- 53m -- 13s ----------------- WT of L.A.N. 11h -- 21m -- 25s WT of 11:30 A.M. 10h -- 51m -- 25s

When, therefore, the watch reads 10h--51m--25s, the deck clocks should be set to 11.30 A.M. and thirty minutes later it will be apparent noon at the s.h.i.+p.

In all these calculations it is taken for granted that the speed of the s.h.i.+p and hence the change in longitude can be gauged accurately. A check on this can be made by comparing the longitude of the A.M. sight with the D.R. longitude of the same time. Any appreciable difference between the two can be ascribed to current. Now, if a proportionate amount of current is allowed for in reckoning the speed of the s.h.i.+p from the time of the A.M. sight to noon, then a proper correction can be made in the net rate of approach of the sun and the corrected time to noon will be very close to the exact time of noon. Of course there will be an error in this calculation but it will be small and the result gained will be accurate enough for ordinary work.

So much for finding the watch time of Local Apparent Noon. Careful navigators carry the process further and get the watch times of 15, 10 and 5 minutes before noon, so that by the use of constants for each one of these times, an accurate check on the noon lat.i.tude can be quickly and easily secured. We have not time in this course to explain how these constants are worked out but it is well worth knowing. The information regarding it is in Bowditch Art. 325, p. 128, and Art. 405, p. 181.

A word about the watch used by the navigator should be included here.

This watch should be a good one and receive as much care, in its way, as the chronometer. It should be wound at the same time every day, carefully handled and, in other respects, treated like the fine time-piece that it is.

While authorities differ on this point, the best practice seems to be not to change the navigator's watch to correspond with the apparent time of each day's noon position. The reason for this is two-fold. First, because constant moving of the hands will have an injurious effect on the works of the watch, and second, because, by not changing the watch, the C-W remains approximately the same, and thus a good check can be kept on both the watch and the chronometer as well as on the navigator's figures in reckoning the times of his various sights.

a.s.sign for night reading the following Arts. in Bowditch: 323, 324, 333.

Also problems similar to the following:

1. At sea, July 28, 1919. Position by observation just found to be Lat.i.tude 44 58' N, Longitude 22 06' W. WT of morning sight 6h-02m-20s. CC 3m 34s slow. Course S 24 W. TZ N 90 E. Speed 9 knots. What will be the watch time of Local Apparent Noon?

2. At sea, August 9th, 1919. Position by observation just found to be Lat.i.tude 38 48' N, Longitude 70 46' W. WT of morning sight 8h-15m-01s A.M.

C-W 3h-56m-32s. CC 3m-43s slow. Course 272. Speed 12 knots. TZ N 90 E. What will be the watch time of Local Apparent Noon?

WEEK VII--NAVIGATION

TUESDAY LECTURE

COMPa.s.s ERROR BY AN AZIMUTH

The easiest and most accurate way to find the error of your compa.s.s is, first, to find the bearing of the sun by your pelorus. If you set your pelorus, so that it will exactly coincide with the course you are steaming as shown by the compa.s.s in your chart house and then get a bearing of the sun by noting where the shadow from the pelorus vane cuts the circ.u.mference, this bearing will be the bearing of the sun by compa.s.s. At the same time, get your true bearing of the sun from the Azimuth Tables. The difference between the two will be the compa.s.s error, marked East or West according to the following rule which put in your Note-Book:

1. Express your Compa.s.s Bearing and your True Bearing by NEW compa.s.s reading.

2. If TZ is to the right of CZ, C.E. is East. Formula: True--Right--East.

3. If TZ is to the left of CZ, C.E. is West. Formula: True--Left--West.

You must now remember that what you have is a Compa.s.s Error, consisting of both Variation and Deviation. To find the Deviation, the Variation and C.E. being given, is merely to apply the rules already given you under Dead Reckoning. For instance, if you had a C.E. of 10 W and a Variation of 4 E, the Deviation would be 14 W.

Put this example in your Note-Book:

LAT 20h 59m 57s Lat. 4 55' N Dec. 10 39' 30" N

s.h.i.+p heading N 11 W. CB of (.) S 88 E. Variation 10 W. What was the s.h.i.+p's true course and Deviation of Compa.s.s on direction s.h.i.+p was heading?

[Ill.u.s.tration]

[Ill.u.s.tration]

CZ 92 (New compa.s.s reading) TZ 80 (New compa.s.s reading) --- CE 12

CE = 12 W Variation 10 W ----- Deviation 2 W

True course being sailed N 23 W or 337.

Let us now work out some of the following examples:

1. L.A.T. 22h--14m--18s Lat. 30 29' S Dec. 17 28' 44" N s.h.i.+p heading S 84 W Compa.s.s Bearing 44 Variation 10 W.

Required T.C. and Deviation on s.h.i.+p's loading.

2. August 29th, 1919. CT 2h 29m 18s A.M. Longitude 120 19' 46" E.

Lat.i.tude 44 14' N. s.h.i.+p heading 98.