Right Ascension is practically celestial longitude. For instance, the
position of a place on the earth is fixed by its latitude and longitude;
the position of a heavenly body is fixed by its declination and right
ascension. But Right Ascension is not measured in degrees and minutes
nor is it measured East and West. It is reckoned in hours and minutes
all the way around the sky, eastward from a certain point, through the
approximate 24 hours. The point from which this celestial longitude
begins is not at Greenwich, but the point where the celestial equator
intersects the ecliptic in the spring of the year, i.e., the point where
the sun, coming North in the Spring, crosses the celestial equator. This
point is called the First Point of Aries. You will frequently hear me
speak of a star having, for instance, a Right Ascension of 5h 16m 32s. I
mean by that, that starting at the celestial meridian, i.e., the
meridian passing through the First Point of Aries, it will take a spot
on the earth 5h 16m 32s to travel until it reaches the meridian of the
star in question.
Roughly speaking then, just as Greenwich Apparent Time means the
distance East or West the Greenwich meridian is from the sun and Local
Apparent Time means the distance East or West your ship is from the sun,
so R.A.M.G. means the distance in time the Meridian of Greenwich is from
the First Point of Aries, measured eastward in a circle. And this
distance is the same as Greenwich Sidereal Time, i.e., Sidereal Time at
Greenwich or the distance in time the meridian of Greenwich is from the
First Point of Aries.
Now, what is the star time that corresponds to local time? It is called
the Right Ascension of the Meridian, which means the R. A. of the
meridian which intersects your zenith. Just as L.A.T. is the distance in
time your meridian is from the sun, so Local Sidereal Time is the R. A.
of your meridian, i.e., the distance in time your meridian is from the
First Point of Aries. Put in your Note-Book:
G.S.T. and R.A.M.G. are one and the same thing.
L.S.T. and R.A.M. are one and the same thing.
G.M.T. + (.).R.A. + (+).C.P. = G.S.T. (R.A.M.G.) If the result is
more than 24 hours, subtract 24 hours.
G.S.T. - (.).R.A. - (+).C.P. = G.M.T.
G.S.T. - W.Lo. = L.S.T.
+ E.Lo.
L.S.T. + W.Lo. = G.S.T.
- E.Lo.
I can explain all these formulas very easily by the following
illustration which put in your Note-Book: (Note to Instructor: If
possible have copies of this illustration mimeographed and distributed
to each student.)
[Illustration: FIRST POINT OF ARIES.]
There is one term I have used which does not appear in the illustration.
It is the Earth's Central Progress ((+).C.P.). The astronomical day
based on the sun, is 24 hours long, as said before. The sidereal day,
however, is only 23h 56m 04s long. This is due to the fact that whereas
the earth is moving in its ecliptic track around the sun while revolving
on its own axis, the First Point of Aries is a fixed point and hence
never moves. The correction, then, for the difference in the length of
time between a sidereal day and a mean solar day is called the Earth's
Central Progress and, of course, has to be figured for all amounts of
time after mean noon at Greenwich, since the Sun's Right Ascension
tables in the Nautical Almanac are based on time at mean noon at
Greenwich.
Now you have a formula for practically all kinds of conversion except
for converting L.M.T. into L.S.T. You could do it by the formula
L.M.T. + W.Lo. = G.M.T. + (.).R.A. + (+).C.P. = G.S.T. - W.Lo. = L.S.T.
- E.Lo. + E.Lo.
But that involves too many operations.
A shorter way, though not so simple perhaps, is as follows: L.M.T. +
Reduction page 2 N.A. for time after local mean noon + (.).R.A. of
Greenwich mean noon +- Reduction page 2 N.A. for Lo. in T. (W+, E-) =
L.S.T.
Note to Instructor:
Explain this formula by turning to page 107 N.A. and work it out by the
formula L.M.T. + Lo. in T (W) = G.M.T. + (.).R.A. + (+).C.P. = G.S.T. -
Lo. in T (W) = L.S.T. Example:
L.M.T. 10h--40m--30s
Lo. in T 4 --56 W +
-------------
G.M.T. 15 --36 --30
(.).R.A. 5 --11 --10
(+).C.P. -- 2 --34
-------------
G.S.T. 20 --50 --14
Lo. W - 4 --56
-------------
L.S.T. 15h--54m--14s
Now Bowditch gets this L.S.T. in still another way. Turn to page 110,
Article 290. There the formula used is L.M.T. + (.).R.A. + (+).C.P. =
L.S.T, and in order to get the correct (.).R.A. and (+).C.P. the G.M.T.
has to be secured by the formula
L.M.T. + W.Lo. = G.M.T.
- E.Lo.
Let us work this same example in Bowditch by the other two methods.
First by the formula
L.M.T. + W.Lo. = G.M.T. + (.).R.A. + (+).C.P. = G.S.T. - W.Lo. = L.S.T.
- E.Lo. + E.Lo.
L.M.T. 22d-- 2h--00m--00s
+ W. Lo. 5 --25
----------------------
G.M.T. 22d-- 7h--25m--00s
(.).R.A. 1 --57 --59
(+).C.P. 1 --13
----------------------
G.S.T. 22d-- 9h--24m--12s
- W. Lo. 5 --25
----------------------
L.S.T. 22d-- 3h--59m--12s
The small difference between this answer and that of Bowditch's is that
the (.).R.A. for 1916 is slightly different from that of 1919. Bowditch
used the 1916 Almanac, whereas we are working from the 1919 Almanac. Now
turn to page 107 of the N.A. and let us work the same example in
Bowditch by the method used here:
L.M.T. 2h - 00m - 00s
Red. for 2h 0 - 20
(.).R.A. 0h 1 - 57 - 59
Red. Lo. 5h - 25m 0 - 53
---------------
L.S.T. 3h - 59m - 12s
The reason I am going so much into detail in explaining methods of
finding L.S.T. is because, by a very simple calculation which will be
explained later, we can get our latitude at night if we know the
altitude of Polaris (The North Star) and if we know the L.S.T. at the
time of observation. Some of you may think that the N.A. way is the
simplest. It is given in the N.A., and in an examination it would be
permissible for you to use the N.A. as a guide because, in an
examination, I propose to let you have at hand the same books you would
have in the chart house of a ship. On the other hand, the method given
in the N.A. is not as clear to my mind as the method which starts with
L.M.T., then finds with the Longitude the G.M.T. That gives you, roughly
speaking, the distance in time Greenwich is from the sun. Add to that
the sun's R.A. or the distance in time the sun is from the First Point
of Aries at Greenwich Mean Noon. Add to that the correction for the time
past noon. The result is G.S.T. Now all you have to do is to apply the
longitude correctly to find the L.S.T., just as when you have G.M.T. and
apply the longitude correctly you get L.M.T. That is a method which does
not seem easy to forget, for it depends more upon simple reasoning where
the others, for a beginner, depend more upon memory. However, any of the
three methods is correct and can be used by you. Perhaps the best way is
to work a problem by two of the three that seem easiest. In this way you
can check your figures. When I give you a problem that involves finding
the L.S.T. I do not care how you get the L.S.T. providing it is correct
when you get it.
Assign for Night Reading in Bowditch the following Arts.:
282-283-284-285. Also the following questions:
1. Given the G.M.T. and the longitude in T which is W, what is the
formula for L.S.T.?
2. Given the L.A.T. and longitude in T which is E, what is the formula
for G.S.T.?
3. Given the L.S.T. and longitude in T which is W. Required G.M.T. Etc.
FRIDAY LECTURE
THE NAUTICAL ALMANAC
For the last two days we have been discussing Time--sun time or solar
time and star time or sidereal time. Now let us examine the Nautical
Almanac to see how that time is registered and how we read the various
kinds of time for any instant of the day or night. Before starting in,
put a large cross on pages 4 and 5. For any calculations you are going
to make, these pages are unnecessary and they are liable to lead to
confusion.
Sun time of the mean sun at Greenwich is given for every minute of the
day in the year 1919 in the pages from 6 to 30. This is indicated by the
column to the left headed G.M.T. Turn to page 6 under Wednesday, Jan.
1st. You can see that the even hours are given from 0 to 24. Remember
that these are expressed in astronomical time, so that if you had Jan.
2nd--10 hours A.M., you would not look in the column under Jan. 2nd but
under the column for Jan. 1st, 22 hours, since 10 A.M. Jan. 2nd is 22
o'clock Jan. 1st, and no reading is used in this Almanac except a
reading expressed in astronomical time. Now at the bottom of the column
under Jan. 1st you see the letters H.D. That stands for "hourly
difference" and represents the amount to be added or subtracted for an
odd hour from the nearest even hour. In this instance it is .2. You note
that even hours 2, 4, 6, etc., are given. To find an odd hour during
this astronomical day, subtract .2 from the preceding even hour. For any
fraction of an hour you simply take the corresponding fraction of the
H.D. and subtract it from the preceding even hour. For instance, the
declination for Jan. 1st--12 hours would be 23 deg. 1.8' or 23 deg.--1'--48", 13
hours would be 23 deg. 1.6' or 23 deg.--1'--36", 12-1/2 hours would be 23 deg. 1.7'
or 23 deg.--1'--42", and 13-1/2 hours would be 23 deg. 1.5' or 23 deg.--1'--30".
Now to the right of the hours you note there is given the corresponding
amount of Declination and the Equation of Time. Before going further,
let us review a few facts about Declination. The declination of a
celestial body is its angular distance N or S of the celestial equator
or equinoctial. Now get clearly in your mind how we measure the angular
distance from the celestial equator of any heavenly body. It is measured
by the angle one of whose sides is an imaginary line drawn to the center
of the earth and the other of whose sides is an imaginary line passing
from the center of the earth into the celestial sphere through the
center of the heavenly body whose declination you desire. Now as you
stand on any part of the earth, you are standing at right angles to the
earth itself. Hence if this imaginary line passed through you it would
intersect the celestial sphere at your zenith, i.e., the point in the
celestial sphere which is directly above you. Now suppose you happen to
be standing at a certain point on the earth and suppose that point was
in 15 deg. N latitude. And suppose at noon the center of the sun was
directly over you, i.e., the center of the sun and your zenith were one
and the same point. Then the declination of the sun at that moment would
be 15 deg. N. In other words, your angular distance from the earth's equator
(which is another way of expressing your latitude) would be precisely
the same as the angular distance of the center of the sun from the
celestial equator. Suppose you were standing directly on the equator
and the center of the sun was directly over you, then the declination of
the sun would be 0 deg.. Now if the axis of the earth were always
perpendicular to the plane of the sun's orbit, then the sun would always
be immediately over the equator and the sun's declination would always
be 0 deg.. But you know that the axis of the earth is inclined to the plane
of the sun's orbit. As the earth, then, revolves around the sun, the
amount of the declination increases and then decreases according to the
location of the earth at any one time with relation to the sun. On March
21st and Sept. 23rd, 1919, the sun is directly over the equator and the
declination is 0 deg.. From March 21st to June 21st the sun is coming North
and the declination is increasing until on June 21st--12 hours--it
reaches its highest declination. From then on the sun starts to travel
South, crosses the equator on Sept. 23d and reaches its highest
declination in South latitude on Dec. 22nd, when it starts to come North
again. This explains easily the length of days. When the sun is in North
latitude, it is nearer our zenith, i.e., higher in the heavens. It can,
therefore, be seen for a longer time during the 24 hours that it takes
the earth to revolve on its axis. Hence, when the sun reaches its
highest declination in North latitude--June 21st--i.e., when it is
farthest North from the equator and nearest our zenith (which is in 40 deg.
N latitude) it can be seen for the longest length of time. In other
words, that day is the longest of the year. For the same reason, Dec.
22nd, when the sun reaches its highest declination in South latitude,
i.e., when it is farthest away to the South, is the shortest day in the
year for us; for on that day, the sun being farthest away from our
zenith and hence lowest down toward the horizon, can be seen for the
shortest length of time.
Put in your Note-Book:
North Declination is expressed +.
South Declination is expressed ---.
Now turn to page 6 of the Nautical Almanac. You will see opposite Jan.
1st 0h, a declination of --23 deg. 4.2'. Every calculation in this Almanac is
based on time at Greenwich, i.e., G.M.T. So at 0h Jan. 1st at
Greenwich--that is at noon--the Sun's declination is S 23 deg. 4.2'.
You learned in the lecture the other day on solar time, that the
difference between mean time and apparent time was called the equation
of time. This equation of time, with the sign showing in which way it is
to be applied, is given for any minute of any day in the column marked
"Equation of Time." You will also notice that there is an H.D. for
equations of time just as there is for each declination, and this H.D.
should be used when finding the equation of time for an odd hour.
Put in your Note-Book:
1. The equation of time is to be applied as given in the Nautical
Almanac when changing Mean Time into Apparent Time.
2. When changing Apparent Time into Mean Time, reverse the sign as given
in the Nautical Almanac.
That is all there is to finding sun time, either mean or apparent, for
any instant of any day in the year 1919. Do not forget, however, that
all this data is based upon Greenwich Mean Time. To find Local Mean Time
you must apply the Longitude you are in. To find Local Apparent Time you
must first secure G.A.T. from G.M.T. and then apply the Longitude.
(Note to Instructor: Make the class work out conversions here if you
have time to do so and can finish the rest of the lecture by the end of
the period.)
So much for time by the sun. Now let us examine time by the
stars--sidereal time. Turn to pages 2-3. There you find the Right
Ascension of the Mean Sun at Greenwich Mean Noon for every day in the
year. You remember that, roughly speaking, the Sun's Right Ascension was
the distance in time the sun was from the First Point of Aries. So these
tables give that distance (expressed in time) for noon at Greenwich of
every day. For the correction to be applied for all time after noon at
Greenwich (i.e., (+).C.P.), use the table at the bottom of the page. For
instance, the (.).R.A. at Greenwich 9h 24m on Jan. 1st would be
(.).R.A. 18h--40m--21s
(+).C.P. 1 33
--------------
18h--41m--54s
Now we must go back to some of the formulas we learned when discussing
star time and apply them with the information we now have from the
Nautical Almanac. If the G.M.T. on April 20th is 4h--16m--30s, what is
the G.S.T. for the same moment? That is, when Greenwich is 4h--16m--30s
from the sun, how far is Greenwich from the First Point of Aries? You
remember the formula was G.S.T. = G.M.T. + (.).R.A. + (+).C.P.
G.M.T. 4h--16m--30s
(.).R.A. 1 --50 -- 6
(+).C.P. 0--42
--------------
G.S.T. 6h--07m--18s
Suppose you were in Lo. 74 deg. W. What would be the R.A.M. (L.S.T.)? You
remember the formula for L.S.T. from G.S.T. was the same relatively as
L.M.T. from G.M.T., i.e.,
L.S.T. = G.S.T. - W. Lo.
+ E. Lo,
Here it would be
G.S.T. 6h--07m--18s
(74 deg. W) - 4 --56 --00
---------------
L.S.T. 1h--11m--18s
Now these are not a collection of abstruse formulas that you are
learning just for the sake of practice. They are used every clear night
on board ship, or should be, and are just as vital to know as time by
the sun.
Suppose you are at sea in Lo. 70 deg. W and your CT is October 20th
6h--4m--30s A.M., CC 2m--30s fast. You wish to get the R.A. of your M,
i.e., the L.S.T. How would you go about it? The first thing to do would
be to get your G.M.T. It is CT--CC.
20d--06h--04m--30s A.M.
--12
------------------
CT 19d--18h--04m--30s
CC --02 --30
------------------
G.M.T. 19d--18h--02m--00s
Then get your G.S.T.
Oct. 19d--18h--02m--00s
(.).R.A. 13 --47 --38.5
(+).C.P 2 --57.7
--------------------
19d--31h--52m--36.2s
--24
--------------------
G.S.T. 19d-- 7h--52m--36.2s
Then get your L.S.T.
G.S.T. 7h--52m--36.2s
W.Lo (--) 4 --40
--------------
L.S.T. 3h--12m--36.2s
The last fact to know at this time about the Almanac is found on pages
94-95. Here is given a list of the brighter stars with their positions
respectively in the heavens, i.e., their celestial longitude or R.A. on
page 94 and their celestial latitude or declination on page 95. These
stars have very little apparent motion. They are practically fixed.
Hence, their position in the heavens is almost the same from January to
December though, of course, their position with relation to you is
constantly changing, since you on the earth are constantly moving.
The relationship between these various kinds of time is clearly
expressed by the following diagram, which put in your Note Book:
[Illustration: Going with arrow, add. Going against arrow, subtract.]
Assign for reading in Bowditch, Articles
294-295-296-297-299-300-301-302-303-304-305-306-307.
If any time is left, have the class work out such examples as these:
1. G.M.T. June 20th, 1919, 5h--14m--39s. In Lo. 68 deg. 49' W. Required
L.S.T., G.S.T., L.M.T., L.A.T.
2. L.M.T. Oct. 15th, 1919, 6h--30m--20s A.M. In Lo. 49 deg. 35' 16" E.
Required L.S.T.
3. L.M.T. May 14th, 1919, 10h--15m.--20s A.M. Lo. 56 deg. 21' 39" W.
Required L.A.T.
4. W.T. April 20th, 1919, 11h--30m--14s C-W 2h--14m--59s CC 4m--30s
slow. In Lo. 89 deg. 48' 30" W. Required G.M.T., G.A.T., L.M.T., L.A.T.,
G.S.T., L.S.T.
5. What is Declination and R.A. on May 15th, 1919, of Polaris, Arcturus,
Capella, Regulus, Altair, Deneb, Vega, Aldebaran?
6. What is the sun's declination and R.A., Time at Greenwich, July 30th:
7h--14m--39s A.M.
4h--29m--14s A.M.
3h--04m--06s
11h--49m--59s
2h--14m--30s A.M.?
SATURDAY LECTURE
CORRECTION OF OBSERVED ALTITUDES
The true altitude of a heavenly body is the angular distance of its
center as measured from the center of the earth. The observed altitude
of a heavenly body as seen at sea by the sextant may be converted to the
true altitude by the application of the following four corrections: Dip,
Refraction, Parallax and Semi-diameter.
Dip of the horizon means an increase in the altitude caused by the
elevation of the eye above the level of the sea. The following diagram
illustrates this clearly:
[Illustration]
If the eye is on the level of the sea at A, it is in the plane of the
horizon CD, and the angles EAC and EAD are right angles or 90 deg. each. If
the eye is elevated above A, say to B, it is plain that the angles EBC
and EBD are greater than right angles, or in other words, that the
observer sees more than a semi-circle of sky. Hence all measurements
made by the sextant are too large. In other words, the elevation of the
eye makes the angle too great and therefore the correction for dip is
always subtracted.
Refraction is a curving of the rays of light caused by their entering
the earth's atmosphere, which is a denser medium than the very light
ether of the outer sky. The effect of refraction is seen when an oar is
thrust into the water and looks as if it were bent. Refraction always
causes a celestial object to appear higher than it really is. This
refraction is greatest at the horizon and diminishes toward the zenith,
where it disappears. Table 20A in Bowditch gives the correction for mean
refraction. It is always subtracted from the altitude. In the higher
altitudes, select the correction for the nearest degree.
You should avoid taking low altitudes (15 deg. or less) when the atmosphere
is not perfectly clear. Haziness increases refraction.
Parallax is simply the difference in angular altitude of a heavenly body
as measured from the center of the earth and as measured from the
corresponding point on the surface of the earth. Parallax is greatest
when the body is in the horizon, and disappears when it is at the
zenith.
[Illustration]
When the angular altitude of the sun in this diagram is 0, the parallax
ABC is greatest. When the altitude is highest there is no parallax. The
sun is so far away that its parallax never exceeds 9". The stars have
practically none at all from the earth's surface. Parallax is always to
be added in the case of the sun.
The semi-diameter of a heavenly body is half the angle subtended by the
diameter of the visible disk at the eye of the observer. For the same
body, the SD varies with the distance. Thus, the difference of the sun's
SD at different times of the year is due to the change of the earth's
distance from the sun.
[Illustration]
The SD is to be added to the observed altitude in case the lower limb is
brought in contact with the horizon, and subtracted if the upper limb is
used. Probably most of the sights you take will be of the sun's lower
limb, i.e., when the lower limb is brought in contact with the horizon,
so all you need to remember is that in that event the SD is additive.
Now at first we will correct altitudes by applying each correction
separately, but as soon as you get the idea, there is a short way to
apply all four corrections at once. This is done in Table 46. However,
disregard that for the moment. Put this in your Note-Book:
Dip is -. Table 14 Bowditch
Refraction is -. " 20 A Bowditch
Parallax is +. " 16 Bowditch
S.D. is +. Nautical Almanac
Observed altitude of Sun's lower limb is expressed (_).
True altitude is expressed -(-)-.
Remember that before an observation is at all accurate, it must be
corrected to make it a true altitude. Remember also that the IE must be
applied, in addition to these other corrections, in order to make the
observed altitude a -(-)- altitude. So there are really five corrections
to make instead of four, providing, of course, your sextant has an IE.
Examples:
1. June 20th, 1919, observed altitude of (_) 69 deg. 25' 30". IE + 2' 30".
HE 16 ft. Required -(-)-.
2. April 15th, 1919, observed altitude of (_) 58 deg. 29' 40". IE - 2' 30".
HE 18 ft. Required -(-)-.
3. March 4th, 1919, observed altitude of (_) 44 deg. 44' 10". IE - 4' 20".
HE 20 ft. Required -(-)-.
Etc.
WEEK IV--NAVIGATION
TUESDAY LECTURE
THE LINE OF POSITION
It is practically impossible to fix your position exactly by one
observation of any celestial body. The most you can expect from one
sight is to fix your line of position, i.e., the line somewhere along
which you are. If, for instance, you can get a sight by sextant of the
sun, you may be able to work out from this sight a very accurate
calculation of what your latitude is. Say it is 50 deg. N. You are
practically certain, then, that you are somewhere in latitude 50 deg. N, but
just where you are you cannot tell until you get another sight for your
longitude. Similarly, you may be able to fix your longitude, but not be
able to fix your latitude until another sight is made. Celestial
Navigation, then, reduces itself to securing lines of position and by
manipulating these lines of position in a way to be described later, so
that they intersect. If, for instance, you know you are on one line
running North and South and on another line running East and West, the
only spot where you _can_ be on _both_ lines is where they intersect.
This diagram will make that clear:
[Illustration]
[Illustration]
Just what a line of position is will now be explained. Wherever the sun
is, it must be perpendicularly above the same spot on the surface of the
earth marked in the accompanying diagram by S and suppose a circle be
drawn around this spot as ABCDE. Then if a man at A takes an altitude,
he will get precisely the same one as men at B, C, D, and E, because
they are all at equal distances from the sun, and hence on the
circumference of a circle whose center is S. Conversely, if several
observers situated at different parts of the earth's surface take
simultaneous altitudes, and these altitudes are all the same, then the
observers must all be on the circumference of a circle and _only one_
circle. If they are not on that circle, the altitude they take will be
greater or less than the one in question.
[Illustration]