practically nothing is actually known in regard to conditions on the
surface of this planet. We do know, however, that Mars more nearly
resembles a miniature of our earth than any other celestial body. The
diameter of Mars is 4,210 miles--almost exactly half the earth's
diameter. The surface area of Mars is just about equal to the total area
of dry land on the earth. Like the earth, Mars rotates about an axis
inclined to the plane of its orbit, and the length of a Martian day is
very nearly equal to our own. The latest determinations give the length
of a Martian solar day as 24h 39m 35s. Fortunately for us, Mars is
surrounded by a very light and transparent atmosphere through which we
are able to discover with our telescopes, many permanent facts.
The most noticeable of these are the dazzling white "polar caps" first
identified by Sir William Herschel in 1784. During the long winter in
the northern hemisphere, the cap at the North pole steadily increases in
size, only to diminish during the next summer under the hot rays of the
sun. These discoveries establish without doubt the presence of vapors in
the Martian atmosphere which precipitate with cold and evaporate with
heat. The polar caps, then, are some form of snow and ice or possible
hoar frost. Outside the polar caps the surface of Mars is rough, uneven
and of different colors. Some of the darker markings appear to be long,
straight hollows. They are the so-called "canals" discovered by
Schiaparelli in 1877. The term "canal" is an unfortunate one. The word
implies the existence of water and the presence of beings of sufficient
intelligence and mechanical ability to construct elaborate works.
Flammarion in France and Lowell in the United States claim the word is
correctly used, i.e., that these markings are really canals and that
Mars is actually inhabited. The consensus of opinion among the most
celebrated astronomers is contrary to this view. Most astronomers agree
that these canals may not exist as drawn--that they are to great extent
due to defective vision. There is no conclusive proof of man-made work
on Mars, nor of the existence of conscious life of any kind. It may be
there but conclusive proof of it is still lacking.
_2. The Stars_
The planets are often called wanderers in the sky because of their ever
changing position. Sharply distinguished from them, therefore, are the
"fixed" stars. These appear as mere points of light and always maintain
the same relative positions in the heavens. Thousands of years ago the
"Great Dipper" hung in the northern sky just as it will hang tonight and
as it will hang for thousands of years to come. Yet these bodies are not
actually fixed in space. In reality they are all in rapid motion, some
moving one way and some another. It is their tremendous distance from us
that makes this motion inappreciable. The sun seems far away from us,
but the nearest star is 200,000 times as far away from us as is the sun.
Expressed in miles, the figure is so huge as to be incomprehensible. A
special unit has, therefore, been invented--a unit represented by the
distance traversed by light in one year. In one second, light travels
over 186,000 miles. In 8-1/3 minutes, light reaches us from the sun and,
in doing so, covers the distance that would take the Vaterland over four
centuries to travel. Yet the nearest star is over four "light years"
distant--it is so far away that it requires over four years for its
light to reach us. When you look at the stars tonight you see them, not
as they are, but as they were, even centuries ago. Polaris, for
instance, is distant some sixty "light years." Had it disappeared from
the heavens at the time Lee surrendered to Grant, we should still be
seeing it and entirely unaware of its disappearance.
Now each star in the heavens is in reality a sun, i.e., a vast globe of
gas and vapor, intensely hot and in a continuous state of violent
agitation, radiating forth heat and light, every pulsation of which is
felt throughout the universe. So closely indeed do many of the stars
resemble the sun, that the light which they emit cannot be distinguished
from sunlight. Some of them are larger and hotter than the sun--some
smaller and cooler. Yet the sun we see can be regarded as a typical star
and from our knowledge of it we can form a fairly correct idea of the
nature and characteristics of these other stars.
Anyone knows that the stars vary in brightness. Some of this variation
is due partly to actual differences among the stars themselves and
partly to varying distances. If all the stars were alike, then those
which were farthest away would be faintest and we could judge a star's
distance by its brilliancy. This is not the case, however. Some of the
more brilliant stars are far more distant than some of the fainter ones.
There are stars near and remote and an apparently faint star may in
reality be larger and more brilliant than a star of the first magnitude.
Vega, for instance, is infinitely farther away from us than the sun, yet
its brightness is more than 50 times that of the sun. Polaris, still
farther away, has 100 times the light and heat of the sun. In fact the
sun, considered as a star, is relatively small and feeble.
_3. Identification of Stars_
Only the brighter stars can be used in navigation. So much light is lost
in the double reflection in the mirrors of the sextant, that stars
fainter than the third magnitude can seldom be observed. This reduces
the number of stars available for navigation to within very narrow
limits, for there are only 142 stars all told which are of the third
magnitude or brighter. The Nautical Almanac gives a list of some 150
stars which may be used, but as a matter of fact, the list might be
reduced to some 50 or 60 without serious detriment to the practical
navigator. About 30 of these are of the second magnitude or greater and
hence easily found. It is not difficult to learn to know 30 or 40 of the
brighter stars, so that they can be recognized at any time. To aid in
locating the stars, many different star charts and atlases have been
published, but most of them are so elaborate that they confuse as much
as they help. The simpler the chart, the fewer stars it pretends to
locate, the better for practical purposes. Also, all charts are of
necessity printed on a flat surface and such a surface can never
represent in their true values, all parts of a sphere. A chart,
therefore, which covers a large part of the heavens, is bound to give a
distorted idea of distances or directions in some part of the sky and
must be used with caution.
There are a few stars which form striking figures of one kind or
another. These can always be easily located and form a starting point,
so to speak, from which to begin a search for other stars. Of these
groups the Great Dipper is the most prominent in the northern sky and
beginning with this the other constellations can be located one by one.
When the groups or constellations are not known, then any individual
star can be readily found by means of its Right Ascension, and
Declination. As you have already learned, Declination is equivalent to
latitude on the earth and Right Ascension practically equivalent to
longitude on the earth, except that whereas longitude on the earth is
measured E. and W. from Greenwich, Right Ascension is measured to the
east all the way around the sky, from the First Point of Aries. With
this in mind, you can easily see that if a star's R.A. is less than
yours, i.e., less than L.S.T. or the R.A. of your Meridian, the star is
not as far eastward in the heavens, as is your zenith. In other words it
is to the west of you. And vice versa, if the Star's R.A. is greater
than yours, the Star is more to the eastward than you and hence to the
east of you. Moreover, as R.A. is reckoned all around a circle and in
hours, each hour's difference between the Star's R.A. and yours is 1/24
of 360 deg. or 15 deg.. Hence if a star's R.A. is, for instance, 2 hours greater
than yours, the star will be found to the east of your meridian and
approximately 30 deg. from your meridian, providing the star is in
approximately the same vertical east and west plane as is your zenith.
When the general east or west direction of any star has been determined,
its north or south position can at once be found from its declination.
If you are in Latitude 40 deg. N. your celestial horizon to the South will
be 90 deg. from 40 deg. N. or 50 deg. S. and to the North it will be 90 deg. + 40 deg. N. =
130 deg. or 40 deg. N. (below the N. pole). The general position of the equator
in the sky is always readily found according to the latitude you are in.
If you are in 40 deg. N. latitude, the celestial equator would intersect the
celestial sphere at a point 40 deg. South of you. Knowing this, the angular
distance of a star North or South of the equator (which is its
declination) should be easily found. Remember, however, that the equator
in the sky is a curved line and hence a star in the East or West which
looks to be slightly North of you may actually be South of you.
Put in your Note-Book:
If the star is west of you its R.A. is less than yours. If east of you,
its R.A. is greater than yours. Star will be found approximately 15 deg. to
east or west of you for each hourly difference between the star's R.A.
and your R.A. (L.S.T.).
Having established the star's general east and west direction, its north
and south position can be found from its declination.
_4. Time of Meridian Passage of a Star_
It is often invaluable to know first, when a certain star will be on
your meridian or second, what star will be on your meridian at a certain
specified time. Here is the formula for each case, which put in your
Note Book:
1. To find when a certain star will cross your meridian, take from the
Nautical Almanac, the R.A. of the Mean Sun for Greenwich Mean Noon of
the proper astronomical day. Apply to it the correction for longitude in
time (West +, East -) as per Table at bottom of page 2, Nautical
Almanac, and the result will be the R.A. of the Mean Sun at local mean
noon, i.e., the distance in sidereal time the mean sun is from the First
Point of Aries when it is on your meridian. Subtract this from the
star's R.A., i.e., the distance in sidereal time the star is from the
First Point of Aries (adding 24 hours to the star's R.A., if necessary
to make the subtraction possible). The result will be the distance in
sidereal time the star is from your meridian i.e., the time interval
from local mean noon expressed in units of sidereal time. Convert this
sidereal time interval into a mean time interval by always subtracting
the reduction for the proper number of hours, minutes and seconds as
given in Table 8, Bowditch. The result will be the local mean time of
the star's meridian passage.
Example:
April 22nd, 1919, A.M. at ship. In Lo. 75 deg. E. What is the local mean
time of the Star Etamin's meridian passage?
R.A.M.S. Gr. 21d--0h 1h--54m--02s
Red. for 75 deg. E (--5h) --49.3
________________
R.A.M.S. local mean noon 1h--53m--12.7s
Star's R.A. 17h--54m--44s
-- 1 --53 --12.7
___________________
Sidereal interval from L.M. Noon 16h--01m--31.3s
Red. for Sid. Int. (Table 8) -- 2 --37.3
___________________
L.M.T. 21 d 15h--58m--54s
Hence, star will cross meridian at 3h--58m--54s A.M. April 22nd.
2. To find at any hour desired what star will cross your meridian, take
the R.A. of the Mean Sun for Greenwich Mean Noon of the proper
astronomical day. Apply to it the correction for longitude in time (West
+, East -) as per table at bottom of page 2, Nautical Almanac, and the
result will be the R.A. of the Mean Sun at local mean noon; i.e., the
distance in sidereal time the mean sun is from the First Point of Aries
when it is on your meridian. Suppose you wish to find the star at 10
P.M. Add 10 sidereal hours to the sun's R.A. just found. The result will
be the R.A. of your meridian at approximately 10 P.M.
Select in the table on p. 94 the R.A. of the star nearest in time to
your R.A. just secured. Subtract the R.A. of the Mean Sun at local mean
noon from the star's R.A. just found on p. 94 of the N.A. and the result
will be the exact distance in sidereal time the star you have just
identified is from your meridian, i.e., the time interval from local
mean noon expressed in units of sidereal time. Convert this sidereal
time interval into a mean time interval by always subtracting for the
proper number of hours, minutes and seconds as per Table 8, Bowditch.
You will then have secured the name of the star desired and the exact
local mean time of the star's meridian passage.
Example No. 2: At sea Dec. 14, 1919. Desired to get a star on my
meridian at 11 P.M. Lo. by D.R. 74 deg. W.
(.).R.A.G.M.N. 17h--28m--26s
Corr. 74 deg. W. (4th - 56m W + ) + 0 --48.6
___________________
(.).R.A. your M. 17h--29m--14.6s
+ 11
___________________
R.A.M. 28h--29m--14.6s
-- 24
___________________
4h--29m--14.6s
R.A. of Star Aldebaran 4h--31m--18.5s
Star R.A. 28h--31m--18.5
(.).R.A. your M. 17 -- 29--14.6
----------------
Sid. Int. from L.M. Noon 11h--02m--03.9s
Red for Sid. Int. (Table 8) -- 1 --48
----------------
L.M.T. 11h--00m--15.9s
Aldebaran, then, is the star and the exact L.M.T. of its meridian
passage will be 11h 00m 15.9s
Note: If the R.A.M. is more than 24 hours, deduct 24 hours. You will
know whether the star is North or South of you by its declination. If
you are in North latitude, the star will be south of you if its
declination is South or if its declination is North and less than your
latitude. If its declination and your latitude are both North and its
declination is greater, the star will be north of you. The same
principle applies if you are in South latitude.
Assign any of the following to be worked in the class room or at night:
1. At sea, November 1st, 1919. In Latitude 40 deg. N., Longitude 74 deg. W.
WT 8h 30m P.M. Observed unknown star about 80 deg. east of my meridian and
25 deg. south of me. What was the star?
2. At sea, December 1st, 1919. CT 10h 45m 01s. CC 20m 16s slow.
In D.R. Latitude 30 deg. N., Longitude 60 deg. 30' W. Observed unknown star
about 60 deg. west of meridian and about 22 deg. S. What was the star?
3. March 15th, 1919. In D.R. Latitude 10 deg. 42' N, Longitude 150 deg.
14' 28" W. CT 5h 14m 28s. CC--2m 10s. Observed unknown star almost
on my meridian and about 28 deg. north of me. What was the star?
4. Aug. 3, 1919, P.M. at ship. In D.R. Latitude 37 deg. 37' N. Longitude
38 deg. 37' W. At what local mean time will the Star Antares be on the
meridian?
5. What star will transit at about 4:10 A.M. on Aug. 3rd, 1919? In
D.R. position Latitude 38 deg. 10' N, Longitude 34 deg. 38' W.
6. At what local mean time will the Star Arcturus transit on July 17th,
1919, in Latitude 45 deg. 35' N., Longitude 28 deg. 06' W.?
WEDNESDAY LECTURE
LATITUDE BY MERIDIAN ALTITUDE OF A STAR--LATITUDE BY POLARIS (POLE OR
NORTH STAR)
To find your latitude by taking an altitude of a star when it is on your
meridian, is one of the quickest and easiest of calculations in all
Navigation. The formula is exactly the same as for latitude by meridian
altitude of the sun. In using a star, however, you do not have to
consult your Nautical Almanac to get the G.M.T. and from that the
declination. All you have to do is to turn to page 95 of the Nautical
Almanac, on which is given the declination for every month of the year,
of any star you desire. The rest of the computation is, as said before,
the same as for latitude by the sun and follows the formula Lat. = Dec.
+- Z.D. (90 deg. - true altitude). As when working latitude by the sun, you
subtract the Z.D. and Dec. when of opposite name and add them when of
the same name. Put in your Note-Book:
Formula: Lat. = Dec. +- Z.D. (90 deg. - true altitude).
At sea, Dec. 24th, 1919. Meridian altitude Star Aldebaran 52 deg. 36' S. HE
20 ft. Required latitude of ship.
Obs. Alt. 52 deg. 36' S
Corr. - 5 08
----------
True Alt. 52 deg. 30' 52" S
- 90 00 00
----------
Z.D. 37 deg. 29' 08" N
Dec. 16 21 00 N
----------
Lat. 53 deg. 50' 08" N
Note to Instructor:
Have class work examples such as the following before taking up Latitude
by Pole Star:
1. At sea, May 5th, 1919. Meridian altitude Star Capella, 70 deg. 29' S. HE
32 ft. Required latitude of ship.
2. At sea, August 14th, 1919. Meridian altitude Star Vega, 60 deg. 15' 45"
N. HE 28 ft. Required latitude of ship.
Etc.
_Latitude by Polaris_ (_Pole or North Star_)
You remember we examined the formula in the N.A. for Lat. by the pole
star when we were discussing sidereal time some weeks ago. We will now
take up a practical case of securing your latitude by this method.
Before doing so, however, it may be of benefit to understand how we can
get our latitude by the pole star. In the first place, imagine that the
Pole Star is directly over the N pole of the earth and is fixed. If that
were so, and imagine for a minute that it is so, then it would be
exactly 90 deg. from the Pole Star to the celestial equator. Now, no matter
where you stand, it is 90 deg. from your zenith to your true horizon. Hence
if you stood at the equator, your zenith would be in the celestial
equator and your true horizon would exactly cut the Pole Star. Now,
supposing you went 10 deg. N of the equator. Then your northerly horizon
would drop by 10 deg. and the Pole Star would have an altitude of 10 deg.. In
other words, when you were in 10 deg. N latitude, the pole star would
measure 10 deg. high by sextant. And so on up to 90 deg., where the Pole Star
would be directly over you and you would be at the North Pole. Now all
this is based upon the Pole Star being in the celestial sphere exactly
over the North Pole of the earth. It is not, however. Owing to the
revolution of the earth, the star appears to move in an orbit of a
maximum of 1 deg. 08'. Just what part of that 1 deg. 08' is to be applied to the
true altitude of the star for any time of the sidereal day, has been
figured out in the table on page 107 of the Nautical Almanac. What you
have to get first is the L.S.T. Find from the table the correction
corresponding to the L.S.T. and apply this correction with the proper
sign to the true altitude of Polaris. The result is the latitude in. Put
in your Note-Book:
To get latitude by pole star, first get L.S.T. This can be secured by
using any one of the three formulas given you in Week III--Thursday's
Lecture on Sidereal Time and Right Ascension. Then proceed as per
formula in N.A.
* * * * *
Note to Instructor:
Spend rest of time in solving examples similar to the following:
1. At sea, Feb. 14th, 1919. CT 13d 21h 52m 33s. CC 1m 14s fast. In
Lo. 72 deg. 49' 00" W. IE + 1' 10". HE 15 ft. Observed altitude Polaris
42 deg. 21' 30" N. Required latitude in.
2. At sea, March 31st, 1919. In Lo. 160 deg. 15' E. CT 7h 15m 19s. Observed
altitude Polaris 38 deg. 18' N. IE + 3' 00". HE 17 ft. Required latitude in.
Etc.
THURSDAY LECTURE
MARC ST. HILAIRE METHOD BY A STAR SIGHT
You have already been given instructions for finding a Line of Position
by the Marc St. Hilaire Method, using a sight of the sun. Today we will
work out the same method by using a sight of a star. Put this in your
Note-Book here and also under I(b) of the formula given you in Week
IV--Friday's Lecture:
Get G.M.T. from corrected chronometer time. With your G.M.T. find the
corresponding G.S.T. according to the formula already given you. With
your G.S.T. apply the D.R. longitude
(- W. Lo.)
----------
(+ L. Lo.)
to get the L.S.T. With the L.S.T. and the star's R.A. subtract the less
from the greater and the result is the star's H.A. at the ship or "t."
In using Sun Azimuth tables always take "t" from the P.M. column. Mark
Azimuth N or S according to the lat. in and E or W, according as to
whether the Star is East or West of your meridian. Then proceed as in
the case of a sun sight. Formula:
(-W. Lo.)
G.M.T. + (.).R.A. + (+)CP = G.S.T. --------- = L.S.T.--Star's R.A.
(+E. Lo.)
(or vice versa if Star's R.A. is greater) = Star's H.A. at ship (t).
Then proceed as in case of sun sight.
Example:
On May 31st, 1919, in D.R. Lat. 50 deg. N, Lo. 45 deg. W, G.M.T. 31d 14h 33m
30s. What was Star's H.A. at ship?
G.M.T. 14h -- 33m -- 30s
(.).R.A. 4 -- 31 -- 44.2
(+).C.P. 2 -- 23
--------------------
G.S.T. 19h -- 07m -- 37.2s
W Lo.-- 3 -- 00 -- 00
--------------------
L.S.T. 16h -- 07m -- 37.2s
Star's R.A.(Spica) 13 -- 20 -- 59
--------------------
Star's H.A. (t) 2h -- 46m -- 38.2s
Now let us work out some examples by this method:
1. Nov. 29th, 1919. CT 30d 2h 14m 39s A.M. CC 3m 14s fast. D.R. position
Lat. 41 deg. 14' N, Lo. 68 deg. 46' W. Observed altitude Star Aldebaran East of
meridian 50 deg. 29' 40". HE 29 ft. Required Line of Position by Marc St.
Hilaire Method and most probable position of ship.
2. Jan. 23rd, 1919. P.M. at ship. CT 3h 45m 40s. Lat. by D.R. 38 deg. 44'
19" N. Lo. 121 deg. 16' 14" E. Observed altitude Star Rigel 28 deg. 59' 20" West
of meridian. IE + 4' 30". HE 42 ft. Required Line of Position by Marc
St. Hilaire Method and most probable position of ship.
Assign for Night Work one or two examples similar to the above.
FRIDAY LECTURE
EXAMPLES: LATITUDE BY MERIDIAN ALTITUDE OF A STAR, LATITUDE BY POLARIS,
MARC ST. HILAIRE METHOD BY A STAR SIGHT
1. At sea, Dec. 5th, 1919. Observed meridian altitude Star Aldebaran 69 deg.
28' 40" S. No IE. HE 26 ft. Required latitude in.
2. At sea, Jan. 20th, 1919. CT 21d 2h 16m 48s A.M. In longitude 56 deg. 29'
46" W. Observed altitude of Star Polaris 48 deg. 44' 30" N. IE + 10' 20". HE
37 ft. Required latitude in.
3. At sea, June 4th, 1919. A.M. at ship. CT 10h 16m 32s. CC 5m 45s fast.
Lat. by D.R. 42 deg. 44' N, Longitude 53 deg. 13' 44" E. Observed altitude of
Star Altair East of meridian, 52 deg. 19' 30". IE--14' 00". HE 56 ft.
Required line of position by Marc St. Hilaire Method and most probable
position of ship.
Etc.
* * * * *
Assign for Night Work the following Articles in Bowditch: 336 through
341, disregarding the formulas.
SATURDAY LECTURE
LONGITUDE BY CHRONOMETER SIGHT OF THE SUN (TIME SIGHT)
You have now learned, first, how to get your latitude by a meridian
altitude of the sun or a star and second, how to get your Line of
Position and most probable fix, including both latitude and longitude,
by the Marc St. Hilaire Method, using for your calculations either the
sun or a star. We are now going to take up a method of getting your
longitude only. This method requires as much, if not more, calculation
than the Marc St. Hilaire Method. Its results, on the other hand, are
far less complete, for while the Marc St. Hilaire Method will give you a
fairly accurate idea of both your latitude and longitude, this method
will, at best, only give you your longitude. Moreover, you can use it
for accurate results only when the sun bears almost due East or West of
you, for that is the best time, as you have already learned, to get a
line of position running due North and South, which is nothing more than
a meridian of longitude. The only reason we explain this method at all
is because it is in common practice among merchantmen and may,
therefore, be of assistance to you, if you go on a merchant ship.
Remember, however, that it belongs to Old Navigation as distinguished
from New Navigation, exemplified by the Marc St. Hilaire Method. It is
undoubtedly being used less and less among progressive, up-to-date
navigators, and will continue to be used less as time goes on. The fact
remains, however, that at present many merchantmen practice it, and so
it will do you no harm to become familiar with the method, too.
This method is based on securing your longitude by a time sight or
longitude by chronometer sight, meaning that at the time the sun bears
as near due East or West as possible, you take a sight of it by sextant
and at the same instant note the time by chronometer. With this
information you proceed to work out your problem and secure your
longitude according to the following formula. Put in your Note-Book:
To find your longitude by chronometer (or time) sight.
1. Take sight by sextant only when the sun bears as near as possible due
East or West. At exact time of taking sight, note chronometer time.
2. Get G.M.T. from corrected chronometer time. Apply Equation of Time to
get the corresponding G.A.T.
3. Correct observed altitude to get T.C.A. Also have at hand Lat. by D.
R. and Polar Distance. (Note: Secure P. D. by subtracting Dec. from 90 deg.,
if Lat. and Dec. are of same name. If Lat. and Dec. are of opposite
name, secure P. D. by adding Dec. to 90 deg..)
4. Add together the T.C.A. the Lat. by D.R. and the P.D. Divide the sum
by 2 and call the quotient Half Sum. From the Half Sum subtract the
T.C.A. and call the answer the Difference.
5. Add together the secant of the Latitude, the cosecant of the P.D.,
the cosine of the Half Sum and the sine of the Difference (Table 44).
The result will be the log haversine of the S.H.A. or L.A.T. It must
always be less than 10. If greater than 10, subtract 10 or 20 to bring
it less than 10.
6. From Table 45, take out the corresponding S.H.A. (L.A.T.), reading
from the top of the page if P.M. at ship, or from bottom of page if A.M.
at the ship.