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COLLEGIATE MATHEMATICS FOE WAK SEEVICE. 



415 



COLLEGIATE MATHEMATICS FOR WAR SERVICE. 

Communications foe this depabtment should be sent to H. Blumbeeg, University of Illinois. 
SOME DRAWINGS AND GRAPHICAL SOLUTIONS IN NAVIGATION. 

By Wm. H. Roeveb. 

1. Introductory Note. — During the past summer the U. S. Naval Auxiliary 
Reserve School, Municipal Pier, Chicago, made arrangements whereby its stu- 
dents were to take a course in navigation at the University of Chicago. As a 
member of the teaching staff at the University during the summer, I offered my 
services as instructor in navigation to these students. Professor Wilczynski and 
I were asked to give the work in nautical astronomy. Most of the students, while 
very intelligent, had only a limited knowledge of mathematics, — not extending 
beyond plane trigonometry, — and no knowledge whatever of astronomy. For 
this reason I at first made frequent appeal to their geometric intuition and solved 
by graphical processes the spherical triangles which arise in the determination of 
time, latitude, longitude, azimuth, and in the problem of great-circle sailing. 
I also drew a number of pictures, illustrating space relations. These pictures 
and graphical processes proved to be so helpful to my students that it seemed 
worth while to those who had charge of the course in Navigation at the University 
of Chicago to have them put in a form whereby they will be available to others. 
Hence this article. It should not be assumed from what has been said, that I 
did not also give the customary methods of computation. 

2. Time Determination. — The determination of time reduces itself to the 
mathematical problem of finding an angle (the hour angle) of a spherical triangle 
of which the three sides (complements of the latitude, altitude and declination) 
are known. To solve graphically such a triangle let us consider the trihedral 
formed by the planes of the sides of the spherical triangle. Fig. 1 represents such 



y 



\tfL 






N, 0< f? \ 












//jo\\ 




M 




^^7- 


7 




\ 



Fig. 1. 



Fig. 2. 



a trihedral in which a, b, c denote the sides of the spherical triangle or the face 
angles of the trihedral, and C denotes the required angle of the spherical triangle 



416 



COLLEGIATE MATHEMATICS FOR "WAR SERVICE. 



or the corresponding dihedral angle of the trihedral. The point represents the 
center of the sphere or the vertex of the trihedral, and N represents any point on 
the edge of the required dihedral angle. At N perpendiculars are erected to the 
edge ON in the faces which meet in ON. These perpendiculars meet the other 
edges of the trihedral in the points L and M . Then, in the triangle represented 
by LMN, the angle at N is equal to the required angle C of the spherical triangle. 
To find C graphically, let us cut the trihedral along the edge ON and open it up, 
or develop it, as shown in Fig. 2. Then N will assume the two positions N\ and 
Ni, which lie at equal distances from on the two positions assumed by the edge 
ON. The perpendiculars erected to these positions at the points N% and N%, 
respectively, meet the new positions of the other edges in L and M, respectively. 
Hence ML, LNi and MN% in Fig. 2 are the true lengths of the sides of the triangle 





Fig. 3. 



Fig. 4. 



represented by LMN in Fig. 1. Thus we are led to the Construction: With 
(Fig. 2) as vertex lay off the angle c equal to the side of the spherical triangle 
which lies opposite to the required angle C. Adjacent to the angle c and on 
either side of it lay off the angles a and b equal to the other two given sides of the 
spherical triangle. Then on the terminal lines of these last two angles take 
points Ni and 2V 2 at equal distances from 0. At these points erect perpendiculars. 
N\L and N%M to these sides, meeting the sides of the angle c in the points L and 
M, respectively. With L as center and LNi as radius draw a circle, and with M 
as center and MNi as radius draw another circle. These two circles intersect 
in a point N 3 , which, together with the points L and M forms the triangle repre- 
sented by LMN in Fig. 1, and hence the angle LN S M of Fig. 2 is the required angle 
C of the given spherical triangle. A check on the construction is furnished by the 
fact that the points N s and must lie on the same perpendicular to the line LM . 



COLLEGIATE MATHEMATICS FOR "WAK SERVICE. 



417 



In navigation, the sides b and c (complements of the latitude and altitude, 
respectively) are always acute, but the side a (complement of the declination) 
may be acute or obtuse. For the case in which the side a is obtuse the point L 
will lie on the produced edge, as shown in Fig. 3, and then the required angle C 
is the supplement of the angle MNL of the triangle MLN. The corresponding 
construction is shown in Fig. 4. 

If the sides a and b of the given spherical triangle are nearly 90°, the points 
L and M of the constructions shown in Figs. 2 and 4, will fall beyond the limits 
of the drawings. In this case the construction explained in § 4 can be used to 
advantage. 

3. Solution of the triangle in the Saint Hilaire Method. — To find the latitude 
and longitude of a ship by the Sumner Method, two lines of position must be 
determined. The Saint Hilaire Method for finding such lines, involves a spherical 
triangle of which two sides and the included angle are known and of which the 
side opposite the given angle is required. 1 To solve graphically this spherical 
triangle, let us consider Fig. 5, in which the edge of the given dihedral angle C 





Fig. 5. 



Fig. 6. 



of the corresponding trihedral is represented as being in a vertical position, i. e., 
perpendicular to the plane (let us say) of a table top. The points in which 
the edges of the trihedral meet this plane are represented by the points N, L, 
and M, of which the first lies on the edge of the given dihedral angle C. The 
center of the sphere, or the vertex of the trihedral, is represented by the point 0. 
Let us now think of this trihedral as cut along its three edges ON, OM, OL and 

1 This side is the computed zenith distance of the observed object. The azimuth is usually 
obtained from an azimuth table, but it can easily be found graphically as shown in the Remark at 
the nd of § 4. 



418 



COLLEGIATE MATHEMATICS FOE WAR SERVICE. 



of its faces as turned down around the lines NM, ML, LN into the plane of the 
table top. They then assume the position shown in Fig. 6. Hence the Con- 
struction: With N (Fig. 6) as vertex lay off the angle C equal to the given angle 
of the spherical triangle. At N erect perpendiculars to the sides of this angle 
and on these take points Oi and Oz at equal distances from N. Through d and 
2 draw lines making with OiN and O2N the angles a and b, respectively, equal 
to the given sides of the spherical triangle, and meeting the sides of the angle C 
in the points L and M, respectively. Then with L as center and L0\ as radius 
draw a circle, and with M as center and MO2 as radius draw another circle. These 
circles meet in a point O3, which with L and M forms the triangle represented in 




Fig. 7. 



Fig. 8. 



Fig. 5 by OLM and of which the angle at is equal to the required side c of the 
spherical triangle. Hence the angle LO3M of Fig. 6 is the required angle c of the 
spherical triangle. A check is furnished by the fact that 3 and N should lie 
on the same perpendicular to LM. 

In the navigation problem the side b (complement of the latitude) is always 
acute, but the parts C and a may be acute or obtuse. In the case where the part 
a is obtuse the point L lies on the produced edge, as shown in Fig. 7, and then the 
required part c is the supplement of the angle LOM in the triangle LOM. The 
corresponding construction is shown in Fig. 8. 

If the parts a and b are nearly 90° the points L and M of Figs. 6 and 8 will fall 
beyond the limits of the drawings, and in this case the construction of § 4 can be 
used. 

4. Constructions available in failing cases of the above. — Constructions which 
have the advantage over the constructions just given, in that all the points re- 



COLLEGIATE MATHEMATICS FOE WAR SERVICE. 



419 



main within the limits of the drawings, will now be given. These constructions 
are perfectly general and may be used whenever the given parts of the spherical 
triangle are less then 180°. These constructions will now be given, but for their 
proof the reader is referred to Loria, Vorlesungen ueber Darstellende Oeometrie, 
Vol. II, p. 6. 

Construction (Fig. 9 for a acute, Fig. 10 for a obtuse). — The three parts a, 
b, c {i. e., the sides of the spherical triangle or the face angles of the corresponding 
trihedral) are laid off from a point as shown in Figs. 9 and 10. A circle T of 
center and any convenient radius is then drawn cutting the sides of the angles 
a, b, c in the points C, A, Bi and B 2 as shown in the figures. From the points 
Bi and B 2 perpendiculars are dropped to the rays OC and OA, respectively, 



A 2 / 





Fig. 9 



Fig. 10. 



cutting these in the points F and E, respectively, and each other in the point G. 
With F as center and FB t as radius a circle Ai is drawn. At a perpendicular is 
erectedto the line BiF cutting the circle Ai in Hi. Then the supplement of the 
angle B x FHi is the required angle C of the spherical triangle of which the sides 
are the angles a, b, c. 

Construction (Fig. 9 for a acute, Fig. 10 for a obtuse). — If a, b, C are the given 
parts of the spherical triangle, the circle T, the perpendicular BiF and the circle 
Ai are drawn as in the above preceding construction. The point Hi is now taken 
on the circle Ai, so that the angle B1FH1 is the supplement of the given angle C. 
From Hi a perpendicular is dropped to FB U meeting it in 6. From G a per- 
pendicular is dropped to OA meeting it in E, and meeting the circle V in B 2 . 
The angle AOB% is then equal to the required side c of this spherical triangle. 

Remark. — In order to find the angle A (which is the azimuth in the nautical 
problem), we have merely to draw the circle A2 of center E and radius EB 2 . At 
G we then erect a perpendicular to B 2 E cutting A2 in H 2 . The supplement of 
the angle B 2 EH% is then equal to the required angle A. 



420 



COLLEGIATE MATHEMATICS FOR WAR SERVICE. 



5. The great-circle-sailing problem. — The problem of great-circle sailing con- 
sists in determining the latitude and longitude of various points Q of the great 
circle which connects two given points A and B on the surface of the earth. The 
points Q are then plotted on a Mercator chart 1 from which the course of the ship 
between consecutive points is determined. To devise a graphical solution of this 
problem let us first think of .the points A and B as both situated in the same hemi- 
sphere (Fig. 11). 

Now think of a plane tangent to the earth at its pole 0' and upon this plane 
let us project the points A and B from the center of the earth, denoting the 
projections by A' and B', respectively. Then the straight line A'B' of this plane 
is the projection of the great circle path AB, and the lines O'A' and O'B' are the 




Fig. 11. 

projections of the meridians of A and B. Let V denote the point of the great 
circle AB which is nearest the pole 0'. Evidently the projection of the point V 
is the foot V of the perpendicular from 0' to A'B' in the triangle O'A'B'. A 
general point Q of the great circle AB projects into a point Q' of the line A'B'. 
The tetrahedron OA'B'O' represented in Fig. 11 is the inverted position of the 
tetrahedron OLMN represented in Fig. 5, so that the parts a, b, C of Fig. 5 cor- 
respond respectively to the co-latitude (90°-La) of A, the co-latitude (90°-Z B ) 
of B and the difference in longitude (equal to the angle B'O'A') of the points 
A and B. Hence the part c of Fig. 5 corresponds to the angular distance AOB 
of Fig. 11. Thus the construction of Fig. 6 determines graphically (in that it 
determines the part c) the angular distance between the two points A and B 
when the latitudes and the difference in longitudes of these points are known. 
In order to find the latitude, longitude and distance from A (or from B) of the 

1 For the definition of Mercator projection see § 7. 



COLLEGIATE MATHEMATICS FOE WAK SEEVICE. 



421 



point V and of the general point Q, we will begin by repeating the construction 
of Fig. 6. In Fig. 12 this much is shown by the heavily drawn lines; the points 
0', A', B', A , b , 0'" of Fig. 12 correspond to the points N, L, M, h 2 , 3 , 
respectively, of Fig. 6. In Fig. 12 the triangle O'A'B' is the true shape of the 
triangle which is shown in Fig. 11 by the same letters. Hence in Fig. 12 the foot 
of the perpendicular from 0' to A'B' is the point V, and any point of A'B' is a 
point Q'. The angles A'0"'V and A'0""Q,' are then the angular distances of 
the points V and Q, respectively, from the point A. Evidently the angle V'O'A' 




(Fig. 12) is the difference in longitude between the points V and A, and the angle 
V'O'Q' is the difference in longitude between V and Q. To find the latitudes of 
V and Q let us first draw a circle T of center 0' and radius O'Oa = O'Ob (equal 
to the distance represented by 00' in Fig. 11). To get the latitude of V draw in 
Fig. 12 a line through 0' perpendicular to O'V and cutting T in V . Then the 
angle O'V'Ov is the latitude of V. Similarly, to get the latitude of Q draw a line 
through 0' perpendicular to O'Q' and cutting T in Q . The angle 0'Q'0 Q is then 
the latitude of Q. This is evident from the fact that the triangles 0'0 V V and 



422 



COLLEGIATE MATHEMATICS FOR WAR SERVICE. 



0'0 g Q' of Fig. 12 are the revolved positions (turned into the plane of the triangle 
O'A'B') of the triangles represented by O'V'O and O'Q'O in Fig. 11. 

Just as we started with Fig. 6 the construction 
of Fig. 12 corresponding to the case where A and 
B are in the same hemisphere, so for the case 
where A and Bare in different hemispheres we 
start the construction with Fig. 8. 

6. Definition of a good picture. — In the absence 
of models, good pictures serve very well to convey 
an adequate notion of space relations. By a good 
picture of a space object is meant a plane repre- 
sentative which, when properly placed, produces 
upon the retinal surface of the eye an image 
which does not differ much from that produced 
by the object itself in a position in which one is 
accustomed to see or think of the object. It can easily be shown that a central 




Fig. 13. 



"circle 



H 



A 

/JP 1 


< \ 

o >^ 

i- \. 

ul X 


/ * 

/ \ 


---- (- — 

\ I 


111 ***"* — ._ 
s — -.. 
E / 
a. / 


JD 


\ 








-Zeffio* 



Fig. 14. 



or parallel (orthographic or oblique) projection in general satisfies this criterion 1 
and therefore such a projection of an object in an accustomed position furnishes a 

1 See for instance, pages 148 and 152 of the article on Descriptive Geometry in Vol. 25 of this 
Monthly. 



COLLEGIATE MATHEMATICS FOE WAR SERVICE. 



423 



good picture. What is here meant by an accustomed position will be made clear 
from the statements which follow. A stereographic projection of a sphere is a cen- 
tral projection for a particular position of the center of projection. 1 However, the 
stereographic projection of the celestial sphere with its circles of reference does 




Fig. 15. 

not produce nearly so good a picture as does an orthographic projection on a 
general plane. This will be evident to the reader when a comparison is made of 
Figs. 13 and 14 which are stereographic and orthographic projections, respectively. 
Hence a stereographic projection, although it is a projection which satisfies the 
criterion of producing the same retinal image as the object itself, does not furnish 
a good picture, because it is projected from a point from which one is not accus- 
tomed to seeing the sphere. 

7. Stereographic and Mercator projections defined. — We have already used the 
terms Mercator projection and stereographic projection. For the purpose of 
defining these, as well as showing a simple relation which exists between them, 
we will make use of the foregoing picture (Fig. 15) which is an orthographic 
projection. 

The stereographic projection of a sphere is the central projection of the points 
of the surface of the sphere upon a diametral plane from a pole of the great circle 

1 For the definition of stereographic projection see § 7. 



424 COLLEGIATE MATHEMATICS FOE WAR SERVICE. 

in which this plane cuts the sphere. In Fig. 15 the plane of projection is the 
equatorial plane, and the center of projection is the south pole. The projection 
of the point P of the sphere is the point P s of the equatorial plane. 

The Mercator projection of a sphere is the development of a right circular 
cylinder, tangent to the sphere along the equator, the points of which are ob- 
tained from those of the sphere in the following manner. The point P of the 
sphere is projected from the center into a point P t of the cylinder (Fig. 15). 
Upon the element of the cylinder which passes through the point P t the Mercator 
projection P m of the point P lies, and its distance u from the equator is equal to 
the Naperian logarithm of the reciprocal (1/r) of the distance (r) of the stereo- 
graphic projection P s from the center of the sphere, the radius of which is taken 
to be unity. 

Thus if of a point P of the sphere, L = latitude and X = longitude, and is 
the co-latitude, so that 

= | - L and hence 0/2 = ir/4 - L/2, 

and if for the stereographic projection P s the polar coordinates are r and X, and 
for the Mercator projection the rectangular coordinates are u and v, then 



and 



U 2 )' 
X = X, 



r = tan "~ = tan 



f , 1 fw L\ 

I u = log, - = — log c tan ^ = — log e tan I 4 — 2 / ' 

\v = \. 

Thus one easily sees the simple relation which exists between these two types 
of map. 

8. Mean sun and equation of time defined. — In the definitions of mean time 
and the equation of time, a good picture showing the relation which exists between 
the mean and apparent suns is very helpful. In the orthographic projection 
shown in Fig. 16 the celestial sphere with the equinoctial and ecliptic are repre- 
sented. In this figure, the vernal equinox is represented by the point r, the 
position which the true sun occupies at perihelion (which occurs about January 
first) is represented by S , and the position which the true sun occupies at any 
particular instant (the position shown in the figure corresponds to about February 
12) is represented by S. Before defining the mean sun, let us first define a fic- 
titious sun which moves in the ecliptic at a uniform rate in the same direction 
as the true sun does and coincides with the true sun at perihelion (and therefore 
also at aphelion). The point F represents the position which this sun occupies 
when the true sun is at S. The mean sun may now be defined as the point 
which moves along the equinoctial at a uniform rate and coincides with the fic- 
titious sun at the vernal equinox (and therefore also at the autumnal equinox). 



COLLEGIATE MATHEMATICS FOE "WAR SERVICE. 



425 



The position which the mean sun occupies when the true sun is at S is repre- 
sented by M . In other words, the right ascension of the mean sun M is equal 
to the mean celestial longitude of the fictitious sun F, these two artificial suns 
being in coincidence at the vernal equinox. Fig. 16 also represents the center 
and the north pole P of the celestial sphere and the hour circles of 8, 
F and M . The hour circles of S and F intersect the equinoctial in the points 




Fig. 16. 



8' and F', respectively. The difference in right ascension between 8 and M, 
i. e., the angle MOS' is called the equation of time. It is the difference between 
mean and apparent solar times. Thus if Z represents the zenith of a particular 
place of the earth at a particular instant, 



and 



t m , = angle ZPM, is the local mean time of this place, 

t a , = angle ZPS, is the local apparent solar time of this place, 

e, = tm — t a = angle MOS', is the equation of time. 1 



9. The Sumner Method. — In describing the Sumner Method of determining 
the position of a ship at sea, a picture representing both the celestial and the 
terrestrial spheres proves to be very helpful. In Fig. 17 such a picture is repre- 
sented in orthographic projection. The positions on the celestial sphere, of the 
pole, zenith and two positions of the sun are represented by P, Z, Si, S2, respec- 
tively. The positions on the terrestrial sphere, of the points immediately below 

1 It should not be forgotten that hour angle (time) and right ascension are measured in opposite 
directions. 



426 



COLLEGIATE MATHEMATICS FOE WAR SERVICE. 



these, are denoted by p, z, s\, s 2 , respectively. Then s is the position of the ship, 
and *i, Si are called the subsolar (or in the case of a star, the substellar) points, 
The meridian of Greenwich is shown on both of the spheres as is also the equator. 
It is thus evident that the terrestrial longitude of a subsolar point is equal to the 
Greenwich apparent time and that the latitude is equal to the declination of the 
true sun. A small circle of the terrestrial sphere which passes through the position 
of the ship (z) and has as pole the subsolar point is called a Sumner circle. It is 




Fig. 17. 



easy to see from the figure that the angular radius of this circle is equal to the 
zenith distance of the sun. The figure shows the two Sumner circles which cor- 
respond to Si and #2 and which intersect in the position z of the ship. 

10. The Gnomonic Chart. — By a gnomonic chart is meant the projection of 
the points of a spherical surface upon a tangent plane to this surface from the 
center of the sphere. A straight line of such a chart corresponds to a great circle 
of the sphere. Such charts are used in solving graphically the problem of great- 
circle sailing, of which one graphical solution has already been given in § 5. In 
fact the projection O'A'B' (Fig. 11) on the plane tangent to the sphere at the 
point 0' is a gnomonic chart of the triangle O'AB of the sphere for the case where 
the plane of projection is tangent to the sphere at the pole 0'. In this case the 
meridians project into straight lines radiating from the point of tangency and the 
parallels of latitude project into concentric circles whose common center is the 
point of tangency. If, however, the plane of projection is tangent to the sphere 
at a point not coincident with a pole, the meridians project into straight lines which 



COLLEGIATE MATHEMATICS. FOE WAK SEBVICE. 



427 



radiate from the point where the axis of the earth pierces the plane of projection, 
and the parallels of latitude project into conies. This is the case because the 
planes of the meridians all pass through the axis, which contains the center of 
projection, and because the projecting rays of a parallel of latitude form a right 
circular cone. Fig. 18 enables one to see this more clearly. If on such a chart a 
sufficient number of meridians and parallels of latitude are represented, it is an 
easy matter to read off from the chart the latitude and longitude of various points 
of the straight line representative of a great-circle path of the sphere. The 
method of doing this as well as that of finding the distance and course (direction 
of sailing) is explained on the great-circle charts which are published by the 
U. S. Hydrographic Office. 

The construction of a gnomonic chart is identical with that of a horizontal 
sun-dial for a place whose latitude is equal to that of the point of tangency of the 
plane of projection of this chart. This can be easily seen by the aid of Fig. 18, 
in which we may now consider the horizontal table top as the dial plate and the 
axis of the sphere as the gnomon, or style, of the dial. The lines representing 
the meridians are then positions of the shadow of the style upon the dial plate, 
and the conies representing the parallels of latitude are the diurnal paths of the 




Fig. 18. 



shadow of a knob (situated at the point 0) of the style on the dial plate. In the 
dialing problem it is required to find the position of the shadow of the style for 
the hours of the day. The diurnal paths of the knob are the conies lying between 
the conies representing the Tropics of Capricorn and Cancer. It is on these 
conies that the equation of time is laid off from the position of the shadow of the 
style at apparent solar noon in order to make possible the use of a sundial for 
the determination of mean time.