illustrates an angular distance on earth's surface. C denotes
the center of earth. Angle ∠ACB is the angular distance from
A to B, or the angle measure of the arc AB.
Any circle on earth's surface whose
center is the same as earth's center is called a great circle.
Great circles describe the shortest distance between points
on a spherical surface, just as straight lines do on a plane surface.
If the angle in the figure is 60°, then the great
circle distance between points A and B is 60 times 60 = 3600 nautical
miles. This great circle path from A to B is the shortest or
most direct path from A to B on earth's surface.
Meridians of longitude are great
circles. However, the equator is the only great circle of
latitude. A corollary to this observation is that if one
travels along a circle of latitude that is not the equator (say for
example at latitude 30° north), the distance traveled will be
greater than along a great circle connecting the same origin and
||Besides earth there is
another sphere of importance in navigation. This one is imaginary, and
is called the celestial
sphere. The celestial sphere is an enormous
sphere whose center is earth. The sun, moon, planets, and
stars—all heavenly bodies—may be imagined to be located on the
surface of this sphere. (I told you to forget Copernicus!)
In relation to the celestial sphere, earth should be regarded
as a single point at the center. The celestial sphere is not
meant as an accurate representation of the universe, but rather as a
conceptual tool for navigational purposes.
Key reference points on earth have their
couterparts on the celestial sphere. Projecting from earth's poles are
a north and south celestial pole. Corresponding to earth's equator is
the celestial equator. Wherever you are on earth, the point
in the sky directly above you is called the zenith. The
zenith is usually abbreviated with the letter Z and the point
opposite the zenith (180° from the zenith) is called the nadir,
abbreviated Z' (pronounced
Celestial meridians (projections of
earth's meridian circles) are called hour circles.
Referring to the figure on the left, if A denotes a heavenly
body (the sun for example), then the celestial latitude of A is called
Declination is the angular distance along the hour circle of A from the
celestial equator. (In the figure P and P' are the celestial poles, and
E the celestial equator.)
Now refer to the figure on the right.
Again P is the celestial pole and Z the observor's zenith.
As in the previous figure, the heavenly body is denoted by A.
The angle ∠ZPA between the observor's celestial meredian and
the hour circle of A is called the local
hour angle of A (abbreviated LHA). Similarly the
angle between the
Greenwich celestial meridian and the hour circle of A is called the Greenwich hour angle
of A (abbreviated GHA).
Declination and Greenwich hour angle are
the coordinates that specify the location of heavenly bodies as a
function of date and time in the same way that latitude and longitude
specify locations on earth. Declinations and Greenwich hour
angles for all navigational bodies are tabled in the Nautical Almanac,
available for purchase in printed form as well as free on-line.
||The horizon is a great
circle tangent to earth, with poles at the zenith and nadir.
The horizon and celestial meridian intersect at the north point and south point.
While the horizon and celestial equator intersect at the east point and west point.
circle, altitude, and azimuth
||For any heavenly body A, the vertical circle of
A is the great circle that passes through Z (the observor's zenith) and
A. The altitude
of A is its angular distance from the horizon measured along the
vertical circle of A. Finally, the azimuth
of A is angle between the local meridian and the vertical circle of A,
measured from 0 to 360 degrees eastward from the north point.
In other words,azimuth refers to the compass bearing from the
observer to A.
The good news is that we are done with
definitions, more-or-less. The bad news is that we now have
to do a little math, but nothing very difficult. According to
my 30+ year old notes, the mathematical starting point in Captain Bob's
short course on celestial navigation was the law of cosines.
I am going to start one step before that with the Pythagorean
Theorem. Nearly everyone can recite the Pythagorean Theorem
by rote, "In a right triangle, the square on the hypotenuse is equal to
the sum of the squares on the other sides." Or, as a formula a2
= b2 + c2. But, if your
life depended on it, could you prove that the theorem is true?
If in doubt, please stare at the drawing below for a few
The law of cosines may be viewed as a
generalization of the Pythagorean theorem for oblique triangles.
Referring to the figure on the left, side a can be labeled in
two parts as d
and a - d.
Applying the Pythagorean theorem, b2
- (a - d)2
... The shared side is the same! Expanding the square on (a -
d) and rearranging obtains a2
-2ad = c2
... Finally, making the substitution d/b = cos(C) produces the law
- 2ab ⋅ cos(C)
The law of cosines may be applied to
solve for any of the three sides of an oblique triangle, provided the
other two sides and the included angle are known. This
equation is all that is needed to derive the two main formulas of
spherical trigonometry, which in turn are used to solve the problem of
Part 2 - Applying
In this section the two formulas of
spherical trigonometry that are needed to solve the celestial
navigation problem will be derived. Recall from Part 1 that a
is any circle on the surface of a sphere whose center is also the
center of the sphere. The angle of intersection of two arcs of great
circles is defined as the angle formed by the tangents to these arcs at
their point of intersection.
The angle measure of an arc of a great
circle is the measure of the angle subtended by the arc at the center
of the sphere. For spherical triangles, this measure is
analogous to the length of a side in plane triangles. Thus sides as well as
angles of spherical triangles have angle measure. The
actual length of the arc corresponding to the side of a spherical
triangle may be easily computed when the radius and angle measure are
three points on the surface of a sphere that are not on a great circle
may be joined by arcs of great circles. The resulting figure is called
a spherical triangle.
This definition is analogous to the assertion that any three points on
a plane that are not on a straight line may be connected by straight
lines, making a plane triangle. The figure on the left
illustrates a spherical triangle with angles A, B, and C, and sides AB,
BC, and CA.
From the standpoint of celestial
navigation, two cases of solving the spherical triangle are important.
They are: 1. Given two sides and the included angle, and 2.
Given three sides. These two cases will be worked out in the
context of the actual navigation problem.
In the figure
above, O represents earth's center. P is the celestial pole.
Z is the observer's zenith, that is the zenith of the
position. B is a celestial body, for example,
the sun. The line PT is tangent to the arc PB at P. Similarly
PS is tangent to arc PZ at P. Thus, angle P is the same as
∠TPS by the definition of spherical angles. In the following
paragraphs, the plane triangle formulas (Pythagorean Theorem and law of
cosines) are applied to this figure to derive the spherical triangle
formulas that are needed.
In this problem the following are known:
The angle ∠PB (recalling that sides of spherical triangles are angles
also), the angle ∠PZ, and the angle P. What we want to know
is the angle ∠ZB, which will give us the zenith distance of the body,
and hence its altitude, since altitude (denoted Hc) is 90° minus
PB is called the co-declination.
It is 90° minus the declination of the body.
Recalling that declination
is the body's distance (in angle measure) north or south of
the celestial equator, and knowing that the equator is 90°
pole, it follows immediately that PB is the co-declination.
Similarly, PZ is called the co-latitude.
It is 90° minus the observer's assumed latitude.
The explanation is analogous to that for co-declination. Finally, angle P is the
local hour angle
of the body B, which may be computed by subtracting the (tabled) Greenwich Hour Angle
(GHA) of the body from the assumed
The drawing above summarizes these
important terms from a different perspective. In a sense this
is the "everything" figure of celestial navigation. ∠BPG
is the Greenwich Hour
of B. ∠ZPG
is the longitude
of Z. ∠ZPB
is the computed Local
Hour Angle (LHA) of B. EB is the declination of B,
PB the co-declination.
ZF is the observer's latitude and
PZ the co-latitude.
BH is the computed altitude
of the body (Hc). Note that Hc is 90° minus ZB, the zenith distance of the body.
(The zenith is 90° above the horizon.)
Now back to the previous spherical
triangles figure (the one with O at the center) ... Examination of this
figure confirms the following correspondences: Arc ZB of the
celestial triangle PBZ is equal to plane angle ∠BOZ.
Similarly, PZ = ∠ZOP,
and PB = ∠POB.
These three equalities establish a correspondence
between plane angles and arcs of the celestial triangle PBZ.
Applying the law of cosines to the plane
triangle OTS obtains:
+ OS2 - 2⋅OT⋅OS ⋅ cos(ZB)
Similarly applying the law of cosines to
the plane triangle PTS obtains:
+ PS2 - 2⋅PT⋅PS ⋅ cos(P) =
Now subtract equation (2) from equation
(1) and slightly rearrange terms to get:
- PT2) + (OS2
- PS2) + 2⋅PT⋅PS ⋅ cos(P) = 2⋅OT⋅OS ⋅ cos(ZB)
We are almost home on the first of the
two spherical triangle formulas. Observe that angles ∠OPT
are right angles. If this is not obvious, think about the
3-dimensional meaning of the drawing, where the point O represents
earth and point P the celestial pole. Since ∠OPT
is a right angle, by the Pythagorean theorem, OT2
- PT2 is OP2.
And since ∠OPS
is a right angle, OS2
- PS2 is also OP2.
Therefore we can substitute OP2 for
the each of the differences and re-write expresssion (3) as:
(4) 2⋅OP2 + 2⋅PT⋅PS ⋅ cos(P) = 2⋅OT⋅OS ⋅ cos(ZB)
Now divide expression (4) by 2⋅OT⋅OS
(OP/OT)⋅(OP/OS) + (PT/OT)⋅(PS/OS)⋅cos(P) = cos(ZB)
Expression (5) might be clearer if it
were exhibited with the numerator on one line, then a horizontal line for
divide and the denominator underneath. However, I don't know
how to do that in HTML at this point. So as written, terms
that will be substituted in the next step are grouped together. Note that
OP/OT is cos(PB). Again, if this is not obvious, think about
the 3-dimensional meaning of the drawing—OT is the hypotenuse of the
right triangle OTP (the right angle is at P). Similarly,
OP/OS = cos(PZ), PT/OT = sin(PB) and PS/OS = sin(PZ). This
takes us to:
cos(ZB) = cos(PB)⋅cos(PZ) + sin(PB)⋅sin(PZ)⋅cos(P)
Equation (6) gives the zenith altitude of the body
ZB, and hence the computed altitude 90° - ZB = Hc. It is the main
formula for solving any spherical triangle, given two sides and the
As explained in the preceding
paragraph, the first of the two spherical triangle formulas gives the
computed altitiude of the body Hc. The next formula will yield
the azimuth of the body. The angle ∠PZB or ∠Z
for short is the bearing toward the body from Z. To obtain an
expression for this angle, first invert expression (6), solving for
cos(P). Again, it would be preferable to display this equation
using a horizontal line for divide, but using what I know it is:
(7) cos(P) = (cos(ZB) - cos(PB)⋅cos(PZ)) / (sin(PB)⋅sin(PZ))
Although expressed in terms of the polar angle P, formula (7) is the
general solution of a spherical triangle whenever the three sides are
known. Thus we are free to transpose sides, PB for ZB, PZ for PB,
and ZB for PZ, obtaining:
(8) cos(P) = (cos(PB) - cos(PZ)⋅cos(ZB)) / (sin(PZ)⋅sin(ZB))
Equation (8) gives the azimuth of the body from the assumed position. To review, PB is the co-declination,
PZ is the co-latitude, and ZB is the zenith distance of the body.
When using equation (6), the first of the spherical triangle
formulas, the zenith distance is unknown, while the local hour angle
(LHA) of the body is computed from assumed longitude and the (tabled)
Greenwich Hour Angle (GHA). On the other hand, when using
equation (8), the second of the spherical triangle formulas, one
computes the zenith distance by subtracting the sextant height (the
observed altitude of the body, including corrections) from 90°.
The declination and assumed latitude remain known, and the
formula is used to compute the zenith angle or azimuth of the body.
Equations (6) and (8) are the two main formulas of celesital navigation.
Part 3 - Line of Position and Beyond
The celestial navigation problem was stated in the fourth paragraph of the introduction (here).
To compute a line of position (LOP) it is necessary to reconcile
a computed estimate of the body's altitude Hc with an observed measure
of its altitude Ho (a
sextant sight). The algebraic sign of the difference between Hc
and Ho indicates whether your true position is closer to the body or
further away from the body than your assumed position. You will
also know that you are somewhere on a line (actually a large circle)
whose azimuth (bearing toward the body) agrees with the bearing
computed from observational data.
Thus a LOP
(solution to the celestial navigation problem) specifies two values: 1)
distance in nautical miles "toward" or "away" from the body, 2) an
azimuth angle or bearing toward the body from your position.
These values are easily transcribed to produce a line on the
navigational chart on which you have plotted your dead reckoning course
navigational sights are taken on the sun (or that was the case when
navigational sights were taken!). However, the moon, planets, and
bright stars (navigational stars) may also be used. The reason
why stars are not used much is that the horizon cannot be seen in
darkness. Therefore the height of stars above the horizon can
only be measured at dawn and dusk, when both the star and horizon are
The Nautical Almanac
tables the GHA of the sun, moon, planets, and Aries, for every second
of the year, as well as various correction factors for which limb of
the sun or moon was sighted, height of the eye, parallax, and so forth.
Aries is the celestial reference point for fixed stars. By
definition, Aries (also called the vernal equinox) is the point of
intersection of the celestial equator and the ecliptic. The Nautical Almanac
tables the sidereal hour angle (SHA) of all navigational stars.
SHA is the angle between the meridian of the star and the
meridian of Aries, measured westward. Astronomers use Right Ascension
(RA), which is the eastward angle and is expressed in units of time
rather than degrees. Since the SHA of fixed stars does not vary
appreciably in the course of a year, the tabled GHA for Aries suffices
to compute the GHA, and hence the LHA of any navigational star.
Part 4 - Summary
The overall theory of celestial navigation was described in Parts 1 and 3 and the main formulas were derived in Part 2. To become proficient at practical celestial navigation also requires taking practice sights with a sextant.
It is most helpful to have a computer or programmable calculator
to carry out the tedious computational steps (especially if using The Long Way).
Celestial navigation software is commonplace now. However,
in the past, when it was truly needed, there was no such thing!
If doing calculations by hand (using tables but not a computer
program) it is easy to make errors. Fortunately such errors
almost always produce absurd results, positions that could not possibly
be correct—thousands of miles away. One of the commonest types
of error is to accidentally invert a sign, substituting plus for minus,
or the reverse.
By the late 1980's I had acquired a programmable pocket computer, the
Radio Shack TRS-80 model PC-4 pictured above. This little
computer had a built-in BASIC language interpreter, and 1 or 2 KB of
memory. I programmed the Line of Position problem, and later,
dead reckoning and fix (intersection of two circles of
position). Eventually I got all the navigational problems to fit
in memory, with a single user interface (if it could be called that).
A facsimile scan of the original Line of Position program from my
old PC-4 notebook together with a contemporary computer BASIC translation of the program and a worked example are reproduced here.
 Errors in this recounting of course notes are mine.