Navigation

There are two navigation problems.

Directed pair-wise Navigation

The classical navigation problem is that of the directed pair-wise navigation. The longitude and latitude of each of two points is given; solve the triangle, including its altitude. This is an SAS problem, with the angle at the North pole.

Consider the triangle with the vertex A at the North pole, the vertex B at our present location, and the vertex C at the destination. Then the side c is the co latitude of our present position; while the side b is the co-latitude of the destination. The side a is the distance between our present position and the destination. The angle A is the difference of the longitudes; i.e., the longitude of the destination less the longitude of our present position. Finally, drop the altitude h -- it is the co-latitude of the northernmost point of the journey -- from the vertex A. It partitions the angle A into A1 and A2 and the side a into a1 and a2, with the one being at the present position and two being at the destination.

Employ the spherical Law of Haversines to find the distance a between our present position and the destination

a = 2 Asin(sqrt((sin((c - b) /2 ))^2 + sin(c) sin(b) (sin(A / 2))^2) * sign(A))

(In passing, we note that the relative error of the Law of Haversines is negligible. Had be employed the Law of Cosines; the magnitude of the relative error would increase as the angle a approaches zero. For instance, at a distance on the surface of the Earth of 38 cm the relative error of the Law of Cosines -- computed in an Excel spreadsheet -- is -0.85%. At 12 cm; it is -16%.)

The resulting expression is symmetric in b and c, as it should be. Employ the spherical Law of Sines to find the bearing B from our present position and the bearing C whence we arrive at our destination

B = Asin(sin(b) sin(A) / sin(a))

C = Asin(sin(c) sin(A) / sin(a))

The altitude, by definition, is perpendicular to the base of the triangle. Employ the spherical Law of Sines to find the altitude h. Then, substitute the angle B from the foregoing

h = Asin(sin(B) sin(c) / sin(pi / 2)) = Asin(sin(b) sin(c) sin(A) / sin(a))

The resulting expression is symmetric in b and c, as it should be. Now, we have a right triangle, with the hypotenuse c, one side h, and the angle opposite B known. Employ the spherical Mollweide's Formulae. But, we have to replace the symbols; because, the Mollweide's Formulae are listed with the angle A as the right angle. The replacement is a --> c, b --> a1, c --> h, B --> A1, and C --> C.

#3. cos(a) = cos(b) cos(c) --> cos(c) = cos(a1) cos(h) yields a1 = Acos(cos(c) / cos(h)) * sign(A)

#2b. cos(C) = cos(c) sin(B) --> cos(C) = cos(h) sin(A1) yields A1 = Asin(cos(C) / cos(h))

Likewise

a2 = Acos(cos(b) / cos(h) * sign(A)

A2 = Asin(cos(B) / cos(h))

In the isosceles special case; that is, if the latitudes of the original position and the destination are the same, the foregoing haversine equation for the distance between the two positions simplifies to

a = 2 Asin(sin(c) sin(A / 2)) when b = c

The rest of the algorithm remains substantially unaltered.

For comparison, d the distance along a parallel of latitude is

d = A sin(c)

Let y be their difference

y = d - a = A sin(c) - 2 Asin(sin(c) sin(A / 2))

In the special case of A = pi, these expressions become

a = 2 c

d = pi sin(c)

y = d - a = pi sin(c) - 2 c

To find its maximum, set its derivative equal to zero

0 = dy / dc = pi cos(c) - 2

Solving for c, we obtain

c = Acos(2 / pi) = 0.8806892354203566672269437825689

Substitution back into y yields the maximum

maximum y = pi sqrt(1 - (2 / pi)^2) - 2 Acos(2 / pi) = 0.66134817512852572571566504324526

Throughout this problem -- and, for that matter, all of Spherical Trigonometry -- each side of the spherical triangle is given as the central (of the sphere) angle. To obtain each of the corresponding linear distances, multiply an angular distance -- expressed in radians -- by rho, the radius of the sphere, Earth, in this case.

For any real x, the sign function is defined as

sign(x) = 1 if x > 0, -1 if x < 0, 0 otherwise.

An Excel spreadsheet, named Distance calculator, is available for the foregoing calculation. It is contained in the distcalcnew.zip file. In this calculator, one enters the coordinates of each of the two points (beginning in row eight) as the pair (longitude, latitude) of quintets of sign (negative for east meridian of longitude or south parallel of latitude), name, degrees, minutes, and seconds, for each point. Then, the calculator computes and displays the solution of the navigation problem to the right, on the same row. For use on an other sphere (than the Earth), the radius may be entered near the upper-right corner of the spreadsheet. Please observe that there are many hidden columns, which may be unhidden; if you are curious.

Historical note

Until approximately 1775, when the marine chronograph became available, it had not been feasible to determine the meridian of longitude of a ship at sea. The strategy for sailing from the departure point A to the destination point B was to sail along the great-circle meridian of longitude to the way-point C -- defined as having the same longitude as A and latitude as B -- then to sail along the small-circle parallel of latitude to the point B. Paradoxically, a small-circle is longer than a great-circle between a given pair of points. With the knowledge of time provided by a chronometer -- a modern quartz-crystal wrist-watch suffices --, one may sail directly from point A to point B, along a great-circle. This path passes through the extreme north-south point D. The aforementioned spherical distance calculator accepts the radius of the Earth -- or any other sphere -- (in any linear units) and the longitude and latitude (in degrees, minutes, and seconds) of the points A and B as input and calculates the distances AC, CB, and AB and the difference AC + CB - AB. It also calculates the bearing (measured clockwise from north) of B wrt (= with respect to) A, the bearing (measured clockwise from north) of A wrt B, and the coordinates of the way-point C and of the extreme-point D. However, the extreme-point D may be hazardously close to the icebergs of the poles. The pair of coordinates of any address within the USA is available from Geocode. An excellent custom-made precision astrolabium (Latin; English: astrolabe) may be ordered from Janus, for a modest fee.

Triangulation

Triangulation is the name given to the problem of locating a point, given its distance -- and, perhaps, its bearing -- from each of a set of points.

Known bearing

The Tacan (= Tactical aircraft navigation) system, which Hoffman manufactured for more than a decade, a set of base stations. The problem is to find the location of an aircraft. By measuring the time between an omni directional pulse and the passage of a rotating beacon from a known base station, the aircraft determines its bearing from that base station. The distance from the base station is half the travel time of a signal sent by the aircraft to the base station, which then is sent back by the transponder at the base station.

There are two weaknesses with the Tacan system. The fairly large width of the sweeping beacon introduces an error into the precision of the determination of the bearing. The transponder at a base station may become saturated with requests from multiple aircraft. An aircraft which does not receive a signal back from the transponder assumes itself to be rather distant from the base station; thus to have a weak signal. To compensate, the aircraft increases intensity and priority of its transmitted signal. If instead the response had not been forthcoming because the transponder was saturated, the load upon the transponder will be increased even more. The transponder will lockup, until all of the aircraft cease their requests, by the pilot of the aircraft either switching off or changing to another base station.

This, again, is an SAS problem. Thus, the algorithm is the same as that of the foregoing. However, now, the known included angle A is at the base station. Hence, the relationship between the longitudes and latitudes and the designation of the variables has to be altered, accordingly. Doing so is left as an exercise for the reader. :-)

Unknown bearing

At radio frequencies, angular measurements, with the precision required for navigation, are not feasible, except if one employs a very large interferometer. But interferometric methods are too cumbersome for navigation. Hence, we have to devise methods employing the measurement of distances, alone.
On a plane surface, the locus of a given distance from a given point, is a circle with the given radius and center. Then a second such circle will intersect the first in two points. A third circle will remove this ambiguity. And a fourth circle is desirable as an error-check.
If either depth or elevation is desired, that is if the problem is in three dimensions, one additional – now sphere – is required.
However, the measurement of distance requires either a reflector (or preferably, a transponder), so one can measure the round-trip distance. The requirement for an active role on the part of the reference stations places them in the undesirable situation of being subject to saturation (exhaustion of their resources).
If clocks were perfect, we could transmit a pulse at an agreed-upon time from the reference stations. Then the time of the reception of the pulse from each station would indicate the distance to each station.
However, your clock drifts. How do you measure distance and the time-of-day, both?
Well, time-of-day is easy, once you know your position, hence the propagation delay from each reference station. Good.
But, how do you measure distance, without knowing your own time?
Well, all you can measure is the difference in the arrival times of the pulses from the reference stations. Then, the locus becomes a hyperbola (in two dimensions) or a hyperboloid of revolution of two sheets (in three dimensions). Observe that we require one additional reference station. Only one of the branches of the hyperbola -- or sheets of the hyperboloid -- Is pertinent, the other being extraneous.
Thus, in three-dimensional space, five reference stations are required, plus additional ones for error-checking, redundancy, and to compensate (at least in part) for an unfavorable pattern of deployment of the reference stations and for the possibly unfavorable viewing angle of the reference stations.
This is the problem that the GPS (= Global Positioning System) navigational device has to solve. The Sonobuoy has the same problem when it is operating in passive mode. In active mode, it needs one less Sonobuoy and the Sonobuoys will employ spheres.
Modern clocks are extremely stable; thus once the local clock is synchronized with the clocks on the GPS satellites, the GPS receiver may switch to employing spheres. The slight drift of the local clock can be corrected from the miss-alignment of the intersection of the redundant spheres.
In summary, the minimal requirements are:

	Active (Time is known)	Passive (Time is unknown)
Two-dimensional	Two reference stations employing circles	Three reference stations employing hyperbolas
Three-dimensional	Three reference stations employing spheres	Four reference stations employing hyperboloids of revolution of two sheets

But, as discussed previously, one more reference station is required to select from among the several intersection points. Additional (redundant) reference stations will improve the precision of the determination of the intersection point.

Where Are We?

Abstract. However, before one can navigate from here to some desired destination, one must know where one is. The most precise determination of ones location may be obtained by the use of a GPS (= Global Positioning System) device -- provided the system and the device are available. Alternatively, one may employ celestial navigation: Given observations of two known celestial objects (or, of the same object, at two different times), find the location of the observer and the direction to the North Pole. Required instruments: a sextant and a marine chronometer (a modern digital wristwatch will do). Also required is an ephemeris of the Sun. From this ephemeris, you would know the declination (latitude) of the Sun each day. You could interpolate to find the declination at the time of each observation. The difference in time between the two observations gives you the angle at the North-Pole.

Formulae. Since we will be employing the spherical Law of Haversines thought the following discussion, let us quote it here:

haversine(a) = haversine(b – c) + sine(b) sine(c) haversine(A)
(sin(a / 2))^2 = (sin((b - c) / 2))^2 + sin(b) sin(c) (sin(A / 2))^2
(sin(A / 2))^2 = (sin(a / 2) + sin((b - c) / 2)) (sin(a / 2) - sin((b - c) / 2)) / (sin(b) sin(c))

and the short-form of the Law of Sines, which we employ only to determine the sign of an angle

sin(A) / sin(a) = sin(B) / sin(b) = sin(C) / sin(c).

We consider each side of a triangle as being positive. We also will be employing the algorithm for the solution of a triangle. "To solve a triangle" means that given three (or more) of the six elements of a triangle, find each of the remaining elements. Finally, for a triangle, the spherical excess E is given by

E = A + B + C – pi = area / rho^2,

where rho is the radius of the Earth. For a spherical-triangle, the value of the spherical excess E is in the interval [0, 4 pi]. However, whenever any of the partitioning triangles of the triangle ANB falls outside of ANB, we assign its angles so that its spherical excess -- and hence area -- would be negative. Then, the sum of the spherical excess of the partitions adds up to the spherical excess of the whole triangle ANB. Please observe that if the spherical excess of any one (or more) of the four triangles should be exactly zero -- as a result of the three vertices being in a straight-line --; the problem becomes meaningless and hence the spreadsheet calculation will error-out -- there is no diagnostic provided.

Combining the foregoing formula, we obtain

SAS
- a = 2 Asin(sqrt(sin((b - c) / 2))^2 + sin(b) sin(c) (sin(A / 2))^2))
- B = 2 Asin(sqrt((sin(b / 2) + sin((c - a) / 2)) (sin(b / 2) - sin((c - a) / 2)) / (sin(c) sin(a)))) * sign(sin(A))
- C = 2 Asin(sqrt((sin(c / 2) + sin((a - b) / 2)) (sin(c / 2) - sin((a - b) / 2)) / (sin(a) sin(b)))) * sign(sin(A))
SSS
- For the angle A, both solutions are valid. Some external condition may select one of them, however.
  - A = 2 Asin(sqrt((sin(a / 2) + sin((b - c) / 2)) (sin(a / 2) - sin((b - c) / 2)) / (sin(b) sin(c)))) acute or obtuse angle
  - A = 2 pi - 2 Asin(sqrt((sin(a / 2) + sin((b - c) / 2)) (sin(a / 2) - sin((b - c) / 2)) / (sin(b) sin(c)))) reflex angle
- B = 2 Asin(sqrt((sin(b / 2) + sin((c - a) / 2)) (sin(b / 2) - sin((c - a) / 2)) / (sin(c) sin(a)))) * sign(sin(A))
- C = 2 Asin(sqrt((sin(c / 2) + sin((a - b) / 2)) (sin(c / 2) - sin((a - b) / 2)) / (sin(a) sin(b)))) * sign(sin(A))

Statement of the problem. Let A be the location of the first celestial object and B that of the second. In this discussion, without loss of generality, we assume that B is westerly of A. Otherwise, interchange these two points. (However, the spreadsheet will handle B easterly of A as an absolute value. Then, the computed longitude of O, as well as the computed value of the bearing of A and that of B will be adjusted accordingly.) Let N be the location of the North Pole. And, let O be the location of the observer. In this discussion, we assume that the point O is interior to the triangle ANB. Following the British, we adapt a left-handed coordinate system: The longitude, measured positive in the westerly direction; the latitude, measured positive in the northerly direction; and the radius, measured positive out from the center of the Earth. As a consequence, the bearing is measured clockwise.

We are given: The coordinates (longitude and latitude) of A and those of B. The observed altitude (elevation) of A and that of B. We want to find: The coordinates of O and the bearing of A and that of B. To do so, we have to solve three spherical triangles; but, actually, solve four as an internal check.

Triangle one NBA. We know the side AN as the co-latitude of A, the side NB as the co-latitude of B, and the included angle ANB as the longitude of B minus that of A. Thus, this is an SAS triangle. We employ the Law of Haversines to find the remaining side BA. Now, the triangle becomes SASS, which we consider as SSS. Then, we finish the solution of this triangle by means of two applications of the Law of Haversines, supplemented by the sign from the short-form of the Law of Sines, to obtain the angles BAN and NBA. The spherical-excess is the sum of the three interior-angles of the triangle, minus two pi. Here, we have translated from spherical-coordinates to a spherical-trigonometry problem. By solving this triangle, we have obtained the distance AB between the two points A and B.

Triangle-two OBA. We know the side AO as the co-altitude (co-elevation) of A, the side OB as the co-altitude (co-elevation) of B, and the remaining side BA as that obtained from the solution of the first triangle NBA. Thus, this is an SSS triangle. We employ the Law of Haversines to find the angle AOB. We obtain its sign as a minus-one if the observer is southerly from the line AN. Then, we finish the solution of this triangle by means of two applications of the Law of Haversines, supplemented by the sign from the short-form of the Law of Sines, to obtain the angles BAO and OBA. The spherical-excess is the sum of the three interior-angles of the triangle, minus two pi. By solving this triangle -- the climax of our problem --, we have found the specified one (of two) intersection point O of the two circles: of radius AO about A and of radius BO about B. The crucial result is the angle AOB.

Triangle-three ANO. We know the side AN as the co-latitude of A. (We already used this same AN for the first triangle.) We know the side OA the co-altitude (co-elevation) of A. (We already used this same AO for the second triangle.) We compute the included angle OAN as the difference of BAN and BAO, which angles we obtained from the solution of the first and second triangles, respectively. Thus, this is an SAS triangle. We employ the Law of Haversines to find the remaining side NO. Then, we finish the solution of this triangle by means of two applications of the Law of Haversines, supplemented by the sign from the short-form of the Law of Sines, to obtain the angles ANO and NOA. The spherical-excess is the sum of the three interior-angles of the triangle, minus two pi. By solving this triangle, we have found the distance NO of the point O from the North Pole.

Triangle four BON. This is an optional solution – as a check. . We know the side NB as the co-latitude of B. (We already used this same NB for the first triangle.) We know the side BO the co-altitude (co-elevation) of B. (We already used this same OB for the second triangle.) We compute the included angle NBO as the difference of NBA and OBA, which angles we obtained from the solution of the first and second triangles, respectively. Thus, this is an SAS triangle. We employ the law of haversines to find the remaining side ON. Then, we finish the solution of this triangle by means of two applications of the Law of Haversines, supplemented by the sign from the short-form of the Law of Sines, to obtain the angles ONB and BON. The spherical-excess is the sum of the three interior-angles of the triangle, minus two pi. By solving this triangle, we again have found the distance NO of the point O from the North Pole.

Clean-up. The angle ANO is the longitude of O minus that of A. Hence, the longitude of O is that of A plus the angle ANO. As a check, the sum of the angles ANO and ONB should equal the angle ANB. The side NO (obtained in triangles three and four – these values should be equal) is the co-latitude of O. The angle NOA is the bearing of A. Minus the angle BON is that of B. As a check, the angles around the observer O should add-up to two pi. Also, the sum of the spherical excesses of the triangles OBA, ANO, and BON should equal the spherical excess of the triangle NBA. Finally, we translate back from spherical-trigonometry to the spherical-coordinates.

An Excel spreadsheet -- entitled where are we -- is contained in the distcalcnew.zip file. The first two rows of calculations are examples. Following a blank row, enter the data as follows: Column #A a minus-one if the observer O is southerly of the line AN; otherwise a plus-one. Columns #B, #C, and #D are the longitude, latitude, and altitude (elevation) of the celestial-object A. They are to be entered in radians. Columns #E, #F, and #G are those of the celestial-object B. All of the remaining columns are computed by the spreadsheet. For each of the four triangles, the (remaining) elements are as labeled. The label E indicates the spherical-excess. Hidden columns #AE, #AF, #AG, and #AH are the aforementioned checks. Finally, the column #AI is the longitude of the observer O and #AJ is the latitude. Column #AK is the bearing of the celestial-object A and #AL is that of the celestial-object B. These results are in radians.

The calculation within the spreadsheet, of necessity, is performed in radians. At present, the input and output values also are in radians. However, the input and output could be in some other form, for instance: 1) hours, minutes, and decimal seconds. 2) decimal degrees. 3) degrees and decimal minutes. 4) degrees, minutes, and decimal seconds. 5) grads. Either obtain the spreadsheet from me and add the necessary conversions yourself or else request me to do so for you.

All of the foregoing is done on a sphere. By convention, we project any three-dimensional point radially to the standard sphere, on which we find the shortest distances -- the great circles are the geodesics --, angles, and surface area.

GPS

A GPS (= Global Positioning System) device provides is easier to use than a sextant and provides our location with a much smaller error. Furthermore, it provides the third coordinate -- the height wrt (= with respect to) the oblate-spheroid (usually called the ellipsoid). Allegedly, however, the latitude is also the ellipsoidal latitude. Since this page is in the spherical trigonometry, we provide a link to a discussion of the ellipsoidal coordinates. We also have written an Excel spreadsheet for this ellipsoidal input. It is contained in the distcalcnew.zip file. In the upper right-hand corner, I have entered the average, equatorial, and polar radii of the Earth in Kilometres. Hence, each height also has to be entered in Km. Likewise, each linear distance is displayed in Km. You may change to any consistent units by entering the three appropriate radii of the Earth or any other oblate spheroid. Caution: the latitude *must* be entered as the ellipsoidal latitude. The corresponding spherical latitude is computed and displayed. The longitude and latitude may be entered in any combination of degrees (or decimal degrees), minutes (or decimal minutes), or seconds (or decimal seconds). You will have to unhide the columns to be able to enter either minutes or seconds.