Position front rate of change of altitude.—Air navigators should be especially interested in the following method of navigation which is entirely independent of the tedious method of finding position by deductions, known as dead reckoning. It is an accepted fact that dead reckoning in the air is far more difficult than at sea, due to the highly variable factor presented by the wind.
Mr. Edward J. Willis has contributed a method of finding position, entirely without the aid of dead reckoning. Furthermore, at sunrise or sunset, position may be found by his method without using dead reckoning or a sextant. It is surprising to find that such an advantageous method is unknown to most navigators. The method has been published since 1928 in a brochure entitled, The Line of Azimuth, which is a supplement to The Methods of Modern Navigation, by the same author. The obscurity of Willis’ method may be due to the lack of descriptive value in the title of the brochure in which it is described. Then again, it may be due to the fact that his equations are derived by means of the calculus, a useful tool which navigators are prone to disregard.
For the navigator who prefers spherical trigonometry to the calculus, it will be shown that Willis’ equations may be derived by means of nothing more than the law of sines, together with one of Napier’s rules, “The sine of any middle part is equal to the product of the cosines of the opposite parts,” remembering to take the complement of the hypotenuse and the angles adjacent thereto previous to applying the rule. The equations obtained may then be converted into functions of the secant and cosecant so that H. O. 211 may be used for their solution.
Willis’ equations for finding latitude and local hour angle are based upon Prestel’s equation,
cos L=dh/d csc Z0,
for finding the latitude by means of the rate of change of altitude of a body. Z0 is the azimuth measured at the equator from the elevated pole. The fractional part of the equation represents the change in altitude corresponding to a change in time, or hour angle. This ratio is known as the rate of change of altitude. By dropping the esc Z0 from the above equation, Prestel’s equation for finding the approximate latitude when the body is on the prime vertical is obtained. The esc Z0 may be dropped since the natural cosecant of an angle within one degree of the prime vertical is practically unity.
Willis uses Prestel’s equation, transposed to make the azimuth the dependent variable:
sin Z0=dH/dt sec L.
Since Z0is the azimuth measured from the equator, the latitude from which the azimuth is measured will always be 0°. The natural secant of 0° is unity; therefore sec L may be dropped from the above equation, giving
sin Z0=dH/dt
So that H. O. 211 may be used for its solution, the equation will be inverted,
csc Z0=dt/dH
csc Z0=time interval in secs. of time/4 /alt. interval in mins. of arc
Prestel’s equation for finding the latitude is accurate theoretically; that is, it would give an accurate latitude if the rate of change of altitude could be obtained for the instant that the body is on the prime vertical. This is practically impossible, with present instruments, as altitude and time must change (considerably for practical reasons) to furnish data for finding the rate of change of altitude, and while altitude and time are changing there will be a resultant change in azimuth.
In Prestel’s equation for finding the latitude, the function of the azimuth was dropped with the understanding that the equation is to be used only when the body is on the prime vertical. If the body is not on the prime vertical the equation is consequently inaccurate, as it no longer contains a correction for azimuth. Since a change in azimuth accompanies a change in time and altitude, Prestel’s equation with the azimuth correction omitted cannot give an accurate solution for latitude, unless an instrument is devised for procuring the rate of change of altitude at the instant the body is on the prime vertical.
Willis’ equation, although derived from Prestel’s equation for finding the approximate latitude, does not give an approximate azimuth, but gives an accurate one, providing the time interval (dt) corresponding to an altitude interval (dH) is taken accurately. The azimuth will be accurate since the rate of change of altitude is the independent variable and consequently any change in the independent variable will directly affect the azimuth, while the independent variable itself is unrestricted. With Prestel’s equation for finding the approximate latitude, azimuth is a part of the independent variable, so, unless the rate of change of altitude could be found exactly when the body is on the prime vertical, the latitude (dependent variable) will be directly affected by the resulting change in azimuth.
A study of tables VII and VIII in H. 0. 200 will be of value, as they show the change of altitude per minute of arc of hour angle and the change of azimuth per minute of arc of hour angle, respectively, at different latitudes. These tables will show at what latitudes and at what time of the day best results may be obtained from the use of Willis’ method. Table VII shows that the rate of change of altitude is zero when the body is on the meridian or when the observer is at the pole, and
[FIGURE 1]
that its maximum value, unity, is attained when the body is on the prime vertical and the observer on the equator. Willis’ method therefore cannot be used in extremely high latitudes, and when the body is on the meridian there is of course no substitute for the conventional meridian altitude method.
From equation (1), the azimuth of a body measured at the equator from the elevated pole may be found. This azimuth would be of little value in the solution of the conventional polar triangle (PMZ, Fig. 2), but its use will greatly simplify a solution of the triangles whose bases lie along the equinoctial (hereafter referred to as the equinoctial triangles).
The equation for finding latitude will now be developed from these triangles. It is evident from Fig. 1 that the distance from observer to body is (90°-H). Knowing the distance from the observer to the body, the distance from the body to the point where the line of azimuth cuts the
[FIGURE 2]
equator is now to be found. From the equinoctial triangle (Fig. 2) containing the declination, the value of c is found by using Napier’s “opposite side” rule, and D as a middle part,
sin D =cos Z0 sin c
sin c = sin D sec Z0
c = arc sin (sin D sec Z0).
The distance from the observer to the point where the line of azimuth cuts the equator is now known, and is
c' = (90° —H) -fare sin (sin D sec Z0).
The equinoctial triangle containing the latitude will now be solved, using L as the middle part and giving
sin L =cos Z0 sin c'.
Substituting the value of c’ gives
sin L = cos Z0 sin [(900-H+arc sin (sin D sec Z0)]
sin L = cos Z0 sin [H+co arc sin (sin D sec Z0)]
sin L = cos Z0 sin [(H-arc sin (sin D sec Z0)]
sin L = cos Z0 sin [H-arc sin sin D/cos Z0]
sin L = cos Z0 sin [H-arc csc csc D/sec Z0]
The equation (2a) is for finding the latitude when L and D have the same sign and L is greater than D. There are three other cases, but since the required space would be too great to show their individual derivation, the equation which applies to these three cases is given. It will be noticed that equation (2b) merely differs from the former equation (2a) by having a plus instead of a minus sign.
csc L= sec Z0 sec [h+ arc csc csc D/sec Z0] 2(b)
Perhaps the simplest method of finding the local hour angle equation by spherical trigonometry is to solve the conventional polar triangle (PMZ) for t, using the law of sines; the result will be the conventional local hour angle equation.
[EQUATION]
Referring to the equinoctial triangle containing the latitude, the value of Z. will be found in terms of Z0 by using Napier’s rule and co Z0 as the middle part,
cos Z0 = sin Zs cos L
sin Zs = sin Z0 sec L.
Since the sine of an angle is identical to the sine of its supplement (180°-Angle):
sin Zn = sin Z0 sec L.
Substituting this value for the sin Z0 in the conventional local hour angle equation and then inverting gives
sin t =cos H sin Z0 sec L sec D,
csc t =sec D sec L/ sec H csc Z0 (3)
Equations (1), (2), and (3) are all that are required for finding position from the rate of change of altitude.
Space will not be consumed in discussing the methods of finding the rate of change of altitude as this subject is thoroughly covered in Willis’ The Line of Azimuth. Briefly, there are two methods of finding this rate. One method, Prestel’s, can only be used with the sun or moon as it requires the time that the body takes to change its altitude equal to its diameter, the diameter being taken from an almanac after the observation. The other method is to take two altitudes and measure the time interval between them. With either method a stop watch will give a more accurate reading of the time interval than either a chronometer or a hack watch. This is because time is recorded instantly on a stop watch by a mere pressure on the stem, while time from a chronometer or a hack watch must be read from a rapidly rotating hand. The stop watch must be accurate and of a type that is either started or stopped by a slight pressure on the stem. Some stop watches have separate starting and stopping devices of a type that cannot be conveniently operated with one hand. The stop watch can scarcely be considered as an extra or unusual instrument since many navigators use it in preference to a hack watch. In either of the above methods the sextant altitude, used to obtain the observed altitude which is a part of Willis’ equations, is the mean of two altitudes, and the G.C.T. corresponding to this altitude is the G.C.T. used in finding the declination and G.H.A. from an almanac.
The observation for the rate of change of altitude is slightly more difficult than the usual single altitude observation. However, it will be appreciated that with Willis’ method position is obtained by direct calculation, without chart or plotting sheet work; while with the single altitude two observations separated by a considerable change in hour angle are required to procure an accurate fix.
The second problem from page 17 of The Line of Azimuth will be reworked by H. 0. 211. This problem is of particular interest as the observation was taken by Mr. Willis at sunset with a stop watch and a chronometer set to G.C.T., as his only navigational instruments. The G.C.T. is required to find the declination and G.H.A. from an almanac.
Notes.—The division of dt by dH (Equation 1) is preferably worked by logarithms. It will be found convenient to paste a 5- place table of logarithms between the two blank pages of H. O. 211, labeled “Star Data from Nautical Almanac.” The symbols X, Y, U, and V, used in working the following problem, are arbitrary, and are used solely to simplify the printing of the problem. It will be noticed that the cologarithm of 4 is a constant for all problems. The division of time by 4 is necessary as values of time and arc must be related, and 1 minute of arc is equal to 4 seconds of time. It is not necessary to use a table of cologarithms as the cologarithm may be readily taken from a logarithm table by mentally subtracting the logarithm from 10 (figure by figure, starting from the right) while taking it from the table. The characteristics of the logarithms and cologarithms need not be used. The following observation was of the sun. When a star, planet, or the moon is the observed body, the time interval obtained from a stop watch adjusted to solar time must be converted to sidereal time. If the use of an extra instrument is considered preferable to the application of a simple correction, a stop watch adjusted to sidereal time is suggested. If the observer is moving, a correction for the run should be made. This correction is given on page 14 of The Line of Azimuth. Since the L.H.A. was greater than 90° it was taken from the bottom of the table.
[FIGURE]
The above observation was taken by Mr. Willis aboard the yacht Oceana. At the time of the observation the yacht’s D.R. position was Lat. 43°-02' N., and Long. 14°-23' E.
Although the D.R. position is not used in Willis’ method, an example is given to show that it could be used to advantage in precomputation work.
Example.—The navigator of a Pan- American air liner flying the great circle from Alameda Airport to Honolulu desires to know the time in seconds that the sun will take to change its altitude equal to its apparent diameter when the “clipper” reaches D.R. Lat. 30° N., and D.R. Long. 143° W., which he expects to reach at G.C.T. 18h of November 17, 1935. This time interval is to be used as a check on the actual time interval obtained by an observation taken by Prestel’s method.
Two altitudes were precomputed by H. O. 211 six minutes on each side of the desired G.C.T. of the observation. The only interpolation made was when taking Hc from the tables.
[EQUATION]
The sun’s diameter (2 S.D.) from the almanac is 32.42'.
[EQUATION]
Explanation.—Dividing the altitude difference by the diameter gives N, which is the number of body’s diameters that are in the altitude difference (e.g., 3.7 diams. in above example).
Dividing the G.C.T. difference by the number of diameters, N, gives the time in seconds that will be required for one diameter to change its altitude by an amount equal to its diameter.
The G.C.T. from which the altitudes were precomputed, 6 minutes on each side of the G.C.T. of the observation, is an arbitrary amount, which may be varied without affecting the result. Six minutes was used as it is one-tenth of an hour, and therefore simplifies interpolating for the declination.
Meridian altitude Sumner line.—A method of finding the meridian altitude Sumner line involving but one simple rule will be derived from Ageton’s equations for the secant-cosecant Marcq Saint-Hilaire method. Although the equation for finding the computed altitude is derived from Ageton’s equations, H. O. 211 is not required for its solution.
Referring to the diagram on page 4 of H. O. 211, it is apparent that for a meridian altitude observation the body M will coincide with X, R and t will therefore equal 0, and K will equal the declination.
From Ageton’s equation (2), on the same page as his diagram, it will be shown trigonometrically that K is equal to D.
[EQUATION]
Sec (D~L) will be substituted for sec (K~L) in Ageton’s equation (3), and since R is 0 and the secant of 0 is unity, R may be dropped from the equation, and it becomes
(3) esc H = sec (D~L).
It is evident that the trigonometric functions are of no more value than of serving to find the complement of (D~L), therefore the computed altitude is merely equal to the complement of (D~L), and our final equation for finding the computed altitude is
[EQUATION]
Problem 2 from page 119 of Cugle’s Practical Navigation (1927 ed.) will be reworked by the above equation and by Dreisonstok’s method, so that a comparison may be made.
It will be noticed that the only rule involved in the new method is a rule common to most navigational methods; if D and L are same name subtract, if different names add. With Dreisonstok’s method the rule for combining D and b is opposite to the rule just given, and in addition it is required to remember that b takes the same name as the latitude. Therefore, there are the equivalent of two rules to Dreisonstok’s method against one by this method.
This problem should be accompanied by the secant-cosecant time and altitude azimuth formula given at the end of this article. An altitude intercept without an accurate azimuth is incongruous. The azimuth of a body on the meridian is either 0° or 180° according to whether the observer is facing the elevated or the depressed pole when taking the observation. In the above problem the body apparently bore south, so the azimuth should have been 180°. Actually the body might have been appreciably off the meridian and consequently its azimuth would be somewhat different from 180°. A calculated azimuth would show this divergence.
The calculated azimuth will complete the required data for laying down an accurate Sumner line. However, since the above methods require the use of the D.R. longitude, as well as the D.R. latitude, it is doubtful if the advantage in using the Sumner line will compensate for the disadvantage of using the D.R. position. The writer prefers the well-known meridian altitude method, which is unhampered by dead reckoning.
The value for He obtained in the above problem may be used, if desired, as a noon constant, by applying the usual corrections with reversed signs.
If the latitude were required instead of the Sumner line, the altitude intercept, a, may be used as a latitude correction.
D.R. Lat. 18°-10.5'
Cor. a + 0.1'
LAT. 18°-10.6' (Computed)
This method, given in H. 0. 208, has the disadvantage of requiring the D.R. latitude, without having the advantage gained from the use of the Sumner line.
[FIGURE 3]
The sign of correction a may be determined either by means of a rough sketch such as Fig. 3, or by means of a rule suggested by Mr. Mclsaacs in “Meridian Altitudes by H. O. 208,” in the March, 1935, issue of the Proceedings. The rule being, “Name H0 plus and H0 minus, then take the lesser from the greater and apply the sign of the greater to the difference.”
Longitude at the equator by H. 0. 211.— The only occasion that the celestial horizon is coincident with a celestial meridian is when the observer is on the equator. Such a condition greatly simplifies finding the local hour angle and consequently the longitude. It is apparent in Fig. 4 that if the declination were zero, ZM and Zq would be congruent, and t would therefore be equal to (90-17), or the zenith distance. Admiral Coutinho made use of this simple fact to find t on his flight from the Cape Verde Islands to St. Paul’s Rocks. Declination is seldom zero, and then only momentarily, so interpolations are required to correct t. Since on the equator zenith distance and hour angle change at the same rate, interpolation is simplified.
[FIGURE 4]
The simplest of interpolations, however, are undesirable, so the following method is recommended. No rules are involved in this method, and the sign of the declination is immaterial.
The value of t at the equator will be found by solving triangle X (Fig. 4), using H as a middle part.
[EQUATION]
Example.—March 29, 1935, crossing equator at G.C.T. 22h, corrected altitude of sun 70°-57.5', find the longitude by H. 0. 211.
[EQUATION]
Time and altitude azimuth by H. 0. 211. —After reading Commander J. B. Olden- dorf’s valued discussion of the writer’s “Longitude by H. O. 211” in the April, 1935, issue of the Proceedings, and Lieutenant Commander W. A. Mason’s “Fallacy of the Time Sight” in the January, 1935, issue, the addition of an azimuth formula to the time-sight formula was considered at least desirable, if not an essential, requisite.
The secant-cosecant time-sight formula given in “Longitude by H. O. 211,”
[EQUATION]
should be complete when supplemented by the standard time and altitude-azimuth formula converted into terms of the secant and cosecant,
[EQUATION]
or by Lieutenant Ageton’s modification of this formula, given in his discussion of “Longitude by H. O. 211” in the August, 1935, issue of the Proceedings,
[EQUATION]
The time sight, taken when the body is on the prime vertical, is an extremely accurate method of finding either L.A.T. or longitude, and should be accompanied by an azimuth formula that will give azimuth to a corresponding degree of accuracy. H. 0. 71 and H. O. 120 are of value in finding the approximate time of the body on the prime vertical in order to know when to take the observation. An azimuth computed from the same data used in working the time sight will show whether or not the body was actually on the prime vertical at the time of the sight. The above azimuth formulas are accurate and are easily solved, while an accurate azimuth obtained from H. O. 71 or from H. O. 120 requires a tedious grill-work interpolation, or some equally laborious method.
If time is desired and a time sight is taken for it when the body is assumed to be on the prime vertical, the value of an accurate azimuth with the time sight formula seems evident. If the azimuth does not show that the body was actually on, or close to, the prime vertical, the sight should not be expected to give accurate time. If longitude is desired and the body is somewhat off the prime vertical, or the D.R. latitude is very much in error, it cannot be found with accuracy from the time-sight formula, but an accurate azimuth will aid in laying down a Sumner line by the “tangent” method.
To illustrate the use of the above formula the time and altitude azimuth problem from page 157 of H. O. 9 (1934 ed.) will be reworked by H. O. 211.
t | 39°-52' E. |
d | 22 -07 S. |
H | 24 -59 |
log CSC t | 19314 |
log sec d | + 3319 |
| 22633 |
log sec H | - 4266 |
log esc Z0 | 18367 |
Z0 | S 40°-56' E |
Time azimuth by H. O. 211.—Since altitude is involved in the formulas, the azimuth obtained from Lieutenant Ageton’s formulas given on page 4 of H. O. 211 might appear to be a time and altitude azimuth. However, it will be noticed that the calculated altitude, and not the observed altitude, is used in his formulas. The observed altitude is not used in a time azimuth, but the use of the calculated altitude is immaterial.
The second time azimuth problem from page 155 of H. O. 9 will be reworked by H. O.211.
t | 56°-04' E. |
d | 7 -38 N. |
L | 2 -16 N. |
log CSC t | 8108 |
log sec d | + 387 |
log esc R | 8495 |
log esc d | 87669 |
log sec R | -24486 |
log esc K | 63183 |
K | 13°-30' N. |
L | - 2 -16 N. |
{K~L) | 11 -14 |
log sec (K~L) | 840 |
log sec R | +24486 |
log esc Hc | 25326 |
log sec R | 8495 |
log sec Hc | -8104 |
log esc Z | 391 |
Z | N 82M9.5 |
Altitude azimuth by H. 0. 211.—The altitude azimuth problem from page 156 of H. O. 9 will be reworked by H. 0. 211.
[EQUATION]
Amplitude by H. O. 211.— The top amplitude problem from page 154 of H.O.9 will be reworked by H.O. 211.
[EQUATION]
Note.—The above azimuth problems were computed entirely without interpolation