A navigator, adrift at sea in a life raft or on the ground, makes a last valiant effort on the autumnal equinox to fix his position using his only available tools—an accurate chronometer set to Greenwich Mean Time (GMT) and a slide rule. He observes the Sun at sunset and times its transit (lower to upper limb) through the horizon. He notes the time that the upper limb of the Sun is on the visible horizon as 0547 GMT (he assumed that 1800 local mean time at Greenwich is the time of sunset at the autumnal equinox). The transit is 127.8 seconds. Where is he and who might he be?

Hint: The single body position fix can be used to determine latitude through the use of the rate of change of elevation (dh/dt) and the knowledge of the azimuth (Z) of a celestial body. Longitude can be determined by observing the time of local sunset in GMT. The equation for latitude (L) is cos(L) = dh/dt is the elapsed time of the transit of the Sun’s disk through the horizon and Z is the Sun’s azimuth (true bearing of the celestial body). Assume the diameter of the Sun is 31.8 arcminutes. On the equinox (vernal or autumnal) the Sun sets due West (270 deg) at all latitudes. Therefore the cscZ term goes to unity. Knowing that the dh/dt term will be in units of arcmin/sec, you need the conversion term of arcmin/4 sec (same as 15 deg/hr) to achieve a dimensionless answer to determine latitude.

a. Commander Richard E. Bird, after his ill-fated Fokker transatlantic flight and forced landing in the sea in an attempt to be among the first to cross the Atlantic from West to East at 49° 30'N latitude and 1° 00'W longitude

b. The navigator, First Lieutenant John J. DeAngelis, of the B-17 crew ferrying Eddie Rickenbacker that missed Canton Island, ran out of fuel, crashed and was rescued with his remaining crew and Rickenbacker near the Ellice Islands in the Pacific during WWII at 7°S latitude and 177° 40'W longitude

c. Fred Noonan, Amelia Earhart’s navigator, in the Lockheed Electra down in the Pacific near the missed destination of Howland Island at 5°N latitude and 17° 45'W longitude

d. The navigator, 2nd Lt. Dp Hays, on the ill-fated B-24 “Lady Be Good” during WWII returning from a bombing mission from Italy to Libya near the crash site 26°N latitude and
13° 12'E longitude

The answer is:

Amelia Earhart and her navigator Fred Noonan, in their Lockheed Electra on their famed around-the-world flight, were lost at sea in their 2,500 mile leg from New Guinea to Howland Island (a speck in the Pacific) near the intersection of the equator and the international date line on July 2, 1937. They were never found—despite intensive searches. Many theories have been advanced and books written on the fate of Amelia Earhart and Fred Noonan. Suffice to quote an old mariner’s saying: “The ocean is so vast and my boat is so small.” The solution is based on a “What if…” account: for this Brain Game, Noonan found himself.

Determining latitude (L) is merely solving cos(L) = dh/dt csZ (from a differential calculus derivation* of the astronomical triangle used in celestial navigation). The azimuth of the Sun has to be accurately known which fortunately is the case on the autumnal equinox; it is 270 deg (in navigation Z is measured from true north clockwise to the body). The cscZ for 270 deg is unity. Solving for latitude (L): cosL = 31.8 arcmin/127.8 x 4 sec/arcmin = 0.995

L = 5°N latitude by slide rule, 5.55°N by calculator

Solving for longitude: The observation was made at 0547 GMT. Knowing that the Sun moves at 15 deg/hour (Earthrate) and sunset occurred at 1800 the previous day at Greenwich (0 deg longitude), Noonan was able to establish his longitude by multiplying the elapsed time between his observation of sunset and sunset at Greenwich by Earth rate: 15 deg/hour (Earthrate) and sunset occurred at 1800 the previous dat at Greenwich (0 deg longitude), Noonan was able to establish his longitude by multiplying the elapsed time between his observation of sunset and sunset at Greenwich by Earth rate: 15 deg/hr x (0547 + 2400 - 1800) = 15 deg/hour x 11 hr 47 min (11.783 hr) = 176.75 deg which is expressed as

176° 45'W longitude

In summary, unique circumstances made it possible to employ this single body position fix (Figure 15 ) solution with only a chronometer and slide rule: it was the autumnal equinox and the Sun was setting over the equator (0 deg declination and due West throughout the world allowing the single body fix equation to reduce simply to one term cos(L) = dh/dt as the cscZ term went to unity.