The Longines Lindbergh Hour Angle is a fascinating watch (it is a reproduction, to scale, of the one designed by Charles Lindbergh himself after his famous ocean flight), but its use is not easy for anyone who has not studied astronomy. All the same, we want to try to explain it, to erase once and for all the many false beliefs about this instrument.
One should know, first of all, that to really use it one would need additional tools, such as a sentant, an almanac and ephemeris tables.
To begin with, let us start with the definition of 'hour angle', which is the quantity that can be measured by the clock and is used by navigators to orient themselves with the Sun or the stars. The hour angle of a point (T) on the celestial sphere is the angular distance between the hour circle passing through the point and the meridian of reference (e.g. the Greenwich meridian). It is one of the two coordinates of a spherical coordinate system known as the hour system. The hour angle is always measured towards the west and varies between 0° and 360°.
The second coordinate of the hour system is declination: the angular distance between a point on the celestial sphere and the celestial equator, measured along the hour circle passing through that point. The declination of a star is measured with the sextant.
In order to calculate the hour angle with a watch, it is first necessary to synchronise the watch with the reference time (let us consider Greenwich Mean Time). The Longines Lindbergh Hour Angle allows this to be done very quickly by turning the disc inside the dial (which shows the seconds scale) until index 60 is positioned at the seconds hand when the hour signal strikes the correct time. The crown is pulled out to the intermediate position and the central disc is rotated so that the seconds hand is at index 60 when the hour strikes (e.g. 00:00). From then on, the seconds hand will be read at this central scale on the dial and will correspond to the exact time. The second operation to be performed is that of correcting the equation of time, which can be done by means of the rotating bezel of the watch, graduated in degrees and minutes (60 minutes per degree) from 1 to 15. Civil time, in fact, differs from real solar time due to the difference between the true solar day and the mean solar day (fixed at exactly 24 hours). The indication of the equation of time is not provided by the clock, so it must be looked up in an almanac. Let us take the example of the Lindbergh Hour Angle instruction manual and say that on the day on which you are calculating the hour angle, the equation of time gives you a value of -4 minutes and 50 seconds: this means that solar time is almost 5 minutes behind civil time, marked by the clock. How does this relate to the calculation of longitude?
Let us take a step back.
The concept behind the Hour Angle clock is based on the fact that the earth is divided into 24 meridians, which can therefore be made to correspond to the 24 hours of the day, just as the distance between two meridians can be expressed in 1 hour or in 15° of longitude (15°=360°/24), which is then the angle by which the earth rotates in an hour. Hence, the use of the clock to facilitate the calculation of the hour angle. We can in fact correspond the subdivision of the dial in hours to a subdivision in degrees: each hour corresponds to 15° of longitude. Similarly, the minute sphere will indicate 15° in one revolution of the dial, corresponding to 60 minutes, so 1° every 4 minutes (4=60/15°). Finally, the seconds sphere makes one revolution of the dial in one minute, corresponding to ¼ of a degree, i.e. 15 minutes of a degree. All these graduations are shown both on the rotating disc central to the dial of his watch, which is the reference for the seconds hand, and on the rotating bezel, which is the reference for the minute ball, and on the main dial, which is the reference for the hour hand.
Let us now return to our example. Once the equation of time is known and the clock is synchronised with Greenwich Mean Time, we need to find out at what time the Sun is at its zenith on the meridian where we are. This can be done empirically by noting the moment at which a stick vertical to the ground presents the minimum shadow. Let us suppose that at local noon (empirically detected) that our timepiece marks 04:37:12 (4 and 37 minutes and 12 seconds, as in the example given in the House instruction booklet). We then proceed as follows.
First of all, we make the correction relative to the minutes of the equation of time (-4): we then rotate the rotating bezel by 4 divisions (counter-clockwise), corresponding to 1° longitude (at 12 o'clock, instead of index 15 on the bezel, we will find index 1).
Secondly, we make the correction relative to the seconds of the equation of time (-50): at index 50 (50 seconds) on the rotating disc central to the dial we read the value 12½', corresponding to the angle by which the Earth rotates in 50 seconds.
Let us now move on to the correspondence between the hour and minute indications and the degrees of longitude: at 4:37, the hour hand indicates on the main dial the hour index 4, i.e. the value in degrees of 60°. At the same time, the minute hand indicates the hour index 37, corresponding on the bezel (which has already been rotated for the correction relative to the equation of time) to the value of 10° and 15 minutes. Finally, let us consider the position of the seconds hand at index 12 on the central rotating disc, equivalent to 3 minutes of degree (red indexes on the same disc).
Adding up all these values, we obtain the hour angle of the sun with respect to the Greenwich meridian:
60° + 10° 15' + 3' + 12½' = 70° 30½' at 4:37 minutes and 12 seconds on a given day of the year (not specified in our example).
Having made the calculations at the local noon of our meridian, the hour angle of the Sun with respect to Greenwch equals the longitude of the place where we are.
As can be seen, the matter is far from simple and the common mistake in the description of this clock is to oversimplify matters by claiming that one can calculate longitude with the clock alone.
