In most of my sailing experience, the only confusion that arose in my mind over direction was in making conversions from true to magnetic to compass. Then I started to study spherical trigonometry and everything I thought I knew about direction went into the wastebasket.

The trouble started when trying to understand the concepts behind amplitude. Amplitude is the angle north or south from the equator to the sun when it is on the celestial horizon (crudely put, at sunrise or sunset). One fact that I hoped to explain was: why, when viewed from 40anddeg; north latitude, would the sun rise north of east all summer long. It seemed to me that because the sun’s declination is never greater than 23 1/2 degrees north, the sun would always be south of my position and therefore should have a bearing of southeast at sunrise.

Spherical trigonometry helped me out of this jam. Using only the most basic knowledge, and a simple trigonometric formula (this can be found in a recent issue of Ocean Navigator, “Triangles in the round,” Issue No. 51), I was able to construct a diagram of the navigational triangle at sunrise. For me, the critical element was to realize that the andquot;horizonandquot; is a circle around our zenith position on the celestial sphere. This circle has a radius of 90 degrees of arc. That means that any line from the zenith to the horizon has a length of 90 degrees of arc.

Since amplitude is measured north or south from the intersection of the horizon and equator, I had to figure out the bearing to this intersection. To do this, I constructed a navigational triangle placing the sun on the equator at its intersection with the horizon (see accompanying figure).

My goal was to solve for the bearing from the zenith to the sun – the azimuth angle (Z). In this triangle, the declination is 0 degrees because the sun is on the equator. The sun’s altitude (H) is also 0 degrees because the sun is on the horizon. Latitude is 40 degrees here, but could be any value. Use the standard formula for determining Z:

Sin 0 degrees is 0, so both sin Dec and sin H are 0. Cos 0 degrees is 1, so the result is:

Z is 90 degrees. Once this is determined, it is relatively easy to continue on and understand amplitude. My confusion with direction starts here. This result seems to say that the bearing from our zenith to a point on the equator is 090 degrees true, or due east. This is a disturbing concept; how could the equator be east? One of the certainties of direction in my mind is that the equator is south of any position in the northern hemisphere.

Experience has taught me that if I sail due east, I will end up at a position which has the same latitude as my starting point. Now I questioned even that. Could this just apply to relatively short distances and not over the course of ocean passages? More concretely, I wondered if I started at New York and sailed 090 degrees true, would I make landfall at Lisbon, Portugal (also roughly 40 degrees N), or on the northwest coast of Africa on my way to the Congo (a point on the equator roughly 90 degrees from New York)? I felt the answer had to be Lisbon, but how could the trigonometry be wrong?

After some deliberation, I now feel confident that I would arrive at Lisbon. If I started sailing at 090 degrees true and maintained that course until landfall, my track would be a rhumbline, or loxodromic track. A loxodromic track crosses each meridian at the same angle – 90 degrees in this case. Doing a Mercator sailing calculation as well as plotting the course on a Mercator chart both confirmed that this was true.

The disturbing trigonometric problem remained: how to explain that the equator has a bearing of due east from New York. The answer lies in the peculiar properties of a sphere. The shortest distance between two points on Earth (or any sphere) lies along a great circle. This is a straight line when drawn on a sphere. Although it follows a straight line, the heading of a vessel following a great circle track is continually changing. Because of these course changes, the vessel starting from New York on a great circle track with an initial heading of 090 degrees will make a landfall in Africa.

The loxodromic track is not a great circle and actually curves across Earth’s spherical surface. (There is an exception to this: the loxodromic track is also a great circle when heading due north or south.) Here is the crux of the problem: on a flat, Mercator-projection chart, the loxodromic track is straight and the great circle track is curved; on the globe, a great circle track is straight and a loxodromic track is curved. While we navigate on flat paper, we live on a sphere. A vessel steering 090 degrees true across the Atlantic Ocean from New York will follow a curved track to Lisbon. A bearing of 090 degrees taken from New York must follow a straight line and that is a great circle which passes through the equator in the Congo. It should be noted that the bearing to that same spot in the Congo is different from different points on this great circle track.

For visual confirmation, I took a piece of paper and wrapped it around a globe so that the paper edge was in constant contact with the globe. On a track from New York to Lisbon the paper edge did not follow 40anddeg; north latitude, but went northeast, then east and finally southeast to reach Lisbon. It was also easy to see that a great circle with its vertex in New York would intersect the equator in the Congo. It passed well to the south of Lisbon.

All this work served only to reiterate what most navigators instinctively know: If you sail due east, you will end up in a spot that is on your original latitude, not on the equator.Cameron Bright