Somebody was commenting here that they thought the sun would rise from the Boston (south) side of the river in the morning, but in fact it rises from the Cambridge (north) side these days. A little more than ten years ago, sitting in my northern-latitude abode watching the sun set into the northwest, I wondered the same confused thing: why does the sun appear to venture into the northern part of the sky? (For reference, the sun’s direct projection on the earth never crosses north of the tropic of cancer, and that is south of here.)

The answer is recognized with the aid of this map, which shows the interesting effect due to the earth’s tilt and its spherical geometry. At every point along the day-night boundary curve, the sun’s rays are orthogonal to the curve and on the plane tangent to the earth’s surface at that point. In other words, the sun appears to rise (set) from (into) the direction orthogonal to the day-night boundary. This means between the spring and autumn equinoxes, the sun rises out of the northeast and sets into the northwest, and between the autumn and spring equinoxes, the sun rises out of the southeast and sets into the southwest.

This reminds me of a rather surprising method for direction-finding without a compass. It goes like this: point a stick in the direction of the sun, at any time during the day, and wait a short moment, then the direction of the new (very short, infinitesimal) shadow from the base of the stick points to the east. Here’s a short “proof”: Let the sun’s original location be called Q, let the base of the stick be called A and let the tip of stick be called B. So the vectors AB and AQ are collinear. At any given moment, the earth’s surface instantaneously rotates into the eastern direction at A, so in the earth’s local reference frame (the one we’re interested in) the sun’s apparent location moves slightly to the west to a new point Q’. Now, the shadow cast by the stick has a tip at a new point C (on the ground). Since A, B, C, Q, Q’ all must be coplanar, and the vector AC has no elevation component (same as QQ’), then the vector AC must be parallel to QQ’, and BQQ’ and BAC being similar triangles, AC must point to the east.

The trick is, you can’t wait too long, or else all the linear approximations and Euclidean geometry break down and you won’t get the right direction.