Okay, it appears nobody else did this. It's a shame, really, as it's such a simple experiment....
Anyway, here's my results: 800 mm, 310 degrees. Using these numbers, the sun should have been above 15.01 -99.7039.
However, the sun really was (in either model) at 15.35 -88.95, about ten degrees farther east than my observations showed. FE predicted that my yardstick would cast a shadow 943 mm long at 293 degrees, about 14 cm longer than my actual value.
MATHS:
Note: The RE maths are way more complicated, and I don't feel like going through them. The following is geometry done on flat planes and thus coincide with what is predicted by FE.
From Points to shadow measurementsConsider a triangle with points A, B, and N at the place of observation, the point directly beneath the Sun, and the north pole respectively. Leg NA is, by definition, parallel to the line of longitude at point A; the same logic applies to leg NB. Leg AB is our target unknown--the total distance between our place of observation and the point directly beneath the Sun.
Let us create a third line, AC, which connects from Point A to Leg NB such that it intersects Leg NB at a right angle. Let us call the point at which Leg AC connects to Leg NB "Point C". This has effectively split our triangle into two right triangles: Triangle NAC and ABC. Leg NB has also been split into Legs NC and CB, though it is important to note that Leg NB still exists as a whole just as our original triangle NAB exists as a whole.
Leg NB is easy to calculate if you know the latitude of Point B, as each degree of latitude is equal to 110.574 kilometers. So,
|NB| = latB * 110.574 km. Leg NA can be calculated in like:
|NA| = latA * 110.574 km.
The angle between Legs NA and NB is simple to calculate, that being the absolute value of the differences in longitude between A and B. Let us call this angle Theta.
Theta = |lonB - lonA|.
The length of Leg AC can be calculated as follows:
|AC| = |NA| * sin(Theta). This is the first value listed in the above examples, labeled 'west' or 'east' of the sun.
Similarly, the length of Leg NC can be calculated:
|NC| = |NA| * cos(Theta).
With |NB| and |NC| now calculated, we can calculate |CB| as follows:
|CB| = |NB| - |NC|. This is the second value listed in the above examples, labeled 'north' or 'south' of the sun.
With both legs of triangle ACB now known, we can calculate the hypotenuse:
|AB| = sqrt(|AC|2 + |CB|2). This is the third value listed in the above examples, labeled 'total horizontal distance' from sun.
Calculating the predicted length of a shadow is now a matter of similar triangles. We know how far away the observation point is from a point directly beneath the sun, and we know how high the sun is from the Earth (roughly 4828 km, according to FE calculations). Our yardstick-shadow triangle is similar to the sun height-distance triangle. Thus, the length of a shadow produced by a 36-inch-tall object can be predicted with the following equation:
|AB| * 914.4 / 4828.
Since Leg NA is, by defintion, the direction 'north' from Point A, the direction of the shadow can be calculated by its deviation from leg NA. Because Leg AB lies entirely between the observation point and the sun, the shadow would be on the other side of A, parallel to Leg AB. To find this deviation, we add together the angles BAC and CAN and subtract this total from 180.
Angle BAC is
tan-1(|CB| / |AC|).
Angle CAN is
90 - Theta.
If Point B is east of Point A (the diagram is in this configuration) the direction of the shadow would be
360 - (180 - (Angle BAC + Angle CAN)).
If Point B is west of Point A (invert the diagram along its vertical axis) the direction of the shadow would be
180 - (Angle BAC + Angle CAN).
Single-equation form:
|AB| = sqrt{[(latA * 110.574 km) * sin(|lonB - lonA|)]2 + [(latB * 110.574 km) - ({latA * 110.574 km} * cos(|lonB - lonA|))]2}From shadow measurements to pointsTo go back the other way (from measurements to point positions) is more of the same. The direction of our shadow is the angle NAB' with B' being a point beyond A from B such that angle B'AB is 180 degrees. Thus,
Angle NAB = 180 - NAB'.
Leg NA remains the same calculation of
|NA| = latA * 110.574 km.
Leg AB is also known, but only for an assumed value of the altitude of the sun over the Earth (I used 3,000 miles, as that is the most common number in FE models). Our yardstick-shadow dimensions form a triangle similar to the sun's altitude-horizontal distance triangle. Thus,
|AB| = lengthshadow * altitudesun / heightyardstick.
With Legs AB and NA known, as well as the angle between them, we can use the Law of Cosines to solve for leg NB:
|NB| = sqrt(|NA|2 + |AB|2 - 2 * |NA| * |AB| * cos(NAB)).
With all three legs known, we can solve for angle ANB also using the Law of Cosines:
ANB = cos-1((|NB|2 + |NA|2 - |AB|2) / (2 * |NB| * |NA|)).
The latitude of Point B is an inverse of what was done above:
lonB = |NB| / 110.574 km.
The longitude of Point B is simply the longitude of Point A subtract angle ANB:
latB = latA - ANB.