Actually on second thought, no, your idea will not work. The distance to celestial bodies are triangulated differently depending on whether a Flat Earth or a Round Earth is assumed. For example, simple triangulation can be used to calculate the sun to be 3,000 miles above the earth or 93 million miles above the earth depending on the earth model we assume.
Tom, you are losing your touch. You are supposed to wait for us to finish the observations and post the results. Then you are supposed to jump in and tell us that we assumed a spherical Earth for the calculations.
On the distance to the sun on an FE, for example: On March 21-22 the sun is directly overhead at the equator and appears 45 degrees above the horizon at 45 degrees north and south latitude. As the angle of sun above the earth at the equator is 90 degrees while it is 45 degrees at 45 degrees north or south latitude, it follows that the angle at the sun between the vertical from the horizon and the line from the observers at 45 degrees north and south must also be 45 degrees. The result is two right angled triangles with legs of equal length. The distance between the equator and the points at 45 degrees north or south is approximately 3,000 miles. Ergo, the sun must be an equal distance above the equator.
What if you aren't at 45° latitude on that day does the math still work?
I will save the time since that is the way that this thread is working. You are right Tom, it does save time waiting for a response that we all know the answer to... it doesn't.
Here is data from our local noon position observation on 21 June...
Vessel Latitude - 28° 01.3' N
Sun Declination (Latitude) - 23° 26.3' N
Observed Altitude of the Sun - 83° 56.9'
Because of the fact that at Meridian Passage (Local Apparent Noon) the body being observed is directly south of your position, it is one of the best time to make an observation. You can use one sight to determine your position. It also happens to be the time that Mr. Rowbatham's observations were made...
The base line in any operation being horizontal and always a carefully measured one, the process becomes exceedingly simple. Let the altitude of the Sun be taken on a given day at 12 o'clock...
Now my distance from the Sun's declination (the point where the Sun would be directly overhead) is is 363 nautical miles (417 statute miles).
Using planar trigonometry the equation would be...
Tangent of the observed = (altitude of the Sun)/(Distance from the Sun's declination)
or
Tan 83° 56.9' = (Altitude of the Sun)/363
or
Altitude of the Sun = Tan 83° 56.9' * 363
or
Altitude of the Sun = 3424 nm (3938 statute miles)
This is off by over 20% from the results that are used for 45° of difference in latitude and a observed height of 45° (21.1% to be exact). That is a significant difference in the altitude of the sun.