NavList:

Frank,

(1) - About working on the geoid, you mentioned "This really isn't relevant at all" ...

Yes and no actually would be my comment here.

- In this specific example, it is not relevant.

- But for other examples, e.g. here and here  it can be relevant.

- And, if we are to be didactic and comprehensive - one of your main qualities, Frank - is it not worth mentioning the geoid then ?

(2) - "But what about refraction?"

A topic I hesitated to mention actually, because you did not specifically address the apparent height above the local horizon of the said poles when seen from one another. So I had set it aside initially.

As a rule of thumb the atmospheric refraction elevates an apparent target by 0.2' every 25 NM. So, from A to B (6 NM) the height elevation would barely reach 3", and from A to C (10 NM) it would barely reach 0.1'. Hence a totally negligible effect here in this specific drill environment. Nonetheless, in different environments it can have some appreciable effect.

(3) " and you will discover comparably large errors in the angles [except at one latitude... anyone?] "

I would then surmise that this would happen at a Latitude where the relative lengths of both 1° of Latitude and 1° of Longitude would exactly match the same ratio observed on a Sphere.

If so, and for WGS84, the Curvature Radius of the North-South direction would be the same as the original equatorial radius, i.e. 6378.137km. That is happening at Latitudes very close from N54°47' and S54°47'.

Only a guess here, which have no time to closely check. 

An other and easier way to check it would be to look up in a dedicated table the Latitude at which the length 1° of Latitude is exactly 111.111 km . But I do not carry such dedicated table.

(4) - Extra topic

Since I just adressed the Curvature Radius towards a specific direction (see the Swiss mathematician Leonhard Euler), it can be a very useful tool to compute distances on an Ellipsoid, and quite good for distances up to 200 NM.

e.g. derived from this reference

Departure N44°/E006°/0 m and Arrival N42°/E002°/0 m

Solution on a 40.000km equator sphere :

GC Distance 212.6 NM (0.061 838 123 Rad) Departure/Arrival Courses 237.0°/234.3°

RL distance 212.6 NM Departure/Arrival Course 235.6° (halfway between GC courses)

Solution on WGS84

Vincenty

Distance 394 575.867 m / 213.05 NM Dep/Arr Courses 237.113 653 4°/234.384 648 7°

Euler

Departure Curvature Radius towards Arrival 6 381 902.392 m

Arrival Curvature Radius towards Departure 6 379 664.755 m

Mean Curvature Radius 6 380 783.376 m

Mean Curvature Radius * spherical geocentric angular separation (at 0.061 838 123 Rad)

394 575.667 m (20 cm off from Vincenty)

Geometrical 3 D from local vertical plans containing the opposite segment end

Departure course 237.113 713°(0.2" off Vincenty)

Arrival Course 234.384 711 (0.2" off Vincenty)

Impressive, no ?

Kermit