Join Date: Nov 2007
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Ah well I am guessing that this is some sort of projection of the celestial sphere as well. I agree that M is the body being sighted but my suspicion is that D is the north celestial pole and C is the zenith, in which case MCD is the celestial triangle as you suggested.
However I get stuck because I think this diagram is relying on conventions that are unknown to me. I'm just assuming north is D and BF is the equator as you suggested, so the line parallel to BF is the declination which is closer to D, therefore north. I'm not sure what the line parallel to AE would represent.
Since D appears to be above the horizon AE, I'm guessing that the latitude is north.
The arc from D through M to H would be the object's meridian. Since the diagram is flattened and I don't know the conventions by which it was constructed, I cannot guess at the azimuth except that it is certainly not north, since then the arc should go between D and E, rather than between D and C.
Since M is above the horizon AE, M is visible.
Alternatively you could assume that AE is the equator so that the line parallel to AE is the declension. Then BF is the horizon, D is the zenith, and the arc is M's vertical circle. I don't like this interpretation because the thing you measure directly is the altitude, which would be very difficult to plot in this case.
In the first version, CX is the vertical circle orthogonally projected, so you could place M on the line at a distance equal to the sine of the altitude you measured. You could do the declension line similarly.
Flybyknight: I feel like this sort of diagram could be very useful---possibly enabling CN without a calculator---but I don't know the conventions by which it is constructed. Do you have any references that would give a complete description?
s/v Laelia - 1978 Pearson 365 ketch
s/v Essorant - 1972 Catalina 27