There are two methods of making astronomical calculations from maps that are accurate enough for the purposes of archaeology. The second is the more complex, but is needed when the use of more than one map is called for.
An important measurement is the azimuth of the sun at rising and at setting. The azimuth, A, is associated with the declination, δ, by the equation:
ø is the latitude of the observer and h is the altitude of the sun.
Some correction is always needed, since the viewed rising and setting points of astronomical objects are not exactly where the equation shows, because of a number of physical distortions.
One is the refraction of light. The sun’s, or moon’s, light is bent by the atmosphere, so that the sun appears to be higher than it is above the horizon. This bending varies under differing conditions, but is about 0.55° at the horizon. Refraction makes the time of rising appear earlier, so decreasing the azimuth, and the setting later, increasing the azimuth.
A correction for parallax, unnecessary for the sun, must always be made for the moon, which is about 0.95°. Parallax works in the apparently opposite way to refraction.
The height of the horizon must also be included. When a long distance is involved, the angular elevation (plus or minus) of the horizon should be reduced to allow for the curvature of the Earth, at a rate of 0.0027961° per statute mile of the distance from the observer to the horizon.
Therefore, the full value of h, when used, is: h = (horizon altitude) – (curvature of the Earth correction) + (parallax correction) – (refraction).
This will give the azimuth when the centre of the disc is on the horizon. For the azimuth of first flash, at the rising, or last instant, at the setting, the value of h must be corrected by 0.25°.
To find the azimuths of sunrise and sunset at an archaeological site; then let δ = Є, using the value of Є for the date of the site, and correct h as described. For summer rising, let δ = + Є. If the azimuth that is found is subtracted from 360°, it will give the summer setting. For winter rising, let δ = – Є. The azimuth will lie between 90° and 180°, but, if subtracted from 360°, the setting azimuth will be given.
Alternatively, when the map coordinates of two points are known, the azimuth of the line joining them can be calculated, and the distance between them. The distance is needed for the working out of the angles of altitude.
Following A. Thom, let λc Lc, be the latitude and longitude of the observer at C and λd Ld be the same coordinates for the observed point D.
Δλ = λd – λc,
ΔL = Ld – Lc, (east longitude reckoned positive),
λm = ½ (λd + λc) = mean altitude.
A = required azimuth measured clockwise from north.
Find tan ɸ from: tan ɸ = K cos λm ΔL/Δλ, which gives ɸ.
Find ΔA from: ΔA = ΔL sin λm and the azimuth of D from C is A = ɸ – ½ΔA.
If the Earth were a sphere, K would be unity. To allow for its being an oblate spheroid, K may be taken to vary from 1.0028 in latitude 50° to 1.0017 in latitude 60°.
The distance CD in statute miles is:
c = CD = 0.01922Δλ/cosɸ or 0.01926Δlcos λm/sin ɸ.
To calculate the apparent angle of altitude of D as seen from C in terms of the distance, c, between them and the amount by which D is higher than C, the curvature of the Earth must be allowed for, as must the refraction that bends the light between D and C. Then:
h = H/c – c(1 – 2k)/2R,
where H = height of D above C,
c = distance of D from C,
R = radius of curvature of the spheroid,
k = coefficient of refraction.
k is usually 0.075 for rays passing over land, and 0.081 for rays passing over the sea.
Let H be given in feet, c in statute miles, and h in minutes of arc. Then:
h = o.65H/c – 0.37c
Since delivering this paper, my confidence in my ability to handle lunar calculations has more than waned. However, I have left them as first observed, so that a more competent mathematician may repudiate, or confirm, the statements. If I have erred, it does not change the main thrust of the argument, or cause to dematerialise the stone axe.
ALAN GARNER, January 1997