AV observation: A solution by DM

There’s probably a simpler way to do it if you know spherical trig. which I don’t. This is how I tried to do it:

The longitudes of A and V need to obtained in the globe with the magnetic poles as the new North and South poles.

In a cartesian system with its x-axis along the direction from the earth’s centre to the 0 N, 0E point and z-axis along the direction from the earth’s centre to the north pole, the coordinates of a location with latitude l and longitude m, are (cos(l)cos(m),cos(l)sin(m),sin(l)). So the longitude is given by arg(x,y).

In the new globe with the magnetic poles as its N and S pole, the 0 N, 0 E shifts to l0,m0. The new unit vectors in terms of the old ones are given by

i’ = cos(l0)cos(m0) i + cos(l0)sin(m0) j + sin(l0) k
j’ = -sin(m0) i + cos(m0) j
k’ = -sin(l0)cos(m0) i – sin(l0)sin(m0) j + cos(l0) k

So the coordinates in the new system,

x’ = cos(l)cos(m)cos(l0)cos(m0) + cos(l)sin(m)cos(l0)sin(m0) + sin(l)sin(m0)
y’ = -cos(l)cos(m)sin(m0) + cos(l)sin(m)cos(m0)

and the new longitude is given by arg(x’,y’).

It resulted in 13400 BCE and 4500 BCE as the bounds for Epoch of Arundhati.

(Note: Above calculations do not take into account ‘proper motions’ of Arundhati and Vasistha, and also effect of ‘Nutation’.)

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