Anthony Ayiomamitis wrote:

[concerning http://www.perseus.gr/Astro-Lunar-Parallax.htm]

Thanks for the feedback Ernie.

Thanks to you and Pete for a fascinating collaborative exercise!

I hope you'll forgive me now for blathering a bit, but it'll lead to a
much better estimate of the Moon's distance derived from your images.

We both got estimates of about 440,000 km, somewhat higher than the true
distance of about 396,000 km, and I mentioned yesterday that most of the
error is because the base of the triangle we're using, the line between
Selsey and Athens, is skewed. The base we *should* be using is a line
that directly faces the Moon.

The situation looks something like this:

.. * Selsey
.. \
.. \ -------- to the Moon
.. \
.. * Athens

The base we should be using,

.. * Selsey ....... |
.. |
.. | -------- to the Moon
.. |
.. * Athens ... |

is the projection of the Selsey-Athens line onto the image plane. Its
length is just the dot product of two vectors: the Selsey-Athens line and
the Earth-Moon line. To find the vectors, we need a common coordinate
system for Athens, Selsey, and the Moon.

I used geocentric equatorial coordinates. The r.a. and dec. of the Moon
are easy enough to find. The coordinates for the cities are just the r.a.
and dec. of the zenith at the time of the observation, which are the local
sidereal time and geographic latitude, respectively.

Convert these to cartesian coordinates in the usual way:

x = cos( dec ) * cos( ra )
y = cos( dec ) * sin( ra )
z = sin( dec )

Subtract the Athens (x, y, z) from the Selsey (x, y, z) and normalize
(divide by the vector length) to get a unit vector pointing from one to
the other. The dot product of this direction vector with the one for
the Moon is the cosine of the angle between them. The length of the
projection we want is the cosine of the difference between this angle
and 90 degrees.

When I did this, I got a length factor of 0.928. Multiplying this by the
chord length distance between Athens and Selsey (2356 km) gives a triangle
base of 2186 km. Using your (probably more careful than mine) estimate of
the parallax angle, 1113.6", yields a distance estimate of 404,897 km, for
an error of only a little more than 2%. That's pretty cool!

Anyway, a nice exercise. Just ask Oriel.

Too bad Gerald's not equipped to appreciate it. It very much has the
flavor of the Ancient Greek efforts to measure the scale of the solar
system.

- Ernie http://home.comcast.net/~erniew