Navigator wrote:

No, I'm trying to get people to think about the relationship between
altitude and period between rise and set for a fixed lunar orbital
period.

Not so much a side-step, as a stumble.

C'mon you can try better than that.

Given that the moon doesn't spend each day jumping up and down like a
bouncing ball, it wouldn't be too far off the mark to say that its altitude
is a function of the the azimuths of its rise and set, and the observer's
latitude. In other words, its apparent path across the sky is largely due to
the rotation of the earth - like that of the sun.

They key thing about the sun is that its altitude for a given azimuth
changes from day to day due to the obliquity of the ecliptic - the plane of
the equator is different from the plane of the earth's orbit around the sun.
This, combined with the earth spinning on its axis, results in the sun's
rise and set azimuths changing daily and producing summer and winter
solstices, and spring and vernal equinoxes. At the equinoxes, the sun is
passing over the equator (the intersection of the equatorial and solar
orbital planes is in the direction of the sun) and, to the observer on
earth, it appears as though the obliquity of the ecliptic is zero, resulting
in the sun's rise/set being exactly due east/west. Anyone who cares to sit
on a hill for a long time will notice that this cycle is yearly.

A notable aspect of the moon's orbit is that it, too, isn't parallel to the
plane of the equator. For the observer, this is manifest as a series of
sun-like solstices and equinoxes, but with a much shorter period - monthly.
Each month, the moon has a northern and southern solstice (or standstill) ,
and twice passes over the equator midway between these, once when going from
the northern to the southern standstill, and vice versa.

The relationship between rise and set azimuths and altitude (at zenith) is
important insofar as the earth doesn't speed up and slow down during its
axial spin. Further, the altitude of the moon is lower when the rise/set
azimuths are closer to south (for the northern hemisphere). In other words,
when moonrise is towards the north, it takes a longer path across the sky
than it does when moonrise is towards the south. If the earth's rotation
speed is constant, then it must follow that the moon takes more time to
traverse the sky with a northerly rise point than with a southerly one. So,
over the period of one month, the time taken for the moon to traverse the
sky varies from day to day.

As I said earlier, if moon rise/set are due east/west, the time to traverse
the sky is about 12 hours. It isn't *exactly* 12 hours because the moon is
orbiting the earth, but it is *about* 12 hours, because it's orbiting the
earth *slowly*. To render some precision to the description: If I stand at
the Greenwich Observatory and note the moon's azimuth at midnight tonight,
and then wait for it to pass through that azimuth tomorrow night, it'll take
nearly an hour for the moon to 'catch up' - nearly 25 hours.

For an east/west rise/set, the nominal time taken to traverse the sky is 12
hours - half the period of the earth's axial rotation. However, since the
moon takes time to catch up, the time to traverse would be longer - I guess
about 12.5 hours (half the earth's rotation period, plus half the moon's
'catch up' time). It would seem that moonrise at the northern standstill
would entail a greater proportion of the moon's catch up time being added
because the moon is visible for a greater portion of the earth's rotation
period, and vice versa for moonrise at the southern standstill. A true 12
hour traverse of the sky by the moon would consequently entail a rise/set a
little to the south of east/west to account for the delay in the moon
catching up to reach the same azimuth.


--
Wally
www.forthsailing.com
www.wally.myby.co.uk