[time-nuts] Calculating time of lunar eclipse

[jimlux]

On 1/21/19 5:50 AM, Magnus Danielson wrote:
> Hi,
> 
> On 2019-01-21 08:07, Mark Sims wrote:
>> While on the subject of the accuracy/reliability of various algorithms 
>> and web pages showing various astronomical data, we had a full moon / 
>> total lunar eclipse in the northern hemisphere.  And not just any full 
>> moon, but a Super Blood Werewolf Zombie Apocalypse full moon (or some 
>> such drivel spouted by all the TV stations).
>>
>> Anyway, I wanted to know when the eclipse was at it's maximum.  Most 
>> web sites gave a time here as 23:12,  some differed by several 
>> minutes.  None gave the time to the second.  That just won't do for a 
>> moon worshiping time-nut, will it?  So, I tricked up a version of Lady 
>> Heather to do a screen dump when the difference of the  sun and moon 
>> azimuth and elevation were at a (180 degree) minimum.  Looks like it 
>> happened at 23:12:04
> 
> Well, the best time as we see it is actually about a second after it 
> actually occurred, due to the time it takes for the light of the moon to 
> reach us. So the time for the actual event and the time for observation 
> becomes notisable different.
> 
> Cheers,
> Magnus

Magnus, this is time-nuts.. "about a second"?

There's a whole bunch of factors that need to be taken into account.

We are close to lunar perigee, where the distance is 357344 km - 1.2 
seconds (approx, see below), were we at apogee in a couple weeks 406555 
km that would add 150 milliseconds (approx).


Here in Southern California, it was about 9PM local, so we were farther 
away from the moon than folks on the US East Coast watching at local 
midnight. Why that's almost 2000km - 6 microseconds.  (Forgive my 
"approximate" here - the intent is to form an 1 sig fig estimate of the 
potential error - if needed, we can discuss making this more precise)

Solid Earth tides (assuming the observer is on land)will lead to a few 
nanoseconds difference as well you might be closer or farther.  For the 
lucky folks watching at moon set or moonrise, this effect will be 
minimized, since the tidal motion is orthogonal to the line of sight. 
For water borne observers, the tidal motion is more complex, and you'll 
have swells to worry about as well.

One should also take into account the density and properties of the 
atmosphere, which changes the propagation speed of light. One should, of 
course, also account for the dispersion - red light travels at a 
different speed than blue - but perhaps the red moon was sufficiently 
monochromatic that this isn't an issue?


As it happens, it was overcast here, so there was no direct viewing 
possible, so my viewing, such as it was, had to be mediated through the 
internet, which has its own set of time lags, some of which we have 
discussed on the list.


In all seriousness, computations of the observing times of astronomical 
events does need to take a variety of factors into account, and light 
speed is but one of them.

In my own professional field of spacecraft telecom, as beamwidths get 
narrower (i.e. going to optical wavelengths instead of RF) knowing the 
precise relative position at a specified time is important.  This is 
biting lots of "cubesat" folks as they move from UHF with 30 degree 
beamwidths to ground station antennas with 1 degree beamwidth.

And, when looking at links to outer planets, with transit time in hours, 
do you need to "point ahead" (earth moves about 30km/sec in its orbit) - 
from Mars at a sort of average 1AU, that's about 0.007 degrees. 
Probably not an issue with a 1 degree beamwidth antenna.

A typical 20cm optical aperture will produce a beamwidth of 1500 km at 1 
AU, a bit less than 1 minute's motion of Earth.

One thing I'll note is that for the most part, folks realize the 
impracticality of computing all this stuff open loop - you do a 
calculation to get you close enough, then use closed loop tracking to do 
the final pointing. But think about this - if I'm pointing my telescope 
back at Earth, finding Earth is easy.  Pointing to the centroid of Earth 
is easy. But my "spot" is only 1500km wide, on a 14,000km wide earth, so 
I need to know "where on the Earth disk" to point.


http://kiss.caltech.edu/workshops/optcomm/presentations/Sburlan.pdf is a 
nice high level presentation on Optical Comm from a few years ago.