[time-nuts] Accurate Thunderbolt position

[Lux, James P]

-----Original Message-----
From: time-nuts-bounces at febo.com [mailto:time-nuts-bounces at febo.com] On Behalf Of Brooke Clarke
Sent: Tuesday, July 14, 2009 3:56 PM
To: Discussion of precise time and frequency measurement
Subject: Re: [time-nuts] Accurate Thunderbolt position

Hi Mark:

0.00001 degrees (five decimal places) = 0.00060' = 0.0360" of arc.
If one second of arc is about 100 feet of longitude then the total distance for 
0.00001 degrees is about 3.6 feet.  To get latitude multiply by COS(your lat).
That way you know how far to move the antenna.

But to check apply the above to the Z12 antenna, i.e, see if you can move the 
Z12 antenna so it's position, after OPUS, is right on a five digit location.

One of my goals is to locate a number of survey monuments so that they are on 
even arc second locations.  i.e. 123:24:40.00000000, 123:24:50.00000000, etc.
It's an iterative process.


--

Harder than you think..

The cos(latitude) is only an approximation, and you're talking about displacements here where not only do you need to take the spheroid into account, but probably even finer scale aspects.


My house moves about 2cm north and 1cm vertically on the average, every year, because of tectonic movement. Over a reasonable period of time, it's entirely possible with typical time-nuts equipment to actually measure this (check out the SCIGN, for instance)

Tidal displacement of the earth's surface is something else you should be taking care of. I believe that crustal deformation is on the order of a few tens of cm.

A GPS guy I know comments that when you start talking down in the sub-meter sorts of accuracies, particularly for absolute measurements, there's a whole raft of factors that are all of the same general magnitude that you need to take into account: tidal deformation, ionosphere, multipath, thermal distortion of your antenna, changes in the cable due to temperature, etc.etc.etc.