NavList:

Jim Rives wrote:

"If 1 arc min = 4 secs of time, then 16x4 = 64 seconds.  Which would make GMT 64 seconds earlier than 2210, or 22h08m56s.  Is this correct? ...... I don't understand the "Approximate error in Longitude: 8d 00.8'  ".  That seems huge!  "

Jim, you pulled the "1 arc min = 4 secs of time" phrase out inapproprately. That comes about due to the fact that the Geographical Position of celestial bodies (that point on the earth's surface where the body is directly overhead) is moving at about a quarter of a nautical mile per second East to West, because of the rotation of the earth

But when you are doing lunars, what you are concerned about is the speed of the Moon against the background of the fixed stars. The Moon is moving at about 12 degrees per day across the sky, so you would expect any lunar distance to a star (near the ecliptic), or Venus in this case, to change by about 12 degrees per day. That is about 30 minutes of arc per hour.

So, if have an error of 16' in your lunar distance, that actually amounts to about 32 minutes error in time. 

Now, suppose the sun is rising directly in the East or setting directly in the West and you take an altitude of the Sun. The resultant position line would be North/South and would sit right along your line of longitude (meridian) - provided your clock was correct. But if your clock was in error by 32 minutes of time, and you took a sighting of the Sun in the East or the West, how far to the East or West of your meridian would the position line sit? NOW you can pull out your "1 arc min = 4 secs of time" and you will see that an error of 32 minutes of time on your clock will correspond to an error in longitude of about 8 degrees.

You are right, that is a huge error. Without looking in detail at your workings, I would suggest you forgot to take into account the semi-diameter of Moon, which would be about 16', or actually corrected for the diameter rather than the semi-diameter.

Geoffrey Kolbe