NavList:

Brad Morris wrote:
"Are you attempting an ocean crossing on a latitude line? Then a minute or so in time either way won't make any real difference, as the objective is to be following a latitude either to the south or north of your intended destination. Hit land and make the appropriate turn to the north or south and follow the coast to your destination."

Brad has apparently lost the forest for the trees. This has NOTHING, absolutely NOTHING, to do with following latitude to one's destination.

Brad continued:
"If on the other hand you are attempting map making (like determining the latitude of a harbor within 5" of arc) then the exact time of meridian passage to the second is critical if you intend to use raw data."

This is quite wrong. And it is easy to see so by simulating a specific case just as Brad suggested --though apparently he didn't try it. Suppose you are at the lighthouse out on Montauk Point (Long Island, New York) and you want to determine your latitude by the meridian passage of the Sun (the numbers for stars will be comparable). And, as per Brad's scenario, you want to determine the latitude to a tenth of a minute of arc (near enough to the 5" he proposed). Let's take a look by simulation and find out just how long the Sun appears to "hang" at its meridian altitude to within a tenth of a minute of arc.

An easy way to do this is to use the well-known web tool at the USNO web site (the successor to the old "ICE" software) located here:
http://aa.usno.navy.mil/data/docs/celnavtable.php
Set the AP (assumed position) to 41°04.3'N, 71°51.4'W. For the date, try May 4, 2014. That's a month from now, so out of sextant range for artificial horizon sights and also closer to the solstice when the sensitivity increases. Brad claimed that the exact time of the sight "to the second is critical". It's easy to see that this is wrong. You can easily determine that the Sun reaches and holds its maximum altitude of 65°00.5' (corrected) for the entire period from 16:43:24 to 16:45:11 UT/GMT. That's nearly TWO MINUTES of time with a change less than a tenth of a minute of arc. But back to Lewis & Clark, if we want the latitude within 0.5' which is about the best that they could hope for, then the Sun is above 65°00.0 and less than its peak altitude of 65°00.5' from 16:41:15 to 16:47:20. That's SIX MINUTES of "hang time". Anytime during that period, the altitude of the Sun will change imperceptibly in an instrument with a precision of +/0.5'. And note that this long hang time also explains why a noon sight ALONE cannot be used as a substitute for a time sight for longitude. For that you need more sights "around" noon to pin down that central time, or you need to take extra sights (with more complicated reduction) in the morning or afternoon. But this peak altitude for Noon Sun works great for latitude, and altitudes by stars, if you can find them in the A.H. work much the same way.

Greg, for latitude by meridian star altitudes, you simply wait for the altitude to stop changing for a couple of minutes, just like Noon Sun. For an example, you might try the same date as above and look at the meridian transit of Mars which occurs around 02:41:25. In fact, Greg, this would be a good real-world example to try. Mars is nice and bright now since it's at opposition this week, and you should have no trouble getting a meridian latitude from it using a mirror artificial horizon.

Brad Morris added:
"Knowing your longitude, you can calculate the exact moment of upper or lower meridian crossing. "

But this is NOT necessary. The observations around noon THEMSELVES provide this time regardless of any uncertainty in longitude and Greenwich time, which Lewis & Clark, of course, suffered from. Greg was SPECIFICALLY asking about Lewis & Clark who certainly did it by the observational process and not the suggested calculation.

Brad concluded:
"I'm sorry if this isn't a crystal clear answer! The issue is that your question isn't quite succinct."

I found his question clear enough. He wanted to know how far off meridian passage you could be. In azimuth, it's just a degree or two. But this doesn't matter since the maximum altitude is judged by observation, not azimuth alignment --and certainly not by timing.

-FER
PS: By the way, OBSERVATORIES with transit instruments certainly did perform the observation by azimuth alignment because they could. When a star reaches due south (or north), it is exactly on the meridian. In traditional observatories, a transit instrument was placed in alignment exactly north-south, both to determine meridian altitudes and also for exact determination of sidereal time, which could be translated to local mean time (for use by customers, like railroads, as we discussed recently). But this is no solution for observers in the field or at sea.

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