# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Raw data for bubble**

**From:**Bill B

**Date:**2007 Mar 22, 17:43 -0400

> From: "Bill Noyce"> I think the original question was, how to predict the time at which > the sun has an azimuth of 90 or 270 degrees, given a known or assumed > latitude. In this case, you have the co-latitude, the co-declination > of the sun, and the azimuth (which is a right angle) as three known > parts of the triangle. (The sun's declination comes from a calendar, > plus perhaps an estimate of longitude to the nearest 15 degrees. You > would have to be more careful if using the moon.) > The formulas will come out something simple like > cos LHA = tan d / tan L and sin h = sin d / sin L Interesting. For Peter to calculate the exact moment the body is at 90/270d he must know his location exactly. While you state co-latitude is a known, it cannot be known exactly until the exact time of transit is known. However delination is a relatively slowly changing variable, so if we can estimate the time of 90/270 to a half hour or so we can derive a good estimate of LHA from AP Lat (LHA = tan d /tan L). With LHA and AP lon we can estimate GHA, from which we can derive estimated time of transit within a few minutes or better. With time of transit, we get a closer approximation of declination and GHA. We could use the direct computation of azimuth from the formula George posted a while back as a sanity check. "Tan Z = sin LHA / (cos LHA sin lat - cos lat tan dec) and the rules for putting Z into the right quadrant, 0 to 360, clockwise from North, are: --If tan Z was negative, add 180 deg to Z. --If hour-angle was less than 180 deg, add another 180 deg to Z" If the computed azimuth is within a degree or so of 90/270d, we are pretty much good to go as we have seen earlier. One could be off the actual time for a given AP by several minutes or more with little damage done unless AP latitude was horribly off. If off more than a degree or so, make an estimate of the change in time needed to move the azimuth to the desire angle and try again. > I think the intent is to observe until the predicted h occurs, and > record the resulting GMT. From that and the matching LHA you compute > GHA and Longitude. Right? And you use the graphical method to > interpolate to hit the desired h exactly. You would have to ask Peter about that. My impression is once he had the time of transit, he computed Hc. Any difference between Hc or Ho was plotted east/west from AP for longitude. > > From the modern perspective, a sight at this special time gives a LOP > like any other; its main advantage is that the LOP runs exactly north > and south if you're at the assumed latitude, so any error in the > latitude has only a tiny effect on the resulting longitude. Exactly. Thanks Bill B --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@fer3.com To unsubscribe, send email to NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---