NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Gary LaPook
Date: 2013 May 4, 02:22 -0700
| (I am attaching a pdf of this post to preserve the format of the table.) This is an example of how the graph of precomputed altitudes is made for use in the standard “landfall” procedure. Fred Noonan most likely created a graph like this but used a different computation table, H.O. 208, while I will utilize the simpler method of H.O. 249. July 2, 1937 Howland Island, 0̊ 48' North, 176̊ 38' West. Declination of the sun at 1800 Z is 23̊ 02.5' North. Greenwich Hour Angle at 1800 Z is 89̊ 02.5' Assumed latitude is 1̊ North Assumed longitude is 177̊ 02.5' West 1800 Z 1820 Z 1840 Z GHA 89̊ 02.5' + 360 449 02.5 Alon -177 02.5 LHA 272̊ 277̊ 282̊ Hc (tab) 2̊ 14' d=0 Z = 67̊ 6̊ 50' d=2 Z = 67 11̊ 26' d= - 4 Z = 67 corr 0 0 0 Hc 2̊ 14' ZN = 67̊ 6̊ 50' ZN=67 11̊ 26' ZN = 67 Ref. + 17' + 8' + 5 Hp 2̊ 31' 6̊ 58' 11̊ 31' By choosing times exactly 20 minutes appart we know that the GHA of the sun will change 5̊ so we can use the same AP and only need to add 5̊ to the LHA. In case you are wondering why the refraction correction was added when the normal marine practice is to subtract for flight navigation it is important to be able to plot the fix as quickly as possible so as much as possible the calculations are done prior to taking the observation. To accomplish this we add the refraction correction to the Hc to calculate Hp (altitude precomputated) so all we have to do after taking the observation is to subtract the Hs from Hp to find the intercept. In marine practice we have to wait until after the observation to then look up the correction and subtract it from Hs to find Ho which we then subtract from Hc. Give it some thought and you will see that either method produces the same intercept. This establishes the Hp at the AP but we need it from the destination so we plot a line through the AP that is parallel to the LOP, 157̊-337̊, then we measure the perpendicular distance from Howland to this line (18 NM = 18') and this is the adjustment we need to determine Hp at Howland. Since the sun is off to the northeast we can see that Howland is closer to the sun than the line so the altitude at Howland must be larger so we add the distance we just measured, 18', to each of the already calculated Hp’s. 1800 Z 1820 Z 1840 Z Hp 2̊ 31' 6̊ 58' 11̊ 31' Adj. + 18' + 18' + 18' Hp 2̊ 49' 7̊ 16' 11̊ 49' We can continue to use the same adjustment until the azimuth changes significantly which took a lot of time since the azimuth didn’t change at all, remaining at 67̊ until 1847 Z and then stayed at 66̊ until 1923 Z and had changed to 63̊, a total change of only 4̊, at the time of Earhart’s last transmission at 2012 Z. We use these numbers to plot points on the graph and draw straight lines between the plotted points which gives us the Hp at intermediate times. The second graph just expands the altitude scale. gl ------------------------------------------------- --- On Fri, 5/3/13, Gary LaPook <garylapook@pacbell.net> wrote:
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