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    Re: A little space navigation
    From: Brad Morris
    Date: 2018 Oct 8, 22:07 -0400
    Hello Dave

    Based on your analysis, can we see the North Pole in the image?  I think the answer is yes when I look at your drawing.  Here's why:

    You have drawn Q as if it is above the Equator.  The north pole P is hidden by the curvature of the Earth, grazing at B.  

    But Q is above Texas, with the southernmost latitude of Texas being at 25.8°.  If we accept 32 degrees of latitude as the point directly below Q, this has the affect of causing the true North Pole rotate towards OQ, to now be visible from Q.  The point P is along our submeridian from the true North Pole.  This is all directly related to the magnitude of the angle P-O-B, again, from your drawing.

    If P-O-B is greater than 23.6°, then we should be able to see the Arctic Circle as it crosses our submeridian.  I don't think this is the case, perhaps you can clarify.  I'm having an internal tussle with this one.

    This is all relevant to finishing the navigation problem.  In CN, the altitude is fixed at the radius of the Earth.  We now know an altitude above a point on the Earth, but where are we in the solar system? 

    Do we wish to remain Lost in Space?  Can we resolve this?


    Danger Will Robinson! Danger!  (The catch phrase of the robot in the sit-com.  That show had a wonderful impression on a space-crazed kid in the 1960s)

    On Mon, Oct 8, 2018, 8:34 PM David Pike <NoReply_DavidPike@fer3.com> wrote:

    I wrote earlier:

    I was rather hoping that someone would tear my geometry to pieces, because I know roughly where I held the camera, and the ratio seemed to reach 45% long before I got down to the 2700km predicted by my maths.  DaveP

    I realised this morning that I’d overlooked the fact that the photo is of a hemisphere, so the USA
    is nearer the camera than the Earth’s periphery.  I roughed something out, but had to leave it, because I was committed elsewhere until this evening.  Below is my revised solution.  Fig I assumes SF to Miami is 45% if disc diameter (as measured by me off my computer) and shows what the camera would see neglecting Earth curvature.  The line from N through MI gives the position of the lens.  Fig II then adjusts for Earth curvature.  The line from Q to the disc’s edge defined B.  The line from U through MI gives the revised position of the lens T.

    Ideally, more iteration is required, but we’re close enough.  The camera is approximately 14,000km above the USA.  This ties in better with my earlier photos of my globe.  I hope you can read the 2B pencil.  It’s a bit dim, but it’s easy to rub out. DaveP


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