Welcome to the NavList Message Boards.


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

Compose Your Message

Add Images & Files
    Re: A little space navigation
    From: David Pike
    Date: 2018 Oct 7, 13:38 -0700

    Brad Morris you wrote:  The orb of the Earth is illuminated, therefore we are between the orbit of the Earth and that of Venus.  The Earth is illuminated over the almost all of the orb, or thereabouts, so the Sun is just about directly behind us.

    From our apparent location over the Earth, we are above the ecliptic.  The southern most latitude in Texas is 25.8°.  This is in conflict with the first paragraph.  The maximum declination of the sun is 23.45 degrees.  It cannot illuminate the orb completely if the sun isn't behind us.
    There isn't a star field visible behind the Earth in the image, so we cannot resolve our position along the Earth's orbit around the sun.  
    We cannot tell our altitude without knowing the magnification of the lens that was used to take the photo.  


    I’m not sure why Venus comes into this.  Perhaps you could explain.  I’m not entirely sure why the fact that the camera’s sub point is outside the ecliptic in going to affect things either.  Agreed it’s best if the Sun is directly behind the camera, but if not exactly so, isn’t there a point as with our Moon after half-Moon when a complete diameter is visible at some angle or other.  Therefore, by moving a ruler over the surface of the photograph, a maximum value can be found to compare with the distance from Miami to San Francisco.  On my screen SF-Mi was 45% of maximum diameter on Frank’s photograph.  From my photos you can see this ratio increases as the camera approaches the Earth. 

    Being a non-expert I’m happy to assume that light travels in straight lines between the Earth’s atmosphere and the front of the camera lens, which is probably the best that money can buy, so no ‘fisheye’ effects, and any magnification should affect all parts of the photo equally and not affect ratios.  Therefore, what’s wrong with taking a series of photos as I did but carefully measuring the distance in Earth radii of the camera above the globe, then after printing these on the same laptop printer combination as Frank’s photo and plotting ratio v distance on a graph, and finding the distance that has the same ratio as Frank’s photo? 

    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




    Browse Files

    Drop Files


    What is NavList?

    Join NavList

    (please, no nicknames or handles)
    Do you want to receive all group messages by email?
    Yes No

    You can also join by posting. Your first on-topic post automatically makes you a member.

    Posting Code

    Enter the email address associated with your NavList messages. Your posting code will be emailed to you immediately.

    Email Settings

    Posting Code:

    Custom Index

    Start date: (yyyymm dd)
    End date: (yyyymm dd)

    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site