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(17B) (19B) (27S) Percentage of Sphere's Surface Visible for Height Above Surface
03-31-2019, 05:13 PM (This post was last modified: 03-31-2019 05:14 PM by ijabbott.)
Post: #1
(17B) (19B) (27S) Percentage of Sphere's Surface Visible for Height Above Surface
Code:
VIS%=HEIGHT*50/(HEIGHT+RADIUS)

For example:

Radius of Earth = 6378 km
Distance of Apollo 17 from Earth's surface at 1972-12-07T10:39Z = 29000 km
Visible percentage of Earth's surface in 1972 "Blue Marble" photograph = 40.99% (41%)

— Ian Abbott
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04-01-2019, 10:51 PM (This post was last modified: 04-02-2019 07:44 AM by StephenG1CMZ.)
Post: #2
RE: (17B) (19B) (27S) Percentage of Sphere's Surface Visible for Height Above Surface
I was just wondering whether this formula holds up if the height were negative (for example, if the GPS shows you in a valley).

I'm thinking in such a case the physics would suggest 100% visible (for a hollow sphere or Earth), or 0% (for an opaque sphere or Earth), rather than a negative value. Geometrically, I think I'd choose 100% - except its the interior of the surface you can see, not the outside. Perhaps - 100%? With the negative flagging an interior view? I'm not sure what result would be preferred.

Stephen Lewkowicz (G1CMZ)
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04-02-2019, 07:40 AM (This post was last modified: 04-02-2019 07:46 AM by ijabbott.)
Post: #3
RE: (17B) (19B) (27S) Percentage of Sphere's Surface Visible for Height Above Surface
For a negative height, the simple formula will give a negative percentage visibility, but I have no idea how that should be interpreted geometrically!

I only posted the formula because it turned out to be so simple and elegant compared to how I was expecting it to turn out.

— Ian Abbott
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04-02-2019, 01:46 PM (This post was last modified: 04-02-2019 03:19 PM by SlideRule.)
Post: #4
RE: (17B) (19B) (27S) Percentage of Sphere's Surface Visible for Height Above Surface
An interesting formulation from 1913
   

as well as an un-referenced web source …
   

BEST!
SlideRule
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