(meteorobs) Camera sky coverage math question

[James Beauchamp]

Just noticed the second number should be 510.072 E^12 Km^2  (510,072,000 km^2)


--- On Mon, 4/4/11, James Beauchamp <falcon99 at sbcglobal.net> wrote:

> From: James Beauchamp <falcon99 at sbcglobal.net>
> Subject: Re: (meteorobs) Camera sky coverage math question
> To: "Global Meteor Observing Forum" <meteorobs at meteorobs.org>
> Date: Monday, April 4, 2011, 9:58 AM
> Very rough method...
> 
> 350 nm = 640 km
> 
> A = Pi * (640,000)^2 = 1.268796 E^12 m^2  (1,286,796
> km^2)
> 
> Earth Surface = 510.072 E^6 Km^2  (510,072,000 km^2)
> 
> Your sky = ~2.5 %, neglecting atmospherics, earth
> curvature, ect.
> 
> Again, very rough, but I bet it would be pretty close to a
> more advanced analysis.
> 
> 
> 
> 
> --- On Mon, 4/4/11, Thomas Ashcraft <ashcraft at heliotown.com>
> wrote:
> 
> > From: Thomas Ashcraft <ashcraft at heliotown.com>
> > Subject: (meteorobs) Camera sky coverage math
> question
> > To: "Global Meteor Observing Forum" <meteorobs at meteorobs.org>
> > Date: Monday, April 4, 2011, 9:41 AM
> > I have a math question.
> > 
> > My all-sky camera can see fireballs in the sky for
> > approximately 350 
> > miles/ 563 km in all directions.  I am wondering
> what
> > percentage of the 
> > Earth's sky this camera can see?
> > 
> > Thanks for your math help in advance.
> > 
> > Thomas  /  New Mexico
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