Very interesting, but one question....
At 45 kilometers, how do you know where to point the telescope to find your subject?
If your target has a permanent address it shouldn’t be to big a problem.
For instance, if the Seals had this available when they were after Osama Ben Laden they could have used this to confirm his presence in the compound after they suspected he was there.
They could have set it up on a mountain that looked down on the compound.
That mirror ball cap sure does the trick.
Individual photons are pretty small.
How can you hear the person yell “say cheese!”?
Simple. You just record everything and sort it out later. /NSA
At 45 kilometers, how do you know where to point the telescope to find your subject?"
Even more interesting is that the "curve of the earth" would prohibit any photography shot from a straight line of sight from 45 km or about 28 miles away of anything but a building about half the height of the Empire State Building.
If you were trying to shoot something 45 km or 28 miles away, your object would be about 520 feet under the horizon line (somewhat less if you held the camera at eye height of five feet) or "under the curve" ... completely outside any direct line of sight, unless you were shooting a building that was taller than 520 feet...
and then you would only see the very top of the building at 28 miles away: http://earthcurvature.com/
https://dizzib.github.io/earth/curve-calc/?d0=28&h0=6&unit=imperial
Earth Curvature Calculator
by NyttNorge.com
Accurately calculate the curvature you are supposed to see on the ball Earth.
Distance.........Curvature
1 mile...........0.00013 miles = 0.67 feet
10 miles.........0.01263 miles = 66.69 feet
50 miles.........0.31575 miles = 1667.17 feet
100 miles .......1.26296 miles = 6668.41 feet
200 miles....... 5.05102 miles = 26669.37 feet
500 miles .......31.5336 miles = 166497.53 feet
1000 miles.......125.632 miles = 663337.65 feet
Explanation: The Earth's radius (r) is 6371 km or 3959 miles, based on numbers from Wikipedia, which gives a circumference (c)of c = 2 * π * r = 40 030 km
We wish to find the height (h) which is the drop in curvature over the distance (d)
Using the circumference we find that 1 kilometer has the angle 360° / 40 030 km = 0.009°.
The angle (a) is then a = 0.009° * distance (d)
The derived formula h = r * (1 - cos a) is accurate for any distance (d)
And what are the odds nothing is in the way? Like a tree trunk or a crowd of people or a truck or an antifa mask etc. Can it see around large objects?