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Monday, September 19, 2016

Flat Earth Follies: The ISS is going too fast to take unblurred pictures!

Flat Earth Claim:

The ISS is going too fast to take unblurred pictures!

The Facts:

This is what comes from people being mathematically illiterate.

The orbital velocity of the ISS is about 17,150 miles per hour or about 4.76 miles per second.  That sure SEEMS fast, golly gee.  But if that is where you stopped in your thought process then you will come to false conclusions because you have ignored critical variables in the question.

The main variables in photographic exposures are lighting, distance, film speed, shutter speed, and aperture.

From this article in PetaPixel we find that they commonly use the 'Sunny 16' rule which says:

On a sunny day set aperture to f/16 and shutter speed to the reciprocal of the ISO film speed/setting for a subject in direct sunlight.

So at f/16, ISO 200 the resulting shutter speed is stated as "often 1/200".

So we take our 4.76 miles per second and divide it by 200 which means the ISS moves approximately 0.0238th of a mile over that 1/200th of a second or about 126 feet.

Let's double check that:  (17150 miles * 5280 feet/mile) per hour / 60 / 60 / 200 - 125.766 feet

[126 feet already doesn't seem nearly as much of a problem]

Ok, so we've moved 126 feet but we're 250*5280 = 1,320,000 feet from the CLOSEST point to our subject but we are 7,529,000 feet from the horizon (almost 6 times further away).

So the question is, assuming we are looking directly towards the Earth, how much does that move in our visual field during that 1/200th of a second.

So the only thing remaining here is understanding how angular size works.  The good news is that it is simply based on a the consideration of a right triangle:

We're going to consider one-half of our motion on one side, and we will double that to get the full result.

So let's call the full 126 feet 'g':

g = 126 feet

So one-half that distance, or from point B to point C making side a = 126/2=63 feet

a = g/2 = 63 feet

and side b is 1,320,000 feet.  What angle α results?

We know from basic trig that

tan(α) = a/b

What this means is that a/b is our slope (simply rise over run).   What the tangent function (tan()) does is convert an angle into a slope and the opposite function is called arctan() which converts a slope back into an angle.  So we can rewrite our equation, by applying arctan() to both sides giving us:

α = arctan(a/b)

Then we substitute 'g' back into it:

α = arctan(g/2/b)

And remember that this is only one HALF the total so we double this angle to get our final angular size equation which relates this angle to the size of the object and the distance.

α = 2*arctan(g/2/b)

Plugging in our values we get:

α = 2*arctan(126/2/1320000)

So we find that this is:

0.005469° or 19.69" (less than 20 arcseconds).  This is a microscopic angle.

Since our total Field of View is on the order of 60 degrees (or 216000 arcseconds) over some 1920 pixels you have some 112.5 arcseconds per pixel.

And while our FOV isn't divided up equally over all the pixels this is good enough to tell us that we are nowhere NEAR blurring, even without getting into the detailed specifics of the lens.


So it seems that the amount of blurring here would almost imperceptible, even at 4K resolution.

And if they were getting any blurring f/16 is a very small aperture, there is plenty of latitude to open up that aperture a bit and increase the shutter speed, to 1/250th, 1/500th, or even 1/1000th of a second.

As you get further and further away from directly below them the distances become much larger and the angular motion less and less.  So, for most of the Earth, you are looking at only a few arcseconds over the exposure time.

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