http://media.npr.org/assets/
I wondered--is that even possible? Or is this a completely Photoshopped image?
First, how big is the ISS? Here's a picture from NASA to give a sense of scale:
How big would a football field look if if were flying at the altitude of the ISS?
Well... here's a bit of math I did to do a quick check on it.
I did a few obvious searches to figure out that the ISS is 109m maximum width and flies at an average distance of 350km. With these two bits of info and a little geometry, you can work out what the subtended linear angle of the ISS would have to be. The geometry is easy...
54.5m is half the ISS width, which you need as the base of the triangle. 278.000005 is the length of the hypotenuse according to Pythagoras.)
To work out the subtended angle, you just compute the arcsin ( 54.5m / 278.000005km ) -- that's about 0.011 degrees.
How did I do that? Once I'd worked out the hypotenuse length (350.0004 km) I just turned to Google Calculator. Here's my query:
[ arcsin ( 54.5m / 350.0004 km ) in degrees ]
That is, this bit of trig computes "what is the subtended angle of the ISS?"
( NOTE: I added "in degrees" at the end because the Google Calculator gives back sin/cos/tan/arcsin... (etc) measurements in radians. But I wanted degrees, because I happen know (from another query) that the sun is 0.53 degrees wide. )
This Google Calculator expression tells us that half the ISS width is 0.008 degrees wide (remember, we divided the image in half in order to do the right-angle trig up above). So the subtended linear angle of the ISS at 350 km overhead is 0.016 degrees.
( NOTE: I added "in degrees" at the end because the Google Calculator gives back sin/cos/tan/arcsin... (etc) measurements in radians. But I wanted degrees, because I happen know (from another query) that the sun is 0.53 degrees wide. )
This Google Calculator expression tells us that half the ISS width is 0.008 degrees wide (remember, we divided the image in half in order to do the right-angle trig up above). So the subtended linear angle of the ISS at 350 km overhead is 0.016 degrees.
So...
... if you measure the photographed width of the ISS in the image, the width of the ISS image should be about 1/50th the width of the sun image. (How do I know what? The subtended linear angle of the sun is 0.53 degrees. Divide 0.53 by 0.01 and you get 53.
I measured it quickly by doing a little copy/pasting on the image (see below), and got it at 1/52cnd... close enough. (That's well within measurement error on my part.) Here's the diagram I drew to measure it. I took the image and drew a light blue line below the ISS that's exactly the width of the ISS. I then drew a bunch of dark blue bars that same width across the radius of the sun.
Each of the dark blue bars is 1 ISS width. Each of those light blue lines is 10 ISS widths across... so it looks to me like the sun is 52 ISS widths wide.
I also looked around a bit, and found multiple OTHER images made by other amateur astronomers, all showing the ISS at more-or-less the same size with respect to the sun.
Overall, the picture checks out. It's internally consistent (that's why I was measuring it) and it's been replicated by other astronomers. So yeah... I believe it.
What a beautiful image!
Search (and measure) on!
I also looked around a bit, and found multiple OTHER images made by other amateur astronomers, all showing the ISS at more-or-less the same size with respect to the sun.
Overall, the picture checks out. It's internally consistent (that's why I was measuring it) and it's been replicated by other astronomers. So yeah... I believe it.
What a beautiful image!
Search (and measure) on!
im needing to check the validity of a pic posted on instagram, how do I do this?
ReplyDeleteAlas, there's no quick and straighforward answer to your question. Every picture has it's own story, and as a consequence, every picture has a different way to validate it. (And truthfully, in many cases you won't be able to do so.)
ReplyDeleteFun post! Thanks for the write up. Fun hack: for large distances the angular diameter is estimated well with just width/distance: 109/350000 = ~ 0.01784° (which is closer to your 0.016 than you might think, read below)
ReplyDeleteWhy this works consider:
To avoid the need to compute the hypotenuse you can also use arctan():
θ = arctan(opposite/adjacent) = arctan(54.5/350000) ~ 0.000155714 rad (0.0089°)
interestingly, when the denominator is large (i.e., the angle is small) you can extremely closely approximate the arctan (in radians) by simply ignoring it: 54.5/350000 = 0.000155714 << neat eh?
the 'usual' Angular diameter formula is: α = 2*arctan(g/2/r) -- where g is the width of the object and r is the distance. You did the same thing with arcsin() (took half the width and doubled the angle)
So for distant objects this simplifies to just g/r in radians -- so we can say 109/350000 rad ~ 0.01784°
Which is slightly larger than your 0.016° estimate only because you truncated the half-angle -- this link shows our full values are the same:
http://www.wolframalpha.com/input/?i=(2*arcsin(54.5%2F350000.4))+in+degrees
One note for those attempting to measure the sun -- the altitude of the ISS has varied from 320km to 415km (currently in 2017 ~400km) -- which is significant enough to affect the apparent angular size so readers should be sure to get the data for the actual time of observation before making estimations.
I like math but even I don't always like doing calculations by hand, so I really like this site for doing quick angular diameter conversions: https://rechneronline.de/sehwinkel/angular-diameter.php