tag:blogger.com,1999:blog-4953008377950396317.post705685882238489837..comments2024-03-28T18:39:59.184-07:00Comments on SearchReSearch: Quick check on the validity of an imageDan Russellhttp://www.blogger.com/profile/13603209997260423532noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-4953008377950396317.post-41851803745546718552017-05-17T14:40:04.767-07:002017-05-17T14:40:04.767-07:00Fun post! Thanks for the write up. Fun hack: fo...Fun 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)<br /><br />Why this works consider:<br /><br />To avoid the need to compute the hypotenuse you can also use arctan():<br /><br />θ = arctan(opposite/adjacent) = arctan(54.5/350000) ~ 0.000155714 rad (0.0089°)<br /><br />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?<br /><br />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)<br /><br />So for distant objects this simplifies to just g/r in radians -- so we can say 109/350000 rad ~ 0.01784°<br />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:<br /> http://www.wolframalpha.com/input/?i=(2*arcsin(54.5%2F350000.4))+in+degrees<br /><br />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.<br /><br />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<br />Dark Starhttps://www.blogger.com/profile/04356850749159919331noreply@blogger.comtag:blogger.com,1999:blog-4953008377950396317.post-73646568494224739092012-08-14T20:03:51.377-07:002012-08-14T20:03:51.377-07:00Alas, there's no quick and straighforward answ...Alas, 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.) Dan Russellhttps://www.blogger.com/profile/13603209997260423532noreply@blogger.comtag:blogger.com,1999:blog-4953008377950396317.post-74201774971661900082012-08-14T18:17:23.885-07:002012-08-14T18:17:23.885-07:00im needing to check the validity of a pic posted o...im needing to check the validity of a pic posted on instagram, how do I do this? Anonymoushttps://www.blogger.com/profile/13438855457355489165noreply@blogger.com