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You can measure Earth with a scale and a shadow.It’s a hot summer day. The sun is beating down mercilessly. You’re walking down the road lost in your own thoughts.Suddenly, you look down at the ground and notice that your shadow is missing!You look up into the sky. The sun is right above your head. (Of course! That's why there are no shadows). It's the Summer Solstice. At your location, the rays from the sun are hitting perpendicular to the ground. Your shadow has literally vanished into a small area under your feet.But the sun can't be overhead everywhere on earth at the same time, right? Two hundred miles from your home, there's another city where the people aren't so lucky. The sun appears a bit off-center in the sky. The shadows are still there.If there's something that cannot change across the world, it is the fact that every town on our planet- big or small - has a clock-tower. This one does too. It stands proudly, right in the middle of the city square, casting a pretty big shadow.You take a measuring tape and find out the length of the shadow. You discreetly give a hundred rupee note to the elderly security guard of the clock-tower, and he tells you the exact height of the building. You divide the second value by the first, feed it into the archtan function in your scientific calculator, and bingo, you get the "angle of the shadow".But you've done something else too. Without even realizing it, you have also calculated another very important, very impossible-to-measure angle.The angle that the two cities subtend on the center of the planet!image source - author’s galleryYou are AB. Sun's rays are falling perpendicular on your head. If you extend this ray into the interior of the planet, as I've done, it would pass through the center of the Earth O.Q is the second city and PQ is the clock-tower in this city. (Note that like every building in existence, the clock-tower is perpendicular to the ground - Angle PQR is 90 degrees. Also, note that in this city, the sun’s rays don’t fall perpendicular to the ground. They fall at an angle, thus causing the shadows.)It casts the shadow RQ.Angle PRQ (shown as the single-stroke-arc) is the 'angle of the shadow'. You just calculated this.Here's the thing - as soon as you calculated angle PRQ, you also automatically calculated angle RPQ (the one represented by double-stroke-arc ). It is 90 degrees minus the first angle. (Why?)Now, shift your focus a bit and look at RPOB. It forms a Zed, doesn't it? And in every Zed, the angles of the two corners are equal!Which means the angle RPQ is equal to angle QOB. Which is the angle that the two cities (Q and B) subtend at the center of the planet!Pause for a moment and marvel the superhuman nature of the feat you've just achieved with nothing but a pen, a paper, and a little bit of mathematics. (Plus don't forget to thank your friend in the other city Q who is going around taking the measurements for you, while you comfortably sit in city B doing your calculations!)But that's not all. In fact, we're just getting started.Forget everything. Just focus on the sector QOB.The ratio of the length QB to the circumference of the entire circle would be equal to the ratio of angle QOB to 360 degrees.Length QB / Length of the entire circle = Angle QOB / 360 degreesIt's logical. The 'total' angle around the centre of the circle is 360 degrees. While QOB is the angle that our arc QB subtends on the center. It's logical that the ratio of these two angles must be equal to the ratio of the arcs that subtend them (Arc QB and the entire circle respectively)You know QB, the distance between cities Q and B (200 miles in this case).You know Angle QOB (You just calculated it above).And you know 360 degrees (duh!)Just rearrange things and you can find out the only unknown in this equation - the circumference of the earth!That's one hell of a mind-fuck!Ladies and Gentlemen, wrap your minds around this - We have just calculated the circumference of the Earth!This is exactly what Eratosthenes did. He was a super-intelligent dude who lived in the Greek town of Cyrene, who once heard the stories that in Egypt there was a well that cast no shadows on one particular day of the year (the Summer Solstice).That gave him this idea.He found a suitable tower in his hometown and calculated the "angle of the shadow" on the noon of that day.He estimated the distance between Egypt and Cyrene to be 5000 stadia (a unit of measurement in those times).He plugged in the numbers and found the circumference of Earth to be 25000 stadia.In today's units, that is 40,000 km.The actual circumference of Earth is 41,670 km. That's an error of less than 4 percent!And he did all of this in 250 BC.If that doesn't blow your mind, as it did mine when I first came across it a decade ago, I don’t know what to say to you.

This question is an opportunity for me to present what people used to do in the past to obtain such approximations. I am, indeed, going wayyyyy back in the past — at the time of Archimedes of Syracuse (of ‘naked Eureka’ fame), approximately 2250 years ago.In his proof that [math]\pi[/math] is a value between [math]\frac{223}{71}[/math] and [math]\frac{22}{7}[/math] (which is another fascinating story of his, requiring its own Quora answer), Archimedes used the following result:[math]\frac{265}{153}<\sqrt{3}<\frac{1351}{780}.[/math]In our decimal notation, that amounts to stating that[math]1.732026144<\sqrt{3}<1.732051282.[/math]The above is an astonishingly accurate estimation, especially considering that this was discovered literally millennia ago.Archimedes, however, stated this result without proof. This fact has puzzled many people for centuries. Did Archimedes prove this result, but decided not to write down his proof (a la Fermat)? Did Archimedes actually write down his proof, but his proof was lost? Or was this result “common knowledge” for the people of that era, perhaps discovered by the Babylonians even before Archimedes?Whatever the case was, people started to try to reconstruct the proof that [math]\frac{265}{153}<\sqrt{3}<\frac{1351}{780}[/math]. Several theories have been proposed. One theory is that this result was deduced by applying the Babylonian method (which Newton later generalized in the Newton-Raphson method) to evaluate square roots. This, essentially, iterates [math]x_{n+1}=x_n/2+k/(2x_n)[/math] to find [math]\sqrt{k}[/math], starting from an initial estimate [math]x_0[/math]. However, it seems difficult to obtain the above fractions [math]\frac{265}{153}[/math] and [math]\frac{1351}{780}[/math] as two consecutive iterates using this method. (We also need to remember that the Babylonians used base sixty for their numbers.)A popular theory, perhaps, is that these numbers were produced using a continued fraction. As explained in Alan Bustany’s answer, if one uses continued fractions, then these two bounds [math]\frac{265}{153}[/math] and [math]\frac{1351}{780}[/math] do appear. However, a better lower bound than [math]\frac{265}{153}[/math], namely [math]\frac{989}{571}[/math], is obtained before the upper bound [math]\frac{1351}{780}[/math] using this method, so if continued fractions were indeed the method used to obtain them, it seems weird to state the lower bound [math]\frac{265}{153}[/math] rather than this better one. Still, continued fractions were certainly known to the ancient Greeks, so some people theorize that these two bounds were produced in this manner.Below is a more plausible explanation of how Archimedes may have obtained his upper and lower bounds for [math]\sqrt{3}[/math]. The reason why this is a more plausible explanation is because our two bounds [math]\frac{265}{153}[/math] and [math]\frac{1351}{780}[/math] are obtained in consecutive iterations.Let me first illustrate the general method, then apply it for [math]\sqrt{3}[/math].Suppose [math]\frac{p}{q}[/math] is an approximation for [math]\sqrt{n}[/math], where [math]n[/math] is not a perfect square. Then [math]\frac{p^2}{q^2}-n[/math] must be a small rational number [math]r[/math]. We can write this down as[math]\dfrac{p^2-nq^2}{q^2}=r.[/math](If [math]r=1/q^2[/math], then the above is the famous Pell equation.) Note that [math]r[/math] cannot ever be zero, otherwise [math]\sqrt{n}[/math] is rational.We now want a better approximation for [math]\sqrt{n}[/math] that makes [math]r[/math] even closer to zero. Suppose we substitute [math]px+nq[/math] for [math]p[/math] and [math]qx+p[/math] for [math]q[/math] on the left hand side of the above equation. The new numerator becomes[math](px+nq)^2-n(qx+p)^2[/math][math]=p^2x^2+2pqnx+n^2q^2-n(q^2x^2+2pqx+p^2)[/math][math]=p^2x^2+2pqnx+n^2q^2-nq^2x^2-2pqnx-np^2[/math][math]=(p^2-nq^2)x^2-n(p^2-nq^2)[/math][math]=(p^2-nq^2)(x^2-n)[/math]But [math]p^2-nq^2=rq^2[/math], so[math](px+nq)^2-n(qx+p)^2=rq^2(x^2-n).[/math]On the other hand, the new denominator is simply [math](qx+p)^2[/math], so[math]\dfrac{(px+nq)^2-n(qx+p)^2}{(qx+p)^2}=\dfrac{rq^2(x^2-n)}{(qx+p)^2}.[/math]If we now substitute [math]x=\frac{p}{q}[/math] in the right hand side of this formula, we obtain:[math]\dfrac{rq^2\left(\left(\frac{p}{q}\right)^2-n\right)}{4p^2}=\dfrac{rq^2\left(\frac{p^2-nq^2}{q^2}\right)}{4p^2}=\dfrac{r^2q^2}{4p^2}=\left(\dfrac{qr}{2p}\right)^2.[/math]This new error is smaller than the old error [math]r[/math] exactly when [math]0<|r|<\left(\frac{q}{2p}\right)^2[/math]. This can easily be arranged.Thus, we have concluded the following: if [math]\frac{p}{q}[/math] is an approximation to [math]\sqrt{n}[/math], then the expression[math]\dfrac{px+nq}{qx+p}[/math] with [math]x=\frac{p}{q}[/math]is a better approximation, provided that the absolute value of [math]\left(\frac{p}{q}\right)^2-n[/math] is less than [math]\frac{q^2}{4p^2}[/math]. Thus, we may iterate this expression until a good estimate is found.Let us apply this for [math]\sqrt{3}[/math]. If we let [math]p=5[/math], [math]q=3[/math] and [math]n=3[/math], the expression [math]\frac{p^2-nq^2}{q^2}[/math] becomes [math]\frac{5^2-3\times 3^2}{3^2}[/math][math]=-\frac{2}{9}[/math], which is our [math]r[/math]. Since [math]|r|[/math] is less than [math]\frac{p^2}{4q^2}=\frac{25}{36}[/math], we may take [math]\frac{5}{3}[/math] as our first approximation to [math]\sqrt{3}[/math]. Then, our iterative function will be:[math]\dfrac{px+nq}{qx+p}=\dfrac{5x+3(3)}{3x+5}=\dfrac{5x+9}{3x+5}.[/math]Substituting [math]x=\frac{5}{3}[/math] in this iterative function should provide us with a closer approximation to [math]\sqrt{3}[/math]:[math]\dfrac{5\frac{5}{3}+9}{3\frac{5}{3}+5}=\dfrac{\frac{25}{3}+9}{10}=\dfrac{25+27}{30}=\dfrac{26}{15}.[/math]It can also be shown that if the first error is negative, then the errors of subsequent iterates will have alternating signs. Thus, [math]\frac{5}{3}[/math] is smaller than [math]\sqrt{3}[/math], because its [math]r[/math] was negative, but [math]\frac{26}{15}[/math] is larger than (and closer to) [math]\sqrt{3}[/math]. The next iterate will be smaller, and even closer to, [math]\sqrt{3}[/math] and the next one will be larger, and even closer to, [math]\sqrt{3}[/math].And do you know what are the next two iterates? They are[math]\frac{265}{153}[/math] and [math]\frac{1351}{780}[/math].These two values should ring a bell… they are precisely the two values that Archimedes said are lower and upper bounds for [math]\sqrt{3}[/math].Of course, if we want even better estimates, we may continue iterating this as much as we want.(Adapted from Archimedes and the Square Root of 3.)

One Trump's tweet probably confirmed that the resolution of American KH-11 spy satellite’s cameras can let American intelligences read license plate number through a spy satellite photo.On 30 Aug 2019, President Trump posted a spy satellite photo in his tweet, mocking Iran's missiles and showing off American advanced spy satellite technology.What's really shocking is how clear this photo is. In this photo, not only can we see the four lightning towers around the circular launch pad, but the length and angle of the shadow of the tower are very obvious, and we can even see the hollow structure of the launch pad itself.What’s more unbelievable is that the photo is not enlarged through post-processing—because the doors of the rocket’s mobile launching and erecting device are clearly visible, and the “empowered by the State” written in Persian on the edge of the circular launch pad can be distinguished.Let's not forget the photo we the public can seew was from Trump's tweet. Trump’s men printed this photo and took a picture of the printed spy satellite photo. It means that the original spy satellite photo is more clearer.(Commercial satellites of Planet and MacArthur also took pictures of it)Then I'll explain why this photo proves how powerful the cameras on spy satellite are. (The calculation below was originally from Marco Langbroek’s post to the SeeSat-L Internet forum. I'll try to explain his calculation in plain language)First, the location of Imam Khomeini Spaceport determined this photo can only be taken by spy satellite not drones.The United States is unable to send a drone flying a thousand miles, through Armenia, Azerbaijan and Iran, with a risk of being shot by Iran, especially when Iran just shot a RQ-4 in Persian Gulf a month before.So the photo can only be taken by a spy satellite. And KH-11 type electro-optical surveillance satellite is able to take such a high-resolution picture.Calculated according to the scale on Google Maps, the diameter of the circular launch pad where the accident occurred is about 60 meters, which occupies about 600 pixels in the Twitter photo, 60/600=0.1, so the resolution of the photo alone is 10 cm per pixel.Like I wrote before, the resolution of the original photo can only be higher.According to the light diffraction formula, and the visibility limit of the observable object caused by the natural diffraction when light propagates, plus parameters visible light wave band 500 nanometers and the lens diameter 2.4 meters, we can calculate that (1.22*5*10^-5)/240 is about 0.05 arc second.Assume that the orbital height of the satellite is 250 kilometers(in fact it's probably higher), 0.05*250000/206265=0.06 (I'll add in reference explaining why 206265 is used here), the limited diffraction resolution of a 2.4-meter lens under vacuum is about 6 cm. And don't forget LEO is not real vaccum and the satellite is probably higher.To sum up, the resolution of the photo is probably very close to the theoretical limit!!!That's why I wrote that it can “let American intelligences read license plate number through a spy satellite photo” in the beginning.Extremely high-resolution camera is not the only information we can get from this Trump's tweet.This launch pad is located in 35.2346° N, 53.9210° E . The direction of the lightning ° at the north and south corners is 192.8 degrees. Thanks to Google Map.If the launch pad is circular, and the azimuth difference between the north and south lightning towers is identified as 3.8°, meaning azimuth angle of the shooting satellite is 192.8+3.8=195.8°. The shooting angle of elevation can be calculated by sinh=b/a, a is 270 and b is 195 and sinh(195/270)=0.787, meaning elevation angle is about 45°.And when we combine calculation results and location of launch pad, we will see it's USA-224 that took this high-resolution picture.Why did I add all of these when a picture?itself can prove how powerful the high-resolution spy satellite camera is?Because USA-224 is already not the most advanced spy satellite. It's believed that after the launch of Block-V KH-11 satellites like USA-290, USA-224 and the ability of it's camera is not so sensitive. So NRO would agree to post some information of this “dated” spy satellite. Just like America released photos of Soviet AC taken by Block-I KH-11 satellite when Block-II KH-11 satellites had served.That's why we can see the phone in Trump's tweet. And it means that we the public still have no idea of howw powerful are the most advanced cameras of spy satellites.We can only be sure that they will be more powerful than the high-resolution camera which took the photo of Iran's launch pad on 29 Aug 2019.Reference:KH-11 satelliteKH-11 Kennen - WikipediaThe KH-11 KENNEN [1] [2] [3] [4] (later renamed CRYSTAL , [5] then Evolved Enhanced CRYSTAL System , and codenamed 1010 [6] and Key Hole [6] ) is a type of reconnaissance satellite first launched by the American National Reconnaissance Office in December 1976. Manufactured by Lockheed in Sunnyvale, California , the KH-11 was the first American spy satellite to use electro-optical digital imaging , and so offer real-time optical observations. [7] A conceptual drawing based upon Hubble Space Telescope (HST) layout. A conceptual drawing based upon Hubble Space Telescope (HST) layout with internal views. Later KH-11 satellites have been referred to by outside observers as KH-11B or KH-12, and by the names "Advanced KENNEN", "Improved Crystal" and "Ikon". Official budget documents refer to the latest generation of electro-optical satellites as Evolved Enhanced CRYSTAL System . [8] The Key Hole series was officially discontinued in favor of a random numbering scheme after repeated public references to KH-7 Gambit , KH-8 Gambit -3, KH-9 Hexagon , and KH-11 satellites. [9] The capabilities of the KH-11 are highly classified, as are images they produce. The satellites are believed to have been the source of some imagery of the Soviet Union and China made public in 1997; images of Sudan and Afghanistan made public in 1998 related to the response to the 1998 U.S. embassy bombings ; [ citation needed ] and a 2019 photo, revealed by President Donald Trump, of a failed Iranian rocket launch. [10] Program history and logistics Edit The Film Read-Out GAMBIT (FROG) served as NRO Program A's competitor to NRO Program B's initial electro-optical imagery (EOI) satellite. [11] After a precursor EOI study with the codeword Zoster, President Nixon on 23 September 1971 approved the development of an EOI satellite under the initial codeword Zaman. [12] In November 1971, this codeword was changed to Kennen, which is Middle English for "to perceive." [13] [14] Data is transmitted through a network of communications satellites ; the Satellite Data System . [5] The initial ground station for the processing of the electro-optical imaging was a secret National Reconnaissance Office facility in Area 58 , later confirmed to be located in Fort Belvoir . [15] [16] In 1999, NRO selected Boeing as the prime contractor for the Future Imagery Architecture (FIA) program, aimed at replacing the KH-11 satellites by a more cost effective constellation of smaller, and also more capable reconnaissance satellites. After the failure of the FIA in 2005, NRO ordered from Lockheed two additional legacy hardware KH-11s. [17] USA-224 , the first of these two, was launched in early 2011 two years ahead of the initial schedule estimate. [18] In January 2011, NRO offered NASA two space optical systems with 2.4 m diameter primary mirrors, similar to the Hubble Space Telescope, yet with steerable secondary mirrors and shorter focal length resulting in a wider field of view. These could either be sparehttps://en.m.wikipedia.org/wiki/KH-11_KennenUSA-224Engineering:USA-224 - HandWikiUSA-224 , also known as NRO Launch 49 ( NRO L-49 ), is an United States reconnaissance satellite . Launched in 2011 to replace the decade-old USA-161 satellite, it is the fifteenth KH-11 optical imaging satellite to reach orbit. Project history and cost After the Boeing -led Future Imagery Architecture program failed in 2005, NRO ordered two more KH-11s, including USA-224. Critics worried that each of these "exquisite-class" [2] satellites would cost more than the latest Nimitz -class aircraft carrier (CVN-77), which had a projected procurement cost of US$6.35 billion as of May 2005 (equivalent to approximately $8B in 2019). [3] [4] USA-224 – the first of these two – was completed by Lockheed $2 billion under the initial budget estimate, and two years ahead of schedule. [5] USA-224 was launched atop a Delta IV Heavy rocket from Vandenberg AFB Space Launch Complex 6 in California . The launch was conducted by United Launch Alliance , and was the first flight of a Delta IV Heavy from Vandenberg. [6] Liftoff occurred on 20 January 2011 at 21:10:30 UTC. [7] Upon reaching orbit, the satellite received the International Designator 2011-002A. [8] The satellite began operating 33 days after its predecessor, USA-161, stopped doing its primary mission. This coverage gap was much smaller than originally feared, thanks to USA-224's earlier-than-planned launch and operational changes to extend the lifetime of USA-161. [5] As the fifteenth KH-11 satellite to be launched, USA-224 is a member of one of the later block configurations occasionally identified as being a separate system. Details of its mission and orbit are classified, but amateur observers have tracked it in low Earth orbit . Shortly after launch it was in an orbit with a perigee of 251 kilometres (156 mi), an apogee of 1,023 kilometres (636 mi) and 97.9 degrees of inclination , typical for an operational KH-11 satellite. [9] By April it was 260 by 987 kilometres (162 by 613 mi) at 97.93 degrees. [10] Imaging of Nahid-1/Safir launch preparation accident A photo tweeted by President Trump which is believed to have been taken by USA-224 On 30 August 2019, President Donald Trump tweeted a classified picture [11] from an intelligence briefing showing the aftermath of an accident that apparently occurred during launch preparations of a Safir rocket at the Imam Khomeini Spaceport a day earlier. [12] [13] [14] According to analysts, the photo is likely to have been taken by USA-224. [15] [16] The opinion is based on a close agreement between the estimated time when the photo was taken (based on the orientation of shadows cast by structures in the photo), and the location of the satellite at that same time, as estimated with tracking data maintained by the amateur satellite watching community. [17] [18] [19] The off- nadir photograph stands out for its high-resolution (estimated by analysts to be 10 cm or less per pixel), sharpness and lack of atmospheric distortion. [15] Before this tweet, the only KH-11https://handwiki.org/wiki/Engineering:USA-224USA-290USA 290USA 290, launched from Air Force Western Test Range, California, USA in 2019.https://in-the-sky.org/spacecraft.php?id=43941calculation and some images are from( I didn't find Marco Langbroek’s original post):一张照片带来的惊人信息川普的推特一直以语出惊人而出名，但他在8月31日发的推特还是再一次刷新人们对他乃至美国的认知。而这一切，仅仅是因为川普在嘲讽伊朗的推特下，配上了一张非常惊人的照片。 这还要从8月29日美国全国广播电台（NPR…https://zhuanlan.zhihu.com/p/81527863?utm_source=wechat_session&utm_medium=social&utm_oi=882999480234971136&s_r=0why use 206265Why do we use the value 206265 in the small angle formula?$D=dx÷206 265$ What does this number mean?https://astronomy.stackexchange.com/questions/36777/why-do-we-use-the-value-206265-in-the-small-angle-formulanewsSurveillance photos reveal apparent explosion on Iranian launch padThis image, tweeted by President Trump on Friday, appears to come from a high-resolution camera on a U.S. government drone or satellite. It shows damage to an Iranian satellite launch pad. A day after commercial satellite images revealed a smoke plume over an Iranian launch pad, President Trump tweeted a remarkably high-resolution aerial photograph of the remote satellite launch facility Friday — apparently from a classified satellite or airborne drone — and denied U.S. involvement in the accident. “The United States of America was was not involved in the catastrophic accident during final launch preparations for the Safir SLV Launch at Semnan Launch Site One in Iran,” Trump tweeted. “I wish Iran best wishes and good luck in determining what happened at Site One.” The Safir is a two-stage Iranian satellite launcher, and the accident was the third time Iran has failed to place a satellite into orbit this year. Trump’s attachment to the tweet of a photograph showing a detailed aerial view of the launch drove imagery analysts and satellite experts into a frenzy. The image Trump tweeted appeared to be taken from a cell phone or other type of camera, showing a digital or printed rendering of a surveillance photograph captured from an intelligence-gathering drone or satellite. In either case, Trump’s disclosure appears to reveal new details about the U.S. government’s normally secret intelligence-gathering capabilities. U.S. presidents can declassify information at their own discretion. The United States of America was not involved in the catastrophic accident during final launch preparations for the Safir SLV Launch at Semnan Launch Site One in Iran. I wish Iran best wishes and good luck in determining what happened at Site One. pic.twitter.com/z0iDj2L0Y3 — Donald J. Trump (@realDonaldTrump) August 30, 2019 Within hours, analysts and amateur satellite trackers joined forces on social media and Internet forums to track down a possible source of the image Trump tweeted. Christiaan Triebert, a visual investigations journalist at the New York Times, deduced from shadows in the image that it was taken between 1:30 and 2:30 p.m. local time, probably on Thursday. Open-source orbital data compiled by amateur trackers indicated that a top secret U.S. government spy satellite flew over the Semnan launch site during that time, according to Michael Thompson, a graduate student at Purdue University studying astrodynamics and spacecraft navigation. Using public data, Thompson calculated that the satellite, officially designated USA-224, passed over the Iranian space center shortly after 2:10 p.m. local time Thursday. Marco Langbroek, an experienced tracker of satellite movements who lives in the Netherlands, confirmed that the USA-224 spacecraft was over the Semnan launch base around that time. In a post to the SeeSat-L Internet forum, where enthusiasts share their satellite sightings, Langbroek wrote the “angle from which the image seems to be taken … matches thehttps://spaceflightnow.com/2019/08/30/surveillance-photos-reveal-apparent-explosion-on-iranian-launch-pad/