Hey Scott, a couple points to remember: 1) the moon orbits in an ellipse and its apogee, perigee and mean distance are all different. Closest perigee is 356,400km. 2) remember the published value is from the centre of Earth, not a point on its surface, which is some fraction of 6,000km closer. So your estimates may be better than you think.
If I was a High School science teacher, I’d try to get together with another school and do this measurement. I think it would be a great project and would teach a lot about these concepts.
A high school math teacher would also be able to make it into a lesson. A history teacher could make it with a bit of a stretch if a scientist comes up who somehow is connected. A geography teacher could do it as a demonstration of how map distances where done in the past.
Ken Smith There are definitely a lot of things that could be done with this. The exciting part is that kids could be shown that they could do real science themselves. I think it would make it exciting for them.
Well, if you know that your cameras are 10 centimeters tall, you can just stack them on top of each other. I dunno why you'd use cameras instead of metersticks, though. ...Wait, you mean _taking pictures_ with the cameras?
This urban myth seems to have become known through an article in "Saturday Review", a magazine, in 1968. An examiner sets the question for a physics student, to measure the building's height with a barometer - expecting him to calculate the pressure difference. The student suggests typing a rope to the barometer, lowering it down the building, and measuring the rope.The examiner thinks that should score zero, the student thinks it's a perfect answer, and suggests many other creative answers. He knows the "real" answer, but objects to the overly dogmatic teachings; the scholasticism. It's also demonstration of creativity; thinking outside the box. The internet meme has a similar story about the Nobel prize-winning physicist Niels Bohr, when he was a student. But that's almost certainly a myth.
Did an entire section on this in high school. We had a classmate who happened to be on a trip to Europe take a photo, and we took a photo from the middle of the US. Knowing the distance apart that the photos were taken, we got the distance to the moon by parallax. Then we stretched a wire "spider web" over the football field using one of the goal posts as a focal point to get a dish antenna in the FM band, and timed the "echo" of a powerful local radio station to measure the speed of light. Fun times!
+CowBones Indeed, hence the slight amount of shade in my comment. Doesnt help that we emailed about possibly re-doing it as a colab but I guess I didn't live far enough away.
Cut would have been simultaneously easier and harder I think. I don’t know why we never followed up on this. What we should try doing is measuring distances to a geostationary satellite.
I love how Apollo 11 left mirrors on the surface to measure the exact distance between Earth and Moon. Not that an amateur could use this system, but it’s still super cool what they were able to do on that first mission. Maybe you could do a video sometime about the scientific discoveries about the moon from the moon landings and what things we still reeeally need to know about the moon.
Those are also great for dealing with lunar landing deniers. Anytime someone in range says the landings were faked I screech "APOLLO RETROREFLECTORS" into whichever ear is less protected or closer until I see blood.
Not gonna argue there, just one thing makes me wonder, though: do we possess a tech that is precise enough to hit a 50x50cm array over 300,000km? Or when they fire a laser, the radius of the laser spot on the moon is much bigger and they just fire in the general direction? If the latter, which I assume it is, is there a video of some sort explaining the math behind it? Would be cool to watch.
By the time the laser beam hits the Moon, it's several kilometres in diameter! However, because the light is of a tightly-controlled wavelength, we only need to receive a few photons back in order to make measurements. Only 1 in several millions of photons will hit the reflector and be fired back toward Earth, where those few are again spread over several kilometres area. By measuring the time, we can calculate the distance to an accuracy of a couple of centimetres. Incidentally, you may be interested to know about the forthcoming new retroreflector. MoonLIGHT, also known by the snappy title "Lunar Laser Ranging Retroreflector Array for the 21st Century" (LLRRA-21) which should be deployed in 2019, and should give 100 times more accurate results.
Actually Apollo 11, 14, and 15 all left retro-reflector arrays, as did the Russian Luna 17 and 21 rover missions. You can see a picture of the Apollo 15 reflector (the largest on the moon) here: en.wikipedia.org/wiki/Apache_Point_Observatory_Lunar_Laser-ranging_Operation#/media/File:ALSEP_AS15-85-11468.jpg As the previous poster pointed out, the beam is a few km across by the time it reaches the moon. Of course, you still have to point at it pretty precisely, as there are a number of effects on range-data quality from mis-pointing. Also, since the moon is a moving target, you have to fire your laser about .01 degrees ahead of where you are looking for the return photons. Now the .01 degrees does not sound like a big angle, but when the field of view of the telescope used is only .016 degrees, you do not have a lot of wiggle room. Plus you have to put your extremely narrow beam (0.0006 degrees(ish)) on a very small portion of the moon, then look for the very small return. It is challenging, to say the least. As the previous poster mentioned, you may only get a small number of photons back for every pulse. In fact, most ILRS stations really only want one photon return from each pulse and attenuate the light to get that. They send thousand(s) of pulses per second, and do not worry too much if they miss a few. From the resultant data you can estimate the distance to the moon to a small number of mm, and probably less.
They had to land something there anyway. You cannot fake a signal coming stationary from the moon. You don't even need paralaxes. Everyone who could receive the signal could just check when the moon went below the horizon and when he lost the signal... Unique time for every location on earth. Unfake-able. the silly thing: for a "fake" that would fool even the casual observer for 5mins, they would have to do so many chunks of the mission anyway, that a fake would barely save them any effort. It would indeed ADD effectively a second space program... Against the massive risk of ending every career and company reputation involved forever. It's all just such complete nonsense....
Fascinating to see the limitations in measurement accuracy imposed by the technological level of the observer, and to imagine what this would've meant for early astronomers doing this with an eyeglass and parchment instead of precision telescopes and computer modelling. It's no wonder it took us so long to get (relatively) accurate measurements for the features of celestial bodies when precision is so closely tied to the quality of the tools at hand.
Is there any way to try this with a single camera at different times of night (e.g. moonrise and moonset), using the fact that you’re a point on a rotating sphere - which means, essentially, two different points when measured at two different times? I guess one problem would be that the moon could move relative to the background stars (does it move fast enough for that to be a problem?), rendering stars useless as reference points. But I wonder if you could instead use a telescope rig that can precisely measure the angle it’s pointing at.
Yeah , that i was thinking too - but you know the moon orbit period and can just substract the difference . I think it can be done from a mountain top with a long bubble level and a good clock knowing just the diameter of the earth
Hello, sorry, I have a question in the video. What are the two points on the moon, I know why choose dark parts, but why those two points? And why two points chose on the moon? Could you please explain it a little clearer? Thanks a lot!
There is a possible source of error that you didn't mention: Your typical camera shoots photos in rectilinear projection, i.e. for viewing on a flat surface in front of your eyes. Imagine standing right in front of a big, flat advertisement poster. You'll notice that the part that's right in front of you will be larger in your vision than the stuff that's on the edges of the poster, because those are further away from you. This means that on the edges, things that take up the same area on the poster than in the center will actually cover a smaller angle. There is also deformation due to perspective - looking at the edge of the poster to your left. A sphere on the poster at that point will look like an ellipse that's higher than it's wide. Going back to the photo camera, this means that on the edge of your picture, the same angle will take up more pixels than in the center. Now, given that the angle of your photos is pretty small the effect may be negligible, but I think it's worth mentioning when you talk about equating pixels and angles.
So how about pinging the retroreflector experiment left by the Apollo mission. How much laser power would you need and how big of a telescope would you need to detect it?
Specialist stuff, cost about a million US dollars. About 3.5m telescope and quite a specialist laser; you need to fire _really_ short bursts. I can't really give a specific power rating, because it's a fast pulsing laser with a small pulse width. Like, 115 mJ per pulse, 20 Hz, which is only 2.3W. You fire quadrillions of photons, and get a handful back. So... universities and stuff can do it, and they do. But it's not really something you can do in most back yards. (I could say lots more... but I'm guessing that sorta answers?)
how do you get the conversion from pixel to degree? That's what confuses me the most. "on the sky we know that is 4.31 degrees" . HOW do we know that???
There are two main ways to do that. One is to look at the camera settings you used to take the picture, which will tell you the andular width of the shot which you can use as a reference for everything else in the shot. The other method is to use visual clues: for instance, the angular diameter of the Moon is close to half a degree with some monthly variation. You can also identify stars in the shot and look up their distances on a star map, which is what Scott did here: he identified two stars from Virgo whose angular separation is known to be 4.31 degrees.
What I did - which gave a crude result, but that's all I was looking for - was attach a 10cm length of construction paper to the end of a tape measure, then push the tape measure out from my eye until the two edges of the construction paper just touched two stars I had taken a picture of as part of my moon picture. That gave me the slope of the angle between the stars, which you can use some inverse trig on to convert to an angle. (In my case I just used the slope, so I didn't even bother converting it to an angle.)
Maybe the inaccuracys are in part because of different atmospheric effects? At 3:30 you show "what the moon sees". From that perspective light hitting the east and west part of the US hit the atmosphere at different angles Wouldn't that cause the light to deviate its course differently for both of you?
Dear Scott, we did the same experiment but with Venus as the callibration "star", we ended up overprojecting the distance to 325.000 miles way back in Dec 2016. We used a telescope to measure and obviously a cheapo! Over long ditances, it is incredibly difficult to point a telescope unless you have precision guides in place. Nevertheless, we DID end up "close" to the accepted values. Probably 25% off, but still incredible that ANYONE can measure the distance to the moon or even other celestial bodies with this method. There are more advanced and better methods, but for the ameteur observer, this is incredible. Please shed more light on the technique and equipment used for this particular achievement. You are Scott the MAN and will always fly safe :-)
Boy, your spherical trig calculations gave me flashbacks to my observational astronomy class in college. The textbook dated from the 1930's and was full of what my professor called "British schoolboy exercises." Every problem set was a new fresh hell. A typical problem: "Prove that, in latitude 45º, the interval between the moment at which a star's azimuth is 90º and the moment of setting is constant."
Proof by contradiction: Circumpolar stars exist for a viewer at 45°N. Circumpolar stars never set. Thus the time between culmination (azimuth 90°) and set (elevation angle 0°) of a circumpolar star is infinite. Thus the earth is flat? I think that's the conclusion I'm supposed to use.
hey Scott. great video. love your work. I have seen my share of FE videos and I can only surmise they steer clear of your channel and VSAUCE and Codys Lab to name a few. KEEP IT UP BUDDY.
Yeah, but you need about a million US $ worth of kit to manage it. They are working great. But you need a 3.5m telescope and stuff, to make use of them. I gave a bit more detail in other replies.
Jackalovski yes, but only if you take the picture with one phone and quickly place the second one on top facing each other to trap the light between them. Making the light confused on with way to go. The light will now hang out in the middle of the two cameras longer, giving you a better exposure.
My budget is whatever I have laying round. I've been gifted a second hand iPhone 7 and the 6 is the works phone. The best I could manage in terms of 'propper hardware' is ducktaping them to a pair of soviet binoculars with my current budget.
Sorry if this is a stupid question but at 5:13 you said left image was high in the sky and right image was down near the horizon. I always thought that when the moon was low on the horizon some factor beyond my knowledge would distort the light making the moon look larger that it actually is but your image shows its smaller?
The apparent size of the Moon from Earth depends *mostly* (bear with me...) on how far away it is. That distance varies from _about_ 406,000 km to _about_ 360,000, every lunar month. In this video, they both measured it at the same time, so that is not an issue. The other effect is an optical illusion. When we see something near the horizon, our brain thinks it's larger than something directly above us. Nobody knows for sure why that happens. But it is not real. If you take a photo, and check, it's the same size. So that does not affect the results. A low Moon certainly _looks_ larger, but if you check using a camera, it is not. It is not a stupid question, because despite many books on the subject, nobody is quite sure why our brains work that way. Hope that helps.
There is a refraction effect that is different than the moon illusion described. This primarily squashes the moon in one direction, but leaves the other direction pretty much unchanged. It's more pronounced the closer to the horizon you get, and if you are ~15° above the horizon it's not really anything to worry about.
Thanks for the responses guys! I had a feeling it had something to do with the way the eyes and brain interpret the signal. My next question is; when will I be able to get Optical Augments so I wont have to worry about Illusions anymore?
When I bought the original PC version of "The Hitchhiker's Guide to the Galaxy" in 1984, inside the box was a pair of "Peril Sensitive Sunglasses", made from opaque black cardboard. What you don't know can't hurt you.
Cool video :) It may not vary by much for a zoomed in photo, but the camera's angle of view is projected onto a sensor that is flat, so pixels / degree is not constant across the entire image.
Probably already mentioned here in a comment...but why not just use the time for one lunar orbit to work it out??? I just used 28 days (close enough) and it comes to 383,215,240.2m which is pretty darn close!
According to flat earth logic, that assumes gravity is real, thus is fake. In the real world, multiple approaches to the same measurement converging on the same value gives you pretty good confidence the value is correct.
ann onn just use the formula r^3=(G*M*T^2)/4pi^2 . That's the standard formula used when balancing centrifugal force and gravity. G is the universal gravitation constant 6.67e-11 kg^-2. M is the mass of earth 5.98e24 kg. T is the time period of one orbit 28 days (use 2419200s). That gives r which is the distance between the center of earth and the center of the moon. For the distance from the surface of earth, subtract the earth radius 6371km. Technically you could also subtract the moon radius to get surface to surface, but I didn't bother with that so my number is just from Earth surface to moon orbit.
Interesting vid Scott, I remember measuring the moon distance with a brush stale and two different sized discs using the small angle approximation with the Open University. amazing stuff thanks for sharing👍🏼
You could use this to determine the diameter of the moon. Then use this as a reference lenght for measuring the shadow cast by a lunar mountain. Then you can measure the height of the mountain using trigonometry.
Don't you have to take into account the optics of the camera? Depending on lens and zoom the distances measured in pixels in one part of the resulting image don't necessarily correlate linearly to other parts in the image.
Hi Scott, I was thinking about the 2nd example with the two muns being compared directly: don't you have to take into account a lens effect of the earths atmosphere, that is stronger for images taken with the mun being close to the horizon because of the longer way of the light passing the atmosphere compared to a picture taken straight up?
Scott Manley Near the horizon it will certainly exaggerate the vertical. Refraction near the horizon changes a lot, even within the arc subtended by the moon. I wonder if you aren't seeing some refraction effect in your parallax measurement due to the low elevation of the east coast photo.
It's probably just caused by inaccuracies stacking up. Remember, Scott's using consumer-grade hardware designed for terrestrial photos, not cameras designed for astronomy. (And I'd imagine that the coordination between him and the other guy, while good, was not up to professional standards of precision.)
Diffraction would act on the starlight as well as moonlight. Errors would creep in if one object was measured near the horizon and another was overhead so that diffraction of the object near the horizon would put in visually in a different location. The Moon and stars were just a few degrees apart.
Lenard Segnitz But from the looks of things the shot from the East coast was very low elevation. At low elevation a few degrees makes a YUGE difference in refraction. Really bigly.
Or you can talk to an amateur radio operator that bounties signals off of the moon. Its called EME (Earth Moon Earth).You can hear the signal coma back at you a few seconds later. The timing matches the round trip distance to the moon and back divided by the speed of light.
Just a little nitpick: July 16th was fairly close to the perigee. According to what I could find, the actual distance on the day was around 369000km, not 384000.
Scott Manley ah ok, I'm an idiot. Of course you didn't use the pictures from before you set things up... Nice coincidence that you ended up on an average distance day then!
i have books from 1840 and 1850. the distances to the sun and the moon and the sizes of many if not all the planets were known. They are essentially spot on with todays numbers, How is it possible that they could measure the size of venus or saturn without presuppositions?
Couldn't you also do parallax observations based on the stars in the background at near moonrise compared to near when it sets? That probably makes figuring out your baseline even harder, but you could do that, right?
The problem is that the Moon moves in its orbit, and thus against the stars, between the two observations. What you would measure is a combination of that movement, which is about six degrees in 12 hours, and the true parallax which is about two degrees. If you can correct for the Moon's motion then in principle you can find the parallax but it would be difficult and probably not very accurate.
Guys, I have a 3 years old who's absolutely fascinated by the moon and stars. You think a regular rifle scope fixed on a camera tripod is good enough for now or should I buy him a dedicated telescope already? I'd also appreciate some links to astronomy videos for young kids if you know any.
Even simple binoculars are great, you can see the Galilean moons, the moon very bright, excellent for the upcoming Lunar Eclipse. You can't see the rings of Saturn however
A decent riflescope will work, but it's not going to be as good as a larger front lens. I carry one with me for that purpose, but mainly for terrestrial viewing. But if you already got the scope, use it, kiddo is only 3 yro.
There are rifle scopes that would be ideal, but they're hardly "regular". They're made for natural-light varmint shooting at night, and have something like a 70mm opening at the front. And they're $tupidly expen$ive. Let the sprout look through a good pair of binoculars instead.
We could use the right angle triangle of a half moon to get how many times further away the sun is that the moon. Not sure what equipment would work best though - the angle would be very close to 90.
Scott - this was pretty damned cool. Could you go a little deeper into it and identify likely sources of error and how they contributed to your final result? When synchronizing with the remote person, other than time, what other factors must you take into account, i.e. focal length of the camera, etc.?
Another fun exercise using only consumer level cameras is to prove Uranus exists and that it is not like other stars. Neptunus is a though a bit beyond the reach of at least cheaper ones. Figuring out when it's best to try is an exercise itself.
Scott Manley Right - my bad - you're taking sidereal rate into account by aligning the stellar background, so it's only the moon's motion relative to the stars. so only about 160km per second. Sorry about that. Did you happen to calculate the exact lunar surface to your station distance to compare? The 385Mm is the mean center to center, but you were probably like 7Mm closer. Update: I get 383.17Mm from Oakland, CA to the Moon's center, which works just fine for the limb.
And to clarify, I was talking about the moon's motion (orbiting earth) effect on your range estimate by paralax. 13.2(ish) degrees per day or about 0.55 degrees per hour.
You could figure out the gravitational constant by the pull of one object on another then use a cubed equals t squared and the time it takes the moon to complete an orbit, then you have the simimajor axis
Great Video And I really like the Idea of measuring astronomical distances with a camera. I was wondering if you added the earths radius to your measurement since the luna distance is define as the distance between the center of earth to the center of moon. Its pretty obvious so i assume you did but I had to ask. Maybe I just missed it in the video or you didnt mention it but any way great video and keep on doing stuff like this
You didn't account for the Moon being in an elliptical orbit. Sometimes it will be closer to Earth, which could help explain the results you were getting. 346,000km is a much better estimate if the Moon is at perigee than if it's at apogee. According to a quick google, the Moon was at perigee on July 13, and will reach apogee on July 27.
He was measuring what it is at the time of the photo being taken. If he wanted he could certainly do this daily and measure the movement around the entire orbit
Could someone explain to me, using simple words, why the distance between the observers is measured along the surface of the earth instead of in a straight line? I mean, the moon is out in space, so the parallax should be the same?
You're using a measure of pixels/degree, which I don't believe is uniform across a lens (and certainly not for two) so perhaps that could have contributed to your inaccuracy. Super cool video btw!
I think you got the math wrong with the second estimation. cos(90-40) implies 50 degree latitude and neither of you were that far north. The full 6378 km distance difference only occurs at the equator (cos(0)). At the north pole (cos(90)) it is zero. So just put in your latitude into the cos function. For San Francisco cos(37.5)*6378=4945 km distance and for upstate for New York 4644 km. Depending on who made the two photos the calculated moon distance is: San Francisco: 412,000 km New York: 387,000 km
I haven't done anything on paper, but thinking a bit I'm pretty sure you can't add the moon latitude with your observing latitude. The right equation would then be cos(42)*6378*cos(7) It's obvious in an extreme example. When the moon would be over the pole ( at 90 degrees off the equator). Then the earth rotation doesn't change the distance at all. But cos(90+anything) would calculate a difference. Any sideway position from the center-center line has a negligible effect as well. Proved by Pythagoras: (6387²+380000²)**0.5=380053
Wait, no error analysis? Would be great to have seen even a rough uncertainty propagation calculation. That's really the heart and soul of real science. Getting close to the real answer might be luck, but if it falls within your uncertainty bounds it's all the more convincing.
(Accel of Gravity)=CauseMass * G/distance^2 (Period of orbit)=(Circumference of Orbit)/Velocity Velocity=sqrt(Accel of Gravity * Distance)=sqrt(CauseMass * G/Distance) Circumference=Tau * distance Period = Tau * Distance / sqrt((earth mass + moon mass)*G/Distance) By knowing the period of the month 27.3 days, the mass of the earth = radius^2 * surface gravity, and radius can be calculated by walking really big triangles to find 5.97x10^24 kg, and the moon mass which is proportional to the distance, we get a really complicated function with one solution which can be plotted if we don't know the moon's mass already, or a really easy direct function for distance given we know that mass, and all within 3 digits of accuracy. I've don't those calculations already, and it works quite well. Only knowing the tides, the month, the size of Earth, and the strength of its pull, we can calculate the distance, mass, volume, and density of the moon and guess its composition based off of the that. In fact, we don't even need the sidereal month directly, just the solar month and the length of a sidereal year, and the sidereal year is MUCH closer to the solar year to the point of near negligibility.
If I remember correctly, (moon mass^3 / (earth + moon mass)^2) = (period^4 * moon's gravity on us^3)/(G*tau^4) If you plot it by hand, since it is technically linear-ish to the moon's mass, there should only be one solution (and there is in this case, at least in the positive reals) or just throw it into a graphing calculator and zoom out way far to find the graph before finally zooming into the X intercept (where the graph is the difference between the two sides of the equation)
Mainly the formula was designed to calculate mass, but from the mass knowing the effect of lunar gravity you could calculate the distance. My best attempts produced a result about 9% off
I'd love to see someone apply this same math to the "flat Earth map" to see how the geometry works out. As I recall, that group has their own figure for the distance to the moon. Then again, I don't remember what they think the stars are. Why did I go down this rabbit hole?
The flat Earth map does not work out. Simple as that. It's patent nonsense. And actually, they have no map. They have no consistent model. They have various insane ideas about what stars are, none of which make any sense. They don't show how things work, they just waffle, gaslight, spam links, insult people, and babble incoherently. Not a single one of them can answer simple questions that rely on high-school maths. Such as, the angle to Polaris, how we precisely predict an eclipse, what the eclipse looks like, or the path of the Moon and Sun. They'll just change the subject - usually pasting pages and pages of utter nonsense and more links. Or they become abusive, or just do not respond. Some are unbelievably stupid and gullible, and some are actually mentally ill; I feel sorry for those. But many are wilfully lying to get popularity, subscribers, and so on. They are the scum of the Earth, snake-oil salesmen, charlatans praying on the most vulnerable in society for their own gain. They are no different to televangelist Jim Bakker, or religious nutcase L Ron Hubbard. Both of those were convicted of fraud; I wonder when the first Flat-Earth TH-camrs will be?
I guess this wont work if i´m halfway across the world from you. Or anyone for that matter. Actually im antipode of New Zealand currently, usually Antipodes of the middle of nowhere in the pacific.
I bet if you calculated the shift in the Moons position due to refraction and dispersion (especially when low down) that might improve your results. Different magnitude depending on the altitude of the Moon and hence the amount of atmosphere refracting. It makes a large difference when Moon rising/setting. As I'm sure you know the setting of the sun is different from pure geometrical expectations due to this very effect.
I assume you took into account where between apogee and perigee the moon was comparing it to the real answer. I suspect the 'distance to moon' numbers are commonly between centoids, whereas you were measuring surface'ish to centroids.... I wonder how much that would eat into the discrepancy.....
Computer vision 101 you need to find the calibration of your cameras first. Raw pixel differences don't preserve relative lengths on most cameras. Can possibly do the calibration after the fact and recalculate.
Posidonius calculated 2100 years ago that the distance is about 370,000 km (2 million stadia), and he didn't use a camera. Pliny wrote: "Posīdōnius nōn minus quadrāgintā stadiōrum ā terrā altitūdinem esse in quam nūbila ac uentī nūbēsque perueniant, inde pūrum liquidumque et inperturbātae lūcis āera, sed ā turbidō ad lūnam uīciēns centum mīlia stadiōrum".
Translation, "Posidonius holds that the height in which mists and winds and clouds reach is not less than 7400 meters from the earth, and that from that point the air is clear and liquid and perfectly luminous, but that the distance between the cloudy air and the moon is 370,000 km"."
It's not so much "the universe" but mostly empty space-time that's flat. And in this case, "flat" just means two parallel lines remain so regardless of distance. If space were curved (and it could be, but if it is the amount is within the margin of error) then given enough distance you'd see the lines either converge or diverge. The overall universe itself, if it does have something we can call a shape, I suspect we'd call a sphere.
Atmospheric diffraction? This is a real problem in astrometry and one of the reasons why high precision astrometry is done from space. (Hipparcos, Gaia) Maybe this is a reason for the difference?
Using google earth to measure the distance between A and B uses the surface of the earth not a state line between. There is a formula to get the exact state line distance but I did not see that in your video. Some of this escapes me so I may have just missed it.
The eclipse on July 27 should be visible from almost everywhere on Earth except North America (and some bits of Russia), to a greater or lesser degree. Best around Europe, Africa, and Australia. In the UK, it'll be at max for an hour, and partial for 2 more hours... the longest one in over 100 years. if you do miss out, there's a *total* lunar eclipse on January 20-21, 2019 - and the Americas are the perfect place to see that.
So I’m a little confused and maybe you can help me out with this… If you are using triangulation using a baseline to measure the distance to the moon, and the sun and the moon are the same apparent size in the sky, wouldn’t this method show that the sun would be the same distance as the moon? Really curious about this so anyone that has a good answer let me know
The angular size of the object doesn't matter here: the key is parallax, ie how much it moves compared to background stars: even if the Moon was 10 times smaller or bigger you'd get the same results. It is possible to use the same method to measure the distance to the Sun, but it is much more difficult for a number of reasons, the most problematic one being that it is so bright it outshines any nearby stars that could be used as reference points.
@@davesims7917 you know its distance by measuring the difference in angular position in the sky from two different points of view. This difference is explained in the video on the graph at 0:25 where it is noted as the dotted angle by the Moon. By measuring this angle and knowing the distance between A and B you can then use trigonometry to work out the value of r which is the distance to the Moon. Notice that the angular size of the Moon isn't needed in this calculation, so if it were bigger or smaller it wouldn't change the value we get. The Sun does have approximately the same angular size as the Moon, but if you succesfully do this experiment with the Sun instead (which as I said is really not easy to do, and other methods are preffered), you'll find that the angle of the corresponding triangle is much smaller, meaning the distance r is much larger (400 times larger in fact). But that doesn't mean the angular size is useless: once you know the distance to the Moon, you can simply multiply it by its angular size to get its real size, which is how we know the Moon is 3400km in diameter.
Most cameras have a clock and timestamp every image. So after syncing watches to Naval time we each took a picture of the watches and could calculate accuracy of camera clocks to within a seconds.
Your video was very useful. But please do a video about SUN's distance from EARTH. I wonder how many kilometers is the distance of the SUN to the EARTH? I would be very happy if you answer me. Best Regards.
I don't know if you took this into account (at 3/4's through the video, I figured you would have mentioned it by now.) His attitude is measured differently than yours. (Sea level differs defending on which coast you're on.)
I loved seeing when the moon appeared near Venus and then near Jupiter, and I wanted to take pictures, but my camera is garbage. I'm guessing in a few days, it will appear near Mars too.
Only if you know how big it is. If you know how far away something is, you can calculate how big it is. If you know how big something is, you can calculate how far away it is. If you don't know either, you need at least two pieces of data. That could be two photos taken at different times, if it's moving and some stuff isn't. Or it can be photos taken from two places, knowing the distance apart. (Or you can bounce a laser off it)
Thanks for having me on the team. It was fun.
Fail! the earth is Flat assuming a globe through your math way off.
@@zechariahdedwards what?? This better be a joke. 😂 They used spherical trig
@@zechariahdedwards no way thats not a joke
Hey Scott, a couple points to remember: 1) the moon orbits in an ellipse and its apogee, perigee and mean distance are all different. Closest perigee is 356,400km. 2) remember the published value is from the centre of Earth, not a point on its surface, which is some fraction of 6,000km closer. So your estimates may be better than you think.
If I was a High School science teacher, I’d try to get together with another school and do this measurement. I think it would be a great project and would teach a lot about these concepts.
A high school math teacher would also be able to make it into a lesson.
A history teacher could make it with a bit of a stretch if a scientist comes up who somehow is connected.
A geography teacher could do it as a demonstration of how map distances where done in the past.
Ken Smith
There are definitely a lot of things that could be done with this. The exciting part is that kids could be shown that they could do real science themselves. I think it would make it exciting for them.
For a high school it's easier to measure satellites since they're closer
It also means planning your lesson around the schedule of the moon rather than the school, not an easy option for many teachers sadly.
A public high school. Pffft, Yea right.
Well, if you know that your cameras are 10 centimeters tall, you can just stack them on top of each other. I dunno why you'd use cameras instead of metersticks, though.
...Wait, you mean _taking pictures_ with the cameras?
Reminds me of a method of determining the height of a building with a barometer.
"This house is 29 barometers tall"
There is a german poem describing more than 10 ways of determining the height of a building with the use of a barometer.
How many Smoots away is the moon? "We'll need a lot of spacesuits..."
that is the most xkcd thing i have ever seen on youtube
This urban myth seems to have become known through an article in "Saturday Review", a magazine, in 1968.
An examiner sets the question for a physics student, to measure the building's height with a barometer - expecting him to calculate the pressure difference. The student suggests typing a rope to the barometer, lowering it down the building, and measuring the rope.The examiner thinks that should score zero, the student thinks it's a perfect answer, and suggests many other creative answers. He knows the "real" answer, but objects to the overly dogmatic teachings; the scholasticism. It's also demonstration of creativity; thinking outside the box.
The internet meme has a similar story about the Nobel prize-winning physicist Niels Bohr, when he was a student. But that's almost certainly a myth.
Did an entire section on this in high school. We had a classmate who happened to be on a trip to Europe take a photo, and we took a photo from the middle of the US. Knowing the distance apart that the photos were taken, we got the distance to the moon by parallax. Then we stretched a wire "spider web" over the football field using one of the goal posts as a focal point to get a dish antenna in the FM band, and timed the "echo" of a powerful local radio station to measure the speed of light. Fun times!
So easy even I could do it. ;)
Cody'sLab Oh you, Cody
You should try it with Venus or Jupiter
Didnt you do it like 3 ago? v=F1cbyhbVpQY
+CowBones
Indeed, hence the slight amount of shade in my comment. Doesnt help that we emailed about possibly re-doing it as a colab but I guess I didn't live far enough away.
Cut would have been simultaneously easier and harder I think. I don’t know why we never followed up on this. What we should try doing is measuring distances to a geostationary satellite.
I love how Apollo 11 left mirrors on the surface to measure the exact distance between Earth and Moon. Not that an amateur could use this system, but it’s still super cool what they were able to do on that first mission. Maybe you could do a video sometime about the scientific discoveries about the moon from the moon landings and what things we still reeeally need to know about the moon.
Those are also great for dealing with lunar landing deniers. Anytime someone in range says the landings were faked I screech "APOLLO RETROREFLECTORS" into whichever ear is less protected or closer until I see blood.
Not gonna argue there, just one thing makes me wonder, though: do we possess a tech that is precise enough to hit a 50x50cm array over 300,000km? Or when they fire a laser, the radius of the laser spot on the moon is much bigger and they just fire in the general direction? If the latter, which I assume it is, is there a video of some sort explaining the math behind it? Would be cool to watch.
By the time the laser beam hits the Moon, it's several kilometres in diameter! However, because the light is of a tightly-controlled wavelength, we only need to receive a few photons back in order to make measurements.
Only 1 in several millions of photons will hit the reflector and be fired back toward Earth, where those few are again spread over several kilometres area.
By measuring the time, we can calculate the distance to an accuracy of a couple of centimetres.
Incidentally, you may be interested to know about the forthcoming new retroreflector. MoonLIGHT, also known by the snappy title "Lunar Laser Ranging Retroreflector Array for the 21st Century" (LLRRA-21) which should be deployed in 2019, and should give 100 times more accurate results.
Actually Apollo 11, 14, and 15 all left retro-reflector arrays, as did the Russian Luna 17 and 21 rover missions. You can see a picture of the Apollo 15 reflector (the largest on the moon) here: en.wikipedia.org/wiki/Apache_Point_Observatory_Lunar_Laser-ranging_Operation#/media/File:ALSEP_AS15-85-11468.jpg
As the previous poster pointed out, the beam is a few km across by the time it reaches the moon. Of course, you still have to point at it pretty precisely, as there are a number of effects on range-data quality from mis-pointing.
Also, since the moon is a moving target, you have to fire your laser about .01 degrees ahead of where you are looking for the return photons. Now the .01 degrees does not sound like a big angle, but when the field of view of the telescope used is only .016 degrees, you do not have a lot of wiggle room. Plus you have to put your extremely narrow beam (0.0006 degrees(ish)) on a very small portion of the moon, then look for the very small return. It is challenging, to say the least.
As the previous poster mentioned, you may only get a small number of photons back for every pulse. In fact, most ILRS stations really only want one photon return from each pulse and attenuate the light to get that. They send thousand(s) of pulses per second, and do not worry too much if they miss a few. From the resultant data you can estimate the distance to the moon to a small number of mm, and probably less.
They had to land something there anyway. You cannot fake a signal coming stationary from the moon. You don't even need paralaxes. Everyone who could receive the signal could just check when the moon went below the horizon and when he lost the signal... Unique time for every location on earth. Unfake-able.
the silly thing: for a "fake" that would fool even the casual observer for 5mins, they would have to do so many chunks of the mission anyway, that a fake would barely save them any effort. It would indeed ADD effectively a second space program... Against the massive risk of ending every career and company reputation involved forever. It's all just such complete nonsense....
"The Twin Hells of Spreadsheets and Spherical Trigonometry"... should have been the true title of this video. :D
stellarfirefly Do I smell another poem about Dante?
See, I don't get that. I *like* spreadsheets & spherical trig.!
Fascinating to see the limitations in measurement accuracy imposed by the technological level of the observer, and to imagine what this would've meant for early astronomers doing this with an eyeglass and parchment instead of precision telescopes and computer modelling. It's no wonder it took us so long to get (relatively) accurate measurements for the features of celestial bodies when precision is so closely tied to the quality of the tools at hand.
Exactly my thoughts when I was watching the video.
I wonder how you could make it more precise without deploying better hardware
Ozone Grif Moar measurements (a whole lot of statistical analysis)
Is there any way to try this with a single camera at different times of night (e.g. moonrise and moonset), using the fact that you’re a point on a rotating sphere - which means, essentially, two different points when measured at two different times?
I guess one problem would be that the moon could move relative to the background stars (does it move fast enough for that to be a problem?), rendering stars useless as reference points. But I wonder if you could instead use a telescope rig that can precisely measure the angle it’s pointing at.
Yeah , that i was thinking too - but you know the moon orbit period and can just substract the difference . I think it can be done from a mountain top with a long bubble level and a good clock knowing just the diameter of the earth
I will indeed *fly safe* now that I know my margins for a moon encounter. Wouldn't wanna crash into the moon on my way to visit aliens.
You strike me as someone who knows where their towel is.
I really like your videos when you explain space stuff. It makes sense!
Also KSP video are always a number one in my book.
"twin hells of spreadsheets and spherical trigonometry" *violent flashbacks*
This is awesome. I don't know what I expected but when you overlayed the photos, it was great seeing the trigonometry in action.
By any chance, would your spreadsheet files be available? This may make a great activity for my science students. Thank you for sharing!
Hello, sorry, I have a question in the video. What are the two points on the moon, I know why choose dark parts, but why those two points? And why two points chose on the moon? Could you please explain it a little clearer? Thanks a lot!
He could have chose any point. He just has to ensure that's the *same point* on both images.
Is there a chance a difference in focal lengths between the two cameras induced an error?
There is a possible source of error that you didn't mention: Your typical camera shoots photos in rectilinear projection, i.e. for viewing on a flat surface in front of your eyes. Imagine standing right in front of a big, flat advertisement poster. You'll notice that the part that's right in front of you will be larger in your vision than the stuff that's on the edges of the poster, because those are further away from you. This means that on the edges, things that take up the same area on the poster than in the center will actually cover a smaller angle. There is also deformation due to perspective - looking at the edge of the poster to your left. A sphere on the poster at that point will look like an ellipse that's higher than it's wide. Going back to the photo camera, this means that on the edge of your picture, the same angle will take up more pixels than in the center. Now, given that the angle of your photos is pretty small the effect may be negligible, but I think it's worth mentioning when you talk about equating pixels and angles.
So how about pinging the retroreflector experiment left by the Apollo mission. How much laser power would you need and how big of a telescope would you need to detect it?
Specialist stuff, cost about a million US dollars. About 3.5m telescope and quite a specialist laser; you need to fire _really_ short bursts. I can't really give a specific power rating, because it's a fast pulsing laser with a small pulse width. Like, 115 mJ per pulse, 20 Hz, which is only 2.3W. You fire quadrillions of photons, and get a handful back.
So... universities and stuff can do it, and they do. But it's not really something you can do in most back yards.
(I could say lots more... but I'm guessing that sorta answers?)
how do you get the conversion from pixel to degree? That's what confuses me the most. "on the sky we know that is 4.31 degrees" . HOW do we know that???
There are two main ways to do that.
One is to look at the camera settings you used to take the picture, which will tell you the andular width of the shot which you can use as a reference for everything else in the shot.
The other method is to use visual clues: for instance, the angular diameter of the Moon is close to half a degree with some monthly variation. You can also identify stars in the shot and look up their distances on a star map, which is what Scott did here: he identified two stars from Virgo whose angular separation is known to be 4.31 degrees.
What I did - which gave a crude result, but that's all I was looking for - was attach a 10cm length of construction paper to the end of a tape measure, then push the tape measure out from my eye until the two edges of the construction paper just touched two stars I had taken a picture of as part of my moon picture. That gave me the slope of the angle between the stars, which you can use some inverse trig on to convert to an angle. (In my case I just used the slope, so I didn't even bother converting it to an angle.)
Maybe the inaccuracys are in part because of different atmospheric effects? At 3:30 you show "what the moon sees". From that perspective light hitting the east and west part of the US hit the atmosphere at different angles
Wouldn't that cause the light to deviate its course differently for both of you?
Dear Scott, we did the same experiment but with Venus as the callibration "star", we ended up overprojecting the distance to 325.000 miles way back in Dec 2016. We used a telescope to measure and obviously a cheapo! Over long ditances, it is incredibly difficult to point a telescope unless you have precision guides in place. Nevertheless, we DID end up "close" to the accepted values. Probably 25% off, but still incredible that ANYONE can measure the distance to the moon or even other celestial bodies with this method. There are more advanced and better methods, but for the ameteur observer, this is incredible. Please shed more light on the technique and equipment used for this particular achievement. You are Scott the MAN and will always fly safe :-)
Boy, your spherical trig calculations gave me flashbacks to my observational astronomy class in college. The textbook dated from the 1930's and was full of what my professor called "British schoolboy exercises." Every problem set was a new fresh hell. A typical problem: "Prove that, in latitude 45º, the interval between the moment at which a star's azimuth is 90º and the moment of setting is constant."
Proof by contradiction: Circumpolar stars exist for a viewer at 45°N. Circumpolar stars never set. Thus the time between culmination (azimuth 90°) and set (elevation angle 0°) of a circumpolar star is infinite.
Thus the earth is flat?
I think that's the conclusion I'm supposed to use.
hey Scott. great video. love your work. I have seen my share of FE videos and I can only surmise they steer clear of your channel and VSAUCE and Codys Lab to name a few.
KEEP IT UP BUDDY.
Astronomical figuratively and literally! Thanks Scott, amazing to see you working with some real measurements and numbers.
what program did you use to overlay the images? (Your first method)
Any program you like. Photoshop works fine, other free programs can do that too.
Photoshop. It has an inbuilt ruler tool too to measure pixels and other measurements on images.
My first Scott Manley video and never looked back since. Thanks for the info Scott!
Do the corner cube reflectors on the Moon still work? I'd rather just blink a laser at them and measure the round trip time. Because lasers.
Yeah, but you need about a million US $ worth of kit to manage it. They are working great. But you need a 3.5m telescope and stuff, to make use of them. I gave a bit more detail in other replies.
How close will i get, trying this with the camera on my iPhone 7? Can I improve the resolution by taping an iPhone 6 and the iPhone 7 together?
Jackalovski yes, but only if you take the picture with one phone and quickly place the second one on top facing each other to trap the light between them. Making the light confused on with way to go. The light will now hang out in the middle of the two cameras longer, giving you a better exposure.
Technically, but only with smart AI to combine them. But the iPhone 7 cam is more than good enough.
with a budget like this you could buy proper hardware too
My budget is whatever I have laying round. I've been gifted a second hand iPhone 7 and the 6 is the works phone. The best I could manage in terms of 'propper hardware' is ducktaping them to a pair of soviet binoculars with my current budget.
Sorry if this is a stupid question but at 5:13 you said left image was high in the sky and right image was down near the horizon. I always thought that when the moon was low on the horizon some factor beyond my knowledge would distort the light making the moon look larger that it actually is but your image shows its smaller?
The apparent size of the Moon from Earth depends *mostly* (bear with me...) on how far away it is. That distance varies from _about_ 406,000 km to _about_ 360,000, every lunar month. In this video, they both measured it at the same time, so that is not an issue.
The other effect is an optical illusion. When we see something near the horizon, our brain thinks it's larger than something directly above us. Nobody knows for sure why that happens. But it is not real. If you take a photo, and check, it's the same size.
So that does not affect the results. A low Moon certainly _looks_ larger, but if you check using a camera, it is not.
It is not a stupid question, because despite many books on the subject, nobody is quite sure why our brains work that way.
Hope that helps.
There is a refraction effect that is different than the moon illusion described. This primarily squashes the moon in one direction, but leaves the other direction pretty much unchanged. It's more pronounced the closer to the horizon you get, and if you are ~15° above the horizon it's not really anything to worry about.
Thanks for the responses guys! I had a feeling it had something to do with the way the eyes and brain interpret the signal. My next question is; when will I be able to get Optical Augments so I wont have to worry about Illusions anymore?
When I bought the original PC version of "The Hitchhiker's Guide to the Galaxy" in 1984, inside the box was a pair of "Peril Sensitive Sunglasses", made from opaque black cardboard.
What you don't know can't hurt you.
Cool video :) It may not vary by much for a zoomed in photo, but the camera's angle of view is projected onto a sensor that is flat, so pixels / degree is not constant across the entire image.
Probably already mentioned here in a comment...but why not just use the time for one lunar orbit to work it out??? I just used 28 days (close enough) and it comes to 383,215,240.2m which is pretty darn close!
According to flat earth logic, that assumes gravity is real, thus is fake.
In the real world, multiple approaches to the same measurement converging on the same value gives you pretty good confidence the value is correct.
@HelmutBemboka Please show your working?
ann onn just use the formula r^3=(G*M*T^2)/4pi^2 . That's the standard formula used when balancing centrifugal force and gravity. G is the universal gravitation constant 6.67e-11 kg^-2. M is the mass of earth 5.98e24 kg. T is the time period of one orbit 28 days (use 2419200s).
That gives r which is the distance between the center of earth and the center of the moon. For the distance from the surface of earth, subtract the earth radius 6371km.
Technically you could also subtract the moon radius to get surface to surface, but I didn't bother with that so my number is just from Earth surface to moon orbit.
Interesting vid Scott, I remember measuring the moon distance with a brush stale and two different sized discs using the small angle approximation with the Open University. amazing stuff thanks for sharing👍🏼
You could use this to determine the diameter of the moon. Then use this as a reference lenght for measuring the shadow cast by a lunar mountain. Then you can measure the height of the mountain using trigonometry.
Don't you have to take into account the optics of the camera? Depending on lens and zoom the distances measured in pixels in one part of the resulting image don't necessarily correlate linearly to other parts in the image.
Yes, and that's probably a big part of my value being off by 10%
Hi Scott, I was thinking about the 2nd example with the two muns being compared directly: don't you have to take into account a lens effect of the earths atmosphere, that is stronger for images taken with the mun being close to the horizon because of the longer way of the light passing the atmosphere compared to a picture taken straight up?
no, because the lens effect doesn't magnify the image it just lifts up the light near the horizon.
Scott Manley Near the horizon it will certainly exaggerate the vertical. Refraction near the horizon changes a lot, even within the arc subtended by the moon. I wonder if you aren't seeing some refraction effect in your parallax measurement due to the low elevation of the east coast photo.
The vertical will be compressed, the horizontal remains roughly the same.
Scott will you upload a video on the new NASA's project Parker?
Please could you share the spreadsheet you made? It would make it a lot easier to try the same thing!
Is the error caused by diffraction in the different paths through the atmosphere?
It's probably just caused by inaccuracies stacking up. Remember, Scott's using consumer-grade hardware designed for terrestrial photos, not cameras designed for astronomy. (And I'd imagine that the coordination between him and the other guy, while good, was not up to professional standards of precision.)
Diffraction would act on the starlight as well as moonlight. Errors would creep in if one object was measured near the horizon and another was overhead so that diffraction of the object near the horizon would put in visually in a different location. The Moon and stars were just a few degrees apart.
Allan Folmersen you talk about 5km error at most
Lenard Segnitz But from the looks of things the shot from the East coast was very low elevation. At low elevation a few degrees makes a YUGE difference in refraction. Really bigly.
Or you can talk to an amateur radio operator that bounties signals off of the moon. Its called EME (Earth Moon Earth).You can hear the signal coma back at you a few seconds later. The timing matches the round trip distance to the moon and back divided by the speed of light.
Just a little nitpick: July 16th was fairly close to the perigee. According to what I could find, the actual distance on the day was around 369000km, not 384000.
pictures were taken on 19th
Scott Manley ah ok, I'm an idiot. Of course you didn't use the pictures from before you set things up... Nice coincidence that you ended up on an average distance day then!
i have books from 1840 and 1850. the distances to the sun and the moon and the sizes of many if not all the planets were known. They are essentially spot on with todays numbers, How is it possible that they could measure the size of venus or saturn without presuppositions?
All experimental science has presuppositions.
What do you mean by that?
Couldn't you also do parallax observations based on the stars in the background at near moonrise compared to near when it sets? That probably makes figuring out your baseline even harder, but you could do that, right?
The problem is that the Moon moves in its orbit, and thus against the stars, between the two observations. What you would measure is a combination of that movement, which is about six degrees in 12 hours, and the true parallax which is about two degrees. If you can correct for the Moon's motion then in principle you can find the parallax but it would be difficult and probably not very accurate.
Good point, I hadn't considered that.
Can someone explain the formula at 6:20? Or at least something to google to learn about it.
Guys, I have a 3 years old who's absolutely fascinated by the moon and stars. You think a regular rifle scope fixed on a camera tripod is good enough for now or should I buy him a dedicated telescope already?
I'd also appreciate some links to astronomy videos for young kids if you know any.
Even simple binoculars are great, you can see the Galilean moons, the moon very bright, excellent for the upcoming Lunar Eclipse. You can't see the rings of Saturn however
A decent riflescope will work, but it's not going to be as good as a larger front lens. I carry one with me for that purpose, but mainly for terrestrial viewing. But if you already got the scope, use it, kiddo is only 3 yro.
There are rifle scopes that would be ideal, but they're hardly "regular". They're made for natural-light varmint shooting at night, and have something like a 70mm opening at the front. And they're $tupidly expen$ive. Let the sprout look through a good pair of binoculars instead.
Just do the good ol' American thing and just shoot the moon with your rifle instead. Science is for pansies.
This would be a great science project for schools next time the moon is near the Pleiades.
We could use the right angle triangle of a half moon to get how many times further away the sun is that the moon. Not sure what equipment would work best though - the angle would be very close to 90.
It's a really really hard measurement to make. At best it can give you confidence that the sun is no closer than "y" times the earth-moon distance.
Scott - this was pretty damned cool. Could you go a little deeper into it and identify likely sources of error and how they contributed to your final result? When synchronizing with the remote person, other than time, what other factors must you take into account, i.e. focal length of the camera, etc.?
Another fun exercise using only consumer level cameras is to prove Uranus exists and that it is not like other stars. Neptunus is a though a bit beyond the reach of at least cheaper ones. Figuring out when it's best to try is an exercise itself.
How will did you synchronize the times of the photographs? You will be off about 4200km (one way or the other) per second of synchronization error.
Huh? The moon moves at about 1km/sec
Scott Manley Right - my bad - you're taking sidereal rate into account by aligning the stellar background, so it's only the moon's motion relative to the stars. so only about 160km per second. Sorry about that.
Did you happen to calculate the exact lunar surface to your station distance to compare? The 385Mm is the mean center to center, but you were probably like 7Mm closer.
Update: I get 383.17Mm from Oakland, CA to the Moon's center, which works just fine for the limb.
And to clarify, I was talking about the moon's motion (orbiting earth) effect on your range estimate by paralax. 13.2(ish) degrees per day or about 0.55 degrees per hour.
You could figure out the gravitational constant by the pull of one object on another then use a cubed equals t squared and the time it takes the moon to complete an orbit, then you have the simimajor axis
Did you calculate the distance between your cameras over the surface of the Earth or "as the mole digs" straight through?
Neither.
I sortof skimmed over your "Twin Hells of Spreadsheets and Spherical Trigonometry" without understanding it.
03:50 "...we have a special type of trigonometry that works on the surface of spheres..." Now, don't you wish the earth was flat? :)
😂😂it is.
Duh. You fire up Kerbal, with realism mods, and you look at it's Peri and Apo. ;)
Duh, you just google it.
Great Video And I really like the Idea of measuring astronomical distances with a camera. I was wondering if you added the earths radius to your measurement since the luna distance is define as the distance between the center of earth to the center of moon. Its pretty obvious so i assume you did but I had to ask. Maybe I just missed it in the video or you didnt mention it
but any way great video and keep on doing stuff like this
Your measurement results are quite consistent. So it should be 340-350k km indeed or it's an unfortunate coincidence?
You didn't account for the Moon being in an elliptical orbit. Sometimes it will be closer to Earth, which could help explain the results you were getting. 346,000km is a much better estimate if the Moon is at perigee than if it's at apogee. According to a quick google, the Moon was at perigee on July 13, and will reach apogee on July 27.
He was measuring what it is at the time of the photo being taken. If he wanted he could certainly do this daily and measure the movement around the entire orbit
Yeah but he was comparing his result to what the average distance is, rather than what the actual distance was at the time the photos were taken.
Could someone explain to me, using simple words, why the distance between the observers is measured along the surface of the earth instead of in a straight line? I mean, the moon is out in space, so the parallax should be the same?
If you know the distance along the curved surface, you can calculate the straight-line distance. That's the part where he used a spreadsheet.
The position of the moon also changes the baseline.
Thank you for all the replies, I must have missed Scott explaining it 🤔
Makes me want to try it. I learn a lot from your videos, thank you
The distance to the Moon that you get off the Internet is the average distance (apogee + perigee) / 2 and is measured from the center of the Earth.
You're using a measure of pixels/degree, which I don't believe is uniform across a lens (and certainly not for two) so perhaps that could have contributed to your inaccuracy. Super cool video btw!
I think you got the math wrong with the second estimation. cos(90-40) implies 50 degree latitude and neither of you were that far north. The full 6378 km distance difference only occurs at the equator (cos(0)). At the north pole (cos(90)) it is zero. So just put in your latitude into the cos function. For San Francisco cos(37.5)*6378=4945 km distance and for upstate for New York 4644 km.
Depending on who made the two photos the calculated moon distance is:
San Francisco: 412,000 km
New York: 387,000 km
Moon at time was above latitude -7 degrees
Then why didn't you write cos(43+7) ?
Because the moon wasn't directly south when it was imaged.
I haven't done anything on paper, but thinking a bit I'm pretty sure you can't add the moon latitude with your observing latitude. The right equation would then be
cos(42)*6378*cos(7)
It's obvious in an extreme example. When the moon would be over the pole ( at 90 degrees off the equator). Then the earth rotation doesn't change the distance at all. But cos(90+anything) would calculate a difference.
Any sideway position from the center-center line has a negligible effect as well. Proved by Pythagoras: (6387²+380000²)**0.5=380053
Wait, no error analysis? Would be great to have seen even a rough uncertainty propagation calculation. That's really the heart and soul of real science. Getting close to the real answer might be luck, but if it falls within your uncertainty bounds it's all the more convincing.
He at least acknowledges and alludes to the measurement errors when he shows the pixels per arcsecond value for 3 different star pairs are different.
(Accel of Gravity)=CauseMass * G/distance^2
(Period of orbit)=(Circumference of Orbit)/Velocity
Velocity=sqrt(Accel of Gravity * Distance)=sqrt(CauseMass * G/Distance)
Circumference=Tau * distance
Period = Tau * Distance / sqrt((earth mass + moon mass)*G/Distance)
By knowing the period of the month 27.3 days, the mass of the earth = radius^2 * surface gravity, and radius can be calculated by walking really big triangles to find 5.97x10^24 kg, and the moon mass which is proportional to the distance, we get a really complicated function with one solution which can be plotted if we don't know the moon's mass already, or a really easy direct function for distance given we know that mass, and all within 3 digits of accuracy. I've don't those calculations already, and it works quite well. Only knowing the tides, the month, the size of Earth, and the strength of its pull, we can calculate the distance, mass, volume, and density of the moon and guess its composition based off of the that. In fact, we don't even need the sidereal month directly, just the solar month and the length of a sidereal year, and the sidereal year is MUCH closer to the solar year to the point of near negligibility.
If I remember correctly, (moon mass^3 / (earth + moon mass)^2) = (period^4 * moon's gravity on us^3)/(G*tau^4)
If you plot it by hand, since it is technically linear-ish to the moon's mass, there should only be one solution (and there is in this case, at least in the positive reals) or just throw it into a graphing calculator and zoom out way far to find the graph before finally zooming into the X intercept (where the graph is the difference between the two sides of the equation)
Mainly the formula was designed to calculate mass, but from the mass knowing the effect of lunar gravity you could calculate the distance. My best attempts produced a result about 9% off
I'd love to see someone apply this same math to the "flat Earth map" to see how the geometry works out. As I recall, that group has their own figure for the distance to the moon. Then again, I don't remember what they think the stars are.
Why did I go down this rabbit hole?
The flat Earth map does not work out. Simple as that. It's patent nonsense.
And actually, they have no map. They have no consistent model. They have various insane ideas about what stars are, none of which make any sense.
They don't show how things work, they just waffle, gaslight, spam links, insult people, and babble incoherently.
Not a single one of them can answer simple questions that rely on high-school maths. Such as, the angle to Polaris, how we precisely predict an eclipse, what the eclipse looks like, or the path of the Moon and Sun.
They'll just change the subject - usually pasting pages and pages of utter nonsense and more links. Or they become abusive, or just do not respond.
Some are unbelievably stupid and gullible, and some are actually mentally ill; I feel sorry for those.
But many are wilfully lying to get popularity, subscribers, and so on. They are the scum of the Earth, snake-oil salesmen, charlatans praying on the most vulnerable in society for their own gain.
They are no different to televangelist Jim Bakker, or religious nutcase L Ron Hubbard. Both of those were convicted of fraud; I wonder when the first Flat-Earth TH-camrs will be?
I guess this wont work if i´m halfway across the world from you. Or anyone for that matter. Actually im antipode of New Zealand currently, usually Antipodes of the middle of nowhere in the pacific.
I bet if you calculated the shift in the Moons position due to refraction and dispersion (especially when low down) that might improve your results. Different magnitude depending on the altitude of the Moon and hence the amount of atmosphere refracting. It makes a large difference when Moon rising/setting. As I'm sure you know the setting of the sun is different from pure geometrical expectations due to this very effect.
Very informative! A great project for amateur astronomy community!
Do a video on how to calculate the heights of the mountains on the moon!
I knew you were in the bay area, but didn't realize it was in Oakland... we should grab a beer some time!
i love these types of videos
I assume you took into account where between apogee and perigee the moon was comparing it to the real answer. I suspect the 'distance to moon' numbers are commonly between centoids, whereas you were measuring surface'ish to centroids.... I wonder how much that would eat into the discrepancy.....
What happens to the equation if you account for the fact that the earth is not 100% spherical, but squashed.
The 21km difference at the poles would not affect a result that is accurate to within some 40,000km or so.
surface of a 3-dimensional globe?! triggered!
twin hells of speadsheets and spherical geometry triggered me into laughter.
are there any non 3 dimensional globes? ;p
Martin Rößner There at least higher dimensional spheres.
TheFLOW1978 true. but not lesser dimension ones :)
Computer vision 101 you need to find the calibration of your cameras first. Raw pixel differences don't preserve relative lengths on most cameras. Can possibly do the calibration after the fact and recalculate.
Posidonius calculated 2100 years ago that the distance is about 370,000 km (2 million stadia), and he didn't use a camera. Pliny wrote: "Posīdōnius nōn minus quadrāgintā stadiōrum ā terrā altitūdinem esse in quam nūbila ac uentī nūbēsque perueniant, inde pūrum liquidumque et inperturbātae lūcis āera, sed ā turbidō ad lūnam uīciēns centum mīlia stadiōrum".
Translation, "Posidonius holds that the height in which mists and winds and clouds reach is not less than 7400 meters from the earth, and that from that point the air is clear and liquid and perfectly luminous, but that the distance between the cloudy air and the moon is 370,000 km"."
I did the same thing with The Traveler in Destiny. You can take screenshots from opposite ends of The Tower and determine how large The Traveler is.
Well? How big is it?
I love this video, but I’m confused as to where the 57.3 came from? Keep up the good work
180/pi
I remember when flat earthers were just a meme...
Years later, they measure the true distance between the Moon and Earth and realize that this video was right.
I know, I was just making a joke.
Now lets just wait for the triggered flat earthers to arrive.
Too much math. Mathematics to flat-Earthers is like garlic to vampires.
True dat.
To be fair to the flat earthers, modern cosmologists believe the UNIVERSE is flat!
Andromeda is flet.
It's not so much "the universe" but mostly empty space-time that's flat. And in this case, "flat" just means two parallel lines remain so regardless of distance. If space were curved (and it could be, but if it is the amount is within the margin of error) then given enough distance you'd see the lines either converge or diverge. The overall universe itself, if it does have something we can call a shape, I suspect we'd call a sphere.
Atmospheric diffraction? This is a real problem in astrometry and one of the reasons why high precision astrometry is done from space. (Hipparcos, Gaia) Maybe this is a reason for the difference?
you can see the video compression distortion you should post the photo used
You could also just take two photos, the whole point of this video is that you can do it yourself!
Using google earth to measure the distance between A and B uses the surface of the earth not a state line between.
There is a formula to get the exact state line distance but I did not see that in your video. Some of this escapes me so I may have just missed it.
MistxTube that's what the whole spreadsheet spherical trig part was about.
There’s supposed to be a lunar eclipse in my area next Friday but the weather report says it’s going to rain. :(
Go above the clouds
The eclipse on July 27 should be visible from almost everywhere on Earth except North America (and some bits of Russia), to a greater or lesser degree. Best around Europe, Africa, and Australia. In the UK, it'll be at max for an hour, and partial for 2 more hours... the longest one in over 100 years.
if you do miss out, there's a *total* lunar eclipse on January 20-21, 2019 - and the Americas are the perfect place to see that.
So I’m a little confused and maybe you can help me out with this…
If you are using triangulation using a baseline to measure the distance to the moon, and the sun and the moon are the same apparent size in the sky, wouldn’t this method show that the sun would be the same distance as the moon?
Really curious about this so anyone that has a good answer let me know
The angular size of the object doesn't matter here: the key is parallax, ie how much it moves compared to background stars: even if the Moon was 10 times smaller or bigger you'd get the same results. It is possible to use the same method to measure the distance to the Sun, but it is much more difficult for a number of reasons, the most problematic one being that it is so bright it outshines any nearby stars that could be used as reference points.
Релёкс84 “If the moon was 10 times bigger or smaller you would get the same results”
So then how is it possible to know the distance of the moon?
@@davesims7917 you know its distance by measuring the difference in angular position in the sky from two different points of view. This difference is explained in the video on the graph at 0:25 where it is noted as the dotted angle by the Moon. By measuring this angle and knowing the distance between A and B you can then use trigonometry to work out the value of r which is the distance to the Moon. Notice that the angular size of the Moon isn't needed in this calculation, so if it were bigger or smaller it wouldn't change the value we get. The Sun does have approximately the same angular size as the Moon, but if you succesfully do this experiment with the Sun instead (which as I said is really not easy to do, and other methods are preffered), you'll find that the angle of the corresponding triangle is much smaller, meaning the distance r is much larger (400 times larger in fact).
But that doesn't mean the angular size is useless: once you know the distance to the Moon, you can simply multiply it by its angular size to get its real size, which is how we know the Moon is 3400km in diameter.
Scott, you are the Irving Welsh of astrophysics.
Cool stuff ... wish I understood spherical trig, nautical navigation and all
Any ham (amateur radio) operator can measure the distance with EME moonbounce radio propagation
Oh lordy this is why you said you were doing spherical trigonometry the other day.
The difficult part was synchronise your watches.
Most cameras have a clock and timestamp every image. So after syncing watches to Naval time we each took a picture of the watches and could calculate accuracy of camera clocks to within a seconds.
We take this ability as a given, but it is, what makes the title of your video true.
reminded me of the :clock in the sky'.didn"t some old timey scientist use transits of the Jovian moons to "synchronise your watches"?
Just make sure Cooper doesn't mess with them in the 5D tesseract.
I know à guy who can help with that. But he went through a Stargate at some point, don't know if you can get in touch.
Your video was very useful. But please do a video about SUN's distance from EARTH. I wonder how many kilometers is the distance of the SUN to the EARTH? I would be very happy if you answer me. Best Regards.
It's more difficult, as Sun is way further.
Eclipses is a good way to get a measure.
So far away. But so easy to see in the sky.@@ThomasKundera
@@theunionofjackbeanie2464 : As it's huge and very bright.
Determining the distance to the moon in 137 easy steps.
Step 1 take 2 photos
step 2 compare positions of moon to estimate difference
Step 3 distance = 57.3 * Baseline / anglular difference.
I find myself compelled to say,
Step 1 Google it
step 2 ???
step 3 profit
But ffs, don't take that seriously :-)
i'd be really concerned if the Moon happened to be in the middle of the US....
btw "my moon" from early july was very blurry.
i.imgur.com/OotmlSf.jpg
Might have been my camera n00bism though.
I don't know if you took this into account (at 3/4's through the video, I figured you would have mentioned it by now.) His attitude is measured differently than yours. (Sea level differs defending on which coast you're on.)
I loved seeing when the moon appeared near Venus and then near Jupiter, and I wanted to take pictures, but my camera is garbage. I'm guessing in a few days, it will appear near Mars too.
Couldn't you do this with a single observer in a single measurement? Just find the moon's width in mils and use the mildot ranging formula.
Only if you know how big it is.
If you know how far away something is, you can calculate how big it is.
If you know how big something is, you can calculate how far away it is.
If you don't know either, you need at least two pieces of data. That could be two photos taken at different times, if it's moving and some stuff isn't. Or it can be photos taken from two places, knowing the distance apart.
(Or you can bounce a laser off it)
I have such a shitty camera on my phone that I could never do this.
I am searching here for flathearh's
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Me too, would like to know how did they end up with 3000 km.
So are you telling us the Moon is getting closer? We need Flash Gordon to do something about that!
The moon is actually getting farther away. About 1.5 inches per year.