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Apollo 16 famously photographed the far side of the Moon, as seen below.

enter image description here

This seems odd to me. Apollo 16 orbited the Moon at an orbit of only 100 km. It seems too close to get this kind of a photograph of the Moon. How was this picture captured with a global view of the Far Side, from Apollo 16?

Anko
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PearsonArtPhoto
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    A (really, really) long selfie stick? (ducks) Good question though. From an 'infinite' distance we would presumably see near 180° of any side of the moon. Do we know how many degrees are seen in that image? – Andrew Thompson Jul 08 '16 at 14:18
  • Do you know the NASA photo number perchance? – Organic Marble Jul 08 '16 at 15:37
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    I'm pretty sure it's AS16-3021 – PearsonArtPhoto Jul 08 '16 at 15:49
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    http://nssdc.gsfc.nasa.gov/imgcat/html/object_page/a16_m_3021.html – Mark Adler Jul 08 '16 at 15:53
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    Note that the Apollo missions operated on the near side during the lunar day, so most of the far side would be dark. In fact, you can already see the terminator in the lower left of the image above, which is only a small part of the far side with a good chunk of near side in the upper right. – Mark Adler Jul 08 '16 at 16:06
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    @MarkAdler In order to allow the maximum amount of daylight for the mission, landings were timed to land on the Moon soon after local dawn. This means the terminator would be to the West of the landing zone and they'd land during the first phase of the Moon. Apollo 16 wasn't far from the "bulls-eye" so the Moon would've been approaching half-full when they landed. As you rightly point out, the photo is not square on to the far-side, but rather is a good bit off to the side. – Oscar Bravo Jul 12 '16 at 14:35
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    Maximizing the daylight has relatively little to do with it. The main reason is shadows. Landing near sunrise allows for good shadows to be visible, which help with depth perception. None of the missions lasted long enough to have daylight even be close to an issue, as the day is 14 days long. I think morning probably helped with thermal considerations as well, the lunar surface wouldn't be as hot after such a little amount of time to heat up. – PearsonArtPhoto Jul 12 '16 at 14:38

4 Answers4

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The photo (frame 3021) appears to have been taken from an approximate altitude of 1180 KM, on the return journey to Earth.

We infer it was taken on the return journey as frame 3005 was taken after trans-Earth injection. And, presumably, by "Revolution: TE" in the image's information.

The image information tells us the photo was taken by the Metric/Mapping Camera. NASA provides details of the The Fairchild Lunar Mapping Camera:

The metric camera has a 3-inch (76-mm) focal length, f/4.5 lens. The format is 4½ x 4½ inches on 5-inch-wide film and the angular coverage is 74° by 74°.

Knowing the camera's angle of view and the radius of the moon, we can calculate the minimum altitude required to see a full disc (a 360° horizon) using the law of sines. The line from our camera to the horizon is a tangent, which gives our second angle, a right-angle.

Minium altitude to see a full discCalculate the altitude given an observer's angle of view and sphere's radius

h+R/sin 90 = R/sin θ
h+R/1 = R/sin θ
h = R/sin θ - R

Moon's equatorial radius: 1738.1 km
Angle of View: 73.74° (calculated from the focal length and format)

h = 1738.1/sin(73.74/2) - 1738.1 = 1158.726 km

Knowing the minimum altitude required, and assuming the image shows the entire photo, we can recreate the camera using 3D software and quickly find its general position in relation to a scale model of the moon by lining up the features.

A more precise method would probably use known distances between features.

Using 3D software we can also see what the image would have looked like had it been taken at an altitude of 300km. (The red highlight indicates the near-side.)

Recreated Lunar Mapping Camera position

The visible footprint viewed from a more familiar angle:

The visible footprint

Leorex
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    Very nice answer! Welcome to the site. – user Jul 08 '16 at 22:24
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    Nice answer! Can you expand a bit on the methods and software you used to produce these visualizations? – E.P. Jul 09 '16 at 23:32
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    I've added how to calculate the minimum altitude required to see a full disc with a given angle of view, and the 'eyeball' method I use to get a general position. I use Blender, though I think most 3D programs would have similar virtual camera and scale features. I have corrected my initial answer, which gave half the approximate altitude (having fed my timesaver Blender script the Moon's radius instead of diameter.) – Leorex Jul 10 '16 at 10:33
  • astounding answer. Just a curiosity, i wonder why called the "metric camera": is it a reference to Napoleon, or does it mean in the sense "measurement"? – Fattie Jul 11 '16 at 13:26
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    @Joe Measurement. It is a precision camera specifically designed for photogrammetry, "constructed so that its geometric characteristics do not change from photograph to photograph and the image is as little distorted as possible." (American Society of Civil Engineers' Glossary of the Mapping Sciences) – Leorex Jul 11 '16 at 21:47
  • This answer nicely complements this answer. – uhoh Sep 21 '19 at 00:21
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The initial Lunar orbit was roughly 100x300 km, with the further point being at the far side. From the mission transcripts, we learn of a request to photograph the Moon at the terminator, at the far side (Because there was no communication with Earth) Using this simple calculator, I find that only 7% of the Moon's surface would have been visible at 300 km. The full map of the Far Side can be seen below. It looks like the far side, and the scale is clearly different, but it doesn't seem to be a factor of 7 that would be required!

enter image description here

Lastly, I've found that 10:30 is supposed to be the direction of Lunar north in the image.

A few things stick out to me. First of all, I believe the entire image is not the far side. In fact, I believe the left most edge is actually the Near Side, as the far side doesn't have many mares, and the line between the two is near perfect.

After doing quite a bit of digging around, I've found the source for this image. It turns out it wasn't taken by an astronaut at all, but rather by an instrument known as the Mapping Metric camera, which was used to map the Moon for Apollo 15-17. The film was retrieved in a short EVA on the way home from the Moon. It seems this was one of the photographs taken on the way home from the Moon, as it is one of the last in the series. You can see the entire series at the Apollo Digital Archive website. The time of the photograph isn't recorded that I can find anywhere. The last photograph before it that has a time is 3000, 1972-04-24T23:32:28, which starts to show the curvature of the Moon. I believe what actually happened is the Lunar return trajectory started on the far side, and came along a course that allowed for much of the dark side to still be seen. Trans-Earth Injection happened at 25 Apr 1972 02:15:33. Likely the photo was taken slightly after TEI, although I can't find the timeline more accurately than that. The photos can be seen to slowly get further away from the Moon, starting at 3000. Before that, they were taken about every 30 seconds. They stopped taking pictures per the standard plan, but at 200:33 they turned the camera back on for a while. That was 10 minutes after the TEI, which seems like it would be in a good place to capture the pictures they returned of the "far side" of the Moon. It was on for about 2 hours, until about 202:55. For certain at some point during that time a "global" view of the Moon would be visible, and towards the earlier portion the far side should be visible. I can't find the altitude vs time, but this now all makes sense!

PearsonArtPhoto
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  • So, the only reason you had a question was because you assumed you were seeing a full half of the surface? Only seeing 7% doesn't mean you won't also see the horizon. – Samuel Jul 08 '16 at 17:53
  • It seems odd still to me that it's a perfect circle, which is a bit that I haven't figured out yet. – PearsonArtPhoto Jul 08 '16 at 17:55
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    This had to be taken from a lot further away than 300 km. That is Mare Crisium on the top right limb, which means half the Moon is indeed visible, because that feature if very large. At the link above that Mark Adler put it is labelled. So it had to be as they were being captured by the Moon's gravity, and they had to have travelled beyond the Moon's orbit, slightly, when that happened. – kim holder Jul 08 '16 at 18:01
  • @kimholder The edge of that feature is visible and it's highly distorted. I'm not sure how you can make the claim that 50% is visible in the OP's photo. – Samuel Jul 08 '16 at 18:07
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    @Samuel the size of Mare Crisium in the photo in relation to the whole disk is the size of it when seen from Earth. Note also that the photo is upside down in the question, it is presented that way for compositional reasons. 10:30 isn't north there, it is more like at 5. – kim holder Jul 08 '16 at 18:32
  • @kimholder No, it's significantly larger in the photo here. Are you looking at the correct Mare? – Samuel Jul 08 '16 at 18:36
  • @kimholder Yes, that's the one I was looking at too. The rotation of the image doesn't change the apparent size ratio of the Mare to the visible surface though. It appears much larger because less than 50% of the surface is visible. – Samuel Jul 08 '16 at 18:42
  • @kimholder I understood what you were saying. I disagree with it, specifically your statement "which means half the Moon is indeed visible". It's very clear that less than 50% of the surface is visible in the photo. – Samuel Jul 08 '16 at 18:52
  • @kimholder Yeah there isn't anything to do to salvage that argument. They'd have to get a tenth of the way back to Earth before close to 50% of the surface became visible. Except, since they were on the far side, that's not what happened. So at no point while on the far side was 50% of the surface visible for a single photo. Not sure how to state it more clearly than that. – Samuel Jul 08 '16 at 19:16
  • @Samuel I deleted a few of my comments to shorten the list. My by eye sense of the size of Crisium was quite off as a proportion, and i was multitasking as i wrote all this - the 'half' part was lazy and based on misunderstandings, then i didn't give it enough attention after that. I was mostly thinking of the 300 km being way too low. You are right. – kim holder Jul 08 '16 at 19:46
  • "odd ... a perfect circle" -- Because the moon and every planet is close to a perfect sphere, every photo with it entirely in the frame will show it as a circle. With fisheye cameras, the earth appears as a circle even when photographed holding the camera only a few feet above the earth -- " http://photo.stackexchange.com/questions/23760/how-are-little-planet-photos-created ". – David Cary Jul 11 '16 at 12:40
  • That's only if you are far enough away. If you are in a plane, for instance, the Earth doesn't look like a circle. – PearsonArtPhoto Jul 11 '16 at 13:02
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enter image description here

So I roughly got into the right position to match the photo taken. The eye altitude is a little too low, so ignore that, but I think the lunar longitude is roughly correct (120° E). I'm very skeptical that this image was taken while at the stated lunar orbit altitudes. Those altitudes are just way too low to have these features appear as they do.

My best guess for when this picture was taken was during the return to Earth, probably not long after the trans-earth insertion burn. At this point the CSM would be on an escape trajectory leaving the moon, so it'd very rapidly gain altitude on the way out. It's also consistent with which hemisphere is visible (eastern), since that would be the departing side of the moon.

The mission report also notes on page 5-8 (pdf pg 48) that the mapping camera was also used "vertically" after trans-earth injection. I would guess that this image is from the mapping camera during that time. Unfortunately I can't really find any information about exactly when and where this occurred.

Another image from a lower altitude:

enter image description here

I apologize for the cropping on this, Google Earth has a fixed horizontal FOV and changing the window width leaves roughly the same horizontal portion of the moon visible. However, I think this image is a better match for the picture in the OP. In this case, approximately 20% of the lunar surface is visible.

Kyle
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    What focal length are you simulating here? – Samuel Jul 08 '16 at 18:03
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    @Samuel Unfortunately I don't know and Google Earth gives me no straight-forward way to adjust it, which is why the eye altitude is bogus for guessing where the module was when the image was taken. Fortunately the focal length won't really matter, since it doesn't change how much of the sphere is visible, just how distorted it is. Zooming in to the lunar orbit altitude would make it obvious that it's too close, since you can't get the right relative spacing between the visible mares. – Kyle Jul 08 '16 at 18:08
  • It looks to me like the visible portion of the moon in the OP's photo is only the upper right quadrant of your photo. Which, if your photo is 50%, it means the OP's photo is in the region of 12% of the surface, which nearly agrees with the OP's answer of 7.4% visible. – Samuel Jul 08 '16 at 18:14
  • @Samuel My photo shows approximately 29% of the lunar surface based on the reported eye altitude. – Kyle Jul 08 '16 at 18:31
  • Even better. The upper right quadrant is about 7.25% of the surface then. – Samuel Jul 08 '16 at 18:32
  • @Samuel I've added a second image that's considerably closer in that I think better matches OP's image. The reported altitude is still significantly higher than 300 km (~186 mi). Center to center, the two mares in the upper right are about 250 miles apart, which is 1/8 the diameter of the moon. From 186 miles, I think they'd take up much more of the visible disk than they do here. – Kyle Jul 08 '16 at 18:42
  • I'm not convinced that they would. I think that the tool you're using has a margin of error large enough (due to the fixed field of view) that we can probably assume the photo was taken from around 300km with a wide angle lens. Occam's razor. – Samuel Jul 08 '16 at 18:48
  • I'm marking this as correct. I think the bottom image is a good match, and it matches the information in my own answer. The FOV doesn't make much of a difference, as Kyle correctly noted, it doesn't change what is visible, only the size of what is visible. – PearsonArtPhoto Jul 08 '16 at 19:23
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https://en.wikipedia.org/wiki/Apollo_16#Launch_and_outbound_trip

When they performed a burn to enter lunar orbit above the far side, they were at ~300km above the surface. I can't find when this photo was taken, but I'd say there's a good chance it was then.

M.A.H.
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    It seems the photo was likely taken at about 76:34 from http://history.nasa.gov/ap16fj/11_Day4_Pt2.htm . I'm guessing that is time since launch. DOI was at 77:39. 300 km is much better than previous, but still seems a bit odd to have such a perfect circular moon from such an altitude. – PearsonArtPhoto Jul 08 '16 at 14:57
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    ?? The outline of a sphere is a perfect circle from any altitude. – Mark Adler Jul 08 '16 at 15:58
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    @PearsonArtPhoto at 74:24, are they talking about someone farting? – kevingreen Jul 08 '16 at 15:59
  • @MarkAdler The problem is getting the whole circle in frame from so close. – kevingreen Jul 08 '16 at 15:59
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    Wide angle lens. – Mark Adler Jul 08 '16 at 16:07
  • @kevingreen is an inventory of lenses brought on the mission available? Fisheye? – uhoh Jul 08 '16 at 16:07
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    @MarkAdler : Wide angle lens for sure. How wide of an angle to see as much as possible? Even at 300km they would be missing most of the surface. I assume that it would still appear spherical from their vantage point. They definitely did it, but what are the details of how? – kevingreen Jul 08 '16 at 16:21
  • @kevingreen It would seem so. – PearsonArtPhoto Jul 08 '16 at 16:23
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    I get that at 300 km you would need a 20 mm focal length wide angle lens for a standard 35 mm film/sensor size camera (I used the shorter 24 mm dimension of the sensor). You can get lenses with shorter focal lengths than that. – Mark Adler Jul 08 '16 at 16:27
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    @MarkAdler: Also, larger film sizes were often used on Apollo, which would increase the focal length allowed for the equivalent field of view. – Jacob Krall Jul 08 '16 at 16:33
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    Yes, the Hasselblad 500EL cameras used on Apollo on orbit and on the surface uses 56 mm x 56 mm film, as compared to the 36 mm x 24 mm standard "35 mm" film size. – Mark Adler Jul 08 '16 at 16:42
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    A 38 mm lens was available for the Hasselblad, which would do the trick. You need a 63° FOV for that picture, and that lens would give the Hasselblad a 73° FOV. – Mark Adler Jul 08 '16 at 16:49
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    My small sensor (APS-C) SLR with my 10 mm wide-angle lens also has a 73° FOV on the short dimension, so I could have taken that picture. I would be happy to go give it a shot. So to speak. – Mark Adler Jul 08 '16 at 17:05