# How did Apollo 16 capture this full photograph of the Moon's far side?

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

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?

• 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? Jul 8, 2016 at 14:18
• I'm pretty sure it's AS16-3021 Jul 8, 2016 at 15:49
• nssdc.gsfc.nasa.gov/imgcat/html/object_page/a16_m_3021.html Jul 8, 2016 at 15:53
• 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. Jul 8, 2016 at 16:06
• @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. Jul 12, 2016 at 14:35

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.

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

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.)

The visible footprint viewed from a more familiar angle:

• Very nice answer! Welcome to the site.
– user
Jul 8, 2016 at 22:24
• Nice answer! Can you expand a bit on the methods and software you used to produce these visualizations?
– E.P.
Jul 9, 2016 at 23:32
• 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.) Jul 10, 2016 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"? Jul 11, 2016 at 13:26
• @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) Jul 11, 2016 at 21:47

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!

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!

• 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. Jul 8, 2016 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. Jul 8, 2016 at 17:55
• 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. Jul 8, 2016 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. Jul 8, 2016 at 18:07
• @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. Jul 8, 2016 at 18:32

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:

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.

• What focal length are you simulating here? Jul 8, 2016 at 18:03
• @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 8, 2016 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. Jul 8, 2016 at 18:14
• @Samuel My photo shows approximately 29% of the lunar surface based on the reported eye altitude.
– Kyle
Jul 8, 2016 at 18:31
• Even better. The upper right quadrant is about 7.25% of the surface then. Jul 8, 2016 at 18:32

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.

• It seems the photo was likely taken at about 76:34 from 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. Jul 8, 2016 at 14:57
• ?? The outline of a sphere is a perfect circle from any altitude. Jul 8, 2016 at 15:58
• Wide angle lens. Jul 8, 2016 at 16:07
• @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? Jul 8, 2016 at 16:21
• 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. Jul 8, 2016 at 17:05