# The Moon's shadow could not possibly look like this — could it?

The NASA Goddard video Tracing the 2017 Solar Eclipse is a visualization of the calculated shape of the umbra during the upcoming August 2017 total solar eclipse, as it moves across North America.

The simulation takes into account the actual measured 3D shape of the Moon from the LRO's laser altimeter, the shape of the Earth is from SRTM (radar map from the space shuttle), and the locations of the Sun, Earth and Moon from the JPL Development Ephemeris.

Question: What is confusing me are the big "corners" on the Moon's shadow. I had thought that the moon was close to an ellipsoid. There are large mountains, mare, and craters for sure, but have these deviations been greatly exaggerated in the animation? The largest fat spot shows an "altitude deficit" of roughly 4% of the Moon's diameter, or about 140 kilometers! Does this turn out to be in fact far less accurate than projecting a simple ellipsoidal moon's umbra?

below: Screen shots from the linked NASA Goddard video Tracing the 2017 Solar Eclipse.

• Even if the Earth and Moon were perfectly spherical, the instantaneous shape of the Moon's shadow on the Earth would be a circle only the Moon and Sun are directly overhead. In the example, the Sun's azimuth is 45.1 degrees, so about halfway up the sky (or halfway down from vertical). The shadow of a spherical Moon on a spherical Earth would be close to a rather elongated ellipse. That's the "smooth limb" curve in the bottom picture. – David Hammen Jul 22 '17 at 21:13
• @AnthonyX I'm not sure you've looked carefully at the figures or watched the video, so I've added a pair of arrows to every image. You can see that the major deviations are completely static throughout the video, independent of any topography. Note that the first image shows the shape over water with no topography at all. The last image is a diagram without topography either. – uhoh Jul 23 '17 at 3:15
• @DavidHammen The profile has completely static bumps, or you can say there are flattened areas between them. The deviation is of the order of 140 kilometers! I've added some arrows - take a look again, especially the profile shown over the ocean. – uhoh Jul 23 '17 at 3:20
• Yeah, something's not right here. WP says 18km range from lowest to highest point on the moon, about 0.005 diameters. Measuring the long diagonal of the "Umbra Shapes" picture, I see ~14 px deviation from smooth limb to true limb on a 700 px diagonal, 0.02 diameters -- 4 times the maximum I'd expect. I guess the curved projection accounts for some of that, but it still seems strange. en.wikipedia.org/wiki/List_of_mountains_on_the_Moon – Russell Borogove Jul 23 '17 at 3:56
• I think what is going on is that the area of totality is only a fraction of the Moon's actual size, but it's surface irregularities are represented on the shadow's edge at approximately their actual size, making them seem bigger in relation. It's to do with the sizes of the Sun and the Moon and their distances. – Anthony X Jul 23 '17 at 5:33

## 3 Answers

The deviations are real and directly related to the size of valleys on the Moon. The simple calculation you do here does only work for a point-like light source that projects the shape of the Moon onto Earth. In case of an eclipse, the light source has almost the same apparent size as the object. For this eclipse, the magnitude is given as 1.03. This is the ratio between the apparent diameters of Moon and Sun.

You can easily imagine what is going on by assuming the Moon to be the same apparent size as the Sun. That is, the Moon can just cover the Sun, but not more. Now, the slightest valley on the Moon would make totality vanish - there would be a (small) amount of Sun still visible.

Taking the actual magnitude of 1.03 into account and putting our observer in the middle of the eclipse (on the red path), there is a ring of 1.5% of the diameter of the Moon "around" the Sun. In the screen shot above uhoh measured a deviation of 4% from a perfect ellipse. These 4% we have to put in relation with this ring of 1.5% of the Moon's diameter, resulting in a valley depth of about 2 km, which is absolutely reasonable. To put it in other words: If there was a valley of 4% / 140 km depth on the Moon, you wouldn't have any totality at all, because Moon wouldn't be able to cover the Sun at any moment.

• They = the creator of the animation? Sure they took it into account, with a point-like sun the zone of totality would have been larger than the US, not just 100 km. – asdfex Aug 30 '17 at 18:19

I'll add this to the other answers, which do a very good job of explaining why this very irregular shape is not inconsistent with the Moon's original much "rounder" shape.

This could be done with some geometry and math, but I was lazy and did a rendering in Blender. I made a spheroid and added some bumps, then added a Sun lamp above and screen far below. I can adjust the half-angle of the sun lamp. This is not intended to be an exact simulation, but once you see the GIF the principle described in words in the other two answers is reinforced.

Note the appearance of sharp corners on the shape of the "umbra" that do not exist on the original shape!

In order to delineate the "umbra" where the light reaches zero, I just turned up the brightness of the Sun to 100.

below: The original shape, moved down into the plane of the screen so that the cross-section shape is clear.

below: Selected frames, showing a nearly square shaped "umbra" with the appearance of sharp corners that do not exist on the original shape!

Here's the Python script I used in Blender to make the shape. Animation is easily scripted also, I just did it manually this time because it was quicker.

import bpy
import numpy as np

halfpi, pi, twopi, fourpi = [f*np.pi for f in [0.5, 1, 2, 4]]

half_nth = 50
nth, nph = 2*half_nth + 1, 200

th  = np.linspace(-halfpi, halfpi, nth+2)[1:-1]

phi = np.linspace(0, twopi, nph+1)[:-1]

n1, n2 = 8, 32

thing = 0.5*(1.+np.cos(np.linspace(0, pi, n2+1-n1)))[:,None]
ramp   = np.zeros_like(th)[:,None]

ramp[half_nth+n1   : half_nth+(n2+1)] = thing
ramp[half_nth-n2   : half_nth-n1+1  ] = thing[::-1]
ramp[half_nth-n1+1 : half_nth+n1    ] = 1.0

nwiggle = 4
awiggle = 0.05

wiggle = 1. + awiggle * np.cos(nwiggle*phi) * ramp

X = wiggle * np.cos(phi)[None,:]       * np.cos(th)[:,None]
Y = wiggle * np.sin(phi)[None,:]       * np.cos(th)[:,None]
Z =          np.ones_like(phi)[None,:] * np.sin(th)[:,None]

verts = [tuple(thing) for thing in zip(X.flatten(), Y.flatten(), Z.flatten())]

faces = []
for ith in range(nth-1):
for iph in range(nph):
v1 =  iph        + nph*ith
v2 = (iph+1)%nph + nph*ith
v3 = (iph+1)%nph + nph*(ith+1)
v4 =  iph        + nph*(ith+1)
faces.append((v1, v2, v3, v4))

bot = [tuple([iph               for iph in range(nph)])]
top = [tuple([iph + nph*(nth-1) for iph in range(nph)])]

faces += bot + top

me = bpy.data.meshes.new('wow')
ob = bpy.data.objects.new('wow', me)

ob.location = (0.0, 0.0, 102.0)
bpy.context.scene.objects.link(ob)
me.from_pydata(verts,[],faces)

bpy.data.objects['wow'].select = False
bpy.data.objects['wow'].select = True

bpy.ops.object.shade_smooth()


The outline of the lunar umbra on Earth (shadow of totality) will be defined by those rays of light emanating from the very edge of the solar disk as seen from Earth which just miss the surface features at the lunar terminator (line between day and night on the Moon's surface). Because the Moon is some 500 times smaller than the Sun (by diameter), those rays are more-or-less converging to a point about 1/500 the distance between Sun and Moon (the ratio of their actual diameters).

As it happens, the Earth-Moon distance is about 1/500 the distance between the Sun and the Earth/Moon system. In the upcoming eclipse, the Moon will be a little closer than "normal" and the "point" of convergence will be "behind" the Earth's surface. The umbra on Earth's surface will be on the order of 100km across.

If you consider a point source at the Sun, as rays pass by objects on the Moon, they will form a shadow on the Earth slightly larger than their true size - if we use the approximation of sizes and distances as 500:1, the shadows will be about 501:500 of their true size. But, the umbra is only a small fraction of the Moon's size. So, as the rays converge down on their way to the Earth's surface, the shadows of lunar surface features at the terminator will overlap and "blur" together to form the odd shape shown in the video.

• @uhoh I'm happy to edit in more accurate values, just guesstimated from the illustration. Feel free to replace the numbers in my answer with true values if you like. – Anthony X Jul 23 '17 at 18:54
• I just noticed the edit. Looks good! – uhoh Jul 29 '17 at 9:28
• I had to choose one of two great answers. You figured it out first though. Thanks for your help! – uhoh Jul 29 '17 at 9:32