As an interesting topic for improving general knowledge about light and shadow casting, I am starting here a discussion about how to calculate the Total Solar Eclipse Umbra and Penumbra sizes, angles of projection, etc, in order to calculate the proportional Sun and Moon sizes and respective distances. As the umbra (the small disc of total shadow where you can not see the Sun during the eclipse) is a conical projection, it assumes a round shape if casting close to 90° angle to Earth's surface, or an ellipsoid when this angle decreases.

There are 8 variable in this project.

1.) The Sun distance

2.) The Moon distance

3.) The Sun diameter

4.) The Moon diameter

5.) The Umbra diameter if a circle

6.) The Umbra dimensions (length and width) if an elipse

7.) The Penumbra dimensions

8.) The Shadow casting Angle on Earth's surface

The Umbra and Penumbra round shape can tell us the conical result of projection, with 5 and 7 we can calculate the proportions between 3 and 4 and between 1 and 2 above. That is pure optics and shadow casting.

**A)** If the Sun is bigger than the Moon (what it is), it will projects a Moon shadow in a conical form, smaller the diameter as far it gets from the Moon, until the distance is such that the cast shadow reach the apex, a very narrow point. If nothing intercepts this cast conical shadow, it just disappears in the space, nobody sees the cone. If anything comes inside such dark cone, it will intercepts and shows the shadow. The apex distance from the Moon is directly proportional to the sizes Sun & Moon and the distance between them.

**B)** If the Sun and the Moon has the same size, it will projects a Moon shadow in a conical form with infinite apex, known as "cylinder". The shadow diameter would be the same as the Moon and will be projected at far distances without changing size.

**C)** If the Sun is smaller than the Moon, it will projects a Moon shadow with an inverted cone, apex on the Sun. The cone would increase in diameter as it distances from the Moon, blocking solar light in vast area, in such way that the entire Andromeda galaxy could be in this inverted cone shadow.

Considering history and records, close to 90° of Sun's position where Umbra happens on total eclipse, the Umbra diameter was never smaller than 120km in diameter.

**Considering (A)** above, the Moon must be way larger than 120km in diameter, the Sun must be way larger than the Moon and they are not close at all. With the same measurement date for the Penumbra, one could calculate the size proportions between Sun & Moon and the proportional distances between them and Earth.

**Considering (B)**, both Moon and Sun must have 120km in diameter in order to cast a same size shadow. The problem here is that the Umbra diameter changes even at 90° (round Umbra), in different eclipses according to factors, one is the actual distance Moon-Earth at the moment of eclipse, another is the angle of shadow casting over Earth surface. Considering only a closer to 90° Sun & Moon position, then both must have 120km in diameter. The killing factor here is that on this situation the only way to have Penumbra is when the same size Sun is far away from the Moon, it can be calculated, measured, simulated on any optical lab or kitchen table. Grab your flashlight, put it face down over a paper, using the flashlight face draw a circle with a pencil, cut this disc of paper (lets call it "blocker"), now projects the flashlight light beam 90° against the wall and use the blocker to block the light, change distances Flashlight-Wall, Flashlight-Blocker and see what happens. No matter the distance from the wall, the blocker shadow diameter will be almost constant. Now cut another blocker half the diameter of the first one and repeat the experience, you will notice the conical shadow projected on the wall, if the flashlight and the smaller blocker is away from the wall, there will be no casting shadow visible defined, as (A). With the blocker same size of flashlight, the Umbra never changes size, and it changes in real life, this option (B) is eliminated.

**Considering (C)**, Cut a blocker twice the diameter of the flashlight and play with it against the wall... it will be impossible, the cast shadow will always be larger than the blocker or the flashlight. If flashlight and blocker are very close, there will be no light projected to the wall at all, no Penumbra. So, this option (C) is eliminated. The only way to have a 120km round shadow Umbra (90°) and larger Penumbra for a 48km diameter Sun is for the Moon to be way from the Sun and larger. It is easy to calculate. With the FE Sun at altitude of 4800km and 48km in diameter, to project a Umbra shadow of 120km in diameter, the blocker (Moon) can be calculated using this formula:

SD = Sun Diameter

MD = Moon Diameter

SPD = Shadow Projected (on Earth) Diameter

H1 = Distance Sun to Cone Apex (behind/above the Sun)

H2 = Distance Sun to Moon

H3 = Distance Earth to Moon

H = H1+H2+H3

First you need to find H1

Suppose:

H2+H3 = 10

SD = 2

SPD = 8

2 = 8 * H1 / (10+H1)

2 * (10+H1) = 8 * H1

8 * H1 / 2 = 10+H1

4 * H1 - H1 = 10

H1 = 10 / 3

H1 = 3.333

Then we have the total cone H as 13.333.

Now you can calculate Moon Diameter (MD) for a certain distance (H3)

MD = SPD * (H2+H1) / H

Suppose all the numbers above the same, and H2 (Sun to Moon Distance) is 3.

MD = 8 * (3+3.333) / 13.333

MD =3.8

So, if the Sun (diameter 2) is 10 from Earth, Moon is 7 from Earth, Shadow is 8, Moon diameter must be 3.8, almost double the Sun.

If H2 = 1, MD = 8 * (1+3.333) / 13.333

MD = 2.6, still bigger than the Sun even being very much closer to the Sun.

So, for a Sun = 48km (SD=48), H2+H3 = 4800km, SPD = 120km

48 = 120 * H1 / (4800+H1)

48 * (4800+H1) = 120 * H1

120 * H1 / 48 = 4800+H1

2.5 * H1 - H1 = 4800

H1 = 4800 / 1.5 = 3200km

Now, calculate diameter of the Moon, in kilometers, by distance from Sun:

MD = 120 * (H2+3200) / 8000

MD = 0.015 * (H2 + 3200)

MD = H2*0.015 + 48

See, no matter the distance of the FE Moon from the Sun, it starts with at least 48km in diameter if very near the Sun, or 120km if very near to Earth. The Moon should increase 15 meters in diameter for every kilometer from the Sun. The FE wiki says Sun and Moon are at the same altitude and size, in that case, a total solar eclipse would obscure the whole Earth. On that Nodal calculation of the Moon performing an angled path under and over the Sun, it becomes clear that the FE Moon would be not at the same altitude as the Sun during a total eclipse, it will be lower, so, to projects a 120km Umbra, the Moon needs to be bigger than the Sun. If FErs publish the numbers, angle of inclination, altitude, etc, one could even calculate the Moon diameter based on the 120km Umbra and Penumbra. It will be interesting to see those FE numbers, mostly to produce the very large penumbra that can cover diferentes continents at once.

Then you need to use the Penumbra numbers to calculate how far is the Sun from the Moon. For example, at the maximum point of the Great American Eclipse in August 2017, the penumbral shadow spanned from Panama all the way to Greenland, covering North America, Central America, and large parts of the northern polar regions. For such gigantic Penumbra, only option (A) is possible. Moon is way bigger than 120km in diameter, very far away from Earth and the Sun is way bigger than the Moon, 400 times farther than the Moon.

Interesting is that if you use the RE numbers, Umbra and Penumbra fits perfectly on the solar eclipses numbers on record, calculating with Sun and Moon distances, angles of projections and ellipsoid sizes if lower than 90° projection, and lines of nodes. The average Umbra conical shadow distance is no more than 380,000 km.

I wonder if FEs could provide some simple calculations to justify their own numbers for Sun/Moon altitude and lines of nodes, to create the same Umbra/Penumbra results.

Suppose FE Moon is farther from the North Pole than the Sun, shadow lower than 90°, and by some way it projects a shadow further yet, the ellipsoid cast shadow would be wider in the farther side from the North Pole, and narrow on the closer side. Also, the farther side on FE will be fully projected on the ground, with good contrast, while on the RE the further side disappears due planet curvature. The same would happen if Sun/Moon aren't well aligned to 90° and over a little different longitude.

Records show (and it is possible to calculate) the Umbra speed is around 1700km/h Eastward, the total eclipse umbra lasts no more than 7 1/2 minutes at a specific point. A Concorde flight was able to fly under Umbra at Mach 2.5 on June 1973 for more than 74 minutes.

https://www.nature.com/articles/246072a0MIR image shadow from the Solar Eclipse from Aug/11/1999:

From ISS on March 29, 2006:

I invite everyone to participate in this exercise of numbers.