Tom, have you reviewed my video where I attempted to explain how the full moon math really works? I sense a new video based on this shadow question.
I haven't looked at that one yet. I'll look at it in a bit.
It actually doesn’t matter what you believe. Your feelings on the matter are irrelevant. You are simply incorrect in your conception of the problem.
As I said earlier, what the observer sees
does matter. It is not irrelevant. model 29 was correct in his previous assessment. At the distance of the moon the earth would seem bigger then the sun, and at the distance where the umbra meets, the sun would seem bigger than the earth. Look at the images Bobby posted above. Look at what happens to the size of the sun and the earth as you get out further than the moon:
How to Calculate the Angular Size of the Sun --
Our sun is enormous compared to the Earth, measuring 109 times the diameter of the planet. When the great distance between the sun and Earth is factored in, however, the sun appears small in the sky. This phenomenon is known as the angular diameter. Astronomers use a set formula to calculate the relative sizes of celestial objects. The size and distance of objects is directly related; while the sun is 400 times larger than the moon, it is also 400 times farther away, making each object appear to be the same size in the sky -- and making solar eclipses possible.
Multiply the distance between the sun and the observer by 2. For example, to find the angular diameter of the sun as it appears on Earth, multiply 93 million miles by 2 to get 186 million.
Divide 865,000 -- the actual diameter of the sun in miles -- by the result from the previous step. The result is 0.00465.
Calculate the arctangent of the result from the previous step. On a scientific calculator, the arctangent function may be listed as either "tan-1" or "atan." The arctangent of 0.00465 is 0.26642.
Multiply the arctangent by 2. This result, 0.533 degrees, is the angular diameter of the sun as it appears on Earth.
Angular Diameter of the Sun from the Earth1.) 93,000,000 (dist from sun to earth) x 2 = 186,000,000
2.) 865,000 (diameter of sun) / 186,000,000 = 0.00465
3.) Arctan (0.00465) = 0.26642
4.) 0.26642 x 2 = 0.533 degrees
Simplified formula
2(Arctan (::Diameter of Sun:: / (::Dist from Sun to Earth:: x 2))) in degrees
2(Arctan (865000 / (93000000 x 2))) in degrees
Output if the above line from
Wolfram Alpha =
0.533 degreesAngular Size of the Sun from the MoonAverage distance between earth and moon = 238900 miles (Source: Google)
Distance from Moon to Sun if Moon is directly behind earth
93,000,000 + 238900 = 93,238,900
2(Arctan (865000 / (93238900 x 2))) in degrees =
0.5315 degreesAngular Size of the Earth from the MoonDist from Earth to Moon = 238,900
Diameter of Earth = 7917.5
2(Arctan (7917.5 / (238900 x 2))) in degrees =
1.899 degreesAt this point, the earth is bigger than the sun.
Angular Size of Sun from the Moon, if the moon were 6x further from the Earth93,000,000 + (6 x 238,900) = 94,433,400
2(Arctan (865000 / (94433400 x 2))) in degrees =
0.5248Angular Size of Earth from the Moon, if the moon were 6x further from the earthDistance from Earth to Moon (x6): 6 x 238,900 = 1,433,400
Diameter of Earth = 7917.5
2(Arctan (7917.5 / (1433400 x 2))) in degrees =
0.3165As we can see, now the sun is bigger.
Now lets get the value for the length of the umbra:
According to
http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit2/eclipses.htmlThe Earth's umbra is ~1.4 Million km long:
About 3.7x the mean Earth-Moon distance.
Angular Size of Sun from the Moon, if the moon were 3.7x further from the Earth93,000,000 + (3.7 x 238,900) = 93,883,930
2(Arctan (865000 / (93883930 x 2))) in degrees =
0.5279Angular Size of Earth from the Moon, if the moon were 3.7x further from the earthDistance from Earth to Moon (x3.7): 3.7 x 238,900 = 883,930
Diameter of Earth = 7917.5
2(Arctan (7917.5 / (883930 x 2))) in degrees =
0.5132The sun is slightly bigger at that distance, answers the problem I asked earlier, and shows that what the observer sees
is relevant.