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Epiphany during the Blood Red Super Full Moon Total Lunar Eclipse, 27 Sep 2015

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posted on Sep, 28 2015 @ 12:08 AM
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According to NASA's Neil DeGrasse Tyson, the Earth is an oblate spheroid and
slightly wider below the equator than above the equator. Like a pear.
Neil DeGrasse Tyson Says Earth is Pear Shaped

As I was filming the Blood Red Super Full Moon Total Lunar Eclipse earlier tonight, it
dawned on me that if the Earth is an oblate Spheroid as Neil insists, how is it that the
Earth does not cast a pear-shaped or oblate shadow during this and every past
eclipse that ever took place?

Rather, the Earth's shadow is a perfect circle that
casts a shadow sized exactly overlaying the moon's circular totality.

Of course, this observation only works if the world is a ball rather than flat.

SO where is the bulge and imperfections in the shadow?




posted on Sep, 28 2015 @ 12:10 AM
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a reply to: imd12c4funn

Just a wild guess here, but I think the reason that the earth doesn't cast a pear shaped shadow on the moon is because the earth's shadow fully engulfs the moon, since the earth is several times bigger



posted on Sep, 28 2015 @ 12:12 AM
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a reply to: imd12c4funn

I was not aware that Mr.Degrasse was a mouth for NASA.



posted on Sep, 28 2015 @ 12:13 AM
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a reply to: imd12c4funn

He doesn't literally mean 'pear-shaped,' he means slightly short of spherical.



posted on Sep, 28 2015 @ 12:14 AM
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a reply to: muse7

But as the shadow moves across the moon until total eclipse is achieved and then moving onward until the shadow no longer casts on the moon, you can see the arch of the shadow and that is where you can see there is no bulge or pear shape to the shadow. It is a perfectly circular arch.



posted on Sep, 28 2015 @ 12:14 AM
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Sorry, that is not an a Epiphany. That is a brainfart. I just hadda say it.



posted on Sep, 28 2015 @ 12:15 AM
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a reply to: imd12c4funn

Rather, the Earth's shadow is a perfect circle that
casts a shadow sized exactly overlaying the moon's circular totality.
No. The Earth's shadow is significantly larger than the Moon. In particular, since this was a "super Moon", when the Moon is closest to the Earth.
www.ldas.org.uk...


SO where is the bulge and imperfections in the shadow?
The "bulge" is too small to be noticeable. An exaggerated diagram:
www.eumetcal.org...
The actual difference is 0.3%

edit on 9/28/2015 by Phage because: (no reason given)



posted on Sep, 28 2015 @ 12:16 AM
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a reply to: imd12c4funn
The Earth's shadow is a lot larger than the Moon. It's also not a tight circle, but a "smudge" that gets progressively darker as you get to its centre: picture

Also, the Earth's oblateness is too small to be clearly evident from its shadow.
edit on 28-9-2015 by wildespace because: (no reason given)



posted on Sep, 28 2015 @ 12:17 AM
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originally posted by: Kandinsky
a reply to: imd12c4funn

He doesn't literally mean 'pear-shaped,' he means slightly short of spherical.


Again, view the arch of the shadow pre and post total eclipse. there is no "slightly short of spherical" to the shadow. it is as if you took a compass to draw the shadow perfectly circular. see what I mean?



posted on Sep, 28 2015 @ 12:18 AM
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a reply to: imd12c4funn




there is no "slightly short of spherical" to the shadow.

Again, the difference is too slight to be visible.



posted on Sep, 28 2015 @ 12:18 AM
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originally posted by: Kandinsky
a reply to: imd12c4funn

He doesn't literally mean 'pear-shaped,' he means slightly short of spherical.


Can we rule out flat...hehe



posted on Sep, 28 2015 @ 12:22 AM
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originally posted by: Xtrozero

originally posted by: Kandinsky
a reply to: imd12c4funn

He doesn't literally mean 'pear-shaped,' he means slightly short of spherical.


Can we rule out flat...hehe


Oh man, we've only just made it through a flat Earth phase. You'll start them off again!



posted on Sep, 28 2015 @ 12:22 AM
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a reply to: wildespace

I'll download my video (still on camera) and use my paint program to overlay a perfect circle on the shadow and I bet it is perfect. As Neil says, the equatorial is high as Mt. Everest above the rest of the earth measured from the center of Earth which would be noticable with an overlay
edit on 28-9-2015 by imd12c4funn because: fix interpretation



posted on Sep, 28 2015 @ 12:25 AM
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originally posted by: Xtrozero

originally posted by: Kandinsky
a reply to: imd12c4funn

He doesn't literally mean 'pear-shaped,' he means slightly short of spherical.


Can we rule out flat...hehe

Actually...No



posted on Sep, 28 2015 @ 12:25 AM
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a reply to: imd12c4funn


Mt. Everest measured from the center of Earth which would be noticable with an overlay
No. It wouldn't.

Draw a circle a meter in diameter. To match the oblateness of the Earth, the bulge would be 3 mm. And, mind you, it is not a 3mm bump, it is the difference between the N-S diameter and the E-W diameter. A gradual curve.

How big a circle did you make in Paint? Six inches? The bulge would be .018". Think you would notice? How sharply defined was the curve?

edit on 9/28/2015 by Phage because: (no reason given)



posted on Sep, 28 2015 @ 12:27 AM
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a reply to: Kapusta

Not sure if Neil is a mouth of NASA, but that is what came out of his mouth.



posted on Sep, 28 2015 @ 12:42 AM
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Take this into consideration... Here is an idea of what they're trying to tell you. See the size of the earths shadow to the moon? Even if it was a little more than slightly, it would still look like a circle since you're only getting a very small part of the circle of the earth on the moon when it starts taking a bite out of it.

See HERE



posted on Sep, 28 2015 @ 12:44 AM
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originally posted by: Phage
a reply to: imd12c4funn




there is no "slightly short of spherical" to the shadow.

Again, the difference is too slight to be visible.


Earth's diameter at equator is 7,926.41 miles.
at the poles is 7901 miles.
a difference of 25.41 miles.

Everest is 29,035 feet, or 5.49905303 Miles, or just over 25% of the difference between the equatorial and meridianal diameters. My program can see that. It's not as minute of a difference as you make it sound.



posted on Sep, 28 2015 @ 12:47 AM
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a reply to: imd12c4funn



a difference of 25.41 miles.
Yes. Like I said, a difference of 0.3%.



It's not as minute of a difference as you make it sound.

In a 6" circle the difference is 18 thousandths of an inch. How big was the circle you drew in paint? How much of the total curve of the shadow were you measuring? A fourth of it? Do you think you could detect a difference of 4 thousandths of an inch on your screen? What's your pixel pitch?


edit on 9/28/2015 by Phage because: (no reason given)



posted on Sep, 28 2015 @ 12:51 AM
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originally posted by: imd12c4funn
a reply to: muse7

But as the shadow moves across the moon until total eclipse is achieved and then moving onward until the shadow no longer casts on the moon, you can see the arch of the shadow and that is where you can see there is no bulge or pear shape to the shadow. It is a perfectly circular arch.


So, you think that the shadow you were seeing on the moon accounts for the enterity of Earths shape?

Thats where your epiphany goes wrong......



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