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

a reply to: Phage

haven't downloaded my camera yet, but I am 100% proficient in my program and will post the results when I complete the calculations which will include the Northern and Southern hemispherical supposed difference proclaimed by Neil.

a reply to: imd12c4funn

but I am 100% proficient in my program and will post the results when I complete the calculations.

You're 100% proficient? Neat. How hard is it to draw a circle with it?

What calculations? Do you dispute that 0.3%? You provided the numbers yourself.

Do you dispute that it translates to 18/1000ths of an inch in a 6" circle? With a typical LCD monitor that translates to about 2 pixels. You are not going to be able to measure that (considering that the curve of the shadow is quite blurred and you are only seeing a portion of the entire shadow), much less see it.

originally posted by: imd12c4funn
a reply to: Kapusta

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

Then why did you make this statement

According to NASA's Neil DeGrasse Tyson, the Earth is an oblate spheroid and 

Can you please enlighten me on that.

Ty

Kap

originally posted by: imd12c4funn
a reply to: Phage

haven't downloaded my camera yet, but I am 100% proficient in my program and will post the results when I complete the calculations which will include the Northern and Southern hemispherical supposed difference proclaimed by Neil.

I'm sure you're an absolute whizz at MS Paint, but two things will make you fail, Firstly, the shadow covering the moon is not the entire Earth shadow. Second, the shadow is falling not on a flat surface but a three dimensional sphere. The curve of the shadow is as much a product of the latter as it is the former.

Tyson's metaphor was intended to help people visualise something. You will not see the difference no matter how good at Paint you are.

Even Neal DeGrasse Tyson says that the pear-shaped Earth is NOT something that can be noticed just by looking at the Earth.

The difference in the width of the earth at its widest point compared to the pole-to-pole "height" of the earth is about 44km (28 miles). That's a 28 mile difference over the Earth's 8000 mile diameter.

That tiny bulge is not something you are going to notice by looking at the Earth or its shadow.

If you drew a circle on paper and the difference between the vertical diameter and horizontal diameter were 0.3%, it'd still look like a perfect circle to the naked eye unless you measured it.

I have a damaged venetian blind slat... the gap is roughly rectangular

when the sunlight passes through the hole in the venetian blind I see a rectangular hole BUT when the light hits the wall some 12 feet away the beam of sunlight is round !


the moon is some +200,000 miles away, I suggest that a tear-drop shaped planet would cast a round or circular shadow on the spherical moon also... or at least the leading edge of a eclipse's shadow would be curved....however a really micro measured (by laser) arc of that eclipse leading edge of light-v-darkness would reveal the irregular curvature


I had an odd epiphany during the solar eclipse I was in while I lived in Phoenix AZ... the 'crescent' in the sky got filtered through all the hundreds of peep-holes between the leaves of some fruit tree in the backyard....
the result was astounding... on the cinder block wall there were hundreds of solar crescents projected on the wall which were very sharp images of the overhead solar eclipse taking place back in the 1990's

...the projections were like that pinhole camera one makes to see the event without burning out your rods & cones in your eye...

Below is a link to video that shows what I believe is a non-oblate spheroid earth.

Lunar Eclipse

originally posted by: imd12c4funn
Below is a link to video that shows what I believe is a non-oblate spheroid earth.

Lunar Eclipse

The calculated bulge is tiny -- a difference of only 28 miles more than total diameter of ~8000 miles.

You would not be able to see such a minor bulge in that video.

a reply to: Soylent Green Is People

yeah but try it with a good hi def video and it is the same.

originally posted by: imd12c4funn
a reply to: Soylent Green Is People

yeah but try it with a good hi def video and it is the same.

Can even a high definition video of the shadow of an eclipse show us that the Earth is 99.7% as short than it is wide?

I'm going to say no, especially considering that the the thing the shadow is being projected onto (the moon) is a non-perfect sphere itself, which would skew any measurements.

Maybe (just maybe) if you had a mega-ultra-high resolution picture that showed the Earth to be ~8000 pixels wide taken with a camera that had an ultra-precision lens that had virtually no distortion, you would THEN maybe be able to count those pixels to discern that it is roughly 28 pixels wider than it is high (say, for example, if the Earth were 7972 pixels high but 8000 pixels wide).

However, even then I doubt you would be able to see the difference with your eyes. You would probably only notice the difference by counting pixels.

a reply to: imd12c4funn

Oh dear. Perhaps in another epiphany you can realize that the circular shadow of the 8,000 mile diameter Earth is not the same as the 2,000 mile diameter Moon.

a reply to: imd12c4funn

Good man for coming through with the video. All too often people say one thing and do another so you're in a different class straight away

Although you are convinced by the empirical evidence, there's just no way for the human senses to differentiate between a very slightly 'oblate' sphere and a perfect sphere - not at this scale. The Earth shadow looks to me to be around ~70 degrees of arc.

I guess you can walk away from this thread with your epiphany intact even though you won't convince anyone else.

here's the video embed:

To expand on my previous post, because the Sun is larger than the Earth, the Earth's shadow is actually conical, getting smaller the further from the Earth you get. My trig is a bit rusty, but I figure that, at 220,000 miles, the shadow (or umbra) is going to be ~4,200 miles in diameter - a bit more than twice the diameter of the Moon. Interestingly, most diagrams I see (like this one) show it to be ~3 times the lunar diameter.

For my demonstration, I'll not only take the lower figure, I'll round it down to an even 4,000 miles - twice the diameter of the Moon.

I drew two figures: one a circle 400x400 pixels, and the other an oblate version 400x398 pixels. The difference of height-to-width on the latter is 0.05% - comfortably greater than the actual 0.03% difference in the Earth's polar and equatorial diameters.

Here they are, eclipsing a 200x200 pixel Moon:




Can you tell which is which?

Tomorrow I'll post the uncropped images with the circles drawn, and let you guess again.

originally posted by: St Udio


I had an odd epiphany during the solar eclipse I was in while I lived in Phoenix AZ... the 'crescent' in the sky got filtered through all the hundreds of peep-holes between the leaves of some fruit tree in the backyard....
the result was astounding... on the cinder block wall there were hundreds of solar crescents projected on the wall which were very sharp images of the overhead solar eclipse taking place back in the 1990's

...the projections were like that pinhole camera one makes to see the event without burning out your rods & cones in your eye...

I had the same experience during an partial solar eclipse in California in the 80's...dozens of crescent shaped shadows on my paper as I was doing the pin-hole thing. (I was under a tree.) It was very cool!

a reply to: Saint Exupery

Here's another:

One of these is a perfect circle, and the other is the stated oblate spheroid shape of Earth -- i.e., having a 0.3% difference in height to width, with the wide part being a bit in the lower hemisphere. Which is which?

originally posted by: Saint Exupery
To expand on my previous post, because the Sun is larger than the Earth, the Earth's shadow is actually conical, getting smaller the further from the Earth you get. My trig is a bit rusty, but I figure that, at 220,000 miles, the shadow (or umbra) is going to be ~4,200 miles in diameter - a bit more than twice the diameter of the Moon. Interestingly, most diagrams I see (like this one) show it to be ~3 times the lunar diameter.

OK, I found the mistake in my math. The correct value for the diameter of the umbra at 220,000 is approximately 6,100 miles, or just over 3 times the diameter of the moon. That means that if I'd drawn the above comparison correctly, there would be even less of a difference between the two images than what I've shown.

Note to imd12c4funn: See what I did, just then? See how easy it is to admit it when you're wrong?