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Sep 10 2017 08:21am
Quote (Santara @ Sep 10 2017 03:45am)
When I am not posting from my phone, I will show you the math. But understand that the sun appearing about the same size in the sky as the moon has everything to do with it being about 400 times larger in diameter and also 400 times further away.


I understand that is what they tell you but show me the trig that a 2000 mile wide object creates a 70 mile focused shadow from parallel sun rays and that when the moon eclipses the sun that event is viewable 40 times as it crosses America without using local perspective. Umbra/penumbra is just an excuse cause my friend launched a balloon during the eclipse and photographed the shadow and it was only about 70 miles wide, no penumbra. Compare this to a lunar eclipse where the entire moon is covered in a shadow, and explain the difference. Also explain why there is no Earthshine seen on the "moon" during an eclipse? And while your at it - explain why the sun and moon can be directly overhead during the day so that the Sun is behind the moon but the sun can light up the front of the moon, need some trig on that one too.



This post was edited by card_sultan on Sep 10 2017 08:22am
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Sep 10 2017 11:03am
Tldt triggered much?

Nice block of text :rofl:
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Sep 10 2017 11:11am
Here's "Dr" Zack getting slammed for using Earth tilt 6 months off from reality:

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Sep 10 2017 11:22am
Quote (card_sultan @ Sep 9 2017 03:49pm)
That's not trig, thats a cartoon



this is not a post, it's just text inside html, c what i did there ?

HERP DERPPPPPPPP

This post was edited by JohnMiller92 on Sep 10 2017 11:23am
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Sep 10 2017 12:19pm
Quote (Santara @ Sep 10 2017 07:11am)
Here's "Dr" Zack getting slammed for using Earth tilt 6 months off from reality:

http://www.youtube.com/watch?v=ho2YB9bLTJ4


that's just a clown video that claims debunk with bad logic.

offers a bad link as proof, eclipses do not prove that Helio math was used and if he did why does NASA have to refer to a retired guy, is he the only guy that can do math?
Also the tilt of the Earth is not wrong from the sun's perspective - it just appears that way because you look from an 3rd party 2 d inversed perspective on paper not IRL
also the 3d calculations were done on a 3d model, he just uses the 2d mercator map projection as a refence

Also you didn't address any of my points, or provided any trig and just posted a silly video.

Smashed!

Quote (JohnMiller92 @ Sep 10 2017 07:22am)
this is not a post, it's just text inside html, c what i did there ?

HERP DERPPPPPPPP


that's not text, its just pixels on a lcd screen

hurrrr durrrrrr

Quote (Ep0ch @ Sep 10 2017 07:03am)
Tldt triggered much?

Nice block of text :rofl:


try not to be so rude when the grown ups are talking, k?

This post was edited by card_sultan on Sep 10 2017 12:21pm
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Sep 10 2017 12:53pm
Weak af kid
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Sep 10 2017 07:09pm
Quote (card_sultan @ Sep 10 2017 01:19pm)
that's just a clown video that claims debunk with bad logic.

offers a bad link as proof, eclipses do not prove that Helio math was used and if he did why does NASA have to refer to a retired guy, is he the only guy that can do math?
Also the tilt of the Earth is not wrong from the sun's perspective - it just appears that way because you look from an 3rd party 2 d inversed perspective on paper not IRL
also the 3d calculations were done on a 3d model, he just uses the 2d mercator map projection as a refence

Also you didn't address any of my points, or provided any trig and just posted a silly video.

Smashed!


You didn't smash shit. I told your dumb ass that I would post the math when I wasn't posting from my phone.

Prove his criticism is wrong.

Anyways, on to the math. First and foremost, the sun's light isn't parallel. It is near parallel, but so close as to be indistinguishable to the naked eye. For a total eclipse to occur, the moon must be at or near its perigee. (The moon's orbit, like the Earth's around the sun, is an ellipse.) At the moon's average distance, a total eclipse is not possible.

For an observer of a total eclipse, the distance from the observer to the moon is approximately 363,000 KM, observing an object with a diameter of approximately 3476 KM. The difference in angle of observation from one side of the moon to the other 0.549°. (This is an isosceles triangle.) For the same observer, the distance to the sun is approximately 149,600,000 KM. Since the sun's diameter is 1,391,400 KM, the trigonometry tells us that the angle of observation from the observer to the edges of the sun is 0.533°. (Also an isosceles triangle.) What this tells us is that it is indeed possible for the sun to hide the moon using known measurements, because the angle to the moon's edge is greater than the angle to the sun's edge. Now for frame of reference, when the moon is at its average distance from the Earth (384,000 KM), the observable angle to the edges is 0.519°, meaning the angle is too small to hide the sun in totality. Anyways, back to your 70 mile (112 KM) umbra. Those .016° of separation between the sun's and moon's angles at a distance of 363,000 KM calculates to 101 KM (63 miles) - which is awfully close to the 112 KM value, and just tells us the distance between the Earth and moon was a little greater than 363,000 KM at the time of the eclipse.

There's your math.

This post was edited by Santara on Sep 10 2017 07:09pm
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Sep 10 2017 07:10pm
Quote (Santara @ Sep 10 2017 07:09pm)
You didn't smash shit. I told your dumb ass that I would post the math when I wasn't posting from my phone.

Prove his criticism is wrong.

Anyways, on to the math. First and foremost, the sun's light isn't parallel. It is near parallel, but so close as to be indistinguishable to the naked eye. For a total eclipse to occur, the moon must be at or near its perigee. (The moon's orbit, like the Earth's around the sun, is an ellipse.) At the moon's average distance, a total eclipse is not possible.

For an observer of a total eclipse, the distance from the observer to the moon is approximately 363,000 KM, observing an object with a diameter of approximately 3476 KM. The difference in angle of observation from one side of the moon to the other 0.549°. (This is an isosceles triangle.) For the same observer, the distance to the sun is approximately 149,600,000 KM. Since the sun's diameter is 1,391,400 KM, the trigonometry tells us that the angle of observation from the observer to the edges of the sun is 0.533°. (Also an isosceles triangle.) What this tells us is that it is indeed possible for the sun to hide the moon using known measurements, because the angle to the moon's edge is greater than the angle to the sun's edge. Now for frame of reference, when the moon is at its average distance from the Earth (384,000 KM), the observable angle to the edges is 0.519°, meaning the angle is too small to hide the sun in totality. Anyways, back to your 70 mile (112 KM) umbra. Those .016° of separation between the sun's and moon's angles at a distance of 363,000 KM calculates to 101 KM (63 miles) - which is awfully close to the 112 KM value, and just tells us the distance between the Earth and moon was a little greater than 363,000 KM at the time of the eclipse.

There's your math.



Fixed
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Sep 10 2017 07:11pm
Quote (Santara @ Sep 11 2017 01:09am)
You didn't smash shit. I told your dumb ass that I would post the math when I wasn't posting from my phone.

Prove his criticism is wrong.

Anyways, on to the math. First and foremost, the sun's light is parallel. It is near parallel, but so close as to be indistinguishable to the naked eye. For a total eclipse to occur, the moon must be at or near its perigee. (The moon's orbit, like the Earth's around the sun, is an ellipse.) At the moon's average distance, a total eclipse is not possible.

For an observer of a total eclipse, the distance from the observer to the moon is approximately 363,000 KM, observing an object with a diameter of approximately 3476 KM. The difference in angle of observation from one side of the moon to the other 0.549°. (This is an isosceles triangle.) For the same observer, the distance to the sun is approximately 149,600,000 KM. Since the sun's diameter is 1,391,400 KM, the trigonometry tells us that the angle of observation from the observer to the edges of the sun is 0.533°. (Also an isosceles triangle.) What this tells us is that it is indeed possible for the sun to hide the moon using known measurements, because the angle to the moon's edge is greater than the angle to the sun's edge. Now for frame of reference, when the moon is at its average distance from the Earth (384,000 KM), the observable angle to the edges is 0.519°, meaning the angle is too small to hide the sun in totality. Anyways, back to your 70 mile (112 KM) umbra. Those .016° of separation between the sun's and moon's angles at a distance of 363,000 KM calculates to 101 KM (63 miles) - which is awfully close to the 112 KM value, and just tells us the distance between the Earth and moon was a little greater than 363,000 KM at the time of the eclipse.

There's your math.


How many hundreds of post will you have to make before you understand that he doesnt give a fuck about any math you or anyone else does unless it fits his troll narrative? Fucking hell man.

This post was edited by Hacintosh on Sep 10 2017 07:12pm
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Sep 10 2017 07:13pm
Quote (Hacintosh @ Sep 10 2017 08:11pm)
How many hundreds of post will you have to make before you understand that he doesnt give a fuck about any math you or anyone else does unless it fits his troll narrative? Fucking hell man.


Need a hug brah?
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