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