Quote (card_sultan @ Sep 10 2017 09:07pm)
For your angle of observation theory to be correct - you need to assume the sun rays are converging but you state its a fact that sun run rays are parallel.
So that's hypocritical first off. Please show me an experiment where the the source of light, the object casting a shadow and the casted shadow are relatively parallel but the shadow is 10x smaller than the object.
Also you didn't address the fact we dont see earthshine on the moon during the eclipse, in other words - why cant we see the moon before the solar eclipse or after it?
Also you failed to explain why when the sun and moon are directly overhead during the day, the front side of the moon is lit up, same yourself the troube and just admit the helio model is complete failure.
Maybe you can also explain why if we are really seeing a distant moon eclipsing a very distant sun - this event was shown by Nasa podcast as a local event viewable 40 times as it crossed america due to local perspective.
FE Crush.
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Quote (Santara @ Sep 10 2017 08:09pm)
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.
Just think for once. How could you possibly see the left and right edges of the sun if the light doesn't converge?
So no, you once again failed to "FE crush."
I didn't address your other points because we haven't finished this one.