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The see the earth is round? No need to, we already know that.
This isn't fake
It proves that the Earth is flat. It is impossible to see the moon 70 miles above Nevada, when it is across from Australia, on a Globe. It is only possible, if it the earth is flat, because as you go higher you begin to see both of them, which are inside the firmament, the greater light and the lesser light.
Yes, the timezone that Australia (the part the moon is over) is in 18 time zones to the East of Nevada, but it's only 6 time zones to the West. When the rocket is facing the moon, the camera is facing West, which is why it can see it and why there's no hint of the sun's location in the picture, as the sun at 9:30 AM is still significantly in an eastward direction.
So, I'm pretty sure the OP is a troll but, if some of you may have gathered, I like geometry. So I'll explain this anyway.
TX_Matt, I'm not sure the timezones matter. The shortest distance from Australia to Nevada is about 8300 miles (westward over the Pacific). This implies that if the moon is directly overhead in Australia, a rocket in Nevada needs to be about 672 miles above the surface to see the moon.
Here S = 8300 miles (the arclength between Nevada and Australia) and S = S1 + S2. D = 238,000 miles (distance from Earth to moon) and R = 3959 miles (radius of Earth). The goal is to find H, the height of the rocket at which point it will see the moon.
Trigonometry tells us via the arclength formula:
and therefore
(1)
and from Pythagoras and trig:
Plug in to Equation (1) and you get
Now, from Pythagoras we also know that
and therefore
.
As per usual, this DOES NOT at all prove the Earth is flat.
Why?
Because the moon is not directly overhead in Australia.
On July 14, 2014, the moon would have been about 60 degrees above the horizon at 3 am. This changes the geometry considerably as shown in this diagram:
The geometry and maths get ridiculously complex at this point and I haven't had time to work it out. Here, α = 60 degrees is the angle the moon is above the horizon. Just by looking at the diagram, it is obvious that H is significantly smaller when the moon is not directly overhead in Australia. Here S =S1+S2+S3.
The stupidest thing about all the flat Earth arguments is they always neglect some obvious fact of reality. In this case, they assume the moon is directly overhead in Australia, which it is not.
Cheers,
LRLRL
****Edit: I have solved the geometry problem in the above diagram. Props to anyone who actually reads this*****
From the above diagram, we first know that
which is the distance to the moon. Substituting variables in using trigonometric relationships:
.
Now, we also know that the sum of the angles in a polygon must equal 360 degrees (or 2*pi radians) and therefore:
.
We also know two other facts about the angles:
So, using the above angle relationships and the arc length formula, we can arrive at an equation for D called Equation (A):
The final thing we need is the relationship for H which we will call Equation (B):
.
Now, unfortunately, using Equation (A) and Equation (B), we can not solve for H analytically. We will have to solve it numerically. I have taken values for H from 0.1 miles to 1000 miles in 0.1 mile increments and solved for D. Then I found the value for H which resulted in D being closest to 238,000 miles.
The final result?
.
Crazy, right? In this model, you would only need to go up about 1.4 miles in order to see the moon on the horizon on July 14, 2014 at 9:30 am in Nevada. This actually makes sense too because the moonset for that day in Las Vegas was at 8:16 AM. This means that at 9:30 AM, the moon would have just set about an hour prior so would only be just below the horizon.
I'm actually shocked at how well my simple geometric model is confirmed by the observations. It also explains how the moon can be so far above the horizon for a rocket which is 73 miles up. I thought there would be a variety of errors because I was assuming a 2D diagram and did not account for the 3D nature of the problem.
***End Edits: Props if you read all that***
Why don't you show me a picture of this "curvature".
I beg your pardon?You're deceiving yourself. Satan has his grip on your neck.
What's this then:Max said:There's only one firmament. That's why I emphasized "The" firmament.
CF's own creation.
I beg your pardon?What's this then:
2 Corinthians 12:2 I knew a man in Christ above fourteen years ago, (whether in the body, I cannot tell; or whether out of the body, I cannot tell: God knowethsuch an one caught up to the third heaven.
1. Fish eye lens
2. Photoshopped
3. Just a picture from Earth, with colors edited.
4. Completely fake.
You have no counter arguments for my and the other guy's arguments for the video? Noted.
It depends on which vector you are talking about. Rotation, revolution, the spin of the galaxy, the galaxcy moving through space.
Does anyone actually believe we're moving right now?
1.5 million miles per day?
Can you see, feel, hear, and sense, of prove, in ANY sort of way that we are moving?.....
I don't want a smart science answer, but have you actually WITNESSED in ANY way to movement of the Earth.
I don't understand where you're coming from, Max.1. Fish eye lens
2. Photoshopped
3. Just a picture from Earth, with colors edited.
4. Completely fake.
You're a "Christian" and don't believe the Bible. Noted.
Stop overthinking it. We are not following an orbit right now. We are stationary.
Sorry, but you are going to get one. Galileo's Principle of Relativity. The acceleration is far too small to be felt, so you don't even need Einstein.
Stop underthinking it. Don't just believe what someone tells you the Bible means.
It was sad that it took so much to convince my acquaintance of the truth. I would suggest that we ignore MaxR1996.
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