theo102

Yes, a sunset on a beach and some birds. I can see the error of my ways now.

Because you pulling thirty degrees out of your arse beats describing flat earth astronomy using trigonometry, right? arctan height/distance = angle above horizon on a that earth, using your height and half the approximate circumference of the earth as the diameter.

The key symbol relating to Egypt and Christianity is the Staff of Asclepius. It relates to the crucifixion (John 3:14) and to the contest between Moses and the Egyptian magicians. This in turn goes back to the Abrahamic covenant and the role of blood sacrifice. In the Christian context the serpent represents the murderous priesthood.

Edom = "red", Edom = Rome, Edom = Esau And I hated Esau, and laid his mountains and his heritage waste for the dragons of the wilderness. Malachi 1:3

How does that relate to your thirty degrees?

Point a and point b are imaginary, they don't have any connection to reality. In reality the sun sun disappears from view when it crosses the actual horizon, not an imaginary one. Of course this can't happen on a flat earth because the horizon is the same for all observers and it's always daytime somewhere.

It wasn't what they thought it was. Awake, O sword, against my shepherd, and against the man that is my fellow, saith YHWH of hosts: smite the shepherd, and the sheep shall be scattered: and I will turn mine hand upon the little ones. Zechariah 13:7

What 30 degree cutoff?

No, when I posted the picture I said that the observer was at the equator. The sun orbiting around the equator is a simple example, if you think that a different orbit would make a difference then how would you describe that orbit? If you can't describe a different orbit then the simple conclusion is the earth is not flat because the sun is not above the horizon at midnight like the model says that it should be.

I use the earth's circumference of approx 25000 miles as the diameter of the flat earth. In my diagram the horizontal distance is half that. When you're calculating the sunrise and sunset distance you can use Pythagoras' Theorem to get the horizontal distance as the diagonal of a square with a side length of one quarter the diameter. 8839^2 = (25000/4)^2 + (25000/4)^2 The visual limit for an observer in a boat on a calm ocean is the simplest horizon to use, there's no 30 degrees to complicate things in that case.

OK, let's use your sun height. h/d = 3800/12500, so arctan h/d = 16.9 degrees above the horizon at midnight. At sunrise and sunset h/d = 3800/8839 -> 23.3 degrees above the horizon. Sphere heads for the win. Can any flat earther explain how the math is wrong, or how the sun can be around seventeen degrees above the horizon at midnight?

Using the FE wiki sun height of 5000 miles and 25000 miles for the diameter (h/d=5000/12500) you get a minimum sun elevation of twenty degrees (at midnight). At sunrise & sunset h/d = 5000/8839 which is 29.5 degrees like you said.

You can calculate the minimum angle using the standard flat earth numbers for the height of the sun and the diameter of the earth. In my picture the observer is at the equator, so d is half the diameter of the earth. angle α = arctan h/d

http://thelivingmoon.com/47john_lear/02files/Neutral_Point.html L1 wasn't where it was predicted to be. If the difference was due to non-sphericity of the Earth or the moon then L1 would change over time.

You don't get a choice. The only difference is that you can't deal with the facts that don't fit into your model of the world.