The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Theory => Topic started by: spherical on April 25, 2019, 10:29:10 PM

The flat earth theory states the Earth's equator line is 40000 km in circumference, the diameter of the equator is then 12738 km, and as the equator is a center line between the North pole and the "ice wall" South pole, then the total diameter of the flat earth is 25477 km, making the ice wall of a circumference of 80038 km. Clear? Okay, the theory also states the Sun has the same diameter of the Moon, around 48 km (30 miles), and both are at the altitude of 4800 km (3000 miles).
Okay, based on what we can analyse and decompose the light received from the sun, we can even detect what temperature and what gases are being converted there, Hydrogen to Helium, we can simulate the fusion in computers, even calculate the amount of energy it needs to produce and based on its size, we can calculate how long it is doing that, and how long it probably will continue doing that. But this is based on our normal universe we know, that follows the same laws of physics.
Now, based on flat earth theory, how the 30 miles across disc sun generates radiation? how long it is being doing it? We can measure the sun light power that hits earth surface and measured around 1kW/m², that is a fact. We can add all the energy from the sun that hits planet earth at a single time or second, that is huge and most of it is invisible infrared. Even so, as the Sun radiates all around it, and Earth is very far away from it, we receive just a tinny bitty small part of the total radiation emitted by the Sun.
Lets calculate it using the flat earth theory numbers, and think that all the Sun's radiation is "coned" down to the flat earth.
(https://i.ytimg.com/vi/8x2Fx9BH7vI/hqdefault.jpg)
Radius of flat earth: 25477 km. Area of a disc is = Radius² x PI = 2,038,216,561 km², or 2 x 10^9 km². Clear?
Each km² is a 1000 x 1000 m, so it is 1 million m², or 1 x 10^6 m².
Now, converting 2x10^9km² to m² = 2x10^9 x 1x10^6 = 2x10^15m². Still clear? Math correct, right?
Now consider that your flat earth theory 33% of earth area actually "coned" illuminated by the sun at any given time, divide this big number by 3, ending up with 6x10^14m², now multiply it by 1kW or 1000 Joules, that the sun pours of energy on the illuminated area, 6x10^14m² x 1x10^3W = 6x10^17W, that is beautiful, simple calculations.
Now, do you know how much energy all our hydroelectric plants, nuclear, coal, oil, wind, etc, on the whole planet generates at any given time?
In 2013 everything together produced 5.67x10^20 Joules, that is during a whole year, that has 31536000 seconds, or 3.1x10^7, so divide the 5.67x10^20 by 3.1x10^7 = 1.82x10^13 Watts per second, or Joules. So, the sun, your little disc of only 30 miles in diameter, 3000 miles up, can generate 4 levels of magnitude more energy that ALL our generation on Earth?
The question is HOW a 30 miles flat disc can generate 33 THOUSAND times more energy than ALL our energy production on the planet, and please, answer that with chemical/scientific explanations, so we could create an artificial Sun and solve all our energy problems. See, according to this, an artificial SUN using the same flat earth magic energy generation, to fulfill ALL our energy needs could generate just double of what we are producing now, to do that, it could have an area 16500 times smaller than the 30 miles diameter flat earth sun.
Area = Radius² x PI. A 30 miles (48km) diameter disc has an area of 1.8 x 10^9 m², dividing it by 16500 = 1.1 x 10^5 m² = radius of 186 meters only... Come on, this is smaller than two city block, LETS DO IT RIGHT NOW !!!! Flat earth theory found the solution for our energy needs. Perhaps a small 8 inch disc made of the same flat earth sun material, can fuel an electric car forever, can you imagine? no more fossil fuel engines? and better, using the fantastic levitation techniques of the flat earth sun, we can make cars flying without any propulsion. Fantastic.
But first, please explain how your flat earth sun works, in technical details.
See, I have a small scientific calculator TI36X and can do wonders with it. Some people have tremendous difficult to understand TAN1 or ebased Logs function, it is not their fault, some animals have no competent cognitive capacity, that is their nature.
(https://www.fondriest.com/environmentalmeasurements/wpcontent/uploads/2014/03/par_solarradiation.jpg)
(https://upload.wikimedia.org/wikipedia/commons/thumb/9/9d/Bp_world_energy_consumption_2016.gif/300pxBp_world_energy_consumption_2016.gif)

I hope you realize that the power from the sun isn't actually even.

I hope you realize that the power from the sun isn't actually even.
Of course, radiation burst and solar flares make it radiation fluctuating, this is why we use average measurements.
Anyhow, care to explain how the flat earth sun generates so huge humongous amount of power in so small form factor?
The oblate spheroid planet scientists and engineers wait the answer with great interest, we could solve the world energy crisis.
It would be pathetic to state: "I have a portable flashlight, 10 cm long, weights 200 g, at dark night it can illuminate 10 square miles ahead as clear as Arizona's daylight, but I will not tell you how it does it, the technology involved, nor what kind of power supply it uses, and you must believe me, otherwise you should consider yourself ignorant".
Waiting your answer about the composition and scientific details about the flat earth sun.
"Science can help to take people from ignorance, but even so, sometimes is difficult to take the ignorance from people."

I hope you realize that the power from the sun isn't actually even.
Of course, radiation burst and solar flares make it radiation fluctuating, this is why we use average measurements.
No, I mean it's not even across the Earth's surface. Sunlight coming at an angle is less radiance than sunlight from directly overhead.

I hope you realize that the power from the sun isn't actually even.
Of course, radiation burst and solar flares make it radiation fluctuating, this is why we use average measurements.
No, I mean it's not even across the Earth's surface. Sunlight coming at an angle is less radiance than sunlight from directly overhead.
This is exactly why we use "average" of 1kW/m². The same average I used to calculate the energy within the cone from the flat earth sun.

I hope you realize that the power from the sun isn't actually even.
Of course, radiation burst and solar flares make it radiation fluctuating, this is why we use average measurements.
No, I mean it's not even across the Earth's surface. Sunlight coming at an angle is less radiance than sunlight from directly overhead.
This is exactly why we use "average" of 1kW/m². The same average I used to calculate the energy within the cone from the flat earth sun.
That average was probably calculated based on spherical area. You can't use it as a reference for flat area.