#### Mysfit

##### Wiki - Sun
« on: September 29, 2018, 09:15:44 PM »
Hello,
I will now look at sorting the Sun page

Original - "The sun is a rotating sphere. It has a diameter of 32 miles and is located approximately 3000 miles above the surface of the earth."

Proposed - The sun is a rotating sphere of super-heated gas (called plasma) which provides warmth and light to the earth. Through some simple mathematics (detailed below), we can show that the sun is approximately 32 miles in diameter and 3000 miles above the surface of the earth.

"The Sun's area of light is limited to an elliptic area of light upon the earth much like the light of a lighthouse is limited to a finite area around it."
I believe a lighthouse focuses light in a cone with mirrors, I am unsure how this is done with a sphere. I looked to my light-bulb for reference (not a sphere but pretty close) and it appears to light the room evenly, although a bit brighter near the ceiling. So, the sun is just illuminating the closer part of the earth more vibrantly or... is not a sphere?

"The rotating light on a lighthouse does not propagate infinitely into the distance. This means that only certain portions of the Earth are lightened at a time."
Oh, my light-bulb thing was spot-on. I was worrying I was getting the wrong end of the stick. So the light is dissipating, but how does a sphere do that so harshly?

"It also describes how night and day arise on a Flat Earth."
No, this explains why some areas are light for a cone, but the movement was not explained. A picture demonstrates this later for the lighthouse cone sphere sun(I am getting hung-up on that, sorry).

"The apparent view of rising and setting are caused by perspective, just as a flock of birds overhead will descend into the horizon as they fly into the distance."
I know smarter folks are working this through on the forums and I am pretty certain that this is not right. I don't recall the sinking ship effect being connected to the sun, which should still be 3000 miles up. I will definitely need help with that.

After that, there is a picture of how the above would work for a mono-pole placeholder for the flat earth map.
I am uncertain the current Flat Earth Society rely on this model, so I am confused as to why it is treated as a representation of the earth. Could this be substituted with the bi-polar one?

I need help with the spherical, but cone sun and some more info on sunsets/sunrises.

#### MegaMan2005

• 19
• Hello, I’m am a Flat Earther.
##### Re: Wiki - Sun
« Reply #1 on: October 04, 2018, 06:45:22 PM »

[/quote]Check the wiki, you may be interested on what you find/discover, or just ask a Moderator or more experienced person in the Flat Earth physics field. Directly, like Email or Messaging.
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#### Mysfit

##### Re: Wiki - Sun
« Reply #2 on: October 04, 2018, 09:29:39 PM »
I have checked the wiki, the point is to make the wiki clearer to prevent repeated questions. I don’t like the repetition in the fora. It’s unhelpful.

As for why I put it here, I think Tom or Pete suggested putting an article here is how to help.
I don’t want to bother folks by just messaging them.

If you can help with the lightbulb problem, it would be much appreciated.

I think the lighthouse simile is the main problem, though. If the wiki could point to a similar everyday object that acts like the sun, folks would get it in a snap.

#### shootingstar

##### Re: Wiki - Sun
« Reply #3 on: January 12, 2019, 11:18:06 PM »
Without actually seeing the light bulb it is hard to comment about that. Are you talking about a light bulb hanging from the ceiling? I'm assuming there is no lamp shade surrounding the bulb. If it is a ceiling mounted light bulb then it will brighter on the ceiling.

You are right about the Sun being a rotating sphere of light. It does emit light evenly all over its surface area. This gives us hints about the internal structure of the Sun. There is no directed light in any particular direction such as you see from a lighthouse. The lightbulb in a light house is directed by using a combination of mirrors and lenses just as the bulb in a car headlight is in order to direct more of the light into the direction it is needed.  Directing light from the Sun into a certain direction is impossible for reasons that I can but hope I don't need to explain.

Light is a form of energy and so (as you probably know) is measured in Watts (Joules per unit time). The total wattage of the Sun is 3.84x10^26 watts and if it was only 3000 miles away as FE Wiki seems to suggest then life on Earth, indeed the Earth itself would not exist as it would have been frazzled before it could even form. It shines with an apparent or visual magnitude of -26.7 making it (obviously) the brightest object in the sky by a long shot.  If we could observe it from a distance of 10pc or 32.6 light years (the standard distance for intrinsic luminosity measurement in all stars) then it would shine at a feeble magnitude of +4.8 meaning it would be barely visible to the naked eye. By using the distance modulus (m-M = 5log10D -5) where D is the distance in parsecs (1 parsec = 3.26 light years) we can confirm its distance of 150 million km since we know it s apparent magnitude (m) and absolute magnitude (M).

If you set D in the above equation to 10, then log10(10) = 1 so you would have (5 x 1) -5 which = 0. Therefore m must equal M.
« Last Edit: January 18, 2019, 08:12:39 AM by shootingstar »

#### Mysfit

##### Re: Wiki - Sun
« Reply #4 on: March 03, 2019, 12:08:23 AM »
Sorry, yes. My lightbulb is unshaded, it is also LED. It can be very bright and only seems to be a tiny bit dimmer directly above it.
I am not worrying about the necessary energy output of the sun atm, as I may look into some alternative physics for that. Sandokahn (sorry if spelled wrong) is bound to have something.
The crux of the query was to find an alternative for the lighthouse simile, as it confuses the ‘eck out of me and probly does so for others too.

#### spherical

• 214
##### Re: Wiki - Sun
« Reply #5 on: May 10, 2019, 04:27:25 PM »
Just some numbers to help the calculations:
The solar energy perpendicular that hits Earth's ground is 1kW/m².
It is also calculated the average energy all lands receive under direct sunlight, due spherical inclination, to be around 300W/m².
The FE has no steep angles like a sphere has, so lets increase this power to 500W/m² on FE as a logical guess.

If you calculate the average FE area of the FE disc that actually receives solar radiation, you can calculate the solar energy radiated from that particular steradian of the Sun, extrapolate for the whole solar sphere radiance energy.

If the covered FE area is "A m²" and the average energy measured on the ground is 500W/m² then the energy necessary to cover such area is "A x 500W".  Considering the Sun a sphere radiating energy all around, you can calculate the exact angle from the FE Sun that covers the stated FE area.

As this area is larger than the 30km (diameter of the FE Sun), more than half of the sphere will be projecting radiation down to the FE area, but on the steep angles the radiation is angled and reduced, so, to facilitate the calculations, lets assume just 1/3 of the FE Sun's "visible disc" is illuminating the FE area. It would represent 1/6 from the whole sphere producing energy hits FE.

I am being good here, since the Isosceles Triangle formed between FE diameter (40000km) and Sun's height (3000km) represents the illumination "cone"  that hits FE with 162° of aperture.

It means we could assume the FE Sun produces 6 times more energy that what FE receives, so the FE Sun could produces "6 x A x 500W".

Due the extreme difficult to understand and calculate the illuminated area over FE, considering now, May 10 Central Florida, I can see the Sun from 6:38am to 8:06pm, it is 13h 28min of solar exposure.  If 12 hours of sun represents 180° of solar coverage over the equator line (15°/h) with an average power of 500W/m², then 13:28h would represent 193° of a circular line around FE North Pole.  That great semicircle means the Sun is illuminating the FE from East to West, reaching the FE Ice Wall, and obviously the FE North Pole.

As only FErs can calculate that, I will only guess a possible 40% of the total FE disc being illuminated by the solar light.

As the distance from North Pole to equator is 10,000 km, the FE disc radius would be 20,000km, what means a total area of (PI*R²) 1256636000 km² or 1.25 E+9 km² or 1.25 E+15 m².   So, 40% of that is 5 E+14 m², applying the formula 6 x A x 500W, result in total Sun's energy radiated as 6 x 5 E+14 x 500W = 15000 E+14 = 1.5 E+18 W , or, 1500 PetaWatt.

Based on the same calculations, A x 500W would be the energy FE receives directly from the Sun, 5 E+14 x 500W = 2500 E+14 = 2.5 E+17 or 250 PetaWatts.

On RE measurements, we know the Sun pours 174 PetaWatts directly over our planet, 30% of it is reflected back to the universe, the 70% rest is absorbed by clouds, oceans and land.  It means around 122 PetaWatts (122 E+15 W).   With all the guessing, my calculations above results in 250 PW, what is just double the actual RE calculations.

See, to adjust the solar radiation over FE from 250PW to 122PW, we need to reduce the land area illuminated by the Sun, from 40% to 20% in the calculations above.  But with only 20% of area, it would be impossible for the Sun to cover 193° over the circle at 38°N, I would NOT be seeing the Sun raise at 6:38am and set at 8:06pm on Central Florida on May 10, the day light would short than that.

Somebody must explain this to me.