The Sun

I am really trying to believe in the flat earth theory, some things i have difficulty with.
The sun is one.
According to the FAQ
Q: "Please explain sunrises/sunsets."

A: It's a perspective effect. Really, the sun is just getting farther away; it looks like it's disappearing because everything gets smaller, and eventually disappears as it gets farther away. 

I know the earth is accelerating upward, and the sun is accelerating at the same speed. Does the sun also speed up and then slow down? I live on a lake and get the awesome view of the sunrise most mornings (The sun doesnt always rise due east for a reason i dont iknow of yet).

So the Sun speeds up and slows down and it also moves back and forth over the earth and it also shuts off its beam of light a lot?

I really think we need to rethink this. There must be a better explanation for the sunrises. Every morning I can see it rise from the side of our disc planet. 

Wouldnt the sun going underneath our disc be a better idea?  And then going underneath and coming back the other side.
I know this dismissed the constant upward theory which i believe.

Really confused.


thank you

Wouldn't work.  If the whole world was night at the same time, then yes, it would be hiding under the disk and even I would have to agree that the Earth was flat.  But it doesn't.  Specifically it lights half the planet at any given time.  One half is ALWAYS lit. 

I am really trying to believe in the flat earth theory, some things i have difficulty with.

Pray tell, why?

But it doesn't.  Specifically it lights half the planet at any given time.  One half is ALWAYS lit.  

Especially since, being the way daylight work, a spotlight cannot explain days on a flat disc.


According to projection of RE days to a flat Earth, days would have to look sort of like:



While not exactly like that, it would be similar.



But according to FE, the sun only does this:

And the only guy to try and make a 3D model of the sun projecting onto a flat disk has failed to do so.  I'm not even sure he's still around.

Yes someone tell me how the sun is a spotlight. A lightbulb doesn't become a spotlight until there is something focusing the light. Unless I'm mistaking, I don't think its a FE position that the sun is in a giant housing that blocks light on the other side.

Yes someone tell me how the sun is a spotlight. A lightbulb doesn't become a spotlight until there is something focusing the light. Unless I'm mistaking, I don't think its a FE position that the sun is in a giant housing that blocks light on the other side.

Think of it as a laser pointer? Thats the first thing I thought of when I read about the sun in the Q&A....laser pointers are great at focusing light into a dot, or large circle if far enough away. Maybe the sun has the same focusing features going on....Please pardon my ignorance 

Quote from: Taylor on April 17, 2010, 07:38:10 AM

Yes someone tell me how the sun is a spotlight. A lightbulb doesn't become a spotlight until there is something focusing the light. Unless I'm mistaking, I don't think its a FE position that the sun is in a giant housing that blocks light on the other side.

Think of it as a laser pointer? Thats the first thing I thought of when I read about the sun in the Q&A....laser pointers are great at focusing light into a dot, or large circle if far enough away. Maybe the sun has the same focusing features going on....Please pardon my ignorance 

A laser pointer is just a highly focused beam of light. They still require lenses and reflectors, hence your example leaves us in the exact same position.

I knocked this one down in another thread.  Their FAQ says the Sun is 32 miles in diameter and is 3000 miles above the Earth.  Well......

OK, 32 miles across and 3000 miles away.  A ratio of 4/375.  You could also as that the distance is 93.75 times greater than the size of the object.  Let's apply this ratio to something else.  Hmmm, I know!  A frisbee.  A regulation frisbee, used in sports, measures 10.75 inches across.  Now, let's apply the ratio.  10.75x93.75=1007.8125  That means to observe the frisbee at roughly the same distance FE'ers say we observe the sun and moon, you would need to be 1007.8125 inches away, or 83.98 ft.  Now 3000 miles equals 15,840,000 ft.  So that means our scale is 83.98/15840000, or 188,617.337:1.  Now, lets apply that to the surface area of the "flat Earth".  Using A=3.14*r2, we can figure out that the area of the flat Earth is 65,563,985,000 square miles.  We reduce that number with our ratio, and we get 347,603.174 square miles.  That's a circle measuring 664 miles across.

Now, you mean to tell me a light the size of a frisbee could even think to illuminate even part of the state of Texas form only 83.98 feet away?  Have fun with that one, guys.

I hope that puts their "spotlight theory" into some better perspective.

They'll just ignore you. Whenever we disprove part of the theory using hard science or mathematics they just act as if it never happened.

http://www.theflatearthsociety.org/tiki/tiki-index.php?page=The+Sun

You dodged my question.  How exactly can an object the size of a frisbee illuminate roughly half the state of Texas from an altitude of less than 90 feet?

Quote from: Tom Bishop on April 19, 2010, 06:28:29 AM

http://www.theflatearthsociety.org/tiki/tiki-index.php?page=The+Sun

You dodged my question.  How exactly can an object the size of a frisbee illuminate roughly half the state of Texas from an altitude of less than 90 feet?

Not to mention we'd all freeze to death.


How do you explain the odd shape that daylight makes on your FE map Tom?

I'm going to put this in a little better perspective.  We're going to use a U.S. quarter for our sun.  Changes are in red.

OK, 32 miles across and 3000 miles away.  A ratio of 4/375.  You could also as that the distance is 93.75 times greater than the size of the object.  Let's apply this ratio to something else.  Hmmm, I know!  A quarter.  A quarter measures .96875 inches across.  Now, let's apply the ratio.  .96875x93.75=90.8203125  That means to observe the quarter at roughly the same distance FE'ers say we observe the sun and moon, you would need to be 90.8203125 inches away, or 7.57 ft.  Now 3000 miles equals 15,840,000 ft.  So that means our scale is 7.57/15840000, or 2,092,470.28:1.  Now, lets apply that to the surface area of the "flat Earth".  Using A=3.14*r2, we can figure out that the area of the flat Earth is 65,563,985,000 square miles.  We reduce that number with our ratio, and we get 31,333.293 square miles.  That's a circle measuring 99.89 miles across.

Now, you mean to tell me a light the size of a quarter could even think to illuminate even part of the state of South Carolina (which is still a little small at only 30,109.47 square miles) form only 7.57 feet away?  Have fun with that one, guys.

With these new calculations something occurred to me.  At a height of 7.57 ft, anyone can attempt this experiment.  Go outside at night.  Measure out a height of 7.57 feet.  Now, hold the brightest god damn light you can find at that height and see just how much ground you can light up.  Hell, I won't even make you use a spot light, use what ever light you want.  You don't even have to make it the size of a quarter, which would really be needed for accurate testing, but oh well.  Now, have your lab partner go until the light your holding no longer illuminates the ground.  I bet he won't make it very far.  But for your sun theory to even come close to holding water, he's need to make it almost 50 miles since roughly half the Earth is in daylight at any given time.  Next time you want to try disproving something, Mr. Bishop, try doing it with more than a link to something I've already torn down.

http://www.theflatearthsociety.org/tiki/tiki-index.php?page=The+Sun

From the link referenced by the wiki:

I don't like the way the sun shines, but it 's very difficult to get a small lightsource so close to ground to illuminate half of the Earth's area properly.

I don't think that this is helping your argument much.

Quote from: http://www.theflatearthsociety.org/forum/index.php?topic=1264.msg8922#msg8922

I don't like the way the sun shines, but it 's very difficult to get a small light source so close to ground to illuminate half of the Earth's area properly.

I don't think that this is helping your argument much.

I totally missed that!  HaHa!  That's exactly my point.  I love this game!

We should get together and modify that rendering/make a new one, of the actual FE and Sun.

We should get together and modify that rendering/make a new one, of the actual FE and Sun.

I'm no good with rendering programs.  But could do some real world experiments.

I knocked this one down in another thread.  Their FAQ says the Sun is 32 miles in diameter and is 3000 miles above the Earth.  Well......

Quote from: Sliver

OK, 32 miles across and 3000 miles away.  A ratio of 4/375.  You could also as that the distance is 93.75 times greater than the size of the object.  Let's apply this ratio to something else.  Hmmm, I know!  A frisbee.  A regulation frisbee, used in sports, measures 10.75 inches across.  Now, let's apply the ratio.  10.75x93.75=1007.8125  That means to observe the frisbee at roughly the same distance FE'ers say we observe the sun and moon, you would need to be 1007.8125 inches away, or 83.98 ft.  Now 3000 miles equals 15,840,000 ft.  So that means our scale is 83.98/15840000, or 188,617.337:1.  Now, lets apply that to the surface area of the "flat Earth".  Using A=3.14*r2, we can figure out that the area of the flat Earth is 65,563,985,000 square miles.  We reduce that number with our ratio, and we get 347,603.174 square miles.  That's a circle measuring 664 miles across.

Now, you mean to tell me a light the size of a frisbee could even think to illuminate even part of the state of Texas form only 83.98 feet away?  Have fun with that one, guys.

I hope that puts their "spotlight theory" into some better perspective.

This mathematics is wrong. You've applied a linear ratio to a measurement of area, which is incorrect. If you double the side length of a square, the area multiplies by four, not by two. You need to square the ratio you calculated for the distance from the Earth to the Sun before applying it to the surface area of the Earth.

Quote from: Sliver on April 18, 2010, 09:17:43 PM

I knocked this one down in another thread.  Their FAQ says the Sun is 32 miles in diameter and is 3000 miles above the Earth.  Well......

Quote from: Sliver

OK, 32 miles across and 3000 miles away.  A ratio of 4/375.  You could also as that the distance is 93.75 times greater than the size of the object.  Let's apply this ratio to something else.  Hmmm, I know!  A frisbee.  A regulation frisbee, used in sports, measures 10.75 inches across.  Now, let's apply the ratio.  10.75x93.75=1007.8125  That means to observe the frisbee at roughly the same distance FE'ers say we observe the sun and moon, you would need to be 1007.8125 inches away, or 83.98 ft.  Now 3000 miles equals 15,840,000 ft.  So that means our scale is 83.98/15840000, or 188,617.337:1.  Now, lets apply that to the surface area of the "flat Earth".  Using A=3.14*r2, we can figure out that the area of the flat Earth is 65,563,985,000 square miles.  We reduce that number with our ratio, and we get 347,603.174 square miles.  That's a circle measuring 664 miles across.

Now, you mean to tell me a light the size of a frisbee could even think to illuminate even part of the state of Texas form only 83.98 feet away?  Have fun with that one, guys.

I hope that puts their "spotlight theory" into some better perspective.

This mathematics is wrong. You've applied a linear ratio to a measurement of area, which is incorrect. If you double the side length of a square, the area multiplies by four, not by two. You need to square the ratio you calculated for the distance from the Earth to the Sun before applying it to the surface area of the Earth.

Isn't that automatically compensated for by the distance between earth and sun doubling causing a squared drop off of light intensity? (inverse square law)

Isn't that automatically compensated for by the distance between earth and sun doubling causing a squared drop off of light intensity? (inverse square law)

No.

Quote from: Sliver on April 18, 2010, 09:17:43 PM

I knocked this one down in another thread.  Their FAQ says the Sun is 32 miles in diameter and is 3000 miles above the Earth.  Well......

Quote from: Sliver

OK, 32 miles across and 3000 miles away.  A ratio of 4/375.  You could also as that the distance is 93.75 times greater than the size of the object.  Let's apply this ratio to something else.  Hmmm, I know!  A frisbee.  A regulation frisbee, used in sports, measures 10.75 inches across.  Now, let's apply the ratio.  10.75x93.75=1007.8125  That means to observe the frisbee at roughly the same distance FE'ers say we observe the sun and moon, you would need to be 1007.8125 inches away, or 83.98 ft.  Now 3000 miles equals 15,840,000 ft.  So that means our scale is 83.98/15840000, or 188,617.337:1.  Now, lets apply that to the surface area of the "flat Earth".  Using A=3.14*r2, we can figure out that the area of the flat Earth is 65,563,985,000 square miles.  We reduce that number with our ratio, and we get 347,603.174 square miles.  That's a circle measuring 664 miles across.

Now, you mean to tell me a light the size of a frisbee could even think to illuminate even part of the state of Texas form only 83.98 feet away?  Have fun with that one, guys.

I hope that puts their "spotlight theory" into some better perspective.

This mathematics is wrong. You've applied a linear ratio to a measurement of area, which is incorrect. If you double the side length of a square, the area multiplies by four, not by two. You need to square the ratio you calculated for the distance from the Earth to the Sun before applying it to the surface area of the Earth.

You're right.  It dawned on me earlier today, so here are the new calculations for everyone.

You claim the Sun is 32 miles across, and that the Earth is 24,900 miles across.  That means the Earth is 778.125 times larger.  Now, if we scale down the Sun to something a little more manageable for out little experiment, let's say, the size of a U.S. Quarter.  The quarter measures 31/32 in across or .96875 inches across.  Now, let's use our multiplier of 778.125 to figure out the size of our scale Earth.  We get 753.809 inches, or for easier reference 62.817 ft.  Now that we have our diameters properly scaled down, again, sorry for the mistake earlier, we can figure out our surface areas.  The quarter's would be A=3.14*0.484375^2, which equals .7367 square inches, and the scaled down Earth would be A=3.14*31.4085^2, which equals 3097.59 square ft.

Now my math on the distance scale was correct, so we're keeping the 7.57 ft from an earlier post.  So, for this experiment, you need a light source no bigger that a quarter to illuminate at least half of a circle measuring 62.817 ft across.  Go try it and get back to me.  Please be sure video tape it so you can prove it's possible.

Now my math on the distance scale was correct, so we're keeping the 7.57 ft from an earlier post.  So, for this experiment, you need a light source no bigger that a quarter to illuminate at least half of a circle measuring 62.817 ft across.  Go try it and get back to me.  Please be sure video tape it so you can prove it's possible.

Not only that, it has to light it up so fucked-up retarded that it looks like the day / night cycle.

Now my math on the distance scale was correct, so we're keeping the 7.57 ft from an earlier post.  So, for this experiment, you need a light source no bigger that a quarter to illuminate at least half of a circle measuring 62.817 ft across.  Go try it and get back to me.  Please be sure video tape it so you can prove it's possible.

I would, but I don't feel the need to convert to standard, manageable units if you can't present the data in such a format. I'm also not going to bother trying to confirm your numerical manipulations, given the unnecessary complexity introduced by your choice of unit system.

I would, but I don't feel the need to convert to standard, manageable units if you can't present the data in such a format. I'm also not going to bother trying to confirm your numerical manipulations, given the unnecessary complexity introduced by your choice of unit system.

You being a lazy ass doesn't discredit his statement.

If you require such spoon feeding:

Now my math on the distance scale was correct, so we're keeping the 2.31m from an earlier post.  So, for this experiment, you need a light source no bigger that a quarter (0.243 m in diameter) to illuminate at least half of a circle measuring 19.15m across.  Go try it and get back to me.  Please be sure video tape it so you can prove it's possible.

Quote from: Tom Bishop on April 19, 2010, 06:28:29 AM

http://www.theflatearthsociety.org/tiki/tiki-index.php?page=The+Sun

You dodged my question.  How exactly can an object the size of a frisbee illuminate roughly half the state of Texas from an altitude of less than 90 feet?

If you can produce a frisbee in which nuclear fusion occurs, you can test that.

Quote from: Sliver on April 19, 2010, 06:44:28 AM

Quote from: Tom Bishop on April 19, 2010, 06:28:29 AM

http://www.theflatearthsociety.org/tiki/tiki-index.php?page=The+Sun

You dodged my question.  How exactly can an object the size of a frisbee illuminate roughly half the state of Texas from an altitude of less than 90 feet?

If you can produce a frisbee in which nuclear fusion occurs, you can test that.

Scientists can't artificially create nuclear fusion because it is so unstable and requires so much fuel to initiate. That fact alone discredits the FE sun.

Scientists can't artificially create nuclear fusion because it is so unstable and requires so much fuel to initiate. That fact alone discredits the FE sun.

Actually, it only discredits the RE sun, which is the only one (to my knowledge) which is claimed to operate through nuclear fusion.

Quote from: 2fst4u on April 20, 2010, 02:40:41 AM

Scientists can't artificially create nuclear fusion because it is so unstable and requires so much fuel to initiate. That fact alone discredits the FE sun.

Actually, it only discredits the RE sun, which is the only one (to my knowledge) which is claimed to operate through nuclear fusion.

How does it discredit the RE sun? The RE sun (Well, RE stars) are all hueg and have had massive explosions and energy start them all. Artificial nuclear fusion is so hard because it is so unstable and requires a lot of energy to start it. RE sun has both of these.

EDIT: Where have you seen that the sun is the only star to use fusion?

How does it discredit the RE sun?

Because the RE sun relies on a process that has never been experimentally validated.