lunar eclipse, round earth explanation

Well I was trying to say that the direct line from the sun to the moon would be going through more atmosphere, therefore the longer wavelength light would be more prevalent since the shorter wavelength gets scattered.  Violet, Blue and Green colors being among the shorter wavelengths would naturally not all make the trip there.  The more light making it to the moon the more wavelengths make it so it moves what you see reflected back as more toward the white area (meaning all colors).  So less "stuff" that can scatter the light means more of the wavelengths can make it through, more stuff, shorter wavelengths get scattered and only the longer ones make it.  So as the moon gets more and more light blocked because of the Earth's shadow and atmosphere the more towards red the moon appears.  So at the beginning of the lunar eclipse the moon moves from appearing white to yellow to orange to red then if it gets completely into the shadow it goes to black. 
I'm not sure what you are talking about with the blue moon thing since saying once in a blue moon doesn't refer to the color the moon appears but to a second full moon in a calendar month.  It only happens somewhat rarely, hence the once in a blue moon figure of speech (on average, close to but less than, once per year).  It never referred to the color, it was once called the betrayer moon.  The term originally was belewe moon.  Belewe originally meant to betray.

Which part of 'to the sides' you have trouble understanding?

as i've just said, "describing something does not make it realistic. granting the ideas of scattering and multiple wavelengths (let's not go there), there is a huge difference between a sunset and a lunar eclipse. for one, double the distance would be covered: we should not see any yellow or orange.

I already explained that the amount of atmosphere that light passes through during sunset and during an eclipse is the same.

secondly, the light reaches it from outside the atmosphere. if we look at the tyndal effect, the wavelengths that reach it should be the ones that scatter.

No, because these wavelengths completely scatter before all the light has passed through the atmosphere, and hit the part of the Earth that is facing the sun at that time creating a blue sky during the day.

for that matter, that's the behavior we'd expect to see the rest of the time: the moon should be blue.

No, because the light has already scattered before it passes through all of the atmosphere.
 
 

the fact 'a blue moon' means a rare occurence says all i need to.

Blue moons are caused by volcanic ash being made of particles that blue can filter through. It has nothing to do with lunar eclipses.

the sun's original path would not reach the moon, i've already shown that: the moon is hidden behind the earth. you rely on light being bent and scattered in the 'atmosphere'.

Which can be demonstrated to be true.

it should be blue."

No, because all the blue light has been scattered by the atmosphere before all the light passes through again, (sorry if I'm repeating myself) This light is scattered toward wherever daytime is, creating a blue sky.

the moon is not in a straight path from the sun. look back a few pages, i provided a diagram that showed that: and while there were some minor mathematical errors, alpha corrected them and his answer made the fact more apparent. for light to reach the moon, it needs to come out the 'sides'.

Which can be proven to occur.

it should be blue."

Perhaps you could provide a diagram that shows this.  Afterall, "describing something does not make it realistic."

i have done the math, and you are completely wrong. explanation is forthcoming. the light from the sun could not reach the moon in any capacity.

I usually try to refrain from making any personal remarks about the things that some flat earthers post. But I don't always refrain from doing so and this is just one time where I will ask "How in the world could you make such a dumb statement as that ?"

Or are you just that dumb on the subject of astronomy ? How in the world could you make that statement ? What is the basis of that ? Of course the light from the sun reaches the moon. Did you ever hear of a thing called "Moonlight" ?  It's the light from the moon, which in turn is reflected light from the sun being reflected back to earth.

Can you please give us some idea for this insane idea of yours. Or are you just a troll or something else ?

Your statement does say "The light from the sun could not reach the moon in any capacity" Does "any capacity" include times other than eclipses when the earth is between the sun and the moon ? Maybe I just mis-read your statement. It is true that light from the sun does not reach the moon during a New Moon. Is that "the capacity" of which you refer ? Otherwise some light from the sun always reaches the moon.

Quote from: JRoweSkeptic on April 05, 2015, 03:54:13 PM

i have done the math, and you are completely wrong. explanation is forthcoming. the light from the sun could not reach the moon in any capacity.

I usually try to refrain from making any personal remarks about the things that some flat earthers post. But I don't always refrain from doing so and this is just one time where I will ask "How in the world could you make such a dumb statement as that ?"

Or are you just that dumb on the subject of astronomy ? How in the world could you make that statement ? What is the basis of that ? Of course the light from the sun reaches the moon. Did you ever hear of a thing called "Moonlight" ?  It's the light from the moon, which in turn is reflected light from the sun being reflected back to earth.

Can you please give us some idea for this insane idea of yours. Or are you just a troll or something else ?

Your statement does say "The light from the sun could not reach the moon in any capacity" Does "any capacity" include times other than eclipses when the earth is between the sun and the moon ? Maybe I just mis-read your statement. It is true that light from the sun does not reach the moon during a New Moon. Is that "the capacity" of which you refer ? Otherwise some light from the sun always reaches the moon.

Just want to point out what I bolded.  This is false.  Light from the sun does reach the moon during a New Moon.  It is just that the light reflected off the moon does not reach the earth during a NEw Moon.  I feel like you know that, however, someone like jrowe might take this statement and run with it.

It is true that light from the sun does not reach the moon during a New Moon. Is that "the capacity" of which you refer ? Otherwise some light from the sun always reaches the moon.

Eh? You might want to re-word this?

"I already explained that the amount of atmosphere that light passes through during sunset and during an eclipse is the same."
no you haven't, and you cannot, because it is impossible to do so. the sun would be at a sunset in those locations, and then the light would continue moving through the same distance of atmosphere again, to leave. that is even more atmosphere. no logical mind could ignore this fact.

it needs to come out the 'sides'.
...
Which can be proven to occur.
thank you for agreeing. my point therefore stands: the moon is lit by the same lightwaves that go out through the sky. if this makes the sky blue, it should do the same for the moon.

Perhaps you could provide a diagram that shows this.  Afterall, "describing something does not make it realistic."
i did, several pages ago. some details were wrong, but alpha corrected them, and came to a more persuasive answer. there is no way for light to go to the moon in a straight line, so the tube effect is not relevant. scattered light only would reach it.

Or are you just that dumb on the subject of astronomy ? How in the world could you make that statement ? What is the basis of that ? Of course the light from the sun reaches the moon. Did you ever hear of a thing called "Moonlight" ?  It's the light from the moon, which in turn is reflected light from the sun being reflected back to earth.
we are talking about lunar eclipses. it is the title of the thread. i was also referring to a specific diagram. i am not the dumb one here, you are simply illiterate.

"I already explained that the amount of atmosphere that light passes through during sunset and during an eclipse is the same."
no you haven't, and you cannot, because it is impossible to do so. the sun would be at a sunset in those locations, and then the light would continue moving through the same distance of atmosphere again, to leave. that is even more atmosphere. no logical mind could ignore this fact.

I don't think there is much of a difference in how much atmosphere is this light passes through during sunset and how much atmosphere is this light passes through during an eclipse.



As you can see, the point where the sun's light touches the edges of the Earth is where sunset occurs. After than, the light is past Earth and there is nothing to cause it to scatter.

it needs to come out the 'sides'.
...
Which can be proven to occur.
thank you for agreeing. my point therefore stands: the moon is lit by the same lightwaves that go out through the sky. if this makes the sky blue, it should do the same for the moon.

No, because the scattering of blue light wavelengths occurs on the "front" of the Earth (where the sun is facing directly) and by the time it reaches the sides, only red light is left. You understand? All the blue light is scattered at the "front" of the Earth creating a blue sky. None of this light reaches the moon because it can only go towards the sky and towards space. By the time the light reaches the "sides", only red/orange light is left. After it passes the sides, where sunset light is created, there is no more atmosphere for it to go through. It goes toward the moon, and turns it red/orange.

Perhaps you could provide a diagram that shows this.  Afterall, "describing something does not make it realistic."
i did, several pages ago. some details were wrong, but alpha corrected them, and came to a more persuasive answer. there is no way for light to go to the moon in a straight line, so the tube effect is not relevant. scattered light only would reach it.

No, blue scattered light would not reach it, because blue scattered light only hits the sky hat faces the sun. After that, there is only red light left

elimist, blue light makes it into the atmosphere: this is a fact. if it didn't, we would not be able to see a blue sky. as such, it does exist, it does not magically vanish.
i would suggest you look at those illustraions more atmosphere has to be crossed to get to a lunar eclipse, than in a sunset. that is a basic fact, because the light causing the lunar eclipse would have had to have caused a sunset as it goes past the edges of the earth. unless you are saying people who see sunsets live on the edge of the atmosphere, there is still the same distance out of the atmosphere to go.
red and orange light, which are not scattered, would not go that far: only scattered light could bend the degree necessary for the light to reach the moon: my diagram showed there is no remotely straight line to the moon. this directly implies light must be scattered in order to get around the earth to make it that far: and that means blue light.

elimist, blue light makes it into the atmosphere: this is a fact. if it didn't, we would not be able to see a blue sky. as such, it does exist, it does not magically vanish.
i would suggest you look at those illustraions more atmosphere has to be crossed to get to a lunar eclipse, than in a sunset. that is a basic fact, because the light causing the lunar eclipse would have had to have caused a sunset as it goes past the edges of the earth. unless you are saying people who see sunsets live on the edge of the atmosphere, there is still the same distance out of the atmosphere to go.
red and orange light, which are not scattered, would not go that far: only scattered light could bend the degree necessary for the light to reach the moon: my diagram showed there is no remotely straight line to the moon. this directly implies light must be scattered in order to get around the earth to make it that far: and that means blue light.

No, the light skimming through the lower atmosphere bends by about 1° and is reddened due to scattering and absorption of the shorter wavelengths. This 1° of refraction is enough to fully illuminate the umbra less than 300,000 km from earth, as calculated here.

You're right that the light passing all the way through the lower atmosphere travels twice as far through it as light seen at the surface, which would tend to make it somewhat redder, but by the time it reaches the Moon, it's mixed with light that didn't pass so close to the Earth and is less heavily filtered, thus, less red. The outer portions of the umbra are illuminated by light that doesn't pass nearly so deeply into the atmosphere, and is both refracted less and progressively whiter.

Here's what a sunrise or sunset looks like from outside the atmosphere. Note the dim blue scattered light (dim compared to the overexposed, reddened direct sunlight).



[Cropped from this image: http://upload.wikimedia.org/wikipedia/commons/3/31/ISS-34_Earth's_atmosphere.jpg]

Quote from: JRoweSkeptic on April 07, 2015, 09:52:30 AM

elimist, blue light makes it into the atmosphere: this is a fact. if it didn't, we would not be able to see a blue sky. as such, it does exist, it does not magically vanish.
i would suggest you look at those illustraions more atmosphere has to be crossed to get to a lunar eclipse, than in a sunset. that is a basic fact, because the light causing the lunar eclipse would have had to have caused a sunset as it goes past the edges of the earth. unless you are saying people who see sunsets live on the edge of the atmosphere, there is still the same distance out of the atmosphere to go.
red and orange light, which are not scattered, would not go that far: only scattered light could bend the degree necessary for the light to reach the moon: my diagram showed there is no remotely straight line to the moon. this directly implies light must be scattered in order to get around the earth to make it that far: and that means blue light.

No, the light skimming through the lower atmosphere bends by about 1° and is reddened due to scattering and absorption of the shorter wavelengths. This 1° of refraction is enough to fully illuminate the umbra less than 300,000 km from earth, as calculated here.

You're right that the light passing all the way through the lower atmosphere travels twice as far through it as light seen at the surface, which would tend to make it somewhat redder, but by the time it reaches the Moon, it's mixed with light that didn't pass so close to the Earth and is less heavily filtered, thus, less red. The outer portions of the umbra are illuminated by light that doesn't pass nearly so deeply into the atmosphere, and is both refracted less and progressively whiter.

Here's what a sunrise or sunset looks like from outside the atmosphere. Note the dim blue scattered light (dim compared to the overexposed, reddened direct sunlight).



[Cropped from this image: http://upload.wikimedia.org/wikipedia/commons/3/31/ISS-34_Earth's_atmosphere.jpg]

Thank you.

elimist, blue light makes it into the atmosphere: this is a fact. if it didn't, we would not be able to see a blue sky. as such, it does exist, it does not magically vanish.
i would suggest you look at those illustraions more atmosphere has to be crossed to get to a lunar eclipse, than in a sunset. that is a basic fact, because the light causing the lunar eclipse would have had to have caused a sunset as it goes past the edges of the earth. unless you are saying people who see sunsets live on the edge of the atmosphere, there is still the same distance out of the atmosphere to go.
red and orange light, which are not scattered, would not go that far: only scattered light could bend the degree necessary for the light to reach the moon: my diagram showed there is no remotely straight line to the moon. this directly implies light must be scattered in order to get around the earth to make it that far: and that means blue light.

Maybe you really should try the tank/tube experiment. What color of light manages to pass through the container? Hint: not blue. Identically, the color of light that has traveled through the atmosphere is not blue.

alpha, i have already shown you the problem in those configurations. you assume light moves into the earth at more of an angle than it actually does, as my calculations showed. one degree would not get you the distance you require, plain and simple.
if the light goes on and mixes with light that was not close to the earth's surface, 1. how was the light bent enough to be able to reach the moon? 2. why is that light still exclusively red or orange?

neimoka, when you can show that the atmosphere in your model even begins to behave like a tube, then you have a point. until then, that experment is irrelevant.

You're saying that light is aware of the shape of a container that it travels in? Use a non-tubular container if you think that's a problem.

edit, here's a pic of a rectangular container.

You're saying that light is aware of the shape of a container that it travels in? Use a non-tubular container if you think that's a problem.

edit, here's a pic of a rectangular container.

i'm saying the atmosphere is not even remotely a container. what does the containing?

Quote from: neimoka on April 07, 2015, 12:52:27 PM

You're saying that light is aware of the shape of a container that it travels in? Use a non-tubular container if you think that's a problem.

edit, here's a pic of a rectangular container.

i'm saying the atmosphere is not even remotely a container. what does the containing?

You moron, Honestly, you can troll harder than this scepti. Obviously the average joe cannot go into space and shine sunlight-grade light at the atmosphere, so this is the alternative and it shows the same effect. At least try to make it seem realistic when you pretend to be stupid 

alpha, i have already shown you the problem in those configurations. you assume light moves into the earth at more of an angle than it actually does, as my calculations showed. one degree would not get you the distance you require, plain and simple.
if the light goes on and mixes with light that was not close to the earth's surface, 1. how was the light bent enough to be able to reach the moon? 2. why is that light still exclusively red or orange?

Are you referring to the calculations in this post: http://www.theflatearthsociety.org/forum/index.php?topic=63323.msg1677529#msg1677529?

If so, that's the pure geometric solution that doesn't consider refraction at all. Your calcs were somewhat correct, except for the erroneous size used for the Sun (430 thousand vs. the more accurate 696 thousand km) and the length of the Earth's umbra where it's the same diameter as the Moon (859 thousand instead of almost one million km) already noted.

At any rate, the Earth's umbra, just considering geometry, is a narrow cone with an interior angle of about 1/2° (the same as the apparent angular size of the Sun from Earth), or half angle of 1/4°. This narrow cone is about about 1.5 million km long, and about three times the diameter of the Moon at the distance the Moon orbits. No one is disputing that.

Since sunlight can be refracted by up to 1°, the most-refracted rays form a cone with a half-angle of 1.25°; the length of this cone is 1/5 the length of the cone with a half-angle of 0.25° since the angle is 5 times greater, or about 300,000 km.

Light that encounters only the higher, less dense parts of the atmosphere is refracted (and filtered) less, so it converges farther away from the earth. It would take about 1/2° of refraction to bring refracted sunlight to the center of the umbra at the distance of the Moon. Keep in mind, though, that the Moon is smaller than the umbra and seldom crosses right through the middle, so even though light may only be refracted into the outer part of the umbra, it can still illuminate the entire orb of the Moon.

You may want to dismiss atmospheric refraction, but it's a well-known, well-understood, and often-observed and measured phenomenon, so you don't get to do that just because it's inconvenient for you.

[Edit] typo

Quote from: JRoweSkeptic on April 07, 2015, 12:40:00 PM

alpha, i have already shown you the problem in those configurations. you assume light moves into the earth at more of an angle than it actually does, as my calculations showed. one degree would not get you the distance you require, plain and simple.
if the light goes on and mixes with light that was not close to the earth's surface, 1. how was the light bent enough to be able to reach the moon? 2. why is that light still exclusively red or orange?

Are you referring to the calculations in this post: http://www.theflatearthsociety.org/forum/index.php?topic=63323.msg1677529#msg1677529?

If so, that's the pure geometric solution that doesn't consider refraction at all. Your calcs were somewhat correct, except for the erroneous size used for the Sun (430 thousand vs. the more accurate 696 thousand km) and the length of the Earth's umbra where it's the same diameter as the Moon (859 thousand instead of almost one million km) already noted.

At any rate, the Earth's umbra, just considering geometry, is a narrow cone with an interior angle of about 1/2° (the same as the apparent angular size of the Sun from Earth), or half angle of 1/4°. This narrow cone is about about 1.5 million km long, and about three times the diameter of the Moon at the distance the Moon orbits. No one is disputing that.

Since sunlight can be refracted by up to 1°, the most-refracted rays, form a cone with a half-angle of 1.25°; the length of this cone is 1/5 the length of the cone with a half-angle of 0.25° since the angle is 5 times greater, or about 300,000 km.

Light that encounters only the higher, less dense parts of the atmosphere is refracted (and filtered) less, so it converges farther away from the earth. It would take about 1/2° of refraction to bring refracted sunlight to the center of the umbra at the distance of the Moon. Keep in mind, though, that the Moon is smaller than the umbra and seldom crosses right through the middle, so even though light may only be refracted into the outer part of the umbra, it can still illuminate the entire orb of the Moon.

You may want to dismiss atmospheric refraction, but it's a well-known, well-understood, and often-observed and measured phenomenon, so you don't get to do that just because it's inconvenient for you.

there are many reasons to dismiss the atmosphere, but it's not necessary. look at the corrected calcukation you can: you have to reduce the distance covered to a third of what it is to allow for a lunar eclipse. are you honestly saying one degree will do that? One degree s barely anything. i have no doubt it would have an effect, but it could not be as much as you're insisting.

elimist, you are the one wasting time here. i have told you that i am not scepti, and i do not like having my theories attributed to someone else. i have explained why the analogy fails, stop misrepresenting.

Quote from: Alpha2Omega on April 07, 2015, 02:40:13 PM

Quote from: JRoweSkeptic on April 07, 2015, 12:40:00 PM

alpha, i have already shown you the problem in those configurations. you assume light moves into the earth at more of an angle than it actually does, as my calculations showed. one degree would not get you the distance you require, plain and simple.
if the light goes on and mixes with light that was not close to the earth's surface, 1. how was the light bent enough to be able to reach the moon? 2. why is that light still exclusively red or orange?

Are you referring to the calculations in this post: http://www.theflatearthsociety.org/forum/index.php?topic=63323.msg1677529#msg1677529?

If so, that's the pure geometric solution that doesn't consider refraction at all. Your calcs were somewhat correct, except for the erroneous size used for the Sun (430 thousand vs. the more accurate 696 thousand km) and the length of the Earth's umbra where it's the same diameter as the Moon (859 thousand instead of almost one million km) already noted.

At any rate, the Earth's umbra, just considering geometry, is a narrow cone with an interior angle of about 1/2° (the same as the apparent angular size of the Sun from Earth), or half angle of 1/4°. This narrow cone is about about 1.5 million km long, and about three times the diameter of the Moon at the distance the Moon orbits. No one is disputing that.

Since sunlight can be refracted by up to 1°, the most-refracted rays, form a cone with a half-angle of 1.25°; the length of this cone is 1/5 the length of the cone with a half-angle of 0.25° since the angle is 5 times greater, or about 300,000 km.

Light that encounters only the higher, less dense parts of the atmosphere is refracted (and filtered) less, so it converges farther away from the earth. It would take about 1/2° of refraction to bring refracted sunlight to the center of the umbra at the distance of the Moon. Keep in mind, though, that the Moon is smaller than the umbra and seldom crosses right through the middle, so even though light may only be refracted into the outer part of the umbra, it can still illuminate the entire orb of the Moon.

You may want to dismiss atmospheric refraction, but it's a well-known, well-understood, and often-observed and measured phenomenon, so you don't get to do that just because it's inconvenient for you.

there are many reasons to dismiss the atmosphere, but it's not necessary. look at the corrected calcukation you can: you have to reduce the distance covered to a third of what it is to allow for a lunar eclipse. are you honestly saying one degree will do that? One degree s barely anything. i have no doubt it would have an effect, but it could not be as much as you're insisting.

elimist, you are the one wasting time here. i have told you that i am not scepti, and i do not like having my theories attributed to someone else. i have explained why the analogy fails, stop misrepresenting.

Alright, alright jroa, I'll stop calling you scepti.

Quote from: Alpha2Omega on April 07, 2015, 02:40:13 PM

Quote from: JRoweSkeptic on April 07, 2015, 12:40:00 PM

alpha, i have already shown you the problem in those configurations. you assume light moves into the earth at more of an angle than it actually does, as my calculations showed. one degree would not get you the distance you require, plain and simple.
if the light goes on and mixes with light that was not close to the earth's surface, 1. how was the light bent enough to be able to reach the moon? 2. why is that light still exclusively red or orange?

Are you referring to the calculations in this post: http://www.theflatearthsociety.org/forum/index.php?topic=63323.msg1677529#msg1677529?

If so, that's the pure geometric solution that doesn't consider refraction at all. Your calcs were somewhat correct, except for the erroneous size used for the Sun (430 thousand vs. the more accurate 696 thousand km) and the length of the Earth's umbra where it's the same diameter as the Moon (859 thousand instead of almost one million km) already noted.

At any rate, the Earth's umbra, just considering geometry, is a narrow cone with an interior angle of about 1/2° (the same as the apparent angular size of the Sun from Earth), or half angle of 1/4°. This narrow cone is about about 1.5 million km long, and about three times the diameter of the Moon at the distance the Moon orbits. No one is disputing that.

Since sunlight can be refracted by up to 1°, the most-refracted rays, form a cone with a half-angle of 1.25°; the length of this cone is 1/5 the length of the cone with a half-angle of 0.25° since the angle is 5 times greater, or about 300,000 km.

Light that encounters only the higher, less dense parts of the atmosphere is refracted (and filtered) less, so it converges farther away from the earth. It would take about 1/2° of refraction to bring refracted sunlight to the center of the umbra at the distance of the Moon. Keep in mind, though, that the Moon is smaller than the umbra and seldom crosses right through the middle, so even though light may only be refracted into the outer part of the umbra, it can still illuminate the entire orb of the Moon.

You may want to dismiss atmospheric refraction, but it's a well-known, well-understood, and often-observed and measured phenomenon, so you don't get to do that just because it's inconvenient for you.

there are many reasons to dismiss the atmosphere, but it's not necessary. look at the corrected calcukation you can: you have to reduce the distance covered to a third of what it is to allow for a lunar eclipse. are you honestly saying one degree will do that?

Yes.

One degree s barely anything. i have no doubt it would have an effect, but it could not be as much as you're insisting.

It's four times the angle the geometric shadow forms, and adds to that angle, so it's five times as steep. It has a lot more effect than you're insisting. It shortens the cone of darkness by a factor of five.

Here's my illustration again (it's diagrammatic, not to scale):

Quote from: JRoweSkeptic on April 07, 2015, 02:44:52 PM

Quote from: Alpha2Omega on April 07, 2015, 02:40:13 PM

Quote from: JRoweSkeptic on April 07, 2015, 12:40:00 PM

alpha, i have already shown you the problem in those configurations. you assume light moves into the earth at more of an angle than it actually does, as my calculations showed. one degree would not get you the distance you require, plain and simple.
if the light goes on and mixes with light that was not close to the earth's surface, 1. how was the light bent enough to be able to reach the moon? 2. why is that light still exclusively red or orange?

Are you referring to the calculations in this post: http://www.theflatearthsociety.org/forum/index.php?topic=63323.msg1677529#msg1677529?

If so, that's the pure geometric solution that doesn't consider refraction at all. Your calcs were somewhat correct, except for the erroneous size used for the Sun (430 thousand vs. the more accurate 696 thousand km) and the length of the Earth's umbra where it's the same diameter as the Moon (859 thousand instead of almost one million km) already noted.

At any rate, the Earth's umbra, just considering geometry, is a narrow cone with an interior angle of about 1/2° (the same as the apparent angular size of the Sun from Earth), or half angle of 1/4°. This narrow cone is about about 1.5 million km long, and about three times the diameter of the Moon at the distance the Moon orbits. No one is disputing that.

Since sunlight can be refracted by up to 1°, the most-refracted rays, form a cone with a half-angle of 1.25°; the length of this cone is 1/5 the length of the cone with a half-angle of 0.25° since the angle is 5 times greater, or about 300,000 km.

Light that encounters only the higher, less dense parts of the atmosphere is refracted (and filtered) less, so it converges farther away from the earth. It would take about 1/2° of refraction to bring refracted sunlight to the center of the umbra at the distance of the Moon. Keep in mind, though, that the Moon is smaller than the umbra and seldom crosses right through the middle, so even though light may only be refracted into the outer part of the umbra, it can still illuminate the entire orb of the Moon.

You may want to dismiss atmospheric refraction, but it's a well-known, well-understood, and often-observed and measured phenomenon, so you don't get to do that just because it's inconvenient for you.

there are many reasons to dismiss the atmosphere, but it's not necessary. look at the corrected calcukation you can: you have to reduce the distance covered to a third of what it is to allow for a lunar eclipse. are you honestly saying one degree will do that?

Yes.

Quote

One degree s barely anything. i have no doubt it would have an effect, but it could not be as much as you're insisting.

It's four times the angle the geometric shadow forms, and adds to that angle, so it's five times as steep. It has a lot more effect than you're insisting. It shortens the cone of darkness by a factor of five.

Here's my illustration again (it's diagrammatic, not to scale):

it's not four times anything, unless you're directly comparing the angles, but the angle is part of the tan operator, it's not going to be four times the size. have you seen how slowly those functions change?

One degree s barely anything. i have no doubt it would have an effect, but it could not be as much as you're insisting.

Quote

It's four times the angle the geometric shadow forms, and adds to that angle, so it's five times as steep. It has a lot more effect than you're insisting. It shortens the cone of darkness by a factor of five.

Here's my illustration again (it's diagrammatic, not to scale):

it's not four times anything, unless you're directly comparing the angles, but the angle is part of the tan operator, it's not going to be four times the size. have you seen how slowly those functions change?

Wrong!

For small angles (these qualify), if the angle is expressed in radians,

tan(angle) ≈ angle 

if the angle is expressed in degrees,

tan(angle) ≈ angleInDegrees * pi/180.

The tangent is directly proportional to the angle for small angles like these.

So...
tan(5 * angle)/tan(angle) =  (5 * angle)/(angle) = 5

for all practical purposes when 5 * angle is < 9° or so[nb]This was a rule of thumb in the slide-rule days.[/nb], which it clearly is here.

It's the cosine that changes slowly at small angles.

[Edit] Fix nested quote.
 

Quote from: neimoka on April 07, 2015, 12:52:27 PM

You're saying that light is aware of the shape of a container that it travels in? Use a non-tubular container if you think that's a problem.

edit, here's a pic of a rectangular container.

i'm saying the atmosphere is not even remotely a container. what does the containing?

Stop obsessing about the container, it is irrelevant. You thought it was somehow important that it's a tube, well it's not and now you're just trying to find something else to whine about while ignoring the actual observation. We are not interested in the container, it is there to store the fluid and nothing else. Put the light in there and make your observations from within as well if you want. What we're interested in here is the fact that blue light does not reach the other end, while red does. Do you deny this?

alpha, it is the sine and cosine which change slowly. what exactly is tan based on, again?

neimoka, are you capable of making an argument without forcing your words into someone else's mouth? i am fully aware of what i am saying. the container is crucial, as it is the only time an end can exist, and that sides can exist. this is not the case in the atmosphere. it has already been shown that there is no straight path from the sun to the moon in the case of the lunar eclipse.

alpha, it is the sine and cosine which change slowly. what exactly is tan based on, again?

Sine(angle) changes most rapidly when the angle passes through zero. Where are you getting your information?

Like tangent, for small angles sine(angle) ≈ angle in radians.

Quote from: JRoweSkeptic on April 08, 2015, 05:07:07 AM

alpha, it is the sine and cosine which change slowly. what exactly is tan based on, again?

Sine(angle) changes most rapidly when the angle passes through zero. Where are you getting your information?

Like tangent, for small angles sine(angle) ≈ angle in radians.

it doesn't matter if it's the most or least rapidly changing point, sine is a tiny function, just like the cosine. 'most' means nothing when all the values are small.

Quote from: Alpha2Omega on April 08, 2015, 06:30:55 AM

Quote from: JRoweSkeptic on April 08, 2015, 05:07:07 AM

alpha, it is the sine and cosine which change slowly. what exactly is tan based on, again?

Sine(angle) changes most rapidly when the angle passes through zero. Where are you getting your information?

Like tangent, for small angles sine(angle) ≈ angle in radians.

it doesn't matter if it's the most or least rapidly changing point,

Of course it does. We're dividing by this number, not adding it. The result will be inversely proportional to the change.

sine is a tiny function, just like the cosine. 'most' means nothing when all the values are small.

"Tiny function." What's that supposed to mean? You do realize that dividing some non-zero number by a tiny number gives a much larger number don't you?

Pout all you want. The fact remains that changing the angle from 0.25° to 1.25° (a difference of "only" 1°, but a factor of 5) changes the tangent of the angle from 0.00436 to 0.0218 (a factor of 5), and thus reduces the length of the shadow by a factor of 5. Recall that you said the refracted-light cone "only" needed to be reduced by a factor of 3 to fully illuminate the umbra at the distance to the Moon.

Is this really so hard for you to grasp, or are you being intentionally obtuse?

neimoka, are you capable of making an argument without forcing your words into someone else's mouth? i am fully aware of what i am saying. the container is crucial, as it is the only time an end can exist, and that sides can exist.

Wrong. Read my previous post; the container is irrelevant. We are not specifically interested where it begins or ends, we are interested in how far light of different wavelenghts propagate in the medium. Neither are the sides of the container in any way relevant to what are the 'sides' where light scatters; light scatters 'to the sides' relative to the original direction of the light beam.

You can do the experiment completely ignoring the container by observing the color directly at the medium if you for some perverse reason want to insist that the container somehow affects the outcome, which it does not.

this is not the case in the atmosphere. it has already been shown that there is no straight path from the sun to the moon in the case of the lunar eclipse.

Irrelevant.

Perhaps I should re-word this;

What we're interested in here is the fact that blue light does not reach the other end, while red does propagate as far in the medium as red does. Do you deny this?

Happy?

Is this really so hard for you to grasp, or are you being intentionally obtuse?

I just cant tell anymore.

Remember this is the guy that burst onto the scene claiming that spherical geometry was a hoax.

Show this to Jrowesceptimatic for a better explanation