Flat Earth Evidence.

Quote from: 11cookeaw1 on September 27, 2013, 10:31:36 AM

Quote from: SeekerOfTruth on September 27, 2013, 08:35:18 AM

One piece of evidence for a flat earth, as you requested:

The power emitted from a blackbody (i.e., the Sun) is related to the temperature according to: P=a*T^4, where P is the power, a is the stefan-boltzmann constant (8*10^-8), and T is the temperature. Now according to round earth science, the temperature of the Sun is 5778K. This gives a power of 6.8*10^7 (check the math yourself). Now, this is really the power from the whole surface, so the corresponding power we receive on Earth from the Sun is MORE than this because the Earth has a smaller surface area than the Sun. Specifically, the power on Earth is 6.8*10^7 times the ratios of surface areas, which gives a power of 8.2*10^11. What temperature does this correspond to? 56582K. Check the math yourself. In Fahrenheit this about 1*10^5, or 100,000 degrees. Go outside, is it that hot? No, it is not. How do the round earthers explain this? They claim it is because only part of the Earth receives the power form the Sun. Which is true, only half the Earth experiences daytime at any given time (in the RE model). Okay, so cut the Earth's surface area in half, and re-do the calculation. What do you get? I will let you check it yourself, so that you can see the truth with your own eyes. In short, the Earth is NOT that hot. We would all be dead at that temperature.

The earth only receives a fraction on the sun's energy.

The calculation of power (read: intensity -- see posts above) takes into account the rate per unit area. Please show me a calculation if you believe I am wrong: I showed you the same courtesy.

Your calculation assumes the earth gets all the suns energy.
Temperature of the sun= 5778K. Ratio of intensity of the Sun compared to the earth = (Semi major axis/ radius the sun)^2=(149600000/695500)^2=215.1^2=46300.
Cross sectional area of the earth = 1/4 of total area of earth. 46300*4=185200.
185200^0.25=20.74 5778/20.74=278K. Considering this doesn't include stuff like greenhouse effects and light reflected back out into space this is close enough to the earth's mean temperature.

Quote from: SeekerOfTruth on September 27, 2013, 10:54:27 AM

Quote from: Cartesian on September 27, 2013, 10:28:40 AM

Since things don't look as simple as you made it sound, you can't use your simple calculation as argument.

Alright, so this is a busy plot, showing lots of information. What part of it disagrees with my calculation? And why?

That you don't take many factors into a account.

Hmm, lets talk about this picture. It shows that light of different wavelengths "on average" have different optical depths, but really this changes noticeably only for UV light (which is a small part of the electromagnetic spectrum). These emphasize this because this regime is where the probability for scattering is non-negligible. But most of the visible light, infra-red, and x-rays survive to the Earth. This SUPPORTS my earlier claim. To the right of the plot, you see how the differing densities of molecules as a function of height correlate. To the left, you see that the atmosphere drops the temperature from 1200K to about 250-300K (depending really on the atmospheric model). 

So does this disprove my earlier statement? No no, look. They assume the temperature BEFORE it enters the atmosphere is 1200K, and drops to 300K (lets say). Just look at the plot. My calculation shows a temperature of 100,000K before it enters the atmosphere. Drop that by 800K and what do you get? Still too hot!

You see this plot, while showing useful information, is unfortunately not relevant to disproving my mis-match in temperature.

Quote from: SeekerOfTruth on September 27, 2013, 10:53:11 AM

Quote from: 11cookeaw1 on September 27, 2013, 10:31:36 AM

Quote from: SeekerOfTruth on September 27, 2013, 08:35:18 AM

One piece of evidence for a flat earth, as you requested:

The power emitted from a blackbody (i.e., the Sun) is related to the temperature according to: P=a*T^4, where P is the power, a is the stefan-boltzmann constant (8*10^-8), and T is the temperature. Now according to round earth science, the temperature of the Sun is 5778K. This gives a power of 6.8*10^7 (check the math yourself). Now, this is really the power from the whole surface, so the corresponding power we receive on Earth from the Sun is MORE than this because the Earth has a smaller surface area than the Sun. Specifically, the power on Earth is 6.8*10^7 times the ratios of surface areas, which gives a power of 8.2*10^11. What temperature does this correspond to? 56582K. Check the math yourself. In Fahrenheit this about 1*10^5, or 100,000 degrees. Go outside, is it that hot? No, it is not. How do the round earthers explain this? They claim it is because only part of the Earth receives the power form the Sun. Which is true, only half the Earth experiences daytime at any given time (in the RE model). Okay, so cut the Earth's surface area in half, and re-do the calculation. What do you get? I will let you check it yourself, so that you can see the truth with your own eyes. In short, the Earth is NOT that hot. We would all be dead at that temperature.

The earth only receives a fraction on the sun's energy.

The calculation of power (read: intensity -- see posts above) takes into account the rate per unit area. Please show me a calculation if you believe I am wrong: I showed you the same courtesy.

Your calculation assumes the earth gets all the suns energy.
Temperature of the sun= 5778K. Ratio of intensity of the Sun compared to the earth = (Semi major axis/ radius the sun)^2=(149600000/695500)^2=215.1^2=46300.
Cross sectional area of the earth = 1/4 of total area of earth. 46300*4=185200.
185200^0.25=20.74 5778/20.74=278K. Considering this doesn't include stuff like greenhouse effects and light reflected back out into space this is close enough to the earth's mean temperature.

You do not take into account the Stefan-Boltzmann constant. P=a*T^4

Quote from: Rama Set on September 27, 2013, 09:27:20 AM

No, power is energy per unit of time.  Intensity is power transferred per unit of area.

Yes yes, you are quite right. The confusion is that it is not really meaningful to discuss power in this context without also addressing the area, so typically the law P=a*T^4 is discussed in terms of "per unit area" without explicitly stating P/cm^2, which is indeed an intensity. The fault is mine of course: when you do this stuff for a while you tend to adopt common colloquial shortcuts in the notation and verbiage, which can be confusing if you're not used to it. This is not an excuse, however, many do not know this, and wrong impressions can be given. To restate, I DO take into account the areas in the calculation.   

I see. Sorry for not looking deeper in to the formula. Upon closer inspection you are calculating the power emitted per unit area. This is different than the power received per unit area which would need to be transformed by the inverse square law I think.

Quote from: Cartesian on September 27, 2013, 10:55:46 AM

Quote from: SeekerOfTruth on September 27, 2013, 10:54:27 AM

Quote from: Cartesian on September 27, 2013, 10:28:40 AM

Since things don't look as simple as you made it sound, you can't use your simple calculation as argument.

Alright, so this is a busy plot, showing lots of information. What part of it disagrees with my calculation? And why?

That you don't take many factors into a account.

Hmm, lets talk about this picture. It shows that light of different wavelengths "on average" have different optical depths, but really this changes noticeably only for UV light (which is a small part of the electromagnetic spectrum). These emphasize this because this regime is where the probability for scattering is non-negligible. But most of the visible light, infra-red, and x-rays survive to the Earth. This SUPPORTS my earlier claim. To the right of the plot, you see how the differing densities of molecules as a function of height correlate. To the left, you see that the atmosphere drops the temperature from 1200K to about 250-300K (depending really on the atmospheric model). 

So does this disprove my earlier statement? No no, look. They assume the temperature BEFORE it enters the atmosphere is 1200K, and drops to 300K (lets say). Just look at the plot. My calculation shows a temperature of 100,000K before it enters the atmosphere. Drop that by 800K and what do you get? Still too hot!

You see this plot, while showing useful information, is unfortunately not relevant to disproving my mis-match in temperature.

At 100 km the temperature is nearly the same as the one on the surface. Your calculation is too simple and yet as proven by others too, it's still wrong.

Quote from: SeekerOfTruth on September 27, 2013, 09:35:59 AM

Quote from: Rama Set on September 27, 2013, 09:27:20 AM

No, power is energy per unit of time.  Intensity is power transferred per unit of area.

Yes yes, you are quite right. The confusion is that it is not really meaningful to discuss power in this context without also addressing the area, so typically the law P=a*T^4 is discussed in terms of "per unit area" without explicitly stating P/cm^2, which is indeed an intensity. The fault is mine of course: when you do this stuff for a while you tend to adopt common colloquial shortcuts in the notation and verbiage, which can be confusing if you're not used to it. This is not an excuse, however, many do not know this, and wrong impressions can be given. To restate, I DO take into account the areas in the calculation.   

I see. Sorry for not looking deeper in to the formula. Upon closer inspection you are calculating the power emitted per unit area. This is different than the power received per unit area which would need to be transformed by the inverse square law I think.

How exactly?

Quote from: Rama Set on September 27, 2013, 11:18:00 AM

Quote from: SeekerOfTruth on September 27, 2013, 09:35:59 AM

Quote from: Rama Set on September 27, 2013, 09:27:20 AM

No, power is energy per unit of time.  Intensity is power transferred per unit of area.

Yes yes, you are quite right. The confusion is that it is not really meaningful to discuss power in this context without also addressing the area, so typically the law P=a*T^4 is discussed in terms of "per unit area" without explicitly stating P/cm^2, which is indeed an intensity. The fault is mine of course: when you do this stuff for a while you tend to adopt common colloquial shortcuts in the notation and verbiage, which can be confusing if you're not used to it. This is not an excuse, however, many do not know this, and wrong impressions can be given. To restate, I DO take into account the areas in the calculation.   

I see. Sorry for not looking deeper in to the formula. Upon closer inspection you are calculating the power emitted per unit area. This is different than the power received per unit area which would need to be transformed by the inverse square law I think.

How exactly?

The intensity of radiation emitted from the sun would be affected by the Inverse Square Law.

http://en.m.wikipedia.org/wiki/Inverse-square_law

The physical law you used to calculate the intensity of the sun's radiation is the intensity emitted, not received.

Quote from: SeekerOfTruth on September 27, 2013, 11:15:16 AM

Quote from: Cartesian on September 27, 2013, 10:55:46 AM

Quote from: SeekerOfTruth on September 27, 2013, 10:54:27 AM

Quote from: Cartesian on September 27, 2013, 10:28:40 AM

Since things don't look as simple as you made it sound, you can't use your simple calculation as argument.

Alright, so this is a busy plot, showing lots of information. What part of it disagrees with my calculation? And why?

That you don't take many factors into a account.

Hmm, lets talk about this picture. It shows that light of different wavelengths "on average" have different optical depths, but really this changes noticeably only for UV light (which is a small part of the electromagnetic spectrum). These emphasize this because this regime is where the probability for scattering is non-negligible. But most of the visible light, infra-red, and x-rays survive to the Earth. This SUPPORTS my earlier claim. To the right of the plot, you see how the differing densities of molecules as a function of height correlate. To the left, you see that the atmosphere drops the temperature from 1200K to about 250-300K (depending really on the atmospheric model). 

So does this disprove my earlier statement? No no, look. They assume the temperature BEFORE it enters the atmosphere is 1200K, and drops to 300K (lets say). Just look at the plot. My calculation shows a temperature of 100,000K before it enters the atmosphere. Drop that by 800K and what do you get? Still too hot!

You see this plot, while showing useful information, is unfortunately not relevant to disproving my mis-match in temperature.

At 100 km the temperature is nearly the same as the one on the surface. Your calculation is too simple and yet as proven by others too, it's still wrong.

Only if you assume the temperature at the top of the atmosphere is 1200K, which my formula says it can't be.

Quote from: 11cookeaw1 on September 27, 2013, 11:08:39 AM

Quote from: SeekerOfTruth on September 27, 2013, 10:53:11 AM

Quote from: 11cookeaw1 on September 27, 2013, 10:31:36 AM

Quote from: SeekerOfTruth on September 27, 2013, 08:35:18 AM

One piece of evidence for a flat earth, as you requested:

The power emitted from a blackbody (i.e., the Sun) is related to the temperature according to: P=a*T^4, where P is the power, a is the stefan-boltzmann constant (8*10^-8), and T is the temperature. Now according to round earth science, the temperature of the Sun is 5778K. This gives a power of 6.8*10^7 (check the math yourself). Now, this is really the power from the whole surface, so the corresponding power we receive on Earth from the Sun is MORE than this because the Earth has a smaller surface area than the Sun. Specifically, the power on Earth is 6.8*10^7 times the ratios of surface areas, which gives a power of 8.2*10^11. What temperature does this correspond to? 56582K. Check the math yourself. In Fahrenheit this about 1*10^5, or 100,000 degrees. Go outside, is it that hot? No, it is not. How do the round earthers explain this? They claim it is because only part of the Earth receives the power form the Sun. Which is true, only half the Earth experiences daytime at any given time (in the RE model). Okay, so cut the Earth's surface area in half, and re-do the calculation. What do you get? I will let you check it yourself, so that you can see the truth with your own eyes. In short, the Earth is NOT that hot. We would all be dead at that temperature.

The earth only receives a fraction on the sun's energy.

The calculation of power (read: intensity -- see posts above) takes into account the rate per unit area. Please show me a calculation if you believe I am wrong: I showed you the same courtesy.

Your calculation assumes the earth gets all the suns energy.
Temperature of the sun= 5778K. Ratio of intensity of the Sun compared to the earth = (Semi major axis/ radius the sun)^2=(149600000/695500)^2=215.1^2=46300.
Cross sectional area of the earth = 1/4 of total area of earth. 46300*4=185200.
185200^0.25=20.74 5778/20.74=278K. Considering this doesn't include stuff like greenhouse effects and light reflected back out into space this is close enough to the earth's mean temperature.

You do not take into account the Stefan-Boltzmann constant. P=a*T^4

I did.

Quote from: Cartesian on September 27, 2013, 11:21:05 AM

Quote from: SeekerOfTruth on September 27, 2013, 11:15:16 AM

Quote from: Cartesian on September 27, 2013, 10:55:46 AM

Quote from: SeekerOfTruth on September 27, 2013, 10:54:27 AM

Quote from: Cartesian on September 27, 2013, 10:28:40 AM

Since things don't look as simple as you made it sound, you can't use your simple calculation as argument.

Alright, so this is a busy plot, showing lots of information. What part of it disagrees with my calculation? And why?

That you don't take many factors into a account.

Hmm, lets talk about this picture. It shows that light of different wavelengths "on average" have different optical depths, but really this changes noticeably only for UV light (which is a small part of the electromagnetic spectrum). These emphasize this because this regime is where the probability for scattering is non-negligible. But most of the visible light, infra-red, and x-rays survive to the Earth. This SUPPORTS my earlier claim. To the right of the plot, you see how the differing densities of molecules as a function of height correlate. To the left, you see that the atmosphere drops the temperature from 1200K to about 250-300K (depending really on the atmospheric model). 

So does this disprove my earlier statement? No no, look. They assume the temperature BEFORE it enters the atmosphere is 1200K, and drops to 300K (lets say). Just look at the plot. My calculation shows a temperature of 100,000K before it enters the atmosphere. Drop that by 800K and what do you get? Still too hot!

You see this plot, while showing useful information, is unfortunately not relevant to disproving my mis-match in temperature.

At 100 km the temperature is nearly the same as the one on the surface. Your calculation is too simple and yet as proven by others too, it's still wrong.

Only if you assume the temperature at the top of the atmosphere is 1200K, which my formula says it can't be.

That's because in your model the ENTIRE output of the sun hits the earth. How can the Sun possibly heat the earth to be hooter then itself.

Quote from: SeekerOfTruth on September 27, 2013, 11:22:12 AM

Quote from: Rama Set on September 27, 2013, 11:18:00 AM

Quote from: SeekerOfTruth on September 27, 2013, 09:35:59 AM

Quote from: Rama Set on September 27, 2013, 09:27:20 AM

No, power is energy per unit of time.  Intensity is power transferred per unit of area.

Yes yes, you are quite right. The confusion is that it is not really meaningful to discuss power in this context without also addressing the area, so typically the law P=a*T^4 is discussed in terms of "per unit area" without explicitly stating P/cm^2, which is indeed an intensity. The fault is mine of course: when you do this stuff for a while you tend to adopt common colloquial shortcuts in the notation and verbiage, which can be confusing if you're not used to it. This is not an excuse, however, many do not know this, and wrong impressions can be given. To restate, I DO take into account the areas in the calculation.   

I see. Sorry for not looking deeper in to the formula. Upon closer inspection you are calculating the power emitted per unit area. This is different than the power received per unit area which would need to be transformed by the inverse square law I think.

How exactly?

The intensity of radiation emitted from the sun would be affected by the Inverse Square Law.

http://en.m.wikipedia.org/wiki/Inverse-square_law

The physical law you used to calculate the intensity of the sun's radiation is the intensity emitted, not received.

Alright, I have to agree this is consistent. But really, isn't this inverse square law just a result of the same reasoning that one uses to explain RE gravity, which is ALSO an inverse square law? Hence wouldn't this only be valid in a RE context?

Jingle Jangle:  The Earth's crust didn't just freeze in place.  It's not a completely static outer shell.  Please read about Subduction:

http://en.wikipedia.org/wiki/Subduction

Quote from: SeekerOfTruth on September 27, 2013, 11:17:36 AM

Quote from: 11cookeaw1 on September 27, 2013, 11:08:39 AM

Quote from: SeekerOfTruth on September 27, 2013, 10:53:11 AM

Quote from: 11cookeaw1 on September 27, 2013, 10:31:36 AM

Quote from: SeekerOfTruth on September 27, 2013, 08:35:18 AM

One piece of evidence for a flat earth, as you requested:

The power emitted from a blackbody (i.e., the Sun) is related to the temperature according to: P=a*T^4, where P is the power, a is the stefan-boltzmann constant (8*10^-8), and T is the temperature. Now according to round earth science, the temperature of the Sun is 5778K. This gives a power of 6.8*10^7 (check the math yourself). Now, this is really the power from the whole surface, so the corresponding power we receive on Earth from the Sun is MORE than this because the Earth has a smaller surface area than the Sun. Specifically, the power on Earth is 6.8*10^7 times the ratios of surface areas, which gives a power of 8.2*10^11. What temperature does this correspond to? 56582K. Check the math yourself. In Fahrenheit this about 1*10^5, or 100,000 degrees. Go outside, is it that hot? No, it is not. How do the round earthers explain this? They claim it is because only part of the Earth receives the power form the Sun. Which is true, only half the Earth experiences daytime at any given time (in the RE model). Okay, so cut the Earth's surface area in half, and re-do the calculation. What do you get? I will let you check it yourself, so that you can see the truth with your own eyes. In short, the Earth is NOT that hot. We would all be dead at that temperature.

The earth only receives a fraction on the sun's energy.

The calculation of power (read: intensity -- see posts above) takes into account the rate per unit area. Please show me a calculation if you believe I am wrong: I showed you the same courtesy.

Your calculation assumes the earth gets all the suns energy.
Temperature of the sun= 5778K. Ratio of intensity of the Sun compared to the earth = (Semi major axis/ radius the sun)^2=(149600000/695500)^2=215.1^2=46300.
Cross sectional area of the earth = 1/4 of total area of earth. 46300*4=185200.
185200^0.25=20.74 5778/20.74=278K. Considering this doesn't include stuff like greenhouse effects and light reflected back out into space this is close enough to the earth's mean temperature.

You do not take into account the Stefan-Boltzmann constant. P=a*T^4

I did.

I do not see it in your calculation...

Quote from: Rama Set on September 27, 2013, 11:25:29 AM

Quote from: SeekerOfTruth on September 27, 2013, 11:22:12 AM

Quote from: Rama Set on September 27, 2013, 11:18:00 AM

Quote from: SeekerOfTruth on September 27, 2013, 09:35:59 AM

Quote from: Rama Set on September 27, 2013, 09:27:20 AM

No, power is energy per unit of time.  Intensity is power transferred per unit of area.

Yes yes, you are quite right. The confusion is that it is not really meaningful to discuss power in this context without also addressing the area, so typically the law P=a*T^4 is discussed in terms of "per unit area" without explicitly stating P/cm^2, which is indeed an intensity. The fault is mine of course: when you do this stuff for a while you tend to adopt common colloquial shortcuts in the notation and verbiage, which can be confusing if you're not used to it. This is not an excuse, however, many do not know this, and wrong impressions can be given. To restate, I DO take into account the areas in the calculation.   

I see. Sorry for not looking deeper in to the formula. Upon closer inspection you are calculating the power emitted per unit area. This is different than the power received per unit area which would need to be transformed by the inverse square law I think.

How exactly?

The intensity of radiation emitted from the sun would be affected by the Inverse Square Law.

http://en.m.wikipedia.org/wiki/Inverse-square_law

The physical law you used to calculate the intensity of the sun's radiation is the intensity emitted, not received.

Alright, I have to agree this is consistent. But really, isn't this inverse square law just a result of the same reasoning that one uses to explain RE gravity, which is ALSO an inverse square law? Hence wouldn't this only be valid in a RE context?

The inverse square law can be measured here on earth.

Jingle Jangle:  The Earth's crust didn't just freeze in place.  It's not a completely static outer shell.  Please read about Subduction:

http://en.wikipedia.org/wiki/Subduction

No I did not say the earth underwent sudden cooling.  I meant the earth burst into existence by the vehicle of divine magic.

I stand completely aware of the entire geologic theory of subduction.  I know how a cycle of change and transformation allegedly exists, influencing the earth's crust.  However, in reality, if that molten earth in the mantle cooled down at all, there would be 100 % kryolite crystal.  This simple fact proves the granite layers were specially created.  No deposition of kryolite ever occurs.  No gradual tectonic movement, heating and cooling, and recycling ever took place.


 Do you also understand that if the granite layer really underwent subduction continually that polonium halos would not appear.  Subduction requires the continuous refreshing of entire plates from molten mantle.  Still no kryolite along those plate areas.

Quartz in large concentrations appears within certain caves.  Even though quartz possesses a low melting point, it does not appear all evenly distributed and melted from thousands of years of mantle contact.  It appears as if no diffusion through molten mass ever occurred.  In essence, the existence of lodes and crystal colonies nullifies the statement that random mantle deposition resulted in the earth's composition.

Notice how clear some crystals come.  Their clarity implies their formation in an environment not littered with conflicting impurities.

Undivine mantle operates on the rules of particle diffusion and general equal distribution.  Miraculous mantle creates solid lodes and crystal colonies, separate and organized.

You have to see what I am saying.  The point stands simply put.

Quote from: MonkeyButz on September 27, 2013, 11:46:44 AM

Jingle Jangle:  The Earth's crust didn't just freeze in place.  It's not a completely static outer shell.  Please read about Subduction:

http://en.wikipedia.org/wiki/Subduction

No I did not say the earth underwent sudden cooling.  I meant the earth burst into existence by the vehicle of divine magic.

I stand completely aware of the entire geologic theory of subduction.  I know how a cycle of change and transformation allegedly exists, influencing the earth's crust.  However, in reality, if that molten earth in the mantle cooled down at all, there would be 100 % kryolite crystal.  This simple fact proves the granite layers were specially created.  No deposition of kryolite ever occurs.  No gradual tectonic movement, heating and cooling, and recycling ever took place.


 Do you also understand that if the granite layer really underwent subduction continually that polonium halos would not appear.  Subduction requires the continuous refreshing of entire plates from molten mantle.  Still no kryolite along those plate areas.

Quartz in large concentrations appears within certain caves.  Even though quartz possesses a low melting point, it does not appear all evenly distributed and melted from thousands of years of mantle contact.  It appears as if no diffusion through molten mass ever occurred.  In essence, the existence of lodes and crystal colonies nullifies the statement that random mantle deposition resulted in the earth's composition.

Notice how clear some crystals come.  Their clarity implies their formation in an environment not littered with conflicting impurities.

Undivine mantle operates on the rules of particle diffusion and general equal distribution.  Miraculous mantle creates solid lodes and crystal colonies, separate and organized.

You have to see what I am saying.  The point stands simply put.

I'm no geologist but I don't think this makes much sense. You are saying that it must be designed because there is so much order but then you use other arguments like your big bang one that say that there wasn't enough order so it must be designed.

You're confusing.

Quote from: Jingle Jangle on September 27, 2013, 12:27:48 PM

Quote from: MonkeyButz on September 27, 2013, 11:46:44 AM

Jingle Jangle:  The Earth's crust didn't just freeze in place.  It's not a completely static outer shell.  Please read about Subduction:

http://en.wikipedia.org/wiki/Subduction

No I did not say the earth underwent sudden cooling.  I meant the earth burst into existence by the vehicle of divine magic.

I stand completely aware of the entire geologic theory of subduction.  I know how a cycle of change and transformation allegedly exists, influencing the earth's crust.  However, in reality, if that molten earth in the mantle cooled down at all, there would be 100 % kryolite crystal.  This simple fact proves the granite layers were specially created.  No deposition of kryolite ever occurs.  No gradual tectonic movement, heating and cooling, and recycling ever took place.


 Do you also understand that if the granite layer really underwent subduction continually that polonium halos would not appear.  Subduction requires the continuous refreshing of entire plates from molten mantle.  Still no kryolite along those plate areas.

Quartz in large concentrations appears within certain caves.  Even though quartz possesses a low melting point, it does not appear all evenly distributed and melted from thousands of years of mantle contact.  It appears as if no diffusion through molten mass ever occurred.  In essence, the existence of lodes and crystal colonies nullifies the statement that random mantle deposition resulted in the earth's composition.

Notice how clear some crystals come.  Their clarity implies their formation in an environment not littered with conflicting impurities.

Undivine mantle operates on the rules of particle diffusion and general equal distribution.  Miraculous mantle creates solid lodes and crystal colonies, separate and organized.

You have to see what I am saying.  The point stands simply put.

I'm no geologist but I don't think this makes much sense. You are saying that it must be designed because there is so much order but then you use other arguments like your big bang one that say that there wasn't enough order so it must be designed.

You're confusing.

I just say everything on earth comes with divine design.  No probabilities appeared in the mix.  I demonstrate how just random heating and cooling never actually occurred.

I reveal how the granite layer does not show signs of having gone through a melting and then cooling period.

My statement lies above simply put and in plain language.  Not necessary for geologic experience. 

No kryolite = no gradual formation from melting and cooling

Polonium 218 halos embedded in granite = instant, flash formation of the earth's crust

I will back myself into a corner by attempting to discuss geology like this. Its not my cup of tea. I did however find an argument that satisfactorily refutes your claims on the interwebs. I think you should answer to those questions before assuming anything. It seems you are just regurgitating this Robert Gentry guys work. Have fun with that.

http://www.talkorigins.org/faqs/po-halos/gentry.html

Still my earlier question stands, order and chaos can't both be indicative of creation. You can't have it both ways.

I read that crap article a long time ago.

He changes the subject about what the Precambrian Layer really is.  He knows exactly how there was no recycling of plates and rocks. Polonium 218 defeats him there.  The absence of kryolite defeats him there as well.  In short, these granites went through no deposition or cooling at all.

He submits a hypothesis (and I stress a loose guess), which I already addressed and refuted thoroughly.  It lies in his alpha decay statements and also his statements about varying half lives, which do not mean jack.  Seriously, I talked about half lives and alpha decay and how it is horse puckey.  External alpha decay would only occur on the surface, not deeply embedded in rock.

Millions of millions of years just get knocked out of the water with my above statements.  Varying half lives of even 17 minutes at most still give me the same ammunition.  I summarized his criticisms in a previous post, honestly I did.

No, I did not just copy and plagiarize Gentry's work and take the credit for myself.  My knowledge comes like a quilt from various sources.

I read that crap article a long time ago.

He changes the subject about what the Precambrian Layer really is.  He knows exactly how there was no recycling of plates and rocks. Polonium 218 defeats him there.  The absence of kryolite defeats him there as well.  In short, these granites went through no deposition or cooling at all.

He submits a hypothesis (and I stress a loose guess), which I already addressed and refuted thoroughly.  It lies in his alpha decay statements and also his statements about varying half lives, which do not mean jack.  Seriously, I talked about half lives and alpha decay and how it is horse puckey.  External alpha decay would only occur on the surface, not deeply embedded in rock.

Millions of millions of years just get knocked out of the water with my above statements.  Varying half lives of even 17 minutes at most still give me the same ammunition.  I summarized his criticisms in a previous post, honestly I did.

No, I did not just copy and plagiarize Gentry's work and take the credit for myself.  My knowledge comes like a quilt from various sources.

Well if there is another RE'r around who has any clue about this stuff then I'd hope they can respond to you. As it is right now, simply reading anything you have to say about it or anything this article says about it is over my head. I'm sure I could figure it out but I'd need time to understand it so that I can give a proper response. It'd be idiotic for me to try to formulate a response about a subject I don't feel comfortable with.

Also, I did not say you copied or plagiarized his work. Regurgitated has a far different meaning.

Alright then,

I read his work a while back and I found out that he spent a lot of time beating around his own bush.  He kept talking about slight differences in information that do not amount to a hill of beans.  I suggest never bantering for so long about your own knowledge and then trying to show intellect at the wrong time.

The article could have been made better.

The presence of the halos still defeat his points.  He tries to find holes in armor which he made himself.  And then tries to place the armor on gentry and says, "I have you."

Everyone has just been deceived by long lists of unrelated information.  Everyone just keeps running into liars out there about polonium and everything else, even the big bang.

http://www.halos.com/faq-replies/creation-halos-stand-unrefuted.htm

The polonium 218 halos could not have been caused by radon 222.  There would have been signs.  All the info just comes as disinfo, because these guys do not get published.  Only Gentry gets published.

http://www.living-water.info/10/polonium_halos.htm

Quote from: SeekerOfTruth on September 27, 2013, 11:45:27 AM

Quote from: Rama Set on September 27, 2013, 11:25:29 AM

Quote from: SeekerOfTruth on September 27, 2013, 11:22:12 AM

Quote from: Rama Set on September 27, 2013, 11:18:00 AM

Quote from: SeekerOfTruth on September 27, 2013, 09:35:59 AM

Quote from: Rama Set on September 27, 2013, 09:27:20 AM

No, power is energy per unit of time.  Intensity is power transferred per unit of area.

Yes yes, you are quite right. The confusion is that it is not really meaningful to discuss power in this context without also addressing the area, so typically the law P=a*T^4 is discussed in terms of "per unit area" without explicitly stating P/cm^2, which is indeed an intensity. The fault is mine of course: when you do this stuff for a while you tend to adopt common colloquial shortcuts in the notation and verbiage, which can be confusing if you're not used to it. This is not an excuse, however, many do not know this, and wrong impressions can be given. To restate, I DO take into account the areas in the calculation.   

I see. Sorry for not looking deeper in to the formula. Upon closer inspection you are calculating the power emitted per unit area. This is different than the power received per unit area which would need to be transformed by the inverse square law I think.

How exactly?

The intensity of radiation emitted from the sun would be affected by the Inverse Square Law.

http://en.m.wikipedia.org/wiki/Inverse-square_law

The physical law you used to calculate the intensity of the sun's radiation is the intensity emitted, not received.

Alright, I have to agree this is consistent. But really, isn't this inverse square law just a result of the same reasoning that one uses to explain RE gravity, which is ALSO an inverse square law? Hence wouldn't this only be valid in a RE context?

The inverse square law can be measured here on earth.

You mean for gravity? Can you explain an experiment (done on Earth) that shows this?

Do not change the topic. We were talking about the inverse square law as it relates to the intensity of the sun's EM radiation as it hits the Earth. You calculated the intensity of the emission from the sun and then made an incorrect transformation of that amount as it strikes the Earth. In reality, the intensity of the Sun's EM radiation would be many orders of magnitude lower as it struck the Earth because it intensity varies as a function of 1/d^2.

Everyone has just been deceived by long lists of unrelated information.  Everyone just keeps running into liars out there about polonium and everything else, even the big bang.

http://www.halos.com/faq-replies/creation-halos-stand-unrefuted.htm

The polonium 218 halos could not have been caused by radon 222.  There would have been signs.  All the info just comes as disinfo, because these guys do not get published.  Only Gentry gets published.

http://www.living-water.info/10/polonium_halos.htm

What signs would their be.

Do not change the topic. We were talking about the inverse square law as it relates to the intensity of the sun's EM radiation as it hits the Earth. You calculated the intensity of the emission from the sun and then made an incorrect transformation of that amount as it strikes the Earth. In reality, the intensity of the Sun's EM radiation would be many orders of magnitude lower as it struck the Earth because it intensity varies as a function of 1/d^2.

Yes , according to RE theory, the inverse square law for radiation holds. What I am saying is that this 1/d^2 law also applies to gravity in RE science. This makes 1/d^2 laws suspect in a FE theory, which disagrees with RE gravity. So I am asking for an experiment to measure this supposed 1/d^2 law for EITHER radiation or gravity, that can be done here on Earth, NOT assuming RE or FE is true, but just to see which one MIGHT be true. What CAUSES a 1/d^2 drop off? Because it seems suspicious...contrived even, to make things work out for RE.

Quote from: Rama Set on September 27, 2013, 07:30:23 PM

Do not change the topic. We were talking about the inverse square law as it relates to the intensity of the sun's EM radiation as it hits the Earth. You calculated the intensity of the emission from the sun and then made an incorrect transformation of that amount as it strikes the Earth. In reality, the intensity of the Sun's EM radiation would be many orders of magnitude lower as it struck the Earth because it intensity varies as a function of 1/d^2.

Yes , according to RE theory, the inverse square law for radiation holds. What I am saying is that this 1/d^2 law also applies to gravity in RE science. This makes 1/d^2 laws suspect in a FE theory, which disagrees with RE gravity. So I am asking for an experiment to measure this supposed 1/d^2 law for EITHER radiation or gravity, that can be done here on Earth, NOT assuming RE or FE is true, but just to see which one MIGHT be true. What CAUSES a 1/d^2 drop off? Because it seems suspicious...contrived even, to make things work out for RE.

Because if you multiply a spheres radius by d you multiply the surface area by d^2.

Quote from: SeekerOfTruth on September 27, 2013, 09:18:21 PM

Quote from: Rama Set on September 27, 2013, 07:30:23 PM

Do not change the topic. We were talking about the inverse square law as it relates to the intensity of the sun's EM radiation as it hits the Earth. You calculated the intensity of the emission from the sun and then made an incorrect transformation of that amount as it strikes the Earth. In reality, the intensity of the Sun's EM radiation would be many orders of magnitude lower as it struck the Earth because it intensity varies as a function of 1/d^2.

Yes , according to RE theory, the inverse square law for radiation holds. What I am saying is that this 1/d^2 law also applies to gravity in RE science. This makes 1/d^2 laws suspect in a FE theory, which disagrees with RE gravity. So I am asking for an experiment to measure this supposed 1/d^2 law for EITHER radiation or gravity, that can be done here on Earth, NOT assuming RE or FE is true, but just to see which one MIGHT be true. What CAUSES a 1/d^2 drop off? Because it seems suspicious...contrived even, to make things work out for RE.

Because if you multiply a spheres radius by d you multiply the surface area by d^2.

Okay, I agree. But why does radiation and gravitation FALL OFF AS 1/d^2. We are talking about the propagation of radiation through space, not the geometry of spheres....

Quote from: 11cookeaw1 on September 27, 2013, 10:09:42 PM

Quote from: SeekerOfTruth on September 27, 2013, 09:18:21 PM

Quote from: Rama Set on September 27, 2013, 07:30:23 PM

Do not change the topic. We were talking about the inverse square law as it relates to the intensity of the sun's EM radiation as it hits the Earth. You calculated the intensity of the emission from the sun and then made an incorrect transformation of that amount as it strikes the Earth. In reality, the intensity of the Sun's EM radiation would be many orders of magnitude lower as it struck the Earth because it intensity varies as a function of 1/d^2.

Yes , according to RE theory, the inverse square law for radiation holds. What I am saying is that this 1/d^2 law also applies to gravity in RE science. This makes 1/d^2 laws suspect in a FE theory, which disagrees with RE gravity. So I am asking for an experiment to measure this supposed 1/d^2 law for EITHER radiation or gravity, that can be done here on Earth, NOT assuming RE or FE is true, but just to see which one MIGHT be true. What CAUSES a 1/d^2 drop off? Because it seems suspicious...contrived even, to make things work out for RE.

Because if you multiply a spheres radius by d you multiply the surface area by d^2.

Okay, I agree. But why does radiation and gravitation FALL OFF AS 1/d^2. We are talking about the propagation of radiation through space, not the geometry of spheres....

Because both are emitted in a radial(read, spherical) pattern away from the source, with approximately equal intensity in all directions.  The amount of energy/force remains constant, but the density of that energy/force changes with distance.

Off topic, but it's nice to have someone on the other side that actually halfway cares about the validity of what either side is saying.  Since I haven't already, I'd like to wish you a good, long stay here.

Quote from: SeekerOfTruth on September 27, 2013, 10:14:35 PM

Quote from: 11cookeaw1 on September 27, 2013, 10:09:42 PM

Quote from: SeekerOfTruth on September 27, 2013, 09:18:21 PM

Quote from: Rama Set on September 27, 2013, 07:30:23 PM

Do not change the topic. We were talking about the inverse square law as it relates to the intensity of the sun's EM radiation as it hits the Earth. You calculated the intensity of the emission from the sun and then made an incorrect transformation of that amount as it strikes the Earth. In reality, the intensity of the Sun's EM radiation would be many orders of magnitude lower as it struck the Earth because it intensity varies as a function of 1/d^2.

Yes , according to RE theory, the inverse square law for radiation holds. What I am saying is that this 1/d^2 law also applies to gravity in RE science. This makes 1/d^2 laws suspect in a FE theory, which disagrees with RE gravity. So I am asking for an experiment to measure this supposed 1/d^2 law for EITHER radiation or gravity, that can be done here on Earth, NOT assuming RE or FE is true, but just to see which one MIGHT be true. What CAUSES a 1/d^2 drop off? Because it seems suspicious...contrived even, to make things work out for RE.

Because if you multiply a spheres radius by d you multiply the surface area by d^2.

Okay, I agree. But why does radiation and gravitation FALL OFF AS 1/d^2. We are talking about the propagation of radiation through space, not the geometry of spheres....

Because both are emitted in a radial(read, spherical) pattern away from the source, with approximately equal intensity in all directions.  The amount of energy/force remains constant, but the density of that energy/force changes with distance.

Off topic, but it's nice to have someone on the other side that actually halfway cares about the validity of what either side is saying.  Since I haven't already, I'd like to wish you a good, long stay here.

Thanks man! There's no reason we can't all be civil, even if we disagree  .

So by your reasoning, if a light source was not emitted spherically, like from a flashlight or something, then the 1/d^2 law would not hold? I find this curious, because the equations do not insist on the shape of the source -- it is not some variable that you input.