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Author Topic: 50 Plus Reasons The Earth Is Not Flat  (Read 225499 times)

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Offline Meg

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Re: 50 Plus Reasons The Earth Is Not Flat
« Reply #1200 on: February 16, 2018, 12:27:09 PM »
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  • Just because you think you are a good, knowledgable christian does not mean you understand anything about Eastern terms, their applications, origins, or reality. How many books have you read on biology, reflexology, energy, martial arts, exercises, and actually put it into practice in the physical world?


    Actually, I was a new ager long ago. I do know something about it.

    Your user name - "Student of Qi." I assume that you are a practitioner of Qi/Chi? 
    "It is licit to resist a Sovereign Pontiff who is trying to destroy the Church. I say it is licit to resist him in not following his orders and in preventing the execution of his will. It is not licit to Judge him, to punish him, or to depose him, for these are acts proper to a superior."

    ~St. Robert Bellarmine
    De Romano Pontifice, Lib.II, c.29

    Offline Meg

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1201 on: February 16, 2018, 12:30:11 PM »
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  • If we are a part of the Church and the Church is the Mystical Body of Christ and the His Body is part of His kingdome, I think it logical to conclude that we are "incorporated" into God and and likewise His Kingdome inside us. Is it that hard to comprehend? It seems a simple matter to me... I do knot deny that Heaven is also a physical place, but it also has a spiritual nature, and the spiritual envelopes the physical in a way i don't think most will ever even partly understand.
     

    If you don't deny that Heaven is a physical place, then why do you accuse flat-earthers for believing as such?
    "It is licit to resist a Sovereign Pontiff who is trying to destroy the Church. I say it is licit to resist him in not following his orders and in preventing the execution of his will. It is not licit to Judge him, to punish him, or to depose him, for these are acts proper to a superior."

    ~St. Robert Bellarmine
    De Romano Pontifice, Lib.II, c.29


    Offline Student of Qi

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1202 on: February 16, 2018, 12:33:32 PM »
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  • Actually, I was a new ager long ago. I do know something about it.

    Your user name - "Student of Qi." I assume that you are a practitioner of Qi/Chi?
    To certain degree, yes, but what you think of as Qi is most likely not what I and many shaolin monks and average martial artists think of it as. Let's take this to another thread, and tell me what you believe "Qi" is.
    Many people say "For the Honor and Glory of God!" but, what they should say is "For the Love, Glory and Honor of God". - Fr. Paul of Moll

    Offline Meg

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1203 on: February 16, 2018, 12:37:23 PM »
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  • To certain degree, yes, but what you think of as Qi is most likely not what I and many shaolin monks and average martial artists think of it as. Let's take this to another thread, and tell me what you believe "Qi" is.

    I could care less about how you have incorporated your new age beliefs into Catholicism. You are hardly the first to do so. But you are wrong to believe that we have to accept you error.

    That "God is within you" is a common theme of new agers.

    Heaven is place. That's what the Church teaches. 

    "It is licit to resist a Sovereign Pontiff who is trying to destroy the Church. I say it is licit to resist him in not following his orders and in preventing the execution of his will. It is not licit to Judge him, to punish him, or to depose him, for these are acts proper to a superior."

    ~St. Robert Bellarmine
    De Romano Pontifice, Lib.II, c.29

    Offline sundance

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1204 on: February 16, 2018, 01:54:26 PM »
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  • Student of IQ said:
    "One of the things I find discouraging with the Flat Earthers is the continual "God feels so much closer then before" argument, which I can't help but think is completely heretical. As Catholics we believe in his Omnipresence, meaning He is everywhere, even inside of us. 
    You say "Where is Heaven?" but when Daniel asked God how far away it was, God replied "The length of a prayer". So, if you say a prayer silently in your head it technically did not go anywhere. Therefore, Heaven is inside YOU and there is no need to look up to the sky and think that is how far away your Creator is from you. He sees all, hears all, knows all. You are connected to God most assuredly if you are in the state of Sanctifying Grace and and even the Holy Bible says tjat you are the Temple of The lord, and no one but you can drive him from that Temple. Remember that.
    Hence, for me this argument is the worst for embracing Flat Earth."

    Catechism  Question:
     
    Where is God?
     
    God is everywhere, but He shows forth His glory especially in Heaven.
     
    Isaiah    40:22
    It is he that sitteth upon the circle of the earth, and the inhabitants thereof are as locusts: he that stretcheth out the heavens as nothing, and spreadeth them out as a tent to dwell in. 


    Offline Neil Obstat

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1205 on: February 16, 2018, 02:06:31 PM »
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  • .
    I have a 1609 Catholic Bible that infallibly defines "A firmament" (Genesis 1:6, "And God said: Let there be a firmament made amidst the waters: and let it divide the waters from the waters") as follows: 

    .
    "Chap. 1.   Ver. 6.   A firmament.  By this name is here understood the whole space between the earth, and the highest stars. The lower part of which divideth the waters that are upon the earth, from those that are above in the clouds."
    .
    Consequently, the firmament is not any solid or hard material but rather it is the zone or space between the earth and the highest stars which modern man would call the atmosphere and outer space. The water that is divided is the separation between liquid water as it occurs on the surface of the earth and vaporous water as it occurs in clouds in the sky. The ancient Bible authors had no way of understanding that the principal limit of water vapor is the earth's atmosphere. But today, we know that even in outer space, water can exist albeit in a very sparse and expanded form, with much distance between each water molecule without confines of ambient air pressure.
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    Offline Neil Obstat

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1206 on: February 16, 2018, 02:07:47 PM »
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  • Heaven is place. That's what the Church teaches.
    .
    Where do you find the Church teaching that "heaven is place?"
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    Offline Student of Qi

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1207 on: February 16, 2018, 02:35:11 PM »
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  • I could care less about how you have incorporated your new age beliefs into Catholicism. You are hardly the first to do so. But you are wrong to believe that we have to accept you error.

    That "God is within you" is a common theme of new agers.

    Heaven is place. That's what the Church teaches.
    You are still wrong to maintain I have anything to do with new agers. Maybe it's your own assoiations with them that makes you like a rock to argue with. If you don't want to hear what I have to say, so be it. But you know nothing of my studies, nor anything of how it has helped ailing people around me. You are presumptuous.
      Secondly, I don't care if you think the same thing(s) I do on this matter and you are again presumtuous for thinking I believe you have to hold the same ideals as me.
      Third, you can't claim I'm imppssing my "error" on others when you don't even actually know what I think. - And I mean this in respect to the subject of qi, since this is what it seems to me that you are refering to.

    If there is something wrong with my statement of heaven or at least a part of it being interior, I retract it. I've already apologielzed previously. But it's probably heterodox at the very least.

    I also contend with you on the "God is within you", as a Catholic you should now that is what the Church teaches. Especially ao in terms of the reception of the Eucharist, because as we change food we eat into our bodily materials, the oppoite happens with the Eucharist. It is in the Catechism that it changes us in to the Body of Christ. So, yes, God is in us both physically and spiritualy.
    However, I get this is not what new agers probably mean by that.
    Many people say "For the Honor and Glory of God!" but, what they should say is "For the Love, Glory and Honor of God". - Fr. Paul of Moll


    Offline Student of Qi

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1208 on: February 16, 2018, 03:07:53 PM »
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  • You say when flat earthers are faced with various questions they just fall silent and can give no explanation.  Now, while its true, flat earthers cannot explain everything about the flat earth model, the one thing they claim beyond doubt is that earth is not a globe.  And for that, we have proof.  Globalists have no proof earth is a globe.  Globalists have no explanation for why there is no curve, how water sticks to a ball, why boats do not disappear behind the curve, even how Scripture describes a flat earth and traditional teaching on the subject.  Yet, you come in here and say we can't explain.  Now, that's weird.  When globalists can address any one of these, we can discuss what flat earthers don't know.  Until then, the ball is in the globalist camp.    
    Getting back on subject...

    The Flat Earthers may unanimously say the Earth is not a globe, but they have not provided mathematics or other such articles that I believe I can agree with without reserve. Neil has provided articles on the function of telecommunications, and St. Ignatious has provided some insight into the use of radio waves and their need for factoring in curves, etc. Flat Earthers have not provided examples other then the lack of horizon and some pieces of scripture that are not explicitly in favor of F.E. even if it COULD be understood that way.
     Also, for me AES' demonstration of the lack of unanimity of the fathers and their understanding/use of scripture (esp. St. Basil) convinces me that scripture is not enough. This especially because the F.Es are saying the Church in antiquity supports F.E. but on the flip side is the bits from Challoner -for example- which make a full contradiction that what the F.E.ers are saying, because the Challoner is approved. That would mean the Church is wrong... and we don't want to go there.
    Many people say "For the Honor and Glory of God!" but, what they should say is "For the Love, Glory and Honor of God". - Fr. Paul of Moll

    Offline kiwiboy

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1209 on: February 16, 2018, 03:36:20 PM »
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  • Getting back on subject...

    The Flat Earthers may unanimously say the Earth is not a globe, but they have not provided mathematics or other such articles that I believe I can agree with without reserve. Neil has provided articles on the function of telecommunications, and St. Ignatious has provided some insight into the use of radio waves and their need for factoring in curves, etc. Flat Earthers have not provided examples other then the lack of horizon and some pieces of scripture that are not explicitly in favor of F.E. even if it COULD be understood that way.
     Also, for me AES' demonstration of the lack of unanimity of the fathers and their understanding/use of scripture (esp. St. Basil) convinces me that scripture is not enough. This especially because the F.Es are saying the Church in antiquity supports F.E. but on the flip side is the bits from Challoner -for example- which make a full contradiction that what the F.E.ers are saying, because the Challoner is approved. That would mean the Church is wrong... and we don't want to go there.
    Globers have not provided a single proof of the curve, or a single proof for water that sticks to the outside of a ball.  They have not explained how entire ocean bodies of water curve.  The rest of their explanations are lengthy but unconvincing on any subject, but mostly because they fail to address the big questions.  As for Challoner, if that is all you took away from the recent discussions, you're dismissing the argument for a very minor piece of evidence that may or may not support your position.  The problem with the glober's arguments is the content.  Sometimes, they look impressive, but when examined closely, they do not work.  

    Offline Neil Obstat

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1210 on: February 16, 2018, 03:49:52 PM »
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  • .
    The following web page gives a general explanation for the phenomenon of refraction over the surface of the earth.
    Factors such as air temperature, air density, water vapor and distance to subject viewed must be considered:
    .
    https://aty.sdsu.edu/explain/atmos_refr/bending.html
    .
    Calculating Ray Bending
    Introduction
    Mirages and other refraction phenomena are the result of the bending of rays in the Earth's atmosphere. But how much do rays bend? This page shows how to estimate the bending from simple information on temperature gradients in the lower atmosphere.

    I'll try to keep the treatment as simple as possible here, so a lot of refinements needed for an accurate calculation will be ignored. My aim is to show the basic physics, not to produce exact numbers. The accuracy of the results will be roughly 10%. (I'll also give the results of more accurate calculations for comparison.)

    This kind of rough approximation is what one of my professors used to call a “bathtub calculation” — the sort of thing you can do in your head in the bathtub, without pencil and paper. (I'll actually not be quite that rough, so that the numerical results will be useful practical guides.)

    By the way: if you're not familiar with the concepts of geometrical optics, I suggest you take a look at my introduction to optics page, before proceeding.

    Basic facts
    Let's start out with some basic facts about the atmosphere. Air is nearly a perfect gas, so its density is accurately given by the ideal gas law, and so is proportional to P/T, where P is the pressure, and T is the (absolute) temperature. (We don't need to worry about the “gas constant” or what units we're using, because only relative changes will be needed here.)

    The density is important, because rays of light are bent toward the denser gas where the density changes from place to place. Air is densest near or at the surface of the Earth, and its refractivity is very nearly proportional to the density.

    The atmosphere is also curved, as it follows the shape of the Earth. In the calculations below, we'll adopt a rough value of 6400 km as the radius of curvature, to make the calculations simple. We'll also assume the density depends only on distance from the center of curvature — a pretty good assumption.

    We'll measure temperatures in Celsius degrees, which are 9/5 the size of Fahrenheit degrees. Just a reminder: temperatures have to be measured from absolute zero, which is −273°C. That means a comfortable outdoor temperature of 25°C (77°F) is nearly 300 K, where the “K” means temperature on the absolute scale, named after Lord Kelvin; the Kelvin scale has degrees the same size as Celsius degrees, but starts at absolute zero instead of at the freezing point of water. I'll use 300 K as a typical temperature, even though it's a little warmer than average.

    Straight lines?
    Everybody says “light travels in straight lines,” but if that were true we wouldn't have atmospheric refraction or mirages. The common saying should be amended by adding “in a homogeneous medium,” but everybody forgets to mention that. (Air isn't homogeneous: it has different densities at different heights above the surface. That's the whole point of this exercise.)

    Even if the air were isothermal (i.e., if it had the same temperature everywhere), it wouldn't have the same density everywhere, because it's in hydrostatic equilibrium: the air at each level in compressed by the weight of the air above it. The higher you go, the less compression; the density is lower at greater heights (which is why it's hard to breathe on mountaintops, and why jet airliners have to be pressurized.) So let's first figure out how to make the atmosphere (locally) homogeneous.

    Obviously, if the density is decreasing upward because of the decreasing pressure, we can counteract this decrease by changing the temperature. In fact, the pressure falls off by about 1 part in 8000 for every meter of increase in height above sea level. We can compensate for this by decreasing the temperature by the same 1 part in 8000 for every meter of height. If the temperature is 300 K, we'll need a temperature decrease with height of 300°/8000 m, or .0375° per meter, to make up for the effect of decreasing pressure. If the temperature is colder than 300 K, a proportionately smaller lapse rate will suffice.

    (For comparison, Wegener (1918 ) calculates a value of 0.034°/m for 273 K; so the “bathtub” estimate is off by less than 10%, as promised above.)

    Rays will be perfectly straight in a layer with this lapse rate ; but such a temperature gradient hardly ever occurs. It's over three times the adiabatic lapse rate (i.e., the gradient assumed by air in free convection), so this large drop in temperature with height can be maintained only near a warm surface. In models that reproduce inferior mirages over water, this value is reached only in the lowest 10 meters or so of the atmosphere.

    An aside
    You may have noticed, when I did the division a couple of paragraphs above, that a length of 8000 m somehow appeared in the denominator. This turns out to be the height the whole atmosphere would have if it had the same density throughout that it actually has at sea level. It's sometimes called “the height of the homogeneous atmosphere,” or the “scale height” of the atmosphere. (Radau calls it the reduced height — a better name.) Although the real atmosphere actually extends much higher than this, it's a useful characteristic length to keep in mind.

    Circulating rays
    Next, let's ask what temperature gradient will bend the rays so they follow the curve of the Earth. This condition divides temperature profiles that produce ducting from those without a duct. As ducts produce certain types of mirages, it is helpful to understand the minimum requirements for a duct.

    Note that a ray concentric with the surface of the Earth is not straight, but it is always “horizontal” — in contrast to the ray discussed in the previous section, which was really straight, and so could be horizontal at only one point. (See the figure below.)

    Some people have worried about how to apply the refraction law to such a horizontal ray of light, because it does not cross any horizontal boundaries between denser and less-dense air in a stratified atmosphere. (In general, the curvature of the ray is different from the curvature of the Earth's surface; but here, we are only concerned with rays that bend just enough to follow the curve of the Earth.) The simplest solution to the “horizontal-ray paradox” is to remind ourselves that “rays” are an unrealistic mental construct: in reality, we always have a beam of light; infinitely narrow “rays” don't really exist. So let's just apply Huygens's principle to such a beam:
    In the figure at the left, the heavy arc denotes the surface of the Earth, with center at C, and the lighter arcs AB and A'B' concentric with it represent the sides of a beam of light propagating at constant height above the surface, from AA' to BB'. We can regard AA' and BB' as wavefronts of the (bending) horizontal beam at two different places; the direction of propagation is perpendicular to the wavefront. (Equivalently, we can say that the direction of propagation of a horizontal beam must always be perpendicular to the local vertical; the radial lines CAA' and CBB' radiating from the center of the Earth are the local verticals at A and B respectively.)

    Evidently, to make the beam follow the Earth's curvature, its lower edge AB must travel more slowly than its upper edge A'B'. The speed of propagation is in fact just inversely proportional to the refractive index, n. As the distance to be traversed by our circulating beam is proportional to the distance R  from the center of the Earth, we require that 1/n (which is proportional to the speed) be proportional to R ; or, in other words, we need the product nR  to be constant, independent of height above the Earth. (This will make the “optical path length” along AB the same as along A'B'.)

    [If you find this argument too superficial to be convincing, you can go read the more rigorous derivations of the nR = constant condition by Biot (1836), by Auer and Standish (2000), or one of the papers cited in my “horizontal-ray paradox” file, such as Bravais (1856) or Thomson (1872). There's also a wonderful mathematical treatment of circulating rays by Kummer (1860); a French translation of it by Verdet (1861) is available on Gallica.]

    To have nR  remain constant with height, n must decrease by 1 part in 6.4 × 106 for every meter of height, as R = 6400 km = 6.4 × 106 m. But the refractivity ( n − 1 ) is only about 1/3200 of n ; so the density of the air [which is proportional to ( n − 1 ), not to n] must fall by about 3200/6.4 × 106 m, or 1 part in 2000 for every meter.

    Now, the decrease in density due to the decrease in pressure with height (1 part in 8000 per meter) is only 1/4 of this, so we need 3 times as much decrease from the temperature gradient, or 3 parts in 8000 per meter. That means the temperature must increase by 3 parts in 8000 of the 300 K, or about 900/8000 of a degree = 0.11° per meter. So a temperature inversion (i.e., increasing upward, instead of the usual decrease) of about 0.11°/m will produce a circulating beam or ray.

    As a check, we can do the arithmetic a little differently. Because the refractive index n decreases by 1 part in a million per degree, and we need a decrease in n of 1 part in 6.4 million per meter, we would need about 1/6.4 of a degree per meter if the pressure stayed constant. But, as the pressure alone gives an effect equivalent to 3/80 of a degree per meter, we really need only ( 1/6.4 − 3/80 ); or, expressing the fractions as decimals, about 0.16 − 0.04 = 0.12°/m. (The lapse rate is, of course, the negative of this value.)

    [For comparison, Wegener (1918 ) gives 0.114°/m as the critical temperature gradient; and J. de Graaff Hunter (1913) gives 0.066°F/foot, which corresponds to 0.116°C/m.]

    Note that −0.12°/m is −120°/km, almost −20 times the lapse rate of 6.5K/km in the Standard Atmosphere. So the slight change of temperature with height in the Standard Atmosphere has hardly any effect on the ray curvature. This justifies, a posteriori, the neglect of temperature effects at the start of the arguments presented on this page.

    Ray curvature in an arbitrary atmosphere
    We can now estimate the ray curvature for any temperature gradient. If a positive lapse rate of 0.034 K/m produces straight rays (zero curvature), and a negative lapse rate of 0.12 K/m makes the rays match the curvature of the Earth, then for every (0.12 + 0.034 = 0.154) K/m change in lapse rate, the ray curvature changes by 1 in units of the Earth's curvature. So the ray curvature for an arbitrary lapse rate  γ K/m will be

    k  = ( 0.034 − γ ) / 0.154 ,
    where we take γ to be positive if the temperature decreases with height, and a positive curvature means a ray concave toward the Earth.

    Example 1: the Standard Atmosphere: In the Standard Atmosphere, the lapse rate is 6.5°/km or  γ = 0.0065 K/m. The numerator of the formula above becomes .034 − .0065 = .0275, so the ratio k is about 1/5.6 or 0.179. In other words, the ray curvature is not quite 18% that of the Earth; the radius of curvature of the ray is about 5.6 times the Earth's radius.

    Example 2: free convection:In free convection, the (adiabatic) lapse rate is about 10.6°/km or  γ = 0.0106 K/m. The numerator of the formula above becomes .034 − .0106 = .0234, so the ratio k is about 1/6.6 or 0.152. In other words, the ray curvature is about 15% that of the Earth; the radius of curvature of the ray is about 6.6 times the Earth's radius. This is close to the condition of the atmosphere near the ground in the middle of the day, when most surveying is done; the value calculated is close to the values found in practical survey work.

    Bear in mind that these values apply only near sea level. At great elevations, the density of the air is less, so the ray curvature is also less. Likewise, these estimates are for sea-level temperatures near 300 K; in cold places, the density of the air is greater, and so is the ray curvature. And remember that these curvature estimates apply only to rays that are horizontal, or nearly so.

    Note: You might have noticed that these values around 1/6 are about the fraction by which the setting Sun is flattened at the horizon. This is not a coincidence; these two quantities are in fact equal, as can easily be shown.

    Historical note
    The method used here to find the ratio of curvatures of the ray and the Earth is essentially the argument employed by Thomas Young in 1821. He also obtained a ratio of 5.6, as in the first example above.

    Copyright © 2002 – 2009, 2012, 2016, 2018 Andrew T. Young
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    Offline kiwiboy

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1211 on: February 16, 2018, 04:18:39 PM »
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  • .
    The following web page gives a general explanation for the phenomenon of refraction over the surface of the earth.
    Factors such as air temperature, air density, water vapor and distance to subject viewed must be considered:
    .
    https://aty.sdsu.edu/explain/atmos_refr/bending.html
    .
    Calculating Ray Bending
    Introduction
    Mirages and other refraction phenomena are the result of the bending of rays in the Earth's atmosphere. But how much do rays bend? This page shows how to estimate the bending from simple information on temperature gradients in the lower atmosphere.

    I'll try to keep the treatment as simple as possible here, so a lot of refinements needed for an accurate calculation will be ignored. My aim is to show the basic physics, not to produce exact numbers. The accuracy of the results will be roughly 10%. (I'll also give the results of more accurate calculations for comparison.)

    This kind of rough approximation is what one of my professors used to call a “bathtub calculation” — the sort of thing you can do in your head in the bathtub, without pencil and paper. (I'll actually not be quite that rough, so that the numerical results will be useful practical guides.)

    By the way: if you're not familiar with the concepts of geometrical optics, I suggest you take a look at my introduction to optics page, before proceeding.

    Basic facts
    Let's start out with some basic facts about the atmosphere. Air is nearly a perfect gas, so its density is accurately given by the ideal gas law, and so is proportional to P/T, where P is the pressure, and T is the (absolute) temperature. (We don't need to worry about the “gas constant” or what units we're using, because only relative changes will be needed here.)

    The density is important, because rays of light are bent toward the denser gas where the density changes from place to place. Air is densest near or at the surface of the Earth, and its refractivity is very nearly proportional to the density.

    The atmosphere is also curved, as it follows the shape of the Earth. In the calculations below, we'll adopt a rough value of 6400 km as the radius of curvature, to make the calculations simple. We'll also assume the density depends only on distance from the center of curvature — a pretty good assumption.

    We'll measure temperatures in Celsius degrees, which are 9/5 the size of Fahrenheit degrees. Just a reminder: temperatures have to be measured from absolute zero, which is −273°C. That means a comfortable outdoor temperature of 25°C (77°F) is nearly 300 K, where the “K” means temperature on the absolute scale, named after Lord Kelvin; the Kelvin scale has degrees the same size as Celsius degrees, but starts at absolute zero instead of at the freezing point of water. I'll use 300 K as a typical temperature, even though it's a little warmer than average.

    Straight lines?
    Everybody says “light travels in straight lines,” but if that were true we wouldn't have atmospheric refraction or mirages. The common saying should be amended by adding “in a homogeneous medium,” but everybody forgets to mention that. (Air isn't homogeneous: it has different densities at different heights above the surface. That's the whole point of this exercise.)

    Even if the air were isothermal (i.e., if it had the same temperature everywhere), it wouldn't have the same density everywhere, because it's in hydrostatic equilibrium: the air at each level in compressed by the weight of the air above it. The higher you go, the less compression; the density is lower at greater heights (which is why it's hard to breathe on mountaintops, and why jet airliners have to be pressurized.) So let's first figure out how to make the atmosphere (locally) homogeneous.

    Obviously, if the density is decreasing upward because of the decreasing pressure, we can counteract this decrease by changing the temperature. In fact, the pressure falls off by about 1 part in 8000 for every meter of increase in height above sea level. We can compensate for this by decreasing the temperature by the same 1 part in 8000 for every meter of height. If the temperature is 300 K, we'll need a temperature decrease with height of 300°/8000 m, or .0375° per meter, to make up for the effect of decreasing pressure. If the temperature is colder than 300 K, a proportionately smaller lapse rate will suffice.

    (For comparison, Wegener (1918 ) calculates a value of 0.034°/m for 273 K; so the “bathtub” estimate is off by less than 10%, as promised above.)

    Rays will be perfectly straight in a layer with this lapse rate ; but such a temperature gradient hardly ever occurs. It's over three times the adiabatic lapse rate (i.e., the gradient assumed by air in free convection), so this large drop in temperature with height can be maintained only near a warm surface. In models that reproduce inferior mirages over water, this value is reached only in the lowest 10 meters or so of the atmosphere.

    An aside
    You may have noticed, when I did the division a couple of paragraphs above, that a length of 8000 m somehow appeared in the denominator. This turns out to be the height the whole atmosphere would have if it had the same density throughout that it actually has at sea level. It's sometimes called “the height of the homogeneous atmosphere,” or the “scale height” of the atmosphere. (Radau calls it the reduced height — a better name.) Although the real atmosphere actually extends much higher than this, it's a useful characteristic length to keep in mind.

    Circulating rays
    Next, let's ask what temperature gradient will bend the rays so they follow the curve of the Earth. This condition divides temperature profiles that produce ducting from those without a duct. As ducts produce certain types of mirages, it is helpful to understand the minimum requirements for a duct.

    Note that a ray concentric with the surface of the Earth is not straight, but it is always “horizontal” — in contrast to the ray discussed in the previous section, which was really straight, and so could be horizontal at only one point. (See the figure below.)

    Some people have worried about how to apply the refraction law to such a horizontal ray of light, because it does not cross any horizontal boundaries between denser and less-dense air in a stratified atmosphere. (In general, the curvature of the ray is different from the curvature of the Earth's surface; but here, we are only concerned with rays that bend just enough to follow the curve of the Earth.) The simplest solution to the “horizontal-ray paradox” is to remind ourselves that “rays” are an unrealistic mental construct: in reality, we always have a beam of light; infinitely narrow “rays” don't really exist. So let's just apply Huygens's principle to such a beam:
    In the figure at the left, the heavy arc denotes the surface of the Earth, with center at C, and the lighter arcs AB and A'B' concentric with it represent the sides of a beam of light propagating at constant height above the surface, from AA' to BB'. We can regard AA' and BB' as wavefronts of the (bending) horizontal beam at two different places; the direction of propagation is perpendicular to the wavefront. (Equivalently, we can say that the direction of propagation of a horizontal beam must always be perpendicular to the local vertical; the radial lines CAA' and CBB' radiating from the center of the Earth are the local verticals at A and B respectively.)

    Evidently, to make the beam follow the Earth's curvature, its lower edge AB must travel more slowly than its upper edge A'B'. The speed of propagation is in fact just inversely proportional to the refractive index, n. As the distance to be traversed by our circulating beam is proportional to the distance R  from the center of the Earth, we require that 1/n (which is proportional to the speed) be proportional to R ; or, in other words, we need the product nR  to be constant, independent of height above the Earth. (This will make the “optical path length” along AB the same as along A'B'.)

    [If you find this argument too superficial to be convincing, you can go read the more rigorous derivations of the nR = constant condition by Biot (1836), by Auer and Standish (2000), or one of the papers cited in my “horizontal-ray paradox” file, such as Bravais (1856) or Thomson (1872). There's also a wonderful mathematical treatment of circulating rays by Kummer (1860); a French translation of it by Verdet (1861) is available on Gallica.]

    To have nR  remain constant with height, n must decrease by 1 part in 6.4 × 106 for every meter of height, as R = 6400 km = 6.4 × 106 m. But the refractivity ( n − 1 ) is only about 1/3200 of n ; so the density of the air [which is proportional to ( n − 1 ), not to n] must fall by about 3200/6.4 × 106 m, or 1 part in 2000 for every meter.

    Now, the decrease in density due to the decrease in pressure with height (1 part in 8000 per meter) is only 1/4 of this, so we need 3 times as much decrease from the temperature gradient, or 3 parts in 8000 per meter. That means the temperature must increase by 3 parts in 8000 of the 300 K, or about 900/8000 of a degree = 0.11° per meter. So a temperature inversion (i.e., increasing upward, instead of the usual decrease) of about 0.11°/m will produce a circulating beam or ray.

    As a check, we can do the arithmetic a little differently. Because the refractive index n decreases by 1 part in a million per degree, and we need a decrease in n of 1 part in 6.4 million per meter, we would need about 1/6.4 of a degree per meter if the pressure stayed constant. But, as the pressure alone gives an effect equivalent to 3/80 of a degree per meter, we really need only ( 1/6.4 − 3/80 ); or, expressing the fractions as decimals, about 0.16 − 0.04 = 0.12°/m. (The lapse rate is, of course, the negative of this value.)

    [For comparison, Wegener (1918 ) gives 0.114°/m as the critical temperature gradient; and J. de Graaff Hunter (1913) gives 0.066°F/foot, which corresponds to 0.116°C/m.]

    Note that −0.12°/m is −120°/km, almost −20 times the lapse rate of 6.5K/km in the Standard Atmosphere. So the slight change of temperature with height in the Standard Atmosphere has hardly any effect on the ray curvature. This justifies, a posteriori, the neglect of temperature effects at the start of the arguments presented on this page.

    Ray curvature in an arbitrary atmosphere
    We can now estimate the ray curvature for any temperature gradient. If a positive lapse rate of 0.034 K/m produces straight rays (zero curvature), and a negative lapse rate of 0.12 K/m makes the rays match the curvature of the Earth, then for every (0.12 + 0.034 = 0.154) K/m change in lapse rate, the ray curvature changes by 1 in units of the Earth's curvature. So the ray curvature for an arbitrary lapse rate  γ K/m will be

    k  = ( 0.034 − γ ) / 0.154 ,
    where we take γ to be positive if the temperature decreases with height, and a positive curvature means a ray concave toward the Earth.

    Example 1: the Standard Atmosphere: In the Standard Atmosphere, the lapse rate is 6.5°/km or  γ = 0.0065 K/m. The numerator of the formula above becomes .034 − .0065 = .0275, so the ratio k is about 1/5.6 or 0.179. In other words, the ray curvature is not quite 18% that of the Earth; the radius of curvature of the ray is about 5.6 times the Earth's radius.

    Example 2: free convection:In free convection, the (adiabatic) lapse rate is about 10.6°/km or  γ = 0.0106 K/m. The numerator of the formula above becomes .034 − .0106 = .0234, so the ratio k is about 1/6.6 or 0.152. In other words, the ray curvature is about 15% that of the Earth; the radius of curvature of the ray is about 6.6 times the Earth's radius. This is close to the condition of the atmosphere near the ground in the middle of the day, when most surveying is done; the value calculated is close to the values found in practical survey work.

    Bear in mind that these values apply only near sea level. At great elevations, the density of the air is less, so the ray curvature is also less. Likewise, these estimates are for sea-level temperatures near 300 K; in cold places, the density of the air is greater, and so is the ray curvature. And remember that these curvature estimates apply only to rays that are horizontal, or nearly so.

    Note: You might have noticed that these values around 1/6 are about the fraction by which the setting Sun is flattened at the horizon. This is not a coincidence; these two quantities are in fact equal, as can easily be shown.

    Historical note
    The method used here to find the ratio of curvatures of the ray and the Earth is essentially the argument employed by Thomas Young in 1821. He also obtained a ratio of 5.6, as in the first example above.

    Copyright © 2002 – 2009, 2012, 2016, 2018 Andrew T. Young
    Too superficial to be unconvincing?  No.  What's unconvincing is the argument itself.  It is full of "if this, then that" none of which is readily provable.  Again, we have to take some guy's word?  And what is this? 
    " So let's first figure out how to make the atmosphere (locally) homogeneous.

    Obviously, if the density is decreasing upward because of the decreasing pressure, we can counteract this decrease by changing the temperature. In fact, the pressure falls off by about 1 part in 8000 for every meter of increase in height above sea level. We can compensate for this by decreasing the temperature by the same 1 part in 8000 for every meter of height. If the temperature is 300 K, we'll need a temperature decrease with height of 300°/8000 m, or .0375° per meter, to make up for the effect of decreasing pressure. If the temperature is colder than 300 K, a proportionately smaller lapse rate will suffice."

    Making the atmosphere locally homogenous doesn't affect other data?  "Decreasing the temperature by the same 1 part of 8000 for every meter of height" doesn't affect outcome?  How do we know either way?  There are many other problems with this article, but for now, the big question remains, is refraction always a present condition?  If arbitrary, and it is, it cannot be relied on to prove the earth is a globe.  This is one of those wordy proofs that fails to prove what they say it does.

    Offline Neil Obstat

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1212 on: February 16, 2018, 04:30:59 PM »
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  • Too superficial to be unconvincing?  No.  What's unconvincing is the argument itself.  It is full of "if this, then that" none of which is readily provable.  Again, we have to take some guy's word?  And what is this?  
    " So let's first figure out how to make the atmosphere (locally) homogeneous.

    Obviously, if the density is decreasing upward because of the decreasing pressure, we can counteract this decrease by changing the temperature. In fact, the pressure falls off by about 1 part in 8000 for every meter of increase in height above sea level. We can compensate for this by decreasing the temperature by the same 1 part in 8000 for every meter of height. If the temperature is 300 K, we'll need a temperature decrease with height of 300°/8000 m, or .0375° per meter, to make up for the effect of decreasing pressure. If the temperature is colder than 300 K, a proportionately smaller lapse rate will suffice."

    Making the atmosphere locally homogenous doesn't affect other data?  "Decreasing the temperature by the same 1 part of 8000 for every meter of height" doesn't affect outcome?  How do we know either way?  There are many other problems with this article, but for now, the big question remains, is refraction always a present condition?  If arbitrary, and it is, it cannot be relied on to prove the earth is a globe.  This is one of those wordy proofs that fails to prove what they say it does.
    .
    Just because you don't understand it doesn't make it objectionable.
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    The article never said that refraction is always a present condition. Your comprehension is lacking. That's the problem.
    .
    The article does not claim that arbitrary refraction proves the earth is a globe. That's a straw man.
    .
    Evidence for the sphericity of the earth is everywhere we look; we don't have to rely on refraction to "prove" it.
    .
    .--. .-.-.- ... .-.-.- ..-. --- .-. - .... . -.- .. -. --. -.. --- -- --..-- - .... . .--. --- .-- . .-. .- -. -.. -....- -....- .--- ..- ... - -.- .. -.. -.. .. -. --. .-.-.

    Offline WholeFoodsTrad

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1213 on: February 21, 2018, 04:27:02 PM »
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  • .
    Nonsense.
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    The Church teaches that "a firmament" in Gen. 1:6 is the whole space between the earth and the furthest stars.
    .
    You lose.
    First, what is your source Neil?  
    "Even a man who is pure in heart and says his prayers by night
    may become a wolf when the wolfbane blooms and the autumn moon is bright."

    Offline Neil Obstat

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    Re: 50 Plus Reasons The Earth Is Not Flat
    « Reply #1214 on: February 21, 2018, 06:01:07 PM »
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  • First, what is your source Neil?  
    .
    I already explained what my source is, a Douay-Rheims bible, 1609 edition (pre-KJV), reprinted by TAN Books. 
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    The Bible's text itself contains no definition of what the sacred author meant by the word(s) the Church translates as "a firmament."
    .
    The Church has seen fit to attribute to that term the meaning she has, because scientific evidence cannot be in contradiction to the revelation of God. The definition I provide is found in the footnote for Genesis 1:6.
    .
    Consequently, while a "hard" or "metallic" substance or the like, might be legitimate interpretation in general, it is in obvious contradiction to the known facts of modern science, therefore "a firmament" as used here in particular cannot have been meant to be "hard" or "metallic" by the sacred author.
    .
    By "a firmament" the author meant a zone obedient to natural law which separates the water above us (which sometimes falls as rain) from the earth below us. Even so, no rain was known to fall from the sky prior to the Great Flood, but the plants received water sustenance from springs that bubbled up from the earth and from a light dew that fell at night. Consequently it's easy for us to understand why the neighbors ridiculed Noah, for they had no idea that water could ever fall as very heavy rain, let alone for 40 days and nights. Furthermore, it underscores the tremendous FAITH of Noah who nonetheless persisted in his great work for 100 years even though there was no natural reason for him to do so from what the cuмulative worldly scientific knowledge could provide at the time.

    .--. .-.-.- ... .-.-.- ..-. --- .-. - .... . -.- .. -. --. -.. --- -- --..-- - .... . .--. --- .-- . .-. .- -. -.. -....- -....- .--- ..- ... - -.- .. -.. -.. .. -. --. .-.-.