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Centaur
Pooh-Bah
Reged: 07/12/04
Posts: 1123
Loc: Chicago
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Jim and Timo, you have mentioned a cosine rule several times. If properly applied, the result would be a tiny shrinkage of the Moon's apparent horizontal diameter as measured by direct angular distance, not a magnification. The cosine should not be inverted.
Due to vertical refraction the Moon near the horizon appears higher than otherwise. Both its left and right limbs appear raised, but remain at their true azimuths. Vertical refraction does not alter the azimuths of the left and right limbs, i.e. the difference in their azimuths remains constant. Due to vertical refraction and not a true increase in altitude, as the Moon’s apparent altitude is increased the constant azimuth lines of the left and right limbs converge toward the zenith. The result is that the direct angular distance between the left and right limbs appears slightly reduced, i.e. not magnified.
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For astronomical graphics, including
monthly wallpaper calendar, visit:
www.CurtRenz.com/astronomical.html
Curt Renz - "Centaur"
Edited by Centaur (05/24/08 11:09 AM)
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Centaur
Pooh-Bah
Reged: 07/12/04
Posts: 1123
Loc: Chicago
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Perhaps the description in the above post can be better understood if you imagine two tiny but bright spaceships hovering just off the surfaces of the left and right limbs of the Moon and close to your horizon. Then let the Moon disappear. Vertical refraction would not alter the spaceships’ azimuths, but convergence of the azimuth lines toward the zenith would make the spaceships appear closer than the true direct angular distance of their separation. Now allow the Moon to reappear between the spaceships. The Moon would appear to have shrunk horizontally by a tiny amount.
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For astronomical graphics, including
monthly wallpaper calendar, visit:
www.CurtRenz.com/astronomical.html
Curt Renz - "Centaur"
Edited by Centaur (05/24/08 11:44 AM)
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photonovore
Moonatic
Reged: 12/24/04
Posts: 2472
Loc: tacoma wa
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Just to add some references which might further explain what is going on in horizon "flattening" of the Sun and the role of magnification (more accurately minification as the magnification is less than one--except "minification" is not really an accepted usage of "minify", which *is* an actual word, so science seems stuck with using "magnification" even when the magnification referred to is less than one--or anything *but* magnified. --ack! english!):
Flattening of the setting Sun
"...calculate the magnification at the horizon, and hence the flattening of the setting Sun." This turns out to be magnification = 5/6 (less than one or technically a reduction or "minification")
Magnification at the Astronomical Horizon
Explains the geometry & math behind calculating horizon "magnification".
Interesting because the "magnification (reduction) occurs on only one axis, the vertical, as has already been stated i believe.
Re; the Moon Illusion, when i am asked about this i always reply that there are no external physical components of the effect, no atmospheric refraction etc, but rather the illusion is entirely psychological, i.e. "between the ears"-- and how *that* works *no* one really knows for sure--we being the mysterious creatures we are and all.. 
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Mardi
4" achromat, ETX-70.
Whitepeak Lunar Observatory Website
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Jim Mosher
sage
Reged: 05/22/06
Posts: 233
Loc: Newport Beach, CA
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Curt,
I believe that if you re-read my postings carefully you will see that we are finally saying the same thing. Normal atmospheric refraction moves the perceived directions of objects (terrestrial or celestial) along lines of constant azimuth (measured relative to the observer’s zenith).
The reason you’re reaching a different conclusion from me is that you’re thinking about light originating at or above the horizon (the great circle at 90° from the observer’s zenith) and being raised above that by refraction. The perceived horizontal angle between such objects is decreased by refraction, just as you say. And I see now that my own photo shows the Moon as it appears above the horizontal (and not at the lower angles one might see looking out, for example, over the ocean), and your reasoning certainly applies to it, since the light from which it is composed originated above the great circle of my horizon.
But I thought (perhaps incorrectly) that the discussion was about what we see when we look horizontally. When the Moon, or stars, or quasars, or spaceships appear at 90° from the observer’s zenith (what I would call horizontal) it is only because refraction has placed them there. In “reality” they are at the observed azimuth but on a little circle below the perceived horizon -- normally by about 0.5°, but possibly much more. That is the open lip of the inverted “goldfish” bowl I was trying to describe. As refraction slides the objects slide up to the equator along the lines of constant azimuth, the perceived horizontal spacing between them increases in exactly the same way you are describe it decreasing for objects starting above the equator and moving still higher.
In general, the amount of horizontal magnification or minification is the ratio of the diameter of the celestial circle (relative to the observer’s zenith) on which the light is perceived to the one on which it originates; and that, in turn is the ratio of Cosines of the two angles above or below the equator. If light is seen at the observer’s equator (offset angle = 0°), but due to refraction it is actually originating at -0.5°, the horizontal magnification will be M = Cos(0°)/Cos(-0.5°) = 1/0.99996 = 1.000 04 (as previously stated). Exactly the same formula applies to each of the cases you mention. For example, for light that originates on the equator, but (due to refraction) is seen 0.5° above it, the magnification is M = Cos(+0.5°)/ Cos(0°) = 0.99996/1 = 0.99996 (i.e., the horizontal diameter is reduced). And, as you say, this is an extremely small effect for the amounts of refraction normally encountered on Earth.
I must apologize again for not having expressed my thoughts (and what I was talking about) more clearly at the outset.
For those who are utterly confused at this point, in addition to the well-known strong vertical flattening, the theory of standard atmospheric refraction predicts an extremely slight apparent horizontal magnification of objects seen at or below the horizon, and an equally slight horizontal compression of objects seen above it. The dividing point between horizontal magnification and minification occurs at the altitude where the refractive bending causes light originating a slight bit below the celestial equator (defined relative to the observer’s zenith) to be perceived an equal angle above it. For the normal amounts of refraction encountered on Earth the effect is too small to be of practical significance, certainly for lunar observing.
-- Jim
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Centaur
Pooh-Bah
Reged: 07/12/04
Posts: 1123
Loc: Chicago
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I understand what you are saying, Jim, when you consider the true Moon to be slightly beneath the horizon. Indeed, the center of that true Moon would be located on a non-great circle parallel to the horizon and beneath it. But it is also slightly further away than it would be if truly centered on the horizon. That makes its true width as seen from that greater distance slightly smaller than when it is actually elevated to the horizon. When raised by refraction so that its center appears on the horizon, the widening between azimuth lines would exactly compensate for the Moon’s greater distance, and make the Moon appear its true width as would be seen when the observer is in the center of the actual ring plane with no atmospheric effect.
Proof is the example of 720 Moons packed together in a ring around the horizon. Their true separations for an observer at the exact center of the ring with no obscuring Earth or atmosphere would be 0.5°. When the ring plane is beneath the observer, yet the refracted Moons appear with their centers precisely on the horizon, they would each appear to be the true width of 0.5°. If any appeared wider, then others would have to appear narrower.
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For astronomical graphics, including
monthly wallpaper calendar, visit:
www.CurtRenz.com/astronomical.html
Curt Renz - "Centaur"
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Centaur
Pooh-Bah
Reged: 07/12/04
Posts: 1123
Loc: Chicago
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To follow up, as the Earth’s rotation causes the Moon to appear to rise even further, the observer would be drawn closer to the Moon, thus making its horizontal angular width even larger. Therefore, the ultimate point is that the Moon’s horizontal angular width is not at its greatest when on the horizon, but instead when it is closer to the zenith.
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For astronomical graphics, including
monthly wallpaper calendar, visit:
www.CurtRenz.com/astronomical.html
Curt Renz - "Centaur"
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Jim Mosher
sage
Reged: 05/22/06
Posts: 233
Loc: Newport Beach, CA
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Quote:
wouldn’t it actually be interesting, that under the altitude of 10 degrees from the horizon, the rising or setting Moon would be effected by paraxial angular magnification - due to layered airmass of the troposphere.
The real answer as to precisely how large the Moon will look, and in what direction it will be seen at a particular moment from a particular point on Earth and under particular atmospheric conditions can in truth be arrived at only by randomly tracing test rays out into space and locating the ones that intercept the Moon’s actual physical position at the moment they cross. It is a very complex problem, but one which our eyes and our telescopes solve effortlessly. And also rather different from the diver face-mask problem.
Tracing the rays through the atmosphere produces two things: (1) an estimate of the total effective airmass along the light path (which can be used to estimate attenuation), and (2) the origin and direction of the rays that exit the Earth’s atmosphere into space.
As has been previously stated, airmass by itself does not produce bending. What produces bending is variation in the density of the air (or more precisely, its “refractive index”) in the direction crossways to the light path, and encountering an inclined interface between two media of differing refractive index (as in the face mask case). The familiar Snell’s law from high school physics is most useful in the latter case (“paraxial” means almost normal -- perpendicular -- to the interface). It is less useful in atmospheric optics where there are, in reality, no sharp interfaces; and where we are also often interested (as in the case of the Moon at the horizon) in rays that are initially traveling parallel to the direction of stratification. Snell’s law does not work at all well in the that case, which is in some ways similar to a ray of light traveling parallel to the axis of an optical fiber (a form of Gradient Index Lens). The guiding principle of the GRIN approach to ray tracing is that the radius of curvature of the ray is equal to the index of refraction divided by the rate of change in the index of refraction transverse to the ray path. In the case of the optical fiber, a ray down the exact center will not be curved, but one parallel to the axis, yet off the center, will encounter denser (higher index) material towards the center, and hence be bent towards it. This traps the ray in a sinusoidal path down the fiber.
In the case of the Earth’s atmosphere, an initially horizontal ray encounters higher index air below and lower index air above, and hence the GRIN approach tells us it immediately bends down (to predict this behavior with Snell’s law one has to approximate the atmosphere as a series of fixed index shells, and the bending doesn’t start until the ray exits the first shell). At ground level on Earth, the basic fall off of air density with height produces a radius of curvature of about 27 000 km. Since this is larger than the radius of the Earth (about 6 37l km), initially horizontal (parallel to surface) rays bend down but don’t hit the surface. The GRIN approach can also properly predict horizontal deflections if there are horizontal variations in density -- something the Snell’s law applied to uniform shells approach has great trouble with.
Deviations in the sea level bending from the value expected for a standard atmosphere are dictated primarily by variations in the vertical temperature gradient. If the temperature increases with height the bending will be stronger; if it decreases with height (giving the higher air a higher than normal index) it will be less. A temperature fall-off of 34 deg/km is sufficient to neutralize the normal bending; and an even steeper fall-off can make the initially horizontal ray bend up towards the sky.
A proper ray-tracing calculation using either Snell’s law (on an imaginary set of idealized layers) or the GRIN approach (on a more realistic smooth atmosphere) should give the same result. They are the basis of the stylized ray diagrams you’ve seen on Les Cowley’s pages. As we’ve been discussing here, they lead to a prediction of vertical flattening, and very little distortion in the horizontal direction. They rarely, if ever, predict that a celestial object will appear magnified (vertically stretched), although terrestrial ones can, as in certain kinds of mirages. Even when you see pictures of sunsets where portions of the Sun appear stretched out vertically, you will almost always find they are just less compressed than the other parts.
So “paraxial angular magnification” below 10° elevation is not an additional effect. It (and all other consequences of Snell’s law) are already included in the ray tracing calculations that predict the vertical flattening described by Mardi.
-- Jim
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Mare Nectaris
Pooh-Bah
Reged: 03/09/08
Posts: 1100
Loc: Toijala, Finland
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Thanks for you patience, Jim and Curt - and the most thorough answers you have provided on this!
Hope you have enjoyed this as much as I have. - Can I ever again see the Moon rise - and not remember all these fishes, bowls, rays, formulae, cosini, tangents or mysteries of the layered troposphere.
Ever since I was a little boy, I have wondered how far the horizon r e a l l y is. Now that I (think) I k n o w, it seems like having lost something mysterious... yet, the horizon continues to escape onto new questions - and perhaps also answers...
Having chased the horizon with you all - - and a bit my own tail on the way I guess - - I finally picture myself sitting with you at the seashore - - as the night starts to fall - - with binoculars raised on my eyes to wonder the rising Moon - -
... only suddenly realizing: I'm actually witnessing the incident of the Moon rising as if I was "looking" through the (tropospherical) binoculars inverted - into the objective - - finally s e e i n g (against the perception) the rising Moon appear in fact slightly "minified"... (thanks for that, Mardi).
Be well all - by the seashore the Moon is gently rising on...
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Share - and you shall have it all
Timo Keski-Petäjä
CtheMoon
Observation shelter KuuMaja (MoonHut)
TAL 250K*Celestron C8-N*SkyWatcher Skymax 150 Pro*TAL1(Mizar)*EQ6 Pro SynScan*Celestron Advanced GT (CG-5 GOTO)*Baader Hyperion Clickstop Zoom 8-24*17 mm UWA-70*TeleVue BIG 2x Barlow*Celestron 2x Barlow Ultima SV Series
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Jim Mosher
sage
Reged: 05/22/06
Posts: 233
Loc: Newport Beach, CA
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Quote:
Therefore, the ultimate point is that the Moon’s horizontal angular width is not at its greatest when on the horizon, but instead when it is closer to the zenith.
Curt,
Again, I must apologize for not understanding properly what the point of the discussion was. Yes. The Moon is roughly 6400 km closer when it is seen at the zenith than when it is seen at the horizon. And (because of this) it grows measurably in horizontal diameter as it rises.
I (apparently mistakenly, again) thought we were talking about whether the presence of the Earth’s atmosphere makes the Moon’s horizontal diameter look larger or smaller than it otherwise would when seen at the horizon. You were arguing, I thought, that horizontal magnification, however small, was logically impossible; while I thought that infinite magnification would be possible on an exotic planet with 90° bending.
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As to your next most recent posting, which again seems to indicate that horizontal magnification of the Moon at the horizon is impossible, the argument that the Moon’s changing distance (you are talking about the changing distance due to the rotation of the Earth?) “exactly” compensates for the change in magnification due to refraction cannot possibly be correct. The effect of the changing distance is a linear effect (near the horizon) that depends on the radius of the Earth and the distance to the Moon. The changing refractive magnification is a quadratic effect related to the density of the Earth’s atmosphere and the amount of bending it produces at a particular observing angle. There is no logical connection between the two that I can see; and if, by chance, they happen to compensate for one set of parameters they can’t possibly compensate for another.
I suppose I should also have mentioned earlier that I have great trouble with the many thought experiments that have been proposed involving a physically moving ring of “moons.” The Moon is, as we all know, a real object at a very definite location (relative to the Earth) in the vacuum of space. The presence of the Earth’s atmosphere changes where and how we see it, but it does not change where it is. With all due respect, attempting to think about the effects of refraction as being somehow equivalent to a physically moving Moon is not likely to produce sensible conclusions.
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To make sure we’re talking about the same thing this time, lets postulate a 6,370 km radius Earth with a 1,740 km radius Moon situated 384,000 km away (center-to-center).
In the presence of an atmosphere producing 0.5° of refractive bending, to see the center of the Moon at his local horizontal, an observer will have to be positioned 0.5°, or about 56 km “back” on the Earth’s “farside” (relative to the Moon’s center). The straight line distance from the observer to the Moon’s center will be roughly 384,109 km.
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The question I have been trying to answer is how does the Moon’s angular horizontal diameter as seen under these circumstances compare to the diameter that would be observed from the same location in space in the absence of the Earth and its atmosphere?
In the direct view (with the Earth removed), the Moon is going to have some angular size (0.519°) which subtends some range of azimuth (also about 0.519°) on a celestial circle at 90.5° from the observer’s zenith (0.5° below the horizontal). As we both seem to agree, when the Earth and its atmosphere are added, the atmospheric bending is going to move the Moon’s left and right limbs up along diverging lines of constant azimuth (on a celestial sphere relative to the observer’s zenith), producing a change in apparent horizontal angular size equal to the ratio of the cosines of the original and final angles from the equator. If the hypothesis that refraction moves things along lines of constant azimuth is correct (and I believe it is) then the formula which I thought we had both just agreed upon applies exactly and is the answer to the question. The horizontal diameter is magnified, in this case, by M = Cos(0°)/Cos(-0.5°) = 1/0.99996 = 1.000 04 when an atmosphere is present (compared to the direct view with no atmosphere).
A larger horizontal magnification would, of course, be observed with a larger refractive bending. For 5° of bending at the horizon, the Moon’s horizontal diameter would be M = Cos(0°)/Cos(-5°) = 1/0.9962 = 1.004 times larger in the refracted view versus in the hypothetical direct view from the same location.
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An alternative way to pose the question, which is not the one I had in mind, but which seems much more closely related to your comments about changing distances, is to say that if we were to remove only the atmosphere, but not the body of the planet, then our view of the Moon would be blocked. So to return the Moon’s center to the horizon (without refraction) the observer would have to move the 56 km to the Earth’s “limb”. The question would then be: is the horizontal diameter of the Moon observed on the horizon of the atmosphere-less planet larger or smaller than the diameter of the Moon actually observed on the horizon of the same planet with its atmosphere?
In the present case, removing 56 km from the distance to the Moon’s center is going to increase its apparent angular size in inverse proportion to the change in distance (384,109/384,053) = 1.000 15 (compared to a direct view from the original, more distant, position). But, as previously demonstrated, from the original position, atmospheric refraction enhanced the horizontal diameter by 1.000 04 compared to the direct view from that position. So this result can be described by saying that the Moon as observed at the horizon from the planet with an atmosphere is 1.000 04 / 1.000 15 = 0.999894 of the size that would be observed for the “same” Moon on the horizon of the “same” planet without an atmosphere. In this case, the changing distance factor overweighs the refractive magnification. So (in this case) you’re right.
On the other hand, we can consider what would happen if we increased the refractive bending angle (at the horizon) to 5.0°. In that case, when the atmosphere is present, to see the Moon in a horizontal direction, the observer would have to be positioned 5°, or about 556 km back on the Earth’s “farside”. From this position, the refraction would enhance the Moon’s horizontal diameter by M = Cos(0°)/Cos(-5°) = 1/0.99619 = 1.0038 compared to a direct view (with no planet or atmosphere). Alternatively, if we remove the atmosphere and rotate the planet 5° to put the Moon’s center back on the horizon of the atmosphere-free planet, then by shortening the distance to the Moon by 556 km we have increased the Moon’s angular diameter (compared to the direct view from the original position) by (384,609/384,053) = 1.0014.
In this case, the Moon seen at the horizon is larger by 1.0038/1.0014 = 1.0023 when an atmosphere is present than when it is not. So it appears to me that the result with both refraction and changing distances can come out either way.
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You can perhaps pose the question in still different ways that will have still different answers; but can we perhaps agree that we’re both right and move on? The practical effect seems much too small to worry about.
-- Jim
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Centaur
Pooh-Bah
Reged: 07/12/04
Posts: 1123
Loc: Chicago
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Quote:
As to your next most recent posting, which again seems to indicate that horizontal magnification of the Moon at the horizon is impossible, the argument that the Moon’s changing distance (you are talking about the changing distance due to the rotation of the Earth?) “exactly” compensates for the change in magnification due to refraction cannot possibly be correct.
...but can we perhaps agree that we’re both right and move on? The practical effect seems much too small to worry about.
No, Jim, in that post I was not talking about changing distance due to the rotation of the Earth. In the post that followed I was talking about that and may have confused you. I tried to describe the situation clearly and succinctly, but apparently failed. Please reread what I wrote. If that fails again, I’ll put it another way. I was referring to the fact that when the true Moon is slightly beneath your horizon, while the center of the refracted Moon is precisely on your horizon, the true Moon is farther from you than it would be for the location at the intersection of the line connecting you with center of the Earth and the perpendicular to that line connecting to the center of the Moon. To you, the true un-refracted Moon seen by X-ray vision would appear smaller than it would to someone with X-ray vision at that other location because you are farther from the Moon. The greater distance can be calculated through the same cosine formula used for the "magnification". However, with refraction placing the image of the center of the Moon precisely on your horizon, it would appear expanded to exactly the same size as it would appear to the X-ray observer at that other location. Yes, you are correct there is an expansion along the constant azimuth lines, but it exactly compensates for the fact that the true Moon was farther from you than it was for that other location. The Moon seen through X-ray vision at that other location would have appeared 0.5° wide and the refracted Moon that seems centered on the horizon for you would also appear to be 0.5° wide. If the other observer saw a ring of 720 Moons, so would you. To both of you, all of the Moons would appear 0.5° wide.
I’ll certainly agree to your final point. 
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