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#11
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Atmospheric thickness (rewrite)
On Sun, 30 Sep 2012 12:05:18 -0700, "W. eWatson"
wrote: It seems what I'm after is misinterpreted. I was talking with a friend about some variable star work he does at his domed observatory. Somehow I asked him at what is the lowest elevation he surveys. He said 30 degrees, about two atmospheres of thickness. He then said," maybe near the horizon, it's 47 thickness." That's what I'm looking for. How many thickness of atmosphere will I see from zero to ninety degrees above the horizon. It depends on your meteorological conditions, particularly when you are near the horizon. First, it is directly proportional to your current air pressure divided by the standard air pressure of 1013.25 millibars. Second, it also depends on how the temperature varies with elevation at your location and time and that dependency gets stronger the closer you get to the horizon. And, third, one should really also account for atmospheric refraction, at least at lower altitudes above the horizon but afaik nobody has ever done such a computation. To summarize: the subject is quite complex, and therefore computed air masses aren't particularly reliable below about 20-30 degrees altitude. |
#12
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Atmospheric thickness (rewrite)
On Sun, 30 Sep 2012 12:05:18 -0700, "W. eWatson"
wrote: It seems what I'm after is misinterpreted. I was talking with a friend about some variable star work he does at his domed observatory. Somehow I asked him at what is the lowest elevation he surveys. He said 30 degrees, about two atmospheres of thickness. He then said," maybe near the horizon, it's 47 thickness." That's what I'm looking for. How many thickness of atmosphere will I see from zero to ninety degrees above the horizon. As an idealized approximation, consider the Earth as a sphere sitting at the origin of the usual x-y-z Cartesian coordinates. Without loss of generality, we can limit ourselves to the z=0 plane. We end up with the circle x^2 + y^2 = R^2 where R is on the order of 4000 miles but you can choose whatever units and accuracy you like. If the atmosphere has a depth of r miles, its boundary forms a concentric circle outside the first with the formula x^2 + y^2 = (R+r)^2 Also without loss of generality, we can put the observer at (0,R), the top of the circle (corresponding to the north pole). Obviously, looking up vertically (90 deg) means looking through r miles of atmosphere. And looking out horizontally (0 deg) means looking through sqrt(2Rr+r) miles. When looking up at an angle e, you look through the atmosphere to the point where the line y = sin e * x + R meets the above circle. Replacing y in equation of the circle produces an equation which is easy enough to solve for x and then compute y. What this doesn't account for is atmospheric density. If we assume the atmosphere is 20 miles thick, then looking horizontally passes through 400+ miles of atmosphere. While this is only 20 times as much as looking up, most of it is much thicker and probably has greater visual impact. So while it is only 20 times as thick, it might well be 47 times the impact. Since the density is a continuous function of altitude, the formula for computing impact is probably an obnoxious integral. It would take into account that as you move out horizontally, the altitude increases slowly and the rate of increase itself increases as the ground level falls away. -- Remove del for email |
#13
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Atmospheric thickness (rewrite)
On 2012-09-30, W. eWatson wrote:
It seems what I'm after is misinterpreted. I was talking with a friend about some variable star work he does at his domed observatory. Somehow I asked him at what is the lowest elevation he surveys. He said 30 degrees, about two atmospheres of thickness. He then said," maybe near the horizon, it's 47 thickness." That's what I'm looking for. How many thickness of atmosphere will I see from zero to ninety degrees above the horizon. You want to look up "air mass" and "zenith distance". Bud |
#15
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Atmospheric thickness (rewrite)
In article , says...
It seems what I'm after is misinterpreted. I was talking with a friend about some variable star work he does at his domed observatory. Somehow I asked him at what is the lowest elevation he surveys. He said 30 degrees, about two atmospheres of thickness. He then said," maybe near the horizon, it's 47 thickness." That's what I'm looking for. How many thickness of atmosphere will I see from zero to ninety degrees above the horizon. Some other useful approximations can be found he http://en.wikipedia.org/wiki/Air_mass_%28astronomy%29 A particularly simple approximation, which yoelds results right down to the horizon which aren't that far off, is: Air mass = 1 / ( cos(z) + (1/40)*exp(-11*cos(z)) ) This formula yields a horizontal air mass (cos(z) = 0) of 40 |
#16
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Atmospheric thickness
On Oct 1, 1:02*am, "W. eWatson" wrote:
On 9/30/2012 11:58 AM, W. eWatson wrote: On 9/30/2012 3:43 AM, wrote: On Sep 24, 4:39 pm, "W. eWatson" wrote: Is there a tool or graph that shows the thickness of the atmosphere at various elevations and accounts for the altitude of the observatory? http://www.denysschen.com/catalogue/density.aspx Looking at what elevation above the horizon? Read what I have to say at my 12:05 pm (rewrite) post. *Elevation, as in elevation above the horizon, or as in az/el. Elevation and altitude tend to be used interchangeably and you didn't initially specific elevation of what. By "thickness" one can assume you meant density. What you seem to want to know is what is the total mass of air between your scope and the object you observe, correct? (This would have to take the scope's aperture into account.) Atmospheric conditions will vary from ground level to the top of the atmosphere, especially when considering directions close to the horizon, and this info might be hard to come by. The conditions are also likely to vary with azimuth. |
#17
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Atmospheric thickness
On Sep 30, 2:58*pm, "W. eWatson" wrote:
On 9/30/2012 3:43 AM, wrote: On Sep 24, 4:39 pm, "W.. eWatson" wrote: Is there a tool or graph that shows the thickness of the atmosphere at various elevations and accounts for the altitude of the observatory? http://www.denysschen.com/catalogue/density.aspx Looking at what elevation above the horizon? You could try this: http://www.ehow.com/how_5172369_calculate-airmass.html |
#18
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Atmospheric thickness (rewrite)
On Oct 2, 12:08*pm, Paul Schlyter wrote:
Your approximation of the atmosphere is called the "homogeneous atmosphere" which is the simplest possible atmospheric model: one then assumes that the air density is constant everywhere up to some "upper surface" where the atmosphere would end (somewhat like the oceans of the Earth). *If the atmosphere was a liquid rather than a mix of gases, this approximation would work well. *However, the "thickness of the atmosphere" would not be as large as 20 miles, but instead about 8 kilometers (= about 5 miles). *That's what you get if you take the ground air pressure (= the weight per unit area of the atmosphere) and divide it by the ground air density and then also by the Earth's acceleration of gravity. The "homogeneous atmosphere" approximation would yield some 40 airmasses in the horizontal direction compared to the vertical direction. These formulae may be useful when you're observing from a high mountaintop. All this is pretense Schlyter and especially when none of you can work with elevations and rotations speeds which is far more meaningful that the harmless topic here.I quite enjoy putting the information out there as few people even consider the difference in latitudinal speeds when view at an elevation of Denver where an unobstructed view from horizon to horizon along a longitude meridian covers 2 degrees of latitude or roughly 224 KM. At the location of Denver (using a hypothetical 1 km elevation),the rotational speed for 40 degrees lat is 1281 km per hour so that the observer can see a location at the horizon South of his meridian rotating at 1297 km per hour and looking North across the same meridian can see a location at the horizon rotating at a speed of 1261 km per hour insofar as difference in latitudinal speeds from horizon to horizon,as seen from 1 degree North and 1 degree South of Denver is roughly 36 km per hour. http://www.ncgia.ucsb.edu/education/...s/table02.html I spoke with a colleague today about errors that are not immediately obvious and it happened before where the atmosphere and horizon trajectories are concerned such as the overlooking of a metric/ imperial conversion - http://en.wikipedia.org/wiki/Mars_Climate_Orbiter The mindnumbing error which prevents All readers here from affirming that the Earth turns once in a 24 hour day is amazing given that when men generally talk through a problem,they enjoy the resolution of a mistake and act to correct the issue.In this era the senseless attempt to explain the Earth's rotation using stellar circumpolar motion,a mistake of enormous proportions,people either act like children or abdicate complete responsibility for correcting something which is plainly and clearly wrong and completely destroys any chance of working with cause and effect between the planetary cycles and terrestrial effects. What is it that is preventing readers from seeing what is true in order to obliterate the principles that are incorrect for as long as this looms in the background,anything you and your empirical colleagues try to promote will count for nothing. |
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