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Atmospheric thickness



 
 
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  #11  
Old October 1st 12, 11:33 AM posted to sci.astro.amateur
Paul Schlyter[_3_]
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Posts: 1,344
Default 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  
Old October 1st 12, 08:12 PM posted to sci.astro.amateur
Barry Schwarz[_2_]
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Posts: 52
Default 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  
Old October 2nd 12, 02:22 AM posted to sci.astro.amateur
William Hamblen[_2_]
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Posts: 236
Default 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

  #14  
Old October 2nd 12, 11:08 AM posted to sci.astro.amateur
Paul Schlyter[_3_]
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Posts: 1,344
Default Atmospheric thickness (rewrite)

In article ,
says...

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



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.


The second simplest approximation to the Earth's atmosphere is the
"isothermal atmosphere": here one assumes that the atmosphere is an ideal
gas which has the same temperature everywhere, and one also neglects the
(small) changes in the Earth's acceleration of gravity at different
altitudes. The result is that both the air pressure and the air density
decreases exponentially with elevation. There's a quantity here called
the "scale height" of the atmosphere, which is the difference in
elevation over which the pressure and temperature decreases by a factor
of e (= the base of the natural logarithms). Interestingly, the scale
height in the isothermal atmosphere approximation is exactly the same as
the "thickness of the atmosphere" in the homogeneous atmosphere
approximation: about 8 kilometers. And the air masses at different
altitudes above the horizon is indeed described with a ghasty integral --
the solution to that integral is called the Chapman function. This
integral lacks an analytic solution, but there are useful approximations
available.


To proceed further, we need some quantities:

T = absolute temperature
k = Boltzmann's constant
m = average molecular weight of the air
g = the Earth's acceleration of gravity

H = scale height = k*T / (m*g)

z = zenith distance
r = radius of curvature of the surface of the Earth

X = r / H

y = sqrt( 0.5 * X ) * abs(cos(z))


With these quantities, a useful approximation to the Chapman function
ch(X,z) is:

ch(X,z) = sqrt(pi/2 * X) * exp(y*y) * erfc(y)

where erfc(y) is the complementary error function: erfc(y) = 1 - erf(y),
which can be approximated as:

0 = y = 8:
a = 1.0606963
b = 0.55643831
c = 1.0619896
d = 1.7245609
erfc(y) = exp(-y*y) * (a + b * y) / (c + d * y + y*y)
ch(X,y) = sqrt(pi/2 * X) * (a + b * y) / (c + d * y + y*y)

8 = y = 100:
f = 0.56498823
g = 0.06651874
erfc(y) = exp(-y*y) * f / (g + y)
ch(X,y) = sqrt(pi/2 * X) * f / (g + y)


There are also series expansions of the erfc function:
http://en.wikipedia.org/wiki/Erfc

A series expansion which is useful in this case is:

erfc(y)*exp(y*y) = (1/(y*sqrt(pi))) * ( 1 + SUM(n=0,1,2,...)((-1)^n * ( 1
* 3 * 5 * ... * (2*n-1) ) / ( (2*y*y)^n ) ) )

which can be written as:

erfc(y)*exp(y*y) = (1/(y*sqrt(pi))) * ( 1 + A )

where

A = SUM(n=0,1,2,...)((-1)^n * ( 1 * 3 * 5 * ... * (2*n-1) )/((2*y*y)^n))

This series diverges for every y, but if y is large enough (i.e. larger
than approx. 4 to 5), only the first few terms need to be included.


If the zenith angle is exactly 90 degrees, then y = 0 and erfc(y) = 1 and
the Chapman function formula can be simplified to:

ch(X,90_deg) = sqrt(pi/2 * X)

which, for common atmospheric conditions, yields:

T = absolute temperature = ( 273 + 15 ) = 288 K
k = Boltzmann's constant = 1.3806488E-23 J/K
m = average molecular weight of the air = 28.97 * 1.66053886E-27 kg
g = the Earth's acceleration of gravity = 9.80665 m/s^2

H = scale height = k*T / (m*g) = 8428 m = 8.428 km


X = 6378 / 8.428

ch(X,90_deg) = sqrt(pi/2 * 6378/8.428) = 34.5

i.e. at the horizon one sees 34.5 air masses if the temperature is +15
deg C.



For zenith angles larger than 90 degrees (i.e. if you look below the
horizon), you get:

ch(X,z) = 2*exp(X*(1-sin(z))) * ch(X*sin(z),90_deg) - ch(X,180_deg-z)

ch(X,z) = 2*exp(X*(1-sin(z))) * sqrt(pi/2 * X * sin(z)) - ch(X,180_deg-z)

These formulae may be useful when you're observing from a high
mountaintop.


The paper describing these calculations was originally published in 1972
by F.L. Smith and Cody Smoth. The paper, titled "Numerical evaluation of
Chapman's grazing incidence integral ch(X, Chi)" is available from the
link below. Unfortunately, it's not for free, you must pay to get a copy:

http://www.agu.org/pubs/crossref/197...19p03592.shtml


Willian Swider extended this to cases where the scale height varies with
altitude. His paper "The determination of optical depth at large solar
zenith angles" can be obtained from this link. Again, this paper is not
for free, you must pay to get it:

http://www.sciencedirect.com/science...3206336490056X



None of these formulae accounts for the refraction in the Earth's
atmosphere. They were intended to be used for studies of the Earth's
ionosphere, where atmospheric refraction of light is insignificantly
small.

  #16  
Old October 2nd 12, 11:31 AM posted to sci.astro.amateur
[email protected]
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Default 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  
Old October 2nd 12, 11:34 AM posted to sci.astro.amateur
[email protected]
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Posts: 9,472
Default 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  
Old October 2nd 12, 07:01 PM posted to sci.astro.amateur
oriel36[_2_]
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Posts: 8,478
Default 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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