Many thanks! -- Joe --- On Mon, 2/9/09, Canopus56 <canopus56@yahoo.com> wrote: From: Canopus56 <canopus56@yahoo.com> Subject: Re: [Utah-astronomy] quick question To: "Utah Astronomy List Serv" <utah-astronomy@mailman.xmission.com> Date: Monday, February 9, 2009, 7:41 PM Joe asked:

Hi, Can anybody tell me right off-hand how many arc-minutes the night sky is? Thanks, Joe

What Daniel said.  My calculator returns a slightly different number than Daniel's:  Total celestial sphere: 148,538,682 arcmin^2 Percent of the total celestial sphere you see across 24 hours at latitude 40N: 88.3% Arcmins^2 visible at 40N latitude in 24 hours: 131,159,656 arcmin^2 - Kurt P.s. - The math: r = 1 radian = 180 degrees / pi = 57.3 degrees r = 1 radian = ( 60 arcmins / deg ) * 180 degrees / pi = 10,800 arcmins / pi r = 1 radian = 10,800 arcmins / pi A = 4*pi*r^2   pi = 3.141 A = 4*pi*(10,800/pi)^2 = 4*(pi/pi^2)*(10,800)^2 = (4*(10,800)^2)/pi =~ 148,538,682 arcmin^2 Appended find excerpts from two post that I did on sci.astro.amateur in 2005.  They also discuss the related question of "how many square degrees does an observer at latitude x see over a 24 hour period?"  From 40N latitude, you see a truncated sphere (truncated at about S40 degs) and not the full sphere.   http://members.csolutions.net/fisherka/astronote/astromath/fracsphere.gif Clear Skies - Kurt =============================== sci.astro.amateur post 8-21-2005 canopus56 (excerpt) The observer, at whatever latitude they are standing, always see 1/2 of a sphere in the local horizon system. As you and others have noted, the area of a sphere is 4*pi*r^2 and where an infinite unitized sphere is used, r = 1 radian = 180 degrees / pi = 57.3 degrees. So, the area of sphere in steradians is: 4*pi*(180/pi)^2 = 4*(pi/pi^2)*180^2 = (4*180^2)/pi = 41253 deg^2 A local observer's view is 1/2 the local horizon system, and, in steradians is always a fractional part of angular area of a total sphere (steradians in square degrees), or 41253 deg^2. So - 1/2* 41253 = 20626.5 deg^2 If the local observer looks at the hemisphere of the night sky using another coordinate system at a rotated angle as the reference - in this case the North Pole of the celestial coordinate system from 40 deg north latitude - that changed angle isn't going to change the physical fact that there are only 20,626.5 degs^2 (steradians) in the hemisphere of the local horizon system. sci.astro.amateur post 8-24-2005 canopus56 Thanks for the further explanations.  I was having problems relating the 1/2*(1+cos(lat)) formula to any geometric construction of the problem.  With the additional tips you provided,  it now makes sense to me. To resummarize - The surface area of a sphere is - S = 4*pi*r^2 There are 41252.96 square degrees on a sphere. At any instant during the night,  you see 1/2 (50%) of the celestial sphere or 20626.5 sq. degs. The observer's local horizon can be vizualized as a layer or plane, titled at the origin of the celestial sphere at the observer's geographic latitude - 40 degs north in my case - that bisects the sphere. Across twenty-four hours of daylight and nighttime, the observer's local horizon rotates 360°. That rotation inscribes the bottom of the celestial sphere with a cone shape.  Think of taking an apple and cutting a cone of the bottom of it such that the sides of the cone have angles equal to your geographic latitude with respect to the equator of the apple.  The "cone" shape is the part of the earth blocked from the observer's view. That sphere with a cut-out cone shape can be abstracted as simple truncated sphere.  Turned upside down, the truncated sphere looks like one of the spherical light globes used in out lighting fixtures that we amateur astronomers hate so-much. E.g. - http://www.thelightingcenter.com/products/*/*/1033 With the truncated sphere model, we can apply (per Brian), Archimedes Hat-Box-Theorem to find the area of the truncated sphere. http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html That area of the truncated sphere is given by: S_cap = 2*pi*r*h Surface area of zone or cap "h" is the length of the line vertically through the origin of the sphere, which at starts the sphere's apex and ends in the truncated layer at one end of the sphere. See - http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html - or for an upside down sphere - http://www.thelightingcenter.com/products/*/*/1033 The answer sought in this thread is "what is the ratio (or fractional part) of the truncated sphere to the entire sphere." or - S_trunc = Fraction * [4*pi*r^2] (See http://en.wikipedia.org/wiki/Solid_angle ) The fractional part seen by an observer across a single day is: Fraction =    [2*pi*r*h] /  [4*pi*r^2] - which solves to: Fraction =    h /  2*r - Our celestial sphere is a unit sphere << http://en.wikipedia.org/wiki/Sphere >>, so r=1 and Fraction =    h /  2 The answer to the question "what fraction of the sky can you see from any latitude" boils down to what is "h" for any latitude in Archimedes Hat-Box-Theorem: http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html - except we are applying the theorem to the "big" part of the upside truncated sphere - http://www.thelightingcenter.com/products/*/*/1033 - and not the small cap shown in - http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html Take a great circle through longitude 0 degrees on the sphere.  From that 2-D planar construction, it can easily be seen that "h" is the sum of cosine of 90 degrees, plus the cosine of the observer's latitude. http://members.csolutions.net/fisherka/astronote/astromath/fracsphere.gif The cosine of 90 degrees = 1, so - h = 1 + cos(latitude) - and Fraction of the sky that is observable across twenty-four hours of darkness and daylight is: Fraction = [1+cos(geolatitude)] / 2 In my case at 40 north degs, that's the 88.3% you guys calculated, or about 36426 sq. degrees. Thanks again for the further explanation of the geometric construction.  The extra tips helped me to piece it together. Sorry to be so dense initially. - Canopus56 P.S. - If you approach the problem by finding the surface area of the small cap cut-off the end of the truncated sphere - that is diagram at Wolfram's website - http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html - then the area of the small cap is - Fraction_small_cap = [1-cos(geolatitude)] / 2 - or 11.7% at 40° north latitude. P.P.S. - Amateur astronomers primarily are concerned with the fraction of the sky that we can see during the night.  Even of the shortest day of the year at 40° north, daylight and nighttime are equally divided - about 12 hours each - and the Earth rotates nearly a full 180° at night.  As a consequence, we might see the full 88% as "dark sky" around the longest day of the year.  Otherwise, night time at 40° north latitude is shorter than 12 hours and the Earth rotates less than 180°.  So, at 40° north, we see some fraction of the night sky between 88% and the 50% of the night sky that can be seen at any one instant during the night. _______________________________________________ Utah-Astronomy mailing list Utah-Astronomy@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/utah-astronomy Visit the Photo Gallery: http://gallery.utahastronomy.com Visit the Wiki: http://www.utahastronomy.com

Kurt: I was suprised buy the high number for a 40N observer but the numbers do check out. We can see 88% of the celestial sphere from here. Having pulled Omega Centari out of the mud in Rush Valley, I would say the not everthing that far south can be see with much satifaction. A better number would be how much can be see at at least 30 degrees elevation, which is high enough for highpower difraction limited viewing with an amateur telescope. The answer to that is 67 percent or 2 thirds. Not too shabby for a stay at home astronomer. DT --- On Mon, 2/9/09, Canopus56 <canopus56@yahoo.com> wrote:

From: Canopus56 <canopus56@yahoo.com> Subject: Re: [Utah-astronomy] quick question To: "Utah Astronomy List Serv" <utah-astronomy@mailman.xmission.com> Date: Monday, February 9, 2009, 6:41 PM Joe asked:

...

Hi, Can anybody tell me right off-hand how many arc-minutes the night sky is? Thanks, Joe

What Daniel said.  My calculator returns a slightly different number than Daniel's: 

Total celestial sphere: 148,538,682 arcmin^2

Percent of the total celestial sphere you see across 24 hours at latitude 40N: 88.3%

Arcmins^2 visible at 40N latitude in 24 hours: 131,159,656 arcmin^2

- Kurt

P.s. - The math:

r = 1 radian = 180 degrees / pi = 57.3 degrees r = 1 radian = ( 60 arcmins / deg ) * 180 degrees / pi = 10,800 arcmins / pi r = 1 radian = 10,800 arcmins / pi A = 4*pi*r^2   pi = 3.141 A = 4*pi*(10,800/pi)^2 = 4*(pi/pi^2)*(10,800)^2 = (4*(10,800)^2)/pi =~ 148,538,682 arcmin^2

Appended find excerpts from two post that I did on sci.astro.amateur in 2005.  They also discuss the related question of "how many square degrees does an observer at latitude x see over a 24 hour period?"  From 40N latitude, you see a truncated sphere (truncated at about S40 degs) and not the full sphere.  

http://members.csolutions.net/fisherka/astronote/astromath/fracsphere.gif

Clear Skies - Kurt

=============================== sci.astro.amateur post 8-21-2005 canopus56 (excerpt) The observer, at whatever latitude they are standing, always see 1/2 of a sphere in the local horizon system. As you and others have noted, the area of a sphere is 4*pi*r^2 and where an infinite unitized sphere is used,

r = 1 radian = 180 degrees / pi = 57.3 degrees.

So, the area of sphere in steradians is:

4*pi*(180/pi)^2 = 4*(pi/pi^2)*180^2 = (4*180^2)/pi = 41253 deg^2

A local observer's view is 1/2 the local horizon system, and, in steradians is always a fractional part of angular area of a total sphere (steradians in square degrees), or 41253 deg^2. So -

1/2* 41253 = 20626.5 deg^2

If the local observer looks at the hemisphere of the night sky using another coordinate system at a rotated angle as the reference - in this case the North Pole of the celestial coordinate system from 40 deg north latitude - that changed angle isn't going to change the physical fact that there are only 20,626.5 degs^2 (steradians) in the hemisphere of the local horizon system.

sci.astro.amateur post 8-24-2005 canopus56

Thanks for the further explanations.  I was having problems relating the 1/2*(1+cos(lat)) formula to any geometric construction of the problem.  With the additional tips you provided,  it now makes sense to me.

To resummarize -

The surface area of a sphere is - S = 4*pi*r^2

There are 41252.96 square degrees on a sphere.

At any instant during the night,  you see 1/2 (50%) of the celestial sphere or 20626.5 sq. degs.

The observer's local horizon can be vizualized as a layer or plane, titled at the origin of the celestial sphere at the observer's geographic latitude - 40 degs north in my case - that bisects the sphere.

Across twenty-four hours of daylight and nighttime, the observer's local horizon rotates 360°.

That rotation inscribes the bottom of the celestial sphere with a cone shape.  Think of taking an apple and cutting a cone of the bottom of it such that the sides of the cone have angles equal to your geographic latitude with respect to the equator of the apple.  The "cone" shape is the part of the earth blocked from the observer's view.

That sphere with a cut-out cone shape can be abstracted as simple truncated sphere.  Turned upside down, the truncated sphere looks like one of the spherical light globes used in out lighting fixtures that we amateur astronomers hate so-much. E.g. - http://www.thelightingcenter.com/products/*/*/1033

With the truncated sphere model, we can apply (per Brian), Archimedes Hat-Box-Theorem to find the area of the truncated sphere. http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html

That area of the truncated sphere is given by:

S_cap = 2*pi*r*h Surface area of zone or cap

"h" is the length of the line vertically through the origin of the sphere, which at starts the sphere's apex and ends in the truncated layer at one end of the sphere. See - http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html - or for an upside down sphere - http://www.thelightingcenter.com/products/*/*/1033

The answer sought in this thread is "what is the ratio (or fractional part) of the truncated sphere to the entire sphere." or -

S_trunc = Fraction * [4*pi*r^2]

(See http://en.wikipedia.org/wiki/Solid_angle )

The fractional part seen by an observer across a single day is:

Fraction =    [2*pi*r*h] /  [4*pi*r^2]

- which solves to:

Fraction =    h /  2*r

- Our celestial sphere is a unit sphere << http://en.wikipedia.org/wiki/Sphere >>, so r=1 and

Fraction =    h /  2

The answer to the question "what fraction of the sky can you see from any latitude" boils down to what is "h" for any latitude in Archimedes Hat-Box-Theorem: http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html - except we are applying the theorem to the "big" part of the upside truncated sphere - http://www.thelightingcenter.com/products/*/*/1033 - and not the small cap shown in - http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html

Take a great circle through longitude 0 degrees on the sphere.  From that 2-D planar construction, it can easily be seen that "h" is the sum of cosine of 90 degrees, plus the cosine of the observer's latitude. http://members.csolutions.net/fisherka/astronote/astromath/fracsphere.gif

The cosine of 90 degrees = 1, so - h = 1 + cos(latitude) - and

Fraction of the sky that is observable across twenty-four hours of darkness and daylight is:

Fraction = [1+cos(geolatitude)] / 2

In my case at 40 north degs, that's the 88.3% you guys calculated, or about 36426 sq. degrees.

Thanks again for the further explanation of the geometric construction.  The extra tips helped me to piece it together. Sorry to be so dense initially.

- Canopus56

P.S. - If you approach the problem by finding the surface area of the small cap cut-off the end of the truncated sphere - that is diagram at Wolfram's website - http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html

- then the area of the small cap is -

Fraction_small_cap = [1-cos(geolatitude)] / 2

- or 11.7% at 40° north latitude.

P.P.S. - Amateur astronomers primarily are concerned with the fraction of the sky that we can see during the night.  Even of the shortest day of the year at 40° north, daylight and nighttime are equally divided - about 12 hours each - and the Earth rotates nearly a full 180° at night.  As a consequence, we might see the full 88% as "dark sky" around the longest day of the year.  Otherwise, night time at 40° north latitude is shorter than 12 hours and the Earth rotates less than 180°.  So, at 40° north, we see some fraction of the night sky between 88% and the 50% of the night sky that can be seen at any one instant during the night.

_______________________________________________ Utah-Astronomy mailing list Utah-Astronomy@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/utah-astronomy Visit the Photo Gallery: http://gallery.utahastronomy.com Visit the Wiki: http://www.utahastronomy.com