Number of square degrees in sky

[callisto]

Constellations are officially areas of the sky and each of the 88 has a
number of square degrees. Some websites give such info, such as

http://www.starshine.com/frankn/astronomy/constellations.asp

Can one of you physics or math guys tell me exactly how many square
degrees are in the whole celestial sphere? Also, how many degrees of
sky can be seen at different latitudes (taking into account the 23.5
degree tilt of the spin axis of the earth). For example I am at 40
degrees north and I cannot see the Southern Cross at any time of the
year but how many square degrees of sky can I see? I presume that under
ideal conditions I could see something on the horizon with a
declination of minus 50 degrees but how does this translate into square
degrees?

On the other hand, people at the equator I think could see the entire
celetial sphere if you ignore atmospheric effects at the horizon.

[Klaatu]

callisto wrote:
<cut> how many square

Post by callisto
degrees are in the whole celestial sphere

<cut>

41253 square degrees
http://www.badastronomy.com/bitesize/bigsky.html

[William Hamblen]

Post by callisto
Can one of you physics or math guys tell me exactly how many square
degrees are in the whole celestial sphere?

The area of a sphere is 4 * pi * r * r with pi being 3.14159265... and
r being the radius. If a sphere has a circumference of 360 degrees it
has a radius of 180 / pi degrees. Plugging that into the formula you
get 4 * 180 * 180 / pi degrees or almost 41253 square degrees.

Post by callisto
Also, how many degrees of
sky can be seen at different latitudes (taking into account the 23.5
degree tilt of the spin axis of the earth). For example I am at 40
degrees north and I cannot see the Southern Cross at any time of the
year but how many square degrees of sky can I see? I presume that under
ideal conditions I could see something on the horizon with a
declination of minus 50 degrees but how does this translate into square
degrees?

The tilt of the Earth's axis doesn't affect how much of the sky you
can see, although it does affect which part of the sky you can see.
The thing that affects how much of the sky you can see is where you
are on the Earth. Because you can't see through the Earth it blocks
part of the sky. The fraction of the sky you can see at any one time
depends on your latitude. At the poles half of the sky is always
below the horizon. At the equator all of the sky is visible over the
course of a day. The formula is ( 1 + cosine( latitude ) ) / 2. At
40 degrees North or South you can see more than 88% of the sky. Which
part of the sky depends on which hemisphere you are in.

Post by callisto
On the other hand, people at the equator I think could see the entire
celetial sphere if you ignore atmospheric effects at the horizon.

Refraction means you can in principle see more than half of the sky at
any one time.

[ixtok]

Post by William Hamblen
The area of a sphere is 4 * pi * r * r with pi being 3.14159265... and
r being the radius. If a sphere has a circumference of 360 degrees it
has a radius of 180 / pi degrees. Plugging that into the formula you
get 4 * 180 * 180 / pi degrees or almost 41253 square degrees.

Why wouldn't it be 360 * 360 or 12960 square degrees?

[Chris L Peterson]

Post by ixtok

Post by William Hamblen
The area of a sphere is 4 * pi * r * r with pi being 3.14159265... and
r being the radius. If a sphere has a circumference of 360 degrees it
has a radius of 180 / pi degrees. Plugging that into the formula you
get 4 * 180 * 180 / pi degrees or almost 41253 square degrees.

Why wouldn't it be 360 * 360 or 12960 square degrees?

Because the area of a sphere isn't the square of the circumference. It
is the square of the radius times 4 times pi.

_________________________________________________

Chris L Peterson
Cloudbait Observatory
http://www.cloudbait.com

[William Hamblen]

Post by ixtok

Post by William Hamblen
The area of a sphere is 4 * pi * r * r with pi being 3.14159265... and
r being the radius. If a sphere has a circumference of 360 degrees it
has a radius of 180 / pi degrees. Plugging that into the formula you
get 4 * 180 * 180 / pi degrees or almost 41253 square degrees.

Why wouldn't it be 360 * 360 or 12960 square degrees?

Because then it would be a square.

[Robert L]

Because degrees of declination getter shorter as you leave the
equator. there are indeed 12960 "holes" in the grid.

[Paul Schlyter]

Post by ixtok

Post by William Hamblen
The area of a sphere is 4 * pi * r * r with pi being 3.14159265... and
r being the radius. If a sphere has a circumference of 360 degrees it
has a radius of 180 / pi degrees. Plugging that into the formula you
get 4 * 180 * 180 / pi degrees or almost 41253 square degrees.

Why wouldn't it be 360 * 360 or 12960 square degrees?

First, 360 * 360 is 129600, not 12960

Well, a sphere isn't a square with a side of 360 degrees. At least
you should have suggested 360 * 180 = 64800 square degrees (since the
circumference of the celestial equator is 360 degrees but the
distance from pole to pole is only 180 degrees). But a sphere isn't
such a rectangle either.

Consider a region in the sky, which is one degree wide in RA and one
degree wide in Decl. What's the area of this region? One square
degree? No, it's rather

cos(Decl) (approximately)

square degrees, where Decl is the declination of the midpoint of this
area. At the celestial equator the area of the region will be one
square degree, but at 60 deg declination its area will only be one
half spherical degree.

This effect will make the area of the celestial sphere considerably
less than 360 * 180 = 64800 square degrees. To find out the area of
the celescial sphere, do like this:

Circumference = 360 degrees

Radius = circumference / (2*pi) = 360/(2*pi) = 180/pi degrees

Area = radius^2 * 4*pi = (180/pi)^2 * 4*pi = 129600/pi
= 41252.9612494193..... square degrees

So your initial guess that the apparent area of the entire celestial
sphere is 360 * 360 = 129600 square degrees was too large by exactly
a factor of pi.

[Sam Wormley]

Post by callisto
Can one of you physics or math guys tell me exactly how many square
degrees are in the whole celestial sphere?

Google is your friend
http://www.google.com/search?q=convert+4+pi+%28radians%5E2%29+to+degrees%5E2

[Odysseus]

Post by callisto
Can one of you physics or math guys tell me exactly how many square
degrees are in the whole celestial sphere?

Well, I'm neither one of those, but it's fairly easy to calculate.
One method starts with the information that there are 4*pi steradians
in the surface of a sphere (you may remember the formula A =
4*pi*r^2) and that there are 180/pi degrees in a radian, giving
4*pi*(180°/pi)^2 = 129600/pi ~= 41253 square degrees.

Another, slightly less direct, approach (also less accurate in
practice, unless you have access to high-precision computation
software) involves solving the right spherical triangle whose legs
are 1° in length (therefore having an area of half a square degree),
and finding the spherical excess. A handy relation exists between
this quantity and the area of a spherical triangle, namely that the
ratio of the excess to a circle equals the ratio of the area to a
hemisphere. Here we have 180°/(atn(sec(1°)) - 45°) ~= 41253.

Post by callisto
Also, how many degrees of
sky can be seen at different latitudes (taking into account the 23.5
degree tilt of the spin axis of the earth). For example I am at 40
degrees north and I cannot see the Southern Cross at any time of the
year but how many square degrees of sky can I see? I presume that under
ideal conditions I could see something on the horizon with a
declination of minus 50 degrees but how does this translate into square
degrees?

The obliquity of the ecliptic makes no difference; it affects the
visibility of the Sun but not the celestial sphere in general. The
only relevant angle is the latitude: from a given northern latitude
L, ignoring the atmosphere and assuming a 'geometrically perfect'
horizon, the portion of the celestial sphere with declination south
of L - 90° is never visible. (You seem to know this already -- you
didn't have to use the 23.5° figure to derive the 'threshold' of -50°
Dec. from your latitude of +40°.) Note also that stars with Decs.
north of 90° - L are circumpolar, i.e. never set.

Anyway, the area of the spherical zone from the pole to declination D
is 2*pi*(1 - sinD) steradians; substituting in terms of latitude as
above this becomes 2*pi*(1 + cosL). So from a latitude of 40°N the
available portion of the sky covers 2*pi*(1 + cos40°)*(180°/pi)^2 =
(64800/pi)*(1 + cos40°) ~= 36427 square degrees. Note that this
represents (1 + cos40°)/2 = 88.3% of the celestial sphere.

Post by callisto
On the other hand, people at the equator I think could see the entire
celetial sphere if you ignore atmospheric effects at the horizon.

Yes. Where L = 0° above, L - 90° = -90° so the southern boundary of
visibility is at the pole itself (right on the horizon), 90° - L =
+90° so nothing but the NCP itself is circumpolar, and (1 + cosL)/2 =
100% of the sky. Of course north and south are perfectly symmetrical
here. Compare with the other extreme, at the (geographic) North Pole:
where L = 90°, L - 90° = 0° so nothing south of the celestial equator
is visible, 90° - L = 0° so everything north of the CE is
circumpolar, and (1 + cosL)/2 = 50% of the sky.

[Paul Schlyter]

Post by Odysseus
Anyway, the area of the spherical zone from the pole to declination D
is 2*pi*(1 - sinD) steradians; substituting in terms of latitude as
above this becomes 2*pi*(1 + cosL). So from a latitude of 400N the
available portion of the sky covers 2*pi*(1 +
cos40_deg)*(180_deg/pi)^2 = (64800/pi)*(1 + cos40_deg) ~= 36427
square degrees. Note that this represents (1 + cos40_deg)/2 = 88.3%
of the celestial sphere.

One easy rule which helps you remember the formula for the area of a
spherical zone is this:

The area of a spherical zone = the area of (the curved part of) a
cylinder with the same radius as the sphere and the same height as
the spherical zone: area = 2*pi*r*h where r=radius and h=height

The height of a spherical zone is of course h = r * (sin(d2)-sin(d1))
where d1 and d2 are the "limiting declinations". If d1 = -90_deg and
d2 = +90_deg then the spherical zone becomes the entire sphere, and
h = 2*r which yields the formula for the area of the entire sphere:
area = 2*pi*r*h = 2*pi*r*2*r = 4 * pi * r^2

Post by Odysseus

Post by callisto
On the other hand, people at the equator I think could see the entire
celestial sphere if you ignore atmospheric effects at the horizon.

Even if we include atmospheric refraction, all of the sky will be
seen from the Earth's equator. A small part around each celestial
pole will even be circumpolar.

If we include "atmospheric effects" in general though, the situation
becomes quite different. Clouds is a common atmospheric effect,
particularly at the equator, and clouds can completely prevent us
from seeing any part of the sky .... :-)

[canopus56]

Post by Odysseus
Anyway, the area of the spherical zone from the pole to declination D
is 2*pi*(1 - sinD) steradians; substituting in terms of latitude as
above this becomes 2*pi*(1 + cosL). So from a latitude of 40°N the
available portion of the sky covers 2*pi*(1 + cos40°)*(180°/pi)^2 =
(64800/pi)*(1 + cos40°) ~= 36427 square degrees. Note that this
represents (1 + cos40°)/2 = 88.3% of the celestial sphere.

I'm not entirely following your math, but it seems to me that you are
falsely equivocating steradians ("square radians") viewed when the
observer looks at the zenith in the local horizon system (alt-az) with
the steradians seen when the observer looks at an angle at the north
pole of the celestial sphere in the celestial coordinate system.

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.

The local observer will see more lines of latitude, in the celestial
coordinate system, near the north horizon than would be implied by a
180 degree horizon, because degrees are smaller near the North
Celestial Pole than they are at the celestial equator.

But that doesn't imply that a square degree at let's say RA 0hr between
Dec 88 and 89 degrees has the same size in steradians (square degrees)
that that same square degree in the local horizon system has at Az 0
deg, Alt 21.5 to 22.5 degs. This can easily be seen using the Cartes
de Ciel planetarium program. Cartes de Ciel allows the user to look at
the North Celestial Pole and to project both grid lines for the
equatorial (celestial) coordinate system and the az-alt (local horizon)
coordinate system at the same time.

I don't know the math to find the equivalent steradians between the
local horizon system and the celestial coordinate system viewed at a
rotated angle.

Finally, you computation of 88% of the celestial sphere being visible
implies that a steradian in the celestial coordinate system between
let's say RA 0hr between Dec 88 and 89 degrees and RA 0hr between Dec 1
and 2 degrees is the same size - which they aren't.

In conclusion, not matter what coordinate system you use, you can only
see 1/2 of the night sky sphere, which in sterdians (square degrees) is
always:

1/2* 41253 = 20626.5 deg^2

Sorry, if I'm not being clear.

- Perplexed at 41 deg North lat - Canopus56

[canopus56]

canopus56 wrote:
<snip all>

P.S. - Some useful links for lurkers -

http://en.wikipedia.org/wiki/Solid_angle
http://en.wikipedia.org/wiki/Steradian

[William Hamblen]

Post by canopus56
Finally, you computation of 88% of the celestial sphere being visible
implies that a steradian in the celestial coordinate system between
let's say RA 0hr between Dec 88 and 89 degrees and RA 0hr between Dec 1
and 2 degrees is the same size - which they aren't.

The area between lines of latitude and longitude one degree apart is
not a square degree. It's always less. Slightly less near the
equator and a lot less near the poles.

[canopus56]

Post by William Hamblen
Because you can't see through the Earth it blocks
part of the sky. The fraction of the sky you can see at any one time
depends on your latitude. At the poles half of the sky is always
below the horizon. At the equator all of the sky is visible over the
course of a day. The formula is ( 1 + cosine( latitude ) ) / 2. At
40 degrees North or South you can see more than 88% of the sky.

Okay, I guess I missed what you were saying. I think I get the
geometry.

Take an infinite celestial sphere with a radius of one radian. The
Earth is a dimensionless point at the middle of the celestial sphere.
Take a plane that represents the observer's horizon in the celestial
sphere that goes through the center. For a 40 deg north observer, tilt
the plane by 40 degrees to represent the horizon seen at the 40 deg
latitude. Now rotate the plane through 360 degrees, representing the
rotation of the earth over twenty-four hours.

The result is there is a 40 degree solid-angle cone that is always
visible - that is it is always above the horizon. There is a 40 degree
solid angle under your feet that you never see.

The 40 degree solid angle you can't see divided by the sphere's 360
degree solid angle (40/360) = 0.1111.

Or the fractional percent of the sphere you can't see is 11.11% and you
can see 88.88% or your 88% of the visible sky.

Since you can see 88% of the visible sky at 40 deg north _across the
entire rotation of the Earth during one twenty-four hour period_
(discounting the effect of daylight), then, using the steradian formula
discussed previously in this thread, the fractional area in square
degrees you can see on anyone day (discounting daylight) is:

88.88% * 41253 deg^2 = 36666 deg^2

I assume, your equation ( ( 1 + cosine( latitude ) ) / 2 ) gives the
inverse of the solid angle you can't see.

Is that right?

But 88% still looks wrong to me.

Doesn't the blocked cone have a solid angle of 2 x 40 degrees or 80
degrees? Seems like that would be the blocked out cone when a plane
tilted at 40 degrees of latitude is rotated through 360 degrees? In
that case, the blocked solid angle is 80 degrees and the fractional
part is 80/360 or 22.22% and the part you can see is:

77.77% * 41253 deg^2 = 32086 deg^2

The guy at 62 deg north has a 124 degree solid angle blocked out by the
Earth, or (124/360) 34.44%. He can see 65% of the sky.

The local horizon of the guy at the North Pole traces out a 180 degree
solid angle across 24 hours, so he sees 180/360 or 50% of the sky.

- Still perlexed and trying to understand the geometry. - Canopus56

P.S. -

To add the effect of daylight -

Your 88% or my 77% of the sky is never seen at 40 deg north because of
daylight. At 40 deg north, on the longest day of the year, there are
about 8 hours 45 minutes of daylight and 15 hours and 15 minutes of
darkness - or about 63% darkness. The "tilted plane" only rotates
through 228.25 degrees ( 15:15 * 15 degrees per hour). You never the
full 360 degree rotation of the plane in darkness and you never see the
full 88% or 77% of celestial sky that passes over your head.

While the guy at 62 degrees north can only see 65% of the sky, on Dec.
24, he has about 5 hours and 10 minutes of daylight, or 18 hours and 50
minutes of darkness, or about 283.25 degrees of rotation. He does not
get to see his full 65% of the sky that rotates over his head each day.


The person at the equator has 100% of the sky pass overhead, but they
are limited to 12 hours of daylight, so he or she really only gets to
see 180 degrees of the sky each day.

The guy at the North Pole, on the longest day of the year sees 1/2 the
sky; on the shortest, he see nothing.

[Brian Tung]

Post by canopus56
Take an infinite celestial sphere with a radius of one radian. The
Earth is a dimensionless point at the middle of the celestial sphere.
Take a plane that represents the observer's horizon in the celestial
sphere that goes through the center. For a 40 deg north observer, tilt
the plane by 40 degrees to represent the horizon seen at the 40 deg
latitude. Now rotate the plane through 360 degrees, representing the
rotation of the earth over twenty-four hours.
The result is there is a 40 degree solid-angle cone that is always
visible - that is it is always above the horizon. There is a 40 degree
solid angle under your feet that you never see.

So far, so good, although you use some non-standard terminology. It's
inappropriate to speak of an infinite celestial sphere with a radius of
one radian; if it's really infinite, it can't have a radius of one
anything, and in any event, a radian is usually a (dimensionless) unit
of angle.

Post by canopus56
The 40 degree solid angle you can't see divided by the sphere's 360
degree solid angle (40/360) = 0.1111.

Ahh, no. First of all, the cone has a *half*-angle of 40 degrees. Its
angle from stem to stern, if you will, is 80 degrees. Why would you
divide the half-angle of the solid angle by the whole of the sphere's
girth?

More importantly, though, if you are to figure out the proportion of sky
that is visible, you must measure things in *solid* angles, not planar
angles, as you've done.

Suppose you have a perfect spherical shell, and you measure out a cap
that has a half-angle of 40 degrees; that is to say, two lines drawn
from the center--one toward the center of the cap, and one toward its
edge--will meet at an angle of 40 degrees.

Now, cut the cap off. What is the surface area of that cap, in relation
to the area of the entire original sphere? It is *not* 40 divided by
360, nor is it even 80 divided by 360.

Archimedes discovered a surprising fact that Paul Schlyter mentioned
earlier in this thread. I wondered if anyone would comment on it, but
I guess it was lost in the middle of his post. Here it is: Suppose you
wrap a cylinder tightly around a sphere, so that the radius of the
cylinder's bases is equal to the radius of the sphere, and the height of
the cylinder is equal to the diameter of the sphere. Now cut a slice
out of both figures, of any thickness, but perpendicular to the axis
of the cylinder. The surface areas of both slices is exactly the same!

One special case is where the slice is cut from the end of the cylinder,
so that you have an endcap of the sphere. In that case, the thickness
of the slice is the height of the endcap. For a 40-degree half-angle,
the thickness would be r*(1-sin(50)). The radius of the cylinder is
r, so the area of both the cylinder's slice and the sphere's slice is
2*pi*r^2*(1-sin(50)). (You must neglect the area of the bases of the
cylinder.)

Since the area of the original sphere was 4*pi*r^2, the fraction of the
sphere represented by the endcap is (1-sin(50))/2, or about 0.117. That
*is* close to 0.111..., but it's not quite right. If the observer were
at latitude 72 degrees, your method would have yielded 72/360, or 0.2,
whereas the correct answer is (1-sin(28))/2, or about 0.345. By the way,
the answer can also be rewritten as

(1-sin(90-L))/2 = (1-cos(L))/2 = sin^2(L/2)

if that's easier to remember.

Another aside: Archimedes's property of the cylinder and the sphere was
part of the way to his formula for the volume of the sphere. All done
in classical Greek times!

[canopus56]

Brian Tung wrote:

This link may help lurkers to visualize Brian's narrative -

http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html

- except in the "special case," Brian is talking about the segment from
the top of the sphere down to 40 deg lat and the top of the cylinder,
down to the same height as 40 deg lat on the sphere.

- Canopus56

Post by Brian Tung
Suppose you have a perfect spherical shell, and you measure out a cap
that has a half-angle of 40 degrees; that is to say, two lines drawn
from the center--one toward the center of the cap, and one toward its
edge--will meet at an angle of 40 degrees.
Now, cut the cap off. What is the surface area of that cap, in relation
to the area of the entire original sphere? <snip>
Archimedes discovered a surprising fact that Paul Schlyter mentioned
earlier in this thread. I wondered if anyone would comment on it, but
I guess it was lost in the middle of his post. Here it is: Suppose you
wrap a cylinder tightly around a sphere, so that the radius of the
cylinder's bases is equal to the radius of the sphere, and the height of
the cylinder is equal to the diameter of the sphere. Now cut a slice
out of both figures, of any thickness, but perpendicular to the axis
of the cylinder. The surface areas of both slices is exactly the same!
One special case is where the slice is cut from the end of the cylinder,
so that you have an endcap of the sphere. In that case, the thickness
of the slice is the height of the endcap. For a 40-degree half-angle,
the thickness would be r*(1-sin(50)). The radius of the cylinder is
r, so the area of both the cylinder's slice and the sphere's slice is
2*pi*r^2*(1-sin(50)). (You must neglect the area of the bases of the
cylinder.)

<snip>

[Odysseus]

Post by canopus56

Post by Odysseus
Anyway, the area of the spherical zone from the pole to declination D
is 2*pi*(1 - sinD) steradians; substituting in terms of latitude as
above this becomes 2*pi*(1 + cosL). So from a latitude of 40°N the
available portion of the sky covers 2*pi*(1 + cos40°)*(180°/pi)^2 =
(64800/pi)*(1 + cos40°) ~= 36427 square degrees. Note that this
represents (1 + cos40°)/2 = 88.3% of the celestial sphere.

I'm not entirely following your math, but it seems to me that you are
falsely equivocating steradians ("square radians") viewed when the
observer looks at the zenith in the local horizon system (alt-az) with
the steradians seen when the observer looks at an angle at the north
pole of the celestial sphere in the celestial coordinate system.

"Equivocating"? I may not have been perfectly clear but I assure you
it wasn't intentional!

A steradian is the same size regardless of what direction you're looking.

Post by canopus56
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.
4*pi*(180/pi)^2 = 4*(pi/pi^2)*180^2 = (4*180^2)/pi = 41253 deg^2

Didn't I (and several others) say just that?

Post by canopus56
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.

I think you're misapprehending the original question, which was not
how much of the sky one can see at any one time (which is indeed
always exactly half, under the usual simplifying assumptions), but
how much of it is *ever* visible during the course of a day or year
from a given location. You should be able to understand,
qualitatively at least, that the amount ranges from half of it, at
the poles, up to all of it, at the equator.

Post by canopus56
The local observer will see more lines of latitude, in the celestial
coordinate system, near the north horizon than would be implied by a
180 degree horizon, because degrees are smaller near the North
Celestial Pole than they are at the celestial equator.

Nonsense. Degrees and square degrees are always the same size,
despite that a square degree isn't the area of a 'square' formed by
lines of latitude and longitude 1° apart anywhere but on the equator.
Don't mistake the spherical coordinate grid for a general means of
measuring distances or areas. This is probably the largest obstacle
to understanding spherical geometry, _viz_ that it's non-Euclidean.
On the surface of a sphere distances are treated as angles because
they're measured along geodesic arcs, not along straight lines as in
plane geometry. The cells in a grid of parallels and meridians
certainly aren't equal in area, but that doesn't mean we have to
throw the definition of area out the window; we just have to use
different formulas to calculate it.

<snip>

Post by canopus56
I don't know the math to find the equivalent steradians between the
local horizon system and the celestial coordinate system viewed at a
rotated angle.

It makes no difference what direction you're looking in, or how the
coordinate system is oriented; spherical areas are invariant under
all concentric rotations. I thought I'd given a fairly complete
derivation of the formulas I used; please point out where you find it
unclear or misleading. On the one hand I'm quite willing to explain
any of my reasoning that didn't come across; on the other I'm quite
open to correction on specifics.

Paul's reply to my post mentioned Archimedes' insight about the
equality in areas of a spherical zone and its projection onto a
circumscribed cylinder: this result is well worth studying, and with
just a little algebra can provide an 'independent' justification for
the (1 + cosL)/2 formula.

Post by canopus56
Finally, you computation of 88% of the celestial sphere being visible
implies that a steradian in the celestial coordinate system between
let's say RA 0hr between Dec 88 and 89 degrees and RA 0hr between Dec 1
and 2 degrees is the same size - which they aren't.

It implies nothing of the kind. Look at it this way: from 40°N
latitude the only part of the celestial sphere that one can *never*
see is the bowl-shaped region within 40° of the SCP. Just from
inspecting a sketch -- ideally on a globe -- is it not at least
plausible that such a bowl has only about an eighth the area of the
whole sphere? Either with a little calculus or by using Archimedes'
cylinder, the proportion can be shown to be (1 - cos40°)/2 = 11.7%
(derivation supplied on request).

[Brian Tung]

Post by Odysseus
"Equivocating"? I may not have been perfectly clear but I assure you
it wasn't intentional!

I'm sure he meant "equating." But you're right: To paraphrase Stein,
(a) steradian is a steradian is a steradian.

[canopus56]

Odysseus wrote:
<snip all>

Odysseus and Brian T. -

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.
Loading Image...

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.

[John Savard]

Post by callisto
Can one of you physics or math guys tell me exactly how many square
degrees are in the whole celestial sphere?

Indeed. 360 times 180 times 2/pi.

John Savard
http://www.quadibloc.com/index.html
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