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{{distinguish|spherical angle}}
While there is not any definite remedy for hemorrhoids, there are nevertheless hemorrhoid treatment [http://www.progressivehealth.com/catalog/Avatrol.htm] tricks accessible to you to avoid the painful onslaught of the disease. It is heavily suggested that we consult a doctor inside order to ensure that the hemorrhoids will not receive worse. It is also important that you go by a treatment regime on the homes to counter the disease.<br><br>Now there are natural [http://hemorrhoidtreatmentfix.com/thrombosed-hemorrhoid thrombosed hemorrhoids] s. Aside from making we feel relieved, it will also strengthen the vein walls and improve vascular circulation. In this means you're preventing internal plus exterior hemorrhoid to occur. It can moreover stop hemorrhoid itching plus bleeding. The natural hemorrhoid treatment is the best thing to employ considering it does not only get rid of the pain, nevertheless also treat the casual source of hemorrhoid. It helps we ease the symptoms of hemorrhoid such as itching burning, swelling plus anal discomfort.<br><br>Another treatment for hemorrhoids is to utilize witch hazel. It's been shown to be successful at helping decrease swelling and helps with healing hemorrhoids which are bleeding. If your piles are extremely swollen, this may undoubtedly enable we today.<br><br>Then there is a high tech way of removing your hemorrhoid. Laser is regarded as the latest tool which is being use inside purchase to receive rid of the hemorrhoid. This is usually applied with those internal plus outside hemorrhoids depending on the severity of the issue.<br><br>Another tip that would offer certain immediate relief to the hemorrhoid issue is talking a good warm bathtub. The bathtub usually sooth the pain we feel, plus if you add a little salt into a bath, regarding a teaspoon or so, plus massage the hemorrhoid this may even further assist with pain relief.<br><br>Three. The next internal hemorrhoids treatment which we can utilize that will help we with your piles is to add a peeled garlic clove into a rectum. The garlic will help we eliminate the hemorrhoid.<br><br>These hemorrhoid treatments as revealed above are considered to become the top methods which have been selected by several persons. However, in the event you have tried these techniques and they cannot help you, then we want to see the doctor, that is really possible that you'll be suggested to try a surgical answer. Although surgery is considered to be an effective answer, nevertheless there are certain risks involved. Besides, it is very very expensive and it takes longer to heal too.
{{refimprove|date=December 2011}}
In [[geometry]], a '''solid angle''' (symbol: '''Ω''') is the two-dimensional angle in three-dimensional space that an object [[Subtended angle|subtend]]s at a point. It is a measure of how large the object appears to an observer looking from that point. In the [[International System of Units]] (SI), a solid angle is a [[dimensionless quantity|dimensionless]] [[unit of measurement]] called a '''[[steradian]]''' (symbol: '''sr''').
 
A small object nearby may subtend the same solid angle as a larger object farther away. For example, although the [[Moon]] is much smaller than the [[Sun]], it is also much closer to [[Earth]]. Therefore, as  viewed from any point on Earth, both objects have approximately the same solid angle as well as apparent size. This is evident during a [[solar eclipse]].
 
== Definition and properties ==
An object's solid angle is equal to the [[area (geometry)|area]] of the segment of a [[unit sphere]], centered at the angle's [[vertex (geometry)|vertex]], that the object covers. A solid angle equals the area of a segment of a unit sphere in the same way a planar [[angle]] equals the length of an arc of a [[unit circle]]. Solid angles are often used in [[physics]] and [[astrophysics]] where they are usually calculated as area divided by distance squared.
 
[[Image:Steradian.svg|thumb|Any area on a sphere, totaling the square of its radius and observed from its center, subtends precisely one [[steradian]].]]
 
The solid angle of a sphere measured from a point in its interior is 4[[Pi|π]] sr, and the solid angle subtended at the center of a cube by one of its faces is one-sixth of that, or 2π/3 sr. Solid angles can also be measured in [[square degree]]s (''1 sr'' = (180/π)<sup>2</sup> ''square degree'') or in fractions of the sphere (i.e., ''fractional area''), ''1 sr'' = 1/4π ''fractional area''.
 
In [[spherical coordinates]], there is a simple formula as
 
: <math>d\Omega = \sin\theta\,d\theta\,d\varphi</math>
 
The solid angle for an arbitrary [[oriented surface]] S subtended at a point P is equal to the solid angle of the projection of the
surface S to the unit sphere with center P, which can be calculated as the [[surface integral]]:
 
: <math>\Omega = \iint_S \frac{ \vec{r} \cdot \hat{n} \,dS }{r^3} = \iint_S \sin\theta\,d\theta\,d\varphi</math>
 
where <math> \vec{r} </math> is the [[vector (geometry)|vector]] position of an infinitesimal area of surface <math> \, dS </math> with respect to point P and where <math> \hat{n} </math> represents the unit vector normal to <math> \, dS </math>. Even if the projection on the unit sphere to the surface S is not [[isomorphic]], the multiple folds are correctly considered according to the surface orientation described by the sign of the scalar product <math>\vec{r} \cdot \hat{n}</math>.
 
==Practical applications==
*Defining [[luminous intensity]] and [[luminance]], and the correspondent radiometric quantities [[radiant intensity]] and [[radiance]].
*Calculating spherical excess ''E'' of a [[spherical triangle]]
*The calculation of potentials by using the [[boundary element method]] (BEM)
*Evaluating the size of [[ligand]]s in metal complexes, see [[ligand cone angle]].
*Calculating the [[electric field]] and [[magnetic field]] strength around charge distributions.
*Deriving [[Gauss's Law]].
*Calculating emissive power and irradiation in heat transfer.
*Calculating cross sections in [[Rutherford scattering]].
*Calculating cross sections in [[Raman scattering]].
*The solid angle of the [[acceptance cone]] of the [[optical fiber]]
 
== Solid angles for common objects ==
 
=== Cone, spherical cap, hemisphere ===
[[Image:Steradian cone and cap.svg|thumb|right|150px|Section of cone (1) and spherical cap (2) inside a sphere. In this figure {{nowrap|''θ'' {{=}} ''A''/2}} and {{nowrap|''r'' {{=}} 1}}.]]
The solid angle of a [[cone (geometry)|cone]] with [[Apex (geometry)|apex]] angle <math>2 \theta \,\!</math>, is the area of a [[spherical cap]] on a [[unit sphere]]
 
:<math>\Omega = 2\pi \left (1 - \cos {\theta} \right) </math>
 
For small ''θ'' in radians such that sin(''θ'')~''θ'', this reduces to the area of a circle π''θ''^2.
 
The above is found by computing the following [[double integral]] using the unit [[Spherical coordinate system#Integration and differentiation in spherical coordinates|surface element in spherical coordinates]]:
 
: <math>\int_0^{2\pi} \int_0^{\theta} \sin\theta' \, d \theta' \, d \phi = 2\pi\int_0^{\theta} \sin\theta' \, d \theta' = 2\pi\left[ -\cos\theta' \right]_0^{\theta} = 2\pi\left(1 - \cos\theta \right)</math>
 
Over 2200 years ago [[Archimedes]] proved, without the use of [[calculus]], that the surface area of a spherical cap was always equal to the area of a circle whose radius was equal to the distance from the rim of the spherical cap to the point where the cap's axis of symmetry intersects the cap. In the diagram opposite this radius is given as:
 
:<math> 2r \sin \left( \frac{ \theta}{2} \right) </math>
 
Hence for a unit sphere the solid angle of the spherical cap is given as:
 
:<math> \Omega = 4\pi \sin^2 \left( \frac{\theta}{2} \right) = 2\pi \left (1 - \cos {\theta} \right) </math>
 
When {{nowrap|''θ'' {{=}} π/2}}, the spherical cap becomes a [[Sphere|hemisphere]] having a solid angle 2π.
 
The solid angle of the complement of the cone (picture a melon with the cone cut out) is clearly:
:<math>4\pi - \Omega = 2\pi \left(1 + \cos {\theta} \right)</math>
 
A Terran astronomical observer positioned at latitude <math> \theta \,\!</math> can see this much of the [[celestial sphere]] as the earth rotates, that is, a proportion
:<math> 2\pi \left (1 + \cos {\theta} \right) </math>
 
At the equator you see all of the celestial sphere, at either pole only one half.
 
A segment of a cone cut by a plane at angle <math>\gamma</math> from the cone's axis can be calculated by the formula:<ref name = Mazonka>{{cite journal| last =  Mazonka| first = Oleg| year = 2012| title = Solid Angle of Conical Surfaces, Polyhedral Cones, and Intersecting Spherical Caps| journal = Cornell University Library Archive| url = http://arxiv.org/abs/1205.1396}}</ref>
 
:<math> \Omega = 2 \left( \arccos \frac{\sin\gamma}{\sin\theta} - \cos\theta \arccos\frac{\tan\gamma}{\tan\theta} \right) </math>
 
===Tetrahedron===
Let OABC be the vertices of a [[tetrahedron]] with an origin at O subtended by the triangular face ABC where <math>\vec a\ ,\, \vec b\ ,\, \vec c </math> are the vector positions of the vertices A, B and C. Define the vertex angle <math> \theta_a \, </math> to be the angle BOC and define <math> \theta_b ,\, \theta_c </math> correspondingly. Let <math> \phi_{ab} \, </math> be the [[dihedral angle]] between the planes that contain the tetrahedral faces OAC and OBC and define <math> \phi_{bc} ,\, \phi_{ac} </math> correspondingly. The solid angle at <math> \Omega </math> subtended by the triangular surface ABC is given by
 
:<math> \Omega = \phi_{ab} + \phi_{bc} + \phi_{ac} - \pi </math>
 
This follows from the theory of [[spherical excess]] and it leads to the fact that there is an analogous theorem to the theorem that ''"The sum of internal angles of a planar triangle is equal to <math> \pi </math>"'', for the sum of the four internal solid angles of a tetrahedron as follows:
 
:<math> \sum_{i=1}^4 \Omega_i = 2 \sum_{i=1}^6 \phi_i - 4 \pi </math>
 
where <math> \phi_i \, </math> ranges over all six of the dihedral angles between any two planes that contain the tetrahedral faces OAB, OAC, OBC and ABC.
 
An efficient algorithm for calculating the solid angle at <math> \Omega </math> subtended by the triangular surface ABC where <math>\vec a\ ,\, \vec b\ ,\, \vec c </math> are the vector positions of the vertices A, B and C has been given by Oosterom and Strackee:<ref>{{cite journal| last = Van Oosterom| first = A| coauthors = Strackee, J| year = 1983| title = The Solid Angle of a Plane Triangle| journal = IEEE Trans. Biom. Eng.| volume = BME-30| issue = 2| pages = 125–126| doi = 10.1109/TBME.1983.325207}}</ref>
 
:<math>\tan \left( \frac{1}{2} \Omega \right) =
  \frac{\left|\vec a\ \vec b\ \vec c\right|}{abc + \left(\vec a \cdot \vec b\right)c + \left(\vec a \cdot \vec c\right)b + \left(\vec b \cdot \vec c\right)a}
</math>
 
where
: <math>\left|\vec a\ \vec b\ \vec c\right|</math>
 
denotes the [[determinant]] of the matrix that results when writing the vectors together in a row, e.g. <math>M_{i1}=\vec a_i</math> and so on—this is also equivalent to the [[triple product|scalar triple product]] of the three vectors;
 
: <math>\vec a</math> is the vector representation of point A, while <math> \, a </math> is the magnitude of that vector (the origin-point distance);
 
: <math>\vec a \cdot \vec b</math> denotes the [[scalar product]].
 
When implementing the above equation care must be taken with the <code>atan</code> function to avoid negative or incorrect solid angles. One source of potential errors is that the determinant can be negative if a,b,c have the wrong [[determinant|winding]]. Computing <code>abs(det)</code> is a sufficient solution since no other portion of the equation depends on the winding. The other pitfall arises when the determinant is positive but the divisor is negative. In this case <code>atan</code> returns a negative value that must be biased by <math>\pi</math>.
 
<source lang="python">
from scipy import dot, arctan2, pi
from scipy.linalg import norm, det
def tri_projection(a, b, c):
  """Given three 3-vectors, a, b, and c."""
  determ = det((a, b, c))
 
  al = norm(a)
  bl = norm(b)
  cl = norm(c)
 
  div = al*bl*cl + dot(a,b)*cl + dot(a,c)*bl + dot(b,c)*al
  at = arctan2(determ, div)
  if at < 0: at += pi # If det > 0 and div < 0 arctan2 returns < 0, so add pi.
  omega = 2 * at
 
  return omega
</source>
 
Another useful formula for calculating the solid angle of the tetrahedron at the origin O that is purely a function of the vertex angles <math>\theta_a ,\, \theta_b ,\, \theta_c </math> is given by
[[Simon Antoine Jean L'Huilier|L' Huilier]]'s theorem as
 
: <math> \tan \left( \frac{1}{4} \Omega \right) =
    \sqrt{ \tan \left( \frac{\theta_s}{2}\right) \tan \left( \frac{\theta_s - \theta_a}{2}\right) \tan \left( \frac{\theta_s - \theta_b}{2}\right) \tan \left(\frac{\theta_s - \theta_c}{2}\right)} </math>
 
where
: <math> \theta_s = \frac {\theta_a + \theta_b + \theta_c}{2} </math>
 
===Pyramid===
The solid angle of a four-sided right rectangular [[Pyramid (geometry)|pyramid]] with [[apex (geometry)|apex]] angles <math>\, a</math> and <math>\, b</math> ([[dihedral angle]]s measured to the opposite side faces of the pyramid) is
: <math>\Omega = 4 \arcsin \left( \sin {a \over 2} \sin {b \over 2} \right) </math>
 
If both the side lengths (''α'' and ''β'') of the base of the pyramid and the distance (''d'') from the center of the base rectangle to the apex of the pyramid (the center of the sphere) are known, then the above equation can be manipulated to give
 
: <math>\Omega = 4 \arctan \frac {\alpha\beta} {2d\sqrt{4d^2 + \alpha^2 + \beta^2}} </math>
 
The solid angle of a right n-gonal pyramid, where the pyramid base is a regular n-sided polygon of circumradius (r), with a
pyramid height (h) is
 
:<math>\Omega = 2\pi - 2n \arctan\left(\frac {\tan{\pi\over n}}{\sqrt{1 + {r^2 \over h^2}}} \right) </math>
 
The solid angle of an arbitrary pyramid defined by the sequence of unit vectors representing edges <math> \{ s_1, s_2, ..., s_n \} </math> can be efficiently computed by:<ref name ="  Mazonka"/>
 
:<math> \Omega = 2\pi - \arg \prod_{j=1}^{n} \left(
    \left[ s_{j-1} s_j \right]\left[ s_{j} s_{j+1} \right] -
    \left[ s_{j-1} s_{j+1} \right] +
    i\left[ s_{j-1} s_j s_{j+1} \right]
  \right)
</math>
 
where parentheses are [[scalar product]] and square brackets is a [[scalar triple product]], and <math>i</math> is an [[imaginary unit]]. Indices are cycled: <math> s_0 = s_n </math> and <math> s_{n+1} = s_1 </math>.
 
=== Latitude-longitude rectangle ===
The solid angle of a latitude-longitude rectangle on a [[globe]] is <math>\left ( \sin \phi_N - \sin \phi_S \right ) \left ( \theta_E - \theta_W \,\! \right)</math>, where <math>\phi_N \,\!</math> and <math>\phi_S \,\!</math> are north and south lines of [[latitude]] (measured from the [[equator]] in [[radian]]s with angle increasing northward), and <math>\theta_E \,\!</math> and <math>\theta_W \,\!</math> are east and west lines of [[longitude]] (where the angle in radians increases eastward).:<ref>{{cite journal| year = 2003| title = Area of a Latitude-Longitude Rectangle| journal = The Math Forum @ Drexel| url = http://mathforum.org/library/drmath/view/63767.html}}</ref> Mathematically, this represents an arc of angle <math>\phi_N - \phi_S \,\!</math> swept around a sphere by <math>\theta_E - \theta_W \,\!</math> radians. When longitude spans 2π radians and latitude spans π radians, the solid angle is that of a sphere.
 
A latitude-longitude rectangle should not be confused with the solid angle of a rectangular pyramid. All four sides of a rectangular pyramid intersect the sphere's surface in [[great circle]] arcs. With a latitude-longitude rectangle, only lines of longitude are great circle arcs; lines of latitude are not.
 
=== Sun and Moon ===
The [[Sun]] is seen from Earth at an average angular diameter of about 9.35{{e|-3}} radians. The [[Moon]] is seen from Earth at an average diameter of 9.22{{e|-3}} radians. We can substitute these into the equation given above for the solid angle subtended by a cone with [[Apex (geometry)|apex]] angle <math>2 \theta \,\!</math>:
 
:<math>\Omega = 2 \pi \left (1 - \cos {\theta} \right) </math>
 
The resulting value for the [[Sun]] is 6.87{{e|-5}} steradians. The resulting value for the [[Moon]] is 6.67{{e|-5}} steradians. In terms of the total celestial sphere, the [[Sun]] and the [[Moon]] subtend ''fractional areas'' of 0.000546% ([[Sun]]) and 0.000531% ([[Moon]]). On average, the [[Sun]] is larger in the sky than the [[Moon]] even though it is much, much farther away.
 
== Solid angles in arbitrary dimensions ==
The solid angle subtended by the full surface of the unit [[n-sphere]] (in the geometer's sense) can be defined in any number of dimensions <math>d</math>. One often needs this solid angle factor in calculations with spherical symmetry. It is given by the formula
 
:<math>\Omega_{d} = \frac{2\pi^\frac{d}{2}}{\Gamma\left(\frac{d}{2}\right)}</math>
 
where <math>\Gamma</math> is the [[Gamma function]]. When <math>d</math> is an integer, the Gamma function can be computed explicitly. It follows that
 
:<math>
  \Omega_{d} = \begin{cases}
    \frac{1}{ \left(\frac{d}{2} - 1 \right)!} 2\pi^\frac{d}{2} & d\text{ even} \\
    \frac{\left(\frac{1}{2}\left(d - 1\right)\right)!}{(d - 1)!} 2^d \pi^{\frac{1}{2}(d - 1)} & d\text{ odd}
  \end{cases}
</math>
 
This gives the expected results of 2π rad for the 2D circumference and 4π [[steradian|sr]] for the 3D sphere. It also throws the slightly less obvious 2 for the 1D case, in which the origin-centered unit "sphere" is the set <math>{ -1 , 1 }</math>, which indeed has a [[measure (mathematics)|measure]] of 2.
 
== References ==
{{Reflist}}
{{commons category}}
 
== External links ==
*Arthur P. Norton, A Star Atlas, Gall and Inglis, Edinburgh, 1969.
*F. M. Jackson, Polytopes in Euclidean n-Space. Inst. Math. Appl. Bull. (UK) 29, 172-174, Nov./Dec. 1993.
*M. G. Kendall, A Course in the Geometry of N Dimensions, No. 8 of Griffin's Statistical Monographs & Courses, ed. M. G. Kendall, Charles Griffin & Co. Ltd, London, 1961
*{{mathworld|urlname=SphericalExcess|title=Spherical Excess}}
*{{mathworld|urlname=SolidAngle|title=Solid Angle}}
 
{{DEFAULTSORT:Solid Angle}}
[[Category:Angle]]
[[Category:Euclidean solid geometry]]

Latest revision as of 09:00, 4 October 2014

While there is not any definite remedy for hemorrhoids, there are nevertheless hemorrhoid treatment [1] tricks accessible to you to avoid the painful onslaught of the disease. It is heavily suggested that we consult a doctor inside order to ensure that the hemorrhoids will not receive worse. It is also important that you go by a treatment regime on the homes to counter the disease.

Now there are natural thrombosed hemorrhoids s. Aside from making we feel relieved, it will also strengthen the vein walls and improve vascular circulation. In this means you're preventing internal plus exterior hemorrhoid to occur. It can moreover stop hemorrhoid itching plus bleeding. The natural hemorrhoid treatment is the best thing to employ considering it does not only get rid of the pain, nevertheless also treat the casual source of hemorrhoid. It helps we ease the symptoms of hemorrhoid such as itching burning, swelling plus anal discomfort.

Another treatment for hemorrhoids is to utilize witch hazel. It's been shown to be successful at helping decrease swelling and helps with healing hemorrhoids which are bleeding. If your piles are extremely swollen, this may undoubtedly enable we today.

Then there is a high tech way of removing your hemorrhoid. Laser is regarded as the latest tool which is being use inside purchase to receive rid of the hemorrhoid. This is usually applied with those internal plus outside hemorrhoids depending on the severity of the issue.

Another tip that would offer certain immediate relief to the hemorrhoid issue is talking a good warm bathtub. The bathtub usually sooth the pain we feel, plus if you add a little salt into a bath, regarding a teaspoon or so, plus massage the hemorrhoid this may even further assist with pain relief.

Three. The next internal hemorrhoids treatment which we can utilize that will help we with your piles is to add a peeled garlic clove into a rectum. The garlic will help we eliminate the hemorrhoid.

These hemorrhoid treatments as revealed above are considered to become the top methods which have been selected by several persons. However, in the event you have tried these techniques and they cannot help you, then we want to see the doctor, that is really possible that you'll be suggested to try a surgical answer. Although surgery is considered to be an effective answer, nevertheless there are certain risks involved. Besides, it is very very expensive and it takes longer to heal too.