Subtended Solid Angles
ISCARE
Hi,
It is a well known fact that the angle subtended by an arc on the
circumference of a circle is half that subtended by it at the centre.
I had a doubt whether this pushes on to 3D, say on a sphere:
i.e. is the solid angle subtended by a patch on the sphere's surface,
half (or is it any other number?) of the solid angle subtended by the
same patch at the centre of the sphere.
Does it make sense to define solid angle this way?
Could you give me good and workable references to solid angles?
Thanks for your help!
Ian Parker
Yes, because if you take a circular cross section of the sphere the
angle at the centre is twice that at the circumference. The same is
true of any other cross section. Hence the same rule must apply to
solid angles. As the angle is solid it will be 4 times the value.
We can verify this by dividing the angle up into strips. The angular
thickness of each strip being twice the size at the centre.
- Ian Parker
serge bouc
Le 14.04.2010 13:40, ISCARE a Žcrit :
any piece of the sphere surface, I don't think there is a fixed ratio
between these two solid angles : consider the case of a circle, drawn on
the sphere. From the center of the sphere, this supports a circular
cone, with half-angle say A at the top. Then the corresponding
solid angle seems to be 4 * pi * (sin(A/2))^2.
But now if you consider this from the point of the sphere "opposite"
to your circle, you have another circular cone, with top half-angle
equal to A/2 (because it is indeed seen from a point on a circle
for which the corresponding angle at the center is A). So the new
solid angle is 4 * pi * (sin(A/4))^2.
The ratio of these two is equal to 4 * (cos(A/4))^2, so it depends
on A.
Ian Parker
Correction. I now don't think this is so. Let us take a small patch on
the surface and view it from a point on the circumference of the
sphere. It is clear that if I draw a great circle between the patch
and the point at which the angle is subtended, the angle at the centre
will always be 0.5 of that at the centre. However suppose I view the
angle subtended at right angles. As I move along the Great Circle
towards the patch the angle at right angles to the GC increases until
eventually it becomes 2Pi.
Sorry about that. We all have to feel our way.
- Ian Parker
gudi
" Integra Curvatura " or integral curvature as KF Gauss called it, is
defined as an isometric invariant, double integral K dA, or dA /(R1*
R2), an inherent surface property, evaluated from the coefficients of
the first fundamental form of surface theory. If the surface is made
of rubber sheet and bent without stretching it retains the same this
value. The spherical images of any patch (surface normal vectors at
boundary) translated/ displaced to a unit sphere center describe a
patch of surface area with so many steredians. The sphere or any
closed simply connected enclosure has 4 Pi steredians solid angle , it
does not depend on subtended solid angle at the surface of a sphere or
any other outer enclosure,quite unlike the limiting rays of 2D angles.
The integral curvature and angle turned in the tangent plane together
add up to a topological constant, mentioned in the Gauss-Bonnet
theorem. Any textbook on differential geometry contains references to
this after treatment of the Egregium theorem.
I should suppose that there is no way of measuring solid angles with
respect to a central datum point as you suggest, the way it is done
analogously in 2D, using any circle arc between two reference rays.
Please note that even in 2D, integral ds/R is an isometric
object,equalling the rotation of the curve tangent. IIRC the text book
by Guggenheimer / Dover publishers mentions these aspects.
Regards
Narasimham
Narasimham
gudi
> On Apr 14, 4:40 pm, ISCARE <lifesjustliket...@gmail.com> wrote:
>
>
>
>
>
> > Hi,
>
> > It is a well known fact that the angle subtended by an arc on the
> > circumference of a circle is half that subtended by it at the centre.
>
> > I had a doubt whether this pushes on to 3D, say on a sphere:
>
> > i.e. is the solid angle subtended by a patch on the sphere's surface,
> > half (or is it any other number?) of the solid angle subtended by the
> > same patch at the centre of the sphere.
>
> > Does it make sense to define solid angle this way?
>
> > Could you give me good and workable references to solid angles?
>
> > Thanks for your help!
>
> defined as an isometric invariant, double integral K dA, or dA /(R1*
> R2), �an inherent surface property, evaluated from the coefficients of
> the first fundamental form of surface theory. If the surface is made
> of rubber sheet and bent without stretching it retains the same this
> value. The spherical images of any patch (surface normal vectors at
> boundary) translated/ displaced to a unit sphere center describe a
> patch of surface area with so many steredians. The sphere or any
> closed simply connected enclosure has 4 Pi steredians solid angle , it
> does not depend on subtended solid angle at the surface of a sphere or
> any other outer enclosure,quite unlike the limiting rays of 2D angles.
> The integral curvature and angle turned in the tangent plane together
> add up to a topological constant, mentioned in the Gauss-Bonnet
> theorem. Any textbook on differential geometry contains references to
> this after treatment of the Egregium theorem.
>
> I should suppose that there is no way of measuring solid angles with
> respect to a central datum point as you suggest, the way it is done
> analogously in 2D, using any circle arc between two reference rays.
> Please note that even in 2D, integral ds/R is an isometric
> object,equalling the rotation of the curve tangent. IIRC the text book
> by Guggenheimer / Dover publishers mentions these aspects.
Even for a continuous 2D curve, total tangent/normal rotation is
reckoned
for the entire curve, the constituent angle is defined instantaneously
or
differentially only at a particular point, the radial rays keep moving
all
the time. So why should it be different in 3D with a fixed inertial
datum?
Narasimham
ISCARE
Let us take a specialized case.
The semicircle : 180* (asterisk for degree) at the centre i.e.
straight line / diameter and 90* anywhere inside.
The hemisphere : solid angle of 2*pi at the centre AND SOLID ANGLE OF
__(?)___ on the surface?
Fill in the (?)
How do we confirm this?
Ilya Zakharevich
On 2010-04-14, ISCARE <lifesjus...@gmail.com> wrote:
>
>
> Hi,
>
> It is a well known fact that the angle subtended by an arc on the
> circumference of a circle is half that subtended by it at the centre.
>
> I had a doubt whether this pushes on to 3D, say on a sphere:
No. See below.
> i.e. is the solid angle subtended by a patch on the sphere's surface,
> half (or is it any other number?) of the solid angle subtended by the
> same patch at the centre of the sphere.
>
> Does it make sense to define solid angle this way?
I do not see any definition in this paragraph.
> Could you give me good and workable references to solid angles?
Solid angle is just a cone. If you want to "measure it as a number":
intersect it with a ball centered at the vertex, and measure the
volume. Or intersect it with the corresponding sphere, and measure
the area. (These two answers are proportional.) Pick up a good unit,
so the answer is more convenient to you.
======
Back to the initial question: proceed infinitesimally. We want a small
arc be visible with angle proportional to its length - with the same
coefficient c at all points of the curve.
In 2D, the fundamental "conservation law" is the inverse distance law:
"intensity" at double the distance is half of the "intensitiy" at a
distance. The "visible angle" of a small object depends on the
distance (as described), and incidence angle: the smaller the
incidence angle, the smaller the visibility angle (the exact
coefficient is sin(angle).
So to have c constant, one needs sin(incidence) to be proportional to
the distance. This is an ODE of the first order on the curve
(depeinding on the coefficient of proportionality); all the solutions
are similar one to another.
At small distance, the incidence angle is small, so it is not a
surprise that the base point is on the solution curve. What IS
surprising is that the relation holds for ANY point of the curve; in
other words, the curve (= a circle) has more symmetries "than expected".
---
Likewise, in 3D, one changes the inverse distance law to inverse
square law. If one assumes the surface to be a surface of revolution,
then one still gets an ODE of the first order on the curve (one to
rotate).
But it is a different ODE, so the solution is not a sphere. Moreover,
the curvature at the base point vanishes (as it is easy to see); hence
the curve is NOT a circle, and has no "extra symmetries". In other
words, the obtained surface is not a sphere. It has the required
property w.r.t. the base point, but with respect to other points on
the surface.
Hope this helps,
Ilya
ISCARE
Hi Ilya,
Thanks for your response.
> > Hi,
>
> > It is a well known fact that the angle subtended by an arc on the
> > circumference of a circle is half that subtended by it at the centre.
>
> > I had a doubt whether this pushes on to 3D, say on a sphere:
>
> No. ?See below.
I am searching for a canonical proof of this statement.
>
> > i.e. is the solid angle subtended by a patch on the sphere's surface,
> > half (or is it any other number?) of the solid angle subtended by the
> > same patch at the centre of the sphere.
>
> > Does it make sense to define solid angle this way?
>
> I do not see any definition in this paragraph.
The imperceptible definition of solid angle is with regard to the
existence of the "corresponding sphere" of which the spherical angle
is a legit cone, with a uniform slant height
>
> > Could you give me good and workable references to solid angles?
>
> Solid angle is just a cone. ?If you want to "measure it as a number":
> intersect it with a ball centered at the vertex, and measure the
> volume. ?Or intersect it with the corresponding sphere, and measure
> the area. ?(These two answers are proportional.) ?Pick up a good unit,
> so the answer is more convenient to you.
>
As I said above, I need to find the solid angle for a family of cones
having 'non-uniform' slant height... as would be for solid angles that
are subtended by a patch on the sphere at an apex point on the surface
of the sphere
(Akin to how it is is the 2D relation)
> ======
>
> Back to the initial question: proceed infinitesimally. ?We want a small
> arc be visible with angle proportional to its length - with the same
> coefficient c at all points of the curve.
>
> In 2D, the fundamental "conservation law" is the inverse distance law:
> "intensity" at double the distance is half of the "intensitiy" at a
> distance. ?The "visible angle" of a small object depends on the
> distance (as described), and incidence angle: the smaller the
> incidence angle, the smaller the visibility angle (the exact
> coefficient is sin(angle).
>
> So to have c constant, one needs sin(incidence) to be proportional to
> the distance. ?This is an ODE of the first order on the curve
> (depeinding on the coefficient of proportionality); all the solutions
> are similar one to another.
>
> At small distance, the incidence angle is small, so it is not a
> surprise that the base point is on the solution curve. ?What IS
> surprising is that the relation holds for ANY point of the curve; in
> other words, the curve (= a circle) has more symmetries "than expected".
>
> ---
>
> Likewise, in 3D, one changes the inverse distance law to inverse
> square law. ?If one assumes the surface to be a surface of revolution,
> then one still gets an ODE of the first order on the curve (one to
> rotate).
>
> But it is a different ODE, so the solution is not a sphere. ?Moreover,
> the curvature at the base point vanishes (as it is easy to see); hence
> the curve is NOT a circle, and has no "extra symmetries". ?In other
> words, the obtained surface is not a sphere. ?It has the required
> property w.r.t. the base point, but with respect to other points on
> the surface.
>
I am unable to dig into these ODE equations. (specifically the
extrapolation from 2D to 3D) Could you help me out with a few concrete
equations?
> Hope this helps,
> Ilya
Just to clarify the question at stake,
in 2d the locus of points such that angle subtended by an arc is
"half" in value with respect to a point O, the centre of said circle
is the circumference of the circle.
In 3D for a sphere, you wish to claim that the locus of points such
that the value of the solid angle subtended at the points is such that
it is half (or a constant value) of that subtended at point O, the
centre of the said sphere is NOT the surface of the sphere. (which
will be evident from claimed broken symmetries in 3D)
Can the claim be fortified using trigonometry?
Could you hint at what kind of surface the locus would be?
Thanks,
Iscare
Ilya Zakharevich
>> Solid angle is just a cone. ?If you want to "measure it as a number":
>> intersect it with a ball centered at the vertex, and measure the
>> volume. ?Or intersect it with the corresponding sphere, and measure
>> the area. ?(These two answers are proportional.) ?Pick up a good unit,
>> so the answer is more convenient to you.
> As I said above, I need to find the solid angle for a family of cones
> having 'non-uniform' slant height...
So apply the definition given above.
> I am unable to dig into these ODE equations. (specifically the
> extrapolation from 2D to 3D) Could you help me out with a few concrete
> equations?
There are only 2 equations, and 2d and 3d differ only by power of sin.
A
_"
_"
_x angle theta
x"
"xx_``````````````_xx"```````````` B
O "xxx______xxx" I
(this assumes fixed-width fonts). You have a curve OA; take any ray
OB intersecting your curve (say, at point I). (On the picture, the
ray is horizontal, since it is easier for my plotting software; in
general, better rotate the picture slightly. You get two rays at I:
the ray IB, and the tangent ray to the arc IA; denote the angle
between them as theta.
So the ODEs are
sin theta = |OI|^(D-1)
with D being 2 and 3 correspondingly.
Hope this helps,
Ilya