3
$\begingroup$

I have a sphere with radius $R$ and $O$ is the origin.enter image description here

Inside sphere there are 3 circles. The small circle in black colour is fixed with it's position defined by and $\alpha$ angle and among other two circles one is great circle and another small circle can rotate by maintaining same distance and they are always parallel to each other. Both blue circles intersect the black great circle at point A and B. The rectilinear distance of A to B can be calculated using this relation: $AB = 2RSin {\theta\over2}$

Orientation of upper blue great circle is defined as $\beta$ which is the angle starting from $z$ axis.The range of $\beta $ can vary from $0^0 -360^0$ and it is a known value. Though my both blue coloured circle is parallel to each other, is it possible to get the lower circle's orientation angle from the centre of sphere that respect $\theta$ as it is the rectilinear distance? We always assume that both blue circle intersect the black circle.

$\endgroup$
  • 1
    $\begingroup$ These are not all great circles. A great circle is the intersection of the sphere with a plane through the sphere's center. Consequently, all great circles have the same diameter -- the diameter of the sphere. Two distinct great circles necessarily intersect in two points. $\endgroup$ – Eric Towers Jul 24 at 22:36
  • $\begingroup$ Is there any other name for those? I can edit. $\endgroup$ – T. an Jul 24 at 22:40
  • $\begingroup$ A circle of a sphere is a circle lying on a sphere. The circles of a sphere come in two types: great circles, described previously, and small circles, which lie on any other plane and therefore have smaller radii than that of the sphere. $\endgroup$ – Eric Towers Jul 24 at 23:18
  • $\begingroup$ @EricTowers Thanks you sir. I am editing according to that. $\endgroup$ – T. an Jul 24 at 23:22
  • $\begingroup$ Shouldn't be $AB=2R \sin \theta/2$ ? $\endgroup$ – Jean Marie Jul 25 at 3:05
1
$\begingroup$

The black circle is the intersection of a plane $P\cdot (0,0,1)=a$ (I don't think you've specified $a$) and the sphere $P\cdot P =1$.

The blue great circle is the intersection of $P\cdot (\cos \beta, 0, \sin \beta) = 0$ with the sphere, and the blue small circle is the intersection of $P\cdot (\cos \beta, 0, \sin \beta) = b$ with the sphere.

The line of intersection between $P\cdot (0,0,1)=a$ and $P\cdot (\cos \beta, 0, \sin \beta) = b$ is fairly obviously $$P(t) = (\frac{b - a\sin\beta}{\cos\beta}, t, a)$$. Intersection with the sphere gives $$t^2 = 1 - a^2 - \left( \frac{b - a\sin\beta}{\cos\beta} \right)^2$$

The angle between two points on the unit sphere is the arccos of their dot product, so the equation to solve is $$\frac{0 - a\sin\beta}{\cos\beta} \frac{b - a\sin\beta}{\cos\beta} + \sqrt{ \left( 1 - a^2 - \left( \frac{0 - a\sin\beta}{\cos\beta} \right)^2 \right) \left( 1 - a^2 - \left( \frac{b - a\sin\beta}{\cos\beta} \right)^2 \right) } + a^2 = \cos \theta$$ which can be simplified to $$\frac{- a\sin\beta(b - a\sin\beta) + \sqrt{(\cos^2\beta - a^2)(\cos^2\beta + 2ab\sin\beta - a^2 -b^2)}}{\cos^2 \beta} + a^2 = \cos \theta$$ which can be further rearranged into a quartic in $\sin\beta$ (WARNING: I may have made some errors while rearranging): $$(1 - \cos^2\theta)\sin^4\beta - 2ab(1 - \cos \theta)(\sin^3\beta - \sin\beta) + (\cos^2\theta -k - 2) \sin^2\beta + ( k + 1) = 0$$ where $k = a^2 b^2 - 2a^2 - b^2 + 2a^2 \cos \theta - \cos^2\theta$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.