All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes
The Shadow in a Sphere (Posted on 2009-08-19) Difficulty: 3 of 5
Consider a hollow sphere of radius R, in which a light source is placed at its centre. A square plate of side length S is held in place within the sphere by a pole of length L units. The square plate's position is then such that the displacement between the centre of the square and the light source is R-L units.

The square plate is also oriented in a way such that an imaginary line drawn perpendicular to the surface of the plate and passing through the plate's centre will pass through the light source.

Determine the surface area of the shadow formed on the spherical shell, due to the square plate.

No Solution Yet Submitted by Chris, PhD    
Rating: 4.0000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Long way around solution, but it works... Comment 7 of 7 |
The solution I came up with is:
Calculating the
dihedral angle of planes that make the lateral faces of the pyramid with the base representing the square plate and the apex being the center of the sphere we find the angles of the shadow on the sphere.
H=R-L -height of pyramid
D=
(S*√2)/2 -semi-diagonal of the square plate.
HL= √(S²/4+(L-R)²) -lateral height of the pyramid
E= √(S²/2+(L-R)²) -lateral edge of the pyramid
AL= (S*HL)/2 -area of one lateral face of the pyramid
HS2= S*HL/E -distance from one corner of the base to the lateral edge of the pyramid(this point being the place where we calculate the dihedral angle, as the line from this point (we name it X) and the corner of the base is perpendicular to the edge of the pyramid which represents the intersections of the planes that are limiting the shadow on the sphere.
H2- line from X to the center of the square plate
β= arcsin(D/HS2) -angle formed by H2 and HS2 which is half of the dihedral angle we are looking for.

             ( √2     √[S²/2+(L-R)²]   )
β=arcsin( ---- * -------------------- )
             (  2      √[S²/4+(L-R)²]   )

Now, using β we can calculate the area of a spherical triangle that is half the area of the shadow, using formula:

Area = R²[(A+B+C)-π] -A, B, C being angles of the spherical triangle, in our case A=B=β and C=2β

<img src="http://img183.imageshack.us/img183/3193/formulaw.jpg" alt="Formula">
<img src="http://img183.imageshack.us/img183/1671/drawingm.th.png" border="0">

  Posted by TheKPAXian on 2009-10-01 16:53:38
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information