Is there a maximal finite depth infinite index irreducible subfactor?

Shadow depth calculation for Dec 21 3PM for any latitude/long on a tilted plane PART 3:?

• Some time ago i posted a question asking on how to calculate the depth of the shadow and it was answered in detailed by Dr. Bob. http://answers.yahoo.com/question/index;_ylt=AtVgK9IhDgqPuwN5u2WfKBbsy6IX;_ylv=3?qid=20111009103722AA8OEPZ Now, my question is how do i calculate the depth of the shadow for Dec 21 for any latitude/longitude on a tilted plane. For instance if i have an object sitting on a plane that is tilted 5 degrees to the south or north or east or west or anything in between, the depth of the shadow would change. thanks.

• Answer:

  You can still use Dr Bob's solution, but first you have to imagine moving your tilted plane to a location where it is horizontal. Let's say you have a plane tilted towards the direction 19 degrees east of north. And let's say the "tilt" is 5 degrees. What you need to do, to bring the plane to a horizontal position, is to start the motion in a direction 19 degrees east of north, but continue along a great circle, for exactly 5 degrees of arc. To calculate where you will end up, consider a spherical triangle with the North Pole at one vertex, the starting location of your tilted plane at another vertex (call it S), and the place where your plane will be horizontal at the third vertex (call it H). You know the starting latitude, that's 90 minus the arc length NS. You know the arc length of side SH needs to be 5 degrees. You know that angle NSH is 19 degrees. That's three "parts" of the spherical triangle. The law of cosines will give you the latitude of H: cos (NH) = cos (NS) cos (SH) + sin(NS) sin(SH) cos (angle NSH) All the quantitites on the right are known. When you've found arc NH, subtract from 90 degrees to get the latitude of H. To find the longitude of H, you could use the law of cosines again: cos (SH) = cos(NH) cos(NS) + sin (NH) sin(NS) cos (angle SNH) Here everything is known except angle SNH, solve for that. When you get it, it's the longitude difference between S and H (should go eastward from S, in the case we've described). Knowing the latitude and longitude of H, find the shadow length on a horizontal plane at H, using Dr Bob's earlier methods. cos (SH) =