# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Zenith Pipe
From: Brooke Clarke
Date: 2004 Dec 20, 09:09 -0800

```Hi Fred:

For a rough approximation there are two main factors.

First is the time determined by the size of the sun.  If the pipe were
very long and very small in diameter then the sun would be visible for
the time it takes the sun to move it's own diameter which is about 2
minutes and 6 seconds.  This can be calculated based on the apparent
diameter of the sun and that it takes 24 hours to make a full circle.

Second is the time determined by the geometry of the pipe.  When the
leading edge of the sun can just see the far wall of the pipe the angle
is 1/1000 radians from the centerline of the pipe.  When the trailing
edge of the sun can just see the near wall of the pipe the angle is
So the included angle is 2/1000 radians.  Multiply that by the suns rate
of 24 hours per 2*Pi radians and you get 24/(1000*Pi hours). The time
determined by the pipe geometry is 27.5 seconds.

The total time the sun shines through the pipe is 2 minutes and 33.5
seconds.

Have Fun,

Brooke Clarke, N6GCE

--
w/Java http://www.PRC68.com
w/o Java http://www.pacificsites.com/~brooke/PRC68COM.shtml
http://www.precisionclock.com

Fried Squash wrote:

>My last question here was answered with such acumen, I
>shall try another...
>
>I am trying to flesh out an analogy to better explain
>the rapidly moving GP of the sun, but have no idea of
>how to do the math to portray it accurately.
>
>Let's say I suspend a pipe, 1cm inside diameter and
>10m long, over the Earth at a point within the
>ecliptic.  At the moment when the Sun's GP coincides
>with my "zenith-pipe", for how long a time will I see
>the light of the Sun making it all the way through the
>pipe?
>
>Of course, I guess I could just go to Home Depot and
>
>Thanks for any takers!
>
>-Jeff
>
>
>
>

```
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