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[TowerTalk] anchor points

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Subject: [TowerTalk] anchor points
From: kurscj@oampc12.csg.mot.com (Chad Kurszewski WE9V)
Date: Mon, 30 Jun 1997 17:07:42 -0500
>>Ah ... a cubic yard is 3 x 3 x 3 or 27 cubic feet.
>>27*.67= 18.09 cubic feet required (using your figure)
>>18.09/3.14=5.76 feet depth required for 1 foot diameter;
>>18.09/6.28=2.88 feet depth required for 2 foot diameter.
>
>Hmm. My back of the envelope computation says something isn't right. If 
>you double the diameter, that quadruples the volume per foot of height. 
>If 5.76 feet is right for a 1 foot hole, then a two foot hole should 
>require 1/4 of that, or roughly 1.5 feet, less about an inch or so.
>
>The volume should be 2*pi*h*r**2.


Well, maybe it should be, but it isn't.

Let's try pi*r^2*h   (no 2* in there)
Area of a circle is pi*r^2, times the height is volume.



>>Not sure whether I'd try to second guess the engineering.
>
>Certainly not with math like that. <grin>

Nor with that math.  :)


p.s.  a 1 foot diameter hole would need to be 23 feet deep to equal 18 ft^3
      a 2 foot diameter hole would need to be 5.76' deep to equal 18 ft^3

---
Chad Kurszewski, WE9V         e-mail:  Chad_Kurszewski@csg.mot.com
The Official "Sultans of Shwing" Web Site:  http://www.QTH.com/sos



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