[OKDXA] MATH QUESTION

[AD5PE]

a^2 + b^2 = c^2, with a=120 and b=44, c = 127.812' or 127' 9" plus a bit.

for the second part create a new triangle, similar to the first - the
included angle (at the tiedown) is the same, therefore both the hypotenuse
and the height (your coax length) are shortened by the same proportion =
57.8/127.8= about .45, so the feedpoint is just past 1/2 way down the slope,
and then length is 45% of the overall height (.45*120=54').

You need 54' of coax for the vertical part plus whatever it takes to get
into the shack, etc.

73,
Jay
AD5PE  

-----Original Message-----
From: okdxa-bounces at mailman.qth.net [mailto:okdxa-bounces at mailman.qth.net]
On Behalf Of K8fu at aol.com
Sent: Wednesday, October 03, 2007 11:44
To: okdxa at mailman.qth.net
Subject: [OKDXA] MATH QUESTION

Guys I need help with a computation..............................
 
I'm trying to figure out the distance between 2 points in a
triangle..........................
 
OK got ur thinkin hats on and calculus calculators at the ready
??.............
 
OK here goes.............................
 
If the top of the tower leg is 120' (A) and the ground level of this leg is
point (B) and the distance from the base of the tower leg to where the
bottom of  the sloper will be tied is 44' (C).....44' is then the distance
between points B  & C................
 
If the sloper is center fed (D) @70' down from points (A) to (C),How much
coax will be needed to get from point (D) to the tower leg which is  the
vertical line from points (A) to (B) I hope this makes sense as I'm not
sure I can express it better............................
 
RIBBIT....................................RIBBIT............................
..
.   BOHICA



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