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VK6BE > TECH 25.09.05 21:37l 24 Lines 1071 Bytes #999 (0) @ WW
BID : B70389VK6BE
Read: GUEST DL1LCA OE7FMI
Subj: Re: calculations
Path: DB0FHN<DB0FOR<DB0SIF<DB0EA<DB0RES<ON0AR<VK6HGR<VK6BBR<VK6ZRT<VK6TJ<
VK6JY
Sent: 050924/1435Z @:VK6JY.#ALY.#WA.AUS.OC #:2690 [Albany] wFBB7 $:B70389VK6BE
From: VK6BE@VK6JY.#ALY.#WA.AUS.OC
To : TECH@WW
Me too. But I am more than 57. The rule remains the same. I calculated the
height of a tower here when I was asked about it by measuring the base
(measured from the base of the tower to a spot a convenient distance from
the base) of the triangle and the angle to the top. Easy using Pythagoras.
Bob VK6BE.
BTW it was 180 feet!
> Does the name Pythagorus ring a bell? "in any rightangled triangle the
> square on the hypotenuse equals the sum of the square on the other 2
> sides." Call the Hypotenuse "a", and the other 2 sides "b" & "c".
> Therefore, a2 = b2 + c2 (2 means squared) Any reasonable calculator will
> give the answer in minutes. So, If you know the height of the support, and
> the length from support to anchor point, then square both of these
> measurements, add them together, then take the square root of the answer.
> That will be the length if the aerial wire or the "hypotenuse". Simple
> huh? Im 57 and stuoill remember this fron the 3rd form High chool days 40+
> yrs ago.
>
> Allen ZL1AJG
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