| |
ZL1AJG > TECH 24.09.05 00:31l 28 Lines 1148 Bytes #999 (0) @ WW
BID : 2C1308ZL1AJG
Read: GUEST DL1LCA OE7FMI
Subj: Re: calculations
Path: DB0FHN<DB0THA<DB0HDF<DB0ERF<DB0FBB<DB0FHK<DB0ACC<DB0GOS<ON0AR<ON0AR<
7M3TJZ<F6CDD<N2BQF<VK6HGR<ZL2BAU<ZL1UX
Sent: 050923/2057Z @:ZL1UX.#20.NZL.OC #:47365 [Hamilton] FBB7.00i $:2C1308ZL1AJ
From: ZL1AJG@ZL1UX.#20.NZL.OC
To : TECH@WW
>>>> LinPac 0.17pre3 : Message sent by VK5KMG : Date 16/09/05 Time 14:33 <<<
> Hi there, I want to work out the length of the hypotenuse of the triangle
> This is in regard to an aerial support system,and its been a while since the
> brain has had to think about these things.
> So if anyone has the data-formulae,handy and can forward it to the address
below
> it will be appreciated.
> Have a good day.
>
>
> 73 de VK5KMG ! Geof
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
Read previous mail | Read next mail
| |
