>>At the posited high wind force, the 5/16" guy
>>will put, say, 10,000 pounds of tension on the equalizer plate, which will
>>rotate to "equalize". Since the holes in the plate are equally spaced about
>>the pivot (anchor) hole, equal force will be exerted on the 3/16" guy, far
>>exceeding its rated capacity.
>
>It would seem to me that the rotational force applied to the smaller guy
>would be far less than the direct force applied to the anchor.
Yes, the rotational force applied to the smaller guy WILL be far less.
The way I visualize it to understand this is to imagine another guy
equalizer plate but with only two guy wires instead of three. If the
hole that goes to the guy anchor was directly in the middle of the two
guywire holes (all three holes in a straight line), then the force
applied to the lower guy would be equal to the top. But since the hole
to the anchor is "set back" from the guywire holes, the forces will be
considerably different. Looking at it another way (the opposite
extreme), if the hole to the guy anchor was set back from the guywire
holes a considerably further distance than normal, say two feet, then
changes in one guywire tension would have VERY little effect on the
other guywires. I don't have the math skills to back this up, but it
seems easy to me to visualize this difference by imagining these
scenarios.
73,
Dave WD5N ... VP5EA ... <harperd@tx.slr.com>
>
>
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