It's a right triangle. Straight trig, but I can never remember which
function. It's the side opposite over the hypotenuse or sine is used.
So, it's the sin of the angle the guy forms with the anchor multiplied
by the guy tension, multiplied by the height of the guy to get the
horizontal force applied by that anchor at ground level..
Assuming I didn't make any mistakes (which requires a leap of faith):
If the guy forms a 45 degree angle with the anchor which is 7 feet tall
and the guy tension is 400#
it becomes sin(45) * 400 * 7 IOW The 7 ft tall post multiplies the
horizontal force at ground level by 7 times.
sin(45) = 0.707 Multiplied by the 400 # of tension = 282.84#, but the 7'
post = 282.84*7 = 1979.9#. That's enough to require quite a chunk of
concrete, or a good back guy which if in line with the guy would only
need to hold 282.8#. However it sounds like he wants an elevated guy
anchor which has to resist an overturning moment of nearly one ton.
Much depends on the soil, but I have a similar arrangement with no back
guys I think I calculated the NW guy anchor weighs close to 17,000#. In
about 13 years the top of that anchor post has now moved a good foot out
of plumb. That anchor sets in a mixture, or layered soil composed of
peat and clay. The other two anchors are in clay and I do not measure
any movement in them.
73
Roger (K8RI)
On 7/16/2015 11:34 AM, Lee George AK4QA wrote:
Does anyone have a the formula for guy posts?
I have a friend that wants tall guy posts (7 feet) for a 100 foot tower so he
can walk under them. I need to show him the stress that is involved in that as
opposed to 2 feet out of the ground.
I've always used the wooden pole rule of thumb; for every foot up you need 3
feet down.
Also, if you have the calculation for the back guy (i.e. earth screws) well my
friends, that would be gravy on my biscuit!
All the best and
73,LeeAK4QA
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73
Roger (K8RI)
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