Tom, W8JI wrote:
>You have the triangle upside down.
To my example:
> Example:
> A 100lbs. of horizontal force restricted by a 45 degree guy arrangement
> would result in horiz. & vert. force components of 70.7lbs each. The same
> loading with a 22.5 degree guy arrangement would result in a horizontal
> component of 38.3lbs and a vertical component of 92.4lbs. The vert. comp
> keeps going up as you approach the tower, but you'll never exceed the 100
> lbs. of force put in the horizontal axis, agreed?? And, at this juncture,
> I kinda forgot what I was arguing about! Hi! So I'll drop it at this
> point! 7 3 Tom K3GM
Tom, W8JI elaborated:
>No way! If that was true, we'd all be using self-support towers with
>tiny base piers for TV broadcast towers!! Our uniform cross section
>towers would stand up hundreds of feet without any strain, because
>the vertical load would never exceed 100 pounds and the horizontal
>load 100 pounds.
>The vertical load in the tower due to side force = side force / tan of
>angle
>The load in the guy line due to side force = side force / sin of the
>guy angle
>Watch how this makes sense......
>With a 90 degree angle the vertical tower load is 100/ infinity or
>zero pounds.
>The guy line load is 100/ 1 or 100 lbs. Tower load is zero.
>With a 45 degree angle, the vertical tower load due to the
>horizontal force is 100/ 1= 100 lbs
>The load in the guy line is 100/ .707 = 141 lbs
>With a 22.5 degree angle the tower load is 100/.4142 = 241.4 lbs
>The guyline load is 100/.3827 = 261.3 lbs
>With a 5 degree angle the tower load is 100/.0875 = 1143 lbs
>The guyline load is 100/.087= 1147 lbs
snip...............
Tom et. al.,
Well....I really hate to say this, but Tom, you're absolutely right. I had
the triangle upside down! Your solution and example is right on, Tom. I
can't believe that got by me. I spent a good part of the day in work trying
to figure how I got turned around. Anyway, excuse me while I get this egg
off my face. After that, I have a full plate of crow to eat!
7 3, Tom K3GM
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