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## Re: [TowerTalk] Calculating forces in guy wires - How

 To: va3pl@cuic.ca, Re: [TowerTalk] Calculating forces in guy wires - How Jim Lux Tue, 25 Nov 2003 12:06:04 -0800
 At 02:31 PM 11/25/2003 -0500, va3pl@cuic.ca wrote: ```How do you calculate forces in guy wires? Here is scenario:``` ```Tower 80 feet high Two Antennas spaced 10 feet totaling 25 sq feet (bottom 15 top 10 sq feet) on 22 feet, 2.5" dia. mast Rotor is Orion 6 feet below top plate Wind 80 M/h``` ```What are the forces in guy wires then? How strong the guy wires I need?``` 73 de Andy - VA3PL You can kind of ball park this, but, really, you need to consult a more definitive reference. You're also missing an essential piece of information, which is where the guys are attached and how far out they are from the base. I'll run through the general calculation, but bear in mind that you shouldn't use these numbers, but use them as an explanation of how the mfr or an engineer will go about calculating. ```25 square feet, 80 mi/hr can be turned into force as: F = Area * mph^2/391 = 25 * 80*80/391 = about 410 pounds.``` Bear in mind, though, that the tower itself will have a significant wind load. Ballparking the tower as a 80 foot high, 1 foot wide rectangle.. 1300 pounds windload. For ballpark, assume that the load is distributed evenly between the top and bottom (so, you'd add another 650 pounds to the 410 pound load at the top..) And, is that 80 mi/hr at the top, or at ground level? Now, say the guy is anchored 40 feet out, and happens to be perfectly in line with the direction of the wind... You can us trigonometry to figure out what the load is in the guy wire (assuming the tower is perfectly rigid and pivots at the base... not necessarily a valid assumption!) The side load is, say, 1100 pounds at the top. since the guy is anchored out at half the height of the tower, the down force on the tower will be twice the side load: 2200 pounds. The tension in the guy will be sqrt(1100^2+2200^2) or 2500 pounds... If this were the whole story (and it isn't.. it's just for an example... get someone to do the real calculations for your system), you'd want a guy that could handle, say, 5000 pounds (for a 2:1 margin) if not a lot more. There are a huge number of traps for the unwary here... 1) The guy probably doesn't attach at the antenna height, so you've got a canteliever sticking up above the guy attachment point, which changes how the forces are arranged 2) Most folks use 3 guys in a "Y", and the forces need to be divided among the guys. 3) Most folks put enough tension in the guys so that they don't go "slack" when the tower gets loaded. That tension has to be added in. 4) The guys are probably not perfectly symmetrical, nor precisely the same distance out, all of which can increase the loads (or decrease them... the point is, you don't know for sure). 5) small changes in the distance of the guy from the base can have huge effects on the required tension (it's that horrible 1/tangent(theta) thing...) Bottom line here: Consult someone who KNOWS (or can calculate) the answers! The mfr of the tower may have cookbook numbers for standard installations, with suitable margins and allowances for installation variations and materials. Jim, W6RMK _______________________________________________ See: http://www.mscomputer.com for "Self Supporting Towers", "Wireless Weather Stations", and lot's more. Call Toll Free, 1-800-333-9041 with any questions and ask for Sherman, W2FLA. ```_______________________________________________ TowerTalk mailing list TowerTalk@contesting.com http://lists.contesting.com/mailman/listinfo/towertalk```
 Current Thread [TowerTalk] Really elementary question, Pete Smith Re: [TowerTalk] Really elementary question, Jim Lux [TowerTalk] Calculating forces in guy wires - How, va3pl Re: [TowerTalk] Calculating forces in guy wires - How, Jim Lux <= Re: [TowerTalk] Calculating forces in guy wires - How, Jim Lux Re: [TowerTalk] Really elementary question, Pete Smith