Towertalk
[Top] [All Lists]

[TowerTalk] Rohn questions

To: <towertalk@contesting.com>
Subject: [TowerTalk] Rohn questions
From: K7LXC@aol.com (K7LXC)
Date: Thu, 9 Apr 1998 01:07:45 EDT
In a message dated 98-03-26 18:45:00 EST, you write:

> Here are my unanswered questions.  These are related to Rohn 45.

      I just checked my outgoing mailbox and they still looked unanswered -
sorry.
>  
>  Since Rohn's wind loads are substantially higher for a house bracketed 70'
>  tower (at 56' and 28') (32 sq. ft. at 70 mph) than a tower with their two
>  recommended guys at 35' and 65' (9.5 sq. ft. at 70 mph assuming tower has 8
> sq.
>  ft. for sidearms and 6 transmission lines coming up the sides), how much 
> extra
>  wind load, if any, would be gained by using 3 guys at 23' increments?
> 
      Kinda inconsistent, eh? I really don't know why the discrepancy unless
they consider a building to be stronger than sets of guys - hi. 

    If you add in the 8.0 sq.ft. for a commercial antenna mount, that makes a
total of 23.9 sq.ft. Still a bunch less than the housebracketed tower. 

     I'm not an engineer so I can't make any comments on adding a set of guys.
I don't think it would change anything because the limiting tower capacity
factor is how much compression the legs can take. If you doubled the number of
guys, that leg factor would still be the same. 
 
>  Concerning subtracting 8 sq. ft. for the side arms, is that 8 sq. ft. a 
> round
>  member spec or a flat member spec?  (If it is the former that would be
>  equivalent to 4.8 sq. ft. of antenna wind load.)

     I've always assumed it's round. 
>  
>  The Rohn notes say that the tower design includes a 1/2" and 7/8" 
> transmission
>  line on each of the three faces.  Ours will only have two (probably .5
inch) 
> so
>  it appears that I can also subtract the surface area of the missing 4 lines
(
> 1
>  1/2" line and 3 7/8" lines), i.e. ( 70' * (1/2 + 3 * 7/8) / 12 ) * .6 ) =
10.
> 9
>  sq.ft.   But that assumes that wind load of the coax along the side is the 
> same
>  as the wind load for an antenna at the top, which probably isn't
equivalent.
>  How much is the right amount, if any, to subtract for the unused coax?
>  
           Your cable calc looks reasonable. Now the tower capacities are
within 12% or so of each other. Don't forget you've still got plenty of
engineering overhead built-in already.

Cheers,  Steve   K7LXC

--
FAQ on WWW:               http://www.contesting.com/towertalkfaq.html
Submissions:              towertalk@contesting.com
Administrative requests:  towertalk-REQUEST@contesting.com
Problems:                 owner-towertalk@contesting.com
Search:                   http://www.contesting.com/km9p/search

<Prev in Thread] Current Thread [Next in Thread>