[TowerTalk] Rotator Questions

Rob VE6TR at telus.net
Mon Jun 12 21:32:05 EDT 2006


Perhaps I am wrong, but the whole idea of wind load on a rotator seems to be a misnomer.  The only time that spec 
would be valid is for side load, not torque, and side load only counts if the rotator is outside of the tower.  Wind load 
will entirely depend on the antenna design and how uneven the wind area is on either side of the mast which will 
produce the torque.  If the antenna had 10 ft^2 on either side of the mast then the torque is 0 yet the wind rating 
would be 20 ft^2.  I know I am simplifying things but torque ratings should be just that... measured in inch pounds or 
the like.  If you want to compare rotators, then look for torque values, not wind load.

My two cents....  73


On Mon, 12 Jun 2006 18:11:14 -0700, Al Williams wrote:

>Not only is it that the M2 rating of 35 sq feet is meaningless, also the 
>rotational torque inertia may even be more important!
>See my previous posting on 4/28/06 at 9:08 am.  There have been no responses 
>to that previous concern.
>----- Original Message ----- 
>From: "Mike Harris" <mike.harris at horizon.co.fk>
>To: "Al Williams" <alwilliams at olywa.net>; "Tower Talk" 
><towertalk at contesting.com>
>Sent: Monday, June 12, 2006 5:27 PM
>Subject: Re: [TowerTalk] Rotator Questions
>> G'day,
>> | The M2 2800 rating of 35 sq feet is comparable to my own rating of 400
>> sq
>> | feet (on an absolute calm day!).  Without a specified wind speed the
>> rating
>> | is meaningless?
>> An interesting point.  Maybe there is someone out there who can offer
>> clarification on this.
>> I found the windloading of the Optibeam OB9-5 somewhat confusing:
>> Windload at 130km/h    578N / 0.72m sq / 7.8 ft sq
>> I received the following from Thomas at OB:
>> <quote>
>> on our web site you should find all the data which are usual, i.e. m2,
>> feet2 and Newton.
>> And we always calculate it at 130 km/h.
>> So the windload on the 9-5 is 578 Newton = 0.72square meters  = 7.8 square
>> feet.
>> I am not doing the calculation, a friend of mine has written a program
>> which is doing the work.
>> The program checks what is the bigger wind resistance, the boom or the
>> elements when pointing into the wind with the maximum surface.
>> <unquote>
>> From this I assume that the antenna at some angle to the wind has a
>> maximum surface area of 7.8 ft sq and at 130km/h offers a windload of 578
>> Newton.
>> Maybe someone can comment upon this interpretation as well.
>> My heavy duty 12 metre crank up/fold over tower has an unguyed and
>> extended "area of headload" rating for "tubular aerials" at 130km/h
>> (80mph) of 24.5 ft sq.  At 100mph 14.4ft sq and 120mph 8.7ft sq.  80mph
>> winds would be very unusual here and higher speeds even more so, an OB9-5
>> on a PST61D on my tower should be pretty solid, even taking into account
>> the wind area of the rotator and mast.
>> We had a blow a couple of weeks ago that was 55kt sustained peaking 75kt
>> that caused a little damage around town.  My 6ft sq log and rotary 20m
>> dipole above didn't complain.  I couldn't telescope the tower if I wanted
>> to 'cos the wind pressure stopped the top section from lowering under
>> gravity.  A pull down rope would seem useful in extreme circumstances.
>> Regards,
>> Mike VP8NO
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