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Total 24 documents matching your query.

1. [TowerTalk] W6NL 40M Moxon (score: 1): Author: "Stan Stockton" <stan@aqity.org>; Date: Mon, 20 Jul 2009 10:50:32 -0500; Has anyone out there used EZNEC to model the Cushcraft 2 element 40 with the W6NL Moxon design? I would appreciate a copy to compare to what I have here. Either I have done something wrong, EZNEC doe; /archives//html/Towertalk/2009-07/msg00390.html (6,834 bytes)
2. [TowerTalk] W6NL 40m Moxon (score: 1): Author: "Joern DK5TT" <dk5tt@opretzka.de>; Date: Fri, 30 Oct 2009 10:20:03 +0100; Hi, I'd like to build the 40m Moxon 2 Element by W6NL. So far I gathered some pdf documents and drawings, but there are still open questions. It seems, there are basically 2 different version, one wi; /archives//html/Towertalk/2009-10/msg00462.html (7,140 bytes)
3. [TowerTalk] W6NL 40m Moxon (score: 1): Author: Kevin Stover <kevin.stover@mediacombb.net>; Date: Fri, 8 Apr 2016 21:18:11 -0500; I need to figure out the wind area of the W6NL 40 Moxon antenna. The antenna starts like as a Cushcraft XM-240 which they say is 5.5 sqft. The Moxon is going to be more but how much? I'm guesstimatin; /archives//html/Towertalk/2016-04/msg00109.html (6,762 bytes)
4. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: Ken K6MR <k6mr@outlook.com>; Date: Sat, 9 Apr 2016 21:51:22 -0700; Ive wondered this myself but never bothered to model it. Here are some rough figures from Yagi Stress: Reflector element without the tee: 2.88 sq ft. Driven element without the tee: 2.76 sq. ft. Refl; /archives//html/Towertalk/2016-04/msg00120.html (9,147 bytes)
5. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: "Joe Subich, W4TV" <lists@subich.com>; Date: Sun, 10 Apr 2016 07:55:37 -0400; I suppose with some trig you could figure the total projected area at various wind angles. One going down and one going up. Would some angle that gives equal projected area from each figure be the ma; /archives//html/Towertalk/2016-04/msg00124.html (10,900 bytes)
6. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: Kevin Stover <kevin.stover@mediacombb.net>; Date: Sun, 10 Apr 2016 09:34:46 -0500; Thanks for the input Ken and Joe. I used 9ft in the ARRL mast calculator and end up with 131 mph survival wind speed on a 3" x .25 wall 87,000 ksi mast. The antennas are an Optibeam 11-5 at 70' and t; /archives//html/Towertalk/2016-04/msg00125.html (12,558 bytes)
7. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: Ken K6MR <k6mr@outlook.com>; Date: Sun, 10 Apr 2016 09:05:56 -0700; Thanks Joe. I was thinking about wind vectors but didnt work it through. With these numbers it should be simple to add a torque compensator to make up for the higher torque generated by the reflector; /archives//html/Towertalk/2016-04/msg00126.html (14,402 bytes)
8. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: "Joe Subich, W4TV" <lists@subich.com>; Date: Sun, 10 Apr 2016 13:37:18 -0400; I used 45 degrees here because the element and boom components were nearly the same. If one had long boom 6 or 10 meter antennas it would probably be necessary to solve for the angle of maximum force; /archives//html/Towertalk/2016-04/msg00127.html (14,327 bytes)
9. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: Ken K6MR <k6mr@outlook.com>; Date: Sun, 10 Apr 2016 11:55:16 -0700; Ok, this even makes mathematical sense (as it should): Assume the wind angle is with respect to the antenna direction. Total area = Element area * COS (angle) + Boom area * SIN(angle) First derivativ; /archives//html/Towertalk/2016-04/msg00128.html (8,739 bytes)
10. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: "Hardy Landskov" <n7rt@cox.net>; Date: Sun, 10 Apr 2016 18:09:23 -0400; Hi All, My 2 cents. I have seen the wind suddenly die--then change direction--then change again. It's like a box of chocolates.... N7RT/4 You don't need to be concerned with equal areas. The force (l; /archives//html/Towertalk/2016-04/msg00129.html (12,902 bytes)
11. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: Dan Maguire via TowerTalk <towertalk@contesting.com>; Date: Sun, 10 Apr 2016 23:26:22 +0000 (UTC); I'd like to add a "me too" to that thank you. Nice mental exercise. I whipped up a little Excel workbook which uses the Solver to find the wind angle that produces the maximum effective area. Here's; /archives//html/Towertalk/2016-04/msg00130.html (8,487 bytes)
12. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: Ken K6MR <k6mr@outlook.com>; Date: Sun, 10 Apr 2016 17:11:12 -0700; All the more reason to be sure that the antenna is torque balanced. Those shifting winds can beat a rotator to death (not a prop motor, but all regular ones). I can loosen the bolts on my reasonably; /archives//html/Towertalk/2016-04/msg00131.html (14,622 bytes)
13. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: Ken K6MR <k6mr@outlook.com>; Date: Sun, 10 Apr 2016 17:14:45 -0700; You are without a doubt the Excel Master :^). I couldnt live (er, at least do any antenna sims) without AutoEZ. It is amazing. Ken K6MR I'd like to add a "me too" to that thank you. Nice mental exerc; /archives//html/Towertalk/2016-04/msg00132.html (9,085 bytes)
14. [TowerTalk] W6NL 40m Moxon (score: 1): Author: "Jim Thomson" <jim.thom@telus.net>; Date: Tue, 12 Apr 2016 21:10:38 -0700; You don't need to be concerned with equal areas. The force (load) on the element will be broadside to the element and will vary with the angle of the wind to the element. The force (load) on the boom; /archives//html/Towertalk/2016-04/msg00166.html (9,827 bytes)
15. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: Ken K6MR <k6mr@outlook.com>; Date: Tue, 12 Apr 2016 21:53:02 -0700; OK Jim, I need that explained. If the wind is at 45 degrees to the boom, how is there not a contribution of force by both the boom and the elements? Ken K6MR You don't need to be concerned with equal; /archives//html/Towertalk/2016-04/msg00168.html (10,412 bytes)
16. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: "Jim Thomson" <jim.thom@telus.net>; Date: Wed, 13 Apr 2016 04:54:56 -0700; SW....heading to the NE, all the forces off all the els will be due east. All the forces off the boom will be to the due north. If you have an equal amount of boom on either side of the mast, and ea; /archives//html/Towertalk/2016-04/msg00169.html (11,537 bytes)
17. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: gboutin@infinichron.com; Date: Wed, 13 Apr 2016 09:16:09 -0300; Ken, This isn't exactly new. I think you'll find what you're after by reviewing K7NV's summary post from back in 1998. http://lists.contesting.com/archives//html/Towertalk/1998-08/msg00499.html <http; /archives//html/Towertalk/2016-04/msg00171.html (10,881 bytes)
18. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: "Joe Subich, W4TV" <lists@subich.com>; Date: Wed, 13 Apr 2016 08:17:04 -0400; SW....heading to the NE, all the forces off all the els will be due With quartering winds (winds not perpendicular to either the boom or mans), you have forces on the mast/tower to the north due to t; /archives//html/Towertalk/2016-04/msg00173.html (13,595 bytes)
19. Re: [TowerTalk] W6NL 40m Moxon (score: 1): Author: Stan Stockton <wa5rtg@gmail.com>; Date: Wed, 13 Apr 2016 08:55:43 -0500; If calculating wind load for the purpose of deciding which rotator to use, I would think the boom length would factor in. Surely a square plate 4'x4' mounted to the mast would put less stress on a ro; /archives//html/Towertalk/2016-04/msg00175.html (15,961 bytes)
20. [TowerTalk] W6NL 40m Moxon (score: 1): Author: "Jim Thomson" <jim.thom@telus.net>; Date: Wed, 13 Apr 2016 08:05:57 -0700; With quartering winds (winds not perpendicular to either the boom or mans), you have forces on the mast/tower to the north due to the boom and forces on the mast/tower to the east due to the elements; /archives//html/Towertalk/2016-04/msg00177.html (11,235 bytes)

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