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*Subject*: Re: [amsat-bb] wind load calculations?*From*: "Murray Peterson VK2KGM" <vk2kgm@xxxxxxxxxxx>*Date*: Tue, 15 May 2001 09:37:42 +1000

Hi Mark, Wind load is calculated by multiplying the stagnation presure by the coefficent of drag (and lift if the shape is likely to produce lift) and by the crossectional aera presented to the wind. The stagnation pressure is the presure due to the unit kinetic energy of the fluid flow. (A fluid is something that flows; it could be gas or liquid) I don't know what unit system you are used to but the calculations are much more simple in the metric system so I will give you a metric example (you can translate it by multiplying by the appropriate conversion factors - otherwise ask me). For Sydney, NSW, Australia (it is probably typical of a lot of places) the design wind speed (wind speed expected to occurr once in every 100 years) is 47 metres per second (170 km/h or 105 MPH). The stagnation pressure is half the density of the fluid (air) times the square of the velocity. ie 0.5 x 1.2 kg/m3 x 47m/s x 47m/s = 1325 Pa (A Pascal is one Newtow per square metre) For the grid dish which is 3 feet x 2 feet that is 0.6 metres x 0.9 metres which is 0.54 m2. If this was a solid dish, and you use it for automatically tracking satellites (it could be aimed in the worst direction for wind loading at a time of high wind) you would use the worst case coefficent of drag for a dish shape which is 1.2. For the open grid dish which is basically a collection of "rods" as far as the wind is concerned, then a coefficent of drag of 1.0 would be appropriate. The dish area used in the calculation would be the area which resists the wind which is probably about two thirds of the area of a solid dish. For the grid dish in Sydney for example:- Aera = 0.54 x 2/3 = 0.36 m2 Drag Co = 1.0 Stagnation Presure = 1325 Pa Wind Load = 0.36 x 1.0 x 1325 = 477 N (48.7 Kg force or 107 lbs.) There are complicated ways of reducing the design wind speed but if you use the basic design wind speed you design will be safe. If you are more that 50 km form the coast you would be unlikely to be subject to tropical cyclones (hurricanes) so unless you have tornadoes in you area the 47 m/s is probably OK in absence of other information. In Australia the design wind speed for areas subject to Tropical Cyclones is 85 m/s which probably covers most of the world worst weather senarios. Regards, Murray Peterson VK2KGM ----- Original Message ----- From: "Mark L. Hammond" <hammond@surrealnet.net> To: "amSAT-BB" <AMSAT-BB@AMSAT.Org> Sent: Monday, May 14, 2001 6:57 AM Subject: [amsat-bb] wind load calculations? > Does anybody have a good link (or explanation) on how to calculate antenna > wind load? I've searched the Handbook, but can't find how to figure > it. And I'm talking for a variety of antenna types--meaning, what is the > windload for a BBQ style dish? (simple as 2'x 3' = 6 sq. ft, more or less > is solid?) Helical antennas for example? > > TIA, > > Mark L. Hammond [N8MH] ---- Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA. To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org

**References**:**wind load calculations?***From:*Mark L. Hammond

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