Need help! Simple question.

"Jim Y" wrote in news:5Acvc.20771$ snipped-for-privacy@bgtnsc05-news.ops.worldnet.att.net:

Well, I don't *have* a Schaum's. I tried figuring it out using the formula F=1/2CApv^2 where F=force, C=coeffecient of drag (I used 0.5), A=cross sectional area (3X12=36), p (rho) is the density of air (forgot value off the top of my head), and v=velocity. Either I made a big mistake making all the values match before computing, or something else is wrong, because the answer didn't seem to make sense.

In the English system of units you have to be careful about mixing lb_f and lb_m. If you have density in slugs/ft^3, you should be ok

Simple Answer: For a first estimate, I use the fabled paratrooper who weighs 180 lb and who drops at 120 MPH terminal. His cross section is about

I easily remember a simple model for air drag F = K A V^2 F force A area V air speed K scaling constant.

Paratrooper:

NOW, we're ready for your antenna section: Area 0.25 sq ft Speed = 100 mph K = 0.00125

F = K A V^2 F = 0.00125 * 0.25 * 100 * 100

Force = 3.125 lb.

If you're smart, you promptly multiply this estimate by a factor of safety. Like an aircraft designer. Use X6 and you won't regret it.

Brian Whatcott Altus OK

Did you convert the 100 mph to ft/sec? V = 100 X 5280 / 60 / 60 = 146.67 ft/sec Wair = .075 lb/cu ft Cylinder L/D = 12/3 = 4, therefore, assume Cp = 1 (not a precise number using the graph) Cylinder, Area = 3" x 12" / 144 "/sq ft = .25 sq ft Gravity, g = 32.174 Force, F = Cp x Wair x Area x (V x V) / (2 x g) F = 1 x .075 x .25 x (146.67 x 146.67) / (2 x 32.174) = 6.3 lb

Hope that helps, Jim Y

Brian Whatcott wrote in news: snipped-for-privacy@4ax.com:

Thanks! Jim Y also replied but his calculation included gravity. I don't know if he understood the problem (force of wind on a vertical cylinder), but thanks to him too for bothering to answer. So if I wanted to actually *measure* the force on the antenna, a 10lb load cell would get me in the ball park.

Try plugging this into Google's web search box:

I used C=1, and the result is about 6 lbf.

Ref

Dan Major wrote in news:Xns94FC5E59D2B29soonerboomergbronlin@68.12.19.6:

Yes, he understood the problem which requires the density of air to be in slugs/ft^2 which is (lbs/ft^2)/g. It is always important to understand the units required for a particular formula and to keep them consistient. The standard form for the formula for drag, a force, is D = (C*p*A*v^

In the answer above note that K must include the drag coefficient, the density of air, and a factor of 2. (In the wonderful way of the internet, the previous post has not yet arrived here.) Note also that the cross sectional area is entered in ft^2, but the velocity is in mph not ft/sec. K must also include the conversion factor for mph to ft/sec for the answer to be correct. (If you correct the 100 mph to 146 ft/sec, the answer changes to 6.7 lbs which is close to Jim Y's answer so I suspect the conversion factor is not in K.)

The drag coefficient is a mater of some judgement and the particular references you happen to work from. At Re of 2E5, I'd use 1.2 rather than the 1 of some of the other answers. If you plug the numbers in, out pops D = 7.5 lbs a little higher than Jim Y's also correct answer of 6.3 lbs.

Charly Coughran wrote in news:Xns94FC82943508Eccoughranucsdedu@132.239.1.221:

Without quoting the entire article, I apologize to Jim Y. With Charly's thorough explaination, looks like *I* didn't understand the problem. I had tried solving it in Imperial (inches/feet/miles/pounds,etc) as well as metric units, but neither answer seemed right. Thanks again to all!

He understood the problem. But a force of 3 to 6 pounds is not what a load cell at a base mount would feel, necessarily. Moment. Leverage. All that good stuff. I know you will keep this in mind.

Brian W

Don't forget that wind is never a steady force and a mast is not rigid. You will have flexing, movement, inertia and acceleration forces in the mast. There will also be variations in wind speed and direction which will add to the mast movements.

John

"John Manders" wrote in news:c9po68$ snipped-for-privacy@newton.cc.rl.ac.uk:

Not to mention that the 100 mph wind is probably a hurricane and the horizontal rain component of the drag will be much greater than the wind drag.

Charly Coughran wrote in news:Xns94FE5886F58A4ccoughranucsdedu@132.239.1.221:

Nah. Here in Central Okla. 100mph is barely an F1 tornado. We also get straight-line winds up to 70-80mph with some storms. Heck, I have sneezes worse than that (bad allergies!).

The explanation that I gave you assumes no factor of safety, it answers your question only. For example, you later stated that 100 MPH wind is not uncommon in your area. You do not design anything for the normal conditions. You take into consideration the extreme that may happen. Then you design the machine to suit worst possible conditions. (Then you get into an argument with management because they say it is too expensive to build and management won't get their bonus that year and their new Ferrari. Of course, you will be held to blame if management over rides your design and the machine fails. Yes, I am being a bit cynical, but it is the truth based on my real life experience.)

In the design of an antenna used in a public area, most states require the seal of a professional engineer. Be careful of what you are doing and this is not one of those designs requiring the seal of a PE.

Jim Y (retired PE - rolling mill machinery)