Satellite dish wind resistance

rough estimate of wind force on dish, stress in mount is dependent on mount geometry

if you want to do the calcs get a copy of an ME handbook (like Mark's; check abebooks.com for used copy)

1/2 * rho *v^2 * Cd *A

rho air density v wind speed Cd object drag coefficient A frontal area

use a consistent set of units & you'll get the wind force on the dish

max force about 2000 lbf (about 10kN)

are you sure the dish structure, dish skin & dish to mount connection can handle this force level?

we're talking about some pretty high winds here

regards Bob

thanks also on my behalf for answering

"Bob K 207" schrieb im Newsbeitrag news: snipped-for-privacy@mb-m03.aol.com...

A thumbnail calculation can get you close enough (you use liberal design margins too...)

A parachutist free falls at 120 kt. Model his cross section as 2 X 6 feet. Drag varies as the square of air speed.

With these data, we compute

1.2m = 40 inch + 8 inch = 4 ft diam Area = pi r sq = 3.14 X 2 X 2 = 12.5 sq ft This is close enough to the jumpers cross section.

The speed requirement is 150 kt so the drag is (150/120) squared = 25/16 = 1.6

The parachute jumpers weight is (say) 170 lb. That's the drag provided by the air at 120 kt.

So at 150kt, the drag is 1.6 X 170 lb = 270 lb.

If you stress for a wind load of 3 X 270 = 800 lb, everyone lives happily ever after.

(If this is a homework problem, you will be a brave student to copy this approach. Only the best schools would give extra credit for this method :-)

Brian whatcott Altus OK

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I think this number is too high. What values did you use for the calc?

Brian W

Brian-

I emailed you outside the news group.

Bob

I agree it seems high, I emailed you the numbers I used

here's a link to an antenna mfr

''%20to%2039''.htm

Thanks for the response: I double checked our respective numbers - my cross section for the parachute jumper was over estimated, so my force estimate was understated. Your coefficient of drag was somewhat high, so your force estimate was overstated.

But you win the cigar on the basis of your solution statement: your answer to one significant place allows a range 1000 to 3000 my answer to one significant place allows a range of 700 to 900 lb: I conclude your answer is more accurate on that basis.

Brian Whatcott

Brian-

Per my email (I assume you got it) I used 2.0 as the Cd.

Upon more searching perhaps 1.2 is closer to correct; 2.0 was my guess at an upper bound based on a square flate plate

My calc actually yielded 1870 (but I fudged it up to 2 kips). Using a Cd of

1.2 would yield 1120 lbf, quite a bit lower than my original conservative SWAG.

regards Bob