w7xman said:
Impressive quick math. Peak (gust only) on KMWN was 105kts. Can you post what that was, just for reference? My math is tired today...
My math was easy too... (except I made an error)
Rules of thumb:
* 17kts gives a dynamic pressure of 1 lb/sq-ft at sea level
- dyn pres is proportional to the square of the wind speed
* density of air at 5Kft is about .85 of that at sea level
- dyn pres is proportional to the density of air
You can get the density at altitudes from
http://www.digitaldutch.com/atmoscalc/
(My error was remembering 16 kts--I just recalculated to get the correct number. The old pressures were 13% high.)
So the table should be:
Approx dynamic pressures at an altitude of 5K ft:
105 kts .. 32 kbs.sq-ft
97 kts ... 28 lbs/sq-ft
90 kts ... 24 lbs/sq-ft
80 kts ... 19 lbs/sq-ft
70 kts ... 14 lbs/sq-ft
60 kts ... 11 lbs/sq-ft
50 kts ..... 7 lbs/sq-ft
In case anyone cares:
dyn_pres = .5 * rho * v**2
... where
... dyn_pres = dynamic pressure in lbs/sq-ft
... rho = density of air = .0024 slugs/cu-ft at sea level
... v = wind velocity in ft/sec
force = dyn_pres * effective_flat_plate_area
... where
... effective_flat_plate_area = real_area * coefficient_of_drag
31 pounds of force per square foot sounds easy, but considering that the average bundled human is about a sqare yard +/- pack = 31lbs/foot^2 x 9 feet^2 = 279 pounds pushing you. That's alot!
The above assumes that the frontal geometric area is close to the effective flat-plate area. (A car with a frontal area of 30 sq-ft and a coeficient of drag of .2 would have an effective area of 6 sq-ft.) Furthermore, a human will be down in the boundary layer which will subject him to a lower wind speed. My guess is that the actual force will be less than 279 lbs.
Doug
