Flying trailers

F-35 shown obsolete on previous posts

   There was a tornado near Dallas, Texas that lifted truck trailers.

   For a  trailer to be lifted, there must be either:

1. a wind such as B, blowing harder above the trailer than below to cause a greater pressure drop and therefore produce lift
2. a wind such as C, that deflects upwards from the ground and produces lift through stagnation pressure

    For B, assuming a trailer weighs 12 000 lbs and is 8 1/2 ft wide, U.S. standard, and 53 ft long, maximum allowed in U.S., the load would be 12 000 / 53 X 8.5 = 30 lbs per square foot, psf.  The magic number for pressure is about 8000in the US. Common System.  For 30 psf, v^2 X 1/800 = 30, or v^2 = 24 000.   If the air speed were to be zero below the trailer, the wind would have to blow at 155 ft/ sec, 100 mph, 160 kph, to produce lift.  If the speed below were 60 mph, 90 fps, the upper speed would have to be 90^2 = 8100 + 24 000 = 32 000 = v^2, v= 180 fps, 120 mph, 192 kph.  With ground drag and other trailers blocking the low wind, it might be achieved, but the trailer would not be a perfect surface, the area would not be fully effective.  Instead one would have to adjust the area downwards by a factor of maybe0.8.  All v would increase by 10 % in that case.
    In C, if the wind is at an angle to ground of 30 degrees, with the same factor of 0.8, the weight density would be 33 psf for v^2 of 26 000, but only 0.5 of velocity would be incident on trailer.  However, there would be a vacuum above the trailer that would match in theory the  pressure force below.  That would reduce the need force by 0.5 but with the wind speed vertical of 0.5 speed, v^2 = 26 000 X 2 = 52 000 orwind speed 230 fps, 160 mph, 260 kph.
   To fly, the trailers would have no inherent aerodynamic lift.  The square corners would prevent circulatory flow and would violate Kutta's hypothesis.  The vertical sides height be 10 ft high.  If the trailer is rolled so the side is 30 deg to the horizontal;
 The sides are 2 x 10 x 0.5, wind angle x 0.87 cosine of lift
 The top an bottom are 8.5 x 2 x .87, wind angle x 0.5 lift cosine
The effective area is 8.7 - 7 = 1.7 ft width.
For 53 ft = 90 ft^2 or 133 psf
Wind speed would be 340 fps, 230 mph, 370 kph.
     The equation of force goes as 2 X (L1 - L2) X sin 2x, where x is the angle of inclination and L1, L2 are the lengths of sides.  This has to br multiplied by the effective adjustment of thearea for edge effects.