ASCE 74 shows the CD on a long cylinder tube would be 1.2 .8 for a short on long flat plates 2.0 and on short plates 1.4
as does the EIA-222-F
"[Force = A x P x Cd x Kz x Gh
A = the projected area of the item
P , Wind pressure (Psf), =.00256 x V^2 (V= wind speed in Mph)
Kz, Exposure Coefficient, = [z/33]^(2/7) 1.0 <= Kz <= 2.58
z = height above average ground to midpoint of the item (IE, antenna, or tower span) in feet.
Gh, Gust response factor = .65+.60/(h/33)^(1/7) 1.0 <= Gh <= 1.25
h = overall height of a tower (used for an antenna mounted at its top) in feet.
Cd = 2.0 for long flat plates and 1.2 for long cylinders. Aspect ratios >=25
Cd = 1.4 for short flat plates and .8 for short cylinders. Aspect ratios <=7
The relationship between drag coefficients for cylinders and flat things is, 1.2/2.0 = .6 or .8/1.4 = .57, in this case less than 2/3
EIA-222-F thinks the wind speed is the "fastest mile basic wind speed" at 33 feet above the ground, not the actual peak sustained wind speed. These values are not the same as the 222-C spec, they are defined by State & County locations, rather that the older wind speed zone maps.
There are no additional site specific exposure factors cited in this spec.]"
Like I said I just used a simple bonehead shortened quickie Frank formula But I've never gotten in trouble by over estimating
I think it can be agreed on that someone missed the mark on their anchorage though with 6 anchor points positioned around the base @ 60 degrees to each other any 2 anchor points should have been sized to withstand the full force of a max wind load for the region with a FOS of 4 to one minimum code and 8to one California std on a 4 or less anchor point it would have been any single anchorage
Here is a link that some may find helpful
http://www.arraysolutions.com/Products/windloads.htm