some numbers for a 45degree 'wing'
*Hoerner assigned Cd 1.03 for a 45degreeangled structure.
*We know that the hypotenuse,C of this rightangled triangle is 3feet.
*and from the Pythagorean Theorem : Asquared + Bsquare = Csquared.
*and @ 45degrees,A = B.
*then,Csquared = 2Asquared.
*and Csquared = 9
*then Asquared = 9/2,or, 4.5.
*and the square root of 4.5 = 2.121feet.
*so the frontal area is 6feet by 2.121feet,or 12.727sqft.
*and CdA = 1.03 X 12.727 = 13.108 sqft.

*On say,a 2014 F150 of Cd 0.402,and 36.0 sqft frontal area,we have CdA= 14.472 sqft.
*CdA of the truck and wing is 13.108 + 14.472 = 27.58 sqft,
*and adding 5% interference drag (as per Hoerner's methodology),we get CdA 28.96 sqft.
*The frontal area of the truck and wing is 48.727 sqft.
*dividing that into the CdA of the 'rig' gives Cd 0.595 for the pickup driving around with the wing on,but not towing anything.
*This would be like driving six 1stgen Honda Insights simultaneously.Ouch!
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