View Single Post 03-04-2017, 01:58 PM #61 (permalink) aerohead Master EcoModder   Join Date: Jan 2008 Location: Sanger,Texas,U.S.A. Posts: 13,450 Thanks: 21,443 Thanked 6,482 Times in 4,051 Posts some numbers for a 45-degree 'wing' *Hoerner assigned Cd 1.03 for a 45-degree-angled structure. *We know that the hypotenuse,C of this right-angled triangle is 3-feet. *and from the Pythagorean Theorem : A-squared + B-square = C-squared. *and @ 45-degrees,A = B. *then,C-squared = 2A-squared. *and C-squared = 9 *then A-squared = 9/2,or, 4.5. *and the square root of 4.5 = 2.121-feet. *so the frontal area is 6-feet by 2.121-feet,or 12.727-sq-ft. *and CdA = 1.03 X 12.727 = 13.108 sq-ft. -------------------------------------------------------------------------- *On say,a 2014 F-150 of Cd 0.402,and 36.0 sq-ft frontal area,we have CdA= 14.472 sq-ft. *CdA of the truck and wing is 13.108 + 14.472 = 27.58 sq-ft, *and adding 5% interference drag (as per Hoerner's methodology),we get CdA 28.96 sq-ft. *The frontal area of the truck and wing is 48.727 sq-ft. *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 1st-gen Honda Insights simultaneously.Ouch! __________________ Photobucket album: http://s1271.photobucket.com/albums/jj622/aerohead2/ 