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Old 03-04-2017, 01:58 PM   #61 (permalink)
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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!

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