Aasc = (0.25) ( 1 . 0 7 8 C L ) = 0.269 CL (3) A similar example using the unit vortex approach (page 22 of reference 4) yields a value of 0 . 2 5 4 C L . The difference of 0.015° (at CL — 1.0) is not surprising in view of the uncertainties of the extrapolation and the fact that reference 4 uses an average value rather than that at the centerline. Better agreement would reasonably be expected for a floor mounted model, since the tunnel shape would then be more reasonable. (4) At M = 0.85, lt/B(l M ) / 2 = 0.0307/[1 (0.85)] = 0.0583 and r = 0.475. Aasc then equals (0.475) (1.078CL) = 0MSCL.