lec10 - 10 PRACTICAL LIFT CALCULATIONS 10.1 Characteristics...

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Unformatted text preview: 10 PRACTICAL LIFT CALCULATIONS 10.1 Characteristics of Lift-Producing Mechanisms At a small angle of attack, a slender body experiences transverse force due to: helical body vortices, the blunt trailing end, and fins. The helical body vortices are stable and symmetric in this condition, and are convected continuously into the wake. The low pressure associated with the vortices provides the suction force, usually toward the stern of the vehicle. The blunt trailing end induces lift as a product of added mass effects, and can be accurately modeled with slender body theory. A blunt trailing edge also induces some drag, which itself is stabilizing. The fins can often be properly modeled using experimental data parametrized with aspect ratio and several other geometric quantities. (Continued on next page) 10.2 Jorgensen’s Formulas 45 As the angle of attack becomes larger, the approximations in the fin and slender-body anal- ysis will break down. The helical vortices can become bigger while remaining stable, but eventually will split randomly. Some of it convects downstream, and the rest peels away from the body; this shedding is nonsymmetric, and greatly increases drag by widening the wake. In the limit of a 90 → angle of attack, vorticity sheds as if from a bluff-body, and there is little axial convection. 10.2 Jorgensen’s Formulas There are some heuristic and theoretical formulas for predicting transverse force and moment on a body at various angles of attack, and we now present one of them due to Jorgensen. The formulas provide a good systematic procedure for design, and are best suited to vehicles with a blunt trailing edge. We call the area of the stern the base area. Let the body have length L , and reference area A r . This area could be the frontal projected area, the planform area, or the wetted area. The body travels at speed U , and angle of attack ∂ . The normal force, axial force, and moment coefficients are defined as follows: F N C N = (135) 1 πU 2 A r 2 F A C A = 1 πU 2 A r 2 M x m C M = 1 πU 2 A r L ....
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This note was uploaded on 02/27/2012 for the course MECHANICAL 2.154 taught by Professor Michaeltriantafyllou during the Fall '04 term at MIT.

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lec10 - 10 PRACTICAL LIFT CALCULATIONS 10.1 Characteristics...

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