# Shock stand-off distance for blunt body

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 August 13, 2003, 10:59 Shock stand-off distance for blunt body #1 Praveen Guest   Posts: n/a There is an empirical formula for calculating shock stand-off distance for a 2-D cylinder and 3-D hemisphere as a function of Mach number. Does anybody have these formulae ? I think these are given in Anderson's Modern Compressible Flow, but I dont have access to that book. I would be grateful if somebody can give me the formulae. Thanks

 August 13, 2003, 22:29 Re: Shock stand-off distance for blunt body #2 Michael Guest   Posts: n/a Hypersonic and High Temperature Gas Dynamics (Anderson, Page 189: 5.4 Correlations for Hypersonic Shock-Wave Shapes Here's a summary: Shock shape is given by the equation: x = R + delta - Rc*cot^2[beta*(sqrt(1+(y^2*tan^2(beta))/Rc^2)-1 )] where: delta/R = 0.143*exp(3.24/Minfinity^2) for sphere-cone 0.386*exp[4.67/Minfinity^2) for cylinder-wedge and Rc/R = 1.143*exp(0.54/(Minfinity-1)^1.2) for sphere-cone 1.386*exp(1.8/(Minfinity-1)^0.75) for cylinder-wedge For cartesion cooridinates (x,y), where delta is the stand off distance and R is the sphere / cylinder radius and beta is the limiting angle to which the curved bow shock is asymtopic. Rc is the radius of curvature at the vertex of the hyperbola. The curves are for a sphere-cone (axisymmetric) and cylinder-wedge (planar) combination. The cone/wedge angle is 14.03624 degrees.

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