How this calculator works
Resolve upstream Mach number normal to the shock and apply the normal-shock conservation equations. The tangential velocity is unchanged. The theta-beta-M relation connects turn and shock angles and can have two attached solutions. The weak branch usually has a smaller pressure rise. The solver finds the maximum allowed turn and rejects larger turns, which require detached-shock analysis. Dimensional state inputs should describe one physically consistent gas state. Heat transfer, viscosity, real-gas chemistry and variable heat capacity are excluded.
Formula
tan(theta) = 2 cot(beta)[M1^2 sin^2(beta)-1]/[M1^2(gamma+cos(2beta))+2]. Mn1 = M1 sin(beta). p2/p1 = [2gamma Mn1^2-(gamma-1)]/(gamma+1). rho2/rho1 = (gamma+1)Mn1^2/[(gamma-1)Mn1^2+2]. T2/T1 = (p2/p1)/(rho2/rho1). Mn2^2 = [1+(gamma-1)Mn1^2/2]/[gamma Mn1^2-(gamma-1)/2]; M2 = Mn2/sin(beta-theta).
Worked example
For gamma = 1.4, M1 = 2 and beta = 45 degrees, the normal upstream Mach is sqrt(2), turn angle is about 14.74356 degrees, p2/p1 = 13/6 and rho2/rho1 = 12/7. The temperature ratio is 91/72.
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Frequently asked questions
What is the strong branch?
It is the larger shock-angle solution for the same upstream Mach and turn, with greater compression and usually subsonic downstream flow.
Why can a turn be rejected?
Beyond the maximum attached turn, no straight attached shock satisfies these equations. A detached flow field needs another model.
Is stagnation pressure conserved?
No. A finite shock is irreversible and loses stagnation pressure, even though stagnation temperature stays constant in this adiabatic model.