- #1

- 5

- 0

tan(θ)= 2cot(β) * (M

^{2}sin2(β)-1) / (M

^{2}(1.4+cos(2β))+2)

This seems like it should be a simple problem, but I can't seem to figure it out.

Note, I am using constant specific heats hence the 1.4 in the equation.

Any hints?

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- Thread starter twmggc
- Start date

- #1

- 5

- 0

tan(θ)= 2cot(β) * (M

This seems like it should be a simple problem, but I can't seem to figure it out.

Note, I am using constant specific heats hence the 1.4 in the equation.

Any hints?

- #2

- 545

- 10

However, "Elements of Gas Dynamics", by Liepmann and Rosko give an approximate relationship between β and θ when the mach number is large:

[itex]\beta ≈ (\gamma +1)/2 * \theta [/itex]

L&R provide a graphical solution to the weak and strong shocks on page 87. This is probably the best method to show the relationship between deflection angle and shock angle. Also, they note that for a specific mach number there is a maximum possible deflection angle, θ.

- #3

- 5

- 0

- #4

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- #5

- 208

- 0

Yep, gogo Newton's method. Gotta love Matlab.

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