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Old 24th May 2006, 22:36
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SR71

Mach 3
 
Join Date: Aug 1998
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I'll have a go.....

Using the relationship for incompressible lift from elliptical wing theory and Prandtl Glauert (or even better von Karman and Tsien) relationships, along with the tables here:

http://books.elsevier.com/companions...le-2/table.htm

you can work out a compressible lift coefficient assuming the lift curve slope of the two dimensional wing is 95% of the theoretical value and an angle of attack of 4 degrees in the cruise.

Suffice to say, lets call it 0.4 - because I can't be bothered to work it out otherwise.

You can then look in some magic tables which plot M_crit (zero sweep) against t/c ratio for various C_L.

The tables in the html file above give t/c ratio for 733 as ~12.9 and sweep as 25 degrees.

M_crit (zero sweep) is thus 0.69 according to my magic table.

Since M_crit (with sweep) is equivalent to M_crit (zero sweep) divided by the cosine of the sweep angle, I'm going for

M_crit (25 degrees sweep) = 0.75.

A good approximation of M_{drag divergence}, assuming that this is the Mach number where the gradient of the C_d versus Mach number curve reaches 0.05, is:

M_{drag divergence} = M_crit*[1.02 + (1-cos(sweep angle))*(0.08)]

This assumes two dimensional wing theory and applies only exactly to a wing of infinite span. It should be good for wings of high aspect ratio.

This gives

M_{drag divergence} ~= 0.77

We know

Mmo = 0.82 and

M_ne should be about 5% higher: ~= 0.86

I have observed Mach Tuck in a 737 sim before but I can't remember what Mach number it manifest itself at.

Someone tell me my guesses are right!



PS: The other good time to see shock waves is in humid conditions at TO which only serves to illustrate Sam_airman's point about how dependent M_crit is on C_L.
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