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Critical Mach Number  
User currently offlineZiggy From United States of America, joined Jun 2001, 178 posts, RR: 0
Posted (12 years 2 months 3 weeks 14 hours ago) and read 9673 times:

I am reading "The Illustrated Guide to Aerodynamics" by H.C. "Skip" Smith. I am trying to learn more about high speed flying and have come acrossed the section on wing sweep and how it can increase an aircrafts critical Mach number. I have tried using models and still cannot understand the example or the author's thinking on how this is done, so if someone could please help and put it into perspective.

The book goes to says that if the wing is swept aft, only a component of the velocity of the air will flow over it chordwise. Another component will flow spanwise along the wing.

Question

Understanding the air will flow over the wing whether it be directly or diagonally. The wing's thickness stays the same no matter which direction the air flows over the wing. And at the highest point on the camber line ( or thickest point on the wing) the air velocity should be at it's peak.

So for example if a unswepted wing has a critical Mach number of .75, the air velocity at the top of the camber line will be 1.0 Mach. Now my question is that if you sweep the wing back 30*, the new critical Mach number is .86. But the camber wasn't changed, only it's highest point was moved rearward a little. So how by just sweeping the wings back, decrease the velocity of the air at the peak of the camber line? (or am I barking up the wrong tree?)

Ziggy  Confused



4 replies: All unread, jump to last
 
User currently offlineBellerophon From United Kingdom, joined May 2002, 583 posts, RR: 59
Reply 1, posted (12 years 2 months 3 weeks 8 hours ago) and read 9666 times:

Ziggy

Let me try and explain as simply as I can, and hope I don’t confuse you more!

So how by just sweeping the wings back, decrease the velocity of the air at the peak of the camber line?

It doesn’t, the speed of the air passing over the peak of the camber line is (broadly) unchanged, but wing sweep changes the angle at which the airflow hits the leading edge of the wing.

I don’t know how good you are at Maths, but we can break down (resolve) the motion of the air across the wing into two components. One component that flows directly across the wing at right angles to the leading edge (chordwise flow), and the other component, which is at right angles to the first component, that flows along the wing from root to tip (spanwise flow).

So on a swept wing, the speed of the air passing over the peak of the camber line at right angles to the leading edge of the wing has been decreased.

Very broadly, it is only the component of velocity across the chord of the wing, (chordwise flow), that is causing the shock wave, and interests us in terms of critical mach number.

In your example of an aircraft with wings swept at 30°, if the aircraft is travelling at V knots, the speed of the air flowing across the chord of the wing has been reduced to V*Cos 30° knots, or around 87% of the aircraft’s forward speed.

If that’s a little difficult to understand, try this. Pick up a ruler and pretend that it’s a wing. Mark a point at the middle of one of the (long) edges, and draw a thick black line backwards from that point, at right angles to the long edge, across the ruler to the other edge.

Hold the ruler between finger and thumb at one end, pretend it is a wing on an imaginary aircraft, and try moving it to different angles of sweep. Can you see that the speed of the airflow over this line will vary as the angle of sweep varies?

The speed of the air flowing along this line will be the same as the aircraft speed when the wing is fully forward (sweep = 0°) and will be zero when the wing is fully aft (sweep = 90°). In between, if the angle of sweep is x° , the chordwise speed will be V*Cos x°.

If we sweep the wings to 30°, whilst still flying the aircraft at Mach 0.75, the spanwise flow of the air will reduce in speed to Mach 0.75 * Cos 30°, roughly Mach 0.65.

Conversely, we could keep the chordwise flow at Mach 0.75, but increase the speed of the aircraft, from Mach 0.75, to (Mach 0.75 / Cos x°).

In your example, the Critical Mach number of the aircraft with an unswept wing was Mach 0.75, which produced a chordwise flow at Mach 1.0 at the peak of the camber point. Sweep the wings by 30°, and you can increase the aircraft speed to (Mach 0.75 / Cos 30°) which comes out to Mach 0.86.

I hope this helps,

Regards

Bellerophon






User currently offlineSeagull From United States of America, joined Jun 2001, 340 posts, RR: 1
Reply 2, posted (12 years 2 months 3 weeks 5 hours ago) and read 9626 times:

Actually, all that stuff about span-wise and normal flow is just good for modeling the situation for calculations. It tells you the "how" without the "why".

Consider instead the problem in the first place -- that of the air in front of the wing not getting any "warning" that the wing is coming along. The idea is to give it a "warning". It turns out that the portion of the wing in advance of the area in question (any point on the wing you want to use) sends out pressure waves radially, so the more forward portion of the wing provides that warning and allows flow to exist as it does subsonically.

I can do it the other way too, but that doesn't explain anything except for a convenient way to approximate the results of the above.


User currently offlineZiggy From United States of America, joined Jun 2001, 178 posts, RR: 0
Reply 3, posted (12 years 2 months 3 weeks 5 hours ago) and read 9623 times:

Thanks Bellerophon, that explanation helped a lot and now starting to see the big picture. But still confused on one aspect.
As states in my book, chordline is a straight line connecting the leading edge, or forward most tip, to the trailing edge. In my humble opinion that still leaves a lot to interpretation, so here how I'm seeing it. With an unswept 0* wing, the chordline would be perpendicular to the leading edge - or how the air would flow over the wing. Now in my thinking, if you swept the wing back 30*. The chordline would also change 30* to stay parallel to the air flow, or the majority of it. I can understand that the 30* sweep would affect a component of it to travel spanwise. I've probably made my dilemma as clear as mud, but if someone could help correct my thinking.

Ziggy  Confused


User currently offlineZiggy From United States of America, joined Jun 2001, 178 posts, RR: 0
Reply 4, posted (12 years 2 months 3 weeks 4 hours ago) and read 9623 times:

Seagull:
Thanks for you reply. So here's a stab in the dark, but this warning would it bend or manipulate the air flow a little as so it would flow more perpendicular to the leading edge? And also would it slow the air flow velocity down a little before it gets to the wing?

Ziggy


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