Mcrit Sweepback
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Mcrit Sweepback
I know there are a couple of threads on this subject. But I still have some issues with it.
Let me try to explain my point of view.
Why does sweepback increase Mcrit? - "Because the effective chord of the wing increases in length." Ok..
I see it like this.
Let say two parcels of air hit a straight wing. One of the parcels go above the wing and the other underneath. (please correct me on the thing Im about to state next) The parcels will meet at the aft side of the wing at the same time?
So the parcel on the upper side have to travel at a higher speed to cover a longer distance in the same time. Hence we get higher dynamic V on top, and the static pressure drops and we have lift.
Still with me?
If two parcels hit a swept wing, the parcel going on top now have an even greater distance to cover?
So The parcel on top is moving even faster. Here is what I don't get. Then the swept wing would lower Mcrit?
Help would be appreciated!
Let me try to explain my point of view.
Why does sweepback increase Mcrit? - "Because the effective chord of the wing increases in length." Ok..
I see it like this.
Let say two parcels of air hit a straight wing. One of the parcels go above the wing and the other underneath. (please correct me on the thing Im about to state next) The parcels will meet at the aft side of the wing at the same time?
So the parcel on the upper side have to travel at a higher speed to cover a longer distance in the same time. Hence we get higher dynamic V on top, and the static pressure drops and we have lift.
Still with me?
If two parcels hit a swept wing, the parcel going on top now have an even greater distance to cover?
So The parcel on top is moving even faster. Here is what I don't get. Then the swept wing would lower Mcrit?
Help would be appreciated!
The idea that the two parcels of air must meet again at the trailing edge has no logical basis. It has been dismissed long ago by most people who know anything at all about aerodynamics.
It might be better to consider what the air must do in order to permit the wing to move through it.
Let's imagine a wing that is 30 cm thick. As it moves forward, the air in contact with the upper surface of the leading edge must move upwards and the air in contact with the lower surface of the leading edge must move downwards to make a space 20 cm thick for the wing to pass through.
If the air is to remain in contact with the wing, then after the thickest section of the wing has passed, the air must move back to close up the gap.
The acceleration needed to achieve this depends upon the amount of time taken for the wing to pass through. This time is determined by the TAS and the chord length.
For any given TAS and wing thickness, increasing the chord length will reduce the acceleration rate that is required. This will reduce the highest local mach number for any given free stream mach number, so MCrit will increase.
So to have a high MCrit we can use very thin wings, and/or very long chord lengths (such as with delta wings) or sweepback the wings.
It might be better to consider what the air must do in order to permit the wing to move through it.
Let's imagine a wing that is 30 cm thick. As it moves forward, the air in contact with the upper surface of the leading edge must move upwards and the air in contact with the lower surface of the leading edge must move downwards to make a space 20 cm thick for the wing to pass through.
If the air is to remain in contact with the wing, then after the thickest section of the wing has passed, the air must move back to close up the gap.
The acceleration needed to achieve this depends upon the amount of time taken for the wing to pass through. This time is determined by the TAS and the chord length.
For any given TAS and wing thickness, increasing the chord length will reduce the acceleration rate that is required. This will reduce the highest local mach number for any given free stream mach number, so MCrit will increase.
So to have a high MCrit we can use very thin wings, and/or very long chord lengths (such as with delta wings) or sweepback the wings.
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Keith - you may well ridicule the simple theory but it is more than adequate for the needs of dcoded (and yours also depends on it to some extent - nature abhors a vacuum and all that) and what you have missed is that dcoded does not see that the lower surface path has also increased with sweep, and that effective camber reduces with sweep for a given section.
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My question is: What is the "cost" of wingsweep?
Reduced CL?
Reduced CL?
Reduced CLmax (proportional to cos(sweep) roughly)
Worse induced drag factor, at low M anyway
Higher wing weight (increased structural 'span')
Increased rolling moment/sideslip sensitivity (Dutch Roll, bigger roll response to sidegust, crosswind landing)
OTOH:
Reduced lift curve slope (less 'g' for a given up gust at a given wing loading/airspeed)
There may be others.
Going back to the original question, the usual explanation is that with a swept wing you can resolve the oncoming airflow into two components: along the span and normal to the LE (or quarter chord). The first has not much effect on Mcrit, but the component normal to c/4 is reduced by cos (sweep) so the airfoil 'sees' a lower Mach Number than freestream.
Cheers
CliveL
Last edited by CliveL; 2nd Feb 2011 at 15:28. Reason: additional comments
BOAC
It was not my intention to ridicule anything, and I do not believe that I have done so. The simple fact is that the idea that two parcels of air, having been separated by the leading edge will meet again at the trailing edge is quite simply untrue.
We could if we wish put arguments about the effects of the inertia of the air resisting longitudinal acceleratrions, beven this would not cause the two particles to meet again.
dcoded
If the explanations so far have not cleared up the matter, you might like to try a simple demonstration.
Take a large carrot (or other generally cylindrical object).
Place the carrot on a worktop and cut vertically downwards across the carrot at 90 degrees close to one end.
About 1/3 of the way along the carrot, cut vertically downwards across it at an angle of about 30 degrees to represent 30 degree sweepback.
About 2/3 of the way along the carrot, cut vertically downwards cross it at an angle of about 60 degrees to represent 60 degree sweepback.
Now compare the cross-sections of the three cuts. These represent what air would experience if it were to flow over the carrot.
The 90 degree cut which represents straight winds will be approximately circular.
The 30 degree cut which represents 30 degree sweepback will appear to be elongated in one direction.
The 60 degree cut which represents 60 degree sweepback will appear to be even more elongated.
The greater the sweepback angle the greater will be the chord length compared to the thickness. This illustrates the way in which sweepback reduces the thickness to chord ratio of a wing.
This reduction in thickness to chord ratio reduce the accelerations experienced by the air flowing over and under the wing. This in turn increases Mcrit.
It was not my intention to ridicule anything, and I do not believe that I have done so. The simple fact is that the idea that two parcels of air, having been separated by the leading edge will meet again at the trailing edge is quite simply untrue.
We could if we wish put arguments about the effects of the inertia of the air resisting longitudinal acceleratrions, beven this would not cause the two particles to meet again.
dcoded
If the explanations so far have not cleared up the matter, you might like to try a simple demonstration.
Take a large carrot (or other generally cylindrical object).
Place the carrot on a worktop and cut vertically downwards across the carrot at 90 degrees close to one end.
About 1/3 of the way along the carrot, cut vertically downwards across it at an angle of about 30 degrees to represent 30 degree sweepback.
About 2/3 of the way along the carrot, cut vertically downwards cross it at an angle of about 60 degrees to represent 60 degree sweepback.
Now compare the cross-sections of the three cuts. These represent what air would experience if it were to flow over the carrot.
The 90 degree cut which represents straight winds will be approximately circular.
The 30 degree cut which represents 30 degree sweepback will appear to be elongated in one direction.
The 60 degree cut which represents 60 degree sweepback will appear to be even more elongated.
The greater the sweepback angle the greater will be the chord length compared to the thickness. This illustrates the way in which sweepback reduces the thickness to chord ratio of a wing.
This reduction in thickness to chord ratio reduce the accelerations experienced by the air flowing over and under the wing. This in turn increases Mcrit.
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Originally Posted by dcoded
I am seeing it clearer now.
PBL
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I thought the best way to summarize the effect of a swept wing would be that it effectively trades span for chord, lengthening the wing for the same chord and thus reducing the effective T/C ratio.
