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#131
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On Thu, 05 Feb 2004 10:22:11 -0500, Todd Pattist
wrote: Jim wrote: I should not have gone farther than just the observation that the inside wing in a stable descending turn is going down while the outside wing is going up ( and the opposite situation in an ascending turn). This is incorrect. Both wings are going down (and both are going forward). They both go down at the same rate, they go forward at different rates (inner wing slower). If you think about it, that means that the inner wing has a higher AOA. Todd Pattist - "WH" Ventus C (Remove DONTSPAMME from address to email reply.) Thank you for clarifying again for me. I sure love this stuff but I sure don't have the background to truly understand it! But I sure love flying gliders! Jim |
#132
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On Thu, 05 Feb 2004 10:22:11 -0500, Todd Pattist
wrote: Jim wrote: I should not have gone farther than just the observation that the inside wing in a stable descending turn is going down while the outside wing is going up ( and the opposite situation in an ascending turn). This is incorrect. Both wings are going down (and both are going forward). They both go down at the same rate, they go forward at different rates (inner wing slower). If you think about it, that means that the inner wing has a higher AOA. I was just reading the Gider Pilot Manual by Ken Steward, pages 68 & 69. There he talks about the larger angle of attack difference in shallow turns than in steep turns, because of the larger difference in the circumference described by both wingtips, and concludes that steep turns are safer than shallow turns. I believe Steward left out an important factor, intentionally or not. Even if the the distance between the inner and the outer circles is larger in shallow turns, they also are much wider, and that has a large impact in the AOA difference. Otherwise, the shalowest turn would have the largest AOA difference, which is clearly not true. So far this thread talked about one bank angle. Would anybody care to compute the AOA delta for the whole range of bank angles? I guess it would be given as a function of the wing span... |
#133
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In article ,
jhandl wrote: On Thu, 05 Feb 2004 10:22:11 -0500, Todd Pattist wrote: Jim wrote: I should not have gone farther than just the observation that the inside wing in a stable descending turn is going down while the outside wing is going up ( and the opposite situation in an ascending turn). This is incorrect. Both wings are going down (and both are going forward). They both go down at the same rate, they go forward at different rates (inner wing slower). If you think about it, that means that the inner wing has a higher AOA. I was just reading the Gider Pilot Manual by Ken Steward, pages 68 & 69. There he talks about the larger angle of attack difference in shallow turns than in steep turns, because of the larger difference in the circumference described by both wingtips, and concludes that steep turns are safer than shallow turns. I believe Steward left out an important factor, intentionally or not. Even if the the distance between the inner and the outer circles is larger in shallow turns, they also are much wider, and that has a large impact in the AOA difference. Otherwise, the shalowest turn would have the largest AOA difference, which is clearly not true. So far this thread talked about one bank angle. Would anybody care to compute the AOA delta for the whole range of bank angles? I guess it would be given as a function of the wing span... I'm figuring bank angle, airspeed, and wingspan are the three factors. What would be really interesting is comparing the results against the airspeeds given in the simple bank angle tables for G loading found in many books, POH's and manuals. If one flies the airspeeds for the bank angles shown on these charts, for each glider (pick some popular ones), what is the bank angle that causes the highest ratio of airspeed between outer wingtip and inner wingtip? Perhaps Bob Hanson at St. Olaf can be convinced to code this up |
#134
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"Gary Boggs" wrote in message ...
The paper that Rich wrote on spin training that is posted on his web site is a must read! Thank you very much Rich. -- Gary Boggs Hood River, Oregon, USA You're welcome, Gary! BTW, I'll be giving a seminar on Stalls & Spins at the NW Aviation Conference in Puyallup, WA on Feb. 21 (4:00 PM) if anyone's interested... Rich http://www.richstowell.com |
#135
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Mark James Boyd wrote:
In article , jhandl wrote: On Thu, 05 Feb 2004 10:22:11 -0500, Todd Pattist wrote: Jim wrote: I should not have gone farther than just the observation that the inside wing in a stable descending turn is going down while the outside wing is going up ( and the opposite situation in an ascending turn). This is incorrect. Both wings are going down (and both are going forward). They both go down at the same rate, they go forward at different rates (inner wing slower). If you think about it, that means that the inner wing has a higher AOA. I was just reading the Gider Pilot Manual by Ken Steward, pages 68 & 69. There he talks about the larger angle of attack difference in shallow turns than in steep turns, because of the larger difference in the circumference described by both wingtips, and concludes that steep turns are safer than shallow turns. I believe Steward left out an important factor, intentionally or not. Even if the the distance between the inner and the outer circles is larger in shallow turns, they also are much wider, and that has a large impact in the AOA difference. Otherwise, the shalowest turn would have the largest AOA difference, which is clearly not true. So far this thread talked about one bank angle. Would anybody care to compute the AOA delta for the whole range of bank angles? I guess it would be given as a function of the wing span... I'm figuring bank angle, airspeed, and wingspan are the three factors. What would be really interesting is comparing the results against the airspeeds given in the simple bank angle tables for G loading found in many books, POH's and manuals. If one flies the airspeeds for the bank angles shown on these charts, for each glider (pick some popular ones), what is the bank angle that causes the highest ratio of airspeed between outer wingtip and inner wingtip? Perhaps Bob Hanson at St. Olaf can be convinced to code this up As a former math teacher, I liked to do a little math about the subject. The result was that the airspeed difference deltaV between the CG and the wingtip as function of bank angle phi, airspeed V, half wingspan b and gravity acceleration g is given by: deltaV = (b*g*sin(phi))/V i.e. speed delta constantly increases with bank angle when the airspeed remains the same (surprising ?), and decreases when airspeed increases at a given bank angle (not surprising). The surprising result for the difference as function of the bank angle should be relativized as it is not aerodynamically correct to compare the deltas at different bank angles and the same airspeed. We know that the load factor increases with increasing bank angle, so lift must increase in the same way. This can be done by increasing either AOA or airspeed, but AOA is limited by the stall angle and similar aerodynamic conditions are better maintained by increasing airspeed. If we express deltaV as a function of bank angle at a given AOA, which may be characterized as the speed V1 at which the glider would fly at this AOA with load factor 1, we get: deltaV = (b*g*sin(phi)sqrt(cos(phi)))/V1 and this has a maximun for cos(phi) = 1/sqrt(3), equivalent to tan(phi) = sqrt(2), or phi = .955 radians = 54.7 degrees. For the AOA delta at bank phi, airspeed V and sink Vz, I find: deltaA = sin(phi)*cos(phi)*b*g*Vz/(V**3) Here again it would be more significative to consider what happens at a given AOA, characterized by the airspeed V1 and sink Vz1 obtained at load factor 1, but as Vz is Vz1*(n**(3/2)) and V is V1*(n**(1/2) when the load factor is n, the effect of the load factor vanishes and the formaula remains the same: deltaA = sin(phi)*cos(phi)*b*g*Vz1/(V1**3) and the maximum is obtained when phi = 45 degrees. |
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