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Supersonic turning radius 
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#1
Nov206, 07:34 AM

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When we take the turn in a car at a high speed, the turning radius is pretty big. Similarly, at supersonic speeds, the turning radius of aircrafts would be huge. Could any one tell me the turning radius of aircrafts as such speeds and the minimum turning radius of any aircraft.....



#2
Nov206, 10:39 AM

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The equation is quite simple (a=v^2/r) so you can just do the solving yourself. I'm not sure what kind of g forces a standard supersonic turn would be, but I wouldn't think more than 2 or 3 g's. So for a plane going 1000 mph (447 m/s) and turning with an acceleration of 2 g's it's turning radius would be 10 km. Going 1400 mph would increase that turning radius to nearly 20 km. You get the idea... 


#4
Nov206, 11:11 AM

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Supersonic turning radius



#5
Nov206, 01:22 PM

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Anyway, the SR71 had a turn radius of about100mi at cruising speed/altitude, but I don't know at what gforce that was. 


#6
Nov206, 11:27 PM

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I remember reading about the SR71 guys. They would reference what state to start their turns over. The turning radius was huge.
Standard G profiles now take them up to 9 g's for a very brief period in training. 


#7
Nov406, 12:37 PM

P: 83

Hi Folks,
MechEngineer has started with the correct basic premise (constant acceleration), but we can go a bit further by considering the limitations of the aerodynamics, and we will even arrive at an equation which includes the load factor (i.e. the number of g's you are pulling during the turn). In general, the maximum turning performance is restricted by aerodynamic, structural, thrust/power and buffet limits. The aerodynamic limit is the most important (i.e. it becomes a central design parameter based on the mission needs) since you can design the strength of the structure and the power of the engines to meet your needs. So the aerodynamic limit describes the minimum turn radius available to the airplane when it is operated at its maximum lift coefficient. The absolute minimum turn radius is achieved at the turning stall speed, and is given by the following equation: Radius(min) = (Vstall^2/g)*(n/SQRT(n^21)) Where: Vstall = Aerodynamic stall speed for the wing+body configuration. g = gravitational constant (i.e. 32.2 ft/sec^2) n = steadystate maneuver load factor (i.e. maximum # of g's you can pull) This is a vast and expansive topic in aircraft design, mostly because of the interaction of the aerodynamics with power available, structural strength, and the number of g's a pilot can withstand. But if you do a little numerical analysis on what happens as the maneuver glevel (n) increases, you will see that the absolute minimum turn radius (if you assume you can pull infinite g's) does indeed reduce to Vstall^2/g. Rainman 


#8
Nov406, 05:23 PM

P: 4,777

Hello Rainman,
I for one would love to see a derivation of this (if it is not too involved) and you have the time. I would expect in a powerful turn that the bottom of the wing along the centerline of the fuselage would be subjected to a strong negative bending moment because of the increased pressure forces distributed along the top of each wing section. It seems like a delta would be better for sharper turns because the wing length is typically reduced, and so would the associated bending moments. This is just me thinking out loud, so please correct me where I am wrong. (and I probably am wrong). Also, could you treat the wing spar as a beam under pure moment, as say a first order approximation. Fixed on a knife edge at the centerline of the wing, and given two forces on each wing some distance, [tex]\delta[/tex] from each wing tip as the applied force. ***Actually, I should have said, the top wing would have a greater force than the bottom wing, because it travels a greater arc and has a higher velocity. I would be interested to know how you account for some of these things. 


#9
Nov506, 02:37 PM

P: 83

Hello Cyrus,
Rainman 


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