# Supersonic turning radius

by banerjeerupak
Tags: radius, supersonic, turning
 P: 123 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.....
PF Gold
P: 2,248
 Quote by banerjeerupak 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.....
Two words: "Centripetal Acceleration"

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...
 P: 15,319 Pilots can do up to about 8gs before risking a black out.
PF Gold
P: 2,248

 Quote by DaveC426913 Pilots can do up to about 8gs before risking a black out.
But is it a good idea to pull 8g's turning around while doing Mach 2?
Mentor
P: 22,303
 Quote by DaveC426913 Pilots can do up to about 8gs before risking a black out.
Just not for all that long...

Anyway, the SR-71 had a turn radius of about100mi at cruising speed/altitude, but I don't know at what g-force that was.
 Sci Advisor P: 5,095 I remember reading about the SR-71 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.
 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^2-1)) Where: Vstall = Aerodynamic stall speed for the wing+body configuration. g = gravitational constant (i.e. 32.2 ft/sec^2) n = steady-state 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 g-level (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
 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, $$\delta$$ 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.
P: 83
Hello Cyrus,
 Quote by cyrusabdollahi I for one would love to see a derivation of this (if it is not too involved) and you have the time.
It is a bit long to try to reproduce in-toto here on the forum. And besides that it is much easier to point you to any of a number of existing books on aerodynamics that includes aircraft performance topics. I learned aircraft performance, stability, and control using Dr. Jan Roskam's books, and his shows the derivation of the specific equation I cited above. But any other book (such as: Introduction to Flight-Anderson, Fundamentals of Flight-Shevell, Fundamentals of Aircraft Design-Nicolai) will also show how the basic equation is arrived at by starting with constant-acceleration equations of motion for an aircraft in turning flight. Most books derive the equation without the "n" in the numerator that is shown in the equation I have cited. Dr. Roskam's derivation shows why his includes this extra load factor and where it comes from. HERE is a paper that shows the derivation, but arrives at a slightly different version of the equation (but still with the V^2/g form). Being that I am a teacher, I think it is important that the student work the derivations through themselves to find out what the variations on the general "V^2/g" form of this equation are used for.

 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.
Yes, very good. The delta planform is superior in turning performance for this reason and also because a delta is inherently more structurally rigid than a conventional wingspan.

 This is just me thinking out loud, so please correct me where I am wrong. (and I probably am wrong).
So far I don't see anything you have said as being wrong! On the contrary, they are right on!

 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, $$\delta$$ from each wing tip as the applied force.
Yes, this is precisely how we teach a first-order approximation for wing bending. It is treated as a beam fixed at a wall (the fuselage) subject to a constant load that varys from wing root out to wing tip (i.e. a load profile). Of course, this first-order (literally) approximation will only give you solutions for static tensile, compressive, and shear stress tensors. It will not tell you anything of the dynamic (frequency) response of the wing's bending and vibration modes. For that you would, of course, need a dynamic model.

 ***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.
And not only that but the upper surface of the wing is also where the greatest lifting loads are generated, as a percentage of the total lifting load of the upper and lower surfaces.

 I would be interested to know how you account for some of these things.
Aerodynamic, structural, and power (thrust) effects on turning performance are analyzed in an integrated metric that we call the "V-n" maneuver diagram. As you will read on this page, the V-n diagram is a composite metric (which is usually calculated and levied as an aircraft-level design requirement) that considers the loads and speed of both maneuvering flight and time-varying atmospheric gusts and the additional loads that they generate on the airframe.

Rainman

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