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How can one calculate the maximum turning angle F-16 can turn in x second flying at corner speed. Is it changed with height or weight?

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To calculate this, you need to look at the energy Maneuver diagram for the aircraft. This diagram depicts, for a specified altitude and gross weight, the excess power per unit pound of weight the aircraft would have at full afterburner at any g-Loading and true airspeed. There would be a different chart for each aircraft configuration, gross weight and Density Altitude combination.

Instantaneous Turn rate is depicted on the vertical scale, true airspeed or mach number is depicted on the horizontal scale. The concave curved lines that go from upper left to far right represent flight conditions of constant G-loading, while the straight lines that radiate from the origin represent constant turn radius. The peak of the flight envelope, (at about Mach 0.7, 20 deg/sec turn rate and 8 Gs, is corner velocity. The lines that start at the left boundary or at the bottom, go upwards, curve to the right and then back down again, and are labeled with values 600, 400, 300, 0, -200, etc. are called Ps, (pronounced P-sub-S), lines. They depict the specific power, or actually, specific excess power - a measure of how much power per pound of aircraft weight the aircraft would have left over to climb or accelerate at the depicted point on the diagram. So at the top of the diagram, at the corner velocity, where the turn rate is near 20 deg/sec, the specific excess power is less than -800. The -800+ Ps curve indicates that the aircraft, if it established flight at those conditions, would be bleeding energy (airspeed or altitude) at a rate in excess of negative 800 ft/sec.

The best performance on the diagram that can be sustained, on the other hand, is the highest point on the zero Ps line, which is at mach 0.8, 7 Gs, and it would produce a sustained turn rate of just over 14 deg/sec.

So, in a level turn, at Mach 0.8, 7 Gs, and 14 deg/sec, the F-16 could turn 180 degrees in about 13 secs.

Of course, if the aircraft performs the turn in a descent, it can make the turn in less time. This is indeed a common training maneuver, to turn 180 degrees in the shortest amount of time possible. It is performed by rolling into 135 degrees of bank at or slightly above corner velocity and pulling at maximum G-load. The maneuver is performed at maximum available G-load, in a constant plane tilted at 45 degrees from the horizontal. Bank angle (starting at 135 degrees), decreases throughout the maneuver, reaching 90 degrees bank after 90 degrees of turn, and 45 degrees bank at the end of the maneuver (after 180 degrees of turn). At 15,000 feet MSL, this will generate almost 20 degrees per second, and can be done in about 10 seconds.

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For an uncoordinated turn, all things are possible, consider a hammerhead, wing over, or Lomcovak! For a constant altitude coordinated turn, weight effects the load and stall which does play a part. The two primary factors are bank angle and rate of turn. As the angle of bank increases the load factor or thrust required for speed will limit the aircraft's ability to bank.

enter image description here

To completely answer your question requires more information such as the weight, load limit, stall speed of the aircraft in the configuration of the turn, atmospheric conditions, etc.

As the angle of bank and turn rate increase, so must the speed and load factor, soon the power available or aircraft structural capability is exceeded. As the aircraft approaches a 90deg bank the vertical lift component approaches zero, the speed and load required to hold altitude approaches infinity.

enter image description here The graph and turn performance of the F-16 are discussed in detail here...

The "FAA's Pilot's Handbook of Aeronautical Knowledge" gives the formulas for rate of turn and turning radius on page 4-34. Chapter 15 of "The Science of Flight" by W.N. Hubin (1992) is the source of the other formulas.

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The variables used are:

  V = true airspeed in knots    
  R = turning radius in feet    
  θ = bank angle in degrees    
  ω = rate of turn in degrees per second
  g = 11.26

  Note: The FAA formulas use deg/sec, while Hubin uses rad/sec.

For example, at 120 knots and a 30° bank angle, the turn radius and rate of turn are:

enter image description here

Hubin goes into greater detail and provides information for other scenarios.

enter image description here

The formula for a standard rate turn (3deg/sec) is... enter image description here

but other velocity's can be plugged in.

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The AOA is 15 degrees and pulling 9G for making the maximum performance turn in level turns in F16. ( 22 degrees /sec)

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  • $\begingroup$ They question is about turn rate, i.e. degrees per second, not AOA or max Gs. $\endgroup$ – Michael Hall Nov 30 at 17:22
  • $\begingroup$ Then 22degrees /sec $\endgroup$ – George Geo Nov 30 at 17:23
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    $\begingroup$ When you are asked for more info in a comment, and you have more to add, don't forget to add it to your answer. (I edited that in for you) $\endgroup$ – KorvinStarmast Nov 30 at 20:48

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