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Rate Of Turn and Turn Radius of an aircraft

  1. Aug 6, 2007 #1
    I have been working on a problem trying to solve the turn diameter of an aircraft in a steady state level turn. I have been using the standard equations for turn radius and rate of turn. The formulas I am using are:

    1) Turn Radius = (Velocity*Velocity)/(11.26 *Tan (bank Angle)) , velocity in Knots, answer in feet.

    2) Rate of Turn = (1092.95*Tan(bank angle))/ Velocity , velocity in Knots, answer in degrees/second

    Where equations 1 & 2 come from:

    The actual formula for the Turn Radius is derived from the mechanics of the problem and free body diagram of the forces acting on the aircraft in a steady state turn at a constant bank angle. The aircraft turns due to the horizontal component of lift. The formula comes from the physics of the problem were we solve for Centripetal Force (the horizontal component of lift in a turn). FAC1 = W Tan (Bank Angle) and since it is also given as FAC2 = mV2/r, and m = W/g, we get FAC2 = WV2/gr, solving for r, we get:

    r = WV^2 / g FAC2, Note: FAC1 = FAC2

    replacing FAC2 with FAC1 = W Tan(Bank Angle or “b” ) we get:

    r = W V^2 / g W Tan(b) = V^2 / g Tan(b), with V in feet per second and radius in feet.

    If we wish to use Knots in place of Feet/Second then converting Ft./sec to Knots gives a conversion factor of 1.6878 (1nm/hr*1.15077948 miles/nm*1 hr/3600 sec = 1.687809904).

    This result must be squared since the velocity will be squared, the result is 2.848702272 which is divided into g=32.17417 ft/s2 gives 11.29 not 11.26 a figure that was calculated using slide rules.

    Note that the value of 11.29 will decrease with altitude and thus using 11.26 is a closer approximation at high altitudes.

    The Rate Of Turn is also a derived formula. Given the circumference of a circle is 2*pi*radius we have the following:

    Time to turn 360 degrees (or 360t) = distance / velocity = 2*pi*radius / velocity, for the radius we use the answer we get for radius of a turn from above and add this to the equation:

    360*time = (2*pi*[ V^2 / g Tan(b)]) / V , this gives the time to make a complete circle. The degrees per second then is found by dividing both sides by 360 so,

    T = pi*V / 180*g*Tan(bank angle), solving all constants (i.e., pi, 180 and g – gravity) we get;

    Rate of Turn = V (in fps) / 1844.92*Tan(bank angle).

    These are standard formulas for Rate of Turn and Turn Radius. My problem is attempting to resolve it with the vector solution converting the polar velocity into the rectangular components for each second of time and summing all the “X” and “Y” distances. One of which should be equal to the Turn Radius times 2.


    Here is my problem. The Rate of Turn is an angular velocity, ω. Angular velocity where the radius and the velocity vector are perpendicular is given by ω = velocity/radius. Using this equation to solve for radius does not give me the same answer I get using equation number 1.

    Last, I tried to use the Rate of Turn value given using equation number 2 in a vector solution to find turn diameter. I made the assumption that initial heading was 0 degrees, 150 knots, bank angle of 45 degrees. I then set up an excel spreadsheet to solve for the vector values of X and Y using the following formulas,

    X= (150knots *Cos (heading angle after 1 second)) *(1.15*5280/3600), answer in feet

    Y= (150knots *Sin (heading angle after 1 second)) *(1.15*5280/3600), answer in feet

    Since I changed my heading angle by the Rate of Turn every second, I would have a new heading angle for each second that went by. Rate of Turn in degrees per second. I would them sum all my X and Y distances. It should make sense that the total distance in the X direction should be zero for 180 degrees of turn. Then the Y distance should sum up to the turn diameter.

    Would you please consider my methods and explain what I am doing wrong if possible.

    Thank You.
     
  2. jcsd
  3. Aug 7, 2007 #2

    AlephZero

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    I suggest you start by working in consistent units (e.g. feet, feet/sec, and radians). When you get that working OK, then go back to your preferred units.

    I haven't checked carefully through every step of what you did, but the general ideas looks OK.

    Try plotting a graph of the X-Y track you calcuated in EXCEL and see what it's telling you. Check if the speed, rate of turn, and turn radius are what you expect.

    If the EXCEL graph looks like you are doing aerobatics not a slow turn, remember the sin() and cos() functions in EXCEL use angles in radians not degrees.
     
  4. Aug 14, 2007 #3
    Solution to Rate of Turn, Radius of Turn Problem

    Thank you for your suggestions. I was using degrees in Excel but did convert them to radians. I broke down the steps I was making towards my solution by setting solutions for each step along side of the appropriate column and was able to identify the problem columns. The problem was one of correcting for the interval of time between each step and summing the vectors found for each step.

    I was attempting to determine the turn radius needed by Cory Lidle during his fateful flight in New York last October. Using the turn radius formula does not allow for adding of a wind vector. The vector solution worked very well for this purpose.

    Thanks again for your suggestions.
     
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