Here let me work out the basic equations in the thread. The inputs we need are airspeed, wind (optional, and I will skip it unless requested), and rate of turn (or angle of bank).
The reason why the plane turns when it is banked is because a component of the force of lift is horizontal. As a consequence of banking, the vertical component of lift is decreased which means the pilot will need to do some combination of adding power and pulling back on the yoke in order to maintain altitude. I'm going to assume the pilot already has that all figured out and is simply maintaining the desired airspeed while turning and descending as needed. Neglecting vertical motion shouldn't have a significant effect on turn radii. The basic formula for relating forces to circular motion is:
Code:
Centripetal Force = m v^2 / r
Using the definitions of sine and cosine, the horizontal component of lift is L sin (theta). Theta is angle of bank, and the use of sine makes sense since if the bank angle is zero, then sin is zero, and there is no horizontal component to the lift. If theta is 90 degrees, then sin(theta) is 1, all of the lift is horizontal, and none of it is vertical. Since the only horizontal force pulling the plane towards the center of the circle is the horizontal component of the lift, we can combine the two expressions above and get:
Code:
L sin (theta) = m v^2 / r
Unfortunately, we haven't figured out what to do with L yet. Since we've already taken care of the math in the horizontal direction, we can work in the vertical direction next. Although the plane may be descending during the turn, it has a constant vertical speed. Therefore, the upward force (vertical component of lift) has to equal the weight of the plane (m * g). Going back to the definitions of sine and cosine, the vertical component of lift is L cos (theta) and the equation for vertical forces becomes:
This makes sense, because when theta is 0 (no bank), cosine is 1 and L = m g. Using algebra techniques, we can combine the above two questions that are in code blocks. The goals are to make the L's cancel away, m's cancel away, and combine the sine and cosine terms into just one term. We can divide the lower equation into the upper equation. Meaning, create a new equation where the left hand side is L sin (theta)
over L cos (theta), and the right and side is m v^2 / r
over m * g. Here is the result:
Code:
sin (theta) / cos (theta) = v^2 / [r g]
Using the definition of tangent, the equation can be simplified to:
Sanity checking that equation, for small angle of banks, increasing bank angle increases tan(theta). This means, as you increase bank angle, you need to either increase speed, decrease turn radius, or decrease gravity in order to maintain stable turning flight, which all seems reasonable to me. Rearranging to isolate r:
Essentially, this is the equation on that Flight Learnings
page @Somedudeintn linked to before. The last step here is to determine the proper value for g so that you can plug in knots and get back the radius in feet. I will convert the well known 9.8 m/s^2 into nautical miles per hour squared:
Code:
9.8 m/s^2 * 0.000539957 nm/m * [3600 s / hr]^2 = 68578.86 nm/hr^2
So if you want the radius of the turn in nautical miles, square the speed in knots, divide by 68578.86, then divide again by the tangent of the bank angle. For example, if I'm turning with a speed of 75 knots and banking at 25 degrees, my radius of turn will be about .175 nm, and the diameter of the turn (how far you should be from the runway centerline to start) will be 0.35 nm. Since there are 6076 feet in a nautical mile we can change that 68578.86 number of a factor of 6076 in order to make the equation spit out feet instead of nautical miles. 68578.86 / 6076 is 11.28, which is almost exactly what @Somedudeintn provided in that link (rounding errors likely caused a very slight discrepancy.) This confirms that the first equation on the linked page is correct.
None of this is useful unless you know how to choose a proper bank angle. The last thing I'll do is relate bank angle to rate of turn. If the plane turns 3 degrees in a second (standard rate), it actually swept out 3 degrees of an arc around the center of the circle after one second. The length of that arc is the angle (in RADIANS) times the radius (commonly referred to as S = theta r). We also can take as given that distance equals speed times time. From the first relationship I just discussed:
Since the arc-length is a distance, and distance is always speed times time, we can combine those two relationships like this:
We know rate of turn is usually provided in degrees per second (e.g., 3 degrees per second). We want to isolate a term that is of the form angle per time, as such:
Technically, that angle is still in radians. What the equation tells us is that the rate of turn, in radians per hour, is the speed in knots divided by the radius in nautical miles. I need to bring back the equation for radius from much earlier in this post. Recall from before:
Combining the above two equations in code blocks:
Code:
angle / t = [g tan(theta)] / v
I will rename "angle / t" ROT, multiply by a factor of 180/pi, and divide by a factor of 3600, to make the units work out:
Code:
ROT = g tan (theta) * 180 / [pi * 3600 * v]
Remember that g is actually 68,578.86. Using the above example, a 75 knot turn at a 25 degree angle of bank results in a ROT of 6.78 degrees per second, well in excess of standard rate. Combining numbers together:
Code:
ROT = 1,091.467 * tan(theta) / v
What angle of bank would be standard rate? Solve for tan(theta):
Code:
tan(theta) = v * ROT / 1,091.467
Using 3 as ROT, 75 as v, the right hand side is .20614. Taking the arc tangent of that value while in degree mode, the angle of bank is 11.6 degrees. The thumb rule for standard rate turns is to take 10% of the speed (7.5) and add half of that (3.75). The sum is 11.25, which is pretty close to 11.6.
Well, in case you were wondering where the equations came from, there it is.
