Increasing bank angle, load factor and rate of turn

I came across this question and I get why the answer is correct but I don’t really understand why by increasing angle of bank from 20 to 30 the load factor has not increased (option B in the image). In theory load factor equals 1/cos(bank angle).

Hope the question is clear. Thanks.

• What's the source? Dec 28, 2022 at 7:00
• @tedder42 Aviation Exam Dec 28, 2022 at 15:22
• @ArisMartínezVentura unless it's a book called "aviation exam" that isn't a source. Presumably an aviation exam for a specific country and rating from a specific website or app or book? Dec 28, 2022 at 20:31
• @quietflyer It is, I’ve added the tag. Dec 29, 2022 at 18:12
• @tedder42 there is indeed a book. It’s called “JAA Test Prep” from Aviation Exam. It’s an ATPL question bank. Dec 29, 2022 at 18:14

You are right, of course. B is also correct. Never put too much stock in the so-called "right" answer to a test question. It is extremely common to find mistakes in these materials.

Here's a bit of conjecture as to how this error may have occurred: in ground school materials, many graphs exist that appear to show that the increase in load factor is negligible at bank angles of 30 degrees or lower. Perhaps the test question was created by someone who was looking at such a graph.

Here is one example of such a graph:

On closer examination, we find a lot of problems with this graph. For one thing, the lines are so wide that small changes in the parameters are hard to see. A more serious defect is that the load factor curve has been shifted downward from where it ought to be if the y value is meant to be zero along the bottom of the graph-- after all, the load factor is 1, not zero, in wings-level linear flight. Also, no scale is given for the load factor curve-- it is clear that it cannot be the same as the scale for the "percent increase in stall speed" graph.

If the lower curve were labeled "percent increase in load factor" rather than simply "load factor", then the scale on the left (y) axis could serve for both curves-- but the "percent increase in load factor" curve would have to be drawn above the "percent increase in stall speed" curve for all points where the bank angle is greater than zero. The "percent increase in load factor" curve would also have to curve upward more sharply than the "percent increase in stall speed" curve. For example, in constant-speed constant-altitude turns, at 20 degrees bank, the stall speed is 1.032 times the wings-level value, representing an increase of 3.2%, while the load factor is 1.065 times the wings-level value, representing an increase of 6.5%. At 30 degrees bank, the stall speed increases by 7.5% while the load factor increases by 15.5%. At 45 degrees bank, the stall speed increases by 18.9% while the load factor increases by 41.4%, and at 60 degrees bank the stall speed increases by 41.4% while the load factor increases by 100%. (All comparisons are to the values for wings-level flight at a constant altitude. Equations: load factor = 1 / (cosine bank angle), and the stall speed scales in proportion to the square root of the load factor.)

The other alternative is that B was supposed to be a "trick" wrong answer just like C and D, where the stated relationship between the values in the 20 and 30 degree-banked turns was reversed from reality, but somewhere along the line "greater than" was inadvertently substituted for "less than".

• D is also correct as you have to increase the angle of attack in a turn, no?
– Ben
Dec 27, 2022 at 21:10
• @Ben no its not, look at it carefully. The lift coefficient (and therefore AoA) will be higher on "A" because of it's higher bank angle. Dec 27, 2022 at 21:43
• All this assumes you are doing a Level (constant Altitude) turn. Clearly, to maintain a level turn, the steeper the bank angle the greater the lift is required to ensure that the component of Lift pointed away from the earth is sufficient to prevent a descent. The greater the Lift, the greater the lift coefficient Dec 27, 2022 at 21:49