The paper discussed factors to complete a 180 degree turn, on instruments, with restrictions on the maximum bank angle (reasonable for instrument flight), followed by no turning (banking) below 100 AGL.

It did not definitively explore all options for a visual return to land on the runway.

BJC

Yes that is true. My understanding of the 45 recommendation is more of a math thing. This paper presents it well:

http://www.campbells.org/BIG_FILES/airplaneImpossibleTurn.pdf
I am curious to know what others think about this math, but my understanding from this paper is that 45 is actually better than higher angles in terms of preservation of altitude. 60 degrees will get you through the turn faster, but you will be closer to the ground afterward, despite having spent less time turning, because your induced drag is higher during the turn at 60 degrees and you have to drop the nose further to maintain a higher speed through the turn than you do at 45, so your form drag is higher also.

Specifically (cut and paste from article - please see attached pdf also):

The Optimum Bank Angle Following the development in Jett3, we consider a simple energy analysis of the optimum conditions for a steady gliding turn to a new heading. In a gliding turn the aircraft trades the potential energy embodied in altitude to overcome drag and maintain velocity above the stall velocity of the aircraft. A larger bank angle in the gliding turn requires a higher rate Figure 1. Teardrop flight path. of descent to maintain steady conditions. Consequently, minimum time in the gliding turn to a new heading yields the optimum turn conditions.

From Fig. (2) we have Lcos φ = 1 2 ρV 2SCL cos φ = W (1) and Fc = Lsin φ = V 2 R W g (2) Thus, combining Eqs. (1) and (2) yields the radius of the turn, i.e. R = V 2 g tanφ (3) Figure 2. Forces in the yz plane acting on an aircraft in a steady state gliding turn. Minimizing the radius of the turn keeps the aircraft close to the end of the runway and thus results in a decreased glide distance after completion of the turn.

The time required to turn thorough a given angle, Ψ, is t = Ψ Ψ˙ (4) and in a steady state turn Ψ =˙ dΨ dt = V R = Ψ t (5) The rate at which the aircraft expends the potential energy available from altitude must equal the energy required to overcome drag. Thus, W dh dt = DV (6) Integrating for steady state conditions yields W h t = DV (7) or h = DV t W (8)

Introducing Eqs. (3) and (5) yields h = D W V 2 Ψ g tan φ (9) In a gliding turn with bank angle φ D W = CD CL cos φ (10) Recalling that V 2 = 2W ρSCL cos φ (11) Eq. (9) is written as h = CD C2 L 4W ρSg 1 cos φ sin φ Ψ (12) The steady state conditions for minimum loss of altitude in a gliding turn to a new heading are obtained by differentiating Eq. (12). The result is dh dΨ = CD C2 L 2W ρSg 1 sin 2φ (13) where we have used sin 2φ = 2 sin φ cos φ to simplify the result. Examining this result shows that for a parabolic drag polar, CD = CD0 + kC2 L, the first term CD C2 L = CD0 C2 L + k (14) is a minimum at CLmax. Thus, the optimum speed for minimum loss of altitude in a gliding turn to a new heading occurs for CLmax, i.e., at the stall velocity. Neglecting the small density change with altitude, the second term, 4W/ρSg sin 2φ, is a minimum for sin 2φ = 1 or φ = 45◦, i.e., the optimum bank angle during a gliding turn to a new heading is 45◦.