The descent angle helps to reduce the lift that the wing needs to provide, in two ways. It leads the aircraft into higher density air, where more absolute lift is possible at the same Mach number. That takes a while, but a more immediate effect comes from pointing the aircraft's nose downwards. Then, part of the gravity-induced forces can be counteracted by drag (D in the sketch below). This is similar to a glider, only with much lower aerodynamic efficiency and steeper angles. Terry reduces thrust and opens the speed brakes to avoid overspeeding, so he can dive as steeply as possible, thus reducing lift requirements.
Now more of the wing's lift L can be employed for turning the aircraft around. If he wants to achieve a constant sink rate, he still has to produce enough lift to balance weight, but reduced by the cosine of the glide path compared to the level flight condition. At 25° nose down, this is a reduction by 10%.
Now you are concerned about his roll angle. Please note that he wants to create as much sideways component of lift as possible, which will look like gravity to all onboard. At 60° roll angle (I think his 80° are a bit extreme), this works out to twice the normal vertical acceleration, and with the dive actually just 1.8 g, so everyone will feel almost twice the force of normal gravity. With the 80° roll, this will not be a stationary turn and not enough lift will remain to counteract gravity, so the aircraft will accelerate downwards during the turn. It will actually fall out of the sky. When already at Mach 0.85, I would be more careful than Terry, because things get ugly very quickly when Mach goes further up.

The term $m\cdot\frac{v^2}{R}$ is the centrifugal force which can be afforded by the horizontal component of the lift and which needs to be as big as possible for tight turns (a smaller radius R helps). This is the lift component that Terry needs to pull the aircraft around.
If we stay with stationary turns (without the "falling out of the sky" part), the maximum bank angle is given by the maximum load factor of the 747. At 400 KEAS, this is just 1.5g, so even with the 25° nose down attitude the maximum bank angle would be 53°. If you fly slightly slower, the load factor goes to 2g (equals 63° in a 25° descent) and tops out at 2.5g at 310 KEAS (68.7° in a 25° descent). Flying at Mach 0.85 in 30.000 ft will produce an equivalent airspeed of 306 kt in normal atmospheric conditions. The precise value varies with the actual flight speed, and the maximum safe bank angle is in the region of 60° to 70°.
I expanded my answer to the original question, and I hope in combination with this I can cover your questions. If not, please keep asking!
EDIT: I see, now you want to know the time to turn around. All I can do is calculate it under the assumption that enough lift is available, regardless of buffeting. In reality, I doubt that even the 48° case will give you a pleasant ride. All results are for the 30.000 ft case, because at 36.000 ft those higher bank angles will not be cleanly flyable due to compressibility effects.
At 60° the airplane flies with 2g, and with 25° nose down still at 1.8g. The turn rate would be 3.77°/s and 180° would be completed after 48 s.
At 70° it flies with almost 3g, and with 25° nose down at 2.65 g which is a little outside of the g limits. The turn rate would be 5.98°/s and 180° would be completed after 30 s (plus the time to roll to 70°, which now starts to become important).
The 80° case becomes really hypothetical, because even with the 25° nose down attitude the load factor will be 5.2g if flown without sideslip and enough lift to prevent overspeeding. The 747 will be close to breaking apart, but maybe a future aerobatic version can fly this maneuver safely. It will result in a turn rate of 12.35° and need 14.6 s for a 180° turn. I would add at least 4 seconds to bank to 80°, but the turn can be completed within 20 s. Hypothetically.
What Terry hinted to when he said he would bank to 80° is not the clean turns I am calculating here, but a dynamic dive where the plane accelerates because not enough lift is left to keep it from falling. To compute this, I would need more aerodynamic data on the 747 and need to employ at least a finite difference algorithm or a lot of paper and pencils.
Just that much: If the 747 is rolled to 80° but the pilot pulls only as many gs as the aircraft can sustain (due to buffeting I doubt that even 2.5 g are possible), the lift will almost exclusively serve to fly a tight turn, but not enough will be left to counteract gravity. Therefore, the aircraft will slip sideways, and directional stability will turn the nose downwards. This will take a while (I am guessing maybe 12 seconds, really a guess), but the aircraft will pick up speed from the beginning of the maneuver. I am a little more scared of flying faster than Mach 0.85 than Terry is, and the various reports of 747s going up to Mach 0.98 support his view that a little overspeeding is OK. It is hard for me to guess how much time it needs to become uncomfortably fast, and then a lot of lift will be needed to pull out of the dive. On the other hand, when the airplane is diving, all it needs is a roll and a pull-out to come out at 180° to the previous course. The pull-out will be performed at a much lower altitude, and creating the gs there will not be a problem.
18 years of flying — typed CE-500, SA-227, B-727, B-747 — last 10 years primarily international on the 747.
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