Below is cut and paste from the comments to the article in the OP's link. Note that he's driving a Mooney.
Bank angle and TAS are the drivers re how long it takes to complete a 180. The formula for rate of turn is 1,091 x (tangent of the bank angle)/TAS. Because the tangent goes up with angle of bank any increase in bank will increase rate of turn and any increase in speed will reduce it. And since the distance between a dead-stick airplane and the ground can be measured in precious seconds, an expeditious rate of turn is desirable. To point out the obvious, more bank and lower TAS are what you want (with an eye on best glide speed and stalling speed as g’s increase). To illustrate: a 180° turn at 30° of bank at 85 KTAS takes 24 seconds to complete; a 180 at 45° takes 14 seconds; and a 180 at 60° can be completed in only 8.1 seconds. Double the bank from 30 to 60 and cut the time needed to head back to the field from 24 to 8 seconds. The steeper bank has the added advantage that the offset from the runway is less on rollout.
Yes, we are all aware that steep-banked turns near the ground invite a stall. A 2 g tug on the yoke will stall my 20E at 90 KIAS; coincidentally, 2 gs are required to hold a plane in a stable 60° bank turn. Not much room for error when you slip below Vy and crank it over to 60° of bank.
Practice
Several iterations (at altitude) in a Mooney 20E yielded average altitude losses in a 180° turn of:
360′ lost at 30° bank,
270′ lost at 45° bank, and
200′ lost at 60° bank.