There was an interesting comment posted the other day that essentially said that a turn to final won't result in an accelerated stall because the aircraft is descending and therefore has a load factor of less than 1 G.  Of course, as long as the aircraft isn't accelerating vertically the situation is the same as for level flight - the load factor will be greater than 1 G.  But what if one did decide to turn the 90-degree turn while keeping the load factor at 1 G?  Certainly any turn at any bank angle can be made at 1 G as long as the aircraft is allowed to accelerate downward.  How much, I asked, so I went to my trusty spreadsheet and did some calculations.  If you turn 90 degrees at a bank angle of 15 degrees (load factor of 1.0 instead of the normal 1.04) and an airspeed of about 90 knots you will exit the turn at a vertical descent rate of 1,000 ft/min MORE than when you entered.  Do it at a bank angle of 30 degrees and the number is about 2,000 ft/min.  at first it seemed to me that it is unlikely that doing this type of thing makes sense, although I suppose one could go from a base leg at level flight and use the turn to make the transition to descending flight.  A 1,000 to 2,000 ft/min descent rate isn't far off for a normal final approach, so maybe it would be a reasonable thing to do as normal practice.  Just a thought.

Gary Casey
ES #157