Some numbers have come up in another thread which caused me some trouble.
I thought I might discuss this scenario in a new thread but with no reference to the source. Please let's just discuss the physics here not the incident.
Scenario: Slow descending turn - possible accelerated stall.
- Stall speed 48kn
- Angle of bank approx 48 degrees
- descent rate >1500 ft/min
- speed during turn 59kn
Let's agree a stall is caused by the angle of attack exceeding the critical angle.
We use speed as an indication of when a stall will occur and we are told that stall speed increases in turns.
Firstly:
Lets consider a 48 degree level turn.
Say vertical component of the total lift is 1 - in order to keep the turn level.
The horizontal component is 1.2 which pulls us around the turn.
The resultant or total lift is 1/cos48 which is 1.5
The stall speed increases with the square root of the load factor √1.5 = 1.22
So the level turn stall speed becomes 1.22x 48kn = 59kn
Which is also the reported airspeed - so in this turn it is possible the airflow was starting to sperate from at least part of one wing.
But, and here is my problem. Everything above assumes a level turn. Or at least is calculated on the special case of a level turn.
This is where my concern starts. To be descending at a steady rate the system must be in equilibrium, that is the weight + lift + air resistance = 0 resultant force
weight is fixed at mass x acceleration due to the earth's gravity. (yes, apparent weight might change with g, but mass is fixed for a given fuel load and the earth always sucks the same.)
Air resistance goes up with the square of speed (from memory) In this case we are looking at the vertical speed and the vertical component of air resistance. 1500ft/min is around 8 m/s. Pretty quick really.
So I am thinking to obtain a steady 1500 ft/m the lift required is much less than the force of 1g. (Yes, technically 1g is an acceleration not a force - but it's convenient.)
Maybe pick a number of 0.8g vertical lift component. In the scenario above this would make the total lift 0.8/cos48=1.2 and the stall speed √1.2 x 48 = 52kn
Maybe 0.7g - which is 49kn. (the other 0.3g would be providing around 200kgf to work against the air resistance which doesn't seem huge at that speed over that projected area?)
These numbers would suggest a stall is unlikely.
Am I missing something here?