BlurE Posted January 28, 2016 Share Posted January 28, 2016 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? Link to comment Share on other sites More sharing options...
diesel Posted January 28, 2016 Share Posted January 28, 2016 All assumes the machine is in balance. In reality its probably over ruddered because of a reluctance to bank. Straight wing is now a swept wing? Chas. Link to comment Share on other sites More sharing options...
Head in the clouds Posted January 28, 2016 Share Posted January 28, 2016 I'd agree that all of the OP assumes the machine is in balance, and the numbers will change for the worse if it isn't - BUT in practical terms and given those numbers and that situation I'd say you could induce the worst-case for possible stall/spin, and still not stall either wing, though at that worst-case position it might be close. The worst-case would be a full control-deflection skid which in many aircraft would, in a left turn, be full left rudder and sufficient right aileron to prevent rolling beyond the 48 degrees bank angle. The above is a bit similar to the classic stall/spin killer on the base/final turn we hear so much about, with the main differences being that the killer turn has the pilot resisting bank, so is rarely at anywhere near 48 degrees, and generally nowhere near a 1500ft/min descent rate. Given both of those the wing is considerably unloaded which makes it quite stall resistant. The OP provided the numbers and my description above is the non-scientific way of considering it i.e. comparing it with real-life things you've tried when 'experimenting' up high - we used to be encouraged to try things out years ago, these days it's all considered stupid or it's illegal - anyway, in real life you have to work pretty hard to stall in a descending situation such as that described, but it's quite do-able if you try hard enough ... the OP suggests, though, that the flying wasn't intended to be greatly unbalanced, so without further information I would think we'd only have to factor in a small amount of imbalance. Link to comment Share on other sites More sharing options...
diesel Posted January 28, 2016 Share Posted January 28, 2016 strange things happen when you get behind the drag curve. Tiger moth or glider. Speed has nothing to do with stalling. I do wonder at all the aircraft that need? 30 degrees up elevator. Chas Link to comment Share on other sites More sharing options...
Head in the clouds Posted January 28, 2016 Share Posted January 28, 2016 strange things happen when you get behind the drag curve. Tiger moth or glider. Speed has nothing to do with stalling. I do wonder at all the aircraft that need? 30 degrees up elevator. Chas Huh? I don't quite get what you're saying here ... Link to comment Share on other sites More sharing options...
P4D Posted January 28, 2016 Share Posted January 28, 2016 Forget the numbers and fly the plane. Maybe your pitot tube was occluded by the down-going wing, maybe your instruments were not calibrated this morning, maybe your bank angle indicator was off by 1%, maybe you were in a momentary skid, maybe you over corrected with rudder for .07 seconds its academic to a successful flight. Instruments are lagging indicators and tell you whats just happened. They cannot anticipate anywhere near as well as you can as PIC. 2 Link to comment Share on other sites More sharing options...
Keenaviator Posted January 28, 2016 Share Posted January 28, 2016 Huh? I don't quite get what you're saying here ... The risk of stalling the elevator? Jabiru's have 18 degrees up elevator if my memory serves me correctly. Are full flying elevators limited to less than 15 degrees? Link to comment Share on other sites More sharing options...
BlurE Posted January 28, 2016 Author Share Posted January 28, 2016 The numbers in the scenario and the conclusion "probably resulted in an accelerated aerodynamic stall" come from the ATSB. In level flight I think the analysis is solid. I am not so sure about the application in a high rate descending turn. Link to comment Share on other sites More sharing options...
SDQDI Posted January 28, 2016 Share Posted January 28, 2016 Forget the numbers and fly the plane. Maybe your pitot tube was occluded by the down-going wing, maybe your instruments were not calibrated this morning, maybe your bank angle indicator was off by 1%, maybe you were in a momentary skid, maybe you over corrected with rudder for .07 seconds its academic to a successful flight.Instruments are lagging indicators and tell you whats just happened. They cannot anticipate anywhere near as well as you can as PIC. While I partly agree with you P4D another part of me knows how much our bodies work against us. For the most part we should be able to avoid stalling without looking at the instruments at all BUT put us in a low level turn with a little wind and all of a sudden our bodies start to take too many queues off the ground and it isn't long before we are flying uncoordinated. I still haven't finished my ll endo but the little I did made me very aware that yes our eyes need to be outside at least 90% of the time BUT it was also super important to do a quick glance often to make sure the ball was centred and the airspeed was healthy because our bodies definitely lie. Link to comment Share on other sites More sharing options...
djpacro Posted January 29, 2016 Share Posted January 29, 2016 - Stall speed 48kn - Angle of bank approx 48 degrees - descent rate >1500 ft/min - speed during turn .......Maybe pick a number ....... Don't need to guess at numbers as the arithmetic is fairly simple and I would expect the ATSB to have done something sensible with the data they have. Why not ask them (eg discussion on their Facebook page)? Link to comment Share on other sites More sharing options...
Ultralights Posted January 29, 2016 Share Posted January 29, 2016 one word when it comes to any turn, any bank angle, and AOA..... BALANCE. keep the aircraft balanced, and you can turn at 60 Deg AOB at 15 deg AOA. if you stall, a recovery will require nothing but a small reduction in back stick, just enough to get the wing back to 15 Deg AOA or less (depending on wing stalling AOA) if the aircraft is unbalanced, then you will have problems.. big ones. 1 Link to comment Share on other sites More sharing options...
kasper Posted January 29, 2016 Share Posted January 29, 2016 And the only thing I would add as a 'practical' flight of this profile is that in slow descending turn I run the stall + 1/3rd rule to account for the non-perfect handling that P4D noted - I am an average pilot and I get things a bit off every now and then. On a 48knt stall that would give me an indicated 62knt target speed or 3knts more than the scenario outlined. As outlined the 59knt target is stall + 23% ... cutting my margin of safety below my personal comfort level but as noted in the calcs probably still under the stall speed ... and I would always target to exit the decent more than 1000ft AGL. Link to comment Share on other sites More sharing options...
Deskpilot Posted January 29, 2016 Share Posted January 29, 2016 If it's a long and slow decent, don't forget carbie heat for safety's sake. Link to comment Share on other sites More sharing options...
Happyflyer Posted January 29, 2016 Share Posted January 29, 2016 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? Your figures are probably fine but don't take into account variable such as turbulence, sink/lift, pilot over correcting with too much back stick, speedo error etc etc. A healthy safety margin in speed and height is always prudent. Link to comment Share on other sites More sharing options...
planesmaker Posted January 29, 2016 Share Posted January 29, 2016 Problem might be the pull out at the bottom of descent. That is when the stall is most likely. 1 Link to comment Share on other sites More sharing options...
nickduncs84 Posted January 29, 2016 Share Posted January 29, 2016 The numbers in the scenario and the conclusion "probably resulted in an accelerated aerodynamic stall" come from the ATSB.In level flight I think the analysis is solid. I am not so sure about the application in a high rate descending turn. I read this report and it's the first thing I thought of too. Makes the whole report seem like amateur hour in my opinion. Link to comment Share on other sites More sharing options...
silvercity Posted January 29, 2016 Share Posted January 29, 2016 one word when it comes to any turn, any bank angle, and AOA..... BALANCE. keep the aircraft balanced, and you can turn at 60 Deg AOB at 15 deg AOA. if you stall, a recovery will require nothing but a small reduction in back stick, just enough to get the wing back to 15 Deg AOA or less (depending on wing stalling AOA) if the aircraft is unbalanced, then you will have problems.. big ones. I agree that balance is important but there are many aircraft that will bite you in the blink of an eye if you stall them in the configuration you describe. Link to comment Share on other sites More sharing options...
OZJohn Posted January 29, 2016 Share Posted January 29, 2016 This is why the FAA is pushing for all aircraft to have AOA instruments. Link to comment Share on other sites More sharing options...
diesel Posted January 29, 2016 Share Posted January 29, 2016 slow decending turn. I teach that as emergency decent. Last time a couple weeks ago in 300 hp C180. Slow down, we were probably 50kts. Etablish turn while pulling in full flap. Trim is hard back. Just enjoy the no load, low speed decent. Safe right down if required and just relax things as you roll out and carry on. Can even leave enough power on to keep jus warm. Quite safe. Chas 1 Link to comment Share on other sites More sharing options...
Keenaviator Posted January 29, 2016 Share Posted January 29, 2016 I don't regard a 48 degree turn onto final with a 1700 fpm decent rate a slow decending turn. 3 Link to comment Share on other sites More sharing options...
Aldo Posted January 29, 2016 Share Posted January 29, 2016 A balanced descending turn (I.e no wing loading) any angle of bank is pretty much impossible to stall but as Planesmaker said you need to be aware of pulling out at the completion of the turn. This is why oval circuits are much safer than rectangular but they are not taught any more unless you are in the military Aldo 1 Link to comment Share on other sites More sharing options...
Flyer Posted January 29, 2016 Share Posted January 29, 2016 BlurE, don't overthink it, just fly it. As a glider guider, I spend a lot of time at bank angles of 45 deg or more, often very close to the stall in a thermal. That is the nature of the beast. As glider pilots, we are taught good stall recognition and recovery techniques as well as good spin recognition and recovery. If you want to do some really good training, jump into a glider. There are a number of good clubs and instructors around that I can pass on if you like. 2 Link to comment Share on other sites More sharing options...
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