Originally Posted by
HarryC
The sims are really games at heart and don't want to make it as difficult as it really is. The model flight sims do have good uses but they also have their limits, I recall one chap in my club telling me he could do rolling circles, turned out he meant on his sim, and he could, I saw him do it, yet in the real world he couldn't fly a level turn never mind a rolling circle!
No plane can do a "proper" turn at 90 degrees of bank, it is physically impossible. By "proper" turn I mean a balanced level turn, not an aerobatic knife edge manouevre. Any plane can do a balanced turn at 90 degrees of bank but it won't be a level turn - a level turn means maintaining height. It must lose height. The wing is what holds a plane up. When banked the lift force is angled away from vertical, that's what makes the plane turn, but there is not enough vertical component so the total lift must be increased. That's why you pull back on elevator, to increase the angle of attack and get more lift, it's also why the stall speed rises when you turn. But at 90 degrees of bank, all the lift force is pointing sideways, there is no vertical component, so no wing lift to hold the plane up. It must either lose height, or the pilot goes knife edge to get lift from the fuz, but for those of you who have no experience of full-size aerobatics, believe me it is most uncomfortable for those not used to it, and the drag rise is a heavy penalty. It is potentially dangerous since it is up elevator and lots of rudder, again the controls for a spin/flick, so should only be done by experienced pilots with a suitable model and safe height.
I totally agree with this post and your previous one.
I will preface this explanation by saying that we are NOT considering use of rudder for any degree of knife edge input, I'll get to that later,
No plane can do a proper 90 degree 'LEVEL' turn by increasing elevator alone.
The math is actually pretty easy to prove this.
To maintain level flight lift must equal weight. IE 1 G of vertical lift must be produced (opposite to gravity)
Wings develop lift perpendicular to their span. IE looking at a plane from behind, the lift is directly up if the wings are level.
As you bank the wings, that lift tilts over to remain perpendicular to the wing.
You gain a turning force but lose vertical component. If you just bank and do nothing else with the controls the plane will descend.
If you increase elevator your lift force increases. At 60 degrees bank you need 2 times the lift (2G) as this will give you 1G of vertical component,
Simple trigonometry. Cosine of 60 is 0.5. 2 x 0.5 = 1
To calculate how much extra lift you need to sustain a level turn at any bank angle use this forumula.
1/(cosine of bank angle)
eg 30 degrees bank = 1.15G
45 degrees bank = 1,41G
60 degrees bank = 2G
75 degrees bank = 3.8G
85 degrees bank = 11.5G
89 degrees bank = 57 G
90 degrees bank = infinity G
I'm pretty sure no plane can pull 57 G let alone infinity G.
a sustained altitude 90 degree bank level turn is impossible using the lift from the wing alone.
Enter rudder and knife-edge flight.. At 90 degrees bank your wings are doing nothing to support the weight of the aircraft.
You can apply rudder which then gives the fuselage an angle of attack to the airflow and there is a lifting force generated by the body.
Depending on power and aircraft design you may be able to sustain level knife edge flight indefinitely.
Getting back to normal bank (no knife edge input) we can calculate stall speed increase quite easily too.
New stall speed at any bank is 1G stall speed x (square root of G)
so for a 60 degree bank the new stall speed will be (square root of 2) = 1.4.
IE 40% more than in level flight.
Increasing lift to produce extra G has the penalty of producing extra drag.
This is usually the limiting factor in light aircraft. Most cessna's and small piston aircraft run out of puff around 2.5G sustained turn.
They can pull more G momentarily but not sustain it due to insufficient power.
now the final part. Structural limit.
Example - (real numbers for a Robin 2160)
A typical light aerobatic plane such as a Robin 2160 has a flight g limit of + 6 and - 3 G.
this means it can pull 6 G every day without damage.
if we know the stall speed we can calculate the maximum manoeuvre speed very easily.
assume stall speed is 55kts.
Stall speed at max G (6 in this example) is:
55kts x (square root of 6) = 135 kts.
this is the maximum manoeuvre speed for this aircraft.
what does it actually mean?
Flying less than 135 kts you cannot pull 6G as the plane will stall before achieving 6G.
IE flying less than 135 kts it is impossible to break the aircraft in flight.
Flying above 135 kts it is possible to exceed 6G if you apply full control deflection. IE you can break the aircraft
in a light aircraft this number is where the green arc on your airspeed indicator become orange (amber)
Flying in the green speed range you can use full control input.
If your speed is in the amber range you must use caution and limit control inputs accordingly.
I doubt most of us know the structural limit of our models.
Also no way to know exact airspeed either unless you use telemetry.
Personally I have dived all my RC planes at full power and then used full elevator to pull out and not one has ever broken.
The principles still apply.