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Engine Out and Enough Altitude to do a 180...

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Old 08-19-2010 | 03:44 PM
  #21  
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Originally Posted by shdw
Load Factor (G) = 1 / Cosine (Bank Angle)
that equation is simplified for a specific case and assumes that you are keeping altitude constant. in a gliding situation, (also in an ag turn) you are not keeping altitude constant so load factor will not increase that much and therefore stall speed will not increase by that much. remember stall is related to critical angle of attack only, and has nothing to do with airspeed. the only time stall is related to a particular airspeed is when you maintain a constant altitude.

when you maintain a constant altitude, lift is held constant so you reach critical angle of attack at a particular airspeed. in accelerated flight, you are producing more lift so you reach critical angle of attack at a higher airspeed. when gliding, lift remains constant so stall speed remains constant. the horizontal component of lift is bought at the expense of altitude so that you can maintain L/Dmax speed.
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Old 08-19-2010 | 03:47 PM
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Originally Posted by mshunter
Good write up, but too much to think about in the heat of the moment. 45 deg will get the average pilot close enough, agreed?
Haha what you can't hold 48.73 degrees of bank...what kind of flight instructor are you?

Yes, 45 will absolutely suffice. Though this write up isn't for 'heat of the moment' calculations. It is for calculations prior to flight by the professional pilot taking off under conditions that may not favor ditching straight ahead. More commonly, it's for glider pilots learning the speed and bank angle for soaring.


Originally Posted by N9373M
Plus there's a tick mark at 45' on the attitude indicator.
Not always. I actually can't recall an aircraft I've flown that had a mark for 45 degrees. Though I know it exists, my aircraft stink I guess.
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Old 08-19-2010 | 03:53 PM
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Originally Posted by N9373M
Plus there's a tick mark at 45' on the attitude indicator.
not every plane has those gyro-thingies.
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Old 08-19-2010 | 04:05 PM
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Originally Posted by SunDog
that equation is simplified for a specific case and assumes that you are keeping altitude constant. in a gliding situation, (also in an ag turn) you are not keeping altitude constant so load factor will not increase that much and therefore stall speed will not increase by that much. remember stall is related to critical angle of attack only, and has nothing to do with airspeed. the only time stall is related to a particular airspeed is when you maintain a constant altitude.

when you maintain a constant altitude, lift is held constant so you reach critical angle of attack at a particular airspeed. in accelerated flight, you are producing more lift so you reach critical angle of attack at a higher airspeed. when gliding, lift remains constant so stall speed remains constant. the horizontal component of lift is bought at the expense of altitude so that you can maintain L/Dmax speed.
Great explanation! That comes from experience...

We do base to final turns every day at ~ 60 degree bank....below the 2g stall speed... Nothing happens....it can be an effective way to bleed energy and manage descent rate
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Old 08-19-2010 | 04:15 PM
  #25  
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Originally Posted by SunDog
in a gliding situation, (also in an ag turn) you are not keeping altitude constant so load factor will not increase that much and therefore stall speed will not increase by that much.
This isn't actually the case, though it seems intuitively so. As long as the descent rate is constant, this formula holds true.

While flying straight and level, lift must equal weight. When banked our total lift must increase to keep vertical lift equal to weight, just like when straight and level. But wait, doesn't lift also equal weight in a steady climb or descent?

In fact it doesn't, but the difference is so minuscule (<10 pounds in a Cessna Vy climb) that it can be assumed negligible. In other words, in a Vy climb less than 10 pounds of the aircrafts weight is being held aloft by the engine thrust. The rest is by lift, thus lift and weight can be assumed equal for any typical steady flight condition.

We conclude that a steady climb or descent (glide) must also have lift equal to weight. Thus, the formula 1 / cosine bank angle holds true because our goal is still the same as level flight -- get vertical lift to equal weight.
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Old 08-19-2010 | 04:43 PM
  #26  
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Originally Posted by shdw
This isn't actually the case, though it seems intuitively so. As long as the descent rate is constant, this formula holds true.

While flying straight and level, lift must equal weight. When banked our total lift must increase to keep vertical lift equal to weight, just like when straight and level. But wait, doesn't lift also equal weight in a steady climb or descent?
descent rates in Ag turns, base to final turns, intial engine out glider turns, lazy 8s, wing overs, etc... don't have to be constant and the rule here doesn't apply.
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Old 08-19-2010 | 05:21 PM
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Originally Posted by ryan1234
descent rates in Ag turns, base to final turns, intial engine out glider turns, etc... don't have to be constant and the rule here doesn't apply.
You are grasping straws. The few seconds for the aircraft's airspeed and descent rate to stabilize during an engine out can and should be ignored.

Whether or not 1 / cosine (bank angle) accurately calculates load factor depends solely (see disclaimer) on the condition of flight: stable versus unstable. If the condition is a stable one, be it climb, level or descent, then the formula holds true. Stability, in this case, refers to a constant rate of climb/descent.

Disclaimer to avoid more straw grasping: Extreme degrees of pitch will result in a far greater quantity of weight being supported by thrust or drag. In these conditions the formula 1 / cosine (bank angle) will not accurately depict load factor.
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Old 08-20-2010 | 04:03 AM
  #28  
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Originally Posted by shdw
You are grasping straws. The few seconds for the aircraft's airspeed and descent rate to stabilize during an engine out can and should be ignored.
No straw grasping here - just pointing out that in real-life the turn may be over before it "stabilizes"
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Old 08-20-2010 | 01:02 PM
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Originally Posted by SunDog
that was my thought too. six thousand? should be able to make a full pattern, maybe twice, from that high.

at a glide ratio of 10:1 you've got 10 miles glide from 6k. even if you loose 1000' in the turn you have something like 8 miles of glide left. of course don't forget the wind when figuring it. if you have an engine failure on takeoff and do the 180, you will be landing with a tailwind.

something to consider: on a lot of airplanes L/D_max (best glide) is better with 5 to 10 degrees flaps. (about the same amount of flap as you can get will full down aileron deflection, or a little less than that) in most training that isn't taught because a little too much flap is a lot worse than a little too little flap.

I suggest that if you fly one airplane a lot get to know that airplane throughout the envelope, not just the normal profiles you use every day. go up high and see how she glides at different speeds, different weights, different flap settings. (or in the sim for the folks flying the big expensive ones)
Hi Sundog.

Ref your statement that some aircraft have a better L/Dmax with some flap out: This does not sound right. Which aircraft are you talking about and and where did you get the data?

Joe
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