Angle of bank affects on descent rate

I’m attempting to do some maths regarding the ‘impossible turn’ (engine failure after takeoff, followed by a 180+ deg turn back to the field to land). I understand how to determine radius, rate of turn, and best glide descent rate; however, I’m unable to determine how descent rate is impacted in a turn at best glide. My only theory is that the equation for load factor might hold some clue as to the percentage increase in VSI. Is there anyway to determine the loss in magnitude of the vertical component in the turn back to the field and the associated increase in VSI (assuming an engine out best glide profile)? Is there a rule of thumb? Is there an equation? Thank you

Yes, of course. It's the square root of 1 over cosine bank angle × airspeed.

That will determine speed increase required to glide while banked, but ...

Turn radius favors lower airspeed. You want to bank but not increase pitch beyond that which maintains proper airspeed for that bank angle. With no power, the aircraft will start a spiral descent.

Try these at altitude. Starting with a 30 degree bank. See which combination of airspeed and bank brings you around with the least loss of altitude.

VSI is not as important as bank angle and airspeed indicator. That's what you need to be focusing on.

But, please, beware. Stalling attempting the "impossible" turn at low altitude may not be survivable. Controlled flight wings level gives one the best chance of survival.

Also, turning from a head wind back into a tailwind (assuming the turn is attempted on take-off) means the ground speed increases 2x the windspeed. If a sudden tail wind gust does not stall you, the ground speed on landing can be higher.

VSI can be calculated by rate of (180 degree turn) divided by altitude loss as a statistic. Once this is determined from flight data, it could serve as a reference while a turn is in progress.

• Thank you for your comment Robert. But, I believe you have provided the equation for increase in stall speed for a level flight turn. My goal is to find the altitude loss, mathematically, in a turn. And then, given the tool of determining rate of turn, determine VSI. The key I think - is determining the resultant loss of vertical component with an AoB. I don’t think my theory of load factor is correct - If you use the extreme of 89 degs AoB as an example Commented May 14 at 14:25
• @RyanBW I would start with Vbg and glide angle at Vbg, which is arc tan glide ratio. In a bank, Vbg increases by square root of 1/cos theta. What one must do is calculate (for gliding) is not only the increase in speed, but also the increase in angle of descent (in a spiral) based on the increase in drag, which is compensated by the increase in vertical gravity component. But, the real key to successfully handing a low level engine out is to immediately initiate the turn at Vbg and best bank angle. Practice makes perfect. Commented May 14 at 14:44
• Alright, I think I get it. I don't think I get this performance though; can you provide additional information regarding the increase in drag? I ask because I get the following - Cessna152 glide performance (1.5 NM per 1,000'), results in an angle of 6.26 deg. 45 deg angle of bank results in an increase of 41% - 8.83 degs. Assuming TAS of 60 knots, radius of turn is 320', and 180 deg turn is 1,000'. Flattening out this half circle, turn back to the field, you lose 155' in 10 secs (VSI 930'/Min). Does this make sense? When I'm practicing the steep spiral, I lose about 700' / turn. Thanks again! Commented May 14 at 18:03
• 700 feet/turn is pretty close to what I got in a 172, which is 350 feet per half turn. Work with IAS, remember at 45 degrees stall speed rises from 50 to 60 knots. Add in more altitude for additional maneuvering as a pure 180 does not align with the departing runway. I wouldn't do a 45 any slower than 70. It'll be close between a 30 at 63 knots and a 45 at 70 knots. You can also try various flap settings. Commented May 14 at 20:31