Minimum Radius Turn back.

[photofly]

Given the situation, where you have a component of motion forward to start with, and have an airframe that's extremely draggy in the vertical direction, your vertical speed isn't going to increase anywhere near that much.

I think your conception of what a falling airplane falls like is conditioned by your extensive experience of how a flying airplane descends while the wings are producing lift.

I can trivially lose a thousand feet in the first three seconds of a spin - that's a rate of descent of 20,000 fpm.

When the drag on a falling body has risen to equal its weight, it will no longer accelerate and has reached its terminal velocity. A falling human body - not even slightly streamlined - has a terminal velocity of (about) 150 knots, that's about 15,000 fpm.

The terminal velocity if of a falling airplane will depend on its orientation, but in a nose down orientation even a draggy airframe will quickly accelerate massively beyond Vne - for reference, 200 knots is 20,000fpm.

I accept that when the direction of motion is a steep descent, airframe drag will provide a braking force that slows the acceleration due to gravity, but there's absolutely no issue getting an airplane to go downwards at many tens of thousands of feet per minute, once you decide to unload the wings and stop producing lift in the traditional manner.

I'm fairly sure that when an airplane actually falls, it does so like most objects, and needs a 'chute (eg. SR22) if you want the rate of descent to be slower than that of most other objects.

[photofly]

What about doing your numbers with a 60 degree bank?

Pick a g budget for the wing, and what g you're prepared to use fighting gravity, and I'll calculate the rest.

[rookiepilot]

What about doing your numbers with a 60 degree bank?

Pick a g budget for the wing, and what g you're prepared to use fighting gravity, and I'll calculate the rest.

1.4

[photofly]

1.4g is your g-budget. Now tell me how much do you want to use fighting gravity?

[rookiepilot]

1.4g is your g-budget. Now tell me how much do you want to use fighting gravity?

Lets assume I wanted a g-load of a 45 degree turn but actually bank 60 in a descending turn....with 2000 feet of altitude to give up. Possible?

[photofly]

Ok let's see....

According to my trigonometry, a coordinated turn at 60° of bank but pulling 1.4g only has a vertical component of 0.7g and a horizontal component of 1.21g.

it takes π . v / a seconds to make a 180, which using appropriate units means that at 70 knots (36m/s) that 180 is going to take 3.14 x 36 / (1.2 * 9.81) = 9.6 seconds.

Your vertical acceleration after subtracting what you're prepared to pull from the wing is 0.3g.

After 9.6 seconds you will have lost 453 feet due to gravitational acceleration and be descending at 5700 fpm.

It occurs to me that that the altitude lost doesn't take into account the energy required to maintain a steady airspeed against the drag in this gliding descending turn. A regular 45° steep gliding turn at about 70 knots requires a rate of descent of about 1200fpm to maintain airspeed, so in your 10 second turn you'll lose another 200 feet.

I conclude that your proposed 180 degree extra-steep-unload-the-wing turn manoeuvre will leave you in a nearly 6000 fpm descent, with a total altitude loss of 653 feet. You can convert that extra downward speed back into some altitude at the end of the manoeuvre, by zoom-climbing.

We can compare this with a 45° bank pulling 1.4g, which at 70 knots takes about 12 seconds to turn 180 degrees, and costs about 240' in altitude, and which leaves you in a 1200fpm descent, only.

Overall your proposed accelerated half-unloaded turn would be faster by a couple of seconds, and a few feet tighter (might be helpful in a canyon), but lose you more altitude.

On reflection I think this analysis is a bit too simple, and here's why: Your 6000 fpm descent represents 60 knots of downward speed. There's no point building up speed (albeit down instead of forward) and then being too timid to increase your g-load - airspeed is airspeed, and if you want to avoid the high drag stalled regime and get more lift from the wing to increase your rate of turn building up speed in a dive would be the way to get that airspeed. So you could arrange to be pulling a lot more than 1.4g at the end of your turn.

Here's one alternative, which is easy to analyze:

Unload the wing to dive straight down to 140 knots, then use that airspeed to pull a steady 4g descending turn. How would that work out?

Accelerating from 70 knots to 140 knots will cost you about 600 feet in altitude. A 4g turn has a horizontal acceleration of about 3.9g, and at 140 knots will take about 5.5 seconds. What would your steady rate of descent be in a 4g gliding turn? Drag scales with square of airspeed (someone can split out induced and parasite drag and work them separately if they want) but I'm going to guess that multiplying the g load from 1.4 to 4, *and* doubling your airspeed from 70 to 140 is going to boost your rate of descent by a factor of 8, or maybe 16, so your steady descent in that turn will be 10,000 or 20,000 fpm. Your 5.5 seconds of turn will cost you 1000-2000 feet, and you still have to recover to a sensible airspeed as you roll out. You can cash in your 140 knots of airspeed to get back 600 feet of altitude in a zoom climb as you slow to 70 knots (if you haven't hit the ground first), so there's that....

[TeePeeCreeper]

I’m impressed Photofly!

Maybe you could challenge this “gem” at the next calculator “bee”!

https://m.youtube.com/watch?v=LOeuDr1d4N8

[AirFrame]

I can trivially lose a thousand feet in the first three seconds of a spin - that's a rate of descent of 20,000 fpm.

When the drag on a falling body has risen to equal its weight, it will no longer accelerate and has reached its terminal velocity. A falling human body - not even slightly streamlined - has a terminal velocity of (about) 150 knots, that's about 15,000 fpm.

The terminal velocity if of a falling airplane will depend on its orientation, but in a nose down orientation even a draggy airframe will quickly accelerate massively beyond Vne - for reference, 200 knots is 20,000fpm.

Okay, I stand corrected... I hadn't done the math to figure out what 20,000fpm was in knots, that is more eye opening. However:

I'm fairly sure that when an airplane actually falls, it does so like most objects, and needs a 'chute (eg. SR22) if you want the rate of descent to be slower than that of most other objects.

By "most other objects" if you mean other objects with a similar cross section relative to the wind, then yes. But I expect in most cases that will be slower than the proverbial steel ball dropped alongside. For example, the Blanik training gliders with spoilers extended will descend nose down (vertical) and not exceed Vne (which must be somewhere below 200kt but I forget the number right now). And there's been one example of a Vari-Eze that had an engine out, the pilot elected to jettison the canopy and bail out (he had a chute) and somehow got it into a deep stall in the process. He realized the plane was descending at about the same rate as his parachute would, so he stayed on board, rode out the impact... And walked away.

I only mention these to clarify that the configuration of the aircraft relative to the wind *does* have an effect. The acceleration isn't unrestricted.