For people who enjoy formulae and numbers - a puzzle

[photofly]

An airplane which normally cruises at 120 knots descends at 600 feet per minute at an identical attitude when in a power-off glide at 60 knots.

What is the effective angle of attack* during cruise at 120 knots?


* Effective AoA is defined to be zero when the coefficient of lift is zero.

[Adam Oke]

$500



***Wrong place for the AvCan 500 joke?***

[Gravol]

It's Sunday morning dude

[Strega]

1.4

[digits_]

1/30 radians or 1.9 degrees nose up.

[Rookie50]

I'm not that smart, but admire those that are.

[photofly]

1/30 radians or 1.9 degrees nose up.

I believe this is correct. Do you want to explain how you got that?

[digits_]

Sure.

Some assumptions:
- Cl is linear in respect to AoA. So Cl = k * AoA with k a constant for our AoA range.
- Since we are looking for effective AoA, we replace the actual plane by a plane where the AoA = degrees pitch up of the airplane (don't have definite proof for this, but otherwise I don't think it can be solved with the data provided)

Situation 1 (in cruise)
Balance of forces in the vertical, perpendicular to flight path:
W = 1/2 r S V1^2 Cl(a) cos(a) with a the AoA and pitch up attitude in this case

Situation 2 (descent)
The flight path of the descent is going down with an angle b
Balance of forces perpendicular to flight path (note, this is not the vertical):

W cos(b) = 1/2 r S V2^2 Cl(a+b)

We know that V2 = V1 / 2

To calculate b:
We descend 600 fpm, so 600 feet per minute. In one minute we fly 1 NM (approximately). 1 NM = 6076 feet (go metric system !). That tells us that
tan(b) = 600 / 6076 = 1/10 => b is approximately 1/10 and cos(b) is approximately 1

Solving situation 1 and 2 for W gives us:
Cl(a) = Cl(a+b) / ( 4 cos(b) )

For a linear Cl we get:

k a = (k a + k b) / ( 4 cos(b) )
4a = a + b
a = b/3

a = 1/30

[5x5]

For anyone else like me, perhaps this link will help you understand the above post.

http://www.kli.org

[iflyforpie]

For anyone else like me, perhaps this link will help you understand the above post.

http://www.kli.org

Qapla'

Personally... I don't think there is enough information and too many assumptions are being made. The effects of prop wash, thrust angle, and thrust-drag coupling are being ignored.

[Strega]

To calculate b:
We descend 600 fpm, so 600 feet per minute. In one minute we fly 1 NM (approximately). 1 NM = 6076 feet (go metric system !). That tells us that
tan(b) = 600 / 6076 = 1/10 => b is approximately 1/10 and cos(b) is approximately 1

The 60kts is not horizontal. it is @ angle b. 600 fpm is a component of this.

[digits_]

To calculate b:
We descend 600 fpm, so 600 feet per minute. In one minute we fly 1 NM (approximately). 1 NM = 6076 feet (go metric system !). That tells us that
tan(b) = 600 / 6076 = 1/10 => b is approximately 1/10 and cos(b) is approximately 1

The 60kts is not horizontal. it is @ angle b. 600 fpm is a component of this.

Yes, hence the "approximately". That angle is so small you can just ignore the difference. But you are right.

You can calculate it exactly by:

sin(b) = 600 / 6076 => b is 0.099 rad, so still basically 1/10

[photofly]

Sure.

Some assumptions:....

That's basically it. You can work it out "in the bath", as follows:

600fpm at 60kts is a 1-in-10 glide descent (about average for a Cessna single), which works out at 6°. By comparison, anyone familiar with an ILS at 3° knows that's 1-in-20.

If you keep the same attitude and change from level flight to a 6° descent your AoA just went up by ... 6°

Halve your airspeed from 120 to 60kts and your lift just dropped to one quarter. Since the lift in regular flight is equal to (or very close to) the weight, you need to increase your effective AoA by a factor of 4.

So your level flight AoA increases by a factor of 4 when you add 6 to it. It must have been 2° - since 2 + 6 = 4 x 2

The "in the bath" answer is 2°.


If you want to be more precise, 600fpm at 60kts is about 5.66°, not 6°.

Also the lift isn't quite the same in the descent. Because the lift vector is tilted forward slightly, you need more lift (just like you need more lift in a turn when it's tilted to the side. But similarly to how you don't need much extra lift in a 6° banked turn you don't need a lot of extra lift in a 6° descent either. And there's now a (small) contribution from the vertical component of the drag. But those are minor effects.

[Posthumane]

Also the lift isn't quite the same in the descent. Because the lift vector is tilted forward slightly, you need more lift (just like you need more lift in a turn when it's tilted to the side. But similarly to how you don't need much extra lift in a 6° banked turn you don't need a lot of extra lift in a 6° descent either. And there's now a (small) contribution from the vertical component of the drag. But those are minor effects.

I don't think you can assume that the lift will be larger in a descent due to the tilting of the lift vector, as the vertical component of drag is not necessarily negligible. The corner case of a 90 degree (vertical) descent has drag countering gravity and zero lift being produced by the wings.

In the 6 degree descent case the vertical component of lift is cos(6*) = 0.9945, so a reduction of 0.55%.
The vertical component of drag is sin(6*) = .1045 or about 10% of the overall drag. If the L/D is approximately 10:1 the vertical component of drag is just over 1% of the lift, greater than the reduction of the vertical component of lift. Therefore, lift created by the wings is slightly lower in the descent compared to level flight. The same is true for a climb where the vertical component of thrust ends up being greater than the reduction of the vertical component of lift, with the corner case being a plane climbing entirely on its prop (or a helicopter).

[photofly]

You're right: in the glide descent the lift will decrease very slightly, not increase. The weight is opposed by the (vector) sum of lift and drag. There's no thrust so the other three forces sum to zero.

So the lift and drag form the two sides of a rectangle whose diagonal is the overall aerodynamic force that opposes the weight. You are correct to say the L/D is 10:1 because the descent is 1-in-10 - it's the same 5.7° (or 6°) angle. So the lift will be 1/sqrt(1.01) of what it was in level flight, or 99.5%.

So the error in the estmated AoA caused by assuming the lift stays the same and ignoring the drag is only about one half of one percent.

[5x5]

Personally... I don't think there is enough information and too many assumptions are being made. The effects of prop wash, thrust angle, and thrust-drag coupling are being ignored.

Agreed. But the most glaring omission has to be the lack of accounting for variable tailwinds.

[Adam Oke]

Agreed. But the most glaring omission has to be the lack of accounting for variable tailwinds.

Hehehe