For people who enjoy formulae and numbers  a puzzle
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For people who enjoy formulae and numbers  a puzzle
An airplane which normally cruises at 120 knots descends at 600 feet per minute at an identical attitude when in a poweroff 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.
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.
Kirk: This is a dangerous mission. Likely, one of us will die. The landing party will be me, Spock, McCoy, and Ensign Ricky.
Ensign Ricky: Aw, crap.
Ensign Ricky: Aw, crap.
Re: For people who enjoy formulae and numbers  a puzzle
$500
***Wrong place for the AvCan 500 joke?***
***Wrong place for the AvCan 500 joke?***
Air to Ground Chemical Transfer Technician turned 3 Bar Switch Flicker and Flap Operator
Re: For people who enjoy formulae and numbers  a puzzle
It's Sunday morning dude
Re: For people who enjoy formulae and numbers  a puzzle
1.4
Rule books are paper  they will not cushion a sudden meeting of stone and metal.
— Ernest K. Gann, 'Fate is the Hunter.
— Ernest K. Gann, 'Fate is the Hunter.
Re: For people who enjoy formulae and numbers  a puzzle
1/30 radians or 1.9 degrees nose up.
Re: For people who enjoy formulae and numbers  a puzzle
I'm not that smart, but admire those that are.

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Re: For people who enjoy formulae and numbers  a puzzle
I believe this is correct. Do you want to explain how you got that?digits_ wrote:1/30 radians or 1.9 degrees nose up.
Kirk: This is a dangerous mission. Likely, one of us will die. The landing party will be me, Spock, McCoy, and Ensign Ricky.
Ensign Ricky: Aw, crap.
Ensign Ricky: Aw, crap.
Re: For people who enjoy formulae and numbers  a puzzle
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
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
Re: For people who enjoy formulae and numbers  a puzzle
For anyone else like me, perhaps this link will help you understand the above post.
http://www.kli.org
http://www.kli.org
Being stupid around airplanes is a capital offence and nature is a hanging judge!
“It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so.”
Mark Twain
“It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so.”
Mark Twain
 iflyforpie
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Re: For people who enjoy formulae and numbers  a puzzle
Qapla'5x5 wrote:For anyone else like me, perhaps this link will help you understand the above post.
http://www.kli.org
Personally... I don't think there is enough information and too many assumptions are being made. The effects of prop wash, thrust angle, and thrustdrag coupling are being ignored.
Geez did I say that....? Or just think it....?
Re: For people who enjoy formulae and numbers  a puzzle
The 60kts is not horizontal. it is @ angle b. 600 fpm is a component of this.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
Rule books are paper  they will not cushion a sudden meeting of stone and metal.
— Ernest K. Gann, 'Fate is the Hunter.
— Ernest K. Gann, 'Fate is the Hunter.
Re: For people who enjoy formulae and numbers  a puzzle
Yes, hence the "approximately". That angle is so small you can just ignore the difference. But you are right.Strega wrote:The 60kts is not horizontal. it is @ angle b. 600 fpm is a component of this.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
You can calculate it exactly by:
sin(b) = 600 / 6076 => b is 0.099 rad, so still basically 1/10

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Re: For people who enjoy formulae and numbers  a puzzle
That's basically it. You can work it out "in the bath", as follows:digits_ wrote:Sure.
Some assumptions:....
600fpm at 60kts is a 1in10 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 1in20.
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.
Kirk: This is a dangerous mission. Likely, one of us will die. The landing party will be me, Spock, McCoy, and Ensign Ricky.
Ensign Ricky: Aw, crap.
Ensign Ricky: Aw, crap.

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Re: For people who enjoy formulae and numbers  a puzzle
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.photofly wrote: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.
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).
"People who say it cannot be done should not interrupt those who are doing it." George Bernard Shaw

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Re: For people who enjoy formulae and numbers  a puzzle
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 1in10  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.
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 1in10  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.
Kirk: This is a dangerous mission. Likely, one of us will die. The landing party will be me, Spock, McCoy, and Ensign Ricky.
Ensign Ricky: Aw, crap.
Ensign Ricky: Aw, crap.
Re: For people who enjoy formulae and numbers  a puzzle
Agreed. But the most glaring omission has to be the lack of accounting for variable tailwinds.iflyforpie wrote:Personally... I don't think there is enough information and too many assumptions are being made. The effects of prop wash, thrust angle, and thrustdrag coupling are being ignored.
Being stupid around airplanes is a capital offence and nature is a hanging judge!
“It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so.”
Mark Twain
“It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so.”
Mark Twain
Re: For people who enjoy formulae and numbers  a puzzle
Hehehe5x5 wrote:Agreed. But the most glaring omission has to be the lack of accounting for variable tailwinds.
Air to Ground Chemical Transfer Technician turned 3 Bar Switch Flicker and Flap Operator