What's my angle, by schooner 69

[AuxBatOn]

I see that this image also shows it's impossible to fly vertical down at that speed. Which makes sense, since the speed would run up unless the propeller started to act as an air brake, or there would be reverse thrust. Which also shows that the graph posted by AuxBatOn is incorrect, especially for the descend area.

The 2 graphs show 2 different relationships. I know that mine is correct from experience, having been straight up and straight down and anywhere in between.

Photofly: here's how I apply the small angle approximation:

L cos(β) + T sin(α+β) - D sin(β) - W =F_v = 0
L cos(β) + T (sin(α)cos(β)+cos(α)sin(β)) - D sin(β) - W = F_v = 0

Since α is small, sin(α)~0 and cos(α)~1

So, L cos(β) + T (0*cos(β)+1*sin(β)) - D sin(β) - W = F_v = 0
L cos(β) + T sin(β) - D sin(β) - W = 0

[Colonel Sanders]

Since α is small, sin(α)~0 and cos(α)~1

That's a superb example of the difference between a
mathematician and a physicist. Meanwhile, an engineer
draws a straight line between the two points

[photofly]

Since α is small, sin(α)~0 and cos(α)~1

I would call that the zero angle approximation. it might be interesting to try a real small angle approximation:

sin α = α

and see where that takes you.
and optionally:

cos α = 1-α^2/2

(α in radians, of course, in both cases)
It's important because at slow airspeeds, α is a non-zero quantity of interest, even if small.

EDIT: oops - missed a factor of 2.

[Skyhunter]

If descending straight down, at constant speed, drag is actually supplying all the lift. Exact opposite of straight up where thrust provides all the lift.

Photo fly, have neither the time nor the desire to go any farther with this, but I will say, in the aircraft, the actually effect does not correspond to your diagram. Even a slower speeds, such as approach configuration at 8.1 deg AOA. I am not a mathematician or an engineer, but I do not a bad job at flying an airplane and noticing what goes on. From experience, you are wrong. You guys argue the math or graphs all you want. From repeated experience, your results do correlate with the real world.

I am officially out of this one

[photofly]

If descending straight down, at constant speed, drag is actually supplying all the lift. Exact opposite of straight up where thrust provides all the lift.

I don't think you mean lift: if the flight path is vertically downwards any lift would be acting horizontally; that's the definition of lift - perpendicular to the direction of flight. But if you mean drag is opposing the weight of the aircraft, then, yes, of course that's true.

Maybe you could be a bit more specific about what you notice in an aircraft that differs? A blanket "you are wrong" isn't very enlightening.

Here's a force diagram for a (nearly) vertical descent, at 100kts still. The thrust is reversed (or equivalently a lot of drag is added) and the AoA is close to zero. It doesn't strike me as unphysical or even particularly controversial:

Screen Shot 2013-12-01 at 1.08.39 PM.png (179.62 KiB) Viewed 1541 times

Could you explain what you don't think's right?

[photofly]

Here's a proof that AoA is stationary (maximum, minimum, or point of inflexion) with respect to angle of climb/descent when, for a given airspeed, thrust is horizontal. It requires no knowledge of how the lift or drag vary with AoA other than that they both be continuous differentiable functions. It's therefore true for any lift/drag curve at all.

As before, the aircraft is rigged so the thrust line T is parallel to the zero-AoA direction. The direction of motion is shown right to left, parallel with the drag D, and β is the angle between the vertical (W) and the lift vector L, i.e. the angle of climb/descent. Without loss of generality (i.e. by choice of units) let W = 1.

See the following figure:

Screen Shot 2013-12-02 at 3.20.00 PM.png (45.16 KiB) Viewed 1447 times

We wish to prove that if α = -β then dα/dβ = 0.

Applying elementary trigonometry to the diagram:

cos β = L + (sin β +D).tan α -- eq.1

Differentiate both sides w.r.t β

-sin β = dL/dβ + tan α.(cos β + dD/dβ) - ( (sin β+D)/cos^2 α ).(dα/dβ) --eq.2

D and L are functions primarily of α but since a value of α fixes a value of β we can also consider them as functions of β.

Using the change of variable rule for differentiation:
dD/dβ = (dD/dα).(dα/dβ) and
dL/dβ = (dL/dα).(dα/dβ)

-sin β = (dL/dα).(dα/dβ) + tan α.cos β + tan α.(dD/dα).(dα/dβ) - ( (sin β+D)/cos^2 α ).(dα/dβ) --eq.3

Therefore:

-sin β - tan α.cos β = (dα/dβ).[ (dL/dα) + tan α.(dD/dα) - ( (sin β+D)/cos^2 α) ] --eq.4

now set α = -β. Then -sin β = sin α, cos β = cos α and the LHS is

sin α - tan α.cos α = sin α - sin α = 0, identically

therefore from eq.4:

(dα/dβ ) . [ some function of D(β) and L(β) ] = 0 --eq.5

If two quantities P and Q satisfy P.Q = 0 then either P = 0 or Q = 0 (or both).
In this case the term in the [ ] in eq.5 is not identically zero for all β, therefore the other part must be.

dα/dβ = 0.

QED.

[AuxBatOn]

So, where did you get those equations from?

[photofly]

W,L,T and D and the angles α and β are defined in the question; the other dimensions in the diagram are derived from them by basic trigonometry and addition, and equation 1 is read straight from the diagram.

Everything else is elementary calculus and algebra, applied to one line to give the next.

[digits_]

But what does that teach us ? It's like saying to an olympic swimmer "you might have finished first, you might have finished last, or something in between".

My algebraic knowledge is a bit rusty, but in eq5, you say the function is not 0 for all beta. It can still be 0 for some values of beta, which means that dα/dβ could be different from zero for certain values. I think there is a piece missing that says that dα/dβ is a continuous function.

[photofly]

But what does that teach us ? It's like saying to an olympic swimmer "you might have finished first, you might have finished last, or something in between".

It's another piece of the puzzle. We were all pretty much agreed that the AoA goes down in both a steep descent and a climb; just arguing over whether the maximum was at level flight or level thrust. It helps me be a bit more clear that it's at level thrust. (For an "ideal" airplane yada yada yada.) But the maths is valid, so there could be cases where it was a local minimum. I think we would actually have that in real life if you were able to fly on the back side of the LD polar, in the stalled region.

My algebraic knowledge is a bit rusty, but in eq5, you say the function is not 0 for all beta. It can still be 0 for some values of beta, which means that dα/dβ could be different from zero for certain values. I think there is a piece missing that says that dα/dβ is a continuous function.

Two points there: I do rely on dα/dβ being finite-valued; that's clearly true at least piece-wise on a real L/D curve but I should have been careful enough to say so.

I was also imprecise to say "D and L are functions primarily of α but since a value of α fixes a value of β we can also consider them as functions of β." - this diagram shows you clearly that α is not a (single-valued) function of β if you include the stalled region: same flight path (nearly horizontal, by coincidence), same airspeed, two possible different values of thrust and two possible different angles of attack:

Screen Shot 2013-12-04 at 11.59.04 AM.png (152.97 KiB) Viewed 1358 times

I do require that L and D both be smoothly parametrized by β so that dD/dβ and dL/dβ exist. Also generally it looks like things are going to get interesting whenever dL/dD = -tan β (can you see why?) and I think that is true for some point near the top of Von MIses' "real" LD polar.

As for "the bit in the [ ]": you could investigate when it becomes zero, and see if it leads to anything interesting. It might tie up with the singularities previously discussed.