What's my angle, by schooner 69
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What's my angle, by schooner 69
Since the mods don't want to let us know where it went (seemed to disappear in the transfer), I still had it in an open window so cut and pasted. I was about to post on it, but will let a few more go and see if there is any interest before I do.
Old question, but maybe time to re-introduce..
Imagine you are flying a certain two-seat red and white bi-plane fitted with a gi-normous engine. You are in level flight at 4500 feet maintaining 100 knots. Throwing caution to the wind, you open the taps and commence a climb still maintaining 100 knots. You are now all trimmed out and stable in your new climb attitude when the guy-in-the-front-seat asks: “Oy! Considering the angle of attack: has it increased, decreased, or remained the same compared to the level flight condition?”
Your answer is…?
PS A certain "Coronel" has to answer last. (;>0)
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photofly
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 10:43 pm
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What's the rate of climb? And what is the thrust to weight ratio of the ginormous engine?
Without that information I don't think you can say. in a steep climb a proportion of the lift of the wing is directed rearwards, so the lift must increase to maintain the weight of the aircraft (just like it does in a steep turn when a proportion of the lift is directed sideways); that means the AoA has to increase.
On the other hand, a proportion of the considerable engine thrust is now directed downwards, decreasing the amount of weight carried by the wing, which means a reduced AoA is required.
Which effect is bigger requires more information to determine.
EDIT: actually I think the one most important piece of information that you need to specify is the percentage power used at level 100kt cruise. I think if you specify that, with some reasonable approximations the question is well posed. The second most important information would be the power-off glide ratio of the Pitts; but one could just assume, say, 8:1.
Last edited by photofly on Sat Nov 23, 2013 11:11 pm, edited 1 time in total.
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FenderManDan
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:09 pm
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The same AoA. A/S is the same.
I am ppl around 150 hr TT
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Schooner69A
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:18 pm
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Photofly: one of your answers is on the right track and all the info required is contained within the preamble. It's a good question for a training school. Great question for Friday night beer calls, also. Lots of yelling and shouting. Maybe not so many correct answers, but lots of yelling and shouting! (;>0)
John
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iflyforpie
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:21 pm
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Let's put a huge engine in that biplane and point it straight up at 100 knots... what is the angle of attack now?
For the regular climb, I think you will find that it will be less than in level flight.... because the thrust is supporting some of the weight of the aircraft.
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photofly
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:27 pm
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Schooner69A wrote:
Photofly: one of your answers is on the right track and all the info required is contained within the preamble. It's a good question for a training school. Great question for Friday night beer calls, also. Lots of yelling and shouting. Maybe not so many correct answers, but lots of yelling and shouting! (;>0)
John
iffp's corner-case analysis is helpful. If the aircraft can maintain a vertical climb at 100kts (approx 10k fpm) then the AoA has clearly decreased to zero.
For a more reasonably powered aircraft, the AoA has increased; the loss in vertical lift through tilting the lift vector rearwards is more significant than the vertical "vectored thrust" component.
Where the crossover between the two lies is an interesting question. I'll write more about that when I've drawn a few sketches.
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photofly
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:33 pm
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Another way to examine the question: did you have to trim forward or back to maintain 100kts?
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Schooner69A
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:41 pm
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iflyforpie must have been at that beer call to which I alluded! When you ascertain the angle of attack required to climb vertically, the answer drops into your lap (so to speak).
Well done, iffp!
Old question, but maybe time to re-introduce..
Imagine you are flying a certain two-seat red and white bi-plane fitted with a gi-normous engine. You are in level flight at 4500 feet maintaining 100 knots. Throwing caution to the wind, you open the taps and commence a climb still maintaining 100 knots. You are now all trimmed out and stable in your new climb attitude when the guy-in-the-front-seat asks: “Oy! Considering the angle of attack: has it increased, decreased, or remained the same compared to the level flight condition?”
Your answer is…?
PS A certain "Coronel" has to answer last. (;>0)
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photofly
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 10:43 pm
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What's the rate of climb? And what is the thrust to weight ratio of the ginormous engine?
Without that information I don't think you can say. in a steep climb a proportion of the lift of the wing is directed rearwards, so the lift must increase to maintain the weight of the aircraft (just like it does in a steep turn when a proportion of the lift is directed sideways); that means the AoA has to increase.
On the other hand, a proportion of the considerable engine thrust is now directed downwards, decreasing the amount of weight carried by the wing, which means a reduced AoA is required.
Which effect is bigger requires more information to determine.
EDIT: actually I think the one most important piece of information that you need to specify is the percentage power used at level 100kt cruise. I think if you specify that, with some reasonable approximations the question is well posed. The second most important information would be the power-off glide ratio of the Pitts; but one could just assume, say, 8:1.
Last edited by photofly on Sat Nov 23, 2013 11:11 pm, edited 1 time in total.
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FenderManDan
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:09 pm
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The same AoA. A/S is the same.
I am ppl around 150 hr TT
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Schooner69A
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:18 pm
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Photofly: one of your answers is on the right track and all the info required is contained within the preamble. It's a good question for a training school. Great question for Friday night beer calls, also. Lots of yelling and shouting. Maybe not so many correct answers, but lots of yelling and shouting! (;>0)
John
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iflyforpie
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:21 pm
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Let's put a huge engine in that biplane and point it straight up at 100 knots... what is the angle of attack now?
For the regular climb, I think you will find that it will be less than in level flight.... because the thrust is supporting some of the weight of the aircraft.
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photofly
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:27 pm
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Schooner69A wrote:
Photofly: one of your answers is on the right track and all the info required is contained within the preamble. It's a good question for a training school. Great question for Friday night beer calls, also. Lots of yelling and shouting. Maybe not so many correct answers, but lots of yelling and shouting! (;>0)
John
iffp's corner-case analysis is helpful. If the aircraft can maintain a vertical climb at 100kts (approx 10k fpm) then the AoA has clearly decreased to zero.
For a more reasonably powered aircraft, the AoA has increased; the loss in vertical lift through tilting the lift vector rearwards is more significant than the vertical "vectored thrust" component.
Where the crossover between the two lies is an interesting question. I'll write more about that when I've drawn a few sketches.
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photofly
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:33 pm
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Another way to examine the question: did you have to trim forward or back to maintain 100kts?
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Schooner69A
Post subject: Re: WHAT'S MY ANGLE?Posted: Sat Nov 23, 2013 11:41 pm
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iflyforpie must have been at that beer call to which I alluded! When you ascertain the angle of attack required to climb vertically, the answer drops into your lap (so to speak).
Well done, iffp!
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Re: What's my angle, by schooner 69
What's that Jewish guy doing in the front seat? 

Re: What's my angle, by schooner 69
A rather quick analysis:
Look at the lift equation. lift is increased, therefore one of the variables must have changed. S is constant, V is stated as constant, rho will change with altitude, but is negligible in this case, so all that is left to change is CL. Since lift is increased, the change in CL must be positive, so what do you have to do to AOA to increase the lift coefficient?
Look at the lift equation. lift is increased, therefore one of the variables must have changed. S is constant, V is stated as constant, rho will change with altitude, but is negligible in this case, so all that is left to change is CL. Since lift is increased, the change in CL must be positive, so what do you have to do to AOA to increase the lift coefficient?
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Re: What's my angle, by schooner 69
It depends...
You have to specify a couple more things. It could easily be assumed
that this aircraft - with its 'ginormous' engine - has a thrust-to-weight
ration greater than 1.
In this case, its flight path could be vertical and the AOA would be
approximately equal to the wings' angle of incidence. Although even
that requires a couple of assumption regarding airfoil and
'engine-mount-angle'.
You have to specify a couple more things. It could easily be assumed
that this aircraft - with its 'ginormous' engine - has a thrust-to-weight
ration greater than 1.
In this case, its flight path could be vertical and the AOA would be
approximately equal to the wings' angle of incidence. Although even
that requires a couple of assumption regarding airfoil and
'engine-mount-angle'.
Re: What's my angle, by schooner 69
ok...... my 2 cents.... first a couple of assumtions.... 0 AOA is the angle required by the wing to make 0 lift (this may not be the case with all airfoils of course). The lift vector of the wing is is parrellel to the normal axis of the aircraft. Again, in reality could be out slightly.. just putting these out so non factors in discussion.
One other point, I am not a mathologist, or a rocket surgeon.
My theory,,, the lift required by the wing in a constant climb (or descent for that matter) is the the cosine of the angle of climb (or descent) multiplied by the lift required fo the same conditions in level flight. For example....a 1000 pound airplane.... at 100 kts in level flight...
At 30 climb, the wing would have to produce approx 860 pounds of lift (cos 30 x 1000_
At 45.... 700
at 60..... 500
at 90.... 0
As noted it is the same for a descent. Based on that less lift will be requred, so will less AOA.
This theory by the way has been verified. With those exact climb and descent angles... with an airplane, with a HUD, and camera, that showed angle of climb (not attitude), g, and AOA. AOA does decrease in a climb or descent compared to level flight. The numbers for g required in a climb and decsent above are correct by personal observation, in aircraft with the thrust to do it. If you are in a 45 deg climb and increase the g from .7 to 1.0 the aircraft will start to pitch up. Not at a great rate, but it will.
I will leave it to the PHD's to argue the why and hows. Hmmm in a decsent.. is drag helping lift the airplane... there is something to ponder hmmm.
One other point, I am not a mathologist, or a rocket surgeon.
My theory,,, the lift required by the wing in a constant climb (or descent for that matter) is the the cosine of the angle of climb (or descent) multiplied by the lift required fo the same conditions in level flight. For example....a 1000 pound airplane.... at 100 kts in level flight...
At 30 climb, the wing would have to produce approx 860 pounds of lift (cos 30 x 1000_
At 45.... 700
at 60..... 500
at 90.... 0
As noted it is the same for a descent. Based on that less lift will be requred, so will less AOA.
This theory by the way has been verified. With those exact climb and descent angles... with an airplane, with a HUD, and camera, that showed angle of climb (not attitude), g, and AOA. AOA does decrease in a climb or descent compared to level flight. The numbers for g required in a climb and decsent above are correct by personal observation, in aircraft with the thrust to do it. If you are in a 45 deg climb and increase the g from .7 to 1.0 the aircraft will start to pitch up. Not at a great rate, but it will.
I will leave it to the PHD's to argue the why and hows. Hmmm in a decsent.. is drag helping lift the airplane... there is something to ponder hmmm.
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Re: What's my angle, by schooner 69
When I did this problem in my head... for assuming a sustained vertical speed of 100 knots... I just took the aircraft as a system and set it to a zero-lift AOA. The wings might be producing some lift, or negative lift, or nothing at all depending on what the fuselage and tail contribute to the system. For thrust lines.... you will have some drift if the thrust is off-axis... so there will have to be some positive or negative lift, and perhaps some yaw introduced to get a pure vertical. I don't think the finer points of that need to be explored.... if it were even possible without empirical data.
In the vertical example, we are in equilibrium and the thrust is equal to both drag and weight. Lift has no force to oppose it, so unless we are changing our attitude (going into the top portion of a loop, let's say) then it must be zero.
Let's take it down to 45 degrees and 100 knots. Thrust and drag are still opposed, and weight is straight down. Because weight is down... we can no longer use thrust to overcome all of it... just the vector that is parallel with the aircraft which is about 70% of the total weight (cos45 * wt). We need lift to take up the slack for the vertical component relative to the aircraft... which is again about 70% of the weight (sin45 * wt).
It goes without saying that in level flight, thrust lines aside, the wings are supporting 100% of the weight of the aircraft. So yes, we can see that as we climb at higher angles, the lift will actually be lower. This also makes sense because it is excess thrust that allows us to sustain a climb, not excess lift.
If the lift is lower, the airspeed the same, the density the same (we aren't climbing very far
), the wing area the same, then the only thing we are changing is Cl. Provided that we leave the flaps alone and aren't flying a swing-wing bi-plane or don't have spoilers.. the only way we can reduce Cl is by reducing angle of attack.
It is important to note that each of these examples is when the aircraft is in equilibrium. Initiating the attitude changes will require an increase in AOA, but that is all. If we kept a high AOA through the maneuver... we would continue into a loop.
In the vertical example, we are in equilibrium and the thrust is equal to both drag and weight. Lift has no force to oppose it, so unless we are changing our attitude (going into the top portion of a loop, let's say) then it must be zero.
Let's take it down to 45 degrees and 100 knots. Thrust and drag are still opposed, and weight is straight down. Because weight is down... we can no longer use thrust to overcome all of it... just the vector that is parallel with the aircraft which is about 70% of the total weight (cos45 * wt). We need lift to take up the slack for the vertical component relative to the aircraft... which is again about 70% of the weight (sin45 * wt).
It goes without saying that in level flight, thrust lines aside, the wings are supporting 100% of the weight of the aircraft. So yes, we can see that as we climb at higher angles, the lift will actually be lower. This also makes sense because it is excess thrust that allows us to sustain a climb, not excess lift.
If the lift is lower, the airspeed the same, the density the same (we aren't climbing very far

It is important to note that each of these examples is when the aircraft is in equilibrium. Initiating the attitude changes will require an increase in AOA, but that is all. If we kept a high AOA through the maneuver... we would continue into a loop.
Re: What's my angle, by schooner 69
No they're not; the rest of your analysis fails.IFFP wrote:Let's take it down to 45 degrees and 100 knots. Thrust and drag are still opposed
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Re: What's my angle, by schooner 69
Aside from minor differences in thrust line vs direction of flight... yes they are.
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Re: What's my angle, by schooner 69
It may already have been answered but here is my take on it...
You were maintaining level flight at 100 knots at x RPM. You were in equilibrium because of your steady rate of change...you're moving forward at 100 knots and up at 0 FPM. Therefore the weight must be equal to your lift and the drag must be equivalent to 100 knots of speed.
Now, you change the airplane to climb at (let's say) 20 degrees to the horizontal. But you're trimmed and maintaining forward movement (relative movement through the air) at 100 knots at x + (let's say) 500 RPM. Yes you're climbing at 0 FPM + y but since your vertical speed is constant, that means that you're climbing at a steady rate (steady rate of change). You're once again in equilibrium. You're also still at 100 knots of forward air speed. Therefore, relative to the wing (which is moving at 100 knots) you haven't changed your angle of attack at all. Your climb performance (increased coefficient of lift) is coming from your engine, which is to be expected, since you are producing more power (500 RPM more).
You were maintaining level flight at 100 knots at x RPM. You were in equilibrium because of your steady rate of change...you're moving forward at 100 knots and up at 0 FPM. Therefore the weight must be equal to your lift and the drag must be equivalent to 100 knots of speed.
Now, you change the airplane to climb at (let's say) 20 degrees to the horizontal. But you're trimmed and maintaining forward movement (relative movement through the air) at 100 knots at x + (let's say) 500 RPM. Yes you're climbing at 0 FPM + y but since your vertical speed is constant, that means that you're climbing at a steady rate (steady rate of change). You're once again in equilibrium. You're also still at 100 knots of forward air speed. Therefore, relative to the wing (which is moving at 100 knots) you haven't changed your angle of attack at all. Your climb performance (increased coefficient of lift) is coming from your engine, which is to be expected, since you are producing more power (500 RPM more).
Re: What's my angle, by schooner 69
Your AOA will decrease for 2 reasons:whoop_whoop wrote:It may already have been answered but here is my take on it...
You were maintaining level flight at 100 knots at x RPM. You were in equilibrium because of your steady rate of change...you're moving forward at 100 knots and up at 0 FPM. Therefore the weight must be equal to your lift and the drag must be equivalent to 100 knots of speed.
Now, you change the airplane to climb at (let's say) 20 degrees to the horizontal. But you're trimmed and maintaining forward movement (relative movement through the air) at 100 knots at x + (let's say) 500 RPM. Yes you're climbing at 0 FPM + y but since your vertical speed is constant, that means that you're climbing at a steady rate (steady rate of change). You're once again in equilibrium. You're also still at 100 knots of forward air speed. Therefore, relative to the wing (which is moving at 100 knots) you haven't changed your angle of attack at all. Your climb performance (increased coefficient of lift) is coming from your engine, which is to be expected, since you are producing more power (500 RPM more).
1- Less G, as mentionned by Skyhunter, to maintain a constant attitude.
2- TAS will increase as you climb, thus reducing your AOA. (100 KCAS level at 1000' is greater than AOA at 100 KCAS at 20 000')
Re: What's my angle, by schooner 69
The differences aren't minor, and you can't discount them.Aside from minor differences in thrust line vs direction of flight... yes they are.
If the thrust was always along the direction of flight, the angle of attack would be constant, all the time. Think about it. The answer to this questions hinges exactly on *how* those two angles differ.
Last edited by photofly on Mon Nov 25, 2013 4:26 pm, edited 1 time in total.
Re: What's my angle, by schooner 69
That's just complete rubbish. Keeping the same indicated airspeed means keeping the same dynamic pressure, so for level flight at both altitudes the AoA is unchanged.AuxBatOn wrote: Your AOA will decrease for 2 reasons:
...
2- TAS will increase as you climb, thus reducing your AOA. (100 KCAS level at 1000' is greater than AOA at 100 KCAS at 20 000')
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Re: What's my angle, by schooner 69
1) I don't think the diminished G force will affect the AoA. The wing is still moving through the air at 100 knots so it will still need a positive AoA (excluding vertical or near-vertical climb scenarios).AuxBatOn wrote:
Your AOA will decrease for 2 reasons:
1- Less G, as mentionned by Skyhunter, to maintain a constant attitude.
2- TAS will increase as you climb, thus reducing your AOA. (100 KCAS level at 1000' is greater than AOA at 100 KCAS at 20 000')
2) I recognize that you're right in that TAS will increase as you get higher but if you maintain 100 KCAS at 1000' and 100 KCAS at FL200, you'll still produce the same amount of lift and still fly the wing at the same AoA (yes you'll need more engine power because of the decreased density and your TAS will be higher).
Re: What's my angle, by schooner 69
I'm fascinated by this "diminished g-force" idea. Can we get the g-force to go zero? or negative? In which case ... free lift. Tell me more.
Re: What's my angle, by schooner 69
At 1G, your AOA will be greater than at 0G (for the same airspeed). Same if you pull 5G, at a given airspeed, your Stall Speed will be greater because your AOA will be greater than at 1G... (remember, critical AOA is constant for a wing form). For a symmetrical wing, if you hold 0G, and the alpha is 0, you should not be producing any lift. If you push to -1G, you will be producing the same amount of life as at 1G, but it will be in the opposite direction. So, when you fly inverted and want to maintain straight and level, you need to push to -1G. Since the lift will be perpendicular to the wing chord, but towards the bottom of the wing, it will be in the up direction.photofly wrote:I'm fascinated by this "diminished g-force" idea. Can we get the g-force to go zero? or negative? In which case ... free lift. Tell me more.
Don't believe me? I recommend you buy an airplane with an AOA indicator and try it out. It works.
Why is it less G in a climb? Let's put it to the extremes... If you want to maintain level flight, you will maintain 1G. Is you want to maintain a pure vertical climb and keep the nose pointed upwards, you will need to push 0G. Anything in the middle and you will have to maintain between 0 and 1G, depending on the climb angle (or descent angle). (cos(climb angle) is the amount of G you need to hold to maintain a certain climb angle).
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Re: What's my angle, by schooner 69
I've got a G meter sitting at home right now waiting to be installed on my airplane. If you tip it vertically, guess what? It reads zero!photofly wrote:I'm fascinated by this "diminished g-force" idea. Can we get the g-force to go zero? or negative? In which case ... free lift. Tell me more.
We typically only measure G force vertically relative to the aircraft. Lift is also only produced vertically... at least the component of the total aerodynamic force that we consider as lift.
Do you really think that the engine will be installed in such a way that the wings will have to produce the same amount of lift to keep the airplane going straight up as they would to keep the plane level?photofly wrote:The differences aren't minor, and you can't discount them.Aside from minor differences in thrust line vs direction of flight... yes they are.
If the thrust was always along the direction of flight, the angle of attack would be constant, all the time. Think about it. The answer to this questions hinges exactly on *how* those two angles differ.
Re: What's my angle, by schooner 69
Rubbish, again. The lift is absolutely not perpendicular to the wing chord; it's perpendicular to the direction of flight.<some semi-correct but not relevant stuff, and then....>
Since the lift will be perpendicular to the wing chord, but towards the bottom of the wing, it will be in the up direction.
True, for any aircraft that has enough thrust to allow the wings to perform at zero load. However most do not; the vertical component of the thrust in a typical climb is *less* than the loss of vertical lift because the wing is tilted back. The wing needs to work harder because it's lift is tilted to help overcome extra drag, and the AoA increases.Is you want to maintain a pure vertical climb and keep the nose pointed upwards, you will need to push 0G.
Re: What's my angle, by schooner 69
Just plain wrong. Lift is perpendicular to the relative airflow, by definition. Go look at your textbooks. Even "From the Ground Up" has that correct. I'm surprised you've forgotten.Lift is also only produced vertically... at least the component of the total aerodynamic force that we consider as lift.
Re: What's my angle, by schooner 69
Semantic.. I should have called it aerodynamic force. The component perpendicular to the direction of flight is the lift.. But the rest of the concept remains valid.photofly wrote:Rubbish, again. The lift is absolutely not perpendicular to the wing chord; it's perpendicular to the direction of flight.
Is you want to maintain a pure vertical climb and keep the nose pointed upwards, you will need to push 0G.
It has nothing to do with thrust. If you maintain 0G, you will be a 0 AOA and not producing any lift (if you were, you would be off of the vertical). If you were flying perfectly, and were able to maintain absolutely 0G the entire time, what would happen is that you would start a tail slide once you reach 0 KCAS. The whole time you are decelerating at 9.81 ft/m2 minus the thrust you produce. The only difference with thrust is the rate at which your speed will change. Once you are in the tail slide, you would be at 180 AOA. As you get slower, the easier it is to reach critical AOA (lift limit) with small deviation (not maintaining exactly 0G). I wish I could upload videos of my HUD recorder here and show you, but it is not a possibility.... It would show you the G, AOA, angle of climb/descent and their relationship with each other...photofly wrote: True, for any aircraft that has enough thrust to allow the wings to perform at zero load. However most do not; the vertical component of the thrust in a typical climb is *less* than the loss of vertical lift because the wing is tilted back. The wing needs to work harder because it's lift is tilted to help overcome extra drag, and the AoA increases.
Re: What's my angle, by schooner 69
AuxBatOn,
I think you're wrong. Just because you see 0 G at the same time as 0 AOA does not mean they are directly related. However, assume you are in level flight (equilibrium) and enter a climb. At the moment you enter a climb, lift>weight, the AOA is also higher and since you are no longer in equilibrium, you experience a G force.
0G, by definition, means you are accelerating towards the earth at 9.8 m/s^2. You could be entering a spin, have an AOA of 45 degrees and have 0 G because you are accelerating towards the earth at g.
Anyways, back to the original question. Assuming constant air density, in a climb, AOA will remain the same, lift=weight, thrust>drag.
Also, G-force is not actually a force. It is an apparent force (like centrifugal force) which is the result of two bodies accelerating at different rates.
I think you're wrong. Just because you see 0 G at the same time as 0 AOA does not mean they are directly related. However, assume you are in level flight (equilibrium) and enter a climb. At the moment you enter a climb, lift>weight, the AOA is also higher and since you are no longer in equilibrium, you experience a G force.
0G, by definition, means you are accelerating towards the earth at 9.8 m/s^2. You could be entering a spin, have an AOA of 45 degrees and have 0 G because you are accelerating towards the earth at g.
Anyways, back to the original question. Assuming constant air density, in a climb, AOA will remain the same, lift=weight, thrust>drag.
Also, G-force is not actually a force. It is an apparent force (like centrifugal force) which is the result of two bodies accelerating at different rates.
Last edited by Bede on Mon Nov 25, 2013 7:14 pm, edited 1 time in total.
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Re: What's my angle, by schooner 69
I think its more like this Colonel.


We can't stop here! This is BAT country!
Re: What's my angle, by schooner 69
Don't play the innocent. It's your aircraft we're arguing about. I blame you.Colonel Sanders wrote:Train wreck
DId you hear the one about the jurisprudence fetishist? He got off on a technicality.
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Re: What's my angle, by schooner 69
Another way of looking at things:
You are in level flight at 10000 feet with a 500 pound bomb on board that is going to be delivered via a sixty degree dive. (Used to be standard procedure in the old days).
After throwing out the speed brakes and remembering to reduce the power to idle, you trim the aircraft for an exact sixty degree dive angle which gives you 450 knots at release altitude. Now, compared to your angle of attack in level flight, is the angle of attack in the dive larger, smaller, or the same as before?
John
You are in level flight at 10000 feet with a 500 pound bomb on board that is going to be delivered via a sixty degree dive. (Used to be standard procedure in the old days).
After throwing out the speed brakes and remembering to reduce the power to idle, you trim the aircraft for an exact sixty degree dive angle which gives you 450 knots at release altitude. Now, compared to your angle of attack in level flight, is the angle of attack in the dive larger, smaller, or the same as before?
John
Re: What's my angle, by schooner 69
Did you miss my symmetrical wing assumption? I am talking about a constant regime, not transitory (like a spin is). In my experience (and I have flown vertically up at 0G to maintain the vertical and at 50+ AoA and at 7.5Gs) and this is what happens. When I am diving at the ground at a 45 degree angle, trying to deliver a MK-82 accurately, I need to push to 0.7Gs to maintain 45 degrees, this decreasing my AoA. You guys are talking theoratically. I am talking from experience.Bede wrote:AuxBatOn,
I think you're wrong. Just because you see 0 G at the same time as 0 AOA does not mean they are directly related. However, assume you are in level flight (equilibrium) and enter a climb. At the moment you enter a climb, lift>weight, the AOA is also higher and since you are no longer in equilibrium, you experience a G force.
0G, by definition, means you are accelerating towards the earth at 9.8 m/s^2. You could be entering a spin, have an AOA of 45 degrees and have 0 G because you are accelerating towards the earth at g.
Anyways, back to the original question. Assuming constant air density, in a climb, AOA will remain the same, lift=weight, thrust>drag.
Also, G-force is not actually a force. It is an apparent force (like centrifugal force) which is the result of two bodies accelerating at different rates.
Going for the deck at corner