Aces High Bulletin Board
General Forums => Aircraft and Vehicles => Topic started by: titanic3 on October 28, 2011, 11:30:18 AM
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Just a question that popped into my head the other day.
1. If a plane had engines capable of exceeding 1:1 thrust to weight ratio, does that mean they can simply point their nose 90 degrees up and climb without fear of stalling out? I would imagine that gravity and less dense air at higher altitudes would eventually cause the plane to stall out. But assume that the atmospheric conditions stays the same, would it still be able to go up and up?
2. Were there any planes during WWII that could do this? Besides the Me-163, because it is a manned rocket, not a real plane.
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A five ounce bird could never carry a three pound coconut! It's a simple question of weight ratios! ;)
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Just a question that popped into my head the other day.
1. If a plane had engines capable of exceeding 1:1 thrust to weight ratio, does that mean they can simply point their nose 90 degrees up and climb without fear of stalling out? I would imagine that gravity and less dense air at higher altitudes would eventually cause the plane to stall out. But assume that the atmospheric conditions stays the same, would it still be able to go up and up?
2. Were there any planes during WWII that could do this? Besides the Me-163, because it is a manned rocket, not a real plane.
1. Yes theoretically they could point their nose up and accelerate indefinitely except for the drop of engine performance at altitude. So if the atmospheric conditions stayed the same, yes, it could go on forever.
2. No WWII plane could. Thrust to weight ratio of 1:1 is pretty impressive and not all modern fighters have it.
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2. Were there any planes during WWII that could do this? Besides the Me-163, because it is a manned rocket, not a real plane.
Me163 cannot do that either. Best sustained it can do is, IIRC, 70 degrees.
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1. Yes theoretically they could point their nose up and accelerate indefinitely except for the drop of engine performance at altitude. So if the atmospheric conditions stayed the same, yes, it could go on forever.
2. No WWII plane could. Thrust to weight ratio of 1:1 is pretty impressive and not all modern fighters have it.
Me163 cannot do that either. Best sustained it can do is, IIRC, 70 degrees.
:aok Thanks for clearing it up guys.
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You don't need lift if your thrust overcomes gravity.
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You don't need lift if your thrust overcomes gravity.
This is slowly coming to me but...does that mean that:
1. Hot Air Balloons and blimps are able to "fly" because their LIFT overcomes gravity, but has no self-thrust, relying on wind/currents? (Or using small propellers/engines)
2. Rockets/Missiles have no lift, but has enough THRUST to "fly".
3. Airplanes produce both LIFT AND THRUST. Being the heaviest of them all.
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A fixed wing aircraft going straight up is not producing any lift. It is relying just as much on pure thrust as a missile.
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Titanic it's only as complicated as it seems because that's the engineering challenge presented by beating gravity - making what you want from what you have, economically. That means we can't just strap a Saturn V engine to anything. It's why modern private planes are so light, IE efficient in lift compared to EG warbirds.
It always comes down to the net forces acting on any body. To fly you have to somehow beat gravity. Whether by throwing stuff out the back the way you'd throw a rock from a small raft on still water, or by moving thru air fast enough that you have a net upwards force from the air moving against the surfaces you designed to produce just that effect.
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1. Yes theoretically they could point their nose up and accelerate indefinitely except for the drop of engine performance at altitude. So if the atmospheric conditions stayed the same, yes, it could go on forever.
If it is a prop plane or helicopter, the thrust will drop off with vertical speed regardless of altitude.
Simple trivia/ physics question.
When a plane is in a steady state climb does it need more lift than if it is flying level?
HiTech
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No. It paradoxically needs less lift since the thrust supports part of the weight.
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The first modern jet to have a T/W > 1 was the F-15 Eagle.
shdo
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So it is unnecessary to push to zero G on the accelerometer in a vertical climb to reduce lift induced drag on the wings to nothing?
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So it is unnecessary to push to zero G on the accelerometer in a vertical climb to reduce lift induced drag on the wings to nothing?
You're describing a zoom climb, not a steady state climb. Reducing drag is good but you need lift for a steady state climb.
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Let me rephrase the question then: will I zoom higher by pushing to Zero G or by leaving in on 1?
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Let me rephrase the question then: will I zoom higher by pushing to Zero G or by leaving in on 1?
In a pure vertical climb your G-meter will read zero. The G-meter reads loads normal to the aircraft centerline. If you're reading 1G on your G-meter in a vertical climb you are actually doing a loop.
In a pure vertical zoom you need to push (and trim) to a slightly negative AoA to counteract lift generated by the typical wing's non-symmetrical airfoil shape. In AH, you also have to deal with CT which is automatically set based only on speed with the assumption it's trimming for level flight. In a vertical climb, CT will actually begin to trim your nose "up" as you decelerate causing you to tend to arc over on your back unless you push/trim nose down.
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This is slowly coming to me but...does that mean that:
1. Hot Air Balloons and blimps are able to "fly" because their LIFT overcomes gravity, but has no self-thrust, relying on wind/currents? (Or using small propellers/engines)
2. Rockets/Missiles have no lift, but has enough THRUST to "fly".
3. Airplanes produce both LIFT AND THRUST. Being the heaviest of them all.
Hot air balloons and blimps (and dirigibles) rise because they are lighter than the surrounding air, hence the terms "lighter-than-air" and "airship" used to describe them. The force that causes them to rise is more accurately described as buoyancy vice "lift" (think submarines here). Just to be clear though, blimps and dirigibles can generate lift as they move forward through the air. The shape of the envelope acts as a wing and can be used to climb, descend, and turn without changing the overall buoyancy of the craft just as submarines do.
A rocket or missile launched vertically from the ground must have thrust greater than its weight. A horizonally flying missile/rocket (think air-to-air or air-to-ground) does generate lift otherwise they would act just like a bullet but just hit the ground at a higher speed. These have small control surfaces (or thrust vectoring) to control AoA and they generate most of their lift by airflow over the body of the missile itself. This lift allows the missile to fly giving it greater range and is used to guide the missile to its target. Even the Saturn V generated lift this way as it arced over after launch.
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Let me rephrase the question then: will I zoom higher by pushing to Zero G or by leaving in on 1?
In a less than vertical zoom climb you'll go higher at 1G because you'll end up in a steady state climb and the lift will add more to your energy gain than the drag will subtract. It should be easy enough for you to test different angles and note the altitude gain vs a pure vertical 0G zoom climb. If you fly at 0G at less than vertical your nose will come down early in an outside half loop.
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Hot air balloons and blimps (and dirigibles) rise because they are lighter than the surrounding air, hence the terms "lighter-than-air" and "airship" used to describe them. The force that causes them to rise is more accurately described as buoyancy vice "lift" (think submarines here). Just to be clear though, blimps and dirigibles can generate lift as they move forward through the air. The shape of the envelope acts as a wing and can be used to climb, descend, and turn without changing the overall buoyancy of the craft just as submarines do.
A rocket or missile launched vertically from the ground must have thrust greater than its weight. A horizonally flying missile/rocket (think air-to-air or air-to-ground) does generate lift otherwise they would act just like a bullet but just hit the ground at a higher speed. These have small control surfaces (or thrust vectoring) to control AoA and they generate most of their lift by airflow over the body of the missile itself. This lift allows the missile to fly giving it greater range and is used to guide the missile to its target. Even the Saturn V generated lift this way as it arced over after launch.
Ah, thanks for the info. But then how come aircraft designers won't just built a plane based on a rocket/missile? I know they tried that with the F-104(I think?) Starfighter, but that was back during the early Cold War, why not try it again but with much lighter materials, much stronger engines, and maybe even turn it into a drone?
I remember reading somewhere that the AIM-9 could pull more Gs than a pilot ever could, so why not make a drone based on the AIM9, just bigger and able to carry its own missile.
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Ah, thanks for the info. But then how come aircraft designers won't just built a plane based on a rocket/missile? I know they tried that with the F-104(I think?) Starfighter, but that was back during the early Cold War, why not try it again but with much lighter materials, much stronger engines, and maybe even turn it into a drone?
I remember reading somewhere that the AIM-9 could pull more Gs than a pilot ever could, so why not make a drone based on the AIM9, just bigger and able to carry its own missile.
To be able to use the missile body for lift you need tremendous speed which means lots of thrust which means lots of fuel or very, very short duration. For most missiles, more than half the weight is fuel and they only burn from a matter of seconds for the AIM-9 up to a couple of minutes for the AIM-54. Yes, they can pull 20 or 30G's but it's impractical to cruise around at Mach 3 to M5. Also, 20 or 30G's sounds impressive but you also have to remember that with the high speed comes a large turn radius and, obviously, a manned aircraft can't pull this many G's without crushing the pilot. They have developed super maneuverable unmanned aircraft that get closer to this kind of maneuverability but then you've got the disadvantages of an unmanned aircraft to consider. Personally, I'd concentrate on improving the maneuverability of the missiles. Imagine not having to point at your target, just fire the missile and let it do the turning.
You're right about the F-104, it's basically an engine with a cockpit and wings bolted on. It was a fast interceptor, capable of Mach 2, and was a beautiful plane but it turned like crap and needed external fuel tanks for decent range and endurance. In the end it had little advantage over more conventional designs considering its limitations and even with a modern engine and materials it would still be a dog in the turn department.
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In a pure vertical climb your G-meter will read zero. The G-meter reads loads normal to the aircraft centerline. If you're reading 1G on your G-meter in a vertical climb you are actually doing a loop.
In a pure vertical zoom you need to push (and trim) to a slightly negative AoA to counteract lift generated by the typical wing's non-symmetrical airfoil shape. In AH, you also have to deal with CT which is automatically set based only on speed with the assumption it's trimming for level flight. In a vertical climb, CT will actually begin to trim your nose "up" as you decelerate causing you to tend to arc over on your back unless you push/trim nose down.
This is exactly what I thought the situation was, many thanks!
But now this:
In a less than vertical zoom climb you'll go higher at 1G because you'll end up in a steady state climb and the lift will add more to your energy gain than the drag will subtract. It should be easy enough for you to test different angles and note the altitude gain vs a pure vertical 0G zoom climb. If you fly at 0G at less than vertical your nose will come down early in an outside half loop.
Confuses me. Are you saying that instead of zooming vertically upwards at zero G, I'd gain more height if I climbed on an incline (lift vector still pointing upwards not downwards) and utilised the lift from the wings? Even with a thrust ratio of less than 1.0?
I gathered that lift induced drag cost quite a penalty at lower speeds?
Sorry to bang on about this but one can usually learn a lot from an apparent contradiction.
Thank you both :salute
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In a less than vertical zoom climb you'll go higher at 1G because you'll end up in a steady state climb and the lift will add more to your energy gain than the drag will subtract. It should be easy enough for you to test different angles and note the altitude gain vs a pure vertical 0G zoom climb. If you fly at 0G at less than vertical your nose will come down early in an outside half loop.
This is not correct.
Lift does not add to your energy in any way.
Lift simply changes the direction of flight.
Lift by definition is a force perpendicular to an objects velocity vector. Gaining energy requires and increase in altitude or an increase in speed. Any excess lift (I.E. more then needed to oppose gravity approximated by the cosine of the climb angle) will cause the airplane to change the direction of it's velocity vector but not increase it.
HiTech
The most efficient energy gain will ALWAYS happen when excess power is the greatest.
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This is exactly what I thought the situation was, many thanks!
But now this:
Confuses me. Are you saying that instead of zooming vertically upwards at zero G, I'd gain more height if I climbed on an incline (lift vector still pointing upwards not downwards) and utilised the lift from the wings? Even with a thrust ratio of less than 1.0?
I gathered that lift induced drag cost quite a penalty at lower speeds?
Sorry to bang on about this but one can usually learn a lot from an apparent contradiction.
Thank you both :salute
Sorry for the confusion. Mace is correct. I wasn't disagreeing with him. My point is that when you aren't vertical you don't want 0G. In other words, be sure to be vertical and don't just use 0G for every zoom climb. Most zoom climbs are at angles less than vertical where 0G will give you a curving flight path that will reduce your altitude gain.
This is not correct.
Lift does not add to your energy in any way.
Lift simply changes the direction of flight.
Lift by definition is a force perpendicular to an objects velocity vector. Gaining energy requires and increase in altitude or an increase in speed. Any excess lift (I.E. more then needed to oppose gravity approximated by the cosine of the climb angle) will cause the airplane to change the direction of it's velocity vector but not increase it.
HiTech
The most efficient energy gain will ALWAYS happen when excess power is the greatest.
My bad I should have said altitude gain not energy gain to distinguish it from total energy. The question was how to zoom higher. You're saying that in a 60 degree zoom climb with fighter A at 0G and fighter B at 1G then fighter A will have more energy even though he's lower. I'm not disagreeing. :salute
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The most efficient energy gain will ALWAYS happen when excess power is the greatest.
Thus the optimal sustainable climb speed is actually the speed and attitude at which excess power is the greatest, because the propeller is producing maximum thrust at this speed as well as parasitic drag and lift induced drag being at their minimum?
Sorry FLS but you are confusing me further. You are saying that it is always better for your overall energy state to climb steadily on a gradient, even though the zero G vertical zoomer gains significantly more altitude? Because I intuitively felt the vertical zoomer would have the least drag and the most thrust especially towards the top? Plus he surely took the shorter route?
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Thus the optimal sustainable climb speed is actually the speed and attitude at which excess power is the greatest, because the propeller is producing maximum thrust at this speed as well as parasitic drag and lift induced drag being at their minimum?
Sorry FLS but you are confusing me further. You are saying that it is always better for your overall energy state to climb steadily on a gradient, even though the zero G vertical zoomer gains significantly more altitude? Because I intuitively felt the vertical zoomer would have the least drag and the most thrust especially towards the top? Plus he surely took the shorter route?
Yes that's correct, to minimize drag climb at 0G. I never said it's always better to climb on a gradient.
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This is not correct.
Lift does not add to your energy in any way.
Lift simply changes the direction of flight.
Lift by definition is a force perpendicular to an objects velocity vector. Gaining energy requires and increase in altitude or an increase in speed. Any excess lift (I.E. more then needed to oppose gravity approximated by the cosine of the climb angle) will cause the airplane to change the direction of it's velocity vector but not increase it.
HiTech
The most efficient energy gain will ALWAYS happen when excess power is the greatest.
....and therein is the reason to have an understanding of how to apply use of the lift vector in air to air combat maneuvering.
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Let me rephrase the question then: will I zoom higher by pushing to Zero G or by leaving in on 1?
I think zooming with G < 1 means you are falling. :headscratch: :uhoh
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I think zooming with G < 1 means you are falling. :headscratch: :uhoh
That's the difference between a vertical zoom climb and any other zoom climb. :aok
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Yes that's correct, to minimize drag climb at 0G. I never said it's always better to climb on a gradient.
Going into 0G minimizes drag by lowering the induced drag component to zero. The parasitic drag is still there and will change with speed and the zoom/dive develops. The general rule is that going 0G minimizes the instantaneous drag and for a few brief moments allow the best rate of building energy at the current speed. The thing is that the current speed will quickly change.
Since in a very steep climb (lets approximate to vertical) the speed is unsustainable and lowers, 0G is a good approximation to the most efficient zoom. In a shallow angle, 0G will send you on a ballistic-like course and at some point speed will not decrease and even start to increase (even before reaching max trajectory hight). From around this point, adding lift (1>G>0) will keep the total drag at a minimum which is sustainable for more than a moment - one essentially transits into a steady climb at the best climb speed. So typically the quickest way to gain energy is by zooming at a steep angle and 0G and then switching to a steady climb once the best climb speed is reached.
Going into a 0G dive momentarily reduces the drag and builds up speed. This is often used in a fight when you only have a short moment available for building energy and there is no time for zooming up and slowing down to a good climbing speed, then accelerating again to maneuvering speeds. This is also a good disengaging maneuver when one is not in immediate danger. You both accelerate, waste less alt per gain in speed and burn less E when leveling vs. pushing down into a vertical dive (in 0G the angle steepens with time so eventually it will reach a vertical dive).
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Going into 0G minimizes drag by lowering the induced drag component to zero.
Induced drag is a component of lift. If lift is created at 0G, you have induced drag. However, at zero G, wing loading is zero, so the load factor does not contribute to the amount of induced drag being created. Only way to eliminate induced drag is to fly at an attitude that reduces the effective angle-of-attack to whatever is required to create a coefficient of lift = zero.
So really, flying at zero G really just removes weight and all of its influences...
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Going into 0G minimizes drag ...
I was keeping it simple because my earlier slightly fuller explanation created some confusion. :D
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Induced drag is a component of lift. If lift is created at 0G, you have induced drag. However, at zero G, wing loading is zero, so the load factor does not contribute to the amount of induced drag being created. Only way to eliminate induced drag is to fly at an attitude that reduces the effective angle-of-attack to whatever is required to create a coefficient of lift = zero.
So really, flying at zero G really just removes weight and all of its influences...
As far as I understand the definitions, G = lift/weight. So zero G means zero lift, and zero lift means no formal induced drag. Of course that is not perfectly accurate locally since at a total of 0 lift some parts of the plane could be producing lift canceled by other parts producing negative lift.
I was keeping it simple because my earlier slightly fuller explanation created some confusion. :D
I know FLS. I was expanding on this for the other readers.
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As far as I understand the definitions, G = lift/weight. So zero G means zero lift, and zero lift means no formal induced drag. Of course that is not perfectly accurate locally since at a total of 0 lift some parts of the plane could be producing lift canceled by other parts producing negative lift.
I know FLS. I was expanding on this for the other readers.
You may be thinking about the lift = weight or that required lift = weight X load factor (G). From that, it would appear that at zero G, lift would = zero but that relationship is only true > 0G. But any time the wing is at an AoA that produces lift, even at zero G, there will be induced drag being created. So, unless the aircraft is at an attitude where the airfoil's lift coefficient is = zero, there will be induced drag present. However, at zero G, the induced drag for that attitude will be minimized. Remember that relative wind over the airfoil is what produces lift, and that's independent of load factor. Flying at 0G merely reduces induced drag to the absolute minimum, for that attitude or AoA.
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You may be thinking about the lift = weight or that required lift = weight X load factor (G). From that, it would appear that at zero G, lift would = zero but that relationship is only true > 0G. But any time the wing is at an AoA that produces lift, even at zero G, there will be induced drag being created. So, unless the aircraft is at an attitude where the airfoil's lift coefficient is = zero, there will be induced drag present. However, at zero G, the induced drag for that attitude will be minimized. Remember that relative wind over the airfoil is what produces lift, and that's independent of load factor. Flying at 0G merely reduces induced drag to the absolute minimum, for that attitude or AoA.
I am missing something in your post stony. Are you simply speaking about the (for lack of better summation) stability AOA generating lift I.E. sum of all torques must be 0 and sum of all forces perpendicular to the velocity vector must be 0?
Because 0g = 0 NET lift generated by the plane and hence 0 NET lift coefficient.
HiTech
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I am missing something in your post stony. Are you simply speaking about the (for lack of better summation) stability AOA generating lift I.E. sum of all torques must be 0 and sum of all forces perpendicular to the velocity vector must be 0?
Because 0g = 0 NET lift generated by the plane and hence 0 NET lift coefficient.
HiTech
If the airfoil in the wing is at an AoA that creates lift, there will be induced drag, regardless of load factor. At 0G, there is still relative wind over the airfoil. Unless the airfoil is at an angle of attack that causes the airfoil Cl to = zero, there is lift being produced, and hence, induced drag. Since load factor = zero, the Cli will be minimized to its smallest value at that AoA, but will exist, theoretically, even if its almost negligible.
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If the airfoil in the wing is at an AoA that creates lift, there will be induced drag, regardless of load factor. At 0G, there is still relative wind over the airfoil. Unless the airfoil is at an angle of attack that causes the airfoil Cl to = zero, there is lift being produced, and hence, induced drag. Since load factor = zero, the Cli will be minimized to its smallest value at that AoA, but will exist, theoretically, even if its almost negligible.
Stoney, are you talking about a local phenomenon or an integrated one? a section of the wing can be producing lift with an associated "induced drag", while other sections produce negative lift such that the total is zero. In that case, I think it is formally considered as part of the global parasitic drag. A wing with a twist to its angle along the length of the wing will do this. Also, at 0 total lift the wings still need to produce some roll torque (one wing positive lift, the other negative) to cancel the torque from the prop.
In any case, the total (integrated) instantaneous drag at a given airspeed is lowest when the plane is practically at 0 (total) lift and hence 0 G. Perhaps one can setup some odd body where this is not perfectly accurate, but I doubt it will apply to any flying plane.
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Stoney, are you talking about a local phenomenon or an integrated one? a section of the wing can be producing lift with an associated "induced drag", while other sections produce negative lift such that the total is zero. In that case, I think it is formally considered as part of the global parasitic drag. A wing with a twist to its angle along the length of the wing will do this. Also, at 0 total lift the wings still need to produce some roll torque (one wing positive lift, the other negative) to cancel the torque from the prop.
In any case, the total (integrated) instantaneous drag at a given airspeed is lowest when the plane is practically at 0 (total) lift and hence 0 G. Perhaps one can setup some odd body where this is not perfectly accurate, but I doubt it will apply to any flying plane.
No, not talking about localized lift effects from the fuselage, H-stab, etc. nor twist or washout. Induced drag is a function of lift. Cl is a function of AoA. Right?
Assume an unswept, constant-chord wing, with no washout. Unless that wing is at an attitude where the lift coefficient of the wing = zero, there must be induced drag being created, regardless of load factor.
Maybe I'm wrong?
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Stoney could you give an example of 0G flight with an AOA that creates lift? I get that the Cl comes from the AOA but isn't the load factor also a function of lift?
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No, not talking about localized lift effects from the fuselage, H-stab, etc. nor twist or washout. Induced drag is a function of lift. Cl is a function of AoA. Right?
Assume an unswept, constant-chord wing, with no washout. Unless that wing is at an attitude where the lift coefficient of the wing = zero, there must be induced drag being created, regardless of load factor.
Maybe I'm wrong?
Stoney.
Lift = 1/2 ro * v^2 * area * cl.
0g Means lift = 0.
Hence Cl must be zero, you are not moving ,in a vacuum or infinitely small.
Am I missing something?
HiTech
0 g
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Stoney.
Lift = 1/2 ro * v^2 * area * cl.
0g Means lift = 0.
Hence Cl must be zero, you are not moving ,in a vacuum or infinitely small.
Am I missing something?
HiTech
0 g
:O HiTech is so smart. :) All I can say to that is wow... just wow! :cheers: :salute
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Maybe I'm looking at this wrong. I know that we use Lift = weight. So, it follows that at zero G, lift = 0. But, if the wing is at an AoA that produces any coefficient of lift, per the lift formula, and there is dynamic pressure (i.e. relative wind over the wing), there is lift being created, right? And, if lift is being created (and I'm talking about negligible amounts here, from a theoretical perspective), there must be induced drag being created, because you can't have one without the other, unless you have a wing with infinite span.
So, the way I figure, the whole "lift = weight" can only exist when G does not equal zero. Now, when G is equal to zero, no amount of lift is required to keep the aircraft "flying", but given that the plane has forward motion in some direction, there must be some lift being produced, unless the attitude of the aircraft in that direction is such that the Cl of the wing = 0.
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Lift can only equal weight at 1G.
If I understand you you're saying that lift from something like the Bernoulli effect could cause some lift and induced drag at 0 AOA? Unless you're taking about absolute vs geometric AOA isn't 0 Cl the definition of 0 AOA?
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Maybe I'm looking at this wrong. I know that we use Lift = weight. So, it follows that at zero G, lift = 0. But, if the wing is at an AoA that produces any coefficient of lift, per the lift formula, and there is dynamic pressure (i.e. relative wind over the wing), there is lift being created, right? And, if lift is being created (and I'm talking about negligible amounts here, from a theoretical perspective), there must be induced drag being created, because you can't have one without the other, unless you have a wing with infinite span.
So, the way I figure, the whole "lift = weight" can only exist when G does not equal zero. Now, when G is equal to zero, no amount of lift is required to keep the aircraft "flying", but given that the plane has forward motion in some direction, there must be some lift being produced, unless the attitude of the aircraft in that direction is such that the Cl of the wing = 0.
:headscratch:
Lift can only equal weight at 1G.
If I understand you you're saying that lift from something like the Bernoulli effect could cause some lift and induced drag at 0 AOA? Unless you're taking about absolute vs geometric AOA isn't 0 Cl the definition of 0 AOA?
:headscratch:
:old: :airplane: :joystick: :banana:
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Ahh I understand your disconnect Stoney.
G loading by definition is Lift / Weight. This applies in any attitude. So there is not any lift (BY DEFINITION) being generated at zero g. If you assume some lift, then we would not be at Zero g.
DISCLAIMER.
The above only is the basic all the minor forces going into the calcs of lift needed such as offset thurst, CG , wing twist ......
HiTech
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I understand it so that if you e.g. push the stick forward and effectively reach a 0G state (ballistic arc), it means that only portion of lift that carries the weight of the aircraft, at that speed, is negated but it does not necessarily mean that all of the lift is negated i.e. induced drag is still created, or it is not at its minimum, unless the wing AoA is at such angle that the lift on both sides of airfoil are equal, but depending on speed that could be even a -G state of the airframe. That would also effetively mean that there is likely a certain speed where the AoA net effect is really 0 as well as the G loading of the airframe is 0 at the same time but only at that certain speed.
-C+
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Lift can only equal weight at 1G.
If I understand you you're saying that lift from something like the Bernoulli effect could cause some lift and induced drag at 0 AOA? Unless you're taking about absolute vs geometric AOA isn't 0 Cl the definition of 0 AOA?
Only if you're talking about the effective AoA (or geometric AoA as you put it).
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Ahh I understand your disconnect Stoney.
G loading by definition is Lift / Weight. This applies in any attitude. So there is not any lift (BY DEFINITION) being generated at zero g. If you assume some lift, then we would not be at Zero g.
DISCLAIMER.
The above only is the basic all the minor forces going into the calcs of lift needed such as offset thurst, CG , wing twist ......
HiTech
I think I might be getting tripped up by the whole "gravitational G" vs. "G load". I tend to forget those aren't the same.
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:headscratch: <leaves the brainy think tank room now> :bolt:
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Only if you're talking about the effective AoA (or geometric AoA as you put it).
I think I mean the opposite. Geometric AOA is aligned with the chord line. Absolute AOA is 0 AOA = 0 Cl. So wouldn't effective AOA be absolute AOA?
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I think I mean the opposite. Geometric AOA is aligned with the chord line. Absolute AOA is 0 AOA = 0 Cl. So wouldn't effective AOA be absolute AOA?
As far as I know AOA is always measured against cord line when plotting CL curves.
Unless the wing is symmetrical, (normally only on planes designed for aerobatics) almost all wings have CL 0 at a negative AOA.
HiTech
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As far as I know AOA is always measured against cord line when plotting CL curves.
Unless the wing is symmetrical, (normally only on planes designed for aerobatics) almost all wings have CL 0 at a negative AOA.
HiTech
FLS, I've never seen the term "geometric" AoA before. Effective AoA is the angle of attack created when incidence and camber are included and compared to the relative wind. Absolute AoA is the angle of attack when compared to the zero-lift AoA of the wing when incidence and camber are included.
@HiTech, yes, since almost all aircraft use cambered airfoils, the zero-lift AoA will be negative.
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Here's a reference for geometric AOA.
http://www.av8n.com/how/htm/aoa.html#sec-raoa-aaoa
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Got it FLS.
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Ahh I understand your disconnect Stoney.
G loading by definition is Lift / Weight. This applies in any attitude. So there is not any lift (BY DEFINITION) being generated at zero g. If you assume some lift, then we would not be at Zero g.
DISCLAIMER.
The above only is the basic all the minor forces going into the calcs of lift needed such as offset thurst, CG , wing twist ......
HiTech
I don't think this statement is technically correct Hitech.
The definition of G loading creates a "divide by zero" mathematical situation at 0g. That doesn't mean that there isn't any lift, just that the denominator is zero.
Think of it this way, if I hold an apple in my hand and ask you to divide it by zero, does that make the apple disappear?
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Baumer G loading is the relationship of acceleration to free fall. No acceleration is 0 G is free fall. If you hold an apple in your hand you are accelerating it to 1 G. If you remove that acceleration the apple has no lift and at 0 G it disappears from the spot it occupied.
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FLS Sorry I was talking about load factor as Hitech posted about.
(http://upload.wikimedia.org/wikipedia/en/math/8/d/5/8d5e0ee2a980b3622cb6833fcfad700b.png)
where:
n= Load Factor
L= Lift
W= Weight
BTW a falling apple (on earth in a normal atmosphere) will have lift and aerodynamic drag as it accelerates towards the ground.
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FLS Sorry I was talking about load factor as Hitech posted about.
(http://upload.wikimedia.org/wikipedia/en/math/8/d/5/8d5e0ee2a980b3622cb6833fcfad700b.png)
where:
n= Load Factor
L= Lift
W= Weight
BTW a falling apple (on earth in a normal atmosphere) will have lift and aerodynamic drag as it accelerates towards the ground.
So was I. I was having fun with the apple. It does disappear. :D Load factor exists only when it is more than 0 G. When there is no acceleration there is no relationship. You're saying there is a problem with the definition of load factor for the case of no load and no load factor. That doesn't seem like a big problem.
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That's fine, but just to be clear, there is lift (and induced drag) on an aircraft in a zero g maneuver.
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I think that was Stoney's point too. I don't think anyone is disagreeing with that. It's probably more of a difference of pilot's perspective vs engineer's perspective and understanding which perspective everyone was coming from. There is stuff going on that to the pilot is negligible. I believe that's why Hitech was specifying 0 net lift and basic forces.
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These are the times I enjoy feeling stupid. :cheers:
BTW please no one divide by zero... I enjoy this world :devil
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I don't think this statement is technically correct Hitech.
The definition of G loading creates a "divide by zero" mathematical situation at 0g. That doesn't mean that there isn't any lift, just that the denominator is zero.
Think of it this way, if I hold an apple in my hand and ask you to divide it by zero, does that make the apple disappear?
This is what I meant when I said the whole lift = weight relationship could only exist at > or < 0 G. And really, that should be "required" lift = weight. Anyway...
Nice to see you Baumer! :salute
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Thanks Stoney, good to see you to. I took a few months off and just started reading the boards again. Perhaps I'll start flying again soon, who knows? :D
I understand where you're coming from FLS, it was just that reading some of the earlier posts I had the impression that some people might be thinking zero gee was also zero induced drag.
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Thanks Stoney, good to see you to. I took a few months off and just started reading the boards again. Perhaps I'll start flying again soon, who knows? :D
I understand where you're coming from FLS, it was just that reading some of the earlier posts I had the impression that some people might be thinking zero gee was also zero induced drag.
That's probably just sloppy writing where minimum induced drag was intended instead of 0 induced drag. Thanks for clearing that up.
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I don't think this statement is technically correct Hitech.
The definition of G loading creates a "divide by zero" mathematical situation at 0g. That doesn't mean that there isn't any lift, just that the denominator is zero.
Think of it this way, if I hold an apple in my hand and ask you to divide it by zero, does that make the apple disappear?
Say WHATTT!?
Where is the divide by zero? Weight is a constant hence the denominator can NOT be zero. And therefore can not be a divide by zero error.
The drop of the apple creates zero lift (unless it is irregular shape or spinning ). Lift would move it side to side not slow its fall.
DEFINITIONS !!!!!!!
Drag = force opposite the direction of travel.
Lift = force perpendicular to the direction of travel.
By simple definitions there can not be any net lift at zero g. Zero g is defined as a zero load factor.
As per your equations.
n = l/w.
or solving for L
N * W = L.
for a zero load factor I.E. N = 0
0 * W = L
Since zero * ANYTHing is zero
L = ZERO
HiTech
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This is what I meant when I said the whole lift = weight relationship could only exist at > or < 0 G. And really, that should be "required" lift = weight. Anyway...
Nice to see you Baumer! :salute
Stoney the lift = weight thing only applies to sustained level flight. In an steady state attitude that is not horizontal to the ground, lift must always be less then weight or the aircraft will not be traveling in a straight line.
HiTech
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Effective AoA is the result of upwash, downwash, prop effects, roll/pitch rates, etc. You guys are really over-thinking this. All that matters to the guy in the cockpit is that minimum drag, max acceleration, min deceleration all happen at zero G. :D
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On the standard force diagram like this.
(http://www.hitechcreations.com/images/stories/dales/lift1.jpg)
I really wish they drew it like this to remove the confusion of the force definitions.
Also the plane is in a climb in the picture below.
(http://www.hitechcreations.com/images/stories/dales/lift2.jpg)
HiTech
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This is what I meant when I said the whole lift = weight relationship could only exist at > or < 0 G.
Stoney, pro tip: when in science or engineering you get a mathematically singular point, it almost certain that you have a mistake somewhere. If the limit behaves badly, i.e. in our case (following your logic) when you approach weight=0, G goes to infinity, you can be CERTAIN that you have a mistake.
You are thinking 0G means weightless. That is not true in the frame of reference of an observer looking at the plane from the outside (i.e. not accelerating). To make things simple: "weight" in our case means the gravitational pulling force acting on the plane, given by weight = mass*g. It is a constant. The little "g" is the earth's gravitational acceleration.
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Stoney, pro tip: when in science or engineering you get a mathematically singular point, it almost certain that you have a mistake somewhere. If the limit behaves badly, i.e. in our case (following your logic) when you approach weight=0, G goes to infinity, you can be CERTAIN that you have a mistake.
You are thinking 0G means weightless. That is not true in the frame of reference of an observer looking at the plane from the outside (i.e. not accelerating). To make things simple: "weight" in our case means the gravitational pulling force acting on the plane, given by weight = mass*g. It is a constant. The little "g" is the earth's gravitational acceleration.
<enters brainy think tank room, notices apparent reference to Einstein's relativity gravity / acceleration relationship theory, leaves brainy think tank room rather quickly> :bolt:
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Mace I agree with you, we are over thinking this and missing the main point that MINIMAL drag occurs at zero G.
Hitech, apples are not symmetrical and will generate lift. This was actually discussed in one of my aerodynamics classes at ASU.
And just to be clear Mass is a constant, Weight is not a constant.
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notices apparent reference to Einstein's relativity gravity / acceleration relationship theory
Nope. Barely Newtons gravity.
Here, have an apple and it will all be clear.
If still not, try these mushrooms.
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Midway this is all basic stuff or I wouldn't be contributing. If you do a little background reading you'll understand it.
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Mace I agree with you, we are over thinking this and missing the main point that MINIMAL drag occurs at zero G.
Hitech, apples are not symmetrical and will generate lift. This was actually discussed in one of my aerodynamics classes at ASU.
And just to be clear Mass is a constant, Weight is not a constant.
And then they are not at zero load factor or zero G.
HiTech
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And then they are not at zero load factor or zero G.
HiTech
So what you're saying is that if there is relative wind over the wing, and the wing is at an AoA that creates lift, the plane cannot be at 0 G, and vice versa?
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So what you're saying is that if there is relative wind over the wing, and the wing is at an AoA that creates lift, the plane cannot be at 0 G, and vice versa?
Can I repeat my earlier question?
Stoney could you give an example of 0G flight with an AOA that creates lift?
...
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(http://www.sciencebuzz.org/sites/default/files/images/vomit%20comet%20trajectory.gif)
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(http://www.sciencebuzz.org/sites/default/files/images/vomit%20comet%20trajectory.gif)
Vomit comet???
BTW good to see ya back,come yak at me in the TA when you have the time!
:salute
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Bell curve. :D
Dude where's my lift?
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Bell curve. :D
Dude where's my lift?
If you pushover at 350kts, achieving 0 g for a few seconds, while the wing is at an angle of attack that produces lift...that would be an example...kind of like Baumer's illustration.
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If you pushover at 350kts, achieving 0 g for a few seconds, while the wing is at an angle of attack that produces lift...that would be an example...kind of like Baumer's illustration.
Hence my question. When during the 25 second 0 G maneuver is the wing producing lift?
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So what you're saying is that if there is relative wind over the wing, and the wing is at an AoA that creates lift, the plane cannot be at 0 G, and vice versa?
Correct.
But obviously this is the NET lift, I.E. the H stab could be producing lift in the opposite the wing. The net lifting force would be zero but wing would not be at zero lift angle. But as we agreed earlier, this was not what we were speaking of.
HiTech
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Hence my question. When during the 25 second 0 G maneuver is the wing producing lift?
If there's 350kts of relative wind over the wing, its making lift, unless the attitude of the plane gives the wing an AoA that results in a Cl of zero.
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Correct.
But obviously this is the NET lift, I.E. the H stab could be producing lift in the opposite the wing. The net lifting force would be zero but wing would not be at zero lift angle. But as we agreed earlier, this was not what we were speaking of.
HiTech
Got it...
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If there's 350kts of relative wind over the wing, its making lift, unless the attitude of the plane gives the wing an AoA that results in a Cl of zero.
I doubt it's going 350 over the top but I understand what you're saying about airspeed. However, in order to have 25 seconds of 0 G it has to have a 0 lift AOA. You can posit If you pushover at 350kts, achieving 0 g for a few seconds, while the wing is at an angle of attack that produces lift but then you're assuming your conclusion in order to prove your conclusion. I think it's just the word "lift" that's creating a problem.
Airflow over the wing creates pressure differences which when in opposition to gravity or radial acceleration we consider "lift". Sometimes the airflow is not creating lift. The difference is the vector of the net forces. It's only lift, by definition, in opposition to acceleration. So airflow over a wing creates pressure differences which increase up to a point with AOA and minimize at 0 AOA (absolute) and exist at 0G but since there is no opposition to acceleration at 0 G there is nothing we can call "lift". So yes I agree that the relative wind over the wing is still creating pressure differences but the 0 G tells us that there is no lift.
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I doubt it's going 350 over the top but I understand what you're saying about airspeed. However, in order to have 25 seconds of 0 G it has to have a 0 lift AOA. You can posit If you pushover at 350kts, achieving 0 g for a few seconds, while the wing is at an angle of attack that produces lift but then you're assuming your conclusion in order to prove your conclusion. I think it's just the word "lift" that's creating a problem.
Airflow over the wing creates pressure differences which when in opposition to gravity or radial acceleration we consider "lift". Sometimes the airflow is not creating lift. The difference is the vector of the net forces. It's only lift, by definition, in opposition to acceleration. So airflow over a wing creates pressure differences which increase up to a point with AOA and minimize at 0 AOA (absolute) and exist at 0G but since there is no opposition to acceleration at 0 G there is nothing we can call "lift". So yes I agree that the relative wind over the wing is still creating pressure differences but the 0 G tells us that there is no lift.
Ok, I capitulate...I think we've beat this one long enough... I don't understand the physics of flight nearly as well as the aerodynamics, so I'll just push the "I believe" button on this one... I just can't see, dynamically, how a wing could stop producing lift when it still has relative wind hitting it at an angle that otherwise, at any other condition, would cause it to produce lift.
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This is from Wikipedia and is consistent with what I was taught in school.
"A fluid flowing past the surface of a body exerts a surface force on it. Lift is the component of this force that is perpendicular to the oncoming flow direction. It contrasts with the drag force, which is the component of the surface force parallel to the flow direction. If the fluid is air, the force is called an aerodynamic force.
Lift is commonly associated with the wing of a fixed-wing aircraft, although lift is also generated by propellers; kites; helicopter rotors; rudders, sails and keels on sailboats; hydrofoils; wings on auto racing cars; wind turbines and other streamlined objects. While the common meaning of the word "lift" assumes that lift opposes gravity, lift in its technical sense can be in any direction since it is defined with respect to the direction of flow rather than to the direction of gravity. When an aircraft is flying straight and level most of the lift opposes gravity. However, when an aircraft is climbing, descending, or banking in a turn, for example, the lift is tilted with respect to the vertical. Lift may also be entirely downwards in some aerobatic maneuvers, or on the wing on a racing car. In this last case, the term downforce is often used. Lift may also be horizontal, for instance on a sail on a sailboat."
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Ok, I capitulate...I think we've beat this one long enough... I don't understand the physics of flight nearly as well as the aerodynamics, so I'll just push the "I believe" button on this one... I just can't see, dynamically, how a wing could stop producing lift when it still has relative wind hitting it at an angle that otherwise, at any other condition, would cause it to produce lift.
I usually learn something from these discussions. :salute
When does the wing stop producing lift when it still has relative wind hitting it at an angle that otherwise, at any other condition, would cause it to produce lift. I'll bite, when does it? If it is at an angle to produce lift then it will. If it isn't producing lift it's not at that angle. The conditions you propose to coexist contradict each other.
Baumer the acceleration tilts along with the lift.
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Baumer: It is also 100% consistent with my diagrams and equations.
HiTech
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So building on HiTech's previous illustration, are these diagrams a fair representation of the forces that have been discussed? Note the length of the vectors are not meant to imply any meaning, just the orientation.
So in a ideal Zero G zoom climb, the situation would be this, with (essentially) no net lift and no lift induced drag:-
(http://i1114.photobucket.com/albums/k526/rwrk2/ZeroGZoomClimbcopy.jpg)
Thus if we altered the above situation slightly with your accelerometer was reading 1 G (acting on the normal to your thrust line) you would get a lift vector at 90 degrees to the wing and eventually loop round inverted because the weight (the effect of gravity on your mass) would still be straight down?
Then would this be the situation in a steady 45 degree 1 G climb:-
(http://i1114.photobucket.com/albums/k526/rwrk2/1GClimb.jpg)
I really have guessed about the orientation of the lift induced drag vector, it just feels about right that it would act in that direction? Obviously the speed would degrade without sufficient thrust (0.707 to maintain this climb angle, is that right?).
You can keep discussing this with equations, just some people can't reason with mathematics and have to resolve it pictorially.
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ALL drag is opposite the velocity vector . I.E the definition of lift induced drag is in the same direction as parasitic drag.
The arrow you have labeled as parasitic drag should be labeled simply drag on both diagrams.
And drag = parasitic + induced.
Thrust is the strange one with it's definition. Because the force from the prop can be lift and thrust at the same time.
These exact definitions become meaningful when you start trying to do the calculations per lift and drag curves which used these conventions when being charted.
HiTech
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Aha! And since the velocity vector need not align with the thrust vector this explains how a wing varies it's lift and also its lift induced drag. Bingo. :banana:
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Then would this be the situation in a steady 45 degree 1 G climb:-
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A steady 45 degree climb must have G<1. It is actually about cos(45)=0.7.
Alternatively, pointing up at 45 degrees and holding 1G will not be steady. The nose elevation will continue to increase.
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A steady 45 degree climb must have G<1. It is actually about cos(45)=0.7.
That's why climbers, when climbing those vertical walls, float? cos(90)=0. Congrats, you've just invented anti-gravity.
When we talk about how many Gs are we pulling, we actually talk about acceleration, change of direction of travel and opposing forces (inertia) expressed in equivalent of G. But G is always the same (same for all practical purposes but in reality it changes just slightly, depends on the position between pole and equator and the distance from body causing the G).
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That's why climbers, when climbing those vertical walls, float? cos(90)=0. Congrats, you've just invented anti-gravity.
Thanks. I have a patent pending on that.
When we talk about how many Gs are we pulling, we actually talk about acceleration, change of direction of travel and opposing forces (inertia) expressed in equivalent of G. But G is always the same (same for all practical purposes but in reality it changes just slightly, depends on the position between pole and equator and the distance from body causing the G).
hmm... I am not as smart as I thought after all, because I have no idea what you are talking about here. G, poles, equator, distance from body? :headscratch:
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That's why climbers, when climbing those vertical walls, float? cos(90)=0. Congrats, you've just invented anti-gravity.
When we talk about how many Gs are we pulling, we actually talk about acceleration, change of direction of travel and opposing forces (inertia) expressed in equivalent of G. But G is always the same (same for all practical purposes but in reality it changes just slightly, depends on the position between pole and equator and the distance from body causing the G).
This is a little incorrect.
If acceleration is 0 in level flight then g's = +1. If g is +1 when upside down at the top of the loop the acceleration would be approximately 64 FPSS. Hence the general term g and acceleration are not really interchangeable.
I understand what you are speaking of when referring to gravitational constants, but what I am referring to , is just general aeronautical definitions , not general physics definitions. The confusion stems from a plane relative axis system vs a world axis system.
G's, when speaking about air planes, is normally referring to the total FORCE in the direction of the lift vector. Not acceleration.
And yes cos(90) = 0, this is why in a vertical climb with no motor and no drag you are experiencing 0 g's even though your total acceleration would be 32fps.
HiTech
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And yes cos(90) = 0, this is why in a vertical climb with no motor and no drag you are experiencing 0 g's even though your total acceleration would be 32fps.
HiTech
Interesting. I thought a zero G situation was that of a freefalling object. So if I'm in a plane climbing straight up, I'll be weightless?
Edit : I really dont get it. Do you define a G in a plane as a force acting straight down on the floor of the craft in level flight? If so it does make sense as depending on the planes attitude, gravity is going to act differently on your plane. Then by that definition, we could say a plane going straight up is experiencing 0g relative to that imaginary line coming perpendicular to the floor. Yet the pilot and airframe will still experience gravity but along a different axis. Does that even make sense?! I'm kinda comfused by this discussion.
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So if I'm in a plane climbing straight up, I'll be weightless?
Only at the top for a moment I think when the thrust runs out of steam.
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Interesting. I thought a zero G situation was that of a freefalling object. So if I'm in a plane climbing straight up, I'll be weightless?
With the correct throttle setting you would be feeling completely weightless/falling, or more correctly decelerating
Then by that definition, we could say a plane going straight up is experiencing 0g relative to that imaginary line coming perpendicular to the floor. Yet the pilot and airframe will still experience gravity but along a different axis. confused by this discussion.
This is correct.
Remember we are not speaking what total physics are happening, but rather what the common term "g loading" refers to.
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Remember we are not speaking what total physics are happening, but rather what the common term "g loading" refers to.
Hmm... Perhaps I got this confused at that beginning.
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I understand what you are speaking of when referring to gravitational constants, but what I am referring to , is just general aeronautical definitions , not general physics definitions. The confusion stems from a plane relative axis system vs a world axis system.
HiTech
Yes, but in real world ie instrumentation combines the two, hence accelerometer (G-meter) is calibrated to +1. This is for good reason because forces acting on airframe and forces acting on (detached) bodies inside the airframe aren't the same and body inside the airframe, at constant speed no matter the direction, will always experience +1G.
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Yes, but in real world ie instrumentation combines the two, hence accelerometer (G-meter) is calibrated to +1. This is for good reason because forces acting on airframe and forces acting on (detached) bodies inside the airframe aren't the same and body inside the airframe, at constant speed no matter the direction, will always experience +1G.
The accelerometer will show zero when traveling vertical.This I have done 100's of times in real life. Even though we call it an accelerometer, it really is a force gauge. The accelerometer works identical to if the bottom of your seat was a bathroom scale. So in the vertical you are only exerting force on your back or the shoulder straps, not on your seat. Hence the bathroom scale will read 0 just like the accelerometer in the plane does.
HiTech
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The accelerometer will show zero when traveling vertical.
HiTech
Single axis G-meter yes.
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Yes, but in real world ie instrumentation combines the two, hence accelerometer (G-meter) is calibrated to +1. This is for good reason because forces acting on airframe and forces acting on (detached) bodies inside the airframe aren't the same and body inside the airframe, at constant speed no matter the direction, will always experience +1G.
You have confused me what you are defining as 1g. Is your definition of 1 g, that an object orbiting the earth is experiencing 1g? I'm not debating, just want to know your G definition.
Because is sounds to me you are just saying everything is accelerated by gravity the same?
(approx, I understand the distance difference)
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I'm just saying that the forces acting on airframe aren't necessarily the same as forces acting on a body (let's say pilot) inside the airframe.
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With the correct throttle setting you would be feeling completely weightless/falling, or more correctly decelerating
This is correct.
Remember we are not speaking what total physics are happening, but rather what the common term "g loading" refers to.
been listening. Now I get it. ;)
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You have confused me what you are defining as 1g. Is your definition of 1 g, that an object orbiting the earth is experiencing 1g? I'm not debating, just want to know your G definition.
Not sure where we went in different directions.
If you park a plane on tarmac horizontally, one axis G meter will show 1G, if you park it on the tail nose up it will show 0G, even though in both cases plane accelerates upwards at 1G relative to free fall, correct?
Or are we talking about another reference frame?
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I'm just saying that the forces acting on airframe aren't necessarily the same as forces acting on a body (let's say pilot) inside the airframe.
Ok I'm lost again, pilot strapped to plane. Pilot is now part of airplane. Force acts on airplane, but not on the pilot?
HiTech
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Ok I'm lost again, pilot strapped to plane. Pilot is now part of airplane. Force acts on airplane, but not on the pilot?
HiTech
Only forces acting on the pilot are weight and airframe.
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Only forces acting on the pilot are weight and airframe.
I have never heard of a force named airframe.
I have heard of forces like lift,drag,thrust and weight.
All of which except weight are transmitted to the pilot threw an airframe.
But I do not believe an airframe is a force.
P.S. at this point we are just speaking of semantics.
HiTech
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:uhoh <enters brainy think tank room> :O <sees heated discussion> :headscratch: <leaves brainy think tank room as fast as possible> :bolt:
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Only at the top for a moment I think when the thrust runs out of steam.
Hitech is describing a special situation where the only thrust is from momentum and only gravity is decelerating you. I believe his point is that you won't feel weight in the decelerating zoom climb the same as you don't feel weight in freefall.
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I have never heard of a force named airframe.
I have heard of forces like lift,drag,thrust and weight.
All of which except weight are transmitted to the pilot threw an airframe.
But I do not believe an airframe is a force.
P.S. at this point we are just speaking of semantics.
HiTech
Of course it's semantics. You know that I meant the force airframe exerts on the pilot, and if we are picky, you can't call it thrust, lift nor drag, since neither affects pilot directly, but only as a vector sum force resulting from others, minus loss incurring during the transfer ie deformation (straps), or even purposely built device such as shock absorbers/acceleration dampeners.
But lets keep it simple, any time you accelerate (relative to free fall) you feel G forces.
And if you answer the question I asked earlier, we're back in discussion, otherwise, we'll just talk by each other.
If you park a plane on tarmac horizontally, one axis G meter will show 1G, if you park it on the tail nose up it will show 0G, even though in both cases plane accelerates upwards at 1G relative to free fall, correct?
Or are we talking about another reference frame?
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That's why climbers, when climbing those vertical walls, float? cos(90)=0. Congrats, you've just invented anti-gravity.
You're just saying that thrust can't be greater than weight but really you know better.
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If I sum this up,
G load is the measure of acceleration of an airframe along a certain axis, going straight up and down, passing through the CG.
In certain flight regime, the instrumentation wont read the gravity as part of the total G load of an airframe. Level it read 1g, inverted -1g and vertical 0g. And everything in-between.
Gravity being universal, it's always felt along a certain axis, which is going from the center of gravity straight down to mother earth.
Soooooo....where does that leave me? I guess that I should avoid nose high turn if I want to optimize my turn rate as gravity will add to my turn instead of fighting me.
But this was interesting.
Now speaking of anti-gravity, when do we the claw back. T'was such a treat! :x
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You're just saying that thrust can't be greater than weight but really you know better.
You're just saying that Apollo astronauts didn't know about cos(90)=0 but you really know better.
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Interesting. I thought a zero G situation was that of a freefalling object. So if I'm in a plane climbing straight up, I'll be weightless?
Edit : I really dont get it. Do you define a G in a plane as a force acting straight down on the floor of the craft in level flight? If so it does make sense as depending on the planes attitude, gravity is going to act differently on your plane. Then by that definition, we could say a plane going straight up is experiencing 0g relative to that imaginary line coming perpendicular to the floor. Yet the pilot and airframe will still experience gravity but along a different axis. Does that even make sense?! I'm kinda comfused by this discussion.
With the correct throttle setting you would be feeling completely weightless/falling, or more correctly decelerating
I'm sure you all seen the pictures and vids of astronauts training for zero-G in parabolic flights in airliners:
(http://3.bp.blogspot.com/_uOc-e3tRrTM/RsI9XloUNjI/AAAAAAAAAJI/PQcCePdaaJM/s320/300.jpg)
And that's the segment of the flight where they experience zero G:
(http://upload.wikimedia.org/wikipedia/commons/2/2f/Zero_gravity_flight_trajectory_C9-565.jpg)
As you can see, 0G begins while still climbing.
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You're just saying that Apollo astronauts didn't know about cos(90)=0 but you really know better.
Are you wondering about the .3 G? Lift is .7 G. Thrust adds .3 G. Pilot and plane are at 1G.
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As you can see, 0G begins while still climbing.
At that point plane is not really climbing, it's in state of inertial fall, similar to satellites orbiting the Earth. Simply said, it's in free fall but not toward the center of the Earth.
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At that point plane is not really climbing, it's in state of inertial fall, similar to satellites orbiting the Earth. Simply said, it's in free fall but not toward the center of the Earth.
Strange definition of climbing there :headscratch:
As long as I see a plane going UP I would call it climbing.
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Strange definition of climbing there :headscratch:
As long as I see a plane going UP I would call it climbing.
In that graphic, the flight path is not illustrated well. Plane does not climb when astronauts experience 0G.
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0 G zoom climb. Weightlessness begins while climbing.
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In that graphic, the flight path is not illustrated well. Plane does not climb when astronauts experience 0G.
It does.
Here's another one:
(http://upload.wikimedia.org/wikipedia/commons/3/35/Parabolic_flight.png)
Or this one, coming from NASA (http://quest.nasa.gov/space/teachers/microgravity/introF.gif)
Plane is still gaining several thousand feet during the inital part of the 0G phase.
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It does.
It's ascending due to being on highly parabolic inertial fall path (orbit), it does not climb in sense of using thrust or lift, otherwise astronauts would end up slamming the walls.
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in the zero G planes, when the zero G phase begin, the pilot push the nose over slowly in a zero G pitch down. If so the graphics are a bit missleading in that the plane will have a more parabolic trajectory instead of this clear cut pitch down at the top and the trajectory will begin the curve when the zero g phase begin. In the graphics, it look like pitch is constant through the first part of this phase.
Taken from wikipedia(I know, I know but it's not ALL BS) : Initially the aircraft climbs with a pitch angle of 45 degrees. The sensation of weightlessness is achieved by reducing thrust and lowering the nose to maintain a zero-lift angle of attack. Weightlessness begins while ascending and lasts all the way "up-and-over the hump", until the craft reaches a declined angle of 30 degrees.
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It's ascending due to being on highly parabolic inertial fall path (orbit), it does not climb in sense of using thrust or lift, otherwise astronauts would end up slamming the walls.
It's ascending due to being on highly parabolic inertial fall path (orbit), it does not climb in sense of using thrust or lift, otherwise astronauts would end up slamming the walls.
Neither I nor the picture made any claims on that matter in any way. We both said it still climbs when the occupants start to experience 0G (which it does), and you said that it does not (which is wrong). :P
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It's ascending due to being on highly parabolic inertial fall path (orbit), it does not climb in sense of using thrust or lift, otherwise astronauts would end up slamming the walls.
It's climbing in the sense that it's gaining altitude.
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Remember we are not speaking what total physics are happening, but rather what the common term "g loading" refers to.
Sorry, completely missed this one. Yes, I didn't notice you guys were talking about load factors and not physics in general. I shouldn't have interrupted and derail a discussion.
:salute
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FLS I meant that if you were flying only with the y-component of a parabolic curve (over time) without any movement on the x-axis, then surely you'd need to balance your thrust very carefully against your weight on the way up, or wait until the thrust began to be overtaken by weight to achieve the same effect. If you had a second accelerometer, fore / aft then it would also have to read zero to give you the feeling of weightlessness, right?
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OK, definitions:
"G", the big capital G in the context of aviation is the acceleration in a direction perpendicular to the movement excluding gravity. In other words, it is the force over mass, where the force include the sum of all forces on the plane except gravity and then taking only the component perpendicular to the velocity.
Since drag is by definition along the velocity direction and thrust is usually close to be aligned with the velocity and we exclude gravity, the only thing that is left is the lift which by definition is perpendicular to the velocity. Hence G=L/M, but expressed in units of earth gravitational acceleration, usually denoted with a small g. So if you use SI or CGS units G=L/(M*g)=L/W where "W" is the weight.
Some finer points:
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* Generally, G is NOT the actual acceleration of the plane in any axis or total since G does not include gravity.
* a stone being thrown is at 0G from the moment it leaves the hand till it lands, even while going up - the only forces that act on it are drag, which is aligned with the velocity and hence irrelevant to G and gravity which is explicitly excluded from G. Dignify all of us by not arguing that the stone is not spherical or is rotating.
* The pilot feels G by how hard the seat is pressed against hit butt, not his back. In a 90 degree climb (maintaining the angle), the seat is only pushing against the pilots back and not his tuches. The vertical acceleration can be whatever, but since it is aligned with the velocity it is still G=0.