drunkenbeagle
Gang Member
I would just talk to a physics professor, and ask him to politely talk with your aeronautics professor about changing the slide.
This is bigger than just semantics to me.
If you know anything about aerodynamics.....
- Thread starter clayfenderstrat
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I would just talk to a physics professor, and ask him to politely talk with your aeronautics professor about changing the slide.
This is bigger than just semantics to me.
Mike H
Well-Known Member
But throwing the ball harder increases it's downward force, which is for all practical purposes, has the same æffect as using a heavier ball.
tgrayson
New Member
Maybe it's my imagination, but it seems that whenever someone leads into an explanation using the word "technically", they're about to state something technically incorrect.
To your instructor's defense, however, almost any statement can be interpreted to be correct as long as you can assume whatever definitions you wish for the words being used. Rather than using "right" or "wrong", I'd throw in the wishy-washy expression "pedagogically unsound", meaning it can cripple your ability to apply your knowledge to new situations.The problem with interpreting the downward force on the horizontal stabilizer as weight is that the "downward" force no longer is downward once the aircraft leaves straight and level flight. The force is perpendicular to the flight path and no longer be parallel with weight during a climb, descent, or bank, unlike weight, which always acts toward the center of the earth. That alone indicates they are not synonymous.
Moreover, saying that the tail down force is weight could lead to all sort of erroneous conclusions, such as the CG location changing with the quantity of downforce, which it cannot do.
Downforce as weight is a deadend notion.
Mike H
Well-Known Member
Downforce contributes to the overall load factor. Calling it weight is a simplification, but for all practical purposes that's what it is.
In a 60 degree (2G) bank, the load factor doubles. Agree? The aircraft isn't actually any heavier, but for all practical purposes, the wing IS supportng twice the weight as in straight & level flight.
Tld
Waffles Brah.
Wow, a lot of people like to talk aerodynamics! here's my two cents, (I am extremely bored today, already flew, now just flying the desk the rest of the day.)
Lift? Weight? Moment? Negative lift? Let's wrap it all up together here...I have never heard negative lift more simply discussed than by the master Rich Stowell....this was how it was explained to me, and if you ever have any questions on aerodynamics, you need to read this guy's stuff....hope this helps
(in terms of the main wing and CP)...The relationship between CG and CP is critical for stability. Whenever CG slips behind CP, lift creates a moment that pitches the airplane towards high angles of attack, high drag, and stalled flight. When CG slides ahead of CP, lift pitches the airplane towards lower angles of attack, away from stalled flight. The latter is more desirable; hence, let's establish an aft CG limit that's ahead of the wings forward most CP point.
If CG moves too far forward, however, our yet-to-be designed pitch control may not be able to raise the nose enough for a good landing. Consequently, we must also set a forward CG limit based on control effectiveness. We're starting to shape the airplane's operating envelope. The CG range thus defined makes our airplane 'nose-heavy'. We now have to find a way to balance the airplane's inherent forward pitching tendency.
Lets add a small stabilizing wing -the Horizontal stabilizer- to our airplane. It's job is to counter the the main wing's nose down pitching moment. Locating this surface aft of the CG means it must produce a small downward force, called negative lift. We now have a large lift force, close to the CG trying to pitch the nose forward, and a small force, far from the CG trying to pitch the nose back. the resulting moments cancel. In addition, placing the horizontal stabilizer within the downwash of our main wing helps it generate the necessary lift.
Now that moments finally balance, its time to recheck our force balance. Thrust still equals drag in steady level flight, but lift must now equal weight plus the negative lift generated by the horizontal stabilizer. Negative lift is counterproductive to lifting the plane skyward. Locating the horizontal stabilizer as far aft as practical minimizes the adverse impact of this tail down force. It's often convenient to treat negative lift as added weight, which allows us to say lift equals weight in steady flight.
And remember, Airplanes operate in a three-dimensional realm. They don't know what's normal or inverted, left or right, up or down. Airplanes are unaware that a horizon exists at all...They only respond to the relative wind acting on them...just sayin'...it's a good thing to keep in mind when discussing aerodynamics
How does that explain it....clear as mud?
Lift? Weight? Moment? Negative lift? Let's wrap it all up together here...I have never heard negative lift more simply discussed than by the master Rich Stowell....this was how it was explained to me, and if you ever have any questions on aerodynamics, you need to read this guy's stuff....hope this helps
(in terms of the main wing and CP)...The relationship between CG and CP is critical for stability. Whenever CG slips behind CP, lift creates a moment that pitches the airplane towards high angles of attack, high drag, and stalled flight. When CG slides ahead of CP, lift pitches the airplane towards lower angles of attack, away from stalled flight. The latter is more desirable; hence, let's establish an aft CG limit that's ahead of the wings forward most CP point.
If CG moves too far forward, however, our yet-to-be designed pitch control may not be able to raise the nose enough for a good landing. Consequently, we must also set a forward CG limit based on control effectiveness. We're starting to shape the airplane's operating envelope. The CG range thus defined makes our airplane 'nose-heavy'. We now have to find a way to balance the airplane's inherent forward pitching tendency.
Lets add a small stabilizing wing -the Horizontal stabilizer- to our airplane. It's job is to counter the the main wing's nose down pitching moment. Locating this surface aft of the CG means it must produce a small downward force, called negative lift. We now have a large lift force, close to the CG trying to pitch the nose forward, and a small force, far from the CG trying to pitch the nose back. the resulting moments cancel. In addition, placing the horizontal stabilizer within the downwash of our main wing helps it generate the necessary lift.
Now that moments finally balance, its time to recheck our force balance. Thrust still equals drag in steady level flight, but lift must now equal weight plus the negative lift generated by the horizontal stabilizer. Negative lift is counterproductive to lifting the plane skyward. Locating the horizontal stabilizer as far aft as practical minimizes the adverse impact of this tail down force. It's often convenient to treat negative lift as added weight, which allows us to say lift equals weight in steady flight.
And remember, Airplanes operate in a three-dimensional realm. They don't know what's normal or inverted, left or right, up or down. Airplanes are unaware that a horizon exists at all...They only respond to the relative wind acting on them...just sayin'...it's a good thing to keep in mind when discussing aerodynamics
How does that explain it....clear as mud?
tgrayson
New Member
Perfect example of a dead-end. Since
load factor = L / W
and you want to say that both lift (L) and weight (W) double, that leads to the calculation (using L and W for their 1 g values)
load factor = (2 * L) / (2 * W)
load factor = L/W = 1
Hence the load factor in a 60 degree bank is 1. Fail.load factor = L/W = 1
The reason that you have a load factor greater than 1 is that weight does NOT increase while Lift does. The *apparent* weight is merely your acceleration as a result of the unbalanced forces.
Mike H
Well-Known Member
Oh, OK :rotfl:
N18040
Well-Known Member
Anybody here ever read the Aviation Instructor's Handbook. It has a section that specifically deals with types of questions to avoid while testing students. It is available here http://www.sportpilot.info/sp/FAA-H-8083-9_Aviation_Instructors_Handbook.pdf Check out pages 6-5 and 6-6. I know I had to learn this stuff before I got my CFI, so it seems that the prof in this situation should have seen this at some point in time.
BobDDuck
Island Bus Driver
Good thing gravity is a constant.
HAHA I beat you to it! While he is at it he can check out negative xfer of learning, this is a prime example. As tgray and mikeh's argument so elegantly pointed out.
For you people studying for your CFI's reading this, negative xfer of learning is always a weird one to explain and give an example for, take notes, this is the perfect example.
Katabat1c_SEA
Well-Known Member
No. Negative (downward) lift can act like weight but it is still lift (negative).
African_Swallow
Well-Known Member
(Hopefully I'm not repeating anything...didn't get to read all posts.)
But...just some food for thought (I'm not an engineer, aerodynamicist, nor do I work for NASA or Scaled):
The propeller is also an airfoil which creates "lift." What do we call this force, the resultant "lift" which is created parallel to the longitudinal axis of the airplane? Would this change with a pusher? How about with a jet engine?
I'm still on the fence with your prof's answer though. Isn't weight the force which acts through the center of gravity toward the center of the earth? Perhaps the resultant tail-down created by the Bernouilli effect of that airfoil does necessitate more lift from the main wing--more than would be needed without the negative angle of incidence of the h-stab. In that case, I would be lead to believe that the "lift" created by the tail-down force would in fact add to what we are calling weight, or the force vector acting through the CG toward the center of the earth. (TGray...help me out here! Just musing
-A.S.
But...just some food for thought (I'm not an engineer, aerodynamicist, nor do I work for NASA or Scaled):
The propeller is also an airfoil which creates "lift." What do we call this force, the resultant "lift" which is created parallel to the longitudinal axis of the airplane? Would this change with a pusher? How about with a jet engine?
I'm still on the fence with your prof's answer though. Isn't weight the force which acts through the center of gravity toward the center of the earth? Perhaps the resultant tail-down created by the Bernouilli effect of that airfoil does necessitate more lift from the main wing--more than would be needed without the negative angle of incidence of the h-stab. In that case, I would be lead to believe that the "lift" created by the tail-down force would in fact add to what we are calling weight, or the force vector acting through the CG toward the center of the earth. (TGray...help me out here! Just musing
-A.S.
Question: If the downward force acting on the tail was perfectly parallel to the weight vector would it even increase weight if you could find some way to measure it? I would assume it wouldn't as it is just countering the rotation force applied from the CP/CG interaction.
I am asking because he is saying, to simplify it, that any force acting down is weight. But since this force isn't acting down directly on the CG it shouldn't increase weight in any way but instead cause rotation. Just like pushing on the end of a perfectly balanced pencil, it would rotate, not increase in weight.
casey
Well-Known Member
no, if you think about it as a w&b problem you still get a force vector that points downward, representing lift.
Your statement that the CG effects the weight of the airplane is a fallacy. If I move a 50 lb box from the front of the airplane to the back I havent changed the weight but I most certainly have changed the CG.
The CG defines where the axis of rotation for the airplane is going to exist. The reason this changes the flying characteristic of the airplane is because the lift is acting in somewhere on the wing, known as the Center of Lift. The distance between the center of lift and the center of gravity is a distance. when you take the force generated at the center of lift (which is a vector) and take the cross product with the vector representing the arm you get a torque (the direction of which is the axis of rotation).
This torque acting alone would cause the airplane to pitch. The magnitude of this force is directly proportional to the distance between the center of lift and center of gravity. Increase the distance, increase the torque.
The horizontal stabilizer counteracts this pitching moment by creating a torque of its own. In order for the torques to cancel, (assuming the CG is between the center of lift and the tail) the lift the horiz stab generates must be in the opposite direction. It is a simple force balance problem (physics 101).
"Weight" only acts through the center of gravity. If I'm flying with the nose below the horizon such that my thrust vector is below the horizontal plane, is my engine creating weight along with thrust? I think not. And before you try to argue that down arrows are only "weight" if they are below the axis of the airplane, consider that physics is the same whether or not i choose to use natural or cartesian coordinates.
Oh, and to the OP, ask your professor if the lift created by the tail contributes to drag during flight (which it does). When he affirms that there is drag created on the horizontal stab, ask him how "weight" is creating this drag, particularly one that can vary with airspeed. Perhaps that will get him to rethink his position.
tgrayson
New Member
You are correct in that the wing must supply "extra" lift to counteract this tail down force....that's why a rear CG reduces drag on the airplane, because it reduces the quantity of taildown force needed.
But its direction is only the same as weight in straight and level flight. It's much better for analysis to keep forces that have different orgins separate, because only then do you know how to take them into account under a variety of circumstances.
For most purposes, though, you can simply add in the taildown force to lift. Since this usually reduces the lift of the main wing, the main wing will have to produce lift greater than weight in order for the net lift of the main wing and horizontal stabilizer to equal the weight of the aircraft.
tgrayson
New Member
When you apply a force to an object, it has two effects:
1) it causes the object to translate in the direction of the force, and
2) it causes the object to rotate if the force does not act through the cg.
This is why when you apply rudder to stop the yaw (rotation) of a multiengine aircraft with one engine out, the rotation stops, but the aircraft still translates due to the lift being created on the vertical stabilizer.
So, yes, the tail down force causes the airplane to rotate, but it also causes it to translate in the direction of the force. When you pull back on the yoke, the taildown force causes the aircraft to accelerate downwards before the increased lift of the main wing overcomes this tendency. This is one reason that fighter jets benefit from a canard; since it's a lifting surface, pulling back on the stick generates a force in the direction you actually want to go, rather than the opposite, resulting in increased agility.
sdfcvoh
This is my Custom Title
The fact that the HStab is an airfoil makes the entire argument. Otherwise the engineers would have just put a heavy piece of lead back there.
I would be tempted to ask the prof if a strategically placed piece of lead would do the same as the HStab. Because that would be referring to the HStab's component as weight.
I would be tempted to ask the prof if a strategically placed piece of lead would do the same as the HStab. Because that would be referring to the HStab's component as weight.
