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I know this is a weird way of thinking about / quantifying this, but I'm trying to figure out how much torque a 747 can apply to itself when it pitches upwards (turning about an axis from wing to wing, with the nose moving upwards). I figure it's equal to the lift times the distance between where lift is applied to the wings and the axis of rotation, but I haven't been able to find this data.

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  • $\begingroup$ There are a great number of forces torquing a 747 at a time, almost entirely cancelling each other out (hence its stability). Are you asking specifically about the torque induced by the elevator's control surfaces, or also about other forces which would tend to counteract it? $\endgroup$ – Sneftel Apr 3 at 8:46
  • $\begingroup$ Well, any good answer would need to consider both torque used to overcome rotational inertia and change the pitch rate, and torque used to overcome aerodynamic damping and maintain a steady pitch rate. Also the "baseline pitch torque" needed simply to maintain linear flight. Or approach the problem from the other end and look at the max elevator deflection and resulting lift coefficient and figure the aerodynamic force resulting and convert to torque by multiplying by distance to CG. $\endgroup$ – quiet flyer Apr 3 at 21:10
  • $\begingroup$ Raises the question are we looking at NET aerodynamic pitch torque in which case the pitch torque used to overcome aerodynamic damping and maintain a steady pitch rotation rate does not count, and likewise the "baseline" pitch torque used to maintain linear flight does not count-- in this case we can only count the pitch torque component that causes a change in the pitch rotation rate. So what is the max possible delta pitch rotation rate? I think you have asked a very hard question perhaps best explored in an accurate simulator. $\endgroup$ – quiet flyer Apr 3 at 21:24
  • $\begingroup$ The question was meant to be "what's the maximum torque it COULD apply to itself, when it's pitching as hard as it can." Total aviation newbie (less than a newbie, not even in the field whatsoever but working on a physics thought experiment). $\endgroup$ – Jake Kurlander Apr 4 at 9:12
  • $\begingroup$ Though it may be too late to edit the question now without invalidating existing answers, for your thought experiment you might want to be consider whether the flight path is constrained to be linear at the instant that the pitch torque is first applied, or whether the flight path is allowed to be already curving at that instant. But, I saw in another comment that you were considering the plane to be rolling along on the runway on the landing gear, so maybe that answers that. $\endgroup$ – quiet flyer Apr 5 at 10:25
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The pitching force is coming from the horizontal tail's downforce resisting the airplane's net nose down pitching forces, so the real question is how much downforce is being applied and how far away is the tail's downforce from the lifting forces holding the plane up.

For a given set of parameters, you need to know the All Up Weight, the Center of Gravity location, and the Neutral Point, which is the location of the (gaseous) fulcrum the tail has to work about. The NP is defined as the longitudinal point at which pitching moments are zero, the "effective" balance point in other words, taking into account the center of lift plus/minus the effects of various pitching moments (airfoil pitching moment, fuselage lift, nacelle lift, thrust effects, flaps and gear hanging down, etc.) which result in the actual net downforce the horizontal tail has to produce to "hold the nose up" in a given steady state flight condition.

If you know the CG location and the all up weight, then the "torque" acting on the fuselage at the Neutral Point is simply all up weight times the distance between the CG and NP.

So if the plane weighed 700,000 lbs, and the CG was 20 feet forward of the NP (just a wild guess; I have no idea the actual number), the "torque" value, the nose-down pitching force acting at the NP, is 14 million foot pounds. If the aerodynamic center of the tail was 120 ft aft of the NP (another wild guess), that works out to 116666 lbs of downforce required for stable level flight back at the tail. Of course if you now pull pitch to raise the nose, you are accelerating the mass about the lateral axis now, so those forces go up temporarily.

enter image description here

The values will change with changes in configuration, weight, speed, etc. Make a change that moves the Neutral Point forward, like adding thrust, and the torque value and downforce required goes down; airplane pitches up from the thrust application because the existing tail downforce is excessive. Drop the gear, which moves the Neutral Point aft, and the torque value and downforce required to counter it goes up; airplane pitches down from the drag of the gear because the existing tail downforce is insufficient.

As far as finding out what is a typical Neutral Point on a 747, good luck with that...

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  • $\begingroup$ I think this answer could be improved by changing the label (and associated references in text) from "neutral point" to "center of pressure" (or "center of lift"). The center of pressure is the point where we can apply the lift vector without having to also add a pitching moment -- exactly what you've done. If the CG is at the "center of pressure" then the tail downforce is zero. The center of pressure moves in flight, thus the preference for using "aerodynamic center" instead, plus a pitching moment. Regardless, the center of pressure is usually not the same as the Neutral Point. $\endgroup$ – quiet flyer Apr 4 at 14:03
  • $\begingroup$ In other words stable flight is often possible with the CG behind the Center of Pressure (but still ahead of the neutral point), thus causing the tail to generate an upforce rather than a downforce. Related ASE answer -- aviation.stackexchange.com/questions/19388/… . See also av8n.com/how/htm/aoastab.html#sec-pitch-equilibrium (last two paragraphs). $\endgroup$ – quiet flyer Apr 4 at 14:06
  • $\begingroup$ Center of pressure just refers to the net median point of lifting forces. Neutral Point also takes into account other pitch moment influences. A low thrust line applies a pitch up moment. Less tail downforce is required. The NP is moved forward. Dropping the gear creates a nose down pitching moment adding to all the other forces to give a new overall moment. This moves the NP forward. GG must always be forward of NP so there is a moment that must be opposed by tail downforce. Stabs on jets move LE down to go slower. If the tail was making up force, the stab would have to move the other way. $\endgroup$ – John K Apr 4 at 14:12
  • $\begingroup$ Regardless of whether tail is making upforce or downforce, lowering the LE of the stab will trim for slower flight. Well, looks like we're heading for another ASE question here, "Is stable flight possible with tail creating upforce, i.e. with Center of Pressure behind CG". Or "is it possible for CG to be behind Center of Pressure but still ahead of Neutral Point, yielding stable flight with the tail generating an upforce?" $\endgroup$ – quiet flyer Apr 4 at 14:17
  • $\begingroup$ Well that one has been beaten to death lol. You could have CG at or aft of the CP but with a strong ND pitching moment induced by some other influence, like floats hanging down, forcing the tail to generate downforce to oppose, you would have some static pitch stability. I've flown several airplanes with and without floats, and the increased pitch stability with floats installed is subtle but noticeable. The NP must remain aft of the CG in all flight regimes so the operating CG range has to take all that into account. $\endgroup$ – John K Apr 4 at 14:22
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Having been a commercial pilot and instructor for over 36 years I deal with this daily. The 747 or any aircraft is essentially balanced at the center of gravity. The positive moment will equal the negative moment on the cg. The aerodynamic force created by the horizontal stabilizer is easily calculable and is based on the lift coefficient, surface area of the airfoil, velocity, air density (rho), The forces can range from mere hundreds to thousands of pounds of force applied to the arm depending on these factors. By the way the cg is generally about 17 to 25% of the mean aerodynamic chord of the 747 measured from the leading edge of the airfoil at the root.

Once the moment (torque) is calculated then pitch acceleration rates based on weight can be determined. Rates of pitch are only of interest to pilots as they relate to controllability and stability so we don’t worry about actual axial acceleration rates. In this regard the farther to the rear the calculated cg is within the acceptable range (in relation to the center of lift) the more controllable and aerodynamically unstable the plane will be due to moments being closer to equal. The farther forward, the more aerodynamically stable and more difficult to control (greater required forces) it will be due to greater nose down moment.

By the way, most planes are designed to fly with some nose down moment due to loading. This creates a more stable aircraft. The cg range is always ahead of the neutral point. In theory, were it not for the necessity to longitudinally stabilize the aircraft very little horizontal stabilizer force would be required. So a small amount of down force is necessary to stay in level flight, by design. This additional downward force adds to the weight of all forces acted upon by the total lift forces.

Tex

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  • $\begingroup$ It seems you are implying that the neutral point is the point where the tail would generate exactly no upload or download in linear flight. Is that really true as a general statement? Could be grounds for another ASE question. My impression is that many aircraft can be stable even when the CG is located is such that the tail would generate an upward force. Of course in such a case the CG still must be ahead of the neutral point or else the aircraft would not be stable. $\endgroup$ – quiet flyer Apr 4 at 12:56
  • $\begingroup$ Could be grounds for another ASE question: is the "neutral point" the point where the tail downforce would go to zero in linear flight? Here's an interesting take on it-- see last two paragraphs-- av8n.com/how/htm/aoastab.html#sec-pitch-equilibrium $\endgroup$ – quiet flyer Apr 4 at 13:04
  • $\begingroup$ @Thetexan is exactly right. The NP is the effective balance point about which the tail must react the moment created by the CG forward. NP is the center of lift but adjusted for other pitching moments that may help or hinder the basic pitching moment of the CG-CL lever. If I have a seesaw that is in balance on its physical pivot but I add an external influence, like a spring pulling down on one end, that tends to tip it one way, I'd have to move the pivot to a new location to make it balance. This is the NP. The NP must ALWAYS be aft of CG to have positive static stability in pitch. $\endgroup$ – John K Apr 4 at 13:54
  • $\begingroup$ "Trim Drag" is in fact the energy consumed by the tail to generate its downforce. I can move the CG closer and closer to the NP to reduce the downforce required and thereby reduce trim drag, but I end up with diminishing stability, and to run an airliner with the too far aft like that I have to use electronics and software to drive the tail surfaces in the background to actively make up for it. $\endgroup$ – John K Apr 4 at 14:06
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I assume you are asking about the NET aerodynamic pitch torque that the 747 can generate. In this case the pitch torque created by tail just to maintain linear flight doesn't count. Nor does the pitch torque created by the tail to overcome "aerodynamic damping" in the pitch axis during a steady-state (constant airspeed, constant bank angle, constant turn rate) turn. All that matters is the maximum possible rate of change of the pitch rotation rate, and the aircraft's moment of rotational inertia in the pitch axis. That is all the information that we need to solve the problem. Any approach based on other information will lead to a much more complicated solution.

But even boiled down to these simple terms, the problem is very challenging. It is very difficult to determine the maximum possible rate of pitch rotation. This is perhaps best determined by experiments in a very accurate simulator, or in the actual aircraft. Caculating the moment of rotational inertia in the pitch axis is trivial compared to determining the maximum possible rate of change of pitch rotation.

The above notes were aimed at the case where the wheels are off the ground. When some of the wheels are on the ground, that introduces another complication-- or perhaps a simplification. When the CG is so far forward that the airplane can only produce a negligible rate of pitch rotation on takeoff at some particular instantaneous airspeed, then the total aerodynamic pitch torque created by the aircraft in that configuration at that airspeed is equal to the pitch torque created by however much weight is still resting on the main wheels at whatever airspeed the aircraft is moving at at that particular instant, times their distance from the CG. What CG position would cause the aircraft to only be able to cause a negligible rate of pitch rotation at takeoff at some particular airspeed, and how much weight is still resting on the main wheels at that instant, is another thing that might perhaps be explored in a good simulator.

The aerodynamic pitch torque the aircraft can create at a typical rotation airspeed would surely not be the maximum aerodynamic pitch torque the aircraft could create in any situation.

For example, to take a really exotic case-- when the aircraft is inverted at the top of a loop at a relatively low airspeed, gravity is helping to curve the flight path, creating a curvature in the relative wind which essentially has the same effect as bending the fuselage like a banana in the opposite direction (with the concave side toward the wheels/ belly), decreasing effective decalage or creating negative effective decalage. This would increase the power of the elevator and tail to create an aerodynamic pitch torque, and create a high rate of change in pitch rotation rate, in the landing gear/ belly direction, while decreasing the power of the elevator and tail to create an aerodynamic pitch torque in the opposite direction. A similar effect was noted in this related answer, which looked at the problem in terms of the elevator's ability (or inability) to keep the wing at a positive-lift angle-of-attack, rather than in terms of the maximum possible rate of change in pitch rotation rotation rate: What has happened to make me experience negative G with the control stick FULL AFT near the top of a loop?

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  • $\begingroup$ The "torque" created by the nose down pitching moment, CG acting around the neutral point, is present continuously. In level flight there is a static nose down pitching moment being opposed by tail down force. If you pull on the stick, you increase the forces but it's a transient condition, same as turbulence. In a 2 G sustained turn, the level flight static forces are simply doubled. $\endgroup$ – John K Apr 4 at 4:51
  • $\begingroup$ If it makes things easier, I'm only trying to figure this out at takeoff, and I'm happy to just get within a factor of two. I was under the impression that much of the torque used to lift the nose of the plane at liftoff is generated by the wings, but now I'm realizing it might be more generated by the tail? $\endgroup$ – Jake Kurlander Apr 4 at 9:15
  • $\begingroup$ At point of rotation it's almost all the leverage applied by the tail to lift the nose, pivoting around the main landing gear, so initially, the lever arm is tail-to-wheel. The "pivot" shifts forward somewhat during the rotation as the wings start generating lift. The peak downforce demand and torque force on the fuselage is about at the point the tail skid contacts the runway.When fuselages get broken on landing, from being "overtorqued" as happened with Beoing's DC-9/717 test a/c, on an Embraer 145 once, it's on landing, when you add landing G force to the leverage and it's just too much. $\endgroup$ – John K Apr 4 at 12:25
  • $\begingroup$ @JohnK let's say we chose the CG is the pivot point to resolve our torques around. This is a valid choice and arguably the one that makes the problem simplest. Strictly speaking in level flight the net pitch torque is zero. Aerodynamic nose-down pitching moment from wing is opposed by aerodynamic nose-up pitching moment generated by tail; weight itself (or G-load) has no effect on anything. $\endgroup$ – quiet flyer Apr 4 at 12:29
  • $\begingroup$ @JohnK I'm not sure neutral point has any effect on anything no matter what point we choose as the pivot point for our problem, I don't think neutral point is really the same as the point where the airplane would fly with zero tail upload or download if the CG were located there. Could be grounds for another ASE question. I think neutral point is the point where the aircraft becomes unstable in pitch if the CG is moved there, and plenty of aircraft are stable even when flying with an upload at the tail. That too could be grounds for another ASE question. $\endgroup$ – quiet flyer Apr 4 at 12:31
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Is it possible to have a stable aircraft (aerodynamically) in a condition where stabilizer updore is necessary for level flight?

No. Not by the definition of static or dynamic stability.

There are two systems that have to be considered. The first is the wing itself. The airfoil has a center of pressure that moves fore and aft based on angle of attack. It also produces a downward force center of pressure due to Bernoulian and Newtonian forces on the bottom of the wing. The upper and lower centers of pressure are not physically opposite of each other and therefore produces either downward or upward torque on the wing itself. This torque is centered on the mean aerodynamic center. The MAC does not move regardless of AOA. It it through this MAC that the center of lift operates.

Now take this wing with Its own system of balances and forces and attach it to a plane with its own balances and forces. The CL will be behind the CG on most aircraft for stability purposes. Only when the CL and CG are at the same point will the NP be near the CL/CG point. Since weight and speed determine necessary AOA, and AOA effects CP, drag, and stabilizer AOA the NP (or point where no stabilizer force is required for level flight) will never remain in any one location but will be at a different point based on a given set of conditions. The odds of the CG and NP ever being and staying the same are improbable.

As to how much torque is required....torque is torque. You can simple calculate the moments of a plane that is balanced. The force required to move a 400,000 lb airplane about its CG will be a matter of F=ma. That will give rate for a given force. Calculate the force capable with a 747 stabilizer (per L=ClrhoV^2*S) and apply it to the force formula and that should be close to the rate or rotational movement around the lateral axis. My angular momentum physics may be a little rusty.

Angle of incidence also plays into this. Most airplanes have a positive AOI. A B727 had an almost 1.5 degree down AOI.
That’s why it always flew nose up compared to other planes

tex

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    $\begingroup$ Welcome to ASE. It looks like at least part of this answer would be better posted to a different question; there are several existing questions on ASE that this answer might fit. Also on ASE, people are welcome to ask a question and then self-answer it, so that's another option if you feel this answer best fits a question that has not been asked yet. But consider this one -- aviation.stackexchange.com/questions/22087/… $\endgroup$ – quiet flyer Apr 5 at 8:37
  • $\begingroup$ "A B727 had an almost 1.5 degree down AOI."-- really? I wonder why. I see another ASE question coming here... $\endgroup$ – quiet flyer Apr 5 at 9:17

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