New answers tagged

2

We "Do" take viscous shear stresses into account when calculating the drag of an airfoil. This is ALWAYS true for subsonic flows. The underlying reason behind this is called D'Alembert's paradox. For incompressible and inviscid potential flow, the drag force is zero on a body moving with constant velocity relative to the fluid. BOTH pressure drag and ...


8

There are plenty of smaller power planes that achieve those numbers; motorgliders. And motorgliders with L/Ds in the high teens and low 20s are pretty efficient cruisers. So the real question is; why aren't all light aircraft made to be like motorgliders? Well, motorgliders have their disadvantages. The long wing span is a problem fitting in on the ...


5

1. Aeroelasticity Unlike a tailplane, a delta wing is more rigid due to its much bigger chord and multiple spars, so control reversal due to aeroelasticity isn't a special concern. Big subsonic jetliners typically lock the outer ailerons at high speeds. Concorde featured a similar function for the outer elevon, but only if $V_{MO}$ is exceeded by 25 knots ...


2

You are basically describing an ejector pump. This would not be very efficient, I'm afraid (I'll try to dig up some comparative figures). The arrangement in this case would be a jet engine used to accelerate an amount of air inside a duct. With a long enough duct, the end result would be a single airmass exiting at (somewhat) uniform speed. Unfortunately, ...


4

That's because Bernoulli's principle breaks down at and above supersonic velocities. In the subsonic realm, when an atom is sped up, it doesn't have much time to spend pushing around a fixed place (it passes by that place quickly). In doing so, the static pressure drops as the speed increases. And in doing so, the neighboring atoms, who are also not doing ...


2

Ground effect only comes into play within about 1 wingspan’s height above a surface and it is largely unnoticeable until approx 1/4 to 1/10 of a wingspan above the surface. While it is true that ground effect greatly increases the lift to drag ratio of the wings, it does not reduce parasite drag nor does it account for the fact that it is more efficient to ...


1

As noted in the answer linked in comments, the practical limit for ground effect is about half the wingspan over a relatively flat, impermeable surface (you'll get better effect from smooth water than from heavy swell, for instance). Flying in ground effect for either economy (reducing power needed to fly) or in an emergency (engine out prevented climbing, ...


1

What height over a fixed plain can a plane begin to take advantage of the ground effect or when air is compressed between the wing and ground? It varies, but, generally speaking, ground effect usually becomes noticeable somewhere around half the plane's wingspan above the ground. So, for a plane with a 50-foot wingspan, ground effect would only be viable ...


8

No, both would reach the ground at the same time, although the one with the higher airspeed would go farther before hitting the ground, assuming atmospheric conditions are invariant. On the other hand, if both aircraft have the same flight path angle, the one with the higher airspeed would have a higher sink rate.


0

Well, both. Lift can be described as a moving wing colliding with air molecules at an angle, the result of the collision is the wing moves one way and the air mass the other, as per momentum physics. Moving the trailing edge, or the entire surface, increases the angle of attack, resulting in more lift at a given speed $V$: $Lift$ = 1/2 × Lift Coefficient ...


5

Thankfully, aerodynamics in the usual flight range is linear. Therefore, there is a gradient of lift over angle of attack and another one over the flap deflection angle. Both are constant over a range of maybe ±15° and can be combined. The angle of attack is referenced to the fixed part of the flight surface and the deflection angle to the moving part ...


2

A trailing-edge control surface, when it deflects, changes the camber of the overall airfoil. More camber means more lift, in whatever direction that airfoil is mounted. In your example, adding up elevator increases the horizontal stabilizer's camber, which increases the downward force it applies. Philosophically, "why" it does this is just, well, that's ...


5

Equating rate of climb with more lift is wrong. In order to climb the aircraft needs more thrust but a bit less lift. Therefore, the first statement is wrong: Adding more power does not create more lift. The only case where this happens is at high speed and the maximum sustainable lift coefficient: More power allows to fly at a higher load factor, but ...


0

Although the benefit is marginal, the exhaust output of a turbo shaft engine (aka a turboprop engine) is usually vectored to the rear of an airplane. This does add to the forward thrust of the plane.


1

The wings don't benefit all that much. Look at a small airplane from head on - how much wing is directly behind the prop? My airplane for example, the prop is 86" diameter. The fuselage is 46" wide - so 20" of wing on either side is behind the prop. The wing itself is 35'6", or 426", so less than 1/10" is in direct prop airflow. The prop pulls the plane ...


4

Provided the bomb has the same ballistic characteristics, the mass of the bomb would be irrelevant to this. Now the weight of the actual MK82 series 500lb bombs may have a distinct effect on the maneuvering characteristics of the attacking airplane during a bombing run as opposed to carrying a load of MK76 ‘Smurf Killers’ to a target.


3

Source: nap.edu Not really no. The velocity off a propeller does not shoot straight back, rather in circular motion (shown above and below). Source: High-Lift Propeller System Configuration Selection for NASA's SCEPTOR Distributed Electric Propulsion Flight Demonstrator, NASA From the conclusion of that NASA report on span-wise propellers generating lift: ...


0

I'd say yes, it's possible. I read about a late pilot testing modifications in his Velocity Canard; in flight, one of the propeller blades was lost, no thrust; instead of pushing the stick, the pilot pull, with the result of airplane entering a flat high speed fall into the ground, impossible to revert. This is also a caution about the not 100% true belief ...


8

The actual mass alone is nearly irrelevant to its a trajectory - what mostly matters is the mass/drag ratio - so if you scale the drag correctly, it feels quite the same - at least from the operational point of view. As one does not drop bombs by sight only, a small adjustment on the computer will do the other trick.


2

Both the airfoil shape and inherent properties of the air contribute to the stall angle of attack. You asked for a theoretical background, but I will list the factors that influence stall because there is no simple formula for it. The most important factor is the suction peak which develops right behind the stagnation point on the upper side of the leading ...


21

Most twin-engine aircraft with counter-rotating propellers have the rotation set up so that the propellers are rotating inward towards the center at the tops of the propeller arc. This configuration reduces the P-Factor effect at slow speed high angles of attack, and eliminates the "critical" engine that is present on multi-engine aircraft where both ...


4

There are two considerations. Two parallel vortices will tend to move upwards if the air flow between them is upwards. You can think of it as each 'blowing' the other upwards. This suggests the P-38 configuration maximises lift. I am not convinced but I don't have access to the software to know either way. Nor did they. The second consideration concerns ...


0

This observation is only partially correct. If you take a generic, General aviation airfoils (moderately high Raynolds number; ie > 5 Million) this observation appears to be correct. Why is this correct? Because most GA airplanes Fly between a limited maximum speed and stall speed which correspond to lift coefficients say 0.01 to 1.5-2.00. CL is (...


0

For an ultralight aircraft with a wing and tail design and a top speed of less than 60 mph, sweep of greater than 15 degrees does not serve a useful purpose. However, some sweep can help solve a discovered center of gravity issue by moving the center of lift rearwards. This was commonly done on bi-planes such as the Tiger Moth well before the supersonic ...


7

There are two primary benefits of wing sweep at ultralight speeds. The first is that it acts as variable dihedral -- the higher your angle of attack, the more dihedral effect you get from sweep. Secondly, when you see ultralights (or hang gliders) with a lot of sweep, it's generally for a tailless design that needs to get the tip fins/rudders and twisted ...


-1

It is not possible to have the rudder deflection because the huge amount of AOA and control law take that in consideration and cancel (fadeaway) the pilot rudder inputs.Activating the MPO the slabs moved max in order to get the nose down to lower AOA and after that we can get responses from the rudder


2

Let's try to derive something. Assume the x-axis points to the right and the y-axis points up. I'm gonna use the subscripts b and s for bottom and side. The incoming velocity vector is decomposed in two components along the axes, so $v_s=v\sin\theta$. Writing the total drag force as a vector: $\vec D=\begin{bmatrix}-D_s\\D_b\end{bmatrix}=\begin{bmatrix}-C_{...


3

In reality, in a strongly- deflected stuck-rudder situation, the pilot would be best advised to forget about the doors and just land the plane, selecting a runway with a strong crosswind component if at all possible. After all, when landing with a strong crosswind, it's not that uncommon to hold the rudder at close to full deflection in the "downwind" ...


0

Now regarding the incompressible and frictionless gas: if it is incompressible then there will be no pressure difference generating lift. If it is frictionless, then there will be no way to interact with the gas. Both conditions are too limiting to work together. Out of the two I would prefer to keep the viscosity because incompressible is like water where ...


0

There will still be airflow over both the body and the wings of the F-16 or any other aircraft when it flies supersonic. Moreover, the F-16 features a blended wing-body which makes it hard to strictly separate the two. The body of the aircraft produces some amount of lift. Your question seems to ask if this lift is required, at supersonic speeds in ...


0

I wouldn’t work out the static pressure at the inside as equal to the static pressure of the undisturbed air away from the plane because the low pressure is forced at the inside of wing due to outside airflow vacuuming the inside air. What you need is to build an internal structure, add resistance to the skin and redo the calculations. Let the inside static ...


0

This may be a good time to dig into the formulas your software uses and try doing them manually. It seems XFLR v5 is doing a few steps at a time. Changing tail incidence indirectly affects the static stability nuetral point by assuming CG will move to new center (of lift) pressure. Moving CG changes the torque arms of the stabilizing and destabilizing ...


1

The answer depends on several factors: Bank angle. The minimum speed rises with the inverse of the cosine of the bank angle. A steep bank angle is preferred for best climb because it helps to fly as close to the core of the thermal as possible. Wing loading. With old designs made from wood and fabric you can get as low as 18 m/s, with modern composite ...


0

Angle of attack is defined as the angle between the chord line of an airfoil and the relative wind. Now the relative wind is the apparent wind flow opposite the direction of that airfoil movement. The relative wind depicted in these diagrams is for the airfoil we call the wing! Put the relative wind vector for the rotating propeller blades opposite their ...


1

Tires. Tires are under very high stress normally due to the compromise between extreme performance requirements and low weight. Tires aren't cheap. By doing a higher-speed-than-necessary takeoff, you're increasing tire wear and inviting tire problems that you just don't need to invite. It would be a very false economy to save a little in what exactly, ...


0

The lift is a force perpendicular on the cord of the wing. As a vector it points up and backwards, towards 11 o’clock. The horizontal component points towards 9 o‘clock and is called drag. The drag is cancelled by the thrust, a force pointing towards 3 o’clock. The vertical component of lift points towards 12 o’clock and cancels the weight (pointing at 6 o’...


3

This is quite a flabbergasting question... As you're standing on the floor of your home, Lift equals Weight. The lift is supplied by the floorboards to your feet. That's static lift. If you tilt up a floorboard and pull a toy car across it, it will be lifted upwards and even fly up after the floorboard ends. That's dynamic lift. Note that the required ...


4

In straight and level flight at an airspeed V, the weight W is balanced by the lift L, and the aerodynamic drag D is balanced by the thrust T. D is much smaller than L. In a passenger liner, D may be 1/12 ... 1/20 of the weight, or so...


3

It is theoretically possible to do a flaps up take off if you have a long enough runway, but why would you want to? An airplane is designed to be most efficient in the air, so the sooner you get there, the better. The lift available to an aircraft is proportional to the area of the wings (and flaps increase that area), however, the drag those wings produce ...


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