# Tag Info

5

If you ever read something along the line that "even the slightest" cause will result in "very bad changes" when before all was stable, you know that the writer was out of their depth. But you have come to the right place. The way you formulate the question it concerns lateral stability. The combined center of all side forces (= lateral ...

2

A quick refresher: the lateral axis runs through the wings and rotates in pitch. The vertical axis runs through the top to bottom and rotates in yaw. The longitudinal axis runs nose to tail and rotates in roll. So we talk about longitudinal stability around the lateral axis. (Ah ha!) So the relevant control surface is the elevator. Placing the center of ...

-1

I think you have a major misconception about stability. Trim and stability are related but they are not the same thing. A stable aircraft will always seek its equilibrium or trimmed state if disturbed. In a well designed aircraft, the equilibrium is reached by the aircraft automatically without pilot input whatsoever. That is the neatness of it. In your ...

0

Other answers have already pointed out that In a steady-state descent, for the simple case where the thrust vector is considered to act parallel to the flight path, lift =weight * cosine (descent angle). Therefore, the larger the descent angle, the smaller the lift vector must be. However, for reasonably small descent angles-- say 20 degrees or less-- ...

2

you need to specify how that changes Lift/Drag ... Bianfable That's the answer. Lift requirement in descent is defined the same way as in a climb: Cosine angle(from horizon) x weight The rest of the vertical lift requirement is from the the vertical drag component. (In a climb it is from the vertical thrust component). When you dump spoilers and flaps, ...

4

Yes. in an unaccelerated descent - or ascent -, lift - the force acting perpendicular to the direction of motion - scales with the cosine of the angle of flight from horizontal. Hence, an aircraft in a 90 degree dive will have no lift. This is independent of airspeed or the method by which lift reduction is achieved. Here's a graph of lift Vs glide ratio

-1

A: The lifts produced by the two aircraft must be the same. As given the vertical and horizontal speeds are constant. This requires all accelerations, including vertical acceleration, to be zero. Therefore the vertical component of lift must equal weight. This statement applies the same to both airplanes in their particular scenarios. However knowledge that ...

6

As you've already learned from the comments, the hard part of getting to space isn't the height, it's the speed. The Saturn V broke the sound barrier less than 30 seconds after launch. So, how is your refueling plane even going to keep up? The only possible answer, at least with modern technology, is to make it a rocket itself. So, now you've got two rockets ...

0

We draw the connection by noting that lift is a force, and that force is pressure/area $L/S = 1/2 * r * v^2 * CL$ L/S is the average pressure difference between the top and bottom of the wing! Now, we rearrange your equation to make a fixed pressure - or rather in this case, a pressure difference - the subject. P2 - P1 = (1/2 * r * v2^2) - (1/2 * r * v1^2) = ...

-1

I think most answers has missed phugoid oscillation which is a key word here. Most general aviation airplanes are designed to be more or less stable in flight once trimmed for the actual situation. This means that if the flight is disturbed somehow, the plane most often returs to a new stable situtation. This goes for all the three inputs: roll, yaw and ...

5

You are not interested in wingtip vortices. For now you are barking up the wrong tree. For that tug idea to work, you need to focus on the trailing vortex which is a result of the whole wing, not the wingtip. How to maximize trailing vortex intensity? Fly slow, use a small aspect ratio and create lots of lift. This means a heavy airplane with a high span ...

7

Don't overthink it too much. Simplify in your mind. Trim is used to pre-set hands-off angle of attack. Static stability forces will focus on regaining the trimmed angle of attack if the plane is displaced from its trim state. Since pilots use airspeed as a proxy for angle of attack, trim sets hands-off airspeed, as far as the pilot is concerned. The best ...

6

The main purpose of the tail surface is to prevent the pitch instability of the wing from affecting the whole aeroplane. This is why it is often called the horizontal stabilizer. In the simplest case, the stabilizer exerts no lift force when the plane is in trim. If the nose lifts, the tail AoA increases and it starts to generate lift. The design is arranged ...

15

But assume, there comes a wind gust, and the angle of attack will increase. This causes the center of pressure to move forward in front of my gravity location. So the aircraft will become instable because the angle of attack will further increase (nose up). The only thing the pilot could do is to use the elevator to trim in pitch. This is not accurate. Are ...

12

I think you misunderstand how it works, and how you would respond to changes in wind. Increases in airspeed impact all flight surfaces, including the elevator, so a change in airspeed due to a gust isn't going to create large changes in pitch. There will be some change, however typically the changes in pitch would balance out with the fluctuations. It's rare ...

-6

The answer is yes. Lift is greater than gravity in a climb. As noted elsewhere, the aircraft climbs because the force of lift acting vertically is greater than the force of gravity. Therefore total lift (of which vertical lift is only a component) is even greater than gravity. The normal procedure for climbing in an aircraft is to increase pitch while ...

4

Admittedly, the 10 degree sloped roadway is an anomaly, and you would never see an actual runway built on this grade. Don’t try to apply any of this logic to a shallower runway without some careful analysis. Here's my take on the possible landing approaches. I think my performance estimates are generally conservative. I think the flight maneuvers are ...

3

Unless the wind is too strong you should try to land uphill. Tailwind will increase the required length of the runway. When you land uphill you have a significantly larger angle between your glide path and the hill slope compared to landing on a horizontal runway. That means the required length of the runway decreases. Both effects work against each other ...

2

Landing downhill/into the wind may result in more wear on the brakes if done often, but would be much safer as an emergency procedure, especially in unfamiliar terrain. A 10 degree downslope on a road would be a 17% down grade, a very rare road indeed. The glide slope of a 172 is around 1:8 which works out to a 7 degree downslope (using sin and arcsin ...

1

If there was no wind, I'd probably land uphill, but in the circumstance you describe I'll take the into-wind direction landing down the hill. This gives the lowest ground speed and therefore the lowest energy state if things don't go so well (you have a 30 knot energy difference between the two landing directions - that's a lot, and a bigger deal all around ...

1

Plugging in some values using a True Airspeed calculator seems to confirm Peter Kampf's conclusion that the "steps" are indeed coinciding with transonic flight at various IAS and altitudes. There for, there is no "over lapping" of the curves, though they appear to be on the graph. One might suggest the flattening of the curve once the ...

5

As an airplane approaches Mach 1, all pressure changes grow with the Prandtl-Glauert factor of $\frac{1}{\sqrt{1-Ma^2}}$. Therefore, the lift curve slope increases so the wing produces more lift at the same angle of attack and dynamic pressure the closer its Mach number is to 1. On wings with thicker airfoils and higher aspect ratio the maximum lift ...

11

what is causing the "corners" that we see at the following points? Compressibility. Close to Mach 1 the lift curve slope increases according to the Prandtl-Glauert rule with the factor $\frac{1}{\sqrt{1-Ma^2}}$. Since the X-axis shows indicated speed, the Mach 1 point moves left with increasing altitude. Technically, those corners should also be ...

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