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Assume an aircraft is flying at an angle of attack of 10 degrees, and that the incidence angle between the aircraft fuselage and the wing as well as the angle between the wing and the engine mount is also 0 degrees, the thrust force is at 10 degrees relative to the incoming velocity. My question is, since there exists an angle between the incoming the airflow and the thrust force, how does the aircraft keeps propelling forward straight?

I mean if I break down the thrust force into its components, then one component can be added in the direction of lift, and other component can be used to counteract the drag. But why would this make the aircraft to keep moving in a straight path? Because aircraft itself doesn't know how the component breakdown should be established, and also it just sees a resultant thrust force. Moreover, theoretically, I can make the component breakdown of the Thrust force using a different co-ordinate system, but still why would the aircraft keep on propelling in the same straight direction?

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  • $\begingroup$ In such a case, Lift is less than Weight. See aviation.stackexchange.com/questions/40921/… , second-to-last paragraph, for an actual equation, not limited to horizontal flight by the way. Surely this question may be a duplicate of some other? $\endgroup$
    – quiet flyer
    Nov 13, 2020 at 14:59
  • $\begingroup$ Let me check out the question you attached. If it is duplicate, what should I do? $\endgroup$
    – Rameez Ul Haq
    Nov 13, 2020 at 16:32
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    $\begingroup$ The plane moves straight because the pilot is telling it to do so. $\endgroup$
    – Jpe61
    Nov 14, 2020 at 19:45
  • $\begingroup$ @RameezUlHaq To arrive at a steady flight state, the aircraft initially needs to accelerate to this stage, so until it is reached, forces are not in equilibrium. Once it has attained the desired state, forces are reset by use of throttle and stick so they are in equilibrium. To move on in a straight path does not require more forces than those which balance out. $\endgroup$
    – Peter Kämpf
    Nov 15, 2020 at 15:58
  • $\begingroup$ You don't break down forces, you sum them up, divide by mass and get the acceleration. And then you sum their moments, divide by moment of inertia and get the angular acceleration. Whenever both sums are zero (because pilot set the controls to make them so), the aircraft flies straight. Breaking down forces is only a mathematical tool to make summing them up easier. $\endgroup$
    – Jan Hudec
    Nov 16, 2020 at 21:39

2 Answers 2

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Let us assume level flight. Then the forces acting on the aircraft are shown in the following sketch (not necessarily to scale):

Forces

The forces are :

  • the total aerodynamic force $ F_A $, which is split into two components: lift $L$ (perpendicular to direction of motion) and drag $D$ (parallel to direction of motion)
  • the weight $W$
  • the thrust $T$, here acting slightly upwards because of the angle of attack $\alpha$

For un-accelerated level flight, the sum of all forces must be zero. This is independent of the coordinate system you use to split the forces. In our example, this is given by these equations:

$$ L + T \sin \alpha = W $$ $$ T \cos \alpha = D $$

At a given angle of attack, these equations will only be true at one particular airspeed. At higher airspeed, the lift would increase resulting in a climb, which then changes the angle of attack. At lower airspeed, the lift would decrease resulting in a descent.

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  • $\begingroup$ What I mean to say is this: Since, in a level flight, all the forces acting on the aircraft are zero. But why does the aircraft propels in the direction opposite to drag? It can move in any direction with constant speed. But why only in the direction opposite to drag? $\endgroup$
    – Rameez Ul Haq
    Nov 13, 2020 at 16:29
  • $\begingroup$ A net force causes a change in momentum (and therefore at fixed mass, velocity), but it does not have anything to do with the current direction of the velocity. The aircraft keeps flying forwards because that is the direction it was flying in before. The net force equal to zero just means no change. $\endgroup$
    – Bianfable
    Nov 13, 2020 at 16:35
  • $\begingroup$ By the way: drag is by definition opposite to the direction of motion (relative wind). $\endgroup$
    – Bianfable
    Nov 13, 2020 at 16:46
  • $\begingroup$ So in order for the pilot to make an aircraft fly in a specific direction at constant speed, he needs to make use of some control surfaces to achieve that desired direction of motion, and as soon as it is achieved, he should immediately alter the control surfaces such that the overall forces on the aircraft again becomes zero? Subsequently, the aircraft will be able to move in that desired direction. Am I correct? $\endgroup$
    – Rameez Ul Haq
    Nov 14, 2020 at 6:38
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    $\begingroup$ @RameezUlHaq Yes, that is exactly what the pilot does: use control surfaces to fly the aircraft and once you are on your desired path, you trim the aircraft to keep the net force to zero without control inputs. $\endgroup$
    – Bianfable
    Nov 14, 2020 at 10:25
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The pilot controls the direction of flight with elevator and throttle. Pitch to the horizon is determined by this combination provided the aircraft is making enough vertical lift to equal its weight.

So, to get level flight, one would reduce AOA and/or throttle until the sum of the vertical lift components equals weight at zero pitch (to the horizon) angle as shown by the lift equation:

$Lift$ = Area x Density x Coefficient (camber/AOA) x V$^2$

Just like any other plane, to climb, add power, to descend reduce power.

We must take note the drag vector in the direction of flight must equal the thrust vector contributing to velocity in a realistic force diagram. The decomposition of the thrust vector is done to show the vertical component contributing to vertical lift. "T-D" illustrates how much of the total thrust vector is available for the creation of airspeed. Graphing Vy and Vx will show a greater value of "T-D" for a given amount of Thrust at Vy.

And here (finally), a vector diagram, drawn to scale. Note how much drag the angled wing lift vector is making.

enter image description here

So one may muse: why not leading edge SLATS for climbing. Moving the Lift vector to vertical makes T-D = ... T!

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    $\begingroup$ "We must take note the drag vector must equal the thrust vector in a realistic force diagram." Almost correct: the drag vector equal the component of the thrust vector parallel to the drag vector (i.e. total thrust must be higher than drag if the engine was mounted at the wrong angle). Normally this little detail is insignificant but since you explicitly mention engine angle, I thought you may want to be slightly more precise. $\endgroup$
    – Sanchises
    Nov 14, 2020 at 9:40
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    $\begingroup$ "Pitch to the horizon"-- seems to mean pitch attitude but could edit to be more explicit. "provided the aircraft is making enough vertical lift to equal its weight."-- seems to imply aircraft will not reach equilibrium until vertical component of lift vector equals weight, which is not exactly true even in horizontal flight, if there is upthrust or downthrust. Maybe answer meant to say "provided the sum of upward vertical force is equal to weight"? $\endgroup$
    – quiet flyer
    Nov 14, 2020 at 13:03
  • $\begingroup$ Re "We must take note the drag vector must equal the thrust vector in a realistic force diagram."-- not exactly, even in horizontal flight, if there is upthrust or downthrust. Rather, for horizontal flight (and only for horizontal flight, we can say the horizontal component of thrust is equal to drag. $\endgroup$
    – quiet flyer
    Nov 14, 2020 at 13:04
  • $\begingroup$ Re "Downthrust is added to most GA aircraft engine mounting, which avoids this clumsy configuration of forces and allows the aircraft to fly more efficiently on its wing."-- arguably, for whatever horizontal cruising airspeed the wing incidence angle (relative to fuselage) is optimized for, zero downthrust would make the most the most efficient use of the thrust vector, by keeping it purely horizontal. The downthrust may be there for other reasons, relating to pitch stability characteristics, compensating for factors such as varying downwash over tail at various thrust settings, etc. $\endgroup$
    – quiet flyer
    Nov 14, 2020 at 13:08
  • $\begingroup$ Could be a good ASE question-- "Presuming that the wing incidence angle is optimized to streamline the fuselage in some specific regime, why design any downthrust into the engine mount?" A quick look over the site suggests that this would not be duplicate of any existing question. $\endgroup$
    – quiet flyer
    Nov 14, 2020 at 13:10

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