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How can a lateral wind gust create a lateral acceleration?

  1. Jun 18, 2016 #1
    Hello everyone! I have some difficulty to understand what momentum actually is, can you help me understand this concept?

    1. The problem statement, all variables and given/known dat

    If a quadrotor of mass $1kg$ flying along the positive $x$-axis at a speed of $4m.s^{-1}$, why would a wind gust providing an effective force of $5N$ along the $y$-axis would give the drone an acceleration of $20m.s^{-2}$?

    Furthermore about moment.

    if this gust provided a moment of $1 NewtonMeter$,

    - why would the drone, which would have vertical thrust of $9.8N$
    (along the $z$-direction) not be able to hover?
    - Why would it gain acceleration on the $y-axis$?

    2. Relevant equations

    Linear momentum is a vector quantity, possessing a direction as well as a magnitude:
    $$\mathbf {p} =m\mathbf {v}$$

    3. The attempt at a solution

    I understand this is the $x$ speed times the effective force of the gust but not what is the physical explanation about it.
  2. jcsd
  3. Jun 19, 2016 #2
    Linear momentum of a system remains conserved if no external force is acting on the system.
    This is just the general form of conservation of linear momentum.
    Suppose external force ia acting along x-axis then linear momentum along y and z axis will remain conserved.The effect of force can only be seen along x-axis.
    Now you can use this information to solve your question.
  4. Jun 19, 2016 #3


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    I assume you are asking why is it not 5ms-2. The short answer is that I do not know. Perhaps it is some tricky aerodynamic effect related to how quadcopter rotor blades work.

    If we ignore that and just treat it as an object flying through a crosswind, we have to take into account the way the forward motion of the object turns the crosswind into a partial headwind. If drag is kv2, the drag was 16k before the crosswind. If the wind has speed w, the drag with crosswind is 5=k(16+w2). We can break that into components, along and across the direction of travel, and combine with the continuing 16k thrust from the rotors to obtain the net horizontal force. But we have two unknowns, k and w, and only one equation.

    Perhaps we should treat the vertical thrust from the rotors as equivalent to a steady vertical upward current of air, speed u, maintaining the craft aloft by drag. I've no idea whether that is valid, and I still don't see it producing the 20ms-2 answer.

    Where did you get the "moment of 1Nm" from? Moment about where?
  5. Jun 19, 2016 #4


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    Is that the exact problem statement word for word?
  6. Jun 19, 2016 #5
    It is known that the forces cause accelerations such that the net force = m*a
    acceleration and force must be in the same direction in this equation.

    When the quadcopter flies into x+ direction, and then accelerates into y+ direction, then the copter originally has 0 velocity into y+ direction. The acceleration and the force must be applied into the y+ direction.

    there are two avenues which could be pursued in this investigation.
    a= 20 m/s^2
    net force=5N
    mass= x

    mass of the quadcopter is reduced during the flightime from 1kg--> 0.25kg'

    the other explanation

    only the force of the wind =5 N
    net force = x + 5
    a= 20m/s^2
    m= 1kg

    The quadcopter produces upward thrust into y+ direction, at 15 N, and there is an upward "wind" which also affects with force upon the quadcopter. Therefore the ##a= 20ms^-2##

    Therefore the net force = 20N

    In my honest opinion I don't really like this kind of "unclear" instructions for word problems. But I suppose both options could be likely solutions.
  7. Jun 19, 2016 #6
    the third choice which comes to mind is that

    the quadcopter falls down vertically and force of gravity + downward wind force + thrust downward cause an acceleration of ##20ms^-2##
  8. Jun 19, 2016 #7
    If the momentum is conserved in x axis (x is horizontal)

    Copter does have momentum because it has mass and constant velocity in the x-axis.

    I suppose the quadcoptr makes a smooth curve downwards (smooth dive)

    Effectively the copter travels like the curved part of a "pizza slice"

    Therr will be angular momentum.

    Then at the end state of motion. Acceleration should be 20m/s^2 in the y direction.
  9. Jun 19, 2016 #8


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    The problem statement and suggested answer appear to be flawed. The force along the y axis is given at 5 N. The mass is giving at 1kg. Acceleration = force / mass = 5N / 1 kg = 1 m/s2. This would be the initial acceleration, since once the quad starts to move in response to the gust, the force diminishes as the quad's velocity approaches the winds velocity.

    A newton meter is a quantity of energy, = 1 joule. Initially the quad has 1/2 kg (4 m / s)28 joules of energy. After the added energy, the quad has 9 joules of energy, so it's moving at 4.24264 m /s.

    A newton meter could be a quantity of energy, = 1 joule.

    The part about the quad not being able to hover doesn't make sense or is lacking information. The quad's reaction to a gusting crosswind could be to lean into the wind, or lean away from the wind, or not to lean at all, depending on the cross profile center of drag versus the center of mass, and depending on any self correction using gyros and accelerometers as inputs to keep the quad level despite gusts. Transition from hover to forward flight reduces the amount of power required for to maintain altitude, due to induced wash, but a speed change from 4 m/s to 4.24264 m/s is not going to make a significant difference in the power required to maintain altitude.
    Last edited: Jun 20, 2016
  10. Jun 20, 2016 #9


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    Or a unit of moment, depending on the geometric relationship between the Newtons and the metres.
  11. Jun 20, 2016 #10


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    Assuming that it is a moment or torque, the angular inertia is not given, so some information is missing.

    Assuming that 1 g is 10 m/s2, then if the quad became inverted and the props continued to produce the same thrust, then the quad would accelerate downwards at 20 m/s2. A vortex could cause a quad to go inverted, but a gusting cross wind would be unlikely to cause a quad to go inverted, it would probably roll, but not 180 degrees.
    Last edited: Jun 20, 2016
  12. Jun 20, 2016 #11

    I was wondering about what is the direction of the windforce upon the drone? What is the direction of the required acceleration of the drone? ##20 \frac{m}{s^2}##

    Is it strictly ##α=90^o## upwards measured from the horizontal x-axis? Intuitively when the positive x-axis is on the right hand direction, then I suppose the positive y- direction is directly upwards.

    I think that
    F =ma
    a = F÷m
    a= (5N) ÷(1kg)
    a = 5 m/(s^2)

    I could be wrong though but I was also wondering why you think that the force of the wind is not constant force? Elaborate please if you know better than I.
  13. Jun 20, 2016 #12


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    The problem statement mentions thrust in the z axis so I would assume z is vertical and x and y are both horizontal.

    Please can the OP check the problem statement has been given word for word as it appears to have errors or typo.
  14. Jun 20, 2016 #13
    Copters and quad drones tend to move forwards by the tilting of the axis of rotors.

    If the quad drone uses all of the thrust (100%) into forward movement. Then there is not usually enough lift force generated in order to stay afloat at the original altitude.
  15. Jun 20, 2016 #14


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    If the drone only has 9.8N of thrust it cannot hover over the same spot ( constant x,y) in a cross wind. To maintain position it must lean so that some of the thrust counters the cross wind. That reduces the vertical component below that required to hover over the same spot.
  16. Jun 20, 2016 #15


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    Helicopters, both real and model, induce pitch and roll movements by using the "cyclic", which changes the angle of attack of each rotor as it rotates about the central axis. The tail rotor is used for yaw movements. The "collective" changes the overall angle of attack for all rotors (models can have negative pitch for inverted maneuvers).

    Most model quad copters have fixed pitch propellers, with adjacent propellers rotating in opposite directions and propellers across from each other rotating in the same direction. As an example, when generating positive lift, the left front and right rear props rotate clockwise (viewed from above), while the right front and left rear props rotate counter-clockwise. Pitch is controlled by varying the speed between front and rear props, roll is controlled by varying the speed between left and right props, and yaw is controlled by varying the speed between the props that rotate clockwise with the props that rotate counter clockwise.

    Repeating my prior post, one way to get 20 m / s2 is for the quad rotor to go inverted (or reverse the direction of the props on a model that can hover or climb inverted), in which case it accelerates downwards, and assuming that 1 g = 10 m / s2.
    Last edited: Jun 20, 2016
  17. Jun 21, 2016 #16
    yes I got mixed up in the direction of the vectors but this is the correct interpretation by CWatters

    If the copter has upward thrust of 9.81N.

    In completely still wind conditions, the copter would maintain its position in the x,y,z coordinates. For the copter to hover in place. I suppose the assumption is that the same coordinates (x,y,z) are being kept for the copter.

    But when the thrust from rotors is kept at the same angle vertically (90deg from horizontal)... Any lateral wind (sidewind) would push the copter into the direction of the force.

    For the copter to maintain the same position in the windy conditions. The thrust needs to be turned into the wind, such that the vertical component still is able to keep the copter in the same z- coordinate (height). And the horizontal component has to be equal and opposite to the windforce.
    This would also mean, that the thrust magnitude would need to increase from the main rotor (in order for altitude to be maintained). This is the same thing as "leaning into the wind" I assume...

    The reason why the copter would acccelerate into the direction of the windforce is simply that the net force in the y-axis = windforce (5N). That is the basic idea in this simplistic modelling. Because in the vertical thrust scenario, there is nothing to counter the windforce in the y-axis. Therefore the net force = windforce, therefore the acceleration occurs into the same direction of the windforce.
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