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Consider the following figure:

Drift due to wind

Initially aircraft is moving along dotted line. But due to wind, aircraft ground track is changed to solid line but body still pointing parallel to dotted line. I've the following questions:

(1) What is sideslip angle in this case?

(2) What is direction of relative wind?

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  • $\begingroup$ The question would be better phrased "if there were no wind, the aircraft would be moving along the dotted line", or alternatively, "the dotted line shows the aircraft's path through the airmass". Phrased such a way (either of those two), the question would be answerable and totally valid for ASE, unless a duplicate of a previous question. Your use of "initially" suggests that the wind has suddenly arisen (i.e. a gust) which complicates things immensely. But I don't think that's what you really intended. (It might be too late to change it now though-- or not-- try it and see-- ) $\endgroup$ Jan 15 at 18:41
  • $\begingroup$ Both your suggested phrases are depicted visually in the diagram. I don't think they need a written explanation to clarify. A picture is worth a thousand words, right?! ;) $\endgroup$ Jan 16 at 1:11
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The answer to the title question is no, sideslip angle and drift angle are not the same thing. (in this figure, or any other...)

What the diagram shows is drift angle, and drift angle only. If you are moving through a uniform airmass that is itself moving relative to the ground, your path across the ground (ground track, or course) will be affected by the movement of the air. The difference between your heading (simply the direction the aircraft is pointing) and your actual ground track is the drift angle. That is the teaching point, and purpose of this figure.

Regarding the numbered questions:

  1. There is not enough information in the diagram to determine this. Sideslip angle is the lateral angle between the relative wind and the longitudinal axis of the aircraft. If the aircraft is in balanced flight, with the balance ball centered in the slip indicator, the sideslip angle will be zero. It doesn’t matter what your heading is, it doesn’t matter which direction the wind is moving, or whether the aircraft may be drifting from its intended course.

  2. The direction of the relative wind is always opposite to the direction the aircraft is traveling through the airmass. In this example, presuming balanced flight, the relative wind is coming directly from the East. If the aircraft was unbalanced, (balance ball out to one side or the other) there would be a sideslip, and relative wind would be striking against the side of the fuselage instead of head-on. That angle is the sideslip angle.

There are different and better illustrations that show sideslip angle if you need more help understanding what it is.

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  • $\begingroup$ Thanks for the answer. Regarding "There are different and better illustrations that show sideslip angle if you need more help understanding what it is", kindly illustrate for my better understanding of sideslip and relative wind. $\endgroup$ Jan 16 at 10:01
  • $\begingroup$ Just enter “side slip angle” in Google and click on the “images” tab. $\endgroup$ Jan 16 at 15:39
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The relative wind (as felt by an imaginary weather vane on the aircraft) is coming from straight ahead. The side slip angle is thus 0.

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Question is (from one perspective), flawed, or at best incomplete. You are dealing with two different frames of reference. 1) the frame of reference of the atmosphere, and 2) the frame of reference of the earth. Every variable or measurement must be identified and labeled to indicate which frame it is being evaluated in. for the sake of argument assume the wind is 5 knots and the aircraft speed is 100 knots.

First, the word "Track" implies the earth frame of reference. but the word Heading implies only the orientation of the aircraft. But let's assume Track can be measured in either frame...

In the Atmosphere, aircraft Heading is 090 Due East, Track (within the atmosphere), is also 090 (due East), Sideslip (difference between Fuselage reference line and relative wind), is Zero. airspeed will not change (100 knots). "Course" is meaningless as it is a concept relative to the earth frame of reference, not the atmosphere. It is actually somewhat the same as "Track".

Relative to the Earth frame of reference (in your diagram): aircraft Heading is still 090, Track (relative to the ground now), is south of due East (about 095), Sideslip (difference between Fuselage reference line and relative wind), is STILL Zero. airspeed is a concept relative to the atmosphere, so it is not directly measurable in the earth frame of reference. Ground speed on the other hand, will be higher (due to vectoral addition of small tailwind component -maybe 103-105 knots). "Course" is just "track" in the frame of reference (across the ground), so it will be to the south a bit (about 95 degrees).

By the way, in reality, if you actually did this (Take off on a runway aligned due East with a left cross wind), what would happen is this: First of all, to keep the aircraft tracking straight down runway centerline, prior to liftoff, you would need to add DOWNWIND (right) rudder to counteract the wind-milling effect of the crosswind on the vertical stabilizer. Without it the aircraft would turn upwind - into the wind. You would also need to add upwind (left) aileron to keep the upwind wing from rising and to prevent the roll away from the wing that the rudder would otherwise cause. As the aircraft lifted off (as the wheels lost contact with the runway), if you kept those cross-controls applied, the aircraft would be flying on a due easterly ground track, in a sideslip, Tracking straight down extended runway centerline, with a true airspeed velocity vector upwind (slightly north) of the runway centerline (about 085-090 degrees). (i.e., it's velocity vector IN THE MOVING ATMOSPHERE Frame). It would be moving slightly north of east through the atmosphere, as the atmosphere is moving south at the same rate, vectorially cancelling each other out to produce a ground track aligned with the runway (due east).

If at this point you neutralize the controls, what will happen is that aircraft will rapidly weathervane into crosswind, removing all sideslip. It will now be point "upwind" (080-085?), but will still be tracking due east along extended runway centerline.

Physics cannot be ignored, nothing (including aircraft) can change it's velocity with a force acting on it. An aircraft flying in a the air does not feel ANY force from "Wind". "Wind" is just the mathematical representation of the motion (velocity) of the entire atmosphere relative to the motion (velocity) of the earth (or the aircraft carrier, or the car you're riding in, or the train you're on, etc. etc. The only force an aircraft feels is from the "relative" wind, the motion of the air across the aircraft skin due to it's motion THROUGH THE AIR, not across the ground. (and the thrust of the engines and it's weight).

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  • $\begingroup$ Your use of heading and track is somewhat misleading. Track is usually a term reserved for the Earth frame. (Magnetic) Heading is defined w.r.t. the Earth's magnetic field, so it does not change between atmosphere and Earth frame of reference. $\endgroup$
    – Bianfable
    Jan 15 at 16:05
  • $\begingroup$ @Bianfable, This was deliberate. This implied meaning of track, mixed with assumptions about heading and ignoring difference between two frames of reference is the cause of the confusion. $\endgroup$ Jan 15 at 16:19
  • $\begingroup$ OK, I think with the last edit it should be clear ;) $\endgroup$
    – Bianfable
    Jan 15 at 16:27
  • $\begingroup$ The question is not flawed, but there are two flaws in this answer: Heading is the same for either frame of reference, and 5 knots direct crosswind would not net you 105 over the ground if you were at 100Kias. Also, crosswind takeoff technique wasn't really a part of the question. Downvoted for these reasons. $\endgroup$ Jan 16 at 2:24
  • $\begingroup$ @Michael, Perhaps "Flawed" is inappropriate or wrong word to use. I have edited it to attempt to be more clear. - but actual numbers are clearly NOT intended to be perfectly accurate. Adding trigonometric formula or extended decimal values would add nothing to intended meaning. $\endgroup$ Jan 16 at 16:04

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