# Is drift and sideslip angle the same in the accompanied figure?

Consider the following figure:

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?

• 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?! ;) Jan 16, 2021 at 1:11
• Relative to what? To the plane it's directly in the line of flight. To the ground it's the vector sum of plane velocity and the (plane and) airmass velocity. Just draw the vectors. So the plane heading will be equal to the drift angle, but in the opposite direction, in this case around 070 and the drift will put one on course at 090. Nov 18, 2022 at 18:33
• The direction the pilot wishes to fly (desired course) really isn't relevant to anything. Still, there may be enough information here to answer all the questions. The question that should have been put to the O.P. before anyone answered, was "do you intend the arrows to be vectors, i.e. does the length of each arrow correspond to the magnitude of the variable involved (groundspeed, wind)?" (Using "vector" in the physics/ engineering sense, not the ATC sense. which is just a direction to steer.) Nov 20, 2022 at 14:49

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.

• 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. Jan 16, 2021 at 10:01
• Just enter “side slip angle” in Google and click on the “images” tab. Jan 16, 2021 at 15:39
• Re "The answer to the title question is no, sideslip angle and drift angle are not the same thing. (in this figure, or any other...)" -- Elsewhere you've defined the drift angle as the angle between the aircraft heading and the ground track (see aviation.stackexchange.com/questions/95979/…). So, by this definition, isn't the drift angle the same as the sideslip angle, any time an aircraft is sideslipping in no-wind conditions? Nov 18, 2022 at 22:38
• @quietflyer, technically yes, but since the goal in most instances is to fly with zero sideslip, and since the illustration doesn't depict a sideslip angle, and because in cross country flight planning 101 and E6B training we aren't taught to factor in a planned side slip angle and adjust heading to achieve it, my answer is correct for all but those who choose to mitigate the effects of wind with rudder alone. Although the definition would then be incorrect because the aircraft really isn't drifting, it's just misaligned. Nov 19, 2022 at 14:14
• I'd submit that if the arrows are intended to be vectors, i.e. if length of groundspeed (or ground track) and wind arrows are intended to be proportional to the groundspeed and wind speed, then there is enough information here to say whether the plane is sideslipping or not. That's the basis of my answer. (But unfortunately the O.P did not clarify whether that was his/her intent or not.) Also, it appears most likely that the O.P. had some confusion about what sideslip means. Nov 20, 2022 at 15:17

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.

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 without 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).

• 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. Jan 15, 2021 at 16:05
• @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. Jan 15, 2021 at 16:19
• OK, I think with the last edit it should be clear ;) Jan 15, 2021 at 16:27
• 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. Jan 16, 2021 at 2:24
• @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. Jan 16, 2021 at 16:04

Since we all seem to be going "enhanced response" on this answer, let's start by interpreting the picture in the question: an aircraft attempting to fly 090 but is being blown off course.

with these parameters:

airspeed: 100 knots
crosswind: 15 knots from the north

Drift angle: arc tan (15/100) = 8.5 degrees
Groundspeed: cosine 8.5 = 100/x
x = 101 knots

If the pilot wanted to fly 090, then the heading would be 081.5 to correct for the crosswind.

Ground speed would be cosine 8.5 = x/100 (because heading is now the hypotenuse) = 99 knots.

One can see that correcting for a 90 degree cross wind does indeed slow groundspeed slightly.

– Jim
Nov 20, 2022 at 14:23
• @Jim relative to the ground point, the plane is "sideslipping", relative to the airmass, it isn't. More reading here. Nov 20, 2022 at 14:49
• He’s crabbing into the wind, but sideslip is zero (for his assumed coordinated flight). But my comment was because you go to great lengths to show that ground speed is decreased when crabbing into the wind. But that proof isn’t what OP was asking about.
– Jim
Nov 20, 2022 at 16:57
• @Jim acknowledge that point. I felt it kind of helps understand the geometry to discuss airspeed and groundspeed for both the drift and to hold 090. Nov 20, 2022 at 17:01

Disclaimer: since flying crosswise to the external, meteorological wind creates no inherent tendency for an aircraft to sideslip, as far as practical piloting is concerned we normally assume sideslip is zero in wind correction problems, in which case we could assume that the arrow labeled "desired course" is also the "airspeed" vector, simply because it is aligned with the aircraft heading. Therefore answering the particular questions asked here, in terms of the particular "givens" presented here (which do not seem to include an assumption of zero sideslip), is a purely theoretical exercise, with no real practical value.

This answer assumes that the all the arrows in the figure are intended to represent velocity vectors, whose length accurately corresponds to the speed (magnitude) of each of the three variables involved.1

The intention of the diagram appears to be that the "track", "wind", and "desired course" arrows could be re-arranged into a closed triangle.2,3 This means that the "track" (i.e. ground track or groundspeed) vector is exactly equal to the vector sum of the "desired course" arrow and the "wind" arrow.

This is a crucial piece of information. It tells us that the "desired course" arrow represents not only the direction that the pilot wishes to track over the ground, but also the direction and speed that the aircraft is actually moving relative to the airmass.

In other words, the "desired course" arrow is also the "airspeed" vector.4

Therefore we can see that sideslip is zero-- the aircraft is exactly aligned with the flight path through the airmass.

The relative wind vector is always exactly equal in magnitude, and opposite in direction, to the airspeed vector. So the direction of the relative wind is from right to left in the diagram. Or in terms of an azimuth, using the usual convention of naming winds in the direction from which they blow, the relative wind direction is 090.

Is drift and sideslip angle the same in the accompanied figure?

No, the sideslip angle is zero, and the drift angle is not zero.

Note that since the final groundspeed vector is a "given", and the wind direction and speed is a "given", and the aircraft heading is a "given", and the final airspeed vector (and thus the final sideslip angle) can be computed, this is not really a question about what happens when an aircraft flies into a new wind field and the pilot makes no control inputs to try to maintain the original heading, zero-sideslip condition, etc. It's already "locked in" to the question that those control inputs were made, if needed. A question about whether any particular control inputs would be needed to maintain these conditions, either as the aircraft transitioned into the new wind field, or after the aircraft had come into equilibrium in the new wind field, would have to be phrased differently.

Footnotes:

1. If this is not actually the intention, then the question contains insufficient information to answer most of the questions, unless the reader is supposed to make some assumptions about what control inputs the pilot is or is not making, etc.

2. If this is really the intention, it would be a kindness to the reader for the illustrator to actually draw the arrows in the form of a closed triangle, so that the reader would not have to guess as to whether the arrows really all match up in this way or not.

3. A very close examination shows that the "wind" arrow is slightly too long to fit in a closed triangle with the other two arrows. This answer assumes that this is not intentional. If it were intentional, the "airspeed" vector would be slightly longer than, and aimed slightly counter-clockwise of, the "desired course" vector. That would change the answers to questions 1) and 2), but not the answer to the title question. Answering questions 1) and 2) for this alternative case is still entirely possible, and will be left as an exercise to the reader! Note that this alternative scenario is only compatible with the "drift angle" label on the diagram, if we define the "drift angle" as the difference between the aircraft heading and the ground track, not the difference between the direction of the flight path through the airmass and the ground track. Sideslip is almost always assumed to be zero in wind correction problems, but in cases where sideslip is not zero, there is arguably some ambiguity around the exact definition of the "drift angle". For more on this, see this related ASE answer.

4. In addition to the argument based on the fact that the arrows appear to be intended to fit together in a closed triangle, we're essentially told in the first sentence under the figure that the "desired course" arrow also represents the airspeed vector before entering the new wind field, and we're not told of any changes in airspeed, power setting, or climb/sink rate, and we're told that the aircraft heading remains constant, and therefore it's only logical that the "desired course" arrow also represents the airspeed vector after coming into equilibrium in the new wind field. But this line of logic is weakened by the fact that the question seems to entertain the possibility that the aircraft may be flying in a slipping condition after coming into equilibrium in the new wind field. Therefore in the context of this question, the best way to find the flight path relative to the airmass is to do the vector subtraction of the wind vector from the ground track vector. A better presentation of the diagram would explicitly include the airspeed vector, i.e. the vector representing the flight path relative to the airmass-- but in labelling the diagram in such a way, the questioner would probably have realized the correct answer to all of his or her questions, before even posting!