Classic airplane vector problem issues

In summary: So the only thing that changes is the relative motion between the airplane and the medium.In summary, the velocity of an airplane flying with a given wind is not affected by the density of the air.
  • #1
Pengwuino
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So there are two vector problems that seem to be classic: someone trying to swim across a river that is flowing and an airplane flying with a given wind. For example, one might say a swimmer is swimming at 1m/s across a river with a .5m/s flow of water. Of course, you do the exercise to find you have to swim at an angle to reach the other side if you wanted to reach the other side without having drifted along the river bank. The other is the aircraft that travels at something like 200m/s North with a 10m/s crosswind and you find the "true speed" using vectors. I've had issues with how this problem is explained.

With the river problem, it is more convincing that this is a realistic problem. However, I have never convinced myself of the airplane being realistic. If you are given the same problem with a very low density gas/atmosphere with said 10m/s crosswind, I can't imagine it would result in the same resultant velocity!

The idea encompassing everything in this problem is the idea that an object "moves with its medium". Taking it to the extreme, imagine a real airplane (in size), a very low density gas with a very high average velocity of the particles, and I can't believe such a method is valid. So the question seems to be what conditions are necessary to facilitate such a simple analysis of velocity vectors?

Hopefully I'm not the only person to think this strange... or maybe I'm missing something obvious.
 
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  • #3
As a Private Pilot, with some stick and rudder, wind correction angle experience, I believe I understand your question.

I would guess, course correction angle is different at 15,500ft MSL,(10kt crosswind), than the same crosswind component at 3000ft MSL. I would seem to agree the "crosswind vectors" are not the same, for different density altitudes.

I fly, or use to fly VFR. This correction angle can be calculated, but, from a "pracitcal" perspective, I was taught to pick a point on the horizon, and do what you need to do, to make the airplane head toward that point.

Hopefully this was of some help. Looking at Russ's post above, LOL, I have had a few nasty "Crab angles" on a few approaches myself, Colorado Springs seemed to have a constant crosswind, "sideslip" was always favorable to setting the plane down sideways, (in a crab).

A lesser dense gas has less force at a given velocity than a greater density gas, correct?
 
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  • #4
As the density decreases, the medium should have a smaller and smaller effect on the airplane's trajectory. I also have seen my fair share of videos of some crazy crosswinds, so I know it has an effect. However, why should its vector component be exactly the same as the wind's velocity? That's what I'm getting at, that's how it's taught. I have always seen this example given with the idea that an object moves with its medium. That's it. Done. No asterisks. It really doesn't make sense that there seemingly is no frequently cited criteria as to where this approximation no longer becomes valid.
 
  • #5
Pengwuino said:
As the density decreases, the medium should have a smaller and smaller effect on the airplane's trajectory... However, why should its vector component be exactly the same as the wind's velocity?
Take the concept of a crosswind out and reexamine:

1. What effect does the density of the air have on a plane flying in no wind?
2. Does a hot air balloon fly at different speeds in different densities of air?

Perhaps the concept of airspeed (and the fact that there is more than one...) would be of some help here: http://en.wikipedia.org/wiki/Airspeed
 
  • #6
The mistake you are making is that you are thinking of this as like the wind blowing against a car or a person who is standing on the ground. In that case, the "density" of the wind affects the friction force the wind applies.

However, in the case of an airplane flying, it is the wind (air) itself that is supporting the airplane. Think, instead of a wind blowing across a toy car, of a toy car moving on a table while the table itself it carried to the side. That is the situation with the airplane and a cross wind.
 
  • #7
Maybe there is a better way to think about this. The airplane is immersed in a medium. All the forces other than gravity come from the medium no matter how thin. Now imaging the medium is stationary and the Earth is moving underneath. It really doesn't matter how fast the Earth is moving underneath. None of the forces changed when we look at the Earth as stationary and the medium as moving as long as we see the airplane as moving along inside the medium.
 
  • #8
HallsofIvy said:
However, in the case of an airplane flying, it is the wind (air) itself that is supporting the airplane. Think, instead of a wind blowing across a toy car, of a toy car moving on a table while the table itself it carried to the side. That is the situation with the airplane and a cross wind.

This is probably the best example that shows what I'm thinking about. The thing I see as the issue here is that the medium is supporting the toy car but the car is forced to move with the medium due to the fact that there is no slippage perpendicular to the motion of the car. Now, assume that instead of tires, there are ball bearings supporting the car. I am more capable of identifying that sort of situation with the airplane since the medium can go around the airplane. Of course, the table doesn't go "around" the car with ball bearings, but the perpendicular motion of the table doesn't contribute exactly to a perpendicular motion of the car. Increasing the coefficient of friction from the ball bearing/table interface would give the car a higher perpendicular velocity (alternatively, make it frictionless and the car's speed is unaffected). In my mind, I can identify the density of the air for the aircraft with the frictional coefficient of the ball bearings. Altering either should change the resultant vector.

@russ: I'm strictly considering the airplane from the Earth (non-rotating) reference frame and I assume the airplanes velocity (with wind not considered) is directly North (let's say). I put an engine on the air plane and the thrust is directed true South. I then consider this moving medium (the wind) and want to see how valid the idea of "moving with the medium" is.
 
  • #9
Pengwuino said:
The thing I see as the issue here is that the medium is supporting the toy car but the car is forced to move with the medium due to the fact that there is no slippage perpendicular to the motion of the car.

When the plane is flying through the medium, there is no appreciable "slippage" perpendicular to the motion of the plane either. Why would you introduce a condition that isn't present?

The air density will affect the speed that the plane travels through the air mass, not the direction. That will alter the vector slightly but most problems assume the corrected air speed.

Pengwuino said:
I then consider this moving medium (the wind) and want to see how valid the idea of "moving with the medium" is.

Very valid. It is as accurate as the measurements of the air mass' and plane's speed and direction.
 
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1. What is the classic airplane vector problem?

The classic airplane vector problem is a physics problem that involves finding the resultant velocity of an airplane when it is flying in the presence of wind. It requires the use of vector addition and trigonometry to solve.

2. Why is the classic airplane vector problem important?

The classic airplane vector problem is important because it has real-life applications in aviation and navigation. Pilots need to understand how to calculate and adjust for wind velocity in order to accurately navigate and reach their destination.

3. What information is needed to solve the classic airplane vector problem?

To solve the classic airplane vector problem, you will need the velocity of the airplane, the direction it is flying, and the velocity and direction of the wind. This information can be represented using vectors.

4. What are some common mistakes students make when solving the classic airplane vector problem?

One common mistake is forgetting to convert the angles given in degrees to radians before using trigonometric functions. Another mistake is using the wrong sign for the direction of the wind velocity, which can result in a wrong answer.

5. Are there any tips for solving the classic airplane vector problem?

Yes, one tip is to draw a diagram and label all the given information and the unknown variables before attempting to solve the problem. Another tip is to break the velocity vectors into their x and y components to make the calculation easier.

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