# The Plane in the Wind Puzzle: Does a Constant Wind Affect the Round Trip Time?

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• SchroedingersLion
In summary, the wind decreases the plane's speed on the way to airport A by the same amount as it increases its speed on the way to airport B, but the retardation on the way to airport A lasts a greater amount of time than the boost on the way to airport B.
SchroedingersLion
Hi everyone.

I came across the following brainteaser:
A plane travels from airport A to airport B and then returns to A from B. There is no wind, both trips follow a straight line, and the plane flies at constant engine speed.
Suppose now that a constant wind is blowing from A to B. Will the required time for the round trip change or will it remain the same?

If you want to think about it first, don't scroll down.
Solution and my question:The given solution is that the round trip with wind takes longer. While the wind decreases the plane's speed on the way to A by the same amount as it increases its speed on the way to B, the retardation on the way to A lasts a greater amount of time than the boost on the way to B.
As an intuitive explanation, one is supposed to imagine the situation in which the wind speed is equal to the engine speed. Then the plane would be able to reach B twice as fast, but would not move from its place on the way back.

I try to write down some basic mechanics to show this, but I embarassingly failed.
I take "engine speed" to mean that the engine runs at a certain power ##P_0##. Let the distance be given by ##s## and the time taken for the trip ##t_0 = t_{A,0}+t_{B,0}=2t_{A,0}##. The power is related to the velocity by ##v_0=P_0 / F_0## with the force that needs to be overcome ##F_0##, so that the needed time can be expressed as $$t_0=2t_{A,0}=2 \frac{s_A}{v_0}=2\frac{s}{P_0}F_{0}.$$
Now, let the wind blow from A to B. Can't I just model it as a constant force ##F_w## applied in that direction?
Then, the force from A to B with wind is ##F_{A,w}=F_0 - F_w##, and the force from B to A ##F_{B,w}=F_0 + F_w##.
Then, applying the formula above, the required time will be
$$t_w = t_{A,w} + t_{B,w} = \frac {s} {P_0} (F_{A,w} + F_{B,w}) = \frac {s} {P_0} *2F_0 = t_0.$$

Where is my error? Needless to say, it is pretty frustrating to study physics for years, obtain an MSc., only to fail on petty brainteasers that are not even meant to require a physics education

Delta2
I think you are over thinking it.

In the rest frame of the air the plane has the same speed ##v## in either direction, so if the wind speed is ##w## what's the speed relative to the ground in each direction?

vanhees71 and SchroedingersLion
SchroedingersLion said:
The given solution is that the round trip with wind takes longer. While the wind decreases the plane's speed on the way to A by the same amount as it increases its speed on the way to B, the retardation on the way to A lasts a greater amount of time than the boost on the way to B.
Doing the same thing with light instead of an airplane was the basis of the Michelson–Morley experiment which should have measured a greater round-trip time depending on the speed of the aether relative to the line connecting A and B. It didn't show this, which falsified the aether-wind kind of thinking that came from Newtonian physics.

The M-M experiment calculations show that it doesn't matter which way the wind was blowing, only the speed of it. So if A is say east of B, a west wind or a north wind will both add the same time to the duration of the round trip, and if the wind is as high as the airplane speed, the round trip cannot be done at all.

SchroedingersLion
OK, one can assume that the plane's engine speed ##v## is always measured relatively to the medium.
In that case, with tailwind, one has ##t_1=\frac {s}{v+w}## and with headwind ##t_1=\frac{s}{v-w} ##.
Adding them, one sees that the resulting time is larger than the time without wind, i.e. ##\frac {2s}{v}##.

No need to talk about power or forces...So, I assume that in my original post, the two equations ##F_{A,w}=F_0-F_w ## and ##F_{B,w}=F_0 + F_w ## are rubbish then.

I don't know that it's rubbish, since you would formally use some sort of force balance equation (cruising speed is the point at which engine thrust equals drag) to show that the airspeed is what they mean by "engine speed". But then I don't immediately see what you'd do with power after that.

SchroedingersLion
Thanks Ibix!

## 1. What is "The Plane in the Wind Puzzle"?

"The Plane in the Wind Puzzle" is a physics-based puzzle that involves manipulating the direction and speed of a plane in order to reach a specific destination. It is often used as a problem-solving exercise in physics and engineering courses.

## 2. How does the puzzle work?

The puzzle involves a plane flying in a straight line with a constant speed, while being affected by wind blowing in a perpendicular direction. The goal is to adjust the plane's angle and speed in order to reach a designated target, taking into account the wind's influence.

## 3. What are the main principles of physics involved in this puzzle?

The puzzle requires an understanding of vector addition, forces, and motion. The plane's velocity and acceleration are determined by the net force acting on it, which is a combination of its own thrust force and the force of the wind.

## 4. Is there a specific strategy or approach to solving this puzzle?

There are many different strategies that can be used to solve this puzzle, depending on the specific scenario. Some common approaches include breaking down the forces acting on the plane into components, using trial and error to adjust the plane's speed and angle, or using mathematical equations to calculate the optimal solution.

## 5. What are the real-world applications of this puzzle?

This puzzle has many real-world applications, particularly in the fields of aviation and meteorology. It can help pilots understand and account for the effects of wind on their flight paths, and it can also be used to simulate and predict weather patterns. Additionally, the problem-solving skills and critical thinking required for this puzzle are valuable in many other scientific and engineering contexts.

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