Airplane with or without wind

[BMAD]

An airplane flies at constant airspeed c directly above a closed polygonal path in a plane, completing one circuit.

 

Prove or disprove that, compared to no wind, the presence of a wind of constant speed and constant direction will increase the time required.

[k-man]

OK, here is an accurate description of the simplified analysis.

Suppose a plane flies from point A to point B that is due west of point A.

A cross (north or south) wind does not impede his westerly progress. If there is a constant north wind, say, the plane will be blown off course and finish its westerly travel at a point that is south of point B. But if for an equal time there is also a constant south wind, the plane will arrive at B, and its travel time will be the same as if there were no cross wind. This assumes the plane maintains a westerly bearing during its flight.

In the OP, the wind is constant, and the course is closed.

Choose points A and B to be extremal on the path in the direction for which the A to B and B to A travel times are equal. This can always be done. Then collapse the path to the line connecting A and B. Now the cross wind component can be ignored, and the analysis of the head/tail wind applies: the component of the wind along AB is the effective magnitude of the wind.

 

I think your simplified analysis still doesn't address the cross wind correctly. Let's say that the closed path is from point A to point B and back in a straight line. It will take longer with the constant cross wind than without any wind.

 

The red arrow is the constant air speed, the blue arrow is the constant wind speed. The sum of the two vectors is the resulting ground speed (purple vector). If without any wind the plane gets from A to B in a unit of time, then with the cross wind present it must change its course to compensate for the wind. As you can see from the picture on the right, the length of the red vector is the same and now the purple vector is pointing toward B, but is a bit short. That's how far the plane will go in the same unit of time under cross wind.

 

The general case can be viewed as the trajectory of the plane relative to a hot air balloon in the air. The hot air balloon will move with the speed of wind in the direction of the wind, so from the hot air balloon frame of reference, the airplane will traverse a trajectory (purple line) different than the one observed from the ground (black line). Each section of the trajectory can be analyzed in the same way by computing a heading vector based on the desired destination and the wind speed. The resulting ground speed vector will determine the time it will take to complete the section of the trajectory. Head/tail component of the wind will slow down/accelerate the plane (not equally as bonanova has explained), while the cross component of the wind will always slow the plane down. As a result, the trajectory relative to the hot air balloon will always be longer (with constant wind) than the ground trajectory.

[bonanova]

This is one of these counter-intuitive questions that asks us to take an

average over some particular quantity, when there is a choice.

Remember the old question about the guy who walks up a hill at a certain speed and asks how fast he must return to average twice that speed? One might guess it's three time his ascent speed, because the average of 1 and 3 is 2. But in fact it can't be done, since he's used up his allotted time during the climb. He would have to return infinitely fast. The point is you must average speeds over the time spent going at different speeds. You can't get the correct result if you average over the distances traveled at different speeds.

For that reason a (constant) wind of any speed will slow an airplane's traversal of a closed course. It's a headwind for parts of the trip and a tailwind for another parts. But the time lost to the headwind can not be recovered from the tailwind.

Here's the proof.

Work out the hill problem first. It's simple to do. Then, inspect the several straight-line paths of the polygonal course. Take into account the vertex angles must sum to 2pi. Even simpler, Just collapse the polygonal path unto a single straight line by ignoring travel perpendicular to the direction of the wind, where it has no effect on the plane's speed. If the polygon is convex, we're done: it's the original problem. If not, it can be split into several pieces of retraced linear paths, each with the same result: the overall transit time is increased by the wind.

Because we are traveling at plane speed plus or minus wind speed for equal distances, we will lose more time to the headwind than we gain from the tailwind. The transit time will be increased by the wind.

[harey]

by ignoring travel perpendicular to the direction of the wind, where it has no effect on the plane's speed

Even here, it increases the travel time.

 

A non-zero component of the plane's speed is used to compensate the wind.

[bonanova]

@harey

You're right. I need to rethink the effect of flattening the polygon.

[bonanova]

OK, here is an accurate description of the simplified analysis.

Suppose a plane flies from point A to point B that is due west of point A.

A cross (north or south) wind does not impede his westerly progress. If there is a constant north wind, say, the plane will be blown off course and finish its westerly travel at a point that is south of point B. But if for an equal time there is also a constant south wind, the plane will arrive at B, and its travel time will be the same as if there were no cross wind. This assumes the plane maintains a westerly bearing during its flight.

In the OP, the wind is constant, and the course is closed.

Choose points A and B to be extremal on the path in the direction for which the A to B and B to A travel times are equal. This can always be done. Then collapse the path to the line connecting A and B. Now the cross wind component can be ignored, and the analysis of the head/tail wind applies: the component of the wind along AB is the effective magnitude of the wind.

[bonanova]

@ k-man: You are exactly right, and all I can say at this point is, Well at least I tried.

 

I conveniently ignored the crucial fact that the vertices all must be reached along the closed path. If that really, really unreasonable constraint were lifted, and all the plane had to do was to start and finish at point A, then ignoring the crosswind would be permissable.

 

However,   in any event, since the OP only asked for it to be shown (or not) that a wind adds to the travel time, and does not require that we say precisely how much, the fact that I swept some of the time penalty under the rug (and a very transparent rug it turned out to be) leaves my flawed analysis no worse that an understatement of the time penalty incurred by the wind. To wit: the main point, namely the presence of a time penalty, has, however ineptly, been established.

[k-man]

@bonanova: You're right that if the wind direction has both head/tail component and cross component relative to the trajectory then analyzing only one or the other will result in correct conclusion that wind adds to travel time. However, the only crucial fact that I was trying to show in my previous post was that there exist closed paths (i.e. straight line A->B and back) and wind directions (i.e. perpendicular to the path) for which ignoring the cross wind component will result in incorrect conclusion - same time with and without wind. In other words, in some cases, the entire time penalty is swept under the rug.

Edited March 25, 2015 by k-man

[bonanova]

Agree totally with your point.

[dgreening]

The Best Answer still bothers me.

 

The OP states that the plane flies "at constant airspeed c directly above a closed polygonal path in a plane".

 

If the desired path over the ground is [east to ] west at a constant "Air speed" and the wind is blowing due north at some velocity, the place will tend to be thrown off course and it will result in some angle [similar to that shown by K-Man above]. As a result the actual distance moved in the E-W direction will be reduced by the resulting angle. 

 

To compensate for this, a pilot must assume a constant "crab angle" in order to stay above the desired line of flight that is choosing a vector that adjusts for the crosswind. The same trig still applies and the "ground speed" will be reduced. 

If the wind direction and speed are constant over the entire course throughout the flight, the ground speed will be reduced similarly on some portion of the return flight.

Edited April 12, 2015 by dgreening

[gavinksong]

The Best Answer still bothers me.

 

The OP states that the plane flies "at constant airspeed c directly above a closed polygonal path in a plane".

 

If the desired path over the ground is [east to ] west at a constant "Air speed" and the wind is blowing due north at some velocity, the place will tend to be thrown off course and it will result in some angle [similar to that shown by K-Man above]. As a result the actual distance moved in the E-W direction will be reduced by the resulting angle. 

 

To compensate for this, a pilot must assume a constant "crab angle" in order to stay above the desired line of flight that is choosing a vector that adjusts for the crosswind. The same trig still applies and the "ground speed" will be reduced. 

If the wind direction and speed are constant over the entire course throughout the flight, the ground speed will be reduced similarly on some portion of the return flight.

 

Yeah, I think post #6 should be the Best Answer.

[bonanova]

Yeah, I think post #6 should be the Best Answer.

 

Agree.

My post was aimed at proving an increased travel time.

kman gave the best analysis.