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This question already has an answer here:

How much does Earth's rotation affect flight times in going East or West, and how could you calculate if zero winds?

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marked as duplicate by mins, David Richerby, Marco Sanfilippo, aeroalias, anshabhi Dec 17 '15 at 12:30

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    $\begingroup$ It's really helpful. For example, if you are in a helicopter and need to travel directly west, you can just hover and let the earth move below you until it gets to the right spot. $\endgroup$ – Tyler Durden Jan 30 '15 at 20:47
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    $\begingroup$ @TylerDurden if you want to go east, though, you have to go faster than 1180 km/hr, and that's at 45 degrees latitude. So I hope you are headed west. $\endgroup$ – fooot Jan 30 '15 at 20:56
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    $\begingroup$ OP, are you asking about the effects of Time of Arrival due to time zone changes? $\endgroup$ – CGCampbell Jan 30 '15 at 21:07
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    $\begingroup$ What do time zones have to do with the earth's rotation or wind? $\endgroup$ – fooot Jan 30 '15 at 21:09
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It does not affect as much as one would think.

The atmosphere moves along with Earth's rotation. An airplane taking off from one place on Earth has to move through the atmosphere which is moving with Earth.

Now you may ask when does it take longer when we travel from Western Europe to Eastern USA, compared to USA to Europe? That depends on Jet Stream:

Jet streams are fast flowing, narrow air currents found in the atmosphere of ... Earth.

Many air routes take advantage of jet streams, as you can see as follows:

Jet Streams

Pilots know about the jet streams and will try to gain benefit of them. Recently, there were some news that a flight from New York to London traveled faster than expected because it was in jet stream (details here and here).


If you want to know why Earth's atmosphere moves with it, please see this.

There is a similar question on Physics.SE.

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It affects flight time the same way that it affects driving time.

Cars measure their speed over the ground based on wheel speed. Airplanes can measure their speed over the ground using things like GPS. Either way, it's referenced to the ground. Technically the ground is moving, but it's not perceptible to us and we still would say zero speed means not moving over the ground.

Airplanes primarily measure their speed through the air (airspeed), and the air tends to move with the ground, so the situation is the same. If the air is not moving the same as the ground, it's called wind, which is the primary influence on flight time. Wind is influenced by the earth's rotation, but the rotation doesn't directly influence flight time.

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    $\begingroup$ More to the point, aircraft are starting from the moving frame of reference (when an aircraft starts its takeoff roll it's moving at the same speed as the rest of the Earth). If you made a direct vertical takeoff in a no-wind situation you would remain directly above your takeoff point until you added thrust in another direction or ran out of fuel. $\endgroup$ – voretaq7 Jan 30 '15 at 21:17
  • $\begingroup$ @voretaq7: Except for Coriolis force, which is, however, negligible for this purpose. $\endgroup$ – Jan Hudec Jan 31 '15 at 11:37
  • $\begingroup$ @voretaq7: That only works by approximation, or above the poles or the equator, and even then you have to apply just the right amount of vertical thrust. The earth is not an inertial frame. $\endgroup$ – Marcks Thomas Jan 31 '15 at 12:15
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This is a physics question, not an aviation question, and might better be posed on Physics StackExchange.

Simply put, the question has an infinite number of answers! Let me explain.

When you are stationary in a helicopter, waiting to take off, with reference to an observer also standing next to the helicopter, your speed is 0.

But with reference to an observer in a static point in space whose frame of reference is the centre of the earth, you are already moving at about 1040MPH. The surface of the earth is also moving at 1040MPH. (BTW, it is itself impossible to define a static point in space since all of space is moving.)

The atmosphere you are about to fly in (assuming zero wind) is also moving at the same speed. If its speed was zero, with reference to the stationary observer in space, then the 1040MPH wind might make flying a little difficult.

You lift into the hover. Now to both observers, you are still doing the same speed because speed is a scalar and you have added no force to change your horizontal speed. To the observer standing near the helicopter, the earth, the helicopter and the air it is flying in are all moving at 0 speed. To the observer in space, they are all moving at 1040MPH.

You now fly West at 100MPH. To the observer standing on the earth, your speed is now 100MPH but to the observer in space, your speed is now 940MPH. You have slowed down!

By picking any arbitrary observer, at any point in space, and doing any speed you choose, you can see that there are an infinite number of answers to the question.

Only by asking the question with reference to a specific observer, can you answer the question.

If you ignore wind and side effects such as the Coriolis effect, the earths rotation has no effect on your journey time since you and the craft you are in must move at the same speed within the same frame of reference, relative to any given observer, together.

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  • $\begingroup$ This is a really good point to explicit the need of frame of reference and it must be written done (so your answer is a good one). But I think you miss one point: in aviation the frame of reference is implicitly either relative to the air or relative to the earth. Assuming zero wind (as asked) means both frame of reference (airmass and earth) are equivalent. $\endgroup$ – Manu H Jan 31 '15 at 9:53
  • $\begingroup$ You're right, and I did say "assuming zero wind". If there is wind, then you simply have to add the scalar component of the wind vector relative to the observer. $\endgroup$ – Simon Jan 31 '15 at 11:52
  • $\begingroup$ my point is that in aviation the frame of reference is implicitly relative to the airmass (or reative to earth) $\endgroup$ – Manu H Jan 31 '15 at 18:17
  • $\begingroup$ "Only by asking the question with reference to a specific observer, can you answer the question" TOTALLY WRONG - the flight time does not depend on the observer. $\endgroup$ – aaa90210 Oct 26 '16 at 1:48
  • $\begingroup$ @aaa90210 Time is not fixed for all frames of reference and observers. Time dilation means that the journey time is dependent on the observer. But that's really esoteric and at the micro scales we're taking about with aircraft flying from A to B, not relevant. My point was simply that the Earth is a non-inertial frame and the last paragraph is the key one. $\endgroup$ – Simon Oct 26 '16 at 6:13

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