<p>In an idealised scenario (ie where the plane is flying due north or south at a and there are no other perturbations to its course apart from the Coriolis effect), yes. And I guess it would be the most efficient course too!</p>

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2016-09-11T12:29:35.830Z
2016-09-12T01:51:50.283Z
2016-09-12T01:51:50.283Z
Semidiurnal Simon
<p>Say a plane moves from 40° North to 40° South, and the plane flies in straight line without considering the Coriolis effect. Will it still reach the destination.</p>

<p>All I mean is whether the deflection caused in the Northern Hemisphere can be compensated for in Southern Hemisphere</p>

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2016-09-08T16:45:48.457Z
Are you assuming an object that has start and end points that lie along the same meridian, or are you considering points that lie along different meridians, say, -120° and -80°?
BillDOe
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2016-09-08T17:34:42.963Z
in short, yes. f=2 omega sin(lat) so if you integrate from -40 to 40, then the answer is zero.
arkaia
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2016-09-09T22:31:09.650Z
I'd love to be able to delete my comment, as the start and end meridians have nothing to do with the question, only the start and end latitudes.  The comment provided by @aretxabaleta is correct. Put another way, the object will be oppositely deflected in one direction as far as it was in the other hemisphere.
BillDOe
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2016-09-10T04:56:45.857Z
A small digression, @aretxabaleta 
The plane flying over a particular latitude say 20°N on its journey from 40°N to 40°S, will have a component of its speed(velocity) equal to rotational speed corresponding to 20° (where it currently is) or 40° (where it started from) due to inertia
turing
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2016-09-10T13:34:09.583Z
@turing, The effect at 20 N is compensated by the effect at 20 S.
arkaia
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2016-09-11T16:26:00.410Z
@aretxabaleta It was a slightly different doubt in my previous comment, I agree it would be compensated, please read my comment again
turing
<earth-rotation><coriolis>
Can Coriolis Effect cancel itself?
turing
# Can Coriolis Effect cancel itself?

(Asked by: username_0 on 2016-09-08)

Say a plane moves from 40° North to 40° South, and the plane flies in straight line without considering the Coriolis effect. Will it still reach the destination.

All I mean is whether the deflection caused in the Northern Hemisphere can be compensated for in Southern Hemisphere

**Tags:** earth-rotation, coriolis

> **Comments:**
> **username_1**: Are you assuming an object that has start and end points that lie along the same meridian, or are you considering points that lie along different meridians, say, -120° and -80°?  
> **username_2**: in short, yes. f=2 omega sin(lat) so if you integrate from -40 to 40, then the answer is zero.  
> **username_1**: I'd love to be able to delete my comment, as the start and end meridians have nothing to do with the question, only the start and end latitudes.  The comment provided by @aretxabaleta is correct. Put another way, the object will be oppositely deflected in one direction as far as it was in the other hemisphere.  
> **username_0**: A small digression, @aretxabaleta  The plane flying over a particular latitude say 20°N on its journey from 40°N to 40°S, will have a component of its speed(velocity) equal to rotational speed corresponding to 20° (where it currently is) or 40° (where it started from) due to inertia  
> **username_2**: @username_0, The effect at 20 N is compensated by the effect at 20 S.  
> **username_0**: @aretxabaleta It was a slightly different doubt in my previous comment, I agree it would be compensated, please read my comment again  

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## 1 answers

### Answer by username_3 on 2016-09-11. Score: 2

In an idealised scenario (ie where the plane is flying due north or south at a and there are no other perturbations to its course apart from the Coriolis effect), yes. And I guess it would be the most efficient course too!


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