This is the result of mapping the straight line path from a point in US to Poland using Distance Measurement Tool.

Also, planes from Asia to US would travel almost over North Pole.

'straight line' distance from Alberta to Poland

Why is the path so curved? I agree that this is a flat representation of a sphere, so I do expect some arc, but I don't think earth has this much curvature.

What am I missing here?

up vote 96 down vote accepted

Just look at the path on the sphere. Here it is in Google Earth:

Earth from space seen from above Greenland, path from Alberta to Poland shown

The path on your map is strongly curved because your map uses a projection with lots of distortion. (The distortion grows without bound towards the poles and this path is getting close to the north pole.)


The distortion is necessary to explain the curvature of this geodesic on the map but the connection between them is subtle. More can be said that is at once useful, informative, and elegant. See whether you agree.

The OP's map uses a Mercator projection. Its salient qualities are that it is

  • Cylindrical: in particular, meridians are vertical lines on the map,

  • Conformal: any angle at which two paths cross on the earth will be correctly rendered on the map, and

  • Loxodromic: any route of constant bearing (on the earth) is rendered as a straight line segment on the map.

These properties make it easy to read some critical information directly off the map. In this context I am most interested in the angles made by any path with each of the meridians it crosses. (These are the bearings measured from the north.) For instance, the path depicted in the question starts in Canada, around 54 degrees latitude, making an angle of about 30 degrees with its meridian.

What we also need to know about a point at 54 degrees latitude is that it is closer to the earth's axis than points along the equator. In fact, it's cos(54) * R from the axis, where R is the earth's radius. (This is essentially the definition of the cosine. It helps to have some familiarity with cosines, so you understand how they behave, but you don't really need to know any other trigonometry at all. I promise. Well, one more thing: the sine of an angle is the cosine of its complement. E.g., sin(32 degrees) = cos(90-32) = cos(58).)

Finally, note that the earth is rotationally symmetric about its axis. This lets us invoke Clairaut's beautiful

Theorem (1743): On a path in any smooth surface of revolution, the product of the distance to the axis with the sine of the bearing is constant if and only if the path is locally geodesic.

Thus, since we are starting off at latitude 54 degrees at an angle of 30 degrees, the product in the theorem equals cos(54) * R * sin(30) = 0.294 * R.

How does this help? Well, consider what would happen if the path were to continue approximately straight on the map. Sooner or later it would rise to a latitude of 73 degrees. Using Clairaut's theorem we can solve for the bearing at this latitude:

cos(73) * R * sin(bearing) = 0.294 * R;

sin(bearing) = 0.294 / cos(73) = 1;

bearing = 90 degrees.

This says that by the time we reach a latitude of 73 degrees, we must be traveling due east! That is, the path, in order to be a geodesic, must curve so strongly that the initial bearing of 30 degrees (east of north) becomes 90 degrees (east of north).

(Of course I found the value 73 degrees by solving the equation cos(latitude) = cos(latitude) * sin(90) = cos(54) * sin(60). To do this yourself you would have to know that (a) sin(90) = 1 (because sin(90) = cos(90-90) = cos(0) = 1) and (b) most calculators and spreadsheets have a function to solve cosines; it's called ArcCos or inverse cosine. I hope you don't view this little detail as breaking my earlier promise about no more trig...)

After doing a few calculations like this you develop an intuition for what Clairaut's Theorem is saying. A path in a surface of revolution (like the earth) can be geodesic (locally shortest or "straight") only when (a) its bearing becomes more parallel to the meridians at points far from the axis and (b) its bearing gets more perpendicular to the meridians at points closer to the axis. Because there is a limit on how perpendicular one can get--90 degrees is it!--there is a limit to how close to the axis you can get. This constant adjustment of bearing (= angle to the meridian) and latitude (= distance to the axis) causes the apparent curvature of geodesics on most maps, especially on those using cylindrical projections, where the meridians and lines of latitude are rendered as vertical and horizontal lines, respectively.

Here are some easy implications of Clairaut's Theorem. See whether you can prove them all:

  1. The equator must be a geodesic.

  2. All meridians are geodesics.

  3. No line of latitude, other than the equator (and the poles, if you want to include them), can be a geodesic. Not even a small part of a line of latitude can be geodesic.

  4. Loxodromes (aka rhumb lines), which are lines of constant bearing, cannot be geodesics unless they are meridians or the equator. Not even a small part of such a loxodrome can be geodesic. In other words, if you sail or fly in a fixed compass direction, then--with a few obvious exceptions--your path is constantly curving!

Point 4 says if you fly from the Canadian Rockies at an initial bearing of 30 degrees east of north, you must appear, relative to north, to be constantly turning (to the right) in order to fly straight; you will never go north of 73 degrees latitude; and if you continue far enough, you will make it to Poland and will be headed roughly 150 degrees east of north when you get there. Of course the details--73 degrees and Poland and 150 degrees--are obtained only from the quantitative statement of Clairaut's Theorem: you can't usually figure out that sort of thing just using your intuitive idea of geodesics.

It is noteworthy that all these results hold on a general spheroid (a surface of revolution generated by an ellipse), not just on perfect spheres. With slight modifications they hold for tori (surfaces of bagels or truck tires) and many other interesting surfaces. (The sci fi author Larry Niven wrote a novel in which a small artificial torus-shaped world is featured. The link includes an image from the novel's cover depicting part of this world.)

  • nice summary...forgot about the Larry Niven book! – user681 Mar 5 '11 at 21:32
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    Great answer, thanks. This may be a good question to address in our FAQ since it touches on many important fundamentals. – scw Mar 8 '11 at 7:21

In this projection (Google Mercator), that's what the great circle arc between those two places looks like.

  • 6
    +1 Why the downvote? This is a perfectly good answer. It's hard to know what else to say. Moreover, it added some insight by recognizing the projection in the map. – whuber Mar 5 '11 at 21:12
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    it would be nice if there were consequences or control on downvotes. – Brad Nesom Mar 6 '11 at 3:08

Just a quick addition:

Also, planes from Asia to US would travel almost over North Pole.

In that direction, they will often use the jet stream. In the other direction they will indeed fly over/close to the poles. Asia-US jetstream

Mercator map with Tissot indicatrix

The Mercator projection distorts at the poles

more info Tissot's Indicatrix

So the steepness is more acute in the latter poles

  • The TI does not directly indicate which way geodesics will curve. High distortion does not imply "acute steepness." For example, on a stereographic projection the opposite (south) pole is infinitely distorted (as on the Mercator); the TI shows circles of unbounded size there; yet all geodesics emanating from either pole will be straight lines on the map and, in fact, the closer a geodesic gets to the south pole the straighter it appears on the map! The most strongly curved geodesic will be the equator, which lies in a region of intermediate (and uniform) distortion. – whuber Mar 6 '11 at 17:49
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    After some thought I appreciate this contribution better: introducing the TI lets us see the nature of the distortion that leads to the curvature of geodesics on the map. The relationship between the TI and the geodesics is subtle: it depends on the rates of change of the TI. Specifically, the circles graphically depict the Euclidean metric, whose components are traditionally written E, F, and G. Their rates of change produce the Christoffel symbols, which in turn tell us the geodesic directions. On a conformal map like this, a geodesic wants to curl away from the big circles. – whuber Mar 7 '11 at 21:32
  • Thanks, comments appreciated - have taught youngsters so keeping is simple as possible - like draw on your hand flat down - now make a fist - lines become curved and longer? - great for explaining contours on a 2D map! – Mapperz Mar 8 '11 at 16:02
  • Just as a comment, if you assume 1 degree between lines of longitude, they are 70 odd miles apart at the equator, and obviously converge at the poles. This is a good site to work out distance, bearing points, great circles etc, etc, etc : – Hairy Dec 2 '11 at 11:10

I saw a very elegant explanation of this phenomenon on Tom MacWright's blog here, with photos of oranges. The explain-it-to-a-5-year-old version: "On a globe, shortest paths are flat, and navigation lines are curvy. Mercator made a map where navigation lines are straight. This made the shortest paths curvy."

It is due to the projection of a 2D plane onto a polorised 2 spheres surface, as the line moves past the poles, it becomes distorted as far as observers of the 2D plane are concerned because the straight line to the destination appears to be a curved ark of a Great Circle, which is a term in mathematics that relates to the greatest circle that can be sliced from a sphere, as long as the circle passes through the center of the sphere. I have slightly modifies the Images provided in other answers by scribbling a line across to illustrate (rather poorly I'm afraid, I'm new to GIMP) The so called polar distortion. I think some similar concept Is behind Gravitational forces, but I'm not a physicist so I couldn't say.

enter image description here

enter image description here

The closer to the Poles a point gets, the less deformed it appears to be when rendered onto a Flat 2D surface, although it still is by a tiny amount. It also depends on the Projection method used, and there are some that are focused on making the quickest route between two points appear to be flat and then round back on the full spherical view.

  • Although much of what you say will be correct from time to time depending on the projection and the context, almost nothing in this answer is generally true. As an example, the familiar Mercator projection provides a counterexample to the assertion about "the closer to the Poles a point gets, the less deformed it appears to be... ." – whuber Jul 20 '15 at 12:19
  • This statement "the closer to the Poles a point gets, the less deformed it appears to be... ." is true for Azimuthal projections but totally incorrect for Mercator projection or any cylindrical projection for that matter. – yanes Dec 3 '15 at 23:39

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