Let's say I'm in the US and I create a Foucault pendulum. Why does its direction change if its axis is not parallel to the rotation axis of the earth? I can understand why it changes if it's at the North or South pole, but Im having trouble picturing it in my head when the pendulum is between the equator and the north/south pole. I can also understand why it doesn't change direction when it's at the equator, its axis is perpendicular to Earth's axis there.

What confuses me is that when I read explanations of why the direction of the pendulum changes, they always say it's because the Earth is rotating beneath the pendulum, but that isn't exactly true right? If I'm between the equator and the north/South pole, the axis isn't parallel to the axis of Earth's rotation.

Btw I'm not a flat earther, I'm just having a hard time picturing this in my head and it's really bugging me. Its making me feel stupid, especially since I'm majoring in physics.

Edit: I think I understand why it changes direction do to the Coriolis effect if I start the pendulum off by pushing it in the direction of the north or south poles (the speed of Earth's rotation is slower moving up towards the north pole and gets faster moving down towards the equator), but what if I start the pendulum by pushing it east or west? The speed of Earth's rotation doesn't change in those directions. Maybe it's just impossible to push it perfectly in the east/west direction and so it will always have components in the north/south directions?

2nd edit: But wait, if the change in direction is due to the Coriolis effect, then shouldn't the direction still change if I'm at the equator and start the pendulum off by pushing it in the north/south direction?

  • $\begingroup$ How do you feel about the Coriolis force? Does it make sense to you that in a rotating reference frame like the Earth a moving object would be deflected to the side? $\endgroup$
    – M. Enns
    Feb 27, 2019 at 2:02
  • $\begingroup$ It makes sense to me if I think about an object moving north or south, but it doesn't make sense to me if I think about an object moving east or west. $\endgroup$ Feb 27, 2019 at 2:23

2 Answers 2


Here are a couple of steps to help visualize.

First I take a simpler case, one where the object is suspended, like the pendulum bob is, but without the swinging back and forth.

Take an airship. An airship is buoyant, just like a ship. So you can think of an airship as an object that is suspended. The airship is free to slide sideways, but it tends to remain at the same altitude above the ground.

First imagine that airship stationary with respect to the ground below. In that state the airship is co-rotating with the Earth. As I mentioned, the airship is free to slide sideways. So: why doesn't the airship slide all the way to the equator? The airship doesn't slide to the equator because the Earth is a bit flattened. The flattening provides the required centripetal force. (To give you an idea: at 45 degrees latitude the ratio of required centripetal force to mass is 1:580 That is: if an object has a mass of 580 kg then the required centripetal force is 1 kg of force.)

Now image the four cases:
- velocity eastward
- velocity westward
- velocity northward
- velocity southward

When the airship has a velocity in eastward direction it is circumnavigating the Earth's axis faster than the Earth itself is rotating. But the amount of centripetal force is the amount for co-rotation motion. Hence the airship will swing wide, which means it will not stay at the same latitude, but it will slide a bit towards the equator.

When the airship has a velocity in westward direction the airship is circumnavigating the Earth's axis slower than the Earth itself. So now there is a surplus of centripetal force, and again the airship will not remain at the same latitude. The centripetal force will pull the airship closer, thus the airship will slide a bit away from the equator.

On the northern hemisphere:
When the airship has velocity in northward direction the airship is giving in to the centripetal force. Now the centripetal force is doing work, and the angular velocity of the airship (wrt the Earth's axis) increases. So the airship will not remain on the same longitude line, but it will slide to the right of it.

On the northern hemisphere: When the airship has velocity in southward direction the airship is moving away from the center of attraction. Now the centripetal force is doing negative work, decreasing the angular velocity of the airship. So the airship will not remain on the same longitude line, but it will slide to the right of it.

In the case of a Foucault pendulum the pendulum wire is doing two jobs: sustaining the swing, of course, but also providing the required centripetal force so that the pendulum bob keeps circumnavigating the Earth's axis.

You can think of the pendulum bob as being subject to a centripetal force, precisely the amount that is required for co-rotating with the Earth. At every moment in its swing the motion is subject to the angular mechanics effects described above.

Of course usually the amplitude of the swing is much larger than the angle associated with providing required centripetal force. For a perspective: the largest Foucault pendulum setup ever is the one that was constructed by Foucault in the Pantheon in Paris; the wire is 67 meters long. For a wire of 67 meters long the required angle corresponds with a displacement of the bob of 11 centimeters. Interestingly, in his reports about his observations Foucault mentions that on rare occasions there was opportunity for very long uninterrupted runs of the pendulum. On occasion the amplitude of the swing had decayed to a mere 10 centimeters or so. Foucault reports that under those circumstances the pendulum still displayed the same rotation of the plane of swing!

I noticed in the comment section a remark: "How do you feel about the "Coriolis force?". Here is my point: it is better to look at the first principles that are involved, and to think the physics through. I warn against treating physics as an exercise in rote learning. If you apply a rule without understanding the rule then you really haven't learned anything.

For more information about the motion of a Foucault pendulum you can visit the Foucault pendulum page on my website


If you're looking for an intuitive way to understand the Foucault pendulum, get a globe and set a pencil tangent to the surface at a specific latitude. The triangle formed by pencil, plus a line from its center to the center of the Earth, defines the plane in which a pendulum is swinging at a given moment.

The plane can be extended out to include points at infinity- like a distant star. Both the distant star and the center of the Earth will remain included in that plane as the Earth turns. Keep in mind that all points at infinity in the plane are still in the original plane even if the plane is transported in a direction perpendicular to the plane, as long as the plane is not tilted in the process.

Turning the globe slowly while gradually adjusting the direction of the pencil (keeping it tangent to the surface) so that the plane of oscillation always includes the center of the globe and any selected distant point will force the pencil to rotate relative to its original direction. When the pencil starts out on the equator, there is no rotation because a distant point simply moves in the original plane. When the pencil starts out at a pole, the pencil basically needs to hold still while the globe rotates under it - in order to keep distant points in its original plane. Play with the pencil and globe, starting with the pencil in intermediate latitudes, and the Foucault pendulum will gradually make sense.


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