In physics, when you shrink a circle in one direction you stretch it in another and get an ellipse One interesting example is when you look at a rotating two bladed wind turbine from an angle instead of directly from the front or behind. The path of the blade tips appears elliptical instead of circular., and moreover the rotation now appears jerky instead of smooth. This must be because what you see of the ellipse perimeter as it approaches -a and a on the major axis is foreshortened in comparison to the circular path, so the blade appears to slow down to account for the lesser distance instead of maintaining its speed. If the circle is squashed then it must either stretch along the x axis beyond -a and a, unlike in your diagram in which the major axis of the ellipse is the same as the diameter of the circle. Or the apparent reduction in speed means this stretching occurs along the time axis rather than any spatial one, and stretched time passes more slowly.
That is an interesting example. The speed of the blades is way too small for any noticeable relativistic effects, so the time dilation doesn't make much of a difference. You are right about why the apparent slowdown at -a and a occurs. Maybe I misunderstood you as to why the circle must stretch beyond -a and a, but neither the area nor the perimeter remain the same after shrinking the circle in the y-direction. It is just a special case of a linear transformation that shows the connection between an ellipse and a circle. The tip of the blade is an excellent example of this. When viewing from an angle, the ellipse we see will have the same semi-major axis as the radius of the circle that we would see if we were looking from the front (assuming orthographic view).
@@MathStuff Yes, I take your last point. The ellipse doesn't stretch beyond the circle. Also there aren't any noticeable relativistic effects. Would you say that the apparent motion of the blade tip obeys Kepler's equal areas law of planetary motion?
@@chrisg3030 Not really. Because the area of any part of a circle, when projected to a plane, gets multiplied by the cosine of the angle between the circle and the projection plane. That means that the rate at which the area is swept out by the blades is constant. The difference to Kepler's law is that this is the area swept out with respect to the center of the ellipse, where in Kepler's law it is with respect to the focal point. Also, with planetary motion, the highest speed is at the vertex that is closest to the focal point to which the gravity pulls, and the lowest speed is at the other vertex. In the case of a blade, the apparent speed is the same at both vertices.
@@MathStuff If the planet, and therefore the focal point to which the gravity pulls, were at the center of the ellipse, then the lowest speed would be at the vertices and the highest in between them on the ellipse. Over a given length of time would the areas of triangles swept out at these two locations be equal?
@@chrisg3030 In order for a focal point to be at the center of the ellipse, the ellipse itself would have to be a circle. In that case, the speed would be constant, and the rate at which the area is swept would also be constant.
In physics, when you shrink a circle in one direction you stretch it in another and get an ellipse
One interesting example is when you look at a rotating two bladed wind turbine from an angle instead of directly from the front or behind. The path of the blade tips appears elliptical instead of circular., and moreover the rotation now appears jerky instead of smooth. This must be because what you see of the ellipse perimeter as it approaches -a and a on the major axis is foreshortened in comparison to the circular path, so the blade appears to slow down to account for the lesser distance instead of maintaining its speed.
If the circle is squashed then it must either stretch along the x axis beyond -a and a, unlike in your diagram in which the major axis of the ellipse is the same as the diameter of the circle. Or the apparent reduction in speed means this stretching occurs along the time axis rather than any spatial one, and stretched time passes more slowly.
That is an interesting example. The speed of the blades is way too small for any noticeable relativistic effects, so the time dilation doesn't make much of a difference.
You are right about why the apparent slowdown at -a and a occurs.
Maybe I misunderstood you as to why the circle must stretch beyond -a and a, but neither the area nor the perimeter remain the same after shrinking the circle in the y-direction. It is just a special case of a linear transformation that shows the connection between an ellipse and a circle.
The tip of the blade is an excellent example of this. When viewing from an angle, the ellipse we see will have the same semi-major axis as the radius of the circle that we would see if we were looking from the front (assuming orthographic view).
@@MathStuff Yes, I take your last point. The ellipse doesn't stretch beyond the circle. Also there aren't any noticeable relativistic effects.
Would you say that the apparent motion of the blade tip obeys Kepler's equal areas law of planetary motion?
@@chrisg3030 Not really. Because the area of any part of a circle, when projected to a plane, gets multiplied by the cosine of the angle between the circle and the projection plane. That means that the rate at which the area is swept out by the blades is constant.
The difference to Kepler's law is that this is the area swept out with respect to the center of the ellipse, where in Kepler's law it is with respect to the focal point.
Also, with planetary motion, the highest speed is at the vertex that is closest to the focal point to which the gravity pulls, and the lowest speed is at the other vertex. In the case of a blade, the apparent speed is the same at both vertices.
@@MathStuff If the planet, and therefore the focal point to which the gravity pulls, were at the center of the ellipse, then the lowest speed would be at the vertices and the highest in between them on the ellipse. Over a given length of time would the areas of triangles swept out at these two locations be equal?
@@chrisg3030 In order for a focal point to be at the center of the ellipse, the ellipse itself would have to be a circle. In that case, the speed would be constant, and the rate at which the area is swept would also be constant.