Thoughts on animations of swing-hinged dissections of a triangle to a square

On the webpage Animations of swing-hinged dissections of a triangle to a square I presented several animations of swing-hinged dissections of an equilateral triangle to a square that appear in the article:

"Designing a Table Both Swinging and Stable", by Greg N. Frederickson,
College Mathematics Journal, volume 39, number 4 (September 2008), pages 258-266.

Those of you who are interested in animations may wonder why I "chopped up" my second and third animations. Why didn't I start rotating all of the pieces at the same time, and stop rotating them all at the same time? There are two reasons. First, I wanted people to be able to focus on various aspects of the motion one at a time, rather than completely overwhelming them. Second, and more importantly, if all pieces moved simultaneously, then the motion wouldn't be correct for a physical object, since some of the pieces would interpenetrate at times during the motion. Here's the explanation why:

The problem is a result of having all of the pieces complete their rotations during precisely the same interval of time. This means that N₅ and N₇ will rotate twice as fast as N₁, N₃, N₄, and N₆, since they must rotate through 360° rather than 180°, because each has two hinges separating them from the immovable piece N₂.

Let's see how the overlap comes about. Consider the square in Figure 6. For simplicity in the description, rotate the whole square 30° counterclockwise, so that the edge of piece N₄ that touches N₅ and N₂ is vertical. Let R be the length of the edge of N₄ that is now vertical, and let r be the length of the edge of N₅ that is now vertical. Note that r is clearly greater than R/2. Now rotate pieces N₁, N_3, N₄, and N₆ by θ, and at the same time rotate N₅ and N₇ by 2θ.

Let's describe the position of the apex of N₅ for θ in the range 0 ≤ θ ≤ 45°: The x-value of the apex will change by –R sin(θ) + r sin(2θ), and its y-value will change by R – R cos(θ) – r + r cos(2θ). Taking derivatives of each value as a function of θ, we get

dx/dθ = –R cos(θ) + 2r cos(2θ)
dy/dθ = R sin(θ) – 2r sin(2θ)

Taking the limit as θ decreases to zero, we get

lim{θ→0} dx/dθ = –R + 2r
lim{θ→0} dy/dθ = 0

Since r is greater than R/2, we see that the apex of N₅ starts off moving horizontally to the right, with it slowly and continuously then shifting direction down and to the right. Unfortunately, immediately after the rotation of pieces begins, N₅ will collide with N₂, because the angle between their adjacent edges in the neighborhood of the apex is 30° below the horizontal, and there cannot be an abrupt shift in direction of 30° as θ continuously increases.

For additional background material on hinged dissections, see:

Hinged Dissections: Swinging & Twisting, by Greg N. Frederickson,
Cambridge University Press, 2002.