As shown in
Figure 1, when a bar of length
l0 is rotating with an angular speed
ω0, the speed of an arbitrary point
P on the bar, relative to the center
C, is (equation before Equation 3 in
Beghi et al., 1991a)):
Obviously, when
λ = 1/2,
P coincides with
C, and
vpc = 0 The next step is most critical since
Beghi et al. (1991a) apply their principle of minimization. In their words (p. 427) (the two equations below are Equation 3 and the one immediately after in
Beghi et al. (1991a)):
“Let us now associate to point
P an additional velocity component
vz(
P)
from which it follows:
”
Let us examine
vz(
P) as a function of the position
λ (in
Beghi et al. (1991a), the equation below Equation 3 contains a typo,
I0 should be
l0). Note that
when
λ = 1/2, and
when
λ = 0 and 1, respectively. That is to say, with the introduction of the additional velocity component in depth, the bar’s midpoint
C will have the largest depth displacement (the direction of which depends on the sign of
vz(
P=
C)), while its two end points
A and
B will remain on the image plane. This means that the bar cannot possibly remain rigid, contradicting the reported percept. It is impossible for a rotating bar to have the same speed everywhere and remain rigid.
I believe that the case has already been made at this point regarding the inadequacy of the minimal relative speed principle. I would like to further comment on the two methods, analytic and trajectory, of the minimal speed difference model that give rise to the same value when computing the depth difference between
A and
B. Although the true depth difference between
A and
B should be zero,
Beghi et al. (1991a) obtain a different value based on their trajectory derivation. They compute this displacement within a time period
t0 =
π/ω0 as follows (Equation 4 of
Beghi et al. (1991a), p. 428) (A time period of
π/ω0, or of the half rotation, is used, because, according to Footnote 2 of
Beghi et al. (1991a), “the rotating segment will return to its initial position after a full rotation” (p. 427). It is unclear, however, what triggers
vz to reverse its direction at half rotation.):
I will now show that this value is in fact the area swept by the bar during this time divided by
l0, but not the depth difference between
A and
B. I will assume, with
Beghi et al. (1991a), that the bar
BCA moves in the positive
z direction. Note that
vz is only a function of position
λ, not of time
t. Within time interval
π/ω0, this area is (let cos
θ = 2
λ − 1):
.
Beghi et al. (1991a) employ a second method, which assumes that point
P moves from
B to
A within time interval
t0 =
π/ω0 with a constant speed
v0 =
l0/
t0. (There is a minor inconsistency in
Beghi et al. (1991a) with regard to which point (
A or
B) is the starting point and which one is the end point. Equation 1 of
Beghi et al. (1991a) clearly indicates that
B is the starting point. Then this is inconsistent with the first equation on p. 428. Regardless of the starting point, the last two equations in Equation 2 (
xp = …,
yp = …) are incorrect.) Specifically,
P’s speed relative to the midpoint
C,
vp, is calculated first (the 3rd equation from the bottom left on p. 428):
then the minimal speed difference principle is applied (the 2nd equation from top right on p. 428):
where
vz(
t) is the
z-component of the motion of point
P. Finally, the depth displacement
is calculated. Although the same displacement is obtained (
π2/8)
l0, the following needs to be clarified.
The
vP2(
t) derived in
Beghi et al. (1991a) is incorrect (the second equation from the bottom left on p. 428). It does not appear to be a typo since follow-up equations are derived from it. It is unclear how, at the end, the z displacement (
π2/8)
l0 is obtained that is consistent with the analytic method. In fact, a
z displacement of (
π2/4)
l0 should have been the correct solution following these equations. The correct
vP2(
t) that gives rise to the
z displacement (
π2/8)
l0 should be:
This can be seen by observing that (I thank Reviewer #1 who pointed this out):
An intuitive way to understand
Equation 10 is that
P has two orthogonal velocity components relative to
C: one along the bar (constant)
π0l0/
π =
v0, the other perpendicular to the bar (rotation)
π0l0(
t/t0 − 1/2).