Passive appendages generate drift through symmetry breaking

Plants and animals use plumes, barbs, tails, feathers, hairs and fins to aid locomotion. Many of these appendages are not actively controlled, instead they have to interact passively with the surrounding fluid to generate motion. Here, we use theory, experiments and numerical simulations to show that an object with a protrusion in a separated flow drifts sideways by exploiting a symmetry-breaking instability similar to the instability of an inverted pendulum. Our model explains why the straight position of an appendage in a fluid flow is unstable and how it stabilizes either to the left or right of the incoming flow direction. It is plausible that organisms with appendages in a separated flow use this newly discovered mechanism for locomotion; examples include the drift of plumed seeds without wind and the passive reorientation of motile animals.

is extracted from numerical simulations of a flow (Re = 45) around a cylinder with splitter plate of length L = 1.0D at turn angles −20 • < θ < 20 • . In a we compare the angular dependence of the drift force for the plate with the drift force of the cylinder. The forces are normalized with their maximum values in order to show that they have a similar shape. In b, the drift force is compared to the force obtained from the model forÃ = C D /4.

Side view
Top view  where the normal force on splitter plates of different lengths changes direction. b, Flow field around cylinder at Re = 45 is shown. Black lines depict streamlines and green lines contours of zero azimuthal velocity. c, Superposition of splitter plates at selected equilibrium angles (a) and the undisturbed flow around cylinder (b) are shown. d, The definition of cylindrical coordinate system. Note that zero angle direction is aligned with thê y direction. e, The half ellipse of back flow region is defined using major semi-axis s 2 and attachment angle θ 0 , which sets the minor semi-axis s 1 . We only consider shapes with different length of s 2 , keeping θ 0 constant. In e, we also show the direction of azimuthal velocity u θ on the left side of the domain. On the right side of the domain, u θ has the opposite direction.
Supplementary Figure 6: Normal force on the plate. a, The problem of a moving plate can be described in a coordinate system where the plate is not translating but only rotating. In this frame of reference, the plate is exposed to incoming free stream with velocity U. The angle between the plate and the incoming flow is θ. The angle θ and flow velocity U can be time dependent. b, The simplified force model applied on a plate, which is divided in two parts; one part is exposed to the free stream U 1 ; and the other part is exposed to free stream U 2 in the opposite direction.  The distance B from the cylinder surface to edge of the BFR is assumed to be known. The ellipse radius in polar coordinates r, the ellipse angle γ and the turn angle θ s are the unknowns.
Supplementary Note 1: Solving for θ = ±θ s In this note, we show how to obtain the non-trivial solutions of Eq. (4) in the main paper and we show that these solutions are always stable. For convenience, the "hats" over dimensionless variables have been omitted. The angle θ s can be obtained by solving a geometrical problem, where the length of the splitter plate (at unknown turn angle θ s ) inside the back flow region (BFR) is given that we know the total length of the splitter plate L and the coefficient k. A sketch of the geometrical problem is shown in Supplementary Fig. 9. At this point we assume that the shape of the back-flow region (BFR) is half of an ellipse (more information in Methods: estimating function B (θ)) and its center is located at a distance y 0 from the center of the cylinder. The half ellipse is defined by semi-axes s 1 and s 2 . The turn angle θ s is defined as the angle between vertical center line of the ellipse (direction of the straight position) and the line going through the splitter plate.
Consider the point where the splitter plate intersects the BFR. Connecting the intersection point with the center of the ellipse yields a triangle ( Supplementary Fig. 9). We are interested in the turn angle θ s in the triangle, in which two sides are known. To complete the problem formulation, we introduce the ellipse radius r and angle γ. In total, we have three unknowns in our problem. We relate the ellipse radius and angle using ellipse equation in polar coordinates r 2 sin 2 γ s 2 1 + r 2 cos 2 γ s 2 2 = 1.
The second equation is sine rule in the constructed triangle where we have used the property of sine function sin (π − γ) = sin γ. Finally, the third and last equation is the cosine rule r 2 = y 2 0 + (0.5 + B) 2 − 2y 0 (0.5 + B) cos θ s .
Here auxiliary variables are y 0 = 0.5 cos θ 0 , s 1 = 0.5 sin θ 0 and s 2 = B max + 0.5 − y 0 . By analyzing the obtained roots of (6) with respect to the original problem, one finds that the root with plus sign corresponds to solution of the posed problem. Therefore the final expression for the turn angle is The explicit expression derived for θ s corresponds to the blue lines in Figs. 1d, 3b and 5c in the main paper. Next, we determine the stability of the skewed positions θ s in a similar fashion as for θ = 0 in the main text. Expanding the torque expression (4) in the main paper around θ = θ s and using the fact that for the skewed solution (1 + k) [B 2 (θ s ) + B (θ s )] − L 2 − L = 0, we arrive with the first-order term ∂T ∂θ θ=θs = sin(θ s ) First, note that sin(θ s ) > 0 if θ s > 0 (and sin(θ s ) < 0 if θ s < 0) for the range −θ 0 < θ s < θ 0 , where θ 0 ≈ 55 degrees (see Methods: estimating parameters B max and k in main paper). Then, from the sketch in Supplementary Fig. 9, we find ∂B ∂θ θ>0 < 0 and ∂B ∂θ θ<0 > 0, which leads to ∂T ∂θ θ=θs < 0.
This shows that the skewed solutions are always stable. Any deviation in positive angle direction will introduce a negative restoring torque and vice versa.