Are Theo Jansen his Linkages Optimal? |16 Dec. 2019|

Theo Jansen, designer of the infamous ‘‘strandbeesten’’ explains in a relatively recent series of videos that the design of his linkage system is the result of what could be called an early implementation of genetic programming. Specifically, he wanted the profile that the foot follows to be flat on the bottom. He does not elaborate on this too much, but the idea is that in this way his creatures flow over the surface instead of the more bumpy motion we make. Moreover, you can imagine that in this way the center of mass remains more or less at a constant height, which is indeed energy efficient. Now, many papers have been written on understanding his structure, but I find most of them rather vague, so lets try a simple example.

To that end, consider just 1 leg, 1 linkage system as in the figure(s) below. We can write down an explicit expression for $p(L,theta)in mathbf{R}^2$ where L=(a,b,c,d,e,f,g,h,i,j,k,l,m)

and

. See below for an explicit derivation. Now, with his motivation in mind, what would be an easy and relevant cost function? An idea is to maximize the ratio p_w/p_h

, of course, subject to still having a functioning leg. To that end we consider the cost function (see the schematic below for notation):

Of course, this is a heuristic, but the intuition is that this should approximate p_w/p_h

. Let

be the set of parameters as given in Jansen his video, then we apply a few steps of (warm-start) gradient ascent: $L_{k+1}=L_k + k^{-2}nabla c(L_k)$ , k=1,2,dots

Can we conclude anything from here? I would say that we need dynamics, not just kinematics, but more importantly, we need to put the application in the optimization and only Theo Jansen understands the constraints of the beach.

Still, it is fun to see for yourself how this works and like I did you can build a (baby) strandbeest yourself using nothing more than some PVC tubing and a source of heat.

Derivation of :
Essentially, all that you need is the cosine-rule plus some feeling for how to make the equations continuous. To start, let e_1=(1,0)

and

(just the standard unit vectors), plus let R(theta)

be a standard (counter-clockwise) rotation matrix acting on $(x,y)in mathbf{R}^2$ . To explain the main approach, when we want to find the coordinates of for example point p_4

, we try to compute the angle theta_2

such that when we rotate the unit vector e_1

with length

emanating from point p_3

, we get point p_4

. See the schematic picture below.

Then we easily obtain p_1=(-a,-l)

and

. Now, to obtain p_4

and

compute the length of the diagonal, n=|p_1-p_3|_2

and using the cosine rule $beta_1=arccosbig( -(2nj)^{-1}(b^2-n^2-j^2) big)$ , $beta_2=arccosbig( -(2nk)^{-1}(c^2-n^2-k^2) big)$ . Then, the trick is to use cos(theta)=langle a,brangle/ (|a|_2|b|_2)

to compute theta_4:=theta_2+beta_1

. To avoid angles larger than

, consider an inner product with e_2

and add the remaining 1/2pi

; $theta_4=arccosbig(n^{-1}langle e_2,(p_1-p_3)rangle big)+1/2pi$ . From there we get theta_2=theta_4-beta_1

. The point p_6

is not missing, to check your code it is convenient to compute p_6=p_3+R(theta_4)ne_1

and check that p_6

agrees with p_1

for all

. In a similar fashion, $gamma_1=arccosbig(-(2db)^{-1}(e^2-d^2-b^2) big)$ , $delta_1=arccosbig(b^{-1}langle e_1,(p_4-p_1)rangle big)$ , p_7=p_1+R(gamma_1+delta_1)de_1

. Then, for p_8

, again compute a diagonal q=|p_5-p_7|_2

plus $alpha_1=arccosbig(-(2qc)^{-1}(d^2-q^2-c^2) big)$ , $alpha_2=arccosbig(c^{-1}langle e_1,(p_1-p_5)rangle big)$ $alpha_3=arccosbig(-(2qg)^{-1}(f^2-q^2-q^2) big)$ , such that for alpha=alpha_1+alpha_2+alpha_3

we have

. At last, $psi_1=arccosbig(-(2gi)^{-1}(h^2-i^2-g^2) big)$ , such that p(L,theta)=p_5+R(psi)ie_1

for