Parameterizing Sphere Linkages

Using iterated suspensions to get nice coordinates on configuration space

Second in a series on spherical configuration spaces of planar linkages, joint work with Aaron Abrams, Dave Bachmann, and Edmund Harriss. The setup is in the first post.

In the previous post we showed that the configuration space $\mathcal{M}^n_L$ of an $n$-rod planar chain pinned at both ends — the locus in $T^n$ where the rod angles satisfy the closure equation $\sum_k e^{i\theta_k} = L$ — is homeomorphic to the $(n-2)$-sphere whenever $n - 2 < L < n$. The argument was inductive: peeling off the last rod identifies $\mathcal{M}^n_L$ as a topological suspension of $\mathcal{M}^{n-1}_d$ at an appropriate sub-base length $d$, so iterating from the base case $\mathcal{M}^2_L = S^0$ gives a tower of $n - 2$ suspensions, $\Sigma^{n-2} S^0 = S^{n-2}$.

What that proof established was the homeomorphism type. What it didn’t give us, at least not directly, was the explicit homeomorphism — a map you could plug into a computer and calculate with. In this post we’ll write one down:

$$\Psi^n_L \;\colon\; S^{n-2} \longrightarrow \mathcal{M}^n_L,$$

an explicit homeomorphism, for each $n \geq 2$ and every $L$ in the allowable range.

We work up by example. After the trivial base case $n = 2$ — just two configurations of the elbow-up/elbow-down kind — we build $\Psi^3_L$ from $\Psi^2_L$ explicitly, by hand. The first nontrivial step is where the geometric and analytic moves of the construction show themselves most clearly: rescaling, smoothing across the suspension poles, the whole story. Once we see the pattern, we articulate the general “add one rod, get one suspension” recipe. Then we apply that recipe to get $\Psi^4_L$ as a parameterization of $\mathcal{M}^4_L \cong S^2$, and gesture at the iteration to higher dimensions.

Each section will follow the same template: build the parameterization in rod angles first — the natural coordinates inherited from the suspension construction — then convert to position coordinates, the locations of the chain’s interior hinges. The position-space form is consistently cleaner: the inverse trig functions that clutter the rod-angle formula collapse into elementary expressions, and smoothness across the suspension poles becomes transparent.

Angle coordinates and position coordinates

There are two natural sets of coordinates for an $n$-rod planar chain pinned at $0$ and $L$:

• Angle coordinates $(\theta_1, \ldots, \theta_n) \in T^n$, the rod angles, with $\mathcal{M}^n_L$ cut out by closure $\sum_k e^{i\theta_k} = L$.
• Position coordinates $(\mathbf{p}_1, \ldots, \mathbf{p}_{n-1}) \in \mathbb{C}^{n-1}$, the locations of the interior hinges, with $\mathbf{p}_k = \sum_{j \le k} e^{i\theta_j}$. The pinned endpoint $\mathbf{p}_n = L$ is fixed, so can be left off.

The two are related by the cumulative sum: $\mathbf{p}_k = \mathbf{p}_{k-1} + e^{i\theta_k}$ (taking $\mathbf{p}_0 := 0$), and going back, $e^{i\theta_k} = \mathbf{p}_k - \mathbf{p}_{k-1}$ — the unit vector from one hinge to the next.

In position coordinates the closure equation has already been folded in: the last position $\mathbf{p}_n = L$ is fixed by the pinning, so it doesn’t appear among the coordinates at all.

In angle coordinates closure can be used to drop one coordinate too. The first $n-1$ angles determine the position of $\mathbf{p}_{n-1}$, and for the chain to close, the last rod must be the unique unit segment from $\mathbf{p}_{n-1}$ to $L$ — so $\theta_n = \arg(L - \mathbf{p}_{n-1})$ is determined by the others. Dropping it gives an embedding $\mathcal{M}^n_L \hookrightarrow T^{n-1}$ (as mentioned in post 1) that we’ll exploit for visualizations.

$\mathcal{M}^2_L \cong S^0$

For the base case, we’ll think of $S^0$ as the discrete two-point set parameterized by a sign variable

$$\sigma \in \{+1,\; -1\}.$$

The two configurations are the elbow-up and elbow-down isoceles triangles with vertices $(0, 0)$, $\mathbf{p}_1$, $(L, 0)$, with the equal sides of length $1$.

Angle coordinates

The triangle’s two equal base angles are both the elbow half-angle

$$\beta(L) \;:=\; \arccos(L/2),$$

so rod 1 makes angle $\sigma\,\beta(L)$ with the positive $x$-axis and rod 2 makes the opposite angle:

$$\boxed{\; \Psi^2_L(\sigma) \;=\; \big(\,\sigma\,\beta(L),\;\; -\sigma\,\beta(L)\,\big). \;}$$

Closure check. $e^{i\sigma\beta} + e^{-i\sigma\beta} = 2\cos\beta(L) = L$ ✓ (using $\cos\beta(L) = L/2$ by definition).

Position coordinates

The single interior hinge sits on the unit circle at angle $\sigma\beta$. Since $\cos\beta = L/2$ and $\sin\beta = \sqrt{4-L^2}/2$,

$$\boxed{\; \Psi^2_L \;\colon\; \sigma \;\longmapsto\; \mathbf{p}_1 \;=\; \tfrac{1}{2}\big(L + i\sigma\sqrt{4 - L^2}\big), \qquad \sigma \in \{+1, -1\}. \;}$$

$\mathcal{M}^3_L \cong S^1$

We build $\Psi^3_L \colon S^1 \to \mathcal{M}^3_L$ by hand from the base case $\Psi^2_d$. The strategy comes straight from the suspension proof of last post: project a $\mathcal{M}^3_L$ configuration onto its last rod’s angle, parameterize the fiber over that angle as a sub-config in $\mathcal{M}^2_d$, and place the result in the right ambient frame.

Angle coordinates

The suspension structure

Fix a target value $\theta$ for the third rod’s angle. With this last rod placed, the second-to-last vertex sits at $\mathbf{p}_2 = L - e^{i\theta}$. The first two rods then span from the origin to $\mathbf{p}_2$ — a $\mathcal{M}^2$-style chain at sub-base length

$$d \;:=\; |L - e^{i\theta}| \;=\; \sqrt{L^2 - 2L\cos\theta + 1}.$$

For this sub-chain to exist we need $d \leq 2$, which restricts $\theta$ to an arc

$$\theta \in [-\alpha,\, \alpha], \qquad \alpha \;:=\; \arccos\!\frac{L^2 - 3}{2L}.$$

For each interior $\theta$, the sub-chain in its own reference frame — base laid along the positive $x$-axis from $(0,0)$ to $(d, 0)$ — is a configuration of $\mathcal{M}^2_d$, parameterized by the sheet sign $\sigma \in \{\pm 1\}$ from before:

$$\Psi^2_d(\sigma) \;=\; \big(\,\sigma\,\beta(d),\;\; -\sigma\,\beta(d)\,\big), \qquad \beta(d) \;:=\; \arccos(d/2).$$

To place the sub-chain in the ambient frame (where it should run from the origin to $\mathbf{p}_2 = L - e^{i\theta}$ rather than to $(d, 0)$), we rotate by

$$R \;:=\; \arg\!\big(L - e^{i\theta}\big).$$

Both frames share the origin, so this is a pure rotation — no translation needed. (This is the geometric payoff of projecting onto the last rod rather than the first.) Rotating each rod-vector by $R$ adds $R$ to its angle, so the rotated sub-config has rod-angles $(R + \sigma\beta,\; R - \sigma\beta)$.

Stacking all three rods, the parameterization sending $(\theta, \sigma)$ to a configuration in $\mathcal{M}^3_L$ is

$$(\theta, \sigma) \;\longmapsto\; \big(\, R + \sigma\,\beta(d),\;\; R - \sigma\,\beta(d),\;\; \theta \,\big),$$

with $d, R, \beta$ defined above.

Closure check. $e^{i(R + \sigma\beta)} + e^{i(R - \sigma\beta)} = e^{iR}(e^{i\sigma\beta} + e^{-i\sigma\beta}) = 2\cos\beta\, e^{iR} = d\, e^{iR}$, using $\cos\beta = d/2$. And $d\, e^{iR} = L - e^{i\theta}$ by definition of $R$. So the rod-vectors sum to $(L - e^{i\theta}) + e^{i\theta} = L$. ✓

Suspension coordinates

The natural domain for our parameterization is the suspension itself — $\Sigma S^0 = [-1, 1] \times S^0$, with the four endpoints $(\pm 1, \pm 1)$ glued in pairs at the two suspension poles $s = \pm 1$. This is the same domain regardless of $L$. We want a map $\Psi^3_L \colon \Sigma S^0 \to \mathcal{M}^3_L$ on this fixed domain.

The second factor lines up for free: $\sigma$ already lives in $S^0$. The first factor needs a rescaling — currently $\theta$ runs over the $L$-dependent arc $[-\alpha, \alpha]$, but we want it to come from $[-1, 1]$. The simplest choice does it:

$$s \in [-1, 1], \qquad \theta \;=\; \alpha\, s.$$

Pulling our parameterization back through this rescaling defines

$$\boxed{\; \begin{aligned} \Psi^3_L \;&\colon\; \Sigma S^0 \to \mathcal{M}^3_L, \\[4pt] \Psi^3_L(s, \sigma) \;&=\; \big(\, R + \sigma\,\beta(d),\;\; R - \sigma\,\beta(d),\;\; \alpha\, s \,\big), \end{aligned} \;}$$

with $\theta = \alpha s$, $d = \sqrt{L^2 - 2L\cos\theta + 1}$, $R = \arg(L - e^{i\theta})$, $\beta = \arccos(d/2)$.

This is a perfectly good homeomorphism from $\Sigma S^0$ to $\mathcal{M}^3_L$. But the domain isn’t a smooth circle: the sheet sign $\sigma$ is a discrete variable, not a smooth coordinate. To get a smooth map from $S^1$ we have to repackage $(s, \sigma)$ in terms of a single circular variable.

Smoothing the parameterization

Introduce a circular variable $t \in [-\pi,\, \pi)$ — the angle on $S^1 = \{e^{it}\}$ — and parameterize the suspension coordinates by

$$s(t) \;:=\; \sin t, \qquad \sigma(t) \;:=\; \operatorname{sgn}(\cos t).$$

As $t$ runs from $-\pi$ to $\pi$:

• $t = -\pi/2$: $s = -1$, the south suspension pole — fiber has collapsed.
• $t \in (-\pi/2,\, \pi/2)$: $\sigma = +1$, $s$ sweeps from $-1$ up through $0$ (equator) to $+1$ on the upper sheet.
• $t = \pi/2$: $s = +1$, the north suspension pole.
• $t \in (\pi/2,\, \pi)$ and $t \in (-\pi,\, -\pi/2)$: $\sigma = -1$, $s$ sweeps from $+1$ back down through $0$ to $-1$ on the lower sheet.

So $t$ traces the upper sheet from south pole to north pole, then the lower sheet back, completing the full circle. Composing with the suspension formula gives $\theta(t) = \alpha\sin t$ and a single-variable map $\Psi^3_L \colon S^1 \to \mathcal{M}^3_L$,

$$\boxed{\; \begin{gathered} \Psi^3_L \;\colon\; t \;\longmapsto\; \begin{pmatrix} R(t) + \sigma(t)\,\beta(t) \\[2pt] R(t) - \sigma(t)\,\beta(t) \\[2pt] \theta(t) \end{pmatrix}, \qquad t \in [-\pi,\, \pi), \\[10pt] \begin{aligned} \alpha \;&=\; \arccos\!\tfrac{L^2 - 3}{2L}, \\ \theta(t) \;&=\; \alpha\sin t, \\ \sigma(t) \;&=\; \operatorname{sgn}(\cos t), \\ d(t) \;&=\; \sqrt{L^2 - 2L\cos\theta(t) + 1}, \\ R(t) \;&=\; \arg\!\big(L - e^{i\theta(t)}\big), \\ \beta(t) \;&=\; \arccos(d(t)/2). \end{aligned} \end{gathered} \;}$$

Or in code:

function psi3(t, L) {
  const alpha = Math.acos((L*L - 3) / (2*L));
  const theta = alpha * Math.sin(t);
  const sigma = Math.cos(t) >= 0 ? 1 : -1;
  const d     = Math.sqrt(L*L - 2*L*Math.cos(theta) + 1);
  const R     = Math.atan2(-Math.sin(theta), L - Math.cos(theta));
  const beta  = Math.acos(d / 2);
  return [R + sigma * beta, R - sigma * beta, theta];
}

By closure, $\theta_3 = \theta(t)$ is determined by the first two rod angles, so we can drop it and equivalently view $\mathcal{M}^3_L$ as a closed curve in the 2-torus. This lower-dimensional embedding is easier to picture than the full curve in $T^3$ — and the savings will matter even more for the next case ($\mathcal{M}^4_L$ as a sphere in $T^3$ rather than $T^4$).

Position coordinates

To convert the rod-angle formula into hinge positions, sum cumulatively: $\mathbf{p}_1 = e^{i\theta_1}$, $\mathbf{p}_2 = e^{i\theta_1} + e^{i\theta_2}$. The simplifications happen because of two complex-exponential identities — exponentials eat inverse trig functions:

• $e^{i\arg(z)} = z/|z|$: the exponential of an $\arg$ is just a normalization.
• $e^{i\sigma\beta} = \cos\beta + i\sigma\sin\beta = \tfrac{1}{2}(d + i\sigma\sqrt{4-d^2})$, using $\cos\beta = d/2$ and $\sin\beta = \sqrt{4-d^2}/2$.

In the rod-angle picture these inverse trig wrappers stay; in the position-space picture they collapse.

$\mathbf{p}_2$. Sum the two halves:

$$\mathbf{p}_2 \;=\; e^{i\theta_1} + e^{i\theta_2} \;=\; e^{iR}(e^{i\sigma\beta} + e^{-i\sigma\beta}) \;=\; 2\cos\beta \cdot e^{iR} \;=\; d\,e^{iR}.$$

Then $e^{iR} = (L - e^{i\theta})/d$ kills the $d$:

$$\mathbf{p}_2 \;=\; L - e^{i\theta}.$$

Geometric reading: $\mathbf{p}_2$ is the right endpoint $L$ minus the third rod’s vector $e^{i\theta}$ — the chain, end backwards.

$\mathbf{p}_1$. From $\mathbf{p}_1 = e^{iR}\,e^{i\sigma\beta}$, multiply the two simplified factors:

$$\mathbf{p}_1 \;=\; \frac{L - e^{i\theta}}{d} \cdot \frac{d + i\sigma\sqrt{4 - d^2}}{2} \;=\; \frac{\mathbf{p}_2}{2} \;+\; \frac{i\sigma\sqrt{4 - |\mathbf{p}_2|^2}}{2\,|\mathbf{p}_2|}\,\mathbf{p}_2.$$

Geometric reading: $\mathbf{p}_1$ is the midpoint of $\overline{0\mathbf{p}_2}$ plus a perpendicular offset — the elbow of two unit rods spanning a base of length $|\mathbf{p}_2| = d$. The factor $i\,\mathbf{p}_2/|\mathbf{p}_2|$ is the unit vector perpendicular to $\mathbf{p}_2$, the offset magnitude is $\sqrt{4-|\mathbf{p}_2|^2}/2$, and $\sigma$ picks the sheet.

Smoothness, transparently

In rod-angle coordinates, smoothness across $t = \pm\pi/2$ would require chasing the sign jump of $\sigma$ through the cusp of $\beta$ and verifying that they cancel. In position coordinates, every step above is elementary except for $\sigma\sqrt{4-|\mathbf{p}_2|^2}$, and that combines into a single smooth signed amplitude. Define

$$\nu(t) \;:=\; \cos t \,\sqrt{\frac{4 - |\mathbf{p}_2(t)|^2}{\cos^2 t}} \;=\; \cos t\,\sqrt{\frac{2L(\cos\theta - \cos\alpha)}{\cos^2 t}},$$

using $4 - |\mathbf{p}_2|^2 = 2L(\cos\theta - \cos\alpha)$ (from the law of cosines and $\cos\alpha = (L^2-3)/(2L)$). The ratio under the radical extends by continuity to its limit at $\cos t = 0$: a Taylor expansion at $t = \pi/2$ gives $4 - |\mathbf{p}_2|^2 \sim L\alpha\sin\alpha\,\cos^2 t$, so the ratio extends smoothly to $L\alpha\sin\alpha > 0$. Outside the poles $\nu(t) = \sigma(t)\sqrt{4 - |\mathbf{p}_2(t)|^2}$, by direct computation: $\cos t \cdot \sqrt{X/\cos^2 t} = \mathrm{sgn}(\cos t)\sqrt{X}$ for $X > 0$. So $\nu$ is smooth everywhere on $\mathbb{R}$, and its zeros at $t = \pm\pi/2$ are linear crossings — exactly what we’d expect for the signed altitude of a triangle whose elbow passes through the base.

The same expression can be rewritten without any apparent singularity at all. Applying the product-to-sum identity to $\cos\theta - \cos\alpha$ and a half-angle substitution gives

$$\nu(t) \;=\; \alpha\sqrt{L}\,\cos t \,\sqrt{\operatorname{sinc}\!\tfrac{\alpha(1+\sin t)}{2}\,\operatorname{sinc}\!\tfrac{\alpha(1-\sin t)}{2}}, \qquad \operatorname{sinc}(x) := \tfrac{\sin x}{x},$$

with $\operatorname{sinc}(0) := 1$ by continuity.

Substituting:

$$\boxed{\; \begin{gathered} \Psi^3_L \;\colon\; t \;\longmapsto\; \begin{pmatrix} \mathbf{p}_1 \\[4pt] \mathbf{p}_2 \end{pmatrix} \;=\; \begin{pmatrix} \dfrac{p}{2} \;+\; \dfrac{i\,\nu}{2\,|p|}\,p \\[10pt] p \end{pmatrix} \\[10pt] \begin{aligned} \alpha \;&=\; \arccos\!\tfrac{L^2 - 3}{2L}, \\ \theta(t) \;&=\; \alpha\sin t, \\ p(t) \;&=\; L - e^{i\theta(t)}, \\ \nu(t) \;&=\; \cos t\,\sqrt{\dfrac{2L(\cos\theta - \cos\alpha)}{\cos^2 t}}. \end{aligned} \end{gathered} \;}$$

Or in code:

function psi3Position(t, L) {
  const alpha = Math.acos((L*L - 3) / (2*L));
  const theta = alpha * Math.sin(t);
  const c     = Math.cos(t);

  // p = L - e^{i theta}
  const p_re  = L - Math.cos(theta);
  const p_im  =   - Math.sin(theta);
  const p_abs = Math.hypot(p_re, p_im);

  // nu(t): smooth signed amplitude. Numerator and denominator both vanish
  // quadratically at cos t = 0; the ratio extends to L*alpha*sin(alpha).
  const num = 2 * L * (Math.cos(theta) - (L*L - 3) / (2*L));
  const nu  = Math.abs(c) < 1e-6
            ? Math.sign(c || 1) * Math.sqrt(L * alpha * Math.sin(alpha))
            : c * Math.sqrt(num / (c * c));

  // p1 = p/2 + (i nu / (2|p|)) * p
  const k = nu / (2 * p_abs);
  return {
    p1: [p_re / 2 - k * p_im, p_im / 2 + k * p_re],
    p2: [p_re, p_im],
  };
}

Click on the circle to drag the configuration around yourself!

The general step

What we just did for $\Psi^3_L$ generalizes. The geometric structure of the construction — project on the last rod’s angle, fix the sub-base length and orientation, parameterize the fiber by the inner config space — is the same at every level of the iteration. Let’s articulate it as a recipe.

Angle coordinates

Suppose, inductively, that we have a smooth parameterization

$$\Psi^n_d \colon S^{n-2} \longrightarrow \mathcal{M}^n_d$$

in angle coordinates, for every $d$ in the corresponding allowable range $n - 2 < d < n$. We want to build $\Psi^{n+1}_L \colon S^{n-1} \to \mathcal{M}^{n+1}_L$ from $\Psi^n$.

Coordinate the parameterizing sphere $S^{n-1}$ by a latitude $\phi \in [-\pi/2, \pi/2]$ and a fiber point $p \in S^{n-2}$ (the inner spherical coordinates), with the fiber collapsing at the two poles $\phi = \pm\pi/2$. We’ll bake the smooth $\sin$-rescaling directly into the recipe — exactly analogous to what we did for $\Psi^3_L$ — so that the suspension axis is from the start an angle, going pole-to-pole through the equator.

The same three helper functions show up. The half-width of the admissible arc for the last rod, when the sub-chain has $n$ rods that need to span up to distance $n$, is

$$\alpha(n, L) \;:=\; \arccos\!\frac{L^2 - n^2 + 1}{2L}.$$

The sub-base length and sub-frame rotation as functions of the last-rod angle $\theta$ are

$$d(L, \theta) \;:=\; \sqrt{L^2 - 2L\cos\theta + 1}, \qquad R(L, \theta) \;:=\; \arg\!\big(L - e^{i\theta}\big).$$

Set $\theta = \alpha\sin\phi$ as we did for the circle: the standard latitude rescaling that takes the equator $\phi = 0$ to $\theta = 0$ and the poles $\phi = \pm\pi/2$ to $\theta = \pm\alpha$. Then for each $(\phi, p) \in S^{n-1}$, the recipe gives

$$\boxed{\; \Psi^{n+1}_L(\phi, p) \;=\; \big(\, R + \Psi^n_d(p),\;\; \alpha\sin\phi \,\big), \;}$$

where $\alpha = \alpha(n, L)$, $\theta = \alpha\sin\phi$, $d = d(L, \theta)$, $R = R(L, \theta)$, and $R + (\theta_1, \ldots, \theta_n) := (R + \theta_1, \ldots, R + \theta_n)$ is componentwise scalar addition.

Position coordinates

Suppose inductively that we have $\Psi^n_d$ in position coordinates: $\Psi^n_d(p)$ outputs a tuple of $n - 1$ interior hinges in $\mathbb{C}$ for a chain pinned at $0$ and $d$ on the real axis (its “sub-frame”).

Coordinate $S^{n-1}$ the same way as before, with the same $\theta = \alpha\sin\phi$. The outermost interior hinge of the new chain is determined by $\theta$ alone — closure forces the last rod to be $e^{i\theta}$, so

$$\mathbf{p}_n \;=\; L - e^{i\theta},$$

the same “endpoint minus last rod” identity that gave us $\mathbf{p}_2$ in the circle case, now at the outer level. Its modulus $|\mathbf{p}_n|$ is the sub-base length $d$, and its direction $\hat{\mathbf{p}}_n := \mathbf{p}_n / d$ is the rotation that places the sub-frame in the ambient one.

The first $n - 1$ interior hinges come from the inner sub-chain. $\Psi^n_d(p)$ produces them in the sub-frame; componentwise multiplication by $\hat{\mathbf{p}}_n$ rotates them into the ambient frame. So:

$$\boxed{\; \Psi^{n+1}_L(\phi, p) \;=\; \big(\, \hat{\mathbf{p}}_n \cdot \Psi^n_d(p), \;\; \mathbf{p}_n \,\big), \;}$$

with $\mathbf{p}_n = L - e^{i\theta}$ and $\hat{\mathbf{p}}_n \cdot (z_1, \ldots, z_{n-1}) := (\hat{\mathbf{p}}_n z_1, \ldots, \hat{\mathbf{p}}_n z_{n-1})$ being componentwise complex multiplication.

Geometric reading: append one new outer hinge, rotate the rest into place. The whole step is a complex subtraction and a complex multiplication — no inverse trig at the recursion level. ($\hat{\mathbf{p}}_n$ is the same rotation that the angle picture writes as $e^{iR}$, just packaged as a unit complex number rather than a scalar angle.)

$\mathcal{M}^4_L \cong S^2$

We’re now ready to apply the general step with $n = 3$ to build the parameterization of the 2-sphere. Coordinate $S^2$ by latitude $\phi \in [-\pi/2, \pi/2]$ and longitude $t \in S^1$ — in the recipe’s notation, $\phi$ is the suspension parameter and $t$ is the inner fiber point, which we feed directly into $\Psi^3_d$ from §3. As with standard spherical coordinates, this chart has the usual singularities at the poles $\phi = \pm\pi/2$, where the longitude $t$ becomes meaningless.

Angle coordinates

The angle-version recipe gives

$$\Psi^4_L(\phi, t) \;=\; \big(\, R + \Psi^3_d(t), \;\; \alpha\sin\phi \,\big),$$

where the outer-level helpers are functions of $\phi$ alone and the inner $\Psi^3_d(t)$ is the §3 formula evaluated at base length $d$, with its own inner helpers as functions of $\phi$ and $t$. Stacking the four rod angles and laying out all helpers in one place:

$$\boxed{\; \begin{gathered} \Psi^4_L \;\colon\; (\phi, t) \;\longmapsto\; \begin{pmatrix} R + R_\text{in} + \sigma\,\beta_\text{in} \\[2pt] R + R_\text{in} - \sigma\,\beta_\text{in} \\[2pt] R + \theta_\text{in} \\[2pt] \alpha\sin\phi \end{pmatrix} \\[10pt] \begin{aligned} \alpha \;&=\; \arccos\!\tfrac{L^2 - 8}{2L}, \\ \theta(\phi) \;&=\; \alpha\sin\phi, \\ d(\phi) \;&=\; \sqrt{L^2 - 2L\cos\theta + 1}, \\ R(\phi) \;&=\; \arg\!\big(L - e^{i\theta}\big), \\ \sigma(t) \;&=\; \operatorname{sgn}(\cos t), \\ \alpha_\text{in}(\phi) \;&=\; \arccos\!\tfrac{d^2 - 3}{2d}, \\ \theta_\text{in}(\phi, t) \;&=\; \alpha_\text{in}\sin t, \\ d_\text{in}(\phi, t) \;&=\; \sqrt{d^2 - 2d\cos\theta_\text{in} + 1}, \\ R_\text{in}(\phi, t) \;&=\; \arg\!\big(d - e^{i\theta_\text{in}}\big), \\ \beta_\text{in}(\phi, t) \;&=\; \arccos(d_\text{in}/2). \end{aligned} \end{gathered} \;}$$

Or in code:

function psi4(phi, t, L) {
  // Outer level
  const alpha = Math.acos((L*L - 8) / (2*L));
  const theta = alpha * Math.sin(phi);
  const d     = Math.sqrt(L*L - 2*L*Math.cos(theta) + 1);
  const R     = Math.atan2(-Math.sin(theta), L - Math.cos(theta));

  // Inner Psi^3_d(t)
  const alpha_in = Math.acos((d*d - 3) / (2*d));
  const theta_in = alpha_in * Math.sin(t);
  const sigma    = Math.cos(t) >= 0 ? 1 : -1;
  const d_in     = Math.sqrt(d*d - 2*d*Math.cos(theta_in) + 1);
  const R_in     = Math.atan2(-Math.sin(theta_in), d - Math.cos(theta_in));
  const beta_in  = Math.acos(d_in / 2);

  return [
    R + R_in + sigma * beta_in,
    R + R_in - sigma * beta_in,
    R + theta_in,
    theta
  ];
}

By closure we can drop the last rod angle and view $\mathcal{M}^4_L$ as a 2-sphere embedded in $T^3$ rather than $T^4$. The savings really pay off here: $T^3$ as a periodic cube is drawable; $T^4$ is not.

The animation above illustrates this parameterization with the domain $S^2$ on the left and its image $\subset T^3$ on the right. Click and drag the point on the domain to explore the parameterization.

Position coordinates

The position-version recipe gives

$$\Psi^4_L(\phi, t) \;=\; \big(\, \hat{\mathbf{p}}_3 \cdot \Psi^3_d(t), \;\; \mathbf{p}_3 \,\big),$$

with $\mathbf{p}_3 = L - e^{i\theta}$ the new outermost interior hinge and $\hat{\mathbf{p}}_3 = \mathbf{p}_3 / |\mathbf{p}_3|$ the rotation onto the ambient frame. The inner $\Psi^3_d(t)$ from §3 produces the two interior hinges of an $\mathcal{M}^3_d$ chain in its sub-frame:

$$\mathbf{p}_2^\circ \;=\; d - e^{i\theta_\text{in}}, \qquad \mathbf{p}_1^\circ \;=\; \frac{\mathbf{p}_2^\circ}{2} \;+\; \frac{i\,\nu_\text{in}}{2\,|\mathbf{p}_2^\circ|}\,\mathbf{p}_2^\circ,$$

with inner helpers

$$\alpha_\text{in} \;=\; \arccos\!\tfrac{d^2 - 3}{2d}, \quad \theta_\text{in} \;=\; \alpha_\text{in}\sin t, \quad \nu_\text{in} \;=\; \cos t\,\sqrt{\dfrac{2d(\cos\theta_\text{in} - \cos\alpha_\text{in})}{\cos^2 t}}.$$

Multiplying by $\hat{\mathbf{p}}_3$ rotates these into the ambient frame. The middle hinge inherits the recurring “endpoint minus last rod” identity, one level in:

$$\mathbf{p}_2 \;=\; \hat{\mathbf{p}}_3 \cdot \mathbf{p}_2^\circ \;=\; \hat{\mathbf{p}}_3\,(d - e^{i\theta_\text{in}}) \;=\; \mathbf{p}_3 \;-\; \hat{\mathbf{p}}_3\,e^{i\theta_\text{in}},$$

using $\hat{\mathbf{p}}_3 \cdot d = \mathbf{p}_3$. Same shape as $\mathbf{p}_3 = L - e^{i\theta}$, but with the ambient-frame “endpoint” $\mathbf{p}_3$ taking the role of $L$ and the rotated unit vector $\hat{\mathbf{p}}_3\,e^{i\theta_\text{in}}$ taking the role of the last rod $e^{i\theta}$. The innermost hinge inherits the elbow formula:

$$\mathbf{p}_1 \;=\; \hat{\mathbf{p}}_3 \cdot \mathbf{p}_1^\circ \;=\; \frac{\mathbf{p}_2}{2} \;+\; \frac{i\,\nu_\text{in}}{2\,|\mathbf{p}_2|}\,\mathbf{p}_2,$$

since rotation preserves both the elbow’s relative geometry and the modulus $|\mathbf{p}_2| = |\mathbf{p}_2^\circ| = d_\text{in}$. Boxed:

$$\boxed{\; \begin{gathered} \Psi^4_L \;\colon\; (\phi, t) \;\longmapsto\; \begin{pmatrix} \mathbf{p}_1 \\[4pt] \mathbf{p}_2 \\[4pt] \mathbf{p}_3 \end{pmatrix} \;=\; \begin{pmatrix} \dfrac{\mathbf{p}_2}{2} \;+\; \dfrac{i\,\nu_\text{in}}{2\,|\mathbf{p}_2|}\,\mathbf{p}_2 \\[10pt] \mathbf{p}_3 \;-\; \hat{\mathbf{p}}_3\,e^{i\theta_\text{in}} \\[6pt] L \;-\; e^{i\theta} \end{pmatrix} \\[10pt] \begin{aligned} \alpha \;&=\; \arccos\!\tfrac{L^2 - 8}{2L}, \\ \theta(\phi) \;&=\; \alpha\sin\phi, \\ \mathbf{p}_3(\phi) \;&=\; L - e^{i\theta}, \\ \hat{\mathbf{p}}_3 \;&=\; \mathbf{p}_3 / |\mathbf{p}_3|, \\ \alpha_\text{in}(\phi) \;&=\; \arccos\!\tfrac{|\mathbf{p}_3|^2 - 3}{2\,|\mathbf{p}_3|}, \\ \theta_\text{in}(\phi, t) \;&=\; \alpha_\text{in}\sin t, \\ \nu_\text{in}(\phi, t) \;&=\; \cos t\,\sqrt{\dfrac{2\,|\mathbf{p}_3|\,(\cos\theta_\text{in} - \cos\alpha_\text{in})}{\cos^2 t}}. \end{aligned} \end{gathered} \;}$$

The recursion peels rods off from the outside in: each level adds an outer hinge of the form (previous level’s “endpoint”) $-$ (rotated unit vector), and the innermost level closes with the elbow formula.

In code:

function psi4Position(phi, t, L) {
  // Outer level
  const alpha = Math.acos((L*L - 8) / (2*L));
  const theta = alpha * Math.sin(phi);

  // p3 = L - e^{i theta}
  const p3_re = L - Math.cos(theta);
  const p3_im =   - Math.sin(theta);
  const d     = Math.hypot(p3_re, p3_im);
  const p3_hat_re = p3_re / d;
  const p3_hat_im = p3_im / d;

  // Inner level (helpers depend on d)
  const c        = Math.cos(t);
  const alpha_in = Math.acos((d*d - 3) / (2*d));
  const theta_in = alpha_in * Math.sin(t);

  // nu_in: smooth signed amplitude (apparent 0/0 at cos t = 0 is removable)
  const num   = 2 * d * (Math.cos(theta_in) - (d*d - 3) / (2*d));
  const nu_in = Math.abs(c) < 1e-6
              ? Math.sign(c || 1) * Math.sqrt(d * alpha_in * Math.sin(alpha_in))
              : c * Math.sqrt(num / (c * c));

  // p2 = p3 - p3_hat * e^{i theta_in}  (complex multiplication then subtract)
  const e_re   = Math.cos(theta_in);
  const e_im   = Math.sin(theta_in);
  const rot_re = p3_hat_re * e_re - p3_hat_im * e_im;
  const rot_im = p3_hat_re * e_im + p3_hat_im * e_re;
  const p2_re  = p3_re - rot_re;
  const p2_im  = p3_im - rot_im;
  const p2_abs = Math.hypot(p2_re, p2_im);

  // p1 = p2/2 + (i nu_in / (2 |p2|)) * p2  (elbow formula in ambient frame)
  const k     = nu_in / (2 * p2_abs);
  const p1_re = p2_re / 2 - k * p2_im;
  const p1_im = p2_im / 2 + k * p2_re;

  return {
    p1: [p1_re, p1_im],
    p2: [p2_re, p2_im],
    p3: [p3_re, p3_im],
  };
}

The animation below shows this map, directly from $S^2$ to the linkage itself. Click and drag the point on the 2-sphere to explore the configuration space.