Quadrotor with cable-suspended payload

This page derives the differential-flatness property of a single quadrotor carrying a point-mass payload via a taut, massless cable, as established by Sreenath, Michael, and Kumar [1] and the companion geometric-control paper of Sreenath, Lee, and Kumar [2]. The flat output is \(y = (x_L,\, \psi)\), with \(x_L \in \mathbb{R}^3\) the payload position and \(\psi \in \mathbb{R}\) the quadrotor yaw angle. The flat-to-state recovery requires \(x_L\) up to the sixth derivative \(y^{(6)}\) and the yaw up to \(\ddot\psi\).

System

Let \(x_L \in \mathbb{R}^3\) be the payload position and \(x_Q \in \mathbb{R}^3\) the quadrotor centre-of-mass position, both in the world frame. Let \(q \in S^2\) be the unit cable vector in the world frame, oriented from the quadrotor to the payload, and \(\omega \in T_q S^2\) the cable angular velocity (so \(\omega \cdot q = 0\)); see Preliminaries for the \(S^2 / TS^2\) conventions. Let \(R \in SO(3)\) be the quadrotor attitude and \(\Omega \in \mathbb{R}^3\) the body-frame angular velocity, so \(\dot R = R\,\widehat{\Omega}\). The collective thrust \(f \in \mathbb{R}_{\geq 0}\) acts along the body \(e_3\)-axis and the moment \(M \in \mathbb{R}^3\) acts in the body frame. Quadrotor mass is \(m_Q\), payload mass \(m_L\), cable length \(\ell > 0\), and quadrotor inertia \(J = J^\top \succ 0\).

The taut-cable kinematic tie between the two positions is

(28)\[x_Q \;=\; x_L \,-\, \ell\, q, \qquad \dot q \;=\; \omega \times q.\]

With this sign convention, \(q \approx -e_3\) in the hanging equilibrium. Under the taut-cable assumption the system is a hybrid rigid body; from [2] its equations of motion are

(29)\[\begin{split}\begin{aligned} (m_Q + m_L)\bigl(\ddot x_L + g\, e_3\bigr) &= \bigl(q \cdot f R\, e_3 \,-\, m_Q\, \ell\, \|\dot q\|^2\bigr)\, q, \\[2pt] m_Q\, \ell\, \dot\omega &= -\, q \times f R\, e_3, \\[2pt] \dot R &= R\, \widehat{\Omega}, \\[2pt] J\, \dot\Omega \,+\, \Omega \times J\, \Omega &= M. \end{aligned}\end{split}\]

The first line of (29) is Newton’s law for the combined system resolved along \(q\) (the only direction in which the cable transmits force), the second is the cable angular dynamics on \(T_q S^2\), and the third–fourth are the rigid-body attitude dynamics. The four scalar control inputs are \((f,\, M) \in \mathbb{R}^4\), so \(\dim u = 4\). The full state \((x_L,\, \dot x_L,\, q,\, \omega,\, R,\, \Omega)\) lives in \(\mathbb{R}^3 \times \mathbb{R}^3 \times TS^2 \times TSO(3)\), which has dimension \(16\) (agreeing with the 8 DOF / 4-degree underactuation count of [1]: \(3 + 3 + 4 + 6 = 16\), since \(\dim TS^2 = 4\) and \(\dim TSO(3) = 6\)).

Flat output

Take

(30)\[y \;=\; (x_L,\, \psi) \;\in\; \mathbb{R}^3 \times \mathbb{R}.\]

Then \(\dim y = 4 = \dim u\), so the dimension-matching necessary condition for differential flatness (see Differential flatness) is satisfied. The remainder of this page constructs the algebraic map from the jet of \(y\) to \((x_L,\, \dot x_L,\, q,\, \omega,\, R,\, \Omega,\, f,\, M)\), confirming that \(y\) is a flat output. Inspection of the recovery below shows that \(x_L\) is needed up to \(y^{(6)}\) and \(\psi\) is needed up to \(\ddot\psi\).

Constructive recovery

Step 1 — cable direction from payload dynamics

The cable direction can be recovered two equivalent ways: from the constrained-Lagrangian first line of (29), or from a Newton–Euler cut on the payload alone with the cable tension made explicit. Both yield the same unit vector \(q\).

Lagrangian form. Group the gravity-augmented payload acceleration with the combined mass into a single vector,

(31)\[A \;:=\; (m_Q + m_L)\bigl(\ddot x_L + g\, e_3\bigr) \;\in\; \mathbb{R}^3.\]

The first equation of (29) reads \(A = \lambda\, q\) for some scalar \(\lambda = q \cdot f R\, e_3 - m_Q\, \ell\, \|\dot q\|^2\), so \(A\) is parallel to \(q\). In the hanging/taut regime the thrust \(f R\, e_3\) has a positive component along \((-q)\) — it pulls the quadrotor away from the payload — so \(q \cdot f R\, e_3 < 0\) and hence \(\lambda < 0\). Combined with \(\|q\| = 1\), this fixes both magnitude and sign:

(32)\[q \;=\; -\,\frac{A}{\|A\|}, \qquad \lambda \;=\; -\,\|A\|.\]

Newton–Euler-with-tension form. Equivalently, isolate the payload and write Newton’s law with the cable reaction \(-T q\) explicit (following [1], eq. 24):

(33)\[m_L\, \ddot x_L \;=\; -\,T\, q \,-\, m_L\, g\, e_3 \quad\Longleftrightarrow\quad T\, q \;=\; -\, m_L\bigl(\ddot x_L + g\, e_3\bigr),\]

where \(T \in \mathbb{R}_{\geq 0}\) is the scalar cable tension. It is convenient to package the cable reaction into a single vector, the tension vector

(34)\[\mathbf{T} \;:=\; T\, q \;=\; -\, m_L\bigl(\ddot x_L + g\, e_3\bigr) \;\in\; \mathbb{R}^3.\]

Taking norms (with \(\|q\| = 1\)) and dividing yields the scalar tension and unit cable direction in closed form:

(35)\[T \;=\; \|\mathbf{T}\| \;=\; m_L\,\|\ddot x_L + g\, e_3\|, \qquad q \;=\; \frac{\mathbf{T}}{T}.\]

Both routes give the same \(q\): the mass prefactor — \((m_Q + m_L)\) in \(A\) from the Lagrangian form, \(m_L\) in \(\mathbf{T}\) from the Newton–Euler form — cancels in the unit-vector normalisation. The scalars differ (\(\lambda\) is the constrained reaction along \(q\) from the combined-system equation; \(T\) is the physical cable tension), and they are related by \(\lambda = -(m_Q + m_L)/m_L \cdot T\). Both \(T\) and \(q\) are algebraic functions of \(\ddot x_L\) alone.

Which form to compute

For implementation, prefer the Newton–Euler form (34). It depends only on the payload mass \(m_L\) and gives the physical tension \(T\) as a free byproduct — useful both as a slack-cable diagnostic (\(T \to 0\) marks the hybrid-mode boundary at which the cable goes slack) and as a sanity-check when comparing to the simulation. The Lagrangian form requires the combined mass \((m_Q + m_L)\) for the same answer and yields \(\lambda\), which has no direct physical meaning and would have to be back-computed from \(f\), \(R\), \(\dot q\) if needed. The derivation chain below uses \(\mathbf{T}\) accordingly.

Differentiating \(\mathbf{T} = T q\) gives, with \(\dot T = (\mathbf{T} \cdot \dot{\mathbf{T}})/T = \dot{\mathbf{T}} \cdot q\),

(36)\[\dot q \;=\; \frac{\dot{\mathbf{T}}}{T} \,-\, \frac{\dot T}{T}\, q \;=\; \frac{1}{T}\Bigl(\dot{\mathbf{T}} \,-\, (\dot{\mathbf{T}} \cdot q)\, q\Bigr).\]

This is the projection of \(\dot{\mathbf{T}}/T\) onto the tangent plane \(T_q S^2\), automatically perpendicular to \(q\) because \(\|q\| = 1\) forces \(q \cdot \dot q = 0\) (see the \(S^2\) tangent-space discussion in Preliminaries). Since from (34), \(\dot{\mathbf{T}} = -\, m_L\, x_L^{(3)}\), so computing \(\dot q\) requires the payload jerk.

Differentiating \(T q = \mathbf{T}\) once more (Leibniz, \(n = 2\)),

\[ \ddot T\, q \,+\, 2\,\dot T\, \dot q \,+\, T\, \ddot q \;=\; \ddot{\mathbf{T}}, \]

and solving for \(\ddot q\),

\[ \ddot q \;=\; \frac{1}{T}\Bigl(\ddot{\mathbf{T}} \,-\, \ddot T\, q \,-\, 2\,\dot T\, \dot q\Bigr). \]

Substituting the tension derivatives \(\dot T = \dot{\mathbf{T}} \cdot q\) and \(\ddot T = \ddot{\mathbf{T}} \cdot q + \dot{\mathbf{T}} \cdot \dot q\) (both consequences of \(T = \mathbf{T} \cdot q\), see also (39)),

\[ \ddot q \;=\; \frac{1}{T}\Bigl(\ddot{\mathbf{T}} \,-\, (\ddot{\mathbf{T}} \cdot q)\, q \,-\, (\dot{\mathbf{T}} \cdot \dot q)\, q \,-\, 2\,(\dot{\mathbf{T}} \cdot q)\, \dot q\Bigr). \]

The middle term collapses into the manifest tangency form. From (36), \(T\, \dot q = \dot{\mathbf{T}} - (\dot{\mathbf{T}} \cdot q)\, q\). Dotting with \(\dot q\) and using \(q \cdot \dot q = 0\) gives \(T\, \|\dot q\|^2 = \dot{\mathbf{T}} \cdot \dot q\), so \((\dot{\mathbf{T}} \cdot \dot q)/T = \|\dot q\|^2\). Substituting:

(37)\[\ddot q \;=\; \frac{1}{T}\Bigl(\ddot{\mathbf{T}} \,-\, (\ddot{\mathbf{T}} \cdot q)\, q \,-\, 2\,(\dot{\mathbf{T}} \cdot q)\, \dot q\Bigr) \,-\, \|\dot q\|^2\, q,\]

where the trailing \(-\|\dot q\|^2\, q\) packages the \((\dot{\mathbf{T}} \cdot \dot q)/T\) residue and enforces the second-order tangency condition \(q \cdot \ddot q = -\|\dot q\|^2\) obtained by differentiating \(q \cdot \dot q = 0\).

The higher derivatives are cleanest via the general Leibniz rule \(\,(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k}\, f^{(k)} g^{(n-k)}\,\) [3]. Applied to \(T\, q = \mathbf{T}\) and solving for \(q^{(n)}\),

(38)\[T\, q^{(n)} \;=\; \mathbf{T}^{(n)} \,-\, \sum_{k=1}^{n} \binom{n}{k}\, T^{(k)}\, q^{(n-k)},\]

which solves for \(q^{(n)}\) once the lower-order \(q^{(0)}, \ldots, q^{(n-1)}\) and the tension derivatives are known. The tension derivatives are obtained by differentiating the definition \(\mathbf{T} = T q\) and projecting onto \(q\) at each order — equivalent to differentiating \(T = \mathbf{T} \cdot q\) (the projection \(\mathbf{T} \cdot q = T q \cdot q = T\), using \(\|q\| = 1\)). The sub-leading \(\mathbf{T} \cdot q^{(k)}\) pieces collapse via the tangency identities \(q \cdot \dot q = 0\), \(q \cdot \ddot q = -\|\dot q\|^2\), etc., leaving:

(39)\[\begin{split}\begin{aligned} \dot T &\;=\; \dot{\mathbf{T}} \cdot q, \\[2pt] \ddot T &\;=\; \ddot{\mathbf{T}} \cdot q \,+\, \dot{\mathbf{T}} \cdot \dot q, \\[2pt] T^{(3)} &\;=\; \mathbf{T}^{(3)} \cdot q \,+\, 2\,\ddot{\mathbf{T}} \cdot \dot q \,+\, \dot{\mathbf{T}} \cdot \ddot q, \\[2pt] T^{(4)} &\;=\; \mathbf{T}^{(4)} \cdot q \,+\, 3\,\mathbf{T}^{(3)} \cdot \dot q \,+\, 3\,\ddot{\mathbf{T}} \cdot \ddot q \,+\, \dot{\mathbf{T}} \cdot q^{(3)}. \end{aligned}\end{split}\]

Substituting \(n = 3\) and \(n = 4\) into (38),

(40)\[q^{(3)} \;=\; \frac{1}{T}\Bigl( \mathbf{T}^{(3)} \,-\, 3\,\dot T\, \ddot q \,-\, 3\,\ddot T\, \dot q \,-\, T^{(3)}\, q \Bigr),\]

(41)\[q^{(4)} \;=\; \frac{1}{T}\Bigl( \mathbf{T}^{(4)} \,-\, 4\,\dot T\, q^{(3)} \,-\, 6\,\ddot T\, \ddot q \,-\, 4\,T^{(3)}\, \dot q \,-\, T^{(4)}\, q \Bigr).\]

The two earlier expressions (36) and (37) are the \(n = 1, 2\) special cases of this recursion, with the tangency conditions on \(T_q S^2\) written explicitly. Tracking the input order: \(\dot{\mathbf{T}} = -m_L\, x_L^{(3)}\), so \(q^{(3)}\) requires \(x_L^{(5)}\) (crackle) and \(q^{(4)}\) requires \(x_L^{(6)}\) (pop) — the deepest payload-jet rung in the recovery.

Step 2 — quadrotor position and its derivatives

Apply \(x_Q = x_L - \ell\, q\) from (28) and differentiate term-by-term:

(42)\[\begin{split}\begin{aligned} x_Q &\;=\; x_L \,-\, \ell\, q, \\ \dot x_Q &\;=\; \dot x_L \,-\, \ell\, \dot q, \\ \ddot x_Q &\;=\; \ddot x_L \,-\, \ell\, \ddot q, \\ x_Q^{(3)} &\;=\; x_L^{(3)} \,-\, \ell\, q^{(3)}, \\ x_Q^{(4)} &\;=\; x_L^{(4)} \,-\, \ell\, q^{(4)}. \end{aligned}\end{split}\]

The differentiation order combines additively with Step 1: each \(x_Q^{(k)}\) requires the payload jet through \(x_L^{(k)}\) for the direct term and through \(x_L^{(k+2)}\) for the cable term. The worst-case rung is \(x_Q^{(4)}\) (quadrotor snap), which needs \(q^{(4)}\) and therefore \(x_L^{(6)}\) — the source of the sixth-order requirement in the flat output.

Step 3 — cable angular velocity and acceleration

The cable kinematics \(\dot q = \omega \times q\) inverts on \(T_q S^2\) by the same BAC-CAB move used for (7) in Preliminaries:

(43)\[\omega \;=\; q \times \dot q \;\in\; T_q S^2.\]

Indeed, \(q \times \dot q = q \times (\omega \times q) = \omega\,(q \cdot q) - q\,(q \cdot \omega) = \omega\), using \(\|q\| = 1\) and the convention \(\omega \cdot q = 0\). Differentiating (43) once,

(44)\[\dot\omega \;=\; q \times \ddot q,\]

since \(\dot q \times \dot q = 0\). Recovering \(\omega\) therefore needs \(\dot q\) — i.e. \(x_L^{(3)}\) — and recovering \(\dot\omega\) needs \(\ddot q\) — i.e. \(x_L^{(4)}\).

The cable angular dynamics in the second line of (29) are not needed to invert \(\omega\): they serve as a consistency relation between \(\dot\omega\) and the moment-arm of the thrust vector across the payload, which is automatically satisfied by the construction in Step 4.

Step 4 — thrust vector

Newton’s law for the quadrotor in the world frame is

\[ m_Q\, \ddot x_Q \;=\; f R\, e_3 \,-\, m_Q\, g\, e_3 \,+\, \mathbf{T}, \]

where \(\mathbf{T}\) is the cable reaction acting on the quadrotor (Newton’s third law: the payload pulls the quadrotor along \(+q\) via the cable, so the force on the quadrotor is \(+T q = +\mathbf{T}\); on the payload it is \(-\mathbf{T}\), as in (33)). Solving for the thrust vector,

(45)\[B \;:=\; f R\, e_3 \;=\; m_Q\bigl(\ddot x_Q + g\, e_3\bigr) \,-\, \mathbf{T}.\]

Both terms on the right-hand side are already in hand: \(\ddot x_Q\) from Step 2, \(\mathbf{T}\) from Step 1. Equivalently, eliminating \(\mathbf{T}\) via \(\mathbf{T} = -m_L(\ddot x_L + g\, e_3)\) and \(\ddot x_Q = \ddot x_L - \ell\, \ddot q\) produces the two combined-system forms

(46)\[B \;=\; m_Q\, \ddot x_Q \,+\, m_L\, \ddot x_L \,+\, (m_Q + m_L)\, g\, e_3 \;=\; (m_Q + m_L)\bigl(\ddot x_L + g\, e_3\bigr) \,-\, m_Q\, \ell\, \ddot q,\]

obtained by adding the quadrotor and payload Newton’s laws (the \(\pm \mathbf{T}\) pieces cancel). The thrust magnitude and direction are then

(47)\[f \;=\; \|B\|, \qquad b_3 \;=\; R\, e_3 \;=\; \frac{B}{\|B\|}.\]

Compared to the standalone-quadrotor flatness map of Quadrotor, \(B\) replaces that page’s \(A = m_Q(\ddot x_Q + g\, e_3)\) — same role (the thrust vector), corrected by the cable reaction \(-\mathbf{T}\). Once \(B\) is computed, every downstream quantity is identical.

Step 5 — attitude, body rate, moment (via the quadrotor recovery)

With \((B,\, \psi)\) in hand, the remaining quantities \((R,\, \Omega,\, \dot\Omega,\, M)\) are recovered by the same machinery as the standalone-quadrotor flatness map of Quadrotor, applied with \(B\) in place of that page’s \(A\). The algebra is not repeated here; the relevant pieces are:

• Attitude \(R\) from \(b_3\) and the yaw-heading hint \(b_{1,d}(\psi) = [\cos\psi,\, \sin\psi,\, 0]^\top\) — see (21)–(22) in Quadrotor. The two derivatives \(\dot R\), \(\ddot R\) follow the recursion of (23)–(24), which feed off \(\dot b_3\), \(\ddot b_3\) via the projector (16)–(18).

• Body rate \(\Omega\) and angular acceleration \(\dot\Omega\) via the skew-projector form (25)–(26) of Quadrotor. The projector form is preferred over the cancellation form for the numerical-robustness reasons unpacked there.

• Moment \(M\) from the body-frame Newton–Euler equation,

  (48)\[M \;=\; J\, \dot\Omega \,+\, \Omega \times J\, \Omega,\]

  identical to (27) of Quadrotor (the rotational dynamics are unchanged by the cable, since the cable applies a pure force at the body-fixed attachment point and exerts no moment about the centre of mass under the point-mass payload assumption).

Tracking the derivative chain: \(B\) depends on \(\ddot x_Q\) (and hence \(x_L^{(4)}\) via Step 2), so \(\dot B\) needs \(x_L^{(5)}\) and \(\ddot B\) needs \(x_L^{(6)}\). Through the recovery, \(\Omega\) requires \(\dot b_3 \sim \dot B\) and hence \(x_L^{(5)}\), and \(\dot\Omega\) (and therefore \(M\)) requires \(\ddot b_3 \sim \ddot B\) and hence \(x_L^{(6)}\) — the deepest payload-jet rung.

Singularity

Two singularities limit the regular region of the flatness map. The first is the slack-cable limit \(\ddot x_L + g\, e_3 = 0\), where \(\mathbf{T}\) in (34) vanishes and the cable direction \(q\) is undefined — physically, the payload is in free-fall and the cable carries no tension (\(T = 0\)). The second is the yaw-aligned limit where \(b_3 \parallel b_{1,d}(\psi)\), identical to the singularity of the single-quadrotor map (Quadrotor). Both must be excluded from the open subset of jet space on which the recovery is smooth.

Theorem

Theorem 2 (Flatness of the quadrotor with cable-suspended point-mass payload)

The quadrotor with taut, massless-cable-suspended point-mass payload described by (29) is differentially flat, with flat output \(y = (x_L,\, \psi)\) as in (30). Every state and every input is an algebraic function of the sixth-order jet of \(x_L\) and the second-order jet of \(\psi\).

Proof. Dimension matching: \(\dim y = 4 = \dim u\). Steps 1–7 above construct the smooth map

\[ \Phi : J^6 \mathbb{R}^3 \times J^2 \mathbb{R} \;\longrightarrow\; \mathbb{R}^{16} \times \mathbb{R}^4, \qquad \Phi\bigl(J^6_t\, x_L,\, J^2_t\, \psi\bigr) \;=\; \bigl((x_L, \dot x_L, q, \omega, R, \Omega),\; (f, M)\bigr), \]

which recovers the full state and input without integrating the dynamics: \(q\) from \(\ddot x_L\) via (35); \(\omega\) from \(x_L^{(3)}\) via (43); \(x_Q\) and its derivatives up to \(x_Q^{(4)}\) from \(x_L\) up to \(x_L^{(6)}\) via (42); the thrust vector \(B = f R\, e_3\) from \(\ddot x_Q\) and \(\mathbf{T}\) via (45); and finally \((R,\, \Omega,\, \dot\Omega,\, M)\) from \((B,\, \psi)\) by the standalone-quadrotor recovery of Quadrotor (Steps 2–4 there) — closing with \(M\) via (48). Smoothness of \(\Phi\) holds on the open subset \(\{\,\ddot x_L + g\, e_3 \neq 0\,\} \cap \{\,b_3 \not\parallel b_{1,d}(\psi)\,\}\) — the former excludes the slack-cable / free-fall configuration, the latter the yaw-aligned configuration in which the desired attitude is not uniquely defined. This is exactly the form (10) on the product jet space, so the system is differentially flat with flat output \(y\).

This recovery is the substrate of the trajectory generation and aggressive-maneuver experiments in [1]; it generalises to multi-lift and flexible-cable variants in [4].

Derivative-order accounting

Recovered quantity	Needs
\(q\)	\(\ddot x_L\)
\(\dot q,\ \omega\)	\(x_L^{(3)}\)
\(\ddot q,\ \dot\omega\)	\(x_L^{(4)}\)
\(q^{(3)}\)	\(x_L^{(5)}\)
\(q^{(4)}\)	\(x_L^{(6)}\)
\(x_Q,\ \dot x_Q,\ \ddot x_Q\)	\(\ddot x_L\) (via \(q\), \(\dot q\), \(\ddot q\))
\(x_Q^{(3)}\)	\(x_L^{(5)}\) (via \(q^{(3)}\))
\(x_Q^{(4)}\) (snap)	\(x_L^{(6)}\) (via \(q^{(4)}\))
\(f,\ b_3\)	\(\ddot x_L\) and \(\ddot x_Q\), i.e. \(x_L^{(4)}\)
\(R\)	\(\ddot x_L,\ x_L^{(4)},\ \psi\)
\(\Omega\)	\(x_L^{(5)},\ \dot\psi\)
\(\dot\Omega\)	\(x_L^{(6)},\ \ddot\psi\)
\(M\)	\(x_L^{(6)},\ \ddot\psi\)

The worst case is the moment \(M\), which closes the chain at \(x_L^{(6)}\) (pop) and \(\ddot\psi\) — two orders deeper in the payload position than the bare quadrotor case of Theorem 1, where snap was sufficient. The two extra orders are exactly the cost of recovering the quadrotor snap \(x_Q^{(4)}\) through the cable constraint \(x_Q = x_L - \ell\, q\).

Jet input

The map \(\Phi\) of Theorem 2 takes a pair of udaan.flatness.Jet objects — one of order 6 for the payload position, one of order 2 for the yaw — packaged into a udaan.flatness.QuadrotorCsPayloadFlats struct:

import numpy as np
from udaan.flatness import (
    Jet,
    QuadrotorCsPayload,
    QuadrotorCsPayloadFlats,
)

# Payload position and its first six time derivatives at t
x_L_jet = Jet(np.stack([
    x_L,            # row 0 — payload position
    x_L_dot,        # row 1 — payload velocity
    x_L_ddot,       # row 2 — payload acceleration (gives q)
    x_L_dddot,      # row 3 — jerk     (gives q̇, ω, ẋ_Q)
    x_L_snap,       # row 4 — snap     (gives q̈, ẍ_Q, f, b3)
    x_L_crackle,    # row 5 — crackle  (gives q⁽³⁾, x_Q⁽³⁾, Ω)
    x_L_pop,        # row 6 — pop      (gives q⁽⁴⁾, x_Q⁽⁴⁾, dΩ, M)
]))
x_L_jet.order   # 6
x_L_jet.dim     # 3

# Yaw and its first two derivatives
psi_jet = Jet(np.array([psi, psi_dot, psi_ddot]))
psi_jet.order   # 2
psi_jet.dim     # 1

flats = QuadrotorCsPayloadFlats(x_L=x_L_jet, psi=psi_jet)
ref, inputs = QuadrotorCsPayload(
    mass=m_Q,
    inertia=J,
    payload_mass=m_L,
    cable_length=ell,
).recover(flats)
# ref.payload_position, ref.payload_velocity,
# ref.position, ref.velocity, ref.acceleration,
# ref.cable_attitude, ref.cable_angular_velocity,
# ref.orientation, ref.angular_velocity, ref.angular_acceleration
# inputs.thrust, inputs.moment, inputs.tension

The returned ref is a udaan.flatness.QuadrotorCsPayloadRefState and inputs is a udaan.flatness.QuadrotorCsPayloadInputs. Truncation (Definition 3) extracts the lower-order jets needed by intermediate stages of the recovery (for example, x_L_jet.truncate(4) is enough to produce \((q,\, \dot q,\, \ddot q,\, x_Q,\, \dot x_Q,\, \ddot x_Q,\, f,\, b_3)\) but not \((\Omega,\, M)\)).

References

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[1] (1,2,3,4)

Koushil Sreenath, Nathan Michael, and Vijay Kumar. Trajectory generation and control of a quadrotor with a cable-suspended load — a differentially-flat hybrid system. In IEEE International Conference on Robotics and Automation (ICRA), 4888–4895. 2013.

[2] (1,2)

Koushil Sreenath, Taeyoung Lee, and Vijay Kumar. Geometric control and differential flatness of a quadrotor UAV with a cable-suspended load. In IEEE Conference on Decision and Control (CDC), 2269–2274. 2013.

[3]

Tom M. Apostol. Calculus, Volume I: One-Variable Calculus, with an Introduction to Linear Algebra. Wiley, 2nd edition, 1967. General Leibniz rule for the $n$-th derivative of a product, §6.17.

[4]

Prasanth Kotaru, Guofan Wu, and Koushil Sreenath. Differential-flatness and control of quadrotor(s) with a payload suspended through flexible cable(s). In Indian Control Conference (ICC), 352–357. 2018.