@@ -12,6 +12,7 @@ in their code.
 
    Equations of Motion v0 <equations_of_motion.rst>
    Equations of Motion v1 <equations_of_motion_v1.rst>
+   Tip-off <tip_off.rst>
    Elliptical Fins <aerodynamics/elliptical_fins.rst>
    Individual Fin <aerodynamics/individual_fins.rst>
    Roll Moment <aerodynamics/roll_equations.rst>
 
@@ -6,3 +6,7 @@ References
 .. [Niskanen] Niskanen, S. (2013). *Development of an Open Source model rocket simulation software*.
 
 .. [Model] Barrowman, James S.. (1970). *Model Rocketry*.
+
+.. [Chou] Chou, P.-C., et al. *Tip-off effect analysis of a vehicle moving along an inclined guideway by considering dynamic interactions*.
+
+.. [Hosken] Hosken, R. W., et al. *Analysis of missile launchers, part Q: tip-off effects in helical rail launchers*.
@@ -0,0 +1,244 @@
+.. _tipoff:
+
+===========================================
+Tip-off: the 3-DOF Single Rail Button Phase
+===========================================
+
+Introduction
+------------
+
+Between the moment the *upper* rail button leaves the launch rail and the moment
+the *lower* button follows it, the rocket is still guided --- but only at one
+point. It slides along the rail while free to pitch and yaw about that remaining
+button. This short interval is what the literature calls **tip-off**, and it sets
+the attitude and angular rate with which the rocket begins free flight.
+
+This document derives the equations of motion used by
+:meth:`rocketpy.Flight.udot_rail2`, the flight phase that models this interval.
+It is enabled with ``Flight(..., use_udot_rail2=True)``; when disabled (the
+default) the simulation transitions straight from the 1-DOF rail phase to the
+generalized 6-DOF equations, exactly as it did before this phase existed.
+
+The three flight phases around rail departure are, in order of the distance
+``d`` travelled from the launch point:
+
+.. math::
+
+   \begin{aligned}
+   \texttt{udot_rail1} \quad & \text{for } d < \ell_1
+     && \text{(both buttons engaged, 1 DOF)} \\
+   \texttt{udot_rail2} \quad & \text{for } \ell_1 \le d < \ell_2
+     && \text{(upper button gone, lower engaged, 3 DOF)} \\
+   \texttt{u_dot_generalized} \quad & \text{for } d \ge \ell_2
+     && \text{(free flight, 6 DOF)}
+   \end{aligned}
+
+where :math:`\ell_1` and :math:`\ell_2` are the ``effective_1rl`` and
+``effective_2rl`` attributes of :class:`rocketpy.Flight` --- the distances at
+which the upper and the lower button reach the end of the rail. Their difference
+is the button-to-button distance, so the phase has zero length for a rocket with
+a single rail button and is skipped in that case.
+
+Frames and conventions
+----------------------
+
+The solver integrates the same 13-element state vector used by the other
+right-hand sides,
+
+.. math::
+
+   \mathbf{u} = [\,x,\ y,\ z,\ v_x,\ v_y,\ v_z,\ e_0,\ e_1,\ e_2,\ e_3,\
+   \omega_1,\ \omega_2,\ \omega_3\,]
+
+with :math:`\mathbf{r} = [x, y, z]` the inertial position of the **center of dry
+mass** (CDM, the tracked point), :math:`\mathbf{v}` its inertial velocity,
+:math:`\mathbf{e}` the attitude quaternion and :math:`\boldsymbol{\omega}` the
+angular velocity in the **body** frame. The matrix
+:math:`\mathbf{K} = \texttt{Matrix.transformation}(\mathbf{e})` rotates body
+components into inertial ones, and :math:`\hat{\mathbf{z}}_b = [0, 0, 1]` is the
+body roll (symmetry) axis.
+
+The relevant mass geometry at time :math:`t`, in the body frame, is the total
+mass :math:`m`, the CDM-to-center-of-mass offset :math:`\mathbf{r}_{CM}`, and the
+inertia tensor about the CDM, :math:`\mathbf{I}`. Shifting the latter to the
+instantaneous center of mass gives
+
+.. math::
+
+   \mathbf{I}_{CM} = \mathbf{I} - m\left(|\mathbf{r}_{CM}|^2 \mathbb{1}
+   - \mathbf{r}_{CM}\mathbf{r}_{CM}^{\mathsf T}\right).
+
+**Rail geometry.** The rail is a line fixed in the inertial frame, set by the
+launch inclination and heading. Its unit vector is the ``attitude_unit``
+attribute of :class:`rocketpy.Flight`; the constrained body point is the lower
+rail button, at the fixed body position :math:`\mathbf{r}_{B}` relative to the
+CDM.
+
+The constraint
+--------------
+
+A free rigid body has six degrees of freedom. The single engaged button removes
+three of them:
+
+#. **The button stays on the rail line.** Its position may only vary along the
+   rail, so the component of its acceleration perpendicular to the rail
+   vanishes. That is **two** scalar constraints.
+#. **Roll is suppressed** by the button in its rail slot:
+   :math:`\dot{\boldsymbol{\omega}} \cdot \hat{\mathbf{z}}_b = 0`. That is
+   **one** more.
+
+Three constraints leave :math:`6 - 3 = 3` degrees of freedom: translation along
+the rail, pitch and yaw --- the 3 DOF this phase is named for.
+
+The forces that enforce them are the unknowns of the problem:
+
+* a **normal reaction** at the button, perpendicular to the rail because a
+  frictionless slot can neither pull nor push along it. Writing an orthonormal
+  body triad :math:`\{\hat{\mathbf{e}}_1, \hat{\mathbf{e}}_2, \hat{\mathbf{n}}_b\}`
+  around the body-frame rail direction
+  :math:`\hat{\mathbf{n}}_b = \mathbf{K}^{\mathsf T}\hat{\mathbf{n}}`, it has two
+  components: :math:`\mathbf{N}_b = \lambda_1 \hat{\mathbf{e}}_1 + \lambda_2
+  \hat{\mathbf{e}}_2`;
+* a **roll reaction moment** :math:`\mu \hat{\mathbf{z}}_b` about the body axis.
+
+Three unknowns, three constraints: a :math:`3\times3` linear system, solved once
+per evaluation of the right-hand side.
+
+The initial conditions are consistent with the constraint. The phase is entered
+from ``udot_rail1``, where the rocket has no angular velocity and its velocity is
+along the rail, so the button's perpendicular *velocity* is already zero.
+Enforcing zero perpendicular *acceleration* therefore keeps it on the rail.
+
+Augmented dynamics
+------------------
+
+The generalized equations of motion assemble a total force
+:math:`\mathbf{T}_{20}` and a total moment about the CDM :math:`\mathbf{T}_{21}`,
+both in the body frame, and solve
+
+.. math::
+
+   \dot{\boldsymbol{\omega}}_{\text{free}} = \mathbf{I}_{CM}^{-1}
+   \left(\mathbf{T}_{21} + \mathbf{T}_{20} \times \mathbf{r}_{CM}\right),
+   \qquad
+   \mathbf{a}_{\text{free}} = \frac{\mathbf{T}_{20}}{m}
+   - \mathbf{r}_{CM} \times \dot{\boldsymbol{\omega}}_{\text{free}}.
+
+Both totals are *sums of external contributions*, which is what makes the
+constraint clean to add: the reaction wrench simply enters the sums,
+
+.. math::
+
+   \mathbf{T}_{20}' = \mathbf{T}_{20} + \mathbf{N}_b,
+   \qquad
+   \mathbf{T}_{21}' = \mathbf{T}_{21} + \mathbf{r}_{B} \times \mathbf{N}_b
+   + \mu \hat{\mathbf{z}}_b,
+
+after which the same two lines apply. Since the solve is linear in the totals,
+the result splits into the free solution plus a response to the unknowns. Using
+:math:`\mathbf{r}_B \times \mathbf{N} + \mathbf{N} \times \mathbf{r}_{CM}
+= (\mathbf{r}_B - \mathbf{r}_{CM}) \times \mathbf{N}` and writing
+:math:`\mathbf{d} = \mathbf{r}_{B} - \mathbf{r}_{CM}` for the center-of-mass-to-button
+vector,
+
+.. math::
+
+   \Delta\dot{\boldsymbol{\omega}} = \mathbf{I}_{CM}^{-1}
+   \left(\mathbf{d} \times \mathbf{N}_b + \mu \hat{\mathbf{z}}_b\right),
+   \qquad
+   \Delta\mathbf{a} = \frac{\mathbf{N}_b}{m}
+   - \mathbf{r}_{CM} \times \Delta\dot{\boldsymbol{\omega}}.
+
+The button is body-fixed, so its acceleration in body components is
+
+.. math::
+
+   \mathbf{A} = \mathbf{A}_{\text{free}} + \Delta\mathbf{a}
+   + \Delta\dot{\boldsymbol{\omega}} \times \mathbf{r}_{B},
+   \qquad
+   \mathbf{A}_{\text{free}} = \mathbf{K}^{\mathsf T}\mathbf{a}_{\text{free}}
+   + \dot{\boldsymbol{\omega}}_{\text{free}} \times \mathbf{r}_{B}
+   + \boldsymbol{\omega} \times
+   \left(\boldsymbol{\omega} \times \mathbf{r}_{B}\right).
+
+The linear solve
+----------------
+
+Collect the unknowns in :math:`\boldsymbol{\chi} = [\lambda_1, \lambda_2, \mu]`.
+The three constraints read
+
+.. math::
+
+   \mathbf{A} \cdot \hat{\mathbf{e}}_1 = 0,
+   \qquad
+   \mathbf{A} \cdot \hat{\mathbf{e}}_2 = 0,
+   \qquad
+   \dot{\boldsymbol{\omega}} \cdot \hat{\mathbf{z}}_b = 0,
+
+and because :math:`\mathbf{A}` and :math:`\dot{\boldsymbol{\omega}}` are linear
+in :math:`\boldsymbol{\chi}`, they form the system
+:math:`\mathbf{J}\boldsymbol{\chi} = -\mathbf{g}_{\text{free}}` with
+
+.. math::
+
+   \mathbf{g}_{\text{free}} =
+   \begin{bmatrix}
+   \mathbf{A}_{\text{free}} \cdot \hat{\mathbf{e}}_1 \\
+   \mathbf{A}_{\text{free}} \cdot \hat{\mathbf{e}}_2 \\
+   \dot{\boldsymbol{\omega}}_{\text{free}} \cdot \hat{\mathbf{z}}_b
+   \end{bmatrix}.
+
+Each column of :math:`\mathbf{J}` is obtained by evaluating the response above at
+one of the three unit inputs :math:`(\mathbf{N}_b, \mu) =
+(\hat{\mathbf{e}}_1, 0)`, :math:`(\hat{\mathbf{e}}_2, 0)`,
+:math:`(\mathbf{0}, 1)`. Solving for :math:`\boldsymbol{\chi}` and substituting
+back gives the constrained accelerations, which override the free ones in the
+returned derivative:
+
+.. math::
+
+   \dot{\boldsymbol{\omega}} = \dot{\boldsymbol{\omega}}_{\text{free}}
+   + \Delta\dot{\boldsymbol{\omega}},
+   \qquad
+   \mathbf{a}_{CDM} = \mathbf{a}_{\text{free}}
+   + \mathbf{K}\,\Delta\mathbf{a}.
+
+The position and quaternion derivatives are the ordinary kinematic ones. In
+particular :math:`\dot{\mathbf{r}} = \mathbf{v}`: the velocity is **not**
+projected onto the rail. The constraint acts at the acceleration level on the
+*button*, and the CDM legitimately acquires a small perpendicular velocity as the
+rocket pitches about that button.
+
+Modelling assumptions and edge cases
+------------------------------------
+
+* **Roll axis.** The constraint suppresses roll about the *body* axis. The rocket
+  travels only the button-to-button distance during this phase, so the tip-off
+  angle is small and the body axis stays close to the rail direction; the
+  difference between body roll and rail-axis roll is of that order. This is
+  consistent with ``udot_rail1``, which freezes rotation entirely.
+* **Radial button offset.** The button is modelled on the rocket axis. The roll
+  constraint is enforced explicitly by :math:`\mu`, so what is lost is only the
+  (small) roll coupling through the button's radial standoff.
+* **Zero-length phase.** A rocket with a single rail button has
+  :math:`\ell_1 = \ell_2`, and the phase is skipped.
+* **Singular system.** Should the geometry make :math:`\mathbf{J}` singular, the
+  step falls back to the unconstrained dynamics and warns.
+
+Expected behaviour
+------------------
+
+The phase reproduces the two effects tip-off is modelled for. With no wind, the
+center of mass sits ahead of the button that the rocket now pivots about, so
+gravity pitches the nose down by a fraction of a degree. With a crosswind, the
+aerodynamic moment turns the rocket into the wind before it is fully free ---
+the weathercock effect --- and the rocket therefore leaves the rail with a small
+angular rate rather than none.
+
+References
+----------
+
+The tip-off phase and its effect on the initial conditions of free flight are
+treated in the launcher dynamics literature: [Chou]_ studies a vehicle moving
+along an inclined guideway with dynamic interactions, and [Hosken]_ analyses
+tip-off effects in rail launchers.
@@ -2030,8 +2030,9 @@ def udot_rail2(self, t, u, post_processing=False):
         to the rail, 2 DOF) plus a roll reaction moment ``mu`` (1 DOF). The
         three unknowns are found from three constraints -- the button's
         acceleration perpendicular to the rail is zero (2) and the roll angular
-        acceleration is zero (1). See ``scratch/pr920_tipoff_derivation.md`` for
-        the full derivation. All reaction quantities are expressed in the "true"
+        acceleration is zero (1). The full derivation is in the technical
+        documentation, :ref:`Tip-off <tipoff>`. All reaction quantities are
+        expressed in the "true"
         body frame (the one used by ``surfaces_cp_to_cdm``, body-z toward the
         nose), so ``r_CM`` and the button position are taken with that sign
         convention -- independent of the internal (point-to-CDM) convention used