PolynomialProfile.md

Polynomial Point-to-Point Profile

Overview & Motivation

A polynomial profile is a closed-form time law that drives one axis from a start state to a goal state over a fixed duration $T$. A cubic matches endpoint position and velocity; a quintic additionally matches endpoint acceleration, so the motion has no acceleration step at either end.

Point-to-point moves are the bread and butter of servo motion: robot joints, CNC and pick-and-place axes, motor position loops. The coefficients come from a handful of algebraic expressions, solved once. Each servo tick is then a fixed number of multiply-adds, with no iteration, no lookup table and a deterministic cycle count.

Mathematical Theory

Boundary Conditions

Let $t \in [0, T]$, $\Delta q = q_f - q_0$, and let $\alpha_0, \alpha_f$ be the start and end accelerations. A polynomial of degree $n$ has $n+1$ coefficients, so a cubic satisfies four conditions and a quintic six:

$$q(0) = q_0,\quad q(T) = q_f,\quad \dot{q}(0) = v_0,\quad \dot{q}(T) = v_f \qquad \text{(quintic also: } \ddot{q}(0) = \alpha_0,\ \ddot{q}(T) = \alpha_f\text{)}$$

The conditions are linear in the coefficients $a_k$ of $q(t) = \sum_k a_k t^k$ and the time basis is fixed, so the solution is a plug-in formula rather than a runtime matrix solve.

Cubic

The start conditions give $a_0 = q_0$ and $a_1 = v_0$. The end conditions leave a $2\times 2$ system with the closed-form solution

$$a_2 = \frac{3,\Delta q - (2 v_0 + v_f),T}{T^2}, \qquad a_3 = \frac{-2,\Delta q + (v_0 + v_f),T}{T^3}$$

The acceleration is linear in $t$ and the jerk is the constant $6 a_3$.

Quintic

The start conditions give $a_0 = q_0$, $a_1 = v_0$, $a_2 = \alpha_0 / 2$. The end conditions give

$$a_3 = \frac{20,\Delta q - (8 v_f + 12 v_0),T - (3\alpha_0 - \alpha_f),T^2}{2,T^3}$$

$$a_4 = \frac{-30,\Delta q + (14 v_f + 16 v_0),T + (3\alpha_0 - 2\alpha_f),T^2}{2,T^4}$$

$$a_5 = \frac{12,\Delta q - 6 (v_f + v_0),T - (\alpha_0 - \alpha_f),T^2}{2,T^5}$$

The acceleration is a cubic matching $\alpha_0$ and $\alpha_f$, hence continuous; the jerk is a quadratic.

Sampling

Position, velocity, acceleration and jerk are four Horner evaluations of the polynomial and its three derivatives:

$$q = a_0 + t\big(a_1 + t(a_2 + t(a_3 + t(a_4 + t,a_5)))\big), \qquad \dddot{q} = 6 a_3 + t,(24 a_4 + 60 a_5, t)$$

The cubic is the same evaluation with $a_4 = a_5 = 0$. The time is first clamped to $[0, T]$.

Rest-to-Rest Peaks

With zero end velocities (and accelerations, for the quintic) the cubic peaks at $1.5,\Delta q / T$ in velocity and $6,\Delta q / T^2$ in acceleration; the quintic at $1.875,\Delta q / T$ and $\tfrac{10}{\sqrt{3}},\Delta q / T^2 \approx 5.77,\Delta q / T^2$.

Complexity Analysis

Operation	Time	Space	Notes
Coefficient solve	$O(1)$	$O(1)$	Closed-form expressions, once at construction or reset
Sample	$O(\text{degree})$	$O(1)$	Four Horner evaluations, at most 14 multiply-adds; no state change, so deterministic
Stored state	n/a	$O(1)$	Six coefficients, the duration and the degree; stack only, no heap

Step-by-Step Walkthrough

Rest-to-rest cubic with $q_0 = 0$, $q_f = 1$, $T = 2$, $v_0 = v_f = 0$, so $\Delta q = 1$ and $a_0 = a_1 = 0$.

Step	Computation	Result
$a_2$	$(3\cdot 1 - 0)/2^2$	$0.75$
$a_3$	$(-2\cdot 1 + 0)/2^3$	$-0.25$
Position	$0.75,t^2 - 0.25,t^3 = 3(t/2)^2 - 2(t/2)^3$	$q(1) = 0.5$, $q(2) = 1$
Velocity	$1.5,t - 0.75,t^2$	$v(1) = 0.75 = 1.5,\Delta q / T$ (the peak)
Acceleration	$1.5 - 1.5,t$	$+1.5$ at $t=0$, $-1.5$ at $t=2$
Jerk	$6 a_3$	$-1.5$, constant

The position is the smoothstep of $t/2$, symmetric about the midpoint, where the velocity peaks. The acceleration jumps from $0$ at rest to $1.5$ at the first instant and from $-1.5$ back to $0$ at the end.

Pitfalls & Edge Cases

• Duration must be positive. The coefficients contain $1/T^k$ terms; a zero or negative duration is rejected at construction and reset.
• Short durations in single precision. Higher-order coefficients grow like $1/T^5$ for the quintic, and terms of opposite sign cancel near the end of the move, losing digits.
• Clamped time holds the endpoint state. Outside $[0, T]$ the sample is the state at $0$ or $T$, so it reports the end velocity and acceleration, not zero, when they are nonzero.
• Acceleration steps in the cubic. Acceleration is not matched, so at rest endpoints it jumps from or to zero, which excites flexible loads. The quintic's jerk is finite but generally nonzero at the ends.
• No limit enforcement. Peaks follow from $\Delta q$, $T$ and the boundary values; inconsistent boundary velocities can overshoot $q_f$ before returning.

Variants & Generalizations

Variant	Key difference
Cubic (this)	Position and velocity matched; constant jerk; acceleration discontinuous at the ends
Quintic (this)	Adds acceleration matching; continuous acceleration at about $25%$ higher peak velocity for rest-to-rest
Path-parameter time law	A 0 to 1 instance maps time to a normalized path coordinate; with $v_0 = v_f = 1$, $T = 1$ the cubic collapses to $s = t$

Applications

• Joint-space moves — robot joints and servo axes between taught points; multi-axis moves use one instance per axis with a common $T$ so all axes arrive together.
• Pick-and-place and CNC positioning — fixed-time axis moves where smoothness matters more than minimum time.
• Path timing — the scalar time law along a geometric path such as an orientation interpolation.

Connections to Other Algorithms

Algorithm	Relationship
Trapezoidal velocity profile	Respects velocity and acceleration limits with a free duration; polynomials fix the duration and let the peaks follow
S-curve profile	Bounds jerk explicitly under limits; the quintic only smooths the endpoints
Cubic spline interpolation	Handles many via points with continuous acceleration; for two knots it reduces to this cubic

References & Further Reading

• Craig, J. J., "Introduction to Robotics: Mechanics and Control," 3rd ed., Pearson, 2005, Chapter 7 (trajectory generation).
• Biagiotti, L. and Melchiorri, C., "Trajectory Planning for Automatic Machines and Robots," Springer, 2008, §2.1.