Exercises — Proportional navigation guidance — N·V_c·λ̇, derivation

The PN law multiplies three things together — gain, closing speed, sightline turn-rate: $a_c=N{V}_c\dot{\lambda }=4\times 800\times 0.015=48{\text{ m/s}}^2.$ Divide by $g$ to read it in "g's": $\frac{48}{9.81}\approx 4.89g.$ What it looks like: a modest, comfortably-flyable command — the sightline is barely drifting.

By definition $V_c=-\dot{R}=-(-1200)=1200\text{ m/s}$. Since $V_c>0$, the range is shrinking — the missile is closing. Why the minus sign: $\dot{R}$ is negative when distance decreases; flipping the sign makes "closing" a positive, intuitive number.

Invert the law $a_c=N{V}_c\dot{\lambda }$ for $\dot{\lambda }$: $\dot{\lambda }=\frac{a_c}{N{V}_c}=\frac{90}{3\times 1000}=0.03\text{ rad/s}.$ What it looks like: the sightline is drifting at $0.03$ rad/s — the missile is turning to erase that drift.

Require $N{V}_c\dot{\lambda }\le a_{c,\max }$: $N\le \frac{a_{c,\max }}{V_c\dot{\lambda }}=\frac{100}{1000\times 0.04}=\frac{100}{40}=\mathrm{2.5.}$ The largest integer $\le 2.5$ is $N=2$ — but recall from the derivation that convergence needs $N>2$, so $N=2$ sits on the stability boundary and $\dot{\lambda }$ never decays. This is a genuine design conflict: the actuator limit and the stability requirement disagree. See Navigation Constant Selection and Actuator Limits. What it looks like: you can't satisfy both here — the honest answer is "no integer $N$ both flies within the limit and converges; you must accept a briefly saturated actuator with $N=3$, or reduce initial $\dot{\lambda }$ with a better launch heading."

From the derivation $\dot{\lambda }\propto R^{N-2}$, so $\frac{\dot{\lambda }}{{\dot{\lambda }}_0}={\left(\frac{R}{R_0}\right)}^{N-2}.$ With $R/R_0=1/20$:

• A ($N=3$): $(1/20{)}^1=\mathrm{0.05.}$
• B ($N=4$): $(1/20{)}^2=\mathrm{0.0025.}$ Missile B's residual LOS rate is $20\times$ smaller — its terminal geometry is far closer to a perfect collision course, so it ends cleaner (smaller miss). The figure below shows both curves collapsing as $R\to 0$.

$a_c=N{V}_c\dot{\lambda }$ inherits the sign of $\dot{\lambda }$ (with $V_c,N>0$). Take $N=4$:

• $t_1$: $a_c=4\times 900\times (+0.02)=+72{\text{ m/s}}^2$ — accelerate toward increasing $\lambda$ (chase the CCW-drifting sightline).
• $t_2$: $a_c=4\times 900\times (-0.02)=-72{\text{ m/s}}^2$ — accelerate the opposite way, toward decreasing $\lambda$. What it looks like: the command flips sign exactly when the drift flips sign — PN is a negative-feedback loop on $\dot{\lambda }$, always pushing to null it. See Feedback Control — Nulling an Error Signal.

Substitute ($V_c=-\dot{R}$, so $a_c=N{V}_c\dot{\lambda }=-N\dot{R}\dot{\lambda }$): $R\ddot{\lambda }+2\dot{R}\dot{\lambda }=-(-N\dot{R}\dot{\lambda })=N\dot{R}\dot{\lambda }.$ Collect the $\dot{R}\dot{\lambda }$ terms: $R\ddot{\lambda }=(N-2)\dot{R}\dot{\lambda }⟹\frac{\ddot{\lambda }}{\dot{\lambda }}=(N-2)\frac{\dot{R}}{R}.$ Integrate both sides in time (left gives $\ln \dot{\lambda }$, right gives $(N-2)\ln R$): $\ln \dot{\lambda }=(N-2)\ln R+C⟹\dot{\lambda }\propto R^{N-2}.$ Condition: as $R\to 0$ we need the exponent positive, i.e. $\boxed{N>2}$. Check $N=3$: exponent $=1>0$, so $\dot{\lambda }\propto R\to 0$. Converges. ✓ The factor of $2$ (the Coriolis term, see Coriolis Term in Polar Coordinate Acceleration) is what makes the boundary $N>2$ rather than $N>1$.

Set $\ddot{\lambda }=0$ so $R\ddot{\lambda }$ drops out: $2\dot{R}\dot{\lambda }=a_T-a_c=a_T+N\dot{R}\dot{\lambda }.$ Collect $\dot{\lambda }$: $2\dot{R}\dot{\lambda }-N\dot{R}\dot{\lambda }=a_T⟹(2-N)\dot{R}\dot{\lambda }=a_T.$ ${\dot{\lambda }}_{\text{ss}}=\frac{a_T}{(2-N)\dot{R}}.$ Interpretation: a maneuvering target leaves a non-zero residual $\dot{\lambda }$ that plain PN can't null — a steady bias. This is exactly the gap that Augmented PN closes by adding a $\frac{N}{2}{a}_T$ feed-forward term. (Note $\dot{R}<0$ while closing, so with $N>2$ the sign of ${\dot{\lambda }}_{\text{ss}}$ tracks $a_T$ sensibly.)

(a) $a_{c,0}=N{V}_c{\dot{\lambda }}_0=4\times 1000\times 0.05=200{\text{ m/s}}^2=200/9.81\approx 20.4g.$ (A steep launch command — the sightline drifts fast initially.) (b) $R/R_0=2000/8000=1/4$, exponent $N-2=2$: $\dot{\lambda }=0.05\times (1/4{)}^2=0.05\times 0.0625=0.003125\text{ rad/s}.$ (c) $a_c=4\times 1000\times 0.003125=12.5{\text{ m/s}}^2=12.5/9.81\approx 1.27g.$ (d) As range collapses, $\dot{\lambda }$ falls fast, so the command drops from $\sim 20g$ to $\sim 1.3g$ — PN front-loads the effort and coasts into a near-perfect terminal geometry. The figure traces $a_c$ versus $R$.

Lateral demand $\propto \dot{\lambda }$ (with $V_c,N$ equal). Ratio: $\frac{a_{\text{PN}}}{a_{\text{pursuit}}}=\frac{{\dot{\lambda }}_0(1/10{)}^2}{{\dot{\lambda }}_0}=(1/10{)}^2=\mathrm{0.01.}$ PN demands $\frac{1}{100}$ of the terminal lateral acceleration — 100× gentler. This is the core reason PN beats pure pursuit: pursuit saves its whole turn for the end, where the airframe has least margin; PN spends it early and arrives relaxed.