Trajectory optimization — minimum gravity loss, minimum drag loss: Worked examples

This page is the "run every case" companion to the parent topic. We take the two loss integrals it derived and drive them through every kind of input a problem can hand you: each flight-path angle regime, the zero and degenerate cases, the limits, a word problem, and an exam twist. Nothing is left as "obvious".

Before we start, three quantities the parent built, restated in plain words so this page stands alone:

The scenario matrix

Every example below is tagged with the cell of this matrix it exercises. Together they cover all of them.

Cell	What it tests	Why it can bite you
A — $γ = 9 0^{\circ}$ (vertical)	$sin γ = 1$ , worst-case gravity loss	Naive "just add $g t$ "
B — $0^{\circ} < γ < 9 0^{\circ}$ (slanted)	partial $sin γ$ , quadrant-I geometry	forgetting to take the sine
C — $γ = 0^{\circ}$ (horizontal)	$sin γ = 0$ , degenerate: gravity loss vanishes	thinking "loss = 0 so this is best"
D — $γ$ varying with $t$	integral is NOT $g \overset{γ}{ˉ} t$ ; must integrate $sin γ$	pulling sine out of the integral
E — drag at a fixed condition	$D / m \cdot Δ t$ evaluation, units	mixing N, kg, s
F — gravity-turn pitch rate	$d γ / d t = - g cos γ / v$ , sign & limits	$cos$ vs $sin$ confusion
G — limiting case $v \to \infty$ or $v \to 0$	turn rate $\to 0$ or $\to \infty$	why the pitch kick is early
H — real-world word problem	full loss budget bookkeeping	translating words to integrals
I — exam twist (mixed sign / trade-off)	comparing two profiles, choosing min sum	optimizing one term only

Example 1 — Cell A: the pure vertical burn (worst case)

Worked example A model rocket burns straight up

Burn $t_{b} = 45 s$ , straight up so $γ = 9 0^{\circ}$ the whole time, take $g = 9.8 m/s^{2}$ constant. Find the gravity loss.

Forecast: guess before reading — will the loss be more or less than $9.8 \times 45$ ?

Write the integral. $Δ v_{grav} = \int_{0}^{45} g sin γ d t$ . Why this step? Loss is defined as an integral; we start from the definition, never a memorized shortcut.
Insert $γ = 9 0^{\circ}$ . $sin 9 0^{\circ} = 1$ , and it never changes, so the integrand is the constant $g$ . Why this step? Vertical flight is the one case where the whole of gravity opposes motion — the opposite-to-velocity component equals the full $g$ .
Integrate the constant. $\int_{0}^{45} 9.8 d t = 9.8 \times 45 = 441 m/s$ . Why this step? A constant times the interval length; nothing subtle.

Verify: units $= (m/s^{2}) (s) = m/s$ ✔. And your forecast: it is exactly $9.8 \times 45$ , not more — because at $9 0^{\circ}$ the sine is at its maximum $1$ , this is the ceiling. Any other angle gives less. See Tsiolkovsky Rocket Equation for where the "ideal" side of the budget comes from.

Example 2 — Cell B: a slanted (quadrant-I) burn

Worked example Same burn, held at

$γ = 4 0^{\circ}$ $t_{b} = 45 s$ , $g = 9.8$ , but now the velocity arrow sits at $4 0^{\circ}$ above horizon the whole time.

Forecast: the answer should be Example 1's $441$ multiplied by what factor?

Same integral, new angle. $Δ v_{grav} = \int_{0}^{45} 9.8 sin 4 0^{\circ} d t$ . Why this step? Only the component of $g$ along the flight direction robs speed; that component is $g sin γ$ . Look at the figure: the vertical gravity arrow splits into a piece along the path ( $g sin γ$ ) and a piece across it ( $g cos γ$ ).

Pull out the constant sine. $sin 4 0^{\circ} \approx 0.6428$ , so $= 9.8 \times 0.6428 \times 45$ . Why this step? $γ$ is fixed here, so $sin γ$ is a number and leaves the integral.
Multiply. $9.8 \times 45 \times 0.6428 \approx 283.5 m/s$ .

Verify: the multiplier is $sin 4 0^{\circ} = 0.6428$ , so $441 \times 0.6428 = 283.5$ ✔ — matches, and it is less than the vertical case, exactly as the geometry promised.

Example 3 — Cell C: the horizontal (degenerate) burn

Example 4 — Cell D: $γ$ changing during the burn (must integrate)

Worked example Pitch program that tips over linearly

Over a $t_{b} = 100 s$ burn the flight-path angle falls linearly from $9 0^{\circ}$ to $0^{\circ}$ : $γ (t) = 9 0^{\circ} (1 - \frac{t}{100}) = \frac{π}{2} (1 - \frac{t}{100}) rad .$ Take $g = 9.8$ constant. Find $Δ v_{grav}$ .

Forecast: the average angle is $4 5^{\circ}$ . A tempting shortcut is $9.8 \times 100 \times sin 4 5^{\circ} \approx 693$ . Will the true integral be higher or lower than that?

Set up the honest integral. $Δ v_{grav} = \int_{0}^{100} 9.8 sin [\frac{π}{2} (1 - \frac{t}{100})] d t$ . Why this step? $γ$ is not constant, so $sin γ$ cannot leave the integral. Sine of an average $\neq =$ average of the sine.
Substitute. Let $u = \frac{π}{2} (1 - \frac{t}{100})$ . Then $d u = - \frac{π}{200} d t$ , so $d t = - \frac{200}{π} d u$ . Limits: $t = 0 \to u = \frac{π}{2}$ , $t = 100 \to u = 0$ . Why this step? Substitution turns a messy composite into a clean $\int sin u d u$ — the standard tool when the inside of the sine is linear in $t$ .
Integrate. $9.8 \int_{π /2}^{0} sin u (- \frac{200}{π}) d u = 9.8 \cdot \frac{200}{π} \int_{0}^{π /2} sin u d u = 9.8 \cdot \frac{200}{π} [- cos u]_{0}^{π /2}$ .
Evaluate. $[- cos u]_{0}^{π /2} = (0) - (- 1) = 1$ , so $Δ v_{grav} = 9.8 \cdot \frac{200}{π} \approx 623.8 m/s$ .

Verify: the naive " $sin$ of average $= 4 5^{\circ}$ " gave $693$ ; the true value $623.8$ is lower ✔ — because sine is concave on $[0, 9 0^{\circ}]$ , so averaging the sine gives less than the sine of the average. This is exactly why you must integrate. Units: $(m/s^{2}) (s) = m/s$ ✔.

Example 5 — Cell E: drag loss at a fixed condition

Worked example Drag near max-Q

At the worst dynamic-pressure moment: $ρ = 0.5 kg/m^{3}$ , $v = 300 m/s$ , $C_{D} = 0.3$ , $A = 2 m^{2}$ , vehicle mass $m = 1.5 \times 1 0^{4} kg$ , and this condition lasts about $Δ t = 15 s$ . Estimate $Δ v_{drag}$ .

Forecast: will it be tens, hundreds, or thousands of m/s? (Gravity losses were hundreds — guess before computing.)

Drag force. $D = \frac{1}{2} ρ v^{2} C_{D} A = \frac{1}{2} (0.5) (300)^{2} (0.3) (2)$ . Why this step? This is the aerodynamic drag formula (see Aerodynamic Drag and Max-Q); $\frac{1}{2} ρ v^{2}$ is the dynamic pressure, $C_{D} A$ the effective area.
Compute. $(300)^{2} = 90000$ ; $\frac{1}{2} (0.5) = 0.25$ ; $0.25 \times 90000 = 22500$ ; $\times 0.3 = 6750$ ; $\times 2 = 13500 N$ .
Drag deceleration. $D / m = 13500/15000 = 0.9 m/s^{2}$ . Why this step? $Δ v_{drag}$ integrates $D / m$ ; deceleration is force over mass.
Times the duration. $Δ v_{drag} \approx 0.9 \times 15 = 13.5 m/s$ .

Verify: units: $N / kg = m/s^{2}$ , times $s$ $= m/s$ ✔. Answer is tens of m/s — far below the gravity losses of Examples 1–4, confirming the parent's point that gravity loss dominates the trade.

Example 6 — Cell F: gravity-turn pitch rate (the sign matters)

Worked example How fast is the velocity arrow turning?

During a gravity turn at $v = 250 m/s$ , $γ = 7 0^{\circ}$ , $g = 9.8$ . Find $d γ / d t$ .

Forecast: positive or negative? Fast or slow (in degrees per second)?

Use the turn law. $\frac{d γ}{d t} = - \frac{g cos γ}{v}$ . Why this step? In a gravity turn the only sideways force is gravity's across-path component $g cos γ$ — no thrust is spent turning, which is why steering loss $\approx 0$ . Note it is $cos$ , the component perpendicular to velocity (contrast the loss, which used $sin$ , the parallel one). The figure in Example 2 shows both pieces.
Insert numbers. $cos 7 0^{\circ} \approx 0.3420$ , so $= - \frac{9.8 \times 0.3420}{250} = - \frac{3.3516}{250}$ .
Compute. $\approx - 0.01341 rad/s$ .
Convert to degrees. $\times \frac{180}{π} \approx - 0.76 8^{\circ} / s$ .

Verify: sign is negative ✔ — $γ$ shrinks, the rocket tips toward horizontal, as it should. Magnitude under $1^{\circ} / s$ ✔, slow, matching the parent's warning that turning is gentle once you're moving.

Example 7 — Cell G: the limits (why the pitch kick is early)

Worked example Same formula at two extreme speeds

Keep $γ = 7 0^{\circ}$ , $g = 9.8$ . Compare the turn rate at a tiny $v = 10 m/s$ (just after liftoff) versus a large $v = 2500 m/s$ (late ascent).

Forecast: which speed lets you turn faster?

Slow case. $d γ / d t = - \frac{9.8 \times 0.3420}{10} = - 0.33516 rad/s \approx - 19. 2^{\circ} / s$ . Why this step? $1/ v$ is large when $v$ is small, so the turn is rapid — a small nudge early bends the path a lot.
Fast case. $d γ / d t = - \frac{9.8 \times 0.3420}{2500} = - 0.0013406 rad/s \approx - 0.076 8^{\circ} / s$ . Why this step? Now $1/ v$ is tiny; the same gravity barely bends the (fast, heavy-with-momentum) velocity arrow.
The limit. As $v \to \infty$ , $d γ / d t \to 0$ : you essentially cannot turn anymore. As $v \to 0^{+}$ , $d γ / d t \to - \infty$ : infinitely sensitive.

Verify: ratio of the two rates $= 2500/10 = 250$ ✔ (turn rate $\propto 1/ v$ ). The slow-case rate is $250 \times$ the fast-case rate. Lesson: you must plant the initial pitch kick while $v$ is small; wait too long and the vehicle is too fast to turn before fuel runs out — the steering-loss alternative would then cost you.

Example 8 — Cell H: real-world loss budget

Worked example Word problem — closing the budget

A launcher has ideal $Δ v_{ideal} = 9.4 km/s$ (from Tsiolkovsky Rocket Equation). Flight analysis gives gravity loss $1.6 km/s$ , drag loss $0.1 km/s$ , steering loss $0.05 km/s$ . What orbital speed is actually delivered, and is it enough for the required orbital velocity of $7.8 km/s$ ?

Forecast: will there be margin or a shortfall?

Bookkeeping equation. $Δ v_{orbit} = Δ v_{ideal} - Δ v_{grav} - Δ v_{drag} - Δ v_{steer}$ . Why this step? This is the parent's conservation of the $Δ v$ ledger — every loss is subtracted from the ideal.
Plug in. $= 9.4 - 1.6 - 0.1 - 0.05 = 7.65 km/s$ .
Compare to requirement. Required is $7.8 km/s$ (see Orbital insertion and required orbital velocity).

Verify: $9.4 - 1.6 - 0.1 - 0.05 = 7.65$ ✔. Delivered $7.65 < 7.8$ required $\Rightarrow$ shortfall of $0.15 km/s$ — the vehicle does not reach orbit. Fix: reduce gravity loss (higher thrust-to-weight, shorter vertical phase) or add propellant.

Example 9 — Cell I: exam twist (choose the minimum-sum profile)

Worked example Which of two pitch profiles is better?

Same $t_{b} = 80 s$ , $g = 9.8$ , two candidate constant- $γ$ phases (idealized):

Profile P (steep): $γ = 7 5^{\circ}$ . Drag near this steep, fast-exit path estimated at $Δ v_{drag} = 40 m/s$ .
Profile Q (shallow): $γ = 3 5^{\circ}$ . Staying low & fast, drag estimated at $Δ v_{drag} = 260 m/s$ .

Which has the smaller total loss $Δ v_{grav} + Δ v_{drag}$ ?

Forecast: steep loses on gravity, shallow loses on drag — guess the winner.

Gravity loss, P. $9.8 \times 80 \times sin 7 5^{\circ} = 784 \times 0.9659 \approx 757.3 m/s$ . Total P $= 757.3 + 40 = 797.3$ . Why this step? Steep $\Rightarrow$ big $sin γ \Rightarrow$ big gravity loss, but the fast climb keeps drag small.
Gravity loss, Q. $9.8 \times 80 \times sin 3 5^{\circ} = 784 \times 0.5736 \approx 449.7 m/s$ . Total Q $= 449.7 + 260 = 709.7$ . Why this step? Shallow $\Rightarrow$ smaller gravity loss, but drag balloons.
Compare totals. $797.3$ (P) vs $709.7$ (Q).

Verify: $797.3 > 709.7$ , so Profile Q wins by about $87.6 m/s$ ✔. Notice neither term alone decides it — you must add both, exactly the see-saw logic. (The real optimum is neither fixed angle but a gravity turn sweeping through them, which beats both.)

Recall Quick self-test

At $γ = 9 0^{\circ}$ , gravity loss over burn $t_{b}$ with constant $g$ is ::: $g t_{b}$ (since $sin 9 0^{\circ} = 1$ ). Why must you integrate $sin γ$ when $γ$ varies rather than use $sin \overset{γ}{ˉ}$ ? ::: Because sine is concave on $[0, 9 0^{\circ}]$ ; the average of the sine is not the sine of the average. In a gravity turn, is the turn rate governed by $sin γ$ or $cos γ$ ? ::: $cos γ$ — the component of gravity perpendicular to velocity. Why apply the pitch kick early? ::: Turn rate $\propto 1/ v$ ; only while $v$ is small can gravity bend the path appreciably.