Case Study: Designing a Crew Re-Entry Capsule From a G-Limit

"A heat shield is the answer to a question you must ask in exactly the right order: first how gently, then how deep, then how hot, then how heavy."

Executive Summary

In the first case study we took a finished vehicle apart. Here we build one on paper. Given a single human requirement — bring four crew home from low Earth orbit without ever exceeding four g — we will design the entry: choose the entry angle, choose the ballistic coefficient, estimate the thermal environment, size the thermal protection, and check the corridor and blackout. Along the way we will discover why a pure-ballistic capsule cannot comfortably meet a gentle-g requirement, and why every crewed capsule since Apollo has therefore flown with lift. This is the reasoning chain that begins every real entry-vehicle design, and the one you will run on your own mission if it returns through an atmosphere.

All performance numbers are representative (Tier 2/3), chosen to make the design logic legible; the emphasis is the order of the reasoning, not three-significant-figure fidelity.

Skills applied

• Inverting the Allen–Eggers relation to turn a g-limit into an entry angle (§7.6).
• Choosing a ballistic coefficient and reading its consequences for deceleration altitude (§7.4).
• Estimating peak heat flux and total heat load from the $v^3$ scaling (§7.2, §7.3).
• Sizing a thermal-protection mass fraction and checking it against the mass budget (§7.3).
• Verifying the entry against the corridor and estimating blackout (§7.5, §7.6).

Background: the requirement

We will use $g_0 = 9.81\ \text{m/s}^2$, scale height $H = 7{,}200\ \text{m}$, and $e = 2.718$.

Phase 1: From the g-limit to an entry angle

Start with the constraint that matters to the crew: peak deceleration. Invert the Allen–Eggers result $a_{\max} = v_E^2 \sin\gamma_E/(2eH)$ for the entry angle that produces exactly $4\ \text{g}$:

$$ \sin\gamma_E = \frac{a_{\max}\,(2eH)}{v_E^2} = \frac{(4\times 9.81)\,(2\cdot 2.718\cdot 7{,}200)}{(7{,}800)^2} = \frac{(39.24)(39{,}139)}{6.084\times10^{7}} = 0.0252, $$

so $\gamma_E = \arcsin(0.0252) \approx 1.45^\circ$. A ballistic capsule must enter at only about $1.45^\circ$ below horizontal to keep the peak at four g.

That result contains a warning. An entry angle of $1.45^\circ$ is shallow — uncomfortably close to the skip-out wall of the corridor (§7.5). We have discovered, from nothing but the g-requirement, the central tension of crewed entry design: low g pushes you toward a shallow entry, and a shallow entry risks skipping out. A purely ballistic capsule is trapped between them.

Phase 2: Escaping the trap with lift, and choosing $\beta$

The escape is lift. A lifting capsule can enter at a safer, steeper angle — say $\gamma_E = 3^\circ$, comfortably captured and well inside the corridor — and then roll its lift vector up to hold the vehicle in the thin upper air, capping the deceleration at $4\ \text{g}$ even though the entry angle would give more ballistically. Lift decouples the g-load from the entry angle, which is exactly why we give the capsule an offset center of mass and a lift-to-drag ratio of $L/D \approx 0.35$, Apollo-and-Dragon class. This one choice resolves Phase 1's trap.

Now fix the ballistic coefficient. Choose a blunt capsule of base diameter $4\ \text{m}$ (frontal area $A = \pi(2)^2 = 12.57\ \text{m}^2$) with a blunt-body drag coefficient $C_d \approx 1.4$. With the $8{,}000\ \text{kg}$ entry mass,

$$ \beta = \frac{m}{C_d A} = \frac{8{,}000}{(1.4)(12.57)} = \frac{8{,}000}{17.6} \approx 455\ \text{kg/m}^2. $$

That is a deliberately low ballistic coefficient (§7.4), and it is a feature: a low-$\beta$ capsule decelerates high in the atmosphere, where the air is thin, so its peak heating happens where there is less gas to heat it — gentler on the crew and gentler on the shield. Compare a warhead at $\beta \sim 10{,}000\ \text{kg/m}^2$, which knifes to low altitude and heats ferociously; our capsule is built to do the opposite.

import math

def entry_angle_for_g(g_limit, v_entry, H=7200.0):
    """Ballistic entry angle (deg) that yields a given peak-g limit (Allen-Eggers)."""
    sin_g = g_limit * 9.81 * (2 * math.e * H) / v_entry**2
    return math.degrees(math.asin(sin_g))

def ballistic_coefficient(mass, cd, area):
    """beta = m / (Cd * A), in kg/m^2."""
    return mass / (cd * area)

print(round(entry_angle_for_g(4, 7800), 2), "deg (ballistic, for 4 g)")
print(round(ballistic_coefficient(8000, 1.4, math.pi*2**2)), "kg/m^2")
# Expected output:
# 1.45 deg (ballistic, for 4 g)
# 455 kg/m^2

Phase 3: The thermal environment

With the geometry set, estimate the heat. We do not chase an exact stagnation flux here (the absolute value depends on nose radius, the peak-heating altitude, and real-gas chemistry); instead we use the scaling of §7.3, $\dot q \propto \sqrt{\rho/R_n}\,v^3$, and anchor it to known LEO entries. A blunt capsule returning from LEO peaks near

$$ \dot q_{\text{peak}} \sim 0.5\ \text{MW/m}^2 \quad (\text{representative, Tier 2}), $$

roughly a third of a lunar return's flux — consistent with $(7.8/11)^3 \approx 0.36$. The total heat load is this flux integrated over the descent; for a LEO capsule it is on the order of tens of $\text{MJ/m}^2$. Because our low-$\beta$, lifting design stretches the descent (§7.4), it trades a modestly lower peak flux for a somewhat larger total load — the trade that sets shield thickness, not peak temperature.

🔧 Engineering Reality: Notice which number drives which decision. The peak flux ($\sim 0.5\ \text{MW/m}^2$) decides the surface material — what can survive the hottest instant without failing. The total load (tens of $\text{MJ/m}^2$) decides the shield thickness and mass — how much heat must be absorbed or blocked over the whole soak. A designer who sizes only for peak flux builds a shield that survives the worst second and then burns through on the long soak; one who sizes only for total load builds a shield that melts at the peak. You must respect both.

Phase 4: Choosing and sizing the shield

Two material strategies fit a LEO crew return:

1. Reusable tiles (Shuttle-style), if the capsule is to fly many times with minimal refurbishment — but tiles are fragile and inspection-heavy.
2. A light modern ablator such as a PICA-class material (Dragon's PICA-X), which is robust, tolerant of the peak flux, and only lightly eroded by a LEO entry, so a single shield can survive several flights.

For a crew capsule the ablator is the pragmatic choice: robust against the peak flux, forgiving of small damage (unlike a tile mosaic), and — at LEO heat loads — reusable a handful of times. Take a representative thermal-protection mass fraction of $\sim 12\%$ of entry mass for a LEO capsule (Tier 2):

$$ m_{\text{TPS}} \approx 0.12 \times 8{,}000\ \text{kg} \approx 960\ \text{kg} \approx 1\ \text{tonne}. $$

Check it against the mass budget. A one-tonne shield out of eight tonnes leaves seven tonnes for structure, crew, seats, life support, parachutes, and propellant — tight but realistic, and squarely in family with real capsules. This is theme #4 made concrete: the shield that saves the crew is also a tonne of payload you launched at the exponential cost of Chapter 3, so every kilogram of it is fought over.

Phase 5: Corridor and blackout check

Finally, verify the entry we designed survives the corridor and account for the blackout.

• Corridor. Our lifting entry at $\gamma_E \approx 3^\circ$ sits well inside a LEO corridor (LEO corridors are wider than Apollo's lunar one because the entry speed is lower and there is no skip-to-space risk below escape speed). Rolling lift up or down lets the guidance hold the trajectory and land precisely, and keeps the peak at the $4\ \text{g}$ we designed for. Had we been forced into the Phase-1 ballistic $1.45^\circ$, we would have been gambling against skip-out; lift bought us margin.
• Blackout. Near peak heating the sheath (§7.6) reaches $n_e \sim 10^{17}\ \text{m}^{-3}$, so $f_p = 8.98\sqrt{10^{17}} \approx 2.8\ \text{GHz}$. An S-band link at $2.2\ \text{GHz}$ is blocked; we therefore plan for a few minutes of blackout, and mitigate it as Dragon and the Shuttle do — relaying upward through a relay satellite (e.g., TDRS), where the leeward plasma is thinnest — so the ground never fully loses the crew.

Design closed: a $4\ \text{m}$, $8{,}000\ \text{kg}$, low-$\beta$ capsule with $L/D \approx 0.35$, a $\sim 1\ \text{tonne}$ ablative shield, entering near $3^\circ$ under lift control to hold $4\ \text{g}$, inside the corridor, with a planned-for and mitigated blackout. Every choice traced back to one human requirement.

Discussion Questions

1. Phase 1 turned a $4\ \text{g}$ limit into a $1.45^\circ$ ballistic entry angle. Show how the number would change for a $6\ \text{g}$ limit, and explain physically why a higher g-tolerance allows a steeper (and thus safer-from-skip) entry.
2. We chose a low ballistic coefficient on purpose. Trace, through $D/m = \rho v^2/(2\beta)$, exactly how a lower $\beta$ moves the peak-heating event to a higher, thinner-air altitude.
3. Why does the peak heat flux drive the choice of surface material while the total heat load drives the shield's thickness and mass? Give an example of a design that would fail if you respected only one.
4. Our shield came out near $12\%$ of entry mass. If a better ablator halved the required areal mass, where in the vehicle would you most want to spend the recovered mass, and why (tie it to earlier chapters)?

Your Turn: Extensions

• Option A (design). Redesign for a robotic capsule that can tolerate $15\ \text{g}$ and needs no lift. Find its ballistic entry angle, and comment on whether it can be a simple ballistic capsule with a fixed shield. Why is a robot cheaper to bring home than a crew?
• Option B (computation). Extend the Phase-2 code so that, given a g-limit and entry speed, it prints the required ballistic entry angle and flags whether that angle is below a "skip risk" threshold you choose (say $2^\circ$). (Do not run it — hand-trace and add # Expected output:.)
• Option C (your mission). If your MDR track returns to Earth (or enters any atmosphere), run this whole chain for it: pick an entry speed, a g-limit, and a mass; derive the entry angle, $\beta$, TPS strategy, and TPS mass; and record the result. If it never enters an atmosphere, write the one-line reason and move the saved shield mass into payload.

Key Takeaways

1. Design entry in order: how gently, how deep, how hot, how heavy. The crew's g-limit comes first and sets the entry angle.
2. A gentle-g requirement forces a shallow entry, which risks skip-out — so crewed capsules fly with lift to decouple g-load from entry angle and buy corridor margin.
3. Choose a low ballistic coefficient so the capsule decelerates high and gently, where the thin air heats it least.
4. Peak flux sets the material; total load sets the thickness and mass. Respect both, or the shield fails at one end of the descent.
5. The shield is ~10–15% of entry mass — a real payload cost (theme #4) that every design fights to minimize.