Sizing a parachute for a target descent rate

A rocket under a parachute reaches terminal velocity almost immediately after deployment — the point where drag exactly balances weight and the descent rate stops changing. That relationship gives us a direct way to size a canopy for a chosen landing speed.

v = √( (2 × m × g) / (ρ × Cd × A) )
v — descent (terminal) velocity, m/s  ·  m — recovered mass, kg  ·  g — 9.81 m/s²  ·  ρ — air density, ≈1.225 kg/m³ at sea level  ·  Cd — drag coefficient  ·  A — parachute reference area, m²

In design, you already know the descent rate you want and need to solve for canopy area instead:

A = (2 × m × g) / (ρ × Cd × v²)

Target descent rates are largely a function of what you're protecting: roughly 15–20 ft/s (4.5–6 m/s) is a common target for a durable, all-up rocket recovering as one piece, while a fragile electronics bay or a rocket recovering in two heavy sections is often sized for a gentler 10–15 ft/s (3–4.5 m/s).

Worked example
Recovered mass: m = 1.2 kg  ·  Target descent rate: v = 5.5 m/s (~18 ft/s)  ·  Cd = 0.8 (flat circular)
A = (2 × 1.2 × 9.81) / (1.225 × 0.8 × 5.5²) = 23.54 / 29.65 ≈ 0.794 m²
For flat circular: D = √(4A / π) ≈ 1.01 m (≈40 in) diameter

Choosing a drag coefficient

Cd depends on canopy shape — but published values for the "same" shape can vary by more than double between sources, because manufacturers and researchers don't always use the same reference area (the flat cut area of the fabric vs. the smaller inflated/projected area). Always check which convention a value assumes before using it.

Canopy typeTypical Cd rangeReference area convention
Flat circular (standard hobby chute)0.75 – 0.9Flat cut area
Hemispherical / dome1.4 – 1.75Projected (inflated) area
Cruciform / cross-form0.6 – 0.85Flat cut area
Toroidal / annular1.4 – 1.6Projected area
Design practice Treat calculated area as a starting point, not a final answer — Cd also shifts with canopy loading (mass per unit area), and simulation tools like OpenRocket use loading-dependent curves rather than a single fixed value. Ground or drop testing the actual descent rate is the only way to confirm a real number for your build.

Sizing an ejection or separation charge

A black powder ejection charge needs to raise the pressure inside a sealed section (payload bay, parachute compartment) enough to overcome friction and shear the nose cone or airframe joint free. The standard approach uses the ideal gas law, treating the charge as a source of a fixed volume of hot combustion gas:

W = (P × V) / (R × T)
W — black powder mass needed  ·  P — target pressure rise  ·  V — free (unfilled) volume of the sealed section  ·  R — specific gas constant for the propellant used  ·  T — combustion gas temperature

For FFFFg black powder specifically — the standard choice for ejection charges — using P in psi, V in cubic inches, R = 22.16 ft·lbf/(lbm·°R), and T = 3307 °R, the equation simplifies (after unit conversion) to a form you can use directly:

W (grams) ≈ (P × V) / 1937
Worked example
A parachute compartment is 4" in diameter and 18" long.
Volume: V = π × (2")² × 18" ≈ 226 in³
Target pressure: P = 15 psi (typical range: 10–15 psi)
W ≈ (15 × 226) / 1937 ≈ 1.7 grams of FFFFg black powder

Target pressure depends on how the section is held together: a friction-fit nose cone typically needs roughly 10–13 psi, while a system using shear pins is sized to the pins' rated shear force instead — divide the total force the charge needs to produce (pressure × cross-sectional area) by each pin's rated shear strength to find how many pins your charge and section size can support.

Shear pin sizing

Shear pins hold the nose cone or payload section on during flight and are designed to snap cleanly when the ejection charge fires, rather than relying on friction alone. Sizing them is a straightforward force balance:

F = P × A(cross-section)
F — total separation force  ·  P — charge pressure, psi  ·  A — cross-sectional area of the airframe, in²
Worked example
4" diameter airframe, charge sized for 12 psi
Cross-sectional area: A = π × 2² ≈ 12.57 in²
Total force: F = 12 × 12.57 ≈ 151 lbf
Using #2-56 nylon shear pins (~35 lbf each): 151 / 35 ≈ 4.3 → use 4 pins

Round down, not up — using more pins than the charge can reliably shear is a common cause of a "no separation" failure that looks identical to an undersized charge, but is actually an oversized pin count.

Shock cord: length and material

A common rule of thumb is a shock cord at least 4–6 times the airframe's length, giving enough separation between the nose cone and body tube that they don't tangle or collide on the way down.

Field note, preventing "zippering": a nylon shock cord attached directly to the airframe wall can cut a long slit down the tube if it goes taut hard during deployment. A short Kevlar leader (12–18") between the motor mount and the main cord, plus a sliding standoff near the coupler, spreads that shock load and is standard practice above L1 airframes.

Every source on this topic agrees on one point Calculated charge sizes assume complete, ideal combustion and no heat loss to the surrounding structure — real charges usually need somewhat more than the calculation suggests. Ground test every new ejection charge configuration before flying it, size charges for drogue and main separately if flying dual-deploy, and never substitute smokeless powder, which produces very different and unpredictable pressures in this application.