Drogue Sizing

Assumptions have been made for some factors (air temperature, pressure, etc.) to create a simple equation which should work for most HPR situations (flights up to about 1 mile, air temperature about 70 degrees F). When actual conditions are significantly different, such as 100 degree temperature or an altitude of 5 miles, then one may need to perform more complex calculations.

In some cases a drogue is not needed. When an airframe will fall horizontally then drag may slow the rocket enough that a drogue is not needed. This can often be the case when an airframe is split into two approximately equal parts. For example the author has a 5 pound rocket that falls at about 45 feet/second in a flat spin when the fin canister and payload sections are separated at apogee. In this case a drogue is not needed. A small one (12” or 18”) may be used to produce a more stable descent, so that the fin canister and payload sections don’t bang into each other.

Some rocket designs don’t allow the airframe to be used to slow a rocket’s descent. In these cases a drogue is needed. For example, rockets whose airframe cannot be split, or those that deploy the drogue and main parachutes from the same location.

A drogue’s purpose is to slow a rocket’s descent, but still allow it to descend rapidly enough to avoid significant drift. Descent needs to be quick, but not too quick that the rocket and/or main parachute will be torn apart when the main parachute is deployed. The author aims for a descent rate of about 50 feet/second.

It’s useful to know how fast a rocket will descend without a parachute under different conditions. For example, a 4” diameter, 5 pound rocket will come in nose first at about 253 feet/second (172 miles/hour). The equation for calculating terminal velocity is shown below.

$$V_t=\sqrt{\frac{2W}{C_d\rho A}}$$

Where:
$V_t$ is the terminal velocity in feet/second
$W$ is the weight in pounds
$Cd$ is the drag coefficient; assume 0.75
$rho$ is the atmospheric density (about 0.0024 at sea level)
$A$ is the surface area in feet²

If a 4” diameter, 5 pound rocket comes in nose first then:

$$A=\frac{3.1415\times 2^2}{144}=0.087f{t}^2$$

$$V_t=\sqrt{\frac{2\times 5}{0.75\times 0.0024\times 0.087}}=253feet/sec$$

If the same rocket has a 60” long airframe and falls horizontally, then:

$$A=\pi rl=3.1415\times 2\times 60=377i{n}^2=2.6f{t}^2$$

$$V_t=\sqrt{\frac{2\times 5}{0.75\times 0.0024\times 2.6}}=46feet/second$$

If a rocket will fall at a reasonable rate without a drogue, then one is not needed unless you want a small (12”-18”) one to add stability without significantly affecting the descent rate. If a drogue is needed then the terminal velocity equation can be rearranged to calculate the desired size of a circular parachute. The simplified equation is shown below, and assumes a desired descent rate of 50 feet/second.

$$D=24\times \sqrt{0.14W}$$

Where:
$D$ is the drogue diameter in inches
$W$ is the weight in pounds without propellant

Example of a 5 pound rocket:

$$D=24\times \sqrt{0.14\times 5}=20$$

So a 20” or 24” parachute would be a good choice.

Submitted by: Dean A. Roth