How do you determine the right size for your parachute? We can apply an equation from the Rocket Equations Page to figure out the parachute needed for your rocket and how fast your rocket will be going during descent.
The rocket, under its parachute, will speed up toward the ground until the drag force on the chute is equal to the weight of the rocket. So to figure this problem out, we need to find equations for drag force and for the rocket's weight and set them to be equal.
We will try to find the diameter of parachute that brings the rocket down slowly enough that the rocket doesn't break when it hits the ground. How fast is that? It depends on how well you built your rocket, of course, along with where it lands - concrete will hurt more than field grass. If you simply drop your rocket from two feet high, it will hit the ground at a little over 7 mph, or approximately 3 m/s. We'd like to keep the speed to about that.
We will use the drag force, or "wind resistance force" equation, which is
FD = ½ r Cd A v2
FD is the drag force
r (Greek letter "rho") is the density of air = 1.22 kg/m3
Cd is the drag coefficient
A is the area of the chute
v is the velocity through the air
Meanwhile, the weight of the rocket, otherwise known as the force of gravity (FG), is computed to be
FG = m g
m is the mass of the rocket
g is the acceleration of gravity = 9.81 m/s2
Let's find when they're equal:
FG = FD
m g = ½ r Cd A v2
...and solving for chute area...
A = (2 m g) / (r Cd v2)
Now the chute area, in terms of the chute diameter, is A = p D2 / 4, so the chute diameter is
D = sqrt(4 A / p).
Combining the two equations above for A & D leads us to the final form of the chute equation as we will use it:
D = sqrt( (8 m g) / (p r Cd v2) )
Let's size a parachute for an Estes Big Bertha.
D = sqrt( (8 m g) / (p r Cd v2) ) = sqrt( (8*0.0623*9.81) / (3.14*1.22*0.75*32) ) = 0.435 m
...which is equal to 17 inches. This explains why the Big Bertha comes with an 18 inch chute.
We'll size a parachute for my LOC V2, which weighs in at 8 pounds even, or 3.6 kg.
D = sqrt( (8 m g) / (p r Cd v2) ) = sqrt( (8*3.6*9.81) / (3.14*1.22*1.5*52) ) = 1.4 m
...which is a 4 foot 8 inch parachute. Pretty big. In reality I'm using a RocketMan R7C, which is about that size.
Note that we can easily find the descent velocity, given the chute diameter, by simply rearranging the above equation to get
v = sqrt( (8 m g) / (p r Cd D2) )
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Updated 8 August, 2000 with corrected velocity equation (thanks, Jack Anderson).