We based our answer on the following design philosophy: Since rocket payload needs to be kept small, the parachute should be designed as small as possible. Also, using a small chute implies a more vertical and faster descent. In other words, using a large parachute usually leads to more horizontal drift due to winds aloft... i.e. more chances to lose the rocket in neighbor McNasty's backyard.
So how small should the parachute be? It depends on how much shock the rocket can take on landing since the faster the descent, the more risk for damage.
Parachute design: how big the chute?
It depends on two factors:
To estimate the max descent speed your rocket can take under parachute, you could do this: drop the rocket (or an old prototype) from height "d" -without the parachute. The velocity reached when it hits the ground is given by
v = squareroot( 2 g d) g = 32.17 ft/sec^2 (English units) or 9.81 m/sec^2 (Metric) d= height in feet (English) or meters (Metric) v = velocity in ft/sec (English) or m/sec (Metric)The maximum drop height from which the rocket can be dropped without breaking determines the max parachute descent velocity. Once you have determined that max velocity, use the following formula to find out the chute's optimal surface area:
2 W S = ----------- rho CD (v^2) where CD = parachute drag coefficient which is approx 0.75 for a chute without holes or slits cut in the fabric; same value in both Metric and English unit systems rho at sea level = 0.00237 sl/ft^3 (English units) and 1.225 Kg/m^3 (Metric) rho near 4000 ft or 1219 m above sea level = 0.00211 sl/ft^3 (English units) and approx. 1.07 Kg/m^3 (Metric) W = weight of the parachute + load, in pounds (English) or Newtons (Metric) v^2 = square of the vertical descent velocity, where v is expressed in ft/sec (English) or m/sec (Metric) S is the parachute's surface area when measured on a flat surface, in ft^2 (English) or m^2 (Metric).
What is the diameter of the chute when it lays flat on the floor (assuming that it is a circular piece of fabric)?
diameter = 2 squareroot of (S / 3.1416)To improve on the stability of the chute during descent it would be a good idea to cut a hole (covering about 10 percent of surface area) at the apex of the canopy. That way the rocket won't oscillate too much and the descent will follow a straight line.
There are other designs to improve stability besides cutting a vent at the apex, see article by Dr. C. W. Peterson in Physics Today, August 1993 The magazine Physics Today can be found at university and college libraries.