The equation for parachute drag is

D = ½ ρ v2 S Cd

Drag force [ lb ] = (1/2) * (density [ slugs / ft3 ] ) * (vel [ ft / sec ] ) 2 * (area [ ft 2 ] ) * (drag coeff [ dimensionless ] )

Where:

Density is a worst-case value of .002377 slugs / ft3 at seal level on an ARDC standard day. Sea level is where the highest value of density will be found, so let’s assume your parachute opens a few inches above sea level. (If you were 2,000 feet higher, the density would only drop by 5%, so let’s keep things simple and just assume density is constant with altitude, as far as low-power rockets go.)

Vel is deployment/opening velocity, which is typically in the range of 0 to 20-30-40 ft / sec for model rockets. Let’s assume a “severe” case where vel = 100 ft / sec (yes, it’s a silly exaggeration).

Area is the cross-section, or profile, or presented area (however you want to describe it) of the size of the parachute as it sweeps through the air. Let’s assume you have a 24 inch parachute. (Note that I did not say 24 inch para*sheet*, which would be a flat pattern, and which would have a smaller diameter when opened and pulled into shape.) If our parachute is 2 ft across then it has 3.14159 ft2 of area….approximately.

Drag coefficient would be anybody’s guess. Just for comparison, some aero data sources state that the Cd of a flat plate (oriented broadside to the oncoming airflow) is around 1.3, and that the Cd of a hemispherical cup (oriented with open side of cup toward the oncoming airflow) is around 1.4. Now, you aren’t going to actually be able to get significantly better than that without some exotic shaping and venting technologies, but just for the sake of estimating the upper limits of a ridiculously over-done model rocket parachute, let’s assume a value of 3 for our Cd.

Put ‘em all together and you get:

D = ½ (.002377) (100)2 (3.14159) (3) = 112 lbs (if I did my math right…and if I didn’t, I don’t want to hear about it!)

If your complete parachute is generating 112 lbs of force, and you have six shroud lines, each shroud line has 19 lbs of tension load pulling straight ahead. Instead, since each shroud line is really pulling inward (to the attachment point where the shroud lines are tied to the swivel clip), say at an angle of around 42 degr (because we are all using shroud lines equal to 1.5 times chute diam, right?), the actual tension load in the shroud line would be around 26 lbs. Toss in a margin of safety of 50% and the design condition would be 39 lbs of tension load in each shroud line.

Now, I used the absolute worst-and-silliest input values that I thought would put an upper bound on the design question. I would expect actual loads to be waaaaay smaller (90-95% smaller). Either way, it looks to me like your 200 lb test material would handle this.

(and why is it that on a "rocket science" website, we can't show exponents properly?)