Making a parachute for Big Daddy

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plummet

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Hi

I'm planning to build an Estes Big Daddy and I'd like to make my own recovery system. I've got some light-weight ripstop (enough to make a parachute something like 22" in diameter) and plan to get a 9" sq nomex, but I'm not sure what grade/thickness of Kevlar I should use.

From reading a few articles, I'm thinking that 1/10" kevlar (300lb breaking strain) sounds about right for both the shock cord and parachute harness, but I wanted to check whether I should be using something stronger for the shock cord (and maybe something lighter for the harness)?

Thanks

Tony
 

powderburner

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In general-

You can use any shroud line material you like that is flexible enough and strong enough to perform the necessary functions without being so bulky or otherwise troublesome to get in the way of recovery system operation.

With a Big Daddy-type design (huge BT, at least compared to most lowpower designs) you should not have any trouble at all packing in these shroud lines. And with a pull-strength like that, it should be plenty strong for shroud line material.

The bizarre thing that I see happening, however, is that if you do even a 'basic' parachute/parasheet with six shroud lines, and if the shroud line attachments are secure enough that they will not tear loose, that means that your recovery device can generate upwards of 1800 lbs of force before things start coming apart. Even if you leave yourself a 50% safety margin, that's still 1200 lbs of force. And if you design a 'nice' parachute with like 10 or 12 shroud lines (parasheets make a much nicer pseudo-chute shape with a few more lines) now you are talking upwards of 2000 lbs (10 lines, 3000 lbs ultimate, 2000 lbs design load with 50% margin). That's a lot of decel force for a rocket that only weighs a few pounds. Something tells me your ripstop nylon might give up at some point before that.

Now, there is absolutely nothing wrong with using the shroud material you describe, especially if it's what you have in your hands and don't want to wait a week or two for delivery of some other material. If it works, it works, and there is nothing wrong with a little extra beef.

The more serious challenge I see is that you will need a solid anchor point for the recovery system to get loads to the rocket airframe, and distributed just a little so it's not a 'point' load that yanks out. You could use another piece of kevlar to tie/knot/wrap/whatever around the MMT core and then route it past the outer edge of the fwd CR (to keep it furthest away from the ejection charge temperatures at the center) and out the front of the rocket. If you want to do a really solid attachment of this tether, you should probably look at some mid power and high power anchor techniques.

If you use the standard Estes folded-tab, it may or may not hold (if you ever get a 2000 lb parachute deploy load) and you will forever have a big ol' wad of obstacle inside the BT that tries to snag the recovery system on the way out. Just don't do it (weak Nike parody).



PS-- I don't know if we said it yet, but welcome to TRF!
 

plummet

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Thanks - sounds like I need to give some good consideration to the mounting.

Regarding shock-cord strength, I suppose the unknown here is the typical and likely maximum forces you'd get on a Big D at opening with a nylon parachute. That would tell me if 200lb is strong enough.
 

plummet

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That's an excellent site, exactly what I need. Thanks for the link.
 

powderburner

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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 parasheet, 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?)
 
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Handeman

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For shroud lines, I use 150 lb. solid core Dacron line. Here's a link where you can get 3,000 ft for $15. It's the 1.4mm diameter and is the same stuff used on window blinds. If you want the $15 price, you have to take whatever color is on sale. Here's a link to the Kite Studio catalog and a sale as of 6/6/09.

The largest chute I've used this on is a 53" golf umbrella that I use as my main chute for dual deploy on my L1 (3 lbs) and L2 (7 lbs) rockets. The chute snaps open pretty quickly on deployment, but the lines has held up great!
 

privateer

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Uh-oh...I made a trash bag replacement chute for the one that departed my Mean Machine on its second flight. I cut a 24" hex and used orange surveyor string for shroud lines. Based on the stock Estes 12" chute, it appeared the shroud line length from edge of chute to the swivel was 1:1 to the chute diameter. So I cut three lines of string at 50" for shrouds (giving about 24" of shroud line from the chute to the swivel, with a little extra for attaching to chute). And, I secured them to the 1 mil thick plastic with a complex layup of scotch tape.

Flights three and four...perfect chute deployment and recovery. It was a down and dirty fix, though; I should read up on making chutes.

It was kind of pleasing to use my own chute.
 

Handeman

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.... Based on the stock Estes 12" chute, it appeared the shroud line length from edge of chute to the swivel was 1:1 to the chute diameter. ....
You are right about the 1 to 1 on the OEM shroud lines. I have not seen a commercial chute with longer then 1 to 1 shroud line lengths, at least nothing below about 6 foot diameter.

My experience has been that 1.5 to 1 is a much better ratio. Especially on parasheets. The longer shrouds allow the parasheet to open better and they seem to have a lot less pendulum action on the way down. A small spill hole also help reduce the pendulum action.
 

hardinlw

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The commonly used value of CD for a flat sheet parachute is 0.75 based on the flat diameter. This compensates for the fact that the inflated diameter is less. For an elliptical, 1.5 is more like it because the elliptical inflates to the nominal diameter.

I worked with a TARC team and they used the 300# kevlar as a leader and 100# kevlar as a shock cord. The larger leader prevented zippering the body tube if the parachute deployed prematurely. The reason they did not use 300# for the entire shock cord was to keep weight down. At the finals, they got a premature deployment 4.2 seconds into the flight (based on altimeter readout) with the rocket going 200 fps and everything survived, no zipper and no broken shock cord.
 
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