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Determination of Parachute Deployment Loads

I sit at the computer with some trepidation in preparing this treatse, knowing that one of our own has recently lost his life to a failure of some sort in his streamliner's parachute deployment at speed. At this time I do not know whether or not it was a failure of rigging, release, or design. However, I feel compelled to take a look at the loads that a parachute can develope when at very high speeds. If nothing else, I hope the analyses I show below will cause some additional thought by those in the safety community and the owners, builders and drivers of these wonderful cars. They are on the cutting edge of automotive performance and it should be treated the same as that of an experimental aircraft with an unknown performance envelope.


What follows is an academic exercise for me. It is not intended to be a definitive answer to anything except my curiosity. To that end, if you find it useful, amusing or down right distasteful,GREAT! You may find errors, wrong assumptions, or other nastiness and I will appreciate you pointing them out to me. Complaints should be backed up by real data, through either analyses of your own or documented information. I am ameniable to changing my thinking iffn there is good reason to. Opinions, by themselves, just don't cut it.

Parachute Failure Modes.

Over the years at both the salt flats (although this experience is limited), and through video, I have seen a number of parachute failures which could have and have had disasterous results. Typically, the parachutes are deployed at speed, followed immediately by the chute's system destruction. Many times the canopy just blows apart, otheres it is the tow or shroud line breaking. Sometimes it is the canopy wildly oscillating or skying. All of these are very dangerous to a car or vehicle at very high speeds. Failure of the parachute systems means that the streamliner cannot be stopped in a normal manner and that it may overrun the course, wrecking in the process. What is the goal here?It is to take some rather broad assumptions which will permit me to make some determinations of the loads and forces that exist when a parachute on a streamliner is deployed and to explain why oscillating or skying occurs.

The "model"

The model I will be working with is simply described. It is a streamliner, at 400 mph, deploying a high speed parachute. The parachute is 4 feet in diameter, has ten shroud lines, with a tow line length of 80 feet. The altitude is that of bonneville as is the temperature. These are all assumptions and cone be changed in the analyses for you own vehicle, should you chose to. The method to be used introduces some new terms to most people so I will expand a bit. Most material failures occur because some rate of change in strain rate has been exceeded. Many items will withstand a slowly applied load or force but will fail when that load is impulsively applied. While impulse does not quite accurately describe the loading method, there are terms and mathematics which do. These are Yank (Y), which is a time derivative of force (F) and jerk (j), the time derivative of acceleration (a). Not surprisingly, there are even terms for higher derivatives, some with names like Snap, Crackle, and Pop. The scenario is the vehicle is at speed, the parachute is deployed backward with an 80 tow line: it all starts with the summation of forces in play.

FT = FD + FN

FT = Total of forces on the parachute system
FD = Aerodynamic force on the parachute
FN = Inertial force in accelerating the parachute system


FD = 1/2 * rho * C d * A * V2


FN = m * a

rho = air density , in slugs, at Bonneville during Speedweek = 0.00186
C d = parachute drag coefficient (1.20 < C d < 2.40; I'll use 1.20)
A = canopy frontal area in square feet
V = vehicle forward velocity in feet / second
m = mass of the parachute and it's rigging
a = acceleration of the rigging to the vehicle speed after deployment, ft/sec/sec

Now applying the new math terms. Remember Y is the time derivative of force (Yank) and is equal to d(F) / dt

d(FT) / dt = d(FD) / dt + d(FN) / dt
d(FD) /dt = d(1/2 * rho * C d * A * V2) /dt
d(FD) /dt = 1/2 * rho * Cd * A * d(V2) /dt
taking the derivative...
d(FD) /dt = 1/2 * rho * Cd * A * {[2 * V] * dV / dt}


d(FN) / dt = d(m * a) / dt
d(FN) / dt = m * d(a) /dt
remember that acceleration is the time rate of change of velocity, plug this into the equation and get
d(FN) / dt = m * d(V/dt) /dt
In doing some internet research, I found a US Army site that delt with parachutes. It stated that parachutes can be inflated in times as little as 0.03 seconds (30 milli seconds) or as much as 0.480 seconds (480 milli seconds). These are from all types of configurations: cross, full canopy and ring. This is a crucial number as will be shown.

Let's designate th etime of parachute release a t-0. Then if we have an 80 tow line attached to the car and it is deployed by forceful ejection backwards then the time to reach the end of that tow line is simple: tow length divided by speed in ft/second.

t = 80 / (400 * 1.467) [note: 1.467 is a conversion factor from mph to ft/sec] = 0.136 seconds or 136 milli seconds.

Not long at all. The reasoning here is that the chute has been displaced backwards and the vehicle is running away from it. In truth, it may be a bit longer than this, but not much. The acceleration time for the components from 0 speed to that of full speed is hard to estimate. It is far shorter that the 136 ms mentioned, but for now, this is the number, delta t, I will use. The velocity change when the tow line is yanked taught is from essentially zero to 400 mph (586.8 fps). So now the Yank due to just pulling the tow line and parachute along for the ride can be figured. Lets assume that the components weight 10 pounds [this is another assumption that can be challenged - go weigh yours and tell me what it is]. For the dt and dVterms, lets switch to delta t and delta V as these are not infitesimal numbers. Also remember that mass is weight (w) divided by gravity (g).

d(FN) / dt = m * d(V/dt) /dt
d(FN) / dt = w /g * delta V / delta t / delta t
d(FN) / dt = 10 / 32.174 * 586.8 / 0.136 / 0.136

Running the numbers for Yank due only to acceleration of the components to vehicle speed gets us a fair number in itself....

d(FN) / dt = 9860.7 lbs/second load rate of change

Doing the same thing for the Yank due to aero drag ....the delta velovity is the same as in the preceding because the force on the canopy is developed from 0 speed to full speed (IMHO). Canopy opening time used is 30 ms ( I used 30 ms for conservativism). rho is 0.00186. A is 12.56 sq ft (ft foot dia chute). Cd is 1.20. You can change these numbers to what fits yur situation.

d(FD) /dt = 1/2 * rho * Cd * A * {[2 * V] * delta V / delta t}
d(FD) /dt = 1/2 * 0.00186 * 1.2 * 12.56 * {[2 * 586.8] * 586.8 / 0.030}

calculating the loads,due to Yank due to aero drag, we find

d(FD) /dt = 321,658 lbs/second

The 2 components of yank add directly so that the total Yank is around 330,000 lb/sec. Of course, hardware starts failing before it reaches this after a second. Some don't fail at all, and this may be attributed to a more compliant tow line, ie stretches, making the time longer. For 10 shroud lines from the canopy to the tow, the yank loads are divided but then the angle of the shroud line to the tow has to be figured into the equation. What does all this prove? Nothing except that vehicle designers and safety inspectors in tech need to really understand the loads being delt with. Cars that are slower than 300 mph probably do not have to worry much, but should look at analysing this anyway. Don't just attach a parachute and hope it will work. Every component in series has to be able to take a large yank load. I hope this helps someone to think about parachutes in a different light. If you don't believe me, find someone you trust to do the modeling and simulations for you. Just do it!

With regards to canopy oscillation, any shroud line that is more taught than others puts an unsymetrical loading on the canopy. This causes it to move off center and this provides an upset air flow situation. It will oscillate causing it to move the back end of the car around. A side wind causes a similar effect. It deforms the canopy on one side resulting in unsymetrical loading resulting in oscillation. Same for skying. Air flow over the top of the car ca deform the top of the canopy causing it to lift. IMHO.

What to do? Well, know your system first. Don't put the new guy on the team on packing the chute or rigging it. Parachute packers are trained people and they need to be. If you can, twist the parachute shroud lines such thet the chute has to unwind before it inflates completely. Add a snubber in line somewhere to take up the load gradually. Pay out the line with the chute fully blossomed early (although this may only help marginally). Do anything to cause the chute from just immediately banging open. And of course, have it mounted in the proper place to provide a line of pull through the cg of the vehicle.

Feel fre to call me an idiot, I don't care, as long as it makes you analyse your system, rather than just installing it. I'll be a fool as long as nobody else gets hurt. Find and repudiate the above, publish it! Do something to ensure you safety! Don't just take my word for it.

Copyright (C) 1998 - 2004, all dates inclusive, L.E. Mayfield - All Rights Reserved

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