Simple Inexpensive Method to test Pilot Chutes

[JohnSherman 1]

Attached please find an Excel Spread Sheet set up with the imbedded math and instructions for how to test a pilot chute.
I call this the "Free Drop Method".
All that is required is a high place from which to drop, a scale to weight the pilot chute and a timer with which to time the decent. Video is a good idea to confirm timing.
If you only wish to directly compare (a Side by side race) pilot cutes they must weigh the same.
If you fill out the spread sheet it will do the math for you and the weights do not need to be the same.
Share your results and comment here.
I have also forwarded copies of this procedure to the PIA tech committee for consideration.

Have fun,

JS

[RiggerLee 63]

This could be fun, but as usual you're a bit off. First you really need to be looking at this once it's reached terminal velocity. It's not fair to use that formula when it's still accelerating. Say I took your PC and Booths. I hold them out five feet off the ground and drop them. They are still accelerating and do not build up enough speed for their individual drag to become a factor. Waa la. I have just proven by your formula that both you're pilot chute and Bills have the same effective square area. This is an extreme example but it holds true. You can not simple divide the distance fallen by time. It will give you the average speed over that distance but it will not tell you what is happening as the speed builds up and they begin to perform differently. Higher would be better. You could drop them from the plane or from under canopy. With more height the acceleration period is less important over all. But an even better method would be to measure and record the rate of decent. Some body help me out. Where do all the new high tech gizmoey wrist mount altimeters stand on this. Which would you recommend for recording the decent rate? Which ones would interface with a laptop and give pretty graphs for people to ooow and aaagh at? A little padding and it should be fine. The old school way would be to hand a weighted flag underneath it on a 100 ft. line. Video it and time from when the flag hits to touch down.

Also, the weight thing. I don't buy the weight them up evenly thing. The PC has to lift it self along with the free bag. A heavy PC has to work that much harder. It's the price you pay for a heavy spring or a heavy metal cap. Basically what I'm saying is that you are unjustly harming the measurement of the lighter PC's such as your own. All drops should be made with the same "cargo weight" and mass of the PC is just their own burden. A reserve canopy weighs X amount. The container requires X amount of force to pull the free bag free. None of this has any sympathy towards a heavier PC. I think 20 lb. would be a good number. How fast does it have to go to support a 20 lb. weight at terminal velocity. In other words how fast will you have to be falling in order to create 20 lb. worth of drag in order to even begin the opening process. And yes I'm just pulling that number out of my ass. It would get way too scary if we tried to apply a realistic number for some of the containers today. But that's a whole nother story.

So let's do the drops from a plane or take the bundle out the door with you if you like and drop it under canopy. Let's make the weight suspended under the PC 20 lb. Let the PC weigh what ever it wants, don't worry about that. And let's see the real speeds that these pilot chutes will need to perform their job.

Who's doing book on this? I say RI will do well. Shermans are ok. And some of the strong PC should do well. Mirage is a bit heavy but it's got the drag. Who's giving odds?

Lee

Lee
lee@velocitysportswear.com
www.velocitysportswear.com

[JohnSherman 1]

Terminal Velocity is best called "Equalized Velocity" it is when the pounds of Dynamic pressure equalizes to your bodies pound for square foot loading. For a human body which has about 6 sq. ft. of surface area at say 180 pound or 30 pound per square foot.
We accelerate until we reach that equalization.

Terminal velocity for a pilot chute is the same. It is when it's drag equals its loading. For a Racer pilot chute its occures about 1/2 a second after release as it weighs about a half pound and has almost 6 Square feet surface area that is equal to .08 pound per sq. ft. Therefor it reaches its "Terminal" almost instantly.
It requires 1/4 a second for the Vector 2 as it weight about a pound. Twice the weight half the drag.
A drop heigth of 5 feet would do if you could measure the time accurately.
As I show above they won't hit the ground at the same time.
Try it, I have and the numbers corrilate well to wind tunnel testing.
I am not doing drops, I am comparing pilot chutes if you are going to drop them side by side as in a race then they must weight the same.
The weight is the drag.

You ask "How fast do you have to go to produce 20 pounds of drag"?
Simple, it depends on how much effective area your pilot chute has.
If your pilot chute has an "Effective Area of 5 Sq. Ft. then you must be going fast enough to develop 4 "Q" or 4 pounds of Dynamic Pressure. 4 x 5 = 20.
If your pilot chute has 2 feet of effective area then you will need 10 pounds of Dynamic pressure or a little longer in free fall befor you bag extracts.

You method will run us into hundres of dollars my method is considerably simpler and cheeper. Additionally you must instrument you dropped load to measure rate of decent.

I'll make the book. You tell me the pilot chute size and type and weight and drop heigth and I will predict the time.