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phoenixlpr

Was: Recommended specs on reserve exit weight, do you go over max?

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MSW is 160 lbs, I was around 200-220 lbs with gear.
I had a soft, stand up landing.



And if you deploy at terminal while tumbling and the canopy blows up how do you think your landing will be? There is an old saying that fits this situation.

“If you are going to be stupid you had better be tough.”

Overloading your last chance at staying alive is stupid. jmo
My idea of a fair fight is clubbing baby seals

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And if you deploy at terminal while tumbling and the canopy blows up how do you think your landing will be? There is an old saying that fits this situation.

“If you are going to be stupid you had better be tough.”

Overloading your last chance at staying alive is stupid. jmo


A canopy does not "know" about the suspended weight on deployment, only the airspeed. So why would it blow up?

I'm aware that I might use it beyond its certification.

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A canopy does not "know" about the suspended weight on deployment, only the airspeed. So why would it blow up?



And where did you learn this interesting "fact"? Why do you think canopies are tested to a given speed and weight? You can blow a canopy up at a lower airspeed if you use enough weight or you can blow it up at a lower weigh if you use enough airspeed.

To produce shock load of 5000 lbs. you can use 325 pounds at 225 mph or 200 pounds at 300 mph. But it takes a combination of both weight and speed.

Before you give advise you should know what you are talking about.
>:(
My idea of a fair fight is clubbing baby seals

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Notice: I have not given any advice.

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To produce shock load of 5000 lbs. you can use 325 pounds at 225 mph or 200 pounds at 300 mph. But it takes a combination of both weight and speed.


???

Size of a canopy is constant. The air resistance of the canopy is constant on a given size. The system: jumper+gear has a given amount of kinetic energy before opening. So it might take longer time to reach, decelerate to new speed of the system. Applied force on the lines and canopy is depending on the area of the canopy, the factor of air resistance and the deployment speed only.

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???
Applied force on the lines and canopy is depending on the area of the canopy, the factor of air resistance and the deployment speed only.



You are mistaken. I will try to explain giving an extreme example: The drag force at a given speed and canopy size is fixed, but it is may also be very high. To deploy a canopy through the air at a continous terminal speed (behind a large plane for instance) would destroy the canopy as the drag force would be so high it would overstress the lines and fabric.

In a normal deployment, this high drag force is quickly absorbed by slowing the jumper as the canopy deploys. A heavier jumper exposes the canopy lines to larger forces because the canopy cannot decelerate as fast.

Since the drag force at terminal on a canopy is so high, I would think that the time it takes to slow from 120-10mph over a deployment would be almost the same whether the jumper is 100lbs or 200lbs. The difference is the forces on the lines.

Seth

I think I got that right!
It's flare not flair, brakes not breaks, bridle not bridal, "could NOT care less" not "could care less".

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In a normal deployment, this high drag force is quickly absorbed by slowing the jumper as the canopy deploys. A heavier jumper exposes the canopy lines to larger forces because the canopy cannot decelerate as fast.



You'd rather start drawing. Forces are not larger just takes more time....

Force of deceleration is the air resistance of the canopy.
Air resistance is depending on the speed, surface and resistance quotient only.

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Hey gang,

The original poster exercised their privilege of removing the original post. No need for me to know the reason, but since the conversation was so constructive, I clipped all of the replies and put them back in here in G&R with a new subject line.
Arrive Safely

John

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Hi phoenixlpr,

About 40 yrs or so ago when I was in Physics it was
F=MA

F= force
M=mass
A=acceleration

A given mass at a certain acceleration equals the forces that will result.

In engineering & physics there is no deceleration, we called it negative acceleration, it is just delta v.

Now, of course, I might be wrong.

Jerry

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so apparently many go over the max.



The fact that many do it does not make it smart or legal. People have died deploying a reserve outside of the design envelope.


5. Reserve Sizing

There are several factors to consider when choosing the size of your reserve. The most important factor to consider is the maximum exit weight limit for a particular size. Many jumpers exceed the maximum weight limit for their main canopies. While this may be foolish, it is not illegal. The maximum exit weight for a reserve is a legal limit. In the United States, it is a violation of federal law to jump a reserve if your exit weight exceeds this limit, and other countries may enforce this limit as well. The maximum exit weight is published on the Warning Label sewn to the tail of every PD Reserve, in the PD Series Ram-Air Reserve Owner’s Manual, and in the product information on our web site at www.performancedesigns.com.

My idea of a fair fight is clubbing baby seals

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Flight characteristic is depending on the suspended weight.



You are right, the suspended weight will affect how a canopy flies.

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IMHO the overloaded reserve blowing up on opening has no base, that's just a myth.



Your opinion is wrong. I have been involved in testing for over 20 years and have seen many canopies that were tested beyond their design limits blow up. This can happen by the use of too much weight or by too much speed. Or it can be a combination of the 2 that create a force higher than the canopy was designed to take.

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Notice: I have not given any advice.



You are giving advice on what will affect a canopy during deployment and you do not understand the basic physics involved.[:/]
My idea of a fair fight is clubbing baby seals

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In Physics it's
F=MA

F= force
M=mass
A=acceleration



Jerry,

I think phoenix would agree (can't argue with Newton), but he is saying that F is basically constant, so a greater M results in a lower A. (Heavier people on a same size canopy just open more slowly).

I think drag forces at high speeds are so great that the canopy is going to decelerate at about the same speed regardless of the weight below it. Then if A is constant a bigger M will result in a bigger F.

Seth


To put some numbers on my point:
Fd=Cv^2
(Drag Force is proportional to square of the velocity)
This means that for a canopy that can lower 220lb person at 10mph would require a force of 32000lb to move it at 120mph.
It's flare not flair, brakes not breaks, bridle not bridal, "could NOT care less" not "could care less".

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Exit weight has alot to do with the rated capatity of a parachute.
I first saw this back in 88 when we were test dropping over fifty thousand pounds for the C17 project. With 12 cargo parachutes: one of these parachutes opened two secounds before the other ones and needless to say it had totally blown its gores: (the other parachutes brought load down but harder then normal).

So with this it goes for cargo parachutes as well as personal parachutes that if you exceed the max limits things will go.

Plus added to this it is against the law to exit with a parachute that is overloaded to this point. Thats why they are published. While some compnaines will test there systems at a higher loading does not mean that you can jump them and be legal.

So anytime you are going to purchase a main/ reserve you should know the legal limit and if you fall witin that max exit weight then either go with another size larger or lay off that soda/ pop before the next jump... (Best answer super size on the canopy size then the drink)...
Kenneth Potter
FAA Senior Parachute Rigger
Tactical Delivery Instructor (Jeddah, KSA)
FFL Gunsmith

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Think about basic physics. You may end up with similar conclusion.



the equation for kinetic energy is of the most basic = .5 mass X velocity squared. Mass is a big part of total energy. (and a overloading jumper weighs more, so V is likely to be higher as well.

Reserves are certified to open within 3 seconds. I'd be surprised if that testing is done at a normal weight. Others here can say with certainty.

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Under TSO-C23b they tested in 2 categories, standard and low speed. In the standard category they needed to produce a load of 5,000 lbs. In the low seed category they needed to produce a load of 3,000 lbs. The first attachment shows the various combinations of speed and weight that will produce the required loads.

The second attachment shows what is required under TSO-C23d.

In all cases the canopy is tested to a given weight and a given speed at deployment.

phoenixlpr

As I posted before, your opinion is wrong.
My idea of a fair fight is clubbing baby seals

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Hi kelpdiver,

The various versions of the TSO req'ments vary somewhat in detail. But for the most part, the common two tests are what are called 'functional' and 'strength' (OK, some folks will argue with me on what is 'common').

Since I know TSO C23b the best; the functional test was to be performed with a 170 lb dummy and the strength test could vary (more weight & less speed; or less weight & more speed); we did ours with a 400 lb dummy at 200 MPH (BTW, we blew two harnesses apart until I redesigned things; out of necessity).

The later versions of the TSO are very similar; but that is it in a nutshell. For more detail, go to the PIA website and look up Technical Standard 135; this is the standard that the PIA committee developed.

Hope this helps,

Jerry

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Hi kelpdiver again,

I forgot to mention that in TSO C23b the functional tests have a req'ment to be open within 3 seconds. The strength tests do not have any opening time req'ment.


Sparky,

Will you quit getting ahead of me when I am trying to post?:)
Jerry



Opps! What an old grouch.:P
My idea of a fair fight is clubbing baby seals

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About 40 yrs or so ago when I was in Physics it was
F=MA

F= force
M=mass
A=acceleration



Drag (physics)


F=m * a is one side of the equation.

Here is the other side:

F = c * A * v^2

where A is the area of the canopy
V is the speed
c is a coefficient

m * a = c * A * v ^ 2

As you see lager acceleration/deceleration belongs to bigger mass,because forces in this system does not depend on the mass of the system.

c is deppending on the shape, air desity....
A the area of the canopy is constant
V is the deployment speed

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