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# A Brain Teaser for Gear Wonks

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We all know that a pilot chute doesn’t drag, after launch, until it reaches the end of the bridle. It can’t drag if it has nothing to drag against. It has to travel from the container some 16 total feet (13 feet of bridle and 3 feet to the top of the PC) to a point where it applies tension to the bridle. That’s the beginning of drag. This causes orientation if required, and provides lift/drag to the deployment bag through the bridle thus extracting the bag.

The question is: What are the forces which transport the reserve pilot chute from the container to the end of the bridle, some 16 feet?

1. The Free Stream Air flow?
2. The force of the spring?
3. Acceleration due to gravity?

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I want two answers! 1 & 2.

The spring gets it a foot or two off the back. Then the pilot chute, even though there's no resistance back through the bridal yet, still has some drag in the high speed air. Like tossing a napkin out a car window at 60 mph. So it seems like it would be both 1 & 2.

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Drag. If you throw a pilot chute into a wind tunnel, it will drag even if it's not connected to a container. It will have its own terminal velocity.

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It does create drag before there is tenson on the bridle, although this is a lot less compared to a pilot chute with a tensioned bridle where the material can form to half a sphere (more or less) which has much higher drag coefficient. This is because it's already exposed to the air rushing past the freefalling jumper, whereas the speed of the pilot chute in relation to the jumper is close to zero at launch.

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JohnSherman

We all know that a pilot chute doesn’t drag, after launch, until it reaches the end of the bridle. It can’t drag if it has nothing to drag against. It has to travel from the container some 16 total feet (13 feet of bridle and 3 feet to the top of the PC) to a point where it applies tension to the bridle. That’s the beginning of drag. This causes orientation if required, and provides lift/drag to the deployment bag through the bridle thus extracting the bag.

The question is: What are the forces which transport the reserve pilot chute from the container to the end of the bridle, some 16 feet?

1. The Free Stream Air flow?
2. The force of the spring?
3. Acceleration due to gravity?

Yep, others already got it, but the key is that your question was not written correctly. You say "a pilot chute doesn’t drag, after launch, until it reaches the end of the bridle." That's incorrect. What's more correct is that the pilot chute doesn't exert force on whatever is on the other end of the bridle until the bridle is taught. The pilot chute itself still experiences drag, just like any non-zero-volume object will in an airstream. EDITED TO ADD: the drag, of course, will be far less than when it reaches the end of the bridle, because the bridle helps form the shape of the PC into one that produces additional drag. But as another poster said, throw an unformed napkin into the wind and it still has drag.

So the answer is that the spring gets the pilot chute into the air stream (by exerting a downward force against the rig/jumper, causing an equal and opposite upward force on the pilot chute), and then the residual energy from that force + the drag in the airstream continues to push the pilot chute upward (relative to the jumper). Acceleration due to gravity is identical for the jumper/rig and the pilot chute.

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initially it would be "None of the above", since it is the FORCE exerted by the jumpers' hand and arm, as the P C Handle is taken and Yanked.....OUT of the pouch...
THAT Muscle power is what Enables the Pilot chute to lay "NEXT TO " the jumper.... for that split second Before the Jumper Falls AWAY from the inflating PC... Think "air anchor "
good question.
No matter WHAT " force " it IS ,, when you NEED it... May it "Be With You "
jmy A 3914
D 12122

All three.

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OH.... Now i realize that you are asking about The reserve Pilot Chute... sorry...

My answer however remains the same....
It's Muscle power..... from the act of Pulling the reserve handle..
IF the deployment begins via AAD..... then it's "computer power "...

Of course immediately AFTER.... it Does become the Other choices listed here...
jmy

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The main force is not acceleration at all, but rather deceleration. The p/c leaves the mass of the falling jumper. Without that mass resistance of the air causes it to rapidly decelerate. The difference in the fall rate of the jumper and the p/c causes the bridle to stretch out.
Always remember the brave children who died defending your right to bear arms. Freedom is not free.

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gowlerk

The main force is not acceleration at all, but rather deceleration.

Either way works, just a different frame of reference, earth vs. jumper
In addition to what John Sherman mentioned, another potential for force involved in pilot chute movement will of course be the unsteady airflow behind the jumper, the burble, where there may be air swirling downwards relative to the jumper. Pilot chutes can thrash around in there, dependent on their mass, speed, balance, drag from different parts of the pilot chute at different angles and extent of inflation.

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Quote

But as another poster said, throw an unformed napkin into the wind and it still has drag.

If I throw a stick into a moving stream of water and it floats away with the flow of the stream is that drag? Maybe it is "going with the flow".

In order to have drag you must have a Force and a Resistance.

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JohnSherman

If I throw a stick into a moving stream of water and it floats away with the flow of the stream is that drag?

Yes. It will remain in a state of rest until a force accelerates it. Chuck a pilot chute into a wind tunnel and it will fly up to the ceiling if it isn't attached to anything. In both cases, the same force is acting.

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JohnSherman

Quote

But as another poster said, throw an unformed napkin into the wind and it still has drag.

If I throw a stick into a moving stream of water and it floats away with the flow of the stream is that drag? Maybe it is "going with the flow".

In order to have drag you must have a Force and a Resistance.

Yes, it's the same thing. The stick in water has nothing holding it back so it floats along freely with the current. No drag there. Likewise, a pilot chute thrown into 120 mph air has nothing holding it back (at first) and floats along freely with the 120 mph air surrounding it. Only when the pilot chute hits the end of the bridle does there become resistance to going with that flow, and therefore drag is produced.

What do I win? A free Racer?

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Could we all agree on a common definition of "drag," preferably one that is used in physics? The one from Wikipedia seems like a good starting point: "forces acting opposite to the relative motion of any object moving with respect to a surrounding fluid."

Is the pilot chute moving with respect to the air? Then it is experiencing drag. But without tension from the bridle, the speed is quickly reduced from 120 mph to almost zero. The time this takes is a function of the force and mass, with the added caveat that the drag force is continuously decreasing together with the speed. The same is true for the stick on the water. The stick is initially moving with respect to the water and drag is what "decelerates" it to match the flow velocity.

In order to have drag, you must have a relative velocity through the surrounding fluid, nothing else (assuming non-zero density, cross section area, and drag coefficient).

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it depends on who is your frame of reference and control volume.

in the eyes of the pilotchute, the jumper is moving away from it, and it is stationary.

from the eyes of the jumper, the pilot shoot is being pulled away from him with the relative wind.

from the eyes of the jumper + pilot chute system, its the differential effect of terminal velocity. the jumper is at terminal velocity, and the PC is not, hence one is traveling towards earth faster. so there is relative velocity between the two that makes the gap.

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The problem with the horizontal wind tunnel and the stream of water analogy is that neither have the gravitational component which is the weight of the object. If we go to a vertical wind tunnel or could run the water up vertically then the gravitational component would come into play and would resist the flow.

Without gravity or something to resist the flow you have no Drag.

To keep the analogy horizontal and in water we could say the air we skydive in is like a still water lake and we are plowing through it in a gravity strength powered submerged body. Put a pilot chute out thru the turbulence into the still water and that will give you a look at the same thing that happens in the air.

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The skydiver and the reserve pilotchute, once deployed, have different terminal velocities.

The reserve pilotchute terminal velocity is lower than that of the skydiver.

After the transient launch period (and all that goes with that..), this differential (or part thereof) results in the launch force applied to the free bag through the bridle. The launch of the free bag does not require the reserve pilotchute to achieve its natural terminal speed whilst connected to the skydiver. I think...

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Hi John, a quite tricky question by the way. Physics again.

If you take the surrounding space as a frame of reference, depending on the force of the reserve spring, the downward speed of the jumper (including the reserve pilot chute just before launch) is more or less (after launch) canceled by the speed acquired from the spring extension. That means that the pilot chute (more or less) stays stationary (with respect to the surrounding space) and "waits" for the jumper to continue his fall until the complete bridle extension occurs.

A little story:
Suppose a person in a fast boat going at 70 mph South on a river flowing very slowly South-North. When the boat passes in front of an observer on the shore, the person using a bow shoots an arrow at 70 mph North. What will see the observer ? According to the Newton laws of relative speed, the arrow will just fall down flat vertically in front of the observer since the two speeds (boat and arrow) cancel each other.

In this thread, what happens to the pilot chute is a bit similar to the arrow of the story. This pilot chute stays stationary. The jumper falling at 174 ft/sec will get the full bridle extension in
16 ft / (174 ft/sec) = 0.092 sec

Therefore, there is a system of forces and speeds equal to zero (more or less) applied on the pilot chute, but for less than a 1/10th of a second.

Note : there is no more acceleration on the jumper since he is at constant speed (terminal velocity)

Learn from others mistakes, you will never live long enough to make them all.

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erdnarob

This pilot chute stays stationary. The jumper falling at 174 ft/sec will get the full bridle extension in
16 ft / (174 ft/sec) = 0.092 sec

Bollocks!

I know that's a simplification, but it's just not true. Taking the air as a frame of reference, the pilot chute is decelerating very rapidly indeed, but it's never going to stop, even with an unlimited bridle length.

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As you said, the pilot chute is decelerating but it is between the time just before launch where it has the same speed than the jumper to the time when the pilot chute spring fully extended cancels that downward speed because of the speed given upward by the spring. The pilot chute then stops for a short duration. In the meantime, the jumper has fallen enough to get the bridle taut.
Learn from others mistakes, you will never live long enough to make them all.

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So when you do a pull test on the ground the pilot chute is accelerated to 120 mph by the spring?

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While I've seen good springs launch the PC to the full extent of the bridle, in my mind the spring is not primarily for that purpose...

The strength of the PC spring is to ensure it clears the burble and gets into clean air to enable the PC material to do its job.

Unfortunately, even with strong springs, this is not ensured and in that case we are then subject to the dynamic behavior of the material involved and the unpredictable currents and eddies of the burble and actions of the jumper.

JW
Always remember that some clouds are harder than others...

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erdnarob

fully extended cancels that downward speed because of the speed given upward by the spring.

Since you've mentioned it twice now I think I'll suggest something different:

It is incorrect to imply that it slows down because it jumps out at 120 mph.

Once the pilot chute catches clean air, with its relatively low mass and high drag, and only dragging out a loose bridle, yes indeed it will decelerate very quickly.

A spring loaded pilot chute can easily have a drag of at 150 lbs at terminal, and have a weight of around 1 lb (plus only light bridle to accelerate too). So once it clears some burble it will have a very high initial deceleration. (150 G in our idealized case of 150 lbs vs. 1 lb)

That deceleration will fall off very quickly as the speed falls off, and of course a pilot chute will have some small terminal velocity on its own. (They don't float like Wile E. Coyote if dropped off a cliff) So it won't travel too far before having to accelerate the d bag out of the container.

So yes a pilot chute will, compared to the speeds of skydivers, "almost come to a stop" due to its very high drag to mass ratio but won't "stop for a short duration" due to having the speed cancelled by springing out.

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A jumper falls at 174 ft/sec which is 54 meter/sec (120 mph)
According my calculation a pilot chute with a total weight of 1 lbs using a spring having a constant of 535 Newton/meter compressed from 78 cm to 3 cm is theoreticaly able to launch at a speed of about 30 meter/sec. This is why I said "depending of the spring". I agree that we still have a gap between 54 and 30 m/sec. However even if the pilot chute doesn't stop completely, it is slowed down at (in this case) at about half the speed of the jumper (with respect to the space). Therefore there is a vertical distance rapidely increasing between the jumper and the pilot chute to allow a fast extension of the bridle. The pilot chute gets also a relative wind of 24 m/sec (54-30) which contributes to shorten the bridle extension time.

My first assumption was qualitative (by comparing the pilot chute to an anchor), here is a more quantitative explanation where figures make it more concrete.

Ref.1 Potential energy of a compressed linear spring = kinetic energy of the pilot chute. Then one can find the speed of the pilot chute when at full extention
Ref.2 The constant (modulus) of a linear spring is the the force to expand or compress it divided by the distance of expansion or compression.

In the actual situation there are other factors like the burble, the container flaps friction and mass.
I agree very much that a strong spring gives the pilot chute a better chance to clear the burble of the jumper. Here for the spring constant I used the figures obtained by compressing my reserve pilot chute on a bathroom scale.
Learn from others mistakes, you will never live long enough to make them all.

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Finally some numbers!
Your k does seem a bit higher than I was expecting, however. You used this procedure to measure the force of a spring compressed 0.75 m and got 41 kg (90 lbs)? The numbers I've seen on this site were around 40 lbs for the Vector II reserve and lower for the rest.

Even with k=535 N/m, x=0.75 m, and m=0.45 kg, I'm getting a velocity of 26 m/s. How did you get 30? PE=1/2*k*x^2, which is 150.4 J, and v=sqrt(2*PE/m). As you said, even this is an overestimate because lots of energy will be lost as the spring makes its way past the fabric and the flaps. You can get a rough idea of the difference by launching the spring vertically and measuring the height that it reaches, and then repeating the same measurement with the spring launched from a closed container. Half the height means half the energy (PE=m*g*h).

However you look at it, the spring does little more than get the pilot chute past the burble. The rest is all drag. I didn't really get what John meant by differences in vertical and horizontal wind tunnels. The drogue chute for an airplane is deployed horizontally and reaches line stretch through exactly the same mechanism. Gravity can also be eliminated (e.g. on the intentional space station), yet the drag experienced by a napkin thrown at a high velocity will still quickly slow it down. Testing a pilot chute at 120 mph in microgravity might be a bit more difficult

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