Kite trampoline / bungee launching a payload

[Roderick Read]

I've been experimenting with aggregate stability of kite networks. ... Kite like to fly in networks.
And I've written this about it.. https://drive.google.com/open?id=0B0qQxFlXu7t-RGpqRk1ZYXNISkU
For the upcoming Airborne Wind Energy book.
Today though... I want to know what thickness of bungee would be needed to ...

launch a satellite from a kite network?

This video will help to frame the question
So , 100 x 5000m long bungees from a ring network of kites at 10 km
F=ma = 10x11000 = 1100N per bungee 1 large kite can pull that...
A ring of 100+ or even 1000+ kites can launch a 1kg satellite...
true or nonsense?
Elastic potential of bungee = .5kx² k =spring constant x = dist stretched =~ 5000m

 

[mfb]

There are no massless bungee cords. You cannot accelerate a single bungee cord to orbital velocity even without payload. Adding more cords just adds mass and force in the same way, so it doesn't help.

 

[Roderick Read]

Thanks for the reply,
My ~reasoning for a multiple bungee central catapult structure is networked load sharing will keep the kites relatively stable.

Just to re clarify the idea is .. A ring of kites supports a cone of bungee which is stretched to ground. As the ground connection is cut ... the bungee cone catapults a payload.

There are no massless bungees: Absolutely.
By using split bridling (multiple lines at top going to fewer to single thicker lines)... Surely multiple upper bridles can support their own weight and that of long lower bungees.

What is the Max (call it vertical) speed of a 5000m elastic line stretched to 10,000m.
Mass distribution is constant on an unloaded line but suspended from a kite it's upper will be more stretched (thinner section ) than it's lower.
The line air resistance of a contracting linear bungee ... No idea.

The initial g kick would likely crush a brick

 

[Roderick Read]

According to http://www.langmanropes.com/langman-en/products/rope-by-industry/industry/shockcord-langman-ropeshttp://www.langmanropes.com/langman-en/products/rope-by-industry/industry/shockcord-langman-ropes
12mm dia bungee has a breaking load of 500kg and weighs 12.8kg/100m
So a 5000m weighs 640kg Which would snap the line. and give 640kg less pull at the bottom than the top.
We'd have to work much shorter unit lengths.
(As an aside I use Dyneema^(registered) line. It has a remarkably long self supporting length (free breaking length) at 378 km)

If this system were to use cascaded bifurcating upper bridling... (1 lowest level line splits to 2 higher lines, split to 4 next higher level lines, ...)
And we stage the release of the elastic at each bridling level... e.g. the lowest level released first...contracts and lifts a payload closer to the first level...
The first level is released re-tensioning the lowest level bungees and furthering their acceleration, the next level is released...

Using this method... Let's not be as ambitious as launching a satellite... the new question might be...
How strong does an Italian mountain have to be to throw a 1kg rock at Iceland?
How much air would be thrown along with the rock?

 

[jedishrfu]

You should try to really simplify this problem by scaling it down and by eliminating the kite structure and the bungee bifurcating network.

Perhaps a ring structure with bungees connected to ring and then determine when you pull back to launch determine how much force is on the ring and from there decide if a kiting system can support that force trying to pull it down.

I can't do the math but there may be some software like Mathematica that can model these kinds of structures or you could use MATLAB, Julia or numerical Python to program a simulation.

https://www.wolfram.com/system-modeler/industry-examples/education/multibody-system-simulation.html

Here's a link to a paper describing a simple bungee cord model for bridge diving:

http://link.springer.com/article/10.1007/BF02915920

which might give you ideas on how to proceed and one using MATLAB to simulate it:

https://www.wpi.edu/Pubs/E-project/...2313-150806/unrestricted/MQP_final_report.pdf

Also you mentioned you were writing a book on these ideas. What is the audience for the book?

What other aspiration do you have here? Are you trying to start a company or develop an invention?

I like the novelty of your ideas but I'm not sure of the practicality considering the fickleness of wind.

Some of it reminds me of the Ringworld physics discussions from the Larry Niven sci-fi book. People wondered could it be built? how stable was it? How to emulate daylight and evening cycles? ...

 

[mfb]

Let's make some rough estimates. What is the total acceleration distance? I guess 10 kilometers is quite optimistic already. If you want to reach 8 km/s (which is not sufficient for getting things to an orbit, but Italy -> Iceland should work), you need an acceleration of 3200 m/s^2, or 320 g. That reduces your 380 km breaking length to ~1 kilometer. Your cable mass has to increase by a factor of e every kilometer, in total by a factor of e¹⁰ = 22,000. Well, it is a bit better as the upper parts can have a smaller acceleration, but your tapering won't be 100% exact, so let's be optimistic and say a factor of 10,000 is sufficient. You start with 1kg/380km and end up with 1kg/38m. Doesn't look bad, but we are not done.
How far can you stretch Dyneema? https://www.dsm.com/content/dam/dsm/dyneema/en_GB/Downloads/Articles/Marlow%20article%20in%20Seahorse%20November%202014.pdf suggests a few percent, but let's be optimistic and say you can get 10%. To get an acceleration length of 10 kilometers by stretching the material, you would need a cable length of 100 km extended to 110 kilometers. That blows up the factor 10000 from above to 10⁴⁰. Oops. And that is not even taking into account if stretching the material would give the right force we need. Longer or shorter acceleration distances do not change that factor, they just change the overall size of the structure.

What else can you do? Well, if stretching the material doesn't work, you have to accelerate the anchor points. How do you accelerate things at a height of 10 kilometers by 320 g? There is no realistic way to do this.

 

[Roderick Read]

You should try to really simplify this problem by scaling it down...

Thanks Jedishrfu,
I like building models before deriving maths ... But maths saves time and injuries... I must get back to developing proofs before playing.
I can probably simulate this with kangaroo physics in grasshopper where I developed my model.
The great thing in grasshopper is the evolutionary parametric design will fit so well with "AI" models.

A bifurcating bridling (or just a progressively thicker single bungee) is essential for the greater load at higher upper levels.
There is an aggregate stabilising effect on individual kites in tether networks. (..from sketchy results of my own tests and simulations)
I work from home as a house-husband-mad-scientist-inventor-dad. You don't get to fund big lab experiment results in this job.
Bridling in kite lattice networks allows lift from more than 1 kite to be used cumulatively.
That's why I reckon on splitting to a greater number of bungee at higher levels.
And why I'm proposing to do some serious lifting with kite networks.

My kites seem bonkers. Many people struggle to believe what they're seeing.
My Daisy stack kites spin round whilst tied to my e-bike, the bike outputs ~150w latest rig on an easy day.
Not a very impressive figure... but potentially scalable because of the soft structure. ( fast rigid wings would output a lot more power)
I share my designs as open source hardware... The book is Springer Airborne Wind Energy 2016.
I have only submitted a chapter. it's in the last round of peer review ... so hopefully my prototype report is included in the book.

 

[Roderick Read]

Your cable mass has to increase by a factor of e every kilometer,
so let's be optimistic and say a factor of 10,000 is sufficient.
You start with 1kg/380km and end up with 1kg/38m.
What else can you do? Well, if stretching the material doesn't work, you have to accelerate the anchor points. How do you accelerate things at a height of 10 kilometers by 320 g? There is no realistic way to do this.

Thanks for taking time to play numbers mfb. YOU STAR!
An ever increasing factor e of material thickness per height sounds right. (My maths has gone shakey and rusty now. Must revise)
You start with 1kg/380km and end up with 1kg/38m. Have I understood that right... It sounds very lightweight for a bungee weight to support 1kg @ 320g bottom of catapult and 1kg/38m top of catapult?
Kites (even at 10km) can still have a huge pull on their line. Literally tonnes of pull.
I'm going to blindly argue that there is a realistic way to do this.

 

[mfb]

You start with 1kg/380km and end up with 1kg/38m.

If the whole cable accelerates in a uniform way, with 1 kg of payload attached. But there is no way to accelerate the whole cable so fast.
If you need the cable to act as spring, that does not work, and you get the 10⁴⁰ figure, which would need more than the mass of Earth (and also lead to ... ehm... space problems).

 

[jedishrfu]

Have you ever read the Somerset Maugham story called The Kite featured in the Quartet of stories?

https://en.wikipedia.org/wiki/Quartet_(1948_film)

and the B/W movie version:



Its good your wife doesn't mind your hobby.

 

[Roderick Read]

If the whole cable accelerates in a uniform way, with 1 kg of payload attached. But there is no way to accelerate the whole cable so fast.
If you need the cable to act as spring, that does not work, and you get the 10⁴⁰ figure, which would need more than the mass of Earth (and also lead to ... ehm... space problems).

Here's a really good factsheet on dyneema.
You really wouldn't want a 10km acceleration distance of dyneema. The forces really would be massive.
The lashback when that stuff snaps is scary. I've seen it throw anchors past my head to twice the length of the rope.
Free breaking length 378 km
Axial tensile strength 3.6* GPa
Axial tensile modulus 116* GPa
Axial compressive strength 0.1 GPa
Axial compressive modulus 116 GPa
Transverse tensile strength 0.03 GPa
Transverse modulus 3 GPa
Transverse compressive strength 0.1 GPa
Elongation at break 3 – 4 %
Work to break 45 – 70 MJ/m3

But best of all if you stretch it really tight...
ACOUSTIC
Sonic velocity (axial) 10000 - 12000 m/s
Sonic velocity (transverse) 2000 m/s
Could you ride a wave induced on the rope at high speed?

In the original question I was looking to stretch a 5km elastic to 10km.
Surely with multiple bungies arrayed around like a troampoline sheet ..
or even arrayed like stacked trampoline sheets ... then we wouldn't need such a long acceleration length.

 

[mfb]

Could you ride a wave induced on the rope at high speed?

Not in any way where the rope would help.

Surely with multiple bungies arrayed around like a troampoline sheet ..
or even arrayed like stacked trampoline sheets ... then we wouldn't need such a long acceleration length.

A shorter acceleration length means even more acceleration, and an even shorter effective breaking length, so your tapering needs to be stronger. The result is the same - you still need the tapering ratio of 1:10000, if everything accelerates uniformly, and some astronomical number if the cables itself act as springs. And instead of 1 kg payload, you suddenly have to accelerate a ton to some astronomical number worth of cables. You gain nothing.

 

[Roderick Read]

Not in any way where the rope would help.
A shorter acceleration length means even more acceleration, and an even shorter effective breaking length, so your tapering needs to be stronger. The result is the same - you still need the tapering ratio of 1:10000, if everything accelerates uniformly, and some astronomical number if the cables itself act as springs. And instead of 1 kg payload, you suddenly have to accelerate a ton to some astronomical number worth of cables. You gain nothing.

I'm sure riding a rope wave vertically (despite being inefficient...) is possible. Have you ever played with a rip stick? replace the wheels with ratcheting pulleys over the rope and it should ride a wave on a taught line.
That aside... (It's kinda fun but a bit nonsense)

What I was asking earlier, I don't think I made it clear...
Using mfb 's logic ... we can come up with an appropriately tapered single line bungee, which can launch launch with a given acceleration.
It is simple to array single line bungees around in a cone shape. For each comparable bungee you would have another 1x the acceleration available.
If 1 x bungee alone is too mad... surely an array of bungees of reasonable size can do the job.

I'm still going to blindly claim that there is a resolution of those parameters (using a large network of kites (or mountains if you insist on solid lifting structure) to stretch multiple bungees) which allows a rock to be flung from Italy to Iceland. 3241km needs ~5.6km/s release

 

[Roderick Read]

What an awful film.
It's more than a hobby now. I'm making kites with intent.
Not sure there is a direction, but there is some purpose.
Latest Test

Have you ever read the Somerset Maugham story called The Kite featured in the Quartet of stories?

and the B/W movie version:

Its good your wife doesn't mind your hobby.

 

[mfb]

I'm sure riding a rope wave vertically (despite being inefficient...) is possible. Have you ever played with a rip stick? replace the wheels with ratcheting pulleys over the rope and it should ride a wave on a taught line.

I don't see how, unless you use techniques that you could use without the wave as well.

Using mfb 's logic ... we can come up with an appropriately tapered single line bungee, which can launch launch with a given acceleration.

I think I made clear in all my posts that it is not possible to reach orbit with that. Not even close.

It is simple to array single line bungees around in a cone shape. For each comparable bungee you would have another 1x the acceleration available.
If 1 x bungee alone is too mad... surely an array of bungees of reasonable size can do the job.

It cannot, for reasons explained above. All the calculations assumed you have the optimal numbers everywhere and it still did not work out.

 

[Roderick Read]

Thanks mfb,
I'm prone to being very optimistic with kites.
Can you provide / do you still have the calculations you used for completeness please?

 

[mfb]

They are all in the thread, the steps in between should be clear.