Free Fall Arrest: Create Mathematical Model for Safe Thrill-Seeker Experience

[David Thomson]

It is recorded that someone has survived a free-fall from an airplane by being lucky enough to land on snow-laden fir trees.
My challenge is this:
Create a mathematical model which describes the forces at play in order to determine the viability of creating a safe and fun 'thrill-seeker' free-fall experience. As a thought experiment, would it be viable to convert a large building into such a facility? the snow laden fir trees would have to be a cleverly designed alternative: perhaps foam trees with fine upper branches and larger boughs or palm or fern fronds to gently decelerate a human participant. The model should allow overall hight of building to be varied and the ratio of free fall height to height of deceleration zone to be varied. Thus the participant takes the lift to the top floor, walks out on a gangway then leaps carefree into the void space below, briefly experiencing the sensation of free-fall before entering the upper surface of a specially designed 'forest' to slow his fall gradually and gently. After the deceleration he will be deposited unharmed to the ground floor (the forest floor). Obviously, deceleration would have to be gradual. I imagine that the forces experienced whilst jumping on a trampoline would be a good starting point to define reasonable comfort, say 2 or 3g?) Specific questions: given the practicalities and limited height available, over what distance should the deceleration take place to avoid forces greater than 3g, would it be a linear deceleration or exponential? how much heat energy must be absorbed for a 100kg participant? Assumptions: terminal velocity will be achieved after a 450 m, 12 second fall. If space is limited and the free-fall height is 50 meters what height is required for the deceleration zone? As kinetic energy would be transfer to heat, would the 'trees' have to be wet to dissipate this energy and avoid heat burns? or would there be a better method so that participants can leap in everyday clothes?

 

[quickquestion]

This seems like a liability waiting to happen. There is too much randomization going on...the twists and turns of the body could easily slip and possibly miss any contact on the branches. Or the body could bounce in an odd way and miss some of the branches.

 

[David Thomson]

Agreed, the canopy would have to work to arrest the fall if the impact was feet first or, like a diver, finger tips down. I think it is still possible to have a very dense system of artificial tree-like branches, twigs, leaves etc to catch all and slow down the movement. Bags of polystyrene beads are used for emergency fall arrest in the construction industry. This would stop a falling person in 100 cm or so or in a shorter distance if landing with a large surface area. I'm thinking of something that would stop a fall by friction over a distance of say 20m. For design purposes the object to be stopped could be approximated to a 100 kg rectanguloid with a downward facing surface area of 1/3 meter square

 

[quickquestion]

Thing of it is, falling into bags of beads has completely different principles than falling into a tree. Falling into beads is similar to falling into a pillow. The first layer offers low resistance, slowly offering more resistance as the pressure increases into deeper and denser layers. This causes a smooth and comfortable deceleration.

The problem with trees instead of trampolines, is that the amount of resistance on the branches is not uniform. Near the edges you get low pain (until you ultimately hit the ground). Low pain and low resistance...(which is bad because it won't slow you down much and you will hit the ground hard.) Near the trunk you get thicker branches as well as less bending = more pain. Basically, you want to hit the middle of the branches which have medium bending and thickness. But if you make the thickness of the branches uniform, you still get non-uniform behavior because of the bending, it will bend more at the end than the middle. Basically it would be gambling with someone's safety, which, depending on the person, may be entertaining for them but not entertaining once you see the legal fees.

For design purposes the object to be stopped could be approximated to a 100 kg rectanguloid with a downward facing surface area of 1/3 meter square

Such approximations will do more harm than good. The human shape is various ellipsoidal and cylindrical shapes, branches are various cylindrical and cone line shapes. This creates numerous chances of surface rolling/surface normal deflection/slipping and unpredictable or billiard style collision scenarios.

 

[David Thomson]

A fir tree is conical and displaces you increasingly away from the trunk as you descend, so after the initial impact on the most slender and flexible of twigs at the absolute top of the tree, you would be directed downwards and sideways towards the more flexible ends of each branch as you descend, each one bending and absorbing energy and depositing you onto the next branch down, each time a little further from the trunk. i was of course thinking of taking the basic design of a tree and optimising its performance and increasing safety. No pain involved. So, using soft foam with a bit of friction on the surface, a bit but not too much. The foam may have a springy metal interior to absorb energy. The whole thing designed to take you from free fall speed to zero over say 20m, absorbing all the kinetic energy smoothly to minimise the deceleration into a gently stopping force. i.e. a composite metal/ foam engineering solution. You might wear safety googles but that's all?

 

[quickquestion]

The position of the body would be very important for this. A flat belly position would be the most optimal, but if they enter a dive-position or feet-first position they could either miss and/or overpower the branches. The branches would have to be a delicate balance of firm but also flexible. If the branches are too firm, you end up with bumps and bruises but less chance of a fatal fall. If the branches are too soft, you end up with no bumps and bruises, (except for the ones at the end of the fall where you break every bone in your body.)

 

[David Thomson]

you are right, its a design issue. It would have to be designed such that someone diving in at any point in the canopy would be delivered safely to ground level. I think that may be possible if there is a circa 20m depth in which to bring the person to a stop. It would be fantastic to create such a material that you jump into fearlessly. A softer landing than water, so absolutely ok to belly flop or dive. The purpose being to experience the sensation of fee fall without any equipment and a guaranteed soft landing.

 

[quickquestion]

Something like rubber would flop too much for the branches. Your idea, having a metal surrounded by soft foam is a good idea, BUT it would have a problem. A thin metal wire is able to bend without much effort, but the problem is once it bends it stays bent, AND when you bend it too many times it breaks. You'd have to replace all of the branches after each participant, unless you can think of a better idea. Depending on the thickness of the metal, you may not be able to notice the structural damage to the beam, it may "appear" just fine, functioning like a diving board after dozens of runs, until one unlucky contestant gets the "hot potato" of it.

 

[Bandersnatch]

Can you apply the SUVAT equations to the problem? Assume the body experiences evenly spread constant 2g net deceleration (i.e. 3g 'felt') from 50m/s terminal velocity. This will give you the minimum height of the deceleration zone (and it's not insignificant).
That'll be the lower bound, since you'll need to increase the height to compensate for the deceleration sometimes being below the specified maximum.

 

[quickquestion]

2 g's of acceleration is about 20 m/s. Terminal velocity is 54/ms but for safety reasons we can round up to 60 m/s. So it would take 3 seconds to go from 60 to 0/ms.
So first second would be a distance of 40 m/s, second second would be 20/ms, 0th second would be 0 m/s. My math could be wrong since I never learned SUVAT, but the distance would be at the minimum about 60 meters.
EDIT: According to SUVAT the distance would be 90 meters, so at minimum it is definitely 60 meters. I just learned SUVAT today so I could be wrong about this. My guess is, the actual distance in the first second is more than 40 meters, because the actual speed isn't 40 m/s until the end of the second. Thus my first calculations are probably off, but I'd say the distance can be no less than 60, and if I did SUVAT right, no less than 90 also.

 

[Bandersnatch]

Take the time as calculated above, and use $0=\frac{1}{2}at^2-V_0t+r_0$. Solve for $r_0$.

 

[quickquestion]

Take the time as calculated above, and use $0=\frac{1}{2}at^2-V_0t+r_0$. Solve for $r_0$.

Wikipedia says r0 is initial position, and r is final position.

 

[Bandersnatch]

Yes. The initial position is the top of the deceleration contraption, and r is the ground level, hence set to 0.

 

[David Thomson]

Thanks. In round numbers, assuming deceleration at 2g ( -20 m/s2) the stopping distance is 100m. in practice it would be best if this was non-linear, i.e. a slower start ramping up to 3g so the average would be 2g.

In practice I think it would be better to enter the de-accleration zone at a speed of 30 m/s and the corresponding stopping distance would reduce to 45m.

So, to make this happen i would need a very tall building and to build a fake forrest of 40 or so closely packed 'trees' each 45m high. Assuming the friction heat is not a problem, i can invite groups of guests to jump off and experience the thrill of free fall for 5 seconds or so with the landing impact being no worse than trampoline type forces.

 

[quickquestion]

In my mind, the way I envisioned it was 2 or 4 very large trees, creating a sort of "net" or trampoline with their shared branches.

40 trees scattered about would probably lead to someone being impaled on the top of one of the trees. Unless you mean the 40 trees are arranged in a large circle and you fall towards the center of the circle.

 

[Bandersnatch]

Or, you could save millions in investment and likely settlements for bodily damage, and purchase a bungee jumping gear.

 

[David Thomson]

I'm thinking that the decelerating ' stuff' is a circle about 30m in diameter absolutely stuffed with a construction of suitable material (doesn't have to be trees) so its homogeneous and you can enter it at any point. It has a tendency to upright you on the way down but does not matter if you land sideways. The forest floor could be pillows just in case.

 

[David Thomson]

bungee jumping is a good alternative there are accidents tho, and its not so good for corporate events. I want 10 people leaping off the gantry together in evening wear and for it to be absolutely safe and foolproof by good design. i.e. test it with crash test dummys thrown in at all sorts of directions. With good design it should be impossible to injure yourself. ?

 

[quickquestion]

I'm thinking that the decelerating ' stuff' is a circle about 30m in diameter absolutely stuffed with a construction of suitable material (doesn't have to be trees) so its homogeneous and you can enter it at any point. It has a tendency to upright you on the way down but does not matter if you land sideways. The forest floor could be pillows just in case.

Well, the radius of the "max safe zone" would have to be the maximum running speed of a human, multiplied by the time it takes to fall from the ceiling+maximum human jump height -45m.

The other problem is, a trampoline uses a net which is a grid. With tree branches, the distance between branches spans farther the farther you go, because it is angle-based. That's a lot of irregularly shaped crevices.

 

[David Thomson]

Hi, Yes. There would be a lot of design and testing involved. It has to go from snow-laden fir trees to something man made and reliable and suitable for 1000's of operations. Maybe I should look elsewhere in nature, sea anemones maybe?

 

[quickquestion]

Take the time as calculated above, and use $0=\frac{1}{2}at^2-V_0t+r_0$. Solve for $r_0$.

Ok so the value I got for r0 is -72? Did I do this right? My math was

0=.5*20*(2.7*2.7)-0+r

I got 2.7 from 54(free fall)/20(2 g's).

 

[quickquestion]

Hi, Yes. There would be a lot of design and testing involved. It has to go from snow-laden fir trees to something man made and reliable and suitable for 1000's of operations. Maybe I should look elsewhere in nature, sea anemones maybe?

That shape form seems more feasible, less gaps and less margin of error. You'd have to find some kind of sturdy foam that doesn't sag at length.

 

[David Thomson]

i got

0=1/2*10*3*3 -(54*3) +r0

so -r0 =45-150

so r0 =100 in round numbers

 

[quickquestion]

i got

0=1/2*10*3*3 -(54*3) +r0

so -r0 =45-150

so r0 =100 in round numbers

The time interval was 3 when I simplified the terminal velocity to 60, but when you use a terminal velocity of 54 it should be 2.7.
Also, v0 is initial velocity, it should be zero. You put (54*3) as v0*t.

 

[David Thomson]

im not sure. I think v0 is the velocity gained by fee fall, i.e. the velocity before deceleration starts?

But i may be confused. Maybe the advisor will be back to sort it out?

 

[Bandersnatch]

Also, v0 is initial velocity, it should be zero. You put (54*3) as v0*t.

No, David has it right.
It shouldn't be 0. It is the velocity at the top of the contraption, i.e. when the person is falling at full speed and begins decelerating.
The way you wrote it, you will get the same numerical result, but with the opposite sign, and its meaning will be different:

Spoiler

Sign convention is + means up, - means down.
Since 0 is the ground level, $r_0$ with a negative sign means below the ground level. The meaning of the equation the way you decided to write it is:
The skydiver has 0 velocity at some distance $r_0$ below ground level and begins accelerating upwards until he reaches the surface, at which point his final velocity is $V=at$, i.e. 54 m/s.
It's more of a catapult than an arrester this way.

 

[quickquestion]

No, David has it right.

It shouldn't be 0. It is the velocity at the top of the contraption, i.e. when the person is falling at full speed and begins decelerating.
The way you wrote it, you will get the same numerical result, but with the opposite sign, and its meaning will be different:

Oh, you're right. Duh moment.

Well in that case, v0*t should equal (.5*at²)*2 so the equation can be simplified to
r0=v0*t/2, or something like that.