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Skydiver falling parallel to the top of the slide at 90 degrees

Please forgive my lack of artistic ability, but here's my question: Consider that a skydiver, without using his parachute, were to fall exactly parallel to a giant curved slide that starts at $90\,^\circ$ perpendicular to the ground and gradually curves until it is parallel to the ground. Can he survive?

My thinking tells me that if I stood at the top of the slide and slid down, making sure to keep contact with the slide, I would (if the top of the slide was high enough) eventually get to almost terminal velocity, yet when the slide starts to curve I would begin to feel an increase in G-force and friction, but no impact and thus would survive.

So then, if I were to jump directly above the slide given that I had enough time to adjust myself to be perfectly aligned with the slide as it started to shallow, (or even better, if I was able to have my body or part of my body scraping the slide) the impact when the slide moves from 90 degrees to 89 degrees would be soft enough for me to survive - and so forth until I'm actually sliding and no longer falling with the slide.

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The answer is Yes and your thinking is correct.

You try to differ between impact and sliding on a curve. In fact the impact is just a sudden large force, while a curved (e.i. circular) motion similarly applies a force, just much smaller but also over a longer period of time.

The key in surviving any fall is to reduce the force on your body at "impact". A pillow does that. A curved slide does that. And they both do it by extending the impact duration. Remember first Newton's 2nd law:

$$\sum\vec F=\frac{d\vec p}{dt}\approx \frac{\Delta\vec p}{\Delta t}$$

Smaller momentum change $\Delta \vec p$ (that would be smaller speed or lighter skydiver) or larger duration $\Delta t$ will reduce the total force. A soft material like a mattress will extend $\Delta t$. And a curved slide will as well, as you explain it yourself, cause the momentum change over a much longer period of time.

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    $\begingroup$ @Dargscisyhp It's actually that, on impact, different parts of the body are accelerated differently. Let's (gruesomely) imagine that you hit the ground at $50{\rm m\,s^{-1}}$ feet first. The soles of your feet hit first, and are slowed swiftly. But there is no slowing force on the rest of your body aside from that which is transmitted upwards. For an instant, the rest of your body is heading towards your feet, and your shins push back because they can't go anywhere. Depending on the elastic constants, this means that there are huge compressive stresses as the shins transmit their .... $\endgroup$ – WetSavannaAnimal Aug 13 '15 at 23:55
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    $\begingroup$ @Dargscisyhp ... decelerating force. They give way and are crushed. Likewise we get stresses in the connective tissue as the force is transmitted upwards; recall that it is the body tissue and nothing else that is decelerating the parts of the body above that tissue. Contrast this with the case where you imagine all the body to be electrically charged with charge density exactly in proportion to the mass density. Then you accelerate the body with a whopping electric field. In this case, all parts of the body are always accelerated at exactly the same rate and there is no stress between ... $\endgroup$ – WetSavannaAnimal Aug 13 '15 at 23:59
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    $\begingroup$ @Dargscisyhp any two parts of the body. You could accelerate arbitrarily fast with this setup and feel nothing unusual. $\endgroup$ – WetSavannaAnimal Aug 14 '15 at 0:00
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    $\begingroup$ @Dargscisyhp See my follow up question and answer here $\endgroup$ – WetSavannaAnimal Aug 14 '15 at 0:38
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    $\begingroup$ My favorite example of this principle is falling off a building by riding the elevator down. $\endgroup$ – imallett Aug 15 '15 at 23:36
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Let's make life easy for ourselves by assuming that the slide is an arc of a circle:

Slide

We also assume the slide is made out of something with a very low friction, so the skydiver maintains a constant speed $v$ all the way round. The reason that using an arc of a circle makes life easy is that the acceleration felt by the skydiver is simply:

$$ a = \frac{v^2}{r} $$

where $r$ is the radius of the circle (the length of the dotted lines in the diagram). The acceleration is shown by the red arrow, and always points towards the centre of the circle.

Let's take the terminal velocity of the skydiver to be 50 m/s, and let's suppose they can survive an acceleration of 10$g$ (98.1 m/s$^2$). If we rearrange the equation to get:

$$ r = \frac{v^2}{a} $$

then to achieve this the slide would have to have a radius of 25m, which is actually quite reasonable.

In practice there would be a few problems of course. You'd have to land in exactly the right place at the top of the slide, and after you'd left the slide you'd still be travelling at 50 m/s but horizontally rather than vertically.

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    $\begingroup$ @MegaMark: I use Google Draw to do the diagrams. It's good for basic diagrams like this one, and it's free! :-) $\endgroup$ – John Rennie Aug 13 '15 at 16:44
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    $\begingroup$ I don't think a circle is the optimal shape. You'd want to use something more like an exponential curve to decelerate the skydiver at a constant rate while their velocity decreases. $\endgroup$ – Schwern Aug 13 '15 at 20:07
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    $\begingroup$ @Schwern: as Chris says, it depends what you're trying to achieve. Using a circle converts vertical speed to horizontal speed with the minimum possible force on the skydiver. The skydiver can then slide along the runway (or whatever) shedding speed at whatever rate suits them. If the aim is to bring the skydiver to rest at the end of the slide then you're correct that the shape of the slide would be different. Actually that would be quite a complicated calculation. $\endgroup$ – John Rennie Aug 14 '15 at 6:07
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    $\begingroup$ Come to think of it, a one-sided "funnel" (on the "opposite" side of the catching-curve-device) would probably be most practical, "easiest" to build. Divers have fairly good control of xz vector these days, so it would more be a matter of hitting that "wall"; it would push you across and then you'd perfectly hit the top of the curve-device, pointing in-to the curve-device, as you must. $\endgroup$ – Fattie Aug 14 '15 at 11:43
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    $\begingroup$ How about a semicircle followed by a short vertical wall, so the person would be launched upward on the other side? The height would be much less than the original fall (due to the energy lost in air friction at terminal velocity) and the person could simply oscillate between the two sides until all the energy had been gradually lost through additional friction. $\endgroup$ – supercat Aug 15 '15 at 18:13
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I have slid down a much smaller version of this at Burning Man. Paha'oha'o was a 30 foot tall volcano art piece which you climbed and then "sacrificed" yourself by dropping into a pit featuring a slide just like you mention. The drop features a 10 foot free-fall, just enough to take your breath away, after which the careful curve of the slide gently catches you and spits you out horizontally... rather fast. Safety Third!

Paha'oha'o at Burning Man 2014. Picture by The Modern Nomad Picture from The Modern Nomad

It is much more intimidating at night, all lit up red and you can't see what you're dropping into.

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  • $\begingroup$ I've been on these in the UK as well with probably a greater total height. I've also (many years ago) been down a water-slide version - very fast, but with the same catching ramp at the bottom. Of course you're nowhere near terminal velocity after a few metres drop, unlike the skydivers. $\endgroup$ – Chris H Aug 14 '15 at 11:02
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Probably the closest to what you are asking about is the story of Ivan Chisov's survival (see Ivan Chisov); but there have been several other similar cases (see for example 10 Amazing Free Fall Survivors).

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    $\begingroup$ Cool. But I don't think the question is whether outlier events are possible, but whether it can be set up to where most if not all people landing correctly on a large slide would survive. $\endgroup$ – Michael Martinez Aug 14 '15 at 21:18
  • $\begingroup$ @MichaelMartinez Precisely. $\endgroup$ – MegaMark Oct 14 '15 at 19:39
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Assuming terminal velocity of 200 km/hour, the scenario seems equivalent to stepping out of a car that's travelling at 200 km/hour. In that case it's not the fall (hitting the road) that kills you, it's the friction (i.e. sliding or tumbling along the road).

There might be a minimum of friction initially (when you're falling parallel to a vertical wall) but that increases towards one gravity as the wall becomes horizontal.

It might work if you timed it with a very long near-vertical to bleed off speed ("bleed" being more than a metaphor); you couldn't land on a sled (because you'd hit the sled at 200 km/hour) but perhaps you could bring a sled with you (instead of a parachute) when you fall, e.g. harnessed to your back, and land on it; or be wearing skis or roller skates.

I'm not sure that 'road rash' is survivable at 200 km/hour.

John's answer has 40m of ramp, if you assume the same deceleration 40m of increasing resistance can stop you safely.

I still think that that sounds as "safe" as coming off a motorbike at 200 km/hour, and sliding along the road for 40 m -- i.e., not safe at all. Presumably someone has already tried it?

My guess is, what if we assume that the skydiver is spherical and already spinning when they touch the wall (to minimize being scraped along the wall)?

Assume a sphere with 1 m radius (2 m diameter), and assume that this doesn't change the terminal velocity. The circumference (2 pi r) is 6 m, at 50 m/sec that's 8 rotations per sec.

8 rotations per sec and a 1 m radius implies a centrifugal force of about 250 g: at 250 g their head would separate from their neck, aren't I right?

Spinning like a human-sized cylinder with a diameter of only 1 m makes that worse (i.e. twice the centrifugal force, up to 500 g): you couldn't keep your arms and legs together.

IMO you need to be in the centre of a large sphere (20 m in diameter would be 1 rotation per second, which is only 4 g of centrifugal force at 1 m from the centre), or not rotate at all i.e. land on wheels or a sled.

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  • $\begingroup$ to your point... a bicycle would work, however it does pose it's own problem of keeping balance through the G forces and speed... How about a suit that is made to shield the friction?... Practically speaking there are a many reasons that this would be an incredible feat... Aligning yourself in free fall to the slide, not falling off the slide while it's still so vertical that even wind could push you sideways, surviving the G forces and friction, and of course... stopping at some point! $\endgroup$ – MegaMark Aug 13 '15 at 23:33
  • $\begingroup$ Maybe not just any bicycle but there is a least one that can do 200 km/hour without losing its tires: google.fr/search?q=rocket%20bike $\endgroup$ – ChrisW Aug 13 '15 at 23:44
  • $\begingroup$ Obviously something must be done to slow you after the ramp converts your velocity to horizontal. John's answer has 40m of ramp, if you assume the same deceleration 40m of increasing resistance can stop you safely. $\endgroup$ – Loren Pechtel Aug 14 '15 at 5:31
  • $\begingroup$ Is there any reason for the ramp to stop once it reaches horizontal? Upward travel can shed velocity very quickly by converting kinetic energy to potential energy, and any kinetic energy converted into potential energy won't get converted into heat. $\endgroup$ – supercat Aug 15 '15 at 18:15
  • $\begingroup$ A terminal velocity of 50 m/sec is about 5 second's-worth of acceleration in a vacuum (given that g has a value of 9.8 m/sec/sec); to lose that speed again (ignoring friction) you'd need to climb (formula is 0.5 * g * t*t) about 120 m. $\endgroup$ – ChrisW Aug 15 '15 at 19:10
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Yes. In fact it would be better to imagine that you skydive towards a "track" that you can strap a "chair" onto, and then the chair is stuck on the track. the acceleration to keep you in a circular orbit of radius $R$ is only $v^2 / R;$ with terminal velocity being about $v \approx 56 \text{ m/s}$ a $1~g$ acceleration will be accomplished by a radius of curvature of about $300 \text{ m}.$ That's about as high as a modest skyscraper (the Empire State Building is I think $400 \text{ m}$ or so), but you can lower it if you're willing to accept, say, a 2g or 4g acceleration.

The real problem is going to be that you need your chair to have brakes as well, otherwise when you get to the ground you'll still be going at terminal velocity.

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  • $\begingroup$ When you say I'll "still be going at terminal velocity", I'm assuming you mean that I will be traveling horizontally at the same speed I was originally falling, but no longer travelling vertically? $\endgroup$ – MegaMark Aug 13 '15 at 16:41
  • $\begingroup$ @MegaMark Indeed. Your acceleration splits into two components, one of which points perpendicular to how you're going, changing your direction, the other points parallel to how you're going, speeding you up or slowing you down. $\endgroup$ – CR Drost Aug 13 '15 at 16:42

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