# Skydiving without a parachute or a wingsuit?

Recently I heard that a skydiver plummets 7600 meters onto a 30m$\times$30m net without a parachute or a wingsuit. I searched online and found a post, which said that a wingsuit is necessary. I have no idea about these technical terms and formulae. However, in this piece of news, no wingsuit but a net. I wonder how it's possible to

1. land in a 30m$\times$30m area (note the existence of wind),
2. survive under the terminal speed 55 m/s?
• Yes I'm skeptical of this too - how it is possible to aim so precisely. I mean, I've no doubt the stunt was truly done, but the guy always has an emergency chute with him and is said to need it many times. So I'm also wondering how many tries he's had to make to hit the net. Presumably his emergency chute can deploy in a few of hundred meters, so with experience I think he could fairly safely decide when his aim was good enough to hit the net (which is pretty big from 300m away) with enough time to abort if not, so he could "fairly safely" simply keep trying the stunt until he got lucky. – WetSavannaAnimal Jul 31 '16 at 13:50
• As for the net: apparently materials engineering is good enough to break the fall of someone like this. – WetSavannaAnimal Jul 31 '16 at 13:51
• Article with video: telegraph.co.uk/news/2016/07/31/… – K7PEH Jul 31 '16 at 17:01
• yahoo.com/news/… – Qmechanic Aug 1 '16 at 11:28

1. Presumably, experienced sky divers become good at steering with their body. Note that they routinely perform quite high-precision maneuvers relative to each other (whilst in the air, before opening a parachute). Small (short) gusts of winds will have limited effect, whilst a constant blow of wind is easily compensated by someone actively aiming. So whilst I certainly can't imagine doing it myself, it's not out of the realm of possibility that a trained expert can land in a designated 30 x 30 m area.

2. Survival is largely a function of force or acceleration. Assuming the net to be specially engineered to provide a constant braking force would allow us to use the formula $2as = v^2$ to estimate the necessary elongation $s = \frac{v^2}{2a} \approx 5 \,\mathrm{m}$ of the net given a maximum tolerable short-time acceleration of (say) $a=30g \approx 300 \,\mathrm{m/s^2}$. This is a minimum value regardless of the actual force/elongation curve and it is only an order of magnitude estimate. But considering the huge elongations possible with e.g. bungee chord, 10 or even 10s of meters of elongation are conceivable in a 30 by 30 meter net. Hence I conclude that (with very careful engineering and optimization), the stunt you describe may just be possible with a human stuntman.

• The height of the net was about 200 ft, from the Sky Diver link by @K7PEH. If he went down it about 3/4 of the way or about s=50 m the acceleration was closer to about 3g. He did need that longer elongation of the net. He didn't seem bothered at all by the deaccelerarion at the end. I thought 30g or much less makes fighter pilots faint, even in a short time. – Bob Bee Aug 1 '16 at 5:47
• Agree, especially with first paragraph. As a pilot, I'm used to people being incredulous about how precisely I can control the aircraft, and I'm sure the same is true of skydivers. Basically, there is a "target". It's the point on the ground that's getting bigger and isn't moving in any direction. If it's moving, you steer toward it until it isn't moving. What makes it wiggly is wind gusts. Civilian pilots can land on pretty skinny runways. Military pilots can land in formation, one on each side. (They don't like to, but they can.) – Mike Dunlavey Aug 5 '16 at 22:01
• @MikeDunlavey People keep incredulous, but one of the goals of science is to unravel the rationale of these counter-intuitive phenomena and put everything in an inferable, manageable and computable framework, I think. – Yai0Phah Aug 13 '16 at 1:17

This article discusses the actual phenomenon - the preparation, and a bit about the steering - that yes it was done with only a fairly normal looking suit without sails, although an oxygen mask was uses at first.

Here's a plot of a ballpark estimation of the trajectory, the math and a short Python script is included below. I adjusted the numbers a bit, estimating the time as 2min 8sec, neglecting the landing wasn't at sea level.

In a short pre-dive interview snipped shown in the video now linked in the article (thanks @K7PEH for the updated info) the diver says that at 25,000 feet he will be falling at about 150 mph (about 240kph, 67 m/s) and near the end where the air is thicker, 120 mph (about 190kph, 54 m/s). The scale height (change in altitude for a $\frac{1}{e}$ change of pressure) of the lower atmosphere is roughly 7600 m (it varies with local temperature) and the density roughly tracks that as well. 25,000 feet is also about 7600 m, that explains the oxygen mask in the beginning.

Wikipedia says that a skydiver in this orientation will reach a terminal velocity of about 200 kph (about 56 m/s).

If the person's mass is 75 kg and they are at terminal velocity, then the gravitational force

$$F_g = -mg$$

will equal the aerodynamic drag force which can be approximated as

$$F_D = \frac{1}{2} \rho v^2 C_D \ A$$.

With a density of air of about 1.2 $kg/m^3$ (it varies with altitude and temperature) and an area of 0.7 $m^2$ , and setting

$$F_g+F_D=0$$

once terminal velocity is reached (no further acceleration) I get a $C_D$ of 0.58, similar to the value of 0.5 used here.

More to read here and here and here.

Since it seems to have happened, it seems to be possible. Extensive preparation by trained and experienced people were needed.

def deriv(xv, t):

x, v = xv

rho  = rho0 * np.exp(-x/h_scale)  # atmospheric density
Fd   = 0.5 * rho * v**2 * CD * A

a    = -g + Fd / m

return [v, a]  # xdot, vdot

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

# these numbers have been fiddled with until the time of 128 sec was reached.
# this is just a ballpark estimat

alt     = 25000. / 3.3   # meters
h_scale = 7600.          # meters
rho0    = 1.22           # kg / m^3
g       = 9.8            # m/s^2
m       = 70.            # kg
CD      = 0.65
A       = 0.75           # m^2

xv0 = [alt, 0.]

t = np.linspace(0, 128, 100)

answer, info = ODEint(deriv, xv0, t,
rtol = 1E-10, atol = 1E-10,
full_output = True )

plt.figure()
plt.subplot(2,1,1)
plt.title('altitude (m)')
plt.plot(t, x)
plt.subplot(2,1,2)
plt.title('velocity (m/s)')
plt.plot(t, v)
plt.show()

• Is the motional equation something like $F_g+F_D=ma$, or an ODE like $\dot x=x^2+c$ where $x$ is proportional to the speed? This equation could be solved explicitly (with something like $t=\arctan(\dotsb)$). – Yai0Phah Aug 2 '16 at 2:12
• @FrankScience yep! It may indeed have a simple analytical solution. Go for it! I was shamelessly lazy here - I had a python IDE already open so I just made the plots quick to see what it looked like. – uhoh Aug 2 '16 at 2:37

If you watch the original Youtube video posted on July 30th, 2016 you would watch the entire 2 minute free-fall and the final landing in the net. Unfortunately, early this morning (July 31st, PDT), the Youtube video is blocked with copyright infringement.

However, there are other videos available if you search and I found one at: Sky Diver video

The article associated with the link describes how this one done using guidance indicators and of course the skill of the sky diver.

Late breaking Physics news on the jump dated August 2nd: Physics Description of Jump, Wired Magazine

## protected by Qmechanic♦Jul 31 '16 at 20:59

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