Mar 20 2012

SuperJump

Felix Baumgartner is attempting to break the world record for the highest skydive ever. He recently completed a test dive from 71,500ft (22km), putting him in the top three. The purpose of the dive was to test out all of his equipment before he attempts the record breaking dive of 120,000ft later in 2012.

The current record is held by Joe Kittinger, a US Air Force Colonel who jumped from 102,800ft in 1960. At that altitude the air is very thin and cold, so the equivalent of a space suit is required to survive the trip up and down.

Baumgartner, a 42 year old Austrian, said after his test jump that the cold was “hard to handle” and that they may need to work on that problem before the big jump.

When discussing the upcoming jump on a recent episode of the SGU we commented that the high altitude jump is very dangerous. Kittinger, in fact, spun out of control at one point and blacked out. The details, however, are a bit different from what we discussed.

Actually,  Kittinger made three jumps as part of Project Excelsior, two test jumps and then the third for the standing world record. During the first jump is when he spun out of control and blacked out. Here is a description from Gregory Kennedy:

As he prepared to jump, Kittinger noted the balloon had climbed to 76,400 feet, more than three miles higher than intended. Standing in the door of the gondola, he pulled the lanyard that activated the barometric release for the main canopy and the timer for the pilot chute. It took three tries to pull the lanyard clear. Unknown to Kittinger, the first pull started the timer. The result of this was the pilot chute deployed only 2.5 seconds after he left the balloon.

Without sufficient airspeed to create adequate dynamic pressure, the pilot chute flopped around in the thin air and eventually wrapped around his neck. He began spinning. At first, Kittinger could correct the spin, but soon he could no longer compensate for it and eventually blacked out. He didn’t regain consciousness until he was floating beneath the reserve parachute several thousand feet above the ground. The pilot parachute had not deployed the drogue because it was tangled around his neck, so the main parachute never deployed. After Kittinger lost consciousness, the spin continued and eventually reached 80 RPM. A barometric release triggered the reserve parachute at 10,000 feet, but due to the spin, at first it tangled around him. Beaupre had planned for such a contingency and installed the reserve parachute pilot chute with a break-away cord that would enable the canopy to clear itself. Thankfully, the reserve cleared itself at 6,000 feet and inflated.

It seems that the spinning out of control had nothing to do directly with the altitude, but rather the early deployment of the pilot chute. However, the thin atmosphere is what made the timing of the chute deployment critical and ultimately resulted in the dangerous spin out.

Our discussion received some feedback from a listener who is an experienced sky diver that led to an interesting physics question. His point is this – wind resistance is independent of altitude because it is determined by by the wind resistance force at terminal velocity. In higher thinner atmosphere terminal velocity is faster, and at lower thicker atmosphere it is slower, but these two factors exactly balance out and the resulting wind resistance will be the same – enough to counteract the accelerating force of gravity. The experience (and danger) to the skydiver will be the same.

I don’t quite buy this, however. My counter point is this – at higher altitude the atmosphere is thinner and terminal velocity is much faster. It is estimated that Kittinger went a top speed of 625.2 mph during his third and highest jump. Typical terminal velocity for a skydiver (depending on various aerodynamic factors, such as position) is about 117-125 mph. A skydiver in a bullet position can go as fast as 210 mph. So Kittinger was going much faster than that.

I agree that once a skydiver reaches a terminal velocity the force they feel is constant, regardless of speed vs atmosphere density. But as they descend into a thicker atmosphere the terminal velocity will decrease. This means that the skydiver not only is maintaining terminal velocity, they are slowing down to the slower terminal velocity. Therefore the forces on them would be greater. I could not find a figure for how fast Kittinger was going when he deployed the parachute, but peak velocity was reached at 90,000 ft, and so he was slowing down from there until his chute deployed at much lower altitude.

The question then becomes – how much greater are the wind resistance forces due to this deceleration and how does that affect the riskiness of the jump.

I have another question as well, related to Kittinger’s first jump where he blacked out, which was related to the high RPM of the spin. If the skydiver does go into a spin at higher altitude and therefore higher velocity, will the increased momentum mean that the spin is likely to be faster, increasing the risk of blacking out?

I don’t have any definitive answers to these questions. I have chatted with a couple of physicists about it, but no one seems to have a final answer.

So I open it up to my astute readers – how does the higher terminal velocity alter the danger and experience for the super high altitude skydiver? What risks is Baumgartner going to face, other than the oppressive cold? I will continue to search for some appropriate expertise in the meantime, but it seems like a fun thought experiment, so have at it.

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19 responses so far

19 Responses to “SuperJump”

  1. petrucioon 20 Mar 2012 at 5:39 pm

    Why not use SI metrics?

    https://www.notatourist.sg/sites/default/files/images/metricvsstandard.jpg

  2. addisontreeon 20 Mar 2012 at 10:42 pm

    One thing that really surprised me about free fall was how easy it was to spin. Once you were in equilibrium (i.e. stable at terminal velocity) any small asymmetry (even an unintentional one) could set you spinning. It was like being the puck on an air hockey table.

    When learning how to sky dive, one of the exercises is to get stable, intentionally initiate a horizontal spin and go back to a neutral position. Even though you are neutral you often continue to spin for more than a revolution or two before coasting to a stop (rotationally speaking). Seems like the thinner air at higher altitudes would only increase the ease of rotation, lengthening the period of time you would coast once a rotation started.

    It’s certainly true that the higher the altitude the more time there is to build up angular momentum. There must be a “terminal RPM” but I would think it would be more than 80 RPM. That “terminal” rate of rotation might be almost impossible to reach if you were only falling 12,000 feet. Falling from 120,000 feet, though, would leave a lot of time for spin to accrue if something went seriously awry.

    (For the SI metrics crowd: “http://www.wolframalpha.com/input/?i=12000+feet+in+meters”.)

  3. Chewon 20 Mar 2012 at 10:43 pm

    Due to the vagaries of the density and temperature of the atmosphere with changing altitude there will be a time, about 6 seconds after reaching maximum velocity, where he should experience more drag deceleration than gravitational acceleration (by about 1.2 m/s², i.e. drag will decelerate him by 11.0 m/s² while gravity is accelerating him by 9.8 m/s²). From then on drag minus gravitational acceleration will fluctuate several times as he transitions through the upper and lower stratosphere and troposphere.

    time s altitude km drag minus gravity m/s²
    41 28 1.2
    55 25 0.6
    70 21 2.8
    78 19 2.7
    82 18 2.9
    242 3 0.2

    So the maximum decelerating force will occur at 18 km.

    These fluctuations are caused by density but also by the temperature, which determines the speed of sound. The coefficient of drag increases as your Mach number gets closer to 1.

  4. BillyJoe7on 21 Mar 2012 at 5:47 am

    petrucio: “Why not use SI metrics?”

    Probably because it’s easier to say 120,000 feet than 36,576 metres. :D
    (It also sounds higher ;) )

  5. ccbowerson 21 Mar 2012 at 12:03 pm

    “It also sounds higher”

    Maybe 3.6576 × 10^13 nanometers sounds more impressive?

  6. tmac57on 21 Mar 2012 at 1:53 pm

    “Maybe 3.6576 × 10^13 nanometers sounds more impressive?”

    How about 22 folds of a piece of paper?

  7. tmac57on 21 Mar 2012 at 2:26 pm

    Maybe more like about 28.35 folds ? My math, she is not so good :(

  8. HHCon 22 Mar 2012 at 2:21 am

    Baumgartner is 42. Kittinger was 32 when he set the record in top shape as part of the NASA Air Force project. Kittinger is an adviser to this 2012 jump.

  9. joe_giorandinoon 23 Mar 2012 at 11:15 am

    If anyone can tell me, what episode of SGU did they talk about HALO/HAHO skydiving? I don’t remember it, so I think I missed that week…I was sure I hadn’t missed an episode in more than a year.

  10. cason 24 Mar 2012 at 10:44 am

    Average forces are not the way to consider this problem. It’s turbulence, which is much higher in the high velocity case. The Reynolds number is very different in the two cases.

  11. dlbraunon 24 Mar 2012 at 4:41 pm

    Hi Dr. Novella,

    I believe your skydiver is correct. I am ( almost ) a pilot. Consider this: With standard temperature, pressure, and no wind: I take off from sea level and adjust the attitude of the aircraft so that the indicated airspeed is 100 kts. I maintain 100 kts until I reach an altitude of 10,000 feet. Indicated airspeed is a measure of the dynamic pressure on the wing. By maintaining 100 kts during the climb means that the dynamic pressure experienced by the wing has remained constant during the climb. As you climb the air pressure decreases. In order for the dynamic pressure experienced by the wing to remain constant the velocity needs to increase. At 10,000 ft 100 kts relative to the ground may be 130 kts. If I shut the engine down and adjust the attitude of the aircraft so that the indicated airspeed is a constant 100 kts the speed of the aircraft will decrease relative to the ground as I descend into the denser air.

    dynamic pressure = 1/2 * rho * V^2

    where

    rho = air density
    V = velocity

    A quarter century ago I was a skydiver. I do not remember any difference in pressure between 4000 and 14000 feet.

    Cheers,

    David

  12. locutusbrgon 26 Mar 2012 at 10:04 am

    Not being a physics expert it is easy for me to speculate about this. I think Steve your though process assumes that that the speed and density quickly interact. The question that is critical for your theory is how quickly does the air density go up as you descend? Despite the greater velocity in the thinner atmosphere, you will not suddenly slow down as you enter the increasingly dense atmosphere. My thought would be gradual variation in velocity as density increases. Probably not noticeable unless there are significant changes or very high velocity. Such as a returning space vehicle. If you were talking about significant slowing in the dense atmosphere I would think there would be heating and cold would not be the problem but heat instead. Again not doing the math leaves me unsure. I still think that the fact that the fall is a cold experience would lead me to belief that the transition to ever ever increasing density and pressure is gradual or undetectable to human senses. If velocity breaks slowly in ever increasing amounts of atmosphere it could possibly be gentle. Using your example, head down diving greatly accelerates the diver but flattening out is not violent. Pulling the chute is. A demonstration of rapid verses slow changes in velocity.

  13. Pabnauon 27 Mar 2012 at 9:05 am

    Hey,

    A couple thoughts from a physics minor. I’m going to avoid the numbers slog involved with deceleration and turbulence which are mentioned earlier and just consider velocity. Remember that velocity and momentum are both dependent on your inertial frame, so its not as though you are innately going to spin faster just because you are moving faster relative to the Earth. In terms of relevance, all it means, is that the air molecules in the atmosphere are traveling faster relative to you. However, this should be taken into account by the amount of force/pressure you are experiencing (instead of thinking of the diver as moving, it may be useful to think of him being in a wind tunnel). The other relevant difference about being high up is that there is less atmosphere to apply drag while you are spinning, this should make it easier to start spinning, harder to stop, and you would tend to spin faster (if you were in a vacuum being pelted by small pellets, you’d inevitably start to spinning and wouldn’t be able to stop).

  14. snufflufikiston 29 Mar 2012 at 11:29 am

    Dr. Novella,

    I think your skydiving friend is correct.

    One point that is important to point out. During a long fall, there is NOT a balance of forces. There is a near balance. From statics, sum of forces = m * acceleration. As the density of air increases during a long fall, the terminal velocity drops. This means that during the course of the fall, the skydiver is experiencing a slow deceleration, from the ~1000km/hr when he first reaches terminal velocity, to the ~200km/hr when he deploys the parachute. The total velocity change is 800km/hr or 222m/s. Imagine he falls for 5 minutes, or 300 seconds. Average deceleration is (222m/s) / (300s) = 0.74 m/s^2 or 7.6% of acceleration due to gravity (9.81m/s^2)

    So in reality, if you average it over the course of the fall, the skydiver is experiencing 1.076N of drag force for every 1N of weight (or if you want it in pounds, a 200lb man would experience an average drag force of 215.1 lbf. This extra drag force is the result of continuously falling through higher density air.

    Now of course, the next question would be: what is the variation of this extra drag force? I’ve only calculated the average. The plot of average drag force is influenced by 3 related things:
    1. atmospheric density as a function of altitude. Highly asymptotic. (http://en.wikipedia.org/wiki/File:Comparison_US_standard_atmosphere_1962.svg)
    2. terminal velocity as a function of atmospheric density. This is an inverse root function. (http://en.wikipedia.org/wiki/Terminal_velocity#Examples)
    3. how is the rate of change of density during the fall affected by the fact that your velocity through the atmosphere is changing too.

    #1. will skew the force to be higher at the lower altitudes because density changes so quickly near sea level.
    #2. will balance out the vertical axis of #1 because you find it’s not rate of change of density, it’s rate of change of the square root of density that matters
    #3. will balance out the horizontal axis of #1 because metres are flying by at an increased rate at the start of the jump. This will squish the long tail of the asymptote while extending the slower changes near sea level because you are traveling slower.

    When you balance these factors out, I’d be surprised to see more than a 3 or 4x difference in “extra” drag force during the course of the fall. Meaning that to average out to 7.6% of the force of gravity, you could have something like 4%-16%, and the changes will be very gradual, occurring over 5 minutes.

    In short, while you are traveling faster at the higher altitudes, the density is lower (of course) but more importantly the rate of change of density is also lower! The deceleration as a function of altitude is faster at lower altitudes, but it’s offset by the fact that your altitude is changing at a slower rate in the lower altitudes. In general, you are experiencing an average 1.076g of drag force (7.6% higher than gravitational force) which causes a slow deceleration. The change in drag force (if any) will be gradual and IMO no more at any given point during the fall than a fraction of the force of gravity.

    I have nothing to back up my answers, but I am a mechanical engineer (nothing to back that up either ;)

  15. toddzon 30 Mar 2012 at 6:13 am

    Not a physics expert, but a skydiver here, so these are just instinctive thoughts coming from my own (below 25,000 feet) experience. My feeling has always been that the risks in these jumps are more about the near-vacuum cold than about any extraordinary difficulty in performing the skydive. I expect that once terminal velocity is reached, deceleration from thickening air will keep him near equilibrium. Common head-down jumpers going 200mph must transition to a belly-to-earth position and decelerate to the 120mph range to safely deploy a parachute, and this deceleration happens within a few seconds upon changing body position. If Felix is in a belly down position the whole time, then I’d think he’d naturally arrive at 120 by pull time.

    As far as spins go, spinning is caused by an uneven body position (or malfunctioning gear) catching air. If that air is thinner but faster than usual, I would think the net effect on the part catching air would be the same and induce the same kind of spin. I don’t see how it would spin up faster or be any more difficult to recover from at high altitude than at low altitude. Any extra speed a spin would achieve would be a function of the time available to accelerate (though as an earlier commenter said, it seems there must be a “terminal spin rate.”) And I suppose an uncorrected spin even at a low rate would have the potential to last longer, and possibly cause more cumulative blood pooling and whatnot.

    In short, my main concerns would be oxygen and protection from cold. If someone gave me a proven spacesuit and a balloon, I’d be next in line for this jump.

  16. BenNon 31 Mar 2012 at 4:28 pm

    Hi Dr. Novella,

    I e-mailed the SGU on this topic, but I also wanted to discuss this with your readers. I apologize for the redundancy.

    I think that you are correct, and that the resistance encountered by the skydiver will vary with altitude. To support this claim, I wrote out the differential equations, researched the appropriate parameter values, and solved numerically. This model takes into account variation of air density, speed of sound, temperature, and gravitational acceleration with altitude, as well as qualitative differences between the troposphere, tropopause, and stratosphere. It also takes into account the way that drag coefficient varies with velocity in transonic/supersonic regimes. There is still plenty of uncertainty in this crude model, though. For example, there is not a whole lot of empirical data out there about the drag coefficient for a human-shaped object falling at transonic speeds in the stratosphere :)

    Here are the details and results:
    http://www.math.unl.edu/~bnolting2/SGUSkyDive.pdf

    The overall message is consistent with the argument you made above. For any particular altitude, an object falling at terminal velocity would experience air drag exactly equal to the force of gravitational acceleration. However, the terminal velocity changes every instant as the skydiver loses altitude. The skydiver’s velocity chases the terminal velocity, but the latter is a moving target.

  17. ianbasseton 01 Apr 2012 at 10:31 am

    Hi Dr. Novella,

    You have asked 2 very tricky questions, so let’s have an approach to the first one.
    How does the higher terminal velocity alter the danger and experience for the super high altitude skydiver?
    We are dealing with a skydiver that is going to jump from an altitude of 120,000 ft, about 32.57 km above sea level. Up there the atmosphere has a lower density than at sea level, in other words “air is thinner”. Also gravity is less up there, so the skydiver would experience less acceleration and as he’s approaching to ‘earth’ the acceleration -gravity- will increase. But as air density is increasing as approaching to the ground, the diver will experience an increase in drag force.

    I think that terminal velocity is “the highest” velocity the diver can reach before loosing control due to an alternating force, perpendicular to the air flow (generated by the Karman vortices). This force could be high enough to make the diver’s vibrate. As the diver’s position is changing, he’d start spinning. Thus terminal velocity maybe an estimation of the maximal velocity the diver can held before changes in body position would lead to spinning. This velocity is function of the diver’s geometry, atmospheric conditions (air density and viscosity) and the Reynolds number.
    http://www.grc.nasa.gov/WWW/BGH/reynolds.html

    From this perspective, exceeding the terminal velocity any changes body position would lead to spinning. The terminal velocity would prevent a catastrophic failure but may avoid the diver from reaching a higher velocity. I’m pretty sure that there is a safety factor included in the estimation of the terminal velocity =).

    The second question: What risks is Baumgartner going to face, other than the oppresive cold?
    As mentioned before there is a risk of spinning. The diver has to maintain certain body position when falling to reach the terminal velocity and during deceleration. Furthermore, there the risk of spinning out increases during changes the body position. The diver is depending on the estimations for maximum velocity and safety factors considered by the engineering staff and his ability and strength to maintain and change the body position.
    Other risks are the low barometric pressure (so oxygen may be supplied), changes in blood pressure due to deceleration, UV radiation, timers and altimeters accuracy and reliability.

    My response is based on the assumption that the diver will “vibrate” as the result of the air flow i.e. vortex shedding. At some point this vibration may increase in amplitude making the diver loose control and spin out =( . Also I’m not considering a lateral air flow.

    Cheers,

    Sebastian

  18. Calli Arcaleon 03 Apr 2012 at 4:47 pm

    Sebastian,

    Regarding the low barometric pressure, not only is oxygen supplied, but he must wear a pressure suit as well. One can tolerate brief exposures to near-vacuum without a pressure suit, but not prolonged exposure, even with oxygen supplied under pressure (as on a fighter aircraft). Supplemental oxygen under positive pressure is sufficient up to about 50,000 ft — the Armstrong Line. Above that line, you need a pressure suit as well. I believe Baumgartner’s suit is a full pressure suit.

  19. DevoutCatalyston 09 Apr 2012 at 9:55 pm

    “…I could not find a figure for how fast Kittinger was going when he deployed the parachute, but peak velocity was reached at 90,000 ft, and so he was slowing down from there until his chute deployed at much lower altitude…”

    In Kittinger’s 1961 book, The Long Lonely Leap, there is no word about his speed when the main parachute deployed (at 18,000 feet), but as for deceleration there is this one tidbit:

    “Fifty Thousand Feet:
    My body has decelerated now from its peak speed of 614 miles per hour true airspeed at 90,000; my speed is now 250 miles per hour.”

    For what it’s worth.

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