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Source - https://www.weather.gov/media/zhu/ZHU_Training_Page/turbulence_stuff/turbulence3/turbulence.pdf

(Scroll down for photos)

I figured out how many G's the cargo jet took by looking at the distance it traveled vertically (500 feet). The turbulence lasted ~10 secs. Most turbulence spikes last half a second or so, so I did some calculating -

9.81 = 1 G (normal gravity)

9.81 x 15 = 147.15 (15 G)

147.15 metres into feet = 482.77 feet.

Within a second the DC-8 would have dropped (nearly) 500 feet under 15 G of force.

Half of its wing ripped off and an engine was missing but it stayed intact enough that it landed safely.

So now we know that turbulence can produce forces in excess of 15 G's, why are newer planes only certified to handle 2.5 G? I saw a Boeing spokesman say the worst turbulence only gets to around 2.5-3 G.

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  • $\begingroup$ You might be interested in avialogs.com/index.php/en/aircraft/usa/douglas/dc-8/…. Specifically, that's an operations manual for a DC-8 (N818CK), issued in 2000, which clearly (page 1-5, document page 33) specifies that the maximum load is 2.5 G flaps up, 2.00 G flaps down. $\endgroup$ – a CVn Nov 12 '18 at 15:01
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    $\begingroup$ Something is off about your calculations... 15 G's for an excursion of 147 meters would happen in about 1 second. Remember, it is 9.81 meters/second/second. 500 feet in 10 seconds is 50 feet/second, the calculation should give you 50/32 = ~1.6g's. Astronauts getting launched on the Soyuz usually only experience about 4 g's... $\endgroup$ – Ron Beyer Nov 12 '18 at 15:32
  • $\begingroup$ You have to remember turbulence isn't smooth. It's going to be sharp ''bumps.'' A bump can last a second, so it is implied within the span of a second the turbulence forced the plane to drop 500 feet. I've played numerous simulators to confirm this is correct. Turbulence isn't like a gentle children's roller coaster ride - it hits the plane in a series of irregular spikes. $\endgroup$ – Willy A Nov 12 '18 at 15:46
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    $\begingroup$ I can't believe that the aircraft moved 500 feet in 1 second, and nothing in the linked PDF seems to suggest that. 500ft/sec is 340mph, so unless the aircraft pointed straight down, I just don't see it. $\endgroup$ – Ron Beyer Nov 12 '18 at 15:49
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    $\begingroup$ Well, if you've got an updraft at one point, there's a downdraft pretty close by. That's one thing my flight instructor made sure to point out to me quite clearly the first time we encountered one while in the air. It's made me go around on more than one occasion when encountering an updraft on final; not because of the updraft per se (that's managable), but because of the downdraft that's out there somewhere nearby, just waiting to hit you with a sledgehammer if you get unwary. $\endgroup$ – a CVn Nov 12 '18 at 16:33
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Before jumping to conclusions, let's look at the numbers:

The duration was about 10 seconds, the altitude excursions was 500 ft. There is no way we can derive a 15g acceleration from that.

It could have been, for example, a 3g acceleration over 2 seconds, resulting in 200 feet excursion, followed by 0.2 seconds coasting after which the pilots levelled off with 2.3g in 2.6 seconds. The resulting excursion would be 500 feet. If the processes then repeated in the other direction, the total time would be approximately 10 seconds, the maximum excursion 500 feet and the maximum acceleration only 3 $g$.

But the fact is we don't know, it's pure speculation.

Another point of discussion is the vertical travel. How was that measured? Very likely it was measured by the altimeter, which works by measuring the static port pressure. How do we know that the static port was measuring static pressure? Very likely it was not, because the turbulence would probably have had a lateral component, as indicated by the 20 degrees left and right rolls. Lateral wind (sideslip) means the static port is exposed to the airstream, and therefore measuring a mix of dynamic and static pressure. It could have been more than 500 ft, it could have been less. We simply don't know.

The most reliable way of determining g forces is by measuring them on-board and recording them (e.g. in the flight data recorder). With the data we have, there is simply no way to know what the g-force were.

In my opinion, it's very unlikely that the aircraft would have encountered 15 g.


The fact that an aircraft is certified for 2.5 g, doesn't mean it can't withstand more, as the crew of China Airlines flight 006 demonstrated on 19 February 1985. They managed to reach 5 g and bend the wing permanently 5 cm upwards. The aircraft was repaired and flew for another 20 years.


Note that the 2.5 g is for load due to manoeuvring, not only for turbulence. The certification specification for large aircraft on the subject of turbulence and gusts has changed several times since the certification of the DC-8.

In 1964, a formula was introduced describing the gust load that the structure aircraft has to deal with. This was added as FAR 25.341. This section has subsequently been updated in 1990, 1996 and 2015.


In addition to changes in structural requirements on the airframe, the origin of mountain wave turbulence in much better understood nowadays. It is taken into account in the operation of the flight.

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  • $\begingroup$ If the DC-8 only felt 3G, why did part of the wing snap off? What about BOAC 911 which experienced +9 G and -4G? $\endgroup$ – Willy A Nov 12 '18 at 16:48
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    $\begingroup$ @WillyA My remark about the 3g was an example to show that you can't conclude anything from the numbers given in the presentation. The claim in the title stating that 15 g was reached and that current certification only require 20% of that amount is based on false assumptions. $\endgroup$ – DeltaLima Nov 12 '18 at 17:30
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It's what they're certified to. The DC-8 was also certified to 2.5 limit/3.75 ultimate but almost all airplanes are stronger than the minimum, and airplanes designed in the late 50s during the slide rule era are even stronger than that because of the need to apply much larger scatter factors in calculations, especially when you are taking fatigue life into account. So the DC-8's structure would be expected to be a lot heavier than a more modern airplane designed using finite element analysis done by computer.

In airframe design, you have competing objectives - on the one hand every pound more than the minimum required is "ballast"; on the other, you have pad the numbers to allow for variability in many forms. The last 50 years have seen computer design being used to shave more and more from the padding to get a lighter airframe that is still safe from the standpoint of a necessarily arbitrary risk profile.

Another is fatigue life. A structure that sees a lot of flexing has to be stronger than the minimum required to take a single application of a load, so that it can take a lot of cycles of an intermediate load before cracks start to form. Fatigue science was nowhere near as well developed in the 50s but it was known to be a problem, so it was accounted for with much larger structural margins than today. A good example is the DC-3, which being designed in the early 30s when the knowledge was very primitive, was made so robust that the fatigue life of the airframe is almost unlimited.

Also, a chunk of wing DID break off, that would have had the effect of partially unloading the rest of the structure. That fact that the wing failed at that location instead of more inboard is probably what saved them.

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