"A pencil thick spider's silk thread is capable of stopping a Boeing-747 in full flight. "
Ten years ago I found this statement somewhere in a book and put the statement on my site without controlling the sources. So now and then people write me to ask where I found this statement and if the Boeing could be stopped with this pencil thick line. Now in 2007 it is much easier to check these statements and the first check was in www.books.google.com . I has already tried to look in my own books at home but failed to retrieve the statement.
Google did found some reference but all the books where from after 1996 and it was almost certain the phrases were copied from this site because the sentence were almost identical. Searching in my spider books for the lost statement also failed.
I had to calculate to question myself.
The strength of a thread depends on more than one feature. One can hang a weight on a given thread and measure the weight before it breaks. But also the elasticity of a thread is part of its strength. Also its weight per volume is an feature that is added to it strength. The less, the better. We will discard this last feature in our calculations for simplification.
I think I was ready to calculate the diameter of spiders silk to stop a Boeing-747 in full flight.
A 747-400 typically takes off at 180 mph (290 km/h), cruises at 565 mph (910 km/h) and lands at 160 mph (260 km/h).
Typical operating empty weight is 180,000 kg.
Sullivan  cited that the breaking stress of the dragline silk for three species of spiders range from 1420 to 1550 million N/m2 with elongation at break ranging from 16 to 30%. The density of this silk is 1300 kg/m3.
I our calculations we will use a breaking stress of 1500 million N/m2 and 30% break elongation and a plane of 60000 kg flying 1080 km/h = 300 m/s.
When we use a thread with a length of 1000 meter the plane must be stopped in 300 meters.
We assume the thread is weightless and is tough enough to stop the plane when stretched completely.
We also assume that the silk thread follows Hooke's law of elasticity that states that the amount by which a material body is deformed (the strain) is linearly related to the force causing the deformation (the stress). stress and strain are linear
We will assume that all kinetic energy of the moving plane will be changed to the potential energy that will be stored in the thread.
The kinetic energy will be calculated with the formula: E (J) = 0.5 m (N) * v ^2 (m^2/s^2) and the elastic potential energy with the formula:
The elastic potential energy is defined as a work needed to compress or expand an elastic body.
The potential energy of a string or spring that has modulus of elasticity under an extension of x is then
This formula is obtained from the integral of Hooke's law:
Elastic Potential Energy is the kind of energy that is stored in a bow, or in a spring or a silken thread that stops a Boeing-747
The energy stored = the work done to stretch the bow, so:
Elastic Energy (joules) = Average Force (newton) x Distance (meters)
Retrieved from " http://en.wikipedia.org/wiki/Elastic_potential_energy "
End intermezzo -------------------------------------------
Elastic Energy (joules = Nm) = 0.5 K (N) x Distance (m)
Diameter = K (N) / breaking stress (N/m2)
Surface = pi / 4 * diameter^2
But how many spiders do we need to make such a thread? We made two calculations. One for a small Araneus diadematus thread of 0.5 um and one for a thick Nephila thread of 10 um. For the calculation of the surface of the thread was assumed square also for simplification.
Réaumur calculated that 27648 female spiders (Araneus diadematus) would be required to produce 1 lb of silk. In the calculation this is compared with the amount of Nephila spiders needed with 10 um thick silk lines. These results are in good agreement and not with the amount of Araneus diadematus spiders! If one uses 11.25 um thickness for a thread the results equal.
We assumed that one spider can give us a thread of 100 meters.
In the calculations we assumed that all energy is transformed into heat (in reality it is about 65%) and calculated the temperature rise assuming a specific heat capacity of water of 1 J/gK.
|distance to stop||s||300||m||300||m||9000||m|
|Diameter thread to stop Boeing-747||5.7||cm||21.4||cm||1.0||cm|
|No of spiders needed on basis of diameter and 10 spiders needed for one km of thread|
|diameter thread um||0.5||10||0.5||10||0.5||10|
|surface thread m2||2.5E-13||1E-10||2.5E-13||1E-10||2.5E-13||1E-10|
|No of threads||1.02E+10||2.56E+07||1.44E+11||3.60E+08||3.41E+08||8.53E+05|
|No of spiders (10/km)||1.02E+11||2.56E+08||1.44E+12||3.60E+09||1.02E+11||2.56E+08|
|No of billion spiders||102||1440||102|
|Number of Nephila spiders needed when assumed 27648 spider produce 1 lb silk|
|Number of spiders / kg||60952||n/kg||60952||n/kg||60952||n/kg|
|Number of spiders Nephila||2.03E+08||2.85E+09||2.03E+08|
|Number of billion spiders Nephila||0.20||0.26||2.85||3.60||0.20||0.26|
|Temprerature rise in the thread just before breaking|
|Specific heat capacity||1||J/gK||1||J/gK||1||J/gK|
|energ per gram||173||J/g||173||J/g||173||J/g|
Excel worksheet with the calculation
To stop a 180,000 kg Boeing-747 flying 1080 km/h (300 m/s) in 300 meters one needs 1,440,000 million Araneus diadematus spiders or 3000 million Nephila spiders to make a one kilometre long thread with a diameter of 21.4 cm. If the plane comes to a stand still and all the passengers are thrown in the cockpit because of deceleration of 150 m/s2 (15 g) the temperature of the thread is 170 Celsius warmer.
It is possible to stop the plane with a pencil thick thread. When the plane is flying at landing speed (80 m/s) and the thread is 30 km long. One needs 102,000 million Araneus diadematus spiders and the temperature rise is the same because the temperature rise is linear related with the breaking force. (To stop the plane at full speed the line should be 500 km long).
Ed Nieuwenhuys, Leo de Cooman, September 2007