Exploring the Mechanics of Spiderman: Is it Possible?...Exploring the Mechanics of Spiderman: Is it...

Exploring the Mechanics of Spiderman: Is itPossible?

Sabrina Mohamed

November 17, 2015

Candidate Number: 001482 - 0011Centre Number: 001482Centre Name: North London Collegiate School

Sabrina Mohamed, Candidate Number 001482 - 0011

1 Introduction

As I intend to study engineering at university, I decided to focus my MathsExploration on an area of mechanics that we would not usually study under theIB syllabus. When researching a topic, I came across the book ‘The Physics ofSuperheroes’ by James Kakalios, which I thoroughly enjoyed as it looked at someof my favourite literary characters from comic books, from a mathematical andphysical perspective. From there, my starting point was clear. Could it reallybe that these fantastical stories could on some level be compatible with thereal world? Spiderman was my choice of superhero not only because he is thestar of some of the most enjoyable comics and films I’ve read and watched, butalso because his superpowers lend themselves to mathematical analysis moresmoothly than any other.Spiderman, known as Peter Parker, a character created by Stan Lee and SteveDitko, was bitten by a radioactive spider which gave him spider-related super-powers. This included super strength, agility and sense perception in proportionto that of a spider, the capability to crawl on walls and ceilings, and the abilityto shoot out elastic webs at will, which he uses to travel. These abilities areclearly extraordinary, and physically impossible, but, if we grant that the bitedid give Parker these qualities, would the rest of the mechanics involved in hisday-to-day escapades be feasible?I aim to model and analyze Spiderman’s motion, deduce whether it is possible,and, using my measurements, see whether I could potentially be Spiderman.On my street, there are lamp posts which, if I were Spiderman, I would use toswing across to move along the road. Using this model, I am going to try andcalculate how I would do so.

2 Forces

The first thing to consider about Spiderman are his webs. These are what allowhim to swing across the city, and if they are not strong enough to hold hisweight, then the mechanics would fall through from the start. If we considerPeter Parker as a point mass, and neglect friction and air resistance, we can saythat the motion when he swings from a web is an example of circular motion.Circular motion is motion around a point in which the moving object is travellingat a constant speed around a circle with a fixed radius. There is a change invelocity as the direction changes since the object moves around a circle, whichmeans it is accelerating, but since the speed is constant, we can deduce thatthe acceleration is directed towards the centre of the circle. This is because theacceleration is directed towards the change in direction, and since the changein direction is around a fixed point, the acceleration is directed towards thatfixed point. Spiderman swings in an arc, using a new piece of webbing whenhe reaches the opposite side to swing in another arc, so his movement can be

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modeled as a series of semicircles each of which is an example of circular motion,where the centre of the circle, or axis, is the point at which the webbing is stuck.

The tension in the webbing is the force directed towards the axis. In circu-lar motion, this is known as the centripetal force. This can be worked out usingNewton’s second law of motion

F = ma

where F is the force, m is the mass of the object and a is the acceleration. Themass in this case is simply Peter Parker’s mass, which is 167 lbs, according toMarvel Comics Database1, which is approximately 75kg (75.7499kg), and theacceleration is the centripetal acceleration. Acceleration is simply the rate ofchange in velocity, and can be written as

a =4v4t

Over some time 4t, the mass will travel around the circumference on the circlefor some angle 4θ, in radians. Using this, we can see that the arc, l is

l = r4θ

since it is the distance travelled around the circumference. We get this sincethe circumference of the circle is 2πr, and the fraction of circumference we wantwould be the angle swept over 2π, which, conveniently, is the same as 4θ. And

1http://marvel.com/universe/Spider-Man_(Peter_Parker)

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since this is simply the distance travelled, we can combine this equation withthe speed, distance and time equation to get

v4t = r4θ

as we equate the two to both equal distance. The speed of an object around anaxis is known as angular displacement, which is

v =r4θ4t

And from this, we can rearrange to find the angular velocity, which is the rateof change of angular displacement, also written as ω:

4θ4t

=v

r

dθ

dt=v

r

ω =v

r

To get from here to the centripetal acceleration however, we still need to findthe change in velocity. Using vectors to find the difference between v at bothends of the arc, as demonstrated in the diagram below, we can use the fact thatthe angle subtended between v1 and v2 is 4θ, and thus, using the green line 4vas a base, and the fact that the magnitude of v1 and v2 are the same, we cansplit the triangle below in half, and use the right angled triangle rule

sin θ =opposite

hypotenuse

to find that

sin

(4θ2

)=

124vv

=4v2v

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which simplifies to

4v = 2v sin

(4θ2

)Dividing both sides by 4θ, we get

4v4θ

=2v sin

(4θ2

)4θ

4v4θ

= vsin(4θ2

)1

× 2

4θ

4v4θ

= vsin(4θ2

)4θ2

As we take the limit as 4θ tends towards 0, since we want to find the rate ofchange over an infinitely small angle, we can find that,

dv

dθ= lim4θ→0

vsin(4θ2

)4θ2

Using L’Hôpital’s rule, which states that

limx→c

f(x)

g(x)= limx→c

f ′(x)

g′(x)

under certain conditions2.We can thus differentiate the above result to find

dv

dθ= lim4θ→0

v

12 cos

(4θ2

)12

= v

Using the formula for angular velocity above, we can get

dv

dt=dv

dθ× dθ

dt

dv

dt=v

1× v

r

2if limx→c f(x) = limx→c g(x) = 0 or ±∞ and limx→cf ′(x)g′(x) exists and g′(x) 6= 0 for all

x with x 6=c

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dv

dt=v2

r

and since acceleration is the rate of change of velocity,

ac =v2

r

Now that we have both the mass and the acceleration, we can deduce that thecentripetal force will equal

F =mv2

r

which would be the tension on the webbing, labelled centripetal force in thediagram above.In the 2002 film, Peter Parker developed his own biological webbing as a resultof his radioactive spider bite, but in the 2012 film and the comics, he createdthe webbing and invents web shooters. Either way, this one piece of science iseasily proven feasible, as the strength of spider silk is similar to that of steel(450-1970 MPa)3, and almost half as strong as materials such as kevlar.

Above is an image of the road I live on. For Spiderman to travel down it usinghis webs, he would have to use the lamp posts (marked with blue lines) toattach his webbing onto, and use that point as the axis of his swing, stickingthe dragline silk used for the actual swing to the post with pyriform, the silkused to stick the different parts of a web together. This would allow him toswing in an arc below the lamp post, and stick another web onto the same pointof the next one along, so would travel in roughly the pattern marked out below.

3Griffiths, J. R.; Salanitri, V. R. (1980, "The strength of spider silk", Journal of MaterialsScience 15 (2): 491–6

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I began with measuring the height and distances of the lamp posts. Conve-niently, the base was almost the same height as I am, as shown in the pictureabove, thus I was able to measure the lamp post in segments. As seen above,the lampost is about five times my height, and since I am 1.65m tall, I haveevaluated the lamp post to be 8.25m tall. I worked out how far apart the lampposts were using trigonometry. As can be seen in the image below, the anglemade by drawing a triangle between two lamp posts is 70º, and since

tan θ =opposite

adjacent

tan 70× 8.25 = 22.67

it can be calculated that the distance is 22.67m.

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The distance between the lamp posts is quite wide, so Spiderman wouldn’t beable to swing to the midpoint using one long string, as that would require the silkstring to be longer than the height of the posts (over 11m). Instead, he wouldhave to use a shorter radius, and let go of his string and let his momentum carryhimself forwards for a bit before attaching webbing to the next lamp post whenit is close enough. This would force him to make his path rely on projectilemotion, looking like this:

where r is the radius of the silk, θ is the angle of the sector made by Spiderman,x is the direct distance between the points he swings between, ` is the arc length,v is the speed and y is the distance he has to travel by projectile motion.If we assume the length of the string from where it is attached to the lamp postto where Peter Parker is holding it to be 6m, to ensure he does not hit any carsor pedestrians while swinging, and his speed to be 10ms−1, as he would be usinggravity to gain speed as he begins swinging, we need to work out the distance

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needed to travel by projectile motion. To do this, we first need to pick an angleat which Parker swings from, the angle marked θ on the diagram. Let us use π

2to begin with. As the angle between the two radii drawn on the diagram is nowa right angle, we can use Pythagoras’ theorem to find x :√

62 + 62 = 8.49m

Since the distance between the posts is approximately 23m apart, this wouldleave 14.51m to be covered by projectile motion. Projectile motion uses the fixedacceleration kinematics formulas to work out the distance, speeds and time ofmotion in an arc. Here, since we are working out equations in the x direction,acceleration is 0. We can thus use the equation

s = ut+1

2at2

to work out the time taken for Parker to travel the distance necessary. Thisgives us

t =s

u

t =14.51

10= 1.45secs

The time taken to travel along ` we can evaluate by using the simple distance,speed and time triangle.

t =`

v

` = θr =6π

2= 9.42

9.42

10= 0.94secs

Thus the total time would be 2.39 seconds to travel between two posts. Thedistance between two posts being about 23m, this means that, provided thelamp posts were equally spaced apart, Spiderman would be travelling at about21mph, which is a fairly decent speed, and would definitely be quicker thansitting in traffic. In order to increase his speed, he would need to shorten thedistance travelled, but even if he used a smaller radius, it would only increasethe distance needed to be travelled by projectile motion, which evens out to beclose enough to the original time that it would not make a significant difference,for example, using a 4m radius would result in the time being 2.36 seconds.There is a limit to how short the diameter of the swing could be due to thedistance required to travel, thus this is not a viable solution. Another methodof increasing his speed would be to increase his weight so that he is heavier,and the force of gravity acting on him would cause him to travel faster when hetakes off. However, carrying extra weight does not seem to be a viable solutionwhen fighting crime. Thus, while this mode of transport is certainly feasible,it is probably not the most convenient or practical method of getting around.

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The additional weight giving him more speed also means that I, being lighterthan Parker at 57kg, would travel slower than he would be able to, thus wouldbe a potentially less efficient Spiderman.With a 6m radius and travelling at 10ms−1, the force exerted on the silk wouldbe

F =75.75× 102

6= 1262.5N

A strand of spider silk ranges in diameter from 2.5µm − 4µm, so we can take3.25 as the median, 3.25 × 10−6m, which is 8.30 × 10−12m2 in area. The forceexerted, 1263.5N, over that area gives 1.52× 1014Nm−1. 450MPa is equivelentto 4.5 × 108Nm−1, so although Parker would need 337778 strands of silk, thesilk would be able to hold his weight. Although this may seem like a lot, sincethe strands are so fine, the rope Spiderman would have to make would be only655.8 strands in diameter, which would be only 2.13mm across.From these calculations, we know that, at least on a simplified level, the featsof Spiderman are possible.

3 Elasticity

However, upon further research I discovered that if this model were used, Spi-derman’s shoulder would be in a lot of pain from being jerked around due tothe rigid nature of the cord. The silk in the example above is rigid, and doesn’tchange length, thus when Parker jumps to swing, his arm would jolt becausethe silk would not give to reduce the impact of when the silk becomes taut.Luckily, spider silk does indeed have the elasticity to ease this, and will give toaccomodate this jerk. This is similar to how a bungee cord extends when used.The primary formula used here is Hooke’s Law, which states that

F = kx

where x is the length of the extention and k is the spring constant, whichdepends on the elasticity of the material. Thus, the centripetal force is not theonly force acting on the rope, or Spiderman. Due to the elasticity of the silk,the radius of the circle will still be fixed, however it will be fixed at the extendedlength. To compensate for this, Spiderman would have to adjust the length ofthe radius so that the extended length of it is correct. Through some simplemultiplication we can find that, if the elasticity of the silk is 24.1± 2%4, Parkerwould have to divide the ideal radius by 1.241 to find the length of silk he needsto put out. This is because the silk would extend by 24.1%, thus using the exactlength would extend to be too long than required. In the above example, where

4Bonino, Mark J (2003), “Material Properties of Spider Silk”,<http://www.lle.rochester.edu/media/publications/documents/theses/Bonino.pdf> ac-cessed [18-08-15]

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he would want a 6m radius, he would need to put out a length of

6

1.241= 4.83m

which would allow the silk to stretch to 6m, saving his shoulder a lot of pain inthe process.Another quality of spider silk is that it contracts when in contact with water5.This is thought to be so that the dew in the morning can help tighten websmade overnight. However, it means that Spiderman would be unable to use hiswebbing as efficiently in wet weather conditions, and would need to accomodatethat in the radius of his silk used. This could be fixed in a similar but oppositeway to the elasticity problem - instead of decreasing the amount of silk he wouldneed to put out a longer line to ensure that it would be the right length withthe contraction.

4 Conclusion

This maths exploration has allowed me to explore mechanics beyond the HigherLevel syllabus, and I’ve learnt a lot about the potential applications of bothprojectile motion and circular motion, fulfilling my aim of learning more aboutthem in a fun and engaging way. It has been an enjoyable project, and from itI have drawn the conclusion that it is indeed possible for Spiderman to travelabout his city the way he does, at a reasonable pace, and thus, if I were to bebitten by a radioactive spider and given powers, I would be able to become theweb-slinging vigilante, given enough practice. Although the actual powers ofSpiderman are fictional and impossible, in terms of the mathematics, it wouldbe realistic, in his world, for him to travel the way he does, with a rational degreeof accuracy. Keeping in mind the sums needed to figure out the optimal lengthsof silk and just when to release himself, however, it would definitely require a lotof practice, dedication, and intuition to be able to so precisely use the webbinglike that. I have learnt, as well as the derivation of centripetal acceleration,the mechanics behind these motions, and have been left with the reflection onhow they can be used beyond this entertaining example. These mechanics areimportant when considering the angles and speeds at which to throw balls whenplaying rounders or, in a much more serious application, rockets and missiles.Mechanics has endless applications which make it, in my opinion, the mostversatile and useful areas of mathematics, and I’m really pleased with what I’vefound in this exploration and how I can use the information in other situationsand problems.

5Plaza, Gustavo R.; Guinea, Gustavo V.; Pérez-Rigueiro, José; Elices, Manuel (2006)."Thermo-hygro-mechanical behavior of spider dragline silk: Glassy and rubbery states". Jour-nal of Polymer Science Part B: Polymer Physics 44 (6): 994.

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5 Bibliography

References

[1] Bonino, Mark J (2003), “Material Properties of Spider Silk”,<http://www.lle.rochester.edu/media/publications/documents/theses/Bonino.pdf>accessed [18-08-15]

[2] Griffiths, J. R.; Salanitri, V. R. (1980), "The strength of spider silk", Journalof Materials Science 15 (2): 491–6

[3] Kakalios, James (2010), ’The Physics Of Superheroes’, Gerald Duckworth& Co. Ltd

[4] Plaza, Gustavo R.; Guinea, Gustavo V.; Pérez-Rigueiro, José; Elices, Manuel(2006). "Thermo-hygro-mechanical behavior of spider dragline silk: Glassyand rubbery states". Journal of Polymer Science Part B: Polymer Physics44 (6): 994.

[5] Marvel Universe Wiki (2015), <http://marvel.com/universe/Spider-Man_(Peter_Parker)> accessed [18-08-15]

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