Introduction
Hello and welcome to the third blog post of FictionPhysics. Kung Fu Panda has been on my mind lately because of the announcement and trailer for Kung Fu Panda 4. When I first watched Kung Fu Panda 3, I was very fascinated by the character Kai, who is a bull wielding a chain with two blades at either end, which I found very cool. Besides the fact that I thought (and still think) that it was a copy of the Blades of Chaos from God of War, I loved the scenes where he used the blades to cause havoc. Furthermore, I recently found out that there is a tiering system for fictional characters which categorises them in terms of strength, ranging from ‘hypoverse level’ to ‘boundless’ (Check VS Battles Wiki for the whole list). In this system, Kai is tier 7-C: Town level, whose description is as follows:
Characters or objects who can destroy a town, or those who can easily harm characters with large town level durability.
The aim of this investigation is to assess Kai’s strength and see whether my conclusions align with the power level assigned to him. We will examine two feats of strength:
1. Him cutting a large building in half with his blades, which can be watched here:
2. Him hurling a large statue of Master Oogway in a circular motion, which can be watched here:
Methodology
A major part of this investigation is the pixel scaling method to find the lengths, which is described below:
As these scenes take place in fictional worlds, we don’t know the sizes of most objects. Due to this, the only way we can find these sizes is by using the few objects that we know the sizes of and comparing them to the objects we don’t know the sizes of. This works by finding a metre: pixel ratio from an object we know the size of and then applying it to the other objects. This is not limited to objects and can be applied to distances.
For both scenarios, I will calculate the strength in terms of explosive power (mass of TNT) as it somewhat fits with how the tiers are described.
The methodologies for both scenarios are discussed in their respective sections.
Feat 1: Slicing through a building
To model this scenario, I will use the concept of Toughness. Toughness is the ability of an object to absorb energy and plastically deform without fracturing, it is a physical quantity measured in J m-3. In this scene, we can see that there is a lot of debris flying off when the building was chopped, suggesting that the parts were touched by the blade were likely all fractured. With this assumption, I will calculate the volume of brick (which was a common material in ancient China) which was in the path of the blade using the blades dimensions. Firstly, I calculated the distance the blade travelled inside the building using pixel scaling-
Here, the Red line represents Kai’s height, which is 7 feet assuming that he is a larger than average bull. The blue line shows the path of destruction of the blade, which was measured to be 8.6 metres.
Next, we need to calculate the size of the cross section of the blade, for which I have chosen the following frame:
Here, Po is about 5 feet tall (red line). The thickness of the blade is shown by the blue line and the length with the black line. The thickness is 0.072 m (7.2 cm) and the length is 0.94 m. Assuming the cross section to rectangular, we get the cross-sectional area of 0.068 m2. We can multiply this by the length to get a total volume of 0.56 m3.
The toughness of standard bricks is about 2 MJ m-3, using the volume we have, we get
Additionally, I took the time it took for Kai to make that cut, which is about 0.37 seconds, meaning that Kai’s power (in the physics sense) was 3 MW, which is enough to power 2000 homes at the same time. Although this seems like a lot, in terms of destructive energy, its worth about a 267 gramme blast of TNT (taking 1 kilogram to be 4.184 MJ) , which is similar to the 300 gramme blast shown in the following video:
Feat 2: playing hammer throw with a massive statue
Hammer throw is a sport which (as the name suggests) involves hurling a heavy object as far away as possible by holding the object with a rope and spinning around on the spot, releasing it at the right moment. The scene is shown above. We can assume that the statue is moving with a circular trajectory, meaning that we can use the equations associated with circular motion. Firstly, in order to keep an object moving in a circle, a centripetal force, Fc, is required at all times. The equation for this
where m is the mass of the object, v is its linear velocity and r is the distance between the axis of rotation and the object in motion. In this case, r is essentially just the length of the chain, which can be calculated using pixel scaling from another frame:
Here, the red line is Kai’s height and the blue line is the length of the chain. Using pixel scaling, we get this to be about 32 metres.
Going back to the scene, there is clearly no way to measure the linear velocity directly. Therefore, we need to use the equation
where ω is the angular velocity, which can be calculated by measuring the angle the statue swept through over a period of time. To find angular velocity, I considered the two frames below.
The first one takes place before the second. From these frames, we can deduce the following top down view:
Here, S is the statue, r is the length of the chain, and A is the angle we need to find. The time interval between the frames is 0.12 seconds. We can take the chain at different points in time to be the sides of an isoceles triangle, meaning that we can apply the cosine rule to it to find A. Before that, we need to measure the length of the third side (labelled ‘?’). This can be done by taking the tangent of the distance between the statues in the two frames- our method restricts us to only being able to measure the straight line distance and not the actual (circular distance). This distance is labelled in the image below
The red line is the tangenetial distance while the green line is the height of the statue. There was virtually no way I could measure this statue through pixel scaling, therefore, I had to resort to using a similar statue in real life. The one I chose was the statue of Guang Yu in Jhinzou, which is kind of similar to the one below
The statue is 58 metres tall and weighs 1197 tonnes, using this to find the metre: pixel ratio, we get the tangential distance to be 56.5 metres, which can then by used in the cosine rule as follows:
Angular velocity is angular displacement per unit time, which can be expressed in this case as
Now, we have enough enformation to calculate the centripetal force that Kai is exerting-
This is equivalent to having about 60,000 elephants trample over you at the same time. Showing that Kai is really strong. In terms of energy, assuming the system is 100% efficient, all the energy Kai inputted would have been transferred to the kinetic energy of the statue. The kinetic energy (which is his energy expenditure) is expressed as follows:
Conclusion
Looking at the numbers, I think that Kai easily fits the criterion of ‘Chacters who can easily destroy a town’ as he strength can be equivalent to one of the largest non-nuclear bombs in history, a handful of which could easily decimate a town. Furthermore, with his blades he should be able to slice buildings in half with ease. He could probably be considered even more powerful when taking into account his other abilities such as being able to trap the souls of other people. I hope that you found this interesting and I will be back next month with another post.
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