
I'm not here, though, to write you why Detective Comics's 1001 releases with Batman (if you want to, follow note(1)), but to seize the opportunity and talk about Batman's physics.
Hanging on the batrope
Since his debut in Detective Comics #27 Batman has been immortalized on the cover in an acrobatic pose: hanging on a rope while grabbing a criminal over the rooftops of Gotham. However, Batman's aerial acrobatics are not limited to just hanging from his rope, but sometimes he launches himself from tall buildings or glides using hang gliders or a particular rigid version of his mantle.If we take the process of the downward fall, this is very well described by the infographic below created by designer Shahed Syed in 2012 on the occasion of the release of the last chapter of Christopher Nolan's Batman's trilogy:

The image below is always inspired by the same trilogy (in particular Batman Begins): it shows a Batman gliding, taken from an article published in the Journal of Physics Special Topics of the University of Leicester(3):

In closing, a curiosity found on reddit: an exercise published in a physics textbook, probably American:

- Calculations are very: 1027 - 27 = 1000, obviously, but with this operation one of the two extremes is not counted, which therefore must be added. So the total of issues from #27 to #1027 is 1001! ↩
- To derive the free fall equations in a simple way, we must combine the conservation equation of mechanical energy with uniformly accelerated motion.
Let's start with the latter: acceleration is defined as the ratio between the difference in speed and the time taken to change the speed. In this case we have an object that falls from above with zero speed and reaches the ground just before impact with a speed $v_i$. So $g=v_i/t$, hence the equation $v_i = gt$.
The conservation of mechanical energy, on the other hand, is given by the equation $$\frac{1}{2} m v_i^2 = mgh$$ where on the left hand there is the expression of kinetic energy and on the right the gravitational potential energy. From this expression we derive that of speed, $$v_i = \sqrt{2gh}$$ and by replacing the previous equation in the latter, the expression to calculate the fall time is obtained. ↩↩ - Marshall, D., Hands, T., Griffiths, I., & Douglas, G. (2011). Trajectory of a falling Batman. Physics Special Topics, 10(1). (pdf) ↩
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