The most unexpected answer to a counting puzzle

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Recommends OuncesBounces · 2020-08 · Mathematics Stack Exchange · on “Proving that $\pi$ can be computed by colliding blocks”

I recently watched 3Blue1Brown's YouTube video "The most unexpected answer to a counting puzzle" [this video] , and the video suggested (at time 2:56) that an inefficient and elegant way to express $\pi$ is: Implement a physics engine, where: Create 2 blocks, and a wall on the left side, while the right side leads towards infinity The second block is bigger, starts out at the right of the smaller one(which has a mass of 1). Set the number of digits you want to compute to a variable N Set the mass of the heavier block to 100 to the power of N-1 Give the second block some initial push. (Doesn't really matter speed) The rules are: All collisions are elastic (no energy is lost, thus the momentum is c

Recommends Pislify · 2020-11 · Stack Overflow · on “Trying to approximate pi digits in C++”

i saw your post you can just use the equation here it is π/4 = 3 tan^-1(1/4) + tan^-1(1/20) + tan^-1(1/1985) or you can do what Harvard does : a physics sim here is a 3blue1brown video [this video]

Recommends Watson · 2024-02 · Mathematics Stack Exchange · on “Interesting and unexpected applications of $\pi$”

Another example: see this interesting collision problem explained by 3Blue1Brown ( [this video] ), inspired by a paper of Gregory Galperin.

Cites MathAdam · 2019-01 · Mathematics Stack Exchange · on “Why does momentum appear to be not conserved in this elastic collision?”

Here's a bit of a fun physics paradox, which I will pose, and then answer below. (These ideas were inspired by Grant Sanderson's fascinating video [this video] on how the digits of $\pi$ are hidden within elastic collisions.) An object is travelling to the right, encounters an immovable barrier, undergoes a perfectly elastic collision, and returns to the left at an equal but opposite velocity. Since the collision is perfectly elastic, kinetic energy and momentum are both conserved. But the momentum after the collision is not equal to the momentum after the collision, because $p \ne -p$ . The net change, $\Delta p$ should be zero, but instead: $$\Delta p = p_f - p_i = -p - p = -2p \ne 0$$ What's u

Cites GSerg · 2019-01 · Stack Overflow · on “Surprisingly different performance of simple C# program”

`calculates PI in a very convoluted way by counting collisions between two object of different mass and a wall` - [this video]

Cites julek.ca · 2020-01 · Stack Overflow · on “Sticky blocks in 2D elastic collisions with VPython”

I'm trying to implement the famous elastic collision problem introduced in this video [this video] . I'm using VPython to simulate the whole thing, however my code isn't running properly for some reason; it seems like when the third collision occurs, the blocks are "colliding" indefinitely (the collision counter keeps increasing), and the blocks are then sticked together. They're even going through the wall instead of bouncing on it and I can't figure out why. Here is my code: from vpython import * import numpy as np import time def compute_pi(number_of_digits): scene = canvas(width=1280, height=720, range=3.8, center=vec(6, 0, 0)) scene.title = 'Computing PI with collisions' # blocks small_block

Cites user818490 · 2020-08 · Mathematics Stack Exchange · on “3blue1brown Playing Pool with Pi Variation.”

I often like doing variations of math puzzles and riddles. For this one though, it's extremely hard for me to find a simulation runner to solve this variation, and this variation also is a lot tougher. Here's a quick refresher of the original problem, as described in 3blue1brown's YouTube video "The most unexpected answer to a counting puzzle" [this video] . The problem includes two sliding blocks and a wall, the second block is coming at the other block at some velocity, while the second one starts out stationary. This is a overly idealistic problem where the floor is friction-less and all collisions are perfectly elastic. When the second block has a mass of a power of 100 in kilograms and the s

Cites Paul Sinclair · 2021-02 · Mathematics Stack Exchange · on “'Mean' appearance of $\pi$...”

Please don't mix up "this proof didn't provide the intuition I was after" with "this proof is unconvincing". The latter suggests that the proof is in error. The former suggests that you haven't understood the proof well enough to figure out the intuition behind it. Which is quite evidently the case here. Anytime $\pi$ shows up [this video] , there is a circle hiding somewhere [LINK] . Here you should be looking for it in that substitution.

Cites Matthew Milone · 2021-03 · Mathematics Stack Exchange · on “Is the decimal component of the square root of 2 (i.e. sqrt(2) - 1) a named constant? Is it broadly significant?”

I was doing some geometry with circles locked on a grid, and I noticed that the distance between a unit circle and the corner of a square that it is inscribed in is of length $\sqrt{2} - 1$ , or roughly $.41421356237$ . Does this number have a widely-recognized name, and does it pop up in seemingly unrelated situations in the way that pi [this video] or the golden ratio [LINK] do? Granted, the primary purpose of such names are to create shorthand, and $\sqrt{2} - 1$ isn't very long to begin with. Still, it's not much less cumbersome than the mathematical expressions that define the first few metallic ratios [LINK] , which are nonetheless named. Furthermore, if we re-scale the geometry to involve

Cites User85825 · 2021-07 · Mathematics Stack Exchange · on “how does the answer of the colliding blocks puzzle change if we add more blocks with mass proportional to each other?”

In this video by 3blue1brown [this video] talks about when you have a block with a mass equal to $\alpha$ and the other block with a mass equal to $\alpha\times100^{\space d-1}$ moving at a constant velocity towards a wall, the amount of collisions will be the smallest whole number with the first $d$ digits of $\pi$ . (with what I understand about the guide lines if I have 2 related questions I must split them into two different questions rather than making it into one, so when I post the other question it will be Here ) My question is when you have $n$ blocks where $n\geq 3$ blocks with masses $\alpha, \alpha \times100^{d-1}, \alpha \times100^{2d-2}, \dots, \alpha \times100^{nd-d-n+1}$ equally s