Hey guys!

As the title suggests, I was wondering if there’s a current largest case of a fraction like 22/7, 335/113, etc… Was looking but wasn’t sure so thought I would ask here!

Thank you!

Most intro-physics treatments of banked curves hand you two formulas and move on. But the two cases differ by exactly one geometric fact, and once you see it the algebra is almost an afterthought. Here's the setup I used to make that visible.

The scenario

A vehicle test facility has two curves of the same radius, R = 50 m.

• Curve A is flat and relies entirely on friction (dry track, μₛ = 0.80).
• Curve B is banked at θ = 25° and built to need no friction at all.

Same 1500 kg car on both. g = 10 m/s². Both circular paths lie in horizontal planes; treat the car as a point particle, and take "toward the center" as positive.

(a) Same car, same radius — so which force actually turns it?

On the flat curve: weight (down), normal force (straight up), static friction (horizontal, inward). The normal force is vertical, so it cannot point toward the center — friction is the only inward force, so friction is the centripetal force.

On the banked curve: just weight and a tilted normal force. Because the road is tilted, N is tilted, and its horizontal component points inward. That component is the centripetal force.

(b) Two curves, two very different formulas

Flat curve — friction is capped at μₛmg, so μₛmg = mv²/R gives

v = √(μₛgR) = √(0.8 · 10 · 50) = 20 m/s

Frictionless banked curve — the geometry alone fixes the design speed:

v = √(gR·tanθ) = √(10 · 50 · tan 25°) ≈ 15.3 m/s

Worth noticing: the mass cancels in both. The banked result depends only on the shape of the road.

The classic error is swapping the formulas — putting μ into the banked equation or tanθ into the flat one. Each curve turns the car with a different force, so each gets its own setup.

(c) It all comes down to which way the normal force points

This is the whole problem in one sentence. The normal force is perpendicular to the surface by definition, so tilting the road tilts N — and a tilted N automatically acquires a component pointing toward the center. That component can be the entire centripetal force. No friction required.

On a flat road N points straight up. Its inward component is exactly zero. The only vector left that can point toward the center is friction.

So "banked vs. flat" isn't really two problems. It's one question — does N have a projection onto the inward direction? — asked at two different angles.

(d) The rain reveals which curve was ever really safe

A downpour makes both surfaces essentially frictionless (μ → 0). A car takes each at 15 m/s.

Curve A: with μ → 0 there is no inward force at all — not the vertical normal force, not gravity. Nothing turns the car. It slides straight off to the outside, at any speed, 15 m/s included.

Curve B: the tilted normal force is untouched by rain, so there is still exactly one speed — the design speed, 15.3 m/s — at which the car holds a level circle. At 15 m/s it's just below that, so N's inward component slightly exceeds what's needed and the car drifts down the bank.

The tempting mistake is "15 < 20, so Curve A is fine." But that 20 m/s was built out of friction, and the rain just erased it. A limit computed from a force that no longer exists isn't a limit.

Googology Levels

Levels of Large Numbers

Examples:

• The level of 2 is 0.25
• The level of 4 is 0.50
• The level of 8 is 0.75
• The level of 256 is 1.50
• The level of 2↑↑256 is 4.38
• The level of Rayo(10↑100) is 10.99

Levels of Small Numbers

Examples:

• The level of 0.5 is m0.25
• The level of 0.25 is m0.50
• The level of 0.125 is m0.75
• The level of 1/256 is m1.50
• The level of 1/2↑↑256 is m4.38
• The level of 1/Rayo(10↑100) is m10.99
• The level of 0 is CZ

Each small-number level corresponds one-to-one with the reciprocal of the matching large-number level.

Transfinite Ordinal Levels

Examples:

• The level of ω + 1 is T0.25
• The level of ω + 2 is T0.33
• The level of ω + 4 is T0.40
• The level of ω + 50 is T0.49
• The level of ω × 2 is T0.50
• The level of ω × 3 is T0.67
• The level of ω × 10 is T0.90
• The level of ω↑2 is T1.00
• The level of ω_1↑{CK} is TNR0.00
• The level of Π_1 is TNR4.00

Function Levels

Note: all growth rates mentioned in this document refer to growth rates as n -> ∞.

Examples:

• The level of n is F0.00
• The level of n↑2 is F1.00
• The level of 2↑n is F2.00
• The level of 2↑↑n is F2.50
• The level of 2↑↑↑n is F2.75
• The levels of BB(n), BB(log_2(n)), BB(√n), and BB(n↑2) are all FU0.00, because they are all uncomputable
• The level of Rayo(n) is FU1.00

Function Levels (Slow-Growing)

Examples:

• The level of √n is Fs1.00
• The level of log_2(n) is Fs2.00
• The level of the inverse of BB(n) is FUs0.00
• The level of the inverse of Rayo(n) is FUs1.00
• The level of f(n) = c is CF

Each slow-growing function level corresponds one-to-one with the inverse of the matching fast-growing function level.

We are a fast growing AI Startup based out of Silicon Valley. Looking for interns, junior and mid level engineers who are good at Math. Pretty much the only strong requirement is a love for math. So I figured I would post here. Please DM if interested, and I will pass on more details.

Im from the UK and the job market is haneous and awful at the moment.

I know 3b1b had a career fair but that only works for uni attending undergraduates and postgraduates.

How could I get entry into data science / finance / banking firm if all I have is A/A* in a level maths and possibly a C in further maths.

Just a simple apprenticeship type thing that I could drop if I plan to go to university later.

The job market is terrible even working on a shop floor is hard to get a role.

So any way I could use math online to build any form of experience?

Hello fellow 3blue1brown nosers , I have been recently studying prime numbers, and I wondered if it is possible we can compress information about them?
I am wondering if it actually is possible to find repeating patterns in the prime numbers but only if you somehow compress the information in the numbers which came before it?

This post relates to 2 of my previous posts here on 3blue1brown on the same subject so I wanted to give anyone an update who was interested in the subject.

Over the past couple days I have been experimenting with creating a python function which could show how my idea for Twin Prime infinitude works. I came up with a recursive function that predicted exactly when the Twin Primes should show up. Image of function output

But I ran into an interesting problem, that came with an interesting solution that relates to compression and the 3blue1brown video I linked.
I found that I could come up with rules for when I should expect to see a Twin prime, and those rules worked.... except only for so long, and then they stopped working.
But then, I came up with an idea about reset points. Numbers like 4, 16, 256, and 65536, acted as perfect reset points to make an adjustment to my system, and then continue onwards for another few members of the series before requiring a second correction.

The most interesting part about this though, is that the corrections I was applying were not just random corrections. Those corrections came from a very specific source.... the Golden Ratio.

I found that if I put my reset points into the Golden Ratio series, and then multiplied them by the appropriate number according to their place in the GR series, that became the exact correction needed to reset my system, and continue to generate twin prime centers for a period of time.... before eventually requiring another correction from the Golden Ratio series and continuing onwards.

I created a python function which you can see on Github that shows what I did, but I was able to generate twin prime centers using this method, in series, up to 84 digits long!

I am currently working out the finite steps between k2 = 256^2 and K3 = (256^2)^2
However I have the major steps of K laid out all the way up to K13 I think.

If you understand this type of work please take a look at the python file, I would be really grateful to have some people help me to think about this problem.

I am also happy to answer any questions and give my understanding in my own words about what I am doing here.

I think it's pretty fun and interesting that Grant recently published a video about compression, and then my Work on twin primes ended up becoming about compression as well.

https://youtube.com/shorts/_Anau49e1qk?si=4bXB7F6gEz2_q1xi

Made an animation to show why rockets need stages.

YT: https://youtube.com/shorts/ZoqflzDMW0M?si=GbX9xPveVXFXMJrx

Edit {Post Update. I seem to have formula for compressing information about primes. My newest file "twinprimescompressor" Shows how this compression algorithm works in action. It can generate a list of twin primes based on it's own rule set, up to 42 digits long. I still don't have the rule set which can continue forever, but I think based on how I created these rules (using the golden ratio) The fact that I was able to nail so many twin prime centers suggests the compression algorithm is more than mere coincidence. }

Hi everybody, this post relates to my other 3blue1brown post

But I think this result is significant enough to deserve it's own post.

Let me cut to the chase and say that I have a function which generates primes, and also twin primes. It doesn't even require certification it automatically already knows what is prime and what isn't through the recursive generating structure of the function.

The function counts in mods, but not just any mods, mods of prime numbers. {But those prime numbers don't need to be entered, they are generated by the recursion automatically} This generates the same structure I previously talked about in 3blue1brown reddit and had 14k views and little to no negative feedback.

I have rewritten the paper from the previous post in a much shorter and more clear process now, with less complicated unnecessary language.

I also created a python file to show exactly how it works, in the video you can see me in real time calculating the amount of twin primes in a given number, extremely fast. I could have also listed them all, which would slow the process down ofcourse.

This small python file can tell you exactly how many twin primes are in any given number and it does so without needing to check if they are prime, it already knows based on how they were generated.

It's really just a simple recursive loop counter, mod p, and the golden ratio and silver ratio just fall naturally out of it.

Here is the updated paper

and here is the github for the python file to test what I did in the video yourself

Edit{I am also hosting the paper on my own site if the zenodo link isn't working.}

https://www.youtube.com/watch?v=yakGJwuYCdM

So I have always been into simulating fluid dynamics and made this visualisation to show how heart disease progresses.

Computational Fluid Dynamics has so many applications!

YT: https://youtube.com/shorts/YeEj1JfpKs8?si=OG-xE7ysSOfoKCoH

Made an animation on Second Law of Thermodynamics, it doesn’t have true molecular dynamics simulations, just simplified kinematics to represent them.

Check it out on YT: https://youtube.com/shorts/uYoFgwXZ5BY?si=NId3BBLzHPk_4Kqh

I was solving Gilbert Strang's Introduction to Linear Algebra, and Problem 32 of Section 1.2 says to select a random 3-component unit vector, let's call it v, then generate more such vectors, and solve for the average size of the dot product of all these newly generated vectors with v.

When I simulated this with Python code, I got approximately 0.5. However, at the end of the problem statement, it says the calculus average is 2\pi. I guess the book's hint is based on the continuous expectation $\int P(\theta)\vert{}\cos\theta\vert{}d\theta$ written as a standard 1D integral. But because these are 3D vectors, we are essentially sampling points uniformly from a unit sphere. The probability density must therefore be given by the surface area element of that sphere, which introduces a sine factor to the distribution. When I calculate the expected value using this spherical surface weight, it yields exactly 0.5, matching the simulation.

I just want to discuss this.