Math in nature?

interrobangzingkapow · 3 months

People here have recommended lots of good resources, but I would also highly recommend the documentary, "The Secret Life of Chaos". It reveals the history and theory behind the very mathematical patterns you mentioned, and how they work in nature. The link I gave is to the preview, but I would definitely recommend the full length movie, given that you seem quite interested in the topic.

bhtp · 3 months

Here you go

doobiedobahbah · 3 months

Very cool!

doobiedobahbah · 3 months

Right here, right at the border of Pennsylvania an Maryland. It gets really wide and deep down here in PA!

doobiedobahbah · 3 months

Yup, that's the spot! I love that little island just offshore that you can swim to to rest.

doobiedobahbah · 3 months

Actually, my physics is rather precise according to d = 1/2 a t² (link), and your 60 foot jump only took about 2 seconds. However, I agree that it definitely SEEMED to take twice as long, because time almost stands still when you're falling that fast.

doobiedobahbah · 3 months

I jumped 4 times. Long story short, the first time went great; second time I got my balls smacked a little, so I protected them with my hands the third time around. My last jump (the one in the pic), I spread my arms like a dumbass and got bruises on them. Water shot up my nose all 4 times. Still fun as hell though; the drop makes your stomach flip, then the adrenaline kicks in as you land, so you don't really feel that much pain.

doobiedobahbah · 3 months

Yeah it kind of does at that angle. My friend also took a video of it at a better angle, but I don't know what she did with it(I'll ask her). We timed how long it took to fall in the video (just under 2 seconds), and that's how we know it's about 50 feet.

doobiedobahbah · 3 months

Cliff jumping on the Susquehanna, from about 50 ft high.

Also, me and my brother when we get bored.

allanvv · 4 months

Impressive! My friend and I designed and built a similar device [video], but it has a vertical row of LEDs that spins in a cylindrical path. Ours uses a synchronous motor from a hard drive, and supplies power to the LEDs using a brush and a slip ring. The video is somewhat washed out, but it is capable of full RGB, with 32 tri-color LEDs and adjustable color depth using a PWM/dithering algorithm I wrote. I can trade off color depth&resolution for more frames in the animation, as it has limited memory.

EGraw · 4 months

I'm working on a small JavaScript-based app that allows for easy manipulation of N'th order polynomial curves using N draggable control points and other dynamic tools. It's really interesting to experimant and play around with, and I'll definitely post it on reddit soon. For the example in this post, I added some constraints such that the x only had odd exponents and the y only had even exponents, thus making the final graph symmetric.

neotheseventh · 4 months

The resize feature was my idea, but I wrote the code that does both.

neotheseventh · 4 months

My only claim to reddit fame is that I came up with the original idea and code for this feature. It was since added to the Reddit Enhancement Suite.

tylertennisman · 5 months

Think visually about the extreme case: where the ellipse is so narrow, it almost looks like a straight line. Now put a large border on it, and observe that it is very close to the shape of two semi-circles connected by straight lines(like an ice hockey rink), and is very obviously not elliptical. This visual makes it clear that the border cannot be fit to an ellipse, unless of course, as other people are pointing out, both ellipses are circles.

musicismath · 6 months

Say you're on a road trip, and you take the amount of time left in your trip and divide it by your current speed. That's approximately how much time you can cut off of your trip for every mph. you speed.

For instance, let's say you have a 60-minute drive, at an average speed of 60 mph. For every (small number) of miles per hour you speed, it will cut off 60/60=1 minute of your trip. It works pretty accurately for small amounts of speeding, obviously you can't go 120 miles per hour and make your trip take 0 minutes!

I figured this out during (guess what!) a LONG road trip.

Bubbasauru · 1 year

I solved it using partial fraction expansion.

bdesham · 1 year

The correct c is log( e⁶ - π⁵ )/log(π) ≈ 4.00000016

dodgepong · 1 year

Also, notice that the curve is a quadratic bezier curve with control points at (0, 1), (0, 0), and (1, 0).

scrabbleigh · 1 year

The only thing wrong with this problem is that the star's height and width are the same. For a normal star shape with rotational symmetry, the star's width is a little larger than its height. If you stare at each line segment in the star, this will become obvious. If you actually want the star's height and width to be equal, then the interior angles of the pentagon cannot all be 108°, as the diagram seems to imply.

dudemanfreak · 1 year

Or division... Someone's gotta be on top!

SkepticalEmpiricist · 1 year

Who'd want to wear something like that? It conveys such a negative message.

h0rror · 1 year

Story of my life.

h0rror · 1 year

[; k=\frac{\frac{x^2+x}{x}+23-x}{6}=\frac{\frac{x(x+1)}{x}+23-x}{6}=\frac{x+1+23-x}{6}=\frac{24}{6}=4\neq 3 ;]

GlueBoy · 1 year

One-eyed for 2 hours

ಠ_ರ

zoroth · 1 year

When you're summing up a geometric series, there's so many different formulas you can use, each with different starting and ending boundaries. For instance:

[;\sum_{i=1}^{\infty}x^i = \frac{x}{1-x}\;\;\;(|x|<1) ;]

[; \sum_{i=0}^{\infty}x^i=\frac{1}{1-x}\;\;\;(|x|<1) ;]

[; \sum_{i=0}^{n}x^i = \frac{1-x^{n+1}}{1-x} ;]

[; \sum_{i=1}^{n}x^i = \frac{x-x^{n+1}}{1-x} ;]

I was always getting them confused and remembering the wrong one or not remembering them at all. Then, I realized that there was a general form with arbitrary limits so that you could just plug in whatever starting and ending boundaries you want, and you can easily get any of the above formulas, plus many more:

[; \sum_{i=a}^{b}x^i = \frac{x^a-x^{b+1}}{1-x} ;]

You can plug in 0, 1, +∞, -∞, or any integer for a and b, and you get the formula you're looking for (assuming x^∞ →0 for |x|<1 and x^-∞ →0 for |x|>1.) This form might be a little harder to memorize, but it's infinitely more useful, in my opinion.

Also, if you're like me and you prefer to derive rather than memorize, here's the derivation for the general formula:

[; \begin{align*}%20\sum_{i=a}^{b}x^{i}=S=x^a\;\;\;+&x^{a+1}+\cdots+x^{b-1}+x^b%20\\%20x\cdot%20S=x^{a+1}+&x^{a+2}+\cdots+x^{b}\;\;\;+x^{b+1}%20\\%20S-xS=x^a\;\;\;+&x^{a+1}+\cdots+x^{b-1}+x^b%20\\%20-(&x^{a+1}+x^{a+2}+\cdots+x^{b}\;\;\;+x^{b+1})%20\end{align*}\\\begin{align*}%20S(1-x)&=x^a-x^{b+1}\\%20\;\;\;\;\;\sum_{i=a}^{b}x^{i}=S&=\frac{x^a-x^{b+1}}{1-x}\\%20\end{align*} ;]

EGraw

TROPHY CASE

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