/sci/ - Science & Math

[Anonymous No.7337389]

What is the shape of this curve, a sine wave or parabolic wave?

[Anonymous No.7337431]

>>7337389
Sine wave. Think of the height of the endpoint of every flame.

[Anonymous No.7337436]

>>7337431
The highest point isn't always a tip

[Anonymous No.7337644]

>>7337389
Sine, because it's made of points that come from a circle (sorry for my very bad english)

[Anonymous No.7337847]

>>7337389
Bumping so I have time to solve this.

[Anonymous No.7337945]

>>7337389
Assuming constant angular difference between adjacent boxes the curve should be parameterized by something like (a* t + cos(t),sin(t)). That is not necessarily a nice curve. Hell if a=1 it's not even differentiable.

[Anonymous No.7337947]

>>7337945
That was a=2. Here is a=1.

[Anonymous No.7338099]

Bumping again. I'm almost done with it.

[Anonymous No.7338102]

>>7337847
>>7338099

You don't need to bump every few hours

/sci/ is a very slow board

[Anonymous No.7338109]

>>7338102
I guess I should have checked the page number before bumping.

[Anonymous No.7338186]

This is a bit of a pain to solve analytically, but it's a very natural question. You could phrase it as: "what is the region swept out by the segment between a moving object and an object which orbits it circularly?" Someone ask the mathexchange people whether that is a well known question in some form. My searches didn't turn up anything. Anything relevant anyway.

[Anonymous No.7338190]

>>7337644
that only applies if the fire walls are all moving in the same direction in a complete circle at the same speed and are the same distance apart

if you vary any of those parameters then you won't have a sine wave anymore

[Anonymous No.7338240]

It's a Curtate Cycloid

[Anonymous No.7338246]

>>7338240
Not obviously. As the anon in the second post pointed out the highest point isn't necessarily a tip of a fire wall. Also the rate and direction of rotation doesn't have any basis in reality here, unlike the cycloid.

[Anonymous No.7338248]

>>7338186
I enjoyed reading your perspective on that

[Anonymous No.7338264]

>>7338186
Space. Swept out describes a void as if there's nothing there. If space can be swept away it simultaneously is replaced with space.

[Anonymous No.7338274]

I spent three hours solving this, and now my scanner won't work.

[Anonymous No.7338320]

>>7337389

>>7337847
>>7338274
My scanner still doesn't work, so I'm just going to have to use my phone camera.

Anyway, here are the parametric equations to the curve enveloping the fire balls. a_up=3.5 (the length of each fire rod), lambda=4*9=36 (the length of each cycle), and n denotes the tile on the ground from which each rod of fire protrudes from.

Here's a link to a graph of it:
https://www.desmos.com/calculator/qkwoeu4l3c

I will be posting my rough work too, so 1/??

[Anonymous No.7338325]

>>7338320
Here's the graph overlaid on the game screenshot.

[Anonymous No.7338328]

>>7338320
How did you go about solving this?

[Anonymous No.7338335]

>>7338325
looks gaussian to me.

[Anonymous No.7338336]

>>7338328
>I will be posting my rough work, so 1/??
Patience, I'm not done posting yet

>>7338320 <---forgot to add that that's post 2/??

Here, are the equations for the locus of the tips of the fire rods. As many anons have stated, this is not the curve that envelopes the fire rods, so it isn'tthe final answer. However, it is useful as an intermediate step.

3/??

[Anonymous No.7338367]

>>7338335
It is not gaussian, but in fact cyclic and technically piecewise (though I bypassed writing it in piecewise notation by using absolute values).

>>7338336
Here is a picture of the tips drawn in for emphasis, as well as a graph overlaid of the equations I posted in >>7338336.

4/??

[Anonymous No.7338372]

>>7338367
Here, I start to work out the equations for the enveloping curves. More to come.

5/??

[Anonymous No.7338376]

>>7338367
If not gaussian than the collective movement is most likely a sinc function. If it was a sine the slope at the beginning and at the top doesn't match.

[Anonymous No.7338377]

>>7338372
6/??

[Anonymous No.7338383]

>>7338376
I will show you that it is not sinc either. Look at my equations in >>7338320. Those are definitely not classical functions.

>>7338377
7/??

[Anonymous No.7338397]

>>7338383
Now we are getting close to the solution. Here is the graph of the equations posted in >>7338383 overlaid on the screenshot. You can see that it fits pretty well in most places, but deviates near the ends. This is because there is no cap for the value of a, when it should be truncated for a>3.5 and a<0. I will show how to do that in the next post.

8/??

[Anonymous No.7338418]

>>7338397
These are the final steps. The first two lines give a function that truncates a for a>a_max=3.5 and a<0. This means that for a<0, T(a)=0, for a>a_max, T(a)=a_max, and for everything in between, T(a)=a.

After that, the rest is just substitution, replacing a in the equations in >>7338336 with T(a(n)).

9/??

[Anonymous No.7338427]

>>7338418
Whoops, I should have said that that was my final post 9/9 instead of 9/??. That's all there is to it. If anyone wants me to clarify anything, let me know.

[Anonymous No.7338438]

Is there no intention to describe the wave when it reaches the furthermost block and then rotate along the underside?

[Anonymous No.7338445]

>>7337436
>>7337431
Even if you only considered the "tips" of the arms, it wouldn't be a pure sine wave. The vertical component of each would form a sine wave, but they are moved in the horizontal direction, as well.

It's a series of evenly (horizontally) distributed circles, the point at the end of each arm is a point on its corresponding circle, at an evenly distributed angle.

The result is not a simple y = f(x) function because the the graph has "loops."

[Anonymous No.7338449]

>>7338438
That would simply be the equations in >>7338336 with y truncated for y>0

[Anonymous No.7338451]

>>7338320
>>7338336
>>7338372
>>7338377
>>7338383
>>7338397
>>7338418
Google sure could have saved you a lot of work
https://en.wikipedia.org/wiki/Cycloid

[Anonymous No.7338460]

>>7338451
My intuition told me it was going to be something like a cycloid, but I wanted to prove it to be the case. Afterall, mathematics is all about proofs, and not just intuition.

[Anonymous No.7338466]

>>7338451
>>7338460
Also, it's not exactly a cycloid, like I showed and explained in >>7338397.

[Anonymous No.7338480]

>>7338367
Forgot to add the picture for this one. It should have been pic related.

[Anonymous No.7338485]

Here's another pic

[Anonymous No.7338489]

It looks like a Normal Distribution.

[Anonymous No.7338500]

Holy shit mate that is some quality posting.

[Anonymous No.7338502]

>>7338489
see >>7338335 and >>7338367

[Anonymous No.7338510]

>>7338372
Ignore the square here, that was something that I forgot to erase. Don't worry, that mistake did not carry over to the rest of my work.

[Anonymous No.7338524]

>>7337389
It seems to be a line centered on every half space that has an angle theta that follows (0, 2PI) at some rate.

So.. vectors at n=1, 2, 3, 4... with magnitude x at angle theta=5n, spinning at 2pi*t/T, where T is the time it takes to make one rotation.

So... 5*n+2*pi*t/T=theta(0)+rate*time=theta(n,t)

Not sure how to program this, but ultimately it will make you a field of lines with whole number spacing that are oriented +5° relative to n-1 that are moving at some rate 2*pi*t/T.

Good luck.

[Anonymous No.7338528]

>the day science and math actually did math

[Anonymous No.7338530]

>>7338528
IKR, we should archive this thread

[Anonymous No.7338655]

>>7338451
Yea Google can save tons of mathematicians a lot of work, but what's the point? Some people enjoy doing math, believe it or not!

[Anonymous No.7338809]

>>7338655
Definitely this. Spending hours working on a math problem and finally solving it is one of the best feelings for me.

[Anonymous No.7339193]

Assuming:
1 unit = 1 block
Flame length = 3.5 units
Flame increments 10 degrees per shift of one block/unit to the right

[Anonymous No.7340114]

>>7338451

[Anonymous No.7340120]

Holy fucking shit, thread of 2015

Can you do a formal write up on this some time?

[Anonymous No.7340126]

post a gif of this

[Anonymous No.7340149]

>>7340126
>of this
???

[Anonymous No.7340156]

>>7340126
>this
what

[Anonymous No.7340162]

>>7340149
>>7340156
of the Mario screenshot you dipshits

[Anonymous No.7340203]

>>7340162
I don't think OP just has gifs lying around of pictures he has.

[Anonymous No.7340229]

>>7340203
It was the super mario maker presentation.
https://youtu.be/b88eKjjQbeg?t=330

[Anonymous No.7340272]

>>7340229
Why do you need a gif if you have the video? Just make it yourself!

[Anonymous No.7340381]

>>7340272
I'm not that guy.
I just happen to know where it's from.
http://www.clipconverter.cc/
Also shit internet

[Anonymous No.7340389]

>>7340381
Oh, well then thanks, my bad for thinking you were him.

[Anonymous No.7340399]

>>7340126

[Anonymous No.7340414]

>>7340399
Is this oc? If so, can you make one with a wavelength of 36 units and with rods that are 3.5 units long? That way, it would match op's picture.

[Anonymous No.7340419]

>>7340126
>1000 hours in Matlab

[Anonymous No.7340430]

>>1000 hours in Matlab

[Anonymous No.7340458]

>1000 hours in Matlab

[Anonymous No.7340487]

>>7340229
Oh, so it was just rods spinning at the same angular speed in staggered starting positions? Then it's obviously going to be a cycloid.

[Anonymous No.7340506]

>>7340414

[Anonymous No.7340510]

>>7340414
>>7340506

[Anonymous No.7340525]

>>7340506
>>7340510
Oh man this is perfect

[Anonymous No.7340618]

>>7340506
>>7340510
>>7338320
https://www.desmos.com/calculator/qkwoeu4l3c
It looks like this is it. It seems to be a perfect fit.

[Anonymous No.7342050]

>>7340458
>1 line in maple

I made a more fluent one :)
cba to add balls, my 300euro potato laptop already took over 5mins to generate 100 frames.

[Anonymous No.7342386]

this thread is the balls

[Anonymous No.7343837]

Is there a name for this kind of a mechanical curve? I propose "The Mario Wave" as the official title. One of the gifs in this thread was named that.

[Anonymous No.7344100]

Given that a bar is about 4 squares long and each one’s angle has an offset of an eighth relative to the previous one, we can easily say that the tip of the <span class="math">n[/spoiler]th bar is given by:
<div class="math">P_n=\left( n+4\cos\frac{n\pi}{16};4\sin\frac{n\pi}{16} \right)</div>
Replace <span class="math">n\in\mathbb{N}[/spoiler] with <span class="math">t\in\mathbb{R}[/spoiler] for the parametric version.

[Anonymous No.7344242]

>>7344100
It is easy to solve for the tips, but as pointed out by >>7337436 , the highest point isn't always the tip. Anyway, the problem has already been solved in this thread:
>>7338320
>>7338325
>>7338336
>>7338367
>>7338480
>>7338372
>>7338510
>>7338377
>>7338383
>>7338397
>>7338418
>>7338427