Good Evening ^_^

@anon How are you ?

alright

I need help with making an equilateral triangle

in Computer Graphics

and... I hit a roadblock

@LittleChild Use three circles.

Or the three third roots of unity.

@PeterTamaroff Would you like to see my method ?

computer graphics

Yes @anon :)

(sidenote: have you seen the mathematica stackexchange? that place is INSANE. they are some dedicated peeps)

I am quite used to back-ticking because we use that a lot when mentioning code on S.O.

I thought I might ask @anon for help :-/

what is the roadblock

Well, Assuming the user clicks at a point on the screen $X_1(200,200)$

Using that as the starting point, construct an equilateral triangle of side 100

I am left with two unknowns when I try the distance of a line formula

@LittleChild Don't you have a compass tool?

a single point and a side length is not enough to uniquely determine an equilateral triangle (and that point is supposed to be a vertex right?). I do not know what you mean by "distance of a line" formula or what the unknowns are.

How can I use a compass in a program ? :-/

@LittleChild Do you know GeoGebra?

should the triangle look like this ▲

I think he's programming something him or herself, Peter

@PeterTamaroff I am trying to do this in a language called Processing

in other words, is the base of the triangle supposed to be horizontal?

@anon Aha.

But maybe he can get ideas.

Length = $\sqrt{(X_2-X1)^{2}-(Y_2-Y_1)^{2}}$

@LittleChild No backslash on the minus.

I was using that.

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Yes, base is supposed to be flat

its the pythagorean formula

@LittleChild now, is the given "starting point" supposed to be a vertex of the triangle? and if so, which vertex?

Given two points , this is how you find the length of a line right ?

@LittleChild the formula I posted, not yours, yes

you'll want to do this with math called trigonometry

I know basic Trig. How do I proceed ? :-)

The side is supposed to be , say , 300 pixels.

the edges of the triangle can be viewed as vectors, which makes finding the other two vertices easier.

um hmm.. and ?

the vector of the left side of the triangle should be (L cos(60 degrees), L sin(60 degrees)) where L is the side length

@anon Wouldn't that be 30 degrees?

ok.

@Lord_Farin the interior angle of an equilateral triangle is ... ?

180

no, that's the sum of the interior angles

60

yes

So, how do I find the other two points ??

@anon I know, but if we look from the top vertex, both other vertices extend downward; if we want the base to be horizontal, we're essentially splitting the angle in half.

And yes, the co-ordinate system is different

it is uh.. 4th Quadrant of cartesian plane

The other vertex being at (L cos(-30 degrees), L sin(-30 degrees)).

@Lord_Farin I am computing the vector using the angle at the base, not at the top vertex

except that Y is considered +ve

@anon I will not further interrupt you, so as to minimise confusion.

Look here: both of you: graphics.comsci.us/images/scoord.gif :-)

@Lord_Farin if you go from the top vertex, you need to do (L sin30, L cos 30), because the triangle you're looking at is "sideways"

stares into the distance

the vertices of the equilateral triangle of side length $L$, starting from the top vertex and going counterclockwise, are $$(x,y),\qquad (x-L\cos60^\circ,y-L\sin60^\circ), \qquad (x+L\cos60^\circ,y-L\sin60^\circ)$$

runs to get pen and paper

jotted :)

Calculated.

@LittleChild You may profit from the identities $\cos 60^\circ = \frac 12, \sin 60^\circ = \frac{\sqrt3}2$.

I remember the standard table for common values

for Sin, Cos and Tan

@LittleChild Good. :)

other 3 are reciprocals

Hi

@Vivian A most pleasant evening to you. :)

Thank you. Nice to talk to you.

I saw a very cool slide before.

I do not how which tool was used.

@Vivian What was it about?

The slide was made like a map and has kind of 3D effect.

When you watch it, it would make you dizzy. But it is cool.

Is there anyone know it?

@Vivian It doesn't ring a bell for me, sorry.

@Lord Farin Thank you so much.

I google so hard, but...

you will like it when you saw it.

jeez

I find this so irrelevant

sorry

@Vivian What are you apologizing for?

Now, proceeding to make a Koch Snowflake

@Peter, you found it is so irrelevant

Are you guys good at fractals ? :)

@anon Ok a little more help

No

Lets say I have a screen of size 640 x 480

@Vivian you realize he was talking about the question he linked to right?

What is the largest KOCH FLAKE I can fit in ?

I mean, basically it will need to be an optimum size for equilateral triangle

measured by what? area or height or sidelength?

side

I will need to position the triangle perfectly

@anon, no. I am just in this room.

so that the fractal does not get drawn out of screen bounds

Koch Flake expands after first iteration and then stays in a finite area right ? @anon

it expands indefinitely and stays in a finite area

the first snowflake must actually be smaller than the screen in order for the subsequent iterations to also be contained within the screen area

Yes !

the sidelength of the triangle should min{width of screen,$\sqrt{3}/2\times$ height of screen}

its base should be positioned 1/4 of the screen size up from the bottom

calculating

The formula for finding other x , y pairs was

x-LCos60,y-LCos60

@anon right ?

Now, if I calculate where the base is supposed to be , how do I calculate the peak vertex ?

peak vertex = (center x-coordinate, base + $\sqrt{3}/2\times$ sidelength)

Wait, adding an image

@anon Lets proceed from here :-)

I found one coordinate which is not the peak vertex

I need other 2 coordinates

btw rounding the sidelength down to an integer isn't necessarily a good idea

415.69

I can get the second coordinate for the base as x + L

112.5 + 415.69

@anon right ?

y will remain 360

Am I correct in my assertion that the inverse of the parobala function (x^2) is not a function but merely a relation? And if we confine the range to only the positive real numbers that in becomes a function?

Do I have this correct?

@Jonn_Underwood Seems OK.

What is left is the PEAK @anon :-)

@PeterTamaroff wanna join in this fractal discussion ? :-)]

Is that true though? That the inverse of x^2 is not a function?

@Jonn_Underwood You're being vague.

A function is not merely a formula.

Understood

You need to specify a domain and a range.

For example $f: \Bbb R\to\Bbb R\; /\; x\mapsto x^2$

That is different from $f: \Bbb R_{\geq 0}\to\Bbb R\; /\; x\mapsto x^2$

As it seems you're trying to say, the latter has a functional inverse, and the former doesn't.

Isn't x^2 the same thing as a parobala?

@anon Hello ?

@Jonn_Underwood Didn't you read what I wrote?

(sorry, am doing other stuff)

@peter tamaroff yes but I didn't understand all the letters and forward slashes?

@anon Is my calculation correct in the pic ?

looks so

I'm on the iPad so if your using HTML it might not be working on my end

@Jonn_Underwood Oh, you need LaTeX.

Here

Ok I think I got it working but when I reload the Page it turns off(I think) and I cant add it to my book makes bar

It still looks like jibberish though

That's the page I have

Every time I switch between my tabs it reloads the page and turns it off

@anon are you free now ?

And I can't bookmark it like the instructions say to

@anon SUCCESS BRO !

brool story co

Drunk ? :D

No more close votes?

@PeterTamaroff @PeterTamaroff, it's called "Similarity Dimension"

@TheSubstitute Link? Made some research?

Interesting!

@anon With your help !

I positioned it, accidentally, in such a way that if you look carefully you will find that there is your photo on the right of the window

looking all indifferent :-))

@TheSubstitute web2.wzw.tum.de/ane/dimensions/subsection3_3_6.html

@PeterTamaroff web2.wzw.tum.de/ane/dimensions/subsection3_3_6.html I'm not sure who was the first to define it, but I came across it when reading "Introducing Fractals: A graphic guide."

Yes, same link

Anyone here ?

@LittleChild good job

@anon I have to do two image processing assignments

for my course

I have no clue what.

I want something mathematical

so it's up to you to figure out what to do?

YES

:)

K

But I want something that has some amount of Math in it

Will you help me out ?

hmm

There is some guy in the course who wrote an algorithm to create a pencil-drawn image out of actual image

What can I possibly do that is as good as his ? :)

@anon You in ??

what did that involve? detecting edges and then "drawing" the edges in "pencil"?

Wanna see the code ?

it's possible I'll help in passing with the math

That is what I need. I can do the coding part :)

takes an image (image.jpg) and creates a sketch version by selecting the most "relevant" points and drawing them according to proximity

description says that

What is the equation for a hyperbola?

@anon Or do you have a better idea ? :)

@Jonn_Underwood do you want a vertical axis of symmetry, or arbitrary?

Verticle axis Symmetry

@LittleChild are there any requirements? like, does it have to take an input image in and output another image or images? what about 3d or animations?

Anything as long it relates to digital media

I can do sound processing also

Sky is the limit :-)

fractals

@anon We can try this

do you work with fourier transforms of images at all?

I am on the way to fourier transform. I am learning derivatives and integration for it right ? :)

on the road map

or the Maldebrot Set

you seem interested in fractals. how about this: someone inputs an image and (in some or other fashion) a "region" (like a mirror or surface in the pic) is selected. the program takes a copy of the image, squishes and bends it appropriately into the selected region, and then iterates this indefinitely. (so it will end up looking like one of those infinite mirror things). one could even specify more than one region to paste onto..

Kaleidoscope thing ?

It is almost impossible to find a mirror in an image

@anon So what can we do ?

Just let me know what to do