Can you flip an astroid inside out so it becomes a circle

Programmatically.

Skip to content
# Can you flip an astroid inside out so it becomes a circle. Programmatically.

###

Can you flip an astroid inside out so it becomes a circle

Programmatically.

no

Why

Just the vectors

I dont see how you can suddenly have deformations from what is not deformed set of coordinates that is an flipped inside out circle

Would it not just be a series of rotations on each curve?

Actually thinking about it a little more I don't think it's possible to do it, at least not without also deforming the shape in some.

Asteroids are too big to be flipped

yeah here you go, just write the program for it

...yea my point exa-

Well i was expecting the code for it

Or the general idea

Of the code.

the code is to rotate it 45 degrees, then multiply all points by -1

Draw circle that goes through 4 points

Delete the inner drawing

... no the idea is that the curve is already a circle and it just needs to be flipped to make the vectors a circle.

Link?

...why not just multiply all points by -1 to begin with.

>...why not just multiply all points by -1 to begin with.

because then it will look the same

no link

you can find it

>no the idea is that the curve is already a circle and it just needs to be flipped to make the vectors a circle

...And the curves make a circle that goes through those 4 points. Skip past the bullshit, cheat by drawing the circle.

yeah they successfully did this to a sphere on youtube without it breaking

What is determining the length of the line here? It's going too fast to see obviously. The line seems to be tangent to the circle at each point. I'm guessing it's something to do with the sin/cos?

>after 25 seconds of research

... not this. Basically moving the 4 sides into the other ends using a vector of some sort

>a vector of some sort

perhaps a magic vector of the non-mathematical sort

actually— do me a favor?

can you figure out if the line given by this parameteric equation with k=4, theta from 0 to pi/2

is equal to the curve of the unit circle when reflected across the line x=1-y

and with R=1, to state the obvious

well, are you winning son?

it seems like youd need to reflect across the four lines y=+/-x+/-1

actually shouldnt be terribly hard to figure out from there