Hi

Hey

I'm going to make a few graphs to accompany what I'm trying to say

In your 3d graphs, which axis is the vertical?

ok

So what you want is that all points in the parabola have the same y value?

Remove the Z axis for a moment. Forget about the height

Here's what the given points look like on the XY plane, exclusively

You can find the slope of that line segment (here it's 1, for simplicity sake)

Treating the vertex (2, 1) as the midpoint of a segment, and the first given point (1, 0) as the starting position, you can find the next point (3, 2)

I just think of it as "the parabola is concave in the -z direction".

Then you can add the Z axis back in, which is pretty simple, since you already have the Z values for the first two points, and the third point (the one we just found) has a Z value equal to the first point's (by definition of a parabola)

I'm not clear on what that graph is supposed to represent.

Even in a plane, there are 5 degrees of freedom for a parabola.

Not when the parabola is bound to one direction along one plane

Did you say that somewhere?

> You may assume that the parabola will always be vertical (continuing downwards).

You asked me to forget about that stuff.

And the first point given is the vertex.

Previously, you said that down refers to the Z-axis.

correct

So what is it supposed to mean in the xy-plane?

in the xy-plane, it's just a linear equation

which is how you find the third point

then you translate the xy-plane to the xz-plane and calculate the quadratic of the parabola

What's a linear equation?

a straight line

What's "it" in the phrase "it's just a linear equation"?

if you remove the Z axis, the points become a straight line alone the xy-plane

with these points, the equation of that straight line is y=x-2

Sorry, but you are not communicating clearly.

@El'endiaStarman: I'm grasping at straws trying to explain in a different way. any thoughts?

hmm

It would help if you answered my direct questions.

Like What's "it" in the phrase "it's just a linear equation"?

The parabola in 3D space will always be in the xz plane (where xy is horizontal and the z axis is up and down), but rotated by some angle with respect to the x-axis.

@El'endiaStarman Really? Where are you pulling this information from?

@feersum It's how I understand Zach's challenge. I think it's what he's going for.

I asked if all the points would have the same y coordinate earlier and got no response.

If the parabola is in the xz-plane, then all of its points have y = 0.

Yeah, they do not necessarily all have y=0. They could

So why don't you tell us what the restriction on the parabola is?

Oh, forgot the other restriction: the parabola is concave down (-z) and is symmetric with respect to the z-axis.

Opening down has no meaning in 3-dimensional space.

How so? There is only 1 "down" in 3D space

@ZachGates More like "down" has been defined here.

That also

Then it should be easy for you to give me a mathematical definition of opening down in 3 dimensions.

@feersum To put it another way, the vertex of the parabola has the maximal z coordinate.

@El'endiaStarman We already discussed that.

That restriction does not even decrease the number of DOFs from 9.

"symmetric with respect to the z-axis" does.

Actually, if the vertex is the apex along the z direction, why doesn't that decrease the DOF?

Symmetry along an axis does not even make sense.

In 3 dimensions, symmetry is over a plane.

Unless you mean rotational symmetry?

If you rotate it "upwards", don't the points near the vertex become maximal? I should investigate this in 2D...

Yeah, they do.

The slope at the vertex is 0, so once you rotate it even the slightest bit, it becomes non-zero, thus there is a point with a greater z-coordinate.

@El'endiaStarman To be brutally honest, you are adding more fuzzy thinking to the discussion, rather than clearing it up.

I'm not sure why you're having such difficulty understanding what Zach is trying to do.

If he's talking about mirroring an "endpoint" (so to speak) across the vertex, like you can do in 2D, there's pretty much only one way for that to work in 3D.

He refuses to clearly state any restrictions that may exist on the parabola.

I have not "refused" anything

If you mirror a point over another point you would get 3 collinear points, which is not what the picture in the sandbox post shows.

Yes, of course.

@ZachGates Can the parabola in question be any parabola, or are there some restrictions on it?

But how else would you describe the derived point?

@feersum The first two given points and the point that is calculated are all collinear in the 2D, xy-plane. The linear equation is y=x-2

I thought the parabola was supposed to be determined by the first 2 points of the input.

Okay, let's try this: the parabola is defined by z = ak^2 + bk + c, where a < 0 and k = (r cos(t), r sin(t), 0).

I think that notation is a bit messed up, but whatever. I hope the point comes across.

@ZachGates I'm asking about the parabola in 3 dimensions.

It's a concave-down (-z) parabola in the xz plane rotated about the z axis.

The first 2 points in the input are supposed to uniquely determine a parabola in 3 dimensions?

And with no further rotation about the vertex.

@El'endiaStarman It doesn't..

@ZachGates, do you understand what I'm trying to say?

yes. we're thinking the same thing

Okay, @feersum, what are the 9 DOFs of an unconstrained parabola?

3 for position, 2 for rotation, 1 for scale...?

Hmm maybe there are only 8? I can't remember how I got that now..

Still more than 6.

What am I missing?

There are at least 2 for shape, 3 for rotation and 3 for translation, is one way to look at it.

Right, yes, 3 for rotation. I was only thinking about a line, essentially.

2 for shape? How?

right, that should be 1

Okay, 7 DOF.

3 are taken by the first position given, the position of the vertex.

Does a parabola have a well-defined axis of symmetry? I think it does.

-b/(2a)

I keep getting 8 when I calculate it another way.

huh

Explain?

Do you agree that a parabola in a plane has 4 DOFs?

Position is 2, scale is 1, and rotation is 1. Yeah.

Never mind, 7 seems to be right.

It's that line where a rotation of 180 degrees around it makes no difference.

@feersum ^ agree?

@El'endiaStarman Yeah, what's your point?

Okay, just wanted to make sure we were on the same page.

3 DOF are used by locating the vertex, and another 2 DOF are used by specifying that this axis of symmetry is parallel (not skew) with the z-axis.

At this point, there are 2 DOF left.

I'd like to hear from Zach exactly what the restrictions on the parabola are instead of making more guesses.

Specifying another point takes care of both (in most cases). It fixes both the third(?) Euler angle and the scale.

Incidentally, I realized that my description earlier is slightly wrong.

Lemme try to rephrase it.

Okay, vertex form in 2D is y = a(x-h)^2 + k.

Let's say the parabola's vertex is at (a,b,c).

If I could easily make an animatable model of this, I think that'd help a lot.

@El'endiaStarman I'm not sure what there is left to discuss.

@ZachGates just needs to add a clear description of the restrictions on the input parabola.

I'm not able to explain any more clearly than I previously have

"Vertical" is your best effort?

x(t) = t*cos(theta)
y(t) = t*sin(theta)
z(t) = (x-a)^2 + (y-b)^2 + c

That's the simpler form, the first version I described.

@El'endiaStarman What on earth is that.

x(t) = t*cos(theta) + a
y(t) = t*sin(theta) + b
z(t) = (x-a)^2 + (y-b)^2 + c

^ that allows for an additional rotation about the axis parallel to the z-axis and passing through the vertex of the parabola.

Some kind of mixture of cartesian and cylindrical coordinates and parametric equations?

@El'endiaStarman (here's a theta for you: θ ) :P

Ah, theta is supposed to be a parameter to determine the parabola?

That's just my way of saying that the "shadow" of the parabola on the xy plane is a line.

In the first version, it must pass through the origin. In the second, not necessarily.

@ZachGates Please answer whether any parabola is allowed, or only certain parabolas; and if the latter, which parabolas are excluded from consideration.

The vertex of the parabola has the maximal z value

That's the only restriction?

All other points on the curve will have a lower z value than the vertex

That restriction alone is not enough to determine the parabola.

@El'endiaStarman had a good suggestion though.

What are we discussing? How to clarify what the parabola is?

That the axis of the parabola should be parallel to the z-axis.

@ThomasKwa In a sense? The challenge currently says that you are given the vertex of a parabola in 3D space and another point on the parabola, then asks you if a third, given point is on the parabola. We're discussing restrictions.

Why do we need to specify that it opens down? Isn't that implied by the other point being below the vertex point?

Yes

I'm not sure why there's any confusion

@El'endiaStarman I think the opening straight down is enough of a restriction

Think about it this way.

Translate all the points so the vertex is at the origin.

Then rotate them so point 2 is in the xz plane.

There's exactly one parabola that has a vertex at the origin and points straight down and contains point 2.

It's x^2=cz

Right?

I agree.

Agreed (I think)

This?

@ZachGates @ThomasKwa @feersum ^

@El'endiaStarman I have no clue, honestly