Parabola between two points in a plane

RiemannLives

Given two points, p1 and p2 (each of which have components x and y), how would I find a set of parametric equations of t for the x and y components of a parabola between them?

(note: this is not 'homework', this is actually going to be used in plotting the path of an animation in a program I am writing)

EG: how to find the equations x = f(t) and y = g(t) that will plot a parabola between the points p1 and p2 in a 2d space.

End

I'm not very mathy, but don't you need three points to uniquely define a parabola? And that's assuming it has a horizontal axis of symmetry.

I guess if you define one point as the root, you could bring it down to two.

RiemannLives

End wrote: »

I'm not very mathy, but don't you need three points to uniquely define a parabola? And that's assuming it has a horizontal axis of symmetry.

I guess if you define one point as the root, you could bring it down to two.

You are right in that 2 points will not identify a unique parabola, but it should just have a constant left in the equation basicially indicating how "stretched" it is normal to the line between the two points.

Lykouragh

by "between" you mean that both points lie on the parabola?

Lykouragh

assuming the above is true, procedurally i would just solve the quadratic y=ax^2+bx+c in the normal way given your two points (use each point to eliminate x and y and generate an equation of two variables and a parameter) and spit out the parametrized solution

offhand i have no idea how to tell a computer to "solve a quadratic" though

let's assume we're parametrizing in c and i'll try to do it by hand though

we have (x1, y1) and (x2, y2)

y1=a(x1)^2+b(x1)+c
y2=a(x2)^2+b(x2)+c

solve for a and b, plug back into y=ax^2+bx+c, to obtain parametrized version solve for x and y in terms of c

Aioua

Now, I don't know my maths, but I just want to clarify one point.
I assume you're talking about having the points equidistant from the center like so:


In which case I would think you could contrive some equasion with a 3rd variable that would determine its steepness.

However, if you're trying to apply a parabola to the points unevenly:


I have a feeling finding an equasion would be nigh impossible, or at least a very different and harder equasion.

Also you could just draw a straight line. It approximates a very, very, very, very shallow parabola

Clipse

Like Aioua said, it's much easier if you assume that the vertex of the parabola projects to the center of the line segment between x1 and x2.

Let x1, x2 be vectors in R^2, take n to be normal to x2-x1 (choice of normal doesn't matter), then

t*(x2 - x1) + x1 + a*t*(t-1)*n

with a being the stretch factor is the family of parabolas you're looking for.

If you can't assume that the vertex of the parabola projects to the center of the line segment... well, good luck!

RiemannLives

Ah, thanks for the help guys. I think I found my answer: Bezier curves.

http://en.wikipedia.org/wiki/B%C3%A9zier_curve

Clipse

RiemannLives wrote: »

Ah, thanks for the help guys. I think I found my answer: Bezier curves.

http://en.wikipedia.org/wiki/B%C3%A9zier_curve

Man, I haven't done anything graphics related in so long I didn't even think about splines. Since you only care about the startpoint and endpoint, you might want to consider stepping up to cubic and using Hermite splines instead, though. They allow you to control the tangent at the startpoint and endpoint, which can be very useful for animation.

RiemannLives

Clipse wrote: »

RiemannLives wrote: »

Ah, thanks for the help guys. I think I found my answer: Bezier curves.

http://en.wikipedia.org/wiki/B%C3%A9zier_curve

Man, I haven't done anything graphics related in so long I didn't even think about splines. Since you only care about the startpoint and endpoint, you might want to consider stepping up to cubic and using Hermite splines instead, though. They allow you to control the tangent at the startpoint and endpoint, which can be very useful for animation.

ah, good point. Especially since I already have the values for the tangent at the start and end pre-calculated (as they are the speed and facing of the object at the start and end points).

Savant

Yeah, like has been said before, you need 3 points to interpolate a parabola. If you want to fit those points exactly, then what you'll do is Polynomial interpolation, which is sort of an extension of the earlier idea of setting up systems of equations, but using matrices instead. You probably only want to do that at the higher end when you aren't fitting data but simply want the exact polynomial through a set of points.

Least squared error type regression and B-splines are better for other types of problems where you want to fit a non-polynomial with a simpler function, depending upon what you are trying to do. B-splines are piecewise polynomial functions, not a single polynomial, by the way.