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I need to get a curve that passes through three points with another requirement.

If I have points P0 P1 and P2 and a value Mu in the range $0..1$ representing where on the curve we are,

mu=0 should result in P0

mu=.5 should result in P1

mu=1 should result in P2

Also:

mu=.25 should result in the halfway point between P0 and P1

mu=.75 should result in the halfway point between P1 and P2

I need a Mu value of $.25$ or $.75$ is halfway along the curve between P0 and P1 or P1 and P2

Unfortunately, everything I've tried to do results in a curve whose midpoint of P0->P1 or P1->P2 is closer to $.255$ than to $.25$

I've tried a few different algorithms -- using this page http://blog.sklambert.com/finding-the-control-points-of-a-bezier-curve/#disqus_thread to figure out how to use P1 to find control points that I can use to make a cubic bezier curve mostly works, but the width of the tangent

I also just tried to straight up calculate the offset between the mid point of a quadratic curve and P1, then add $offset*abs((1-mu)^2)$ to the resulting bezier curve, which has a result pretty close, but also wrong. The bezier algorithm weights the samples towards P1, instead of evenly distributing them.

Is there a way to do what I want?

Thanks, Ben

Is there a way to do this?

Arnaldo
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Ben Andersen
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  • Of course you need a smooth curve, right? If not, two consecutive segments would be a simple solution. And I also suppose you wouldn't like a piecewise defined curve, but you should state that explicitly. Is a Bézier curve mandatory? – Intelligenti pauca Dec 22 '16 at 07:36
  • but what if the midpoint is not on the curve ? – mercio Dec 22 '16 at 09:20
  • @Ben Andersen You need to define halfway. Is it " approximately in between" ? – Narasimham Dec 22 '16 at 10:58
  • I'm not sure if this belongs in the original question. I'm using this to calculate curved motion blur based on the acceleration of a point through XYZ space. If I have a point starting at (1,0,0) rotating around Y 90 degrees per frame, I would expect mu=.75 on Frame 1 to match exactly with mu=.25 on Frame 2, since the curves should have the same acceleration and the mid points of the two curve segments should match.

    @Aretino - I do need a smooth curve, yeah. I don't mind calculating the two curves separately as long as the curve is continuous through P1.

     – Ben Andersen Dec 23 '16 at 00:36
  • Also you allow any smooth curve pass through the given points, Right? – Narasimham Dec 24 '16 at 16:42

1 Answers1

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Having $\mu=\frac14$ result in the point half way between $P_0$ and $P_1$ feels like a unit speed parametrization of that first half of the curve. If you do the same for the second half, you will get a discontinuity at $P_1$ unless that point itself is half way between $P_0$ and $P_2$.

Personally I'd take a projective view on this. Three points on a conic define a projective scale, so I'd consider something like the circle through these three points (which becomes the line through them if they are collinear). Usually a projective scale is defined in terms of $0,1,\infty$ instead of $0,\frac12,1$. But the point for $\mu=\infty$ can be constructed like this: construct tangents in $P_0$ and $P_2$. Connect the point where these intersect with $P_1$. That line will intersect the circle in a second point, which is the point for $\mu=\infty$. For the collinear case, using the tangents like this won't work, but there is an alternate way to construct a harmonic throw on a straight line. If $P_1$ is the regular euclidean midpoint between $P_0$ and $P_2$ then this point for $\mu=\infty$ would indeed be at infinity, and you get a regular constant speed parametrization of the line.

I've created a blog post to illustrate this approach. That may help you decide whether it satisfies your requirements or not. If you want to do some computations yourself, here is a formula which is the result of a quick Sage computation:

x = ((x0^2*x1 + y0^2*x1 - 2*x0*x1^2 - 2*x0*y1^2 - 2*x0^2*x2 - 2*y0^2*x2 + 6*x0*x1*x2 - 2*x1^2*x2 + 4*y0*y1*x2 - 2*y1^2*x2 - 2*x0*x2^2 + x1*x2^2 - 2*y0*x1*y2 + 4*x0*y1*y2 - 2*x0*y2^2 + x1*y2^2)*mu^2 + (-x0^2*x1 - y0^2*x1 + 3*x0*x1^2 + 3*x0*y1^2 + x0^2*x2 + y0^2*x2 - 6*x0*x1*x2 + x1^2*x2 - 2*y0*y1*x2 + y1^2*x2 + 3*x0*x2^2 - x1*x2^2 + 2*y0*x1*y2 - 6*x0*y1*y2 + 3*x0*y2^2 - x1*y2^2)*mu - x0*x1^2 - x0*y1^2 + 2*x0*x1*x2 - x0*x2^2 + 2*x0*y1*y2 - x0*y2^2)/((-x0^2 - y0^2 + 4*x0*x1 - 4*x1^2 + 4*y0*y1 - 4*y1^2 - 2*x0*x2 + 4*x1*x2 - x2^2 - 2*y0*y2 + 4*y1*y2 - y2^2)*mu^2 + (-2*x0*x1 + 4*x1^2 - 2*y0*y1 + 4*y1^2 + 2*x0*x2 - 6*x1*x2 + 2*x2^2 + 2*y0*y2 - 6*y1*y2 + 2*y2^2)*mu - x1^2 - y1^2 + 2*x1*x2 - x2^2 + 2*y1*y2 - y2^2)
y = ((-2*y0*x1^2 + x0^2*y1 + y0^2*y1 - 2*y0*y1^2 + 4*y0*x1*x2 - 2*x0*y1*x2 - 2*y0*x2^2 + y1*x2^2 - 2*x0^2*y2 - 2*y0^2*y2 + 4*x0*x1*y2 - 2*x1^2*y2 + 6*y0*y1*y2 - 2*y1^2*y2 - 2*y0*y2^2 + y1*y2^2)*mu^2 + (3*y0*x1^2 - x0^2*y1 - y0^2*y1 + 3*y0*y1^2 - 6*y0*x1*x2 + 2*x0*y1*x2 + 3*y0*x2^2 - y1*x2^2 + x0^2*y2 + y0^2*y2 - 2*x0*x1*y2 + x1^2*y2 - 6*y0*y1*y2 + y1^2*y2 + 3*y0*y2^2 - y1*y2^2)*mu - y0*x1^2 - y0*y1^2 + 2*y0*x1*x2 - y0*x2^2 + 2*y0*y1*y2 - y0*y2^2)/((-x0^2 - y0^2 + 4*x0*x1 - 4*x1^2 + 4*y0*y1 - 4*y1^2 - 2*x0*x2 + 4*x1*x2 - x2^2 - 2*y0*y2 + 4*y1*y2 - y2^2)*mu^2 + (-2*x0*x1 + 4*x1^2 - 2*y0*y1 + 4*y1^2 + 2*x0*x2 - 6*x1*x2 + 2*x2^2 + 2*y0*y2 - 6*y1*y2 + 2*y2^2)*mu - x1^2 - y1^2 + 2*x1*x2 - x2^2 + 2*y1*y2 - y2^2)

Here is how I came up with that formula. I concentrated on cross ratios, with the circle itself playing only a minor role. Along a line, the cross ratio $(P_0,P_2;P_1,Q)$ is $\infty$ for $Q=P_0$, $0$ for $Q=P_2$ and $1$ for $Q=P_1$. To make this fit in with your values of $0,\frac12,1$, you'd apply a (real) Möbius transformation to $\mu$ to map it to $\frac{\mu}{1-\mu}$. The cross ratio of four points on a given conic is the same as the cross ratio of these points seen from a fifth point. The ideal circle points $I=[1:i:0]$ and $J=[1:-i:0]$ lie on every circle. So we can characterize $Q$ by

$$(P_0,P_2;P_1,Q_\mu)_I=(P_0,P_2;P_1,Q_\mu)_J=\frac{\mu}{1-\mu}$$

Plugging in the equation for a cross ratio you'd get

$$\frac{[I,P_0,P_1][I,P_2,Q_\mu]}{[I,P_2,P_1][I,P_0,Q_\mu]}=\frac{\mu}{1-\mu}$$

where the square brackets denote determinants of homogeneous coordinates. Cross multiplication turns this into

$$(1-\mu)[I,P_0,P_1][I,P_2,Q_\mu]-\mu[I,P_2,P_1][I,P_0,Q_\mu]=0$$

This restricts $Q_\mu$ to a line. The homogeneous coordinates of that line can be computed by using $[I,P_2,Q_\mu]=\langle I\times P_2,Q_\mu\rangle$ and so on.

$$(1-\mu)[I,P_0,P_1](I\times P_2)-\mu[I,P_2,P_1](I\times P_0)$$

Doing the same for $J$ instead of $I$, i.e. applying a complex conjugation to the above, will lead to the description of a second line. Intersecting these two lines will give homogeneous coordinates of $Q_\mu$.

MvG
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  • Great, thanks, I'll take a look at graphing a circle between the points, then I should be able to use polar coordinates to pick different mu values, right?

    I have a hard time figuring out how to convert some of the explanations I see on the internet to python. Like this: http://klein.math.okstate.edu/~wrightd/INDRA/MobiusonCircles/node4.html

     – Ben Andersen Dec 23 '16 at 01:00
  • @BenAndersen: Personally I'd not use polar coordinates, but stereographic projection. With polar coordinates, you again have to decide what to do in case $P_1$ is not in the middle between $P_0$ and $P_2$. If you use the point for $\mu=\infty$ as center of a stereographic projection, the projected points will have $P_1$ centered between $P_0$ and $P_2$. – MvG Dec 23 '16 at 11:08
  • @BenAndersen: As a Christmas present I've worked out some formula for you. Explanation can come later, but this might help you use my idea. – MvG Dec 23 '16 at 15:35
  • @BenAndersen I've just created a blog post illustrating my approach, to give you something to toy with. – MvG Dec 30 '16 at 13:10
  • Wow, this is really helpful! That algorithm works really well. There's no way I'd have gotten that on my own.

    I made a mistake earlier when I said that P1 would be equidistant between P0 and P2, since it usually won't be. mu=.5 should still result in P1 though. I do notice that if P1 is not equidistant between P0 and P2, that mu=.25 and mu=.75 aren't equidistant between P0/P1 or P1/P2. I think that means there would be a break point at P1 for unit speed though, which is okay.

     – Ben Andersen Jan 09 '17 at 02:55
  • If P0 and P1 are very near each other and P2 is very far, then .75 is very very close to P1 at the moment, which is problematic :-( – Ben Andersen Jan 09 '17 at 06:38
  • @BenAndersen: If the break in speed is okay, you can pick any curve and do a piecewise constant speed parametrization. Circular arc would be one candidate again. Polar coordinates might work, but would fail badly if the points become collinear. Formulas would become more ugly as you'd get square roots all over the place. – MvG Jan 09 '17 at 07:05
  • Well, I got this working, but unfortunately using the circle method doesn't work if there is a sharp change in direction, like near the crest of a sine wave. I think I need to use some sort of parabolic arc, but I'm not sure it's possible to do that and have F(.25) give me 25% of the distance along the arc (exactly half way between P0 and P1).

    I'm not sure what you mean by a piecewise constant speed parameterization though.

    I was thinking that maybe I could do something like find the mid point of the straight line, then intersect with the curve, but I'm not sure how to do that either.

     – Ben Andersen Jan 10 '17 at 23:11
  • In general a curve is a function $f:[0,1]\to\mathbb R^2$, and it has constant speed if $\int_a^b\lVert\dot f(t)\rVert=(b-a)\cdot\text{const.}$, i.e. if length along the curve is proportional to the change in parameter value. In your case it would be constant speed in $[0,0.5]$ and in $[0.5,1]$ but the factor of proportionality would change at $0.5$. A parabola isn't uniquely defined by three points; even if you have four points there are two to choose from. Which one do you use? Computing arc lengths of conics other than circles is extremely tricky, see http://math.stackexchange.com/a/1123448 – MvG Jan 11 '17 at 07:39