This post is part of my Game Math Series.
While processing data for skeletal animations, we are usually faced with a series of discrete samples of positions and orientations. The positional samples are typically stored as a series of 3D vectors, and the orientational samples are typically stored as a series of quaternions. The most straightforward way to interpolate between positional samples is using piece-wise lerp (linear interpolation), and the counterpart for orientational samples is using piece-wise slerp (spherical linear interpolation). For more information on slerp, please see my previous post on quaternion basics.
The samples are sometimes too far apart, and we can see the visual artifact of discontinuous change in the first-order derivative of interpolation, i.e. the interpolation is not smooth at sample points.
In this post, I will present to you a technique for interpolating orientational samples in a smooth fashion called circular blending. I leaned about this technique from the MAT 351 class at DigiPen, taught by professor Matthew Klassen.
Let’s say we are given a series of quaternions:
Let and denote the two quaternions we are trying to interpolate between, and let denote the interpolation parameter (). Also, let denote the interpolating curve between and .
If we are just using the straightforward slerp approach, we get:
This is a curve, meaning the curve is only continuous up to the zeroth-order derivative, i.e. the curve itself. The first-order derivative is generally not continuous at sample points using this approach.
Circular blending gives us a nice curve, which means the curve is continuous up to the first-order derivative. It is difficult to visualize quaternions, so I will use a 2D analogy to explain how circular blending works and how to mathematically work it out.
In order to interpolate between two quaternions and using circular blending, we also need the two neighboring samples and . Let the figure below represent the four sample points:
If we just use piece-wise slerp, this is what the curve will look like:
We can easily see the abrupt change of slope at sample points.
To prepare for circular blending between and , we draw two circles; one passes through , , and ; the other one passes through , , and . Let us denote these two circles and , respectively. Also, let us denote the arcs on these circles going from to as and , with and .
The formula for circular blending between and is as follows:
It is as simple as taking the slerp between the two arcs connecting and . The arc fully contributes to the slope at , and the arc fully contributes to the slope at .
So why does this give us a curve? Let’s add the curve between the next pair of samples, and . We need another sample in order to draw the circle .
Notice how the arc fully contributes to the slope of at . The arc is part of the same circle as the arc , so the slope is continuous at the sample point .
Now let’s look at how we can find these circles and the desired arcs.
Details & Derivation
Given three points, , , and , we would like to find a circle that passes through these points. Let denote the center of this circle. Also, we would like to find the parameterized arc that goes from to , where and .
Let denote the vector from to , and let denote the vector from to . Let denote the midpoint between and , and let denote the midpoint between and .
If we draw the bisectors of and , they should pass through the center of the circle, because a bisector of a line segment connecting two points on a circle always passes through the center of the circle. The bisectors are perpendicular to their corresponding vectors and . Let the direction vectors of the two bisectors be denoted as and .
To find and , we use the formulas:
where denotes the projection of the vector onto the vector .
Now we have the parameterized formula for the two bisectors, and :
The center of the circle is at the intersection of these two bisectors, so we need to find the parameter pair that satisfies:
If we rearrange the equation, we get:
Remember that we are working with quaternions, so the vectors , , and are all 4D vectors. This means that we have four equations for two unknowns, which is more than enough. All we have to do is to pick two equations and use Cramer’s Rule to solve for . Beware that the two equations you choose might not have a solution, i.e. you get a zero determinant when applying Cramer’s Rule; so be sure pick two equations that do not give you a zero determinant when solving for . We can obtain by plugging either or back into its corresponding bisector equation.
Now that we have the center of the circle , the last step is to find the parameterized arc where and . We aim to find the arc in the following form:
where is the angle between and , so ; is the radius of the circle; and form an orthonormal basis of the plane that contains the circle.
Finding is easy. It is the unit vector pointing from to :
As for finding , we first find the unit vector pointing from to :
and then we can find by taking out from its parallel component to :
Visually, here’s the whole picture:
We are done! We have found the circle that passes through the three points, as well as the parameterized arc on the circle that satisfies and .
One last thing. You might wonder what we should do if the three points are collinear. There’s no way we can find a circle with finite radius that passes through three collinear points! Remember that we are working with unit quaternions here. Three different unit quaternions would never be collinear because they lie on three different spots on the 4D unit hypersphere, just as three different points on a 3D unit sphere would never be collinear. So we are good.
Finally, let’s look at a video comparing the results of piece-wise slerp and circular blending in action.