Linear, cubic and spherical interpolation

Lerp, the equal-power crossfade, cubic Hermite interpolation between samples, and slerp along the great circle between two directions.

September 5, 2026

8 minutes read

1492 words

Contents11

Given two values and a fraction, interpolation names a value in between. Which value depends on what the two are: numbers on a line, samples of a curve, or directions. Each case has its own rule, and the wrong rule gives an answer that lies between the inputs and is still wrong.

Assumed knowledge

Arithmetic, and the sine and cosine of an angle from angles and polar form . The last two sections use the length of a vector, unit vectors and the dot product, all of which matrices and vectors builds up. The cubic section uses the derivative of a polynomial.

Ground covered

Linear interpolation between two numbers. The crossfade, and the law that holds the power constant when the two signals are unrelated. Reading a signal between its samples, and why a straight line dulls the treble. The cubic through four samples and the coefficients it needs. Why a straight line between two directions leaves the sphere, and the rule that stays on it.

Linear interpolation

Between two numbers $a$ and $b$ , the value a fraction $t$ of the way from $a$ to $b$ is

\operatorname{lerp}(a, b, t) = a + t\,(b - a) = (1 - t)\,a + t\,b. \tag{1}

At $t = 0$ the result is exactly $a$ , at $t = 1$ exactly $b$ , and it moves at a steady rate between the two. The same formula works entry by entry for vectors.

Crossfades

A crossfade is equation (1) applied to two signals, with $t$ rising from 0 to 1 over some time. One signal fades out as the other fades in. For a signal that is being replaced by a slightly different version of itself, equation (1) is the whole answer.

For two unrelated signals it has a flaw. Halfway through, each is at half size. Power goes as the square of size, so each contributes a quarter of its power and the total is half. That is a dip of three decibels in the middle of the fade. The cure is to pick gains whose squares add to one.

y = \cos\!\big(\tfrac{\pi}{2}\,t\big)\,a + \sin\!\big(\tfrac{\pi}{2}\,t\big)\,b, \qquad \cos^2 + \sin^2 = 1 \tag{2}

Equation (2) is the equal-power crossfade. At $t = \tfrac12$ both gains are $1/\sqrt2 = 0.707$ , each signal contributes half the power, and the total holds. A dry and wet control on an effect is this formula with $t$ as the knob. The untreated sound and the reverberated sound are unrelated enough for the equal-power law to be the right one.

Between samples

A digital signal is known only at whole-sample positions. Delaying it by $2.3$ samples means reading it at a position that does not exist, so a value has to be manufactured from the neighbours. Equation (1) between the two samples on either side works, and it dulls the treble. Averaging two adjacent samples is a small low-pass filter whose strength depends on $t$ , so a delay that changes slowly makes the tone flutter. Passing a curve through four neighbours instead of a line through two removes most of that.

Cubic Hermite interpolation

Name the four samples around the target $y_{-1}, y_0, y_1, y_2$ , at positions $-1, 0, 1, 2$ . Ask for a cubic $p(\mu)$ on $0 \le \mu \le 1$ that passes through $y_0$ and $y_1$ , with the slope at each end matching the neighbours of that end.

p(0) = y_0, \qquad p(1) = y_1, \qquad p'(0) = \tfrac12\,(y_1 - y_{-1}), \qquad p'(1) = \tfrac12\,(y_2 - y_0) \tag{3}

Those slopes, each half the difference of the samples on either side, are the Catmull-Rom choice.¹ Write $p(\mu) = c_0 + c_1\mu + c_2\mu^2 + c_3\mu^3$ . The two conditions at $\mu = 0$ give $c_0$ and $c_1$ at once. The two at $\mu = 1$ are two equations in $c_2$ and $c_3$ , and solving them gives the other pair.

c_0 = y_0, \quad c_1 = \tfrac12\,(y_1 - y_{-1}), \quad c_2 = y_{-1} - \tfrac52\,y_0 + 2\,y_1 - \tfrac12\,y_2, \quad c_3 = \tfrac12\,(y_2 - y_{-1}) + \tfrac32\,(y_0 - y_1) \tag{4}

The cubic is evaluated from the inside out, which needs three multiplications rather than six.

p(\mu) = \big((c_3\,\mu + c_2)\,\mu + c_1\big)\,\mu + c_0 \tag{5}

Take the samples $1, 2, 4, 8$ and $\mu = \tfrac12$ . Equation (4) gives $c_0 = 2$ , $c_1 = 1.5$ , $c_2 = 1 - 5 + 8 - 4 = 0$ and $c_3 = 3.5 - 3 = 0.5$ . Equation (5) then gives $p = 2.8125$ , a little below the straight-line value of 3. The curve bends upwards, and the chord of a curve that bends upwards lies above it.

The four samples of the worked example, and the interval between the middle pair read again at a larger scale. At the halfway position the straight line reads 3 and the cubic of equations (4) and (5) reads 2.8125.

Directions and the chord

Take two unit vectors $a$ and $b$ , which are directions rather than positions, at an angle $\Omega$ to each other. Their linear midpoint $\tfrac12(a + b)$ has length

\big\lVert \tfrac12 (a + b) \big\rVert = \cos\frac{\Omega}{2} < 1, \tag{6}

because $\lVert a + b \rVert^2 = 2 + 2\cos\Omega = 4\cos^2(\Omega/2)$ . A straight line drawn from one tip to the other cuts through the inside of the sphere. Dividing by the length puts the point back on the sphere, and then the angle swept out is no longer proportional to $t$ . The point hurries in the middle and dawdles at the ends. Over a sweep of 120 degrees in eleven steps, the projected point advances 5.8 degrees at each end and 19.1 degrees in the middle. An even sweep would advance 12 degrees at every step. Where the direction is something a listener can hear changing, that unevenness is audible.

Eleven equal steps along the chord, pushed back onto the arc, against eleven equal steps of angle. The pushed points advance 5.8 degrees at each end of the sweep and 19.1 degrees in the middle, where an even sweep of 120 degrees advances 12 degrees at every step.

Spherical linear interpolation

The rule that stays on the sphere and turns at a steady rate is slerp.²

\Omega = \arccos(a \cdot b), \qquad \operatorname{slerp}(a, b, t) = \frac{\sin\big((1 - t)\,\Omega\big)}{\sin\Omega}\,a + \frac{\sin(t\,\Omega)}{\sin\Omega}\,b \tag{7}

The derivation is short. The wanted point lies in the plane of $a$ and $b$ , at angle $t\Omega$ from $a$ and $(1-t)\Omega$ from $b$ , so it is some combination $\alpha a + \beta b$ . Dotting that combination with $a$ and then with $b$ gives two equations, $\alpha + \beta\cos\Omega = \cos(t\Omega)$ and $\alpha\cos\Omega + \beta = \cos((1-t)\Omega)$ . Solving the pair uses the identity $\cos(t\Omega) - \cos\Omega\cos((1-t)\Omega) = \sin\Omega\,\sin((1-t)\Omega)$ , which is the expansion of $\cos(\Omega - (1-t)\Omega)$ , and returns the weights in equation (7).

Equation (7) has unit length for every $t$ . It sweeps the angle at a constant rate, reaching $t\Omega$ at fraction $t$ . It is exactly $a$ at $t = 0$ and exactly $b$ at $t = 1$ . What has to be guarded is $\sin\Omega$ in the denominator. When $a$ and $b$ are nearly the same direction that sine is nearly zero, and equation (1) followed by division by the length is used instead. The arc is then almost straight, so the substitution makes no visible difference.

Equation (7) does not care how many entries the vectors have. Two directions in six dimensions are interpolated the same way, along the great circle of the five-dimensional sphere that contains them.

Choice of rule

between	rule	why
two versions of one signal	linear, equation (1)	exact at both ends, and there is no power dip to fix
two unrelated signals	equal-power, equation (2)	holds the total power constant
samples of a signal	cubic Hermite, equations (4) and (5)	keeps the treble a straight line would dull
two directions	spherical, equation (7)	stays on the sphere at a steady rate

Figure sources

One script draws both figures. It evaluates equations (4), (5) and (7) directly and prints the two numbers the page quotes, so a wrong coefficient cannot reach the page unnoticed. It needs NumPy and Matplotlib.

interpolation.py

uv run --with numpy --with matplotlib python3 interpolation.py