[Solved] How to calculate interpolated point at a specific time `t` between two waypoints?

Hello all,
I am trying to figure out manually, the point p at which the layer will be present at a given time t. Consider I have a star layer and it’s origin is animated from (0, 0) to (60, 0). The interpolation type is linear. So according to the Synfig code, the in-tangent and out-tangent both will be equal to 60. But these needs to be scaled by a factor of 3.
So finally, to calculate the hermite curve, we need 4 control points(link).
So, for the hermite curve of x-axis:
P0 = 0
P1 = 60/3 = 20
P2 = 60/3 = 20
P3 = 60
Hence the curve will be written as:
P(t) = (1 - t)3 * 0 + 3(1 - t)2t * 20 + 3(1-t)t2 * 20 + t3 * 60
where 0 < t < 1

Let us take t = 0.5:
P(0.5) = 22.5
This is the answer we get after evaluating the curve. But intuitively and also the value at t = 0.5 in Synfig UI is 30.

Could anyone explain where am I wrong here, or where am I making the mistake. I have been stuck on this for quite a while. Any help would be much appreciated! :smile:
I am attaching the .sif file here: star_check.sif (2.8 KB)

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One more view regarding the above:
When the interpolation type is linear or when the interpolation is TCB but it is on the first waypoint(link), does Synfig still use the hermite curve to evaluate the expression?
I could not find anything else than hermite curve being used… Am I missing somewhere to look?
-Anish

@blackwarthog pointed, that Synfig uses Hermite curve, described here - Hermite Curve Interpolation

The formula is:

  x = ( 2*t*t*t - 3*t*t + 1 )*p0
    + (   t*t*t - 2*t*t + t )*p1
    + (   t*t*t -   t*t    )*p2
    + (-2*t*t*t + 3*t*t    )*p3

where p0, p3 - spline points;
p1, p2 - tangents.

When t = 1/2:

  x = ( 2/8 - 3/4 + 1  )*p0
    + ( 1/8 - 2/4 + 1/2)*p1
    + ( 1/8 - 1/4      )*p2
    + (-2/8 + 3/4      )*p3

and with p0 = 0, p1 = 20, p2 = 20, p3 = 60:

  x = ( 2/8 - 3/4 + 1  )*0
    + ( 1/8 - 2/4 + 1/2)*20
    + ( 1/8 - 1/4      )*20
    + (-2/8 + 3/4      )*60

  x = 20*8 - 20/8 + 60/2 = 30
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Thanks a lot @blackwarthog and @KonstantinDmitriev!
Now things are getting much clear than before. I guess now interpolation in the lottie-exporter will be much more similar to that of Synfig's actual interpolation. As I was using bezier curves before.
:smile:

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