I'm trying to implement color-to-grayscale method from this paper. And they use Verlet's integration as:

$$L^*(t+\Delta t)=\frac{F(t)}{m}\Delta t^2+2L^*(t)-L^*(t-\Delta t),$$

for computing new value of luminance. But they not say how to compute new value of luminance for $t=0$. According to this Wikipedia link I think it could be as:

$$L^*(\Delta t)=\frac{F(0)}{2m}\Delta t^2+L^*(0)$$

Is it correct?

  • 1
    $\begingroup$ At time zero, no integration has happened yet, right? So there's some initial state before doing any integration; isn't that described in the paragraph starting with "The instantaneous force applied to a particle..." $\endgroup$
    – Guest
    Apr 14, 2018 at 18:32

1 Answer 1


The function varies as you increase the time step ∆t , t will be zero from the start, So at ∆t =0, the initial condition L*(0) is the gray scale value obtained by converting the color image into L* a* b* color space.

  • $\begingroup$ I don't think that $\Delta t$ should be set to 0. Because $\Delta t$ represents time difference between current value and next value, so if i want to compute new value after 0.01 seconds pass, i would set $\Delta t$ to 0.01. I use fixed timestep so $\Delta t$ is constant. You are right that $L^*(0)$ is obtained from the image. $\endgroup$
    – lived
    Dec 4, 2016 at 14:47

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