up vote 1 down vote favorite

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?

share|improve this question
  • 1
    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..." – Guest Apr 14 at 18:32
up vote 0 down vote

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.

share|improve this answer
  • 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. – lived Dec 4 '16 at 14:47

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.