Trajectory Timestep

[hblumberg]

It seems that the timestep of the outputted trajectories is the minimum of:

• the timestep specified when executing the kernel (the dt argument to execute)
• the smallest timestep of the fieldset

Is this correct? Thank you!

[erikvansebille]

Yep, that is correct. If outputdt < dt, then the particles wouldn't have moved/changed, so it would nog make sense to output again.

[hblumberg]

Thanks so much for confirming!