batch slow down .mov speed

Question

So I have a simple problem, taking a large number of .mov files and slowing them down to half speed. For the life of me I can't seem to find a simple solution to my simple problem. I don't need to rename them, change the frame rate, or anything fancy, just slow down a large number of .mov files to half speed. I've looked into software and avisynth to maybe help, but nothing seem to do what I need. Any suggestions?

Answer 1

I had the same problems and solved it by remuxing and changing the frame rate to half. This can be done using tsMuxer, which has a command line interface. Downside is it can not output in MOV format, but in m2ts or ts. I wrote a simple python script to find all MOV files in the current directory, generate the input file to tsMuxer with framerate according to input and run it. Probably exists better ways, but this is one way.

import os
import sys

allFilesAndFolders = os.listdir(".")
cwd = os.getcwd()
fps = sys.argv[1]
filename = "Slower.META"

for entry in allFilesAndFolders:
    if entry.endswith(".MOV"):

        if os.path.isfile(filename):
            os.remove(filename)

        with open(filename, "a+") as myFile:
            myFile.write("MUXOPT --no-pcr-on-video-pid --new-audio-pes --vbr  --vbv-len=500\n")
            myFile.write("V_MPEG4/ISO/AVC, \"" + cwd + "\\" + entry + "\", fps=" + fps + ", insertSEI, contSPS, ar=As source, track=1\n")

        os.system("\"C:\Program Files (x86)\\tsMuxeR_2.6.12\\tsMuxeR.exe\" " + filename + " " + entry.split(".")[0] + "_slow.m2ts")