0

As a simple solution to avoid dead air, I'd like to overlap each pair of consecutive songs in a playlist by ramping them down/up.

In effect this would be a poor man's DJ solution.

I wouldn't mind if it's entirely non-interactive, just so long as it is suitable for dinner parties, dance events, and the like. Indeed, the DJ software applications on the app store all seem to require an operator. Does an automatic solution exist, one that takes no more input than: 1. playlist, and 2. duration of overlap ?

0

In iTunes playback preferences (iTunes menu > Preferences… > Playback), you can toggle crossfading, and set it to take anywhere from 1 to 12 seconds.

enter image description here

|improve this answer|||||
  • Funny how closely the feature you point to matches what I was seeking. This covers the "dinner" scenario perfectly, but not quite the "dance" scenario. If a song has a very long introduction, it would be nice to either specify a starting point or to trim it. I know QuickTime Pro used to have a trimming feature. I'm sure I could trim through iMovie then extract the audio track. A slimmer feature would be nice. That's a question for another time. – Calaf Nov 1 '16 at 5:24
1

There is also a trimming function within iTunes that works separately from the crossfade solution proposed by @timothymh. This would work better for your dance scenario. If you right-click a song and select Get Info, you get an info window about the song. Under the Options tab, you can set a start and stop point for each song.

As an example, here I've set this song to start 20 seconds from the beginning.

enter image description here

|improve this answer|||||
  • And I can enter the starting and stopping times in boxes? What luxury! I was about to give up on iTunes and seek a way to parse a list of the filenames and starting times to generate one gigantic output using ffmpeg. But seriously.. I wish the crossfade and the start/stop times were parameters of a playlist, not global parameters. – Calaf Nov 2 '16 at 4:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .