I don’t know how I got to this state, but my Ardour project is about 24 hours long, yet there are no tracks that go for more than a few minutes. I think the END marker was moved by accident at some point. I have attached a picture.
On the pic you can see the “Tempo” extends to about 24 hours. It makes the UI pretty hard to use esp. when zooming an panning.
How can I bring the whole thing back to the length of the longest track (a few minutes)?
Check your locations window, pretty sure the END marker is in there IIRC and you can edit the time there, probably the quickest way to start. Just make sure there isn’t a random region that far down.
SeabladeThanks for your help Seablade. I checked the locations window too, but I don’t see the END marker there.
Yes, the END marker is at the end of the longest track. I placed the red cursor at the END marker where the longest track ends, and clicked “select all after edit point” and hit delete. I repeated on both tracks. It does not bring the whole project length back to the duration of the longest track yet. Here is another pic.
If you moved the markers manually once, they will not be automatically updated them anymore.
I think you are right Robin. I moved the markers by hand unintentionally. I found I can pan with the arrow keys, so that makes up for the unusable summary view.
I saw this in another post: Session:set_end_is_free(true);
But I get this error: Error: [string "Session:set_end_is_free(true);"]:1: attempt to call a nil value (method 'set_end_is_free')
Is there any hope for this? Can I edit the session XML file by hand to remedy this? I see in it that the end-is-free appears to be true already:
<Session version="6000" name="cade300" sample-rate="48000" end-is-free="1" session-range-is-free="1" id-counter="8065"
name-counter="1" event-counter="2294" vca-counter="1">
It’s
Session:set_session_range_is_free (true)
Yes, but if it’s already true, then there is a different issue.
The range should expand when you move regions after the current end (but it won’t ever shrink by itself).
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