Delaying a note in a chord?

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1

Is there a way to delay a note in a chord?
for example:

#this code:

play_chord [:c, :e, :g], attack: 0.05, amp: 2, release: 2
sleep 3

#into this:
play :c, release: 2
sleep 0.1
play :e, release: 2
sleep 0.1
play :g, release: 2

#but better of course...
2 
c = (chord :c4, :major)
use_synth_defaults release: 2

for n in c
  tick(:delay)
  time_warp (line 0,1).look
    play n
  end
end

Just a starting point, needs some tweaking.

2 Likes
3

Hi
“better” is debatable, your solution works ok. I don’t think you can arpeggiate the play_chord function, but this works:

crd = chord(:c4, :major)

puts crd

3.times do
  play crd.tick, release: 0.5
  sleep 0.1
end

PD-Pi

1 Like
4

This is a bit more flexible, allowing you to specify different times per note in an array:


define :arpeggiate do |thesenotes, thesedelays|
  (1..thesenotes.length).each do |i|
    play thesenotes.ring[i-1]
    sleep thesedelays.ring[i-1]
  end #loop
end #define


arpeggiate (chord :c4, "m7"), [0.125, 0.25, 0.375, 0.5]
1 Like
5

This is great @HarryLeBlanc . I took the liberty of simplifying this - for both myself and OP - and added the built in rotate and invert methods :wink:

define :arpej do |notes, rel=0.3, del=0.2|
  notes.length.times do |i|
    play notes[i], release: rel
    sleep del
  end
end

arpej (chord :c4, :maj, invert: 1).rotate(2)

PD-Pi

1 Like
6

I like!
But why don’t you run into an off-by-1 problem with this? Isn’t the first array element 0?

7

Yes, you’re right, but it’s because I’m using n.length.times. Your example is more sophisticated, using a range (1…thesenotes.length).each, which starts counting at 1. My arg i initialises at zero, as normal. For comparison:

define :arpeggiate do |thesenotes, thesedelays|
  (0..thesenotes.length-1).each do |i|
    puts i
    play thesenotes.ring[i], release: 0.4
    sleep thesedelays.ring[i]
  end #loop
end #define


arpeggiate (chord :c4, '6*9'), [0.125]

Sorry for muddying the waters :smiley:
PD-Pi

8

No, on the contrary, you cleared away some underbrush. You’re right, that’s so much simpler.
Ruby is slick!

1 Like
9

Woah, these are all such cool and interesting ways to think about this problem!! Thanks so much everyone :smiley: