Hi,

I'm still at the beginning of PD and learning bit by bit in college. Often we're sent away with only bits of information and then with the help of the famous Johannes Kreidler manual I dig around to make sense out of earlier tutorials.

At the moment I'm working on Arrrays and would like to know if with this simple patch (taken from JK's manual) I should only get clicks (or hisses). I would of thought that if I speak it should play back my voice etc, or have I misunderstood something?

Big thanks in advance.

note : I've normally opened the array on a subpatch as told [pd subpatch], I hope that's correct? The patch is just the standard patch from the manual .

http://www.pdpatchrepo.info/hurleur/sampler.pd

Thanks Nau, how do I use that formula? I looked at the first subject about using () or [] etc but couldn't come up with anything that made sense.

Help (and thanks).

Maybe you could simply resize the array loading the audio file using the file's length (in samples) outputted by a [soundfiler] object :

http://puredata.hurleur.com/sujet-3991-recive-sample-lenght-soundfiler-table

to convert duration (seconds) to length (in samples, just do something like

[bang]
|
[t b b]
| \
| \
| [samplerate~]
| /
[2( / <-- your value in seconds
| /

• |
  [88200 \ <-- the result

Nau

Big thanks Nau. As you can see I changed the number box to a print object. The number box just gives a + when the number is too big.

[bang]
|
[t b b]
| \
| \
| [samplerate~]
| /
[2( / <-- your value in seconds
| /

• |
  [print] <-- the result

The result is always rather strange when large. Ex : 60 seconds = 2.646e+06? Could you explain what the 'e+06' means, or maybe it's of no importance?

Thanks.

Hi joesh,

What values are you putting in?

I noticed that you were originally working out in milliseconds but nau's examples are all in seconds. Make sure that you're converting correctly between (1000 milliseconds to a second).

2.646e+06 stands for 2.646 * 10^6 . It's scientific notation for expressing very large or small numbers.

The output of your algorithm will give you the number samples (at a certain samplerate) for the inputted number of seconds.

Running at a samplerate of 44.1KHz means that for every 1 second there for 44100 samples.

In a 60 second period you'll have 44100 * 60 or 2646000 samples.

If you're working milliseconds make sure you divide your sample rate by 1000, as in this case there would 44.1 samples for every 1 millisecond.

Hope that helps.

Hey,

I'm looking for a way to make the array size dependent on the length of the recording instead of giving the array a fixed length that I can fill with data. Is there a way to do so?

Hey Elden
i use [writesf~] and than have it loaded into an array from disk with the appropriate array-size.

you can do so with the -resize flag:

[read -resize $1 array_name(
|
|
[soundfiler]

also nice to know: use the -maxsize <big_number (e.g. 1.5e+08)> flag for using arrays bigger than standard size.

jo

ah ok. seems to fit my needs... and is writesf~ always overwriting the same file?

Every time you start recording with [writesf~], you need to send the [open <filename>( message. Once you stop, it forgets the file. So it will only overwrite the file if you keep opening the same file.

You would want to send [writesf~] a [stop( message. (I'm finding that other ways out of the program don't always close the file cleanly. [Especially not crashing it!] )

And then send a new filename in the next [open <filename>( message if you want to record something else...

Hmm, I shouldn't have to leave pd to play back.

[open foo.wav(
|
| [rewind(
| /
[readsf~]
|
| [number] [better limit the range of values!!!]
| /
[*~ .4]
|
[dac~]

?