i have this proplem with buffer~ en groove~'s sample length.

i made a patch were i can replace or record audio into buffer. when i replace an audio file buffer~ bangs its info to info~ when done reading. immediatly the samplelength is taken from info~ and sent to groove. this works allright. 

but now when i did this and i want to record a file from audio in in that buffer instead groove~(or buffer~ i don't know) takes the previous sample length (from the file i replaced in buffer~ before) as a maximum. (i time the milliseconds between start and stop recording bangs (with timer~) and send that to groove~'s right inlet to set the sample lenght). so if i wan't to record something wich takes longer then that sample lenght it will stop looping @ that previous sample length. but if i record something wich is shorter it adapts perfectly. but then if i want to record something longer it stays @ that previous maximum length.

if i don't replace a sample in the buffer from opening the patch but only record samples it works perfectly, it just takes the maximum sample length i did set in buffer(10000ms) when i reach it.

and when i replace anonther file in this buffer wich takes longer it also adapts nicely.

i have this proplem with buffer~ en groove~'s sample length.
i made a patch were i can replace or record audio into buffer. when i replace an audio file buffer~ bangs its info to info~ when done reading. immediatly the samplelength is taken from info~ and sent to groove. this works allright. 
but now when i did this and i want to record a file from audio in in that buffer instead groove~(or buffer~ i don't know) takes the previous sample length (from the file i replaced in buffer~ before) as a maximum. (i time the milliseconds between start and stop recording bangs (with timer~) and send that to groove~'s right inlet to set the sample lenght). so if i wan't to record something wich takes longer then that sample lenght it will stop looping @ that previous sample length. but if i record something wich is shorter it adapts perfectly. but then if i want to record something longer it stays @ that previous maximum length.
if i don't replace a sample in the buffer from opening the patch but only record samples it works perfectly, it just takes the maximum sample length i did set in buffer(10000ms) when i reach it.
and when i replace anonther file in this buffer wich takes longer it also adapts nicely.
what could i be doing wrong?


buffergroove-sample-length-problem

right before recording newly into buffer:

send message "clear" & "size 60000" to the [buffer] and "60000" to the [record] object *for 60 seconds time) aswell as "reset" to [groove]. while recording have a timer running simultaneously, which should be stopped when recording stops and send its timed value to teh groove object loop max input then. am i clear?

it works here now, but my patch is momentarily such a mess, that you wouldns see thru if i posted it.

i had the same problem. 
right before recording newly into buffer:
send message "clear" & "size 60000" to the [buffer] and "60000" to the [record] object *for 60 seconds time) aswell as "reset" to [groove]. while recording have a timer running simultaneously, which should be stopped when recording stops and send its timed value to teh groove object loop max input then. am i clear?

Thanks alot! Clear as can be. I'll try it out in a moment.

good luck! i'd be glad to supply you an example patch as soon as i cleaned up my undocumented spaghetti mess :)

it works fine. thanks a-lot. you helped me out big time? how much (kilo/)megabytes of ram would a minute recording take?

any other format equivalently (2channel/44.1khz/16bit = 10 mb, 2 channel 44.1.khz/24bit = 15 mb etc...simple math huh?)

just try it out: record one channel with above sample rate and bitdepth for one minute and see how large it gets, i should be about right about the size - us an [sfrecord] object for that, tutorials tell how exactly that works.

1 min as 1-channel 44.1 kz/16bit: 
Approx. 5 MB, i believe.
any other format equivalently (2channel/44.1khz/16bit = 10 mb, 2 channel 44.1.khz/24bit = 15 mb etc...simple math huh?)

400 buffers of 30 seconds should be enough ;).

thanks. 
400 buffers of 30 seconds should be enough ;).


that's nasty, do you run a supercomputer?

http://store.apple.com/1-800-MY-APPLE/WebObjects/AppleStore.woa/wa/RSLID?nnmm=cart&sf=wHF2F2PHCCCX72KDY

it's not like i have a supercomputer:
http://store.apple.com/1-800-MY-APPLE/WebObjects/AppleStore.woa/wa/RSLID?nnmm=cart&sf=wHF2F2PHCCCX72KDY


oops, that was a shopping cart item you posted... nevertheless i assume you have at least 3gb of ram.

oops, that was a shopping cart item you posted... nevertheless i assume you have at least 3gb of ram.
i have only 2, good enough maybe :)

i have 2 gigs to. divided it with 5mb and you get 400. just to joke around, 200 will be good enough to i guess :).

hehe, don't know were anybody would need that for but it would be a hell of a patch :).

Mac Pro
Part Number: Z0EM
Two 3.2GHz Quad-Core Intel Xeon
32GB (8x4GB)
Mac Pro RAID Card
1TB 7200-rpm Serial ATA 3Gb/s
1TB 7200-rpm Serial ATA 3Gb/s
1TB 7200-rpm Serial ATA 3Gb/s
1TB 7200-rpm Serial ATA 3Gb/s
NVIDIA Quadro FX 5600 1.5GB
Two 16x SuperDrives
Apple Mighty Mouse
Apple Keyboard + User's Guide
Accessory kit
$18,899.00

Quote: nit wrote on Thu, 13 November 2008 15:56

----------------------------------------------------

Haha. I love the game "let's build a mac pro that's more expensive than a car". I told my friends about this once and they didn't believe me so I had to show them.

It's kind of crazy. I guess if you are rendering a commercial CG animated movie maybe you need something like that.

Quote: nit wrote on Thu, 13 November 2008 15:56
----------------------------------------------------
> Mac Pro
> Part Number: Z0EM
> Two 3.2GHz Quad-Core Intel Xeon
> 32GB (8x4GB)
> Mac Pro RAID Card
> 1TB 7200-rpm Serial ATA 3Gb/s
> 1TB 7200-rpm Serial ATA 3Gb/s
> 1TB 7200-rpm Serial ATA 3Gb/s
> 1TB 7200-rpm Serial ATA 3Gb/s
> NVIDIA Quadro FX 5600 1.5GB
> Two 16x SuperDrives
> Apple Mighty Mouse
> Apple Keyboard + User's Guide
> Accessory kit
> $18,899.00
>

It's kind of crazy. I guess if you are rendering a commercial CG animated movie maybe you need something like that.


> 1 min as 1-channel 44.1 kz/16bit: Approx. 5 MB, i believe. any other

> format equivalently (2channel/44.1khz/16bit = 10 mb, 2 channel

> 44.1.khz/24bit = 15 mb etc...simple math huh?)

its twice as that, as all audio is 32-bit float in memory.

Btw. the size message to buffer is clearing the buffer, a clear message 

is not necessary. But as always, the order of execution is crucial, and 

jayrope schrieb:
> 1 min as 1-channel 44.1 kz/16bit: Approx. 5 MB, i believe. any other
> format equivalently (2channel/44.1khz/16bit = 10 mb, 2 channel
> 44.1.khz/24bit = 15 mb etc...simple math huh?)

Btw. the size message to buffer is clearing the buffer, a clear message 
is not necessary. But as always, the order of execution is crucial, and 
can be controlled with t ...

-- 
Stefan Tiedje------------x-------
--_____-----------|--------------
--(_|_ ----|-----|-----()-------
-- _|_)----|-----()--------------
----------()--------www.ccmix.com


stefan, i meant the recoring filetype, which one can freely choose from a variety of 16, 24, 32 int and float types by sending for instance "samptype int24" to teh recording [sfrecord] object.

Quote: stefantiedje wrote on Fri, 14 November 2008 13:02

> > 1 min as 1-channel 44.1 kz/16bit: Approx. 5 MB, i believe. any other

> > format equivalently (2channel/44.1khz/16bit = 10 mb, 2 channel

> > 44.1.khz/24bit = 15 mb etc...simple math huh?)

> its twice as that, as all audio is 32-bit float in memory.

> Btw. the size message to buffer is clearing the buffer, a clear message 

> is not necessary. But as always, the order of execution is crucial, and 

and what would be the right order do you think?

Quote: stefantiedje wrote on Fri, 14 November 2008 13:02
----------------------------------------------------
> jayrope schrieb:
> > 1 min as 1-channel 44.1 kz/16bit: Approx. 5 MB, i believe. any other
> > format equivalently (2channel/44.1khz/16bit = 10 mb, 2 channel
> > 44.1.khz/24bit = 15 mb etc...simple math huh?)
>
> its twice as that, as all audio is 32-bit float in memory.
>
> Btw. the size message to buffer is clearing the buffer, a clear message 
> is not necessary. But as always, the order of execution is crucial, and 
> can be controlled with t ...
>
> Stefan
>
> -- 
> Stefan Tiedje------------x-------
> --_____-----------|--------------
> --(_|_ ----|-----|-----()-------
> -- _|_)----|-----()--------------
> ----------()--------www.ccmix.com
>
>
----------------------------------------------------

and what would be the right order do you think?


try the order out yourself - and keep in mind that any object further to the right executes earlier.

buffer~/groove~ sample length problem