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    Isadora connection with Blacktrax

    How To... ?
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    • DusXD
      DusX Tech Staff @Jamie Nesbitt
      last edited by

      @jamie-nesbitt

      I don't think it supports standards like OSC.

      I remember this being an issue for users previously.

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      • J
        Jamie Nesbitt
        last edited by

        I spoke with them and they said they can send dmx. Would that work?

        WolandW markM 2 Replies Last reply Reply Quote 1
        • WolandW
          Woland Tech Staff @Jamie Nesbitt
          last edited by Woland

          @jamie-nesbitt

          If it can send DMX, maybe you could convert the DMX data to MIDI notes or MIDI Show Control via an intermediary piece of software or hardware.

          Just a thought, I'm not familiar with Blacktrax.

          Best wishes,

          Woland

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          • markM
            mark @Jamie Nesbitt
            last edited by

            @jamie-nesbitt said:

            I spoke with them and they said they can send dmx. Would that work?

            With a DMX input device like the ENTTEC DMX USB Pro, yes.

            Does it send Art Net? We'll have ArtNet actors for v2.6 which will come out very soon.

            Best Wishes,
            Mark

            Media Artist & Creator of Isadora
            Macintosh SE-30, 32 Mb RAM, MacOS 7.6, Dual Floppy Drives

            J 1 Reply Last reply Reply Quote 3
            • SkulptureS
              Skulpture Izzy Guru
              last edited by

              The official protocol is:

              https://rttrp.github.io/RTTrP-...

              But I think it can send out artnet. I can double check. 

              Graham Thorne | www.grahamthorne.co.uk
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              • DusXD
                DusX Tech Staff
                last edited by

                The only problem I see with either DMX or ArtNet is the resolution. 
                With only 256 values, it may not be as exact as you would like.
                Once in Isadora you have much finer numerical control, so if you are for example controlling the placement of some Izzy map elements, you will be able to do this very smoothly.
                It is just the measures received from Black Trax will be 'steppy', so you will likely want to smooth / filter them out.



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                • FredF
                  Fred
                  last edited by

                  Blacktrax is a pretty pricey system, I would guess there are some high expectations for a project that can afford it. I would seriously look at using a media system that has the protocol integrated, there is even a touch designer addon for blacktrax. Apart from the steppy motion you will get losing resolution by using DMX, Isadora does not have the best set of 3d tools, which you would need to do tracking and mapping on moving objects. Once you do have a 3d viewport where you can manipulate objects and images, setting the FOV of the virtual camera correctly is not really possible in Isadora. Isadora is great for many things, but every software has its downfalls and it may be more sensible to use something more capable in this case.

                  http://www.fredrodrigues.net/
                  https://github.com/fred-dev
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                  • J
                    Jamie Nesbitt @mark
                    last edited by

                    @mark How difficult (programming wise) would it be for Isadora to receive their protocol ? 

                    1 Reply Last reply Reply Quote 0
                    • J
                      Jamie Nesbitt @Fred
                      last edited by

                      @fred If Isadora could receive their protocol would it be less steppy?

                      FredF 1 Reply Last reply Reply Quote 0
                      • FredF
                        Fred @Jamie Nesbitt
                        last edited by

                        @jamie-nesbitt 

                        In the protocol they use a long for most of the variables, but a float for acceleration and velocity:

                        double has 52 mantissa bits + 1 hidden bit: log(253)÷log(10) = 15.95 digits

                        float has 23 mantissa bits + 1 hidden bit: log(224)÷log(10) = 7.22 digits

                        DMX has 7 bit values, so yes it would be less steppy, but I believe that is only half your problem.

                        http://www.fredrodrigues.net/
                        https://github.com/fred-dev
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                        J 1 Reply Last reply Reply Quote 0
                        • J
                          Jamie Nesbitt @Fred
                          last edited by

                          @fredThanks for the help!

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