Comments by Paul

  • How to make a Balboa robot balance, part 2: inertial sensors

    How to make a Balboa robot balance, part 2: inertial sensors

    - 27 March 2017

    That's a reasonable idea. I have never actually seen it fail to initialize the sensor. If it did occasionally fail, I might change that line as you suggested, but only after looking into the source of the problem and determining that there is nothing I can do about it. Our examples are not intended to be robust against communication failure, so if you need something more robust, you might also want to look into re-initializing the sensor if it fails during loop() and consider enabling the AVR's watchdog timer.

  • Building a Raspberry Pi robot with the A-Star 32U4 Robot Controller

    Building a Raspberry Pi robot with the A-Star 32U4 Robot Controller

    - 4 October 2016

    Thanks for the comment! I have updated the command as you suggested.

    -Paul

  • Paul's dead reckoning robot

    Paul's dead reckoning robot

    - 19 August 2015

    For anyone interested in building a robot like this, our new A-Star 32U4 robot controller includes most of the electronics shown here, on a single PCB. So if you use that, you can skip the breadboard and directly wire your batteries, motors, and sensors to the controller.

  • Paul's dead reckoning robot

    Paul's dead reckoning robot

    - 20 July 2015

    I have not made a wiring diagram (need to learn Fritzing!), but I still have the robot, so if there are any connections that are not clear in my post, I would be happy to check them for you.

  • Math on the Pololu website!

    - 27 June 2014

    Ben wanted me to post the details of the Gaussian integral, so here we go.

    It's simpler to think about the full integral from ``-oo`` to ``+oo``. Let's give that a name:

    ``lambda = int_-oo^oo e^(-x^2) dx``.

    Then, instead of trying to compute ``lambda``, we work on ``lambda^2``:

    ``lambda^2 = ( int_-oo^oo e^(-x^2) dx )^2 = int_-oo^oo int_-oo^oo e^(-x^2) e^(-y^2) dx dy``.

    We note that ``e^(-x^2) e^(-y^2) = e^(-(x^2+y^2))`` and rewrite in polar coordinates:

    ``lambda^2 = int_0^oo e^(-r^2) 2pi r \ dr``.

    Luckily, that integral is easy, since ``d / (dr) pi e^(-r^2) = - 2 pi r \ e^(-r^2)``:

    ``lambda^2 = - [ pi e^(-r^2) ]_0^oo = pi``;

    ``lambda = sqrt(pi)``.

    Finally, since the Gaussian is an even function, our desired integral is just half of that value:

    ``int_0^oo e^(-x^2) dx = sqrt pi / 2``.

  • Pocket-sized USB charger adapter

    Pocket-sized USB charger adapter

    - 6 March 2014

    Thanks for the nice feedback, and please let us know when you post more projects!

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