Comments by Paul
about 3 hours ago
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.
4 October 2016
Thanks for the comment! I have updated the command as you suggested.
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.
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.
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``.
6 March 2014
Thanks for the nice feedback, and please let us know when you post more projects!