## Pololu Blog » User Profile: Paul »

# Comments by Paul

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

- about 3 hours agoThat'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

- 4 October 2016Thanks for the comment! I have updated the command as you suggested.

-Paul

## Paul's dead reckoning robot

- 19 August 2015For 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

- 20 July 2015I 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 2014Ben 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

- 6 March 2014Thanks for the nice feedback, and please let us know when you post more projects!