|
ACS37200LLXTR-150B3 Current Sensor Large Carrier -150A to +150A, 3.3V |
|
Breakout for JST PH-Style Connector, 4-Pin Male Top-Entry, with SH |
|
12V, 125mA Isolated Step-Up Voltage Regulator U5Z9F12 |
|
QTRXL-MD-01RC-S Reflectance Sensor: 1-Channel, 8mm Wide, RC Output, Long Range, Side-Entry Connector |
|
Magnetic Encoder Pair Kit with Top-Entry Connector for 20D mm Metal Gearmotors, 20 CPR, 2.7-18V |
|
Breakout for JST PH-Style Connector, 6-Pin Male Top-Entry, with SH |
|
ACS37200LLXTR-150B5 Current Sensor Compact Carrier -150A to +150A, 5V |
|
2.5-9V Adjustable Step-Up/Step-Down Voltage Regulator S9V11E2A |
|
Glideforce GF01-120501-1-66 Micro Linear Actuator with Limit Switches: 12V, 2.2kgf, 28mm/s, 10mm Stroke |
|
ACS37200LLXTR-100B5 Current Sensor Compact Carrier -100A to +100A, 5V |
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!