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ACS37041KLHBLT-010B5 Current Sensor Micro Carrier -10A to +10A, 5V |
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JST SH-Style Cable, 4-Pin, Female-Female, 40cm |
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ACS37800KMACTR-030B3-I2C Power Monitor Carrier with Secondary I²C Isolation (Soldered Terminal Block) |
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ACS37030LLZATR-020B3 Current Sensor Compact Carrier -20A to +20A, 3.3V |
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ACS37041KLHBLT-010B3 Current Sensor Compact Carrier -10A to +10A, 3.3V |
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Glideforce GF01-120501-1-66 Micro Linear Actuator with Limit Switches: 12V, 2.2kgf, 28mm/s, 10mm Stroke |
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CT433-HSWF70MR TMR Current Sensor Compact Carrier -70A to +70A, 3.3V |
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JST SH-Style Cable, 2-Pin, Single-Ended Female, 12cm |
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JST SH-Style Cable, 2-Pin, Single-Ended Female, 30cm |
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JST SH-Style Cable, 2-Pin, Female-Female, 25cm |
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!