Example calculations
Voltage to acceleration example:
”With the 3.3V supply, the X output reads 2.06V. What acceleration does this correspond to?”
At 3.3V, the 0g point is approximately 1.66V
2.06V – 1.66V = +0.40V with respect to the 0g point
At 3.3V, if sensitivity is 333mV/g, 0.40 / 0.333 = 1.20g
Therefore the acceleration in the X direction is +1.20g
Acceleration to voltage example:
”I am powering the DE-ACCM3D with 2.0V. What voltage will correspond to an acceleration of -0.5g?”
At 2V, the 0g point is approximately 1.00V
If sensitivity at 2V is 195mV/g, -0.5 * 0.195 = -0.0975V with respect to the 0g point.
1.00V – 0.0975V = 0.903V
Therefore you can expect a voltage of approximately 0.903V when experiencing an acceleration of -0.5g.
Voltage to tilt example:
“With a Vcc of 3.0V, and the accelerometer oriented flat and parallel to ground in my robot, Yout is 1.50V.
When my robot goes uphill, Yout increases to 1.67V. What is the slope of the hill?”
1.67V – 1.50 = +0.17V with respect to the 0g point.
With a sensitivity of 300mV/g, 0.17 / 0.300 = 0.567g
Sin
-1
(0.567) = 34.5º
The slope of the hill is 34.5º in the Y axis
Tilt to voltage example:
“I am making an antitheft device that will sound an alarm if it is tilted more than 10º with respect to ground in
any direction. I have measured the 0g bias point to be 1.701V, and I want to know what voltage to trigger the
alarm at. I am using the onboard 3.3V source.”
Sin(10º) = 0.1736 so acceleration with a tilt of 10º will be 0.1736g
0.1736g * 0.333V/g = 0.058V with respect to the 0g point
1.701 + 0.058 = 1.759V
1.701 – 0.058 = 1.643V
Sound the alarm when the voltage reaches more than 1.759V or less than 1.643V.