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I have acceleration data from a sensor. X Y & Z. I move the senor in the Y axis. Mostly in a straight line. So I ignore x & z.

From the sensor documentation 5.2.1 Acceleration output：

ax=((AxH<<8)|AxL)/32768*16g(g is Gravity acceleration，9.8m/s2)

ay=((AyH<<8)|AyL)/32768*16g(g is Gravity acceleration，9.8m/s2)

az=((AzH<<8)|AzL)/32768*16g(g is Gravity acceleration，9.8m/s2)

The data is in (m/s2)

I need a simple calculation that java or C# can take easily. I want to write something in code that calculates the acceleration over time to maximum velocity and average velocity. I need a "speed" value that I can display. For Ex. Max speed 12MPH and Average Speed 8MPH.

In this data the device was moved from the zero point to about 6 inches away less than 1 second.

Time(s) Acc X   Acc Y   Acc Z
48.547  0.4756  0.0864  1.2207
48.563  0.2051  0.2651  1.3350
48.563  0.0044  0.6621  1.3140
48.578  -0.2876 1.0117  1.4292
48.578  -0.0732 1.5586  1.4653
48.594  -0.0659 1.8984  1.3447
48.594  -0.2344 2.4453  1.4043
48.641  -0.2690 3.2148  1.3677
48.656  -0.4072 3.0083  1.4995
48.656  -0.2573 3.2700  1.3545
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The equations at the start look like they're converting sensor data (pairs of 8 bit integer values) into floating point numbers. They're irrelevant to your problem.

What you need to do is :

$$v_k = v_{k-1}+a_k(t_k-t_{k-1})$$

For each time $t_k$ and with $t_0$ and $v_0$ the initial time and velocity.

To find the average velocity over the whole time you can do :

$$u=\frac {\sum v_k(t_k-t_{k-1})}{\sum(t_k-t_{k-1})}$$

The sum on the bottom simplifies to $t_N-t_0$.

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  • $\begingroup$ For clarification tk = time interval, ak = acceleration interval $\endgroup$ – Steve Coleman Jan 29 '18 at 19:24
  • $\begingroup$ How can I solve for this if I don't know what VK is on the right side?vk=vk−1+ak(tk−tk−1) $\endgroup$ – Steve Coleman Jan 29 '18 at 19:33
  • $\begingroup$ You should know the initial $v_0$ (probably zero ?) and can in turn calculate each $v_k$ from that, so $v_1,\,v_2,\,v_3$ and so on. $\endgroup$ – StephenG Jan 29 '18 at 23:10

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