Do you use the magnitude equation to get speed from an accelerometer?

A guy suggested to me that getting speed from an accelerometer required the use of this equation:

$\text{speed} = \sqrt{x^2 + y^2 + z^2}$

This does not make any sense to me, all that you would get from this equation would be the magnitude in $m/s^2$ of the acceleration in the $x, y$ and $z$ axes. Am I correct? Or is my reasoning flawed?

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A link to the original claim would be nice. –  Chris White Mar 21 '13 at 3:57
I added the link into the question for you, mathisnotmyforte, but just remember for the future that when you have additional information to add to a question, edit it in, don't leave it in a comment. –  David Z Mar 21 '13 at 6:09

Basically yes.

1) Acceleration is a change over time in velocity. Since velocity has units of distance per unit time (like meters per second), acceleration has units of distance per unit time per unit time (like $\mathrm{m}/\mathrm{s}^2$). So the accepted answer is mistaken on that point.

2) Just to make sure everyone is on the same page with notation: From reading the question, and better yet the accepted answer to the original question, one can see that all throughout, $x$, $y$, and $z$ are referring to components of acceleration, not speed (or position, as a physicist would assume given those names). I imagine that mistaken answerer just used "speed" and "acceleration" interchangeably - unfortunately.

3) But the answer has the right idea. The quantity $$\sqrt{x^2 + y^2 + z^2}$$ is indeed the magnitude of the vector with components $x$, $y$, and $z$, assuming those components are in orthogonal positions (which they are). If all three are accelerations in $\mathrm{m}/\mathrm{s}^2$, then the result will be too. If the device is otherwise not moving (or even if it is moving, but at a constant speed and in an unchanging direction), then the only acceleration it will feel will be due to the Earth's gravity, which (assuming we are not inside the mantle or as far away as the Moon or something) is pretty constant at $9.8~\mathrm{m}/\mathrm{s}^2$. Thus one would hope the square-rooted quantity we calculated comes out to $9.8$. If it doesn't, you need to tweak the calibrations more.

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You're correct. That equation is for the magnitude of the acceleration vector, not the speed.

Measuring speed using just an accelerometer can be tricky. You essentially have to integrate the measured acceleration by multiplying that acceleration (minus gravity) by the time between measurements and summing them for three components of the velocity, then finding the magnitude with that equation. I don't have experience with doing this, but if you're serious about measuring speed, you could combine that measurement with GPS coordinates using something like a Kalman filter.

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Thanks for answering my question. I'm going to use this paper as a guide –  mathisnotmyforte Mar 21 '13 at 4:18