# Accelration Direction

1. Apr 30, 2007

### IMK

Hello,
Could someone help me with this maths problem please?
I am working on a little project involving a Tri-Axis accelerometer and I need to be able to calculate the g in a particular direction when the devices installed orientation is initially unknown.

Thus when I power the device on I have three force values let us call these the initial values. ix, iy, iz. (From these initial values we have a sense of orientation also a static total g of 1.0 = sqrt( ix2 iy2 iz2 ))

Now I accelerate the device (keeping the orientation) in a straight line so as to get a g of 1.1 (static 1g plus 0.1 acceleration) this gives me a new set of calibration g values that we can call cx, cy, cz.

What I need please is a formula for calculating the g in any direction from the calibration line/direction.

Say if I move the device forward in the same direction as the calibration direction at 0.1g then the formula/function returns 0.1, if I moved it backward the formula/function returns -0.1. and if I acceleration the device at 90 degrees to the calibration direction the formula/function returns zero.

I have tried a few ideas based on dot products but I have become stuck, any help would be much appreciated.
Added the static/initial values and a set of calibration values:

ix = 0.34
iy = -0.74
iz = -0.68

cx = 0.54
cy = -0.58
cz = -0.84

Last edited: Apr 30, 2007
2. May 1, 2007

### Integral

Staff Emeritus
I am confused by your use of the symbol g, generally this is used to represent the acceleration due to gravity. As far as your accelerometer is concerned, g would be a constant acceleration in the direction of the center of the earth. You seem to be using it as your system acceleration. Could you please clarify just what your g is?

3. May 1, 2007

### IMK

Hello, Integral and many thanks for your reply. Well I guess to be correct I should uses acceleration rather then g. But as the specs on the devices always use g, then for this case I think the terms g and acceleration are interchangeable.

Anyway I am finding some of the math for project a little tricky at times so if you have some input it would be more than appreciated.

Many thanks again IMK

4. May 1, 2007

### uart

I'm pretty sure that what you're looking to find is the difference between the accelerometer reading ("a") and the reference vector ("c") projected into the reference direction.

That is,

$$\frac { ( \vec{a} - \vec{c} ) \cdot \vec{c} } { \parallel \vec{c} \parallel }$$

5. May 1, 2007