Removal of 1 g from accelerometer

[mark2468]

Hi.

I was wondering if it is possible to remove the 1g (sqrt(x^2+y^2+z^2)) from an accelerometer when in static positions.

When the device is tilted to different positions the total g-force does not equal 1 and differs quite a bit from position to position.

I would like to be able to have the device at a particular angle which gives a reading of 0,0,0 as the device is static (angle not important) and when linear acceleration is experienced thay it just shows on the relevant axis.

Is there a formula that can achieve this?

thanks.

 

[Archosaur]

It can likely be "zeroed", or calibrated in a way that you redefine the force of gravity to be zero g's. But that's not really what you're asking about, and I hate to disappoint, but, as Einstein would tell you, rest in the presence of a gravitational field and linear acceleration are indistinguishable. There is no way to teach an accelerometer to learn the difference between sitting on a table and accelerating through space in a rocket ship at 9.8 m/s^2. However, robotics engineers have done some clever things by adding gyroscopes and other components to their setup.

 

[mark2468]

I'm not really to worried about rest and linear acceleration as i have a method that determines the position. What I really want is to have a reading of 0,0,0 at rest and then if it accelerates let's say for example in the z-plane that the reading might look something like 0,0,0.2 or 0,0,-0.7. Can this be done.

Thanks.

 

[K^2]

If you have orientation of the accelerometer, yes. What you need to know is which way the true vertical points relative to the X, Y, and Z of accelerometer.

 

[mark2468]

with the values like:

0.6, 0.6, 0.53.

The magnitude is 1 so therefore could just subtract each element from itself to get:

0, 0, 0.

but if there acceleration in z direction which gives something like:

0.6, 0.6, 0.6, or 0.6, 0.6, 0.8. and to get a normalized value:

0,0,z.

How can I solve for z?