scale factor problem

Hi, I've received a new accelerometer with a scale factor declared as: 800 mV/g @ 1.5g. I googled and searched the forum but still can't fully be sure I understand the accelerometer and its functioning, so please correct me where I'm wrong:

One thing that very helped me to get some understanding, was this post: where you see this: giving the calculation of: accelerometer = cos (theta) * gravity theta = acos (accelerometer / gravity)

Formula is easy and beautiful, and this is my first question please: what 'gravity' means there? Suppose the robot owning this accelerometer is on Earth surface, then this 'gravity' means 9.8m/s^2?

Next how to relate this to my accelerometer board output please? I apply a 5V to its board and it has a 10bit ADC, so if I receive: x(LSB) * 5(V)/1024(LSB) now to get rid of V dimension I use the scale factor: x(LSB) * 5(V)/1024(LSB) * (1/{800mV/g}) get rid of 'm': x(LSB) * 5(V)/1024(LSB) * (1000/{800V/g}) or x(LSB) * 5(V)/1024(LSB) * (10/{8V/g}) finally: x*(25/4096)g

questions: 1. but this finall scale factor is a very small number! seems wrong :( 2. in data sheet the scale factor is declared: 800 mV/g @ 1.5g how to take to account the 1.5g? I can imagine when the vertical upward is 1g and vertical downward is -1g, but can't imagine 1.5g. 3. Am I right with the g unit there? As long as I know, this is an embedded number like 9.8m/s^2 on Earth surface. So finally, having 'g' in both accelerometer and gravity in this formula: theta = acos (accelerometer / gravity) we remain with a scalar number and everything stands good at the end of the story, correct?!

Please help to make this clear!

For every g of accekeration, the output will be 800mV. However, the device can't output negative voltages (nor can the AVR accept tem), so zero g is offset to half the supply voltage. So +1g is 800mV plus half the supply voltage.

Using a 5V analogue reference, each 1 count from the A/D is just under 5mV, so 800mV will give you around 180 counts.

Does that help?

Thank you, it sure helps and made some more clear, but some questions remain...

So +1g is 800mV plus half the supply voltage.

So, if I have 5v ref. then +1g is 2.5v+08v = 3.3v?

Supposing I guessed right, then:

Using a 5V analogue reference, each 1 count from the A/D is just under 5mV, so 800mV will give you around 180 counts.

means that for +1g I get some 180 + 500 = 680, as: 2500mV/5= 500.

Am I right with my guesses?

Those values are for a stationary object or one at constant velocity. If it is accelerating in any axis, you need to know about that.

well by calculation way I got from you, the acceleration will be counted from 0 to 680 right?

I wouldn't go with guesses - most common hobbyist accelerometers don't operate off 5V, but usually only about 3V or 3.3V.

So, your zero g point is at 1.5 to 1.65V, so using the latter, 0g reads about 337 from the A/D.

800 mV equates to about 164 counts, assuming a 5V reference.

So, +1g reads about 500, and -1g reads about 173, assuming a 3.3V accelerometer supply, and a 5V analogue reference.

Thank you, I feel better now :D