kievari
Member level 2
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:
MAV-blog : Accelerometer to pitch and roll
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!
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:
MAV-blog : Accelerometer to pitch and roll
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!