0
\$\begingroup\$

We are using a MEMS tri-axial sensor which has an Accelerometer,Magnetometer and Gyrometer. We have also calibrated the Accelerometer and Magnetometer.This sensor package is used in a borehole application. We have calculated the deviation of the borehole using the Accelerometer. Now we are stuck with the direction calculation free from rotation (i.e if the sensor rotates, the direction should not change).Is it possible to calculate a rotation free direction using the three inputs we have (Accelerometer, Magnet, Gyro)?

Please find the attached image of Mechanical orientation of sensor.enter image description here

\$\endgroup\$
  • \$\begingroup\$ Does a magnetometer actually work inside a steel casing? \$\endgroup\$ – WhatRoughBeast Aug 1 '14 at 10:51
  • \$\begingroup\$ Not really sure I understand the mechanical arrangement well enough to answer your question. Can you give us a paragraph about where the sensor is and how it's moving? Is it cam-following the borehole wall? is there an eccentric rotation -- i.e., is there a centripetal acceleration you need to worry about? \$\endgroup\$ – Scott Seidman Aug 1 '14 at 14:29
  • \$\begingroup\$ @WhatRoughBeast It is not steel its aluminium casing. \$\endgroup\$ – Novity Naseer Aug 2 '14 at 4:11
1
\$\begingroup\$

Hi: If the probe is not moving you don't need the gyro, the 3-axis magnetometer and the 3-axis accelerometer are all you need to determine the orientation of the probe.

It turns out that the Earth's magnetic field in Northern California is stronger in the vertical direction than in the horizontal direction so the magnetometer by itself can not determine direction. The companies that make integrated sensor ICs have app notes that go into detail on how to use the accelerometer to correct the magnetic readings and may even have the firmware for that. http://www.prc68.com/I/Sensors.shtml#EMF

\$\endgroup\$
  • \$\begingroup\$ @ Brooke Clarke : The probe will be a free fall from the top and it tends to rotate and deviate from its path. \$\endgroup\$ – Novity Naseer Aug 4 '14 at 4:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.