3
\$\begingroup\$

Is there a device which acts like an accelerometer but only measures absolute velocity? I could integrate the accelerometer's output but errors quickly accumulate and cause problems. Preferably the device is available in a MEMS based chip and is low cost. I could use a GPS device (and am currently, with success), but this is expensive, bulky, and requires a GPS link, which is not available in all circumstances.

\$\endgroup\$
  • 1
    \$\begingroup\$ How would such a thing even work? It needs an external reference of some sort, like GPS or other positioning system. \$\endgroup\$ – endolith Apr 14 '11 at 1:43
  • \$\begingroup\$ @endolith Simple, its almost like measuring voltage with a single probe (no ground reference). Wait, not so simple :-) \$\endgroup\$ – Kellenjb Apr 14 '11 at 2:41
  • 1
    \$\begingroup\$ I presume you mean without mechanical contact with a surface such as a trundle wheel and rotary encoder? Or a surface with stripes on it? That gives me a thought, do you mean like a laser mouse? \$\endgroup\$ – Martin Apr 14 '11 at 9:13
  • 1
    \$\begingroup\$ Perhaps if you describe your application, there might be some usable solution. As others clearly pointed out, you can't simply measure velocity in a self-contained manner. \$\endgroup\$ – Toybuilder Apr 14 '11 at 12:16
10
\$\begingroup\$

There is no such thing as absolute velocity; velocity is always relative to something. Velocity cannot be measured internally to a chip (as can acceleration). You must have external input, such as radio references a la GPS or sensors detecting changes in other phenomena external to the chip and its vehicle.

\$\endgroup\$
1
\$\begingroup\$

As has already been said, there is no such thing as absolute velocity, however STM make a 3D motion sensing chipset (iNEMO). Details of the evaluation kit here

\$\endgroup\$

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.