I apologize for what is likely an ignorant question, but I am fairly new to electronics and have little confidence. I am working on a personal project where I am using a servo motor to move a rod back forth 90 degrees (not exactly, but this analogy should suffice), and I am interested in measuring the instantaneous (angular?) acceleration (I want to use it to calculate torque). Surprisingly, I have come across trouble while researching online for a sensor to do this. My understanding now is that an accelerometer is only capable of measuring linear acceleration (which means the object can't be rotating, right?), while a gyroscope only measures angular velocity when an object is spinning about an axis. Technically, the rod in my project isn't spinning-it's just rotating back and forth between two angles (0 and 90) on the XY plane. Is there a simply way/sensor to do this? More broadly, am I completely misunderstanding the uses of gyroscopes and accelerometers? I hope I explained the issue properly...

Edit: Sorry about the confusion. I am using a servo to rotate a rod back and forth and need to measure the instantaneous torque. While the rod is moving, there will be times at which a force acting against the rod/servo is applied, which means that the torque shouldn't remain constant between experiments.

I thought of a couple of ways to do this, and one idea I had was using calculating the torque using: torque=I (moment of inertia)*angular acceleration. To get the angular acceleration however, I figured I need a sensor. I thought an accelerometer might work, but don't know how I would implement it in my system or if it would even work.

  • \$\begingroup\$ You can directly measure torque for not much money omega.com/googlebase/… \$\endgroup\$ Commented Dec 9, 2014 at 1:41
  • \$\begingroup\$ Oops Actally quite pricey! \$\endgroup\$ Commented Dec 9, 2014 at 1:44

2 Answers 2


A gyro measures angular velocity, and is relatively insensitive to other movement. You would need to differentiate to get angular acceleration. An accelerometer measures acceleration. Period. It will respond to reorientation with respect to gravity (as gravity is an acceleration). It will respond to linear acceleration. If offset from the axis of rotation, it will respond to centripetal acceleration, which will be related to angular velocity (though if motion is sinusoidal, you will need to keep track of tangential accelerations).

If you can orient the accelerometer to be right on the axis of rotation, you can easily track the orientation with respect to gravity (that is, Z), but not so easily in the xy plane (if that is the earth plane in your system)

Your situation may be easier to use some sort of absolute or relative encoder, which will give you angular position without worrying about drifts and offsets associated with integration.

With your improved description, you might want to die toy measure torque! http://www.omega.com/googlebase/product.html?pn=TQ513-514-FOOTMOUNT&gclid=Cj0KEQiAtZWkBRC9ibSfhoKEyLYBEiQA5fDxkcDLSo3_jC7_lsm5amhIy02A4MfjEriBgxXRaW5BxcYaAgOy8P8HAQ

  • \$\begingroup\$ I'm not sure I'm understanding you. Would an encoder be capable of giving me acceleration? Also, I guess I'm not understanding accelerometers either. The rod only rotates (from a fixed point) back and forth-it isn't moving at all in the z-axis. The distance its covering is like a quarter of a circle (~90 degrees), would the angular acceleration be easy to derive here? \$\endgroup\$ Commented Dec 9, 2014 at 1:03
  • \$\begingroup\$ Start by telling us what you're trying to accomplish, and then we can suggest how to best accomplish it. An encoder would provide position, and you could process to get acceleration. What, dpecifically, do you need to measure, and why? \$\endgroup\$ Commented Dec 9, 2014 at 1:12
  • \$\begingroup\$ Scott-sorry about that. I included some more information that should clear up your question in the OP, but me know if you need more (I appreciate you taking time out of your day to help me regardless). \$\endgroup\$ Commented Dec 9, 2014 at 1:23
  • \$\begingroup\$ You could use a gyro and differentiate, or an encoder and differentiate twice. You could also try motor current, which may be proportionap to torque \$\endgroup\$ Commented Dec 9, 2014 at 1:44
  • \$\begingroup\$ Is the sensor that you linked the actual sensor? Because the rest seem very expensive... \$\endgroup\$ Commented Dec 9, 2014 at 2:24

Will the servo be driven by a sine wave or otherwise?
If it's a sine wave (in angle vs time.) then it's easy to measure the period, and calculate the angular velocity and angular acceleration vs time. (a couple of derivatives.)
If you are driving it otherwise, then again you can still calculate the angular velocity and acceleration with the derivative. What kind of signal are you sending to the servo? And how is it measuring angular position? with a pot?

As Scott suggested an angular encoder will give you the angle vs time.
The other thing I've seen is a capacitve sensor with two half moon shapes. One rotates relative to the other... like old air gap variable capacitors.

  • \$\begingroup\$ The servo is actually controlled through PWM. The width of the pulse essentially tells it which angle it should go to. It measures the angular position with a potentiometer (5K linear taper pot). If I derive twice through the potentiometer, do you think it would be accurate at all? \$\endgroup\$ Commented Dec 9, 2014 at 3:48
  • \$\begingroup\$ Also, I can't find anything about capacitive sensors giving acceleration. Are they reliable for acceleration? \$\endgroup\$ Commented Dec 9, 2014 at 4:09
  • \$\begingroup\$ The capacitive thing measures position again... not acceleration. If you are sending it signals that tell it what position to have at what time, then as long as you are not over driving it.. (asking for more than it can give.) then it seems to me you already know the acceleration. You could make a velocity transducer from a coil and permanent magnet... I'm not sure how that would look... it might not be very linear. \$\endgroup\$ Commented Dec 9, 2014 at 13:25

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