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I am trying to compute the angle of a leg instantaneously regarding an initial position using accelerometer readings.

As I approached this, I decided I needed to record an averaged gravity vector and compute some angle between my current reading and that.

Theoretically it seems plausible: the reading of gravity is different on each position of the leg, so in stationary positions the readings should converge to a vector I can use to find the angular displacement of the leg, with respect to the reading at the start.

However, I have tried several combinations: the angle between both vectors using only the Z and Y components (using this); compute the difference vector between both gravity and current, and do atan2(dY,SQRT(dX^2+dZ^2)); compute atan2(dY,dz)... (where dY and dZ are the subtraction of the current and gravity vector for the Y and Z components, respectively)

None of these solutions seem to be working, so I am wondering if I can even do this.

Does anyone have a solution for this?

ravemir
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  • You can get the tilt angles with the accelerometers only. You cannot get "the angle of the leg" regarding an initial position. You either need gyros or a magnetometer for the latter. – Ali Mar 29 '13 at 20:23
  • @ravemir- were you able to find the solution? I am facing the same problem. Any help is welcome. – nia Mar 03 '14 at 17:57
  • It has been over 6 months since I delivered my thesis, and after a while discussing this with Hoan, I decided to use the compass for direction and focus on finishing the rest of the work. But I had a hunch at one point I could do this using Principal Component Analysis (see [my thesis'](https://fenix.tecnico.ulisboa.pt/publico/department/theses.do?thesisID=2353642465657&contentContextPath_PATH=/departamentos/dei/lateral/dissertacoes&method=showThesisDetails&selectedDepartmentUnitID=1911260507896&locale=en_EN) references 25 and 35). This would be a good place to start. Feel free to discuss! :) – ravemir Mar 03 '14 at 22:56

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Assuming the device is tied to the leg with the screen facing the same direction as of the user then you can calculate the angle as the angle between the surface of the screen and the East-North plane. I just give you the idea here you have to do some more work to know if the angle is below or above the East-North plane. The angle can be calculate as 

Math.acos(event.values[3] / norm of event.values[3]);

Thus if the user is standing then the angle will be 90 or -90 depending your convention. And when the leg move up the absolute of the angle will be decreasing. One the thigh is parallel to the East-North plane, ie the screen is laying flat and the leg moves up, you will get the same angle as the leg moves down, so you need to use a little more info to figure this out.
If the screen is orthogonal to the facing direction of the user then you have to calculate the angle between the East-North plane and the plane orthogonal to the screen. I let you figure this out, it probably the above equation with event.values 0 or 1 depending on whether the device is in Portrait or Landscape.

Hoan Nguyen
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  • Theoretically it seems plausible, though: the reading of gravity is different on each position of the leg, so in stationary positions the readings should converge to a vector I can use to find the angular displacement of the leg. Could you elaborate a little bit further on why it doesn't work? – ravemir Mar 30 '13 at 15:53
  • I do not understand what do you mean by "reading of gravity", do you mean it as the value of say z-coordinate. Because to me gravity is gravity and it is approximate 9.8 – Hoan Nguyen Mar 30 '13 at 18:08
  • What I mean is that if the phone is stationary, the accelerometers read the value of gravity, distributed along its axis. Tilting the phone slightly will also provide a reading of gravity, but this time distributed differently along the axis. Isn't it possible to determine the change in positioning through the both these values? – ravemir Mar 30 '13 at 18:13
  • I understand what you are trying to do, using component of the accelerometer values to calculate the angle. But if you think about it, it's boil down to the same thing as finding the device position thus coordinate through accelerometer alone, which is not possible. – Hoan Nguyen Mar 30 '13 at 18:37
  • At the same position if the phone just rotate you will get different coordinates. – Hoan Nguyen Mar 30 '13 at 18:38
  • Even if I just need the angle change, are you saying I can't get it because of the rotational forces? – ravemir Mar 30 '13 at 18:42
  • No, when you rotate the device coordinate system changes and therefore it return different coordinates for the same vector. – Hoan Nguyen Mar 30 '13 at 18:45
  • Precisely: but since the vector is (in some cases) the same, isn't it possible to take both vectors and determine the tilt change by calculating the angle between them? – ravemir Mar 30 '13 at 18:47
  • How do you know the change is the result of rotation or changing position? – Hoan Nguyen Mar 30 '13 at 18:49
  • I assume it because I know my code will be ran from within a pocket during a walk. During a walk, a phone will change orientation (rotate) and move at the same time, but during a specific moment during the walk, the phone will move very little and just rotate in the air (for example, the time where the foot touches the ground and the leg passes beneath the torso) – ravemir Mar 30 '13 at 18:56
  • Ok, I see you really want just the rotation of the device. – Hoan Nguyen Mar 30 '13 at 19:04
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    If assuming one coordinate is fixed then it is very easy but I don't think the result is accurate due to rotational acceleration. – Hoan Nguyen Mar 30 '13 at 19:17
  • Thanks. Your answer is similar to others I had researched but it made me understand it a little bit more. According to Wikipedia (sometimes enough of a reliable source) using atan is preferable to using arccos or arcsin due to precision issues. I had used the atan2 version a while back, with similar results: atan2(y/x). But the angles oscillate between roughly 150 and -150 then, which seems kind of off since the legs shouldn't extend that much upwards during a regular walk. – ravemir Apr 02 '13 at 10:01