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I have just started learning how to fuse the data measured by the gyro and accelerometer to estimate the attitude of an IMU in Matlab. I have no problem estimating or understanding the Euler angles calculated from at set of gyro data. But when I calculate the Euler angles from the accelerometer data, the rotation always seem to go in the opposite direction to that of the gyro's.

(note:the order of rotation is X(Phi) -> Y(Theta) -> Z(Psi))

To troubleshoot, I ran a very simple experiment where I first fixed the x-axis parallel to the ground with the y-axis pointing straight down. Then I gradually rotated the object around the x-axis positively (CCW) by roughly ~+40°. All the signs and direction of rotation follow the right-hand-rule and right-hand-curl rule.

For clarity I attach the following drawing to associate sections of the accelerometer data to the state of rotation (note: +x-axis points out of the page):

z1

The accelerometer data looks correct. ay stays pointing downward for the first few seconds and logs a value of -1g before rotating. As the object frame rotates around the x-axis (positively or CCW), ay decreases in magnitude as az increases its.

I expect the Phi-delta measured by gyro should start with 0 deg and end with ~+40 deg because the rotation is positive (CCW), if we ignore the noise, drift, and accumulation of error due to numerical integration - the calculated graph pretty much agrees.

Then with the accelerometer, I expect the Phi to go from -90 deg to -45 deg. The slope should be positive because, again, the rotation is identical and is positive (CCW). Phi-delta is positive, whereas the Phi values should remain negative because the frame position is still CW or negative away from the horizontal orientation. But when I calculate the Phi, I get the complete opposite:

enter image description here

I am not sure what is wrong in my thought process and hope someone can point out or correct that for me.

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    $\begingroup$ See also: why do accelerometers seemingly have a left-handed coordinate frame $\endgroup$
    – Ben
    Oct 25, 2022 at 17:06
  • $\begingroup$ @Ben I think what the answers provided in the link were trying the say but didn't say so explicitly is that the accelerometer measures the normal force exerted on the the body which otherwise would be in free fall. If one holds an accelerometer in place with z pointing up, it isn't the gravity that it's measuring but the upward normal force exerted by the whatever electromechanical structure internal to the accelerometer - hence it registers +1g. Vice versa, if +z points down, the normal force against gravity is in the -z direction, which explains why we get -1g instead. $\endgroup$
    – KMC
    Oct 25, 2022 at 22:16

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Welcome to Robotics, KMC!

One thing I noticed is that your equation gives $\phi=\arctan2(-a_y,-a_z)$, but you're calculating it as $\phi=\arcsin(-a_y/\cos(\theta))$. If your sensor is oriented as depicted then I would expect $a_x$ to be around zero, which would make $\theta = \arcsin(a_x)$ also around zero. This, then, would make $\cos(\theta)$ around 1, which would leave your calculation as basically $\phi \approx \arcsin(-a_y)$.

Since your y values are ranging from about -1 to about -0.75, I would then expect your values as-calculated to be around $\arcsin(-(-1)) = +90^\circ$ to about $\arcsin(-(-0.75)) = 48.6^\circ$, which seems to be pretty close to what you're getting in your plots.

Just for sanity, because your starting values are easy, I took your other formula, $\phi = \arctan2(-a_y, -a_z)$ and evaluated it at the start. You've got $a_y \approx -1$ and $a_z \approx 0$, which gives $\phi = \arctan2(-(-1),-0)$, or $\phi \approx 90^\circ$. This seems odd, because in your drawing we can see that the rotation should be -90, like you mention in your post.

So then I went back to your equations in matrix form:

$$ \left[\array{ a_x \\ a_y \\ a_z \\ }\right] = \left[\array{\sin(\theta) \\ -\cos(\theta)\sin(\phi) \\ -\cos(\theta)\cos(\phi)}\right] $$

If I divide row 2 by row 3 then I get the following:

$$ \frac{a_y}{a_z} = \frac{-\cos(\theta)\sin(\phi)}{-\cos(\theta)\cos(\phi)} $$

The $-\cos(\theta)$ cancels on top and bottom, leaving

$$ \frac{a_y}{a_z} = \frac{\sin(\phi)}{\cos(\phi)} $$

Which reduces to $$ \frac{a_y}{a_z} = \tan(\phi) $$

and then solve for $\phi$ with $\arctan$:

$$ \phi = \arctan2(a_y, a_z) \\ $$

I'm not sure where the negative signs came from in your $\phi=\arctan2(-a_y,-a_z)$ but if you do this without the negative signs then I think your numbers come out correct. I wouldn't continue to use the version that relies only on $a_y$ and $a_x$ because $a_x$ is not really a valid input for the $\phi$ calculation.

I'd also recommend (as always) using the Madgwick filter for anything to do with IMU handling. It looks like your units are $\times10^6$, too, so I'd double-check that you're in proper engineering units before trying to push them through the filter.

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  • $\begingroup$ yea I don't follow the atan2(-ay,-az) either so I always rely my calculation on the more intuitive matrix form instead. But I think my problem is that I wrongfully think that the accelerometer measures gravity but in practice it measures the normal force. So all I have to do is to negate {ax,ay,az} to get the correct Euler angles and the slope. $\endgroup$
    – KMC
    Oct 25, 2022 at 22:26

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