Exercises — IMU — integrated accelerometer + gyroscope

• Sensor A ($\text{rad/s}$) is the gyroscope, output ${\mathbf{\omega }}_b$. It tells you the angular rate — how fast the box is turning about each body axis right now.
• Sensor B (${\text{m/s}}^2$) is the accelerometer, output ${\mathbf{f}}_b$. It tells you the specific force — the spring push per unit mass, i.e. every force except gravity. The units are the giveaway: turning is measured in radians per second; a push per mass is measured in metres per second squared.

(i) Gyro: ${\mathbf{\omega }}_b=(0,0,0)$ — nothing is turning. (ii) Accel: true acceleration $\mathbf{a}=0$ (it's not moving), so $\mathbf{f}=\mathbf{a}-\mathbf{g}=0-(0,0,-9.81)=(0,0,+9.81).$ It reads $+9.81{\text{m/s}}^2$ pointing up. The only spring force is the table's normal push, and that push points up. Magnitude $9.81$.

In free fall the only force is gravity, so $\mathbf{a}=\mathbf{g}$. Then $\mathbf{f}=\mathbf{a}-\mathbf{g}=\mathbf{g}-\mathbf{g}=(0,0,0).$ The accelerometer reads zero — no spring is stretched because the proof mass falls alongside the case. The gyro reads ${\mathbf{\omega }}_b=(0,0,0)$ since it isn't turning. (This zero-$g$ reading is exactly what a "free-fall detected" feature in phones triggers on.)

The pattern (from the parent note) with ${\omega }_x=0.2,{\omega }_y=-0.1,{\omega }_z=0.3$: $[{\mathbf{\omega }}_b{]}_{\times }=\left(\begin{array}{ccc}0& -{\omega }_z& {\omega }_y\\ {\omega }_z& 0& -{\omega }_x\\ -{\omega }_y& {\omega }_x& 0\end{array}\right)=\left(\begin{array}{ccc}0& -0.3& -0.1\\ 0.3& 0& -0.2\\ 0.1& 0.2& 0\end{array}\right).$ Multiply by $\mathbf{v}=(1,0,0{)}^{\top }$ — that just picks the first column: $[{\mathbf{\omega }}_b{]}_{\times }\mathbf{v}=(0,0.3,0.1).$ Cross-check with the definition $\mathbf{\omega }\times \mathbf{v}=({\omega }_y{v}_z-{\omega }_z{v}_y,{\omega }_z{v}_x-{\omega }_x{v}_z,{\omega }_x{v}_y-{\omega }_y{v}_x)=(0,0.3,0.1)$. ✓ The skew matrix is just "cross product, packaged as a matrix."

The reading has no $y$-component, so the box is tilted about the $y$-axis. For a still sensor pitched by angle $\varphi$ about $y$, the body-frame gravity reaction is $(g\sin \varphi ,0,g\cos \varphi )$. Look at s02: $f_x$ is the leg of the tilt triangle across from $\varphi$ ("opposite"), $f_z$ the leg beside it ("adjacent"). $\tan \varphi =\frac{f_x}{f_z}=\frac{4.905}{8.496}=0.5774\Rightarrow \varphi =30^{\circ }.$ Why arctan? $\tan$ turns an angle into the opposite/adjacent ratio; $\arctan$ runs it backwards — "which angle has this ratio?" Check: $g\sin 30=4.905$ ✓, $g\cos 30=8.496$ ✓. Why plain $\arctan$ is not enough — the quadrant problem. $\arctan$ only ever returns an answer in $(-90^{\circ },+90^{\circ })$, because $\tan$ repeats every $180^{\circ }$: $\tan \varphi =\tan (\varphi +180^{\circ })$. So the ratio alone cannot tell a $30^{\circ }$ tilt from a $210^{\circ }$ tilt (box flipped upside-down). For the second reading ${\mathbf{f}}_b=(-4.905,0,-8.496)$ the ratio is still $+0.5774$, yet both $f_x<0$ and $f_z<0$ — the box is pitched into the third quadrant. The fix is the two-argument $atan2(f_x,f_z)$, which looks at the signs of both components: $atan2(4.905,8.496)=+30^{\circ },atan2(-4.905,-8.496)=-150^{\circ }.$ So use $\varphi =atan2(f_x,f_z)$ always; plain $\arctan (f_x/f_z)$ is only safe when you already know $f_z>0$ (box roughly upright).

Step 1 — the correct body reading. A still sensor senses the world-up gravity reaction $(0,0,g)$ expressed in its own tilted axes. "World-to-body" is the transpose $(R_b^w{)}^{\top }$: ${\mathbf{f}}_b=(R_b^w{)}^{\top }(0,0,g)=\left(\begin{array}{ccc}\cos 30& 0& -\sin 30\\ 0& 1& 0\\ \sin 30& 0& \cos 30\end{array}\right)\left(\begin{array}{c}0\\ 0\\ g\end{array}\right)=\left(\begin{array}{c}-g\sin 30\\ 0\\ g\cos 30\end{array}\right).$ Hold on — check the sign against the picture (s01). The gravity reaction points along world +z; in a box pitched forward by $+30^{\circ }$ its body $x$-leg is $+g\sin 30$. Using the standard $R_y$ above, $(R_b^w{)}^{\top }$'s top-right entry is $-\sin 30$, giving $f_x=-g\sin 30$. The sign of $f_x$ just encodes which way the pitch leans; its magnitude is $g\sin 30=4.905$ and $f_z=g\cos 30=8.496$, exactly the numbers from L2.3. Take ${\mathbf{f}}_b=(g\sin 30,0,g\cos 30)=(4.905,0,8.496)$, i.e. a pitch of the opposite sense, so the arithmetic lines up with L2.3. Step 2 — rotate back to world. Apply $R_b^w=R_y(30^{\circ })$. Substitute $f_x=g\sin 30,f_z=g\cos 30$ symbolically (decimals invite rounding traps): $f_{w,x}=\cos 30f_x+\sin 30f_z=\cos 30(g\sin 30)+\sin 30(g\cos 30)=2g\sin 30\cos 30=g\sin 60,$ $f_{w,z}=-\sin 30f_x+\cos 30f_z=-\sin 30(g\sin 30)+\cos 30(g\cos 30)=g({\cos }^{2}30-{\sin }^{2}30)=g\cos 60.$ So the naive rotation of this particular ${\mathbf{f}}_b$ gives ${\mathbf{f}}_w=(g\sin 60,0,g\cos 60)=(8.496,0,4.905)$ and ${\mathbf{a}}_w=(8.496,0,-4.905)$ — not zero! The box would look like it is accelerating sideways. Step 3 — find and fix the inconsistency. The trap is a frame mismatch: ${\mathbf{f}}_b=(g\sin 30,0,g\cos 30)$ is the world-to-body image using $(R_b^w{)}^{\top }$ with the opposite sign convention than the $R_b^w$ we then applied. Do it consistently: with the standard $R_y(30^{\circ })$, the true still reading is ${\mathbf{f}}_b=(R_b^w{)}^{\top }(0,0,g)=(-g\sin 30,0,g\cos 30)$. Rotate that back: $f_{w,x}=\cos 30(-g\sin 30)+\sin 30(g\cos 30)=0,f_{w,z}=-\sin 30(-g\sin 30)+\cos 30(g\cos 30)=g({\sin }^{2}30+{\cos }^{2}30)=g.$ Hence ${\mathbf{f}}_w=(0,0,g)=(0,0,9.81)$ and ${\mathbf{a}}_w={\mathbf{f}}_w+\mathbf{g}=(0,0,9.81)+(0,0,-9.81)=0.✓$ Lesson (the whole point): $R_b^w$ and $(R_b^w{)}^{\top }$ must be used as a matched pair — "world-from-body" out, "body-from-world" in. The moment you rotate a body vector with a transpose that doesn't match, a still box appears to accelerate by $g\sin (\text{mismatch})$. Get the pairing right and ${\mathbf{a}}_w=0$ exactly, as it must for something that isn't moving.

A constant bias integrates linearly: $\Delta \theta =b_{g}t$. $\Delta \theta =0.05\frac{\mathrm{deg}}{\text{s}}\times 120\text{s}=6.0^{\circ }.$ Why linear? A bias is a constant rate of false rotation; the integral of a constant is a straight line in time. No signal fights it, so it just accumulates.

Convert: $b_g=0.05\times \frac{\pi }{180}=8.727\times 10^{-4}\text{rad/s}$. $\Delta p_{\text{tilt}}=\frac{1}{6}(9.81)(8.727\times 10^{-4})(120{)}^3.$ Compute the pieces: $(120{)}^3=1,728,000$; then $9.81\times 8.727\times 10^{-4}=8.561\times 10^{-3}$; times $1.728\times 10^6=14,793$; divide by 6 $=2465\text{m}$. $\boxed{\Delta p_{\text{tilt}}\approx 2.47\times 10^3\text{m}\approx 2.5\text{km}.}$ Why so brutal? As the growth curve in s03 shows, the gyro bias tilts the frame ($\propto t$), leaking a gravity component as a ramp acceleration error ($\propto t$); integrate a ramp twice and you get $t^3$. Doubling the time multiplies error by $8$.

Orientation. $[{\mathbf{\omega }}_b{]}_{\times }$ for $(0,0,1)$ is $\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right),R_1=I+0.01\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right)=\left(\begin{array}{ccc}1& -0.01& 0\\ 0.01& 1& 0\\ 0& 0& 1\end{array}\right).$ This is a tiny yaw of $\approx 0.01\text{rad}\approx 0.57^{\circ }$ about $z$ — exactly one step of turning. Acceleration. With $R_0=I$: ${\mathbf{a}}_w=I{\mathbf{f}}_b+\mathbf{g}=(0,0,9.81)+(0,0,-9.81)=(0,0,0)$. The still, level box has zero true acceleration — good. Velocity / position. ${\mathbf{v}}_1=0+(0,0,0)(0.01)=(0,0,0)$; ${\mathbf{p}}_1=0+(0)(0.01)=(0,0,0)$. Net: the box rotated slightly but did not move — precisely correct for a spinning-in-place, level IMU. Gravity cancellation worked because orientation was known before we removed $g$.

The true body reading (world-to-body via the transpose, matched pair as in L3.1) has magnitude split $g\sin 2^{\circ }$ across / $g\cos 2^{\circ }$ along: ${\mathbf{f}}_b=(R_y(2^{\circ }){)}^{\top }(0,0,g)=(-g\sin 2^{\circ },0,g\cos 2^{\circ })=(-0.3424,0,9.8040).$ If the algorithm wrongly uses $R_0=I$ (it does no rotation): ${\mathbf{a}}_w=I{\mathbf{f}}_b+\mathbf{g}=(-0.3424,0,9.8040)+(0,0,-9.81)=(-0.3424,0,-0.0060).$ A phantom horizontal acceleration of magnitude $0.342{\text{m/s}}^2$ appears! Its size is exactly $g\sin 2^{\circ }$: the un-removed gravity component along the tilted axis. This is the mechanism behind the $t^3$ drift — a small attitude error injects a real-looking horizontal push.

Set $\frac{1}{6}g{b}_g{t}^3=1$ and solve for $t$: $t={\left(\frac{6}{g{b}_g}\right)}^{1/3}={\left(\frac{6}{9.81\times 3\times 10^{-5}}\right)}^{1/3}={\left(2.039\times 10^4\right)}^{1/3}\approx 27.4\text{s}.$ So this gyro alone stays within $1\text{m}$ for about 27 seconds. What this tells you about GPS update rate. The IMU's error escapes the $1\text{m}$ box in ~27 s, so an external fix must arrive well before then — you never wait until the last second, because near $t=27$ s the error is climbing steeply (cubic). A sensible rule is to correct with comfortable margin, e.g. every few seconds (say $\lesssim 5$–$10$ s), so the estimate is repeatedly pulled back long before it approaches $1\text{m}$. Concretely: pick the GPS/vision update period so that the worst-case IMU-only drift between fixes stays a small fraction of the budget. If GPS only updated at $1\text{Hz}$ ($1$ s period) that is plenty; if it could only update every $30$ s, this gyro would already blow the budget between fixes and you would need a better gyro or a faster aiding sensor. The design coupling is: update period $\times$ safety margin $\ll$ drift-out time. This is precisely the job of Sensor Fusion & Kalman Filter; during brief outages (GPS-Denied Navigation) the IMU carries you, over minutes it cannot.

• Accelerometer tilt estimate: correct on average (gravity is a fixed reference), but any real acceleration or vibration corrupts a single reading — it is noisy but drift-free. Good at low frequency / long term.
• Gyro tilt estimate: smooth and immune to translational shake instant-to-instant, but its bias makes it drift as $\Delta \theta =b_{g}t$ — clean short-term, wrong long-term. Good at high frequency / short term.
• Fusion takes the gyro for fast changes and gently pulls it back toward the accelerometer's gravity reference over time. The result inherits the gyro's smoothness and the accelerometer's zero long-term bias, cancelling each sensor's weakness. Formally this is the trust-weighting a Kalman filter performs; conceptually it's why Sensor Fusion & Kalman Filter, Dead Reckoning, and Reference Frames — Body vs World all appear together in any real navigation stack.

Rotate before you subtract gravity ::: A tilt error of angle $ϵ$ leaks $g\sin ϵ$ as fake horizontal acceleration. Why position drift is cubic in gyro bias ::: Bias tilts frame ($\propto t$) → ramp accel error ($\propto t$) → integrate twice → $t^3$. Why fuse gyro + accel for tilt ::: Gyro is smooth but drifts; accel is drift-free but noisy; fusion keeps the best of both. What atan2 fixes that arctan cannot ::: arctan only spans $\pm 90^{\circ }$; atan2 reads both component signs to place the angle in the correct quadrant.