IMU — integrated accelerometer + gyroscope: Worked examples

Everything here rests on three facts from the parent note. Let us restate them in plain words so nothing is used before it is earned:

First, a plain-words note on position error. When we dead-reckon, we estimate where the box is by adding up (integrating) its motion. The estimate drifts away from the truth; we call that gap the position error and write it $Δ p$ (here $p$ = position, $Δ$ = "the error in"). It is measured in metres. Keep this meaning in mind — it appears in the drift bullet and again in Examples 7 and 8.

Here $a$ = true acceleration (how the box speeds up), $g$ = the gravity vector (points down, so $g = (0, 0, - 9.81)$ if $z$ is up), and $f$ = "specific force" = the push per kilogram that the sensor's internal spring feels. Keep the world axes fixed: $x$ = East, $y$ = North, $z$ = Up.

The scenario matrix

Every IMU problem is one (or a mix) of these cells. The examples below are tagged with the cell they cover.

Cell	What is special	Example
A. Rest, level	$a = 0$ , $ω = 0$ , no tilt	Ex 1
B. Rest, tilted (sign of tilt)	still, but rotated — gravity splits across axes	Ex 2
C. Degenerate: free fall	spring force is zero	Ex 3
D. Linear acceleration, level	true motion, $g$ must be subtracted	Ex 4
E. Pure rotation (gyro only, on axis)	orientation changes, no translation	Ex 5
F. Coupled: accelerate while tilted	must rotate before removing $g$	Ex 6
G. Limiting behaviour: drift over time	error growth $t^{2}$ vs $t^{3}$	Ex 7
H. Real-world word problem	drone in a tunnel, GPS-denied	Ex 8
I. Exam twist: sign trap	negative reading, which way is "up"?	Ex 9
J. Off-axis rotation (centripetal)	sensor not at the spin axis	Ex 10

Example 1 — Cell A: rest and level

Worked example Motionless, perfectly flat

An IMU sits still on a table, axes aligned with the world. What do the gyro and accelerometer read?

Forecast: Guess before reading. Is the accelerometer reading zero, $+ 9.81$ , or $- 9.81$ ? In which axis?

Gyro: the box is not turning, so $ω_{b} = 0$ . Why this step? Angular rate is literally "how fast you spin"; a still box spins at zero.
Accelerometer: true acceleration $a = 0$ , so $f = a - g = 0 - (0, 0, - 9.81) = (0, 0, + 9.81)$ . Why this step? $f = a - g$ ; subtracting a downward vector adds an upward one. Physically, the table's normal force is the only spring push, and it points up.

Answer: $ω_{b} = (0, 0, 0)$ rad/s, $f_{b} = (0, 0, + 9.81)$ m/s².

Verify: magnitude $∣ f_{b} ∣ = 9.81$ = one $g$ . Units: m/s². The reading points up — this is the free "which way is up" calibration mentioned in the parent note. ✓

Example 2 — Cell B: rest but tilted (the sign of tilt matters)

Read the figure below first. The lavender box is the tilted IMU. The dark grey arrow is the true up-reaction $f$ that gravity forces the table to supply — it always points straight up in the world. The coral and mint arrows are the sensor's own tilted body $x$ and $z$ axes. Notice how the single grey arrow casts shadows (dashed lines) onto the two tilted axes: the coral shadow is the $g sin θ$ that leaks into body- $x$ , the mint shadow is the $g cos θ$ that stays along body- $z$ . The butter arc marks the $3 0^{\circ}$ tilt. This picture is the derivation in steps 1–2.

Figure — IMU — integrated accelerometer + gyroscope

In the world frame the sensor still only feels the up-reaction to gravity: $f_{w} = (0, 0, + 9.81)$ . Why? It is still; the physics is identical to Ex 1 in the world frame.

Express that world vector in the body's tilted axes using $R^{⊤}$ (world → body, from the definition box). For $3 0^{\circ}$ about $y$ : $f_{b} = R^{⊤} 00 9.81 = g sin θ 0 g cos θ = (4.905, 0, 8.496) .$ Why this step? Look at the figure: as the box tips forward, part of the "up" push now falls along the body's $x$ -axis (the coral shadow). The $sin$ measures how much leaks sideways; $cos$ how much stays vertical.

Sign check: tilt the other way, $θ = - 3 0^{\circ}$ , and $sin$ flips sign → $f_{b} = (- 4.905, 0, 8.496)$ . The sign of the $x$ -component tells you which way you tipped. Why this matters? This is exactly how the IMU derives a tilt angle: $θ = arctan (f_{x} / f_{z})$ .

Verify: $∣ f_{b} ∣ = 4.90 5^{2} + 8.49 6^{2} = 9.81$ ✓ (rotation preserves length). And $arctan (4.905/8.496) = 3 0^{\circ}$ ✓.

Example 3 — Cell C: free fall (degenerate, spring force zero)

Example 4 — Cell D: linear acceleration, level

Worked example Car flooring it

A level IMU in a car accelerates East at $3$ m/s². It is not tilting. What does the accelerometer read in the body frame (still aligned with world)?

Forecast: Two things push the spring now — the seat pushing you forward, and the ground holding you up. Predict both components.

True acceleration $a = (3, 0, 0)$ . Why? Speeding up eastward at $3$ m/s².
Specific force: $f = a - g = (3, 0, 0) - (0, 0, - 9.81) = (3, 0, + 9.81)$ . Why this step? The horizontal $3$ comes from real acceleration; the vertical $+ 9.81$ is still gravity's reaction.

Answer: $f_{b} = (3, 0, 9.81)$ .

Verify: To recover true acceleration we run it back: $a = f + g = (3, 0, 9.81) + (0, 0, - 9.81) = (3, 0, 0)$ ✓. This is the round trip Dead Reckoning relies on. Magnitude of the reading $= 3^{2} + 9.8 1^{2} = 9 + 96.24 = 10.26$ m/s², slightly more than $g$ — the car "feels heavier" because you're being pushed. ✓

Example 5 — Cell E: pure rotation (gyro only, on the axis)

Read the figure below first. The lavender dot is the IMU, sitting exactly on the vertical spin axis (the dashed slate line). The mint circular arrow shows the spin $ω_{z}$ . Because the sensor is on the axis, it never travels along a circle — it just pivots in place, so there is no circular (centripetal) acceleration. The only force its spring feels is still the up-reaction to gravity (the grey arrow, unchanged from Ex 1). Contrast this with Ex 10, where the dot is moved off the axis and a new inward pull appears.

Gyro reads $ω_{b} = (0, 0, 0.5)$ rad/s, constant. Why? Constant spin about $z$ .

Angle turned $= \int_{0}^{4} ω_{z} d t = 0.5 \times 4 = 2$ rad $\approx 114. 6^{\circ}$ . Why this step? Integrating a constant rate is just rate × time — the simplest case of the strapdown attitude update.

Accelerometer still reads $(0, 0, 9.81)$ — no linear acceleration. Why? The sensor is exactly on the vertical spin axis, so it has no circular motion and gravity is not tipped into other axes. (Move it off the axis and this changes — that is Ex 10.)

Answer: turned $2$ rad; accel unchanged.

Verify: $0.5 \times 4 = 2$ rad, and $2 \times 180/ π = 114.5 9^{\circ}$ ✓. This uses only Reference Frames — Body vs World: orientation changed, position did not. ✓

Example 6 — Cell F: accelerate while tilted (the coupled case)

Read the figure below first. In the fixed world axes (slate) the true acceleration $a_{w}$ (coral, pointing East) and the up-reaction to gravity $- g$ (mint, pointing up) add tip-to-tail to give the total specific force $f_{w}$ (grey diagonal). The lavender dashed axes are the tilted body frame: the very same grey vector, when read off the tilted axes, splits into the mixed-up numbers $(- 3.173, 0, 9.496)$ — that is what the device outputs. The picture shows why you must rotate the grey arrow back onto the world axes before the mint gravity part will cancel cleanly.

Compute $f_{b}$ . With $R^{⊤}$ (world → body) for $+ 3 0^{\circ}$ about $y$ : $f_{b} = (2 cos 30 - 9.81 sin 30, 0, 2 sin 30 + 9.81 cos 30) = (- 3.173, 0, 9.496) .$ Why this step? This is what the physical device outputs — mixed up because the body is tilted.

Correct recovery: rotate to world first with $R$ (body → world), then subtract gravity. $f_{w} = R f_{b} = (2, 0, 9.81), a_{w} = f_{w} + g = (2, 0, 0) .$ Why this step? Only after rotating are the axes world-aligned, so gravity is purely in $z$ and cancels cleanly.

Naive shortcut (wrong): read $f_{b}$ as world and subtract $(0, 0, 9.81)$ : $(- 3.173, 0, 9.496) - (0, 0, 9.81) = (- 3.173, 0, - 0.314) .$ Why it's wrong? You never rotated, so gravity was still smeared across $x$ and $z$ . You now "see" a phantom $- 3.173$ m/s² eastward acceleration.

Answer: correct $a_{w} = (2, 0, 0)$ ; naive method gives the garbage $(- 3.173, 0, - 0.314)$ .

Verify: $R R^{⊤} (2, 0, 9.81) = (2, 0, 9.81)$ so step 2 round-trips exactly ✓. The naive $x$ -error is $2 cos 30 - 9.81 sin 30 = 1.732 - 4.905 = - 3.173$ ✓. Lesson: attitude → rotate → subtract $g$ → integrate, always.

Example 7 — Cell G: limiting behaviour, drift over time

Read the figure below first. The horizontal axis is elapsed time; the vertical axis is how many metres your position estimate has drifted. The coral curve is the accelerometer error, a gentle parabola ( $t^{2}$ ). The lavender curve is the gyro-tilt error, a steeper cubic ( $t^{3}$ ). The mint dashed line marks the crossover moment where the cubic overtakes the parabola — before it the accel error dominates, after it the gyro error runs away. The dots are the four numbers we compute in steps 1–2.

Accel error grows as $Δ p_{a} = \frac{1}{2} ε_{a} t^{2}$ .

$t = 10$ : $\frac{1}{2} (0.02) (100) = 1.0$ m.

$t = 100$ : $\frac{1}{2} (0.02) (10000) = 100$ m. Why this step? A constant acceleration error integrated twice gives a $t^{2}$ position error.

Gyro-tilt error grows as $Δ p_{g} = \frac{1}{6} g b_{g} t^{3}$ .

$t = 10$ : $\frac{1}{6} (9.81) (1.745 \times 1 0^{- 4}) (1000) = 0.2853$ m.

$t = 100$ : $\frac{1}{6} (9.81) (1.745 \times 1 0^{- 4}) (1 0^{6}) = 285.3$ m. Why this step? Bias tilts the frame linearly, leaking a ramp of acceleration; integrating a ramp twice gives $t^{3}$ .

Crossover: at $t = 10$ s accel error (1 m) beats gyro (0.29 m); by $t = 100$ s the gyro cubic (285 m) has crushed the accel quadratic (100 m). Why? $t^{3}$ eventually overtakes $t^{2}$ no matter how small $b_{g}$ is — this is why GPS-Denied Navigation must fuse with other sensors.

Verify: ratios — accel error scales ×100 from $t = 10$ to $t = 100$ (i.e. $1 0^{2}$ ) ✓; gyro error scales ×1000 (i.e. $1 0^{3}$ ) ✓. See Sensor Fusion & Kalman Filter for the cure.

Example 8 — Cell H: real-world word problem

Worked example Drone through a tunnel (GPS-denied)

A drone flies at constant $12$ m/s East into a tunnel where GPS drops out. Its IMU has an effective accel error of $0.05$ m/s². The tunnel is $600$ m long. How far off is the position estimate when it exits, and how long was it inside?

Forecast: Constant velocity means true acceleration is zero — so where does error even come from?

Time in tunnel: $t = 600/12 = 50$ s. Why? Distance / speed, constant velocity.
True acceleration is $0$ , but the sensor has a residual error $ε_{a} = 0.05$ m/s² which dead reckoning cannot know about, so it integrates that error as if it were real motion. Why? The IMU has no outside reference (GPS is gone), so it trusts every number it reads — error included.
Position error $Δ p = \frac{1}{2} ε_{a} t^{2} = \frac{1}{2} (0.05) (50)^{2} = \frac{1}{2} (0.05) (2500) = 62.5$ m. Why this step? Same $t^{2}$ law as Ex 7 — a constant acceleration error integrated twice grows as $t^{2}$ , accumulating even during "boring" constant-velocity flight.

Answer: the drone is inside for $t = 50$ s, and its exit position estimate is off by $Δ p = 62.5$ m.

Verify: $600/12 = 50$ ✓, $0.5 \times 0.05 \times 5 0^{2} = 62.5$ ✓. Units: $(m/s^{2}) (s^{2}) = m$ ✓. A $62.5$ m error over a $600$ m tunnel ( $\sim 10%$ ) is why real drones fuse Dead Reckoning with vision the moment GPS returns. ✓

Example 9 — Cell I: exam twist, the sign trap

Worked example Which way is up?

On a test you're told a stationary IMU reads $f_{b} = (0, 0, - 9.81)$ m/s². A student concludes "it's accelerating downward at $g$ ." What is actually going on?

Forecast: Is the box in free fall, upside down, or is the student wrong about the convention?

If it were truly stationary and $z$ pointed up, a reading of $- 9.81$ in $z$ is impossible — rest gives $+ 9.81$ (Ex 1). Why? $f = a - g$ ; at rest $a = 0$ so $f = - g = (0, 0, + 9.81)$ .
The only consistent reading is that the sensor's body $z$ -axis points down (box is flipped over). Then the up-reaction lies along $- z_{body}$ , giving $- 9.81$ . Why? The accelerometer always reports the up-reaction; if your $z$ -axis is inverted, so is the sign.
Recover truth: with the flipped orientation $R$ (here $R = diag (1, - 1, - 1)$ , a body → world flip), rotate $f_{b}$ to world → $(0, 0, + 9.81)$ , then $a_{w} = f_{w} + g = 0$ . Truly at rest. Why? Rotating first removes the sign confusion — the same lesson as Ex 6.

Answer: the box is upside down but stationary; it is not accelerating. The student fell for reading the body-frame sign as world truth.

Verify: flipping to world, $R f_{b} = diag (1, - 1, - 1) (0, 0, - 9.81) = (0, 0, + 9.81)$ ; then $a_{w} = (0, 0, 9.81) + (0, 0, - 9.81) = 0$ ✓. The sign of a resting reading only tells you which way your axis points, never the motion. ✓

Example 10 — Cell J: off-axis rotation (the case Ex 5 warned about)

Worked example Sensor mounted away from the spin axis

The IMU spins about the world $z$ -axis at $ω = 0.5$ rad/s, but is mounted at radius $r = 0.2$ m out from the axis. The axis itself does not translate. What extra force does the accelerometer feel, and in which direction?

Forecast: In Ex 5 (sensor on the axis) the accel was untouched. Off the axis, does it now feel something — and is it inward or outward?

A point going in a circle of radius $r$ at rate $ω$ has a centripetal acceleration $a_{c} = ω^{2} r$ , pointing toward the centre (the axis). Why this step? To stay on a circle the mass must be pulled inward; that inward pull is the only way to keep bending its straight-line motion.
Magnitude: $a_{c} = ω^{2} r = (0.5)^{2} (0.2) = 0.05$ m/s², directed inward (toward the axis). Why? Plug the numbers into $ω^{2} r$ .
The accelerometer reports specific force $f = a - g$ . The true acceleration now has this inward centripetal part plus the up-reaction, so the horizontal channel reads $0.05$ m/s² (the spring must push the mass inward to keep it circling, so the reported specific force has magnitude $0.05$ in the horizontal plane), while $z$ still reads $+ 9.81$ . Why? The spring must supply the inward pull; specific force is that spring push per mass.

Answer: an extra horizontal $0.05$ m/s² centripetal reading appears; vertical stays $9.81$ . Total horizontal-plane magnitude $0.05$ m/s².

Verify: $ω^{2} r = 0.25 \times 0.2 = 0.05$ ✓. This is why a naive strapdown that assumes the sensor sits on the rotation axis mis-reads spins — a lever-arm correction is needed, closely related to the Coriolis Effect terms in rotating frames. ✓

Recall Quick self-test

A resting accelerometer reads $- 9.81$ in body $z$ — is it accelerating? ::: No. Its $z$ -axis points down; it's stationary but flipped. Why does a $t^{3}$ error eventually beat a $t^{2}$ error? ::: Because for large $t$ , $t^{3}$ grows faster than $t^{2}$ regardless of coefficients. In Ex 6, why did the naive method invent a phantom acceleration? ::: It skipped the rotation, so gravity stayed smeared across the body axes and leaked into $x$ . Off the spin axis, what extra reading appears and how big for $ω = 0.5$ , $r = 0.2$ ? ::: Centripetal $ω^{2} r = 0.05$ m/s² toward the axis. What single word summarises the fix for all coupled cases? ::: Rotate (into the world frame) before subtracting gravity.