Mechanization equations — integrating IMU to get position, velocity, attitude: Exercises

Quick symbol reminder, so line one is readable by anyone:

Definition The cast of characters

$f^{b}$ = specific force the accelerometer feels, written in body axes (the axes painted on the vehicle).
$ω_{ib}^{b}$ = angular rate the gyro feels, body vs. inertial (non-rotating star-fixed) space, in body axes.
$ω_{nb}^{b}$ = angular rate of the body relative to the nav frame, in body axes. This is what actually turns the vehicle relative to North-East-Down; it equals the gyro rate minus the parts caused by Earth spin and by travelling over the curved Earth. In a static, short-run flat-Earth problem $ω_{nb}^{b} \approx ω_{ib}^{b}$ .
$C_{b}^{n}$ = the rotation matrix ("attitude") that turns a body-axis vector into a nav-frame (North-East-Down) vector.
$[ω \times]$ = the skew-symmetric matrix built from a vector $ω = (ω_{x}, ω_{y}, ω_{z})$ so that multiplying by it does exactly the same thing as the cross product $ω \times ()$ : $[ω \times] = 0 ω_{z} - ω_{y} - ω_{z} 0 ω_{x} ω_{y} - ω_{x} 0, [ω \times] u = ω \times u .$
$v^{n} = (v_{N}, v_{E}, v_{D})$ = velocity in nav frame; $g^{n}$ = plumb-bob gravity, pointing Down.
$ϕ$ = latitude (angle North of the equator), $λ$ = longitude (angle East around the axis), $h$ = height.
$Ω_{e} = 7.292 \times 1 0^{- 5} rad/s$ = Earth's spin rate.

L1 — Recognition

Problem 1.1

A stationary IMU lies flat on a bench (NED, Down $= + z$ ). What does each accelerometer axis read, and what do the gyros read?

Recall Solution 1.1

Gyros: $ω_{ib}^{b} \approx 0$ (ignoring the tiny $Ω_{e}$ Earth-spin term the note flags for high-grade gyros). Accels: $f^{b} = (0, 0, - 9.81)$ . The bench pushes up, and in NED the up-direction is $- z$ , so the felt (specific) force sits on Down as $- 9.81$ . It is not zero: the accelerometer cannot feel gravity, only the contact push.

Problem 1.2

Name the three "layers" of mechanization in the order they must be computed, and state the one-word reason attitude comes first.

Recall Solution 1.2

Order: Attitude → Velocity → Position. Reason: rotation. The accelerometer's force lives in body axes; you need $C_{b}^{n}$ (attitude) to rotate it into nav axes before gravity can be added and velocity integrated.

Problem 1.3

Which term in the velocity equation "adds gravity back," and why is adding it necessary at all?

Recall Solution 1.3

The term $+ g^{n}$ . Necessary because the accelerometer physically cannot sense gravitation (gravity pulls every atom equally, so no internal strain). Software must reinsert it.

L2 — Application

Problem 2.1

Flat-Earth, no rotation. A sled reads $f_{x} = 3 m/s^{2}$ on nav- $x$ (gravity handled off-axis). Starting from rest at the origin, find $v_{x}$ and $x$ at $t = 6 s$ .

Recall Solution 2.1

Flat-Earth velocity law with no Coriolis, no on-axis gravity: $\overset{v}{˙}_{x} = f_{x} = 3$ . Why integrate once? Acceleration is the rate of change of velocity, so velocity is the running total (integral) of acceleration: $v_{x} (t) = \int_{0}^{t} f_{x} d τ = 3 t \Rightarrow v_{x} (6) = 18 m/s$ . Why integrate again? Velocity is the rate of change of position, so position is the running total of velocity: $x (t) = \int_{0}^{t} v_{x} d τ = \frac{1}{2} (3) t^{2} = 1.5 t^{2} \Rightarrow x (6) = 1.5 (36) = 54 m$ . That two-step "integral of an integral" is the whole spirit of mechanization: rates in, positions out.

Problem 2.2

Compute the skew-symmetric matrix $[ω \times]$ (defined in the symbol reminder) for $ω = (0, 0, 0.2) rad/s$ and verify it reproduces $ω \times u$ for $u = (1, 0, 0)$ .

Recall Solution 2.2

Plugging $ω_{x} = 0, ω_{y} = 0, ω_{z} = 0.2$ into the skew template: $[ω \times] = 0 0.2 0 - 0.2 00 000 .$ Multiply by $u = (1, 0, 0)^{⊤}$ : gives $(0, 0.2, 0)^{⊤}$ . Cross product $(0, 0, 0.2) \times (1, 0, 0) = (0 \cdot 0 - 0.2 \cdot 0, 0.2 \cdot 1 - 0 \cdot 0, 0) = (0, 0.2, 0)$ . ✔ They match — that is exactly why the skew matrix replaces the cross product in $\dot{C}_{b}^{n} = C_{b}^{n} [ω_{nb}^{b} \times]$ .

Problem 2.3

At latitude $ϕ = 4 5^{\circ}$ , height $h = 0$ , the vehicle flies due East at $v_{E} = 200 m/s$ . The prime-vertical radius of curvature is $R_{N} = 6.389 \times 1 0^{6} m$ . Find the longitude rate $\dot{λ}$ in rad/s.

Recall Solution 2.3

Why this formula? Longitude $λ$ is an angle measured around the Earth's spin axis. As you fly East you trace the circle of constant latitude (a "parallel"). Its radius is not the full Earth radius but the shrunken $(R_{N} + h) cos ϕ$ , because at latitude $ϕ$ the parallel is a small circle whose radius is the big circle's radius times $cos ϕ$ . An angle swept equals arc-length travelled divided by the circle's radius, so $\dot{λ} = \frac{East arc-rate}{parallel radius} = \frac{v _{E}}{( R _{N} + h ) c o s ϕ} .$ Plug in: $\dot{λ} = \frac{200}{6.389 \times 1 0 ^{6} \cdot c o s 4 5 ^{\circ}} = \frac{200}{6.389 \times 1 0 ^{6} \cdot 0.70711} \approx 4.427 \times 1 0^{- 5} rad/s .$ The $cos ϕ$ makes longitude circles shrink toward the poles, so the same East speed spins you around longitude faster the higher your latitude. See Radii of Curvature of the Earth Ellipsoid.

L3 — Analysis

Problem 3.1

A gyro has a constant bias $ε = 0.005 rad/s$ about the East axis. The vehicle is static and initially level. Estimate the attitude (tilt) error after $t = 20 s$ , in degrees.

Recall Solution 3.1

A static-vehicle tilt error accumulates linearly: $δ θ = εt = 0.005 \times 20 = 0.1 rad$ . Convert: $0.1 \times \frac{180}{π} \approx 5.7 3^{\circ}$ . The mechanization believes it rotated when it did not, because it integrates the fake rate.

Problem 3.2

Continue Problem 3.1. This 0.1 rad tilt misrotates gravity. Estimate the spurious horizontal acceleration it injects, using $g = 9.81$ .

Recall Solution 3.2

A tilt $δ θ$ tips the gravity vector so a component $g sin (δ θ)$ spills onto the horizontal channel: $δ a = g sin (δ θ) = 9.81 sin (0.1) \approx 9.81 \times 0.09983 = 0.9794 m/s^{2} .$ This is the dominant INS error path: a small attitude error leaks nearly the whole of $g$ into where it doesn't belong.

The figure makes the geometry concrete. True gravity (red) points straight Down. The mechanization thinks Down is the dashed white axis, tilted by $δ θ$ (blue arc). When it removes "gravity" along its tilted axis, a yellow horizontal slice $g sin δ θ$ is left uncancelled — that leftover is the fake horizontal acceleration you computed.

Figure — Mechanization equations — integrating IMU to get position, velocity, attitude

Problem 3.3

Why does a velocity error from an accelerometer bias $b$ grow like $t$ , while a position error from the same bias grows like $t^{2}$ ?

Recall Solution 3.3

Velocity is one integral of acceleration: $\int_{0}^{t} b d τ = b t$ (linear). Position is one further integral: $\int_{0}^{t} b τ d τ = \frac{1}{2} b t^{2}$ (quadratic). Each integration multiplies by "time," so each layer of the chain adds one power of $t$ to the error — this is why INS drift accelerates and needs GNSS correction.

L4 — Synthesis

Problem 4.1

Level IMU, no rotation, flat-Earth. Accels read $f^{b} = (0.02, 0, - 9.81)$ m/s $^{2}$ (a small $x$ -bias of $0.02$ ). With $g^{n} = (0, 0, + 9.81)$ , find $\dot{v}^{n}$ , then $v_{N}$ and North position after $t = 100 s$ from rest.

Recall Solution 4.1

Since attitude is identity (no tilt, no rotation) $C_{b}^{n} f^{b} = f^{b}$ . Using the flat-Earth velocity core $\dot{v}^{n} = C_{b}^{n} f^{b} + g^{n}$ : $\dot{v}^{n} = f^{b} + g^{n} = (0.02, 0, - 9.81) + (0, 0, 9.81) = (0.02, 0, 0) .$ Down cancels (static vertical channel ✔). Only the North bias survives. $v_{N} (100) = 0.02 \times 100 = 2 m/s$ . North position $= \frac{1}{2} (0.02) (100)^{2} = 100 m$ . A $2 mg$ -scale bias walked us 100 m off in under two minutes.

Problem 4.2

Combine a gyro bias with the accel channel. Static, level, gyro bias $ε = 0.002 rad/s$ about East for $t = 50 s$ . Find (a) the tilt, (b) the leaked horizontal acceleration $g sin δ θ$ at $t = 50$ s treating tilt as instantaneously that value, and (c) roughly bound the horizontal velocity error by assuming acceleration ramps linearly from $0$ to that value.

Recall Solution 4.2

(a) $δ θ = εt = 0.002 \times 50 = 0.1 rad$ . (b) $δ a_{m a x} = 9.81 sin (0.1) \approx 0.9794 m/s^{2}$ . (c) Tilt grows linearly, so leaked accel $\approx g sin (εt) \approx g εt$ for small angles. Integrate: $δ v \approx \int_{0}^{50} g εt d t = g ε \frac{t ^{2}}{2} = 9.81 \times 0.002 \times \frac{2500}{2} = 24.525 m/s .$ The velocity error grows like $t^{2}$ here because the bias feeds a tilt that itself grows with $t$ — a compounding, faster-than-linear leak.

L5 — Mastery

Problem 5.1

A pilot proposes: "For a 30-second drone hop under 1 km, skip Earth-rotation and transport terms." Quantify both terms you'd be dropping. (i) The Coriolis/Earth-rotation part $2 Ω_{e} v$ for $v = 30 m/s$ . (ii) The transport-rate part $ω_{e n}^{n}$ : use its dominant component $v_{E} / (R_{N} + h)$ for the same $30 m/s$ East and $R_{N} = 6.389 \times 1 0^{6} m$ , and turn it into an acceleration $ω_{e n} v$ . Compare both against a typical accel-bias-driven acceleration of $0.02 m/s^{2}$ . Is dropping them justified?

Recall Solution 5.1

(i) Coriolis magnitude bound: $2 Ω_{e} v = 2 \times 7.292 \times 1 0^{- 5} \times 30 = 4.375 \times 1 0^{- 3} m/s^{2}$ . (ii) Transport rate: $ω_{e n} \approx \frac{v}{R _{N} + h} = \frac{30}{6.389 \times 1 0 ^{6}} = 4.696 \times 1 0^{- 6} rad/s$ . As a frame-correction acceleration $ω_{e n} v = 4.696 \times 1 0^{- 6} \times 30 = 1.409 \times 1 0^{- 4} m/s^{2}$ . Compare to bias-driven $0.02 m/s^{2}$ : Coriolis is about $4.6 \times$ smaller, and the transport term is over $100 \times$ smaller still. Over $30 s$ the larger of the two (Coriolis) contributes at most $\frac{1}{2} (4.375 \times 1 0^{- 3}) (30)^{2} \approx 1.97 m$ of position error; the transport term's is a few centimetres. Verdict: dropping Earth-rotation and transport is justified for this short, slow, low-accuracy hop (the flat-Earth 80/20 core). It would not be justified for a long-range, high-grade system where $Ω_{e}$ dominates. See Coriolis and Centrifugal Effects and Strapdown vs Gimbaled INS.

Problem 5.2

Design check. You want horizontal position error under $10 m$ after $60 s$ of pure inertial coasting (no GNSS), dominated by a single accel bias $b$ (assume level, no tilt). What is the maximum allowable $b$ ?

Recall Solution 5.2

Position error from constant bias: $δ r = \frac{1}{2} b t^{2}$ . Set $δ r = 10$ , $t = 60$ : $b = \frac{2 δ r}{t ^{2}} = \frac{2 \times 10}{3600} = 5.556 \times 1 0^{- 3} m/s^{2} .$ In "milli-g" ( $1 mg = 9.81 \times 1 0^{- 3} m/s^{2}$ ): $b \leq 0.566 mg$ . So you need an accelerometer better than about half a milli-g — a real navigation-grade requirement, and exactly why INS-GNSS fusion exists to bound this drift.

Recall One-line summary of the whole ladder

An IMU feels rate and specific force; software rotates, adds gravity, and integrates — and every integration turns a small bias into a fast-growing error, which is why real systems fuse with GNSS.