Question bank — Mechanization equations — integrating IMU to get position, velocity, attitude

False. It reads $-9.81{\text{m/s}}^2$ (in NED, Down $=+z$, the reading is opposite the support). The table pushes it up, and specific force ${\mathbf{f}}^b$ is the contact force, so a static reading is $+g$ magnitude opposite the support — zero happens only in free fall.

False. It reads $0$. Gravity accelerates every part of the proof mass equally, so nothing squeezes the sensor — there is no specific force to detect during free fall.

False. Step it out: a gyro bias $\epsilon$ tilts attitude linearly, $\delta \theta =\epsilon t$. A tilt of $\delta \theta$ misrotates gravity so a horizontal acceleration $a\approx g\delta \theta =g\epsilon t$ leaks into the level channel. Integrating that once gives velocity $\propto t^2$ and again gives position $\propto t^3$ — so position error from a gyro bias grows like $t^3$, worse than the $t^2$ from an accelerometer bias.

False. The accelerometer output ${\mathbf{f}}^b$ lives in body axes. You need $C_b^n$ (attitude) to rotate it into nav axes (${\mathbf{f}}^n=C_b^n{\mathbf{f}}^b$) before you can add gravity, so attitude must come first.

False. It is plumb-bob (apparent) gravity = gravitational attraction minus centrifugal acceleration from Earth's spin. That is why it points along the local vertical (Down), not exactly at Earth's centre.

True. For short baselines the $2{\mathbf{\omega }}_{ie}^n\times {\mathbf{v}}^n$ and ${\mathbf{\omega }}_{en}^n\times {\mathbf{v}}^n$ corrections are negligible; the flat-Earth trio (in NED) ${\dot{\mathbf{r}}}^n={\mathbf{v}}^n,{\dot{\mathbf{v}}}^n=C_b^n{\mathbf{f}}^b+{\mathbf{g}}^n,{\dot{C}}_b^n=C_b^n[{\mathbf{\omega }}_{ib}^b\times ]$ is the 80/20 core. The NED convention fixes every sign in that trio.

False. Finite 3-D rotations do not commute; orientation lives on the curved manifold $SO(3)$. That is exactly why we propagate a matrix/quaternion (${\dot{C}}_b^n=C_b^n[\mathbf{\omega }\times ]$, using the skew matrix defined above) instead of summing angles.

True (it is a convention). In NED the "Down" axis $v_D$ points toward Earth, so positive downward velocity decreases height; the sign is purely from the choice of axis direction. (In ENU it would be $\dot{h}=+v_U$.)

The device measures specific force ${\mathbf{f}}^b$, not acceleration. Integrating raw ${\mathbf{f}}^b$ while static injects a fake $9.81{\text{m/s}}^2$; you must rotate to nav axes and use the full ${\dot{\mathbf{v}}}^n=C_b^n{\mathbf{f}}^b+{\mathbf{g}}^n$ so gravity is added back first.

You need the rate relative to the nav frame, ${\mathbf{\omega }}_{nb}^b={\mathbf{\omega }}_{ib}^b-C_n^b({\mathbf{\omega }}_{ie}^n+{\mathbf{\omega }}_{en}^n)$, and then form its skew matrix $[{\mathbf{\omega }}_{nb}^b\times ]$. Using ${\mathbf{\omega }}_{ib}^b$ raw makes attitude rotate with the Earth and the moving frame.

It is a fictitious (frame) correction, not a real force. It appears only because we express Newton's law in a rotating/curving nav frame; there is no physical push behind it.

Missing the $\cos \varphi$: correct is $\dot{\lambda }=v_E/[(R_N+h)\cos \varphi ]$, where $v_E$ is the East velocity component and $R_N$ the transverse radius. Longitude circles shrink toward the poles, so the same eastward speed sweeps more longitude at high latitude.

They use different radii of curvature: latitude rate uses the meridian radius $R_M$, longitude rate uses the transverse radius $R_N$. The ellipsoid curves differently north–south vs east–west (more sharply north–south), so $R_M\ne R_N$ — see Radii of Curvature of the Earth Ellipsoid.

No — velocity is one integration of acceleration error, position is two. Position error accumulates faster ($\propto t^2$ from accel bias) than velocity error ($\propto t$).

You add ${\mathbf{g}}^n$: the equation is ${\dot{\mathbf{v}}}^n=C_b^n{\mathbf{f}}^b+{\mathbf{g}}^n$, so the $+{\mathbf{g}}^n$ term is where gravity re-enters. The accelerometer already missed true gravitational acceleration (it can't feel it), so gravity is put back in software rather than removed.

Gravity ${\mathbf{g}}^n$ is naturally expressed along the local vertical (a nav-frame direction). To add two vectors they must share a frame, so the body-frame force is rotated by $C_b^n$ first (${\mathbf{f}}^n=C_b^n{\mathbf{f}}^b$).

Coriolis arises from differentiating a velocity observed in a rotating frame — it is inherently a first-derivative-of-velocity effect. Attitude and position kinematics don't involve that cross term the same way (see Coriolis and Centrifugal Effects — Coriolis is $-2\mathbf{\omega }\times \mathbf{v}$, largest when velocity is perpendicular to the rotation axis).

A small attitude tilt $\delta \theta$ leaks the full $g$ into the horizontal channel as $g\sin \delta \theta \approx g\delta \theta$. Since $g\approx 9.81$, even a $0.1^{\circ }$ tilt injects a larger horizontal acceleration than most accelerometer biases.

Euler angles suffer gimbal lock (singularities) and their rates blow up near $\pm 90^{\circ }$ pitch. The DCM/quaternion stays well-behaved everywhere — but you must give it a correct initial condition (an initial alignment $C_b^n(0)$ or unit quaternion) and, because numerical integration drifts off the constraint, renormalize each step (orthonormalize the DCM, or scale the quaternion back to unit norm). See Direction Cosine Matrix and Quaternions.

Because the local "down" direction physically re-tilts as you travel over a curved Earth. The nav frame rotates to keep pointing down, and that rotation must be removed from the gyro measurement.

In strapdown the sensors are bolted to the body, so all the frame rotations (attitude layer) must be done numerically at high rate; a gimbaled platform mechanically holds the frame, doing the rotation in hardware (see Strapdown vs Gimbaled INS).

Arc length $s=R\theta$, so $\theta =s/R$ and $\dot{\theta }=\dot{s}/R$. Northward speed $v_N$ is the meridian arc rate, so $\dot{\varphi }=v_N/(R_M+h)$.

External aids (like GNSS) reveal the growing INS error, and because bias-driven error has a known time signature ($t$, $t^2$, $t^3$), the filter can attribute drift to the underlying bias and correct it (see Kalman Filter for INS-GNSS Integration — the filter blends a drifting-but-smooth INS with a noisy-but-bounded GNSS fix).

The longitude rate $\dot{\lambda }=v_E/[(R_N+h)\cos \varphi ]$ diverges because $\cos \varphi \to 0$ — longitude is ill-defined at the pole. The latitude rate $\dot{\varphi }=v_N/(R_M+h)$ stays perfectly finite (the meridian radius $R_M+h$ is a normal positive length there), so only the east-west channel is singular; wander-azimuth or a different frame fixes it.

It should be $0$. The upward support force ($-9.81$ Down in ${\mathbf{f}}^b$) rotates into nav axes and exactly cancels the added $+9.81$ Down gravity ${\mathbf{g}}^n$, so the two terms in ${\dot{\mathbf{v}}}^n=C_b^n{\mathbf{f}}^b+{\mathbf{g}}^n$ sum to zero.

${\dot{C}}_b^n=C_b^n[0\times ]=0$, so orientation is held constant — correct for a non-rotating body, and it also means any zero-rate bias still silently tilts the frame.

No. Even at low speed, Earth rate ($7.29\times 10^{-5}\text{rad/s}$) and transport effects accumulate over long time, so long-duration navigation needs the full Earth terms regardless of speed.

It grows as $\frac{1}{2}b{t}^2$ without bound — the unaided INS drift signature. This is why bounded external aiding (GNSS) is required for long missions.

No — it reads apparent gravity, which is gravitational attraction minus the equatorial centrifugal term ($\approx 0.034{\text{m/s}}^2$ smaller). The mechanization ${\mathbf{g}}^n$ uses this apparent value (see Gravity Model and Geoid).