Exercises — Inertial navigation — accelerometer measures non-gravitational specific force

WHAT: apply $\mathbf{f}=\mathbf{a}-\mathbf{g}$ in the z-direction. WHY: the sensor reads specific force, never $\mathbf{a}$ directly. $f_z=a_z-g_z=0-(-9.81)=+9.81{\text{m/s}}^2$ Reads $+9.81{\text{m/s}}^2$ (upward). The spring must push up to hold the mass against gravity, and that push is the reading.

In free fall the only force is gravity, so the true acceleration equals gravity: $\mathbf{a}=\mathbf{g}=(0,0,-9.81)$. $f_z=a_z-g_z=-9.81-(-9.81)=0$ Reads $0$ — weightlessness. The spring has nothing to push against, so it reads nothing. (This is the Equivalence Principle in sensor form.)

From L1.1, at rest $\mathbf{a}=0$ so $\mathbf{f}=-\mathbf{g}$. Since $\mathbf{g}$ points down, $\mathbf{f}$ points up. The reading always mirrors the contact force the spring supplies, and at rest that force fights gravity → upward.

Step 1 (WHAT/WHY): downward acceleration means $a_z=-3$ (negative because z is up). Step 2: gravity $g_z=-9.81$. Step 3: $f_z=a_z-g_z=-3-(-9.81)=+6.81{\text{m/s}}^2$ Reads $+6.81{\text{m/s}}^2$, still upward but less than $9.81$. You feel lighter — the spring supports less because the mass is partly "falling away" from it.

$f_z=a_z-g_z=39.24-(-9.81)=49.05{\text{m/s}}^2=5g$ The spring must both support the weight ($1g$) and accelerate the mass ($4g$), so the reading is $5g$. This is why a "$4g$ burn" feels like $5g$ of force on your chest.

Motion is horizontal, so gravity has no x-component: $g_x=0$. Braking = acceleration opposite to motion, so $a_x=-6{\text{m/s}}^2$. $f_x=a_x-g_x=-6-0=-6{\text{m/s}}^2$ Reads $-6{\text{m/s}}^2$ (pointing backward). You lurch forward against the seatbelt; the belt's contact force is what the sensor equivalently registers.

WHAT/WHY: a level turn has two acceleration ingredients — vertical ($a_z=0$, altitude constant) and horizontal centripetal $a_x=v^2/R$ pointing toward the turn centre. Vertical: $f_z=a_z-g_z=0-(-9.81)=+9.81$. Horizontal: $g_x=0$, so $f_x=a_x=\frac{v^2}{R}=\frac{200^2}{3000}=\frac{40000}{3000}=13.33{\text{m/s}}^2$. Magnitude: $|\mathbf{f}|=\sqrt{f_z^2+f_x^2}=\sqrt{9.81^2+13.33^2}=\sqrt{96.24+177.7}=\sqrt{273.98}\approx 16.55{\text{m/s}}^2$ Load factor: $n=|\mathbf{f}|/g=16.55/9.81\approx 1.69g$. The seat must both hold the pilot up and curve their path — the vector sum exceeds $g$, so they feel heavier (a "$1.69g$ turn").

WHY tan? We have the two legs of a right triangle: the horizontal leg $f_x$ (opposite the angle from vertical) and the vertical leg $f_z$ (adjacent). The ratio opposite/adjacent = $\tan \varphi$ encodes how steeply $\mathbf{f}$ tilts from straight down. To recover the angle from that ratio we invert tan → arctan. $\tan \varphi =\frac{f_x}{f_z}=\frac{13.33}{9.81}=1.359$ $\varphi =\arctan (1.359)\approx 53.7^{\circ }$ A coordinated turn banks the aircraft so its wings are perpendicular to $\mathbf{f}$; here that is about $53.7^{\circ }$ from level.

WHAT went wrong: true acceleration is $a_z=f_z+g_z=9.81+(-9.81)=0$, so the real position stays $0$. But the buggy code integrates $\tilde{a}=f_z=9.81$ constant. Double integration of a constant $\tilde{a}$ from rest: $\Delta z_{\text{error}}=\frac{1}{2}\tilde{a}{t}^2=\frac{1}{2}(9.81)(10{)}^2=490.5\text{m}$ Nearly half a kilometre of drift in 10 seconds from one forgotten $\mathbf{g}$. This is the $\frac{1}{2}g{t}^2$ blow-up (see Dead Reckoning and Error Drift).

Consistency check (WHY): at rest $\mathbf{a}=0$, so $\mathbf{f}=-\mathbf{g}$, meaning $|\mathbf{f}|$ must equal $g$. $|\mathbf{f}|=\sqrt{0^2+4.905^2+8.496^2}=\sqrt{24.06+72.18}=\sqrt{96.24}=9.81{\text{m/s}}^2=g✓$ Consistent with rest. Tilt: the reading's z-component is the part aligned with true-up; the horizontal part is the tilt. $\cos \theta =\frac{f_z}{|\mathbf{f}|}=\frac{8.496}{9.81}=0.866\Rightarrow \theta =\arccos (0.866)=30^{\circ }$ The unit is tilted $30^{\circ }$ from horizontal. (A gravity vector splits into slope-parallel and slope-normal parts — accelerometers alone can sense static tilt precisely because they read $-\mathbf{g}$ at rest.)

In orbit the only force on the body is gravity, so the true acceleration equals the local gravitational acceleration: $\mathbf{a}=\mathbf{g}(r)$, with $|\mathbf{g}(r)|=8.11$. $|\mathbf{f}|=|\mathbf{a}-\mathbf{g}|=|\mathbf{g}-\mathbf{g}|=0$ Reads $0$ — orbit is perpetual free fall, so the sensor is weightless despite the centre accelerating at $8.11{\text{m/s}}^2$ toward Earth. A tiny non-zero reading would come from non-gravitational forces — atmospheric drag, solar radiation pressure, thruster firings — because those are contact/applied forces the spring can feel. Accelerometers in orbit are therefore drag sensors, not gravity sensors.

WHAT/WHY: at the loop bottom the centre is directly overhead, so centripetal acceleration is upward: $a_z=+v^2/R$. $a_z=\frac{180^2}{1200}=\frac{32400}{1200}=27.0{\text{m/s}}^2$ Specific force z-component: $f_z=a_z-g_z=27.0-(-9.81)=36.81{\text{m/s}}^2$ Load factor: $n=f_z/g=36.81/9.81\approx 3.75g$. The seat must supply both the $27$ upward to curve the path and $9.81$ to hold weight — the two add at the loop bottom, pinning the pilot down at $\approx 3.75g$.

$f_z=a_z-g_z=-27.0-(-9.81)=-17.19{\text{m/s}}^2$ Reads $-17.19{\text{m/s}}^2$ — negative means the seat is pushing the pilot downward (i.e. the pilot presses up into the seat, inverted). Load factor magnitude $|f_z|/g=17.19/9.81\approx 1.75g$, but it is a "negative-g relative to the seat" situation: the required inward force exceeds gravity, so the restraint pushes the pilot toward the loop centre. All-cases check: at the bottom the terms added ($+3.75g$); at the top they partly cancelled in the vertical sum but the sign flipped — the sensor faithfully tracks the contact force in every quadrant of the loop.