Inertial navigation — accelerometer measures non-gravitational specific force: Worked examples

This page is a drill of every case the accelerometer equation $f = a - g$ can throw at you. If you have not yet seen where that equation comes from, build it first in the parent topic.

Before any numbers, let us agree on the picture and the sign rule — nothing below uses a symbol we have not pinned down.

Recall The equation, in words

The accelerometer reads specific force $f = a - g$ . Here $a$ is the true acceleration of the box (how its velocity actually changes), and $- g$ is "gravity subtracted back out". Component by component: $f_{x} = a_{x} - g_{x}$ and $f_{z} = a_{z} - g_{z}$ . With our convention $g_{x} = 0$ and $g_{z} = - g$ , so $f_{x} = a_{x}$ and $f_{z} = a_{z} + g$ .

Look at the figure: the true acceleration $a$ (blue) and minus-gravity $- g$ (green, always pointing up with length $g$ ) are added tip-to-tail; their sum is the orange reading $f$ . Every example below is just this triangle with different blue arrows.

The scenario matrix

Each row is a case class. The examples that follow are labelled with the cell they cover.

#	Case class	What is special	Example
C1	$a_{z} > 0$ (accelerate up)	reading exceeds $g$	Ex 1
C2	$a_{z} < 0$ (accelerate down, still moving)	reading below $g$ , stays positive	Ex 2
C3	$a_{z} = - g$ exactly (free fall)	zero reading — degenerate	Ex 3
C4	$a_{z} < - g$ (thrown down harder than gravity)	reading goes negative	Ex 4
C5	Pure horizontal $a_{x} > 0$ , $a_{z} = 0$ (forward)	two non-zero components, tilt one way	Ex 5
C6	Combined $a_{x} \neq = 0, a_{z} \neq = 0$ (banked/climbing turn)	full vector, magnitude & tilt	Ex 6
C7	Limiting values ( $R \to \infty$ , and $a_{x} \to \infty$ )	what dominates the reading	Ex 7
C8	Real-world word problem (rocket liftoff)	translate words → vector	Ex 8
C9	Exam twist (reading given, find motion)	invert the equation	Ex 9
C10	Pure horizontal $a_{x} < 0$ , $a_{z} = 0$ (braking)	tilt the other way	Ex 10

We use $g = 9.81 m/s^{2}$ throughout, and round to two decimals.

Examples

Worked example Ex 4 — Pushed down

harder than gravity (Cell C4) A rocket sled aims its thrust to push a payload straight down at $a_{z} = - 15 m/s^{2}$ (faster than free fall). Reading?

Forecast: can a reading be negative? What would that mean physically?

True acceleration: $a = (0, - 15)$ . Why? Given: the vehicle accelerates downward at $15$ , more than $g$ .
Apply: $f_{z} = a_{z} + g = - 15 + 9.81 = - 5.19 m/s^{2}$ . Why? Gravity alone can only supply $9.81$ of downward pull; the extra $5.19$ must come from the spring pulling the mass down — a downward contact force.
Result: $f_{z} = - 5.19 m/s^{2}$ (i.e. $5.19$ downward).

Verify: the sign flipped because $∣ a_{z} ∣ = 15 > g$ . Physically the mass is being dragged down by the case, so the reading points down. Boundary check: at $a_{z} = - 9.81$ it would be exactly $0$ (Ex 3), and going more negative pushes the reading negative ✓.

Worked example Ex 5 — Pure horizontal acceleration, forward (Cell C5)

A car on a flat road accelerates forward at $a_{x} = + 4 m/s^{2}$ ; no vertical motion, $a_{z} = 0$ . What does a 2-axis accelerometer read, and in which direction does it tilt?

Forecast: will the reading be purely horizontal, purely vertical, or slanted?

True acceleration: $a = (+ 4, 0)$ . Why? Forward speeding up, level ground.
Components: $f_{x} = a_{x} = 4$ and $f_{z} = a_{z} + g = 0 + 9.81 = 9.81$ . Why? The spring still supports weight ( $f_{z} = g$ ) and must shove the mass forward ( $f_{x} = 4$ ).
Magnitude: $∣ f ∣ = 4^{2} + 9.8 1^{2} = 16 + 96.24 = 112.24 = 10.59 m/s^{2}$ . Why the square root? $f$ is a vector; its length is the Pythagorean combination of its perpendicular parts.
Tilt angle from vertical: $θ = arctan (f_{x} / f_{z}) = arctan (4/9.81) = 22.1 9^{\circ}$ toward $+ x$ . Why $arctan$ ? We ask "which angle has this ratio of horizontal-to-vertical?" — $arctan$ undoes the tangent to give that angle.

Verify: $∣ f ∣ > g$ as expected (extra horizontal shove), and the reading leans backward relative to the acceleration — this is why a hanging pendulum swings back when you accelerate. Units all $m/s^{2}$ ✓.

Worked example Ex 6 — Aircraft in a banked, climbing situation (Cell C6)

A plane simultaneously accelerates forward-horizontally at $a_{x} = + 6 m/s^{2}$ and climbs with vertical acceleration $a_{z} = + 2 m/s^{2}$ . Find the reading vector, its magnitude, and tilt.

Forecast: bigger or smaller than the car in Ex 5?

True acceleration: $a = (6, 2)$ . Why? Both components given.
Components: $f_{x} = 6$ , $f_{z} = 2 + 9.81 = 11.81$ . Why? Horizontal part passes straight through; vertical part adds the ever-present $g$ .
Magnitude: $∣ f ∣ = 6^{2} + 11.8 1^{2} = 36 + 139.48 = 175.48 = 13.25 m/s^{2}$ .
Tilt from vertical: $θ = arctan (6/11.81) = 26.9 4^{\circ}$ .

Verify: both $f_{x}$ and $f_{z}$ exceed the car's values (extra climb raised $f_{z}$ above $g$ ), so magnitude is larger — consistent. This total is the load factor a pilot feels: $∣ f ∣/ g = 13.25/9.81 = 1.35$ g's. ✓

Worked example Ex 7 — Limiting behaviour (Cell C7)

Two limits. (a) In a coordinated level turn, $f_{x} = v^{2} / R$ , $f_{z} = g$ , where $R$ is the turn radius defined above. What happens to the reading as $R \to \infty$ (straighter and straighter)? (b) What dominates $∣ f ∣$ as forward acceleration $a_{x} \to \infty$ ?

Forecast: (a) does the reading approach $g$ or $0$ ? (b) does the vertical $g$ still matter?

(a) Set $f_{x} = v^{2} / R$ . Why? Centripetal acceleration of a circular path is $v^{2} / R$ , horizontal.
Take $R \to \infty$ : $f_{x} = v^{2} / R \to 0$ , so $∣ f ∣ = (v^{2} / R)^{2} + g^{2} \to 0 + g^{2} = g$ . Why? An infinite radius is a straight line — no centripetal demand — so only weight support remains.
(b) For large $a_{x}$ : $∣ f ∣ = a_{x}^{2} + g^{2}$ . As $a_{x} \to \infty$ , the constant $g^{2}$ becomes negligible, so $∣ f ∣ \approx ∣ a_{x} ∣$ and the tilt $arctan (a_{x} / g) \to 9 0^{\circ}$ (reading lies nearly flat/horizontal).

Verify: limit (a) correctly recovers the straight-and-level answer of $g$ (matches Ex 5 with $a_{x} = 0$ ). Limit (b): if you accelerate insanely hard, gravity is a rounding error and $f$ points along your acceleration ✓.

Worked example Ex 8 — Real-world word problem: rocket liftoff (Cell C8)

A rocket lifts off vertically. At burnout its engines give a true upward acceleration of $a_{z} = + 40 m/s^{2}$ . Its onboard accelerometer (z-up) is used by the Strapdown Inertial Navigation System. (i) What does it read? (ii) The navigation computer must recover true $a$ — show it works.

Forecast: will the computer add or subtract $g$ to get back to $40$ ?

Reading: $f_{z} = a_{z} + g = 40 + 9.81 = 49.81 m/s^{2}$ . Why? Spring supports weight ( $9.81$ ) plus provides the $40$ upward push.
In g's: $49.81/9.81 = 5.08$ g's felt by the crew. Why divide by $g$ ? To express the reading as a multiple of normal gravity.
Computer recovers $a_{z} = f_{z} + g_{z} = 49.81 + (- 9.81) = 40 m/s^{2}$ . Why add $g_{z} = - 9.81$ ? The equation $a = f + g$ says add the modeled gravity vector back; here that model contributes $- 9.81$ in $z$ .

Verify: step 3 returns exactly the $40$ we started with — the loop is self-consistent. Forgetting the $+ g$ term would leave $49.81$ and, per Dead Reckoning and Error Drift, blow the trajectory apart. ✓

Worked example Ex 9 — Exam twist: reading given, find the motion (Cell C9)

An accelerometer (z-up, single axis) outputs $f_{z} = 7.00 m/s^{2}$ . It is known the vehicle moves purely vertically. What is its true acceleration, and describe the motion.

Forecast: is the vehicle accelerating up or down?

Invert the equation: $a_{z} = f_{z} - g = 7.00 - 9.81 = - 2.81 m/s^{2}$ . Why subtract $g$ ? From $f_{z} = a_{z} + g$ , solving for $a_{z}$ gives $a_{z} = f_{z} - g$ .
Interpret sign: $a_{z} < 0$ ⇒ true acceleration points down at $2.81 m/s^{2}$ . Why? Negative $z$ -component means the velocity is changing downward (either falling faster, or rising while slowing).
Sanity on regime: $∣ a_{z} ∣ = 2.81 < g$ , so this is Cell C2 territory (below $g$ , positive reading) — occupants feel lighter but not weightless.

Verify: plug back $f_{z} = a_{z} + g = - 2.81 + 9.81 = 7.00$ ✓ — matches the given reading exactly. A reading between $0$ and $g$ always means downward true acceleration of magnitude less than $g$ . ✓

Worked example Ex 10 — Pure horizontal acceleration, braking (Cell C10)

The same car as Ex 5 now brakes, decelerating at $a_{x} = - 4 m/s^{2}$ ; still level, $a_{z} = 0$ . Find the reading, its magnitude, and which way it tilts — compare with Ex 5.

Forecast: same magnitude as Ex 5? Same tilt direction?

True acceleration: $a = (- 4, 0)$ . Why? Braking while moving forward means the velocity change points backward, so $a_{x} = - 4$ .
Components: $f_{x} = a_{x} = - 4$ and $f_{z} = a_{z} + g = 0 + 9.81 = 9.81$ . Why? Horizontal part passes straight through (now negative); vertical part still supports weight.
Magnitude: $∣ f ∣ = (- 4)^{2} + 9.8 1^{2} = 16 + 96.24 = 112.24 = 10.59 m/s^{2}$ . Why? Squaring erases the sign, so the length equals Ex 5's.
Tilt from vertical: $θ = arctan (f_{x} / f_{z}) = arctan (- 4/9.81) = - 22.1 9^{\circ}$ — i.e. $22.1 9^{\circ}$ toward $- x$ (the opposite side from Ex 5).

Verify: identical magnitude to Ex 5 but mirror-image tilt — the reading leans forward now, which is why you're thrown against the seatbelt when braking. The sign of $a_{x}$ flips the sign of the horizontal component and hence the tilt direction, completing the horizontal quadrant. ✓

Recall Quick self-test

A sensor reads exactly $0$ . What is the vehicle doing? ::: In free fall — true acceleration equals $g$ (Cell C3). A sensor reads $- 5.19$ (downward). What must be true of $∣ a_{z} ∣$ ? ::: $∣ a_{z} ∣ > g$ : the case is dragging the mass down harder than gravity (Cell C4). As turn radius $R \to \infty$ , what does $∣ f ∣$ approach in a level turn? ::: $g$ — a straight path has no centripetal demand (Cell C7). Braking and accelerating at the same rate give the same reading magnitude — what differs? ::: The tilt direction flips (forward vs backward lean); magnitude is identical (Ex 5 vs Ex 10).

Connections

Newton's Second Law — every inversion here starts from $m a = F_{c} + m g$ .
Equivalence Principle — the zero reading in Ex 3.
Strapdown Inertial Navigation System — where Ex 8's $a = f + g$ is integrated.
Gravity Model (WGS-84 / J2) — supplies the $g$ the computer adds back.
Dead Reckoning and Error Drift — what goes wrong if you skip that step.
Gyroscope and Attitude Determination — provides the orientation so $g$ is resolved into the right axes.