Visual walkthrough — Inertial navigation — accelerometer measures non-gravitational specific force

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Step 1 — The one gadget: a mass on a spring inside a box

WHAT. An accelerometer is a small ball (the proof mass) tethered by a spring inside a sealed box. The box is bolted to whatever we want to track — a car, a rocket, an elevator. The only thing the sensor ever reports is how hard the spring is pushing or pulling the ball.

WHY start here. Every symbol we will meet is just an arrow acting on this ball. If we can draw the arrows correctly, the algebra is forced. There is nothing to memorize.

PICTURE. The ball hangs in the middle. Two things can act on it, and we will meet them one at a time in the next steps.

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Step 2 — Arrow #1: gravity pulls on everything

WHAT. Gravity tugs the ball straight down. We write this pull as $m\mathbf{g}$.

Let us earn every symbol:

• $\mathbf{g}$ (bold letter) is the gravity arrow: it points down and its length is $g\approx 9.81$ (in ${\text{m/s}}^2$). The bold means "this is an arrow — it has a direction," while plain $g$ is just its length.
• $m\mathbf{g}$ means "stretch the gravity arrow by the mass $m$." A heavier ball is pulled harder — that is the whole content of multiplying by $m$.

WHY name it separately. Because gravity has a secret property we exploit later: it pulls on every atom of the ball at once, from the inside. Nothing has to touch the ball to make gravity act. Hold that thought.

PICTURE. One magenta arrow, straight down, labelled $m\mathbf{g}$.

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Step 3 — Arrow #2: the spring, the only thing we can read

WHAT. If the ball tries to drift off-centre, the spring pushes it back. We call this push the contact force ${\mathbf{F}}_{\text{c}}$ ("c" for contact — something physically touches and pushes the ball).

• ${\mathbf{F}}_{\text{c}}$ (bold, so it's an arrow) is whatever the spring must supply to keep the ball centred in its box.
• Unlike gravity, this force acts by touching — the spring is physically hooked to the ball. That difference is the entire secret of the sensor.

WHY it is the star. The sensor cannot see the gravity arrow. It can only measure the spring's push, ${\mathbf{F}}_{\text{c}}$. So whatever equation we build, the readable quantity must be ${\mathbf{F}}_{\text{c}}$.

PICTURE. The violet spring-arrow ${\mathbf{F}}_{\text{c}}$ alongside the magenta gravity arrow. Two arrows, two different origins: one from inside (gravity), one from touch (spring).

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Step 4 — Add the arrows with Newton's law

WHAT. Newton's Second Law says: the total force on the ball equals its mass times its true acceleration. "True acceleration" $\mathbf{a}$ is how the ball's velocity is really changing, as seen by someone floating outside in space (an inertial observer). We add our two arrows:

$\underset{\begin{array}{c}\text{mass}\times \text{true}\\ \text{acceleration}\end{array}}{\underbrace{m\mathbf{a}}}=\underset{\begin{array}{c}\text{spring}\\ \text{push}\end{array}}{\underbrace{{\mathbf{F}}_{\text{c}}}}+\underset{\begin{array}{c}\text{gravity}\\ \text{pull}\end{array}}{\underbrace{m\mathbf{g}}}$

Term by term:

• $m\mathbf{a}$ — the net result: mass times the real change in motion.
• ${\mathbf{F}}_{\text{c}}$ — the readable spring push (Step 3).
• $m\mathbf{g}$ — the invisible gravity pull (Step 2).

WHY add, not subtract? Because forces are arrows placed head-to-tail: the net arrow $m\mathbf{a}$ is simply the two input arrows summed. Newton's law is nothing more than "arrows add up to the net."

PICTURE. The two arrows drawn head-to-tail; their sum is the green net arrow $m\mathbf{a}$.

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Step 5 — Solve for what the spring reads

WHAT. We want the spring's arrow alone, because that is all the sensor knows. Move $m\mathbf{g}$ to the other side, then divide by $m$:

$m\mathbf{a}-m\mathbf{g}={\mathbf{F}}_{\text{c}}⟹\frac{{\mathbf{F}}_{\text{c}}}{m}=\mathbf{a}-\mathbf{g}$

Name the left side the specific force $\mathbf{f}$ (force per unit mass):

$\boxed{\mathbf{f}=\mathbf{a}-\mathbf{g}}$

Term by term:

• $\mathbf{f}$ — sensor reading (spring push per kilogram).
• $\mathbf{a}$ — the true acceleration we actually want for navigation.
• $-\mathbf{g}$ — the gravity arrow, subtracted. It vanishes from the reading precisely because it never touched the spring.

WHY this is the whole point. The sensor hands us $\mathbf{f}$. To get true acceleration we must undo the subtraction: $\mathbf{a}=\mathbf{f}+\mathbf{g}$. Someone (the navigation computer) has to remember gravity and add it back by hand.

PICTURE. Rearranged arrows: $\mathbf{f}$ shown as $\mathbf{a}$ with the gravity arrow removed.

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Step 6 — Case at rest: why it reads $+g$ upward

WHAT. The box sits on a table. The ball isn't moving, so the true acceleration is zero: $\mathbf{a}=0$. With $\mathbf{g}=(0,0,-g)$,

$\mathbf{f}=\mathbf{a}-\mathbf{g}=0-(0,0,-g)=(0,0,+g)$

WHY not zero? Even though nothing is moving, the spring still has to push up to stop the ball from sinking onto the box floor. That upward push is a real reading: $+g$ pointing at the sky.

PICTURE. Ball at rest; spring pushes up with length $g$, gravity pulls down with length $g$, they cancel so the ball is still — but the spring reading is the up-arrow.

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Step 7 — Case free fall: why it reads exactly zero

WHAT. Drop the whole box. Now the only force is gravity, so the ball accelerates at exactly $\mathbf{g}$: that is $\mathbf{a}=\mathbf{g}$. Then

$\mathbf{f}=\mathbf{a}-\mathbf{g}=\mathbf{g}-\mathbf{g}=0$

WHY zero. Ball, spring, and box all fall together at the same rate. The ball never drifts off-centre, so the spring never has to push. No push ⇒ zero reading. This is weightlessness, and it is the Equivalence Principle in sensor form: free-fall is indistinguishable from floating in deep space.

PICTURE. Everything falling at once; the spring is relaxed, the ball is centred, reading $=0$.

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Step 8 — Case accelerating: elevator going up

WHAT. The elevator (z-axis up) accelerates upward at $2{\text{m/s}}^2$, so $\mathbf{a}=(0,0,+2)$. Gravity is unchanged, $\mathbf{g}=(0,0,-9.81)$:

$f_z=a_z-g_z=2-(-9.81)=11.81{\text{m/s}}^2$

Term by term:

• $a_z=+2$ — the extra upward acceleration the elevator forces on the ball.
• $-g_z=+9.81$ — the support the spring already supplied at rest (Step 6).
• Sum $=11.81$ — the spring works harder: you "feel heavier."

WHY it adds. The spring must do two jobs now: hold the ball up against gravity ($9.81$) and shove it upward to match the elevator's acceleration ($2$). Two jobs ⇒ two contributions added.

PICTURE. A taller spring-arrow ($11.81$) versus the resting arrow ($9.81$); the difference bar is the felt "extra heaviness."

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The one-picture summary

All three cases in one strip: at rest the reading is $+g$, accelerating up it is $g+a$, in free fall it is $0$. Below each case the same equation $\mathbf{f}=\mathbf{a}-\mathbf{g}$ produces the number — the gravity arrow is always the same; only $\mathbf{a}$ changes.

Recall Feynman retelling of the whole walkthrough

Picture a ball hanging by a spring inside a shoebox. Gravity tugs the ball from the inside — it pulls on every bit of the ball at once, so the spring never feels it directly. The spring only wakes up when the ball tries to drift off-centre, and then it pushes back; that push is the only thing the box can report.

Sit the box on a table: gravity would drag the ball down, so the spring squeezes up to hold it. The box says "$+g$, upward!" — even though nothing is moving. Ride it up an elevator: now the spring has to hold the ball up and shove it along, so it pushes harder — you feel heavier, and the box reads $g$ plus the extra. Drop the whole box: ball, spring, and box all plunge together, the ball stays perfectly centred, the spring goes slack — the box says "$0$, I feel nothing," even while it's screaming toward the floor. That's why astronauts float.

The one lesson: the box can never feel gravity. So to get the true acceleration, the computer must always remember gravity and add it back: $\mathbf{a}=\mathbf{f}+\mathbf{g}$. Forget it, and every second the pretend-position runs away like $\frac{1}{2}g{t}^2$ — see Dead Reckoning and Error Drift.

Recall

A still accelerometer (z up) reads what? ::: $+g\approx 9.81{\text{m/s}}^2$ upward. A free-falling accelerometer reads what? ::: $0$ (weightlessness). Elevator accelerating up at $2{\text{m/s}}^2$ reads what on z? ::: $11.81{\text{m/s}}^2$. What must the computer do to $\mathbf{f}$ before integrating? ::: Add $\mathbf{g}(\mathbf{r})$: $\mathbf{a}=\mathbf{f}+\mathbf{g}$.

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Connections

• Newton's Second Law — Step 4's starting arrow-sum.
• Equivalence Principle — Step 7, free-fall reads zero.
• Strapdown Inertial Navigation System — where $\mathbf{a}=\mathbf{f}+\mathbf{g}$ is integrated.
• Gravity Model (WGS-84 / J2) — supplies the $\mathbf{g}$ we add back.
• Dead Reckoning and Error Drift — what happens if you forget it.
• Gyroscope and Attitude Determination — supplies the orientation to rotate $\mathbf{f}$ into the right frame.