3.5.13Guidance, Navigation & Control (GNC)
Inertial navigation — accelerometer measures non-gravitational specific force
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WHAT is being measured?
HOW: derive specific force from Newton's 2nd law
WHY derive? So the "subtract gravity" rule is forced on us, not memorized.
Consider the proof mass . Two categories of force act on it:
- Gravity (a body force acting on every atom).
- Contact / applied force (the spring / electrostatic pickoff that holds it in the case).
Newton's second law in an inertial frame, with the true acceleration of the mass:
Solve for the contact force per unit mass — that is exactly what the spring reads:
Why this step? (solve for $\mathbf{f}$) The spring only ever supplies ; gravity never touches the spring. So the readable quantity is .
The two "surprising" special cases (Forecast-then-Verify)
Forecast: What does an accelerometer read (a) sitting still on a table, (b) in free fall?
Case A — resting on a table. True acceleration . Gravity (down). It reads upward — the spring is pushing up to support the mass. Verified: a still accelerometer reads , not zero.
Case B — free fall. Only gravity acts, so . Reads zero — this is "weightlessness". Astronauts float because their accelerometer of a body feels , even while accelerating toward Earth. This is Einstein's equivalence principle in sensor form.
The full navigation chain (WHY this note matters)
Inertial navigation integrates sensor output twice to get position:
\;\xrightarrow{\int dt}\; \mathbf{v} \;\xrightarrow{\int dt}\; \mathbf{r}$$ - If you forget to add $\mathbf{g}$, you integrate a wrong acceleration → position error grows like ==$\tfrac12 g\,t^2$== (drift explodes in seconds). - Because $\mathbf{g}=\mathbf{g}(\mathbf{r})$ depends on position, and you need position to know $\mathbf{g}$... this couples the equations — a real navigation loop. ![[3.5.13-Inertial-navigation-—-accelerometer-measures-non-gravitational-specific-force.png]] > [!example] Worked: elevator accelerating upward > An elevator accelerates upward at $2\,\text{m/s}^2$. What does a floor-mounted accelerometer (z-axis up) read? > > - **Step 1:** true acceleration $\mathbf{a}=(0,0,+2)$. *Why?* Elevator physically accelerates up. > - **Step 2:** gravity $\mathbf{g}=(0,0,-9.81)$. *Why?* Always points down regardless of motion. > - **Step 3:** $f_z = a_z - g_z = 2 - (-9.81) = 11.81\,\text{m/s}^2$. *Why?* Spring must both support weight ($9.81$) and accelerate the mass ($2$). > - **Interpretation:** you "feel heavier" — exactly the extra $2\,\text{m/s}^2$ the reading shows. > [!example] Worked: aircraft in level cruise, banked turn > A plane flies a level, coordinated turn (constant altitude). Vertical accel $a_z=0$, but centripetal accel horizontal $a_x = v^2/R$. > - $f_z = 0 - (-g) = +g$ (still supporting weight). > - $f_x = a_x - 0 = v^2/R$. > - Magnitude $|\mathbf{f}| = \sqrt{g^2 + (v^2/R)^2} > g$: this is the **load factor / g-force** pilots feel. *Why?* The seat (contact force) must both hold the pilot up and curve their path. --- ## Common mistakes (Steel-man + fix) > [!mistake] "The accelerometer measures acceleration, so a still device reads 0." > **Why it feels right:** the name says *accelero*meter, and intuitively "not moving = no acceleration = zero." **The flaw:** it measures specific force $\mathbf{f}=\mathbf{a}-\mathbf{g}$, not $\mathbf{a}$. At rest $\mathbf{a}=0$ but $\mathbf{g}\neq0$, so it reads $+g$ upward. > **Fix:** always ask "what force is the spring supplying?" That is the reading. > [!mistake] "In free fall the accelerometer reads $g$ downward." > **Why it feels right:** the object is accelerating downward at $g$, so surely the sensor shows $g$. **The flaw:** in free fall the *only* force is gravity, and gravity is invisible to the spring — no contact force → reads **0**. > **Fix:** free fall ⇒ $\mathbf{a}=\mathbf{g}$ ⇒ $\mathbf{f}=0$. > [!mistake] "Just integrate the raw sensor output twice to get position." > **Why it feels right:** position is the double integral of acceleration. **The flaw:** raw output is $\mathbf{f}$, not $\mathbf{a}$. You must add $\mathbf{g}(\mathbf{r})$ first. > **Fix:** $\mathbf{a}=\mathbf{f}+\mathbf{g}$, *then* integrate. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine you're holding a ball on a spring inside a shoebox. When you sit still, the spring has to squeeze up a little to hold the ball against gravity — the spring is "working," so the box says "I feel something!" But if you drop the whole box, the ball, the spring, and the box all fall together, so the spring doesn't have to push at all — the box says "I feel *nothing*," even though it's zooming toward the ground. That's why astronauts float: everything falls together, so their inner "spring" feels zero. The box can never feel gravity directly — someone (the computer) has to *remember* gravity is there and add it back. > [!mnemonic] > **"A = F + G"** → *"**A**ll **F**light needs **G**ravity added back."* > And for the sign: **"At rest, points to the sky"** (a still accelerometer reads $+g$ *upward*, opposite to gravity). --- ## Active recall #flashcards/physics What physical quantity does an accelerometer actually measure? ::: Specific force $\mathbf{f}$ — the non-gravitational (contact) force per unit mass, not acceleration. Write the fundamental accelerometer equation. ::: $\mathbf{f} = \mathbf{a} - \mathbf{g}$, hence $\mathbf{a} = \mathbf{f} + \mathbf{g}$. Why is gravity invisible to an accelerometer? ::: Gravity acts equally on the proof mass and the case, so the spring (contact force) never has to counter it. Only contact forces are read. What does a stationary accelerometer (z up) read? ::: $+g \approx 9.81\,\text{m/s}^2$ upward (spring supports the weight; $\mathbf{a}=0$, $\mathbf{f}=-\mathbf{g}$). What does an accelerometer read in free fall? ::: Zero — the only force is gravity, which the spring can't feel. This is weightlessness / the equivalence principle. In inertial navigation, what must you do to sensor output before integrating? ::: Add the modeled gravity vector $\mathbf{g}(\mathbf{r})$: $\mathbf{a}=\mathbf{f}+\mathbf{g}$. An elevator accelerates up at $2\,\text{m/s}^2$; what does its z-accelerometer read? ::: $f_z = 2-(-9.81)=11.81\,\text{m/s}^2$. What happens to position error if you forget to add gravity? ::: It grows like $\tfrac12 g t^2$ — drift explodes within seconds. --- ## Connections - [[Equivalence Principle]] — free-fall reading zero is gravity ≡ acceleration. - [[Newton's Second Law]] — the derivation's starting point. - [[Strapdown Inertial Navigation System]] — where $\mathbf{a}=\mathbf{f}+\mathbf{g}$ is integrated. - [[Gyroscope and Attitude Determination]] — needed to rotate $\mathbf{f}$ into the nav frame before adding $\mathbf{g}$. - [[Gravity Model (WGS-84 / J2)]] — supplies $\mathbf{g}(\mathbf{r})$. - [[Dead Reckoning and Error Drift]] — consequence of integration. ## 🖼️ Concept Map ```mermaid flowchart TD ACC[Accelerometer: mass on spring] PM[Proof mass] SF[Specific force f] CONTACT[Contact force Fc: spring] GRAV[Gravity mg] N2[Newton 2nd law: ma = Fc + mg] EQ[f = a - g] NAV[Nav computer adds g] A[True acceleration a] CaseA[Case A: at rest reads +g] CaseB[Case B: free fall reads 0] ACC -->|contains| PM PM -->|displacement gives| SF CONTACT -->|per unit mass equals| SF GRAV -->|invisible to| SF N2 -->|solve for f| EQ EQ -->|so a = f + g| NAV NAV -->|recovers| A A -->|integrated for| NAV EQ -->|predicts| CaseA EQ -->|predicts| CaseB ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, accelerometer ka naam sunke lagta hai ki ye "acceleration" measure karta hai — par asal me ye **specific force** measure karta hai, yaani ek chhoti si mass (proof mass) ko spring kitni force se pakad ke rakh raha hai, gravity ko chhod ke. Iska formula simple hai: $\mathbf{f} = \mathbf{a} - \mathbf{g}$. Yahan sabse bada twist ye hai ki **gravity accelerometer ko dikhti hi nahi**, kyunki gravity proof mass aur box dono ko barabar kheenchti hai, to spring ko gravity se ladna hi nahi padta. > > Isliye do mazedaar cases yaad rakho. Agar accelerometer table pe rakha hai aur bilkul still hai, to $\mathbf{a}=0$, par reading zero nahi aayegi — reading aayegi $+g$ (upar ki taraf), kyunki spring mass ko upar support kar raha hai. Aur agar aap accelerometer ko free fall me chhod do, to reading **zero** ho jaayegi — yahi weightlessness hai, isiliye astronauts float karte hain. Ye Einstein ka equivalence principle hai, sensor ki bhasha me. > > Navigation ke liye ye baat critical hai: computer ko sensor se $\mathbf{f}$ milta hai, aur true acceleration nikalne ke liye usko **manually gravity add karni padti hai** — $\mathbf{a} = \mathbf{f} + \mathbf{g}$. Fir usko do baar integrate karke velocity aur position nikalte hain. Agar aap galti se gravity add karna bhool gaye, to error $\tfrac12 g t^2$ ke rate se badhta hai — matlab kuch hi seconds me position bilkul galat ho jaayegi. Isiliye INS (inertial navigation system) ke andar hamesha ek accurate gravity model (jaise WGS-84) chalta hai. ![[audio/3.5.13-Inertial-navigation-—-accelerometer-measures-non-gravitational-specific-force.mp3]]