Gyroscope — angular velocity measurement, bias, noise: Worked examples

Before anything else, a quick unit reminder so no symbol is unearned.

Definition The symbols we use, in plain words

$ω_{true}$ — how fast the body actually rotates, in $°/ s$ (degrees per second).
$\tilde{ω}$ — the raw gyro reading (the "tilde" mark, $\sim$ , just means "measured, not true"). From the parent's model, $\tilde{ω} = ω_{true} + b + n$ : the true rate plus bias plus noise.
$b$ — bias: the fake rotation the gyro reports when perfectly still, also $°/ s$ .
$n$ — the fast white-noise jitter on each reading, $°/ s$ .
$ARW$ — Angle Random Walk, quoted in $°/ h$ (degrees per square-root-hour). It is the coefficient in front of $t$ .
$Q$ — the noise power spectral density, the "strength" of the white noise. The parent showed the integrated noise has variance $Q t$ , and $ARW = Q$ — so $Q$ and $ARW$ are two names for the same noise level (one squared, one not).
$\hat{θ}$ — our estimated angle after integrating the reading.
$Δ θ$ — the error in that estimate.
$t$ — elapsed integration time.

The scenario matrix

Every gyro-error question is one of these cells. The examples below are tagged with the cell they cover, and the map figure below indexes each cell by letter on the time axis.

#	Case class	What changes	Covered by
A	Positive bias, integrate	$b > 0$ , ramp	Ex 1
B	Negative bias, integrate	$b < 0$ , sign of error	Ex 2
C	Zero bias, pure noise	$b = 0$ , only ARW	Ex 3
D	Small- $t$ limit	$t \to 0$ : which term wins?	Ex 4
E	Large- $t$ / limiting	$t \to \infty$ : which term wins?	Ex 4
F	Exact crossover	bias line = noise line	Ex 5
G	Nonzero true rate + bias	$ω_{true} \neq = 0$ , separate signal from error	Ex 6
H	Full realistic case	true rate + bias + noise together	Ex 7
I	Real-world word problem	drone, unit soup, decision	Ex 8
J	Exam-style twist	Allan slope / averaging trap	Ex 9

Figure — Gyroscope — angular velocity measurement, bias, noise

The figure above is the map for the whole page: the horizontal axis is elapsed time $t$ in seconds and the vertical axis is angle error in degrees. The straight red line is bias error ( $\propto t$ ), the curved blue line is noise error ( $\propto t$ ). They cross once, at the yellow dot. The little labelled markers D, F, E on the time axis are literally the matrix cells above: D at $t = 1$ s (small- $t$ ), F at the crossover $t^{⋆}$ , E far to the right (large- $t$ ). Cells A/B/C sit on the two curves themselves (pure ramp / pure wander), and G–J are variations built on top. Left of the cross, noise looks bigger; right of the cross, bias wins forever. Keep this picture in your head.

Worked Examples

Cell A — Positive bias integrates to a ramp

Worked example Example 1 — a steady tilt

A stationary gyro has bias $b = + 0.03 °/ s$ and negligible noise. You integrate the raw reading for $t = 90 s$ . What is the heading error?

Forecast: Guess — bigger than $1°$ or smaller? Same sign as $b$ or opposite?

Write the model at rest. With $ω_{true} = 0$ , the reading is $\tilde{ω} = b$ . Why this step? At rest the only thing the gyro reports is its own bias — that is the definition of bias.
Integrate. $\hat{θ} = \int_{0}^{t} b d τ = b t$ . Why this step? A constant integrated over time is just (constant × time) — the area of a rectangle of height $b$ and width $t$ .
Plug numbers (one-unit rule: bias uses seconds). $Δ θ = 0.03 \times 90 = 2.7°$ . Why this step? Both numbers are already in $°/ s$ and $s$ , so $°/ s \times s = °$ — units cancel cleanly.

Verify: Height $0.03$ , width $90$ → rectangle area $2.7$ . Sign is $+$ (drifts the same way as the bias). Units: $\frac{°}{s} \cdot s = °$ . ✓

Cell B — Negative bias: the error flips sign

Cell C — Zero bias, only noise (Angle Random Walk)

Worked example Example 3 — pure jitter

A very well-calibrated gyro has $b \approx 0$ but $ARW = 0.4 °/ h$ . Expected angle uncertainty after $t = 1 h$ ? And after $9 h$ ?

Forecast: Going from $1$ h to $9$ h — does the error grow $9 \times$ ? $3 \times$ ? Something else?

Use the noise growth law (one-unit rule: noise uses hours — and here $t$ is already in hours). $σ_{θ} = ARW \cdot t_{h}$ . Why this step? With no bias, the only error is integrated white noise. The parent showed integrated white noise of density $Q$ has variance $Q t$ , so its standard deviation is $Q t = ARW t$ — a $t$ spread.
At $t = 1$ h. $σ_{θ} = 0.4 \times 1 = 0.4°$ . Why this step? $ARW$ is defined per $hour$ , so plugging $t$ in hours needs no conversion.
At $t = 9$ h. $σ_{θ} = 0.4 \times 9 = 0.4 \times 3 = 1.2°$ . Why this step? $9 = 3$ , not $9$ . The random increments partly cancel, so a $9 \times$ longer run gives only a $3 \times$ bigger spread.

Verify: Ratio $\frac{1.2}{0.4} = 3 = 9$ . ✓ This is the signature of a random walk — time up by $k^{2}$ , error up by $k$ .

Cells D & E — the small- $t$ and large- $t$ limits

Worked example Example 4 — who wins when

$t$ is tiny vs huge? A gyro has $b = 0.01 °/ s$ and $ARW = 0.5 °/ h$ . Compare bias error and noise error at (D) $t = 1 s$ and (E) $t = 3600 s$ (1 h).

Forecast: At $1$ second, which term is bigger — the ramp or the wander? Does that stay true at 1 hour?

Apply the one-unit rule up front. Bias error $= b t_{s}$ (seconds). Noise error $= ARW t_{h}$ with $t_{h} = t_{s} /3600$ (hours). Convert both target times once: $t = 1$ s $\Rightarrow t_{h} = 1/3600$ ; $t = 3600$ s $\Rightarrow t_{h} = 1$ . Why this step? You cannot compare two numbers whose $t$ 's are in different units — convert once, carry both.
Cell D, $t = 1$ s.
- Bias: $0.01 \times 1 = 0.01°$ .
- Noise: $0.5 \times 1/3600 = 0.5 \times \frac{1}{60} = \frac{1}{120} \approx 0.00833°$ . Why this step? Both are now in degrees, so we can compare directly: bias $0.01°$ is slightly larger than noise $0.00833°$ . So even at $1$ s the bias is already (barely) ahead — the winner near $t = 0$ is not automatic; it depends on the coefficients, which is exactly why we compute the crossover in Ex 5.
Cell E, $t = 3600$ s (1 h).
- Bias: $0.01 \times 3600 = 36°$ .
- Noise: $0.5 \times 1 = 0.5°$ . Why this step? Over a long time the ramp $t$ crushes the $t$ wander — bias becomes $72 \times$ bigger.

Verify: D: $0.01 > 0.00833$ , bias barely wins. E: $36/0.5 = 72$ , bias dominates hugely. Units are degrees throughout. ✓ The general limit law: as $t \to \infty$ , bias always wins because $t$ beats $t$ ; near $t \to 0$ the winner depends on the actual coefficients (see Allan Variance Analysis).

Cell F — the exact crossover time

Worked example Example 5 — where the two lines cross

Same gyro as Ex 4: $b = 0.01 °/ s$ , $ARW = 0.5 °/ h$ . At what elapsed time $t^{⋆}$ do bias error and noise error become equal?

Forecast: From Ex 4, bias already led at $1$ s. So do you expect $t^{⋆}$ to be below $1$ s?

Set the two errors equal (work in seconds, convert the noise time inline). $b t = ARW t /3600$ (both sides in degrees, $t$ in seconds). Why this step? "Crossover" literally means the two error curves have the same height — that is one equation in one unknown $t$ .
Square both sides — and note what squaring can introduce. $b^{2} t^{2} = ARW^{2} \frac{t}{3600}$ . Why this step? Squaring removes the awkward square root. Caution: squaring can create extraneous roots (solutions that satisfy the squared equation but not the original). So we must, at the end, keep only roots with $t > 0$ — a negative $t$ is physically meaningless (time cannot run backwards) and would make the original right-hand side $t /3600$ undefined. We will check this.
Cancel one $t$ and identify all roots. Rewrite as $b^{2} t^{2} - \frac{ARW ^{2}}{3600} t = 0$ , i.e. $t (b^{2} t - \frac{ARW ^{2}}{3600}) = 0$ . The two roots are $t = 0$ and $t^{⋆} = \frac{ARW ^{2}}{3600 b ^{2}}$ . Why this step? Factoring shows every solution explicitly. $t = 0$ is the trivial crossing (both errors are $0$ at the start, not a meaningful crossover); there is no negative root here, so nothing extraneous survives — but we checked, as the rule demands.
Plug numbers into the meaningful root. $t^{⋆} = \frac{0. 5 ^{2}}{3600 \times 0.0 1 ^{2}} = \frac{0.25}{0.36} \approx 0.694 s$ . Why this step? $3600 \times 0.0001 = 0.36$ ; then $0.25/0.36$ .

Verify: Both roots are $\geq 0$ , so no extraneous solution to discard. $t^{⋆} \approx 0.694 s < 1 s$ , consistent with Ex 4 where bias already led at $1$ s. Below $t^{⋆}$ noise leads, above it bias leads — matches the yellow dot in the scenario-map figure. ✓

Cell G — real rotation plus bias: separating signal from corruption

Worked example Example 6 — a genuine turn, contaminated by bias

A turntable truly rotates at $ω_{true} = + 5 °/ s$ for $t = 20 s$ . The gyro has bias $b = + 0.1 °/ s$ and negligible noise. What angle does the gyro report, and what is the true angle?

Forecast: Will the reported angle be bigger or smaller than the true one, given $b > 0$ ?

True angle. $θ_{true} = ω_{true} t = 5 \times 20 = 100°$ . Why this step? With constant true rate, orientation is just rate × time (all in seconds, so units give degrees).
Reported angle. The reading is $\tilde{ω} = ω_{true} + b$ , so $\hat{θ} = (ω_{true} + b) t = (5 + 0.1) \times 20 = 102°$ . Why this step? Bias adds on top of the true signal; integration passes both through.
The error. $Δ θ = \hat{θ} - θ_{true} = 102 - 100 = 2° = b t$ . Why this step? The true part subtracts out exactly, leaving only the bias ramp — confirming error depends on $b$ , not on how fast you really turned.

Verify: $b t = 0.1 \times 20 = 2°$ matches $102 - 100$ . ✓ Key insight for cell G: the error is independent of the true rate; a fast turn and a slow turn accumulate the same bias error over the same time.

Cell H — the full realistic case: true rate + bias + noise

Worked example Example 7 — everything at once

Now use the complete model $\tilde{ω} = ω_{true} + b + n$ . A turntable truly rotates at $ω_{true} = + 5 °/ s$ for $t = 20 s$ , with bias $b = + 0.1 °/ s$ AND noise $ARW = 0.6 °/ h$ . Give the true angle, the expected reported angle, the deterministic error, and the $1 σ$ noise spread on top.

Forecast: Does the noise change the expected reported angle, or only its spread?

Integrate each term separately (integration is linear). $\hat{θ} = \int_{0}^{t} (ω_{true} + b + n) d τ = signal ω_{true} t + bias ramp b t + random walk \int_{0}^{t} n d τ$ . Why this step? The three effects don't interfere; splitting the integral lets us handle each with the law we already know.
Signal (seconds). $ω_{true} t = 5 \times 20 = 100°$ . Why this step? This is the answer we actually want — the true orientation.
Bias ramp / expected error (seconds). $b t = 0.1 \times 20 = 2°$ . Why this step? Same as Cell G — bias adds a deterministic $+ 2°$ , so the expected reading is $100 + 2 = 102°$ .
Noise spread $1 σ$ (one-unit rule: convert $20$ s $= 20/3600$ h). $σ_{θ} = ARW t_{h} = 0.6 20/3600 \approx 0.0447°$ . Why this step? $E [\int n d τ] = 0$ , so noise contributes no shift to the mean — only a $\pm$ wobble of size $σ_{θ}$ .

Verify: Expected reading $= 100 + 2 = 102°$ (noise mean is $0$ , so it does not move this). Noise scatter $\approx 0.0447°$ is tiny next to the $2°$ bias — confirming again that over $20$ s bias dominates noise. Units: signal, ramp, and spread are all in degrees. ✓ This is the complete realistic picture: a wanted signal $+$ a growing bias ramp $+$ a small random wobble.

Cell I — full real-world word problem

Worked example Example 8 — the GPS dropout

A delivery drone loses GPS and flies on gyro-only heading for $3 min$ . Its gyro: bias $b = 0.02 °/ s$ , $ARW = 0.6 °/ h$ . Mission rule: abort if the heading error can plausibly exceed $5°$ . Does it abort?

Forecast: $3$ minutes feels short. Bias-only, does it stay under $5°$ ?

Convert time once (one-unit rule). $t = 3 min = 180 s$ ; for the noise term $t_{h} = 180/3600 = 0.05 h$ . Carry both from here on. Why this step? Bias uses seconds, ARW uses hours — prepare both up front so no step mixes them.
Bias error (deterministic). $Δ θ_{bias} = b t_{s} = 0.02 \times 180 = 3.6°$ . Why this step? Constant bias → ramp, exactly Ex 1's logic. This is a fixed offset, present with certainty.
Noise spread $1 σ$ (random). $σ_{θ} = ARW t_{h} = 0.6 \times 0.05 = 0.6 \times 0.2236 \approx 0.134°$ . Why this step? Noise is the $t$ contribution and is random, so it is a standard deviation, not a fixed number. One $1 σ$ of noise is only $0.134°$ .
State the risk policy explicitly, then apply it. The heading error is a fixed bias ramp of $3.6°$ plus a random noise term of mean $0$ and standard deviation $0.134°$ . We adopt a conservative $3 σ$ worst-case bound (covers $\approx 99.7%$ of outcomes — standard for a safety abort decision): $Δ θ_{max} = certain b t_{s} + worst-case noise 3 σ_{θ} = 3.6 + 3 \times 0.134 = 3.6 + 0.402 \approx 4.00°.$ Why this step? The abort rule says "can plausibly exceed $5°$ " — "can plausibly" means we must bound the worst realistic case, not the average. Adding $3 σ$ of noise onto the certain bias ramp is that bound. (If the policy had said "on average", we would use the mean $3.6°$ ; if "always/absolute worst", there is no hard cap on a Gaussian, so $3 σ$ is the accepted engineering convention.)

Verify: Worst-case $\approx 4.00° < 5°$ → do NOT abort, but with only a $1.0°$ margin, and it is almost entirely bias. If bias were even $0.028°/ s$ the $3 σ$ bound would breach $5°$ . Fix in the real system: fuse an accelerometer/magnetometer to reset heading (an Inertial Measurement Unit (IMU) does this via a Kalman Filter). Units all resolve to degrees. ✓

Cell J — exam-style twist (the Allan-slope trap)

Worked example Example 9 — "averaging longer must help, right?"

An exam states: "A gyro's Allan deviation is $0.3 °/ h$ at averaging time $τ = 10 s$ , sitting on the $- 1/2$ (ARW) slope. Predict $σ_{A}$ at $τ = 40 s$ . Then explain why the same prediction fails at $τ = 4000 s$ ."

Forecast: On a $- 1/2$ slope, quadrupling $τ$ changes $σ_{A}$ by what factor?

Use the $- 1/2$ slope law. On a log-log plot, slope $- 1/2$ means $σ_{A} \propto τ^{- 1/2}$ . Why this step? In the ARW region, longer averaging reduces error like $1/ τ$ — that is what the negative slope encodes.
Scale from the known point. $\frac{σ _{A} ( 40 )}{σ _{A} ( 10 )} = (\frac{40}{10})^{- 1/2} = 4^{- 1/2} = \frac{1}{2}$ . Why this step? Ratios on a power law only need the ratio of $τ$ 's — the $τ$ units cancel, so no seconds-to-hours conversion is needed here.
Compute. $σ_{A} (40) = 0.3 \times \frac{1}{2} = 0.15 °/ h$ . Why this step? Half the error, as promised by the slope.
The trap at $τ = 4000$ s. Between the $- 1/2$ region and long $τ$ , the curve first flattens to a slope- $0$ floor — that flat minimum is the bias instability you can never average away. Only past the flat region does it turn UP to slope $+ 1/2$ , which is rate random walk (see Random Walk & Wiener Process). Either way, applying the $- 1/2$ law would predict $0.3 \times (4000/10)^{- 1/2} = 0.3/20 = 0.015 °/ h$ , but the real value is larger, not smaller. Why this step? Once you leave the ARW region, averaging longer stops helping (flat floor) and eventually HURTS (rate random walk, slope $+ 1/2$ ) — extrapolating the $- 1/2$ line is the classic mistake.

Verify: $σ_{A} (40) = 0.15 °/ h$ , halved as expected. The naive extrapolation gives $0.015 °/ h$ , which the problem tells us is wrong because the process changed regimes. ✓

Recall Quick self-test (cover the answers)

Bias $0.03°/ s$ for $90$ s gives what error? ::: $2.7°$ (Cell A, ramp $b t$ ). Same but bias is $- 0.03°/ s$ ? ::: $- 2.7°$ — same size, opposite direction (Cell B). ARW $0.4°/ h$ : error at $9$ h vs $1$ h grows by what factor? ::: $9 = 3 \times$ , so $1.2°$ (Cell C). As $t \to \infty$ , which error always wins, bias or noise? ::: Bias, because $t$ beats $t$ (Cell E). In the full model, does noise shift the expected reported angle? ::: No — noise is zero-mean; only bias shifts the mean, noise adds spread (Cell H). In the drone abort problem, what bound did we use for the random noise? ::: A $3 σ$ worst-case bound added onto the certain bias ramp (Cell I). On an Allan $- 1/2$ slope, quadrupling $τ$ does what to $σ_{A}$ ? ::: Halves it ( $4^{- 1/2}$ ) (Cell J). What does the FLAT (slope-0) region of an Allan curve represent? ::: Bias instability — the floor you cannot average away (Cell J).

Related building blocks: Random Walk & Wiener Process (where $t$ comes from), Coriolis Force and Sagnac Effect (how $ω$ is sensed), Attitude Estimation / Dead Reckoning (where these errors bite).