Graphical Integration in Motion Analysis — Question 8

Question 8

A constant accelerometer bias accumulates into position error

An accelerometer reports $a_m=+2.2\ \mathrm{m/s^2}$ during 0–2 s and $a_m=-1.8\ \mathrm{m/s^2}$ during 2–4 s. Its only error is an unknown constant bias $b$, defined by $a_m=a+b$. Independent measurements give $v(0)=v(4)=1.0\ \mathrm{m/s}$ on the same east-positive axis. Position starts at $x(0)=0$; all values are exact model data.

1. Use the total reported acceleration area and the independent velocity change to determine $b$. Write the corrected acceleration history.

2. Reconstruct the corrected velocity graph and calculate $x(4)$ and distance traveled.

3. A logger integrates the raw readings using the correct initial velocity and position. Find its final velocity and position errors, and their signs. Explain why correcting only its final velocity does not correct its already accumulated position.

4. For an arbitrary duration $T$ under this constant-bias model, derive the velocity and position errors. Would the endpoint-velocity condition alone identify a time-varying bias function?

Question 8 - Solution

A constant accelerometer bias accumulates into position error

(a) The independent endpoint velocities constrain one total area.

The reported area is $2(2.2)+2(-1.8)=0.8\ \mathrm{m/s}$, while the true change is zero. Hence $$0=0.8\ \mathrm{m/s}-b(4\ \mathrm{s}),\qquad\boxed{b=+0.20\ \mathrm{m/s^2}}.$$ The true accelerations are $\boxed{+2.0\ \mathrm{m/s^2}}$ and $\boxed{-2.0\ \mathrm{m/s^2}}$. A positive bias means the reported value exceeds the true value in both stages.

(b) Integrate the corrected velocity graph.

Velocity rises linearly from 1 to 5 m/s and then falls to 1 m/s. Each 2 s trapezoid has area $2(1+5)/2=6\ \mathrm{m}$. Thus $$\boxed{x(4)=12\ \mathrm{m}},\qquad \boxed{D=12\ \mathrm{m}}.$$ Velocity stays positive, so unsigned and signed areas agree.

(c) Raw integration creates a growing velocity error.

The logger obtains $v_m(2)=5.4$ and $v_m(4)=1.8\ \mathrm{m/s}$. Its position estimate is $$x_m(4)=2(1+5.4)/2+2(5.4+1.8)/2=13.6\ \mathrm{m}.$$ The final errors (reported minus true) are $\boxed{+0.8\ \mathrm{m/s}}$ and $\boxed{+1.6\ \mathrm{m}}$. The position error is the area under the velocity error over the whole run. Changing only a single endpoint value does not remove that area; the history or its integrated bias correction must be corrected.

(d) Integrate the error, with zero initial errors.

$$e_v(T)=\int_0^T b\,dt=bT,\qquad e_x(T)=\int_0^Tbt\,dt=\frac{bT^2}{2}.$$

A constant acceleration error $b$ gives $$\boxed{e_v(T)=bT},\qquad\boxed{e_x(T)=\tfrac12bT^2}.$$ These formulas assume correct initial velocity and position and a fixed bias. With a time-varying bias, endpoint velocity determines only $\int_0^T b(t)\,dt$. Many bias histories share that area but have different weighted areas $\int_0^T(T-t)b(t)\,dt$, so the position correction would not be uniquely identified.