Robot Control Resolution: 25-Comp-B8 Computer Integrated Manufacturing, May 2015, Q2

Question 2: Robot Control Resolution (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Part (a)	Part (b)
Axis range $R$	0.7 m	1.2 m
Storage capacity $n$	12 bits	to be determined
Required control resolution $CR$	to be determined	$\le 0.5\text{ mm}$

Find. (a) The control resolution of the 12-bit telescoping-arm axis. (b) The minimum bit storage capacity that gives a control resolution no coarser than $0.5\text{ mm}$ on the $1.2\text{ m}$ slide.

Approach. An $n$-bit binary register addresses $2^n$ distinct storage states, which divide the axis range into $2^n-1$ equal control increments; control resolution is the axis range divided by that number of increments, $CR = R/(2^n-1)$. Part (a) evaluates this directly; part (b) inverts it, solving for the smallest integer $n$ that makes $CR \le 0.5\text{ mm}$.

(a) State the control-resolution relation and substitute. With $n=12$ bits, the register has $2^{12}=4096$ addressable states, giving $2^{12}-1 = 4095$ equal increments across the $0.7\text{ m}$ range: $$CR = \frac{R}{2^{n}-1} = \frac{0.7\text{ m}}{4095} = \boxed{1.709\times10^{-4}\text{ m} = 0.1709\text{ mm}}$$ This is the smallest increment of arm extension the controller can distinguish and command.
(b) Convert the target resolution to a required number of increments. The controller needs $2^{n}-1$ increments no coarser than $CR = 0.5\text{ mm} = 0.0005\text{ m}$ across the $1.2\text{ m}$ range, so $$2^{n}-1 \ \ge\ \frac{R}{CR} = \frac{1.2}{0.0005} = 2400$$
Find the smallest integer $n$ satisfying the inequality. Testing successive bit counts: $2^{11}-1 = 2047$ (insufficient, $2047 < 2400$); $2^{12}-1 = 4095$ (sufficient, $4095 \ge 2400$). Therefore $$n_{\min} = \boxed{12\text{ bits}}$$ which gives an actual control resolution of $1.2/4095 = 0.293\text{ mm}$, comfortably finer than the required $0.5\text{ mm}$ (an 11-bit register would only achieve $1.2/2047 = 0.586\text{ mm}$, which fails the specification).

Quantity	Value
(a) Control resolution, 12-bit / 0.7 m axis	0.1709 mm
(b) Minimum storage capacity for ≤0.5 mm resolution on 1.2 m slide	12 bits (actual CR = 0.293 mm)

Topic: Robot control resolution and control memory sizing.

Key relations: $CR = R/(2^{n}-1)$, where $R$ is the total axis range and $n$ the number of storage bits; equivalently the number of controllable increments is $2^{n}-1$ (the $2^n$ addressable states include the zero position, which is not itself an increment).

Why this works: a digital control memory can only represent a finite, discrete set of positions along any axis; the finer the desired positioning increment relative to the total travel, the more distinct states — and hence more storage bits — the controller must be able to address. Control resolution is one of two components of a robot's overall spatial resolution (the other being mechanical inaccuracies such as backlash and gear/linkage compliance); accuracy is generally taken as roughly one-half the control resolution, while repeatability is about $\pm$ this half-resolution around the commanded point.

Common pitfall: dividing the range by $2^{n}$ instead of $2^{n}-1$ (off-by-one from forgetting that $2^n$ states bound only $2^n-1$ intervals), and rounding a required bit count DOWN instead of up — the number of bits must always be rounded up to the next integer that satisfies the inequality, since a fractional bit cannot be built.

Source: Groover, Automation, Production Systems, and Computer-Integrated Manufacturing, 4th ed., Ch.8 (Industrial Robotics) § Robot Accuracy and Repeatability.

Question 2 of 6: Robot Control Resolution

Question 2: Robot Control Resolution (20 marks)