Adding Vectors by Components — Question 7

Question 7

Near cancellation exposes rounding errors

Two vectors are $\vec A=(1000,1)$ m and $\vec B=(-999,1)$ m. Treat these as exact components. A display rounds each component to the nearest 10 m before addition.

1. Find the exact resultant, magnitude and direction.

2. Find the result obtained from the rounded component display. Why is it misleading?

3. If each original component instead has an independent hard error bound of $\pm0.1$ m, bound each resultant component and decide whether the true resultant can vanish.

Question 7 - Solution

Near cancellation exposes rounding errors

(a) Preserve the small residual.

$$\boxed{\vec R=(1000-999,1+1)=(1,2)\,\mathrm m},\quad R=\sqrt5\,\mathrm m,\quad\theta=63.435^\circ$$ $\vec R=(1,2)$ m, magnitude $\sqrt5\approx2.236$ m, direction $\tan^{-1}(2)\approx63.435^\circ$ from east. The large opposite horizontal terms nearly cancel but do not cancel exactly.

(b) Rounding can erase the answer.

$$\vec R_{\mathrm{rounded}}=(1000,0)+(-1000,0)=\vec0\ne\vec R$$ The display shows $(1000,0)$ m and $(-1000,0)$ m, whose sum is zero. That eliminates both the residual length and its direction. Small relative errors in large input vectors can produce large relative errors in their nearly canceled sum. Keep guard digits until after addition.

(c) Propagate hard bounds by interval addition.

$$|\delta R_i|\le|\delta A_i|+|\delta B_i|\le.2\,\mathrm m\Rightarrow R_x\in[.8,1.2],\ R_y\in[1.8,2.2]\quad\mathrm m$$ $R_x\in[0.8,1.2]$ m and $R_y\in[1.8,2.2]$ m because each sum can err by at most 0.2 m. Neither interval contains zero, so the resultant cannot vanish and stays in quadrant I. This bound does not use a root-sum-square statistical assumption.