Adding Vectors Geometrically — Question 10

Question 10

A stretched image can spoil a correct vector construction

An original calibrated drawing represents 8.0 m east followed by 6.0 m north at 4.0 m per paper centimetre. Thus the original leg arrows are 2.0 cm horizontally and 1.5 cm vertically. The resultant is the diagonal from the first tail to the final head.

A scanner doubles every horizontal distance but leaves every vertical distance unchanged. It copies the text “4.0 m per cm” unchanged. A student measures the resulting picture with a ruler and uses that old scale in every direction. The physical displacements themselves have not changed.

1. Find the correct physical resultant magnitude and direction, and the original paper diagonal length.

2. Find the scanned paper diagonal length and its angle above the horizontal. What magnitude and direction would the student’s method report? Compare with the physical answer.

3. Can one new scalar “metres per centimetre” calibration repair this distorted picture for every direction? Give a geometric reason using the two leg calibrations.

4. Describe two valid repairs: one that restores an undistorted drawing and one that reconstructs the physical triangle from the known leg data. Why would a uniformly enlarged image have been a different case?

Question 10 - Solution

A stretched image can spoil a correct vector construction

(a) Use the undistorted right triangle.

$$\boxed{R=10\,\mathrm m},\quad\theta=\tan^{-1}(6/8)=36.870^\circ,\quad\ell=\sqrt{2^2+1.5^2}=2.5\,\mathrm{cm}$$ The original paper triangle preserves the physical aspect ratio. Its diagonal is 2.5 cm, which corresponds to 10 m at 4 m/cm.

(b) Unequal stretching changes both the measured length and angle.

$$\ell_s=\sqrt{4^2+1.5^2}=\tfrac{\sqrt{73}}2\,\mathrm{cm},\quad R_s=2\sqrt{73}\,\mathrm m,\quad\theta_s=\tan^{-1}(1.5/4)=20.556^\circ$$ The old calibration reports approximately 17.09 m at $20.556^\circ$, both incorrect. It even exceeds the physical route length of 14 m.

(c) No single uniform scale can satisfy both axes.

$$k_x=8/4=2\,\mathrm{m/cm},\qquad k_y=6/1.5=4\,\mathrm{m/cm}\ne k_x$$ One scalar calibration cannot simultaneously satisfy these two different axis scales, nor can a length correction alone repair the wrong angle.

(d) Repair the geometry before using graphical lengths and angles.

$$(x_s,y_s)\mapsto(x_s/2,y_s),\qquad\text{uniform scale: }\ell'=q\ell,\ k'=k/q$$ Halve scanned horizontal distances to restore the original drawing, or redraw the known 8 m and 6 m legs at one common scale. Uniform enlargement preserves angles and only requires inverse adjustment of the scalar calibration.