Chapter 4.3, Problem 91E

To determine

To calculate:The difficulty in acquiring a target according to fits law, the difficulty ID

  $ID=\frac{\log \left(\frac{2A}{W}\right)}{\log 2}$

Answer to Problem 91E

The clicking 5mm wide target is 1.231 times difficult than wide target .

Explanation of Solution

Given information:

The difficulty in acquiring a target according to fits law, the difficulty ID

  $ID=\frac{\log \left(\frac{2A}{W}\right)}{\log 2}$

Formula used:

First we find the ID with 5mm and then of 10mm after get these two values divide the term 5mm to 10mm.

Calculation:

Consider the ID,

  $ID=\frac{\log \left(\frac{2A}{W}\right)}{\log 2}$

Rewrite the expression as:

  $\begin{array}{l}ID=\frac{\log \left(\frac{2A}{W}\right)}{\log 2}\\ I{D}_{5mm}=\frac{\log \left(2\times \frac{100}{5}\right)}{\log 2}\\ I{D}_{10mm}=\frac{\log \left(2\times \frac{100}{10}\right)}{\log 2}\end{array}$

Divide both the statements :

  $\begin{array}{c}\frac{I{D}_{5mm}}{I{D}_{10mm}}=\frac{\frac{\log \left(2\times \frac{100}{5}\right)}{\log 2}}{\frac{\log \left(2\times \frac{100}{10}\right)}{\log 2}}\\ =\frac{\log \left(2\times \frac{100}{5}\right)}{\log \left(2\times \frac{100}{10}\right)}\\ =\frac{\log 40}{\log 20}\\ =1.231\end{array}$

Thus,The clicking 5mm wide target is 1.231 times difficult than wide target .