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Difficulty of a Task The difficulty in "acquiring a target"(such as using your mouse to click on an icon on yourcomputer screen) depends on the distance to the target andthe size of the target. According to Fitts's Law, the index ofdifficulty (ID) is given by$$\mathrm{ID}=\frac{\log (2 A / W)}{\log 2}$$where $W$ is the width of the target and $A$ is the distance tothe center of the target. Compare the difficulty of clickingon an icon that is 5 $\mathrm{mm}$ wide to clicking on one that is10 $\mathrm{mm}$ wide. In each case, assume that the mouse is100 $\mathrm{mm}$ from the icon.

$=1.231$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 3

Logarithmic Functions

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University of Michigan - Ann Arbor

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Lectures

01:22

Difficulty of a Task The d…

00:54

The difficulty in "ac…

00:34

Draw rays to scale to loc…

01:32

Draw rays to scale to loca…

06:26

A simple magnifier of foca…

02:33

A person with a near-point…

01:30

A field mouse trying to es…

01:14

You view an object by hold…

02:58

Water drop magnifier. You …

01:16

Solve each problem.Fir…

01:19

When dots are placed on a …

05:06

Because of a calculator&#x…

03:22

Make a rough graph of line…

01:06

How far should you hold a …

01:49

To focus a camera on objec…

03:54

02:57

A magnifier with a focal l…

02:56

Use a proportion to solve …

03:11

You measure the focal leng…

02:41

so to solve this equation, all we have to do is plug in, are given values for the with and so four for a first. With we have this is equal to five millimeters five millimeters. And so plugging this in we get that I d is equal to was equal to the log of of two times 100 because the distances on hundreds of two times 100 is 200. And so we have 200 divided by five, 200 divided by five gives us 40. So we have 40. So we have log of 40. It's right this as longer 40 and then we have This is divided by log of to Well, so the first solution we have is we plug this in plugging in log of 40 over log of to this gives us 5.32 So So this means that I d is equal to 5.3 to And so now, for for our our second with here this is equal to is equal to 10 millimeters and so So we have two times 100 because it distance remains the same. So we have log of two times 100 is 200 then we're dividing that by 10. So 200 divided by 10 gives us 20. So instead of log of 40 in the numerator, we have long of 20 over, log of to And so now if we want to, we want to solve this. Well, all we have to do is plug this into our calculator and what we get is a value that I d is approximately or 0.32 And so since these air approximates, I will use the approximation sign the approximation sign. And so these are our two solutions.

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