Dotting The Dot Map, Revisited. A. Jon Kimerling Dept. of Geosciences Oregon State University

Similar documents
Median and IQR The median is the value which divides the ordered data values in half.

CEE 522 Autumn Uncertainty Concepts for Geotechnical Engineering

PH 425 Quantum Measurement and Spin Winter SPINS Lab 1

Example: Find the SD of the set {x j } = {2, 4, 5, 8, 5, 11, 7}.

Number of fatalities X Sunday 4 Monday 6 Tuesday 2 Wednesday 0 Thursday 3 Friday 5 Saturday 8 Total 28. Day

( ) = p and P( i = b) = q.

NCSS Statistical Software. Tolerance Intervals

Statistics 511 Additional Materials

Properties and Hypothesis Testing

Chapter Vectors

THE SOLUTION OF NONLINEAR EQUATIONS f( x ) = 0.

FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING. Lectures

CS / MCS 401 Homework 3 grader solutions

Lecture 6 Chi Square Distribution (χ 2 ) and Least Squares Fitting

Sample Size Determination (Two or More Samples)

The Pendulum. Purpose

U8L1: Sec Equations of Lines in R 2

The picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled

Kinetics of Complex Reactions

4.3 Growth Rates of Solutions to Recurrences

Lecture 6 Chi Square Distribution (χ 2 ) and Least Squares Fitting

SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES

7-1. Chapter 4. Part I. Sampling Distributions and Confidence Intervals

CS 332: Algorithms. Quicksort

ECE 8527: Introduction to Machine Learning and Pattern Recognition Midterm # 1. Vaishali Amin Fall, 2015

RADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify

MID-YEAR EXAMINATION 2018 H2 MATHEMATICS 9758/01. Paper 1 JUNE 2018

Axis Aligned Ellipsoid

Continuous Functions

A quick activity - Central Limit Theorem and Proportions. Lecture 21: Testing Proportions. Results from the GSS. Statistics and the General Population

REGRESSION (Physics 1210 Notes, Partial Modified Appendix A)

Chapter 8: STATISTICAL INTERVALS FOR A SINGLE SAMPLE. Part 3: Summary of CI for µ Confidence Interval for a Population Proportion p

Power and Type II Error

Confidence Intervals for the Population Proportion p

Nonlinear regression

Example 2. Find the upper bound for the remainder for the approximation from Example 1.

Price per tonne of sand ($) A B C

is also known as the general term of the sequence

Section 9.2. Tests About a Population Proportion 12/17/2014. Carrying Out a Significance Test H A N T. Parameters & Hypothesis

Parameter, Statistic and Random Samples

Economics 250 Assignment 1 Suggested Answers. 1. We have the following data set on the lengths (in minutes) of a sample of long-distance phone calls

Introducing Sample Proportions

If, for instance, we were required to test whether the population mean μ could be equal to a certain value μ

ANALYSIS OF EXPERIMENTAL ERRORS

We will conclude the chapter with the study a few methods and techniques which are useful

CALCULUS BASIC SUMMER REVIEW

Estimation for Complete Data

Injections, Surjections, and the Pigeonhole Principle

11 Correlation and Regression

Response Variable denoted by y it is the variable that is to be predicted measure of the outcome of an experiment also called the dependent variable

(3) If you replace row i of A by its sum with a multiple of another row, then the determinant is unchanged! Expand across the i th row:

Analysis of Algorithms -Quicksort-

CONFIDENCE INTERVALS STUDY GUIDE

a is some real number (called the coefficient) other

CS 332: Algorithms. Linear-Time Sorting. Order statistics. Slide credit: David Luebke (Virginia)

1 Inferential Methods for Correlation and Regression Analysis

CS284A: Representations and Algorithms in Molecular Biology

The Random Walk For Dummies

Polynomial Functions and Their Graphs

Inferential Statistics. Inference Process. Inferential Statistics and Probability a Holistic Approach. Inference Process.

Mathematical Notation Math Introduction to Applied Statistics

WHAT IS THE PROBABILITY FUNCTION FOR LARGE TSUNAMI WAVES? ABSTRACT

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER / Statistics

multiplies all measures of center and the standard deviation and range by k, while the variance is multiplied by k 2.

Bertrand s Postulate

6.1. Sequences as Discrete Functions. Investigate

Probability, Expectation Value and Uncertainty

Fourier Series and the Wave Equation

Discrete probability distributions

PROPERTIES OF AN EULER SQUARE

Chapter 6 Sampling Distributions

Infinite Sequences and Series

6 Sample Size Calculations

Castiel, Supernatural, Season 6, Episode 18

Mathematics for Queensland, Year 12 Worked Solutions to Exercise 5L

Discrete Mathematics for CS Spring 2008 David Wagner Note 22

17 Phonons and conduction electrons in solids (Hiroshi Matsuoka)

Chapter 10: Power Series

Final Examination Solutions 17/6/2010

A statistical method to determine sample size to estimate characteristic value of soil parameters

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =

WORKING WITH NUMBERS

XT - MATHS Grade 12. Date: 2010/06/29. Subject: Series and Sequences 1: Arithmetic Total Marks: 84 = 2 = 2 1. FALSE 10.

KNOWLEDGE OF NUMBER SENSE, CONCEPTS, AND OPERATIONS

Principle Of Superposition

(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is

Analysis of Algorithms. Introduction. Contents

Random Variables, Sampling and Estimation

Chapter 8: Estimating with Confidence

CHAPTER 8 FUNDAMENTAL SAMPLING DISTRIBUTIONS AND DATA DESCRIPTIONS. 8.1 Random Sampling. 8.2 Some Important Statistics

Because it tests for differences between multiple pairs of means in one test, it is called an omnibus test.

Mechatronics. Time Response & Frequency Response 2 nd -Order Dynamic System 2-Pole, Low-Pass, Active Filter

Chapter 4. Fourier Series

Final Review for MATH 3510

Statistical Pattern Recognition

UNIVERSITY OF TORONTO Faculty of Arts and Science APRIL/MAY 2009 EXAMINATIONS ECO220Y1Y PART 1 OF 2 SOLUTIONS

A New Recursion for Space-Filling Geometric Fractals

Appendix: The Laplace Transform

September 2012 C1 Note. C1 Notes (Edexcel) Copyright - For AS, A2 notes and IGCSE / GCSE worksheets 1

4.1 SIGMA NOTATION AND RIEMANN SUMS

Transcription:

Dottig The Dot Map, Revisited A. Jo Kimerlig Dept. of Geoscieces Orego State Uiversity

Dot maps show the geographic distributio of features i a area by placig dots represetig a certai quatity of features where the features are most likely to occur.

Whe makig a dot map, we first select a dot diameter ad the determie the dot uit value. dot diameter dot uit value

The dot uit value is always greater tha oe. dot uit value > 1

Selectig the dot diameter is a subjective decisio.

Selectig the dot uit value is doe by trial ad error or usig the Mackay omograph. J. Ross Mackay, 1949 om o graph ou a graph, usually cotaiig three parallel scales graduated for differet variables, desiged to solve a equatio; also called a aligmet graph or calculatio graph.

Usig the Mackay omograph to fid the dot uit value

Mark the selected dot diameter

Draw a lie from the origi to your mark

Fid where the lie crosses the zoe of coalescig dots

Draw a vertical lie dow to the x-axis

Note the umber of dots per square cetimeter. 52

The, fid the map area of the regio that will have the highest desity of dots 1.2 sq cm WHEAT

Multiply the map area by the umber of dots per sq. cm. from the omograph to fid the umber of dots to place i that area WHEAT 52 dots / sq cm * 1.2 sq cm = 62 dots

Divide the quatity i the desest area by the umber of dots to fid the dot uit value WHEAT 62,000 acres 1997 STATE TOTAL Washigto: 2,422,506 Idaho: 1.410,978 Orego: 882,862 Each dot represets 1000 acres 62,000 acres / 62 dots = 1000 acres / dot

Issues with the Mackay omograph 5 4.5 4 pt 3.5 pt 3 pt 2.5 pt mathematically imprecise zoe of coalescece 2 pt 1.5 pt poit size rage limits??? 1 pt

Let s look carefully at the omograph A 2 pt dot with diameter 0.0706 cm has a area of 0.00391 cm 2, so 128 dots without overlap would have a aggregate area of oly 0.50 cm 2 o overlap! 2 pt

Let s look carefully at the omograph The MacKay omograph is a graph of : dot area dots per sq cm = aggregate area of dots, with a zoe of coalescig dots overlaid. 2 pt

We eed a theoretically soud mathematical basis for dot coalescece.

Modelig dot coalescece usig the Uificatio Equatio from probability theory is the aswer!

( ) ( ) ( ) ( ) ( ) ( ) < < < + < < < = = + + = k... j i k j i k j i j i j i i i i i E... E E E P... E E E P E E P E P E P 3 2 1 1 0 0 1 U The Dot Coalescece Model A 1 cm 2 square has a probability P of 1.0. The area of each of dot (Ei) i proportio to the area of the square is its probability p = P(E i ). The Uificatio Equatio Sice the dots i the square are the same size, E i = E j = = E ad p = P(Ei) = P(Ej) = = P(E)

( ) ( ) < < < + < < < = = + + = k... j i k j i j i i i p... p p p p P 1 3 2 0 0 1 U = = i p p 0 Summatios are umbers of dot combiatios The umber of dot combiatios i each summatio are foud by: ( )! k k!! ( ) = U i p P 0 is the aggregate dot area i cm 2 The Uificatio Equatio

So P 2 3 + 1 U i= 0 ( p) = p 2!! p +! ( 2)! 3! ( 3)! p... + ( 1) k!! ( k)! p k There is a term i the series for each dot added, but I trucated the series at k = 10 without affectig the results for up to: P U i= 0 ( p) = 0. 95 with up to 1,000 poits.

( ) ( ) ( ) ( ) ( ) 10 10 3 2 0 10 10 1 3 3 2 2 p!!!... p!!! p!!! p p P i + + = = U Or doig the factorials ( ) ( ) 10 10 3 2 0 800 628 3 10 9 8 7 6 5 4 3 2 1 1 6 2 1 2 1 p,, ) )( )( )( )( )( )( )( )( )( (... p ) )( ( p ) ( p p P i + + = = U So

Usig the equatio with ESRI dots ESRI dots are i 0.5 Postscript poit icremets from 0.5 to 11 poits, although sizes smaller tha 1 or larger tha 5 poits would ot ormally be used for dot mappig. Give that 1 Postscript poit = 0.353 mm, the followig table gives values of p (dot areas) for this dot size rage. Poit Size p (cm 2 ) Poit Size p (cm 2 ) 1.0 0.000978 3.5 0.011988 1.5 0.002202 4.0 0.015658 2.0 0.003914 4.5 0.011988 2.5 0.006116 5.0 0.024466 3.0 0.008808

Ruig the equatio with ESRI dots i 10 dot icremets gave the results plotted o this graph.

Usig the graph Example: Havig half the 1 cm 2 area covered by 2 pt dots takes 177 radomly placed dots

This compares with 128 dots from the MacKay omograph. 2 pt

Amout of dot overlap is aother measure of coalescece. Percet Overlap = p P U i= 0 ( p) 100

Ruig the equatio with ESRI dots i 10 dot icremets gave the results plotted o this graph. equal overlap poits

Equal overlap poits defie lies of costat dot overlap a more precise form of zoe of coalescig dots. equal overlap poits

Testig the model

Step 1. Geerate lots of radom umbers from 0.0 mm to 10.0 mm. #iclude <stdio.h> #iclude <stdlib.h> mai() { FILE*ofil; char oame[80]; it seed; double r; /* radom value i rage [0,1) */ it cout; seed = 10000; /* choose a seed value */ srad(seed); /*iitialize radom umber geerator*/ pritf("eter output file ame: "); scaf ("%s",oame); if((ofil = fope(oame,"wt")) == NULL) { pritf("error: caot ope output file\"); retur 1; } for(cout=1; cout<=3000; ++cout) { r = 10.0*(rad() / ((RAND_MAX)+(1.0))); fpritf(ofil,"%lf\",r); } retur 0; } /*mai*/ 9.977417 3.888855 3.605957 6.149292 0.285339 7.028809 0.999451 6.518555 2.563782 6.928406 0.412292 2.995300 2.835388 5.465698 1.372375 6.512756 4.749756 7.342224 7.826538 6.076050 3.758240

Step 2. Split the umbers ito two Excel colums ad make a scatterplot.

Step 3. Import the scatterplot ito Freehad ad chage the dots to the desired diameter.

Step 4. Wrap the dots aroud the four edges.

Step 5. Import the graphic ito Photoshop ad crop the edge dots.

Step 6. Use the histogram tool to fid the proportio of the square covered by dots.

I made these dot proportio measuremets for the umber of dots predicted at differet poit sizes for 10-50% dot overlap.

How close did the measured aggregate areas match those predicted from the Uificatio Equatio?

I tested the equatio for 10% overlap of 3 pt dots 55 radom dots.

I also made a Dot Selectio Guide based o aggregate dot area.

What is the problem with radom dot placemet? Cartographers place dots maually i a pseudoradom fashio!

Let s look at pseudo-radomess i terms of maximum allowable overlap of idividual dots 200 dots are placed i each square

Idividual dot overlap is computed from the les area equatio. Area Les = d 2R 1 2 2 1 2 2 2R cos d 4R d

Graphig the les area equatio shows the oliear relatioship betwee the spacig of dot ceters ad the proportio of a dot overlapped by the les. example, 2 dots spaced at 1.4 times their radius will overlap by 0.2 (20% of the dot area)

This iformatio is the basis for a pseudo-radom dot geerator. 1. Select a dot radius ad maximum dot overlap. 2. Geerate the first radom dot positio. 3. For succeedig dots, compute the distace betwee the dot ceter ad all other dot ceters. 4. If all distaces are greater tha d, add the dot to the array of dots otherwise, discard the dot positio ad geerate aother radom positio. 5. Repeat util the umber of dots you eed is geerated, or util a maximum umber of tries is reached.

The procedure I wrote a C program to ru the procedure 30 times for 10%, 20%, ad 30% maximum idividual dot overlaps, i steps of 25 or 50 dots with a maximum of 32,000 tries for each ru. Oce I had a pseudo-radom dot positio array, I could check the distace of each dot agaist all other dots, ad compute the les area if a dot was less tha a dot diameter away. Summig the les areas gave the total dot overlap area, ad hece the aggregate proportio or percetage covered by dots, assumig that there were o triple dot overlaps.

1 pt dots maximum aggregate area reached i less tha 32,000 tries

What is happeig with mutually exclusive dot placemet? rigid radom packig for 1 pt dots i a 1 x 1 cm square

Do the graphs for 1.0 pt to 2.5 pt dots look similar?

Aggregate area equatios Mutually exclusive dots: Area = p Area = p 2!! Totally radom dots: 2! 3 p + p... +! ( 2)! 3! ( 3) ( 1) 10 10!! ( 10)! p 10 My guess is that the equatio for itermediate pseudo-radom dots is a liear combiatio of the two boudig equatios above.

A geeral aggregate area equatio Area = kp + ( 1 k) ( p 2!! p 2 +! ( 2)! 3! ( 3)! p 3... + ( 1) 10 10!! ( 10)! p 10 ) or Area = p + ( 1 k) ( 2!! p 2 +! ( 2)! 3! ( 3)! p 3... + ( 1) 10 10!! ( 10)! p 10 ) where k rages from 0 (totally radom) to 1 (mutually exclusive).

Guesses as to what k is proportioal to? 1. k = 1.0 - les proportio 2. k = d/2 totally radom mutually exclusive

The secod possibility for k fit the data better about twice as good a fit.

Average values for a umber of tries of 30 rus of creatig pseudo-radom are plotted o these graphs. 1 pt dot example

From these values we ca make a pseudoradom dot selectio guide.

Pseudo-radom placemet of 1 pt dots with less tha 5,000 tries

Pseudo-radom Dot Selectio Guide for 1 pt dots with variable maximum dot overlap

mappigceter.esri.com Other Resources