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https://educationlearnacademy.in/ncert-solutions-class-11-maths-pdf/ | 1,721,052,705,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00109.warc.gz | 190,541,124 | 65,796 | Free NCERT Solutions Class 11 Maths PDF Download
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NCERT Solutions Class 11 Maths
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NCERT Solutions Class 11 Maths
NCERT Solutions for class 11 Maths all chapters are given below updated for new academic session 2020-2021. Download NCERT Solutions 2020-21 for other subjects also. If you are having any suggestion for the improvement, your are welcome. The improvement of the website and its contents are based on your suggestion and feedback.
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NCERT Sols for Class 11
NCERT Sols for Class 11 Maths, Physics, Chemistry and Biology are given below to study online or download. Notes and Study material for class 11 Physics, Chemistry, Biology, Business Studies and Physical Education are also given to download.
How To Prepare Class 11 Maths All Chapters
Chapter 1: Sets
Set is the foundation of today’s Maths. In Chapter 1 Sets of Class 11, we will learn sets and their representation. It includes the definition as well as properties of various type of sets. A set which does not contain any element is called the empty set or the null set or the void set. Similarly, a set which is empty or consists of a definite number of elements is called finite. Otherwise, the set is called infinite. We should know about the operations of set like Union, Intersection, Adjoint and Disjoint Sets, Equal and equivalent set. Concepts of subsets, universal sets and Venn Diagram is widely used in Class 11 Maths. The Miscellaneous Exercise of Chapter 1 Class 11 Maths describes the applications of the set and its theorems.
Chapter 2: Relations And Functions
Relations and Functions of Class 11 Maths prepare a solid based for Class 12 Maths Chapter 1 Relations and Functions. In Class 11, we will learn about Cross Product of two or three sets to get ordered pair. This fact helps us to understand the concepts of Relations and Functions in Exercise 2.2 and Miscellaneous of Class 11 Maths. Do you know why students face difficulties in Class 12 Maths Chapter 1? The only reason is that they have not done the concepts of Chapter 2 of Class 11 Maths properly. Class 11 Maths Chapter 2 is the base of Class 12 Chapter 1 Relation and Functions.
Chapter 3: Trigonometric Functions
The first exercise ofNCERT Solutions Class 11 Maths Chapter 3 Trigonometric Functions explains how to convert Radians to Degree or Degree to Radians. Students are more comfortable with degree measure so far. In higher classes, most often we use radians measure only. That is why now start practising in radian rather than degree. Exercise 3.2 explains how the signs of trigonometric functions change from one quadrant to others. All the formulae in Exercise 3.3, will be frequently used in most of the chapters of NCERT Solutions Class 11 Maths. If someone is not able to grasp these formulae, he will face the problem in Class 12 Maths and Physics.
Chapter 4: Principle Of Mathematical Induction
The Principle of Mathematical Inductions tells us if an expression is true for two consecutive natural numbers, then it will be true for all. To verify this, we take it is true for some positive integer k and try to prove that it is also true for k + 1. In this way, we try to prove for two consecutive positive integers. PMI is useful for not only in Class 12 Maths but in higher studies also. There are two sections of questions from 1 to 18 and from 19 to 24. In Exams, question papers consist of at least one question from each section.
Chapter 5: Complex Numbers And Quadratic Equations
In NCERT Solutions Class 11 Maths Chapter 4 Quadratic Equations, if the discriminant is negative, we were not able to find the solutions. We will deal here the same situation but getting the solutions. Now, we can find out the roots of equations even though the D is negative. It is become possible just because of Complex Numbers. In Class 11 Chapter 5 Complex numbers, we can find the real as well as imaginary roots of any quadratic equation. Square roots of a complex number also can be obtained in the supplementary section. Class 11 Maths Chapter 5 Miscellaneous Exercise is most important to know Complex numbers completely.
Chapter 6: Linear Inequalities
NCERT Solutions Class 11 Maths Chapter 6 Linear Equalities, we shall study of linear inequalities in one and two variables only. Representing on a number line, finding solutions sets as Real Numbers, Integers and Problems based on word problems are the common topics of the chapter. Exercise 6.3 is the main exercise for the exams point of view. Its weightage is maximum with respect to other exercises. This exercise is also helpful in Class 12 Maths Chapter 12 LPP. This chapter is easier to score good enough in the CBSE exams.
Chapter 7: Permutations And Combinations
NCERT Solutions Class 11 Maths1 Maths Chapter 7, we shall learn some basic counting techniques. It will enable us to answer so many questions without actually listing or arranging elements. Permutations will be useful to know the number of different ways quickly. As a first step, we shall know to the learning of Permutations tricks. Class 11 Maths Chapter 7 Miscellaneous Exercise questions need the application of both Permutation and Combination for answers. These topics are helpful for the calculation of Probability in Class 11 Maths Chapter 16.
Chapter 8: Binomial Theorem
NCERT Solutions Class 11 Maths Chapter 8 Binomial Theorem contains two exercises. Exercise 8.1 has simple sums on Binomial Expansions with two terms. The applications of general terms are in exercise 8.2. Just like the other chapters, miscellaneous of this chapter is also scoring for exams or tests. In 11th Maths, we know the formula of expansion of 3 or 4 powers but using Binomial Theorem we can expand the expression with any number of powers. In Class 11 Maths, there are only natural numbers as indices. Binomial allows the negative powers also but not in Class 11 CBSE Syllabus 2020-21.
Chapter 9: Sequences And Series
NCERT Solutions Class 11 Maths Chapter 9 Sequence and Series is the next step of Class 10 Maths Chapter 5 A P. We have done A P in Class 10, so about A.P. only a few sums are there in Class 11 Maths. Now we shall study about the arithmetic mean, geometric mean, the relationship between A.M. and G.M. and based questions. We also learn about special series in forms of the sum to n terms of consecutive natural numbers, sum to n terms of squares of natural numbers and sum to n terms of cubes of natural numbers. Class 11 Maths Chapter 9 needs more practice to know the concepts well.
Chapter 10: Straight Lines
NCERT Solutions Class 11 Maths Chapter 10 Straight Lines provides brief recall of 2 – D from earlier classes. In the supplementary part, we will learn about the shifting of origin, the slope of a line and angle between two lines. Exercise 10.1 includes simple questions, mostly based on the concepts of class 10 Maths. In the next exercises, we know about lines parallel to the axis, point-slope form, slope-intercept form, two-point form, intercept form and Normal form. Obtaining the distance of a point from a line is scoring for exams. This chapter is a little bit difficult as compared to others. So it needs more focus in practice to get good marks in tests or exams.
Chapter 11: Conic Sections
NCERT Solutions Class 11 Maths Chapter 11 Conic Sections, we shall study about curves like circles, ellipses, parabolas and hyperbolas. Just try to know who named these curves? Why is the study of these curves under the heading Conics? The 11 Maths Exercise 11.1 explains the terms and equations related to circle only whereas Exercise 11.2 for Parabola. Exercise 11.3 and 11.4 of Class 11 Maths explains about Ellipse and Hyperbola, respectively. We should know about the foci, vertices, length of the major & minor axis, eccentricity and the length of the latus rectum of the Ellipse as well as Hyperbola. In modern time, these curves are in fields such as planetary motion, design of telescopes, and so many other fields.
Chapter 12: Introduction To Three Dimensional Geometry
In Class 10 Maths we have learnt about the distance between the two points in 2 D plane. Here, in Class 11 Maths Chapter 12, we can find the same distance in 3 D plane. At the beginning of the chapter, the basics of quadrant octant are described. Whatever we have done in class 10 Maths Exercise 7.2, the same thing has to be repeated with the concept of 3 D geometry. Only one thing is different, that is Section Formula. In NCERT Solutions Class 11 Maths, we divide a line or line segment into a particular ratio internally, but in class 11 Maths internally and externally both are applicable.
Chapter 13: Limits And Derivatives
The concept of Chapter 13 Limits and Derivatives of Class 11 Maths is new for the students. Here, they learn how to find the value of a function using limits. Derivatives using First Principle is one of the good topics for exams. We should try to do almost all the derivatives using the First Principle. The concepts of LHL and RHL is useful for testing a function whether its Limits exist or not. After doing Chapter 13 of Class 11 Maths, everyone should do the Class 12 Maths Exercise 5.2, 5.3 and 5.4 to understand better the concept of Derivatives.
Chapter 14: Mathematical Reasoning
In mathematical language, there are two kinds of reasoning – inductive and deductive. Chapter 14 of Class 11 Maths follows deductive reasoning. Here, we will learn about a SENTENCE which is called a mathematically acceptable statement if it is either true or false but not both. While dealing with statements, we usually denote them by small letters p, q, r like letters. This Chapter also deals with the negation of a statement; Compound statements with connecting word and decision making final statements.
Chapter 15: Statistics
Exercise 15.1 of Class 11 Maths contains the questions of Deviations about the mean, mode and median. We will deal here with all the measures of central tendencies like Mean Deviation and Range. The calculation will include both grouped and ungrouped data series. Class 11 Maths Chapter 15 will discuss the questions for desecrate and continuous frequency distribution. Coefficient of variation using Variance and Standard Deviation based questions frequently come in the exams. Shortcut method to find variance and the standard deviation is equally useful for exams.
Chapter 16: Probability
Probability is always termed as a scoring topic in all the classes. It is scoring also in CBSE Board exams. Class 11 Maths Chapter 16 is the base for the Class 12 Maths Chapter 13 Probability. If you want to score well in class 12 probability, must do by learning all the concepts in grade 11. Exercise 16.1 is based on Sample Space and simple ideas of Probability. Next exercises include events which are Exclusive, Exhaustive or concepts of sets. Miscellaneous Exercise, containing good questions for practice, is vital for the class test or final exams.
The solutions by us provided are well explained keeping in mind the minutest of details and covering all the aspects of the book. The answers are created in such a manner that even a slow grasper student can also easily grab the concepts and intelligent student can mange out his time well.
NCERT is the standard book followed by 90% of the CBSE (Central Board of Secondary Education) and many state board schools across India. All the solutions provided resonate well with the latest NCERT books and cover the entire 16 chapters.
As we all know preparation for classes such as XI and XII is incomplete if you have not done NCERT Books. The curriculum designed by NCERT (National Council of Education Research and Training) plays an important role in preparation as all other help books and books from other authors are designed taking the pattern and syllabus from NCERT text books.
Mathematics is just not a cup of tea for everyone but we make sure that it becomes one. Our aim is just not to make the students pass the School/Board examination with flying colors but also to make them competent enough to crack top Entrance exams as well such as IIT/JEE, AIEEE and many more.
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1. Which book is best for class 11 maths?
NCERT Solutions Class 11 Maths Textbook is best and enough for CBSE Class 11 Exam preparation. RD Sharma class 11 Textbook is best reference book for Class 11 students.
2. How many chapters are there in class 11 maths?
There are 16 chapters starting from Sets to Probability. educationleaenacademy.com has included all the chapters Answers in this page.
3. Is class 11 Maths tough?
No. If you have very good foundation in lower classes, Maths will be very easy. CBSE has designed curriculum based on NCERT guidelines. These NCERT Textbooks are very good in presenting concepts and application based on these concepts. educationleaenacademy.com recommends you to study introduction part of each and every chapter get the firm grip on the concepts.
4. Write down all the subsets of the following set: {1, 2, 3}.
The subsets of {1, 2, 3} are
Φ,
{1},
{2},
{3},
{1, 2},
{2, 3},
{1, 3}
{1, 2, 3}
5. In a group of students 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Let U be the set of all students in the group.
Let E be the set of all students who know English.
Let H be the set of all students who know Hindi.
∴ H ∪ E = U
Accordingly, n(H) = 100 and n(E) = 50
n( H U E ) = n(H) + n(E) – n(H ∩ E)
= 100 + 50 – 25
= 125
Hence, there are 125 students in the group.
6. If A = {-1, 1}, find A × A × A.
A = {-1,1},
Therefore,
A×A
={-1,1}×{-1,1}={(-1,-1),(-1,1),(1,-1),(1,1)}
and
A×A×A
={(-1,-1),(-1,1),(1,-1),(1,1)} × {-1,1}
={(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)}
7. Write the relation R = {(x, x^3): x is a prime number less than 10} in roster form.
R = {(x, x^3): x is a prime number less than 10}
The prime numbers less than 10 are 2, 3, 5, and 7.
∴ The roster form R = {(2, 8), (3, 27), (5, 125), (7, 343)}
8. मान ज्ञात कीजिए: sin〖765°〗
sin〖765°〗
=sin(2×360° + 45°)
〖=sin 45°〗
[∵ पहले चतुर्थांश में sin धनात्मक होता है]
=1/√2
9. Prove the following by using the principle of mathematical induction for all n ∈ N: 〖10〗^(2n-1)+1 is divisible by 11.
Let the given statement be P(n), therefore,
P(n):〖10〗^(2n-1)+1 is divisible by 11.
For n = 1, we have
〖10〗^(2-1)+1=11,which is divisible by 11.
So, P(1) is true.
Let P(k) be true for some positive integer k, such that
P(k):〖10〗^(2k-1)+1 is divisible by 11.
Let 〖10〗^(2n-1)+1 =11m … (1)
Where, m is any natural number.
Now, to prove that P(k + 1) is true. i.e.
P(k+1):〖10〗^(2k+1)+1 is divisible by 11.
Consider 〖10〗^(2k+1)+1
=〖10〗^(2k-1+2)+1
=〖10〗^2.〖10〗^(2k-1)+1
=〖10〗^2.(11m-1)+1
[From the equation (1),〖10〗^(2n-1)=11m-1]
=100.(11m-1)+1
=1100m – 100 + 1
=1100m – 99
=11[100m – 9],which is divisible by 11.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the Principle of Mathematical Induction, statement P(n) is true for all natural numbers.
10. Solve the following equation: 2x^2 + x + 1 = 0.
The given quadratic equation is 2x^2 + x + 1 = 0.
On comparing the given equation with 〖ax〗^2+bx+c=0,
we obtain a = 2, b = 1, and c = 1
Therefore, the discriminant of the given equation is given by
D = b^2-4ac
= 1^2-4×2×1
=-7
Therefore, the required solutions are
x = (-b±√D)/2a=(-1±√(-7))/(2×2)
= (-1±√7.√(-1))/4
= (-1±√7 i)/4 [∵ √(-1)=i]
11. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is x + 2.
Since both the integers are smaller than 10, therefore
x + 2 < 10 ⇒ x < 10 – 2 ⇒ x < 8 … (i) Also, the sum of the two integers is more than 11. ∴ x + (x + 2) > 11
⇒ 2x + 2 > 11
⇒ 2x > 11 – 2
⇒ 2x > 9
⇒x>9/2
⇒x>4.5 …(ii)
From (i) and (ii), we obtain that the value of x can be 4,5,6 or 7. .
Since x is an odd number, x can take the values, 5 and 7.
Hence, the required possible pairs of numbers are (5, 7) and (7, 9).
12. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Each signal requires the use of 2 flags.
There will be as many flags as there are ways of filling in 2 vacant places _ _ in succession by the given 5 flags of different colours.
The upper vacant place can be filled in 5 different ways by any one of the 5 flags following which, the lower vacant place can be filled in 4 different ways by any one of the remaining 4 different flags.
Thus, by multiplication principle, the number of different signals that can be generated is 5 × 4 = 20
13. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, 115 … 995.
Here, first term, a = 105 and common difference, d = 5
Now,a_n=a+(n-1)d
⇒ 995 = 105+(n-1)×5
⇒ 890=(n-1)×5
⇒ 178=(n-1)
⇒ n=179
S_n = n/2 [2a+(n-1)d]
⇒ S_179 = 179/2 [2×105+(179-1)×5]
⇒ S_179 = (179)[550]
= 98450
Hence, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.
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NCERT Solutions Class 11 Maths
Posted by
NirbhaySingh
Hello friends, I am Hemant, Technical Writer & Co-Founder of Education Learn Academy. Talking about education, I am a student. I enjoy learning things related to new technology and teaching others. I request you that you keep supporting us in this way and we will continue to provide new information for you. :) | 4,909 | 19,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-30 | latest | en | 0.918589 |
https://waseda.pure.elsevier.com/ja/publications/mathematical-analysis-of-three-dimensional-geometric-tolerances-o | 1,643,143,191,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00583.warc.gz | 650,647,687 | 9,613 | # Mathematical analysis of three-dimensional geometric tolerances of mechanical products - analysis under maximum material condition
Naoki Satonaka, Akifumi Yoshida, Nobuhiro Sugimura, Yoshitaka Tanimizu, Koji Iwamura
## 抄録
Dimensional tolerances and geometric tolerances of three-dimensional product shapes are very important for manufacturing process control and quality control of mechanical products. The objective of the present research is to develop systematic methods for planning and analysis of the geometric tolerances of the three-dimensional mechanical products. The tolerance zones of the geometric features have been investigated, in the previous paper, based on the definitions of the geometric tolerances, and the parameters have been defined to describe the deviations of the geometric features within the tolerance zones. Mathematical methods are discussed, in the present paper, to estimate the statistical deviations of the positions and the orientations of the geometric features, based on the deviation parameters and the relationships between the datum features and the target features. In particular, emphasis is given to the analysis of the statistical deviations of the geometric features under the MMC (Maximum Material Conditions), which specifies the interaction between the dimensional tolerances and the geometric tolerances. The proposed method is applied to the analysis of the deviations of the geometric features, to which both the dimensional tolerances and the geometric tolerances are required.
本文言語 English 313-318 6 Published - 2005 はい 3rd International Conference on Leading Edge Manufacturing in 21st Century, LEM 2005 - Nagoya, Japan継続期間: 2005 10 19 → 2005 10 22
### Other
Other 3rd International Conference on Leading Edge Manufacturing in 21st Century, LEM 2005 Japan Nagoya 05/10/19 → 05/10/22
• 産業および生産工学
## フィンガープリント
「Mathematical analysis of three-dimensional geometric tolerances of mechanical products - analysis under maximum material condition」の研究トピックを掘り下げます。これらがまとまってユニークなフィンガープリントを構成します。 | 444 | 2,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-05 | longest | en | 0.81017 |
https://www.lmfdb.org/L/8/825e4/1.1/c1e4/0/3 | 1,638,157,173,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00345.warc.gz | 969,636,798 | 8,009 | # Properties
Label 8-825e4-1.1-c1e4-0-3 Degree $8$ Conductor $463250390625$ Sign $1$ Analytic cond. $1883.32$ Root an. cond. $2.56664$ Motivic weight $1$ Arithmetic yes Rational yes Primitive no Self-dual yes Analytic rank $0$
# Origins of factors
## Dirichlet series
L(s) = 1 − 3-s + 2·4-s + 5·5-s + 7-s − 9·11-s − 2·12-s + 6·13-s − 5·15-s − 5·16-s − 9·17-s + 22·19-s + 10·20-s − 21-s − 11·23-s + 10·25-s + 2·28-s + 6·29-s + 7·31-s + 9·33-s + 5·35-s − 12·37-s − 6·39-s + 41-s − 4·43-s − 18·44-s − 15·47-s + 5·48-s + ⋯
L(s) = 1 − 0.577·3-s + 4-s + 2.23·5-s + 0.377·7-s − 2.71·11-s − 0.577·12-s + 1.66·13-s − 1.29·15-s − 5/4·16-s − 2.18·17-s + 5.04·19-s + 2.23·20-s − 0.218·21-s − 2.29·23-s + 2·25-s + 0.377·28-s + 1.11·29-s + 1.25·31-s + 1.56·33-s + 0.845·35-s − 1.97·37-s − 0.960·39-s + 0.156·41-s − 0.609·43-s − 2.71·44-s − 2.18·47-s + 0.721·48-s + ⋯
## Functional equation
\begin{aligned}\Lambda(s)=\mathstrut &\left(3^{4} \cdot 5^{8} \cdot 11^{4}\right)^{s/2} \, \Gamma_{\C}(s)^{4} \, L(s)\cr=\mathstrut & \,\Lambda(2-s)\end{aligned}
\begin{aligned}\Lambda(s)=\mathstrut &\left(3^{4} \cdot 5^{8} \cdot 11^{4}\right)^{s/2} \, \Gamma_{\C}(s+1/2)^{4} \, L(s)\cr=\mathstrut & \,\Lambda(1-s)\end{aligned}
## Invariants
Degree: $$8$$ Conductor: $$3^{4} \cdot 5^{8} \cdot 11^{4}$$ Sign: $1$ Analytic conductor: $$1883.32$$ Root analytic conductor: $$2.56664$$ Motivic weight: $$1$$ Rational: yes Arithmetic: yes Character: induced by $\chi_{825} (1, \cdot )$ Primitive: no Self-dual: yes Analytic rank: $$0$$ Selberg data: $$(8,\ 3^{4} \cdot 5^{8} \cdot 11^{4} ,\ ( \ : 1/2, 1/2, 1/2, 1/2 ),\ 1 )$$
## Particular Values
$$L(1)$$ $$\approx$$ $$3.675119453$$ $$L(\frac12)$$ $$\approx$$ $$3.675119453$$ $$L(\frac{3}{2})$$ not available $$L(1)$$ not available
## Euler product
$$L(s) = \displaystyle \prod_{p} F_p(p^{-s})^{-1}$$
$p$$\Gal(F_p)$$F_p(T)$
bad3$C_4$ $$1 + T + T^{2} + T^{3} + T^{4}$$
5$C_4$ $$1 - p T + 3 p T^{2} - p^{2} T^{3} + p^{2} T^{4}$$
11$C_4$ $$1 + 9 T + 41 T^{2} + 9 p T^{3} + p^{2} T^{4}$$
good2$C_2^2$ $$( 1 - T^{2} + p^{2} T^{4} )^{2}$$
7$C_2^2:C_4$ $$1 - T - T^{2} - 17 T^{3} + 64 T^{4} - 17 p T^{5} - p^{2} T^{6} - p^{3} T^{7} + p^{4} T^{8}$$
13$C_4\times C_2$ $$1 - 6 T + 23 T^{2} - 60 T^{3} + 61 T^{4} - 60 p T^{5} + 23 p^{2} T^{6} - 6 p^{3} T^{7} + p^{4} T^{8}$$
17$C_2^2:C_4$ $$1 + 9 T + 14 T^{2} - 147 T^{3} - 1001 T^{4} - 147 p T^{5} + 14 p^{2} T^{6} + 9 p^{3} T^{7} + p^{4} T^{8}$$
19$D_{4}$ $$( 1 - 11 T + 67 T^{2} - 11 p T^{3} + p^{2} T^{4} )^{2}$$
23$C_2^2:C_4$ $$1 + 11 T + 38 T^{2} + 125 T^{3} + 821 T^{4} + 125 p T^{5} + 38 p^{2} T^{6} + 11 p^{3} T^{7} + p^{4} T^{8}$$
29$D_{4}$ $$( 1 - 3 T + p T^{2} - 3 p T^{3} + p^{2} T^{4} )^{2}$$
31$C_2^2:C_4$ $$1 - 7 T + 38 T^{2} - 329 T^{3} + 2725 T^{4} - 329 p T^{5} + 38 p^{2} T^{6} - 7 p^{3} T^{7} + p^{4} T^{8}$$
37$C_2^2:C_4$ $$1 + 12 T + 17 T^{2} - 360 T^{3} - 2879 T^{4} - 360 p T^{5} + 17 p^{2} T^{6} + 12 p^{3} T^{7} + p^{4} T^{8}$$
41$C_2^2:C_4$ $$1 - T - 10 T^{2} - 229 T^{3} + 1679 T^{4} - 229 p T^{5} - 10 p^{2} T^{6} - p^{3} T^{7} + p^{4} T^{8}$$
43$D_{4}$ $$( 1 + 2 T + 7 T^{2} + 2 p T^{3} + p^{2} T^{4} )^{2}$$
47$C_2^2:C_4$ $$1 + 15 T + 53 T^{2} + 15 T^{3} + 484 T^{4} + 15 p T^{5} + 53 p^{2} T^{6} + 15 p^{3} T^{7} + p^{4} T^{8}$$
53$C_4\times C_2$ $$1 - T - 52 T^{2} + 105 T^{3} + 2651 T^{4} + 105 p T^{5} - 52 p^{2} T^{6} - p^{3} T^{7} + p^{4} T^{8}$$
59$C_2^2:C_4$ $$1 - 10 T + 101 T^{2} - 990 T^{3} + 10961 T^{4} - 990 p T^{5} + 101 p^{2} T^{6} - 10 p^{3} T^{7} + p^{4} T^{8}$$
61$C_2^2:C_4$ $$1 + 20 T + 129 T^{2} + 520 T^{3} + 4001 T^{4} + 520 p T^{5} + 129 p^{2} T^{6} + 20 p^{3} T^{7} + p^{4} T^{8}$$
67$C_2^2:C_4$ $$1 + 5 T - 27 T^{2} + 385 T^{3} + 6524 T^{4} + 385 p T^{5} - 27 p^{2} T^{6} + 5 p^{3} T^{7} + p^{4} T^{8}$$
71$C_2^2:C_4$ $$1 - 7 T - 2 T^{2} - 569 T^{3} + 8925 T^{4} - 569 p T^{5} - 2 p^{2} T^{6} - 7 p^{3} T^{7} + p^{4} T^{8}$$
73$C_2^2:C_4$ $$1 - 6 T - 57 T^{2} + 130 T^{3} + 4761 T^{4} + 130 p T^{5} - 57 p^{2} T^{6} - 6 p^{3} T^{7} + p^{4} T^{8}$$
79$C_2^2:C_4$ $$1 - 24 T + 177 T^{2} - 392 T^{3} + 225 T^{4} - 392 p T^{5} + 177 p^{2} T^{6} - 24 p^{3} T^{7} + p^{4} T^{8}$$
83$C_2^2:C_4$ $$1 - 21 T + 223 T^{2} - 2625 T^{3} + 30136 T^{4} - 2625 p T^{5} + 223 p^{2} T^{6} - 21 p^{3} T^{7} + p^{4} T^{8}$$
89$C_2^2:C_4$ $$1 + 16 T + 257 T^{2} + 2868 T^{3} + 35165 T^{4} + 2868 p T^{5} + 257 p^{2} T^{6} + 16 p^{3} T^{7} + p^{4} T^{8}$$
97$C_2^2:C_4$ $$1 - 3 T + 12 T^{2} - 865 T^{3} + 11511 T^{4} - 865 p T^{5} + 12 p^{2} T^{6} - 3 p^{3} T^{7} + p^{4} T^{8}$$
$$L(s) = \displaystyle\prod_p \ \prod_{j=1}^{8} (1 - \alpha_{j,p}\, p^{-s})^{-1}$$ | 2,636 | 4,578 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-49 | latest | en | 0.249396 |
https://skhane.org/pyramid-angles-creating-ancient-measurements-1/ | 1,685,855,214,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00314.warc.gz | 561,964,230 | 41,490 | # Great Pyramid Angles creating Ancient Measure {1}
An investigation regarding: W: 4.14644179350061703…
The Ancient Gematria Values denoted as Gem:
GP Decimal Slant Angle 51.77956794514995499…{Apex to a base-side centre}
GP Quoin Slant Angle 41.92109147699452…{Apex to a base corner}
The time interval between one opposition and the next is just about 780 days for Mars, 399 days for Jupiter, and 378 days for Saturn. Given that the GP “face on” is a triangle and there being 180 Degrees in a triangle, we deduct the GP Slant Angle of. 51.77956794514995499… twice from 180 Degrees, to obtain 76.440864109700090016…. the Tan of this being 4.14644179350061703 … denoted as W that as a guide is the Greek Foot of 10.08 ÷ 2.431 The forthcoming number {2} I will, at this time, be extended merely to show the numerous options, given that all are interconnected, I will employ, in order, {A} 46,656, the Cubic Inches in one Cubic Yard {B} Gem Founder of the City 1225 {C} 5280 Feet, the Mile {D} Gem Divine Circle 1224 {E} 364 Maya Temple steps {F} The Stonehenge Sarsen Circle Outer Diameter of 104.2724571428…Feet {G} The Egyptian Foot of 1.14545…{H}
The Maya Earth Year of 364.9122807017…{1,872,000 ÷ 5,130} {I} the 13th PWS member 1536 {J} Gem Simon Peter 1925 {K} The British Admiralty Nautical Mile of 6080 Feet {L} the 16th PWS member 2048 {M} Gem Mary 192 {N} 1.777…which is the Square Root of the Rhind Papyrus fraction {256:81} of 3.160493827{O} Gem Abraxas, Mithras {both Solar “gods”} {P} The Maya “Super Number” 1,366,560 {Q} Gem Lord Jesus Christ 3168
It needs be borne in mind that the “Rhind” of 3.160493827…is 3.111…x 1.015873…The Eye of Horus Fraction {64:63} that multiplied by 1.25 = 1.269841…the GP Tan {480:378} The 3.111…x 18 = 56 the Aubrey holes likewise 3.111…is 1.25 x 2.4888…which is the Eye of Horus Fraction {64:63} multiplied by 2.45 this is the GP ratio 756:480 of 1.575 x 1.555…obviously the 3.111…÷ 2 The Musical Scale Pythagorean Limma {256:243} of 1.05349794238683127…is one-third of 3.160493827…Multiplying the Ancient Comma of Pythagoras {531,441:524,288} twice by the “Limma” 1.0534979423868312…obtains exactly 1.125, which is 9 ÷ 8 hence there is ample scope regarding 1.125.the Square Root of the “Comma” component 531,441 is 729 that multiplied by the “Limma” of 1.05349794238683127…= 768 the 6th PWS {Plato’s World-Soul} member hence and the 729 x 3.160493827…= 2304 the 16th PWS member.
The GP so-called “King’s Chamber” Wall height being 230.4 Inches otherwise 19.2 Feet, the 19.2 x 20 = 384 the 1st PWS member and 19.2 x 10 = 192 Gem Mary that multiplied by 31.666…= 6080 Feet the British Admiralty Nautical Mile , the 31.666…for example, is the 3.111…x 10.17857142…which is 5.9375 Royal Egyptian Cubits of 1.714285…otherwise 31.666…x 12.6 = 399 days for Jupiter and the 10.17857142…x 1.8666…= 19 the Metonic Cycle the 1.8666…is 1.777…x 1.05 the 1.777…is the Square Root of: 3.160493827…the Rhind Papyrus fraction {256:81}. I could endure forever and a proverbial day because the entire are fully, faultlessly interconnected, however I purely proposed to demonstrate this mathematical interconnection prior to commencing the calculations, with the purpose of assuring the reader that when they discover length divided by a Musical Scale, otherwise Time divided by Volume and additional apparently irrational mathematical procedures it simply demonstrates that the entire are interlinked. Such as the Platonic Precession Cycle of 25.920 Years is 0.3 x 86,400 the Seconds in 24 Hours and 86,400 is 1.98347107438016…x 43,560 the Square Feet in one Acre subsequently the Mile of 5280 Feet divided by this 1.98347107438016…= 2,662
This 2,662 multiplied by the Earth Polar Radius of 3,949.714285…Miles = 10,514,139.428571…that divided by the Stonehenge Sarsen Circle Outer Diameter of 104.2724571428…Feet obtains 100,833.333…which multiplied by the commencing Platonic Precession Cycle of 25.920 Years obtains exactly 2,613,600,000 that divided by the Cubic Inches in one Cubic Yard 46,656 = 56,018.518518…that multiplied by the Ancient, much-employed number 54 = 3,025,000 that multiplied by 0.0144 = 43,560 Otherwise 56,018.518518…is the Musical Scale Pythagorean Minor Third {32:27} of 1.185185…x 47,265.625 that multiplied by 0.1024 = 4840 the Square Yards in one Acre the 9th PWS member being 1024 the 19th being 2592, hence 0.1024 x 253,125 = 25,920 “Square One” The 253,125 is Gem Abraxas, Mithras {both Solar “gods” } namely 365 x 693.49315068…that multiplied by the 399 days for Jupiter then by the Maya Calendar number 5256 obtains exactly 1,454,355,000 which is 239,203.125 “Sea” Miles of 6080 Feet and 239,203.125 that multiplied by the 16th PWS member 2,048 = 489,888,000 which is 360 x 1,360,800 the Maya Long Count
The reader will note that on numerous occasions there will be cyclic digits emerging, most of the Ancient Units of Measure will contain the number 124578 as a cyclic in various orders such as the Royal Egyptian Cubit of 1.714285…otherwise the Long Geographical Royal Cubit of 1.73787428571…or else the Ring of Brogar base perimeter of 1,071.428571…Feet. This guarantees they will produce more Units of Measure when multiplied by the number 7 or multiples of, for example the Long Geographical Royal Cubit of 1.73787428571…x 7 = 12.16512 which is 10 x 1.216512 the Roman Remen, likewise the Polar Furlong of 617.142857…{625 ft.} x 7 = 4,320 the number 124578 added obtains 27 obviously 9 x 3 likewise 124 plus 578 = 702 obviously 7 plus 2 = 9 this being one reason why the number 3 is so important otherwise 702 is 3 x 234 which is 3 x 78 which is 3 x 26
Hence the Long Geographical Royal Cubit of 1.73787428571…x 355.113636…= 617.142857…The 355.113636…is 206.6798941…Egyptian Cubits of 1.71818…and this 206.6798941…is 162.76041666…x 1.269841…the GP Tan {480:378} the 162.76041666…is exactly 117.1875 x 1.3888…The 117.1875 x 49.152 = 5,760 the GP height in Inches {480 Feet} and the 49.152 is 32.3232…Greek Cubits of 1.52064 and the 32.3232…x 188.1 = 6,080 Feet the British Admiralty Nautical Mile, the 49.152 is much more than this I show at this point Otherwise, the Greek Stade {600 ft.} of 601.3636…is the 355.113636…multiplied by exactly 1.69344 that divided by the Arabian Hashimi Foot of 1.064448 = 1.59090…that is 1.3888…Egyptian Feet of 1.14545…
The 1.59090…x 30.8955428571…= 49.152 and the 30.8955428571…is the Roman Cubit of 1.4598144 x 21.164021…that multiplied by 9 x 399 days for Jupiter obtains exactly 76,000 which divided by the Sothic Cycle 1,460 = 52.05479452…that multiplied by 11.21894736842…= 584 the Maya Synodic period of Venus and 11.21894736842…x 121,808.219178082…= 1,366,560 the Maya “Super Number” the 121,808.219178082…multiplied by Gem Abraxas, Mithras 365 = 44,460,000 that is the Maya Temple steps 364 x 122,142.857142…this I leave as is to indicate the aforementioned cyclic number 124578 rearranged is forever present The 364 being the product of Maya Temples with 91 steps on each of their four sides obtains:91 x 4 = 364 that added to the top platform itself counting as 1 obviously obtains 365
As I assume is obvious there is no end to the interconnections, likewise the number 7 is prominent in the case of Ancient Units of Measure, however the number being the number 3, simply add, for example 25,920, 86,400, 43,560, 46,656 and so on the products will always be multiples of the number 3, likewise whilst a number ends with a number 5 or zero it is divisible by 5 for example, employing Pi as exactly 3.14159265358979323846265 this being 5 x 0.62831853071795864769253 obviously Pi x 0.2 = 0.628318530717958647692528…This indicated merely as an exercise regarding the impact of the number 5, likewise the number 7 together with the greatest mathematically enmeshed number 3 As regards the 5 one example being Gem Tree of Life 1625 is 5 x 325 which is 5 x 65
…………………………………..
W: is: 4.14644179350061703
Following is several examples of W that I extracted from a list of approximately 150 calculations employing Ancient and Modern-day Units of Measure.The British Admiralty Nautical Mile of 6080 Feet multiplied twice by Gem: Founder of the City: 1225 obtains 9,123,800,000
{30} The Belgic Foot of: 1.0861714285… ÷ W then multiplied by 9,123,800,000 = 2,390,003,616 this being 364 x 399 x 16,456 that divided by 119 = 138.285714…one-seventh of 968 that multiplied by 1.125 = 1,089 this is one-ninth of 9,801 hence 1,089 reversed The 138.285714…divided by the Royal Egyptian Cubit of 1.714285…= 80.666…this is 1.777…x 45.375 that multiplied by 256 = 11,616 this is the GP Tan {480:378} of 1.269841…x 3024 x 3.025 that multiplied by 1,600 = 4,840 the Square Yards in one Acre otherwise, 3.025 x 14,400 = 43,560 the Square Feet in one Acre
{31} The Standard Geographic Athenian Foot of 1.042724571428… ÷ W then multiplied by 9,123,800,000 = 2,294,403,471.36 this being 364 x 399 x 15,797.76 that divided by 119 = 132.75428571… one-seventh of 929.28 that multiplied by 1.125 1.125 = 1,045.44 The 132.75428571… divided by the Royal Egyptian Cubit of 1.714285…= 77.44 this is 1.777…x 43.56 the Square Feet in one Acre 43,560 obviously reduced
{32} The Sacred Foot of 1.3824 ÷ W then multiplied by 9,123,800,000 = 3,041,822,784 this being 364 x 399 x 20,944 that divided by 119 = 176 one-seventh of 1,232 that multiplied by 1.125 = 1,386 that multiplied by 981.8181…= 1,360,800 the Maya Long Count, the 981.8181…is the Royal Egyptian Cubit of 1.714285…x 572.7272…the Egyptian Stade {500ft.} otherwise 981.8181…is 1.428571…Egyptian Stades of 687.2727…{600 ft.} The 1.428571…x 7 = 10 The 176 divided by the Royal Egyptian Cubit of 1.714285…= 102.666…this is 1.777…x 57.75 that multiplied by 24 = 1,386
This {33} I will extend since the options, as with additional Units of measure are numerous {33} The Polar Foot {Root} of 9.795918367346÷ W then multiplied by 9,123,800,000 = 21,554,866,666.666… this being 364 x 399 x 148,412.698412… one-seventh of 1,038,888.888… that multiplied by 1.125 = 1,168,750 that is 1,090.8333…x 1,071.428571…the Ring of Brogar circular base and the 1,090.8333…x 19,760,000 = 21,554,866,666.666…
The 19,760,000 is the Maya Temple steps 364 x 54,285.714285…that is 31.23684764309…x 1,737.87428571…the Silbury Hill circular base The 31.23684764309…x 18,707.8736842…= 584,375 The 18,707.8736842…multiplied by the Metonic Cycle 19 then by the Greek Cubit of 1.52064 obtains exactly 540,510.879744 that is exactly 55,177.1523072 Polar Feet {Root} of 9.795918367346… and 55,177.1523072 multiplied by Gem Tree of Life 1625 = 89,662,872.4992 which is the 1,038,888.888 …multiplied by exactly 86.306508288 that divided by the Roman Cubit of 1.4598144 obtains exactly 59.1215625 that due to the end digit 5 possesses ample prospects
The 148,412.698412… divided by the Royal Egyptian Cubit of 1.714285…= 86,574.074074… this is 1.777…x 48,697.91666…one-twelfth of 584,375 as prior The 148,412.698412…is the GP Tan {480:378} of 1.269841…x 116,875 The 584,375 is 9.795918367346…x 59,654.94791666…that multiplied by the 20th PWS member 2916 = 173,953,828.125 which is Gem Simon Peter 1925 x 90,365.625 this being the below 2,677.5 x 33.75 that multiplied by 1.6 = 54 the much-employed Ancient number The 116,875 is 9.795918367346…x 11,930.9895833…that multiplied by the 6th PWS member 768 = 9,163,000 which is Gem Simon Peter 1925 x 4,760 that is the 1.777…x 2,677.5 which is Gem Divine Circle 1224 x 2.1875 that multiplied by 25.6 = 56 the Stonehenge Aubrey Holes and we understand from previous that the 1.777… is the Square Root of the Rhind Papyrus fraction {256:81} of 3.160493827
W: x 63,136.5428571…= 261,792 that is the Lunar, Yin-Yang number 1080 x 242.4 that multiplied by 2,378.61386…= 576,576 this being the GP height of 5760 {Inches} multiplied by 100.1 which is 10 x 10.01 as we will search presently, likewise Gem Eagle, Pneuma {Breath} 576 The 2,378.61386…x 989,901 = 2,354,592,240 that is 1,730.3 x 1,360,800 the Maya Long Count, the 1,730.3 is 4.1176470588235294… x 420.21571428…that multiplied by 150.24793388429…regains the commencing 63,136.5428571…and 150.24793388429…x 2.431 = 365.2527272…a passable Earth Year
The 989,901 x 18.97474…= 18,783,121.5 and the 18.97474…x 3,960 = 75,140 which divided by the 364 Maya Temple steps obtains 206.428571…which divided by the Royal Egyptian Cubit of 1.714285…= 120.4166…that multiplied by 48 = 5,780 this being the GP height of 5760 plus 20 Otherwise, the commencing 63,136.5428571…÷ 206.428571…= 305.8517647058823…that is the 4.1176470588235294…x 74.2782857142…that is exactly 43.329 Royal Egyptian Cubits of 1.714285…and this 43.329 x 6,000 = 259,974 The 63,136.5428571…x 4.1176470588235294…= 259,974 The 63,136.5428571…divided by the 4.1176470588235294… then multiplied by Gem Founder of the City 1225 obtains exactly 18,783,121.5 as we observed prior The 259,974 is the 364 Maya Temple steps multiplied by 714.2142857…that is the 74.2782857142…x 9.615384…that multiplied by Gem YHWH 26 = 250
Several more will be revealed, consequently they require no explanation, consequently the remaining calculations will be brief as feasible since within Part {1} I indicated many avenues to travel, many Units of Measure to employ whilst employing W: the number {1} will show the easiest method to find W: however at the outset the 4.1176470588235294…contains a cyclic 1176470588235294 which contains, in order, Gem: Only/Begotten Son: 1176 Gem 1764 “Fixed” to 1176 {explained within part {1} } …Gem City of god 882…Gem Church 294 and the GP measurement 235.2 indicated as 2352 these three likewise “Fixed” together and the 294 multiplied obtains: 8821176,…17642352
In that case Gem: Church 294 ÷ 4.1176470588235294…= 71.4 along with 714 x 715 = 510,510 that is Gem Divine Circle 1224 x 417.08333…that multiplied by 4.615384…= 1925 Gem Simon Peter, he who landed the 153 “Fishes” upon the Sea of Galilee and 153 x 8 = 1224, the numbers 1 to 17 add to 153 and 17 x 72 = 1224 which most appropriately is also Gem “Fishes” The 4.615384…multiplied by Gem: YHWH 26 = 120 otherwise Gem Tree of Life 1625 x 4.615384…= 7,500 The Maya Long Count of 1,360,800 is 4.615384…x 294,840 now we espy Gem Church 294 as the first three digits therefore 294,840 is 294 x 1,002.857142…that divided by W then multiplied by Gem Founder of the City 1225 = 298,350 which divided by the Maya Temple steps 364 = 819.6428571…that multiplied by 1.49333…= 1224, the 1.49333…x 2025 = 3,024 Feet the base perimeter of the GP {Great Pyramid of Giza}
The 510,510 is the 819.6428571…x 622.8444…that is the 1.49333…x 417.08333…as we observed prior, due to the cyclic 857142 within the 1,002.857142…and the cyclic 428571 within the 819.6428571…this indicates they are ratios of numerous Units of Measure The 510,510 is the product of the first 7 primes 2 x 3 x 5 x 7 x 11 x 13 x 17 = 510,510 and the sum of the prime factors of 714 equalled the sum of the prime factors of 715 in other words 714 = 2 x 3 x 7 x 17 likewise 715 = 5 x 11 x 13 as well as 2 + 3 +7 + 17 = 29 accompanied by 5 + 11 + 13 = 29 {The Man Who Loved Only Numbers: Author Paul Hoffman}
Consequently, the numbers I employ are not straightforward numbers, there is much more to them than, the Gem {Gematria Values} are of mathematical significance. Likewise, 510,510 is 10.01 x 51,000 obviously one-third of 153,000, regarding the GP “missing” Pyramidion I calculated the base side length to be 9.85359375 Feet that I obtained via dividing the 10.01 by the, most appropriate, Eye of Horus fraction {64:63} of 1.015873…that multiplied by 1.25 = 1.269841…the GP Tan {480:378}, likewise the Pyramidion 9.85359375 Feet divided by 1.575 = 6.25625 Feet, the Pyramidion height hence the Pyramidion is the entire Pyramid in miniature since 756 Feet divided by 480 Feet obtains the identical 1.575
W is 4.14644179350061703
{1} The Greek Foot of 10.08 is W x 2.431
{2} The Royal Egyptian Cubit of:1.714285… is W x 0.41343537…that multiplied by Gem 1224 x Gem 1225 = 619,905 which multiplied by the Rhind Papyrus fraction {256:81} of 3.160493827…then divided by the Mile of 5280 Feet obtains 371.061728395…which multiplied by 46,656 = 17,312,256 that is the Maya Third Sun 312 x 55,488 that is 153 x 1.777…x 204 which is the Stonehenge Aubrey Holes 56 x 3.6428571…that multiplied by 830.1176470588235294…= 3024 Feet the GP base perimeter The 830.1176470588235294… is the prior obtained 4.1176470588235294… x 201.6 and the 3.6428571…x 336 = 1224 Gem Divine Circle.
The magnificent Grand Gallery entombed within the GP has a height of 336 Inches equally the number 366, not the Inches but the number 336 x 9 = 3024 the GP base perimeter being 3024 Feet, the 336 is 12 x 28 Feet hence the GP 3024 is 28 x 108 the Lunar Yin-Yang number being 1080, the 28 x 2 = 56 Stonehenge Aubrey Holes the 1080 being 20 x 54 the Ancient, much employed number moreover 56 x 54 = 3,024. The above 201.6 will obtain: 3024, 8064, 10,080, 12,096, 24,192, 362,880, 4838.4, 5040, 72,576, 9072, 96,768, 108,864, 20,901,888, 479,001,600 {=12!} 195,955,200,000,000 the Nineveh Constant and much more we will see all these Units of Measure as we progress
{3} The Egyptian Cubit of:1.71818… is W x 2.4132730015…that multiplied by 165.335625 obtains the 399 days for Jupiter and 165.335625 multiplied by the 13th PWS member 1,536 = 253,955.52 which is the Maya Temple steps 364 x 697.68 that is Gem Divine Circle 1224 x 0.57 that multiplied by 33.333…= 19 the Metonic Cycle, the 33.333…x 3 = 100
{4} The Roman Cubit of 1.4432727…is W x 2.872944049414… that multiplied by 127.016842105263…= 364.9122807017…the Maya Earth Year {1,872,000 ÷ 5,130} and the 127.016842105263…multiplied by the 399 days for Jupiter obtains exactly 50,679.72 that is 575.8658181…x 88.00612642717905…which is the 127.01684210526…x 1.44327272… the Roman Cubit The 575.8658181…is 335.16 Egyptian Cubits of 1.71818…and the 335.16 is W x 80.83075 that is the straightforward product in {1} below 0.24171875 x 334.4 that, obviously multiplied by the Greek Foot of 1.0022727…will regain the 335.16
A change of tack is required
{5} The Longer Greek Foot of:1.01376 is W x 4.090161175722…that multiplied by 299.254710857142…= 1224, the 299.254710857142…is 2.869930555…x 104.2724571428…Feet, the Stonehenge Sarsen Circle Outer Diameter and the 2.869930555…multiplied by the Earth Polar Radius of 3,949.714285…Miles obtains 11,335.40571428…which divided by the Ring of Brogar base perimeter of 1,071.428571…Feet obtains exactly 10.579712 that is 8.6967592592…Roman Remen of 1.216512 The 8.6967592592…multiplied by the Platonic 25,920 = 225,420 which is Gem Divine Circle 1224 x 184.1666…which is 1.777…x 103.59375 that multiplied by 256 = 26,520 that divided by the 364 Maya Temple steps obtains 72.857142…that multiplied by 2.1 = 153 the “Fishes”
{6} The Shorter Greek Foot of:1.0368 is W x 3.99926870515… that multiplied by 306.055954285714…= 1224, Gem Divine Circle the 306.05595428571…is 2.93515625 x 104.2724571428…Feet, the Stonehenge Sarsen Circle Outer Diameter and the 2.93515625 multiplied by the Earth Polar Radius of 3,949.714285…Miles then divided by the Ring of Brogar base perimeter of 1,071.428571…Feet obtains exactly 10.82016 which is the {5} Longer value Greek Feet of 1.01376 x 10.673295454…that is 9.3179563492…Egyptian Feet of 1.14545…
And the 9.3179563492…multiplied by 3960 = 36,899.10714285…which divided by the Silbury Hill Circular base of 1,737.87428571…Feet obtains 21.2323224103009259259…that, due to the cyclic 925 within, multiplied by the Platonic 25,920 = 550,341.796875 that multiplied by the 4th PWS member 512 = 281,775,000 which is 3,024,000 x 93.179563492…that multiplied by W: = 386.3636…that multiplied by Gem Lord Jesus Christ 3168 = 1,224,000 due to the cyclic 36 within this 386.3636…it means it is a ratio of numerous Units of Measure It is nigh-on unfeasible to obtain a specific technique to achieve an identical product, however one alternative being
{5} The Longer Greek Foot of:1.01376 is W x 4.090161175722…that multiplied by 89.217090726817…= 364.9122807017…the Maya Earth Year, the 89.217090726817…x 50,274 = 4,485,300.0192 The 50,274 x 1.01376 = 50,965.77024 which is the GP 3024 Feet multiplied by 16.85376 that is the Longer value Greek Foot of 1.01376 x 16.625 which multiplied by 365.714285…= 6080 the British Admiralty Nautical Mile, the 365.714285…is the Royal Egyptian Cubit of 1.714285…added to the 364 Maya Temple steps
Besides this 365.714285…multiplied by 360,000 Degrees then divided by the Mile of 5280 Feet obtains 24,935.064935…that multiplied by 54.80475 = 1,366,560 the Maya “Super Number” the 54.80475 divided by Gem Abraxas, Mithras 365 obtains exactly 0.15015 that multiplied by the “Sea” Mile of 6080 Feet obtains 912.912 which is the GP Pyramidion factor 10.01 x 91.2 that is the “Sea” Mile of 6080 Feet divided by 66.666…one-third of 200
{6} The Shorter Greek Foot of:1.0368 is W x 3.99926870515… that multiplied by 91.2447518796992…= 364.9122807017…the Maya Earth Year, the 91.2447518796992…x 49,156.8 = 4,485,300.0192 The 49,156.8 x 1.0368 = 50,965.77024 that divided by the prior “near-miss” Earth Year of 365.714285…= 139.359528 which divided by the Shorter Value Greek Foot of 1.0368 = 134.413125 that multiplied by 256 = 34,409.76 which is 30,240 plus 3024 plus 1,145.76 that is Gem Mary 192 x 5.9675 that multiplied by the 256 = 1,527.68. The Greek Cubit being 1.52064, the 1,527.68 minus an increased 1,520.64 = 7.04 the Stonehenge upright Pillars measurement hence 7.04 will obtain: 21.12, 35.2, 633.6, 105.6, 1760, 5280, 19,008, 3168 and so on all Units of Measure mainly pertaining to Stonehenge
The 4,485,300.0192 is the 1st PWS member 384 x 11,680.4688 that divided by the Arabian Hashimi Foot of 1.064448 = 10,973.263888…that multiplied by the GP 5760 = 63,206,000 which divided by the “Sea” Mile of 6080 Feet then multiplied by the 399 days for Jupiter obtains exactly 4,147,893.75 which is Gem Founder of the City 1225 x 3,386.03571428…that multiplied by 1.16647153751…= 3,949.714285…Miles the Earth Polar Radius and the 1.16647153751…x 1,049.318359375 = 1224 Gem Divine Circle, The 1,049.318359375 multiplied by the 9th PWS member 1024 = 1,074,502 that, merely to end, multiplied by 4.1743058823529411…will regain the commencing 4,485,300.0192 and 4.1743058823529411…is W: x 1.00672 that plus 0.00704 = 1.01376
The 0.00704 will obtain 1760, 5280, 3168, 1.056, 21.12, 633.6, 3960,.7920, and so on all Units of Measure The 4.1743058823529411…is 4.1176470588235294… x 1.01376 prior I indicated the following “In that case Gem: Church 294 ÷ 4.1176470588235294… = 71.4”
……………………….
Otherwise, the straightforward process being as follows:
{1} The Greek Foot of 1.0022727…is W x 0.24171875
{2} The Roman Remen of 1.202727… is W x 0.2900625
{3} The Greek Cubit of 1.50340909…is W x 0.362578125
{4} The Roman Cubit of 1.4432727…is W x 0.348075
{5} The Greek Digit of 0.06264204545…is W x 0.015107421875
Prior to continuing we could, if so required, employ the “Sea” Mile of 6080 the Cubic Inches in one Cubic Yard 46,656 and so on, since the {1} produced 0.24171875 x 6080 x 46,656 = 68,567,990.4 which is the Maya Temple steps 364 x 188,373.6 that divided by the Maya Long Count of 1,360,800 = 0.138428571… that multiplied by the Greek Foot of 1.0022727…= 0.1387431818… due to the cyclic 18 within this has ample scope Otherwise, the 0.138428571…x 753.25820433436…= 104.2724571428…the Stonehenge Sarsen Circle Outer Diameter and the 753.25820433436…multiplied by the 399 days for Jupiter obtains 300,550.023529…that is W multiplied by exactly 72,483.84 which divided by the GP base perimeter of 3024 Feet then by the Rhind Papyrus fraction {256:81} of 3.160493827…= 7.58410714285…that multiplied by the prior 0.138428571…x 313,600 = 329,235.192 this is Pi x 286.929696…x 365.242139799…a tolerable Earth Solar Year that multiplied by 360,000 Degrees then divided by the Mile of 5280 Feet obtains 24,902.87316814…Miles, a tolerable Earth Equatorial Circumference.
The 286.929696…x 3960 = 1,136,241.6 which is Pi x Pi x 3.1666…x 1.46 x 24,900.93954196…whereas the Maya Earth Year component 5,130 x 3 x Phi = 24,901.54308686…hence a slight difference The 1.46 obviously the Sothic Cycle 1460 reduced and 1460 is a multiplied of the Maya 365, 584 the Maya 5256 being 1460 x 3.6 the Maya 18,980 is 1460 x 13, the Maya “Super Number” 1,366,560 is 1460 x 936 that multiplied by 2,000 = 1,872,000 the Maya Earth Year component and so on The 3.1666…x 6 = 19 the Metonic Cycle otherwise 3.1666…x 126 = 399 days for Jupiter, hence the “Sea” Mie of 6080 Feet is 3.1666…x 1,920 Gem Mary is 192 {Mary, Mare, Sea?}
Hence not one of the employed numbers has “magicked” itself out of the proverbial this air, the whole is interconnected, we could also employ Gem Founder of the City 1225 along with the “Sea” Mile of 6080 Feet to obtain different Units of Measure, yet more than enough examples are provided to indicate the interconnections, hence we move progress.
W being: 4.14644179350061703…
{6} The Shorter Greek Foot of:1.0368 is W x 0.25004571428…one-seventh of 1.75032 that is exactly 1.7265625 Longer Greek Feet of 1.01376 and this 1.7265625 x 258,048 = 445,536 which is Gem Divine Circle 1224 x 364 Maya Temple steps and the 258,048 is the GP base perimeter of 3024 Feet multiplied by 85.333…which multiplied by 36 = 3072 the 21st PWS member
{7} The Longer Roman Foot of:0.9732096 isx 0.2347095771428…one-seventh of 1.64296704 that is 1.620666… Longer Greek Feet of 1.01376 the 1.620666…x 1.5 = 2.431 that we recognise as the W product of the Greek Foot of 10.08 ÷ 2.431 = W
{8} The GP Tan {480:378} of 1.269841… is W x 0.30624842529…that multiplied by the “Sea” Mile of 6080 Feet then by Gem Founder of the City 1225 = 2,280,938.271604938…that divided by the Rhind Papyrus fraction {256:81} of 3.160493827…= 721,703.125 which is 710,426.513671875 x 1.015873…the Eye of Horus fraction {64:63}
{9} The Greek Foot of: 1.0022727… is W x 0.24171875 that is 0.365 x 0.662243150684…that multiplied by the 399 days for Jupiter then by the Maya Calendar 5256 obtains exactly 1,388,819.25 which is Gem Founder of the City 1225 x 1,133.73 that is the Maya Temple steps 364 x 3.1146428571…that divided by the Royal Egyptian Cubit of 1.714285…= 1.816875 which is the {10} 0.2900625 x 6.263736…that multiplied by the Maya Calendar First Sun 676 = 4,234.285714…this divided by the Earth Polar Radius of 3,949.714285…Miles obtains 1.0720486111…that multiplied by 5,529.6 = 5,928 which is the Maya Third Sun 312 x 19 the Metonic Cycle and the employed 5,529.6 is 460.8 plus 5,068.8 the Greek Mile, the 26th PWS member being 4608
{10} The Roman Remen of:1.202727… is W x 0.2900625 as indicated within {9}
{11} The Greek Cubit of: 1.50340909… is W x 0.362578125
{12} The Roman Cubit of:1.4432727… is W x 0.348075 that is the {11} 0.362578125 x 0.96
{13} The Greek Digit of:0.06264204545… is W x 0.015107421875 that
{14} The Common Egyptian Cubit of:0.98181… is W x 0.2367857142…one-seventh of 1.6575 that multiplied by 684 = 1,133.73 as within {9} the 684 x 8.888…= 6080 the “Sea” Mile and the 8.888…x 648 = 5760 the GP height
{15} The Common Egyptian Cubit of:1.47272… is W x 0.35517857142…one-seventh of 2.48625 that multiplied by the {14} 8.888…x 51.3 = 1,133.73 as within {9} the 51.3, obviously the Maya Earth Year component {1,872,000 ÷ 5,130} 5,130 reduced I ascertain no valid reason to continue this section since as I mentioned preceding the impact of the number 5 operates such as {16} the 0.3315 is 0.1625 x 2.04 that multiplied by 1.43 = 2.9172 as within {18} the 1.43 x 2,215.384615…= 3168 and the 2,215.384615…x 2.6 = 5760
{16} The Sacred Foot of:1.374545… is W x 0.3315
{18} The Egyptian Cubit of:1.728 is W x 0.4167428571…one-seventh of 2.9172
{20} The Shorter Greek Cubit of:1.512 is W x 0.36465
{21} The Longer Egyptian Foot of:1.152 is W x 0.27782857142…one-seventh of 1.9448
{22} The Roman Cubit of:1.4598144 is W x 0.35206436571428…one-seventh of 2.46445056
Examples regarding the absence of the end digit being a 5, the {21} produced 1.9448 is the {22} Roman Cubit of 1.4598144 x 1.33222415123456790…that multiplied by the Platonic 25,920 x 4 = 138,125 which is Gem Tree of Life 1625 x 85 that multiplied by 14.4 = 1224 Gem Divine Circle.
The {22} produced 2.46445056 is the “Sumerian” Foot of 1.056 x 2.33376 and this is the Arabian Hashimi Foot of 1.064448 x 2.192460317…that divided by the Rhind Papyrus fraction {256:81} of 3.160493827…= 0.69370814732142857…and this multiplied by 458,752 = 318,240 this being the Maya Temple steps 364 x 874.285714…that multiplied by 1.4 = 1224 Gem Divine Circle. Consequently, there being no interruption in the connectivity since the entire are totally interlinked
{23} The Shorter Roman Cubit of:1.45152 is W x 0.350064
{24} The Arabian Hashimi Foot of 1.064448 is W x 0.2567136
{25} The Standard Geographical Megalithic Rod of: 6.95149714285… is W x 1.67649697959183…that multiplied by Gem Founder of the City 1225 then by Gem Divine Circle 1224 = 2,513,739.5712 that divided by the Roman Mile of 4,866.048 = 516.5875
Regarding {23} The Shorter Roman Cubit of:1.45152 is W x 0.350064 this increased to 35,00,640 is 3,453,125 Longer Value Greek Feet of 1.01376, the 3,453,125 being 2.431 x 1,420,454.5454…which is 1,240,079.3650793…Egyptian Feet of 1.14545…and 1,240,079.365079…multiplied by 3 x 399 days for Jupiter obtains exactly 1,484,375,000 Likewise, 1,420,454.5454…is exactly 390.625 x 3,636.3636…ample scope here The 390.625 x 512 = 200,000, the 4th PWS member being 512 due to the cyclic 63 within this 3,636.3636…it will obtain numerous, however in this case it is one-eleventh of 40,000
Common Units of Measure
W being: 4.14644179350061703
Gem: Founder of the City 1225 multiplied by the British Admiralty Nautical Mile of 6080 Feet obtains exactly 7,448,000
{A} Common Mycenaean Canonical Feet of 0.905142857… ÷ W = 0.2182938775510204…that multiplied by 7,448,000 = 1,625,852.8 which is the Maya Temple steps 364 x 4,466.6285714…one-seventh of 31,266.4 that multiplied by the 399 days for Jupiter then divided by the Maya Long Count of 1,360,800 = 9.167617283950…which multiplied by the Cubic Inches in one Cubic Yard 46,656 = 427,724.352 that is 3168 x 135.014 that minus 33.638 = 101.376 obviously the Longer value Greek Foot of 1.01376 increased moreover 33.638 x 90 = 3,027.42 which minus the GP base perimeter of 3024 feet obtains 3.42 that multiplied by 1,777.777…= 6080 Feet, the British Admiralty Nautical Mile I am certain I need not clarify the properties of the 1,777.777…
This {B} calculation process will be extended
{B} Assyrian/Germanic 1.0861714285…÷ W = 0.261952653061… that multiplied by 7,448,000 = 1,951,023.36 that multiplied by the 399 days for Jupiter then divided by the Maya Long Count of 1,360,800 = 572.0593185185…that multiplied by the Ancient, much-employed number 54 = 30,891.2032 that multiplied by the {D} Megalithic Yards {S: Geographic} of 2.7802742857142…= 85,886.017911734857142…which is 1.0861714285…x 79,072.24923481481…this being the Rhind Papyrus fraction {256:81} of 3.160493827…multiplied by exactly 25,018.953859453125
Then Gem Abraxas, Mithras 365 x 25,018.953859453125 = 9,131,918.158700390625 which is the prior 1,951,023.36 x 4.6805785855379…that multiplied by the Mile of 5280 Feet then by 65,536 = 1,619,620,982.4 which divided by the Maya “Super Number” 1,366,560 = 1,185.18102564102…this multiplied by Gem YHWH 26 then by the Greek Cubit of 1.52064 then by the Metonic Cycle 19 obtains exactly 890,303.4353664 which divided by the {C} Long Geographical Royal Cubit of 1.73787428571…= 512,294.498333…that is the prior 9,131,918.158700390625 divided by 17.825524553571428…and this multiplied by the {D} Megalithic Yards {S: Geographic} of 2.780598857142…x 1225 obtains exactly 60,717.900672 which divided by the 6080 then multiplied by the 399 days for Jupiter obtains 3,984.6122316 that multiplied by 3.142857…x 2 obtains 25,046.1340272 which is exactly 1.0058125 x 24,901.39467067…Miles, a tolerable Earth Equatorial Circumference
The 1.0058125 multiplied by the PWS 4th member 512 = 514.976 this being 101.376 plus 413.6 which is 302.4 plus 111.2 that multiplied by 30 = 3,336 which is the Maya Third Sun 312 plus 3024 Otherwise, Gem Tree of Life 1625 plus 1,000 = 2625 that divided by the 399 days for Jupiter then multiplied by the 514.976 = 3,388 which is 3024 plus 364 Maya Temple steps the 514.976 divided by the 101.376 = 5.07986111…that multiplied by 131,040 = 665,665 this being Gem Weapons of god 665 x 1,001 the GP Pyramidion factor 10.01 increased likewise 665 x 9.142857…= 6080 the “Sea” Mile the 131,040 is the Maya Temple steps 364 x 360, and the 9.142857…will obtain 640 Acres on one Square Mile, 4,840 Square Yards in one Acre, 43,560 Square Feet in one Acre likewise 9.142857…will obtain 192, 384, 352, 512, 1024,1056, 1152, 1536, 1728, 1760, 2112, 2048, 2368, 3072, 4096, 5280, 5760, 6080, 704, 768, 86,400, 25,920, 272, etc.
Likewise: 9.142857… denoted as S will obtain
S x 117.1875 = 1,071.428571…Ring of Brogar circular base
S x 190.08 = 1,737.87428571…Silbury Hill circular base
S x 11.4048 = 104.2724571428…Stonehenge Sarsen Circle Outer Diameter
S x 0.76032 = 6.95149714285…Standard Geographical Megalithic Rod
S x 0.1188 = 1.0861714285…Assyrian/Germanic Canonical Unit of Measure
S x 0.297 = 2.715428571…Astronomical Megalithic Yard
S x 79.2 = 724.1142857…Egyptian Furlong
S x 76.032 = 695.149714285…Egyptian Geographic Stade
S x 63.36 = 579.29142857…Egyptian Geographic Stade
S x 0.108 = 0.987428571…Roman Polar Foot {Geographical}
S x 432 = 3,949.714285…Miles Earth Polar Radius
S x 432.98181…= 3,958.69090…Miles Earth Equatorial Radius
S x 0.1188 = 1.0861714285…Drusian Foot
S x 0.162 = 1.481142857…Polar Cubit {Tropical}
S x 64.8 = 592.4571428…Polar Stade {600 ft.}
S x 67.5 = 617.142857…Polar Furlong {625 ft.}
S x 0.57024 = 5.21362285714…One Roman Pace
S x 0.12 = 1.09714285…Sumerian Root
S x 0.1056 = 0.9654857142…Roman Foot Root {Canonical}
S x 4.5625 = 41.714285…Maya STU
S x 0.1875 = 1.714285…Egyptian Royal Cubit
S x 0.2376 = 2.1723428571…Geographic Megalithic Cubit
{C} The Long Geographical Royal Cubit of 1.73787428571…÷ W = 0.419124244897… that multiplied by 7,448,000 = 3,121,637.376 that multiplied by the 399 days for Jupiter then divided by the Maya Long Count of 1,360,800 = 915.29490962962…that multiplied by the Platonic 25,920 = 23,724,444.0576
{D} Megalithic Yards {S: Geographic} of 2.780598857142…÷ W = 0.6705987918367… that multiplied by 7,448,000 = 4,994,619.8016 which is the Maya Temple steps 364 x 13,721.4829714285…that one-seventh of 96,050.3808 that multiplied by the 399 days for Jupiter then divided by the Maya Long Count of 1,360,800 = 28.162920296296…that multiplied by the Square Feet in one Acre 43,560 = 1,226,776.80810666…which is 1.777…x 690,061.95456 that is the commencing 2.780598857142…x 248,170.2647569444…that multiplied by the 16th PWS member 2,304 = 571,784,290 this being Gem Simon Peter 1925 x 297,030.8 which is the Square Yards in one Acre 4,840 x 61.37 that is the GP reduced 30.24 plus the GP 10.01 plus 21.12 the Persian Cubit
{E} Royal Athenian Geographic of 1.042724571428… ÷ W = 0.2514745469387… that multiplied by 7,448,000 = 1,872,982.4256 which minus the Maya Earth Year { 1,872,000 ÷ 5,130 } component 1,872,000 = 982.4256 this being the W:{10.08 ÷ 2.431} component 10.08 x 97.46285714…that is one-seventh of exactly 682.24 that minus a reduced Greek Mile {5068.8} of 506.88 = 175.36 that minus 103.68 = 71.68 which is the W: 10.08 plus 30.24 plus 30.24 plus 1.12 that multiplied by 2,700 = 3024 Feet the GP base perimeter
{F} Roman Pace of 5.21362285714… ÷ W = 1.2573727346… that multiplied by 7,448,000 = 9,364,912.128 which divided by the Stonehenge Sarsen Circle Outer Diameter of 104.2724571428…Feet obtains 89,811.9444…that multiplied by 36 = 3,233,230 which is the 6080 x 531.78125 that multiplied by 256 = 136,136 which is Alexander Thom Megalithic Yard of 2.72 x 50,050 which is the GP Pyramidon factor 10.01 x 5,000
{G} Astronomical Megalithic Yard of 2.715428571…÷ W = 0.6548816326… that multiplied by 7,448,000 = 4,877,558.4 which divided by the Earth Polar Radius of 3,949.714285…Miles obtains 1,234.914236111…that multiplied by 72.7272…will regain the {F} 89,811.9444… The 72.7272…x 79.2 = 5760 the GP height. Otherwise, we need but add to 614.5454… 72.7272…thereby obtaining 687.2727…the Egyptian Stade {600 ft.} The 614.5454…for example, being Egyptian Feet of 536.5079365…that multiplied by the 399 days for Jupiter obtains 214,066.666…one-third of 642,200 which multiplied by 2.916509538461… will regain the {F} 1,872,982.4256 And this 2.916509538461…multiplied by the Maya First Sun: 676 obtains exactly 1,971.560448 which is 195.5913142857…Greek Feet of 10.08 and 195.5913142857…divided by the Astronomical Megalithic Yard of 2.715428571… = 72.0296296…due to the cyclic 296 within there is ample scope regarding 72.0296296…
At commencement I indicated: {1} The Greek Foot of 10.08 ÷ W = 2.431
Gem Divine Circle 1224 is the 72.0296296…x 16.993006… that multiplied by the 2.431 = 41.31 which is the 10.08 x 4.0982142857…that multiplied by 298.666…= 1224, the 298.666…is one-third of 896 that is the Greek Foot of 10.08 x 88.888…which multiplied by 34.02 = 3024 Feet the GP base perimeter and this 34.02 plus 7.29 regains the 41.31 the Square Root of the Comma of Pythagoras {531,441:524,288} component: 531,441 is 729
Ancient Sites
W being: 4.14644179350061703
When drafting these essays, I often question, as I have for many a year, how do I explain the interconnectedness to the reader in a transparent, succinct manner, however I realize full well this is unfeasible since every one of the numbers, for this is all they truly are, “numbers” are totally, flawlessly interlinked. Today we have no tangible perception why this is so, nonetheless without seeking, we will not find, hence I maintain my search for the truth of “Number Codification” codified by whom, likewise during which time-period. Obviously, I realize the numbers 1 to 10 are all that is required, likewise a number cannot be created without utilizing, as I dub it, the Numerical Alphabet parallel to the Letters of the Alphabet. Nonetheless the following ought to suffice, taking into consideration the numbers I will employ, similarly the manner via which they are employed, are, not in the slightest, the exclusive options. Therefore, I will expand this section afterwards, as I have mentioned prior, I will ease off as well as limit the calculations, if at all feasible. In addition to, I genuinely hope the reader recognizes my reasoning, since I realize the reader is more than competent to continue calculating in their own style
{G} The Stonehenge Sarsen Circle Outer Diameter 104.2724571428…Feet. is W x 25.1474546938…that multiplied by 6080 x 1224 x 1225 = 229,253,048,893.44 that divided by the Square Feet in one Acre 43,560 = 5,262,925.824 this divided by the GP base Area {756 x 756} of 571,536 Square Feet obtains 9.2083890148652…that multiplied by 2.730928792569…will regain the 25.1474546938The 2.730928792569…multiplied by the 399 days for Jupiter obtains 1,089.64058823529…that multiplied by 1.079254951308…= 1176 Gem Only/Begotten on
Prior to commencing I indicated: “In that case Gem: Church 294 ÷ 4.1176470588235294 …= 71.4”
The 1,089.6405882352…÷ 4.1176470588235294…obtains exactly 264.627 that multiplied by the 1.079254951308…obtains 285.6 which multiplied by 12,960 = 3,701,376 this is the GP base perimeter of 3024 Feet multiplied by the 1224 Gem Divine Circle. I am certain the 12,960 requires no explanation
{I} The Stonehenge Sarsen Circle Inner Diameter 97.32096 Feet is W x 23.470957714285…that multiplied by the Silbury Hill base perimeter 1,737.87428571…. Feet then by Gem: 1225 obtains exactly 49,967,227.994112 which is 49,289,011.2 Longer Value Greek Feet of 1.01376 The 49,289,011.2 being 46,304,761.904761…Arabian Hashimi Feet of 1.064448 and this 46,304,761.904761…divided by the Rhind Papyrus fraction {256:81} of 3.160493827…= 14,651,116.0714285…that divided by the Ring of Brogar perimeter. 1,071.428571…Feet obtains exactly 13,674.375 this is the Maya Temple steps 364 x 37.56696428571…that multiplied by 1,792 = 67,320 which is the commencing 23.470957714285…x 2,868.22552447
The cyclic 552447 within provides a clue since 2,868.22552447…multiplied by the Maya Calendar number 18,980 = 54,438,920.45454…which is 47,526,041.666…Egyptian Feet of 1.14545…and 47,526,041.666… divided by the Stonehenge Sarsen Circle Inner Diameter 97.32096 Feet obtains 488,343.32980959…that multiplied by 5,839.2576 = 2,851,562,500 this divided by the Cubic Inches in one Cubic Yard 46,656 then multiplied by the Roman Cubit of 1.4598144 x 532.6704545…= 47,526,041.666…as prior This 532.6704545…multiplied by the 5,839.2576 = 3,110,400 which is the Platonic 25,920 x 120 Whilst the 49,967,227.994112 produces an exact number of Longer Value Greek Feet of 1.01376 there are additional Units of Measure that can be employed since via continuously, adding the 1.01376 we will obtain: 5068.8, 608.256, 9.12384, 1.216512, 1.52064, 364953.6, 456192, 4866.048, 729.9072, 97.32096, 1.4598144, and so on, all Units of Measure
Continuing the Ancient Sites
W being: 4.14644179350061703
{J} the Ring Of Brogar perimeter. 1,071.428571…Feet is W x 258.3971088435…that multiplied by 1224 x 1225 = 387,440,625 which is Gem Tree of Life 1625 x 238,425 which is Gem Simon Peter 1925 x 123.857142…which is 71.26933396464……Geographic Royal Cubits of 1.73787428571…
The 71.26933396464…divided by the 399 days for Jupiter then multiplied by 2,169,078,912 will regain the 387,440,625 and this 2,169,078,912 is the GP base Area of 571,536 Square Feet obtains 3,795.174603…which is exactly 3,735.875 x 1.015873… the Eye of Horus fraction {64:63} which multiplied by 1.25 = 1.269841…the GP Tan {480:378} that multiplied by 2.4888…= 3.160493827…the Rhind Papyrus fraction {256:81} that multiplied by 1,200.81696428571…will regain the 3,795.174603…The 1,200.81696428571…multiplied by W: 4.14644179350061703… = 4,979.1176470588235294…which is the prior employed 4.1176470588235294… x 1,209.2142857…that multiplied by 0.9930555… will regain the 1,200.81696428571…that multiplied by 1,792 = 2,151,864 which is the 399 days for Jupiter multiplied by 5,393.142857…which divided by the Earth Polar Radius in Miles of 3,949.714285…Miles obtains 1.3654513888…that multiplied by the 1,792 x 21.6 = 52,852.8 which is the GP Pyramidion factor 10.01 x 5,280 Feet the Mile, the 1,792 x 1.6875 = 3024 Feet the GP base perimeter and the 1.6875 x 256 = 432 hence 1.6875 will obtain numerous
And the 0.9930555…x 26.1818…= 26 or 0.9930555…x 3,190.153846…= 3168 accordingly this 3,190.153846…x 26 = 82,944 a Musical Scales number that is a multiple of 62,208, 41472, 27648, 20736, 13824, 18432, more Musical Scales number likewise Units of Measure 82,944 is also a multiple of, 9216, 6912, 5184 ,6144, 3456, 2592, 2304, 10368, all PWS members. The 26.1818…divided by the Egyptian Cubit of 1.71818…= 15.238095…that divided by the Rhind Papyrus fraction {256:81} of 3.160493827…= 4.82142857…which is 2.8125 Royal Egyptian Cubits of 1.714285…and the 2.8125 multiplied by the 14th PWS member 2048 = 5760 the GP height, the 5th PWS member being 576 likewise Gem Eagle, Pneuma {Breath} being 576
The 4.82142857…x 11.2 = 54 the Ancient, much employed number the 11.2 x 270 = 3024 Feet the GP base perimeter hence the 4.82142857…will obtain 1080, 2160, 378, 432, 5130, 594, 729, 756, 86,400, 12,960, 25,920, and so on, simpler being 4.82142857…x 1.8666…= 9 and the 1.8666…x 30 = 56 Aubrey Holes Likewise, the 26.1818…x 14.666…= 384 the 1st PWS member and the 14.666…x 360 = 5280 Feet the Mile, the 56 multiplied by the 54 = 3,024 Feet, the GP base perimeter
Likewise, a Cubit attributed to Sir Isaac Newton: 2.0736 x 40,000 = 82,944 moreover 82,944 x 1,584 = 131,383,296 Feet, the NC {Numerical Canon} Earth mean Circumference otherwise the NC Moon Circumference of 35,831,808 Feet is 82,944 x 432, likewise 82,944 will obtain: 93,312,000, 4,561,920,000, 479,001,600 {=12!} 195,955,200,000,000 the Nineveh Constant and so on. Equally 82,944 is 30240 plus 3024 plus 19,440 which is 3024 plus 3024 plus 3024 plus 10,368, then again, the 19,440 is 3024 plus 4320 plus 12,096 the Roman Remen {Canonical} is 1.2096 that obtains: 2.4192, 362,880, 4838.4, 3024, 72,576, 9.6768, 1.45152, 399,168,000, 79,833,600, 108,864 and so on
{1} 12,096 is the Persian Cubit of 21.12 x 572.7272…the Egyptian Stade {600 ft.}
{2} 12,096 is the “Sumerian” Foot of 1.056 x 11,454.5454…which is 10,000 Egyptian Feet of 1.14545…
{3} 12,096 is Gem Lord Jesus Christ 3168 x 3.8181…that multiplied by 6.857142…will regain the prior 26.1818…the 6.857142…will be seen below Otherwise, the GP height of 5760 is 3.8181…x 1,508.571428…one-seventh of 10,560 Then again, the 3.8181…is 2.222…Egyptian Cubits of 1.71818…and 2.222…x 4.5 = 10
{4} 12,096 is the Mile of 5280 Feet multiplied by 2.29090…that multiplied by 4.317460…= 9.89090…the Megalithic Foot and 4.317460…is 1.3660714285…x 3.160493827…the Rhind Papyrus fraction {256:81} we previously realize the connection between 3.160493827…as well as the Eye of Horus fraction, the GP Tan, likewise the Musical Scale the Pythagorean Limma {256:243} of 1.0534979423868…is one-third of 3.160493827…The 1.3660714285…x 896 = 1224 Gem Divine Circle, the 896 x 13.5 = 12,096
Of additional significance, the 4.317460…x 914.8235294117…= 3,949.714285…Miles, the NC {Numerical Canon} Earth Polar Radius in addition the 914.8235294117…is W: 4.14644179350061703… x 220.6285714…that is the Stonehenge Sarsen Circle Outer Diameter of 104.2724571428…Feet multiplied by 2.11588541666…that multiplied by the Roman Cubit of 1.4598144 obtains exactly 3.0888 which is the prior 896 divided by 290.080290…that multiplied by the 2.431 = 705.185185…which is the Musical Scale Pythagorean Minor Third {32:27} of 1.185185…x 352.592592…that due to the cyclic 592 within, multiplied by the Platonic 25,920 = 9,139,200 which is the 12,096 x 755.555…one-ninth of 6,800 hence 755.555…x 1.62 = 1224 Gem Divine Circle
Otherwise, the 705.185185…multiplied by the Ancient, much-employed number: 54 = 38,080 that is the 12,096 x 3.148148…which multiplied by 388.8 = 1,224, the 388.8 being 0.15 x 2592, otherwise 3.148148…is the Musical Scale: the Pythagorean Limma {256:243} of 1.0534979423868…multiplied by exactly 2.98828125 which is the GP ratio {756:480} of 1.575 x 1.89732142857…that multiplied by 645.12 = 1224 the 645.12 is the Shorter value Roman Cubit of 1.45152 x 444.444…one-ninth of 4,000 hence 12.96 x 444.444…= 5760 the GP height {Inches} or the base perimeter of 3024 Feet is 444.444…x 6.804 that multiplied by 1,777.777… regains the 12,096, the Square Root of the Rhind Papyrus fraction {256:81} is 1.777…Then again, the 3.148148…x 666 x 20,000 x 3.142857…= 131,790,476.190476…which is the Mile of 5280 Feet multiplied by 24,960.317460…that is the prior {3} highlighted 6.857142…x 3,640.04629629…that resembles the Maya Earth Year {1,872,000 ÷ 5,130} of 364.9122807017…
Therefore, the Maya 5130 multiplied by the 3,640.04629629…obtains exactly 18,673,437.5 that divided by Gem Tree of Life 1625 = 11,491.3461538…that multiplied by the Maya Long Count of 1,360,800 then by Gem YHWH 26 = 406,573,020,000 this being the 12,096 x 33,612,187.5 which is the 3,640.04629629…multiplied by exactly 9,234 that is 9 x 1026 that plus 4,104 = 5,130 the 4,104 is the 3024 plus 1080 Then again, the Platonic 25,920 x 3,640.04629629…= 94,350,000 that divided by the Maya Long Count of 1,360,800 = 69.334215167…that multiplied by the 399 days for Jupiter obtains 27,664.3518518…which divided by the prior 705.185185…= 39.229910714285…that multiplied by the GP base perimeter of 3024 Feet obtains 118,631.25 that multiplied by 283.333…will regain the 33,612,187.5
This 283.333…x 42.69176470588…= 12,096 in addition the 42.691764705882…is W: 4.14644179350061703… x 10.296 that multiplied by the 56 Aubrey Hole increased to 5,600 = 57,657.6 which is the GP 5760 x 10.01 And the 42.691764705882…is the prior 4.1176470588235294… x 10.368
W is 4.14644179350061703
We can employ, Gem Founder of the City 1225 with the Musical Scale Pythagorean Limma {256:243} of 1.0534979423868…with 729 which is the Square Root of 531,441 a Comma of Pythagoras component {531,441:524,288} along with the Stonehenge Sarsen Circle Outer Diameter of 104.2724571428…Feet, the 364 Maya Temple steps to obtain 2,431,000 the product of the Greek Foot of 10.08 divided by W being 2.431
{K} the Silbury Hill base perimeter 1,737.87428571…. Feet is W x 419.1242448979…that multiplied by 1225 = 513,427.2 which multiplied by the Musical Scale Pythagorean Limma {256:243} then by 729 = 394,312,089.6 this being the Maya Temple steps 364 x 1,083,274.9714285…that divided by the Stonehenge Sarsen Circle Outer Diameter of 104.2724571428…Feet obtains 10,388.888…and this multiplied by 234 = 2,431,000 the 234 x 3,234 = 756,756 which is the GP base-side length of 756 Feet multiplied by 1,001 this being 100 x 10.01 that I divided by the Eye of Horus fraction {64:63} of 1.015873…to obtain exactly 9.85359375 Feet, the GP “missing” Pyramidion base-side length
{L} the Earth Polar Radius 3,949.714285…Miles is W x 952.5551020408…that multiplied by 1225 = 1,166,880 which multiplied by the Musical Scale Pythagorean Limma {256:243} then by 729 = 896,163,840 this being the Maya Temple steps 364 x 2,461,988.571428… that divided by the Stonehenge Sarsen Circle Outer Diameter of 104.2724571428…Feet obtains 23,611.111… and this multiplied by 102.96 = 2,431,000 The 102.96 multiplied by the 56 Aubrey Holes increased to 5,600 = 576,576 which is the GP height of 5760 {Inches} multiplied by 100.1 this being 10 x 10.01
{M} the Earth Equatorial Radius 3,958.69090…Miles. is W x 954.72 that multiplied by 1225 = 1,169,532 which multiplied by the Musical Scale Pythagorean Limma {256:243} then by 729 = 898,200,576 this being the Maya Temple steps 364 x 2,467,584…that divided by the Stonehenge Sarsen Circle Outer Diameter of 104.2724571428…Feet obtains 23,664.77272…and this multiplied by 102.7265306122… = 2,431,000 The 102.7265306122…multiplied by Gem Simon Peter 1925 = 197,748.571428…which divided by the {K} Silbury Hill base perimeter 1,737.87428571…. Feet obtains 113.787615740740…that multiplied by the Roman Mile of 4,838.4 = 550,550 which is the {L} 100.1 x 5,500
Resuming the Ancient Units of Measure
W is 4.14644179350061703
We can employ the 1225 with GP Tan {480:378} of 1.269841…with the Rhind Papyrus fraction {256:81} of 3.160493827…with the Eye of Horus fraction {64:63} of 1.015873…along with the GP “missing” Pyramidon base-side length of 9.85359375 Feet
{26} The Assyrian/Germain Canonical of : 1.0861714285… is W x 0.2619526530612…that multiplied by 1225 = 320.892 this multiplied by the GP Tan of 1.269841…then by Rhind Papyrus fraction of 3.160493827…= 1,287.8440446796002…which is Eye of Horus fraction of 1.015873…x 1,267.72148148…and this multiplied by the Pyramidon base-side length of 9.85359375 Feet obtains 12,491.6124666…that multiplied by 194.61058422097…= 2,431,000 The 194.61058422097…x 1,872.2722685333…= 364,364 which is the Maya Temple steps 364 x 1,001 the 1,872.2722685333…x 5760 = 10,784,288.266752 which is 9,918.79777…x 1,087.2576 that is 1,001 x 1.0861714285… The Assyrian/Germain Canonical
The 9,918.79777…is 1.777…x 5,579.32375 that multiplied by 256 = 1,428,306.88 which is Gem Tree of Life 1625 multiplied by exactly 878.95808 this multiplied by 1,065.340909…= 936,390 which is 935.4545…x 1001 The 1,065.340909…x 7.6032 = 8,100 and the 7.6032 is 7 x 1.0861714285…the 7.6032 will obtain 145.98144, 15.2064, 608.256, 4,561,920,000, 4866.048, 91.2384, 97.32096, 106.4448, 11,404,800, 121.6512, 190.08, 364953.6, 729.9072, 479,001,600 {=12!} and considerably more, all being Units of Measure some increased and some decreased.
The 935.4545…is 816.666…Egyptian Feet of 1.14545…and the 816.666…x 1.5 = 1,225 The 935.4545…is 544.444…Egyptian Cubits of 1.71818…and the 544.444…x 2.25 = 1225, so on and so forth. Hence, I do consider the time is satisfactory to decrease the lengthy calculations at that point the reader can, if so desired, verify the Units of Measure that are able to be produced. I will employ Gem: Founder of the City 1225 with a few more numbers that interconnect. For those to whom Gematria Values that I denote as Gem: do not have a place in calculations as well as have no bearing on Modern-Day Units of Measure, let alone Ancient Units of Measure, this needs to be addressed prior to commencing.
The Gem Founder of the City 1225 is 49 x 25 which is a component of the conversion ratio: Greek to Roman Measurements 25:24 that obtains: 1.041666…hence the Greek Foot of 10.08 is 9.6768 x 1.041666…= the Shorter value Roman Foot being 0.96768, the Roman Furlong of 608.256 x 1.041666…= 633.6 the Greek Furlong. Equally 1225 is 7 x 175 a component of the Longer to Shorter Unit of Measure ratio 176:175 that obtains 1.00571428…which is the Square Root of 1.0114612244897…that multiplied by the 1225 = 1,239.04 which is the 1.005714285…x 1,232 this is the Greek Foot of 10.08 multiplied by 122.222…that multiplied by 43.2 = 5280 Feet the Modern-day Mile. The 1,239.04 is 176 x 7.04 Feet the Stonehenge upright Pillars measurement and 7.04 will obtain 10.56, 21.12, 3168, 5280, 633.6, and so on
Gem Simon Peter 1925 x 2.7428571…= 5280, the 2.7428571…is the prior 1.00571428…x 2.7272…which multiplied by 5280 = 14,400 Gem Lord Jesus Christ 3168 x 1.666…= 5280 Feet, the Mile, the 1.666…is 1.6 x 1.041666…{25:24} yet howbeit 3168 is 1925 x 1.64571428…that will obtain 5760, 1152, 1728, 2304, 3456, 4608, 41,472, 6336, 6912, 86,400, 10,368, 12,096, 13,824, 184,32, 20,736, 24,192, 2,7648, 25,920, 362,880, and much more. All PWS members, Units of Measure albeit several being increased, Musical Scales numbers and so on
Likewise, 1925 x 1224 = 2,356,200 which is 153 x 15,400 that is the Maya Temple steps 364 x 42.307692…that multiplied by 124.8 = 5,280 Feet, the Mile the Maya First Sun 676 is the 124.8 x 5.41666…that multiplied by 300 = 1625 Gem Tree of Life various cultures from the Sumerians onward venerated a Tree of Life, as did the Maya. The 1625 is Gem YHWH 26 x 62.5 that multiplied by 21,772.8 = 1,360,800 the Maya Long Count the 21,772.8 is 3168 x 6.87272…a reduced Egyptian Stade {600 ft.} of 687.2727…
Gem Divine Circle 1224 is 8 x 153 Gem the “Fishes” the numbers 1 to 17 add to 153 and 17 x 72 = 1224, therefore are we to understand that since Simon Peter 1925 landed 153 Gem: “Fishes” upon the Sea of Galilee that he gained 153, if so 1925 x 153 = 294,525 which is the 1224 x 240.625. Prior we see “Gem Lord Jesus Christ 3168 x 1.666…= 5280 Feet, the Mile” this 240.625 is the 1.666…x 144.375 that multiplied by 13.333…= 1925 the 13.333…divided by the prior indicated 1.041666…{25:24},obtains 12.8 that multiplied by 95.625 = 1224 and the 95.625 x 35.497159090…= 3168 Gem Lord Jesus Christ The 35.497159090…x 131.9444…= 4,683.6529356060… as we observe below
The 3168 x 0.0608 = 192.6144 which is 131.9444…x 1.4598144 the Roman Cubit, the 0.0608 which is the British Admiralty Nautical Mile of 6080 Feet obviously decreased and 6080 is Gem Mary 192 x 31.666…that divided by the prior 13.333…= 2.375 which multiplied by 2,560 = 6080, or else 31.666…x 12.6 = 399 days for Jupiter
A few “Divine Circles” being the Earth Solar Year, the Earth Equatorial Circumference, the Stonehenge Sarsen Circle, and the Platonic Precession Cycle of 25,920 Years. Hence 1224 divided by the Stonehenge Sarsen Circle Outer Diameter of 104.2724571428…Feet obtains 11.7384785353…that multiplied by the 399 days for Jupiter obtains 4,683.6529356060…this being Pi x 4.08181…x 365.2423911231…tolerably near the two forthcoming Earth Solar Years
In addition to this 365.2423911231…multiplied by 360,000 Degrees then divided by the Mile of 5280 Feet obtains 24,902.89030384…Miles a tolerable Earth Equatorial Circumference The 4.08181…multiplied by Gem Abraxas, Mithras {Both Solar “gods”} namely 365 = 1,489.86363…that multiplied by the Yards in one Mile 1760 = 2,622,160 which is Gem 3168 x 827.70202…that divided by the 4.08181…= 202.777…that multiplied by 1.8 = 365
The Maya Earth Year being 1,872,000 ÷ 5,130 = 364.9122807017…
And exactly 2,200,770,000 divided by the Maya 5256 ÷ 364.9122807017…= 1,147.4422089041095…this is 365.2421989187444823444437169844…x Pi
The 1,872,000 days divided by the 13 Baktun Cycle obtains 144,000 the Maya Long Count calendar measures 1,872,000 days The 1,147.4422089041095…multiplied by the Maya Calendar 5,256 x 9,999 x = 60,303,531,543.75 which is the 2,200,770,000 x 27.401105769230…that multiplied by Gem Tree of Life 1625 x 256 = 11,398,860 which is exactly 1,874.8125 “Sea” Miles of 6080 Feet and 1,874.8125 multiplied by the 13th PWS member 1,536 = 2,879,712 which is the Maya Earth Year component: 1,872,000 x 1.538307692…that multiplied by 3,952.3952…= 6080 Feet The 3,952.3952…multiplied by the 9,999 = 39,520,000 which is the Maya Third Sun 312 x 126,666.666…that multiplied by 3.15 = 399,000 obviously the 399 days for Jupiter increased
Whereas to obtain an improved approximation the 1,872,000 divided by 5,125.366 “Tropical Years” obtains 365.24220904419…that multiplied by 6,025,508.40375 regains the 2,200,770,000 which is 361,968.75 British Admiralty Nautical Miles of 6080 Feet and 361,968.75 x 5.1717…= 1,872,000. The 5,125.366 is but the Maya 5,130 minus 4.634 which is the GP base perimeter of 3024 Feet reduced to 3.024 plus 1.61 that multiplied by 3,600 = 5,796 which minus the GP height of 5760 {Inches} plus 36 obviously 3,600 ÷ 100. Therefore, the 1,872,000 divided by the {now abridged} Pi obtained Earth Year of 365.2421989187…= 5,125.3661420882647764…hence the 5,125.366 is in the vicinity but not entirely precise whilst employing Pi
The 5.1717…multiplied by 99 = 512 the 4th PWS member equally 5.1717…multiplied by exactly 991.03756640625 = 5,125.366 obviously the 991.03756640625 multiplied by the 6080 will regain the 6,025,508.40375 The 5.1717…multiplied by 3.7125 = 19.2 Feet the Wall height of the GP so-called King’s Chamber” otherwise 230.4 Inches the 1st PWS member 384 being 20 x 19.2 the 16th PWS member being 2304 and the 3.7125 obtains: 594 Carnac Menhirs, 237,600, 13,1383,296, 11,404,800, 479,001,600 {=12!} 4,561,920,000 the area of Atlantis {minus the Central Shrine} and so on
The 2,200,770,000 divided by the Maya Long Count of 1,360,800 = 1,617.2619047…which is the Rhind Papyrus fraction {256:81} of 3.160493827…x 511.711774553571428…this multiplied by 40,754.370850202…= 20,854,491.428571 …Miles, the Earth Polar Radius. The 40,754.370850202…multiplied by the 399 days for Jupiter then by Gem: YHWH 26 obtains exactly 422,785,843.2 that is the 2,200,770,000 x 0.1921081454218…that multiplied by 156,804,862.5 = 30,123,491.328
The 1,617.2619047…x 3,402 = 5,501,925 which is the Maya Temple steps 364 x 15,115.17857142…one-seventh of exactly 105,806.25 which is Gem Tree of Life 1625 x 65.11153846…that multiplied by 29.5646523716…= 1925 Gem Simon Peter and 151,666.666……divided by this 29.56465237166…obtains the Maya Earth component 5,130 that multiplied 3 x Phi = 24,901.54308686…another Earth Equatorial Circumference that multiplied by the Mile of 5280 Feet then divided by 360,000 Degrees produces an Earth Year of 365.2226319406…The 151,666.666…is one-third of 455,000 that , for example, is 1250 x 364 the Maya Temple steps so on and so forth
The 156,804,862.5 is 136,656 x Pi x 365.2421989187444823444437169844…obviously the 136,656 is the Maya “Super Number” 1,366,560 divided by 10 And the 30,123,491.328 is the Maya 5256 x 26 x 220.43299473…that multiplied by the Maya Earth Year of 364.9122807017…then by the 399 days for Jupiter then by the Maya 18,980 obtains exactly 609,163,935,744 that is the 2,200,770,000 divided by 276.7958195286…which divided by Pi then multiplied by 4.14545…obtains 365.24292444998…another Earth Solar Year version that is the Maya Earth Year component 1,872,000 divided by 5,125.3559608827769…therefore, once more the exact 5,125.366 is in the vicinity but not entirely precise whilst employing Pi
The 4.14545…x 1,466.666…= 6080 Feet, the “Sea” Mile and the 1,466.666…x 3.6 = 5280 Feet, the “Land” Mile Yet by all accounts the Ancients had no knowledge of true Pi the GP base perimeter is 3024 Feet which is 1.5 x 672 Feet however within my website essay “Pi in the Rhind Papyrus” I wrote:
“Employing the GP 672 Feet as the Diameter of a Circle, hence 672 Feet divided by 9 = 74.666…subsequently 672 minus 74.666…= 597.333…that multiplied by 597.333…= 356807.111…the Ahmes Square Area. The GP Circle Diameter of 672 Feet is obviously 336 Feet Radius, therefore 336 Feet x 336 Feet x Pi = 354673.24421967329…the GP Circle Area.
However, actual Pi divided by the Rhind Papyrus 3.160493827…= 0.994019550549…and this multiplied by the 356807.111…= 354673.24421967329… the same GP Circle Area Hence the 597.333…x 597.333…= 356807.111…this is the 354673.24421967329…x 0.994019550549…that multiplied by the Rhind Papyrus fraction 3.160493827…= Pi
Of interest is the 597.333…x 9.6428571…= 5760 the GP height in Inches and the 9.6428571…is 5.625 Royal Egyptian Cubits of 1.714285…the GP ratio is 756:480 = 1.575 this is the 5.625 ÷ 3.571428…which multiplied by 0.48 = 1.714285…
{A} Pi divided by the Pythagorean Limma {256:243} of 1.05349794238…then multiplied by the Ahmes produced Square Area of 356807.111…= 1064019.73265901…= which is 3 x 354673.24421967329… the GP Circle Area
The 1.05349794238…is 1.777…x 0.592592…that due to the cyclic 592 within multiplied by the Platonic Precession of 25,920 = 15,360 the 13th PWS member is 1536 and this 15,360 divided by the Pythagorean Limma obtains 14,580 the 12th PWS member is 1458, the 19th is 2592. Then again, the 15360 divided by the NC {Numerical Canon} Earth mean Circumference of 24883.2 Miles obtains 0.617283950…that multiplied by the Cubic Inches in one cubic Yard 46,656 = 28,800, the highlighted 1728 x 27 = 46,656, the 14th PWS member is 1728. The 1.777…is the Square Root of the Rhind Papyrus fraction of 3.160493827…
The fact I am attempting to express being that it creates no distinction which number is divided by, multiplied by or additional mathematical procedures by further numbers, since the entire are totally, flawlessly interrelated. Hence whilst the reader may observe a division of time by, for example, an Ancient Unit of Measure, however there is a legitimate explanation, all number is circular, the “Whole” is one large numerical circle
With regards to the commencing 672 Feet as written within “Employing the GP 672 Feet as the Diameter of a Circle” The 672 x 4.5 = 3024 Feet, the GP base perimeter, likewise the awe-inspiring Grand Gallery “entombed” within the GP is 28 Feet otherwise 336 Inches, the 672 is 24 x 28. And the number 336 {not the Inches} the “number” 336 x 2 = 672
Until the next essay | 22,250 | 63,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-23 | latest | en | 0.768929 |
http://topcoder.bgcoder.com/print.php?id=1365 | 1,716,769,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00578.warc.gz | 26,048,708 | 2,116 | ### Problem Statement
A simple way to compress a string is to encode repeated consecutive substrings as a counter followed by the substring. For example, if X represents a substring, and the string contains a sequence "XX...X", we can compress the sequence as "[DX]", where D is the number of times X is repeated (D is a single digit, i.e., 1 <= D <= 9). X itself might contain some compressed substrings as well. For example, the string "CABABABABABABC" can be compressed as "C[6AB]C" or "C[2[3AB]]C". You are given a String toUncompress. Uncompress toUncompress and return the result. The uncompressed string will contain only uppercase letters ('A'-'Z').
### Definition
Class: SimpleCompressor Method: uncompress Parameters: String Returns: String Method signature: String uncompress(String toUncompress) (be sure your method is public)
### Constraints
-The return value will contain between 1 and 1000 characters, inclusive.
-The return value will contain only uppercase letters ('A'-'Z').
-toUncompress will contain between 1 and 50 characters, inclusive.
-toUncompress will contain only uppercase letters ('A'-'Z'), digits ('1'-'9'), and brackets ('[' and ']').
-toUncompress will be a properly compressed string.
-In each occurrence of "[DX]", D will be a single digit, between 1 and 9, inclusive.
-In each occurrence of "[DX]", X will be a non-empty string.
### Examples
0)
`"C[6AB]C"`
`Returns: "CABABABABABABC"`
1)
`"C[2[3AB]]C"`
`Returns: "CABABABABABABC"`
2)
`"CO[1N]TEST"`
`Returns: "CONTEST"`
3)
`"[2[2AB]]"`
`Returns: "ABABABAB"`
4)
`"AAAAAAAAAAAAAAAAAAAAA"`
`Returns: "AAAAAAAAAAAAAAAAAAAAA"`
#### Problem url:
http://www.topcoder.com/stat?c=problem_statement&pm=6599
#### Problem stats url:
http://www.topcoder.com/tc?module=ProblemDetail&rd=10061&pm=6599
Mike Mirzayanov
#### Testers:
PabloGilberto , brett1479 , Olexiy
Recursion | 508 | 1,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-22 | latest | en | 0.743785 |
https://prepinsta.com/deloitte/aptitude/pipes-and-cisterns/quiz-1/ | 1,620,789,332,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991693.14/warc/CC-MAIN-20210512004850-20210512034850-00086.warc.gz | 491,333,967 | 33,012 | Prime
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# Deloitte Pipes and Cisterns Questions and Answers Quiz-1
Question 1
Time: 00:00:00
Tap A and Tap B can fill a water tank in 3 and 4 respectively.
A tap C can empty the tank in 6 hours. If all the three taps are opened
together, then the tank will be filled in?
2(2/5) ℎ
2(2/5) ℎ
5(2/7) ℎ
5(2/7) ℎ
3(3/5)ℎ
3(3/5)ℎ
5(1/7)ℎ
5(1/7)ℎ
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Question 2
Time: 00:00:00
Two large pipes X & Y can separately fill a cistern in 20 minutes
and 30 minutes. Together, they can fill the cistern in:
1 hour 31 minutes
1 hour 31 minutes
12 minutes
12 minutes
2 hours 12 minutes
2 hours 12 minutes
2 hours
2 hours
none of the above
none of the above
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Question 3
Time: 00:00:00
A large water tank can be filled by a pump A in 10 minutes, and by B
can fill the same tank in 20 minutes. If both the pumps work together to
fill the tank, how long will it take to fill the tank.
1hrs 10 minutes
1hrs 10 minutes
20 minutes
20 minutes
4minutes
4minutes
9 hrs.
9 hrs.
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Question 4
Time: 00:00:00
A water container was 3/5 full. Ravi poured 4 buckets from it and then
poured two buckets back in the container. The container is 1/2 full now.
How many buckets can be filled if the entire container is emptied?
32
32
20
20
44
44
40
40
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Question 5
Time: 00:00:00
A pipe can fill a petrol tank in 20 hrs but due to some leakage it got
emptied in 15 hrs. If the Find the time in which the tank will be completely full.
70
70
69
69
60
60
20
20
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Question 6
Time: 00:00:00
One pipe can fill a tank four times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
180 min
180 min
144 min.
144 min.
126 min
126 min
114 min
114 min
none of the above
none of the above
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1/20 + 1/30 -1/8 = -1/24
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Question 7
Time: 00:00:00
The pumps can fill water in 20 tanks in 2 hrs in 1 day. How many tanks
can be filled in 2 days working 4 hours a day.
7
7
3
3
5
5
2
2
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Question 8
Time: 00:00:00
If 3 pipes named as A, B, C. Pipes A and B can fill a tank in 6 and 5 hours
respectively. Pipe C can empty it in 18 hours. If all the three pipes are turned
on together, then the tank will be filled in
12/90 hrs.
12/90 hrs.
7/45 hrs
7/45 hrs
14/45 hrs.
14/45 hrs.
12/45 hrs.
12/45 hrs.
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0/0
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0% | 1,101 | 3,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-21 | longest | en | 0.83475 |
https://codedump.io/share/3bxC1Wv3UDW6/1/flattening-the-key-of-a-rdd | 1,477,186,107,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719136.58/warc/CC-MAIN-20161020183839-00116-ip-10-171-6-4.ec2.internal.warc.gz | 835,106,475 | 8,655 | Armand Grillet - 2 months ago 4x
Scala Question
# Flattening the key of a RDD
I have a Spark RDD of type
`(Array[breeze.linalg.DenseVector[Double]], breeze.linalg.DenseVector[Double])`
. I wish to flatten its key to transform it into a RDD of type
`breeze.linalg.DenseVector[Double], breeze.linalg.DenseVector[Double])`
. I am currently doing:
``````val newRDD = oldRDD.flatMap(ob => anonymousOrdering(ob))
``````
The signature of anonymousOrdering() is
`String => (Array[DenseVector[Double]], DenseVector[Double])`
.
It returns
`type mismatch: required: TraversableOnce[?]`
. The Python code doing the same thing is:
``````newRDD = oldRDD.flatMap(lambda point: [(tile, point) for tile in anonymousOrdering(point)])
``````
How to do the same thing in Scala ? I generally use
`flatMapValues`
but here I need to flatten the key.
If I understand your question correctly, you can do:
``````val newRDD = oldRDD.flatMap(ob => anonymousOrdering(ob))
// newRDD is RDD[(Array[DenseVector], DenseVector)]
``````
In that case, you can "flatten" the `Array` portion of the tuple using pattern matching and a `for`/`yield` statement:
``````newRDD = newRDD.flatMap{case (a: Array[DenseVector[Double]], b: DenseVector[Double]) => for (v <- a) yield (v, b)}
// newRDD is RDD[(DenseVector, DenseVector)]
``````
Although it's still not clear to me where/how you want to use `groupByKey`
Source (Stackoverflow) | 370 | 1,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2016-44 | longest | en | 0.700276 |
http://2012.igem.org/wiki/index.php?title=Team:Colombia/Modeling/Paramterers&direction=prev&oldid=296104 | 1,597,222,705,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738888.13/warc/CC-MAIN-20200812083025-20200812113025-00460.warc.gz | 699,681 | 14,427 | # Team:Colombia/Modeling/Paramterers
## Parameters of the equations
When we want to model a biological process, it is necessary to write the differential equation that model the system which requires a number of constants. If the real value for the constants are unknown, the system can not have any biological sense. These entries are called parameters and they are crucial elements in the mathematical model made in this project.
There are three possible ways to find this parameters:
i) Literature. There are a lot of studies trying to find biological parameters, such as basal rate of protein production, kinetic constants, Monod's constants, etc. Moreover, there are so many biological systems and only a few of them have been characterized. Thus, it is difficult to find the parameters that are needed.
ii) Experimental way. If an experiment is made using the biological system of interest, it is possible to find the parameters for the equations that models the whole system. For this project, it was necessary to model the biological system first. Thus, experiments couldn't be performed to find the constant for the differential equations.
iii) Screening of parameters. Sometimes we don't have the exact number that we need, but we have a rank where it could be or the parameter for a similar biological system, then we can perfom a screening of parameter, where we try to find the value that perfectly fits the reponse of our circuit.
## How did we do it?
As said above finding the right set of parameters for a system is a headache for all the synthetic biologists. That is why we developed a standard procedure that can be use any synthetic biology mathematical model. It has four main steps. But before starting to work in it we need to find as much information in literature as possible.
For our particular case, ee found all the CI promoter box parameters, we normalized all the degradation terms with the cell division time and for the other we found acceptable ranges. We used a lot of resources and methods to approximate the limits for this ranges. Here we show an explanation:
a.Basal levels of the proteins:
Determining a rank which this value could be, we described mathematically a very simple process. Suppose there is a usual differential equation:
Where αo is the basal level, γ is the protein parameter for degradation and P is the protein concentration. Thus, we can assume that in steady state conditions we have that dP/dt= 0:
But γ is approximately 1/T, then:
Establishing the rank, we assumed T as φ and we looked for a normal rank of concentration of a protein in a cell [1]. Another important aspect is that there were constitutive proteins and inducible proteins. Thus, we divided the basal levels in two class: i) one for basal proteins and ii) one for constitutive proteins. Finally, following ranges were defined:
Basal proteins range (LuxR, LuxI, HipB and Salicylic Acid):
Constitutive proteins range (HipA7,Sensor, Chitinase, Chitoporin and CBP):
By the other hand, the basal production of HipA has to be constitutive and inducible because the cell has to have three stages. The first stage is always sleep (constitutive) due to HipA effect. Once an impulse of chitin or 3-OH-PAME is presented, it is going to awake the cell due to concentration decrease of HipA . This is the second stage of the cell. The third stage is when cell is slept again which it is achieve through a positive control (inducible). The basal concentration used in the equations was the constitutive one because it is going to have more effect in the response.
The chosen unit of time was the bacteria life time because it represents when the proteins are diluted and decrease in the cell. We can establish easily the protein degradation using values related to life time of E.Coli. We defined the protein degradation for almost all values as follows:
Again, we found a special case. Degradation of HB is greater, then we assigned it a value of 4/φ.
c. Reaction constant:
This value can vary depending of the described process. Fortunately, we found in literature how to approximate this value [2]. However, we took the approximation and turned it into units described above.
d. Hill coefficients k:
We looked for some reference in different data sources and we found the coefficient of CI (Nitzal et al.2005). Then, we decided to approximate a rank taking this value as a reference.
e. Hill coefficients n:
We based on the value "n" found for CI( NItzal et al.2005). In this case, there was an analysis that took in account the number of molecules that interfere with gene activation.
f. Maximum cell production (β):
Considering the concentration of Rubisco, which is about 20 times the average concentration and rate of production of a protein, [3], we assign the limits for this range as follows.
CI PARAMETERS
All the parameters for CI activation system were found in the literature. (Nitzal et al.2005). We only made a change of units.
Once the ranges of the parameters were set, we proceed to develop the standard method. Here we show the steps proposed with the obtained results for Pest-busters as an example.
Step 1: Objective function: Define your desired behavior
First of all, it is necesary to define the desired response, in this case the Salicylic Acid. Thus, we stablished some ranges where concentration supposed to be. This response depends on the chitin or 3-OH-PAME impulse. Besides, the system is preferable to respond if the signal is long and intense. On the other hand, we don't want our cells to be "awake". The figure below shows this:
Figure 1. Objective function and desired behavior
Step 2. Optimization of parameters: A point within an area
Within the parameter space, there are many sets of them that make the system behaves just like expected; and these can be represented as an area. The main objective is to find this area. Suppose you only have two parameters. The plane represents the parameter space and the green area is where the system works.
Figure 2. Parameter space for a system
Now we want a point within these area to begin the screening. It seems easy to identify in two dimensions, but this process can take a lot of time (weeks or even months) when it is a space of more than 10 parameters. So, one way to achieve this process is making an optimization.
Figure 3. A point in parameter space for a system
A process of optimization takes a function and found maximum or minimum values changing some variables of it. The changing variables are named optimization variables and function is called objective function. If we have some limitations, it is possible to add restrictions to the system, and the function will be optimized without breaking them.
In this case, the objective function is a minimum difference of squares between the points of the expected behavior and the response of the equations with a set of optimization variables. When the distance between these points is minimum, the parameters are found.
This optimization process was done discretizing the differential equations and putting them as restrictions. The same procedure was made for the stable state concentration where the differential equation must be equal to zero (when there is no signal). This algorithm was programmed in a specialized software.
Unfortunately the firsts results were not satisfactory and a new code it is being developed.
Step 3: Sensitivity anlysis
In this step the importance of each parameters in the main outputs behavior (Our case: Salicylic Acid, the toxin HipA7 and the antitoxin HipB) is tested. The objective is to find how much each parameter affect the desired response.
This test considered the following stages: i.) Establish the ranges of the parameters (see above), ii) Determine appropriate division for the ranges, iii) Iterate each parameter while leaving the others fixed in the MATLAB code. iv) See how the difference between the steady state's concentrations and the concentrations during the impulse of the pathogen changed with the change of each parameter.
Note: the exact value for each parameter is not important. It is important their relevance and how they change the response.
Here is an example:
Parameter: LuxI Maximal production rate
Range: 0-22
Step size: 0.5
The figure above shows how the HipB, HipA7 and Salicylic Acid concentration change during the presence of the pathogen is significantly influenced by the maximal production rate of LuxI protein. We make this procedure for all the parameters in both systems (Ralstonia and Fungus). The results are shown below:
Fungus: The sensitivity analysis was performed for the system with the detection module for fungus. The following table shows the different parameters involved in the system with the fungal detection system and how they affect the final behavior
Here we present some of the graphics found in the analysis. The pictures below show two parameters that do not have significant influence in the desired response:
We also found some parameters that have a significant influence in one of the main outputs but were not sensitive to the rest:
Finally, we found some parameters that affect the final response of all the main outputs of interest:
Ralstonia
Below, it is exposed the results obtained from sensitivity test for Ralstonia. The next table identifies those parameters that make a relevant change in the model from those that do not. The main outputs tested were the HipB, HipA7 and the salicylic acid change in concentration due to the impulse.
The following figures show the results obtained for almost all the relevant parameters. Like in the previous case we found some parameters that had no effect at all in the response and others than only had effect in one of the main outputs.
We also found some interesting results, there were some parameters that had a significant influence in a small range, but when the rest of the range was evaluated the response was not sensitive to these parameters:
Finally we found parameters that had significant influence in the response of the three main outputs and in the complete range of the parameter. Here we present some of the most interesting behaviors of these kind of parameters:
Using this results it is possible to know the parameters that affect the response in a major way. Although it is still unknown the exact value of the parameter, it is possible predict how the system is supposed to response. Then, we could proceed to the final step.
Step 4: Screening
If we have a set of parameters, why do we do a screening?
In the second step we found a set of parameters that give the expected response of the system. But this is only a point. What does happen with the rest of points of the area? Does the optimization method “know” if these parameters are real or have biological sense? As long as we specified one single response for the parameters, we don’t know if there are other possible responses of valid solutions. Do they exist?
To answer these questions we performed a screening of the parameters. Beginning at the resulting point of the optimization, we started to move in all directions in a fixed range and then we examined when the acceptable area ends.
How much do we move depends on the sensitivity analysis. If the response is not very sensitive to the parameter, we take longer steps than the case which the response is more sensitive.
There are some parameters that can me modified experimentally like the maximal rate production but most of them depend on molecular details, are unchangeable and can only be known with experiments. The final goal of these standard procedure is to set the changeable parameters in such a way that the cross sectional area of the area of parameters that has desired response is maximized.
Suppose you only have two parameters, one that can be changed and one that does not. If you set the first parameter in a black dot (see figure below), a small change of the uncontrollable parameter would take the system out of its working area. The optimum solution is to set the first parameter in the red line, so the probability of going out of the working area is smaller.
``` -Note: Due to long computational time, the code is being parallelized and we expect to run the tests shortly.
``` | 2,516 | 12,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-34 | latest | en | 0.930288 |
http://mathhelpforum.com/advanced-statistics/83832-sample-data-closer-simple-linear-model-exponential-model-print.html | 1,529,669,023,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864391.61/warc/CC-MAIN-20180622104200-20180622124200-00437.warc.gz | 201,531,234 | 2,835 | # Sample data is closer to a simple linear model or an exponential model?
• Apr 14th 2009, 10:59 PM
Lok
Sample data is closer to a simple linear model or an exponential model?
A guy wants to know if this data sample is closer to a simple linear model or to an exponential model.
The table below represents 4 machines. We want to know the cost in function of the age of the machine. So X:age and Y:maintenance cost
Age 1 | 1.7 | 2 | 2.25
cost 0.5| 0.6 | 1.2 | 2
Either linear
Y = B_0 + B_1X
or Exponential
Y = ae^{bx}
Any tips to get me started?
• Apr 15th 2009, 12:58 AM
mr fantastic
Quote:
Originally Posted by Lok
A guy wants to know if this data sample is closer to a simple linear model or to an exponential model.
The table below represents 4 machines. We want to know the cost in function of the age of the machine. So X:age and Y:maintenance cost
Age 1 | 1.7 | 2 | 2.25
cost 0.5| 0.6 | 1.2 | 2
Either linear
Y = B_0 + B_1X
or Exponential
Y = ae^{bx}
Any tips to get me started?
Fit the data to each model and look at r^2.
• Apr 15th 2009, 01:11 AM
Lok
How do I fit the data to the models? :(
• Apr 15th 2009, 01:18 AM
mr fantastic
Quote:
Originally Posted by Lok
How do I fit the data to the models? :(
Use technology. eg. TI-89 calculator, suitable software. It can even be done by hand (but very tedious). | 422 | 1,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-26 | latest | en | 0.828887 |
http://solvemaths.in/contest-question-11/ | 1,566,364,846,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315809.69/warc/CC-MAIN-20190821043107-20190821065107-00298.warc.gz | 176,503,719 | 21,364 | A magician has 5 hats, 3 green and 2 white in a closed box, as known to all the audience. He ask three persons to stand in a queue and closed their eyes. Magician picks 3 hats from the box and puts one hat onto each person’s head. Then magician ask first person to open eyes and look at the other two person and based on that judge the color of his own hat. First person couldn’t answer confidently. Then magician ask second person to open eyes and look on the other two person and based on that judge the color of his own hat. Second person also couldn’t conclude. Then, magician ask third person. However even without opening the eye, the third person guess correctly the color of hat on his head. What was the color of hat on third person? Explain your answer.
Aptitude
A magician has 5 hats, 3 green and 2 white in a closed box, as known to all the audience. He ask three persons to stand in a queue and closed their eyes. Magician picks 3 hats from the box and puts one hat onto each person’s head. Then... | 234 | 1,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-35 | longest | en | 0.982236 |
http://mathhelpforum.com/calculus/47064-triple-integral-using-spherical-coordinates.html | 1,480,702,596,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540409.8/warc/CC-MAIN-20161202170900-00507-ip-10-31-129-80.ec2.internal.warc.gz | 176,958,622 | 12,217 | # Thread: Triple Integral Using Spherical Coordinates
1. ## Triple Integral Using Spherical Coordinates
Hi Everyone,
I'm having a little problem with the following question that I was hoping someone could help me with.
Basically I need to find the triple integral of (9 - x^2 -y^2) over a region, H, using spherical coordinates where H is the solid hemisphere
x^2 + y^2 + z^2 <= 9,
and z>=0.
Now in terms of actually performing the triple integration, I have no problems, it is getting the spherical coordinates that is really bugging me.
H should be in the following form
H = {p,θ, φ| a<= 0 <= b, α<= θ<= β, c<=φ<=d}
Now I know how to find p(really a and b). In this case a = 0 and b = 3.
But I don't know how to find c, d, α, and β. I would really need a step-by-step guide (something like "Spherical Coordinates for dummies" )
If anyone could let me know how to do this I would be most grateful.
Also, if someone could provide me with another example of the process just to ensure that my grasp on the method is secure, then that would just be gravy.
Eg.
E is the area that is to be integrated over, where E lies between the spheres
x^2 +y^2+z^2 = 1 and x^2 +y^2 +z^2 =4.
(here a = 1, b=2)
Thanks in advance to anyone who helps.
2. Originally Posted by woody198403
Hi Everyone,
I'm having a little problem with the following question that I was hoping someone could help me with.
Basically I need to find the triple integral of (9 - x^2 -y^2) over a region, H, using spherical coordinates where H is the solid hemisphere
x^2 + y^2 + z^2 <= 9,
and z>=0.
Now in terms of actually performing the triple integration, I have no problems, it is getting the spherical coordinates that is really bugging me.
H should be in the following form
H = {p,θ, φ| a<= 0 <= b, α<= θ<= β, c<=φ<=d}
Now I know how to find p(really a and b). In this case a = 0 and b = 3.
But I don't know how to find c, d, α, and β. I would really need a step-by-step guide (something like "Spherical Coordinates for dummies" )
If anyone could let me know how to do this I would be most grateful.
Also, if someone could provide me with another example of the process just to ensure that my grasp on the method is secure, then that would just be gravy.
Eg.
E is the area that is to be integrated over, where E lies between the spheres
x^2 +y^2+z^2 = 1 and x^2 +y^2 +z^2 =4.
(here a = 1, b=2)
Thanks in advance to anyone who helps.
As you correctly stated, the region projected onto the xy plane $(z=0)$ gives you the limits $0\leq\varrho\leq3$
In this case, $\varphi$ can only be the angle between the positive z and the xy plane [since $z\geq0$]. Thus, $0\leq\varphi\leq\frac{\pi}{2}$
So now we have two sets of limits.
Now, lets find the limits for $\vartheta$.
Since the region is a complete circle, $\vartheta$ must be $2\pi$. This is because 1 full revolution around a circle is $2\pi$ radians.
Thus, we see that our integral is $\int_0^{2\pi}\int_0^{\frac{\pi}{2}}\int_0^3\varrho ^2\sin\varphi\,d\varrho\,d\varphi\,d\vartheta$
Does this somewhat make sense?
--Chris
3. Originally Posted by Chris L T521
Thus, we see that our integral is $\int_0^{2\pi}\int_0^{\frac{\pi}{2}}\int_0^3\varrho ^2\sin\varphi\,d\varrho\,d\varphi\,d\vartheta$
Chris, I think you forgot to put the function to be integrated in the region ;p ...
Originally Posted by woody198403
Basically I need to find the triple integral of (9 - x^2 -y^2) over a region, H, using spherical coordinates where H is the solid hemisphere
x^2 + y^2 + z^2 <= 9,
But it should be no problem after finding the integration limits.
Use these to get your function in spherical coordinates:
$x = r \cos \vartheta \sin \varphi$
$y = r \sin \vartheta \sin \varphi$
$z = r \cos \varphi$
4. Originally Posted by wingless
Chris, I think you forgot to put the function to be integrated in the region ;p ...
But it should be no problem after finding the integration limits.
Use these to get your function in spherical coordinates:
$x = r \cos \vartheta \sin \varphi$
$y = r \sin \vartheta \sin \varphi$
$z = r \cos \varphi$
I overlooked something... >_>
Thanks for catching that!
--Chris
5. Awesome, thanks so much. However, there is one other thing with this question that I just wanted to check which is whether this is correct?
$\int_0^{2\pi}\int_0^{\pi/2}\int_0^3$ (9 - (p^2 sin^2 φ cos^2 θ) - (p^2 sin^2 φ sin^2 θ)) dp dφ dθ??
im sure that "- (p^2 sin^2 φ cos^2 θ) - (p^2 sin^2 φ sin^2 θ))" is correct but im uncertain about what do do with the 9 here (only because im working with spherical coordinates).
6. I was thinking that it could be
$\int_0^{2\pi}\int_0^{\pi/2}\int_0^3$ ((p^2 cos^2 φ) - (p^2 sin^2 φ cos^2 θ) - (p^2 sin^2 φ sin^2 θ)) dp dφ dθ
because if you look at x^2 + y^2 + z^2 <= 9, and let x=y=0, z= + or - 3, so z^2 = 9
and
z = p cos φ; so
z^2 = p^2 cos^2 φ
which I believe comes to
$\int_0^{2\pi}\int_0^{\pi/2}\int_0^3$ p^2(cos 2φ) dp dφ dθ??
am I way off here, or is this correct? Thanks. | 1,553 | 4,973 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-50 | longest | en | 0.962182 |
https://ucsd-cse230.github.io/fa21/lectures/09-monads.html | 1,642,632,348,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301592.29/warc/CC-MAIN-20220119215632-20220120005632-00698.warc.gz | 648,134,579 | 10,812 | # Functors and Monads
## Abstracting Code Patterns
a.k.a. Dont Repeat Yourself
### Lists
data List a
= []
| (:) a (List a)
### Rendering the Values of a List
-- >>> incList [1, 2, 3]
-- ["1", "2", "3"]
showList :: [Int] -> [String]
showList [] = []
showList (n:ns) = show n : showList ns
### Squaring the values of a list
-- >>> sqrList [1, 2, 3]
-- 1, 4, 9
sqrList :: [Int] -> [Int]
sqrList [] = []
sqrList (n:ns) = n^2 : sqrList ns
## Common Pattern: map over a list
Refactor iteration into mapList
mapList :: (a -> b) -> [a] -> [b]
mapList f [] = []
mapList f (x:xs) = f x : mapList f xs
Reuse map to implement inc and sqr
showList xs = map (\n -> show n) xs
sqrList xs = map (\n -> n ^ 2) xs
## Trees
Same “pattern” occurs in other structures!
data Tree a
= Leaf
| Node a (Tree a) (Tree a)
### Incrementing the values of a Tree
-- >>> showTree (Node 2 (Node 1 Leaf Leaf) (Node 3 Leaf Leaf))
-- (Node "2" (Node "1" Leaf Leaf) (Node "3" Leaf Leaf))
showTree :: Tree Int -> Tree String
showTree Leaf = ???
showTree (Node v l r) = ???
### Squaring the values of a Tree
-- >>> sqrTree (Node 2 (Node 1 Leaf Leaf) (Node 3 Leaf Leaf))
-- (Node 4 (Node 1 Leaf Leaf) (Node 9 Leaf Leaf))
sqrTree :: Tree Int -> Tree Int
sqrTree Leaf = ???
sqrTree (Node v l r) = ???
## QUIZ: map over a Tree
Refactor iteration into mapTree! What should the type of mapTree be?
mapTree :: ???
showTree t = mapTree (\n -> show n) t
sqrTree t = mapTree (\n -> n ^ 2) t
{- A -} (Int -> Int) -> Tree Int -> Tree Int
{- B -} (Int -> String) -> Tree Int -> Tree String
{- C -} (Int -> a) -> Tree Int -> Tree a
{- D -} (a -> a) -> Tree a -> Tree a
{- E -} (a -> b) -> Tree a -> Tree b
## Lets write mapTree
mapTree :: (a -> b) -> Tree a -> Tree b
mapTree f Leaf = ???
mapTree f (Node v l r) = ???
## QUIZ
Wait … there is a common pattern across two datatypes
mapList :: (a -> b) -> List a -> List b -- List
mapTree :: (a -> b) -> Tree a -> Tree b -- Tree
Lets make a class for it!
class Mappable t where
gmap :: ???
What type should we give to gmap?
{- A -} (b -> a) -> t b -> t a
{- B -} (a -> a) -> t a -> t a
{- C -} (a -> b) -> [a] -> [b]
{- D -} (a -> b) -> t a -> t b
{- E -} (a -> b) -> Tree a -> Tree b
### Reuse Iteration Across Types
Haskell’s libraries use the name Functor instead of Mappable
instance Functor [] where
fmap = mapList
instance Functor Tree where
fmap = mapTree
And now we can do
-- >>> fmap (\n -> n + 1) (Node 2 (Node 1 Leaf Leaf) (Node 3 Leaf Leaf))
-- (Node 4 (Node 1 Leaf Leaf) (Node 9 Leaf Leaf))
-- >>> fmap show [1,2,3]
-- ["1", "2", "3"]
## A Type to Represent Expressions
data Expr
= Number Int -- ^ 0,1,2,3,4
| Plus Expr Expr -- ^ e1 + e2
| Minus Expr Expr -- ^ e1 - e2
| Mult Expr Expr -- ^ e1 * e2
| Div Expr Expr -- ^ e1 / e2
deriving (Show)
## Some Example Expressions
e1 = Plus (Number 2) (Number 3) -- 2 + 3
e2 = Minus (Number 10) (Number 4) -- 10 - 4
e3 = Mult e1 e2 -- (2 + 3) * (10 - 4)
e4 = Div e3 (Number 3) -- ((2 + 3) * (10 - 4)) / 3
## EXERCISE: An Evaluator for Expressions
Fill in an implementation of eval
eval :: Expr -> Int
eval e = ???
so that when you’re done we get
-- >>> eval e1
-- 5
-- >>> eval e2
-- 6
-- >>> eval e3
-- 30
-- >>> eval e4
-- 10
## QUIZ
What does the following evaluate to?
quiz = eval (Div (Number 60) (Minus (Number 5) (Number 5)))
A. 0
B. 1
C. Type error
D. Runtime exception
E. NaN
## To avoid crash, return a Result
Lets make a data type that represents Ok or Error
data Result v
= Ok v -- ^ a "successful" result with value `v`
| Error String -- ^ something went "wrong" with `message`
deriving (Eq, Show)
## EXERCISE
Can you implement a Functor instance for Result?
instance Functor Result where
fmap f (Error msg) = ???
fmap f (Ok val) = ???
When you’re done you should see
-- >>> fmap (\n -> n ^ 2) (Ok 9)
-- Ok 81
-- >>> fmap (\n -> n ^ 2) (Error "oh no")
-- Error "oh no"
## Evaluating without Crashing
Instead of crashing we can make our eval return a Result Int
eval :: Expr -> Result Int
• If a sub-expression has a divide by zero return Error "..."
• If all sub-expressions are safe then return Ok n
## EXERCISE: Implement eval with Result
eval :: Expr -> Result Int
eval (Number n) = ?
eval (Plus e1 e2) = ?
eval (Minus e1 e2) = ?
eval (Mult e1 e2) = ?
eval (Div e1 e2) = ?
## The Good News
No nasty exceptions!
>>> eval (Div (Number 6) (Number 2))
Ok 3
>>> eval (Div (Number 6) (Number 0))
Error "yikes dbz:Number 0"
>>> eval (Div (Number 6) (Plus (Number 2) (Number (-2))))
Error "yikes dbz:Plus (Number 2) (Number (-2))"
## The BAD News!
The code is super gross
## Lets spot a Pattern
The code is gross because we have these cascading blocks
case e1 of
Error err1 -> Error err1
Ok v1 -> case e2 of
Error err2 -> Error err2
Ok v1 -> Ok (v1 + v2)
but look closer … both blocks have a common pattern
case e of
Error err -> Error err
Value v -> {- do stuff with v -}
1. Evaluate e
2. If the result is an Error then return that error.
3. If the result is a Value v then further process with v.
## Lets Bottle that Pattern in Two Functions
• >>= (pronounced bind)
• return (pronounced return)
(>>=) :: Result a -> (a -> Result b) -> Result b
(Error err) >>= _ = Error err
(Value v) >>= process = process v
return :: a -> Result a
return v = Ok v
NOTE: return is not a keyword
• it is the name of a function!
## A Cleaned up Evaluator
The magic bottle lets us clean up our eval
eval :: Expr -> Result Int
eval (Number n) = Ok n
eval (Plus e1 e2) = eval e1 >>= \v1 ->
eval e2 >>= \v2 ->
Ok (v1 + v2)
eval (Div e1 e2) = eval e1 >>= \v1 ->
eval e2 >>= \v2 ->
if v2 == 0
then Error ("yikes dbz:" ++ show e2)
else Ok (v1 `div` v2)
The gross pattern matching is all hidden inside >>=
Notice the >>= takes two inputs of type:
• Result Int (e.g. eval e1 or eval e2)
• Int -> Result Int (e.g. the processor takes the v and does stuff with it)
In the above, the processing functions are written using \v1 -> ... and \v2 -> ...
NOTE: It is crucial that you understand what the code above is doing, and why it is actually just a “shorter” version of the (gross) nested-case-of eval.
## A Class for >>=
The >>= operator is useful across many types!
• like fmap or show or toJSON or ==, or <=
Lets capture it in a typeclass:
class Monad m where
-- (>>=) :: Result a -> (a -> Result b) -> Result b
(>>=) :: m a -> (a -> m b) -> m b
-- return :: a -> Result a
return :: a -> m a
## Result is an instance of Monad
Notice how the definitions for Result fit the above, with m = Result
instance Monad Result where
(>>=) :: Result a -> (a -> Result b) -> Result b
(Error err) >>= _ = Error err
(Value v) >>= process = process v
return :: a -> Result a
return v = Ok v
## Syntax for >>=
In fact >>= is so useful there is special syntax for it.
e1 >>= \v1 ->
e2 >>= \v2 ->
e3 >>= \v3 ->
e
you can write
do v1 <- e1
v2 <- e2
v3 <- e3
e
or if you like curly-braces
do { v1 <- e1; v2 <- e2; v3 <- e3; e }
## Simplified Evaluator
Thus, we can further simplify our eval to:
eval :: Expr -> Result Int
eval (Number n) = return n
eval (Plus e1 e2) = do v1 <- eval e1
v2 <- eval e2
return (v1 + v2)
eval (Div e1 e2) = do v1 <- eval e1
v2 <- eval e2
if v2 == 0
then Error ("yikes dbz:" ++ show e2)
else return (v1 `div` v2)
Which now produces the result
>>> evalR exQuiz
Error "yikes dbz:Minus (Number 5) (Number 5)" | 2,524 | 7,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-05 | latest | en | 0.693253 |
https://www.mathworks.com/matlabcentral/cody/problems/233-reverse-the-vector/solutions/192181 | 1,506,333,867,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690591.29/warc/CC-MAIN-20170925092813-20170925112813-00162.warc.gz | 824,209,765 | 11,577 | Cody
# Problem 233. Reverse the vector
Solution 192181
Submitted on 17 Jan 2013 by Guangxi Liu
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
This solution is outdated. To rescore this solution, log in.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1; y_correct = 1; assert(isequal(reverse(x),y_correct))
``` ans = 1 ```
2 Pass
%% x = -10:1; y_correct = 1:-1:-10; assert(isequal(reverse(x),y_correct))
``` ans = 1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 ```
3 Pass
%% x = 'able was i ere i saw elba'; y_correct = 'able was i ere i saw elba'; assert(isequal(reverse(x),y_correct))
``` ans = able was i ere i saw elba ``` | 225 | 707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-39 | latest | en | 0.807762 |
https://stonespounds.com/206-4-pounds-in-stones-and-pounds | 1,619,013,480,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039544239.84/warc/CC-MAIN-20210421130234-20210421160234-00341.warc.gz | 607,457,120 | 4,683 | # 206.4 pounds in stones and pounds
## Result
206.4 pounds equals 14 stones and 10.4 pounds
You can also convert 206.4 pounds to stones.
## Converter
Two hundred six point four pounds is equal to fourteen stones and ten point four pounds (206.4lbs = 14st 10.4lb). | 76 | 268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-17 | latest | en | 0.761024 |
https://gamedev.stackexchange.com/questions/19924/planar-shadow-matrix-and-plane-b-value | 1,656,493,039,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00311.warc.gz | 315,447,366 | 61,820 | # Planar shadow matrix and plane b value
As you know, we need plane and light factor to calculate a shadow matrix.
The problem is that when I set the plane as D3DXPLANE p(0, -1, 0, 0.1f), the shadows by directional light are correctly rendered, but the shadows by point light are not rendered.
However, if I use D3DXPLANE p(0, 1, 0, 0.1f), the situation is reversed, shadows by directional light are not drawn, the shadows by point light are ok.
I cannot understand why it happens. Is it normal or am i missing something? Please explain to me why this happens.
Code
//main.cpp
ID3D10EffectTechnique* oldTech = meshes[i]->technique;
D3DXPLANE groundPlane(0, -1, 0, 0.01f);
D3DXVECTOR4 light;
if(lightType != 2) //2 here means point light
{
light.x = lightDirection.x;
light.y = lightDirection.y;
light.z = lightDirection.z;
light.w = 0;
}
else
{
light.x = lightPosition.x;
light.y = lightPosition.y;
light.z = lightPosition.z;
light.w = 1;
groundPlane.b = 1;
}
D3DXMATRIX W = meshes[i]->GetWorldMatrix();
D3DXMATRIX matWorld = W * S;
meshes[i]->render(&matWorld);
//Meshes here are drawn by ID3DX10Mesh,
//but I tested other code with LPD3DXMESH, it has the same problem.
meshes[i]->technique = oldTech;
{
PS_Input output = (PS_Input)0;
output.pos = mul(float4(input.pos, 1), matWorld);
output.pos.y = 0.1;
output.pos = mul(mul(output.pos, matView), matProjection);
output.normal = mul( input.normal, (float3x3)matWorld );
output.color = input.color;
output.worldPos = (float3)mul(input.pos, matWorld);
output.tex = input.tex;
return output;
}
float4 PS_Shadow( PS_Input input) : SV_Target
{
return float4(0,0,0,0.5f);
}
DepthStencilState StencilIncreaseOnce {
DepthEnable = true;
DepthFunc = LESS_EQUAL;
StencilEnable = true;
FrontFaceStencilFunc = EQUAL;
FrontFaceStencilPass = INCR;
FrontFaceStencilFail = Keep;
};
BlendState BlendAlpha {
AlphaToCoverageEnable = FALSE;
BlendEnable[0] = TRUE;
SrcBlend = SRC_ALPHA ;
DestBlend = INV_SRC_ALPHA ;
SrcBlendAlpha = ONE;
DestBlendAlpha = ZERO;
};
{
pass P0
{
SetDepthStencilState( StencilIncreaseOnce, 0x0 );
SetBlendState( BlendAlpha, float4( 0.0f, 0.0f, 0.0f, 0.0f ),
Your code looks mostly correct to me. One thing that's a little weird is the output.pos.y = 0.1; in your VS. The reason this concerns me is that when you're using a point light, the shadow matrix is a projection matrix, and that means that after going through it your points are going to come out with w != 1.0. So you can't just set y to 0.1 to make them come out at y = 0.1...you'd have to set y to w * 0.1.
Another thing that might be an issue is a possible mismatch in the convention of lightDirection. I believe the function expects a vector pointing toward the light, but some people prefer to define it in terms of the direction the light is traveling, which the opposite. But if this was the probem I'd expect to see the shadows appearing to the lit side of the object instead of the shadowed side, since the whole vector would be negated and not just the y component. | 843 | 3,010 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-27 | latest | en | 0.670012 |
http://stackoverflow.com/questions/11623264/python-data-structure-index-start-at-1-instead-of-0/11623319 | 1,369,361,519,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704132729/warc/CC-MAIN-20130516113532-00023-ip-10-60-113-184.ec2.internal.warc.gz | 257,790,020 | 12,393 | # Python Data structure index Start at 1 instead of 0?
I have a weird question: I have this list of 64 numbers that will never change:
``````(2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 128)
``````
I need a data structure in Python that will allow me to accsess these numbers using a 1-64 index as opposed to the standard 0-63. Is this possible? Would the best way to accomplish this be to build a dictionary?
-
Just insert a `0` at the beginning of the structure:
``````(0, 2, 4, 6, 8, ...)
``````
-
+1 for a simple solution w/o unnecessary complications – Levon Jul 24 '12 at 2:42
I think there is no other solution that will work well with 1-based slices. – Kay Jul 24 '12 at 2:50
Wow, can't believe it was that simple.. I was gonna use a dictionary. – NASA Intern Jul 24 '12 at 2:59
You could use a dictionary, or you could simply subtract one from your index before accessing it.
Also, I note that your 64 numbers are in a simple arithmetic progression. Why store them at all? You can use this:
``````def my_number(i):
return 2*i
``````
If the list you showed was actually an example, and the real numbers are more complicated, then use a list with a dummy first element:
``````my_nums = [0, 2, 4, 6, 8, ....]
``````
Then you can get 2 as `my_nums[1]`.
-
No, those are the actual numbers. – NASA Intern Jul 24 '12 at 2:58 Why would you store the numbers from 2 to 128? Just double your index, and you have the number... – Ned Batchelder Jul 24 '12 at 3:06 Well, these numbers are ID's from canvas objects. I cant find away to change them. It's rather complicated – NASA Intern Jul 24 '12 at 3:17
You could override the item getter and make a specialized tuple:
``````class BaseOneTuple(tuple):
__slots__ = () # Space optimization, see: http://stackoverflow.com/questions/472000/python-slots
def __new__(cls, *items):
return tuple.__new__(cls, items) # Creates new instance of tuple
def __getitem__(self, n):
return tuple.__getitem__(self, n - 1)
b = BaseOneTuple(*range(2, 129, 2))
b[2] == 4
``````
-
Uhhhh, im not nearly good enough in Python yet to do this... thanks tho! – NASA Intern Jul 24 '12 at 3:00
You could use `range(2, 129, 2)` to generate the numbers in the range 1 - 128 in increments of 2 and convert this list into a tuple if it's not going to change.
``````t = tuple(range(2, 129, 2))
def numbers(n):
return t[n-1]
``````
Given the global tuple `t`, function `numbers` could retrieve elements using a 1-based (instead of 0-based) index.
- | 896 | 2,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2013-20 | latest | en | 0.896457 |
https://jp.mathworks.com/matlabcentral/answers/1683579-goal-seek-excel-function-in-matlab?s_tid=prof_contriblnk | 1,675,185,821,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499888.62/warc/CC-MAIN-20230131154832-20230131184832-00063.warc.gz | 354,657,920 | 25,781 | # Goal seek excel function in MATLAB.
21 ビュー (過去 30 日間)
Alexander Guillen 2022 年 3 月 29 日
Thank you in advance for any guidance or help. I am not sure if it will be hard to understand what I am trying to achieve but here I go.
I am aware of a function in excel called goalseek which allows me to vary column T ( which is temperature) to column p ( total pressure) in order to make this column match p_target so in the end column p will be a column of just 1 in it . I can do this in excel but would like learn how to do this in matlab. My code is below. I would appreciate any guidance, help or links that maybe answer my question.
clear all; close all; clc
p_target = 1; % bar
T= 300:5:330;
x1 = linspace(0,1,7);
x2 = 1-x1;
% Calculating the pressure 1 and 2
% Water- Antoine parameters
a = 4.6543; b = 1435; c = -64.848;
p1_pure = Antoine(a,b,c,T);
p1 = transpose(p1_pure.*x1);
% Benzene: Antoine parameters
a = 4.01814; b = 1203; c = -53.226;
p2_pure = Antoine(a,b,c,T);
p2 = transpose(p2_pure.*x2);
% Total pressure
p = p1+p2;
y1 = p1./p;
y2 = p2./p;
OUTPUT = [p1,p2,y1,y2]
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Translated by | 409 | 1,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-06 | latest | en | 0.740754 |
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# Maximum likelihood decoding of space time codes
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#### aregawi
##### Newbie level 2
Dear all,
I am currently doing my Master's thesis on 'coding and decoding of perfect space time block codes: analysis and performance simulation'.
The point is I wanted to reduce the complexity of ML decoding by conditional optimization principle that the order of complexity will reduce, for example,3x3 antenna dimension the constellation size the power of 6, which the order of 9 to transmit 9-HEX symbols.Can any one give me how i can estimate by ML, assuming that I transmitted 4-QAM symbols, after gray mapping symbowise or bitwise error probability. I want to implement it in matlab.I know the principle that after sending the symbols in a rayleigh fading channel,I will decode the one as correct pattern, let (x1,x2,x3,x4) out of all possible patterns of all these 4-symbols,which minimizes the error found by subtracting the code book of each pattern from what is received. But after finding the minimum error,how can I take the pattern out of it and compare it with the one the patterns at hand.I found it my approach as it has a drawback of accuracy and simplicity.
I will be happy if any one can give me a clue on this
(I needed i may send the exact problem)
Looking forward to receiving an earlier reply
with best regards,
Last edited:
if your looking towards using Matlab, then if you have the communication toolbox there is a function called vitdec, which will decode it. Which QAM are you talking about 8,16... The ML decoder uses a sliding window and gives the most reliable output.
If you look at the communication toolbox most of what you have mentioned already exists. Including modulation.
Childa909
### Childa909
Points: 2
Re: Maximum likelihood decoding of space time block codes
Dear Kalyanasv,
Lot of thanks for your earliest response of my problem.You recommended me to use a function vitdec(Viterbi Decoder) in the matlab toolbox, which suitable for Trellis codes. But here is the core of my question: my codes are space time block codes which are to be transmitted in the form of Alamouti blocks (the code being formed by cyclic arrangement of a certain pattern).
As they are in the form of independent blocks,it is possible to optimize of the the block elements with respect to the rest prior to maximizing the ML log function thereby that optimized value will be used to estimate the elements of the other blocks.Over all the order of complexity will be reduced than the conventional ML.I will use 9-HEX symbols for 3x3 antenna dimension and 16-QAM for 4x4. Here you the attache pdf :View attachment LOWCOMPLEXITY ESSENTIALLY MAXIMUM LIKELIHOOD DECODING OF PERFECT.pdf
I want to evaluate the performance of the conditionally optimized ML of 3x3 shown and do for 4x4 in the same way
With great thanks
Status
Not open for further replies. | 649 | 2,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-50 | latest | en | 0.901353 |
https://bumpercarfilms.com/qa/quick-answer-what-happens-if-two-bodies-collide-elastically.html | 1,620,625,600,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989030.87/warc/CC-MAIN-20210510033850-20210510063850-00574.warc.gz | 174,824,597 | 8,007 | # Quick Answer: What Happens If Two Bodies Collide Elastically?
## Do objects stick together in an inelastic collision?
People sometimes think that objects must stick together in an inelastic collision.
However, objects only stick together during a perfectly inelastic collision.
Objects may also bounce off each other or explode apart, and the collision is still considered inelastic as long as kinetic energy is not conserved..
## What are the 3 types of collision?
There are three different kinds of collisions, however, elastic, inelastic, and completely inelastic. Just to restate, momentum is conserved in all three kinds of collisions. What distinguishes the collisions is what happens to the kinetic energy.
## What happens to the energy when two bodies collide Inelastically?
Perfectly inelastic collision In such a collision, kinetic energy is lost by bonding the two bodies together. This bonding energy usually results in a maximum kinetic energy loss of the system. … In this example, momentum of the system is conserved because there is no friction between the sliding bodies and the surface.
## Is momentum conserved when two bodies collide?
Momentum is conserved in the collision. … Momentum is conserved for any interaction between two objects occurring in an isolated system. This conservation of momentum can be observed by a total system momentum analysis or by a momentum change analysis.
## When two balls at same temperature collide what is conserved?
When two balls at the same temprature collide, some fraction of their KE appears in other forms of energy, like heat energy , second energy. Hence neither temparture, nor velocity or KE will remain conserved. The only quantity which will remain conserved.
## Is momentum conserved in the real world?
Momentum is not conserved if there is friction, gravity, or net force (net force just means the total amount of force). What it means is that if you act on an object, its momentum will change. This should be obvious, since you are adding to or taking away from the object’s velocity and therefore changing its momentum.
## Is momentum conserved when a ball hits a wall?
Clearly, the momentum of the ball is changed by the collision with the wall, since the direction of the ball’s velocity is reversed. It follows that the wall must exert a force on the ball, since force is the rate of change of momentum.
## When two bodies collide elastically then the quantity conserved is?
Both momentum and kinetic energy are conserved quantities in elastic collisions. Suppose two similar trolleys are traveling toward each other with equal speed. They collide, bouncing off each other with no loss in speed. This collision is perfectly elastic because no energy has been lost.
## When two bodies collide the momentum is?
Momentum is of interest during collisions between objects. When two objects collide the total momentum before the collision is equal to the total momentum after the collision (in the absence of external forces). This is the law of conservation of momentum.
## Which of the following is not an example of perfectly inelastic collision?
In ball bearing striking another ball bearing momentum of the balls system is conserved. Therefore it is not an example of perfectly inelastic collision.
## Is momentum conserved when a ball bounces?
conservation of momentum: The amount of momentum in a system remains the same after a collision. elastic collision: A collision in which all of the momentum is conserved. For example, a ball that bounces back up to its original height. … momentum: Mass in motion.
## Which will stop first when these two bodies collide truck or bicycle?
Answer Expert Verified Lets assume that the two vehicles are travelling at the same speed, but due to the big difference between the mass, the truck will have smaller change in its velocity (Newtons 3rd Law of Motion) and so most likely will the one to stop first. bike !! | 784 | 3,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-21 | latest | en | 0.936731 |
https://www.scribd.com/document/356503747/Quick-Reference-Table-txt | 1,544,664,566,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824338.6/warc/CC-MAIN-20181213010653-20181213032153-00622.warc.gz | 998,198,831 | 46,687 | You are on page 1of 6
# Quick Reference Table - AutoLisp
AutoLisp Quick Reference Table.
(+ <NUMBER> <NUMBER> ...)
Returns the sum of two or more numbers.
(- <NUMBER> <NUMBER> ...)
Returns the difference of two or more numbers.
(* <NUMBER> <NUMBER> ...)
Returns the product of two or more numbers.
(/ <NUMBER> <NUMBER> ...)
Returns the quotient of two or more numbers.
(= <NUMBER> <NUMBER> ...)
Returns T if the numbers are equal, nil if not.
(/= <NUMB1> <NUMB2>)
Returns T if the numbers are not equal, nil if they are.
(< <NUMB1> <NUMB2>)
Returns T if <NUMB1> is less than <NUMB2>.
(<= <NUMB1> <NUMB2>)
Returns T if <NUMB1> is equal to or less than <NUMB2>.
(> <NUMB1> <NUMB2>)
Returns T if <NUMB1>is greater than <NUMB2>.
(>= <NUMB1> <NUMB2>)
Returns T if <NUMB1> is equal to or greater than <NUMB2>.
(~ <NUMBER>)
Returns the bitwise of <NUMBER>.
(1+ <NUMBER>)
Returns the <NUMBER> increased by one.
(1- <NUMBER>)
Returns the <NUMBER> decreased by one.
(abs <NUMBER>)
Returns the absolute value of <NUMBER>.
(and <EXPR> ...)
Returns the logical AND of a list of expressions.
Returns nil if expressions are nil.
(angle <PT1> <PT2>)
Returns the angle in radians between two points.
(angtos <ANGLE>[<MODE><PRECIS.>])
Returns the angle as a string. Mode 0=Degrees,1=Deg Min Sec,
(append <EXPR> ....)
Returns a number of lists as one list.
(apply <FUNCTION> <LIST>)
Executes function with the arguments given as a list.
(ascii <STRING>)
Returns ascii value of first character of a string.
(assoc <ITEM><ALIST>)
Searches <ALIST> for <ITEM>.
(atan <NUM1> [<NUM2>])
Returns the arctangent of <NUM1> or <NUM1>/<NUM2>.
(atof <STRING>)
Returns the conversion of a string into a real.
(atoi <STRING>)
Returns the conversion of a string into an integer.
(atom <ITEM>)
Returns T if <ITEM> is not a list.
(Boole <FUNC> <NUMB1> <NUMB2)
Returns Boolean bitwise truth comparisons.
(entdel <ENAME>) Deletes the entity <ENAME> or undeletes it if it has already been deleted. (eq <EXPR1> <EXPR2>) Returns T if expressions are bound to same object.. (chr <NUMBER>) Returns the string represented by an ascii value.) Evaluates any number of lists as arguments. (cdar <LIST>) Returns second element of the first element of list. (entnext [<ENAME>]) Returns the first non-deleted entity after <ENAME>. (entget <ENAME>) Returns <ENAME> from the database as a list. Use for polylines or blocks. (cond (<TEST1><RESULT1>). (equal <EXPR1> <EXPR2>) Returns T if the expressions are equal... (defun <SYM><ARG LIST><EXPR>. (cddr <LIST>) Returns all but first element of last element list. (close <FILENAME>) Closes a file.) Executes an AutoCAD command. (caddr <LIST>) Returns the third element of a list. (entmod <ENAME>) Updates <ENAME> in the database. (cdr <LIST>) Returns all but the first element of a list. (cons <NEW FIRST ELEMENT><LIST>) Contructs a new list. (command <ARGS>..) Defines an AutoLISP function. (boundp <ATOM>) Returns T if <ATOM> has a value bound to it. (eval <EXPR>) Evalutes an expression. NOT for blocks or polylines (entupd <ENAME>) Updates <ENAME> in the database. (cadr <LIST>) Returns the second element of a list. (cos <ANGLE>) Returns the cosine of an angle. (entlast) Returns the last non-deleted entity in the database. (entsel [<PROMPT>]) Selects a single entity by a point pick. (car <LIST>) Returns the first element of a list.<NUMB>'s must be integers. .. (caar <LIST>) Returns first element of the first element of a list. (distance <PT1> <PT2>) Returns the distance between two points. (cadar <LIST>) Returns last element of the first element of a list..
. (getdist [<PT>][<PROMPT>]) Pauses for input of distance at optional start point. (getvar <varname>) Retrieves the value of an AutoCAD system variable. (graphscr) Flips from text screen to graphics screen. (getangle [<PT>][<PROMPT>]) Pauses for input of an angle at optional starting point <PT>. (inters <pt1><pt2><pt3><pt4>[<onseg>]) Returns the intersection point of two lines. If <ONSEG> is greater than 0 the intersection may lie past the endpoints of the lines. (grread <track>) Allows AutoLISP to read AutoCADs input devices.. (float <NUMBER>) Converts <NUMBER> to a real. (fix <NUMBER>) Converts <NUMBER> to an integer. (foreach <NAME><LIST><EXPR>) Evaluates <EXPR> for each element in the <LIST> and assigns to <NAME>. (itoa <number>) Returns the conversion of an integer to a string. (grclear) Clears the graphics screen. (getcorner [<PT>][<PROMPT>]) Getpoint function . (getorient [<PT>][<PROMPT>]) Returns angle in radians from current orientation. (if <testexpr><thenexpr>[<elseexpr>]) Conditionally evaluates expressions.) Defines an "anonymous" function. (getkword [<PROMPT>]) Requests a keyword from the user. (gcd <NUM1> <NUM2>) Returns the greatest common denominator. (grtext <box><text><highlight>]) Write to text portion of graphics screen. (getenv [<VARIABLE>]) Returns string value assigned to a system variable. (lambda <arguments><expr>. (getreal [<PROMPT>]) Pauses for input of a real number. (getpoint [<PT>][<PROMPT>]) Pauses for input of an angle at optional point <PT>.(exp <NUMBER>) Returns E to the power <NUMBER>. (getstring [<CR>][<PROMPT>]) Pauses for input of a string. (findfile <FILENAME>) Searches all paths to find <FILENAME>. .draws triangle as crosshairs move. (grdraw <from><to><color>[<highlight>]) Draws a vector between two points. (expt <BASE> <NUMBER>) Returns <BASE> to the power <NUMBER>.
(member <expr><list>) Searches <LIST> for first occurence of <EXPR> and returns the remainder of the list..) Returns the logical bitwise OR of a list of numbers. (menucmd <string>) Switch between pages in an AutoCAD menu.(last <list>) Returns the last element in <LIST>.. (open <FILENAME><MODE>) Opens a file. (load <filename>) Loads an AutoLISP program in memory. (length <list>) Returns the number of elements in <LIST>.. (numberp <ITEM>) Returns T if <ITEM> is an integer or a real.. (polor <PT><ANGLE><DISTANCE>) Returns point at <ANGLE> and <DISTANCE> from <PT>. (nth <n><LIST>) Returns the "nth" element of <LIST>. (not <ITEM>) Returns T if the expression is nil.) Evalutes each expression sequentially.. (min <NUMBER><NUMBER>.) Returns the logical bitwise AND of a list of numbers. .) Returns the logical OR of a list of expressions. (or <EXPR>. (lsh <numb1><numbits>) Returns the logical bitwise SHIFT operator of <NUM1> by <NUMBITS> (mapcar <function><list1>. (progn <EXPR>. (log <number>) Returns natural log of <NUMBER> as a real. (minusp <ITEM>) Returns T if the item is negative. (null <ITEM>) Returns T if <ITEM> is bound to nil. (max <number><number>. (print <EXPR>[<FILENAME>]) Same as prin1 except a new line is printed before <EXPR> and a space is printed after. (prin1 <EXPR>[<FILENAME>]) Prints <EXPR> on the screen or to <FILENAME>..<listn>) Executes <FUNCTION> on the elements of <LIST1> through <LIST2>... (logand <number><number>. (princ <EXPR>[<FILENAME>]) Same as prin1 except control character are printed without expansion.. (osnap <PT><MODE-STRING>) Returns a point by applying the osnap command.... (list <expr>) Returns the list of a number of expressions. (listp <list>) Returns T if the item is a list. (logior <number><number>.) Returns the maximum of the numbers given.) Returns the smallest of the numbers given...
. (strcat <STRING1><STRING2>. (ssname <SS><INDEX>) Returns the name of the <INDEX>th entity in selection set <SS>. (set <SYM><EXPR>) Sets the value of a sysmbol name.) Concatenates a series of strings. (rem <NUM1> <NUM2> . (sslength <SS>) Returns the number of entities in selection set <SS>... (subst <NEWITEM><OLDITEM><LIST>) Substitutes <NEWITEM> for <OLDITEM> in <LIST>. If <WHICH> is not nill the returned string will be all lower case.. (setvar <VARNAME><VALUE>) Sets a system variable to <VALUE>.. (tblnext <NAME> [<rewind>]) Scans symbol table for next entity.) Evaluates each expression <NUMBER> times..) Sets the value of <SYM1> to <EXPR1>. (ssadd [<ENAME>][<SS>]) Adds entities to selections sets. or <ENAME> in <MODE>. (read-line [<FILENAME>]) Returns a string from the keyboard buffer or from <FILENAME>. (reverse <LIST>) Returns a list reversed. (ssmemb [<ENAME>][<SS>]) Returns <ENAME> if <ENAME> is a member of selections set <SS>. (ssdel [<ENAME>][<SS>]) Deletes entities from selections sets. (sqrt <NUMBER>) Returns the square root of <NUMBER>. (read <STRING>) Returns the first LIST obtained from <STRING>. (read-char [<FILENAME>]) Returns a character from the keyboard buffer or from <FILENAME>.) Returns the remainder of <NUM1>/<NUM2>.. (substr <STRING><START><LENGTH>) Returns a substring of <STRING> begining at <START> of <LENGTH> characters..(prompt <MSG>) Displays <MSG> on the screen PROMPT area. (strcase <STRING>[<WHICH>)] Returns <STRING> as upper case if <WHICH> is nill. (redraw [<ENAME>][<MODE>]) Redraws either entire drawing. (repeat <NUMBER> <EXPR> . (quote <EXPR>) Returns <EXPR> unevaluated. (sin <ANGLE>) Returns the sine of an angle. (strlen <STRING>) Returns the number of characters in a string. (setq <SYM1><EXPR1>. (ssget [<MODE>][<PT1><PT2>]) Returns a selection set. (rtos <NUMBER> [<MODE><PRECIS>]) Returns a real number as a string.
(zerop <ITEM>) Returns T if <ITEM> evaluates to zero. (*error* <STRING>) User-definable error function.. (terpri) Prints a new line on the screen.. (trace <FUNCTIONS>. (type <ITEM>) Returns the type of <ITEM>.. (write-line <STRING> [<FILENAME>]) Writes <STRING> to the screen or to <FILENAME>.. (textscr) Flips from graphics screen to text screen. (trans <PT><FROM><TO>[<DISP>]) Translates a point <PT> from one coordinate system to another. (while <TESTEXPR><EXPR>) Evaluates <TESTEXPR> and if not nill executes <EXPR>.) Clears the trace floag for <FUNCTIONS>. (write-char <NUM> [<FILENAME>]) Writes the character represented by the ASCII value of <NUM> to the screen or to <FILENAME>.(tblsearch <NAME><SYM>[<setnext>]) Searches entire symbol table for <NAME>. (vmon) Enables AutoLISP virtual function pager. (ver) Returns AutoLISP version number.) Debugging aid that returns the last function name. (vports) Returns a list of currently active viewports. . (untrace <FUNCTIONS>. | 2,807 | 10,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-51 | latest | en | 0.597853 |
https://www.2020vw.com/23/adding-and-subtracting-with-decimals-worksheet/ | 1,709,572,094,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476464.74/warc/CC-MAIN-20240304165127-20240304195127-00192.warc.gz | 612,741,023 | 66,311 | Worksheet
# Adding And Subtracting With Decimals Worksheet: A Comprehensive Guide
As a professional writer, I understand the importance of providing helpful content that can aid learners in their education. In this article, I will discuss everything you need to know about adding and subtracting with decimals worksheet, including the main content, frequently asked questions, pros, and tips.
Adding and subtracting with decimals can be challenging for learners, but it is an essential skill they must acquire to excel in their academics. In this section, we will explore the step-by-step process of adding and subtracting decimals.
Step 1: Write the decimals in columns, lining up the decimal points.
Step 2: Fill in any missing places with zeros.
Step 3: Begin by adding or subtracting the digits in the ones place value.
Step 4: Move to the tenths place, and add or subtract the digits.
Step 5: Continue the process until you reach the last decimal place value.
Related Articles
It is important to note that learners must practice regularly to master this skill.
• What is the importance of adding and subtracting with decimals?
Adding and subtracting with decimals is an important skill that learners must acquire to excel in their academics and daily life.
• What are some common mistakes learners make when adding and subtracting with decimals?
Some common mistakes include forgetting to line up the decimal points, forgetting to fill in the missing places with zeros, and miscalculating the answer.
• How can learners improve their skills in adding and subtracting with decimals?
Learners can improve their skills by practicing regularly, seeking help from their teachers or peers, and using online resources.
• What are some real-life applications of adding and subtracting with decimals?
Some real-life applications include calculating money, measuring ingredients in cooking, and recording stock prices in finance.
• What are the key concepts learners must understand when adding and subtracting with decimals?
Learners must understand place value, decimals, and basic arithmetic operations.
• What are some effective strategies for learners struggling with adding and subtracting with decimals?
Some effective strategies include breaking down the problem into smaller parts, using visual aids, and seeking help from their teachers or peers.
• What are some common misconceptions about adding and subtracting with decimals?
Some common misconceptions include thinking that decimals are less important than whole numbers and assuming that adding and subtracting with decimals is too difficult to learn.
• How can parents support their children in learning how to add and subtract with decimals?
Parents can support their children by providing additional resources, encouraging regular practice, and seeking help from their child’s teacher.
## Pros
Adding and subtracting with decimals is an essential skill that learners must acquire to excel in their academics and daily life. It helps them understand basic arithmetic operations, improve their problem-solving skills, and prepares them for more advanced mathematics.
## Tips
Some tips for learners include practicing regularly, seeking help from their teachers or peers, breaking down the problem into smaller parts, using visual aids, and using online resources.
## Summary
In conclusion, adding and subtracting with decimals worksheet is a crucial skill that learners must acquire to excel in their academics and daily life. By following the step-by-step process, seeking help from their teachers or peers, and practicing regularly, learners can improve their skills and become confident in their abilities. | 677 | 3,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-10 | latest | en | 0.928138 |
https://library.soton.ac.uk/mathsandstats/mean-median-and-mode | 1,726,632,150,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00615.warc.gz | 321,076,872 | 15,098 | # Mean, Median and Mode: Maths and Stats
### Mean, Median and Mode
Measures of central tendency can help describe and summarise quantitative data, which in turn helps us understand it better. The mean, median and mode are all examples of measures of central tendency.
#### Mean
The arithmetic mean is the arithmetic average of all the values in the data set, and is calculated by adding all the values in the data set together and then dividing by the total number of values.
The Greek letter µ (pronounced 'mew') is used to denote the mean of a population, whereas (pronounced 'x bar') is used to denote the mean of a sample. Therefore, we can write
and
for the formulas of the population and sample mean respectively, where N is the number of data points in the population and n is the total number of data points in the sample.
Observe the following data set:
7, 12, 5, 18, 5, 9, 10, 9, 12, 8, 12, 16
which has twelve values (that is, n=12). The mean value is therefore calculated to be:
#### Median
The median is the middle value of the data set, provided that the set has been ordered numerically. In general, for a data set of size n, the median is the
-th
value.
For example, observe the data set we had before:
7, 12, 5, 18, 5, 9, 10, 9, 12, 8, 12, 16
To find the median, we need to order it:
5, 5, 7, 8, 9, 9, 10, 12, 12, 12, 16, 18
Now, since there are twelve values (remember, n=12), we will need to find the -th value, or the 6.5th value - which is simply the number between the 6th and 7th values. Here, we have the sixth value as 9 and the 7th as 10, and therefore the median value is:
Notice that, like with the mean, this median value does not actually appear anywhere in our dataset. That's perfectly fine, sometimes this happens.
#### Mode
The mode is the value which appears the most frequently in the data set. For this reason, it is possible for data sets to have more than one mode in it. If there exist two modes in a data set, then it is said to be 'bimodal'; if there exist more than two, then it is said to be 'multi-modal'.
For example, observe the following observations:
7, 12, 5, 18, 5, 9, 10, 9, 12, 8, 12, 16
Here, we can see that the mode of this data set is 12, as it occurs the most frequently. | 631 | 2,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-38 | latest | en | 0.916303 |
https://medium.com/fictional-mathematics/time-funnels-2ca8ef4820b5?source=collection_home---5------0--------------------- | 1,555,709,190,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578528058.3/warc/CC-MAIN-20190419201105-20190419223105-00290.warc.gz | 498,197,804 | 23,564 | # Time Funnels
[Continuation of the time trajectory-theme from here.]
F
Funnels
In considering time tunnels, we may want to think of how they are unlike tunnels. One such would be that the event-objects, event-lines and event-points — the class of Eventuals — that participate in a tunnel are not passive in the sense we think of the constituents of a material tunnel.
We may think of the Eventuals as constraints as much as potentials. When Eventual turns Actual the Eventual also becomes a restraint upon all futures involving the tunnel and near-by, bundled tunnels. It will often be the case that several things are potential, but that they may not all come to pass. A couple of parents may at most have children two times in the same year, and most likely not. No matter how many children are born, the number of times this happens will not exceed two in a year, hardly ever one in a year.
We can thus except a lot of possible scenarios from happening when we know something has happened. In the everyday real world this is trivial, but for time tunnels, it must not be.
If Eventuals are constraints, we can consider an intersection of several eventlines as on the one hand increasing the odds for certain scenarios, on the other hand to decrease certain other scenarios. We can think of the tunnel as funneling.
Points and Objects
Difference between point and object:
1. Event-points would be the singular event as singular event. An intersection, a crossing or forking of paths.
2. Event-objects would be an event as a system of points. A “body”. | 340 | 1,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-18 | latest | en | 0.971245 |
https://www.exodusbooks.com/math-u-see-primer-student-text/68449/ | 1,638,110,660,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00326.warc.gz | 860,184,735 | 9,103 | # Math-U-See Primer - Student Text
Publisher: Math-U-See
Item: 68449
Consumable Workbook
Current Retail Price: \$36.00
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Focus: Have Fun Introducing Math
Prerequisite: The student is beginning to count and quantify.
Number Recognition and Counting from 0-9 Writing Numerals Geometric Shapes: Rectangles Geometric Shapes: Circles Geometric Shapes: Triangles Place Value: Units and Tens Place Value: Hundreds Unit Bars or Weaning from the Green Addition: Introduction and Symbolism Addition: +1 Counting to 20 Addition: 2+2 and 3+3, Vertical Geometric Shapes: Squares Addition: 4+4, 5+5 Skip Counting by 2 Addition of Tens Skip Counting by 10 Addition of Hundreds Solving for an Unknown Skip Counting by 5 Tally Marks Addition: Making 10 Skip Count to Find Area Telling Time with Minutes Telling Time with Hours Telling Time with Minutes and Hours Subtraction: Introduction and Symbolism Subtraction: -1
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# GMAT Test M07 Master Thread - All discussions
Author Message
GMAT Club team member
Joined: 16 Mar 2009
Posts: 115
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15 May 2009, 02:04
Kudos [?]: 996 [0], given: 19
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2761
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Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
### Show Tags
11 Jul 2010, 03:37
This was the first test where I had to switch to review mode as I was getting short of time.
Some of the questions were really time consuming.
Number of PS questions were quite > number of DS questions.
This could be the reason.
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### Show Tags
21 Sep 2012, 06:20
Hi Buneul,
A train traveling at a certain constant speed takes 30 seconds longer to travel 2 kilometers than it would take to travel 4 kilometers at 120 kilometers per hour. At what speed, in kilometers per hour, is the train traveling?
Is this question correct? ( Q 11)
I some how spent 30 minutes and still not able to see how 80 can be the answer,
it must be 48 right??
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Math Expert
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Posts: 41872
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### Show Tags
21 Sep 2012, 06:30
shankar245 wrote:
Hi Buneul,
A train traveling at a certain constant speed takes 30 seconds longer to travel 2 kilometers than it would take to travel 4 kilometers at 120 kilometers per hour. At what speed, in kilometers per hour, is the train traveling?
Is this question correct? ( Q 11)
I some how spent 30 minutes and still not able to see how 80 can be the answer,
it must be 48 right??
Answer is indeed C. Check the screenshot below:
Attachment:
Train.png [ 26.69 KiB | Viewed 1164 times ]
P.S. Please post the questions in the respective forums.
_________________
Kudos [?]: 128639 [0], given: 12181
Re: GMAT Test M07 Master Thread - All discussions [#permalink] 21 Sep 2012, 06:30
Display posts from previous: Sort by
# GMAT Test M07 Master Thread - All discussions
Moderator: Bunuel
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 981 | 3,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-43 | latest | en | 0.876529 |
https://www.convert-measurement-units.com/convert+Centiare+to+Square+micrometer.php | 1,713,924,818,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00490.warc.gz | 638,752,085 | 13,739 | Convert ca to µm² (Centiare to Square micrometer)
## Centiare into Square micrometer
numbers in scientific notation
Direct link to this calculator:
https://www.convert-measurement-units.com/convert+Centiare+to+Square+micrometer.php
# Convert Centiare to Square micrometer (ca to µm²):
1. Choose the right category from the selection list, in this case 'Area'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Centiare [ca]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Square micrometer [µm²]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '379 Centiare'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Centiare' or 'ca'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Area'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '27 ca to µm2' or '59 ca into µm2' or '88 Centiare -> Square micrometer' or '58 ca = µm2' or '46 Centiare to µm2' or '15 ca to Square micrometer' or '51 Centiare into Square micrometer'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(34 * 75) ca'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '379 Centiare + 1137 Square micrometer' or '92mm x 12cm x 60dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.608 901 219 926 9×1029. For this form of presentation, the number will be segmented into an exponent, here 29, and the actual number, here 1.608 901 219 926 9. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.608 901 219 926 9E+29. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 160 890 121 992 690 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 900 | 3,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-18 | latest | en | 0.848891 |
https://questions.examside.com/past-years/jee/question/ptwo-dice-a-and-b-are-rolled-let-the-numbers-obtained-on-jee-main-mathematics-trigonometric-functions-and-equations-pwwbhip1zkf9vfhh | 1,716,493,650,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058653.47/warc/CC-MAIN-20240523173456-20240523203456-00204.warc.gz | 410,879,047 | 53,107 | 1
JEE Main 2023 (Online) 12th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Two dice A and B are rolled. Let the numbers obtained on A and B be $$\alpha$$ and $$\beta$$ respectively. If the variance of $$\alpha-\beta$$ is $$\frac{p}{q}$$, where $$p$$ and $$q$$ are co-prime, then the sum of the positive divisors of $$p$$ is equal to :
A
48
B
31
C
72
D
36
2
JEE Main 2023 (Online) 11th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Let $$S=\left\{M=\left[a_{i j}\right], a_{i j} \in\{0,1,2\}, 1 \leq i, j \leq 2\right\}$$ be a sample space and $$A=\{M \in S: M$$ is invertible $$\}$$ be an event. Then $$P(A)$$ is equal to :
A
$$\frac{47}{81}$$
B
$$\frac{49}{81}$$
C
$$\frac{50}{81}$$
D
$$\frac{16}{27}$$
3
JEE Main 2023 (Online) 10th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Let a die be rolled $$n$$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $$\frac{k}{2^{15}}$$, then $$\mathrm{k}$$ is equal to :
A
15
B
60
C
30
D
90
4
JEE Main 2023 (Online) 10th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that $${2^N} < N!$$ is $${m \over n}$$, where m and n are coprime, then $$4m-3n$$ is equal to :
A
12
B
6
C
8
D
10
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Class 12 | 568 | 1,555 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-22 | latest | en | 0.756865 |
https://www.sanfoundry.com/dc-machines-questions-answers-wave-winding/ | 1,723,722,422,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641291968.96/warc/CC-MAIN-20240815110654-20240815140654-00834.warc.gz | 735,768,180 | 18,929 | # DC Machines Questions and Answers – Wave Winding
This set of DC Machines Multiple Choice Questions & Answers (MCQs) focuses on “Wave Winding”.
1. What will be the value of “Yf + Yb” for a wave winding?
a) Equal to Yc
b) Half of the Yc value
c) Double of the Yc value
d) Four times Yc value
Explanation: In the wave winding, as the number of coil-sides is double the number of segments, the top coil-side of the second coil will be numbered as (1+2*Yc). After numbering other coil sides,
1 + 2*Yc – Yf = 1+ Yb
So Yf + Yb = 2Yc.
2. For a progressive wave winding Yc = ______
a) 2C/P
b) 2(C+1)/P
c) 2(C-1)/P
d) 2C/(P+1)
Explanation: Starting at segment 1 and after going through P/2 coils or Yc (P/2) segments, the winding should end in segment 2 for progressive winding or segment (C) for retrogressive winding. That is mathematically,
Yc (P/2) = (C+1)
Yc = 2(C+1)/P
3. Number of parallel paths in wave winding are ______
a) Equal to P
b) Equal to P/2
c) 2
d) Depends on other parameters
Explanation: In wave winding all coils are divided into 2 groups- all coils carrying clockwise current are series connected and so are all coils with counter-clockwise current- and these 2 groups are in parallel because the winding is closed. Thus, a wave winding has 2 parallel paths irrespective of number of poles.
4. What is the spacing between the brushes for a wave winding when a machine is 6-pole DC armature with 16 slots having 2-coil sides per slot and single-turn coils.
a) 4 segments
b) 8 segments
c) 16 segments
d) 12 segments
Explanation: Only 2 brushes are required in this case as the number of poles in wave winding is equal to 2. So, spacing between the brushes is equal to total number of segments i.e. total slots divided by 2. Spacing between brushes = C/A = 16/2 = 8 segments.
5. What is the relation between conductor current and armature current in wave winding?
a) Ic = Ia
b) Ic = 2Ia
c) Ic = 4Ia
d) Ic = Ia/2
Explanation: the number of parallel paths in the in a wave winding is equal to 2. So, armature current will get divided equally into total number of conductors/paths. Conductor current in a wave wounded machine is half of the Ia.
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6. For a conductor current equal to 4mA, Current carried by a particular brush in a 2-pole machine will be _____
a) 16mA
b) 8mA
c) 2mA
d) 10mA
Explanation: Conductor current in a wave wounded machine is half of the Ia. So, Ia= 8mA. All positive and all negative brushes are respectively connected in parallel to feed the external circuit. Thus, IBRUSH = Ia /(P/2). Solving we get Brush current = 8mA.
7. Equalizer rings are needed in the wave winding.
a) True
b) False
Explanation: The armature coil forms 2 parallel paths under the influence of all pole-pairs so that the effect of the magnetic circuit asymmetry is equally present in both the parallel paths resulting in equal parallel-path voltages. Thus, equalizer rings are not needed in wave winding.
8. For a wave winding when a machine is 6-pole DC armature with 16 slots having 2-coil sides per slot and single-turn coil, Yf value is ____
a) 5
b) 3
c) 2
d) 7
Explanation: Ycs = 16/6 = 2 slots (nearest lower integral value)
Yb= 2*2+1 = 5
Yc= 2(16-1)/6 = 5 segments
Yf = 2Yc – Yb = 5.
9. Wave winding machines are used in ______ currents applications.
a) High
b) Moderate
c) Low
d) Can be used anywhere
Explanation: Lap winding machine has the advantage of large number of parallel paths and lower conductor current and is therefore used in low voltage and high current applications. Wave winding has fixed number of parallel paths so, wave wounded machine is used in low currents application.
10. For a wave wounded machine number of brushes for small, large machines respectively is ________ _________
a) 2, 2
b) 4, 2
c) 2, P
d) Both values depend on the given conditions
Explanation: For a small wave wounded machine number of parallel paths are 2, thus 2 brushes are used. For a large machine total number of brushes is equal to the total number of poles. The spacing between adjacent brushes is C/P commutator segments.
Sanfoundry Global Education & Learning Series – DC Machines.
To practice all areas of DC Machines, here is complete set of
1000+ Multiple Choice Questions and Answers
.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] | 1,214 | 4,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-33 | latest | en | 0.923329 |
http://at.metamath.org/ileuni/pm3.12dc.html | 1,606,912,401,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141708017.73/warc/CC-MAIN-20201202113815-20201202143815-00466.warc.gz | 8,011,619 | 3,500 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > pm3.12dc Structured version GIF version
Theorem pm3.12dc 864
Description: Theorem *3.12 of [WhiteheadRussell] p. 111, but for decidable propositions. (Contributed by Jim Kingdon, 22-Apr-2018.)
Assertion
Ref Expression
pm3.12dc (DECID φ → (DECID ψ → ((¬ φ ¬ ψ) (φ ψ))))
Proof of Theorem pm3.12dc
StepHypRef Expression
1 pm3.11dc 863 . . . 4 (DECID φ → (DECID ψ → (¬ (¬ φ ¬ ψ) → (φ ψ))))
21imp 115 . . 3 ((DECID φ DECID ψ) → (¬ (¬ φ ¬ ψ) → (φ ψ)))
3 dcn 745 . . . . . 6 (DECID φDECID ¬ φ)
4 dcn 745 . . . . . 6 (DECID ψDECID ¬ ψ)
5 dcor 842 . . . . . 6 (DECID ¬ φ → (DECID ¬ ψDECIDφ ¬ ψ)))
63, 4, 5syl2im 34 . . . . 5 (DECID φ → (DECID ψDECIDφ ¬ ψ)))
7 dfordc 790 . . . . 5 (DECIDφ ¬ ψ) → (((¬ φ ¬ ψ) (φ ψ)) ↔ (¬ (¬ φ ¬ ψ) → (φ ψ))))
86, 7syl6 29 . . . 4 (DECID φ → (DECID ψ → (((¬ φ ¬ ψ) (φ ψ)) ↔ (¬ (¬ φ ¬ ψ) → (φ ψ)))))
98imp 115 . . 3 ((DECID φ DECID ψ) → (((¬ φ ¬ ψ) (φ ψ)) ↔ (¬ (¬ φ ¬ ψ) → (φ ψ))))
102, 9mpbird 156 . 2 ((DECID φ DECID ψ) → ((¬ φ ¬ ψ) (φ ψ)))
1110ex 108 1 (DECID φ → (DECID ψ → ((¬ φ ¬ ψ) (φ ψ))))
Colors of variables: wff set class Syntax hints: ¬ wn 3 → wi 4 ∧ wa 97 ↔ wb 98 ∨ wo 628 DECID wdc 741 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-in1 544 ax-in2 545 ax-io 629 This theorem depends on definitions: df-bi 110 df-dc 742 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 669 | 1,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-50 | longest | en | 0.206552 |
https://programsandcourses.anu.edu.au/2021/course/engn1217 | 1,718,506,034,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00152.warc.gz | 409,974,736 | 12,253 | • Offered by School of Engineering
• ANU College ANU College of Engineering and Computer Science
• Course subject Engineering
• Areas of interest Materials Science, Mathematics, Physics, Engineering, Mechatronics More...
• Course convener
• Dr Kiara Bruggeman
• Mode of delivery In Person
• Offered in Second Semester 2021
Introduction to Mechanics (ENGN1217)
Course has been adjusted for remote participation in 2021. Some on-campus activities are available. Attendance at these where possible is encouraged.
This course introduces the fundamental principles in mechanics. Structural design applications of a variety of problems are developed throughout the course using examples that elucidate the theory of mechanics. The primary aim of this course is to provide a solid foundation for students in the field of mechanical engineering. Specific topics include:
• Statics of particles; forces in a plane and in space
• Equivalent systems of forces; Cartesian vector and vector operations
• Equilibrium of rigid bodies; including free body diagrams
• Simple trusses
• Distributed forces; centroids, centres of gravity and moments of inertia
• Static indeterminacy and friction
• Internal forces; including shear and moment diagrams
## Learning Outcomes
Upon successful completion, students will have the knowledge and skills to:
Upon completion of this course, students will have the knowledge and skills to-
1. Identify and manipulate forces and their resultants in one- two- and three dimensions.
2. Recognise and classify moments and couples created by forces.
3. Employ mechanical equilibrium and free body diagrams to solve mechanical statics problems, including bending moment diagrams.
4. Acquire skills for testing the bending of a beam and the construction of a bridge model subject to testing.
5. Analyse and demonstrate the stability conditions of mechanical equilibrium.
6. Define and evaluate the fundamentals of mechanical testing of materials (tension, compression, shear).
Professional Skills Mapping:
Mapping of Learning Outcomes to Assessment and Professional Competencies
## Indicative Assessment
Online assignments (16%); Mid semester exam (14%); Bridge competition (20%); Final exam (40%); Lab assignments (10%)
The ANU uses Turnitin to enhance student citation and referencing techniques, and to assess assignment submissions as a component of the University's approach to managing Academic Integrity. While the use of Turnitin is not mandatory, the ANU highly recommends Turnitin is used by both teaching staff and students. For additional information regarding Turnitin please visit the ANU Online website.
A standard workload of approximately 10 hours per week, including 4-5 contact
hours (lectures, tutorials, laboratories) as well as time for preparation, assignments
## Prescribed Texts
Engineering Mechanics STATICS, 13th edition, R.C. Hibbeler
(No other text)
## Assumed Knowledge
Students are assumed to have achieved a level of knowledge of mathematics comparable to at least ACT Mathematics Methods or NSW Mathematics or equivalent.
## Areas of Interest
• Materials Science
• Mathematics
• Physics
• Engineering
• Mechatronics
• Materials
• Mechanical
• Renewable Energy
• Biotechnology
• Energy Change
## Fees
Tuition fees are for the academic year indicated at the top of the page.
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If you have been offered a Commonwealth supported place, your fees are set by the Australian Government for each course. At ANU 1 EFTSL is 48 units (normally 8 x 6-unit courses). More information about your student contribution amount for each course at Fees
Student Contribution Band:
2
Unit value:
6 units
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## Offerings, Dates and Class Summary Links
ANU utilises MyTimetable to enable students to view the timetable for their enrolled courses, browse, then self-allocate to small teaching activities / tutorials so they can better plan their time. Find out more on the Timetable webpage.
The list of offerings for future years is indicative only.
Class summaries, if available, can be accessed by clicking on the View link for the relevant class number.
### Second Semester
Class number Class start date Last day to enrol Census date Class end date Mode Of Delivery Class Summary
5350 26 Jul 2021 02 Aug 2021 14 Sep 2021 29 Oct 2021 In Person N/A | 1,032 | 4,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.890727 |
http://www.talkstats.com/search/3773852/ | 1,669,599,307,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00848.warc.gz | 99,641,302 | 7,306 | # Search results
1. ### Compare regression coefficients
For reference, you can take a look at constructing a Bland-Altman plot in Excel.
2. ### Is ANOVA right for me?
This is an example of, as staassis mentioned above, a repeated-measures ANOVA, which differs from a standard one-way ANOVA. For a quick primer on repeated measures ANOVA, check out this tutorial: https://www.statology.org/one-way-repeated-measures-anova/
3. ### Simple Probability
For anyone struggling with a similar probability problem, you might find this tutorial helpful: http://www.statology.org/probability-using-sample-spaces/
4. ### Sum of two normal distributions
You might find this brief tutorial helpful on combining random variables: http://www.statology.org/combining-random-variables/ Basically, if you have random variables X and Y, then when you add X and Y, you can form a new random variable Z. The mean of Z is μZ = μX + μY The variance of Z is... | 223 | 940 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-49 | latest | en | 0.751302 |
https://blogmymaze.wordpress.com/2012/12/01/flower-of-life-and-labyrinths-with-multiple-arms/ | 1,576,116,898,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540534443.68/warc/CC-MAIN-20191212000437-20191212024437-00176.warc.gz | 293,301,435 | 23,711 | Feeds:
Posts
## Flower of Life and Labyrinths with Multiple Arms
In my last post I have shown how one-arm labyrinths can be inscribed into the Flower of Life. Now I will also consider labyrinths with muliple arms. I begin with the labyrinth with three arms and three circuits that has already been introduced by Erwin in several previous posts (see also related posts below).
Figure 1: The Labyrinth with three arms and three ciruits
It is possible to inscribe the Ariadne’s Thread of this labyrinth into the original Flower of Life using the same previously described method by following a pathway along the lentiform segments.
Figure 2: The three-arm three-circuit labyrinth in the Flower of Life
However the resulting Ariadne’s Thread is strongly distorted. This is due to the fact that the center is too narrow. The innermost circuit therefore is made up mostly of parallel vertical lines. This makes it difficult to realize that it is a circuit at all.
The solution is to size-up the center so that the innermost circuit comes to lie on the second concentric hexagon. As shown in figure 3 this enables us to give a hexagonal shape also to the innermost circuit. By this, the entire labyrinth is far better recognizable in its intended form, although on a hexagonal layout.
Figure 3: Four circles for three circits with a larger center
In order to inscribe multiple-arm labyrinths into the hexagonal grid, a bigger area is needed. This must have a diameter that is one circle greater than the number of circuits of the labyrinth. Let’s examine this with a larger labyrinth. And, given the hexagonal grid, why not inscribe the Ariadne’s Thread of one of my labyrinth designs with six arms. Figure 4 shows my labyrinth KS 3-3 of the category of my Cascading Serpentine labyrinths with six arms and 11 circuits.
Figure 4: Labyrinth of the type KS 3-3 on the hexagonal grid from the Flower of Life
In order to inscribe the Ariadne’s Thread of this labyrinth into the hexagonal grid, an area with a diameter of 12 circles is needed (11 circuits + 1 additional circle for sizing-up the center).
The Flower of Live provides a hexagonal grid. This grid can be extended without limits. And it is possible to inscribe labyrinths with a greater number of arms and circuits on larger areas covered with this grid.
Related Posts:
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This site uses Akismet to reduce spam. Learn how your comment data is processed. | 593 | 2,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-51 | longest | en | 0.927981 |
https://edurev.in/course/quiz/attempt/7326_Test-Communication-System-6/4fa1915b-dc55-44de-bf73-d3659753793e | 1,653,548,935,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00010.warc.gz | 281,694,354 | 43,747 | # Test: Communication System- 6
## 25 Questions MCQ Test Basic Electronics Engineering for SSC JE (Technical) | Test: Communication System- 6
Description
Attempt Test: Communication System- 6 | 25 questions in 50 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Basic Electronics Engineering for SSC JE (Technical) for Electrical Engineering (EE) Exam | Download free PDF with solutions
QUESTION: 1
Solution:
QUESTION: 2
Solution:
QUESTION: 3
### The Nyquist sampling interval, for the signal sin c(700t) + sin c(500t) is
Solution:
QUESTION: 4
The Nyquist rate for the signal x(t) = 2 cos (2000 π t) cos (5000 π t), is
Solution:
QUESTION: 5
A signal occupies a band 5 kHz to 10 kHz. For proper error free reconstruction at what rate it should be sampled
Solution:
QUESTION: 6
Which one of the following pulse communication systems is digital
Solution:
QUESTION: 7
A signal is sampled at 8 kHz and is quantized using 8-bit uniform quantizer. Assuming SNRqfor a sinusoidal signal, the correct statement for PCM signal with a bit rate of R is
Solution:
QUESTION: 8
What are the three steps in generating PCM in the correct sequence
Solution:
QUESTION: 9
In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor
Solution:
QUESTION: 10
In a PCM system each quantization level is encoded into 8 bits. The signal to quantisation noise ratio is equal to
Solution:
QUESTION: 11
If the number of bits per sample in a PCM system is increased, the improvement in signal-to-quantization noise ratio will be
Solution:
QUESTION: 12
The main factor that determines the accuracy of a reconstructed PCM signal is the
Solution:
QUESTION: 13
The quantization noise of a PCM system depends on
Solution:
QUESTION: 14
Companding is used
Solution:
QUESTION: 15
Companding is used in PCM to
Solution:
QUESTION: 16
Which of the following pulse modulations are digital
1. PCM
2. Differential PCM
3. PWM
Select the correct answer using the codes given below:
Solution:
QUESTION: 17
"Slope overload" occurs in delta modulation when the
Solution:
QUESTION: 18
To overcome slope overload problem, which type of integrator is used in delta modulation
Solution:
QUESTION: 19
In delta modulation, the slope overload distortion can be reduced by
Solution:
QUESTION: 20
Which one of the following modulation technique is most affected by noise
Solution:
QUESTION: 21
In case of data transmission, which one of the following systems will give the maximum probability error
Solution:
QUESTION: 22
If S represents the carrier synchronization atthe receiver and r represents the bandwidth efficiency, then the correct statement for the coherent binary PSK is
Solution:
QUESTION: 23
If the probability of a message is 1/4, then the information in bits is
Solution:
QUESTION: 24
The capacity of a channel is given by the
Solution:
QUESTION: 25
At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by
Solution:
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code | 741 | 3,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-21 | latest | en | 0.867264 |
https://iss.oden.utexas.edu/?p=projects/galois/analytics/gpu-sgd | 1,660,480,828,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572033.91/warc/CC-MAIN-20220814113403-20220814143403-00623.warc.gz | 312,507,090 | 3,904 | # Matrix Completion
## Problem Description
This benchmark implements Stochastic Gradient Descent (SGD). In particular, the benchmark uses SGD to complete unknown entries of a sparse matrix R. The sparse matrix represents a bipartite graph, with one set of nodes represent movies, while the other set represents users. The edge connecting a movie node to a user node denotes that the user has rated the movie, with the edge label representing the rating assigned.
The goal of the benchmark is to build two matrices, A (1xm) and B (1xu), such that the matrix product AB approximately equals R, the original ratings matrix described above. Each entry of A and B is a vector of real numbers and is called a latent vector.
After A and B have been calculated, they can be used to predict unknown entries of R.
## Algorithm
First, the entries of the latent vector are initialized to random numbers. Then the ratings are normalized.
Then for every edge in the graph R, we apply stochastic gradient descent as follows:
```function gradient_update(edge) {
Mlv = edge.src.lv
Ulv = edge.dst.lv
diff = edge.rating - dotproduct(Mlv, Ulv)
update(Mlv, diff)
update(Ulv, diff)
}
```
The key problem in a parallel implementation is to ensure that the gradient_update procedure has exclusive access to the end-points of the edge. Although this is conceptually simple, the benchmarks show that different locking strategies lead to different schedules, which in turn lead to very wide differences in performance. This is especially the case when applied to scale-free graphs. | 326 | 1,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-33 | latest | en | 0.905477 |
https://metanumbers.com/18331 | 1,632,179,322,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057119.85/warc/CC-MAIN-20210920221430-20210921011430-00163.warc.gz | 435,037,388 | 7,316 | # 18331 (number)
18,331 (eighteen thousand three hundred thirty-one) is an odd five-digits composite number following 18330 and preceding 18332. In scientific notation, it is written as 1.8331 × 104. The sum of its digits is 16. It has a total of 2 prime factors and 4 positive divisors. There are 17,512 positive integers (up to 18331) that are relatively prime to 18331.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 16
• Digital Root 7
## Name
Short name 18 thousand 331 eighteen thousand three hundred thirty-one
## Notation
Scientific notation 1.8331 × 104 18.331 × 103
## Prime Factorization of 18331
Prime Factorization 23 × 797
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 18331 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 18,331 is 23 × 797. Since it has a total of 2 prime factors, 18,331 is a composite number.
## Divisors of 18331
1, 23, 797, 18331
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 19152 Sum of all the positive divisors of n s(n) 821 Sum of the proper positive divisors of n A(n) 4788 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 135.392 Returns the nth root of the product of n divisors H(n) 3.82853 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 18,331 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 18,331) is 19,152, the average is 4,788.
## Other Arithmetic Functions (n = 18331)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 17512 Total number of positive integers not greater than n that are coprime to n λ(n) 8756 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2098 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 17,512 positive integers (less than 18,331) that are coprime with 18,331. And there are approximately 2,098 prime numbers less than or equal to 18,331.
## Divisibility of 18331
m n mod m 2 3 4 5 6 7 8 9 1 1 3 1 1 5 3 7
18,331 is not divisible by any number less than or equal to 9.
## Classification of 18331
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (18331)
Base System Value
2 Binary 100011110011011
3 Ternary 221010221
4 Quaternary 10132123
5 Quinary 1041311
6 Senary 220511
8 Octal 43633
10 Decimal 18331
12 Duodecimal a737
20 Vigesimal 25gb
36 Base36 e57
## Basic calculations (n = 18331)
### Multiplication
n×y
n×2 36662 54993 73324 91655
### Division
n÷y
n÷2 9165.5 6110.33 4582.75 3666.2
### Exponentiation
ny
n2 336025561 6159684558691 112913177645364721 2069811459417180700651
### Nth Root
y√n
2√n 135.392 26.3671 11.6358 7.12258
## 18331 as geometric shapes
### Circle
Diameter 36662 115177 1.05566e+09
### Sphere
Volume 2.58016e+13 4.22262e+09 115177
### Square
Length = n
Perimeter 73324 3.36026e+08 25923.9
### Cube
Length = n
Surface area 2.01615e+09 6.15968e+12 31750.2
### Equilateral Triangle
Length = n
Perimeter 54993 1.45503e+08 15875.1
### Triangular Pyramid
Length = n
Surface area 5.82013e+08 7.25926e+11 14967.2
## Cryptographic Hash Functions
md5 bd6a06f41bae67be959e5b44addc880f 38cd1ff8b4cf414d2b33ccb1403a2f293189d7cd 1d2282fc5b426785e341cceb103f1eee4727295beb9f3af811d5d7c8565c2ee2 97663397e138793a31b9b49fd0629495eaaaa9db185049c80a05e625aacee320628575cfeb6005ce394fbeb5ab629eeb555e8078964408b034438478eec99c91 0937fd1eec8f645103d4f4c6dfebd373cae3910e | 1,439 | 4,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2021-39 | latest | en | 0.822383 |
http://en.sourcecodeaplikasi.info/python-is-is-or-better-for-number-comparison/ | 1,726,207,811,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00507.warc.gz | 10,708,249 | 13,391 | Python: Is ‘Is’ or ‘==’ Better for Number Comparison?
Posted on
Python is a popular programming language that offers a wide range of functionalities. One of the most common scenarios while working with Python is performing number comparison operations. There are two operators available for this task, ‘Is’ and ‘==’, but which one is better?
Do you want to know the answer? Looking for an accurate solution for number comparison in Python? If yes, then keep reading this article. We will discuss the differences between the ‘Is’ and ‘==’ operators in detail and help you pick the right one for your use case.
Whether you are a beginner or an experienced programmer, understanding the difference between these two operators can be beneficial to create clean and efficient code. The topic of number comparison may seem simple, but it can be confusing without proper knowledge of these operators. So, don’t miss out on the opportunity to learn more about the comparison options in Python.
By reading this article, you will gain a better understanding of how the ‘Is’ and ‘==’ operators work, as well as their advantages and disadvantages. Moreover, we will explore some use cases where each operator is more suitable. Don’t hesitate to read through the entire article to make an informed decision about the right operator for your project.
“Is It Better To Use “Is” Or “==” For Number Comparison In Python? [Duplicate]” ~ bbaz
Python: Is ‘Is’ or ‘==’ Better for Number Comparison?
Introduction
As a Python developer, you may have encountered a situation where you need to compare two numbers. Python offers two operators for that purpose: ‘is’ and ‘==’. While both seem to do the same thing, they actually differ in their functionality.
The ‘is’ Operator
Judging from the name, the ‘is’ operator might seem like the better choice for comparing two numbers. This operator is used to check if two variables point to the same object in memory. In other words, it checks if two variables are identical.
The ‘==’ Operator
The ‘==’ operator, on the other hand, checks if two variables have the same value. It does not check if the values are stored in the same memory location or if they are the same object.
Table Comparison
Operator Functionality Example
is Checks if two variables point to the same object in memory x is y
== Checks if two variables have the same value x == y
When to Use ‘Is’
The ‘is’ operator should be used when you want to check if two variables are identical. For example, if you want to check if two variables are pointing to the same list or tuple:
`x = [1, 2, 3]y = [1, 2, 3]z = xprint(x is y) # Falseprint(x is z) # True`
When to Use ‘==’
The ‘==’ operator should be used when you want to check if two variables have the same value. For example, if you want to check if two integers or floats are equal:
`x = 5y = 5.0print(x == y) # True`
Opinion
In most cases, the ‘==’ operator is the better choice for comparing two numbers. This is because we typically care about the values of the numbers, not the objects themselves. However, there are some situations where the ‘is’ operator can be useful, such as when working with mutable objects like lists and dictionaries.
Conclusion
Both the ‘is’ and ‘==’ operators have their own unique functionality when it comes to comparing two numbers in Python. While the ‘==’ operator is generally the better choice, there are certain situations where the ‘is’ operator can come in handy. As a Python developer, it’s important to understand the differences between the two so you can choose the right one for your specific use case.
Thank you for visiting our blog and reading our article about number comparison in Python using ‘is’ versus ‘==’. We hope that this article has been helpful in clarifying the differences between these two operators and when to use them.
As discussed in the article, ‘is’ should be used when comparing object identity, while ‘==’ should be used when comparing object values. It’s important to understand this distinction in order to avoid unexpected results in your code.
Python is a versatile programming language with a wide range of applications, from web development to scientific computing. It’s known for its clean syntax and ease of use, making it a popular choice among developers of all skill levels.
We encourage you to continue learning and exploring the capabilities of Python, and to stay up-to-date with the latest developments in the language through our blog and other resources. Thank you again for your interest in Python, and we look forward to seeing you again soon! | 985 | 4,604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.925671 |
xxqiuo.moneyflorida.info | 1,701,459,977,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100304.52/warc/CC-MAIN-20231201183432-20231201213432-00802.warc.gz | 1,195,957,749 | 118,432 | .
.
(ii) Find the difference between sales of grapes and oranges. We are reading the temperature so we start from where we meet the line and travel across.
Step 2 - Click on "Draw" to find the pie chart for a given data.
.
In the graph on the right, percents are used to label the data. . Dec 21, 2022 · In math, the pie chart calculator helps you visualize the data distribution (refer to frequency distribution calculator) in the form of a pie chart.
.
March 8, 2021 corbettmaths. He has released all his music. You can make a pie chart by hand using a mathematical compass, pencil, and some colored pencils or markers.
. Steps for Constructing a Pie Chart.
Now, we shall represent these angles within the circle as different sectors.
A pie chart is a type of graph in which a circle is divided into sectors that each represents a proportion of the whole.
A pie chart is a circular chart. Simply input the variables and associated count, and the pie chart.
. Pie chart is circle divided to slices.
CONTENT.
In other words, a pie chart gives us a visual representation of the numerical.
Dec 14, 2019 · Step by step guide to solve Pie Graph problems.
Choose the pie chart option and add your data to the pie chart creator, either by hand. If Smithville Community Theater has \$1,000 to spend this month, how much will be spent on. 1.
Missing from session. . . He has released all his music. . Each student voted only once.
.
. .
The Pie Graph or Circle Graph Pie Graph – Example 1:.
.
Ask students what the circular image looks like.
Year 6.
00 4. | 361 | 1,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-50 | latest | en | 0.914442 |
http://www.agileconnection.com/article/learning-gnu-make-functions-arithmetic?page=0%2C3 | 1,455,051,435,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701157472.18/warc/CC-MAIN-20160205193917-00172-ip-10-236-182-209.ec2.internal.warc.gz | 252,871,311 | 15,789 | # Learning GNU Make Functions with Arithmetic
[article]
We can avoid recursion in the special case of division by 2; we define the halve function to be the opposite of double:
` halve = \$(subst xx,x, \ \$(filter-out xy x y, \ \$(join \$1,\$(foreach a,\$1,y x))))`
By now you've seen all the functions used as part of halve; work through an example, say \$(call halve,\$(five)) to see how it works.
So, the only tricky thing to do is turn a number entered by the user into a string of x's. The encode function does this by chopping out a substring of x's from a predefined list of x's:
` 16 := x x x x x x x x x x x x x x x input_int := \$(foreach a,\$(16), \ \$(foreach b,\$(16), \ \$(foreach c,\$(16),\$(16))))) encode = \$(wordlist 1,\$1,\$(input_int))`
Here we are limited to being able to enter numbers up to 65536. Once we've got the number in the encoding, only available memory limits the size of integers we can work with.
To really show off this library here's an implementation of a reverse polish notation calculator written entirely in GNU Make functions:
` stack := push = \$(eval stack := \$\$1 \$(stack)) pop = \$(word 1,\$(stack))\$(eval stack := \$(wordlist 2,\$(words \ \$(stack)),\$(stack))) pope = \$(call encode,\$(call pop)) pushd = \$(call push,\$(call decode,\$1)) comma := , calculate = \$(foreach t,\$(subst \$(comma), ,\$1),\$(call \ handle,\$t))\$(stack) seq = \$(filter \$1,\$2) handle = \$(call pushd, \ \$(if \$(call seq,+,\$1), \ \$(call plus,\$(call pope),\$(call pope)), \ \$(if \$(call seq,-,\$1), \ \$(call subtract,\$(call pope),\$(call pope)), \ \$(if \$(call seq,*,\$1), \ \$(call multiply,\$(call pope),\$(call pope)), \ \$(if \$(call seq,/,\$1), \ \$(call divide,\$(call pope),\$(call pope)), \ \$(call encode,\$1)))))) .PHONY: calc calc: ; @echo \$(call calculate,\$(calc))`
You'll need to be using GNU Make 3.80 or later for this to work. The operators and numbers are passed into GNU Make in the calc variable separated by commas. For example,
` make calc="1,3,-,3,21,5,*,+,/"`
Will output 54. Clearly, that's not what GNU Make was designed for, and I don't have space here to explain how the calculator transforms its input into calls to our plus, minus,multiply and divide functions, but I hope I've shown you the power of GNU Make functions.
Now go read chapter eight of the GNU Make manual to learn about the other functions that I have not touched on here.
Appendix
Here's the complete commented Makefile:
`# input_int consists of 65536 x's built from the 16 x's in 1616 := x x x x x x x x x x x x x x xinput_int := \$(foreach a,\$(16),\$(foreach b,\$(16),\$(foreach c,\$(16),\$(16)))))# decode turns a number in x's representation into a integer for human# consumptiondecode = \$(words \$1)# encode takes an integer and returns the appropriate x's# representation of the number by chopping \$1 x's from the start of# input_intencode = \$(wordlist 1,\$1,\$(input_int))# plus adds its two arguments, subtract subtracts its second argument# from its first iff this would not result in a negative result.plus = \$1 \$2subtract = \$(if \$(call gte,\$1,\$2), \ \$(filter-out xx,\$(join \$1,\$2)), \ \$(warning Subtraction underflow))# multiply multiplies its two arguments and divide divides it first# argument by its secondmultiply = \$(foreach a,\$1,\$2)divide = \$(if \$(call gte,\$1,\$2), \ x \$(call divide,\$(call subtract,\$1,\$2),\$2),)# max returns the maximum of its arguments and min the minimummax = \$(subst xx,x,\$(join \$1,\$2))min = \$(subst xx,x,\$(filter xx,\$(join \$1,\$2)))`
## Pages
### About the author
AgileConnection is a TechWell community.
Through conferences, training, consulting, and online resources, TechWell helps you develop and deliver great software every day. | 1,128 | 4,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2016-07 | latest | en | 0.697405 |
https://www.shaalaa.com/question-bank-solutions/determine-the-ap-whose-fifth-term-is-19-and-the-difference-of-the-eighth-term-from-the-thirteenth-term-is-20-nth-term-ap_267528 | 1,680,258,365,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00440.warc.gz | 1,076,686,563 | 9,833 | # Determine the AP whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20. - Mathematics
Sum
Determine the AP whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.
#### Solution
We know that,
The first term of an AP = a
And, the common difference = d.
5th term, a5 = 19
Using the nth term formula,
an = a + (n – 1)d
We get,
a + 4d = 19
a = 19 – 4d .......(1)
Also,
13th term – 8th term = 20
a + 12d – (a + 7d) = 20
5d = 20
d = 4
Substituting d = 4 in equation 1,
We get,
a = 19 – 4(4)
a = 3
Then, the AP becomes,
3, 3 + 4 , 3 + 2(4),…
3, 7, 11,…
Concept: nth Term of an AP
Is there an error in this question or solution?
#### APPEARS IN
NCERT Mathematics Exemplar Class 10
Chapter 5 Arithematic Progressions
Exercise 5.3 | Q 5 | Page 52
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https://www.open-mpi.org/community/lists/users/2013/09/22626.php | 1,464,351,964,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049276759.73/warc/CC-MAIN-20160524002116-00211-ip-10-185-217-139.ec2.internal.warc.gz | 1,032,854,990 | 6,778 | Open MPI User's Mailing List Archives
| Home | Support | FAQ | all Open MPI User's mailing list
Subject: Re: [OMPI users] OMPI_LIST_GROW keeps allocating memory From: Max Staufer (max.staufer_at_[hidden]) Date: 2013-09-09 03:59:01 I am still working on a small example that shows the problem, our problematic call is part of a fairly extensive framework so its not easy to post that part, but see below. As you can see the subroutine is recursive and will be calling itself again depending on the outcome posted here. The MPI_ALLREDUCE of dum(3) is the part that causes the ompi_free_list to grow. Is there an MCA parameter to limit the groth of the ompi_free_list ? Max ----------- RECURSIVE SUBROUTINE setup(l,n,listrank) ! ! USE dagmgpar_mem IMPLICIT NONE INTEGER :: l,n INTEGER, OPTIONAL :: listrank(n+1:*) INTEGER :: nc,ierr,i,j,k,nz LOGICAL :: slcoarse INTEGER, POINTER, DIMENSION(:) :: jap REAL(kind(0.0d0)), POINTER, DIMENSION(:) :: ap LOGICAL, SAVE :: slowcoarse REAL(kind(0.0d0)) :: fw,eta,dum(3),dumsend(3) #ifdef WITHOUTINPLACE REAL(kind(0.0d0)) :: dumbuffer(3) #endif CHARACTER(len=13) :: prtint REAL (kind(0.0d0)) :: fff(1) ! nn(l)=n nlc(1)=n IF (n > 0) THEN nlc(2)=dt(l)%ia(n+1)-dt(l)%ia(1) ELSE nlc(2)=0 END IF ngl=nlc IF (l==2) slowcoarse=.FALSE. slcoarse = 2*nlcp(1) < 3*nlc(1) .AND. 2*nlcp(2) < 3*nlc(2) IF( l == nstep+1 .OR. l == maxlev & .OR. ( ngl(1) <= maxcoarset) & .OR. ( nglp(1) < 2*ngl(1) .AND. nglp(2) < 2*ngl(2) & .AND. ngl(1) <= maxcoarseslowt ) & .OR. ( slowcoarse .AND. slcoarse ) & .OR. nglp(1) == ngl(1) ) THEN nlev=l dumsend(3)=-1.0d0 ELSE dumsend(3)=dble(NPROC) END IF dumsend(1:2)=dble(nlc) #ifdef WITHOUTINPLACE dumbuffer = dum CALL MPI_ALLREDUCE(dumbuffer,dum,3,MPI_DOUBLE_PRECISION, & MPI_SUM,ICOMM,ierr) #else CALL MPI_ALLREDUCE(dumsend,dum,3,MPI_DOUBLE_PRECISION, & MPI_SUM,ICOMM,ierr) #endif ngl=dum(1:2) IF (dum(3) .LE. 0.0d0) nlev=l slowcoarse=slcoarse ... > Yes, the number of elements each freelist accepts to allocate can be bounded. However, we need to know which freelist we should act upon. > > What exactly you means by "MPI_ALLREDUCE is called in a recursive way"? You mean inside a loop right? > > George. > > > On Sep 8, 2013, at 21:36 , Max Staufer wrote: > >> I will post a small example for testing. >> >> It is interesting to note though that this happens only >> >> when MPI_ALLREDUCE is called in a recursive kind of way. >> >> Is there a possibility to limit the OMPI_free_list groth, via an --mca parameter ? >> >> >> >> >> >> >> >> Date: Sun, 08 Sep 2013 14:51:44 +0200 >> From: Max Staufer >> To: users_at_[hidden] >> Subject: [OMPI users] OMPI_LIST_GROW keeps allocating memory >> Message-ID: <522C72E0.9000301_at_[hidden]> >> Content-Type: text/plain; charset=ISO-8859-15 >> >> Hi All, >> >> using ompi 1.4.5 or 1.6.5 for that matter, I came across an >> interesting thing >> >> when an MPI function is called from in a recusivly called subroutine >> (Fortran Interface) >> the MPI_ALLREDUCE function allocates memory in the OMPI_LIST_GROW functions. >> >> It does this indefinitly. In our case OMPI allocated 100GB. >> >> is there a method to limit this behaviour ? >> >> thanks >> >> Max >> >> _______________________________________________ >> users mailing list >> users_at_[hidden] > _______________________________________________ > users mailing list > users_at_[hidden] | 1,185 | 3,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-22 | longest | en | 0.444205 |
http://math.stackexchange.com/questions/109485/antipodal-map-of-complex-projective-space | 1,469,501,505,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824570.25/warc/CC-MAIN-20160723071024-00120-ip-10-185-27-174.ec2.internal.warc.gz | 158,319,207 | 17,855 | # antipodal map of complex projective space
Let $CP(n)$ be the complex projective space with Fubini-Study metric with diameter $=\frac{\pi}{2}$. Fix a point say $p\in CP(n)$; my question is what is the set of points of maximum distance to the point $p$? (The hint given in the class is $CP(n-1)$.) I can't figure out why. Could anyone help me please? Or give any reference with a detailed study of this kind of properties?
-
Since $CP(n)$ is homogeneous, you may start by assuming that $p$ is your favorite point, e.g. the point with homogeneous coordinates $[1:0:0:\cdots:0]$. – Jim Belk Feb 15 '12 at 3:47
What's your definition of $CP(n)$ and the Fubini-Study metric? – Jason DeVito Feb 15 '12 at 4:46
@JasonDeVito, the $CP(n)$ should be written as $\mathbb{CP}^n$ the complex $n$-dimensional projective space. The Fubini-Study metric is the metric you get from Riemannian submersion form the round sphere of dimension $2n+1$ by Hopf fibration. – Sun Feb 15 '12 at 15:13
1. The (identity component) of the isometry group of $S^{2n+1}$ is $SO(2n+2)$. The subgroup of isometries which preserve the Hopf fibration is $U(n+1)\subseteq SO(2n+2)$ (standard embedding). The subgroup $SU(n+1)$ inside of this, thus preserves the Hopf fibration, so descends to an isometry of $\mathbb{C}P^n$. $SU(n+1)$ acts transitively on $S^{2n+1}$, and hence transitively on $\mathbb{C}P^n$. (This is an outline of the proof of what Jim said in his comment). So, as Jim mentioned, we might as well focus on the point $p=[1:0:...:0]$.
2. Given any geodesic $\gamma$ in $\mathbb{C}P^n$ starting at $p$, and given any preimage $\tilde{p}$ of $p$ in $S^{2n+1}$, there is a unique geodesic $\tilde{\gamma}$ starting at $\tilde{p}$ which projects down to $\gamma$ and is everywhere orthogonal to the Hopf circles. In my head, I'm thinking of $\tilde{p}=(1,0...,0)\in S^{2n+1}$ where I'm thinking of $S^{2n+1} = \{(z_0,...,z_n)\in\mathbb{C}^{n+1}\mid \sum |z_i|^2=1\}$.
3. A geodesic $\gamma$ stops minimizing distance when $\tilde{\gamma}$ is half way around the sphere and the set of all such stopping points on $S^{2n+1}$ is $\{(0,z_1,...,z_n)\in S^{2n+1}\}$, diffeomorphic to $S^{2n-1}$.
4. The Hopf actions preserves this $S^{2n-1}$ and becomes $\mathbb{C}P^n$ in the quotient. | 746 | 2,259 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2016-30 | latest | en | 0.84131 |
https://www.physicsforums.com/threads/rms-velocity-of-molecules-in-a-mixture.982603/ | 1,718,527,755,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00490.warc.gz | 841,645,794 | 17,031 | # RMS velocity of molecules in a mixture
• Saptarshi Sarkar
In summary, the conversation discusses the relationship between average kinetic energy and RMS velocity for a single monoatomic gas and a mixture of gases. The first equation states that for a single gas, the average kinetic energy is equal to half the mass times the square of the RMS velocity. The second equation, corrected by the participants, shows that the RMS velocity of the mixture is equal to the square root of the average of the squares of the individual RMS velocities. This leads to the conclusion that if the absolute temperature of the two gases is equal, the ratio of their RMS velocities will be 2.
Saptarshi Sarkar
Homework Statement
A vessel at temperature T contains equal number of molecules of two gases whose masses are m and 4m respectively. What is the RMS velocity of molecules in the mixture?
Relevant Equations
##v_{rms} = \frac {\sqrt {3k_bT}} m##
I know that for a single monoatomic gas with RMS velocity ##v_{rms}## , $$\frac 1 2mv_{rms}^2 = \frac 3 2k_bT$$ where ##m## is mass of a single molecule, ##k_b## is Boltzmann constant and ##T## is temperature of the gas.
For a mixture of gas, I know that the average kinetic energy after mixing the gases will be equal to the sum of average kinetic energy of the two constituent gases before mixing.
##\frac 1 2mv_{rms_1}^2 + \frac 1 24mv_{rms_2}^2 = \frac 1 25mv_{rms}^2##
But, I have no idea how to use this to find the RMS velocity of the mix.
Saptarshi Sarkar said:
the average kinetic energy after mixing the gases will be equal to the sum of average kinetic energy of the two constituent gases before mixing.
I don’t think you mean that.
It might be the average of the averages, which fits better with the equation you wrote, but not the sum.
What does your first equation tell you about the relationship between the two individual rms speeds?
haruspex said:
I don’t think you mean that.
It might be the average of the averages, which fits better with the equation you wrote, but not the sum.
What does your first equation tell you about the relationship between the two individual rms speeds?
I did a mistake and wrote the formula wrong, it should be ##v_{rms} = \sqrt \frac {3k_bT} m##
From this I can say that if the absolute temperature of the two gases be equal, then ##\frac {v_{rms_1}} {v_{rms_2}} = \sqrt \frac {4m} m = 2##
Saptarshi Sarkar said:
I did a mistake and wrote the formula wrong, it should be ##v_{rms} = \sqrt \frac {3k_bT} m##
From this I can say that if the absolute temperature of the two gases be equal, then ##\frac {v_{rms_1}} {v_{rms_2}} = \sqrt \frac {4m} m = 2##
Right, so consider one molecule of each. What is the RMS speed of the pair?
haruspex said:
Right, so consider one molecule of each. What is the RMS speed of the pair?
Should it be ##\frac 3 2v_{rms_1}##?
Saptarshi Sarkar said:
Should it be ##\frac 3 2v_{rms_1}##?
How do you find the RMS of two numbers?
haruspex said:
How do you find the RMS of two numbers?
By taking the root of the average of the square. So, will it be ##\sqrt \frac {5v_{rms_1}^2} 2##?
Saptarshi Sarkar said:
By taking the root of the average of the square. So, will it be ##\sqrt \frac {5v_{rms_1}^2} 2##?
Yes.
Saptarshi Sarkar
## 1. What is RMS velocity?
RMS velocity, or root mean square velocity, is a measure of the average speed of molecules in a gas or liquid mixture. It takes into account the velocities of all the molecules present, and gives a more accurate representation of the overall speed than other measures such as average velocity.
## 2. How is RMS velocity calculated?
RMS velocity is calculated by taking the square root of the average of the squared velocities of all the molecules in the mixture. This can be expressed as:
V(rms) = √(1/N * Σv^2)
where N is the number of molecules and v is the velocity of each individual molecule.
## 3. What factors affect the RMS velocity of molecules in a mixture?
The RMS velocity of molecules in a mixture is affected by the temperature, mass of the molecules, and the nature of the gas or liquid. As temperature increases, the RMS velocity also increases. Heavier molecules will have a lower RMS velocity compared to lighter molecules at the same temperature. Additionally, the type of gas or liquid will also affect the RMS velocity as different molecules have different masses and properties.
## 4. Why is RMS velocity important?
RMS velocity is important because it helps us understand the behavior of gases and liquids at a molecular level. It is used in various equations and models to predict properties such as pressure, diffusion, and viscosity. It also helps in understanding the kinetic energy and heat transfer within a mixture.
## 5. How does the RMS velocity of molecules in a mixture relate to kinetic theory of gases?
The kinetic theory of gases states that the average kinetic energy of gas molecules is directly proportional to the temperature of the gas. RMS velocity is related to kinetic energy as it is the measure of the average speed of molecules. Therefore, the higher the RMS velocity, the higher the kinetic energy and temperature of the gas or liquid mixture.
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1K | 1,428 | 5,755 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-26 | latest | en | 0.924178 |
http://mathhelpforum.com/calculus/113180-value-cosec-print.html | 1,529,558,952,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00183.warc.gz | 198,974,801 | 2,974 | # Value of cosec
• Nov 8th 2009, 08:07 AM
scofield131
Value of cosec
Excuse the silliness of this question. It's been a wee while since I've done maths and I think I've forgotten something really basic.
Anyway, I've got to find the value of y= - cosec^2(pi.x) where x = 1/3
So far, I've gotten down to:
= - 1/sin(60)
Which, when banged into the calculator, = 1/ (square root)3/2 [if that makes sense].
But, I'm conscious of the squared sign in the cosec. Does that mean anything?
• Nov 8th 2009, 08:14 AM
e^(i*pi)
Quote:
Originally Posted by scofield131
Excuse the silliness of this question. It's been a wee while since I've done maths and I think I've forgotten something really basic.
Anyway, I've got to find the value of y= - cosec^2(pi.x) where x = 1/3
So far, I've gotten down to:
= - 1/sin(60)
Which, when banged into the calculator, = 1/ (square root)3/2 [if that makes sense].
But, I'm conscious of the squared sign in the cosec. Does that mean anything?
$\displaystyle cosec^2(\pi \, x) = \frac{1}{sin^2(\pi \, x)}$
As you rightly mentioned $\displaystyle sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
Therefore it becomes $\displaystyle \frac{1}{\left(\frac{\sqrt3}{2}\right)^2}$
I got an overall answer of $\displaystyle \frac{2}{3}$
• Nov 8th 2009, 08:19 AM
scofield131
That makes sense. Thanks :]
• Nov 8th 2009, 08:32 AM
pencil09
sorry double post....
• Nov 8th 2009, 08:36 AM
pencil09
Quote:
Originally Posted by e^(i*pi)
$\displaystyle cosec^2(\pi \, x) = \frac{1}{sin^2(\pi \, x)}$
As you rightly mentioned $\displaystyle sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
Therefore it becomes $\displaystyle \frac{1}{\left(\frac{\sqrt3}{2}\right)^2}$
I got an overall answer of $\displaystyle \frac{2}{3}$
the last one (result), i think it should be $\displaystyle \frac{4}{3}$
since $\displaystyle \frac {1}{\left(\frac{\sqrt3}{2}\right)^2} =\left (\frac{2}{\sqrt3}\right)^2 =\frac{4}{3}$ | 653 | 1,937 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-26 | latest | en | 0.888164 |
https://www.coursehero.com/file/6857188/Lab-1/ | 1,490,513,978,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189130.57/warc/CC-MAIN-20170322212949-00454-ip-10-233-31-227.ec2.internal.warc.gz | 892,528,374 | 60,375 | # Lab 1 - Percent Error= 1.3201 x 9.8 m/s^2/9.8 m/s^2=1.3201...
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Christopher Dunomes Lab Report #1 PHYSICS 102L February 12, 2010 Objective: To measure the acceleration due to gravity using an incline plane. Theory: If the glider starts from rest, the distance it travels is given by formula Data: d(m) t(s) t2(s) T3(s) t(ave) T2(ave) 0.3 .4176 .4175 .4175 .4175 .1743 0.4 .4959 .4950 .4939 .4949 .2449 0.5 .5682 .5680 .5670 .5677 .3223 0.6 .6356 .6398 .6380 .6380 .4070 0.7 .7062 .7042 .7052 .7052 .4973
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Calculations: Acceleration do to gravity is G=2 x .8x1.3201/.2=10.5608 m/s2
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Unformatted text preview: Percent Error= 1.3201 x 9.8 m/s^2/9.8 m/s^2=1.3201 Conclusion: If the glider starts from rest, the distance it travels is given by the formula therefore the above theory is proven true. The acceleration due to gravity is 10.5608m/s 2 with a percent error of 1.3201%. Sources of error that may cause this percent error are inaccurate measure of time, inaccurate measure of distance, and inaccurate measurements of height....
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## This note was uploaded on 03/26/2012 for the course PHYS 101 taught by Professor Dues during the Spring '12 term at Emory.
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Lab 1 - Percent Error= 1.3201 x 9.8 m/s^2/9.8 m/s^2=1.3201...
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Ask a homework question - tutors are online | 520 | 1,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-13 | longest | en | 0.853324 |
https://publish.illinois.edu/illinoisblj/2009/10/12/statistical-sampling-weighing-costs-versus-precision-in-providing-taxpayer-guidance/ | 1,638,723,538,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00022.warc.gz | 526,141,762 | 27,013 | # Statistical Sampling: Weighing Costs versus Precision in Providing Taxpayer Guidance
I. Introduction
In the months preceding elections in the United States it is difficult to avoid statistical sampling, as polling projections are everywhere. Only a sample is used to make these projections because it would take too much time and be too expensive to determine how every voter will vote. [1] Statistical sampling has many others uses as well, including being used as evidence in a trial [2] or being used to estimate how much a taxpayer owes the government on their tax return. [3] As with elections, to determine the exact result for a tax return, every item in the population would need to be investigated. As a population gets larger, this gets more time consuming and more expensive, especially when the information is collected by experts, lawyers, and accountants. Furthermore, each additional item of the population collected will not result in a proportionate change in the precision of the estimate, because the precision of an estimate varies inversely with the square root of the sample size. [4]
II. Statistical Sampling in Trials and Usage by the IRS
Statistical sampling has not always been used as evidence in trials or in preparing tax returns. Early courts were skeptical of statistical sampling estimates and did not admit these estimates. [5] However, modern courts have begun to accept sampled evidence in a wide variety of contexts, including mass torts cases. [6] Similarly, the Internal Revenue Service (“IRS”) utilized statistical sampling in performing tax audits as early as 1964 [7], but it has taken the IRS more time to provide taxpayers guidance for using statistical sampling in preparing tax returns. [8] Some evidence that the IRS is allowing increased use of statistical sampling for taxpayers, is that the IRS provided guidance on using statistical sampling for substantiating meal and entertainment expenses that are excepted from the 50% disallowance rule under Code section 274(n) in 2004 [9], and provided guidance for calculating qualified production activities for the Domestic Manufacturing Deduction in 2007. [10]
The IRS has yet to provide similar guidance for calculating the Research and Development Tax Credit (“R&D Tax Credit”) [11], however recent court decisions have allowed taxpayers to use estimates in calculating their R&D Tax Credits. In Union Carbide and Subsidiaries v. Commissioner, the United States Tax Court accepted estimates based on extrapolations “as a close approximation of all of the qualified research activities.” [12] Similarly, the Fifth Circuit in U.S. v. McFerrin held that employee testimony and estimates may be used to substantiate qualified research expenditures, against arguments by the IRS. [13] As the IRS has not yet provided guidance to taxpayers for using statistical sampling in calculating the R&D Tax Credit [14], and they may in the future [15], the R&D Tax Credit provides a good context for examples that follow.
A major benefit for a taxpayer or for a party in a trial who uses statistical sampling, is the costs that can be saved by using a sample rather than using the entire population. [16] This is especially true when there is a large sample and when amounts are being calculated by expert witnesses, lawyers, or accountants. There are additional benefits as well. For example, additional valuable information can be gained by using the resources available to determine a carefully drawn smaller sample or to collect more information on each item in the sample. [17] Also, there may be drawbacks from using an entire large population, as one recording the entire sample results may get tired or bored enough to start recording information incorrectly. [18]
Even if the costs to calculate a tax deduction or a credit equaled the cost to calculate that deduction or credit, benefits are induced by the presence of the deductions and credits. [19] For example, as the name suggests, the R&D Tax Credit was added to encourage research and development in the United States and as part of the American Jobs Creation Act of 2004, the Domestic Manufacturing Deduction was added to encourage increasing the quality of manufacturing and jobs in the United States. [20] A study of the effectiveness of the R&D Credit has shown a positive impact on R&D activity and “[t]here is significant evidence that nations and states that adopt an R&D tax credit will experience an increase in R&D investments.” [21] If the incentive to participate in these activities was cheaper and easier to calculate, it follows that more people would consider using them.
III. Precision vs. Costs to Increase Precision
The R&D Tax Credit is a difficult credit to calculate because it requires intrusive examinations to determine how many of the costs of a particular research project qualifies as a research expense for the credit. [22] For example, qualified research expenses include qualified wages paid to engineers. [23] It may not be difficult to determine how much a company paid its engineers by looking at payroll detail, but it is more difficult to determine how much of an engineer’s wages qualify as a research expense. This is the case because qualified research expenses, as defined within I.R.C. § 41, which outlines how the R&D Tax Credit is calculated, do not include all wages. [24] Even a twenty minute phone conversations with each engineer, to determine wages that qualify, will add up quickly when you take into account that the engineer could be continuing to conduct research instead, and the costs paid those conducting the interviews. When a company does extensive research and development and has multiple locations with multiple engineers, it adds up even faster.
The precision of an estimate calculated from a sample varies inversely with the square root of the sample size. [25] Therefore, in the example above, if ten engineers were originally interviewed, in order to double the precision the taxpayer would be required to interview forty engineers. [26] Similarly, to increase the precision of a sample by a factor of ten, it would require interviewing one hundred engineers. [27] Adding more numbers to this example, if a sample of ten determines that the mean percentage of time engineers spend doing qualified research is 60%, and you can be 95% sure that the mean of the population falls between 40% and 80%, to be 95% sure that this amount is between 50% and 70%, one would have to have to sample forty engineers. [28] To further increase precision so that you can be 95% sure that the percentage was between 59% and 61% would require interviewing four hundred engineers. [29]
Although the “longstanding” [30] rule developed in Cohan v. Commissioner is that absolute certainty is not required and that close approximations are acceptable when calculating deductions [31], it would be difficult to argue that such a wide range would be acceptable. This would be especially true when it is possible to calculate a more precise number. Using a sample to claim a deduction or credit of \$60, that a taxpayer is 95% sure that is between \$40 and \$80, does not appear to be a close approximation. However, it would be easier to argue that if it was determined that the deduction or credit was \$60 with 95% certainty that that the deduction or credit was between \$59 and \$61, that \$60 is a close approximation. However, if it costs the taxpayer \$1 to determine that they are 95% sure the deduction or credit is between \$40 and \$80, and \$400 to determine that they are 95% sure the deduction or credit is between \$59 and \$61; it is not worth it for the tax payer to calculate the deduction or credit at all if such a high degree of precision is required.
IV. A Compromise is Needed to Make Statistical Sampling Effective
When you have a range of how much tax liability exists, the IRS will always want the taxpayer to pay more and the taxpayer will always want to pay less. When precision is not very high, this difference may be large. Consider if instead of the example above that used tens of dollars, a credit of tens of millions was being calculated. In continuing to provide guidance to what extent statistical sampling is acceptable, the IRS should take into account how much can be saved by using statistical sampling. While they have a legitimate concern over requiring tax returns that are precise, the IRS should realize that the money saved could go elsewhere. Even if a deduction or credit fails to net as much revenue for the government, the presence of the deductions and credits encourage other activities for the benefit of the United States. [32] If it costs the taxpayer more to collect the information needed to calculate a potential benefit, the taxpayer may not participate in the potentially beneficial activity at all. [33]
V. Conclusion
Statistical sampling allows for substantial savings when making conclusions about populations. At the same time, there comes a point when asking for increased precision may cost more than it is worth to have this precision. Tax deductions and credits may be difficult to calculate, but rather than render them worthless to taxpayers or get rid of them completely, statistical sampling should be encouraged when calculations would otherwise be too difficult to calculate.
[1] Robert M. Lawless, Jennifer K. Robbenault, & Thomas S. Ulen, Empirical Methods in Law (forthcoming 2010) (manuscript at 188, released to students).
[2] Id at 208.
[3] Rev. Proc. 2004-29, 2004-20 I.R.B. 918.
[4] Hans Zeisel & David H. Kaye, Sampling, in Prove it with Figures 108-109 (1997).
[5] Lawless, Robbenault, & Ulen, supra note 1, at 208.
[6] Id.
[7] Rev. Proc. 64-4, 1964-1 C.B. 644.
[8] Will Yancey, Sampling for Income Tax and Customs, http://www.willyancey.com/sampling-income-tax.html#cases (last visited Oct. 11, 2009).
[9] Rev. Proc. 2004-29, 2004-20 I.R.B. 918.
[10] Rev. Proc. 2007-35, 2007-23 I.R.B. 1349.
[11] Yancey, supra note 8.
[12] Union Carbide Corp. and Subsidiaries v. Comm'r., 97 T.C.M. (CCH) 1207, 110 (2009).
[13] U.S. v. McFerrin, 570 F.3d 672, 679 (5th Cir. 2009).
[14] Yancey, supra note 8.
[15] Mary Batcher, Statistical Sampling in Tax Filings: New Confirmation from the IRS, Tax Executive (2004), available at http://www.thefreelibrary.com/_/print/PrintArticle.aspx?id=143304208.
[16] Lawless, Robbenault, & Ulen, supra note 1, at 208.
[17] Id at 191-192.
[18] Mary Batcher, Statistical Sampling: A Potential Win for Business Taxpayers, Tax Executive (2001), http://www.thefreelibrary.com/Statistical+sampling:+a+potential+win+for+business+taxpayers-a095446866.
[19] Ross Gitell & Edinaldo Tebaldi, Are Research and Development Tax Credits Effective? The Economic Impacts of a R&D Tax Credit in New Hampshire, Public Finance and Management (2008), http://findarticles.com/p/articles/mi_qa5334/is_1_8/ai_n29431949/.
[20] American Jobs Creation Act of 2004, Pub. L. No. 108-357, 118 Stat. 1418.
[21] Gitell & Tebaldi, supra note 19.
[22] Batcher, supra note 15.
[23] I.R.C. § 41(b)(2)(A)(i) (2008).
[24] Id.
[25] Zeisel & Kaye, supra note 4.
[26] Id.
[27] Id.
[28] Id.
[29] Id.
[30] U.S. v. McFerrin, 570 F.3d at 679.
[31] Cohan v. Comm'r, 39 F.2d 540, 544 (2d Cir. 1930).
[32] Gitell & Tebaldi, supra note 19.
[33] Mary Batcher, supra note 18. | 2,681 | 11,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-49 | latest | en | 0.934264 |
anglejournal.com | 1,713,507,698,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817289.27/warc/CC-MAIN-20240419043820-20240419073820-00849.warc.gz | 91,942,003 | 11,498 | Finance has always been built on mathematics. For example, in his book Summa de Arithmetica, Luca Pacioli, born in 1445, described the double-entry bookkeeping system through which merchants have controlled their businesses over the past half-millennium. Incidentally, Pacioli was a personal friend of Leonardo da Vinci – they once shared a house in Florence and drawings by da Vinci appear in Pacioli’s book Divina Proportione.
Mathematical models for finance evolved radically from about 1900, when Bachelier applied Brownian motion as an underlying process to derive option prices. Since then the use of probability theory and partial differential equations has boomed, to the extent that Merton and Scholes were awarded a Nobel Prize in 1997 for the contribution of mathematical finance to global economics.
Today banking systems, credit card operations, pension funds, and insurance companies amongst many others rely on such models to make decisions about all manner of financial transactions. As has been highlighted in recent years, the risk is that, while those developing these sophisticated models may understand them perfectly, those basing their decisions on them may not.
The effectiveness of work in this area hinges on how accurate the models are, and how effectively they are used
Nowadays, both finance and mathematics have grown into vast edifices, but they continue to be closely interwoven. Mathematics – or, more accurately, the mathematical disciplines – have a diverse, and it is no exaggeration to say fundamental, role to play in many quite different aspects of finance. This article singles out just three aspects: (i) so-called ‘mathematical finance’ (ii) hedge funds, and (iii) retail financial services.
Rocket Science
The first of these, mathematical finance, is concerned with building financial instruments for investment, and controlling risks: designing such things as financial options and derivatives. Through such instruments merchants can protect themselves from risk (e.g. by agreeing to buy raw materials in the future at a specified price, so protecting themselves from unpredictable price changes), and traders can seek to make money (e.g. if the price of the raw material in fact falls over the intervening period, by buying at the new reduced price and simultaneously selling at the agreed price).
This area uses advanced mathematics such as stochastic differential calculus and measure theory. It has been described as one of the few areas where cutting-edge mathematical research has a direct and immediate impact in a practical area. The nature of the mathematics involved in this area explains why it attracts physicists (the same equations describe the evolution of some physical systems and some financial systems), and hence why the phrase ‘rocket science’ has sometimes been attached to it.
The 2011 film Margin Call saw Kevin Spacey play an embattled finance executive, struggling to grasp the complexities of the algorithms his traders were employing daily on the floor
A characteristic of this area is that it has, in the main, been built on economic models of the financial system. The efficient market hypothesis is important in this area. This says that it is impossible to produce a trading strategy (a strategy for buying and selling financial products) which outperforms the market (except randomly) because any information indicating that a particular stock is over or under-valued is already taken account of in the stock’s price. Despite its importance in some areas, the efficient market hypothesis is not an uncontroversial idea, and some people do seem able to outperform the market with a consistency beyond that suggested by chance. But it is clearly almost true – which is why outperforming the market is so hard. A classic book in this area (and there are a great many) is Hull's "Options, Futures, and Other Derivatives"1.
The effectiveness of work in this area hinges on how accurate the models are, and how effectively they are used. In particular, if they are based on shaky premises then there are clearly unquantified risks. The well-known collapse of Long Term Capital Management2 arose as a consequence of a failure to take into account the fact that perhaps there would not always be buyers and sellers willing to take the other side of a trade. The sub-prime crisis was based on perfectly sound mathematical models but if, despite the fact that your model tells you an applicant for a mortgage has a high risk of defaulting on the repayments, you go ahead and make the loan, you cannot really blame the mathematics.
The Turner report ... chooses its words carefully: it does not ask ‘was the mathematics wrong?’, but asks was there ‘misplaced reliance on sophisticated mathematics?’
Furthermore, the models in this area are highly, and increasingly, sophisticated. There is a real risk that while those developing the models may understand them perfectly, higher management may not, with obvious dangers. By way of illustration, the Turner report3 which describes the events leading up to the recent financial crisis, chooses its words carefully: it does not ask ‘was the mathematics wrong?’, but asks if there was ‘misplaced reliance on sophisticated mathematics?’ That is, did those who bore the ultimate responsibility for deciding to use the models really know what they were doing? These aspects are discussed in Hand (2009)4, but to put things into context, the excellent and informed review by Reinhart and Rogoff (2009)5 examines the history of financial crashes over several centuries (with some wonderful data!), from long before sophisticated mathematical models were available.
The second area listed above was that of hedge funds. In fact, the phrase ‘hedge fund’ is not a well-defined concept, as it covers a very wide range of quite distinct kinds of trading activity. Some base their trading strategies on perceived fundamental value of the stocks being traded, others use subjective views on how the market is behaving, others are based on objective statistical models of price time series, and so on. Investors in hedge funds are constrained by regulation, but typically include pension funds, university endowments, and high net worth individuals, but not the public.
Hedge fund managers usually have their own money invested in their funds, so aligning what they do with what their customers want. This need not be the case with banks, where the aims of customers and shareholders might not be aligned. Most hedge funds seek an absolute return, a positive return on investments, regardless of whether the market is going up or down. Hedge funds are a relatively recent development, beginning in about the 1960s. An entertaining history of hedge funds is given by Mallaby (2010)6. One class of hedge funds is based on ‘systematic trading’: the use of mathematical and statistical models to predict market behaviour and make trading decisions with little or no human intervention. In contrast to the models of the financial mathematicians, described above, the models used in these organisations are empirical models, based on intensive statistical analysis of past behaviour of the financial markets. Some commentators have expressed concerns that if many hedge funds adopt similar strategies, then the correlation between the way they behave will introduce instabilities into financial markets – and have attributed the so-called ‘quant quake’ of 2007 to this cause.
Although statistics is often taught in mathematics departments, the two disciplines are rather different ... one might, with equal justification regard statistics as a part of computer science
You may notice the use of the phrase ‘mathematical and statistical models’ above, and the distinction made between 'mathematics' and ‘the mathematical disciplines’. Although statistics is often taught in mathematics departments, the two disciplines are rather different, and few statisticians nowadays regard their discipline as a part of mathematics. Certainly statistics has mathematics at its base, as do physics and engineering, but statistics is no more a part of mathematics than are those disciplines. In particular, the discipline of statistics has been revolutionised over the past fifty years by the development of computing, so that one might, with equal justification regard statistics as part of computer science.
Some hedge funds base their trading strategies on perceived fundamental value of the stocks being traded. Others use subjective views on how the market is behaving, or base trading decisions on objective statistical models of price time series
The key point is that statistics starts with the question and the data, and seeks to apply formal methods of inference to find structures and relationships, and to extract understanding and meaning. Mathematics, in contrast, is fundamentally concerned with deduction about given abstract objects and their relationships. This, in fact, is a commonly made distinction between probabilists and statisticians: the former start with the mathematics, and try to deduce what the data would look like; the latter begin with the data, and try to work out what kind of structure would have generated it. This explains the opening comment of the preface of David Williams’ marvellous book Weighing the Odds7 ‘Probability and Statistics used to be married; then they separated; then they got divorced; now they hardly ever see each other.’
Data Crunching
That small detour serves as an introduction to the third area mentioned at the beginning: the retail financial services sector. Retail banking or consumer banking refers to the sorts of transactions in which you and I engage every day. It covers such things as credit and debit cards, mortgages, car finance, personal insurance, store cards, personal loans, and so on. And it will be immediately obvious that one of the characteristics of this area is that it involves large, even massive, data sets. Many organisations in the sector carry out billions of transactions each year. So, in this area, we are really talking about statistics rather than mathematics.
Statistical algorithms ... far outweigh anything that a human could do: they make more accurate decisions
The aim is to build models of behaviour to answer a variety of questions: will a loan applicant make the scheduled repayments on time; is a credit card owner running into financial difficulties; is that anomalous card transaction evidence of fraud; is that pattern of mortgage applications suggestive of something suspicious? Anderson8 gives an introduction to the area.
Almost universally, the models in this area are empirical models: they are not based on any underlying theory (e.g. from the psychology of behavioural finance) but are entirely data driven. A characteristic will be included in a model if the data analysis shows that it leads to improved prediction, regardless of whether there are theoretical reasons which might lead one to expect it to be predictive.
That last sentence should be slightly qualified. Such a characteristic will be included only if legal restrictions allow it (e.g. the US Equal Credit Opportunity Act of 1974). Typically, anti-discrimination legislation precludes certain characteristics, such as sex, race, colour, and religion, from being included in these so-called credit scorecards.
The models in this area have increased in sophistication since initially introduced in the 1960s, concurrently with the advent of the computer. Initial scepticism about whether statistical algorithms could be as accurate in their decision-making as humans soon gave way to a recognition that they could. Since then, decades of research and refinement, coupled with the growth of massive databases, now means that such systems far outweigh anything that a human could do: they make more accurate decisions.
The models can be very elaborate. They often take the form of logistic regression trees, in which the population is divided into subgroups, with a distinct logistic regression model being built in each segment. They may involve hundreds of variables.
Sex cannot be included in scorecards because gender, per se, is irrelevant to propensity to repay, and one would not want to base decisions on prejudice. But, in general, women are lower risks than men
An important area of work in this area is the evaluation of the scorecards. This is because the domain is characterised by so-called population drift: changes in the nature of the population of people applying for or using the financial products arising from changing economic conditions (e.g. making people less willing to take out loans), changing competitive environment (e.g. often because globalisation encourages international competition or because non-finance players, such as supermarkets, enter the financial arena), or changing technology (e.g. the advent of internet banking, and more recently mobile phone cash transfers).
To illustrate just one of the difficulties, imagine deciding that a loan scorecard’s performance has degraded, so that one wishes to construct a new one. The available data will be the descriptive characteristics of those who sought a loan in the past, along with the outcome (e.g. whether they defaulted or not) of those who were previously given a loan. But this is not the population to which the new model will be applied: we wish to calculate a creditworthiness score for all applicants, not merely those that the old method would have accepted. Thus we have a distorted population from which to build a new model. Coping with such selection bias is a non-trivial problem – to the extent that James Heckman was awarded the 2000 Nobel Prize for Economics for his efforts to tackle it.
Another conceptual and methodological challenge in this area is posed by the legal constraints on allowed characteristics mentioned above. This is described in more detail in Hand (2012)9 but the essence of the challenge is as follows, illustrated for the case of a loan. Sex cannot be included in scorecards because gender, per se, is irrelevant to propensity to repay, and one would not want to base decisions on prejudice. But, in general, women are lower risks than men, so that this exclusion means that women are being forced to pay a higher rate than men. Roughly speaking, the solution to the problem is to construct a scorecard using all the predictive characteristics one can find, so that one’s estimate of the probability of failing to repay is the most accurate estimate one can get, but then to base one’s decision on the model excluding the prohibited variables. This leads to a decision which includes that aspect of gender which is predictive of default probability, but excludes all other (irrelevant, prejudiced-based) aspects.
As illustrated above, the range of interaction between the mathematical disciplines and finance is large, spanning a number of different kinds of application domains. What is common, however, is that as time progresses, all of these areas are requiring an ever higher level of mathematical and statistical sophistication. It would be quite impossible to run a banking system, credit card operation, pension fund, insurance company, or other financial operation without mathematical and statistical tools right at the heart of the enterprise. | 2,921 | 15,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-18 | latest | en | 0.953916 |
https://www.sumproduct.com/blog/article/power-pivot-principles/ppp-az-abs | 1,713,423,806,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817200.22/warc/CC-MAIN-20240418061950-20240418091950-00167.warc.gz | 924,848,691 | 9,113 | Please note javascript is required for full website functionality.
# Power Pivot Principles: The A to Z of DAX Functions – ABS
15 June 2021
In our long-established Power Pivot Principles articles, we are starting a new series on the A to Z of Data Analysis eXpression (DAX) functions. So how about take a brake and read about ABS
The ABS Function
When the ABS function is applied it returns the absolute value of a number, generated by a formula, measure or parameter. The absolute value is the numerical value excluding sign (i.e. the notation that determines whether it is positive or negative). You might think of it as a useful way to determine the distance between two points, ignoring direction.
The syntax is very simple:
=ABS(number)
To illustrate the concept:
For those who may get an error back from the ABS function, it is possible that that the number you used was actually a text value: ensure that the number is numeric.
Example
Imagine you are an analyst that wishes to assess how accurate forecast sales were against the revenue actually achieved. You might record a list as follows:
Your task is to identify the overall discrepancy between forecast and actual sales, regardless of sign. This is not to be performed on a record by record basis, but in aggregation. Therefore, you would first import the data (let’s imagine the table is called Sales) into the Data Model and create two measures to represent both financial fields:
Actual Revenue := SUM(Sales[Actual Sales])
and
Forecast Revenue := SUM(Sales[Forecast Sales])
The absolute discrepancy could then simply be measured using,
Absolute Forecast Error := ABS([Actual Revenue] – [Forecast Revenue])
This would generate a non-negative number, regardless of which measure was the larger (if either).
Come back next week for our next post on Power Pivot in the
Blog section. In the meantime, please remember we have training in Power Pivot which you can find out more about here. If you wish to catch up on past articles in the meantime, you can find all of our Past Power Pivot blogs here. | 430 | 2,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-18 | latest | en | 0.917145 |
https://wattsupwiththat.wordpress.com/2018/02/10/the-global-warming-cure/ | 1,540,171,537,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514443.85/warc/CC-MAIN-20181022005000-20181022030500-00089.warc.gz | 829,631,191 | 127,368 | # The global warming cure
Guest fiction by Sheldon Walker
Graph 1
Imagine that scientists have developed a cure for global warming. But there is a catch. The cure can only be used one time, and it will only last for 10 years.
Warning: the following story is not real. It is fiction, science fiction. However, most of the “facts” that are presented here are completely true. See if you can identify fact from fiction.
There are many questions. Should we use the cure now? Should we delay using the cure until global warming gets worse? Is the cure dangerous? Does the cure have side-effects?
Scientists present the facts to the citizens of Earth.
• The average global warming rate from 1970 to 2017 was about 1.80 degrees Celsius per century.
• The 95% confidence interval for the average global warming rate was about 1.50 to 2.10 degrees Celsius per century (calculated using a linear regression with a correction for autocorrelation).
• The 95% confidence interval does not include zero, so the average global warming rate is statistically significant.
• As well as being statistically significant, the average global warming rate is significant in other ways. We can see the effects of global warming in the real world. It is changing the environment in a number of ways.
• The cure will make the average global warming rate equal to about zero degrees Celsius per century, for 10 years.
• The citizens of Earth are warned that to reduce the average global warming rate from 1.80 to 0.00 degrees Celsius per century, will not be easy. They must be prepared for some hardships. Scientists will be dealing with a new technology, which must be powerful to overcome the significant effects of global warming. But scientists are generally optimistic that the cure can be implemented.
Is this cure a waste of time? Some people believe that the cure is just putting off the inevitable. Others believe that it will give us a 10 year window to develop a different cure.
A citizen in the crowd shouts out a question, “Can you give us any idea what we are going to face, when the cure is implemented?”.
A scientist from the back of the stage steps forward, and says “As a matter of fact, we can”.
The crowd goes completely quiet, and waits for the scientist to speak.
The scientist starts speaking, “from about 2002 to 2012, there was a phenomenon on the Earth known as ‘The Slowdown’. It resulted in an average global warming rate of about zero degrees Celsius per century for 10 years. Almost exactly the same as the cure. Many of the people here will have lived through ‘The Slowdown’. Most people probably didn’t even realise that it was happening. Some people deliberately ignored ‘The Slowdown’. Several groups of people, known as ‘Warmists’ and ‘Alarmists’, tried to pretend that there was no ‘Slowdown’. I guess that the people in those groups must be feeling pretty stupid now.
____________________
As I said at the start of this story, most of the “facts” that are presented in this story are completely true. The existance of the global warming cure is the only fiction in the story. All of the information about ‘The Slowdown’ is true.
If you look at Graph 1 (at the top of this story), then you will see ‘The Slowdown’. The blue line is the GISTEMP global monthly temperature series. The red line is a LOESS Smooth of the GISTEMP data. It uses a local regression size of 10 years to do the smoothing.
The red line looks flat from about 2004 to 2011. Flat means a warming rate of close to zero degrees Celsius per century. I usually claim that the slowdown went from 2002 to 2012. The reason that the LOESS Smooth is not flat for that entire date interval, is that it uses a local regression which includes data to the left and right of the point being plotted. So that when the point for 2002 is plotted, it is calculating a regression over 1998 to 2007.
If you plot a linear regression just over the date interval from 2002 to 2012, then you will get a warming rate of close to zero degrees Celsius per century, for the 10 year period. I wish that we had a “cure” that powerful.
## 279 thoughts on “The global warming cure”
1. Bruce Cobb says:
We don’t need a “cure” for a non-problem.
• Urederra says:
It ain´t broke.
• Doug Huffman says:
Don’t fix it if it ain’t broke lest one fix until broke or broke.
• Greg Goodman says:
Firstly I would not base any claims on “a smooth”. What is that supposed to mean? It is not scientific. Smoothing is a visual effect. Scientists and engineers apply filters and then specify what the filter is supposed to do, eg. remove high frequencies and keep lower frequency changes.
If you look at what the LOESS filter does it treats the end sections differently to the bulk of the data in order to get results which go to the end of the data. With lengthy windows 10 years on such short data that means that middle is getting a different treatment to the rest. And this is just the bit which is showing a different result. Is that part of the reason, I don’t know, but it is very poor data processing.
So that when the point for 2002 is plotted, it is calculating a regression over 1998 to 2007. ….. It uses a local regression size of 10 years to do the smoothing.
Really ? How does it do that for the last year of the data : where it still produces a result?
You can not even define the frequency response of this “filter” since it is not consistent across the span of the data. How can you be sure that what you see is not due to artefacts of the varying performance of the filter? You can’t.
I usually claim that the slowdown went from 2002 to 2012.
Well at least you say it is just a “claim”. When I look at the graph you are clearly stretching it bolster your claim. It is clearly not flat in either 2002 or 2012. Why does the analysis start in 1995, the dataset goes back way further? What is that about?
Any rigorous filter with a ten year window will not produce a result for the first and last 5 years. If you remove that you really don’t have must left to comment on.
I agree that there was a pause , probably considerably longer than 5 or 6 years shown by this analysis, if presented honestly.
I would suggest you learn a little about filters instead of applying arbitrary “smoothers”.
• Mark - Helsinki says:
Its fitting that GISS Temp is used in fiction posting.
Greg called GISS Temp data, lol. Nope, with the stations they use in analysis at GISS Temp, the result is output not data.
GISS is clearly an estimation, there is no data in GISS final product.
• michael hart says:
“It is not scientific. Smoothing is a visual effect.”
Yup. William M Briggs tried to explain it more forcefully:
Do not smooth times series, you hockey puck!
http://wmbriggs.com/post/195/
Words that stick in the memory:
“Unless the data is measured with error, you never, ever, for no reason, under no threat, SMOOTH the series! And if for some bizarre reason you do smooth it, you absolutely on pain of death do NOT use the smoothed series as input for other analyses!”
• R. Shearer says:
Yes, the premise is somewhat ridiculous.
• J Mac says:
Agree!
Global Warming is occurring, naturally.
The Global Warming Pause from 2002 to 2012 occurred, naturally.
When the Holocene Interglacial ends, Global Cooling will occur, with brutal frigid consequences….. naturally!
• Crispin in Waterloo says:
Victor Borge had an uncle who invented a cure for which there was no disease. Sadly, he later caught the cure and died.
• Mike McMillan says:
He is greatly missed.
Victor Borge, that is. Not the uncle.
2. The only cure is to get the science right and the misperceived problem goes away all by itself.
• Latitude says:
…as opposed to never getting one single prediction right
Has there ever been a “science” that’s persisted like this…when it’s never been right in almost 100 years?
3. Sweet Old Bob says:
And using GISTAMP for the chart …..
what fun !
4. Germonio says:
why are you wasting your time worrying about noise? Suppose that the temperature is given by a
function
f(t) = alpha * t +noise
then the trend over any sufficiently short period can be almost arbitrarily large or small (plus both
positive and negative). All that means is that the noise fluctuates faster than your time period. Given
random fluctuations you would expect to find ten year slowdowns in temperature fairly regularly. It doesn’t
mean anything except that we are dealing with a noise system.
• Dave Fair says:
A system that does not respond to rapid increases in CO2 concentrations. Theory trumped by reality, anyone?
• Germonio says:
the response time is thousands of years due to the thermohaline currents in the deep ocean. There is no reason to expect a rapid response from a system with a 5000 year memory.
• Dave Fair says:
Then I’ll get right on it over the next thousand years or so, Germonio.
• hunter says:
Geronimo,
You are arm waving, and looking pretty silly.
Hansen declared the climate catastrophe had started in 1988.
We are bombarded by daily claims that the climate catastrophe is happening now.
So now, after 30 years of failed climate predictions you claim the catastrophe is hiding in the oceans.
Please continue this great comedy bit.
Thanks.
• “There is no reason to expect a rapid response from a system with a 5000 year memory.”
What happened to the argument of what else could it be? Wasn’t the late 20th C warming due to an event 5000 years ago?
• Richard M says:
Germonio, what if the cause is the thermohaline circulation itself? Then what we are seeing now is perfectly natural and will change as the currents change.
• icisil says:
IMO this guy is not funny at all, but his shifty eyes, sly grin and “yeah, yeah… that’s it” remind me of climate explainers.
• Gary Pearse says:
Geronimo, ditto on the warming too! Steve McIntyre (a statistician) showed that the procedure used by Mann was such that noise ALWAYS came out a hockey stick! Fancy that.
5. nn says:
The solution, not for global warming, but rather catastrophic anthropogenic global warming, is the progressive normalization of selective-child (a.k.a. Pro-Choice), as well as other orientations and behaviors that debase human life and are antithetical to human fitness. The Choice is a wicked solution, to albeit a hard problem.
• I have no idea what you are babbling about.
CAGW is a fairy tale backed up by
absolutely NO SCIENCE.
Our planet has had more CO2 in the air than today
for almost all of it’s 4.5 billion years, and there was
never any runaway warming (CAGW), or else we would
not be here debating the fairy tale !
Please tell us what language you write in so we can translate
your comment. I would guess it is Confusedglish.
• Sara says:
Well, nn, if we chose YOUR solution, you might not be alive now.
Try thinking about that for a moment.
• R. Shearer says:
Maxine Waters protested because her mother couldn’t have an abortion.
• Leo Smith says:
so a win win really?
6. stronginva says:
I was thinking about the small temperature differences alleged to be from global warming the other day when I drove my daughter the short distance to her school and the air temperature was 2 degrees F different at the school. Yes, we adjusted easily.
• stronginva:
You are confused:
+ 2 degree F. is no big deal
+2 degrees C. is the beginning of the end
of life on our planet as we know it.
Many people make the same mistake.
• Alan Tomalty says:
Prove that +2C is catastrophic?
• Crispin in Waterloo says:
Richard, that was good.
• John Harmsworth says:
Catastrophic of course, but only for the world. If she stays between home and school she should be ok.
7. Russ R. says:
We already have a cure for the next three years, and hopefully more. And we saved money on the cure, versus spending more, to get get a bigger problem. The “problem” is not a real problem, it is a theoretical one. No one notices a tenth or two per decade, one way or another. It is a global measurement, and we live in one spot at a time. It is not only insignificant, it is undetectable by a single individual.
8. Latitude says:
• Tom Dayton says:
Latitude, your temperature graph stops in 1895. And it’s only from a single spot in Greenland.
• Latitude says:
oh dang, I meant to post the one from Miami
• Red94ViperRT10 says:
Regardless of the actual time used, it certainly appears to indicate that temperature has DECLINED over the same time period that CO2 concentration in the air has INCREASED. Where’s your Greenhouse Theory (sorry, should be more properly called Conjecture) now?
• zazove says:
Red you are talking about a few years? You appear to be with Sheldon (and icisil below it seems) – there is no such thing as noise – unless the two lines are straight that disproves everything, including Arrhenius (1896).
9. icisil says:
Seems like it actually flat-lined to 2014 until the 2014-16 El Nino.
Meanwhile CO2 did this during the same period
• Mark - Helsinki says:
To Add El Nino caused a spike in geometric CO2 growth as measured at mauna loa.
NOAA noted this.
To cause a spike of a well mixed gas in the atmosphere, the CO2 release by the oceans during El Nino dwarfed human emissions over a decade combined at least.
Plus human emissions have stalled since 2014.
10. Ack says:
A cure like the slaughter of 40,000 elephants?
11. Stop warmer nights ?
Stop the greening of the Earth ?
No one in their right mind
would want to stop the good news
from adding Co2 to the air.
Based on real science, of course,
not wild guess computer games.
Climate blog:
http://www.elOnionBloggle.Blogspot.com
12. Using the GISTemp data for the explanatory chart is a real Show-Stopper for me. After I saw that the rest was fiction writ large.
• It still comes down fo the scientific facts about the prroperties of the gas co2.
We need two things, the red blue team in the usa, with lots of govt properganda, the good stuff, plus a epa court case to prove that co2 is a good gas & essential for all life on earth.
Trumps govt must buy advertasing space in the mainly left wing papers to get the message out.
The green industry is big business. Only big govt. Can effectly fight it.
Mje
• Sheldon Walker says:
Could all of the people who don’t like GISTEMP please tell me what I should use as an alternative.
As far as I can see, most of the Land-Ocean temperature series are very similar.
I have a choice, use GISTEMP and have skeptics not believe me, or use UAH or RSS and have warmists not believe me.
I cannot make everybody happy. So tell me a temperature series to use.
• jim says:
The only temperature record on Earth that is mostly consistent over its entire lifetime and goes back centuries is CET ( Central England Temperatures) kept as ‘hard copy’ record by the MET office.
Given all the fiddling, statistical manipulation of every other series of ‘numbers’ that advertise themselves as temperature records it can justifiably be used as the only global record.
No anomalous anomalies required.
• Mark - Helsinki says:
Because GISS temp has no data in its final product.
Are you aware of the coverage of stations used on their analysis Sheldon? The smearing? C’mon.
Take all data sets and average them, balloon sat & surface at least.
• Mark - Helsinki says:
GISS also consistently adjust the monthlies from 1880 onwards every year, hundreds of adjustments are made every year
• Mark - Helsinki says:
and lets not forget the “data” GISS uses has been raped and pillaged before they even get it.
Any problems highlighted on the DATA, Schmidt at GISS promptly and consistently points people to NOAA for any raised issues. More than once I read a tweet from Schmidt that was more or less a tacit admission he thinks NOAA’s adjustments are bollocks
• Richard M says:
The only reasonable data is UAH 6.0. It is the only data that covers 95% of the planet and also avoids UHI and extremely poor areal coverage from the oceans. It’s true that warmists don’t like it. So what.
The only other possibility is hadsst3 but like I said the areal coverage of the oceans is poor.
• Bellman says:
The only temperature record on Earth that is mostly consistent over its entire lifetime and goes back centuries is CET
Much as I love the CET, the idea that it is more reliable and less “manipulated” than every global series is questionable to say the least. But if you insist on using it as a proxy for global temperatures, note that from 1975 to present it shows warming at the rate of 2.28C / century, somewhat faster than GISTEMP.
Over the last 10 years the rate has shot up to 8.86C / century, but of course that’s just noise.
• Bellman says:
Sorry, I messed up the blockquote in that previous comment. The first paragraph is quoting jim, the rest is my response.
[Fixed. -mod]
• Sunsettommy says:
Sheldon Walker writes,
“I have a choice, use GISTEMP and have skeptics not believe me, or use UAH or RSS and have warmists not believe me.”
The difference is that GISS has been adjusted so many times, that the cooling from the 1940’s to the 1970’s have changed from a .50 cooling trend to zero trend. Their adjustments are not properly validated.
Meanwhile UAH changes from one version to another are documented and presented in a science publication. Aside from any possible small corrections, it is accepted.
• Bellman says:
Meanwhile UAH changes from one version to another are documented and presented in a science publication. Aside from any possible small corrections, it is accepted.
As are GISS and RSS changes.
• Sunsettommy says:
Bellman, did you bother to notice how GISS/NOAA adjust OVER and OVER…….., While UAH and RSS doesn’t?
No they don’t publish their reasons in a paper or you would have pointed it out. Satellite data are not adjusted over and over.
That is the crucial difference you miss.
• Martin Smith says:
Sunsettommy wrote: “Satellite data are not adjusted over and over.”
UAH is on version 6.0. That qualifies as over and over.
• Sunsettommy says:
Martin,
Now you are being dishonest since the “adjustments” for UAH have little to do the data themselves but on other factors that effect the data which they explained in the paper for Version 6.0
GISS adjust the adjusted data over and over, not even close to the original raw data anymore.
It is clear you didn’t look at the link I gave you, since GISS clearly has no credible justification to eliminate the well known cooling trend of about .5C from the 1940’s to the 1970’s, which is now about zero today. They DOUBLED the warming trend from 1880 in their latest changes.
The obvious eludes you apparently.
• Martin Smith says:
Sunsettommy wrote: “Now you are being dishonest since the “adjustments” for UAH have little to do the data themselves but on other factors that effect the data which they explained in the paper for Version 6.0”
Now you are accusing me of being dishonest, but did you inform your readers that satellites don’t even measure temperature? They don’t. The temperature data is computed from the data recorded by the satellites using models, and those models have been adjusted many times, which means the temperature data they compute has been adjusted many times. So if one of us is being dishonest, it’s not me.
• Sunsettommy says:
See how Martin tries to mislead by suggesting that I am talking about temperature data when I simply said DATA only.
Meanwhile I made the point that their adjustments are not on the data themselves (which Martin never shows to be wrong) as Dr. Spenser over the years publish what the adjustments are and why they made them: Here is a simple list from Wikipedia showing the adjustments:
UAH version Main adjustment Trend correction Year
A Simple bias correction 1992
B Linear diurnal drift correction -0.03 1994
the rest of the listed changes are in the link,
https://en.wikipedia.org/wiki/UAH_satellite_temperature_dataset
For the most part they are corrections of Satellites situation they are in, the changes of their orbits, the aging status of their sensors and so on.
” The satellite series is not fully homogeneous – it is constructed from a series of satellites with similar but not identical instrumentation. The sensors deteriorate over time, and corrections are necessary for satellite drift and orbital decay.”
Martin stated,
“Now you are accusing me of being dishonest, but did you inform your readers that satellites don’t even measure temperature? They don’t. The temperature data is computed from the data recorded by the satellites using models, and those models have been adjusted many times, which means the temperature data they compute has been adjusted many times. So if one of us is being dishonest, it’s not me.”
He has responded to what I wrote about UAH satellite:
“Now you are being dishonest since the “adjustments” for UAH have little to do the data themselves but on other factors that effect the data which they explained in the paper for Version 6.0″
Here is the published paper on Version 6.0 by Dr. Spencer:
Abstract 3
Version 6 of the UAH MSU/AMSU global satellite temperature dataset represents an
extensive revision of the procedures employed in previous versions of the UAH datasets. The
two most significant results from an end-user perspective are (1) a decrease in the global-average
lower tropospheric temperature (LT) trend from +0.14 C/decade to +0.11 C/decade (Jan. 1979
through Dec. 2015); and (2) the geographic distribution of the LT trends, including higher spatial
resolution, owing to a new method for computing LT. We describe the major changes in
processing strategy, including a new method for monthly gridpoint averaging which uses all of
the footprint data yet eliminates the need for limb correction; a new multi-channel (rather than
multi-angle) method for computing the lower tropospheric (LT) temperature product which
requires an additional tropopause (TP) channel to be used; and a new empirical method for
diurnal drift correction. We show results for LT, the mid-troposphere (MT, from
MSU2/AMSU5), and lower stratosphere (LS, from MSU4/AMSU9). A 0.03 C/decade reduction
in the global LT trend from the Version 5.6 product is partly due to lesser sensitivity of the new
LT to land surface skin temperature (est. 0.01 C/decade), with the remainder of the reduction
(0.02 C/decade) due to the new diurnal drift adjustment, the more robust method of LT
calculation, and other changes in processing procedures.”
and,
“The global coverage by polar-orbiting satellites provides an attractive vantage point from
which to monitor climate variability and change. Average air temperature over relatively deep
atmospheric layers can be monitored, with minimum cloud contamination, using passive
microwave radiometers operating in the 50-60 GHz range which measure thermal microwave
emission from molecular oxygen that is proportional to temperature. The temperature of such
bulk atmospheric layers relate directly to heat content and thus to the rate at which heat may be
accumulating in the atmosphere due to enhanced greenhouse gas forcing and other climate
changes
You are quickly being exposed as a troll here since once again you fail to answer honestly. You never showed where I was wrong either, just a word game is what you are now playing.
Meanwhile notice that Martin suddenly drop the GISS TEMPERATURE adjustments that are shown to be inappropriate manipulations to create a huge false warming trend, that the RAW data never shows. I think I know why……………….
• Martin Smith says:
Sunsettomy wrote: “Meanwhile notice that Martin suddenly drop the GISS TEMPERATURE adjustments that are shown to be inappropriate manipulations to create a huge false warming trend, that the RAW data never shows. I think I know why…”
I’ll ask you again: If you have any evidence that any adjustment to any dataset is fraudulent or incorrect, just post the evidence. So far, you are doing everything you can to avoid providing any evidence. It’s easy for you to do another Gish Gallup like the one you did above, but the facts remains: Satellites don’t measure temperature. The computer models that compute temperature from the satellite radiance data have been adjusted many times, because they were wrong. That means the temperature data inferred from the satellite data have been adjusted many times. In fact, more times than the actual surface thermometer data. And I think nobody gets to review Spencer’s computer models. Yes?
If you have any evidence that any adjustment to any dataset is fraudulent or incorrect, just post the evidence.
• Martin Smith says:
Sunsettommy wrote: “Meanwhile I made the point that their adjustments are not on the data themselves (which Martin never shows to be wrong) .
You are doing a Gish Gallup, tommy. First, the satellite data are adjusted. They have to be adjusted for orbital changes, instrument degradation, and so on. Second, the satellite data are not used directly, because they are not temperatures. Temperature data can only be inferred from the satellite data using a complicated computer model, which you don’t trust when it is a peer-reviewed climate model but you do trust when it is Spencer’s black box temperature computation model. Third, the UAH dataset, which you accept, and the RSS dataset, which you reject, both begin with the same satellite dataset. They then use different, proprietary computer models to compute their respective temperature datasets. Their datasets disagree with each other, by quite a lot. However, the RSS dataset agrees with all the surface temperature datasets, and the UAH dataset does not. Yet you demand that climate science be based on the UAH dataset and only the UAH dataset. You demand that all the other datasets be rejected, and that only your cherrypicked UAH dataset is the one true dataset. And you hold this position despite all the scientific evidence being against you. This is the current summary of all the physics and chemistry you are rejecting so that you can hold on to your belief that UAH TLT version 6.0 is the one true temperature dataset: http://www.ipcc.ch/report/ar5/wg1/
Don’t use the Gish Gallup.
13. Sara says:
Nah. You have all got it all wrong. You have no VISION. You must follow the trail of Elon Musk! Look to the Future!! Get on the rocket ship and GO!
Why do you think NASA is spending so much effort to find a twin to Earth, not just in the Goldilocks zone, but with the same compounds signature in its atmosphere?
You poor souls have no vision! Space is the solution! Once humans have left the planet, warming will become a non-issue!
/snrc
14. TonyL says:
OK, it is a story. You tell us about it.
The existance of the global warming cure is the only fiction in the story.
Then you say:
As well as being statistically significant, the average global warming rate is significant in other ways. We can see the effects of global warming in the real world. It is changing the environment in a number of ways.
How can you see the effects of global warming in the real world? How is the environment changing?
Can anybody tell me what is changing in the real world due to warming.
More hurricanes?
More floods?
More droughts?
Sea level rising to flood our coastal cities?
Anything?
Sheldon Walker writes fiction. GISSTEMP indeed!
• Latitude says:
…but but I read it in the news…and there was a movie
• Sheldon Walker says:
TonyL,
I think that my comment “We can see the effects of global warming in the real world. It is changing the environment in a number of ways.” can be justified.
– temperature changes
– seasons starting at different times
– birds migrating at different times
– changes to the Great Barrier Reef
– etc
– I personally don’t believe that there has been any significant change in hurricanes,
– There may be small changes in flooding and droughts (I am not an expert, so I am not sure).
– Sea level has been rising for a long, long time. Long before AGW. AGW may have increased it a bit, but we are still talking millimetres per year.
– Polar bears are doing much better than warmists claim. With possibly 50,000 polar bears, you have to expect one to die occasionally.
So when you say “Sheldon Walker writes fiction”, is that a compliment?
Sheldon Walker also writes facts.
If not GISTEMP, then what?
If I use GISTEMP, then skeptics won’t believe me.
If I use UAH or RSS, then warmists won’t believe me.
In general, I am trying to convince warmists, since skeptics already know what is right (you didn’t see me write that, ok).
• Latitude says:
• TonyL says:
A reasonable response deserved a reply.
– temperature changes
Here in the US, the warmest time was the mid 1930s. After that, it cooled for over three decades until the 1970s, to levels seen ~1900-1920. Then it warmed again. The warm 1930s and the cooling to the 1970s have been conveniently removed from GISSTEMP. The progressive changes to the historical record By GISS are well known. Australia has similar issues with their record as amply shown here and at Jo Nova’s site.
– seasons starting at different times
Planting times have not changed. USDA (US Department of Agriculture) hardiness zones have changed +5.0 deg F. across the US since 1990! Not even remotely plausible. USDA is fully in the tank with CAGW. Heaven help any gardener who relies on the new USDA zone charts. More broadly, the US wheat belt and corn belt in the midwest are right where they always have been. If the new USDA chart was right, planting times would have changed, but they have not.
– birds migrating at different times
Much ink has been spilled about CAGW affecting bird migration. Always from the usual alarmist sources. Like hurricanes and tornadoes, the claims hold up right until you examine the data.
– changes to the Great Barrier Reefs
The Australian Research Council Centre of Excellence for Coral Reef Studies has been called out for their alarmism. The CAGW party is not all fun and games. A lot of people make their living off the reef. The alarmist message apparently cut deeply into tourism in some areas and caused some real hardship. These games have consequences.
“If I use GISTEMP, then skeptics won’t believe me.”
“If I use UAH or RSS, then warmists won’t believe me.”
I see your point. This is a problem and the debate is polarized.
More broadly, I see where you are coming from. There is *so much* alarmist stuff out there, it is almost impossible not to accept that at least some of it must be real. Sorting out all the claims as fact or fiction is more than a full time job, and most of us have better things to do.
All in all, you take a fair enough position, although the alarmists and skeptics can and will argue the details forever.
• jim says:
As I have said above, use CET, its the only reliable, consistent and generally unabused temperature record out there.
• Red94ViperRT10 says:
Sheldon, I agree with you, conditionally… Much of this is apocryphal, but (forgive me for taking these out of order, I’m stream-of-consciousness responding as they occur to me)… I recall reading in the Little House On The Prairie books, that in upstate New York (the series became popular with the book on living in Ohio, but prior to that the Ingalls family lived in upstate New York, I read all of the books, not just the popular ones) it was time to cut ice for the icehouse when a bucket of water thrown into the air hit the ground as ice. At that time, such a condition occurred virtually every year. Today, not so much. So this addresses the “-temperature changes…” though you can’t tell it from the tortured data.
Birds migrating earlier, or birds altering migration or even ranges at all – I’ll give you that one, https://www.sciencedaily.com/releases/2012/02/120223142642.htm, though the sample size, both number of species and years in the study, may be too small.
-Changes in growing season… the EPA seems to agree with you, though I have not taken a dive in the data, so it could be bollocks https://www.epa.gov/climate-indicators/climate-change-indicators-length-growing-season
-Great Barrier Reef: It has taken quite a hit https://www.nytimes.com/2016/11/29/world/australia/great-barrier-reef-coral-bleaching.html but is it dead yet? Probably not: https://wattsupwiththat.com/2016/05/08/the-great-barrier-reef-is-it-dead-yet/ and in any event, this proves it has happened before and the reef has recovered.
And as for the rest of that, I agree, hurricanes “Accumulated cyclone energy globally has experienced a large and significant downward trend…” https://fabiusmaximus.com/2017/09/14/what-you-need-to-know-about-the-science-of-hurricanes/ floods, droughts, (I’ll leave those last 2 for others to research) … no increase found in the data (though the Warmunist crowd has shown, with the torturing of temperature records, that all data is malleable), polar bears are doing quite nicely thank you very much, and while sea level has been rising for at least 300 years, the rate of the rise may actually be in decline https://wattsupwiththat.com/2018/02/04/the-fantasy-of-accelerating-sea-level-rise-just-got-hosed/
So in summary, Sheldon knows whereof he speaks, don’t cut him short too quick. The thing is, and has been my point ever since I first heard of it, how much is attributable to human activity? And how much of it can we correct? Emissions (if they have anything at all to do with climate anyway) once emitted can hardly be vacuumed up and pretend they never existed, can they? So IF there is climate change, all we can do is figure out how to adapt. And based on the last 20 years, not much of that is necessary, is it?
• Chris says:
TonyL said: “– birds migrating at different times
Much ink has been spilled about CAGW affecting bird migration. Always from the usual alarmist sources. Like hurricanes and tornadoes, the claims hold up right until you examine the data.”
Can you provide a link with that refutation?
15. Frank says:
Sheldon: Rising GHGs slow down the rate at which the atmosphere radiatively cools to space. Conservation of energy demands that must cause warming somewhere on the planet. And that will produce other changes in the net flux at the TOA. The planet will warm until the net imbalance is zero and a new steady state GMST exists.
The slowdown in warming could have been caused by a temporary increase in the chaotic ocean currents that exchange heat between the deep ocean and surface. In chaotic systems, this is called internal variability or unforced variability or NOISE. Or, it could have been caused by a change in the radiative flux across the TOA – say by a less active sun, volcanic aerosols (or possibly more clouds). That would be forced variability, naturally forced not anthropogenically forced. We have no evidence that the slowdown can be accounted for by natural variability, but our data on clouds is highly dubious.
El Nino is associated with a slowing of heat exchange with the deep ocean: a slowing of upwelling off equatorial South American and a slowing of downwelling in the Western Pacific Warm Pool. El Ninos typically rise in roughly a half year (and fall in another half year), but their size and rate (0.4 K in your graph, 0.8 K/yr rise) shows that slow downs in upwelling and downwelling could easily have negated the roughly 0.02 K/yr of warming expected during the slowdown.
ARGO came online during the slowdown and showed that the ocean as a whole warmed during the slowdown, but the top layer of ocean experienced less warming. So the slowdown appears to have been caused by a change in heat flux in the ocean, not across the TOA.
We don’t know how to speed up the exchange of heat between the deep ocean and the surface, so we can’t use that as a permanent cure for rising GHGs. Perhaps such a change will occur for most of this century and reduce warming. Perhaps such a change can last for centuries and was responsible for the LIA. (So far, solar and volcanic explanation have been proposed, but don’t appear to be adequate.) We can’t count on unforced variability to cure GW.
We can (and probably will) cure global warming by placing aerosols in the upper atmosphere to reflect incoming SWR.
• Sheldon Walker says:
Hi Frank,
your comment was well written, and I agree with almost everything that you said.
I am happy to accept that “The Slowdown” was caused by ocean cycles.
• Alan Tomalty says:
You guys are promoting a religion The religion of global warming. There are no valid stats nor any valid theory to support that CO2 absorbs infrared(IR) at the frequencies that the IR is emitted by the surface of the earth. IT DOES NOT. CO2 only absorbs the IR emitted frequencies at temperatures of -80C The only possible place the CO2 could do this is in the Antarctic. All studies have shown that the Antarctic is not warming. Also Before NASA stopped measuring water vapour in 2009, all measurements did not show any increase in water vapour in the atmosphere. Global warming is a flawed concept and can be shown to be false on any level you want to talk about. Skeptics like me relish the debate but warmists never will debate the science cause the science is not there.
• Frank says:
Sheldon wrote: “I am happy to accept that “The Slowdown” was caused by ocean cycles.”
Great. What does this tell us about climate change caused by rising GHGs? ie, What is the significant of this slowdown if we have identified the right cause.
• JohnKnight says:
I got a story . .
Imagine if you will, that (roughly) 97% of scientists were convinced that human CO2 emissions were going to cause catastrophic warming effects, and all the nations of the world agreed it was a grave threat. And there were promising “cures” in the way of aerosols to be sprayed high in the atmosphere, and the threat was declared the number one threat to American national security . . but the promising “cures” were not used . . not so much as tested.
Lots’ of people in America swore they saw stuff being sprayed up there, but it was just a strange coincidence . . that this warming was the number one national security threat, and the promising “cures” involved spraying things up there, and many people sorta hallucinated that very thing going on . . but it wasn’t, because all the people in positions of authority sort of spaced out, and just didn’t.
• Richard M says:
Actually, it is likely the strong warming in the late 20th century was caused by ocean cycles (+PDO and +AMO). The slowdown happened when they went neutral (-PDO and +AMO). Expert cooling to start in a few years when the AMO goes negative too. This is all on top of the millennial cycle which is also related to the oceans (THC speed). When this will change is difficult to assess.
The net is it leaves very little room for GHGs to have much effect. This is probably due to negative water vapor feedback at high altitudes. This was seen in the data as described by Miskolczi’s work and Gero/Turner 2011.
• Frank says:
Alan Tomalty: “You guys are promoting a religion The religion of global warming. There are no valid stats …”
On a planet where an El Nino event can warm temperature 0.4 K in one half a year (0.8 K/yr) because of the unforced variability that exists in our weather and climate, it is simple-minded to look for proof of the effects of rising CO2 in short term temperature records that are have been warming at a rate of 0.015-0.020 K/yr for nearly the last 50 years. It is idiotic to look at much shorter periods due to chaotic fluctuations
Alan continued: “[Nor] any valid theory to support that CO2 absorbs infrared(IR) at the frequencies that the IR is emitted by the surface of the earth. IT DOES NOT.”
Quantum mechanics is a carefully validated theory that describes the interaction between CO2 (and other GHGs) and the thermal IR emitted by the surface of the Earth.
You can see an infrared spectrum of the Earth taken from space below. The effect of CO2 is obviously. Now there is no need to mislead others
Alan continued: CO2 only absorbs the IR emitted frequencies at temperatures of -80C The only possible place the CO2 could do this is in the Antarctic.
You can take an infrared spectrum of CO2 in the laboratory at room temperature with an ordinary spectrometer. There are plenty of spectra on the internet.
Alan wrote: “Global warming is a flawed concept and can be shown to be false on any level you want to talk about. Skeptics like me relish the debate but warmists never will debate the science cause the science is not there.”
People like you hurt the skeptical cause because you are the source of such nonsense. No warmist would debate because you spout a bunch of nonsense. It is a natural reaction for humans to reject things that conflict with deeply held belief and to persist in believing any evidence that supports you position. However, the Internet is full of wrong information and you can find “evidence” to support anything – even the existence of Podestra’s child sex ring at the Cosmic Ping Pong Pizzeria in DC. However, the is a science blog. Readers are supposed to know how to tell science from pseudo-science and fake news. Since you don’t, consider asking questions about what evidence exists.
• Tom Dayton says:
Frank is right. Sheldon, here is what the temperature change looks like when you remove the variations (noise) from ENSO, volcanoes, and solar variation: https://tamino.wordpress.com/2018/01/20/2017-temperature-summary/ And here is a summary of the trends: https://tamino.wordpress.com/2018/02/06/global-warming-rates-by-request/
And here is the umpteenth explanation that global warming is the relentless trend, not the noise: https://tamino.wordpress.com/2018/01/31/global-warming-the-relentless-trend/
• Sheldon Walker says:
Hi Tom,
we posted at about the same time, and you can see from my reply to Frank that I agree with almost everything that he said.
I respect Tamino’s statistical expertise, but I have to take him in small doses.
When you adjust a temperature series for ENSO, volcanoes, and solar radiation, are you removing their effects, or adding your own prejudices. You need to have God like powers to be able to do this correctly. What if the volcanic eruption wasn’t a normal one?
You may have missed my comment about Tamino’s relentless trend, in my article called “A possible compromise on global warming slowdowns and pauses”. Here is a quote:
<<>>
Tamino writes some poetic nonsense about the recent warm temperatures. In case you don’t know what “a highest high born” is, it is referring to the record high temperature in 2016.
[…, a highest high born of the unholy marriage of extreme fluctuation and relentless trend.]
What Tamino fails to mention, is that the “unholy” marriage is between a human and a mouse. The mouse’s name is “relentless trend”, and the human’s name is “extreme fluctuation”. Tamino would like you to think that the mouse and the human are equal partners in the marriage. But reality proves that they are not.
<<>>
• Tom Dayton says:
Sheldon, it is obvious you have not bothered to read the paper linked from that post by Tamino—the paper that describes the method by which those effects were removed. No god-like powers required. Just math. Not even slightly contentious math. Your accusations from ignorance would get you an F if you were one of my graduate statistics students. Or one of my undergrads. Or even a high school student. Read the assignment. Do the homework. Argue from knowledge, not profound ignorance.
• Sheldon Walker says:
Hi Tom,
it is the weekend, and I just handed in my science fiction story. No homework until next week.
I admit that I didn’t study the first 2 Tamino articles that you mentioned. I glanced at them, and realised that I had read them when they first came out. I wrote an article for WattsUpWithThat about the 3rd Tamino article, so I think that I understand that one quite well.
From your reaction, I assume that you are a Tamino fan. Am I right?
I looked back at the first 2 Tamino articles, and even looked at the link to Foster & Rahmstorf.
“Global Warming Rates (by request)” sounds too much like you can request the Global Warming Rate that you want, and Tamino will produce it for you. I know that it doesn’t really mean that (or does it? :).
Describing the method used to adjust the temperature series for ENSO, volcanoes, and solar radiation, does not give the full picture. They rely on various indexes (the multivariate el Niño index or MEI, the aerosol optical thickness data from Sato et al, and total solar irradiance (TSI) data from Fröhlich). Are these indexes accurate? I don’t know. Many people question the accuracy of the various temperature series. Are we adjusting garbage temperature anomalies with garbage indexes?
I like my temperature series to be as unadulterated as possible. I can feel more confident about them that way
You may be interested in some extra-curricular study that I am doing. I am studying the “Multiple Tamino Problem” (this can apply to Time and Space). Imagine that Tamino does a study correcting a temperature series for ENSO, volcanoes, and solar radiation. He finds a result which supports warmists, and publishes his results. Six months later Tamino does another study correcting a temperature series for ENSO, volcanoes, and solar radiation. This time he finds a result which supports skeptics. Does he publish his result? If he doesn’t publish these results, nobody will know. So he doesn’t risk being exposed.
Because of the “Multiple Tamino Problem”, all of Tamino’s results must be divided by 2. It is the only fair way of solving this problem.
.
• Tom Dayton says:
Sheldon, many people in comments to your posts on WUWT and at Open Mind have explained this to you, but I’ll try again. When people say “there is inadequate evidence to support the existence of a slowdown in global warming,” what they mean, and what they endlessly explain, is that the long-term warming trend has been very (not perfectly) consistent, where “long-term trend” means the trend “averaged” across the short-term variations, which are called “noise” in contrast to the “trend.”
Any mathematical smoothing, including the LOESS smoothing you did for this post, is a reduction of the noise–“averaging” out the noise–to reveal the trend. There is an infinite number of ways to reduce the noise to reveal the trend. Your LOESS procedure reduces all sources of noise both known and unknown, but due to its lack of any external information about any of the sources of the noise, it is a blunt instrument. In contrast, Tamino’s approach that I linked for you, smooths out specifically, only, and intentionally only the three particular types of noise from ENSO, volcanoes, and sun. Tamino’s approach uses external sources of information about those types of noise, so it is more powerful in reducing those three particular types of noise. Tamino’s approach leaves untouched all the other sources of noise. But then applying smoothing (e.g., LOESS) to Tamino’s results, reduces all other sources of noise, just as your LOESS smooth did.
• Tom Dayton says:
No, Sheldon, I am not a “Tamino fan.” I am a PhD-trained scientific research methodologist with heavy emphasis on quantitative methods. Versus you, who is proud that you have not even bothered to try to understand the method of Foster and Rahmstorf. Revel in your ignorance.
• jim says:
Mr Dayton Phd etc etc, what a crock of sh*t.
Use real temperature data, not ‘tamino’ anomalous anomalies’ . Use a proper temperature series ( CET).
Now find the ‘relentless trend’.
• Sheldon Walker says:
Hi Tom,
I sense that you are slightly annoyed with me. That’s fine, I can be an annoying person (I have an odd sense of humour, a bit like Monty Python – hence the Multiple Tamino Problem). But I am a genuine person, and my attempt to find out what is happening with AGW is sincere.
You say that many people tell me that there is inadequate evidence to support the existence of a slowdown in global warming. Yes, they do. Many people also tell me to believe in God. And some of them are priests, who are the “experts” in religion. Do I listen to them? Yes, I do. But then I make up my own mind.
Notice that many of the people who tell me that there is inadequate evidence to support the existence of a slowdown in global warming, are warmists. They have a prejudice, which I take into account. They are supporting their favourite soccer club, and they refuse to believe that their goalie isn’t the best in the world. So the issue of trust comes into who I believe.
Do you honestly expect me to believe that noise (randomness) caused a 10 year interval with a warming rate of zero degrees Celsius per century. And of course, it did this while fighting an average warming rate if 1.8 degrees Celsius per century. I must get me some of that noise, it is magical stuff.
You may have read elsewhere, how I believe in AGW. I accept all of the basic science. I am not so sure about CAGW, or what we should do about AGW. I have been attacked, often in a nasty way, by warmists, since before 2009. Even though I believe in AGW. It is partly that nastiness that makes me wary of warmists, and question their motives.
I do not have all of the answers. I am working hard to try and find the answers. I have some skills, but lack other skills. I am not a statistics expert, and have never claimed to be one. I don’t like ignorance. I looked in Foster and Rahmstorf and gave you the names of the indexes that they used (the multivariate el Niño index or MEI, the aerosol optical thickness data from Sato et al, and total solar irradiance (TSI) data from Fröhlich). You have a funny definition of ignorance.
• zazove says:
“Do you honestly expect me to believe that noise (randomness) caused a 10 year interval with a warming rate of zero degrees Celsius per century. And of course, it did this while fighting an average warming rate if 1.8 degrees Celsius per century.”
Despite the fact the premise of your question has been torn to shreds, burnt to cinders and then flushed down the toilet by several individuals much smarter than us – if it was not noise what was it?
• Frank says:
Tom: I’m not a big fan of what Tamino is doing in the posts you link. The temperature perturbation associated with a particular ENSO index is purely a curve fitting exercise. It assumes that the temperature perturbation associated with a particular ENSO state can be predicted by multiplying an ENSO index by a factor determined by linear regression from a modest number of events. That factor contains uncertainty that Tamino ignores. Different ENSO indices provide different factors with different uncertainties. Most involve SSTs in certain regions and perturbation in GMST is what Tamino is trying to project. The whole process has a element of circularity. GMST is rising. There is an ENSO index that involves only atmospheric pressure (Darwin and Tahiti?). That index would be more independent than the one Tamino uses and atmospheric pressure globally is conserved, unlike rising like GMST.
If Tamino did his analysis of volcanic and solar effects using W/m2, that would be great. An imbalance of 1 W/m2 can produce an initial warming rate of 0.2 K/yr assuming all of the heat goes into a mixed layer of 50 m and the atmosphere. But that analysis quickly gets tricky as warming develops and the planet begins to radiate more energy to space due to rising temperature. That process is controlled by the climate feedback parameter, the reciprocal of climate sensitivity (expressed in K/(W/m2) rather K/doubling). And a significant amount of heat is convected below the mixed layer. So Tamino is treating changes at the TOA due to solar and volcanos as arbitrary indices (like an ENSO), not as power fluxes. He does regressions here too and the coefficients have uncertainty. He doesn’t justify the magnitude of those coefficient in terms of physics.
If Tamino showed us the sum of all of the uncertainties introduced when removing the influence of these phenomena, the result would still be a trend with significant ambiguity that would admit the possibility of a slowdown. There are now dozens of papers providing possible explanations for a slowdown, so many climate scientists appear to disagree with Tamino’s simplistic analysis. However, the need for an explanation diminished with the adoption of ERSST4. Our ability to accurately measure temperature change over a single decade is limited. Unlike Sheldon, I find decadal change fairly meaningless.
It might be that the recent state of the PDO or AMO is amplifying or suppressing the effects of ENSO, making Tamino’s corrections too large or too small. I personally prefer to focus on the unadjusted trend and confidence interval for the longest period possible and only draw conclusions from that. CO2 has been rising about 1.5 to 2.0 ppm/yr for the last 40 years (and aerosols have remained stable) producing warming at a rate of 1.5-2.0 K/century.
In previous comments to earlier posts, I have encouraged Sheldon to pay more attention to noise and the confidence interval around trends. ENSO is noise. The slowdown is “real” (whether caused by noise, ENSO, solar activity or volcanos), but doesn’t require an explanation if it could be due to noise – any of these sources of noise.
• Sheldon Walker says:
Tom,
I have an interesting question for you. It is about noise.
If you get cooled to zero degrees Celsius, does it matter (in terms of how cold you are), whether you were cooled by a random process or a deterministic process?
Does the answer to this question have any relevance to the slowdown?
In other words, if we know that you got cold, do we say that you didn’t get cold, if it was caused by randomness.
• Sheldon Walker says:
zazove,
you said:
“Despite the fact the premise of your question has been torn to shreds, burnt to cinders and then flushed down the toilet by several individuals much smarter than us – if it was not noise what was it?”
You have just declared yourself to be “not very smart” (you said “several individuals much smarter than us).
So how do you know that the people smarter than you, are telling the truth?
Stupid people tend to believe stupid things, call other intelligent people stupid, and believe stupid experts. And you have declared yourself stupid
• zazove says:
No Sheldon it is called humility. I am quite adequately intelligent to follow their arguments as well as yours. It seems it is you who cannot follow theirs.
• zazove says:
Let me correct that. I think you are able to understand their arguments but you are choosing not to due to a fixated bias.
• Sheldon Walker says:
Hi zazove,
You say: “Despite the fact the premise of your question has been torn to shreds, burnt to cinders and then flushed down the toilet by several individuals much smarter than us”
Telling me that my premise has been torn to threads, does not indicate what was wrong with my premise”.
Can you tell me in your own words what you think is wrong with my premise? That would at least give me the chance to correct your mistake.
• zazove says:
• Sheldon Walker says:
zazove,
you tell me that I am wrong. But you won’t tell me why I am wrong.
How can I argue against that?
It is not a matter of being worthy, it is a matter of providing adequate information.
• Red94ViperRT10 says:
@Tom Dayton February 10, 2018 at 7:08 pm “…Tamino’s approach…smooths out specifically, only, and intentionally only the three particular types of noise from ENSO, volcanoes, and sun…” And what about all those forcings (call them “noise” if you like, whatever floats your boat) Tamino (nor you) ever even considered? What about AMO? What about orbital obliquity? What about heat from the combustion of fossil fuels? What about solar wind? What about GCRs? (Svensmark says solar wind and GCRs are directly related, but what if they’re not?) What about solar UV? What about things I haven’t even thought of? i.e., what about not only the known unknowns, but also the unknown unknowns? So, from a high school physics perspective, FAIL!!! Tamino hasn’t proven anything.
• Bellman says:
Jim
Use real temperature data, not ‘tamino’ anomalous anomalies’ . Use a proper temperature series ( CET).
Now find the ‘relentless trend’.
Here you go.
Trend since 1975 is 2.28 C / century.
I think you could argue that there has been a slowdown or even a drop in temperatures in CET, but only if you acknowledge that temperatures were increasing much more than global temperatures up to the end of the century. Of course it’s difficult to be sure give the much greater noise in CET than global temperatures.
• Richard M says:
Tamino ignores medium and long term ocean cycles. His analysis is therefore worthless. This is the typical bias I’ve seen from global warming activists. They deny the oceans and their 1000x stronger heat capacity are the real determining factor in climate.
• paqyfelyc says:
@Bellmand.
“Trend since 1975 is 2.28 C / century.”
This swarm rather looks like to me that you can rule out, at 95% confidence, that a trend exist at all.
16. The global warming cure — take two doses of reality, and call me in the morning.
• Frank says:
Take four decades of warming at a rate of 0.15-0.20 (95% confidence interval) and say something intelligent.
• Take four decades of warming at a rate of 0.15-0.20 (95% confidence interval) and say something intelligent.
You say this as though it is reality, but that’s not reality, Frank. That’s a sugar-coated pill at best, and bad medicine, at worst.
Four decades ! — at 0.80 degrees total ! That’s a pretty small pill. How about four million years, with cycles within a range far greater than this, multiple times over. I know that’s a pretty big pill to swallow, by comparison, but I am confident that you can do it.
• Dave Fair says:
You need at least 6 decades (preferably more) to account for identified cyclical fluctuations in numerous climatic metrics, Frank.
• Frank, I guess I couldn’t tell if you were being sarcastic or not. I read it as serious. But I’m thinking now, maybe not. Typing does not convey the subtleties of spoken verbal play sometimes.
• Frank says:
Robert: I was quite serious when I suggested taking four plus decade of 0.15-0.20 K/decade, and you replied with something intelligent:
Robert replied: “Four decades ! — at 0.80 degrees total ! That’s a pretty small pill. How about four million years, with cycles within a range far greater than this, multiple times over. I know that’s a pretty big pill to swallow, by comparison, but I am confident that you can do it.”
Why 4 million years? Why not 400 million years? Or 40,000 years? We think orbital mechanics and continental drift and changing GHGs could have been responsible of change over these long period. I look to the Holocene, the last 10,000 years for guidance. There probably hasn’t been an 0.8 K warming in a half century in the whole Holocene – if you consider that polar amplification means that the change in GMST is probably only half the change in ice cores.
Furthermore we know that our planet’s hasn’t fully equilibrated in response to the change in GHGs. The current forcing is around 2.5 W/m2, but the current imbalance is only 0.7 W/m2 according to ARGO. So we are about 70% of the way to an new equilibrium of +1.2 degC
Yes. 0.8 degC isn’t a particularly big quantity and it is somewhat uncertain. But we appear to have the potential to double or triple the total radiative forcing from aGHGs in the next century. That would be a big pill to shallow. Possibly not as big as the pill of doing without fossil fuels sooner than necessary. Which pill should we take?
The answer may depend on how much richer we expect our descendants to be a century from now. Certainly it would have been idiocy for people in 1918 to spend a lot of taxpayer money back then trying to make our current world a better place.
• Frank says:
David Fair said: “You need at least 6 decades (preferably more) to account for identified cyclical fluctuations in numerous climatic metrics, Frank.”
When dealing with chaotic phenomena (not periodic phenomena) like weather and climate, the is no period long enough to ensure that the average behavior you have experienced so far represents the long term average. Even the 100 centuries of stable Holocene climate does not guarantee this.
• Dave Fair says:
Just a partial response, Frank: Historical fluctuations may give one guidance as to the significance of recent fluctuations.
17. Bellman says:
The scientist starts speaking, “from about 2002 to 2012, there was a phenomenon on the Earth known as ‘The Slowdown’. It resulted in an average global warming rate of about zero degrees Celsius per century for 10 years. Almost exactly the same as the cure.
Yet the rate of warming since 1970 increased from 1.69C /century to 1.72C / century. Or in other words no statistical difference. How exactly is this evidence of a cure?
18. Joel O'Bryan says:
The Alarmists already have offered a “cure” for Global Warming, it is called Socialism. And along with socialism, group identity will prevail and individual freedoms will be limited.
And then when socialism doesn’t work (because it’s an illusory problem), the next cure won’t be offered, it’ll be forced on population.as they’ll have given up the means to resist in the first step. And it’s be called Marxist-Communism, or maybe with a few modifications, just neo-Marxism.
19. u.k.(us) says:
I know the cure for a hangover, and yes, there is a catch.
Still trying to perfect the work-around.
• joelobryan says:
I know a cure for aging, but there’s a catch.
• u.k.(us) says:
Good one :)
20. Alan Tomalty says:
http://jvarekamp.web.wesleyan.edu/CO2/FP-1.pdf
I came upon the above Absorption study to see what scientists say about the magic of CO2 absorption of infrared rays. Well in the introduction you will see the global warming PR so as to placate the peer review censors. At 1st I thought that this study would refute the concept that CO2 is a poor absorber of IR but the farther you read the more you realize they say no such thing. The last 2 graphs say it all. They indicate that CO2 is a worse absorber than I thought.
• TonyL says:
Great Stuff.
They built a Near-IR spectrometer out of junk they scrounged up. The first attempt did not work, so they scrounged up some more junk and got things working. Near IR spectra tend to be high-order harmonics of primary absorption bands in the traditional IR range.
It is an interesting idea they had, though.
• Frank says:
Alan: The first two sentences of the abstract you cite:
“This experiment explored the absorptivity of four peaks, 1437, 1955, 2013, and 2060 nanometers, in the near-IR (NIR) absorption spectrum of CO2. The NIR absorption bands in CO2 can contribute up to 30% of the total solar heating in the MESOSPHERE.
Our climate depends mostly on the troposphere. The mesosphere is irrelevant!. So say something intelligent about the troposphere, one needs to recognize the fact that GHGs both absorb and emit thermal IR. Thermal IR emitted by the surface is modified by absorption and emission of radiation by GHGs. That situation is handled by the Schwarzschild equation.
It is human nature to latch onto stray facts that confirm our deeply held beliefs: The Earth is flat. The sun goes around the Earth. God is perfect so the orbits of the planets must involve the most perfect geometric shape, circular. The rate of evaporation of oceans is depends on surface temperature. When you encounter evidence that contradicts these beliefs, you reject it and can find somewhere on the Internet that agrees with your belief. This is called confirmation bias. Why else would you pay attention to the above website with a single unpublished paper?
One answer is that we do live in a highly polarized society that makes far too many of us act like used car salesmen, including the media.
If you want to know something about how radiation interacts with the atmosphere, I suggest Grant Petty’s textbook for meteorology students (ca \$40), A First Course in Atmospheric Radiation. There is not a word about climate change in this book.
21. mairon62 says:
In 1990, the IPCC predicted 5.0 degrees Celsius of warming per/century for their “business as usual” scenario for CO2 emissions…where is the predicted warming? The “prediction” was real, but the “reality” of said prediction is a “fairy tale”. And don’t go trying to move the goalpost now…you said “+5.0 deg C”.
• Frank says:
Obviously you have never read what the IPCC actually said in 1990. Its is trivial to look up, so why not do so before shooting your mouth off and confusing others.
https://www.ipcc.ch/ipccreports/far/wg_I/ipcc_far_wg_I_full_report.pdf
“under the IPCC Business-as-Usual (Scenario A) emissions of greenhouse gases, a rate of increase of
global mean temperature during the next century of about 0 3°C per decade (with an uncertainty range of 0.2°C to 0 5°C per decade), this is greater than that seen over the past 10,000 years This will result in a likely increase in global mean temperature of about 1°C above the present value by 2025 and 3 °C before the end of the next century. The rise will not be steady because of the influence of other factors.”
The warming rate since 1990 has been 0.17 to 0.20 K/decade with an uncertain of about +/-0.03 K/decade. As this post shows, the rise was not steady. Reality is that observations have been at the optimist end of the IPCC’s projections and about 70% of their central estimate for warming.
No one has moved the goalposts. The field goal attempt glanced off right post. Do it go through? What we can say is that the IPCCs central estimates have been too high, and their worst case scenarios absurd.
22. I don’t deny that the average global warming rate from 1970 to 2017 was about 1.80 degrees Celsius per century.
But I don’t deny that less than 25,000 years ago northern Illinois was covered by more than mile of glacial ice. I don’t deny that less than 12,000 years ago mammoths were flash frozen in Siberia with fresh grasses still in their stomachs. I don’t deny that 1,000 years ago Vikings in Greenland were cropping barley so they could use the grain to make beer. I don’t deny that the Thames River froze solid to such an extent that during 26 separate winters from 1408 until 1814, Londoners were able to hold a Frost Fair on the ice.
The issue of climate is directed towards an end goal that is not an honest nor a useful one. The future of humanity does NOT depend upon the sea being at a certain level. It does not depend upon the average temperature being a certain number of degrees . It does not depend upon the level of CO2 falling. It does however depend upon the health of the biosphere in which we live and through which we depend to be able to grow our food.
Not a single one of the concerns being expressed by the advocates of “climate changed” is focused on the health of the overall biosphere iby itself. Where their research report also expresses a policy recommendation, it is always to the end effect of far bigger government, far higher taxes, far less personal freedom, far fewer choices, far more restrictions. It is always a means to an ideological end. That tells me the “social construct” of the research environment. It is always interventionistic, if not socialistic. Their “science” is just as compromised as any ever sponsored by tobacco companies.
• Latitude says:
…and somehow they convinced a lot of people that a tiny little bitty bit warmer is a bad thing
• Urederra says:
… a bad thing all over the world, including countries like Canada, Finland or Norway.
23. Red94ViperRT10 says:
So riddle me this, Batman… this whole Anthropogenic Greenhouse effect, are these deep thinkers claiming that human emissions are adding CO2 to the atmosphere, so that the total molar mass of our atmosphere is increasing? In which case, one would/should expect an increase in temperature of the air at sea level because the pressure has increased while the total heat in the air has remained constant (thermodynamics, you know), and there is nothing we could do about that anyway (because thermodynamics, you know), nor should we worry about it since the resulting increase of temperature will produce an increase in the ΔT of the Earth-black body relation, and thus increase heat loss to space and thereby maintaining equilibrium; QED, problem solved. Or is this AGW wetdream claiming that the the total molar mass of the atmosphere remains constant but the CO2 spewings of industrial activity is shifting the percentages of the atmosphere? I want a scientific answer to this, the Ideal Gas Law and comparing that result to that natural-log equation with the unknown-and-unknowable (and possibly mythical) sensitivity constant, even if the actual numbers only support millionths of a degree because those same numbers only produce a thousandth of a millimeter Hg Δair pressure. And then do another column comparing total heat of atmospheric air, which will require data on the amount of moisture in the air at each temperature reading. Where does one find that data? (Could it be that even if the temperature increase is real, it has been accompanied by a decrease in moisture in the air, which produces a Δheat = 0 ?) Don’t tell me it’s a difference that makes no difference, we already know that because we can see from the data (“homogenized” though it may be) that natural variations prior to any anthropogenic atmospheric contribution were just the same, if not more variable, as anything occurring now. Does anyone have time to play around with that? Cuz I sure don’t.
• Latitude says:
• Red94ViperRT10 says:
Wow, such data exists… where/how do I get that as a table of numbers? I can do the psychrometrics, so I could take the table of global temperature, lay it alongside the table of relative humidity, and produce a time series of global total heat… What do you suppose that graph would look like? Does anyone have time to do that? Willis? Bueller? Bueller? (some of you will get the reference)?
• Richard M says:
Nice chart and it specifically shows the negative feedback from increasing CO2.
• Toneb says:
http://onlinelibrary.wiley.com/doi/10.1029/2010JD014192/abstract;jsessionid=6CC471BF5E0E3A8F605EF06C895B9E87.f04t01
“Here we consider whether this result holds in other reanalyses and what time scale of climate fluctuation is associated with the negative specific humidity trends. The five reanalyses analyzed here (the older NCEP/NCAR and ERA40 reanalyses and the more modern Japanese Reanalysis (JRA), Modern Era Retrospective-Analysis for Research and Applications (MERRA), and European Centre for Medium-Range Weather Forecasts (ECMWF)-interim reanalyses) unanimously agree that specific humidity generally increases in response to short-term climate variations (e.g., El Niño). In response to decadal climate fluctuations, the NCEP/NCAR reanalysis is unique in showing decreases in tropical mid and upper tropospheric specific humidity as the climate warms. All of the other reanalyses show that decadal warming is accompanied by increases in mid and upper tropospheric specific humidity. We conclude from this that it is doubtful that these negative long-term specific humidity trends in the NCEP/NCAR reanalysis are realistic for several reasons. First, the newer reanalyses include improvements specifically designed to increase the fidelity of long-term trends in their parameters, so the positive trends found there should be more reliable than in the older reanalyses. Second, all of the reanalyses except the NCEP/NCAR assimilate satellite radiances rather than being solely dependent on radiosonde humidity measurements to constrain upper tropospheric humidity. Third, the NCEP/NCAR reanalysis exhibits a large bias in tropical upper tropospheric specific humidity. And finally, we point out that there exists no theoretical support for having a positive short-term water vapor feedback and a negative long-term one.”
• Richard M says:
Toneb, nice example of bias. The establishment climate cartel simply cannot accept reality and once again twist in knots trying to ignore reality. There is theoretical support. They simply deny it.
Any increase in warming of the surface will lead to enhanced convection which in turn leads to more condensation at higher altitudes. That will lower the amount of water vapor. Denying this kind of basic physics is why climate science is obviously anti-science.
• John harmsworth says:
This is a fantastic set of graphs! For relative humidity to fall to this extent on a worldwide basis is absolutely incredible. I cannot believe that the tiny observed temperature changes can explain this. Can someone provide a mechanism to explain this? The reduction in humidity contrasted to the tiny increase in temperatures represents a significant reduction in enthalpy. This is not global warming. This is global cooling.
• Toneb says:
“Toneb, nice example of bias. The establishment climate cartel simply cannot accept reality and once again twist in knots trying to ignore reality. There is theoretical support. They simply deny it.”
Nope the opposite of “bias” as it was a meta-study.
The one it counters as being an outlier is but one.
But again that is just bau for “contrarians”.
There is no “theoretical support”.
Link to it if there is (other than much further on in warming when surface winds have diminished, and hence evap – not the case now).
“simply cannot accept reality and once again twist in knots trying to ignore reality”
You really don’t get the irony of that do you?
That you think the majority and the knowledgeable (experts) are all wrong and the ideologically motivated know better.
“they simply deny it” … and it gets even more ironically contrary.
And of course it is not *you* that “ignores reality”
You see why it is impossible to communicate with mind-sets such as yours.
To say that the consensus is science denies anything is bizarre …. it is a ‘consensus’ because no one has found fault, while all the while studying the subject. Like any science.
It hasn’t been arrived at via a vote.
The study I linked to involved 5 data sets.
The ones that you “ignore” number 4.
The ones I don’t ignore number 4.
my ‘reality’ numbers 4.
If nothing else just basic common sense.
Again:
“We conclude from this that it is doubtful that these negative long-term specific humidity trends in the NCEP/NCAR reanalysis are realistic for several reasons. First, the newer reanalyses include improvements specifically designed to increase the fidelity of long-term trends in their parameters, so the positive trends found there should be more reliable than in the older reanalyses. Second, all of the reanalyses except the NCEP/NCAR assimilate satellite radiances rather than being solely dependent on radiosonde humidity measurements to constrain upper tropospheric humidity. Third, the NCEP/NCAR reanalysis exhibits a large bias in tropical upper tropospheric specific humidity. And finally, we point out that THERE EXISTS NO THEORETICAL SUPPORT for having a positive short-term water vapor feedback and a negative long-term one.” (my caps)
24. No problem … I’ll draw the line
25. Leo Smith says:
Now for fact..
– Global warming, such as it may be, is way below natural ‘noise’
– Humanity depends on fossil fuel for its very existence
– The measures proposed to combat ‘climate change’ are enormously expensive, and will threaten human lives by making energy enormously expensive.
– The measures proposed to combat ‘climate change’ are almost completely ineffective at reducing CO2.
– The evidence that global warming is a problem at all is almost non existent.
– the evidence that CO2 is the driver of such climate change as there is, is flawed.
The overwhelming outcome of cost benefit analysis is to do nothing – business as usual.
The overwhelming consensus amongst governments, ‘green energy’ companies and academics is to divert huge quantities of taxpayers money into the pockets of governments, ‘green energy’ companies and academics to produce virtue signalling solutions to faux problems created by scaremongering research studies.
Quelle surprise.
• zazove says:
You just expressed your opinions, quite a different thing from stating “facts” which surely require a little supporting evidence. Not that there anything at all wrong with that expressing your opinions. Be aware though that facts don’t change wheras opinions do.
• Urederra says:
Past temperatures also change. See GISTEMP, HadCRUT and other datasets.
• Leo Smith says:
Of course your position is typical for New Socialist, raised on the concept that the Truth is a Cultural Construct.
I hope you never fly in an airliner: Having to trust to the mere opinion of an engineer, that it will, in fact, fly.
The underlying truth of the world does not, one assumes, change.
In that you are correct.
Where your arrogance blinds you, is that you think that you are in possession of it already.
The evidence is all there. You only have to be able to think for yourself and view it objectively.
Its a shame that you will never achieve that simple feat. But will always have to rely on someone you hope is an authority, to think for you.
It is not necessary that most human beings ever grow up beyond looking up to authority figures.
It is however dangerous when they look up to false ones.
• Toneb says:
“It is not necessary that most human beings ever grow up beyond looking up to authority figures.”
A splendidly missed point and avoidance of common sense.
Some authority figures are there merely due to said authority. Others, as in science are not a ‘who’ but an ‘it’.
To question the ‘who’ or ‘it’ is fine and indeed necessary, but when you decide that the ‘it’ is wrong on the basis of judging the thing it knows better than you as being wrong, whether because (as it must be via that logic) it is incompetent and or fraudulent. Then we have to bring in motivations that must be over-riding common sense.
I suggest that most, as evidenced here, do not look at the real science but just that viewable on Blogs, (or even that published here, FI, in regard to the deceptive Alley graph) which self evidently have an ‘axe’ to grind. Err, any one/thing coming to it with that attitude are not being ‘scientific’.
• paqyfelyc says:
facts do NOT require supporting evidence, unless hidden or contested.
Most facts are just self-evident.
Now YOU, zazove, may claim that such and such are not “facts” but mere opinions, but you must then … just do it! since you didn’t, what you wrote is useless.
• zazove says:
“you think that you are in possession of it already”
There: our opinions have differed again. Attacking me personally is not going to persuade me, finding out what my opinions are then refuting them, if you can bothered, may.
You expressed your obviously closely-held opinion that:
“the evidence that CO2 is the driver of such climate change as there is, is flawed”,
but then neglect to substantiate it. By doing so you are asking all of us to accept that the theory is “flawed” on your authority – because of your expressed opinion. Sorry, but when it comes to overturning a theory…what’s your evidence? Is there any evidence out there that is not already in your possession?
Btw, who claims CO2 is “the” only driver? If you cannot name them, then my opinion sir is that your opinion is not even wrong.
Rudi.
• paqyfelyc says:
@zazove
“Btw, who claims CO2 is “the” only driver? If you cannot name them, then my opinion sir is that your opinion is not even wrong.”
IPCC mission was to assess
Human-induced climate change,
The impacts of human-induced climate change,
This obviously do NOT preclude that CO2 is the only driver, and indeed in IPCC work many other drivers are mentioned … and discarded as being of lesser to no importance. The conclusion of IPCC are just about CO2, or CO2 equivalent. Nothing about land usage change, water usage, urban heat, or whatever. In this respect, methink it is fair to say that CO2 is “the” only driver for the IPCC, “the” one that rules them all.
But I guess you will easily discard this as mere opinion.
26. Steve says:
I like it….it would make a great movie…after the scientist speaks up all the zombies suddenly wake up and turn on greenies – nothing like finding out you been lied to…
27. Stevek says:
often the “cure” is worse than the “disease”
28. Peta of Newark says:
It don’t matter – a or The Cure for Global Warming, Climate Change or whatever you want to call it.
It Is Not A Problem with the outside real world.
It is all in people’s heads. If you cure cAGW, something else will come along to replace it.
Just recognise what is described in this post = chat about what you, me or ‘the other person’ sees in pictures of the choreography of dancing faeries
Why?
Have a listen to the link coming up. Just the first 2 mins and 15 seconds.
Be brave. Do not just dismiss the guy. Do not play the ‘Big Strong He Man’
Most of all, be utterly honest with yourself in recognising any part of what he’s talking about.
If you can’t, you have got what he’s talking about and there in is The Problem.
Watch/listen here.
There’s a whole series of the guy raving away, trying to sell his book and video series and hey why not, everyone’s gotta live.
He does mention the cure in somewhere in his sales blurb and its very very simple..
Recognise that sugar (carbohydrate) in our diet is NOT a replacement for saturated fat.
That’s all. That’s it.
Period
29. John Dowser says:
A straw man article: no serious person ever claimed that 10 year intervals of any rise, pause or decline would be significant, desired or feared. Thus a solution for any decade of pause would simply not be interesting besides the possibility to understand the mechanics of that pause as to apply it our understanding of longer term fluctuations and cycles. It’s okay to have fiction but it’s more like a non-point made by suggesting a non-starter scenario. As such it’s more theatrical preaching ones belief… (welcome to climate politics then!)
• zazove says:
Sheldon has doubled down on this fiction about 4 times in the last week. Despite the obvious flaws he ain’t letting go.
• Reg Nelson says:
Phil Jones said fifteen years was significant and then backtracked on his assertion:
“Yet he (Phil Jones) insisted that 15 or 16 years is not a significant period: pauses of such length had always been expected, he said.
Yet in 2009, when the plateau was already becoming apparent and being discussed by scientists, he told a colleague in one of the Climategate emails: ‘Bottom line: the “no upward trend” has to continue for a total of 15 years before we get worried.””
—————
No wonder rational, objective people distrust Climate Science and see it for what it is: Political Propaganda.
30. Extreme Hiatus says:
I think a cure for Restless Leg Syndrome would be a better goal.
• Bruce Cobb says:
When my legs are restless, I go for a walk. Problem solved!
31. petermue says:
Letz me say it with the words of George Carlin: Leave nature alone!
So called climate scientists should take off their fingers from things they don’t understand, and most notably from undemanded actions involving the whole mankind.
This is not a question of side effect or cure but a question of overriding each single human’s own decision.
Personally I hate those people wanting to play God on planetary levels. This is terrorism.
There’s no reason and there’s no need.
• Sara says:
I have to agree with petermue. I have spent the last three days shoveling (so far) a totla of 14 inches of global warming off my sidewalk and helping my neighbors shovel their global warming into the streets for the global warming plows to pick up. The original forecast for my area was 6″ to 10″, then 8″ to 12″, then back to 6 to 10, and now it is 14 inches and growing.
I think the cure for the nonsense is to drag every one of these bozos out of their comfy offices, away from their computers and simulations, into the real world of high humidity and heavy snowfalls, and put shovels into their hands. Under supervision, of course, so that they don’t start doing equations in the snow or taking breaks for a nip of brandy.
You have to have a sense of humor about the silliness the represent. If you don’t, if you can’t point at them and giggle while you’re deep in your mug of hot chocolate with or without marshmallows, and some turtle cheesecake at your elbow, then you’re lost.
#makethemshovel!!
32. Martin Smith says:
Sheldon Walker, given that pauses like the one you believe is there are expected, what do you think your alleged pause means, scientifically?
• Richard M says:
But remember, historic proxy data shows century scale increases are also to be expected. So what does any century scale increase means scientifically?
• Martin Smith says:
I believe you are referring to one rapid increase that was explained, ad the explanation for that increase is known not to be operating now. Nevertheless, the current “century scale increase” is caused by burning fossil fuels, and nothing else, so Walker can’t hide from the question by using your alleged increase(s), because it was not caused by burning fossil fuels. The question remains: Given that climate science expects and predicts “pauses” like the one Walker alleges, if we assume the pause is really a pause, what is Walker’s point? What does he think it means scientifically?
• Richard M says:
No Martin, there are more than one century scale increases. Look at the chart for the GISP2 ice cores displayed elsewhere on this thread. Several century scale warming trends.
Sorry, but it is obvious you don’t want to understand the truth. Yes, Walker is fooling himself but not in the way you think. Satellite data shows no warming for the 18 years prior to the recent El Nino blip. And, we can see in global SSTs that the El Nino blip has now dissipated and we are right back where we were.
• Martin Smith says:
Richard, you can’t measure global average temperature using an ice core from the center of Greenland. Nor can you use this old diversion: “Satellite data shows no warming for the 18 years prior to the recent El Nino blip. And, we can see in global SSTs that the El Nino blip has now dissipated and we are right back where we were.”
Satellite data is for the mid-troposphere, which is 15,000 feet above sea level. No one lives there, I think you know. But, again, even if the pause you and Walker allege is a real pause, the question remains: Given that climate science expects and predicts “pauses” like the one Walker and you allege, if we assume the pause is really a pause, what is Walker’s point? What does he think it means scientifically?
By the way, the two satellite datasets disagree with each other. You seem to have ignored that. You can’t.
• paqyfelyc says:
@Martin Smith
“Given that climate science expects and predicts “pauses” like the one Walker and you allege, ”
Oh. Who, and when, expected and predicted such pauses ? The only relevant paper I know of, was written AFTER the onset of the pause, and ruled out that it could last more than 15 years. It lasted 18 years.
But I guess you know of a paper from before 1998, making such prediction: I’ll be happy if you shared it with us.
“Nevertheless, the current “century scale increase” is caused by burning fossil fuels, and nothing else, ”
Again, who wrote that? Who believes he KNOWS so much about the climate, that he can rule out anything else ? How this scientist explains the past climate events, and who believes him?
Edward Lorenz wrote a seminal paper, explaining that this feat (attributing current increase to burning fuels) was just impossible, but I guess you know of a paper refuting his statement (Nobody’s perfect, you don’t have to trust Lorenz just because who he is). Again, could you please share it with us?
• Richard M says:
Martin and Tom, RSS is no longer satellite data. It is a hybrid of mixed sources. Sorry you haven’t figured that out. As for satellite data not being surface data that is a positive. It means it is not infected with UHI, infilling, homogenization, etc.
I know you “want” to believe the surface data is true but that is far from being certain. On top of that, you have this thing called science which doesn’t let tropospheric data venture far from the surface. It’s called the adiabatic lapse rate. Unless you’ve encountered some kind of recent gravitational change in our planet you’d be wise to avoid claiming a long term difference.
As for all the surface data sets agreeing with each other, they should. They are all mostly based on the same data source (GHCN) and use many of the same techniques. Hardly meaningful.
Finally, the Greenland ice cores also agree with several other proxies. What, you really thought there was only one????? Amusing.
• Martin Smith says:
Sheldon Walker, Richard M, and (who rejoices in the name) paqyfelyc, I’m calling your bluff. If you believe there was a statistically significant pause, you must address this question: If the pause you all allege is a real pause, the question remains: Given that climate science expects and predicts “pauses” like the one you allege, if we assume the pause is really a pause, what is Walker’s point? What does it mean scientifically?
Tamino has shown that the alleged pause was not a new signal but just the noise of variation, and such pauses are expected, so, despite the proof to the contrary, assume Walker’s analysis is robust. What does it mean scientifically?
[??? .mod]
• paqyfelyc says:
@Martin Smith
I still wait for you to show us “that climate science expects and predicts “pauses” “. This is not a given for me. Just show us the paper from before the pause expecting and predicting a pause (just a single one, not even plural).
As for your second sentence “Tamino has shown … What does it mean scientifically?”, that’s easy
This scientifically means that what you call “the noise of variation” is at least as strong as the alleged signal, that we have no way to sort this signal out of the noise, that the signal may actually doesn’t exist at all, and even if the signal exist we will never know (and of course won’t be able to assess its strength) as we have no way to rule out the null hypothesis “this is just noise without underlying signal” (*). Which is just a basic skeptic position!
So you are telling us that Tamino turned skeptic. Just fine, he wouldn’t be the first, but pretty unbelievable. Are you sure?
(*) don’t take my word for truth, just ask some statistician
• Martin Smith says:
paqyfelyc wrote: “I still wait for you to show us “that climate science expects and predicts “pauses” “. ”
paqyfelyc, let’s not argue about facts that aren’t controversial. Here is one proof: http://theconversation.com/the-climate-hiatus-doesnt-take-the-heat-off-global-warming-40686
Here is another:
https://skepticalscience.com/global_warming_still_happening.html
Here is the best one:
https://skepticalscience.com/going-down-the-up-escalator-part-1.html
Then paqyfelyc tried to put this up as the meaning of the alleged pause: “This scientifically means that what you call “the noise of variation” is at least as strong as the alleged signal, that we have no way to sort this signal out of the noise,”
That doesn’t make sense. If it did make sense, the global average temperature wouldn’t be rising so fast, as predicted by climate science, and as observed by NASA: https://climate.nasa.gov/vital-signs/global-temperature/
That signal you tried to say is matched by the alleged pause is actually very strong, and the temperature rise is much faster than natural global warming. You haven’t shown there is any meaning to your alleged pause, so, Sheldon Walker, Richard M, and paqyfelyc, I am still calling your bluff. By claiming there was a statistically significant pause, you must address this question: Given that climate science expects and predicts “pauses” like the one you allege, if we assume there really was a statistically significant pause, what is does it mean scientifically?
• Richard M says:
Martin, since you ignored the 18 year pause in satellite data I showed you earlier, it is evident that you will deny anything that does not support your belief system. Here it is again.
Note that the only way the pause ended was by a 4 year period driven by El Nino caused increases in sea surface temperatures. We can now see that warming has completely disappeared. Hence, there is no reason to believe the pause will not reassert itself in the coming years.
Your response was unscientific nonsense which I pointed out and thinking you refuted that graph is also nonsense. You haven’t called anyone’s bluff. You are just fooling yourself.
• Martin Smith says:
Richard M wrote: “Martin, since you ignored the 18 year pause in satellite data I showed you earlier, it is evident that you will deny anything that does not support your belief system.”
I did not ignore a pause, nor have I denied a pause. I’m sure you won’t apologize for your false accusations, so I WILL ignore those. I am still calling your bluff. By claiming there was a statistically significant pause, you must address this question: Given that climate science expects and predicts “pauses” like the one you allege, if we assume there really was a statistically significant pause, what is does it mean scientifically? If there was a pause, it doesn’t damage the AGW case because |AGW projections show pauses like the one you are alleging. So what is your point?
BTW, note again that the the two satellite datasets disagree, and the one you are touting disagrees with all other temperature datasets.
• paqyfelyc says:
@Martin Smith
So you use the words “predict” and “expect”, when someone explains in 2015 what happened starting 1998. Don’t expect that you can convince anyone, with such bad use of words.
All you provided are links to “statistical shit happens, that’s how I explain AFTERWARD my prediction didn’t materialized, but it will, I promise, unless it doesn’t, which may happen from time to time again”. They don’t provide a figure of the probability of this happening, not even a guesstimate, just “may” and “likely”. That’s that you call science, really? Soothsaying can do better, do you call it science, too? Why not?
“Then paqyfelyc … That doesn’t make sense.”
As I said: don’t trust me, just ask some statistician, or some signal theorist : what does this means, when noise overrides a signal you are not even sure exist (nobody told you he sent you a signal, you just are sure there is a signal in some noisy natural data so you are trying to sort it out ).
“That signal you tried to say is matched by the alleged pause”
I guess you mean “That signal you tried to say is matched by the natural downward variation so that a pause appeared but that’s an illusion” (otherwise, the sentence means just nothing). But that’s not what “I” said, that’s what Tamino said, according to you. If you don’t believe it, well… that’s OK for me.
If you asked ME, I would just quote Edward Lorenz: mathematics of Chaos prevents us to know nothing about a man-made signal. May be it is there, maybe not, there is just no scientific way to tell. This is not a science question, there is no way to prove either that it really exists, or that it really doesn’t exist (and of course, you cannot quantify it at all, either).
“temperature rise is much faster than natural global warming.”
Make you mind. This contradict your previous claim, that natural variations have strength enough to keep up with man-made, so that a pause appears.
AND
you claim to know enough about natural global warming to tell it is “much slower” than current temperature rise. Extraordinary claim, that require extraordinary evidence. Methink that it would be worth of both several Nobel Prize and a Field medal in Mathematics, and would dwarf the achievements of, well, list them ALL past geniuses. Not only that, but this knowledge is worth Trillions each year: Just tap on it for a small part, and you’ll have all the money you need to fund whatever you fancy, including Energiewende for the whole world if you will.
• Martin Smith says:
paqyfelyc: “As I said: don’t trust me, just ask some statistician,…”
I am still calling your bluff. By claiming there was a statistically significant pause, you must address this question: Given that climate science expects and predicts “pauses” like the one you allege, if we assume there really was a statistically significant pause, what does it mean scientifically? If there was a pause, it doesn’t damage the AGW case because AGW projections show pauses like the one you are alleging. So what is your point?
Please understand, I’m not arguing with you about whether there was a pause. Obviously, I agree with Tamino’s analysis, not Sheldon Walker’s, because Tamino has proved Walker’s analysis is wrong. So this discussion is not about proving or disproving the pause. We are here assuming, for the sake of argument, that there was a pause. Then given the assumed pause, what do you think it means, given that climate science says such pauses must occur?
• paqyfelyc says:
@Martin Smith
“I am still calling your bluff.”
Well, I already showed my cards, did you fail to notice? Let try again, some other way (the very same card, just at another angle)
Given that climate science expects and predicts “pauses” like the one you allege, if we assume there really was a statistically significant pause, what does it mean scientifically? If there was a pause, it doesn’t damage the AGW case because AGW projections show pauses like the one you are alleging. So what is your point?
So, you have a theory that is just immune to critics whatever happens. Warming Happens? Theory predicted it will. Warming don’t happen? Theory predicted, that, too. Zero falsifiability.
“what does it mean scientifically?” It means the theory is simple NOT science.
Science make serious prediction, serious hypothesis that can be tested, that is, can be proven false, so we can get rid of false hypothesis and try new one and make some progress. A theory that is right whatever happens can bring you some comfort, but I advise you turn to a proper religion instead. No lack of them, proven by centuries or millennia to bring solace and good enough advices to humankind about what to do or not.
• Martin Smith says:
paqyfelyc wrote: “So, you have a theory that is just immune to critics whatever happens.”
No, paq, I didn’t put forth any theory. Natural variation is simply part of the projections produced by global climate models. The outputs they produce include “pauses,” which are called natural variation. You can’t deny there is natural variation, paq. To do so, you would have to demand that AGW produce a monotonically increasing graph. You can’t do that. But, ok. You have asserted that global climate models do not show slowdowns/pauses in their outputs. You haven’t proved that. You demand proof of everything I say, although everything I say is public knowledge, but now you are claiming, without proof of any kind that global climate models do not show pauses from natural variation. They do, of course, but let’s assume they don’t. So now we are assuming (1) there was a pause, and (2) AGW must be monotonically increasing.
We know that your alleged pause began with an extreme El Nino, and we know that the years you include in your pause were dominated by La Ninas with a few weak El Ninos. And, we know there was a lot of volcanic activity that threw up a lot of aerosol pollution, and there was a lot of aerosol pollution thrown up by China, India, and the burnin g of forests in Indonesia and Brazil. I assume you are also claiming (again without proof), that the extreme El Nino followed by mostly La Ninas and weak El Ninos and the volcanic activity, and the air pollution from China, India, Indonesia, and Brazil, had nothing at all to do with causing your “pause.”
Then you really have to give us some idea of what caused your pause, paq. I have now given you every assumption you demand, without asking you to prove any of them. So you really do have to say what your pause means scientifically. You really do have to explain what caused it.
• paqyfelyc says:
@Martin Smith
“No, paq, I didn’t put forth any theory.”
Don’t call it “theory” if you will. Doesn’t change anything. This “whatever you want to call it you put forth” is not science, period.
“Natural variation is simply part of the projections produced by global climate models.”
Nope. These variation are models variation, not natural variation. Nobody knows enough about natural climate variations to model them correctly. You even call them “noise”! Sorry, but Nature signal is THE signal, not noise, however noisy it may look.
“The outputs they produce include “pauses,” which are called natural variation.”
Call them “Arthur” if you will, that won’t allow them to run a round table society, any more than calling them “natural variation” won’t stop them to only be GIGO model variations, very far from natural variation as you admitted against your intention (see below).
” You can’t deny there is natural variation, paq. ”
I don’t, quite the opposite, I accuse warmists of denying them, by retro-calculating them as the difference between what happened in reality and the assumed effect of GHG. A wicked procedure called “begging the question”, that you find in manuals of rhetorical bad tricks, not in science manuals. A proper science would first understand natural variations, and produce a convincing reproduction of past climate with no free parameters to fit, before trying to assess human effect on them.
“To do so, you would have to demand that AGW produce a monotonically increasing graph. You can’t do that. ”
Strawman. I demand that AGW theory produce a normal graph, with proper error bar according to what is known about natural variation and the system, so that we can check against reality. I demand that the graph be just that: a graph, a single one, not an “ensemble” from which one will randomly be right afterward, as a Blind Texas Sharpshooter using a machine gun with bullets of hundreds of manufacturers could randomly hit the target and claim the manufacturer makes good bullets. I demand that the graph be specific enough (i.e. that the target is small enough) that hitting the target is not a random event, so that the check really means something about the model.
I don’t demand that this graph be monotonous. Just that it is drawn before the events, and that it fits them.
Is that too much to ask, for people who call themselves scientists?
Just for instance, physicists recently used 5 sigmas significance to be content that the signal they observed really was a signal and not some random event. By contrast, you proudly link to some work that, after some data torturing to extract unpredicted “noise”, “proved” that you cannot rule out that the Texas sharpshooter actually hit the mark…that is, 0 sigma significance.
“You have asserted that global climate models do not show slowdowns/pauses in their outputs.”
I didn’t. I am pretty sure they do experience slowdowns/pauses, and even worse, that they actually give wildly diverging, completely useless result. And they have to, if they did the job properly, since they deal with a chaotic system. Trouble is, such result is useless. So modelers hide this mess under some average, after discarding the most unbelievable result (iceball or Venus Earth within a century), and voilà.
“You haven’t proved that. ”
Irrelevant, as I said, I don’t believe what you think I asserted, pretty much the opposite.
Besides, the burden of proof lies upon those who push the theory (or whatever you want to call it…), not the one like me.
“You demand proof of everything I say, although everything I say is public knowledge”
So much public knowledge, that you failed to produce a single evidence. How is that possible?
“but now you are claiming, without proof of any kind that global climate models do not show pauses from natural variation. ”
I didn’t claim that.
“They do, of course, but let’s assume they don’t. So now we are assuming (1) there was a pause, and (2) AGW must be monotonically increasing.”
Not “we”. Only you. The important thing for me is an increase according to prediction, not that the increase be monotonic. Yet another strawman of yours.
“We know that your alleged pause began with an extreme El Nino, ”
No, we don’t “know” that, because this is plainly false. If you had checked, you would know that
1) the pause wasn’t calculated starting a convenient point, as warmists do, but backward from the current day to any point in the past showing a zero trend.
2) and the resulting point was not at El nino, but before.
proof here : https://wattsupwiththat.com/2016/02/06/the-pause-hangs-on-by-its-fingernails/
(notice that the writer PREDICTED in his very first sentence that the pause will end soon; he made no effort, used no statistical trick to extend it or manufacture a new one, as your Tamino did to disappear it. Says much about intellectual honesty of each)
So you should stop trust people who told you this lie.
What we DO know, is that it took the recent extreme El Nino to stop the pause, and that it will resume if some decent la Nina occurs, which is not impossible, is it?
” I assume you are also claiming (again without proof), that the extreme El Nino followed by mostly La Ninas and weak El Ninos and the volcanic activity, and the air pollution from China, India, Indonesia, and Brazil, had nothing at all to do with causing your “pause.”
You assume a lot. Actually, all your talk is just that: GIGO assumptions all the way down.
I just claim that you find a 18 year of zero trend, that the theory didn’t predict beforehand. Quite simple, and quite easy to disprove if it were false: just show me a paper from ~2000 or before, warning that warming could go through a decade or two of stalling, according to theory.
I also claim, after reading your excuse, that you admit that the theory (or whatever you call it) cannot cope with ENSO, volcanic activity, air pollution, forest fire, etc. I guess you have shorter time listing what it cope with. In the endless list of factor having an effect on climate, whatever happens, you will always find some excuse afterward to explain why the reality was different from prediction, you will always find some epicycle to add to have climate still revolving around CO2. You call this science?
“Then you really have to give us some idea of what caused your pause, paq.”
What caused the pause? the very same thing that caused climate to go through up and down all along recorded history. That is, the simple mathematical fact that, this is just the way a chaotic system behave without any external reason, and the empirical finding that climate is such a chaotic system.
Be aware that is just what Edward Lorenz wrote, and nobody still didn’t dare write a refuting paper. Climatastrophologists just ignored what he wrote. Do you?
” I have now given you every assumption you demand,”
No you didn’t. You still owe me a paper from ~2000 or before (before the pause), predicting that some pause could happen for a decade or two.
“without asking you to prove any of them. ”
Oh. So, when you wrote (just for instance) ” you are also claiming (again without proof)” or “You haven’t proved that. “, that’s not a way to ask for proof?
Anyway, all I wrote is self evident, proven, or at least not claimed false by anyone (I admit this is no proof of validity, but is the best we currently have)
“So you really do have to say what your pause means scientifically. You really do have to explain what caused it.
Which I already did, several time. Let’s do this again (sigh…)
The pause means that climate does what climate always did: up, down, and stall. The pause means, if predicted possible by the AGW theory, that the theory is not even wrong: impossible to check whether it is true or not; not science. The pause means, if not predicted by the AGW theory, that the theory is wrong.
You’ll notice that answering your claims and demands resulted in an awfully long comment. A pretty redundant one, moreover, has I just essentially repeated what I previously wrote to you. Please be a good faith person, and refrain from that, from now on. Stop making assumption, using strawman, repeating blatantly false claims, and ignoring answers already given to you. Make the effort to read and understand what has already been written before. Please.
• Dave Fair says:
+1, paqyfelyc. But wasted on true believers; facts do not matter in political debates.
• Martin Smith says:
paqyfelyc wrote: “Nope. These variation are models variation, not natural variation. Nobody knows enough about natural climate variations to model them correctly” …and: “Strawman. I demand that AGW theory produce a normal graph, with proper error bar according to what is known about natural variation and the system, so that we can check against reality” …and “You still owe me a paper from ~2000 or before (before the pause), predicting that some pause could happen for a decade or two.”
You are saying the same thing 3 different ways, but you are wrong. The Global Climate Models do have decade+ long natural variations. This is how it works. A GCM includes submodels for all of the physical systems listed as the column headings in Tables 9.A.1 and 9.A.2 beginning on page 854 of the document linked here: https://www.ipcc.ch/pdf/assessment-report/ar5/wg1/WG1AR5_Chapter09_FINAL.pdf
These are the physical subsystems we understand well. Not completely, but quite well enough to model them. The physical subsystems are: Atmosphere, Aerosol, Atmosphere Chemistry, Land Surface, Ocean, Ocean Bio-Geo Chemistry, and Sea Ice. When all these submodels are coupled, they each affect the model output and the result is what you call natural variation. That is, it is the interaction of all these real-world physical systems that cause natural variation. Did you really think natural variation was some separate force we don’t understand? It’s not. It is an emergent property of the interaction of all the physical subsystems, and the natural variation shown in the models’ outputs matches well the natural variation we see in the observed data.
So, none of you has addressed the issue, which remains: We know that your alleged pause began with an extreme El Nino, and we know that the years you include in your pause were dominated by La Ninas with a few weak El Ninos. And, we know there was a lot of volcanic activity that threw up a lot of aerosol pollution, and there was a lot of aerosol pollution thrown up by China, India, and the burnin g of forests in Indonesia and Brazil. I assume you are also claiming (again without proof), that the extreme El Nino followed by mostly La Ninas and weak El Ninos and the volcanic activity, and the air pollution from China, India, Indonesia, and Brazil, had nothing at all to do with causing your “pause.”
Then you really have to give us some idea of what caused your pause, paq. I have now given you every assumption you demand, without asking you to prove any of them. So you really do have to say what your pause means scientifically. You really do have to explain what caused it.
Aside:
And you wrote: “1) the pause wasn’t calculated starting a convenient point, as warmists do, but backward from the current day to any point in the past showing a zero trend.” Wow! That’s the most obvious admission of cherrypicking I have ever seen. Note that (a) your calculation ALWAYS includes the extreme El Nino year 1998, because without it, there is no zero trend (FTR: the trend is still positive even with the El Nino), and (b) you also stop the calculation when there is a large rise. You are admitting to choosing both the starting point and the ending point, and you cherrypick these points to get a zero trend. You can’t do that.
• Martin Smith says:
Dave Fair wrote: “But wasted on true believers; facts do not matter in political debates.”
Dave, will you answer the question then? paqyfelyc refuses. He tried this: “The pause means that climate does what climate always did: up, down, and stall,” which is tautological and ignores completely the anthropogenic CO2 contribution to both the atmosphere and the oceans. We know that your alleged pause began with an extreme El Nino, and we know that the years you include in your pause were dominated by La Ninas with a few weak El Ninos. And, we know there was a lot of volcanic activity that threw up a lot of aerosol pollution, and there was a lot of aerosol pollution thrown up by China, India, and the burning of forests in Indonesia and Brazil. I assume you are also claiming that the extreme El Nino followed by mostly La Ninas and weak El Ninos and the volcanic activity, and the air pollution from China, India, Indonesia, and Brazil, had nothing at all to do with causing your “pause.”
Then what caused your pause, Dave. What your pause means scientifically? You really do have to explain what caused it.
• paqyfelyc says:
@Martin Smith
February 14, 2018 at 2:41 am
“You are saying the same thing 3 different ways, ”
Nice you noticed, at last! Now just answer, if you please: produce a decent scientific graph from ~2000 or before hinting at a possible “pause” for a decade or two. Still waiting.
“These are the physical subsystems we understand well. Not completely, but quite well enough to model them. etc. ”
Sure, but not enough to get rid of the “butterfly effect”, so the model result are just crap. And actually not even godlike understanding would be enough.
“We know that your alleged pause blah blah blah”
Now, this is enough.
To prove Dave Fair (February 13, 2018 at 10:59 pm) right in implying you belong true believers, impervious to simple facts, along with flat-Earthers, creationists and fake-moon-landing conspiracy theorists? You are doing a good job at it, then.
I however already explained you. I’ll do a last time:
Even when fully determinists, perfectly cause-consequence related, systems in chaotic mode, from which climate was the archetype, ALWAYS experience wild variations WITH NO EXTERNAL CAUSE AT ALL
Got it? Is that clear, now?
Stop pretending a cause is required.
Stop demanding the production of some other cause to accept the failure of your favorite (CO2 for you; other people have other obsessions) to explain things.
STOP IT
NOW
FOREVER
And start reading basic literature about chaos theory. Starting wikipedia. You’ll then answer yourself your questions, and facepalm you saying “damn, how could I be so stupid to trust this damn not-even-wrong, not scientific, AGW theory”.
That’s why I didn’t asked you to explain why, despite much more CO2 than previously though was dumped into the atmosphere, the pause happened. I know you cannot. I know this is mathematically impossible. And I know you ask silly rhetorical question because you wrongly believe you and Tamino know something. Neither of you do.
PS Your side note is laughable, and just prove, again, how scientifically weak you are. Cherry picking is when you intend to prove a theory out of a cases that agree while disregarding cases that disagree, which is just what YOU do. Pointing at a disagreeing case to disprove a theory in not only allowed, it is just the way it works, and that’s not cherry-picking.
• Martin Smith says:
paqyfelyc wrote: “Even when fully determinists, perfectly cause-consequence related, systems in chaotic mode, from which climate was the archetype, ALWAYS experience wild variations WITH NO EXTERNAL CAUSE AT ALL”
Sorry, but that’s a complete non sequitur. We’re not talking about external causes for the natural variation. Please do not change the subject. You have to explain what is causing your alleged pause, because the global climate models do show pauses like the one you allege. Here is the explanation. Please address it:
The Global Climate Models do have decade+ long natural variations. This is how it works. A GCM includes submodels for all of the physical systems listed as the column headings in Tables 9.A.1 and 9.A.2 beginning on page 854 of the document linked here: https://www.ipcc.ch/pdf/assessment-report/ar5/wg1/WG1AR5_Chapter09_FINAL.pdf
These are the physical subsystems we understand well. Not completely, but quite well enough to model them. The physical subsystems are: Atmosphere, Aerosol, Atmosphere Chemistry, Land Surface, Ocean, Ocean Bio-Geo Chemistry, and Sea Ice. When all these submodels are coupled, they each affect the model output and the result is what you call natural variation. That is, it is the interaction of all these real-world physical systems that cause natural variation. Did you really think natural variation was some separate force we don’t understand? It’s not. It is an emergent property of the interaction of all the physical subsystems, and the natural variation shown in the models’ outputs matches well the natural variation we see in the observed data.
So, none of you has addressed the issue, which remains: We know that your alleged pause began with an extreme El Nino, and we know that the years you include in your pause were dominated by La Ninas with a few weak El Ninos. And, we know there was a lot of volcanic activity that threw up a lot of aerosol pollution, and there was a lot of aerosol pollution thrown up by China, India, and the burnin g of forests in Indonesia and Brazil. I assume you are also claiming (again without proof), that the extreme El Nino followed by mostly La Ninas and weak El Ninos and the volcanic activity, and the air pollution from China, India, Indonesia, and Brazil, had nothing at all to do with causing your “pause.”
Then you really have to give us some idea of what caused your pause, paq. I have now given you every assumption you demand, without asking you to prove any of them. So you really do have to say what your pause means scientifically. You really do have to explain what caused it.
Aside:
And you wrote: “1) the pause wasn’t calculated starting a convenient point, as warmists do, but backward from the current day to any point in the past showing a zero trend.” Wow! That’s the most obvious admission of cherrypicking I have ever seen. Note that (a) your calculation ALWAYS includes the extreme El Nino year 1998, because without it, there is no zero trend (FTR: the trend is still positive even with the El Nino), and (b) you also stop the calculation when there is a large rise. You are admitting to choosing both the starting point and the ending point, and you cherrypick these points to get a zero trend. You can’t do that.
[?? .mod]
• paqyfelyc says:
@Martin Smith
OMG, are you hopeless? Now you see “non sequitur” in a simple statement of fact…
Just read wikipedia about chaotic systems (and “non sequitur”, too, that wouldn’t hurt). I am not to teach basics to a man who have rather protect his ignorance with pedant misuse of formulae like “non sequitur” and “sherry-picking”, and false assumptions he refuses to let go.
Come back when enlighten, or come back when confused but at least conscious and eager to learn, then I could help.
• Martin Smith says:
paqyfelyc, this is what you wrote: “Even when fully determinists, perfectly cause-consequence related, systems in chaotic mode, from which climate was the archetype, ALWAYS experience wild variations WITH NO EXTERNAL CAUSE AT ALL”
It is a non sequitur. You are claiming that natural variation is not caused because the global climate is chaotic. Your statement is false. Chaos does not mean “without cause.”
I understand chaos, paq. You can’t appeal to chaos theory to refute the point of this discussion. The Global Climate Models do have decade+ long natural variations, what you call pauses. These are projected by the GCM’s. This is how it works. A GCM includes submodels for all of the physical systems listed as the column headings in Tables 9.A.1 and 9.A.2 beginning on page 854 of the document linked here: https://www.ipcc.ch/pdf/assessment-report/ar5/wg1/WG1AR5_Chapter09_FINAL.pdf
These are the physical subsystems we understand well. Not completely, but quite well enough to model them. The physical subsystems are: Atmosphere, Aerosol, Atmosphere Chemistry, Land Surface, Ocean, Ocean Bio-Geo Chemistry, and Sea Ice. When all these submodels are coupled, they each affect the model output and the result is what you call natural variation. That is, it is the interaction of all these real-world physical systems that cause natural variation. Did you really think natural variation was some separate force we don’t understand? It’s not. It is an emergent property of the interaction of all the physical subsystems, and the natural variation shown in the models’ outputs matches well the natural variation we see in the observed data.
You can’t refute the point by calling me stupid. Nor can you ignore the evidence. GCM’s do predict there will be periods where there is little or no temperature rise, even declines are possible, so proving there was a pause (which neither you nor Sheldon Walker, nor Richard M, nor Dave Fair has done, does not argue against AGW. It isn’t evidence in your favor at all — UNLESS you can say what caused the alleged pause. Can you?
• paqyfelyc says:
@Martin Smith
stop this nonsense. You use your mind in the most perverse way, finding the most stupid, nonsensical interpretation of what I wrote, to use it as strawman. Where do you think this attitude will get you?
A non sequitur is about two statements, and a false claim that validity of one says something about of the other. I made a single statement. No “non sequitur” possible.
“I understand chaos, paq.” Well, all what you wrote is proof of the opposite: no one understanding chaos would make the kind of demand for a “cause” that you do. Do you know why I mentioned Ed Lorenz? Someone understanding chaos and what it means for climate, would.
Not understanding chaos is not a trouble per se. You can still learn. I just warn you: understandingh as a price. downgrading part of what you think is “knowledge” into mere “belief” outside of science. Are you ready for that?
• Martin Smith says:
paq, your claim is a non sequitur. You are claiming that natural variation is not caused because the global climate is chaotic. Your conclusion does not follow from the facts. The global climate IS chaotic. No one disputes that. But being chaotic does not mean the natural variation is not caused by the interactions of the many physical subsystems we do understand fairly well.
You can’t appeal to chaos theory to refute the point of this discussion. The Global Climate Models do have decade+ long natural variations, what you call pauses. These are projected by the GCM’s. This is how it works. A GCM includes submodels for all of the physical systems listed as the column headings in Tables 9.A.1 and 9.A.2 beginning on page 854 of the document linked here: https://www.ipcc.ch/pdf/assessment-report/ar5/wg1/WG1AR5_Chapter09_FINAL.pdf
These are the physical subsystems we understand well. Not completely, but quite well enough to model them. The physical subsystems are: Atmosphere, Aerosol, Atmosphere Chemistry, Land Surface, Ocean, Ocean Bio-Geo Chemistry, and Sea Ice. When all these submodels are coupled, they each affect the model output and the result is what you call natural variation. That is, it is the interaction of all these real-world physical systems that cause natural variation. Did you really think natural variation was some separate force we don’t understand? It’s not. It is an emergent property of the interaction of all the physical subsystems, and the natural variation shown in the models’ outputs matches well the natural variation we see in the observed data.
You can’t refute the point by calling me stupid. Nor can you ignore the evidence. GCM’s do predict there will be periods where there is little or no temperature rise, even declines are possible, so proving there was a pause (which neither you nor Sheldon Walker, nor Richard M, nor Dave Fair has done, does not argue against AGW. It isn’t evidence in your favor at all — UNLESS you can say what caused the alleged pause. Can you?
• Richard M says:
Martin Smith claims: ” We know that your alleged pause began with an extreme El Nino”
So what. We also know that the El Nino was followed by a 3 year La Nina which completely balanced out any effect of the El Nino on the trend. Why do warmists continue to prove they are math illiterates? Just check out the trend after all of the ENSO years. It is actually more negative.
http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1997/to:2015/trend/plot/uah6/from:2001/to:2015/trend
• Martin Smith says:
Richard M wrote: “So what. We also know that the El Nino was followed by a 3 year La Nina which completely balanced out any effect of the El Nino on the trend. Why do warmists continue to prove they are math illiterates? Just check out the trend after all of the ENSO years. It is actually more negative.”
Richard M is using UAH data, which (a) shows the temperature at 15,000 feet, not sea level, and (b) disagrees with not only the RSS satellite data for temperature at 15,000, but also disagrees with all the surface temperature datasets. RSS and UAH start with the same data from the satellites and arrive at different results. The RSS result agrees with the surface temperature datasets, and the UAH result, which Richard M is using here, disagrees with the surface temperature datasets.
So Richard M is, by implication, claiming that the UAH result is correct, and the |RSS result is incorrect. Richard M is also claiming by implication, that all the surface temperature datasets (the ones measured by thermometers) are wrong. Richard M’s claim that there was a pause is seen to be false when he uses RSS or any of the surface temperature datasets.
The question remains, Richard. Suppose there was a pause. The occurrence of a pause does not argue against AGW, for there have been many pauses, and yet the observed rise in global average surface temperature continues to match the expected rise predicted by AGW theory: https://skepticalscience.com/graphics.php?g=47
So what is the meaning of your alleged pause?
• Richard M says:
Martin Smith: “If there was a pause, it doesn’t damage the AGW case because |AGW projections show pauses like the one you are alleging.”
Santer et al in trying to support your beliefs actually found out that pauses over 17 years in length were out side the 95% confidence level. So, an 18 year pause is scientifically meaningful. And, if you simply remove years with ENSO effects (let’s call it noise). then you will see a period of over 20 years with no warming.
How long a period of no warming do you need before you question your belief system???? I have a feeling the answer is infinite.
• Martin Smith:
“If there was a pause, it doesn’t damage the AGW case because AGW projections show pauses like the one you are alleging.”
Please show us ANY series of actual model runs, their date-time-group report with all relevant temperatures and temperature graphics, and ALL inputs, that do actually show:
any 5 year pause within a 150 year run, CO2 increasing exactly as it has been, from the conditions found in 1997. NO forced assumptions on CO2.
any 10 year pause in 150 years of projected global average temperatures, same conditions
any 15 year pause in 150 years of projected global average temperatures, same conditions
any 20 year pause in global average temperatures after the same 150 years projection.
IF your claim is correct, there will be a very large number of 5 year flat temperature projections as CO2 increases.
IF your claim is correct, there will be a large number of 10 year pauses as well, but fewer than above. If your models are correct, these pauses will begin randomly as some of the 5-year pauses continue, and some are stopped by cooling (as we had from 1945-1975), and some by a renewal of heating.
Same with 15 pauses: Fewer than the 10 years group, but several. You claim 18 year pauses have been found, so several (3 or 4) shorter 15 year pauses must be easy to find!
IF your claim is correct, there MUST occur at least two or more “random” 20 years pauses. After all, you claim to have evidence the models have already produced these results, do you not?
• Martin Smith says:
Richard M wrote: “Santer et al in trying to support your beliefs actually found out that pauses over 17 years in length were out side the 95% confidence level. So, an 18 year pause is scientifically meaningful.”
Richard, I have asked several times now for you to say what that meaning is. Nevermind that you are still using ONLY UAH data while ignoring ALL the other datasets which disagree with UAH while all agreeing with each other. There have been many pauses: https://skepticalscience.com/graphics.php?g=47
You claim your alleged 18 year pause is scientifically meaningful. What, exactly, is that meaning?
• paqyfelyc says:
@Martin Smith
You are not talking to me. You didn’t even read me. You just build some strawman out of the most stupid interpretation you could imagine out of what I wrote, pushing aside anything that could bother you (this is cherry picking, you know?), so as to have the upper hand. This strawman you call “paq”, but actually is just yourself, as you failed to notice.
Well, you are doing yourself no favor. You are just downgrading yourself in this mudwrestling of your ego against your homunculus, that your ego always win, of course. Why don’t you try to upgrade your play? into an upper league where you could lose, but with panache, and even a loss would actually be a victory for you, because you would learn something? Why don’t you address YOUR current shortcoming, instead of trying to exploit the fake one you assigned to me (which is not actually me, just your homunculus, but is makes no difference AFAYAC) ?
“You can’t refute the point by calling me stupid”.
Oh. I see. You are afraid of being stupid. Which answer previous question. You shouldn’t. It takes a clever mind, to fool oneself in such a way you do. If you weren’t so afraid, you would quit aforementioned mudfight.
Fear usually drives you right were you don’t want to go, your fear of being stupid is what makes you appear so (in your own mind). BTW, to repeat oneself again and again is great way to appear stupid, you know…
I am no calling you stupid. I am calling you ignorant, and ignorant you proved yourself. You obviously do not know why I mentioned Ed Lorenz, and your explanation about chaos was not about chaos, as you think is was, but about a system made of many parts, which is very different and barely overlapping. The climate is not a car. So, don’t try again to prove you know about chaos; well, at least not before you actually do, which would require some reading first. Etc.
Teach less, Learn more. You can do it, I promise you are not stupid. I promise you can become a skeptic (Just wondering if you have any sense of humor…)
“Nor can you ignore the evidence. ”
You mean, the evidence you never produced? Fairly simple to ignore, unfortunately.
• Martin Smith says:
paqyfelyc, please stop attacking me. Address the point. We are here simply assuming the alleged pause is real. We know that global climate models show pauses in their outputs, so what is the meaning of this alleged pause? Here is a good paper on the subject: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
Assuming the pause exists, and given that there have been many pauses (see here: https://skepticalscience.com/graphics.php?g=47), why are you saying THIS most recent pause means AGW is wrong, when none of the other pauses before it (see the link) meant AGW is wrong?
• paqyfelyc says:
@Martin Smith
You repeat yourself.
You still only provide papers from 2010 and after. Still waiting for the prediction you claim exist. Remember: a prediction is made BEFORE having knowledge of the predicted, not after
Alternatively, you could simply recognize you have no evidence of such prediction, and you can only provide papers made afterward. Is that SO hard for you?
You again invent my saying something. Why don’t you simply read what I wrote, instead, for once (no evidence you ever did, so far)?
But, let’s pretend I didn’t made myself perfectly clear. Why not. Let’s try again, then.
I am not saying AGW is wrong. I am saying it is not-even-wrong: nothing can disprove it. The theory begins with huge target hard to miss, and even if it miss a target, no problem, you just need to invoke afterward some volcano, pollution, meteorite, ENSO, whatever, or even some fancy statistical treatment of the poor available data, to explain any discrepancy. “With four parameters, I can fit an Elephant, with five I can have him wiggle his trunk”, and they have HUNDREDS of parameters to choose from.
YOU are, involuntarily, fueling my assertion, actually, when you provide papers “this event we didn’t predict, doesn’t disprove the theory, because [insert made-up afterward excuses]”. QED.
Not falsifiable. Not reproducible. Not predictive. Not operative. How do you call it?
• Martin Smith says:
paqyfelyc wrote: “You still only provide papers from 2010 and after. Still waiting for the prediction you claim exist.”
I’m sorry, are you now also cherrypicking years you will accept papers from? Cherrypicking data as you are is bad enough, but now you refuse to consider papers written since 2010. Sorry, that’s not on. This paper shows that the models that include ENSO do show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
And, sorry again, but the observed data do show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47), so (a) the paper demonstrates that the climate models that do include ENSO do show pauses, and (b) the other link shows that there have been many of what you call pauses.
paqyfelyc also wrote: “I am not saying AGW is wrong. I am saying it is not-even-wrong: nothing can disprove it.”
That’s an interesting topic, but it has nothing to do with this topic, which is this. Assuming the pause exists, and given that there have been many pauses (see link above), if you are not saying your alleged pause invalidates AGW, what are you claiming the pause means? What is the significance of the pause, given that global average temperature is rising as AGW predicts? The only explanation for the temperature rise we have seen over the last 100+ years is AGW. The way you disprove AGW is to propose a different theory that accounts for all the data and that makes better predictions about the future than AGW. I don’t think you have an alternative theory, but here we are only discussing your claim that your cherrypicked pause has scientific significance. What is that significance?
• Richard M says:
Martin Smith: “Nevermind that you are still using ONLY UAH data …”
This gets to the nut of the issue. When faced with uncomfortable facts Martin’s only response is denial. I already answered his made up problems with UAH yet he still spews nothing but denial. It is obvious that Martin’s opinion is driven by what he wants to believe. No facts will be allowed to intrude.
• Martin Smith says:
Richard M wrote: “Santer et al in trying to support your beliefs actually found out that pauses over 17 years in length were out side the 95% confidence level. So, an 18 year pause is scientifically meaningful.”
Richard, I have asked several times now for you to say what that meaning is. You have stated again that your alleged pause is scientifically meaningful. This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47), so (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many of so-called pauses before this one you claim is scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• Richard M says:
Martin, when you look at model produced pauses or pauses during warming periods you will find they are normally caused by ENSO. They usually end with a simulated period of cool La Nina or start with a simulated El Nino or both. That is no doubt true of Santer et al.
Now look at this UAH data.
http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1997/to:2015/trend/plot/uah6/from:2001/to:2015/trend
It actually starts with La Nina conditions and ends at the start of an El Nino. Exactly the opposite of what is needed for the models to produce a pause. Get it? Where are model based pauses of this length that aren’t the result of ENSO (or simulated volcano)?
Welcome to reality. This is what is known as falsification.
• Martin Smith says:
Richard M wrote: “Now look at this UAH data.”
Richard M is trying to deceive you. You can’t get away with this, Richard. Your post is wrong for two reasons. First, you only use the UAH dataset. Why? You cherrypicked the UAH data alone because it is the only dataset that allows you to make your claim there was a pause. If you include the NASA GISS, NOAA NCDC, HadCRUT4, and RSS datasets, you can’t.
Second, your graph has not removed the ENSO effect. You must remove it, because your graph begins with an extreme El Nino anomaly in 1998. Your token inclusion of 1997 doesn’t change this. You can discover for yourself how to remove the ENSO effect. This is not how its done, but do your computation again without that one El Nino datapoint. Where did your alleged pause go?
• Richard M says:
Martin is now ignoring several things I already pointed out. The 1998 El Nino has no effect on the trend since it is surrounded by several La Nina events. This is obvious in the graph I showed him where I also provided a trend line starting in 2001.
http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1997/to:2015/trend/plot/uah6/from:2001/to:2015/trend
UAH is the best data we have. All the rest have huge problems that Martin doesn’t want to accept. None of the surface data has widespread coverage, it’s infected by UHI, it uses poor technology and it is constantly adjusted by biased folks. RSS throws out data and replaces it with something from models or the very same lousy surface data. Pure junk.
The bottom line is Martin has denied anything that prevents him from holding on to his current beliefs. He even ignores obvious facts that have been pointed out to him multiple times.
The pause is based on the best data we have and it has clearly shown that the models do not work. No model produces an 18 year pause that isn’t caused by timely placed ENSO events. If Martin cared about the truth this would be meaningful, but instead all is can do is spew ad homs.
• Martin Smith says:
Richard M wrote: “The 1998 El Nino has no effect on the trend since it is surrounded by several La Nina events.”
That’s astonishing, Richard. You are saying that the highest temperature in the data has no effect on the average of the data, because it has colder temperatures on either side of it. That’s false. Remove the temperature for 1998 and recompute the trend.
Richard M wrote: “UAH is the best data we have.” “RSS throws out data and replaces it with something from models…”
That’s also astonishing, Richard. UAH and RSS start from the same dataset. That data is NOT temperature data. Temperature data is implied from the satellite data by passing the satellite data through a model. The reason the UAH model output disagrees with the RSS model output is that they use different models. The RSS model outputs data that agree with NASA GISS, NOAA NCDC, and HadCRUT4. The UAH model outputs data that disagree with NASA GISS, NOAA NCDC, and HadCRUT4. So you have no basis for claiming the UAH data is “the best data we have.” Your deception fails.
• Richard M says:
Martin Smith: ” You are saying that the highest temperature in the data has no effect on the average of the data, because it has colder temperatures on either side of it. That’s false. Remove the temperature for 1998 and recompute the trend.”
That is exactly what I did, Martin. I provided you a link. Did you even look at it? Of course, I didn’t JUST remove the warm outlier event, I also removed the cold ones. That’s why it balanced out. When that was done the trend was actually lower. Here are both trends yet again.
http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1997/to:2015/trend/plot/uah6/from:2001/to:2015/trend
So, we have now established beyond any doubt that you will lie to maintain your belief system. As for RSS I suggest you read up on what they did in version 4. Your description is wrong again. Or, is it another intentional lie? Here’s what they say:
“To account for changes in observation times, the RSS group used a number of different approaches and models to try and estimate what the temperature would have been if the measurement time remained constant. This involves a combination of satellite observations (when different satellites captured temperatures in both morning and evening), the use of climate models to estimate how temperatures change in the atmosphere over the course of the day, and using reanalysis data that incorporates readings from surface observations, weather balloons and other instruments.”
Martin, as you can see your description was wrong. RSS now uses made up data to replace observations. Why do you insist on lying? Do you actually think lying makes you look smarter?
I no longer plan to respond to your lies. It is obvious you aren’t interested in the truth. Trying to educate someone who is not interested in the truth is a waste of time.
• Martin Smith says:
Richard M wrote: “That is exactly what I did, Martin. I provided you a link.”
No, you didn’t. You chose 2001. That’s not the same interval. I said leave out the data point for 1998. Start at 1997 but leave out 1998. You didn’t do that. The reason you didn’t do that is because you knew it would prove you are wrong. So what did you do? Did you choose to start at 1999, the year after the one you have to delete? No. Why? Because that also proves you are wrong. Try it. Start at 1999. Redo the computation. The trend rises. And if you started at 1997 but left out 1998, it would rise more steeply. What you have done now is prove that you are trying to deceive. Your deception fails.
You have multiple problems in your method. You aren’t removing the effects of ENSO. You aren’t removing the effects of volcanic activity. You aren’t removing the effects of changes in solar irradiance. You can see the effects of doing all of these here: https://skepticalscience.com/global-warming-stopped-in-1998-intermediate.htm
Skip down to the section titled: “Removing El Niño and other Exogeneous Factors.” Note that See the animation. All the datasets are shown, even your preferred UAH. And there is no pause. Why? Because the exogenous factors we can remove have been removed. You didn’t do that. Your method is wrong.
Richard M wrote: “Martin, as you can see your description was wrong.”
Nope. Your understanding is incorrect. Your cherrypicked quote (without attribution) indicates the RSS data has been corrected for a known error, which, you apparently mean, has not been corrected in the UAH data. How can you claim the UAH data “the best data we have” while admitting that it contains a known error?
Richard M wrote: “I no longer plan to respond…”
Of course not. You have been caught out. You must leave the field.
• Richard M says:
Martin Smith: “I said leave out the data point for 1998. ”
Oh I just had to laugh at this. You want only the warm ENSO year removed at the start of the trend but not the cold ENSO years. LMAO. That’s like telling your bank you want only the payments removed from your bank balance but not the deposits and saying “see, I wasn’t overdrawn”. I really didn’t think anyone could be this inane.
Martin, your comments are nothing but pure nonsense. If you think you’ve made any points you are kidding yourself. All you have done is proven beyond any doubt that you are scientifically illiterate.
• Martin Smith says:
Richard M wrote: “You want only the warm ENSO year removed at the start of the trend but not the cold ENSO years.”
Removing a single data point from your alleged pause shows changes your flat trend to an increasing trend. That’s the point. You saw that and reduced the length of your period. Not 1 year; not 2 years, not 3 years, but 4 years. You did that to prevent your readers from seeing the truth. Your alleged pause depends on the extreme El Nino anomaly in 1998. Take that out, and you have nothing.
• Richard M says:
OMG, you aren’t just a troll, you are a clueless troll. You don’t even have a clue what ENSO is or how it works. You obviously think removing just an El Nino event is a valid scientific argument. LOL. My analogy above was perfect. You only want payments removed from your back account.
On the chance that you really are just this naive, let me educate you. ENSO events are whenever the tropical Pacific is either .5 C above (El Nino) or below (La Nina) average. Got that???? Do you understand was “below” means? If you remove JUST the above events you aren’t correcting for ENSO, you are producing scientific gibberish. In fact, let’s flip my analogy above, let’s only remove large deposits from your bank account. Would you think your bank balance was valid if the bank did that???? Well, that is exactly what you want to do.
What we had over the 1997-2001 time period was one large deposit (1998 El Nino) and one large payment (1999-2001 La Nina). They essentially canceled each other out as far as the planet’s energy balance. You can remove all of them or none of them. Anything else is nonsense. That was what the two lines in my link showed. Either way gives you a valid trend and they are almost identical.
I do have to take some of the blame here. I thought anyone coming on here to discuss climate would have a small understanding of subject. Martin is completely illiterate when it comes to climate. He didn’t have a clue what ENSO was (probably still doesn’t).
PS. I used my usual troll bait earlier. If I want to determine if a person is a troll I tell them I am leaving. If they are a troll they will jump up and down and claim victory. Martin took the bait eagerly and showed his true colors. :))
• Martin Smith says:
Richard M wrote: “You don’t even have a clue what ENSO is or how it works. You obviously think removing just an El Nino event is a valid scientific argument.”
No, Richard. I told you from the start, and you ignored: “This is not how its done, but do your computation again without that one El Nino datapoint.”
Richard M wrote: “If you remove JUST the above events you aren’t correcting for ENSO,”
That’s what I said above, Richard. You ignored it. The point is that if you remove one data point from your zero trend line, it reverts to the upward trend that is there in the entire dataset. You can’t do that with any other data point in your cherrypicked interval except 1998, which was an extreme El Nino anomaly. It was the warmest year by far at that point, and it remained the warmest year until, what was it, 2013? 2014? Your entire alleged pause.
Richard M wrote: “What we had over the 1997-2001 time period was one large deposit (1998 El Nino) and one large payment (1999-2001 La Nina). They essentially canceled each other out as far as the planet’s energy balance. You can remove all of them or none of them. Anything else is nonsense.”
No, they didn’t. But your miscue is partially right. You must remove all of them. These actual scientists did remove all of them. See the animation of the result in the section titled: “Removing El Niño and other Exogeneous Factors.” https://skepticalscience.com/global-warming-stopped-in-1998-intermediate.htm
Note that all the datasets are shown, including your preferred UAH. No pause in any of them. Your attempted deception fails.
• Richard M says:
Martin, as I already explained to you several ways, removing just one El Nino event from a trend results in gibberish. What happens if you removed the 1999-2001 La Nina? Yes, that is one event. That would turn the trend into a strongly declining trend. Of course, that would also be gibberish but it demonstrates that your point is silly.
Only removing the outlier data that supports your bias is nonsense and the fact you would even suggest it shows you don’t understand the first thing about science. To get a ****legitimate*** trend you need to balance out the climate noise. Removing part of the noise and leaving another part that we know are related is beyond silly. The fact you keep harping on this anti-science nonsense is informative.
What is really funny is you then refer to the Skeptical Science propaganda which just happens to end with the 2010 El Nino but doesn’t include the following La Nina event. Talk about anti-science. What happens if you compute the trend without the 2010 El Nino? Oh yeah, the trend goes negative. The fact you don’t even see how this completely destroys your arguments is hysterical. Your claim of “no pause” is precisely due to cherry picking ONLY the 2010 El Nino and not including the La Nina event that naturally follows. Let’s look at the same data with the La Nina event.
http://www.woodfortrees.org/plot/uah6/from:1997/to:2012/plot/uah6/from:1997/to:2012/trend/plot/uah6/from:2001/to:2012/trend
Martin, to put gently you are making a complete fool of yourself. Did you even know that El Nino and La Nina events generally come as pairs? Yes, they are physically linked. You would do well to understand Bjerknes feedback and why El Nino conditions spawn La Nina conditions.
• Martin Smith says:
Richard M misses the point: “as I already explained to you several ways, removing just one El Nino event from a trend results in gibberish.”
That’s incorrect, Richard. Forget that these are global average temperature anomalies. They are just data points, one for each year beginning in 1997 and ending in 2015. The number for 1998 is of dubious value because it is extremely high. Using all the data points, the trend line is flat. But if you remove one number, the one for 1998, the trend line rises kind of dramatically. And if you then add all the other data points for all the years you left out, you see that even if we also add the extreme anomaly for 1998, the rising trend line approaches the rising trend line for your 1997-2015 trend line without the 1998 number. That’s the point.
Richard M then wrote: “What is really funny is you then refer to the Skeptical Science propaganda which just happens to end with the 2010 El Nino but doesn’t include the following La Nina event.”
Richard, that’s because the paper was written in 2010 and published in 2011, before your following La Nina event. When it is included, and when all the years following it are included, and when all the years prior to your cherrypick 1997 are included, your attempted deception fails even worse. And Skeptical Science didn’t write the paper. Two real scientists wrote the paper. One real statistician and one real climate scientist: Grant Foster and Stefan Rahmstorf. Here is their paper, which you apparently still have not read: http://iopscience.iop.org/article/10.1088/1748-9326/6/4/044022/meta
And here is the discussion of their paper, which you apparently have still not understood: https://skepticalscience.com/foster-and-rahmstorf-measure-global-warming-signal.html
And finally, Richard M wrote: ” Did you even know that El Nino and La Nina events generally come as pairs?”
No, they don’t. See figure 6: http://onlinelibrary.wiley.com/doi/10.1002/2013EF000165/full
• Martin Smith says:
Richard M wrote: “Martin, as I already explained to you several ways, removing just one El Nino event from a trend results in gibberish.”
Let’s test your theory, Richard. We will leave out the El Nino in 1998, and since you claim an El Nino must be paired with a La Nina, we will leave out the La Nina in 1997. Recompute: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
The trend is up. Your deception fails.
• Martin Smith says:
Richard M wrote: “Martin is working very hard to try and hold on to his belief system.”
Richard, my belief system has nothing to do with the truth or falsehood of scientific propositions. And I do understand Bjerknes feedback.
1. You say the data points in your graph of UAH TLT data from 1997 to 2015 cannot be extracted and looked at as a simple sequence of data points having nothing to do with temperature in the troposphere. But that’s exactly what your tool does. See it again: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
You enter a start year and an end year. The tool then loads the data for the selected years and performs a simple linear regression, which will always compute a straight line through the data. It doesn’t matter what the data is, SLR always computes a straight line through the data. So if we remove the data point for 1998, which is way high abnormally, and which is only the 2nd year in your 18 year period, and then run the tool again, it changes the line it computes quite a lot. Instead of being flat, it rises quite a lot. That is because the El Nino in 1998 was an extreme anomaly, and yes, we can remove one data point. And yes, I know your tool is displaying monthly data, but we can remove the entire 12 months. We can do that; the result really is meaningful. It disproves your claim because without the extreme anomaly of 1998, there was no pause.
2. You say El Nino must be paired with a La Nina for removal from the data, but your tool doesn’t care. In fact, you violated your own rule by starting with a weak La Nina in 1997.
3. But there can be more than one La Nina between El Ninos, and there can be no La Ninas between El Ninos, so when you say they must be removed from the data in pairs, you are making that up.
4. Neither El Ninos nor La Ninas begin on January 1st and end on December 31st. They begin and end any time, but your tool only selects data by year, so it doesn’t support what you are trying to argue.
5. What you are trying to argue is that ENSO and PDO caused your alleged pause and are causing what is known as AGW. But that can’t be correct. ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system. Nor do they prevent energy from leaving the system. Therefore, while they certainly do contribute to natural variation in the global average temperature, they cannot cause an upward trend or a downward trend in global average temperature.
6. About 93% of the energy from the sun that reaches earth and is not reemitted back to space is absorbed by the oceans and by melting of the icecaps. Your UAH TLT data only measures what happens in the lower troposphere, which accounts for less than 7% of the energy from the sun that reaches earth and is not reemitted back to space. That process of 93% of the energy being absorbed by the oceans and by melting ice did not stop during your alleged pause. It didn’t slow down. There is evidence that it increased. Therefore, you cannot say (even if your alleged pause in UAH TLT is statistically significant) that it means global warming paused.
7. When there is an El Nino where an abnormal amount of energy is released from the Pacific Ocean around the equator, or when there is a La Nina and the reverse happens, it doesn’t mean the oceans are releasing or absorbing that amount of energy everywhere. In fact, most of the ocean areas continue to absorb that 93% of the energy reaching earth that is not reemitted back to space. In fact, the oceans continue to absorb energy even during El Nino, just not where El Nino is.
8. You say this paper (https://skepticalscience.com/foster-and-rahmstorf-measure-global-warming-signal.html)
by Foster and Rahmstorf is wrong (I forget what you called it), but they are both professional scientists, one a statistician, the other a climate scientist, and their paper was peer reviewed and publish. You provided no argument that shows their analysis is wrong.
9. Foster and Rahmstorf 2011 removes the ENSO effect from the entire UAH TLT dataset, and from all the other datasets as well. They also remove the effects of variations in aerosols and variations in solar irradiance. Your tool doesn’t do any of this. Your tool uses data that you admit contain these noise causing effects. You even imply that the alleged pause depends on ENSO. But ENSO can’t cause warming or cooling. It only exchanges energy between the ocean and the lower atmosphere.
10. Finally, after all this, you still have not stated what the alleged pause means scientifically. If we assume the pause was statistically significant and scientifically meaningful, as you demand, then what is the scientific meaning of the pause?
• Richard M says:
Martin spews more nonsense … “The number for 1998 is of dubious value because it is extremely high. Using all the data points, the trend line is flat. But if you remove one number, the one for 1998, the trend line rises kind of dramatically. ”
First of all it isn’t a single data point. Each month is a data point. Second, as I already told you, if you leave out the La Nina data points (those numbers are just as “dubious” as the El Nino numbers) the line falls dramatically. Neither one of those situations is reality. The result of leaving them out is completely and totally meaningless.
The fact you keep repeating nonsense only solidifies the fact you are completely illiterate when it comes to science. This is exactly the same as leaving out your bank deposits in my analogy. Would you accept a bank balance where they left out your deposits???? Maybe one of them was big and of “dubious” value. What would it mean? Would it tell you anything at all?
This is not rocket science.
Now it gets even more humorous. Martin claims El Nino and La Nina events don’t often come in pairs. Here is the NOAA data. Let’s take a look.
http://origin.cpc.ncep.noaa.gov/products/analysis_monitoring/ensostuff/ONI_v5.php
1983 El Nino
1984-85 La Nina
1987-88 El Nino
1989 La Nina
1995 El Nino
1996 La Nina
1998 El Nino
1999-2001 La Nina
2005 El Nino
2006 La Nina
2007 El Nino
2008-09 La Nina
2010 El Nino
2011-12 La Nina
2015-16 El Nino
2017 La Nina
Even clueless Martin should see the pattern.
Facepalm.
• Martin Smith says:
Richard M wrote: “Martin spews more nonsense …”
You can’t avoid the point forever…
Richard M wrote: “Martin, as I already explained to you several ways, removing just one El Nino event from a trend results in gibberish.”
Let’s test your theory, Richard. We will leave out the El Nino in 1998, and since you claim an El Nino must be paired with a La Nina, we will leave out the La Nina in 1997. Recompute: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
The trend is up. Your deception fails.
• Richard M says:
Martin is working very hard to try and hold on to his belief system. However, we are making progress. His latest attempt:
“Let’s test your theory, Richard. We will leave out the El Nino in 1998, and since you claim an El Nino must be paired with a La Nina, we will leave out the La Nina in 1997.”
The reason I told Martin to look up Bjerknes feedback is because that is the mechanism that leads to a La Nina following an El Nino. The fact they are physically linked is why you can’t just eliminate the El Nino from a trend. If the El Nino had not happened then it is highly probable that neither would the La Nina. That would have eliminated the “dubious” high anomalies but it also would have eliminated all those “dubious” low anomalies associated with the La Nina. You would have been left a flat line right on the trend line and we’d have the same pause.
Now Martin wants to eliminate the La Nina BEFORE the super 1998 El Nino. At least he now understands they are usually paired. And, if he would have just taken the time to understand ENSO and Bjerknes feedback maybe he wouldn’t have made such a silly request. D-K in action.
Martin wants to remove a small La Nina with the super El Nino because he could see that if you remove the big La Nina you end up with the same pause. Sorry Martin, I live in the real world. Cause and effect. Just like a big wave has high crests and low troughs and a little wave has small ones. You don’t see waves with high crests and small troughs.
BTW, this is also why the 2011 paper you linked to is laughable, anti-science nonsense. Stopping a trend in 2010 is more gibberish. If the authors were interested in science they would have ended the trend BEFORE the El Nino or after the paired La Nina that followed. Clearly, they were not doing science.
• Martin Smith says:
Richard, my belief system has nothing to do with the truth or falsehood of scientific propositions. And I do understand Bjerknes feedback.
1. You say the data points in your graph of UAH TLT data from 1997 to 2015 cannot be extracted and looked at as a simple sequence of data points having nothing to do with temperature in the troposphere. But that’s exactly what your tool does. See it again: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
You enter a start year and an end year. The tool then loads the data for the selected years and performs a simple linear regression, which will always compute a straight line through the data. It doesn’t matter what the data is, SLR always computes a straight line through the data. So if we remove the data point for 1998, which is way high abnormally, and which is only the 2nd year in your 18 year period, and then run the tool again, it changes the line it computes quite a lot. Instead of being flat, it rises quite a lot. That is because the El Nino in 1998 was an extreme anomaly, and yes, we can remove one data point. And yes, I know your tool is displaying monthly data, but we can remove the entire 12 months. We can do that; the result really is meaningful. It disproves your claim because without the extreme anomaly of 1998, there was no pause.
2. You say El Nino must be paired with a La Nina for removal from the data, but your tool doesn’t care. In fact, you violated your own rule by starting with a weak La Nina in 1997.
3. But there can be more than one La Nina between El Ninos, and there can be no La Ninas between El Ninos, so when you say they must be removed from the data in pairs, you are making that up.
4. Neither El Ninos nor La Ninas begin on January 1st and end on December 31st. They begin and end any time, but your tool only selects data by year, so it doesn’t support what you are trying to argue.
5. What you are trying to argue is that ENSO and PDO caused your alleged pause and are causing what is known as AGW. But that can’t be correct. ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system. Nor do they prevent energy from leaving the system. Therefore, while they certainly do contribute to natural variation in the global average temperature, they cannot cause an upward trend or a downward trend in global average temperature.
6. About 93% of the energy from the sun that reaches earth and is not reemitted back to space is absorbed by the oceans and by melting of the icecaps. Your UAH TLT data only measures what happens in the lower troposphere, which accounts for less than 7% of the energy from the sun that reaches earth and is not reemitted back to space. That process of 93% of the energy being absorbed by the oceans and by melting ice did not stop during your alleged pause. It didn’t slow down. There is evidence that it increased. Therefore, you cannot say (even if your alleged pause in UAH TLT is statistically significant) that it means global warming paused.
7. When there is an El Nino where an abnormal amount of energy is released from the Pacific Ocean around the equator, or when there is a La Nina and the reverse happens, it doesn’t mean the oceans are releasing or absorbing that amount of energy everywhere. In fact, most of the ocean areas continue to absorb that 93% of the energy reaching earth that is not reemitted back to space. In fact, the oceans continue to absorb energy even during El Nino, just not where El Nino is.
8. You say this paper (https://skepticalscience.com/foster-and-rahmstorf-measure-global-warming-signal.html)
by Foster and Rahmstorf is wrong (I forget what you called it), but they are both professional scientists, one a statistician, the other a climate scientist, and their paper was peer reviewed and publish. You provided no argument that shows their analysis is wrong.
9. Foster and Rahmstorf 2011 removes the ENSO effect from the entire UAH TLT dataset, and from all the other datasets as well. They also remove the effects of variations in aerosols and variations in solar irradiance. Your tool doesn’t do any of this. Your tool uses data that you admit contain these noise causing effects. You even imply that the alleged pause depends on ENSO. But ENSO can’t cause warming or cooling. It only exchanges energy between the ocean and the lower atmosphere.
10. Finally, after all this, you still have not stated what the alleged pause means scientifically. If we assume the pause was statistically significant and scientifically meaningful, as you demand, then what is the scientific meaning of the pause?
• Richard M says:
Martin comes back with a whole plethora of alarmist lies as his response. Sorry, pasting nonsense from Skeptical Science doesn’t hack it. Stick to the question at hand. I’ll give you one example but I could destroy every one of your points.
“ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system.”
Sure they do. They do it every single time. They are not driven by some magically consistent process. They can differ every time as the 1998 and 2016 El Nino events made perfectly obvious. It may all even out eventually but that could take 1000s of years. If you fell for that ridiculous nonsense then I feel sorry for you. You don’t appear to have the capability to think logically.
Now back to the topic at hand.
“You say the data points in your graph of UAH TLT data from 1997 to 2015 cannot be extracted and looked at as a simple sequence of data points having nothing to do with temperature in the troposphere.”
Future data points are in fact dependent on what happened in the past. I explained that to you and if you actually understood Bjerknes feedback you wouldn’t make sure silly claims. We are looking for a trend over time. And there are most certainly dependencies between what occurs over time. It is called auto-correlation. You would do well to learn a little science. You cannot pull out one years worth of data when that year most definitely led to what occurred in subsequent years.
I’ll say it again. THIS IS NOT ROCKET SCIENCE. This is very simple. Your objections are obviously due to the fact you are in complete denial. Continuing to deny basic science only makes you look foolish.
• Martin Smith says:
I (Martin Smith) wrote: “ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system.”
Richard M tries to respond: “Sure they do. They do it every single time. They are not driven by some magically consistent process. They can differ every time as the 1998 and 2016 El Nino events made perfectly obvious. It may all even out eventually but that could take 1000s of years. If you fell for that ridiculous nonsense then I feel sorry for you. You don’t appear to have the capability to think logically.”
Richard, your statement about differing have nothing to do with ENSO and PDO adding to or taking energy from the system. ENSO and PDO transfer energy from the oceans to the atmosphere and from the atmosphere to the oceans. The atmosphere and the oceans are both inside the system. Therefore, transferring energy from one to the other does not add energy to the system, nor does it take it out of the system.
Not only did you not “demolish” my point, you didn’t even address it.
Richard M wrote: “Now back to the topic at hand.”
…and then he did not get back to the topic at hand. These are the points. Address them if you will:
1. You say the data points in your graph of UAH TLT data from 1997 to 2015 cannot be extracted and looked at as a simple sequence of data points having nothing to do with temperature in the troposphere. But that’s exactly what your tool does. See it again: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
You enter a start year and an end year. The tool then loads the data for the selected years and performs a simple linear regression, which will always compute a straight line through the data. It doesn’t matter what the data is, SLR always computes a straight line through the data. So if we remove the data point for 1998, which is way high abnormally, and which is only the 2nd year in your 18 year period, and then run the tool again, it changes the line it computes quite a lot. Instead of being flat, it rises quite a lot. That is because the El Nino in 1998 was an extreme anomaly, and yes, we can remove one data point. And yes, I know your tool is displaying monthly data, but we can remove the entire 12 months. We can do that; the result really is meaningful. It disproves your claim because without the extreme anomaly of 1998, there was no pause.
2. You say El Nino must be paired with a La Nina for removal from the data, but your tool doesn’t care. In fact, you violated your own rule by starting with a weak La Nina in 1997.
3. But there can be more than one La Nina between El Ninos, and there can be no La Ninas between El Ninos, so when you say they must be removed from the data in pairs, you are making that up.
4. Neither El Ninos nor La Ninas begin on January 1st and end on December 31st. They begin and end any time, but your tool only selects data by year, so it doesn’t support what you are trying to argue.
5. What you are trying to argue is that ENSO and PDO caused your alleged pause and are causing what is known as AGW. But that can’t be correct. ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system. Nor do they prevent energy from leaving the system. Therefore, while they certainly do contribute to natural variation in the global average temperature, they cannot cause an upward trend or a downward trend in global average temperature.
6. About 93% of the energy from the sun that reaches earth and is not reemitted back to space is absorbed by the oceans and by melting of the icecaps. Your UAH TLT data only measures what happens in the lower troposphere, which accounts for less than 7% of the energy from the sun that reaches earth and is not reemitted back to space. That process of 93% of the energy being absorbed by the oceans and by melting ice did not stop during your alleged pause. It didn’t slow down. There is evidence that it increased. Therefore, you cannot say (even if your alleged pause in UAH TLT is statistically significant) that it means global warming paused.
7. When there is an El Nino where an abnormal amount of energy is released from the Pacific Ocean around the equator, or when there is a La Nina and the reverse happens, it doesn’t mean the oceans are releasing or absorbing that amount of energy everywhere. In fact, most of the ocean areas continue to absorb that 93% of the energy reaching earth that is not reemitted back to space. In fact, the oceans continue to absorb energy even during El Nino, just not where El Nino is.
8. You say this paper (https://skepticalscience.com/foster-and-rahmstorf-measure-global-warming-signal.html)
by Foster and Rahmstorf is wrong (I forget what you called it), but they are both professional scientists, one a statistician, the other a climate scientist, and their paper was peer reviewed and publish. You provided no argument that shows their analysis is wrong.
9. Foster and Rahmstorf 2011 removes the ENSO effect from the entire UAH TLT dataset, and from all the other datasets as well. They also remove the effects of variations in aerosols and variations in solar irradiance. Your tool doesn’t do any of this. Your tool uses data that you admit contain these noise causing effects. You even imply that the alleged pause depends on ENSO. But ENSO can’t cause warming or cooling. It only exchanges energy between the ocean and the lower atmosphere.
10. Finally, after all this, you still have not stated what the alleged pause means scientifically. If we assume the pause was statistically significant and scientifically meaningful, as you demand, then what is the scientific meaning of the pause?
• Richard M says:
Martin’s response; “The atmosphere and the oceans are both inside the system. Therefore, transferring energy from one to the other does not add energy to the system, nor does it take it out of the system.”
It appears Martin doesn’t realize that when energy moves from the oceans to the atmosphere it next is radiated to space. In other words, it does take energy out of the system.
I really didn’t realize anyone could be as dense as Martin appears to be. He clearly has no idea how the climate system works. And then copying the same long refuted Skeptical Science lies is hysterical. Martin couldn’t even respond with anything new.
I’ve explained everything quite clearly to Martin and it is now clear his only response will be denial of basic science.
• Martin Smith says:
Richard M wrote: “It appears Martin doesn’t realize that when energy moves from the oceans to the atmosphere it next is radiated to space. In other words, it does take energy out of the system.”
No, Richard. We know how much energy is radiated into space, because we measure it. The increased CO2 blocks the energy preventing it from leaving the system. Yes, some of it does leave, but that’s part of the amount that is already emitted. ENSO and the PDO are both part of the system, and they are cycles, so they don’t add energy to the system or subtract energy from the system. Now please address the points, if you will:
1. You say the data points in your graph of UAH TLT data from 1997 to 2015 cannot be extracted and looked at as a simple sequence of data points having nothing to do with temperature in the troposphere. But that’s exactly what your tool does. See it again: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
You enter a start year and an end year. The tool then loads the data for the selected years and performs a simple linear regression, which will always compute a straight line through the data. It doesn’t matter what the data is, SLR always computes a straight line through the data. So if we remove the data point for 1998, which is way high abnormally, and which is only the 2nd year in your 18 year period, and then run the tool again, it changes the line it computes quite a lot. Instead of being flat, it rises quite a lot. That is because the El Nino in 1998 was an extreme anomaly, and yes, we can remove one data point. And yes, I know your tool is displaying monthly data, but we can remove the entire 12 months. We can do that; the result really is meaningful. It disproves your claim because without the extreme anomaly of 1998, there was no pause.
2. You say El Nino must be paired with a La Nina for removal from the data, but your tool doesn’t care. In fact, you violated your own rule by starting with a weak La Nina in 1997.
3. But there can be more than one La Nina between El Ninos, and there can be no La Ninas between El Ninos, so when you say they must be removed from the data in pairs, you are making that up.
4. Neither El Ninos nor La Ninas begin on January 1st and end on December 31st. They begin and end any time, but your tool only selects data by year, so it doesn’t support what you are trying to argue.
5. What you are trying to argue is that ENSO and PDO caused your alleged pause and are causing what is known as AGW. But that can’t be correct. ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system. Nor do they prevent energy from leaving the system. Therefore, while they certainly do contribute to natural variation in the global average temperature, they cannot cause an upward trend or a downward trend in global average temperature.
6. About 93% of the energy from the sun that reaches earth and is not reemitted back to space is absorbed by the oceans and by melting of the icecaps. Your UAH TLT data only measures what happens in the lower troposphere, which accounts for less than 7% of the energy from the sun that reaches earth and is not reemitted back to space. That process of 93% of the energy being absorbed by the oceans and by melting ice did not stop during your alleged pause. It didn’t slow down. There is evidence that it increased. Therefore, you cannot say (even if your alleged pause in UAH TLT is statistically significant) that it means global warming paused.
7. When there is an El Nino where an abnormal amount of energy is released from the Pacific Ocean around the equator, or when there is a La Nina and the reverse happens, it doesn’t mean the oceans are releasing or absorbing that amount of energy everywhere. In fact, most of the ocean areas continue to absorb that 93% of the energy reaching earth that is not reemitted back to space. In fact, the oceans continue to absorb energy even during El Nino, just not where El Nino is.
8. You say this paper (https://skepticalscience.com/foster-and-rahmstorf-measure-global-warming-signal.html)
by Foster and Rahmstorf is wrong (I forget what you called it), but they are both professional scientists, one a statistician, the other a climate scientist, and their paper was peer reviewed and publish. You provided no argument that shows their analysis is wrong.
9. Foster and Rahmstorf 2011 removes the ENSO effect from the entire UAH TLT dataset, and from all the other datasets as well. They also remove the effects of variations in aerosols and variations in solar irradiance. Your tool doesn’t do any of this. Your tool uses data that you admit contain these noise causing effects. You even imply that the alleged pause depends on ENSO. But ENSO can’t cause warming or cooling. It only exchanges energy between the ocean and the lower atmosphere.
10. Finally, after all this, you still have not stated what the alleged pause means scientifically. If we assume the pause was statistically significant and scientifically meaningful, as you demand, then what is the scientific meaning of the pause?
• Richard M says:
Martin babbles more nonsense. Now he claims the energy buried deep in the PWP was going to leave the climate system without an El Nino. Does he even have a clue how silly this is?
Here is a picture of the SSTs during the recent El Nino. Maybe Martin will tell us what happened to first cause the significant increase in temperature and then to cause all that energy to disappear.
Of course, real scientists do know what happened. Energy buried deep in the PWP was spread out over vast parts of the Pacific raising the average temperature and increasing evaporation. Almost all of that energy eventually was radiated to space and is now gone.
This is a cooling event. Different El Nino events will lead to more or less cooling. The atmosphere is temporarily warmed as the energy is transported to space. Hence, more or stronger El Nino events will definitely cause changes in the “system”.
This is true of all ocean cycles. They are not perfect cycles and will affect the system to varying degrees.
As for CO2 blocking some of that energy. Of course it does. Very little though and even that little gain is offset by an almost equal loss due to a reduction in high altitude water vapor.
• Martin Smith says:
Richard M wrote: “Martin babbles more nonsense. Now he claims the energy buried deep in the PWP was going to leave the climate system without an El Nino.”
I didn’t write that, Richard. You made it up. ENSO and PDO move energy from the sea into the air ad from the air into the sea. These processes do increase and decrease the global average temperature, but they do not add energy to the system or remove energy from the system. They just move the energy aroud in the system. Now, address these points if you can. If you can’t, just say you can’t:
1. You say the data points in your graph of UAH TLT data from 1997 to 2015 cannot be extracted and looked at as a simple sequence of data points having nothing to do with temperature in the troposphere. But that’s exactly what your tool does. See it again: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
You enter a start year and an end year. The tool then loads the data for the selected years and performs a simple linear regression, which will always compute a straight line through the data. It doesn’t matter what the data is, SLR always computes a straight line through the data. So if we remove the data point for 1998, which is way high abnormally, and which is only the 2nd year in your 18 year period, and then run the tool again, it changes the line it computes quite a lot. Instead of being flat, it rises quite a lot. That is because the El Nino in 1998 was an extreme anomaly, and yes, we can remove one data point. And yes, I know your tool is displaying monthly data, but we can remove the entire 12 months. We can do that; the result really is meaningful. It disproves your claim because without the extreme anomaly of 1998, there was no pause.
2. You say El Nino must be paired with a La Nina for removal from the data, but your tool doesn’t care. In fact, you violated your own rule by starting with a weak La Nina in 1997.
3. But there can be more than one La Nina between El Ninos, and there can be no La Ninas between El Ninos, so when you say they must be removed from the data in pairs, you are making that up.
4. Neither El Ninos nor La Ninas begin on January 1st and end on December 31st. They begin and end any time, but your tool only selects data by year, so it doesn’t support what you are trying to argue.
5. What you are trying to argue is that ENSO and PDO caused your alleged pause and are causing what is known as AGW. But that can’t be correct. ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system. Nor do they prevent energy from leaving the system. Therefore, while they certainly do contribute to natural variation in the global average temperature, they cannot cause an upward trend or a downward trend in global average temperature.
6. About 93% of the energy from the sun that reaches earth and is not reemitted back to space is absorbed by the oceans and by melting of the icecaps. Your UAH TLT data only measures what happens in the lower troposphere, which accounts for less than 7% of the energy from the sun that reaches earth and is not reemitted back to space. That process of 93% of the energy being absorbed by the oceans and by melting ice did not stop during your alleged pause. It didn’t slow down. There is evidence that it increased. Therefore, you cannot say (even if your alleged pause in UAH TLT is statistically significant) that it means global warming paused.
7. When there is an El Nino where an abnormal amount of energy is released from the Pacific Ocean around the equator, or when there is a La Nina and the reverse happens, it doesn’t mean the oceans are releasing or absorbing that amount of energy everywhere. In fact, most of the ocean areas continue to absorb that 93% of the energy reaching earth that is not reemitted back to space. In fact, the oceans continue to absorb energy even during El Nino, just not where El Nino is.
8. You say this paper (https://skepticalscience.com/foster-and-rahmstorf-measure-global-warming-signal.html)
by Foster and Rahmstorf is wrong (I forget what you called it), but they are both professional scientists, one a statistician, the other a climate scientist, and their paper was peer reviewed and publish. You provided no argument that shows their analysis is wrong.
9. Foster and Rahmstorf 2011 removes the ENSO effect from the entire UAH TLT dataset, and from all the other datasets as well. They also remove the effects of variations in aerosols and variations in solar irradiance. Your tool doesn’t do any of this. Your tool uses data that you admit contain these noise causing effects. You even imply that the alleged pause depends on ENSO. But ENSO can’t cause warming or cooling. It only exchanges energy between the ocean and the lower atmosphere.
10. Finally, after all this, you still have not stated what the alleged pause means scientifically. If we assume the pause was statistically significant and scientifically meaningful, as you demand, then what is the scientific meaning of the pause?
• Richard M says:
Martin now states: “ENSO and PDO move energy from the sea into the air ad from the air into the sea. These processes do increase and decrease the global average temperature, but they do not add energy to the system or remove energy from the system. ”
That is complete and total nonsense. We know exactly HOW the energy is removed from the system during an El Nino. The warm water from the PWP spreads out over the Pacific and through radiation and evaporation that energy is moved into/through the atmosphere and eventually to space. It is 90+% a one way street. There is only a small path for back radiation to warm the water. It is so small it would be a round off error. Sorry but the energy is not moved around. It is moved out.
So please inform the scientific world exactly how you recapture this energy and move it back into the ocean.
You are so far out in left field that all I can do is chuckle at your nonsensical claims.
• Martin Smith says:
Richard M writes: “… and through radiation and evaporation that energy is moved into/through the atmosphere and eventually to space.”
Nope. Now you are denying the greenhouse effect. Stop making up nonsense and address these points or admit you can’t:
1. You say the data points in your graph of UAH TLT data from 1997 to 2015 cannot be extracted and looked at as a simple sequence of data points having nothing to do with temperature in the troposphere. But that’s exactly what your tool does. See it again: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
You enter a start year and an end year. The tool then loads the data for the selected years and performs a simple linear regression, which will always compute a straight line through the data. It doesn’t matter what the data is, SLR always computes a straight line through the data. So if we remove the data point for 1998, which is way high abnormally, and which is only the 2nd year in your 18 year period, and then run the tool again, it changes the line it computes quite a lot. Instead of being flat, it rises quite a lot. That is because the El Nino in 1998 was an extreme anomaly, and yes, we can remove one data point. And yes, I know your tool is displaying monthly data, but we can remove the entire 12 months. We can do that; the result really is meaningful. It disproves your claim because without the extreme anomaly of 1998, there was no pause.
2. You say El Nino must be paired with a La Nina for removal from the data, but your tool doesn’t care. In fact, you violated your own rule by starting with a weak La Nina in 1997.
3. But there can be more than one La Nina between El Ninos, and there can be no La Ninas between El Ninos, so when you say they must be removed from the data in pairs, you are making that up.
4. Neither El Ninos nor La Ninas begin on January 1st and end on December 31st. They begin and end any time, but your tool only selects data by year, so it doesn’t support what you are trying to argue.
5. What you are trying to argue is that ENSO and PDO caused your alleged pause and are causing what is known as AGW. But that can’t be correct. ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system. Nor do they prevent energy from leaving the system. Therefore, while they certainly do contribute to natural variation in the global average temperature, they cannot cause an upward trend or a downward trend in global average temperature.
6. About 93% of the energy from the sun that reaches earth and is not reemitted back to space is absorbed by the oceans and by melting of the icecaps. Your UAH TLT data only measures what happens in the lower troposphere, which accounts for less than 7% of the energy from the sun that reaches earth and is not reemitted back to space. That process of 93% of the energy being absorbed by the oceans and by melting ice did not stop during your alleged pause. It didn’t slow down. There is evidence that it increased. Therefore, you cannot say (even if your alleged pause in UAH TLT is statistically significant) that it means global warming paused.
7. When there is an El Nino where an abnormal amount of energy is released from the Pacific Ocean around the equator, or when there is a La Nina and the reverse happens, it doesn’t mean the oceans are releasing or absorbing that amount of energy everywhere. In fact, most of the ocean areas continue to absorb that 93% of the energy reaching earth that is not reemitted back to space. In fact, the oceans continue to absorb energy even during El Nino, just not where El Nino is.
8. You say this paper (https://skepticalscience.com/foster-and-rahmstorf-measure-global-warming-signal.html)
by Foster and Rahmstorf is wrong (I forget what you called it), but they are both professional scientists, one a statistician, the other a climate scientist, and their paper was peer reviewed and publish. You provided no argument that shows their analysis is wrong.
9. Foster and Rahmstorf 2011 removes the ENSO effect from the entire UAH TLT dataset, and from all the other datasets as well. They also remove the effects of variations in aerosols and variations in solar irradiance. Your tool doesn’t do any of this. Your tool uses data that you admit contain these noise causing effects. You even imply that the alleged pause depends on ENSO. But ENSO can’t cause warming or cooling. It only exchanges energy between the ocean and the lower atmosphere.
10. Finally, after all this, you still have not stated what the alleged pause means scientifically. If we assume the pause was statistically significant and scientifically meaningful, as you demand, then what is the scientific meaning of the pause?
• Richard M says:
OMG Martin, you don’t even have the slightest clue about energy flows. Your statement:
“Nope. Now you are denying the greenhouse effect. Stop making up nonsense and address these points or admit you can’t:”
Is hilarious. You would have 100% of the energy trapped by GHGs???? Sorry, that qualifies as “not even wrong”. If this is your level of understanding I can only chuckle. Once again the D-K effect in action.
The Earth receives ~240 w/m2 of energy from the sun each and every day. If GHGs trapped all of it, the planet would be an inferno in weeks. Of course, anyone who understands the energy budget of the planet also knows the Earth also radiates just as much energy back into space. The GHE from a doubling of CO2 only captures at most 3.7 w/m2. When you get an El Nino the planet’s surface and atmosphere warm up which means even more energy is radiated to space.
ROTFLMAO.
• Martin Smith says:
Richard M wrote: ” You would have 100% of the energy trapped by GHGs?”
Richard, you know what you wrote above is false because you know I wrote this: “We know how much energy is radiated into space, because we measure it.” So you deliberately misrepresented what I wrote, and you you did that because you are still trying to deceive the readers. Fortunately, I think they’re all gone now. It’s just you and me, and your argument has been proved wrong. We know the amount of energy emitted from earth because we measure it with satellites. You like satellites.
Richard M wrote: “Of course, anyone who understands the energy budget of the planet also knows the Earth also radiates just as much energy back into space.”
You are still denying the greenhouse effect. If your statement above were true, the earth would not be warming. But it IS warming, much faster than it would warm naturally. Actually, were it not for AGW, earth would be cooling now, but let that pass. Earth is not emitting “just as much energy back into space,” because: Global Warming continues.
Richard M wrote: “When you get an El Nino the planet’s surface and atmosphere warm up which means even more energy is radiated to space.”
You can say that for 1998, because 1998 was an extreme El Nino anomaly. In fact, that’s why you cherrypicked 1998 to be at the start of your 1997-2015 time period. But, Richard, when the atmosphere warms because of an El Nino, the atmosphere warms because of the El Nino. That means the energy is trapped in the atmosphere. When the atmosphere warms, the energy is still there. You are trying to imply the energy is gone. It’s not. It’s still there. So now you are arguing against your own argument. Here is why:
After 1998, there are 2 weak El Ninos and 2 moderate El Ninos in your cherry picked interval. There are no other strong El Ninos. But there are 3 weak La Ninas in your cherrypicked interval, 2 moderate La Ninas, and 4 strong La Ninas. http://ggweather.com/enso/oni.htm
So your cherry picked time interval contains 5 El Ninos (only 1 strong one right at the start) and 9 La Ninas (4 strong ones). 15 of the years in your cherrypicked interval were complete or partial La Nina years. That’s 15 out of 18. And you chose to start in 1997 because of the extreme El Nino in 1998.
But as you are arguing now, more energy leaves earth during an El Nino, which means more energy stays on earth during La Nina years. Less energy than your mythical balanced energy budget is radiated back to space during La Nina years, and most of the years in your cherrypicked interval are La Nina years. And we know that during La Nina years, that extra energy that is not being radiated to space is being absorbed by the oceans. That’s warming, Richard. It is warming that doesn’t show in your cherrypicked UAH TLT data, because you are deliberately ignoring the 93% of the energy that goes into the oceans.
And, as Foster and Rahmstorf 2011 showed, when you remove the effects of El Niño/southern oscillation, volcanic aerosols and solar variability, the warming is still there; there was no pause.
http://iopscience.iop.org/article/10.1088/1748-9326/6/4/044022/meta
That proves your argument is wrong. But you still haven’t addressed these points. You say you have, but you have not:
1. You say the data points in your graph of UAH TLT data from 1997 to 2015 cannot be extracted and looked at as a simple sequence of data points having nothing to do with temperature in the troposphere. But that’s exactly what your tool does. See it again: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
You enter a start year and an end year. The tool then loads the data for the selected years and performs a simple linear regression, which will always compute a straight line through the data. It doesn’t matter what the data is, SLR always computes a straight line through the data. So if we remove the data point for 1998, which is way high abnormally, and which is only the 2nd year in your 18 year period, and then run the tool again, it changes the line it computes quite a lot. Instead of being flat, it rises quite a lot. That is because the El Nino in 1998 was an extreme anomaly, and yes, we can remove one data point. And yes, I know your tool is displaying monthly data, but we can remove the entire 12 months. We can do that; the result really is meaningful. It disproves your claim because without the extreme anomaly of 1998, there was no pause.
2. You say El Nino must be paired with a La Nina for removal from the data, but your tool doesn’t care. In fact, you violated your own rule by starting with a weak La Nina in 1997.
3. But there can be more than one La Nina between El Ninos, and there can be no La Ninas between El Ninos, so when you say they must be removed from the data in pairs, you are making that up.
4. Neither El Ninos nor La Ninas begin on January 1st and end on December 31st. They begin and end any time, but your tool only selects data by year, so it doesn’t support what you are trying to argue.
5. What you are trying to argue is that ENSO and PDO caused your alleged pause and are causing what is known as AGW. But that can’t be correct. ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system. Nor do they prevent energy from leaving the system. Therefore, while they certainly do contribute to natural variation in the global average temperature, they cannot cause an upward trend or a downward trend in global average temperature.
6. About 93% of the energy from the sun that reaches earth and is not reemitted back to space is absorbed by the oceans and by melting of the icecaps. Your UAH TLT data only measures what happens in the lower troposphere, which accounts for less than 7% of the energy from the sun that reaches earth and is not reemitted back to space. That process of 93% of the energy being absorbed by the oceans and by melting ice did not stop during your alleged pause. It didn’t slow down. There is evidence that it increased. Therefore, you cannot say (even if your alleged pause in UAH TLT is statistically significant) that it means global warming paused.
7. When there is an El Nino where an abnormal amount of energy is released from the Pacific Ocean around the equator, or when there is a La Nina and the reverse happens, it doesn’t mean the oceans are releasing or absorbing that amount of energy everywhere. In fact, most of the ocean areas continue to absorb that 93% of the energy reaching earth that is not reemitted back to space. In fact, the oceans continue to absorb energy even during El Nino, just not where El Nino is.
8. You say this paper (https://skepticalscience.com/foster-and-rahmstorf-measure-global-warming-signal.html)
by Foster and Rahmstorf is wrong (I forget what you called it), but they are both professional scientists, one a statistician, the other a climate scientist, and their paper was peer reviewed and publish. You provided no argument that shows their analysis is wrong.
9. Foster and Rahmstorf 2011 removes the ENSO effect from the entire UAH TLT dataset, and from all the other datasets as well. They also remove the effects of variations in aerosols and variations in solar irradiance. Your tool doesn’t do any of this. Your tool uses data that you admit contain these noise causing effects. You even imply that the alleged pause depends on ENSO. But ENSO can’t cause warming or cooling. It only exchanges energy between the ocean and the lower atmosphere.
10. Finally, after all this, you still have not stated what the alleged pause means scientifically. If we assume the pause was statistically significant and scientifically meaningful, as you demand, then what is the scientific meaning of the pause?
Best of luck in your future endeavors.
• Richard M says:
LMAO. Now Martin tries to blame me for his nonsense comment. Then, he repeats his inane claim.
“when the atmosphere warms because of an El Nino, the atmosphere warms because of the El Nino. That means the energy is trapped in the atmosphere. ”
No, it doesn’t. It means energy is continually transferred from the oceans into the atmosphere and then the atmosphere into space. The oceans are very warm and it can take several years for this to happen. You can clearly see it recent with SSTs. It took a full 4 years but now all the energy has disappeared from the last El Nino.
Next comes this gem:
“We know the amount of energy emitted from earth because we measure it with satellites.”
Yes we do, and because of that we know the amount of energy radiated to space goes up during El Nino events just like I told you. It is not magically trapped in the atmosphere as you just claimed. Yes, the atmosphere warms during this time because much of the energy is latent heat which must first be transferred into the atmosphere through condensation before it can be radiated away.
Then Martin says “But as you are arguing now, more energy leaves earth during an El Nino, which means more energy stays on earth during La Nina years. ”
That is correct. I’m simply relating facts to you. This is basic science. Then …
“And we know that during La Nina years, that extra energy that is not being radiated to space is being absorbed by the oceans. That’s warming”
That’s one theory. In reality La Nina events are no different than any other non El Nino period. It is just a period with stronger trade winds and therefore more upwelling cold water takes place. I do believe the Earth is warming during La Nina events because there will probably be fewer clouds and thus more solar absorption (notice no greenhouse effect is needed). This is still open science.
Now you appear to accept that El Nino and La Nina events cause warming and cooling just like I told you. You even indicate that more La Nina events will lead to warming by saying:
“That’s warming”
Isn’t that what I told you right at the start of this discussion? I told you that ENSO can affect the global atmospheric temperature. You denied it. Now, it appears you are starting to catch on. However, you seemed to think that you can just count El Nino events and La Nina events and gain knowledge. Nope, it’s more difficult. El Nino events tend to be much stronger than La Nina events. And, many other processes are also taking place.
Now, keep in mind we were discussing the global atmospheric temperature data. This is not the same as the global energy which would also need to measure the energy in the oceans (which we cannot measure accurately). However, one good proxy for the oceans is the SSTs. Unfortunately, we don’t have a good satellite measure there either.
I recently read a paper that reconstructed SSTs in one area for the last few centuries. They found the SSTs have been increasing for 400 years. So, why would anyone think humans are now the cause? You would need to understand the mechanism for the natural warming and then subtract that from human warming. Can you do that, Martin? Of course not and neither can anyone else.
So, what we know is the oceans have been warming for a long time and it is likely that has led to natural warming of the atmosphere just like happens with El Nino. We also know during the last 20 years that, except for temporary warming and cooling from ENSO, the atmosphere stopped warming despite what climate models project.
Like it or not that means there is no evidence to support the claims that CO2 emissions will warm the planet.
• Martin Smith says:
Richard M wrote: “Like it or not that means there is no evidence to support the claims that CO2 emissions will warm the planet.”
That’s a direct denial of the greenhouse effect, without which, the planet would be more than 30C colder than it is now and there would be no human race. So you are still denying basic physics. Nevertheless, your conclusion has absolutely nothing to do with the subject of this thread. Stop avoiding the subject.
Richard M wrote: “So, what we know is the oceans have been warming for a long time and it is likely that has led to natural warming of the atmosphere just like happens with El Nino.”
Very good, Richard, but that is a point not in dispute. The oceans absorb about 93% of the energy that reaches earth and is not radiated back to space. That process is accelerating because of the rise in Anthropogenic CO2., which blocks more and more energy from radiating back to space. Instead, it reflects it back down to the surface, where the oceans absorb more ad more of it. So you argued for a point that was not in dispute.
By the way, you have proved there was no pause, since most of the years in your cherrypicked time interval were La Nina years and you have concluded the oceans absorb more heat in La Nina years.
Richard M wrote: “In reality La Nina events are no different than any other non El Nino period.”
You can’t get away with that. You have previously argue here that La Ninas are opposites of El Ninos; they come in pairs, and they must paired in the statistical analysis. That’s wrong, but it is what you have already claimed (see below), so you can’t also argue against your own argument.
Now please address the subject of this thread, which is explained below. You continue to avoid the real issue.
1. You say the data points in your graph of UAH TLT data from 1997 to 2015 cannot be extracted and looked at as a simple sequence of data points having nothing to do with temperature in the troposphere. But that’s exactly what your tool does. See it again: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
You enter a start year and an end year. The tool then loads the data for the selected years and performs a simple linear regression, which will always compute a straight line through the data. It doesn’t matter what the data is, SLR always computes a straight line through the data. So if we remove the data point for 1998, which is way high abnormally, and which is only the 2nd year in your 18 year period, and then run the tool again, it changes the line it computes quite a lot. Instead of being flat, it rises quite a lot. That is because the El Nino in 1998 was an extreme anomaly, and yes, we can remove one data point. And yes, I know your tool is displaying monthly data, but we can remove the entire 12 months. We can do that; the result really is meaningful. It disproves your claim because without the extreme anomaly of 1998, there was no pause.
2. You say El Nino must be paired with a La Nina for removal from the data, but your tool doesn’t care. In fact, you violated your own rule by starting with a weak La Nina in 1997.
3. But there can be more than one La Nina between El Ninos, and there can be no La Ninas between El Ninos, so when you say they must be removed from the data in pairs, you are making that up.
4. Neither El Ninos nor La Ninas begin on January 1st and end on December 31st. They begin and end any time, but your tool only selects data by year, so it doesn’t support what you are trying to argue.
5. What you are trying to argue is that ENSO and PDO caused your alleged pause and are causing what is known as AGW. But that can’t be correct. ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system. Nor do they prevent energy from leaving the system. Therefore, while they certainly do contribute to natural variation in the global average temperature, they cannot cause an upward trend or a downward trend in global average temperature.
6. About 93% of the energy from the sun that reaches earth and is not reemitted back to space is absorbed by the oceans and by melting of the icecaps. Your UAH TLT data only measures what happens in the lower troposphere, which accounts for less than 7% of the energy from the sun that reaches earth and is not reemitted back to space. That process of 93% of the energy being absorbed by the oceans and by melting ice did not stop during your alleged pause. It didn’t slow down. There is evidence that it increased. Therefore, you cannot say (even if your alleged pause in UAH TLT is statistically significant) that it means global warming paused.
7. When there is an El Nino where an abnormal amount of energy is released from the Pacific Ocean around the equator, or when there is a La Nina and the reverse happens, it doesn’t mean the oceans are releasing or absorbing that amount of energy everywhere. In fact, most of the ocean areas continue to absorb that 93% of the energy reaching earth that is not reemitted back to space. In fact, the oceans continue to absorb energy even during El Nino, just not where El Nino is.
8. You say this paper (https://skepticalscience.com/foster-and-rahmstorf-measure-global-warming-signal.html)
by Foster and Rahmstorf is wrong (I forget what you called it), but they are both professional scientists, one a statistician, the other a climate scientist, and their paper was peer reviewed and publish. You provided no argument that shows their analysis is wrong.
9. Foster and Rahmstorf 2011 removes the ENSO effect from the entire UAH TLT dataset, and from all the other datasets as well. They also remove the effects of variations in aerosols and variations in solar irradiance. Your tool doesn’t do any of this. Your tool uses data that you admit contain these noise causing effects. You even imply that the alleged pause depends on ENSO. But ENSO can’t cause warming or cooling. It only exchanges energy between the ocean and the lower atmosphere.
10. Finally, after all this, you still have not stated what the alleged pause means scientifically. If we assume the pause was statistically significant and scientifically meaningful, as you demand, then what is the scientific meaning of the pause?
Best of luck in your future endeavors.
• Richard M says:
Martin claims:
“That’s a direct denial of the greenhouse effect ”
which once again demonstrates he is scientifically illiterate. Here’s an analogy. We know when water is placed out in air with a RH of less than 100% that it will evaporate. When I tell you that all of water in the world is not going to evaporate and leave the oceans dry, am I denying the basic physics that water evaporates?
The answer is of course no. But that is essentially the only thought that crossed your feeble mind when I mentioned adding CO2 to the atmosphere is not causing warming.. Unlike you, I am open to our atmosphere being more complex than your simplistic mind thinks.
Martin then repeats this silly propaganda:
“Anthropogenic CO2., which blocks more and more energy from radiating back to space. Instead, it reflects it back down to the surface, where the oceans absorb more ad more of it. ”
LOL. CO2 does not “block energy” nor “reflect energy”. It absorbs energy within certain frequency domains and transfers that energy to other molecules through kinetic energy transfers well over 99.999% of the time. In addition the oceans only absorb LW radiation to a depth of a few microns. This may cause warming of the skin but also leads to enhanced evaporation which is a cooling process removing heat from the ocean.
You should try to learn to understand the science.
Now, CO2 also occasionally gains energy through kinetic transfers and radiates that energy with a little less than half directed back to the surface. Adding more CO2 molecules with increase this process. Of course, CO2 is also radiating energy towards space and adding more CO2 molecules also increases this process. I doubt Martin is even aware of this fact.
All things being equal the net result of these actions would be a warming of ~1 C per each doubling of CO2. Hardly worth any concern. Your climate priests tell us we’ve already seen this much warming. But, all things are not equal and this is where the debate rages.
Your climate priests claim this warming is more than tripled mainly through positive water vapor feedback. What real scientists tell us is this is nonsense and some scientists even believe this value is reduced significantly due to negative water vapor feedback.
Here’s how it works. Increases in CO2 leads to increases in IR radiation to the surface. The holy greenhouse effect. This radiation has a high probability of striking a water molecule which then increases the probability of evaporation. Since water vapor is lighter than air this will also enhance convection. Basic physics. Speeding up convection will drive convecting air columns higher in the troposphere which will lead to more of that water vapor being condensed into clouds and rain. Basic physics.
You end up with more clouds reflecting solar energy and less high altitude water vapor which makes the air denser and also enhances convection.
Did you get this? Less water vapor, the most important greenhouse gas, right at the place where it has the most effect. Both of these are negative feedbacks.
Now, did you see me denying the greenhouse effect at any time? Nope. All I did was show the system is far more complex than you imagined. All I had to do is use simple physics and the net effect could be no warming at all or even cooling. However, since convection occurs most during the day, we should see cooler days and some warming at night. This also means warmer winters and cooler summers.
Now it appears Martin agrees with me that ENSO can affect the global temperature. He says:
“By the way, you have proved there was no pause, since most of the years in your cherrypicked time interval were La Nina years and you have concluded the oceans absorb more heat in La Nina years. ”
As I told you you cannot simply use time by itself to understand the climate effects of ENSO. However, if we assume you are right then you should also be aware that from 1977-1998 we has more El Nino influence than La Nina influence. That would mean the warming over that period is exaggerated. So instead of a 20 years warming period followed by a 20 year pause, you end up with 40 years with a very low warming rate. Doesn’t exactly help your case. And, now we get into the AMO and PDO which have been predominately in their respective warm phases over that 40 year period and the millennial cycle that has seen warming since the LIA.
Next Martin spews this lie:
“You have previously argue here that La Ninas are opposites of El Ninos; they come in pairs, and they must paired in the statistical analysis. ”
I did no such thing. I explained that they generally come in pairs. I never said always. However, when they do come in pairs then you need to keep that association. You cannot pull out one year and have anything representing reality. And, they aren’t really opposites. It is far more complex than that which is why your simplistic thinking always fails.
• Martin Smith says:
Richard M makes yet another ad hominem attack: “[Martin Smith] is scientifically illiterate.”
Richard, this is your belief. It is what you wrote: ““Like it or not that means there is no evidence to support the claims that CO2 emissions will warm the planet.”
Your statement can only be true if the physics of the greenhouse effect are wrong. But those physics are correct. You are denying the greenhouse effect.
Richard M denied his own statements: “I did no such thing.”
Yes, you did. The proof is in the thread above. Please stop avoiding the subject. Here it is. You have not addressed these points:
1. You say the data points in your graph of UAH TLT data from 1997 to 2015 cannot be extracted and looked at as a simple sequence of data points having nothing to do with temperature in the troposphere. But that’s exactly what your tool does. See it again: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
You enter a start year and an end year. The tool then loads the data for the selected years and performs a simple linear regression, which will always compute a straight line through the data. It doesn’t matter what the data is, SLR always computes a straight line through the data. So if we remove the data point for 1998, which is way high abnormally, and which is only the 2nd year in your 18 year period, and then run the tool again, it changes the line it computes quite a lot. Instead of being flat, it rises quite a lot. That is because the El Nino in 1998 was an extreme anomaly, and yes, we can remove one data point. And yes, I know your tool is displaying monthly data, but we can remove the entire 12 months. We can do that; the result really is meaningful. It disproves your claim because without the extreme anomaly of 1998, there was no pause.
2. You say El Nino must be paired with a La Nina for removal from the data, but your tool doesn’t care. In fact, you violated your own rule by starting with a weak La Nina in 1997.
3. But there can be more than one La Nina between El Ninos, and there can be no La Ninas between El Ninos, so when you say they must be removed from the data in pairs, you are making that up.
4. Neither El Ninos nor La Ninas begin on January 1st and end on December 31st. They begin and end any time, but your tool only selects data by year, so it doesn’t support what you are trying to argue.
5. What you are trying to argue is that ENSO and PDO caused your alleged pause and are causing what is known as AGW. But that can’t be correct. ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system. Nor do they prevent energy from leaving the system. Therefore, while they certainly do contribute to natural variation in the global average temperature, they cannot cause an upward trend or a downward trend in global average temperature.
6. About 93% of the energy from the sun that reaches earth and is not reemitted back to space is absorbed by the oceans and by melting of the icecaps. Your UAH TLT data only measures what happens in the lower troposphere, which accounts for less than 7% of the energy from the sun that reaches earth and is not reemitted back to space. That process of 93% of the energy being absorbed by the oceans and by melting ice did not stop during your alleged pause. It didn’t slow down. There is evidence that it increased. Therefore, you cannot say (even if your alleged pause in UAH TLT is statistically significant) that it means global warming paused.
7. When there is an El Nino where an abnormal amount of energy is released from the Pacific Ocean around the equator, or when there is a La Nina and the reverse happens, it doesn’t mean the oceans are releasing or absorbing that amount of energy everywhere. In fact, most of the ocean areas continue to absorb that 93% of the energy reaching earth that is not reemitted back to space. In fact, the oceans continue to absorb energy even during El Nino, just not where El Nino is.
8. You say this paper (https://skepticalscience.com/foster-and-rahmstorf-measure-global-warming-signal.html)
by Foster and Rahmstorf is wrong (I forget what you called it), but they are both professional scientists, one a statistician, the other a climate scientist, and their paper was peer reviewed and publish. You provided no argument that shows their analysis is wrong.
9. Foster and Rahmstorf 2011 removes the ENSO effect from the entire UAH TLT dataset, and from all the other datasets as well. They also remove the effects of variations in aerosols and variations in solar irradiance. Your tool doesn’t do any of this. Your tool uses data that you admit contain these noise causing effects. You even imply that the alleged pause depends on ENSO. But ENSO can’t cause warming or cooling. It only exchanges energy between the ocean and the lower atmosphere.
10. Finally, after all this, you still have not stated what the alleged pause means scientifically. If we assume the pause was statistically significant and scientifically meaningful, as you demand, then what is the scientific meaning of the pause?
Best of luck in your future endeavors.
• Richard M says:
Poor Martin didn’t like being called a science illiterate yet he has proven it over and over again. Sorry, but telling the truth is not an ad hominem. It simply is.
The rest of your last comment is nothing but more lies. I explained everything to you in simple to verify scientific terms. Obviously, you couldn’t understand a word of it. It appears you aren’t just scientifically illiterate, you are also too slow to grasp simple concepts..
Thanks for proving once again that climate true believers are among the densest people on the planet.
• Martin Smith says:
Richard M wrote: “Poor Martin didn’t like being called a science illiterate…”
Enough of your ad hominem and dissembling. Address the subject of this discussion. The points listed below. If you can’t, then Foster and Rahmstorf 2011 proves there was no pause: http://iopscience.iop.org/article/10.1088/1748-9326/6/4/044022/meta
Here are the points you must address:
1. You say the data points in your graph of UAH TLT data from 1997 to 2015 cannot be extracted and looked at as a simple sequence of data points having nothing to do with temperature in the troposphere. But that’s exactly what your tool does. See it again: http://www.woodfortrees.org/plot/uah6/from:1997/to:2015/plot/uah6/from:1999/to:2015/trend
You enter a start year and an end year. The tool then loads the data for the selected years and performs a simple linear regression, which will always compute a straight line through the data. It doesn’t matter what the data is, SLR always computes a straight line through the data. So if we remove the data point for 1998, which is way high abnormally, and which is only the 2nd year in your 18 year period, and then run the tool again, it changes the line it computes quite a lot. Instead of being flat, it rises quite a lot. That is because the El Nino in 1998 was an extreme anomaly, and yes, we can remove one data point. And yes, I know your tool is displaying monthly data, but we can remove the entire 12 months. We can do that; the result really is meaningful. It disproves your claim because without the extreme anomaly of 1998, there was no pause.
2. You say El Nino must be paired with a La Nina for removal from the data, but your tool doesn’t care. In fact, you violated your own rule by starting with a weak La Nina in 1997.
3. But there can be more than one La Nina between El Ninos, and there can be no La Ninas between El Ninos, so when you say they must be removed from the data in pairs, you are making that up.
4. Neither El Ninos nor La Ninas begin on January 1st and end on December 31st. They begin and end any time, but your tool only selects data by year, so it doesn’t support what you are trying to argue.
5. What you are trying to argue is that ENSO and PDO caused your alleged pause and are causing what is known as AGW. But that can’t be correct. ENSO and PDO are cycles. They don’t add energy to the system, nor do they take energy out of the system. Nor do they prevent energy from leaving the system. Therefore, while they certainly do contribute to natural variation in the global average temperature, they cannot cause an upward trend or a downward trend in global average temperature.
6. About 93% of the energy from the sun that reaches earth and is not reemitted back to space is absorbed by the oceans and by melting of the icecaps. Your UAH TLT data only measures what happens in the lower troposphere, which accounts for less than 7% of the energy from the sun that reaches earth and is not reemitted back to space. That process of 93% of the energy being absorbed by the oceans and by melting ice did not stop during your alleged pause. It didn’t slow down. There is evidence that it increased. Therefore, you cannot say (even if your alleged pause in UAH TLT is statistically significant) that it means global warming paused.
7. When there is an El Nino where an abnormal amount of energy is released from the Pacific Ocean around the equator, or when there is a La Nina and the reverse happens, it doesn’t mean the oceans are releasing or absorbing that amount of energy everywhere. In fact, most of the ocean areas continue to absorb that 93% of the energy reaching earth that is not reemitted back to space. In fact, the oceans continue to absorb energy even during El Nino, just not where El Nino is.
8. You say this paper (https://skepticalscience.com/foster-and-rahmstorf-measure-global-warming-signal.html)
by Foster and Rahmstorf is wrong (I forget what you called it), but they are both professional scientists, one a statistician, the other a climate scientist, and their paper was peer reviewed and publish. You provided no argument that shows their analysis is wrong.
9. Foster and Rahmstorf 2011 removes the ENSO effect from the entire UAH TLT dataset, and from all the other datasets as well. They also remove the effects of variations in aerosols and variations in solar irradiance. Your tool doesn’t do any of this. Your tool uses data that you admit contain these noise causing effects. You even imply that the alleged pause depends on ENSO. But ENSO can’t cause warming or cooling. It only exchanges energy between the ocean and the lower atmosphere.
10. Finally, after all this, you still have not stated what the alleged pause means scientifically. If we assume the pause was statistically significant and scientifically meaningful, as you demand, then what is the scientific meaning of the pause?
Best of luck in your future endeavors.
• Richard M says:
Martin, why should I address long refuted nonsense from a cartoonist’s website when you have proven you won’t understand my comments. Are you 12 years old? If you can’t understand what I explained to you above then there is zero chance you will understand when I explain anything else that relates to climate.
BTW, I already answered your point 10. Since you keep asking I assume you aren’t even reading my comments except for the first sentence or two. Act like a troll and you will be treated like a troll.
• Martin Smith says:
Richard M wrote: “Martin, why should I address long refuted nonsense from a cartoonist’s website when you have proven you won’t understand my comments.”
Richard, every one of your posts in this thread has been a Gish Gallup. You have used the Gish Gallup because you CAN’t address the subjects of this thread, which are: (a) Foster and Rahmstorf 2011 have shown that when El Niño/southern oscillation, volcanic aerosols and solar variability are removed from the datasets (including your cherrypicked UAH TLT version 6.0, there was no pause/hiatus/slowdown. Foster and Rahmstorf 2011 proves that the global warming signal is still there; and (b) If we assume, despite having seen the proof that there was no pause, that there was a pause, what is the scientific significance of that pause?
Foster and Rahmstorf 2011: http://iopscience.iop.org/article/10.1088/1748-9326/6/4/044022/meta
I believe you did not. But if you put your answer at the end of one of your Gish Gallup’s it is possible I missed it. If so, I won’t apologize, because your purpose in using so many Gish Gallups is to hide responses you can’t support with facts, so you can say later, as you have now said: “I already answered your point 10.”
So I believe you have not addressed either point (a) or point (b) above. I know you have not addressed point (a). If you have addressed point (b) and I missed it, I do apologize, but please cut and paste your answer here, because I can’t find it anywhere. And please address point (a).
BTW, what scientists do in their spare time has nothing to do with the truth of their science.
• Martin Smith says:
If you have any evidence that any adjustment to any dataset is fraudulent, post your evidence. I have only ever seen you post evidence that data has been adjusted. But in each case, of course, the adjustment is a correction. Correcting errors in data is not “tampering.” So if you have any evidence that any adjustment to any dataset is fraudulent, post your evidence.
• co2islife says:
Tony Heller over at Real science has done an exhaustive job demonstrating the relationship. The tell tail sign of fraud is that the temperature “adjustments” are not random, but almost all make temperatures not linear. CO2s absorption of LWIR isn’t linear, it is logrythmic, so the physics don’t support the physics, but linear temperatures work better in the linear CO2 vs Temp model pushed by the iPCC.
• Martin Smith says:
co2islife wrote: “Tony Heller over at Real science has done an exhaustive job demonstrating the “adjustments.”
No, he hasn’t. He posts evidence that data have been adjusted, and then he calls it fraud.
co2islife wrote: “linear temperatures work better with a linear co2 vs temperature model being pushed by the IPCC.”
The IPCC is not pushing “a linear co2 vs temperature model.”
• Martin Smith says:
Sunsettommy wrote: “Martin, the “adjustments” happens over and over.”
Evidence of adjustment is not evidence of fraud. When an error is discovered, if a way to correct the error can be found, the data is adjusted to correct the error. Again: If you have any evidence that any adjustment to any dataset is fraudulent or incorrect, post the evidence.
• co2islife says:
More Evidence Congress Needs to Investigate the Field of Climate “Science”
Study Finds Temperature Adjustments Account For ‘Nearly All Of The Warming’ In Climate Data A new study found adjustments made to global surface temperature readings by scientists in recent years “are totally inconsistent with published and credible U.S. and other temperature data.”…The new study doesn’t question the adjustments themselves but notes nearly all of them … Continue reading
https://co2islife.wordpress.com/2017/07/06/more-evidence-congress-needs-to-investigate-the-field-of-climate-science/
• Martin Smith says:
co2islife wrote: “More Evidence Congress Needs to Investigate the Field of Climate “Science””
So, you can’t produce any evidence that any adjustment to any dataset is fradulent or incorrect.
• Martin Smith says:
co2islife wrote: “This post has a graphic of the linear IPCC Model:”
No it doesn’t. The IPCC doesn’t do climate modeling.
• co2islife says:
More on the subject:
4 Graphs That Demonstrate Why The IPCC Climate Models Will NEVER Be Accurate
Y = mX + b One of the most basic statistical techniques used in science is the linear regression. The linear regression defines the relationship between the independent variable (cause) and the dependent variable (effect). The mathematical relationship is Y = mX + b, where Y is the dependent variable, m is the slope of the … Continue reading
https://co2islife.wordpress.com/2017/05/13/4-graphs-that-demonstrate-why-the-ipcc-climate-models-will-never-be-accurate/
33. Bruce Cobb says:
Looks to me like we need a global cooling “cure”:
34. Alan Tomalty says:
https://www.skepticalscience.com/The-Physical-Chemistry-of-Carbon-Dioxide-Absorption.html
I finally found one of the warmists scientific explanations of CO2 warming. They are few and far between cause remember the IPCC has always depended on computer models. It makes for fascinating reading mainly because of his unproven assumptions (CO2 promotes forcing of more water vapour, 400ppm is enough to increase temp… etc) which all warmists have. But particularly interesting is that even though he is a PhD in chemistry he makes the following statement. “These conclusions are not based on the earth’s temperature history or upon complex computer programs (which I certainly believe to be of great importance ”
The fact that a PhD in chemistry would rely on any computer simulated climate model simply boggles my mind. The human race is in trouble if we are producing PhDs who place great importance on computer climate models. They are the reason why we have got into this CO2 mess.
He also makes a number of mistakes
1) “It turns out that at equilibrium all matter emits radiation the distribution of which is determined only by its temperature.”
The above is false because no blackbody behaves the same as any other blackbody they all have a different emissivity coefficient.
2) “So the very fact that we can report a temperature for a given place at a given day, and we routinely do that at any place on any day, means that the earth at that time and place is close enough to being in thermal equilibrium that we are justified in talking about its temperature.”
The above logic boggles the mind. There can be temperature changes of 10 C in one hour at any place on the earth. So how can their be an equilibrium? If he is talking about a global equilibrium even that is not possible if his GHG theory is correct. Because runaway global warming or global warming of any kind is not an equilibrium.
3) “What we believe, and it has been borne out by many studies, is that in general we can do two different things: we can make a number of measurements, reach conclusions based upon those measurements and then average the conclusions, or we can average the measurements and reach conclusions based upon that average measurement. For example we could measure the temperatures at a very large number of places on the earth and 1. Use Planck’s law to calculate the energy radiated at each point and then average or, 2. Average the temperature and use that temperature with Planck’s law to calculate the radiated energy. What has been found is that the final results are essentially the same.”
The above paragraph is laughable. In the 1st case you can only reach conclusions if you plot all the measurements and then take averages or trends or use stats for confidence inertvals …etc. You cannot average conclusions. In the 2nd case you had better have a wide coverage of the area you are measuring with very many measuring stations and no outside extraneous effects like urban heat islands( which by the way all GISS measurements are contaminated with).
4) “The earth’s spatially averaged temperature, when averaged over a year, comes out to be 288K. The balance: energy in = energy out for the earth for a year results because if the energy in from the sun (corrected for albedo) during a year were greater than the energy radiating out from the earth then the earth’s average temperature would rise and, according to Planck’s Law, the earth would radiate more energy and reestablish the balance. ”
The above paragraph seems to support the idea that we have nothing to worry about. If the earth automatically readjusts its IR emittance to rebalance the temperature equilibrium then what in the hell is he promoting?
5) ” Thus the one piece of information that is lacking at this point is NL/V of carbon dioxide in the atmosphere. ”
What he really meant was N/V. The L is the wave length of the frequency of the IR.
6) “Because one half of the reemitted radiation comes back to the earth ”
A statement with no supporting evidence.
7) ” this flux is equal to one half the Planck flux in the absorbing interval multiplied by one minus the diffuse, broadband transmittance. ” ………”The energy leaving at the final time equals the energy entering at that time (for the reason discussed previously) and, because we know by how much that energy is increased by the carbon dioxide greenhouse effect, we know by how much the earth’s temperature is increased by that effect.”
The above assumes that all of the extra flux collides with N2 or the O2 molecules and that none of it escapes to the upper atmosphere and then to space. it doesnt consider the role of clouds nor of water vapour through its own absorption of the extra heat flux. He further ignores condensation.
8) “When the calculation is done, as in GWPPT6, the conclusion is that the earth’s temperature is currently rising by 0.014 degrees per year because of the greenhouse gas effect of the additional carbon dioxide entering the earth’s atmosphere.”
Soooooooooooo what 1.4 degrees in 100 years would be a good thing. Much more plant vegetation and stuff to eat. I cant for the life of me figure out what is scary in all this? And that assumes that he is correct in the temperature rise. I predict that in the next 100 years we will probably see a fall in average temperatures rather than a small rise that he is predicting. Why did politicians buy into a non scary rise of 1.4 degrees in a century?
• 1. “because no blackbody behaves the same as any other blackbody they all have a different emissivity coefficient”
“Blackbody” means emissivity is 1.
2. “There can be temperature changes of 10 C in one hour at any place on the earth. So how can their be an equilibrium?”
He is talking about local thermodynamic equilibrium. In the quote, you cut off the lead-in sentence:
“This can be seen immediately when you realize that to say that something is “at thermal equilibrium” means that it has a temperature and vice versa.”
LTE is indeed the requirement for having a defined temperature.
3. “You cannot average conclusions.”
The “conclusion” is the implied radiative flux. You can average that. All they are saying is that you get the same answer whether you calculate the flux and then average, or average the temperature and calculate the flux. Well, you do.
4.” If the earth automatically readjusts its IR emittance to rebalance the temperature equilibrium”
He’s talking about the temperature rising to rebalance flux equilibrium.
5. “What he really meant was N/V”
No, he means NL/V. N is transmittance per unit wavelength, so you have to integrate over avelengths.
6. “A statement with no supporting evidence.”
He is simply saying that re-emission is not directional. On average half goes up, half down.
7. “all of the extra flux collides with N2 or the O2 molecules”
No, it’s just heat balance; calculating the effect of what didn’t get out. There is no implication that it is a radiative flux that could “collide”. Not that thermal can.
8. “non scary rise of 1.4 degrees in a century”
That is just the CO2 component. But they quote the current annual rate; it is likely to increase. And even if not, we can’t maintain 1.4°/century for ever.
• Alan Tomalty says:
1. There is really no such thing as a blackbody. So we are really talking grey bodies that all have different emissivity coefficients. I used the term blackbody because everyone else does.
2. There can be an equilbrium with an enclosed system with no heat entry or escape. Other than that equilibriums dont exist especially when talking about the weather.
3.You will never get the same answer with those 2 methods. The temperatures you add up and average will always be from a different refence point than the flux you calculate.A further problem is that it would be impossible to calculate this flux at any point of time because you would have to have a million different measurements of the flux on the earths surface.
4. Th earth would not radiate more temperature. You would get an increase in evaporation from the oceans with an increase in water vapour and eventually an increase in condensation.
5) The author defined N/V as the concentration. N cannot be both the number of molecules and the transmittance per unit wavelength.
6. And you also mean that half goes each sideways? Obviously clouds and water vapour absorptions and n2 collisions and O2 collisions rear their ugly heads.
7.At the same time that this flux contained within the CO2 is in the atmosphere some of the CO2 is colliding with N2orO2 molecules. So you cant really get a heat balance equation that is to any degree accurate.
8.
He did not talk about water vapour so we cant argue in this thread about the details of any temperature change that might make. But at 1.4 degrees per century maybe in 5 centuries we might start worrying about overheating but believe me we will have a mini ice age long before that.
35. Martin Smith says:
I think those of you arguing that there is/was an 18 year pause have painted yourselves into a corner. Assuming the alleged pause exists, and given that there have been many pauses before it (see here: https://skepticalscience.com/graphics.php?g=47), why are you saying THIS most recent pause means AGW is wrong, when none of the other pauses before it (see the link) meant AGW is wrong?
• paqyfelyc says:
The “Pause” argument emerged when real world temperature exited the error bar band of IPCC. Obviously, since IPCC had painted this band on previous pauses that were known at the time, that didn’t allow to question the same discrepancy.
https://wattsupwiththat.com/2017/03/31/a-ground-breaking-new-paper-putting-climate-models-to-the-test-yields-an-unexpected-result-steps-and-pauses-in-the-climate-signal/
Funny, isn’t it? Looks like true AGW believers are now arguing that the view Tamino mocked as a fool trick of crackpots deniars, is actually correct. I grab my popcorn and watch the fight from afar.
Note the similarity with those older questions (don’t bother to answer, that’s not the place)
1) Assuming the alleged warming exists, and given that there have been many bigger, faster, warming before it [insert graph at different time scale: millennia, tens of millennia, millions of years…] , why are you saying THIS most recent warming means AGW is the only explanation, when none of the other before required it?
2) Assuming the alleged warming exists, and given that there have been many periods of higher temperature before it [insert graph at different time scale: millennia, tens of millennia, millions of years…] , why are you saying THIS most recent warming means endangers the Planet, when none of the other before did ?
• Martin Smith says:
paqyfelyc wrote: “Funny, isn’t it? Looks like true AGW believers are now arguing that the view Tamino mocked as a fool trick of crackpots deniars, is actually correct. I grab my popcorn and watch the fight from afar.”
Tamino didn’t walk step-wise warming. Tamino proved that your alleged pause is not statistically significant. This new paper doesn’t change that.
paqyfelyc wrote: “1) Assuming the alleged warming exists, and given that there have been many bigger, faster, warming before it [insert graph at different time scale: millennia, tens of millennia, millions of years…] , why are you saying THIS most recent warming means AGW is the only explanation, when none of the other before required it?”
We don’t have to assume alleged warming exists. The alleged warming has been observed. The alleged warming is also apparent in the satellite data you prefer. AGW is the only possible explanation for the most recent warming because the other warmings you refer too that occurred over “millennia, tens of millennia, millions of years,” required millennia, tens of millennia, and millions of years to happen, while the current warming has occurred over 100+ years. That means the causes of those much slower warmings can’t have caused the most recent warming. The only cause that explains the data is AGW.
paqyfelyc wrote: “2) Assuming the alleged warming exists, and given that there have been many periods of higher temperature before it [insert graph at different time scale: millennia, tens of millennia, millions of years…] , why are you saying THIS most recent warming means endangers the Planet, when none of the other before did ?”
No one says the current warming endangers the planet. The current warming endangers the human race (and other species). The reason the previous warm periods with “higher temperature” didn’t endanger the human race is the human race didn’t exist yet, when those warmer warm periods occurred.
Note that I addressed your points immediately, while you have yet to address the one Sheldon Walker raised by arguing that the alleged pause was statistically significant: This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47), so (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many of so-called pauses before this one you claim is scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• No one says the current warming endangers the planet. The current warming endangers the human race (and other species). The reason the previous warm periods with “higher temperature” didn’t endanger the human race is the human race didn’t exist yet, when those warmer warm periods occurred.
?? Not true.
The previous warm periods actually killed exactly what species at what time?
If sea level rise is 2 mm per year, how is the human race threatened?
If man’s release of CO2 increases plant growth (food, fuel, fodder, and feed are increased) by 12 – 27% over previous years, exactly how is the human race threatened by warmer temperatures with greater CO2?
Last month’s global average temperature was less than 0.3 degrees above that reported for the mid-1970’s. What harm has this actually caused to date?
Artificially limiting CO2 emissions by artificially increasing fossil fuel prices and artificially restricting fossil fuel use worldwide will absolutely kill millions of innocents worldwide, will save no lives, and will harm 6 billions more economically and physically. Yet that guaranteed harm to many for many years has only a possibility of potentially reduce future temperature rises by less than 1/2 of one degree. Why do you call for those deaths and that harm?
Is it for the trillions in carbon futures trading the global bankers seek?
Also: You showed a graphic from skeptical science claiming evidence of many previous pauses. But then you also claim the models predict those pauses – but the graphic is merely plots of past actual temperatures stair-stepping upwards. The plots models predicting future pauses are not shown.
• Martin Smith says:
RACookPE1978 wrote: “?? Not true.The previous warm periods actually killed exactly what species at what time?”
You misread. No one is saying AGW will kill the human race. AGW endangers the human race. It already is endangering people in many places with bigger, longer, and drier droughts, bigger, longer, and deeper floods, bigger and wetter storms, sea level rise, etc. And yes, some species will die because AGW happens too fast for them to evolve. Coral is one possibility.
Note that I answered your question immediately, while you have yet to address the one Sheldon Walker raised by arguing that the alleged pause was statistically significant: This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47), so (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many of so-called pauses before this one you claim is scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• Not true.
Abstract
The question of how climate model projections have tracked the actual evolution of global mean surface air temperature is important in establishing the credibility of their projections. Some studies and the IPCC Fifth Assessment Report suggest that the recent 15-year period (1998-2012) provides evidence that models are overestimating current temperature evolution. Such comparisons are not evidence against model trends because they represent only one realization where the decadal natural variability component of the model climate is generally not in phase with observations. We present a more appropriate test of models where only those models with natural variability (represented by El Nino/Southern Oscillation) largely in phase with observations are selected from multi-model ensembles for comparison with observations. These tests show that climate models have provided good estimates of 15-year trends, including for recent periods and for Pacific spatial trend patterns.
Well-estimated global surface warming in climate projections selected for ENSO phase | Request PDF. Available from: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase [accessed Feb 15 2018].
They selectively picked certain models, then selectively analyzed only those models that artificially pre-created (simulated) only those future ENSO periods that were “in phase” with the selected models, then analyzed the result to see if a “short pause” was observed.
Sort of like selecting 75 answers from 77 replies from 2 questions out of 5 asked in one survey from 3500 self-selected survey results from 13,500 members to claim 97% of all scientists believe CAGW is occurring!
• Martin Smith says:
RACookPE1978 wrote: “They selectively picked certain models, then selectively analyzed only those models that artificially pre-created (simulated) only those future ENSO periods that were “in phase” with the selected models, then analyzed the result to see if a “short pause” was observed.”
Obviously, models that don’t include ENSO can’t be used to predict a “pause” caused in large part by ENSO. And obviously, given the complaining by you all that the climate models did not predict your alleged “pause,” only those models that used ENSO phasing that coincided with the real ENSO had to be used to see if their outputs would produce your alleged “pause.”
RACookPE1978 wrote: “Sort of like selecting…”
No, it was exactly what you all demanded, as explained above.
• Martin Smith says:
RACookPE1978 wrote: “If sea level rise is 2 mm per year, how is the human race threatened?”
See south Florida for an example.
RACookPE1978 wrote: “If man’s release of CO2 increases plant growth (food, fuel, fodder, and feed are increased) by 12 – 27% over previous years, exactly how is the human race threatened by warmer temperatures with greater CO2?”
Increasing temperature caused by increasing anthropogenic CO2 already is endangering people in many places with bigger, longer, and drier droughts, bigger, longer, and deeper floods, bigger and wetter storms, sea level rise, etc.
RACookPE1978 wrote: “Last month’s global average temperature was less than 0.3 degrees above that reported for the mid-1970’s. What harm has this actually caused to date?”
RACookPE1978 wrote: “Why do you call for those deaths and that harm?”
Now that renewable energy is competitive with oil and coal, taxing CO2 at the source with a revenue-neutral carbon tax that is redistributed to all citizens equally, will not cause the deaths and harm you are implying it would cause. BTW, oil and gas have indeed caused a lot of deaths and harm.
Note that I answered your questions immediately, while you have yet to address the one Sheldon Walker raised by arguing that the alleged pause was statistically significant: This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47), so (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many of so-called pauses before this one you claim is scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• Martin Smith says:
paqyfelyc wrote: “The “Pause” argument emerged when real world temperature exited the error bar band of IPCC.”
I don’t think the real world temperature exited the error bar, but if we assume it did, that doesn’t mean AGW is wrong. The error bar is based on probability. It says the global average temperature will remain within the error bar with 95% probability, but it CAN go outside the error bar range, below or above the limits. It is currently inside the error bar.
• paqyfelyc says:
@Martin Smith
“I don’t think the real world temperature exited the error bar, but if we assume it did, that doesn’t mean AGW is wrong. ”
Yes, once again you are saying that AGW can never be wrong, that it is not science. I praise your constancy.
In physics, they use several sigma to validate a theory. It must had been be 5 sigma outside a random event to validate recent breakthrough.
You are saying AGW doesn’t even need a single sigma.
Not science. QED
• Martin Smith says:
paqyfelyc wrote: “…once again you are saying that AGW can never be wrong, that it is not science.”
I said no such thing. AGW is the only explanation for the current warming that accounts for all the data. There is no viable alternative explanation that accounts for even most of the data. If you disagree, start a discussion about your alternative theory of global warming. In the meantime, please address this point, raised by Sheldon Walker by arguing that the alleged pause was statistically significant:
This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47).
So, (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many of so-called pauses before this one you claim is scientifically meaningful.
Then state the scientific meaning of your alleged pause.
36. Martin Smith says:
Richard M wrote: “Santer et al in trying to support your beliefs actually found out that pauses over 17 years in length were out side the 95% confidence level. So, an 18 year pause is scientifically meaningful.”
Richard, I have asked several times now for you to say what that meaning is. You have stated again that your alleged pause is scientifically meaningful. This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47), so (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many of so-called pauses before this one you claim is scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• paqyfelyc says:
@Martin Smith
“I’m sorry, are you now also cherrypicking years you will accept papers from? ”
I’m sorry, are you still again building some strawman argument, as a diversion from your failure to deliver?
Still waiting for the prediction you claim exist. Remember: a prediction is made BEFORE having knowledge of the predicted, not after
(I Don’t even believe I had to write that, and have to repeat that…but I did! )
So far, you provided no prediction. I predict you won’t. Never.
Actually, notorious deniar Mann even wrote a paper saying you cannot, as this was unpredictable, according to him.
https://wattsupwiththat.com/2018/02/10/the-global-warming-cure/#comment-2741695
https://wattsupwiththat.com/2018/02/10/the-global-warming-cure/#comment-2742899
(the middle, right after my quoting you that miss quotation mark)
https://wattsupwiththat.com/2018/02/10/the-global-warming-cure/#comment-2743452
(second last paragraph)
What don’t you understand in this?
“The only explanation for the temperature rise we have seen over the last 100+ years is AGW. ”
Blatantly false, and doubly so
False, because the number of possible explanation is very long (sun, clouds, microbes, ocean…), and we don’t have 100+ years of data to rule them out.
False, because so-called GHG forcing just don’t match with warming. Speaking of the pause: human GHG emission accelerated, so much so that the period accounted for ~1/3 of the total. You need to add lots of unexplainium, feedback etc. to have GW and AGW match. Occam advise to stick to unexplainium alone, in such case, since you need it anyway.
“The way you disprove AGW is to propose a different theory that accounts for all the data and that makes better predictions about the future than AGW. ”
QED. That is the kind of reasoning to support God interventions, Cupidon, witchcraft, conspiracy theory, alien abduction, etc. since we lack other theories to explain miracles, falling in love, streak of (mis)fortune, many dramatic events, strange disappearance … You just placed AWG on par with these. And you were right, All these beliefs are alike. Not falsifiable, not science, fallacies based.
Science, on the other hand, don’t need to have a better theory to drop a failed one. Physicists just dropped a theory about Dark matter, but they still have no better.
In science, the way you disprove a theory is just to expose it shortcoming. Period. If you already have an alternative theory, just fine; if you don’t, you proceed to find another. Well, if you need it, of course.
• paqyfelyc says:
@Martin Smith
“I’m sorry, are you now also cherrypicking years you will accept papers from? ”
I’m sorry, are you still again building some strawman argument, as a diversion from your failure to deliver?
Still waiting for the prediction you claim exist. Remember: a prediction is made BEFORE having knowledge of the predicted, not after
(I Don’t even believe I had to write that, and have to repeat that…but I did! )
So far, you provided no prediction. I predict you won’t. Never. The fun is, even is you did, Mann already disagree that this prediction was even valid, as, according to him, it was just not predictable.
Funny, isn’t it?
• paqyfelyc says:
February 12, 2018 at 4:46 am
February 13, 2018 at 9:24 am
February 13, 2018 at 8:50 pm (second last paragraph)
What don’t you understand in this? Why do you still repeat the very same question, already answered ?
• Martin Smith says:
No. Neither you nor Sheldon Walker, nor Richard M, nor anyone else here has ever addressed this point: You have stated that your alleged pause is statistically significant and scientifically meaningful. This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47).
So, (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many so-called pauses before this one that you claim is both statistically significant and scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• paqyfelyc says:
You are acting like a child who, when asking “2+2= ?” and answered “4”, says “no, you didn’t adress my question, so I repeat again, 2+2= ?”
Not sure what can be done for you.
May be, to begin with, state the scientific meaning of your question?
• paqyfelyc says:
“The only explanation for the temperature rise we have seen over the last 100+ years is AGW. ”
Blatantly false, and doubly so
* False, because the number of possible explanation is very long (sun, clouds, microbes, ocean…), and we don’t have 100+ years of data to rule them out.
* False, because so-called GHG forcing just don’t match with warming. Speaking of the pause: human GHG emission accelerated, so much so that the period accounted for ~1/3 of the total. You need to add lots of unexplainium, and assumptions about the inside working of the climate blackbox (feedback etc.) to have GW and AGW match. Occam advise to stick to unexplainium and assumptions about the inside working of the climate blackbox, in such case, since you need them anyway.
• Martin Smith says:
paqyfelyc wrote: “the number of possible explanation is very long (sun, clouds, microbes, ocean…), and we don’t have 100+ years of data to rule them out.”
Those aren’t explanations of global warming. You have simply listed items that affect global climate. The sun and the ocean are both accounted for in AGW, and their contributions are fairly well understood, so again you can’t claim the established science of the climate roles played by the sun and oceans is wrong without proposing an alternative explanation for those roles that fits the data. The affect of clouds on global warming is more of an unknown; there are both positive and negative feedbacks caused by clouds, but these are feedbacks, paq, not original drivers. If there is increasing cloud cover it is because warmer air holds more water. The cosmic ray idea is understood well enough to know that they are not affecting cloud cover enough to make a difference in global temperature. Microbes have caused major climate changes in the past, and they will again, possibly, if we don’t stop AGW, but we know they are not causing the current warming.
paqyfelyc wrote: “so-called GHG forcing just don’t match with warming.”
Yes it does. Please stay on topic: Neither you nor Sheldon Walker, nor Richard M, nor anyone else here has ever addressed this point: You have stated that your alleged pause is statistically significant and scientifically meaningful. This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47).
So, (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many so-called pauses before this one that you claim is both statistically significant and scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• paqyfelyc says:
@Martin Smith
Now you are ridiculous. YOU put some topic on the table, not me. I just answered you, and when I did, you accuse me of being out of topic…
If the topic you put on the table was out of topic, what is the topic, then?
• Martin Smith says:
paqyfelyc wrote: “what is the topic, then?”
This is the topic: Neither you nor Sheldon Walker, nor Richard M, nor anyone else here has ever addressed this point: You have stated that your alleged pause is statistically significant and scientifically meaningful. This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47).
So, (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many so-called pauses before this one that you claim is both statistically significant and scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• paqyfelyc says:
Don’t trunk my sentence. My question wasn’t “what was the topic?” , it was “If the topic you put on the table was out of topic, what is the topic, then?”
You didn’t answer. I want to know how the topic could simultaneously be the topic (when you claim something), and not the topic (when you get answer to your claim).
• Martin Smith says:
paqyfelyc wrote: “If the topic you put on the table was out of topic, what is the topic, then?”
I didn’t put a topic on the table. I have continuously tried to put this discussion BACK on topic AFTER your attempts, and the attempts of Richard M and RACookPE1978, to divert the topic to something else. THIS IS THE TOPIC: Neither you nor Sheldon Walker, nor Richard M, nor anyone else here has ever addressed this point: You have stated that your alleged pause is statistically significant and scientifically meaningful. This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47).
So, (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many so-called pauses before this one that you claim is both statistically significant and scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• paqyfelyc says:
“The way you disprove AGW is to propose a different theory that accounts for all the data and that makes better predictions about the future than AGW. ”
QED. That is the kind of reasoning to support G0d interventions, Cupidon, w!tchcraft, consp!racy theory, al!en abduction, etc. since we lack other theories to explain miracles, falling in love, streak of (mis)fortune, many dramatic events, strange disappearance … You just placed AWG on par with these. And you were right, All these beliefs are alike. Not falsifiable, not science, fallacies based.
Science, on the other hand, don’t need to have a better theory to drop a failed one. Physicists just dropped a theory about Dark matter, but they still have no better.
In science, the way you disprove a theory is just to expose it shortcoming. Period. If you already have an alternative theory, just fine; if you don’t, you proceed to find another. Well, if you need it, of course.
• paqyfelyc says:
“The way you disprove AGW is to propose a different theory that accounts for all the data and that makes better predictions about the future than AGW. ”
QED. That is the kind of reasoning to support supernatural interventions and illuminati plots, since we lack other theories to explain miracles, falling in love, streak of (mis)fortune, many dramatic events … You just placed AWG on par with these. And you were right, All these beliefs are alike. Not falsifiable, not science, fallacies based.
Science, on the other hand, doesn’t need to have a better theory to drop a failed one. Physicists just dropped a theory about Dark matter, but they still have no better.
In science, the way you disprove a theory is just to expose it shortcoming. Period. If you already have an alternative theory, just fine; if you don’t, you proceed to find another. Well, if you need it, of course.
• Martin Smith says:
paqyfelyc wrote: “That is the kind of reasoning to support supernatural interventions and illuminati plots, since we lack other theories to explain miracles, falling in love, streak of (mis)fortune, many dramatic events …”
There is nothing to respond to there. It reads like gibber to me. AGW is the only theory that explains the current global warming. It is the only explanation for the current warming that accounts for all the data. It’s projections match observations quite well. No other possible explanation we know of comes close. That’s why AGW is the theory used for explaining the current global warming. The way you disprove AGW is by finding a better theory.
I await your proposal. In the meantime, please address this point: Neither you nor Sheldon Walker, nor Richard M, nor anyone else here has ever addressed this point: You have stated that your alleged pause is statistically significant and scientifically meaningful. This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47).
So, (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many so-called pauses before this one that you claim is both statistically significant and scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• paqyfelyc says:
“AGW is the only theory that explains the current global warming. It is the only explanation for the current warming that accounts for all the data. It’s projections match observations quite well. No other possible explanation we know of comes close. That’s why AGW is the theory used for explaining the current global warming. The way you disprove AGW is by finding a better theory.”
That’s a lot of claims, none of them proven, most of them most of them easy to prove false, all of them worthy to be forever marked on the hall of shame with a mention “do not ever says this again”. The question is, is it worth the effort to show you wrong? Do you have good faith enough? Won’t you rather answer “this is off topic” to discard inconvenient evidence, or turn your claim into a new version untouched by the evidence the first version was false? That’s what you did every time so far.
So, then again, is it worth the effort? YOU tell me.
Let’s start with a simple one “No other possible explanation we know of comes close. ”
Who dare bring a https://en.wikipedia.org/wiki/Argument_from_ignorance Argument from ignorance in a scientific debate? You do. And you stick to it, again and again. Stop that. Forever.
Marked on the hall of shame with a mention “do not ever says this again”
• Martin Smith says:
paqyfelyc wrote again: “That’s a lot of claims, none of them proven…”
The statement you are denouncing, which is presented again below, is a statement of fact:
AGW is the only theory that explains the current global warming. It is the only explanation for the current warming that accounts for all the data. It’s projections match observations quite well. No other possible explanation we know of comes close. That’s why AGW is the theory used for explaining the current global warming. The way you disprove AGW is by finding a better theory.
If you have an alternative theory that explains all the data, post it now.
• paqyfelyc says:
@martin Smith
“AGW is the only theory that explains the current global warming. It is the only explanation for the current warming that accounts for all the data. It’s projections match observations quite well. No other possible explanation we know of comes close. That’s why AGW is the theory used for explaining the current global warming. The way you disprove AGW is by finding a better theory.”
That’s a lot of claims, none of them proven, most of them most of them easy to prove false, all of them worthy to be forever marked with a mention “do not ever say this again”. The question is, is it worth the effort to show you wrong? Do you have good faith enough? Won’t you rather answer “this is off topic” to discard inconvenient evidence, or turn your claim into a new version untouched by the evidence the first version was false? That’s what you did every time so far.
So, then again, is it worth the effort? YOU tell me.
Let’s start with a simple one “No other possible explanation we know of comes close. ”
Who dare bring the fallacy “I don’t know any other possible explanation, so the witch is guilty” in a scientific debate? You do. And you stick to it, again and again. Stop that. Forever. The choice is simple: either the theory does get proper support, and this argument is useless. Or it doesn’t… This argument may bu useful for a cult or a superstition, but for this very reason it is unsuitable as far as science is concerned and weaken your case, not strengthen it.
Marked with mention “do not ever say this again”.
• Martin Smith says:
paqyfelyc wrote: “That’s a lot of claims, none of them proven…”
What I wrote is factual and correct. AGW is the only theory that explains the current global warming. It is the only explanation for the current warming that accounts for all the data. It’s projections match observations quite well. No other possible explanation we know of comes close. That’s why AGW is the theory used for explaining the current global warming. The way you disprove AGW is by finding a better theory.
I await your proposal. In the meantime, please address this point: Neither you nor Sheldon Walker, nor Richard M, nor anyone else here has ever addressed this point: You have stated that your alleged pause is statistically significant and scientifically meaningful. This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47).
So, (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many so-called pauses before this one that you claim is both statistically significant and scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• paqyfelyc says:
You didn’t anwser.
Is it worth the effort to you prove you wrong?
Well, actually you did. I pointed at a most blatant fallacy, you didn’t even deny it is a fallacy, but you stick to it no matter what.
If you do that for a fallacy, no amount of evidence will ever convince you. It is much more easy for you to cherry-pick papers supporting your claims and discard all contradicting evidence, than accept changing your mind. Even a paper from Mann you discarded, although it just mitigated a claim without impact on the AGW theory.
You succeeded in proving right Dave Fair February 13, 2018 at 10:59 pm
It is a detector of people like Martin Smith, ready to use any fallacy to support their claims.
• Martin Smith says:
paqyfelyc wrote: “You didn’t anwser (sic).”
I did answer; you didn’t. But that is actually my point. None of you has even attempted a serious explanation of why you think you alleged pause is important. Whether there was or was not a pause, there isn’t one now. If there was a pause, it can’t be evidence against AGW.
I have continued to reply to your missives because each one gives me a new opportunity to prove this point: You can’t provide a serious explanation of what your alleged pause means in AGW. Your real purpose is to manufacture doubt about the accuracy of global climate models by implying that, because their outputs don’t show the alleged pause when the ENSO submodel is not included, global climate models can’t be trusted.
That’s your real purpose. It is not science; it is not skepticism. It is nefarious.
• paqyfelyc says:
Act V, scene IX
(Thunderclap. Martin Smith, previously dressed as an ordinary man, rip open his dress, exposing his red cassock and holy cross previously concealed, and gleam in triumph. paqyfelyc falls on her knee stunned in awe; for the rest of the scene, will makes herself smaller and smaller, tamed. )
Martin Smith (talking quickly and very loud, with frantic gestures, expressive face, pointing of finger, etc. Walking speedily all aver the scene to show complete possession of the space and domination):
“aha ah (evil laughter), Nobody expect Spanish Inquisition! I am not the ordinary man you thought!
I tricked you, evil witches, and it duly worked as I intended.
I have continued to reply to your missives because each one gives me a new opportunity to prove this point: any proof of whatever, I can weave hand at, and pretend it doesn’t exist, or is not serious, or off topic, as I see fit. So none of you has even attempted a serious explanation of why you think your alleged fossil is important.
And lo, behold, I am mind-reader, too. You didn’t expect that, either, did you? I know your real purpose! You intent to smear the Holy Bible, by implying that, just because at first reading it doesn’t seem to mention fossils when the ENGEL submodel is not included, the Gospel cannot be trusted.
That’s your real purpose. It is not science; it is not skepticism. It is nefarious.”
(paqyfelyc cowers away of the stage as sneakily as possible. Martin Smith stand up alone at the center of the scene, in full control, under bright light, with a gloriole.
Curtain closes)
Audience blows in laughter and delight, and stand-up clapping for the fabulous performance.
• Martin Smith says:
paqyfelyc continued his ad hominem attack but remains unwilling, as are Sheldon Walker and Richard M, to answer the obvious question resulting from the issue Walker raised:
You have stated that your alleged pause is statistically significant and scientifically meaningful. This paper shows that the models that include ENSO show pauses like your alleged one: https://www.researchgate.net/publication/264348479_Well-estimated_global_surface_warming_in_climate_projections_selected_for_ENSO_phase
The observed data show that there have been many pauses like the one you allege (see here: https://skepticalscience.com/graphics.php?g=47).
So, (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many so-called pauses before this one that you claim is both statistically significant and scientifically meaningful.
Then state the scientific meaning of your alleged pause.
• Martin Smith
So, (a) the paper demonstrates that the climate models that include ENSO show pauses, and (b) the other link shows that there have been many so-called pauses before this one that you claim is both statistically significant and scientifically meaningful.
No, you’ve linked to that animated gif of a self-called “escalator” at the SS (Skeptical Science) site several times now. it’s “information” has not changed.
It claims to show “many pauses” in the past. The SS gif does not do that. What that animation does show very well are six separate short periods ranging from 2-1/2 years to 12 years – averaging less than 8 years each for the total of 45 years before 2015 – of oscillating past recorded temperatures between the 1970’s and 2000’s as temperatures gradually rose after they fell from 1945 to 1975. Now, global atmospheric CO2 levels continuously INCREASED from 1945 through 2018. But global average temperatures DECREASED from 1945 through 1975 – a feature the SS animated gif blocked out – and then INCREASED between 1975 and 1998.
Thus, the SS animated gif only displays what no one is disputing: Global average temperatures and global average CO2 levels both increased over a 23 year span between 1975 and 1998. The SS gif selectively HIDES the decline before the unique 30 year period in earth’s history when temperatures fell but CO2 rose at the same time.
The single, long, enduring 18 year “pause” as we found it by BACKTRACKING from each date after 2003 (when it began to be noticeable) is up to 2, 3 and up to 5 TIMES longer than the six short short stairsteps conveniently highlighted in the SS animated gif. The gif does not establish anything but our own points that contradict the CAGW theory:
– CO2 increased for more than 30 years between 1940 and 1975, and global average temperatures fell.
– CO2 increased for 20+ years between 1975 and 1998, and global average temperatures rose.
– CO2 increased even MORE for 20 years between 1998 and 2018, and global average temperatures remained near-steady. (The pause, until the El Nino rise of 2016-2017, and now today’s GAT of +0.28 above the mid-1970’s baseline of 0.0.)
37. paqyfelyc says:
@mod
I wrote several comment that never appeared, I think they went to the moderation buffer because of some trigger word. I reformulated and the modified comment appeared, so don’t bother to save those previous comments, as they would double | 63,404 | 281,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-43 | longest | en | 0.933782 |
http://nrich.maths.org/public/leg.php?code=-68&cl=3&cldcmpid=4821 | 1,472,285,461,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982298875.42/warc/CC-MAIN-20160823195818-00142-ip-10-153-172-175.ec2.internal.warc.gz | 187,784,105 | 10,852 | # Search by Topic
#### Resources tagged with Visualising similar to Ratio Pairs 3:
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### Square It
##### Stage: 1, 2, 3 and 4 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### Square it for Two
##### Stage: 1, 2, 3 and 4 Challenge Level:
Square It game for an adult and child. Can you come up with a way of always winning this game?
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##### Stage: 4 and 5 Challenge Level:
This is an interactive net of a Rubik's cube. Twists of the 3D cube become mixes of the squares on the 2D net. Have a play and see how many scrambles you can undo!
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
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Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
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Can you find a rule which connects consecutive triangular numbers?
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##### Stage: 3 and 4 Challenge Level:
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Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
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##### Stage: 3 Challenge Level:
Every day at noon a boat leaves Le Havre for New York while another boat leaves New York for Le Havre. The ocean crossing takes seven days. How many boats will each boat cross during their journey?
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A game for 2 players
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### Instant Insanity
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Show that all pentagonal numbers are one third of a triangular number.
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Show that among the interior angles of a convex polygon there cannot be more than three acute angles.
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##### Stage: 4 Challenge Level:
The triangle OMN has vertices on the axes with whole number co-ordinates. How many points with whole number coordinates are there on the hypotenuse MN?
### Muggles Magic
##### Stage: 3 Challenge Level:
You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.
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##### Stage: 3 Challenge Level:
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### Clocking Off
##### Stage: 2, 3 and 4 Challenge Level:
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### There and Back Again
##### Stage: 3 Challenge Level:
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Can you mark 4 points on a flat surface so that there are only two different distances between them?
### Baravelle
##### Stage: 2, 3 and 4 Challenge Level:
What can you see? What do you notice? What questions can you ask?
### Partially Painted Cube
##### Stage: 4 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Route to Infinity
##### Stage: 3 and 4 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next? | 2,504 | 10,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2016-36 | longest | en | 0.899474 |
http://www.gicgac.com/FirstSet/KIDS/Practice_Free_Ratio_to_Fraction_Questions_online.html | 1,601,510,829,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402130531.89/warc/CC-MAIN-20200930235415-20201001025415-00007.warc.gz | 162,377,267 | 4,066 | # Learn and Practice Free Ratio to Fraction questions - Kids Study
## Practice Free Ratio to Fraction Questions online
By Practicing the below question set :
The student will get the basic and advanced knowledge on Ratio to Fraction.
The student will learned the priciples and methedology of Ratio to Fraction.
The student will evaluate the knowledge level on Ratio to Fraction.
Student can practice number of times the question till he is clear.
Student can provide the review / feedback / suggestion on the questions which will help to others.
The questions can be practiced by All the standards / Grades.
The question contains the simple to complex level difficulties so that student can improve the knowledge.
## You can get answer of the below sample questions by going through the above Question sets.
Write the Ratios as Fraction: 8 : 9
Write the Ratios as Fraction: 9 : 16
Write the Ratios as Fraction: 5 : 11
Write the Ratios as Fraction: 3 : 13
Write the Ratios as Fraction: 11 : 15
Write the Ratios as Fraction: 17 : 20
Write the Ratios as Fraction: 21 : 31
## Practice Other Selected Questions
### Click here to Practice Free Find fraction by Object Questions onlineClick here to Practice Free Find the Elapsed Time Questions onlineClick here to Practice Free Fraction to Percentage Questions onlineClick here to Practice Free Fraction Word problem Questions onlineClick here to Practice Free Gartitude Questions onlineClick here to Practice Free GAT Math Questions onlineClick here to Practice Free GED Math Questions onlineClick here to Practice Free GMAT Math Questions onlineClick here to Practice Free Greatest Common Factor questions onlineClick here to Practice Free Identify 3D Shapes Questions onlineClick here to Practice Free Identify Equation Questions onlineClick here to Practice Free Inside Outside questions online
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Mail: Support@gicgac.com | 515 | 2,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-40 | latest | en | 0.855019 |
https://studylib.net/doc/25485058/unit-3 | 1,675,114,646,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499829.29/warc/CC-MAIN-20230130201044-20230130231044-00201.warc.gz | 572,267,929 | 11,378 | # unit 3
```Unit-3
PRODUCTION PLANNING
AND PROCESS PLANNING
PRODUCT PLANNING
• Product Planning is the ongoing process of
identifying
and
articulating
market
requirements that define a product’s feature
set.
• Product planning serves as the basis for
promotion.
EXTENDING ORIGINAL PRODUCT
INFORMATION
VALUE ANALYSIS
•
Value, as defined, is the ratio of function
to cost. Value can therefore be increased by
either improving the function or reducing
the cost
LACK OF PRODUCT PLANNING
•
•
•
•
Unsatisfied customer
Durability deteriaration
Lack of quality
Loss of brand name., Etc
PROCESS PLANNING
ROUTING
SCHEDULING
DISPATCHING
PROCESS PLANNING – INPUT INFO.
•
•
•
•
•
•
•
PRODUCT
PROCESS
CAPACITY
ORDERS
DUE DATES
RESOURCES
ETC.,
STEPS IN PROCESS PLANNING
• Analyzing environment
• Establishing objectives or goals
• Seeking necessary Information
QUANTITY DETERMINATION –
BATCH PRODUCTION
• Economic batch quantity (EBQ), also called "optimal batch
quantity" or economic production quantity, is a measure used
to determine the quantity of units that can be produced at
minimum average costs in a given batch or production run.
• Economic Production Quantity model (also known as the EPQ
model) is an extension of the Economic Order Quantity
model. The Economic Batch Quantity model, or production
lot-size model, is similar to the EOQ model in that an optimum
is to be calculated for the batch quantity to be produced.
EBQ
In working with this EBQ model, principal assumptions are:
•
The demand (D) is known and constant within a certain period of time
•
The unit cost of the inventory item (U) is constant
•
The annual holding-cost per unit (Ch) is constant
•
The setup-cost per batch (C) is constant
•
The production time (tp) is known and constant
•
there is one kind of product
•
There is no interaction with other products
•
The aspect of time does not play a role, just the setup time does
•
The setup cost is constant and does not act upon the batch quantity.
Variables
•
K = setup cost
•
D = demand rate
•
F = holding cost
•
T = cycle length
•
P = production rate
Formula: Sqrt(2x annual demandx setup costs)/(inventory carrying cost per unit)
``` | 564 | 2,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-06 | longest | en | 0.880199 |
https://www.numberempire.com/2199472 | 1,601,526,817,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402130615.94/warc/CC-MAIN-20201001030529-20201001060529-00536.warc.gz | 869,532,784 | 6,812 | Home | Menu | Get Involved | Contact webmaster
# Number 2199472
two million one hundred ninety nine thousand four hundred seventy two
### Properties of the number 2199472
Factorization 2 * 2 * 2 * 2 * 11 * 12497 Divisors 1, 2, 4, 8, 11, 16, 22, 44, 88, 176, 12497, 24994, 49988, 99976, 137467, 199952, 274934, 549868, 1099736, 2199472 Count of divisors 20 Sum of divisors 4649256 Previous integer 2199471 Next integer 2199473 Is prime? NO Previous prime 2199433 Next prime 2199479 2199472nd prime 35917069 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 1000011000111110110000 Octal 10307660 Duodecimal 8a0a14 Hexadecimal 218fb0 Square 4837677078784 Square root 1483.0616979748 Natural logarithm 14.603727889524 Decimal logarithm 6.3423184376369 Sine -0.84097104248004 Cosine 0.5410801287333 Tangent -1.5542449220022
Number 2199472 is pronounced two million one hundred ninety nine thousand four hundred seventy two. Number 2199472 is a composite number. Factors of 2199472 are 2 * 2 * 2 * 2 * 11 * 12497. Number 2199472 has 20 divisors: 1, 2, 4, 8, 11, 16, 22, 44, 88, 176, 12497, 24994, 49988, 99976, 137467, 199952, 274934, 549868, 1099736, 2199472. Sum of the divisors is 4649256. Number 2199472 is not a Fibonacci number. It is not a Bell number. Number 2199472 is not a Catalan number. Number 2199472 is not a regular number (Hamming number). It is a not factorial of any number. Number 2199472 is an abundant number and therefore is not a perfect number. Binary numeral for number 2199472 is 1000011000111110110000. Octal numeral is 10307660. Duodecimal value is 8a0a14. Hexadecimal representation is 218fb0. Square of the number 2199472 is 4837677078784. Square root of the number 2199472 is 1483.0616979748. Natural logarithm of 2199472 is 14.603727889524 Decimal logarithm of the number 2199472 is 6.3423184376369 Sine of 2199472 is -0.84097104248004. Cosine of the number 2199472 is 0.5410801287333. Tangent of the number 2199472 is -1.5542449220022
### Number properties
Examples: 3628800, 9876543211, 12586269025 | 748 | 2,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-40 | latest | en | 0.632166 |
https://ioflood.com/blog/python-absolute-value/ | 1,718,703,344,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00548.warc.gz | 271,979,617 | 15,362 | Python Absolute Value | abs() Function and Alternatives
# Python Absolute Value | abs() Function and Alternatives
Ever felt lost trying to find the absolute value of a number in Python? Just like a compass guiding you through the wilderness, Python’s `abs()` function can help you find the absolute value. It’s a simple, yet powerful tool that can make your coding journey much smoother.
This comprehensive guide will walk you through the steps of using the `abs()` function in Python, helping you understand its basic and advanced uses, potential issues you might encounter, and alternative methods.
So, buckle up and get ready to delve into the world of Python absolute values!
## TL;DR: How Do I Find the Absolute Value in Python?
To find the absolute value in Python, you use the `abs()` function. Here’s a simple example:
``````number = -10
absolute_value = abs(number)
print(absolute_value)
# Output:
# 10
``````
In this example, we’ve used the `abs()` function to find the absolute value of `-10`, which is `10`. The `abs()` function is a powerful tool in Python that can handle both positive and negative numbers with ease. If you’re intrigued and want to learn more about the `abs()` function, including advanced usage scenarios and potential pitfalls, keep reading. This comprehensive guide has got you covered!
## Python’s abs() Function: A Beginner’s Guide
Python’s `abs()` function is a built-in function used to return the absolute value of a number. The absolute value of a number is its distance from zero, regardless of the direction. In other words, it’s the non-negative value of a number. Let’s see it in action:
``````number = -15
absolute_value = abs(number)
print(absolute_value)
# Output:
# 15
``````
In this example, we use the `abs()` function to find the absolute value of `-15`, which returns `15`. Notice how regardless of the negative sign, `abs()` returns the positive counterpart. This is the basic use of the `abs()` function.
The `abs()` function is a handy tool when you’re dealing with operations where you need the magnitude of a number, but the direction (positive or negative) is not important. It’s simple to use and works seamlessly with both positive and negative numbers.
However, it’s important to remember that the `abs()` function only works with numbers. If you try to use it with a data type that is not a number, Python will raise a TypeError. For example:
``````string = 'I am not a number'
try:
absolute_value = abs(string)
except TypeError:
print('TypeError: bad operand type for abs(): str')
# Output:
# TypeError: bad operand type for abs(): str
``````
In this example, we tried to find the absolute value of a string, which resulted in a TypeError. This is one of the potential pitfalls you need to be aware of when using the `abs()` function.
## Handling Complex Numbers with abs()
Python’s `abs()` function isn’t limited to just integers and floating-point numbers; it can also handle complex numbers. A complex number is a number that can be expressed in the form `a + bj`, where `a` and `b` are real numbers and `j` represents the square root of `-1`. In the context of complex numbers, the `abs()` function returns the magnitude of the complex number.
Let’s see an example:
``````complex_number = 3 + 4j
absolute_value = abs(complex_number)
print(absolute_value)
# Output:
# 5.0
``````
In this example, `3 + 4j` is a complex number. When we pass this complex number to the `abs()` function, it returns the magnitude of the complex number, which is `5.0`. The magnitude is calculated as the square root of the sum of the squares of the real and imaginary parts (`sqrt(a^2 + b^2)`).
This advanced use of the `abs()` function opens up a whole new range of possibilities. However, it’s important to understand that the `abs()` function treats real and complex numbers differently. For real numbers, it returns the number without the sign, while for complex numbers, it returns the magnitude. As you delve deeper into Python, you’ll find this function to be a powerful tool in your arsenal, especially when dealing with mathematical computations involving complex numbers.
## Exploring Alternatives: The math.fabs() Function
While the `abs()` function is the most common method for finding the absolute value in Python, it’s not the only one. Python also provides the `math.fabs()` function as part of its math module. The `math.fabs()` function returns the absolute value of a number as a floating-point number.
Here’s an example of how to use the `math.fabs()` function:
``````import math
number = -7.5
absolute_value = math.fabs(number)
print(absolute_value)
# Output:
# 7.5
``````
In this example, we’ve used the `math.fabs()` function to find the absolute value of `-7.5`, which returns `7.5`. Notice that even though the input was a float, the `math.fabs()` function still successfully returned the absolute value.
However, it’s important to note that the `math.fabs()` function only works with real numbers. If you try to use it with a complex number, Python will raise a TypeError.
MethodWorks with IntegersWorks with FloatsWorks with Complex Numbers
abs()YesYesYes
math.fabs()YesYesNo
As you can see, both `abs()` and `math.fabs()` functions have their advantages and disadvantages. While `abs()` is more versatile and can handle complex numbers, `math.fabs()` returns a float for all inputs. Your choice between the two will depend on the specific requirements of your code.
## Troubleshooting Common Issues with Absolute Values in Python
When finding absolute values in Python, you might encounter a few common issues. One of the most common errors is the `TypeError`. This error occurs when you try to find the absolute value of a non-numeric data type, such as a string or a list. Let’s look at an example:
``````string = 'Hello, Python!'
try:
absolute_value = abs(string)
except TypeError:
print('TypeError: bad operand type for abs(): str')
# Output:
# TypeError: bad operand type for abs(): str
``````
In this code block, we tried to find the absolute value of a string, which resulted in a `TypeError`. Python’s `abs()` function only works with numeric data types, so trying to use it with a string or other non-numeric data types will result in an error.
Another common issue is using the `math.fabs()` function with complex numbers. While the `math.fabs()` function works perfectly with integers and floats, it doesn’t support complex numbers. Here’s an example:
``````import math
complex_number = 5 + 2j
try:
absolute_value = math.fabs(complex_number)
except TypeError:
print('TypeError: can't convert complex to float')
# Output:
# TypeError: can't convert complex to float
``````
In this example, we tried to find the absolute value of a complex number using `math.fabs()`, which led to a `TypeError`. This is because `math.fabs()` can only handle real numbers and not complex numbers.
When encountering these common issues, the solutions are usually straightforward. For `TypeError` with `abs()`, make sure the input is a numeric data type. For issues with `math.fabs()`, ensure the input is a real number. Understanding these potential problems and their solutions will make your Python coding journey smoother and more enjoyable.
## Python Number Data Types: Integer, Float, and Complex
To fully grasp the concept of finding the absolute value in Python, it’s essential to understand Python’s number data types: integer, float, and complex numbers.
### Integer Numbers
Integers in Python are whole numbers that can be positive, negative, or zero. For example, `5`, `-3`, and `0` are all integers. When you use the `abs()` function with an integer, it simply returns the positive counterpart of the number. Here’s an example:
``````integer_number = -8
absolute_value = abs(integer_number)
print(absolute_value)
# Output:
# 8
``````
In this example, we’ve used the `abs()` function to find the absolute value of `-8`, which is `8`.
### Float Numbers
Floats in Python are real numbers that contain a decimal point. For example, `3.14`, `-0.01`, and `0.0` are all floats. The `abs()` function works with floats just like it works with integers, returning the positive counterpart of the number. Here’s an example:
``````float_number = -7.5
absolute_value = abs(float_number)
print(absolute_value)
# Output:
# 7.5
``````
In this example, we’ve used the `abs()` function to find the absolute value of `-7.5`, which is `7.5`.
### Complex Numbers
Complex numbers in Python are numbers that have a real and an imaginary part, expressed as `a + bj`. For example, `3 + 4j` is a complex number. When you use the `abs()` function with a complex number, it returns the magnitude of the number. Here’s an example:
``````complex_number = 3 + 4j
absolute_value = abs(complex_number)
print(absolute_value)
# Output:
# 5.0
``````
In this example, we’ve used the `abs()` function to find the magnitude of the complex number `3 + 4j`, which is `5.0`.
Understanding these number data types and how the `abs()` function works with each of them is fundamental to mastering the concept of finding the absolute value in Python. Whether you’re working with integers, floats, or complex numbers, Python’s `abs()` function has got you covered!
## Expanding Horizons: Absolute Values in Real-World Applications
Finding absolute values is not just a theoretical concept, but a practical tool used in various real-world applications, especially in mathematical computations and data analysis. For instance, in data analysis, absolute values can help calculate the magnitude of change or difference between two data points, disregarding the direction of the change. Here’s a simple example:
``````change = -15
magnitude_of_change = abs(change)
print(magnitude_of_change)
# Output:
# 15
``````
In this example, we have a change of `-15`. Using the `abs()` function, we can find the magnitude of this change, which is `15`, regardless of whether it was a decrease or increase.
Beyond just the `abs()` function, Python offers a rich set of mathematical functions in its math module, which can be a great next step to explore. For example, the `math.sqrt()` function can be used to calculate the square root of a number, and the `math.pow()` function can be used to raise a number to a power.
Additionally, Python’s capabilities for data analysis extend far beyond just mathematical functions. Libraries such as Pandas, NumPy, and Matplotlib offer powerful tools for data manipulation, analysis, and visualization. These libraries, combined with Python’s built-in functions like `abs()`, can be a powerful toolkit for any data analyst or scientist.
## Further Resources for Python Functions
To deepen your understanding of Python and its applications, consider exploring these resources:
## Recap: Python’s Absolute Value
In this comprehensive guide, we’ve explored how to find the absolute value in Python using the `abs()` function. This built-in function is a versatile tool that works with integers, floats, and even complex numbers. Here’s a quick recap of what we covered:
• Basic Use of abs(): The `abs()` function returns the positive counterpart of a number, effectively giving us its absolute value. It’s a simple and straightforward method, as seen in this example:
``````number = -10
absolute_value = abs(number)
print(absolute_value)
# Output:
# 10
``````
• Advanced Use of abs(): We’ve seen that the `abs()` function can also handle complex numbers, returning the magnitude of the complex number.
• Alternative Approach: Python also provides the `math.fabs()` function, which can be used to find the absolute value of a number as a floating-point number.
• Troubleshooting: We discussed common issues that might arise when using these functions, such as `TypeError`, and provided solutions for them.
• Beyond the Basics: We also touched on the real-world applications of finding absolute values in Python, particularly in mathematical computations and data analysis.
MethodWorks with IntegersWorks with FloatsWorks with Complex Numbers
abs()YesYesYes
math.fabs()YesYesNo
Whether you’re a beginner just starting out or an intermediate programmer looking to brush up on your skills, understanding how to find the absolute value in Python is an essential skill. It’s our hope that this guide has helped shed light on this concept and provided you with the tools you need to navigate Python’s absolute value functions with ease. | 2,794 | 12,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-26 | latest | en | 0.796053 |
https://www.equationsworksheets.net/solve-equations-by-graphing-worksheet/ | 1,709,516,119,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00059.warc.gz | 763,867,065 | 15,343 | # Solve Equations By Graphing Worksheet
Solve Equations By Graphing Worksheet – The aim of Expressions and Equations Worksheets is for your child to be able to learn more efficiently and effectively. The worksheets include interactive exercises as well as problems determined by the sequence of operations. With these worksheets, kids can grasp both simple as well as complex concepts in brief amount of period of time. These PDF resources are completely free to download and can be used by your child in order to practice math equations. They are useful for students between 5th and 8th Grades.
These worksheets can be used by students in the 5th through 8th grades. These two-step word problems are constructed using fractions and decimals. Each worksheet contains ten problems. These worksheets are available both online and in printed. These worksheets can be used to practice rearranging equations. Alongside practicing restructuring equations, they can also aid students in understanding the characteristics of equality and reverse operations.
These worksheets are designed for fifth and eight grade students. They are ideal for students who have trouble calculating percentages. You can select from three different kinds of problems. It is possible to work on one-step problems that include decimal or whole numbers or utilize word-based strategies to do fractions or decimals. Each page contains 10 equations. The Equations Worksheets are used by students in the 5th to 8th grade.
These worksheets are a great way for practicing fraction calculations and other concepts in algebra. Some of the worksheets let users to select from three different kinds of problems. It is possible to select the one that is word-based, numerical or a mixture of both. The problem type is also vital, as each will present a different problem kind. Each page contains ten problems, making them a great resource for students from 5th to 8th grade.
The worksheets will teach students about the relationship between variables as well as numbers. These worksheets allow students to test their skills at solving polynomial equations, and discover how to use equations to solve problems in everyday life. If you’re in search of an educational tool that will help you learn about expressions and equations, you can start by exploring these worksheets. These worksheets can teach you about the different kinds of mathematical issues as well as the many symbols that are used to describe them.
These worksheets could be beneficial for students in their beginning grades. These worksheets will help them develop the ability to graph and solve equations. They are great for practice with polynomial variables. They will also help you learn how to factor and simplify these variables. There are many worksheets available to teach kids about equations. Working on the worksheet yourself is the most efficient way to master equations.
There are a variety of worksheets for teaching quadratic equations. There are several worksheets on the different levels of equations for each stage. These worksheets are a great way to solve problems to the fourth level. After you’ve completed the required level and are ready working on other types of equations. Then, you can work on solving the same-level problems. It is possible to, for instance solve the same problem in a more extended form. | 620 | 3,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-10 | longest | en | 0.961127 |
http://www.math.wisc.edu/wiki/index.php?title=Math_567_--_Elementary_Number_Theory&oldid=985 | 1,531,972,392,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590493.28/warc/CC-MAIN-20180719031742-20180719051742-00156.warc.gz | 499,218,174 | 8,469 | # Math 567 -- Elementary Number Theory
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MATH 567
Elementary Number Theory
MWF 1:20-2:10, Van Vleck B119
Professor: Jordan Ellenberg (ellenber@math.wisc.edu) Office Hours: Weds, 2:30-3:30, Van Vleck 323.
Grader: Silas Johnson (snjohnson3@wisc.edu) Office Hours: Thurs, 1:15-2:15, Van Vleck 522. Silas's page, with problem set solutions.
Math 567 is a course in elementary number theory, aimed at undergraduates majoring in math or other quantitative disciplines. A general familiarity with abstract algebra at the level of Math 541 will be assumed, but students who haven't taken 541 are welcome to attend if they're willing to play a little catchup. We will be using William Stein's new (and cheap) textbook Elementary Number Theory: Primes, Congruences, and Secrets, which emphasizes computational approaches to the subject. If you don't need a physical copy of the book, it is available as a free legal .pdf. We will be using the (free, public-domain) mathematical software SAGE, developed largely by Stein, as an integral component of our coursework. There is a useful online tutorial. You can download SAGE to your own computer or use it online.
Topics include some subset of, but are not limited to: Divisibility, the Euclidean algorithm and the GCD, linear Diophantine equations, prime numbers and uniqueness of factorization. Congruences, Chinese remainder theorem, Fermat's "little" theorem, Wilson's theorem, Euler's theorem and totient function, the RSA cryptosystem. Number-theoretic functions, multiplicative functions, Möbius inversion. Primitive roots and indices. Quadratic reciprocity and the Legendre symbol. Perfect numbers, Mersenne primes, Fermat primes. Pythagorean triples, Special cases of Fermat's "last" theorem. Fibonacci numbers. Continued fractions. Distribution of primes, discussion of prime number theorem. Primality testing and factoring algorithms.
Course Policies: Homework will be due on Fridays. It can be turned in late only with advance permission from your grader. It is acceptable to use calculators and computers on homework (indeed, some of it will require a computer) but calculators are not allowed during exams. You are encouraged to work together on homework, but writeups must be done individually.
Many of the problems in this course will ask you to prove things. I expect proofs to be written in English sentences; the proofs in Stein's book are a good model for the level of verbosity I am looking for.
Grading: The grade in Math 567 will be composed of 40% homework, 20% each of three midterms. The last midterm will be take-home, and will be due on the last day of class. There will be no final exam in Math 567.
Syllabus: (This may change as we see what pace works well for the course. All section numbers refer to Stein's book.)
• Sep 3-10: Prime numbers, prime factorizations, Euclidean algorithm and GCD (1.1-1.2)
• Sep 13-17: The integers mod n, Euler's theorem, the phi function (2.1-2.2)
• Sep 20-24: Modular exponentiation, primality testing, and primitive roots (2.4-2.5)
• Sep 27-Oct 1: Public-key cryptography and RSA (3.1-3.4)
• Oct 4 - 8: Algebraic numbers
• Oct 6: First midterm exam
• Oct 11-15: Quadratic reciprocity (4.1-4.4)
• Oct 18-22: Finite and infinite continued fractions (5.1-5.3)
• Oct 25-29: Continued fractions and diophantine approximation (5.4-5.5)
• Nov 1-5: Diophantine equations I: Pell's equation and Lagrange's theorem
• Nov 8-12: Elliptic curves (6.1-6.2)
• Nov 10: Second midterm exam
• Nov 15-19: Applications of elliptic curves (6.3-6.4)
• Nov 22-Dec 3: Diophantine equations II: Fermat, generalized Fermat, and probabilistic methods
• Dec 6-15: advanced topic TBD: maybe a look at the Sato-Tate conjecture?
Homework: Homework is due at the beginning of class on the specified Friday. Typing your homework is not a requirement, but if you don't already know LaTeX I highly recommend that you learn it and use it to typeset your homework. I will sometimes assign extra problems, which I will e-mail to the class list and include here.
• Sep 13 (note this is Monday, not Friday!): 1.1, 1.3, 1.5, 1.7 (use SAGE), 1.8, 1.14.
Problem A: Use SAGE to compute the number of x in [1..N] such that x^2 + 1 is prime, for N = 100, N = 1000, and N = 10000. Let f(N) be the number of such N.
a) Can you formulate a conjecture about the relationship between f(N) and N/log N?
b) What if x^2 + 1 is replaced with x^2 + 2? Can you explain why x^2 + 2 appears less likely to be prime? (Hint: consider x mod 3.)
c) Prove that f(N) is at most (1/2)N+1. (Hint: consider x mod 2.)
d) Give as good an upper bound as you can for f(N).
Note that, despite the evident regularities you'll observe in this problem, we do not even know whether there are infinitely many primes of the form x^2 + 1! You would become very famous if you proved this.
• Sep 17: 2.6 (the formulation of numerical evidence should be done by Sage if you've got Sage working, and by calculator if not; you can use an online tool like this to test whether a number is prime.) 2.8,2.9,2.11,2.12,2.14,2.19
.
• Sep 24: 2.15, 2.16 (note that I presented part a) of this in class), 2.20, 2.23, 2.26.
Problem A: Prove that if n=pq, with p,q prime, then n is not a Carmichael number.
• Oct 1:
Book problems: 2.31 (I will give a hint for this problem later in the week.) 3.4,3.5,3.6
Problem A. Prove that there are infinitely many primes p such that 2 is not a primitive root in Z/pZ. We break this up into steps. Problem A.1. Prove that, if x is an element of Z/nZ, then x^2 is not a primitive root. Problem A.2. Prove that there are infinitely many primes p such that 2 is a square in Z/pZ. Hint: suppose there are only finitely many such primes p_1, .. p_r, and define N = (p_1 .. p_r)^2 - 2. Where can you go from here...? Problem A.3. Give a list of five primes p such that 2 is not a primitive root in Z/pZ (you can use the method of this proof or any other.)
Problem B. Prove that 24 is the largest integer n such that every element of (Z/nZ)^* is a root of x^2-1.
• Oct 8:
Problem A. Give a prime factorization of the Gaussian integer 7+9i. Problem B. We showed in class that Z[i] satisfies a reduction theorem: if n and d are Gaussian integers, then there exists integers q and r such that n = qd = r and Norm(r) < Norm(d). But (by contrast with the case of Z) this d may not be unique. In some contexts it is better to be able to choose r uniquely, even if this means letting r have norm greater than Norm(d).
B.1. When d = 1+2i, show that, for each n in Z[i], there is a unique pair (q,r) in Z[i] such that n = qd+r and r is contained in the set {0,1,2,3,4}. For instance, i can be written as i(1+2i) + 2, so we say i reduces to 2 mod (1+2i).
B.2. Show that if n is an integer in Z, the reduction of n mod (1+2i) is equal to its reduction mod 5.
Problem C. Let's try to figure out how to define "phi(d)" for a Gaussian integer d. Suppose S is a set of Gaussian integers such that every n in Z[i] can be written uniquely as qd+r, with q in Z[i] and r in S. (So for instance when d=1+2i, we showed in problem B that we can take S to be {0...4}. It would also be OK to take S to be {1..5} or {0,i,2i,1+i,1+2i}. In fact, it turns out that S has to be a set of size Norm(d) (I might or might not prove this in class; if not, feel free just to accept it.)
Now define phi(d) to be the number of elements s of S such that s and d are coprime.
C.1. Compute phi(1+2i) and phi(3). C.2. Prove that the value of phi(d) does not depend on the choice of S. C.3. Prove that every n in Z[i] which is coprime to 3 satisfies n^phi(3) = 1 mod 3; that is, Euler's theorem holds in this case. (You can prove this by direct computation; of course, if you want, you are welcome to prove that Euler's theorem holds for Z[i] in general, adapting the proof in Stein's book or the one we gave in class.)
Problem D. Express 50005 as the sum of two squares. | 2,255 | 7,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-30 | latest | en | 0.856501 |
https://codedump.io/share/s3pAmWJ1PoxM/1/calculate-data-attributes-and-output-into-html | 1,527,459,661,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870470.67/warc/CC-MAIN-20180527205925-20180527225925-00503.warc.gz | 537,421,755 | 8,796 | user2931494 - 1 year ago 69
jQuery Question
# Calculate data attributes and output into HTML
I want to calculate all
`<select>`
data attributes and output the total into a
`<span>`
.
Here are the selects:
``````<select name="product-1" id="product-1">
<option value="0" data-price="0">0 (0.00\$)</option>
<option value="10" data-price="52.39">10 (52.39\$)</option>
<option value="20" data-price="61.93">20 (61.93\$)</option>
</select>
<select name="product-2" id="product-2">
<option value="0" data-price="0">0 (0.00\$)</option>
<option value="10" data-price="57.92">10 (57.92\$)</option>
<option value="20" data-price="77.81">20 (77.81\$)</option>
</select>
<select name="product-3" id="product-3">
<option value="0" data-price="0">0 (0.00\$)</option>
<option value="10" data-price="64.63">10 (64.63\$)</option>
<option value="20" data-price="84.07">20 (84.07\$)</option>
</select>
``````
Here is the span:
``````<span id="total">4.99</span>
``````
Here is the jQuery:
`````` \$(document).ready(function() {
delivery = 4.99, \$("select").change(function() {
total = \$(this).find("option:selected").data("price") + delivery, \$("#total").html(total.toFixed(2))
})
});
``````
The problem:
I can't figure it out, how to sum all of the
`<select>`
attributes into one total. With my current code, it only output's the last changed option.
The easiest way to do this is to loop through the selected options of each `select` when a value is changed and keep a running total of the selected prices. Try this:
``````\$(document).ready(function() {
var delivery = 4.99;
\$("select").change(function() {
var total = delivery;
\$('select option:selected').each(function() {
total += parseFloat(\$(this).data('price'));
});
\$("#total").html(total.toFixed(2))
})
});``````
``````<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<select name="product-1" id="product-1">
<option value="0" data-price="0">0 (0.00\$)</option>
<option value="10" data-price="52.39">10 (52.39\$)</option>
<option value="20" data-price="61.93">20 (61.93\$)</option>
</select>
<select name="product-2" id="product-2">
<option value="0" data-price="0">0 (0.00\$)</option>
<option value="10" data-price="57.92">10 (57.92\$)</option>
<option value="20" data-price="77.81">20 (77.81\$)</option>
</select>
<select name="product-3" id="product-3">
<option value="0" data-price="0">0 (0.00\$)</option>
<option value="10" data-price="64.63">10 (64.63\$)</option>
<option value="20" data-price="84.07">20 (84.07\$)</option>
</select>
<span id="total">4.99</span>``````
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download | 819 | 2,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-22 | latest | en | 0.479338 |
https://proprogramming.org/python-program-reverse-number/ | 1,611,260,399,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527850.55/warc/CC-MAIN-20210121194330-20210121224330-00734.warc.gz | 532,442,605 | 12,945 | # Python Program to Reverse a number
Here we will write a program in Python to Reverse a number using while loop. Lets start with the steps:
### Step-wise Solution:
1. Take the value of the integer and store in a variable.
2. Using a while loop, get each digit of the number and store the reversed number in another variable.
3. Print the reverse of the number.
4. Exit.
Now lets write the program for the same.
## Python Program to Reverse a number:
```#Program to reverse a given number
n=int(input("Enter number: "))
rev=0
while(n>0):
dig=n%10
rev=rev*10+dig
n=n//10
print("Reverse of the number:",rev)```
### OUTPUT:
```Enter number: 124
Reverse of the number: 421```
Comment for any suggestion or concerns. | 183 | 721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-04 | latest | en | 0.65005 |
https://www.studymode.com/essays/Freezing-Point-Depression-And-Boiling-Point-67935286.html | 1,621,286,621,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991870.70/warc/CC-MAIN-20210517211550-20210518001550-00450.warc.gz | 1,051,226,008 | 25,766 | # Freezing Point Depression And Boiling Point
Topics: Freezing-point depression, Van 't Hoff factor, Physical chemistry Pages: 24 (491 words) Published: February 15, 2015
COLLIGATIVE
PROPERTIES:
FREEZING POINT
DEPRESSION AND
BOILING POINT
ELEVATION
DAY 1 – 04 FEBRUARY 2015
Colligative Properties
Depends on the NUMBER of solute, not on the nature of
solute particles
Freezing Point Depression
Boiling Point Elevation
Vapor Pressure Lowering
Osmotic Pressure
Electrolyte and Nonelectrolytes
Electrolytes
•Separates in water forming a
solution that conducts electric
current
•IONIC COMPOUNDS
Non- electrolytes
• does not allow the flow of an electric current
• COVALENT COMPOUNDS
Freezing
Point
Depressio
n
Freezing Point
Depression
Adding a solute to a solvent decreases freezing point of the solvent Amount of
solute
Freezing Point of
solvent
Tf° (pure solvent) > Tf (solution) Freezing point of pure
solvent – 0o
Tf = Tf° - Tf
Freezing Point
Depression
Formula (non-electrolytes):
Temperature
Freezing Point Depression constant
(1.860 C /m)
Molality (mol/kg)
Freezing Point
Depression
Formula (electrolytes):
ΔTf = i Kf m
ΔTf
i
freezing point temperature
van ‘t Hoff factor (sum of
subscripts)
Kf
m
Freezing Point Depression
constant (1.860 C /m)
Molality (mol/kg)
Freezing Point
Depression
EXAMPLES
CaCl2
NaCl
Na
1
Cl
+
i =2
Ca
1
ΔTf
Cl2
+
1
i =3
2
Kf
i
Na PO
3
4
freezing point temperature
Na
(PO )
van ‘t Hoff
3 factor (sum of4
subscripts)
3
+
1
Freezing Point Depression
constant (1.860 C /m)
m
i =4
Molality (mol/kg)
PRACTICE
What is the new freezing point of 200 g
of water (Kf = -1.86 oC) if195 g of
sucrose (C12H22O11) are added to it?
∆Tf = -5.301°C
PRACTICE
What is the molality of barium sulfate
(Ba2SO4) with a freezing a point of
1.12°C?
Molality = 0.20 m
PRACTICE
A solution of 3.39 g of an unknown compound in
10.00 g of water has a freezing point of 7.31°C.
The solution does not conduct electricity. What
is the molar mass of the compound?
Molar mass = 86.26 g/mol
Boiling
Point
Elevation
Boiling Point Elevation
Temperature difference between a solution’s boiling point and a pure solvent’s boiling point
For nonelectrolytes:
Amount of
solute
Boiling Point
Elevation
Tb (solution) > Tb° (pure solvent) Boiling Point of pure
solvent - 100°C
Tb = Tb° + Tb (+)
Boiling Point Elevation
Formula (non-electrolyte):
Boiling point of elevation
Molal boiling elevation constant
(0.5120 C /m)
Molality (mol/kg)
Boiling Point Elevation
Formula (electrolytes):
ΔTb = i Kb m
ΔTb
freezing point temperature
i
van ‘t Hoff factor (sum of
subscripts)
Kb
Molal Boiling Point Elevation
m
Molality (mol/kg)
constant (0.5120 C /m)
PRACTICE
What will be the boiling point of an
aqueous solution containing 55.0 g of
glycerol, C3H5(OH)3, and 250 g of water?
Kb(H2O)= 0.512 °c/m
∆Tb = 1.23°C
PRACTICE
What is the boiling point of a solution
containing 34.3 g of magnesium nitrate
dissolved in 0.107 kg of water?
∆Tb = 3.3°C
PRACTICE
A solution containing 28.4 g of sodium
bromate dissolved in water. Find the
mass of the solvent if the boiling point of
the solution is 8.5°C.
Mass of solvent = 0.034 kg | 1,045 | 3,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-21 | latest | en | 0.797803 |
https://mtp.tools/converters/data-storage/terabit-to-exabit-calculator | 1,590,522,712,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391309.4/warc/CC-MAIN-20200526191453-20200526221453-00098.warc.gz | 466,220,532 | 5,539 | # Terabit to Exabit calculator (Tb to Eb)
Convert terabits to exabits (Tb to Eb) by typing the amount of terabits in the input field below and then clicking in the "Convert" button. If you want to convert from exabits to terabits, you can use our exabit to terabit converter.
## Formula
Formula used to convert Tb to Eb:
F(x) = x / 1000000
For example, if you want to convert 1 Tb to Eb, just replace x by 1 [Tb]:
1 Tb = 1 / 1000000 = 0.000001 Eb
## Steps
1. Divide the amount of terabits by 1000000.
2. The result will be expressed in exabits.
## Terabit to Exabit Conversion Table
The following table will show the most common conversions for Terabits (Tb) to Exabits (Eb):
Terabits (Tb) Exabits (Eb)
0.001 Tb 0.000000001 Eb
0.01 Tb 0.00000001 Eb
0.1 Tb 0.00000010000000000000001 Eb
1 Tb 0.000001 Eb
2 Tb 0.000002 Eb
3 Tb 0.000003 Eb
4 Tb 0.000004 Eb
5 Tb 0.000005 Eb
6 Tb 0.000006 Eb
7 Tb 0.000007 Eb
8 Tb 0.000008 Eb
9 Tb 0.000009 Eb
10 Tb 0.00001 Eb
20 Tb 0.00002 Eb
30 Tb 0.00003 Eb
40 Tb 0.00004 Eb
50 Tb 0.00005 Eb
60 Tb 0.00006 Eb
70 Tb 0.00007 Eb
80 Tb 0.00008 Eb
90 Tb 0.00009 Eb
100 Tb 0.0001 Eb
A terabit is a unit of measurement for digital information and computer storage. The prefix tera (which is expressed with the letter T) is defined in the International System of Units (SI) as a multiplier of 10^12 (1 trillion). Therefore, 1 terabit is equal to 1,000,000,000,000 bits and equal to 1,000 gigabits. The symbol commonly used to represent a terabit is Tb (sometimes as Tbit).
A exabit is a unit of measurement for digital information and computer storage. The prefix exa (which is expressed with the letter E) is defined in the International System of Units (SI) as a multiplier of 10^18 (1 quintillion). Therefore, 1 exabit is equal to 1,000,000,000,000,000,000 bits and equal to 1,000 petabits. The symbol commonly used to represent a exabit is Eb (sometimes as Ebit).
## FAQs for Terabit to Exabit converter calculator
### What is Terabit to Exabit converter calculator?
Terabit to Exabit converter is a free and online calculator that converts Terabits to Exabits.
### How do I use Terabit to Exabit converter?
You just have to insert the amount of Terabits you want to convert and press the "Convert" button. The amount of Exabits will be outputed in the input field below the button.
### Which browsers are supported?
All mayor web browsers are supported, including Internet Explorer, Microsoft Edge, Firefox, Chrome, Safari and Opera.
### Which devices does Terabit to Exabit converter work on?
Terabit to Exabit converter calculator works in any device that supports any of the browsers mentioned before. It can be a smartphone, desktop computer, notebook, tablet, etc. | 806 | 2,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-24 | latest | en | 0.79578 |
https://erlang.org/pipermail/erlang-questions/2011-November/062551.html | 1,623,686,650,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00116.warc.gz | 229,394,720 | 2,629 | # [erlang-questions] [Erlang-Question] Is Erlang good for Matrix manipulation and AI related algorithms
Barco You barcojie@REDACTED
Thu Nov 17 09:09:21 CET 2011
```Hi Peer,
Thank you for your links, but the formula is seemingly not right!
The number of paths to [n,m,k] should equals to (n+m+k)! / (n!*(m+k)!).
I write down a snippet of erlang cold as following. Please correct me if I
did it in an inefficient way.
-module(lattice).
-export([build/1]). %%One list as argument specifying the dimensions of
the lattice
build([]) ->
[];
build([H|T]) ->
build([H|T], []).
build([H|[]], D) ->
build_innerlist(H, D, [] );
build([H|T], D) ->
Bu = fun(X) -> build(T, [X|D]) end, %%Any better way here?
build_outterlist(H, Bu, []).
build_innerlist(0, _D, L) ->
L;
build_innerlist(N, D, L) when N > 0 ->
build_innerlist(N-1, D,
[factorial(N+lists:sum(D))/(factorial(N)*factorial(lists:sum(D))) | L]).
build_outterlist(0, _X, L) ->
L;
build_outterlist(N, X, L) when N > 0 ->
build_outterlist(N-1, X, [X(N)|L]).
factorial(0) ->
1;
factorial(N) when N > 0 ->
factorial(N,1).
factorial(0, V) ->
V;
factorial(N, V) ->
factorial(N-1, N*V).
Eshell V5.8.4 (abort with ^G)
1> lattice:build([2,2,2]).
[[[3.0,6.0],[4.0,10.0]],[[4.0,10.0],[5.0,15.0]]]
On Thu, Nov 17, 2011 at 12:33 AM, Peer Stritzinger <peerst@REDACTED> wrote:
> Forget what I said about Catalan Numbers (these count paths staying
> below the diagonal).
>
> What you need is this: http://mathworld.wolfram.com/LatticePath.html
>
> From this you could easily derive that the number of paths in a n x m
> x k cuboid is:
>
> (n+m+k)! / (n!*m!*k!)
>
> Well that was easier than I thought.
> -- Peer
>
> On Tue, Nov 15, 2011 at 6:19 AM, Barco You <barcojie@REDACTED> wrote:
> > Hi Peer,
> > Could you please show me the one-line operation in erlang? Thank you!
> >
> > Regards,
> > Barco
> >
> > On Tue, Nov 15, 2011 at 1:18 PM, Peer Stritzinger <peerst@REDACTED>
> wrote:
> >>
> >> On Thu, Nov 10, 2011 at 6:39 AM, Barco You <barcojie@REDACTED> wrote:
> >>
> >> > I hope to know there are how many paths across a specific point in a
> >> > cubic
> >> > lattice if we walk from the origin to the far-most diagonal point.
> >>
> >> It is not difficult to derive a closed form for this number, then
> >> It'll be a fast O(1) operation in one line in any language.
> >>
> >> Cheers,
> >> -- Peer
> >
> >
>
-------------- next part --------------
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``` | 868 | 2,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-25 | latest | en | 0.737489 |
http://stackoverflow.com/questions/3108497/data-structure-and-algorithms-for-a-directed-cyclic-graph-f/3121358 | 1,435,955,049,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375096209.79/warc/CC-MAIN-20150627031816-00120-ip-10-179-60-89.ec2.internal.warc.gz | 253,203,249 | 17,691 | # Data structure and algorithms for a directed cyclic graph (F#)
I'm trying to analyse an application where the assembly references should be a directed-acyclic-graph, but aren't. There is also a related problem of sub-assemblies referencing different versions of one sub-sub-assembly (think Escher...)
What I want to do is analyse each assembly-subassembly pair and build up a picture of where things are wrong.
I need some guidance on what would be a good data structure for this. I'm not too sure that I can build up an immutable one, but I don't mind having it mutable internally then transformed to immutable at the end.
The other part of the question is what kind of algorithms I should use for filling the data structure, and also afterwards for 'analysing' the problems.
-
You can just use NDepend, it analyzes your assemblies and detects dependency cycles.
If you really want to do this yourself, I'd use QuickGraph to model the dependency graphs, it also includes graph algorithms, like topological sort.
-
Thanks, the situation is actually slightly more complicated that my question implies. We have a home-made system which 'tracks' the dependencies, but it is patently not up-to-date. My aim is to find all the places it is not working. So your second suggestion looks to be more useful for me. – Benjol Jun 25 '10 at 6:17
I don't mind having it mutable internally then transformed to immutable at the end.
You may well find it easier to use immutable data structures throughout. In particular, you can easily represent a graph as a `Map` from source nodes to sets of destination nodes. For a topological sort, you want efficient access to the source nodes of a destination node so you may want to augment your graph with another `Map` going in the opposite direction.
I just implemented this in F# and the topological sort is just 12 lines of code... :-)
-
Jon, I'm in need of a topological sort function. Any chance you can share? – Cameron Taggart Jan 17 '14 at 18:06
I wrote an article about it here: fsharpnews.blogspot.co.uk/2010/10/… – Jon Harrop Jan 17 '14 at 21:04
What you want to do is called "Topological sorting". Wikipedia has a good overview:
http://en.wikipedia.org/wiki/Topological_sort
- | 507 | 2,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2015-27 | latest | en | 0.943196 |
https://wiki.analytica.com/index.php?title=StudentT | 1,568,983,532,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574018.53/warc/CC-MAIN-20190920113425-20190920135425-00199.warc.gz | 717,938,605 | 7,567 | # Student's t-distribution
(Redirected from StudentT)
Release: 4.6 • 5.0 • 5.1 • 5.2
The Student's t-distribution describes the deviation of a sample mean from the true mean when the samples are generated by a normally distributed process. It is a continuous, unbounded, symmetric and unimodal distribution.
The statistic
t = (m - u)/(s*Sqrt(n))
where m is the sample mean, u the actual mean, s the sample standard deviation, and n the sample size, is distributed according to the Student's t-distribution with n - 1 degrees of freedom. The parameter, «dof», is the degrees of freedom. Student's t-distributions are bell-shaped, much like a normal distribution, but with heavier tails, especially for smaller degrees of freedom. When n = 1, it is known as the Cauchy distribution. For efficiency reasons, when a latin-hypercube sampling method is selected, psuedo-latin-hypercube method is used to sample the Student-T, which samples from the T-distribution, but does not guarantee a perfect latin spread of the samples.
## Functions
### StudentT( dof, over )
The distribution function. Use this to specify that a chance variable or uncertain quantity has a Student's t-distribution with «dof» degrees of freedom.
Use the optional «over» parameter to create independent and identically distributed quantities over one or more indexes.
### DensStudentT(x, dof, over)
The probability density at «x», given by
$p(x) = { {\Gamma\left({ {d+1}\over 2}\right)}\over {\sqrt{\pi d} \Gamma\left( d\over 2 \right) }} \left( 1 + { x^2 \over d} \right)^{-{ {d+1}\over 2} }$
where $d$ is «dof», and $\Gamma(x)$ is GammaFn.
### CumStudentT(x, dof, over)
The cumulative density up to «x», i.e., the probability that the outcome is less than or equal to «x».
$F(x) = { {\Gamma\left({ {d+1}\over 2}\right)}\over {\sqrt{\pi d} \Gamma\left( d\over 2 \right) }} \int_{\infty}^x \left( 1 + { t^2 \over d} \right)^{-{ {d+1}\over 2} } dt$
### CumStudentTInv(p, dof, over)
The inverse cumulative probability function, aka quantile function. This is value x at which the area under the probability density graph falling at or to the left of x is «p».
## Statistics
When 0<dof<=1, all moments are undefined.
The theoretical statistics (i.e., in the absence of sampling error) when dof>1 are as follows.
• Mean = Mode = Median = 0
• Variance = $\left\{\begin{array}{ll} \infty & \mbox{when } 1 < dof \leq 2 \\ dof / (dof-2) & \mbox{when } dof>2\end{array}\right.$
• Skewness = 0, when dof>3.
• Kurtosis = $\left\{\begin{array}{ll} \infty & \mbox{when } 2 < dof \leq 4 \\ 6 / (dof-4) & \mbox{when } dof>4\end{array}\right.$
## Parameter Estimation
If you want to estimate the parameter from sample data X indexed by I, you can use the following estimation formula provided that Variance(X, I) > 1:
«dof» := 2*Variance(X, I)/(Variance(X, I) - 1) | 842 | 2,848 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-39 | latest | en | 0.730526 |
https://www.gradesaver.com/textbooks/math/geometry/geometry-common-core-15th-edition/chapter-9-transformations-get-ready-page-541/3 | 1,713,232,660,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817036.4/warc/CC-MAIN-20240416000407-20240416030407-00550.warc.gz | 712,584,257 | 13,352 | ## Geometry: Common Core (15th Edition)
$\triangle RTS$
Let's first list the corresponding congruent angles: $\angle A ≅ \angle R$ $\angle B ≅ \angle T$ $\angle C ≅ \angle S$ Therefore, $\triangle ABC ≅ \triangle RTS$. | 63 | 219 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-18 | latest | en | 0.512278 |
https://www.studiestoday.com/download-book/ncert-class-6-maths-symmetry-174039.html | 1,669,838,939,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710771.39/warc/CC-MAIN-20221130192708-20221130222708-00597.warc.gz | 1,040,419,795 | 15,898 | # NCERT Class 6 Maths Symmetry
Read and download NCERT Class 6 Maths Symmetry chapter in NCERT book for Class 6 Mathematics. You can download latest NCERT eBooks for 2022 chapter wise in PDF format free from Studiestoday.com. This Mathematics textbook for Class 6 is designed by NCERT and is very useful for students. Please also refer to the NCERT solutions for Class 6 Mathematics to understand the answers of the exercise questions given at the end of this chapter
## Symmetry Class 6 Mathematics NCERT
Class 6 Mathematics students should refer to the following NCERT Book chapter Symmetry in standard 6. This NCERT Book for Grade 6 Mathematics will be very useful for exams and help you to score good marks
### Symmetry NCERT Class 6
Symmetry
13.1 Introduction
Symmetry is quite a common term used in day to day life. When we see certain figures with evenly balanced proportions, we say, “They are symmetrical”. These pictures of architectural marvel are beautiful because of their symmetry.
Suppose we could fold a picture in half such that the left and right halves match exactly then the picture is said to have line symmetry (Fig 13.1). We can see that the two halves are mirror images of each other. If we place a mirror on the fold then the image of one side of the picture will fall exactly on the other side of the picture. When it happens, the fold, which is the mirror line, is a line of symmetry (or an axis of symmetry) for the picture.
The shapes you see here are symmetrical. Why?
When you fold them along the dotted line, one half of the drawing would fit exactly over the other half.
How do you name the dotted line in the figure?
Where will you place the mirror for having the image exactly over the other half of the picture?
The adjacent figure 13.2 is not symmetrical. Can you tell ‘why not’?
Do This
Take a piece of paper. Fold it in half.
Spill a few drops of ink on one half side.
Now press the halves together.
What do you see?
Is the resulting figure symmetric? If yes, where is the line of symmetry? Is there any other line along which it can be folded to produce two identical parts?
Try more such patterns.
Inked-string patterns
Fold a paper in half. On one half-portion, arrange short lengths of string dipped in a variety of coloured inks or paints. Now press the two halves. Study the figure you obtain. Is it symmetric? In how many ways can it be folded to produce two identical halves?
List a few objects you find in your class room such as the black board, the table, the wall, the textbook, etc. Which of them aresymmetric and which are not? Can you identify the lines of symmetry for those objects which are symmetric?
Please refer to attached file for NCERT Class 6 Maths Symmetry
NCERT Class 6 Maths Knowing our Numbers
NCERT Class 6 Maths Mensuration
NCERT Class 6 Maths Algebra
NCERT Class 6 Maths Ratio and Proportion
NCERT Class 6 Maths Symmetry
NCERT Class 6 Maths Practical Geometry
NCERT Class 6 Maths Whole Numbers
NCERT Class 6 Maths Playing with Numbers
NCERT Class 6 Maths Basic Geometrical Ideas
NCERT Class 6 Maths Understanding Elementary Shapes
NCERT Class 6 Maths Integers
NCERT Class 6 Maths Fractions
NCERT Class 6 Maths Decimals
NCERT Class 6 Maths Data Handling
NCERT Class 6 Maths Answers to all Chapters
NCERT Class 6 Maths Brain Teasers | 792 | 3,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2022-49 | latest | en | 0.911806 |
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# Propositional logic
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### Transcript of "Propositional logic"
1. 1. Rushdi Shams, Dept of CSE, KUET, Bangladesh 1 Knowledge Representation Propositional Logic Artificial Intelligence Version 2.0 There are 10 types of people in this world- who understand binary and who do not understand binary
2. 2. Rushdi Shams, Dept of CSE, KUET, Bangladesh 2 Propositional Logic
3. 3. Rushdi Shams, Dept of CSE, KUET, Bangladesh 3 Introduction Need formal notation to represent knowledge, allowing automated inference and problem solving. One popular choice is use of logic. Propositional logic is the simplest. Symbols represent facts: P, Q, etc.. These are joined by logical connectives (and, or, implication) e.g., P Λ Q; Q R Given some statements in the logic we can deduce new facts (e.g., from above deduce R)
4. 4. Rushdi Shams, Dept of CSE, KUET, Bangladesh 4 Syntactic Properties of Propositional Logic If S is a sentence, S is a sentence (negation) If S1 and S2 are sentences, S1 S2 is a sentence (conjunction) If S1 and S2 are sentences, S1 S2 is a sentence (disjunction) If S1 and S2 are sentences, S1 S2 is a sentence (implication) If S1 and S2 are sentences, S1 S2 is a sentence (bi-conditional)
5. 5. Rushdi Shams, Dept of CSE, KUET, Bangladesh 5 Semantic Properties of Propositional Logic S is true iff S is false S1 S2 is true iff S1 is true and S2 is true S1 S2 is true iff S1is true or S2 is true S1 S2 is true iff S1 is false or S2 is true i.e., is false iff S1 is true and S2 is false S1 S2 is true iff S1 S2 is true and S2 S1 is true
6. 6. Rushdi Shams, Dept of CSE, KUET, Bangladesh 6 Truth Table for Connectives
7. 7. Rushdi Shams, Dept of CSE, KUET, Bangladesh 7 Model of a Formula If the value of the formula X holds 1 for the assignment A, then the assignment A is called model for formula X. That means, all assignments for which the formula X is true are models of it.
8. 8. Rushdi Shams, Dept of CSE, KUET, Bangladesh 8 Model of a Formula
9. 9. Rushdi Shams, Dept of CSE, KUET, Bangladesh 9 Model of a Formula: Can you do it?
10. 10. Rushdi Shams, Dept of CSE, KUET, Bangladesh 10 Satisfiable Formulas If there exist at least one model of a formula then the formula is called satisfiable. The value of the formula is true for at least one assignment. It plays no rule how many models the formula has.
11. 11. Rushdi Shams, Dept of CSE, KUET, Bangladesh 11 Satisfiable Formulas
12. 12. Rushdi Shams, Dept of CSE, KUET, Bangladesh 12 Valid Formulas A formula is called valid (or tautology) if all assignments are models of this formula. The value of the formula is true for all assignments. If a tautology is part of a more complex formula then you could replace it by the value 1.
13. 13. Rushdi Shams, Dept of CSE, KUET, Bangladesh 13 Valid Formulas
14. 14. Rushdi Shams, Dept of CSE, KUET, Bangladesh 14 Unsatisfiable Formulas A formula is unsatisfiable if none of its assignment is true in no models
15. 15. Rushdi Shams, Dept of CSE, KUET, Bangladesh 15 Logical equivalence Two sentences are logically equivalent iff true in same models: α ≡ ß iff α╞ β and β╞ α
16. 16. Rushdi Shams, Dept of CSE, KUET, Bangladesh 16 Deduction: Rule of Inference 1. Either cat fur was found at the scene of the crime, or dog fur was found at the scene of the crime. (Premise) C v D
17. 17. Rushdi Shams, Dept of CSE, KUET, Bangladesh 17 Deduction: Rule of Inference 2. If dog fur was found at the scene of the crime, then officer Thompson had an allergy attack. (Premise) D → A
18. 18. Rushdi Shams, Dept of CSE, KUET, Bangladesh 18 Deduction: Rule of Inference 3. If cat fur was found at the scene of the crime, then Macavity is responsible for the crime. (Premise) C → M
19. 19. Rushdi Shams, Dept of CSE, KUET, Bangladesh 19 Deduction: Rule of Inference 4. Officer Thompson did not have an allergy attack. (Premise) ¬ A
20. 20. Rushdi Shams, Dept of CSE, KUET, Bangladesh 20 Deduction: Rule of Inference 5. Dog fur was not found at the scene of the crime. (Follows from 2 D → A and 4. ¬ A). When is ¬ A true? When A is false- right? Now, take a look at the implication truth table. Find what is the value of D when A is false and D → A is true ¬ D
21. 21. Rushdi Shams, Dept of CSE, KUET, Bangladesh 21 Rules for Inference: Modus Tollens If given α → β and we know ¬β Then ¬α
22. 22. Rushdi Shams, Dept of CSE, KUET, Bangladesh 22 Deduction: Rule of Inference 6. Cat fur was found at the scene of the crime. (Follows from 1 C v D and 5 ¬ D). When is ¬ D true? When D is false- right? Now, take a look at the OR truth table. Find what is the value of C when D is false and C V D is true C
23. 23. Rushdi Shams, Dept of CSE, KUET, Bangladesh 23 Rules for Inference: Disjunctive Syllogism If given α v β and we know ¬α then β If given α v β and we know ¬β then α
24. 24. Rushdi Shams, Dept of CSE, KUET, Bangladesh 24 Deduction: Rule of Inference 7. Macavity is responsible for the crime. (Conclusion. Follows from 3 C → M and 6 C). When is C → M true given that C is true? Take a look at the Implication truth table. M
25. 25. Rushdi Shams, Dept of CSE, KUET, Bangladesh 25 Rules for Inference: Modus Ponens If given α → β and we know α Then β
26. 26. Rushdi Shams, Dept of CSE, KUET, Bangladesh 26 References Artificial Intelligence: A Modern Approach (2nd Edition) by Russell and Norvig Chapter 7 http://www.iep.utm.edu/p/prop-log.htm#H5
1. #### A particular slide catching your eye?
Clipping is a handy way to collect important slides you want to go back to later. | 1,825 | 5,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2015-32 | latest | en | 0.837101 |
http://study.com/academy/topic/ny-regents-quadrilaterals-help-and-review.html | 1,511,508,442,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807146.16/warc/CC-MAIN-20171124070019-20171124090019-00486.warc.gz | 277,362,650 | 23,622 | # Ch 5: NY Regents - Quadrilaterals: Help and Review
The Quadrilaterals chapter of this NY Regents Exam - Geometry Help and Review course is the simplest way to master quadrilaterals for the NY Regents Exam. This chapter uses simple and fun videos that are about five minutes long, plus lesson quizzes and a chapter exam to ensure students learn the essentials of quadrilaterals.
## Who's it for?
Anyone who needs help learning or mastering geometry material for the NY Regents Exam will benefit from taking this course. There is no faster or easier way to learn high school geometry. Among those who would benefit are:
• Students who have fallen behind in understanding theorems and proofs, measurements of area or angles, or other quadrilaterals topics
• Students who struggle with learning disabilities or learning differences, including autism and ADHD
• Students who prefer multiple ways of learning math (visual or auditory)
• Students who have missed class time and need to catch up
• Students who need an efficient way to learn about quadrilaterals for the NY Regents Exam
• Students who struggle to understand their teachers
• Students who attend schools without extra math learning resources
## How it works:
• Find videos in our course that cover what you need to learn or review.
• Press play and watch the video lesson.
• Refer to the video transcripts to reinforce your learning.
• Test your understanding of each lesson with short quizzes.
## Why it works:
• Study Efficiently: Skip what you know, review what you don't.
• Retain What You Learn: Engaging animations and real-life examples make topics easy to grasp.
• Be Ready on Test Day: Use the Quadrilaterals chapter exam to be prepared.
• Get Extra Support: Ask our subject-matter experts any quadrilaterals question. They're here to help!
• Study With Flexibility: Watch videos on any web-ready device.
## Students will review:
This chapter helps students review the concepts in a quadrilaterals unit of a standard high school geometry course. Topics covered include:
• Definition and properties of various quadrilaterals, including kites and squares
• Measurements for the area of parallelograms, rhombuses, rectangles and trapezoids
• Division of line segments into equal parts
• Heron's formula
Final Exam
Chapter Exam
### Earning College Credit
Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 543 | 2,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-47 | latest | en | 0.898546 |
https://www.instructables.com/community/How-possible-would-it-be-to-build-a-pedal-powered-/ | 1,606,804,440,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141652107.52/warc/CC-MAIN-20201201043603-20201201073603-00027.warc.gz | 655,007,848 | 13,023 | 186Views5Replies
# How possible would it be to build a pedal-powered mobile toast shop? Answered
I am thinking to build a bike-with trailer that powers one or more toasters, by charging a battery then releasing that power. Do the electrical among you believe this is practical? Could there be enough lekky? What considerations should I have? Also - a lot of instructions for this on the web are based on US power supply. Does anyone have any links to UK based instructions.
Tags:
## Discussions
The forums are retiring in 2021 and are now closed for new topics and comments.
So... if I was charging batteries (I think running them live off a dynamo, although it sounds cool would be impractical / hard to serve people)
Lets keep it simple.
How long pedalling would it take to charge a battery that would keep one toaster going for 1 hour?
Lemonie is right - to power a toaster you would need to generate about 1kW of human power. But can a human put out that much power - even pedalling hard? The Wikipeida page on Human Power says that a human can exert under half a horsepower for a short time and a Hp is about 750W and so it's looking like you could exert yourself and generate about 300-350 watts. This is going to make what my son calls 'Golden Toast' but what I prefer to call slightly dried bread!
Toasting uses a lot of energy, let's say for example a cheap 2-slice @ 1Kw
A car battery would run toaster elements, but for how long?
50/60 quid standard-ish battery, 12V 60Ah.
1Kw at 12V is 83A, so it'd do you ~ 40 min of toasting.
How much toast, you can work out from that.
An inverter would convert the 12V supply to 240VAC, but you might want 2 batteries and a 24V inverter?
L
As a side note... it doesn't necessarily have to have batteries. You could get to a place, drop down a prop/stand that lifts the back wheel then you can sit on the bike and run the toasters by pedaling, converting to 'lekky' with the dynamo. Toasting the bread as your customers watch :)
I really doubt it would be practical to charge batteries to power the toasters, If you charge the batteries from a power supply (not your bike) you could just use the bike to top up the batteries. But the toasters will drain your batteries really good, so then you have to work extra hard to charge them by pedalling... Do you have a link to the US instructions, Usually they are really easy to run off UK power... | 576 | 2,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-50 | latest | en | 0.965752 |
https://mathematica.stackexchange.com/tags/sound/hot | 1,721,290,965,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00090.warc.gz | 338,047,172 | 24,101 | # Tag Info
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### Weird Export of .M4A format
I found the same problem with version: 11.3.0 for Mac OS X x86 (64-bit) (March 7, 2018) The documentation for M4A says that MMA follows the ISO/IEC 13818-7 standard and that "a few sampling ...
• 5,250 | 1,382 | 5,838 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-30 | latest | en | 0.847946 |
https://metanumbers.com/25163 | 1,638,979,303,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363515.28/warc/CC-MAIN-20211208144647-20211208174647-00164.warc.gz | 466,315,666 | 7,277 | # 25163 (number)
25,163 (twenty-five thousand one hundred sixty-three) is an odd five-digits prime number following 25162 and preceding 25164. In scientific notation, it is written as 2.5163 × 104. The sum of its digits is 17. It has a total of 1 prime factor and 2 positive divisors. There are 25,162 positive integers (up to 25163) that are relatively prime to 25163.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 17
• Digital Root 8
## Name
Short name 25 thousand 163 twenty-five thousand one hundred sixty-three
## Notation
Scientific notation 2.5163 × 104 25.163 × 103
## Prime Factorization of 25163
Prime Factorization 25163
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 25163 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.1331 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 25,163 is 25163. Since it has a total of 1 prime factor, 25,163 is a prime number.
## Divisors of 25163
2 divisors
Even divisors 0 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 25164 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 12582 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 158.628 Returns the nth root of the product of n divisors H(n) 1.99992 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 25,163 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 25,163) is 25,164, the average is 12,582.
## Other Arithmetic Functions (n = 25163)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 25162 Total number of positive integers not greater than n that are coprime to n λ(n) 25162 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2778 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 25,162 positive integers (less than 25,163) that are coprime with 25,163. And there are approximately 2,778 prime numbers less than or equal to 25,163.
## Divisibility of 25163
m n mod m 2 3 4 5 6 7 8 9 1 2 3 3 5 5 3 8
25,163 is not divisible by any number less than or equal to 9.
## Classification of 25163
• Arithmetic
• Prime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Prime Power
• Square Free
## Base conversion (25163)
Base System Value
2 Binary 110001001001011
3 Ternary 1021111222
4 Quaternary 12021023
5 Quinary 1301123
6 Senary 312255
8 Octal 61113
10 Decimal 25163
12 Duodecimal 1268b
20 Vigesimal 32i3
36 Base36 jez
## Basic calculations (n = 25163)
### Multiplication
n×y
n×2 50326 75489 100652 125815
### Division
n÷y
n÷2 12581.5 8387.67 6290.75 5032.6
### Exponentiation
ny
n2 633176569 15932622005747 400912567530611761 10088162936772783742043
### Nth Root
y√n
2√n 158.628 29.3036 12.5948 7.58844
## 25163 as geometric shapes
### Circle
Diameter 50326 158104 1.98918e+09
### Sphere
Volume 6.67384e+13 7.95673e+09 158104
### Square
Length = n
Perimeter 100652 6.33177e+08 35585.9
### Cube
Length = n
Surface area 3.79906e+09 1.59326e+13 43583.6
### Equilateral Triangle
Length = n
Perimeter 75489 2.74173e+08 21791.8
### Triangular Pyramid
Length = n
Surface area 1.09669e+09 1.87768e+12 20545.5
## Cryptographic Hash Functions
md5 7e4ddecd5564e9fe8a83e41821797ca3 9de21030f0e59c89f67b94b9116d25acb17277b4 1ec6cb932d05a872096776c6f65f5253dae0550b49f52ebe35cfaa3bb101d418 d54b8b2db8482aa043b6c8570cc6ad7210e38e9d4b6c91bbdbef7bc80037be1676f2994a812153e38650b69f62656fd084f9f9ed3ec2f8a2c8cb026695467ff9 6d7500e4fb5058f428f9ab3fb711123e12fa2312 | 1,450 | 4,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-49 | latest | en | 0.825507 |
https://www.thestudentroom.co.uk/showthread.php?t=105156 | 1,534,602,515,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213689.68/warc/CC-MAIN-20180818134554-20180818154554-00442.warc.gz | 983,978,299 | 41,129 | You are Here: Home >< Maths
Graph Question Help watch
1. Hi can somebody please tell me what to do on this question?
Part of the graph y=2x^3-1 is drawn
By drawing a suitable line solve the equation 2x^3-2x-3=0
2. (Original post by lucasharrison)
Hi can somebody please tell me what to do on this question?
Part of the graph y=2x^3-1 is drawn
By drawing a suitable line solve the equation 2x^3-2x-3=0
Draw the line, y= -2x-2
solve for x, by finding factors in > 2x^3-2x-3=0
3. (Original post by Vijay1)
Draw the line, y= -2x-2
solve for x, by finding factors in > 2x^3-2x-3=0
no.
draw the line y = 2x + 2. where they intercept, is the x value you want.
4. (Original post by chewwy)
no.
draw the line y = 2x + 2. where they intercept, is the x value you want.
Silly me
5. yes it does, around 1.42
6. (Original post by chewwy)
yes it does, around 1.42
so it does, not an obvious solution > 1.43211
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Chat with other maths applicants | 510 | 1,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-34 | latest | en | 0.908111 |
https://tutorial.eyehunts.com/python/sort-list-of-tuples-by-first-element-python-example-code/ | 1,713,322,013,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817128.7/warc/CC-MAIN-20240417013540-20240417043540-00008.warc.gz | 533,303,302 | 28,950 | # Sort list of tuples by first element Python | Example code
Using the sorted() function or in-place sort are the ways to Sort a list of tuples by the first element in Python. Both methods need to use the key keyword.
Note: `key` should be a function that identifies how to retrieve the comparable element from your data structure. In your case, it is the second element of the tuple, so we access `[1]`.
## Example Sort list of tuples by first element Python
Simple example code sort by the first element in the tuple.
Using sorted method
This method will not modify the original list of tuples.
``````a = [(2, 8), (0, 4), (1, 7)]
# sort by first element in tuple
res = sorted(a, key=lambda tup: tup[0])
print(res)
``````
Output:
Do in-place sort
This method will change the original data.
``````a = [(2, 8), (0, 4), (1, 7)]
a.sort(key=lambda tup: tup[0])
print(a)
``````
Output:
[(0, 4), (1, 7), (2, 8)]
Using the `itemgetter()` function from the `operator` module:
``````from operator import itemgetter
my_list = [(3, 'apple'), (1, 'banana'), (2, 'cherry')]
sorted_list = sorted(my_list, key=itemgetter(0))
print(sorted_list)
``````
Using a list comprehension:
The list comprehension then creates a new list that contains only the second elements of the sorted tuples.
``````my_list = [(3, 'apple'), (1, 'banana'), (2, 'cherry')]
sorted_list = [tup for _, tup in sorted(my_list)]
print(sorted_list)``````
Comment if you have any doubts or suggestions on this Python tuple topic.
Note: IDE: PyCharm 2021.3.3 (Community Edition)
Windows 10
Python 3.10.1
All Python Examples are in Python 3, so Maybe its different from python 2 or upgraded versions. | 462 | 1,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-18 | latest | en | 0.655044 |
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_10&diff=prev&oldid=76455 | 1,620,796,840,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991252.15/warc/CC-MAIN-20210512035557-20210512065557-00535.warc.gz | 130,506,877 | 11,226 | # Difference between revisions of "2013 AMC 8 Problems/Problem 10"
## Problem
What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$
## Solution 1
To find either the LCM or the GCF of two numbers, always prime factorize first.
The prime factorization of $180 = 3^2 \times 5 \times 2^2$.
The prime factorization of $594 = 3^3 \times 11 \times 2$.
Then, to find the LCM, we have to find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is $3^3, 5, 11, 2^2$). Multiply all of these to get 5940.
For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. $3^2 \times 2$ = 18.
Thus the answer = $\frac{5940}{18}$ = $\boxed{\textbf{(C)}\ 330}$.
## Similar Solution
We start off with a similar approach as the original solution. From the prime factorizations, the GCF is $18$.
It is a well known fact that $\gcd(m,n)\times \operatorname{lcm}(m,n)=|mn|$. So we have, $18\times \operatorname{lcm} (180,594)=594\times 180$.
Dividing by $18$ yields $\operatorname{lcm} (180,594)=594\times 10=5940$.
Therefore, $\dfrac{\operatorname{lcm} (180,594)}{\gcd(180,594)}=\dfrac{5940}{18}=\boxed{\textbf{(C)}\ 330}$. | 464 | 1,394 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2021-21 | latest | en | 0.778814 |
https://nl.mathworks.com/matlabcentral/answers/59287-converting-binary-to-decimal | 1,566,549,606,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318243.40/warc/CC-MAIN-20190823083811-20190823105811-00400.warc.gz | 573,696,633 | 18,534 | ## converting binary to decimal
on 19 Jan 2013
### Image Analyst (view profile)
i have a value as S=-5
i converted it to binary by
Cw1=dec2bin(typecast(int16(S),'uint16'),15)
now how to convert again to -5
plz assist
### Image Analyst (view profile)
on 19 Jan 2013
Why are you using uint16 when you have a negative number?????
Anyway,...
S = -5;
Cw1=dec2bin(typecast(int16(S),'uint16'),15)
S_recovered = bin2dec(Cw1)
maxUint16 = intmax('uint16')
maxInt16 = intmax('int16')
if S_recovered > maxInt16
S_recovered = int16(S_recovered - double(maxUint16) - 1)
end
S_recovered will be -5 as a signed 16 bit integer, which is what I assume you want.
### Walter Roberson (view profile)
on 19 Jan 2013
typecast(bin2dec(Cw1), 'int16')
FIR
### FIR (view profile)
on 21 Jan 2013
Walter i get
typecast(bin2dec(Cw1), 'int16')
ans =
0 0 -160 16623
Walter Roberson
### Walter Roberson (view profile)
on 21 Jan 2013
typecast(uint16(bin2dec(Cw1)),'int16') | 318 | 952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-35 | latest | en | 0.648374 |
https://physics.stackexchange.com/questions/595411/two-questions-on-the-nuclear-spin | 1,653,610,034,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00752.warc.gz | 515,937,644 | 66,472 | # Two Questions on the nuclear spin
To those who are familiar with nuclear physics, I have two questions of understanding:
1. Why does it happen that nuclei have a spin greater than 1 (e.g. I=8 at 90Nb)?
2. How can we infer the parity of a nucleus?
My ideas:
1): I have the nuclear shell model in mind. If the levels for protons and neutrons are filled one after the other. In the case of "even-even" nuclei, all levels are fully occupied, so the total spin adds up to 0. If we now have an odd number of protons and neutrons, then according to this model one level would only be half full. If the spins are parallel, this adds up to I=1. But how do I get I=8 (or other integer values)?
To 2): My idea about "even-even" nuclei: The total spin is 0, so the spin wave function is antisymmetric. Since according to the Pauli principle the total wave function must be antisymmetric, it follows that the local wave function is symmetric, i.e. the parity is positive.
Is this argumentation correct?
What about "odd-odd"-nuclei or "even-odd"-nuclei? How can we conclude parity here?
I would be very grateful, if someone could help me. Sorry if my text is not understandable in some places, English is not my mother tongue.
Note that a the total spin of a nucleus includes both the intrinsic spins and the orbital angular momenta of the nucleons. For a simpler example, compare oxygen-16 (which, like all stable even-even nuclei, has ground state spin-parity $$0^+$$) with oxygen-17, which has $$J^P=5/2^+$$. A simple shell-model-ish description would put the ninth neutron in oxygen-17 in an empty orbital around an oxygen-16 core. That ninth orbital must have even parity, since O-16 and O-17 have the same parity, which means that the orbital angular momentum of the ninth neutron is even. But it can't be in an $$s$$-wave, $$\ell=0$$ orbital, because then there is not enough angular momentum to reach $$J=5/2$$. Apparently the ninth neutron in oxygen-17 has orbital angular momentum $$\ell=2$$, and its spin and orbital angular momenta are aligned.
Of course, the wave function for an oxygen nucleus is much more complicated than a core and single orbital. But this should give you an idea about how the quantum numbers have to work out.
$$^{90}$$Nb has 41 protons and 49 neutrons. In the non-deformed single particle shell model according to Wikipedia nuclear shell article (I don't have my textbookds with me), the single particle state for both proton 41 and neutron 49 is $$g\frac{9}{2}$$. That means the total contribution (particle spin plus orbital angular momentum) has a quantum number, $$j=\frac{9}{2}$$. The $$g$$ is the orbital state and has quantum number $$\ell = 4$$. That means the parity of each single particle will be positive (+). Therefore, the total ground state parity is predicted to be +.
The total angular momentum quantum number will be between the sum and the difference of the individual particle $$j$$ so $$|\frac{9}{2}-\frac{9}{2}|\le \mathcal{I} \le\frac{9}{2}+\frac{9}{2} \\ 0 \le \mathcal{I} \le 9 .$$ Consequently, 8+ is totally consistent with the single particle shell model due to the orbital angular momentum contribution of each particle.
To summarize
1. You must look at the total angular momentum quantum number of the extra particle(s) and use the q.n. combination rule to find possible total q.n.
2. You must look at the $$\ell$$ of each extra (odd) particle to determine its parity from $$\left(-1\right)^{\ell}$$. Standard spectroscopic notation of $$s,~p, ~d,~f, ~g, ...$$ has $$\ell= 0,1,2,3,4,...$$ Final parity is the product of the odd particle parities. | 909 | 3,617 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-21 | longest | en | 0.906526 |
https://de.scribd.com/document/415272337/Minor05-Ans-Dlp-NEET18-Pmtcorner-in | 1,600,708,232,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400201826.20/warc/CC-MAIN-20200921143722-20200921173722-00412.warc.gz | 357,387,978 | 84,750 | Sie sind auf Seite 1von 8
# DISTANCE LEARNING PROGRAMME
## LEADER TEST SERIES / JOINT PACKAGE COURSE
TARGET : PRE-MEDICAL 2018
Test Type : Unit Test Test # 05 Test Pattern : NEET-UG
TEST DATE : 03 - 09 - 2017
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 1 3 3 3 4 1 4 1 2 2 4 3 2 1 2 2 2 1 1
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 2 1 2 3 4 2 1 1 4 4 1 1 1 2 4 3 3 3 1 2
Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 2 2 1 3 3 4 3 4 3 4 3 3 1 4 1 2 2 3 1 2
Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans. 2 2 2 4 4 3 2 1 3 4 1 4 2 1 2 4 1 3 1 3
Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans. 1 2 2 3 3 3 3 1 2 3 3 2 4 4 3 1 2 1 2 3
Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans. 2 2 3 4 3 3 2 1 2 4 2 2 4 3 4 4 4 2 3 4
Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Ans. 3 3 3 4 3 2 3 1 1 3 2 3 3 2 2 4 2 3 2 3
Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Ans. 1 3 1 2 1 3 4 2 3 2 4 4 1 1 2 4 1 2 1 4
Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans. 1 3 3 4 3 1 3 2 2 3 2 3 3 2 2 2 2 3 4 3
HINT – SHEET
1. In ground to ground projectile motion, KE R
decreases first (in upward dirn) then increases S ˆi Hjˆ
2
(in downward dirn) because at maximum height
particle has only horizontal velocity. Work done by gravity = F.S
2. Change in KE = P dt mg
u 2 sin 2
= –mgH =
4 2g
(3t 2 2t 1)dt
2
6. V 2g(1 cos )
= 46 J U ˆ U ˆ U ˆ
7. F i j k
1 x y z
3. U Kx2 U x2
2 F (kiˆ kj)
ˆ
4. Power = F.U
S (x 2 x1 )iˆ (y 2 y1 ) ˆj ˆi 2 ˆj
5. F mg( ˆj)
W F S 3k
0999DMD310317005 LTS/HS-1/8
Target : Pre-Medical 2018/NEET-UG/03-09-2017
8. U x2 1
O + mgL = mv 2L O
U (2)2 4 2
2
U' (10 2) 144 VL 2gL
U' = 36U at lowest point :-
KE1 P 2 / 2m 1 m 2 mv 2L
9. TL mg
KE 2 P 2 / 2m 2 m 1 L
TL = 3 mg
5
= 21. W FS
1
23. Vector form of Bio-savart law but need to
10. Angle between acceleration and momentum is
calculate at origin.
less than 90°. So acceleration has two
24. Both exduce magnetic field equal in magnitude
component one is ar to momentum
as current will be divided.
(centripetal) and another is parallel to
momentum (tangential) which increases speed. B1
12. aCP V2 so 1
B2
13. i = 0
f = 20 R
f=5 25. r< B=0
2
=?
R
f R>r>
i t 50 2
2
0i R 2
Number of revolution
25
B = 2 r 4r , r>R
2
15. By COLM 1
mv0 + 0 = (m + M)V B i current density
r
m So as 4
V V0 ...(1)
mM 26.
By CONE
1 1
(m M)V2 kx 2 ...(2) 1mm
2 2
by solving (1) and (2) N
B = 0 I
L
m 2 v 20 N × 2r = L (r = radius of wire)
x
(m M)k
N 1
so
16. Vmax = µrg L 2r
18. TA – mg = m2r 0 I
T1 = mg + m2r ...(1) B=
2r
TB = T2 = m2r ...(2)
TC + mg = m2r 4 107 5 103
B=
T3 = m2r – mg ...(3) 2 0.5
TD = T4 = m2r ...(4) B = 20 × 10–4 T
T1 > T2 > T3 and T2 = T4 B = 2 × 10–3 T
19. Tmax = m2max r 27. F i ( B)
20. By CONE
LTS/HS-2/7 0999DMD310317005
= i × 2r × B 32. M = NI A
= i2rB M = N × I × nR2
= 2irB
L2
28. As particle is released from rest so it will more M=N×I×N×
4n 2 N2
along electric field since || B so Fm = 0 so it L = N × 2nR
will move in a straight line (v || B) L 1
R= M
2nN N
29. Since v || B
For Mmax N = 1 and area of circle is maximum
So, f m = 0
fe = q 33. P = M B
So where M is aloy area vector so P is max.
q in position I and minimum in IV.
a=
m
n = 0 I
36. T = 2n MB
Since speed brone H
2v0 = y = const. = 0
T I M
So, 2v01 = 2x 2y
T0 m
4v 20 2x 20 T1
2m
x x at
T1 2T0
q
3v 0 t
m FC
3v 0 m
t 37.
q
FA
dx
F1 F2 µ 0 I 2
30.
2nd
F12 F22
Fnet
Q
f1 µ I2
dQ = dx Fnet 2 0
2nd
I = (dQ)f
dm = (I) (A) 2µ 0 m P
38. Baxis =
2 4 r 3
dm (dQf)(x )
µ0 m P
Q Bequi =
2 4n r 3
dm f x dx
0 if 0 or 180º
Q 3 so PQ1 and PQ2
f
3 µ0M
31. A charge particle can accelerated but speed 39. B =
4 r 3
can't be charged while a close loop may
experience a sorce in non uniform magnetic 4 107 M
field. 4× 10–5 =
4 (6.4 103 )3
0999DMD310317005 LTS/HS-3/7
Target : Pre-Medical 2018/NEET-UG/03-09-2017
M = 1.05 × 1023 A-m 2
t 0.875 8n1 1
Accordy to fleming left hand rule.
t 0.5 2 n 1 1
Ans. 2
40. = 90° 48. K = AeEa/RT K1 = AeEa1 /RT
f K2 = AeEa2 /RT K3 = AeEa3 /RT
so f = = I B sin
2
f = 1 × 3 × 10–5 K1 K 2 3
= 3 × 10–5 N/m K = on putting all values in this
K3
42. Circle due to maximum area
43. Z = MBsin equation.
M = NIA 2
= 10 × 5 × 12 × 10–4 = (a1+a2–a3)
3
= 6 × 10–2 A-m2, B = 0.3 T
as = 0 2
= 180 80 50 = 140 KJ/mol
Z=0 3
49. [A]t = [A]0 – Kt = 1 – 0.001 × 10 × 60 = 0.4M
44. G
R [B]t = 0.001 × 10 × 60 = 0.6M
Rg = 12, Ig = 3mA 50. A 3B
P° O Pt = P° + 2x
VR P°–x 3x
Rg = P P
Rg RH x t 0
2
VR After long time
RH = Rg
Ig P
O 3P° = P or P° =
RH = 5988 3
q2 P
45. KEmax Pt
m x= 3
so, it is minimum for deutron. 2
46. K = 10–14
P
1 d(A) 1
K(A)2 K = n 3
3 dt t P 3Pt P
3 6
d(A)
= 3×10–14×(0.5)2
dt
= 7.5 × 10–15 M/ P
1
7.5 10 15 K = n 3
1.25 10 16 M / sec t P P
=
60 t
2 2
1 1 1
47. Kt = n1 n1 1 2P
n 1 At A0
K = t n 3 P P
t
1 2 n1 1
Kt 0.5 = 83.314 75
n 1 A n0 1 51. K1 = A.e R500 K2 = A.e R500
1 8n1 1 K2
8.314103
Kt 0.875 = e 8.314500
n 1 A0n1 K1
LTS/HS-4/7 0999DMD310317005
K2 2.303 a
2.303 log 2 59. t= log
K1 K ax
K2
10 0.868 2.303 5
K1 or t = 3
log 34.07s
15 10 3
= 7.38
60. Radioactive decay follows first order kinetics.
1 dCB dCB Therefore,
52. K1 .CA = 2 K1 CA
2 dt dt 0.693 0.693
Decay constant () =
= 2×2×10–3×2 t1/ 2 5730
= 8×10–3Ms–1
Given, R0 = 100 R = 80
1 d NH 3 1 d H 2 O
53.
4 dt 6 dt 2.303 [R]0 2.303 100
and t = log log
d H 2O [R] 0.693 80
6 5730
×3.6×10–3 = 5.4×10–3mol L–1s–1
dt 4
54. If a= 0 2.303 5730
Ea / RT t= × 0.0969 = 1845 years
K = Ae 0.693
and K = Ae0 = A 2.303 [A1 ]
Hence, K becomes independent of T. 61. K = (t t ) log [A ]
2 1 2
[X]
55. Rate of reaction = 2.303 1.45 2.303
t K= log 0.2169
[X] = Xi–Xf = 0.50 – 0.38 = 0.12M (1600 800) 0.88 800
= 6.24 × 10–4s–1
0.12
Rate = = 2×10–4Ms–1 62. Average rate during the time interval 30 – 60sec.
10 60
(C2 C1) (0.17 0.31) 0.14
Rate 1.02 10 4 mol L1s1 Rate =
56. [N2O5]= 3 mol L1 t2 t1 60 30 30
K 3.4 10 –5 s–1
= 4.67 × 10–3 mol L–1s–1
57. Rate = K[A]x [B]y
From exp.(1), 63. Arrhenius equation, K = Ae Ea / RT
5×10–4 = K[2.5×10–4]x [3×10–5] y ...(i) Given equation is
From exp.(2), K = (4.5×1011s–1) e–28000 K/T
4×10–3= K[5×10–4]x [6×10–5]y ...(ii) Comparing both the equations, we get
Dividing (ii) by (i), Ea 28000K
–
4 10 3 RT T
= 2x . 2y = 8
5 104 Ea = 28000 K×R = 28000 K × 8.314 JK–1mol–1
From exp.(3), = 232.79 KJ mol–1
1.6×10–2 = K[1×10–3]x [6×10–5]y ...(iii) 64. For endothermic reaction, the minimum
Dividing (iii) by (ii), amount of activation energy will be more than
H.
1.6 10 2
= 2x = 4 Ea
4 10 3 65. log K = logA–
or x = 2, y = 1 2.303RT
Hence, order with respect to A is 2 and with 35 103
respect to B is 1. log K = log(1.45×1011) –
2.303 2 573
d[IO ] [I ]1 [OCl ]1 log K = 11.16 – 13.26 = –2.1
58. K
dt [OH ]1 Taking antilog, K = 7.94 × 10–3s–1
order of reaction = 1 + 1 – 1 = 1
0999DMD310317005 LTS/HS-5/7
Target : Pre-Medical 2018/NEET-UG/03-09-2017
Let at equilibrium, mole of SO2 and SO3 be
40
66. n(SO3) = 0.5 same then –
.80
2SO32SO2+O2 PSO3 PSO2
0.5 0 0 2
3.2 KP =
P
SO3
## 0.5–2x 2x x = 3.2 gm = 0.1 mol 2
32 P P
SO 2 O2
0.1 1 1
x = [O2] = 1000 = 1 mol/L
100 or PO2 = = 0.29 atm
K P 3.5
x 1 72. By adding all the three reactions, we will get
ROR = = 0.1 mol L–1 min–1
t 10 target reaction. On adding reaction, there
67. 3A B + 2C equilibrium constant will be multiplled
8 0 0 73. PCl3 Cl 2
PCl 5
8–3x x 2x
8 – 3x = 5 1 0 0
x = 1 mol/lit 1–
x 1 2 P 2P
ROR = KP =
t 10
= 0.1 mol/lit/min 1 1 1 2
## 68. 2NO(g) Br2(g )
2NOBr(g) KP 1
= ;
P p
3P 2P P
P
4 4 4 74. PCl3 Cl 2
PCl 5
Initial moles a 0 0
PT = P 3P 2P P P After moles a–0.1 0.1 0.1
4 4 4
0.1 0.1
2
2P P KC = 1 1 V =1L
4 P P a 0.1
4 4
KP = 2
P 9 4 1 1
4 0.01
0.0414 =
a 0.1
P
1 a = 0.3415 mole
KP
75. NH 3(g) H 2 S(g)
NH 4 HS(s)
69. G° for change = 2 G NO2 G N2O4 Pressure at equilibrium P P
= 2 × 12.39 – 23.49 total pressure at equilibrium =
= 1.29 KCal 2P = 1.12 atm
G° = –2.303 RT log KP 1.12
1.29×103 = –2.303 × 2 × 298 log KP P= atm
2
KP = 0.1132 atm 1 1
KP = PNH3 PH2S
KP n
RT
70. KC KP = 1.12 1.12 0.3136 atm2
2 2
2 = (0.0821×T)1 76. A 2(g) 3B2(g )
2AB3(g)
T = 24.36K
at t = 0 8 0 0
71. 2SO3
2SO 2 O 2 at eqm 8–a a/2 3a/2
LTS/HS-6/7 0999DMD310317005
pOH = –log[OH–] = –log[3×10–3]
[A 2 ][B2 ]3
thus KC = [AB ]2 = –log[3×10–3]
3 = –[0.4771 – 3]
= 2.52
a
also 2 a=4 pH = 14 – 2.52 = 11.48
2
83. A weak acid and strong base salt (NaX)
4 2 6 Kw
and [AB3] = ; [A2] = and [B2]= 10 14
1 1 1 Kh = K = Ch2 0.1 × h2
a 10 5
2 63 h2 = 10–8 h = 10–4
KC= = 27 mol2L–2
42 % hydrolysis = 10–4×100 = 0.01%
77. Keq =
Kf
20
10 84. pH = pKa + log [NaCN]
Kb Kb [HCN]
Kb = 0.5 [NaCN]
78. Ka = 10–4 = Co 2 % = 2% 9.3 = 9.3 + log Ka = 5×10–10
[HCN]
2
= 0.02 [NaCN] pKa [log 5 10log]
100 log =0
[HCN] pKa 9.3
K 10 4 1
Co = 2a = [NaCN]
0.02 0.02 4 =1
[HCN]
85. pH > 7
+
[H ] = Co It means solution is basic
It is only possible when the salt contains strong
1 base.
= 0.02 0.005 5 10 –3
4
pH = –log[H+] Ka
87. (C = Constant for equimolar solution)
= –log[5×10–3] C
= –[log5 – 3log10]
= 2.3 1 K a1
79. At T = 30°C 2 K a2
Kw
Ka = 1 K a2 2 K a1
55.5
Ka 1 12 K a 2 22 K a1
0.018 18 :1000
K w 55.5 88. pH = 12, pOH = 14–12 = 2.
[OH–] = 10–2M
80. H3BO3 B(OH)3
82. pH of HCl = 2 Mg 2 2OH –
Mg(OH)2
[H+] = 10–2 M 0.5×10–2M 10–2M
pH of NaOH = 12
[OH–] = 10–2 M w B 1000
Molarity =
HCl + NaOH NaCl + H2O mB V
Initial mili mol
100 × 10–2 200 × 10–2 w B 1000
0.5×10–2 =
1 2 0 0 58 200
Final mili mol 0 1 1 1 w B = 58 × 2 × 0.5 × 10–2
= 0.58 gm 0.60 gm
1
[OH–] from NaOH = 3 10 –3 89. S = 1.43×10–3 gL–1 Mw (AgCl) = 143
300
0999DMD310317005 LTS/HS-7/7
Target : Pre-Medical 2018/NEET-UG/03-09-2017
## 1.43 10 –3 123. NCERT Pg. No. # 77 ; Para-1
= 124. NCERT Pg. No. # 70 ; Para-2
143
128. NCERT Pg. No. # 74 ; Para-4
S = 10 mol L–1
–5
## Ksp = S2 130. NCERT Pg. No. # 82 ; table-5.3
= (10–5)2 = 10–10 M2 131. NCERT Pg. No. # 81 ; Para-2
132. NCERT Pg. No. # 83 ; Para-4
90. A 2 2B
AB2
80%
133. NCERT Pg. No. # 83 ; Para-4
0.8S 2×0.8S 135. NCERT Pg. No. # 83 ; Para-4
Ksp = (0.8×S)(2×0.8S)2 139. NCERT Pg. No. # 83 ; Para-4
8×10–5 = 2.048S 3 2S
3
141. NCERT Pg. No. # 74 ; Para-4
142. NCERT Pg. No. # 89 ; Para-1
8 10 –5 143. NCERT Pg. No. # 89 ; Para-1
S3 =
2 146. NCERT Pg. No. # 86 ; Para-3
S = 40 × 10–6
3
148. NCERT Pg. No. # 86 ; Para-3
S= 40 10 –6
3 149. NCERT Pg. No. # 85 ; Para-1
= 3.4 × 10–2M 150. NCERT Pg. No. # 85-86 ; Para-1-2
= 0.034M 151. NCERT Pg. No. # 85 ; Para-2
91. NCERT Pg. No. # 69 ; Para-3 152. NCERT Pg. No. # 91 ; Para-4
92. NCERT Pg. No. # 69 ; Para-4 153. NCERT Pg. No. # 91 ; Para-2
94. NCERT Pg. No. # 69 ; Para-2 155. NCERT Pg. No. # 112 ; Para-2
95. NCERT Pg. No. # 72 ; Para-1 156. NCERT Pg. No. # 91 ; Para-2
96. NCERT Pg. No. # 71 ; Table-1 159. NCERT Pg. No. # 114 ; Para-1
97. NCERT Pg. No. # 75 ; Para-2 165. NCERT Pg. No. # 182 ; Para-1
99. NCERT Pg. No. # 71 ; Para-2 166. NCERT Pg. No. # 184 ; Para-2-3
100. NCERT Pg. No. # 70 ; Para-2 167. NCERT Pg. No. # 184 ; Para-2-3
102. NCERT Pg. No. # 71 ; Para-1 168. NCERT Pg. No. # 187 ; Para-2-3
103. NCERT Pg. No. # 73 ; Para-2 169. NCERT Pg. No. # 187 ; Para-4
104. NCERT Pg. No. # 73 ; Para-2 170. NCERT Pg. No. # 195 ; Para-5
105. NCERT Pg. No. # 74 ; Para-4 171. NCERT Pg. No. # 199 ; Fig.-11.4
106. NCERT Pg. No. # 73 ; Fig. 5.4 172. NCERT Pg. No. # 199 ; Para-2
107. NCERT Pg. No. # 79 ; Fig. 5.7 173. NCERT Pg. No. # 200 ; Para-2
111. NCERT Pg. No. # 77 ; Para-2 174. NCERT Pg. No. # 198 ; Para-11.3
115. NCERT Pg. No. # 77 ; Para-2 175. NCERT Pg. No. # 198 ; Para-2
116. NCERT Pg. No. # 77 ; Para-2 176. NCERT Pg. No. # 200 ; Para-3
118. NCERT Pg. No. # 74 ; Para-4 177. NCERT Pg. No. # 209 ; Para-2
121. NCERT Pg. No. # 90 ; Para-1 178. NCERT Pg. No. # 210 ; Para-2-3
122. NCERT Pg. No. # 77 ; Para-2 179. NCERT Pg. No. # 208-210
180. NCERT Pg. No. # 214 ; Para-2
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## Transcription
1 Contents By Kamran Ahmed Lecture # 7 Antenna polarization of satellite signals Cross polarization discrimination Ionospheric depolarization, rain & ice depolarization The polarization of an electromagnetic wave is defined as the orientation of the electric field vector. Recall that the electric field vector is perpendicular to both the direction of travel and the magnetic field vector. The polarization is described by the geometric figure traced by the electric field vector upon a stationary plane perpendicular to the direction of propagation, as the wave travels through that plane. is also describe as the "direction of vibration" on the radio wave. It depends the orientation of elements of an antenna, when you set elements vertical, it generates vertical-polarized radio wave similarly when you set as horizontal, it generates horizontal-polarized. In the case of YAGI antenna, the direction of Electronic-Field is same as the direction of its elements. Radio stations have to set as a same direction of polarization for communication each other. Types of An electromagnetic wave is frequently composed of (or can be broken down into) two orthogonal. This may be due to the arrangement of power input leads to various points on a flat antenna, or due to an interaction of active elements in an array, or many other reasons. The geometric figure traced by the sum of the electric field vectors over time is, in general, an ellipse as shown in Figure 2. Under certain conditions the ellipse may collapse into a straight line, in which case the polarization is called linear.
2 In the other extreme, when the two components are of equal magnitude and 90 0 out of phase, the ellipse will become circular as shown in Figure 3. Thus linear and circular polarization are the two special cases of elliptical polarization. Linear polarization may be further classified as being vertical, horizontal, or slant. and its types makes the beam more concentrated FSS satellites use horizontal and vertical polarization, whereas DBS satellites use leftand right-hand circular polarization To use the channels that are available for satellite broadcast as efficiently as possible, both horizontal and vertical polarization (and left- and right-hand circular polarization) can be applied simultaneously per channel or frequency. In such cases the frequency of one of the two is slightly altered, to prevent possible interference Horizontal and vertical transmissions will therefore not interfere with each another because they are differently polarized. This means twice as many programs can be transmitted per satellite Consequently, via one and (almost) the same frequency the satellite can broadcast both a horizontal and a vertical polarized signal (H and V), or a left- and right-hand circular polarized signal (LH and RH).
3 Radio stations have to set as a same direction of polarization for communication each other. When you try to hear the vertical-polarized wave with horizontal- polarized antenna, what will be happened? A theory tells it is impossible to receive. In fact, although it is possible, It becomes very difficult (very weak less than -20dB ). This is due to:- The radio waves do not travels with pure-polarized condition, and There is no real antenna that has pure-polarized character. Anyway, you should to adjust the polarization for better communication. Is Circular better choice for satellite? Circular-polarization (CP) is another choice when you could not decide the polarization of your choice. CP is the special style of polarization, the direction of Electric-Field rotates one times par one cycle. The CP antenna can receive both horizontal and vertical polarized radio wave, even in the direction of slant-polarized. CP is very popular technique for satellite communication both commercial and amateur satellite systems. Antenna Table 1 shows the theoretical ratio of power transmitted between antennas of different polarization. These ratios are seldom fully achieved due to effects such as reflection, refraction, and other wave interactions, so some practical ratios are also included. The sense of antenna polarization is defined from a viewer positioned behind an antenna looking in the direction of propagation. The polarization is specified as a transmitting, not receiving antenna regardless of intended use. We frequently use "hand rules" to describe the sense of polarization. The sense is defined by which hand would be used in order to point that thumb in the direction of propagation and point the fingers of the same hand in the direction of rotation of the E field vector.
4 For example, referring to Figure 4, if your thumb is pointed in the direction of propagation and the rotation is counterclockwise looking in the direction of travel, then you have left hand circular polarization. The polarization of a linearly polarized horn antenna can be directly determined by the orientation of the feed probe, which is in the direction of the E-field. In general, a flat surface or sphere will reflect a linearly polarized wave with the same polarization as received. A horizontally polarized wave may get extended range because of water and land surface reflections, but signal cancellation will probably result in "holes" in coverage. Reflections will reverse the sense of circular polarization. For a linearly polarized antenna, the radiation pattern is taken both for a co-polarized and cross polarized response. The polarization quality is expressed by the ratio of these two responses. The ratio between the responses must typically be great (30 db or greater) for an application such as cross polarized jamming For general applications, the ratio indicates system power loss due to polarization mismatch. For circularly polarized antennas, radiation patterns are usually taken with a rotating linearly polarized reference antenna.
5 Manual Switching The CP antenna reduces QSB so it might be better for comfortable operation, but the CP antenna is bigger and more complicated than the simple linear-polarized antenna. Also the big and complicated antenna will be expensive. 3dB loss will be a problem with some limited conditions. There is another choice. Setup a pair of vertical/horizontal polarized independent antenna and switch them at your shack. You select where either is better during its pass. This is the theory of "Divercity" reception of satellite signal Applied for geo-stationary satellites Horizontal polarization = parallel to the equatorial plane Vertical polarization = parallel to the Earth's axis angle at earth station r = local gravity direction k = the direction of the wave propagation p = unit polarization vector f = k x r, normal to the reference plane x = the angle between the reference plane (r and k) and the polarization vector Depolarization The electric field E 1 is depolarized after going through a depolarizing medium. The result is, as shown in the figure, an orthogonal (E 12 ) component may be generated. E 11 is called the co-polar component and E 12 is called the cross-polar component. This phenomenon can cause interference. Cross-polarization discrimination (XPD) One measure to quantify the effects of polarization is called the cross-polarization discrimination (XPD) Cross-polarization discrimination observations - rain depolarization Looking at XPD as a function of the co-polar attenuation (A), it can be concluded that: XPD degrades at a given co-polar attenuation as the frequency decreases XPD degrades with increasing co-polar attenuation XPD for the Vertical wave is better than that for Horizontal XPD for the Vertical and the Horizontal waves are better that the Circular
6 XPD and co-polar attenuation A θ -> the elevation angle in degrees τ > the polarization tilt angle τ = 45 for circular polarization Ionospheric effects Faraday s effects The rotation of a linearly polarized wave due to the earth s magnetic field is called the Faraday s effect. It is proportional to the 1/f2 factor. Ionospheric scintillation Due to the refractive index variations in the ionosphere caused by local concentrations of ionization. It is also proportional to the 1/f2 factor.
### EMG4066:Antennas and Propagation Exp 1:ANTENNAS MMU:FOE. To study the radiation pattern characteristics of various types of antennas.
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https://amruthapublicschool.relents.in/tag/2c2a7e-pythagoras-theorem-history | 1,686,212,443,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00062.warc.gz | 124,696,760 | 5,719 | # pythagoras theorem history
Euclid, in his book The Elements, presents a proof of the Turns out … Much as known about Pythagoras, although many historical facts were not written down about him until centuries after he lived. He was highly involved in the religious sect and founded his own religious movement called Pythagoreanism (Machiavelo, 2009). The problem he faced is explained in the Sidebar: Incommensurables. This may be the original proof of the ancient theorem, which states that the sum of the squares on the sides of a right triangle equals the square on the hypotenuse (. The history of the Pythagorean theorem can be divided as: knowledge of Pythagorean triples, the relationship among the sides of a right triangle and their adjacent angles, and the proofs of the theorem. He was a Greek mathematician who revolutionized areas such as geometry and architecture, as well as philosophy. The Pythagorean Theorem was invented by Pythagoras of Samos. B.C.E. Thus, not only is the first proof of the theorem not known, there is also some doubt that Pythagoras himself actually proved the theorem that bears his name. The semicircles that define Hippocrates of Chios’s lunes are examples of such an extension. This is used when we are given a triangle in which we only know the length of two of the three sides. Finally, the Greek Mathematician stated the theorem hence it is called by his name as "Pythagoras theorem." He perhaps was the first one to offer a proof of the theorem. If a = 3, and b = 4, we could then solve for c. 32 + 42 = c². tiles shown here is typical of those seen Pythagoras was a Greek philosopher who made important developments in mathematics, astronomy, and the theory of music. Although the theorem has long been associated with Greek mathematician-philosopher Pythagoras (c. 570–500/490 bce), it is actually far older. Instead, in the margin of a textbook, he wrote that he knew that this relationship was not possible, but he did not have enough room on the page to write it down. Premium Membership is now 50% off! So according to the History Channel, the plan for the tunnel was based on the Pythagorean Theorem. In the 17th century, Pierre de Fermat(1601-1665) investigated the following problem: For which values of n are there integral solutions to the equation. If a is the adjacent angle then b is the opposite side. Pythagoras theorem was introduced by the Greek Mathematician Pythagoras of Samos. Pythagorean Theorem History. expressed as a length that can be measured with a ruler divided He did not leave a proof, though. The Pythagorean theorem says that the area of a square on the hypotenuse is equal to the sum of the areas of the squares on the legs. They called It was probably independently discovered in several different cultures. Discoveries like the one of the musical trilogy (tonic, dominant and subdominant) determining for the understanding as much […] The Pythagorean OrderO Students at the Pythagorean School had to live by strict rules since they all strove to create pure minds and bodies.O They all thought that the mind could be purified through studying Geometry, … triangle, thus giving an angle of exactly 90 degrees. The Pythagoreans wrote many geometric proofs, but it is Pythagoras theorem history 1. The pattern of. into fractional parts, and that deeply disturbed the Apparently, Euclid invented the windmill proof so that he could place the Pythagorean theorem as the capstone to Book I. Discoveries like the one of the musical trilogy (tonic, dominant and subdominant) determining for the understanding as much […] It qualities. The Pythagorean Theorem. Although the Pythagorean theorem bears his name, the discoveries of the Pythagorean theorem and that the square root of 2 is an irrational number were most likely made after his death by his followers. Can you figure out the method of proof used in the figure below? In this picture, the area of the blue square added to the area of the red square makes the area of the purple square. Visual demonstration of the Pythagorean theorem. -500 BCE. (See Sidebar: Euclid’s Windmill.) The theorem now known as Pythagoras's theorem was known to the Babylonians 1000 years earlier but he may have been the first to prove it. Greece, and did much traveling through Egypt, learning, among Later in Book VI of the Elements, Euclid delivers an even easier demonstration using the proposition that the areas of similar triangles are proportionate to the squares of their corresponding sides.
Problem And Solution Essay, Mtg Arena Cash Tournaments, Razer Blade 15 2018, Totino's Canadian Bacon Pizza Discontinued, American Walnut Vs Black Walnut, Sonic Advance 2 Release Date, How To Cook Buckwheat, White Linkedin Logo, | 1,041 | 4,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-23 | longest | en | 0.984269 |
http://forums.wolfram.com/mathgroup/archive/2004/May/msg00019.html | 1,718,577,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00391.warc.gz | 12,891,866 | 8,448 | Re: FindMaximum doesn't work?
• To: mathgroup at smc.vnet.net
• Subject: [mg47952] Re: FindMaximum doesn't work?
• From: suomesta at yahoo.com (Carol Ting)
• Date: Sun, 2 May 2004 04:50:50 -0400 (EDT)
• References: <c6l8p2\$iv6\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```My bad. I copied it manually and sure enough a couple of typos
sneaked in and messed things up.
Here is the function which I tried to apply FindMaximum to:
Test[c_]:=-10 + (1/(530604*c))*((10 - 185*c + Sqrt[5]*Sqrt[20 +
19660*c + 6641*c^2])^2*(-41 - 134*c + Sqrt[5]*Sqrt[20 + 19660*c +
6641*c^2]))/;(c<=1.4<6.2)
If you plot the function you will see it's decreasing in c with the
maximum at 1.4. But unless I choose an initial guess of c less than
2.6 Mathematica just throws my guesses back at me. Again, the old
examples:
In[57]:=FindMaximum[test[c],{c,5}]
Out[57]:={-0.895005,{c->5.}}
Any one has an idea of what's going on? Thanks!
Carol
"DrBob" <drbob at bigfoot.com> wrote in message news:<c6l8p2\$iv6\$1 at smc.vnet.net>...
> Show us the code you REALLY used, or we can't help you. If test[c] is
> defined the way you say in the post, it (a) has no global maximum and (b)
> has no local maximum at 1.4. Even if you define it as follows, to enforce
> the limits you mention, it still has no maximum at 1.4, and it doesn't have
> the values at 5 and 1.4 that you show in the post. Nor is it decreasing in
> c, near 2 or 5.
>
> test[c_] /; 1.4 ? c < 6.2 :=
> ((10 - 185c + ((5^0.5)(20 + 19660c + 6641c^2))^2)(-41 -
> 134c + (5^0.5)(20 + 19660c + 6641c^2)))/(530604c)
> FindMaximum[test[c], {c, 5}]
> (error message)
> {7.77508372302857*^10, {c -> 5.}}
>
> FindMaximum[test[c], {c, 2}]
> (error message)
> {3.009388196803684*^9, {c -> 2.}}
>
> DrBob
>
> www.eclecticdreams.net
>
>
> -----Original Message-----
> From: Carol Ting [mailto:suomesta at yahoo.com]
To: mathgroup at smc.vnet.net
> Subject: [mg47952] FindMaximum doesn't work?
>
> Hi List,
>
> I have one function which is defined only for a certain range of its
> independent variable:
>
> test[c_]:=((10-185c+((5^0.5)(20+19660c+6641c^2))^2)(-41-134c+(5^0.5)(20+1966
> 0c+6641c^2)))/(530604c)
> 1.4<=c<6.2
>
> This function is decreasing in c so the maximum should be at 1.4.
> However, when the initial guess I give is greater than 2.6,
> FindMaximum only returns the values at that point.
>
> For example,
>
> In[57]:=FindMaximum[test[c],{c,5}]
> Out[57]:={-0.895005,{c->5.}}
>
> But when I give
>
> In[58]:=FindMaximum[test[c],{c,2}]
> Out[58]:={2.09973,{c->1.4}}
>
> Then it works.
>
> My program uses a While loop to generate piecewise functions (I only
> put the piece containting the global maximum here). The maximum
> condition is needed to decide whether to terminate the loop so I
> cannot always stop the execution and keep throwing in different
> guesses.
>
> Did I do anything wrong, or is this a bug? Thanks a lot!
>
> Carol
```
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# Computer Programming
Assignment
## Due Date: 09 May 2016
1. Write a program that prints the numbers between 1and 25 along with their square roots in a
tabular form.
2. Define a function that takes two integer arguments, if second argument is greater than first the
function returns the sum of the arguments otherwise their product is returned.
3. Write a function integerPower (base, exponent) that returns the value of
base exponent
For example, integerPower( 3, 4 ) = 3 * 3 * 3 * 3. Assume that exponent is a positive, nonzero
integer and that base is an integer. The function integerPower should use for or while to control
the calculation. Do not use any math library functions.
4. Define a function calcTax that takes the value of a product as a double value. It then return the
sale value after adding tax to it @ 15%.
5. Write a function that displays at the left margin of the screen a solid square of asterisks whose
side is specified in integer parameter side. For example, if side is 4, the function displays the
following:
****
****
****
****
6. Modify the function created in Ex # 4 to form the square out of whatever character is contained
in character parameter fillCharacter. Thus, if side is 4 and fillCharacter is #, then this function
should print the following:
####
####
####
####
7.
An integer is said to be a perfect number if the sum of its factors, including 1 (but not the
number itself), is equal to the number. For example, 6 is a perfect number, because 6 = 1 + 2 +
3. Write a function perfect that determines whether parameter number is a perfect number.
8. One of the most popular games of chance is a dice game known as "craps" which is played in
casinos and back alley worldwide. The rules of the game are straight
forward:
A player rolls two dice. Each die has six faces. These faces contain 1, 2, 3, 4, 5 and 6 spots.
After the dice have come to rest, the sum of the spots on the two upward faces is calculated. If
the sum is 7 or 11 on the first roll, the player wins. If the sum is 2, 3 or 12 on the first roll
(called "craps"), the player loses (i.e., the "house" wins). If the sum is 4, 5, 6, 8, 9 or 10 on the
first roll, then that sum becomes the player's "point." To win, you must continue rolling the
dice until you "make your point." The player loses by rolling a 7 before making the point.
9. Write a function that takes an integer value and returns the number with its digits reversed. For
example, given the number 7631, the function should return 1367. Make sure that the number
entered is a 4 digit number. If not, re-prompt the user to enter it again.
10. Simulate the rolling of dice. Roll the dice 1000 times and draw the results in the
histogram as given below.
No.
1
2
3
4
5
6
form of a
Frequency
+++**************
++****************
+*************************
+***************
++********************
********************
## Where + represents 100 and * represents 1.
11. Write a program that declares an integer array of 10 elements. Initialize the array using an
initializer list. The program then prints the elements of the array in a tabular form.
12. Write a program that declares an integer array of size 10 and scans the values of array elements
from user. It then calculates the sum of all the elements of the array.
13. Write a program that declares a float array of size 10 and scans the values of array elements
from user. It then adds a value of 5.5 to even indexed elements and prints the array on screen.
14. Use an integer array to solve the following problem. Read in 10 numbers, each of which is
between 10 and 100, inclusive. As each number is read, validate it and store it in the array only
if it is not a duplicate of a number already read. After reading all the values, display only the
unique values that the user entered.
15. Write a program that declares an integer array of size 10. Initialize the array elements with
random numbers in the range 1-20 and graph the information as a histogram as shown:
Element Value Histogram
0
3
***
1
5
*****
2
3
...
8
11
********
***********
Note:
Plagiarism is strictly prohibited in this assignment; Copy of even a single question will
lead to rejection of whole assignment and will result in 0 marks.
This assignment should be sent via email, Email id is waqasdocument@gmail.com. Email
Subject should be Computer Programming Assignment 3 Reg# Otherwise your
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# OpenGL Programming/GLStart/Tut3
## Tutorial 3: Drawing PrimitivesEdit
Primitives are basic shapes that you can easily draw. It can be a triangle, a square, or even a single point. In this lesson we are going to learn how to draw some basic primitives including: points, triangles, and different polygons.
### Immediate ModeEdit
The easiest way to do drawing in OpenGL is using the Immediate Mode. For this, you use the glBegin() function which takes as one parameter the “mode” or type of object you want to draw.
Here is a list of the possible modes and what they mean:
GL_POINTS Draws points on screen. Every vertex specified is a point. Draws lines on screen. Every two vertices specified compose a line. Draws connected lines on screen. Every vertex specified after first two are connected. Draws connected lines on screen. The last vertex specified is connected to first vertex. Draws triangles on screen. Every three vertices specified compose a triangle. Draws connected triangles on screen. Every vertex specified after first three vertices creates a triangle. Draws connected triangles like GL_TRIANGLE_STRIP, except draws triangles in fan shape. Draws quadrilaterals (4 – sided shapes) on screen. Every four vertices specified compose a quadrilateral. Draws connected quadrilaterals on screen. Every two vertices specified after first four compose a connected quadrilateral. Draws a polygon on screen. Polygon can be composed of as many sides as you want.
When you are done drawing all the vertices of the specific primitive, you put in the next line the glEnd() function which ends drawing of that primitive.
#### Drawing PointsEdit
Lets do an example with the simplest mode, the GL_POINTS. When drawing points using OpenGL, the default size of the points is 1 pixel wide and high. This would be very hard to see when you run the program. To edit the size of the point you want to draw, you use the glPointSize() function which takes as one parameter the size of the point you want.
Now in the Render() function, before you write the glBegin() code, we will set the point size to be 10 pixels in size:
``` glPointSize(10.0f);
```
After that, any drawing of points will be drawn 10 pixels wide and high. Now write the glBegin() function with the GL_POINTS mode parameter. Then after that specify the vertices you want to use using the glVertex3f() function. For this example, we want the upper – right corner (1.0,1.0,0.0) and the lower – left corner (-1.0,-1.0,0.0) of the screen to have a point. After drawing those two points, make sure to end the drawing with the glEnd() function:
``` glBegin(GL_POINTS); //starts drawing of points
glVertex3f(1.0f,1.0f,0.0f);//upper-right corner
glVertex3f(-1.0f,-1.0f,0.0f);//lower-left corner
glEnd();//end drawing of points
```
Here is the whole render function for your reference followed by a sample output. The whole code for this section of the tutorial can be found in the downloadable files. This example is called “points”:
```void Render()
{
//clear color and depth buffer
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glTranslatef(0.0f,0.0f,-4.0f);//move forward 4 units
glColor3f(0.0f,0.0f,1.0f); //blue color
glPointSize(10.0f);//set point size to 10 pixels
glBegin(GL_POINTS); //starts drawing of points
glVertex3f(1.0f,1.0f,0.0f);//upper-right corner
glVertex3f(-1.0f,-1.0f,0.0f);//lower-left corner
glEnd();//end drawing of points
}
```
#### Drawing a Line LoopEdit
I would like to cover every single mode in the glBegin() function, but it would take up too much time and space. So I will cover the advanced ones and most likely cover the other ones when they are deemed useful.
A line loop requires at least two vertices. Every vertex specified after that is connected to the vertex specified before it and the first vertex specified. So if we put a vertex to the left of the window, to the right of the window, on top and on the bottom, we would have a rotated square:
Lets do this diagram in OpenGL. First we use the glBegin() function and pass the parameter of GL_LINE_LOOP to tell OpenGL we are going to start drawing a line loop. Then we pass the four vertices to create the rotated square. The first vertex being to the left of the window (-1.0f,0.0f,0.0f), second one being at the bottom of the window (0.0f,-1.0f,0.0f), third at the right of window (1.0f,0.0f,0.0f), and the fourth and last being at the top of the window (0.0f,1.0f,0.0f). Then we make sure to put in glEnd() to tell OpenGL we are done drawing the line loop:
``` glBegin(GL_LINE_LOOP);//start drawing a line loop
glVertex3f(-1.0f,0.0f,0.0f);//left of window
glVertex3f(0.0f,-1.0f,0.0f);//bottom of window
glVertex3f(1.0f,0.0f,0.0f);//right of window
glVertex3f(0.0f,1.0f,0.0f);//top of window
glEnd();//end drawing of line loop
```
Following is the whole Render() function followed by a sample output. Look for the project file called “lineLoop” on the downloadable file for this chapter:
```void Render()
{
//clear color and depth buffer
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glTranslatef(0.0f,0.0f,-4.0f);//move forward 4 units
glColor3f(0.0f,0.0f,1.0f); //blue color
glBegin(GL_LINE_LOOP);//start drawing a line loop
glVertex3f(-1.0f,0.0f,0.0f);//left of window
glVertex3f(0.0f,-1.0f,0.0f);//bottom of window
glVertex3f(1.0f,0.0f,0.0f);//right of window
glVertex3f(0.0f,1.0f,0.0f);//top of window
glEnd();//end drawing of line loop
}
```
#### Drawing TrianglesEdit
Triangles are composed of three vertices. For this example we are going to use the regular GL_TRIANGLES mode to draw two triangles side by side.
First we want a triangle to the left. So we need three vertices on the left side representing one triangle and three vertices on the right side of the window representing the second triangle. Note that you don’t need two glBegin() functions to draw two triangles. Since the mode GL_TRIANGLES is plural, it can handle more than one triangle in between one glBegin() and glEnd() function call:
Note that I wrote on the bottom – right corner of the diagram the coordinates of the vertices. Here is the code to draw these two triangles:
``` glBegin(GL_TRIANGLES);//start drawing triangles
glVertex3f(-1.0f,-0.25f,0.0f);//triangle one first vertex
glVertex3f(-0.5f,-0.25f,0.0f);//triangle one second vertex
glVertex3f(-0.75f,0.25f,0.0f);//triangle one third vertex
//drawing a new triangle
glVertex3f(0.5f,-0.25f,0.0f);//triangle two first vertex
glVertex3f(1.0f,-0.25f,0.0f);//triangle two second vertex
glVertex3f(0.75f,0.25f,0.0f);//triangle two third vertex
glEnd();//end drawing of triangles
```
Following is the whole Render() function for this example followed by a sample output. To view the whole code, please see the “triangle” project folder in the downloadable files:
```void Render()
{
//clear color and depth buffer
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glTranslatef(0.0f,0.0f,-4.0f);//move forward 4 units
glColor3f(0.0f,0.0f,1.0f); //blue color
glBegin(GL_TRIANGLES);//start drawing triangles
glVertex3f(-1.0f,-0.25f,0.0f);//triangle one first vertex
glVertex3f(-0.5f,-0.25f,0.0f);//triangle one second vertex
glVertex3f(-0.75f,0.25f,0.0f);//triangle one third vertex
//drawing a new triangle
glVertex3f(0.5f,-0.25f,0.0f);//triangle two first vertex
glVertex3f(1.0f,-0.25f,0.0f);//triangle two second vertex
glVertex3f(0.75f,0.25f,0.0f);//triangle two third vertex
glEnd();//end drawing of triangles
}
```
#### Drawing PolygonsEdit
Polygons consist of at least three vertices that, when connected, make up a shape. In this example we are going to be using the GL_POLYGON mode to draw a six – sided shape.
The GL_POLYGON mode allows you to draw a shape with any number of sides as you want. Since GL_POLYGON is a singular word (meaning no “S” at the end of the word), you can only draw one polygon between a glBegin() and glEnd() function call. Also, the last vertex you specify is automatically connected to the first vertex specified. And of course, since a polygon is a closed shape, like the triangle, the shape will be filled with your specified color (which for now is blue). Here is the diagram for our six – sided shape:
(note: this can't be used for concave polygons)
Here is the code to draw this polygon:
``` glBegin(GL_POLYGON);//begin drawing of polygon
glVertex3f(-0.5f,0.5f,0.0f);//first vertex
glVertex3f(0.5f,0.5f,0.0f);//second vertex
glVertex3f(1.0f,0.0f,0.0f);//third vertex
glVertex3f(0.5f,-0.5f,0.0f);//fourth vertex
glVertex3f(-0.5f,-0.5f,0.0f);//fifth vertex
glVertex3f(-1.0f,0.0f,0.0f);//sixth vertex
glEnd();//end drawing of polygon
```
Following is the whole Render() function followed by a sample output. Look for the whole code in the “polygon” project folder on the included downloadable files:
```void Render()
{
//clear color and depth buffer
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glTranslatef(0.0f,0.0f,-4.0f);//move forward 4 units
glColor3f(0.0f,0.0f,1.0f); //blue color
glBegin(GL_POLYGON);//begin drawing of polygon
glVertex3f(-0.5f,0.5f,0.0f);//first vertex
glVertex3f(0.5f,0.5f,0.0f);//second vertex
glVertex3f(1.0f,0.0f,0.0f);//third vertex
glVertex3f(0.5f,-0.5f,0.0f);//fourth vertex
glVertex3f(-0.5f,-0.5f,0.0f);//fifth vertex
glVertex3f(-1.0f,0.0f,0.0f);//sixth vertex
glEnd();//end drawing of polygon
}
```
### Display ListsEdit
Using the OpenGL immediate mode has several drawbacks. For example, vertex data must be transferred on the fly to the graphics memory whenever drawing. This can be improved by using Display Lists. Display Lists essentially take a glBegin()/glEnd() command sequence and store that on the graphics card side in an efficient manner.
(todo: elaborate on that later, see [1] for some infos - glGenLists(), glNewList(), glEndList(), glCallList(), glDeleteLists())
### Vertex Arrays and Vertex BuffersEdit
Display Lists also have drawbacks. Once compiled, Display Lists are static, and cannot be changed. Also, vertices shared by several primitives still need to be represented and transformed multiple times when drawing. This is very inefficient.
These problems are addressed by Vertex Arrays and Vertex Buffers. The idea is simply to use arrays to store vertex data. Using glDrawArrays(), primitives using these vertices can be drawn.
For storage of data in video memory, Vertex Buffers can be used (glGenBuffer(), glBindBuffer(), glBufferData(), glDeleteBuffers()). Two types of Vertex Buffers exist:
• GL_ARRAY_BUFFERs hold actual vertex data.
• GL_ELEMENT_ARRAY_BUFFERs hold indices to vertices stored in a separate GL_ARRAY_BUFFER and, thus, allow reusing vertices in several primitives.
(todo: elaborate on that later, example) | 2,957 | 10,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2015-22 | latest | en | 0.858103 |
http://www.jiskha.com/display.cgi?id=1361412191 | 1,498,524,719,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320873.54/warc/CC-MAIN-20170626235612-20170627015612-00025.warc.gz | 550,392,318 | 4,177 | # chem
posted by .
For the reaction I2(g) + Br2(g) 2 IBr(g), Kc = 280. at 150.°C. Suppose that 0.570 mol IBr in a 2.50-L flask is allowed to reach equilibrium at 150.°C. What are the equilibrium concentrations of IBr, I2, and Br2?
• chem -
M = mols/L = 0.570/2.50L = about 0.228M
........I2 + Br2 ==> 2IBr
I.......0....0......0.228
C.......x....x.....0.228-2x
E.......x....x.....0.228-2x
Substitute the E line into Kc expression and solve.
• chem -
do u have to use the quadratic to find x?
• chem -
what do i do after i get
.051984+4x^2 - .912/(x^2)
• chem -
.051984+4x^2 - .912x=280(x^2)
• chem -
then subtract the 4x^2? so
.051984-.912x=276x^2
now what?
• chem -
To here is ok.
.051984+4x^2 - .912x=280(x^2)
Just rearrange to
0.05918 + 4x^2 - 0.912x = 280x^2
Then 280x^2-4x^2 + 0.912x - 0.05198 = 0 and
276x^2+0.912x-0.05198 = 0 and use the quadratic formula. I have x = 0.0122 or close to that.
### Answer This Question
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### Related Questions
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Post a New Question | 404 | 1,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-26 | latest | en | 0.823153 |
http://planetmath.org/1131constructionofcauchyreals | 1,521,852,339,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649508.48/warc/CC-MAIN-20180323235620-20180324015620-00426.warc.gz | 241,085,039 | 7,380 | # 11.3.1 Construction of Cauchy reals
The construction of the Cauchy reals $\mathbb{R}_{\mathsf{c}}$ as a higher inductive type is a bit more subtle than that of the free algebraic structures considered in \autorefsec:free-algebras. We intend to include a βtake the limitβ constructor whose input is a Cauchy sequence of reals, but the notion of βCauchy sequence of realsβ depends on having some way to measure the βdistanceβ between real numbers. In general, of course, the distance between two real numbers will be another real number, leading to a potentially problematic circularity.
However, what we actually need for the notion of Cauchy sequence of reals is not the general notion of βdistanceβ, but a way to say that βthe distance between two real numbers is less than $\epsilon$β for any $\epsilon:\mathbb{Q}_{+}$. This can be represented by a family of binary relations, which we will denote $\mathord{\sim_{\epsilon}}:\mathbb{R}_{\mathsf{c}}\to\mathbb{R}_{\mathsf{c}}\to% \mathsf{Prop}$. The intended meaning of $x\sim_{\epsilon}y$ is $|x-y|<\epsilon$, but since we do not have notions of subtraction, absolute value, or inequality available yet (we are only just defining $\mathbb{R}_{\mathsf{c}}$, after all), we will have to define these relations $\sim_{\epsilon}$ at the same time as we define $\mathbb{R}_{\mathsf{c}}$ itself. And since $\sim_{\epsilon}$ is a type family indexed by two copies of $\mathbb{R}_{\mathsf{c}}$, we cannot do this with an ordinary mutual (higher) inductive definition; instead we have to use a higher inductive-inductive definition.
Recall from \autorefsec:generalizations that the ordinary notion of inductive-inductive definition allows us to define a type and a type family indexed by it by simultaneous induction. Of course, the βhigherβ version of this allows both the type and the family to have path constructors as well as point constructors. We will not attempt to formulate any general theory of higher inductive-inductive definitions, but hopefully the description we will give of $\mathbb{R}_{\mathsf{c}}$ and $\sim_{\epsilon}$ will make the idea transparent.
###### Remark 11.3.1.
We might also consider a higher inductive-recursive definition, in which $\sim_{\epsilon}$ is defined using the recursion principle of $\mathbb{R}_{\mathsf{c}}$, simultaneously with the inductive definition of $\mathbb{R}_{\mathsf{c}}$. We choose the inductive-inductive route instead for two reasons. Firstly, higher inductive-recursive definitions seem to be more difficult to justify in homotopical semantics. Secondly, and more importantly, the inductive-inductive definition yields a more powerful induction principle, which we will need in order to develop even the basic theory of Cauchy reals.
Finally, as we did for the discussion of Cauchy completeness of the Dedekind reals in \autorefsec:RD-cauchy-complete, we will work with Cauchy approximations (\autorefdefn:cauchy-approximation) instead of Cauchy sequences. Of course, our Cauchy approximations will now consist of Cauchy reals, rather than Dedekind reals or rational numbers.
###### Definition 11.3.2.
Let $\mathbb{R}_{\mathsf{c}}$ and the relation $\mathord{\sim}:\mathbb{Q}_{+}\times\mathbb{R}_{\mathsf{c}}\times\mathbb{R}_{% \mathsf{c}}\to\mathcal{U}$ be the following higher inductive-inductive type family. The type $\mathbb{R}_{\mathsf{c}}$ of Cauchy reals is generated by the following constructors:
• β’
rational points: for any $q:\mathbb{Q}$ there is a real $\mathsf{rat}(q)$.
• β’
limit points: for any $x:\mathbb{Q}_{+}\to\mathbb{R}_{\mathsf{c}}$ such that
$\forall(\delta,\epsilon:\mathbb{Q}_{+}).\,x_{\delta}\sim_{\delta+\epsilon}x_{\epsilon}$ (11.3.2)
there is a point $\mathsf{lim}(x):\mathbb{R}_{\mathsf{c}}$. We call $x$ a Cauchy approximation.
• β’
paths: for $u,v:\mathbb{R}_{\mathsf{c}}$ such that
$\forall(\epsilon:\mathbb{Q}_{+}).\,u\sim_{\epsilon}v$ (11.3.3)
then there is a path $\mathsf{eq}_{\mathbb{R}_{\mathsf{c}}}(u,v):u=_{\mathbb{R}_{\mathsf{c}}}v$.
Simultaneously, the type family $\mathord{\sim}:\mathbb{R}_{\mathsf{c}}\to\mathbb{R}_{\mathsf{c}}\to\mathbb{Q}_% {+}\to\mathcal{U}$ is generated by the following constructors. Here $q$ and $r$ denote rational numbers; $\delta$, $\epsilon$, and $\eta$ denote positive rationals; $u$ and $v$ denote Cauchy reals; and $x$ and $y$ denote Cauchy approximations:
• β’
for any $q,r,\epsilon$, if $-\epsilon, then $\mathsf{rat}(q)\sim_{\epsilon}\mathsf{rat}(r)$,
• β’
for any $q,y,\epsilon,\delta$, if $\mathsf{rat}(q)\sim_{\epsilon-\delta}y_{\delta}$, then $\mathsf{rat}(q)\sim_{\epsilon}\mathsf{lim}(y)$,
• β’
for any $x,r,\epsilon,\delta$, if $x_{\delta}\sim_{\epsilon-\delta}\mathsf{rat}(r)$, then $\mathsf{lim}(x)\sim_{\epsilon}\mathsf{rat}(r)$,
• β’
for any $x,y,\epsilon,\delta,\eta$, if $x_{\delta}\sim_{\epsilon-\delta-\eta}y_{\eta}$, then $\mathsf{lim}(x)\sim_{\epsilon}\mathsf{lim}(y)$,
• β’
for any $u,v,\epsilon$, if $\xi,\zeta:u\sim_{\epsilon}v$, then $\xi=\zeta$ (propositional truncation).
The first constructor of $\mathbb{R}_{\mathsf{c}}$ says that any rational number can be regarded as a real number. The second says that from any Cauchy approximation to a real number, we can obtain a new real number called its βlimitβ. And the third expresses the idea that if two Cauchy approximations coincide, then their limits are equal.
The first four constructors of $\mathord{\sim}$ specify when two rational numbers are close, when a rational is close to a limit, and when two limits are close. In the case of two rational numbers, this is just the usual notion of $\epsilon$-closeness for rational numbers, whereas the other cases can be derived by noting that each approximant $x_{\delta}$ is supposed to be within $\delta$ of the limit $\mathsf{lim}(x)$.
We remind ourselves of proof-relevance: a real number obtained from $\mathsf{lim}$ is represented not just by a Cauchy approximation $x$, but also a proof $p$ ofΒ (11.3.2), so we should technically have written $\mathsf{lim}(x,p)$ instead of just $\mathsf{lim}(x)$. A similar observation also applies to $\mathsf{eq}_{\mathbb{R}_{\mathsf{c}}}$ andΒ (11.3.3), but we shall write just $\mathsf{eq}_{\mathbb{R}_{\mathsf{c}}}:u=v$ instead of $\mathsf{eq}_{\mathbb{R}_{\mathsf{c}}}(u,v,p):u=v$. These abuses of notation are mitigated by the fact that we are omitting mere propositions and information that is readily guessed. Likewise, the last constructor of $\mathord{\sim_{\epsilon}}$ justifies our leaving the other four nameless.
We are immediately able to populate $\mathbb{R}_{\mathsf{c}}$ with many real numbers. For suppose $x:\mathbb{N}\to\mathbb{Q}$ is a traditional Cauchy sequence of rational numbers, and let $M:\mathbb{Q}_{+}\to\mathbb{N}$ be its modulus of convergence. Then $\mathsf{rat}\circ x\circ M:\mathbb{Q}_{+}\to\mathbb{R}_{\mathsf{c}}$ is a Cauchy approximation, using the first constructor of $\mathord{\sim}$ to produce the necessary witness. Thus, $\mathsf{lim}(\mathsf{rat}\circ x\circ m)$ is a real number. Various famous real numbers $\sqrt{2}$, $\pi$, $e$, β¦ are all limits of such Cauchy sequences of rationals.
Title 11.3.1 Construction of Cauchy reals
\metatable | 2,140 | 7,217 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 77, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-13 | latest | en | 0.885566 |
https://www.physicsforums.com/threads/local-linear-approximation-vs-linearization.721937/ | 1,508,413,475,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823282.42/warc/CC-MAIN-20171019103222-20171019123222-00533.warc.gz | 977,565,891 | 13,899 | # Local Linear Approximation vs Linearization
1. Nov 10, 2013
### Lebombo
Are Local Linear Approximation, Linear Approximation, and Linearization all the same thing?
Question is, I learned about something called Local Linear Approximation in Calc 1. Now in Calc 2, the topic of Linearization from Calc 1 was mentioned. But I never did anything that was referred to as Linearization. The closest sounding topic was Local Linear Approximation. So are they the same exact topic just referred to with different names?
2. Nov 10, 2013
### HallsofIvy
Staff Emeritus
Perhaps a slight difference. When you have a "local linear approximation", what is "local" is already decided. That is, you have a specific point at which you approximate the curve by a line. The "linearization" of a function requires a choice of the point at which you will linearize. "Linearization" of the same function at two different "x" values will give two different linear functions.
Once you have chosen that point at which to linearize the function, then linearization about that point is the local linear approximation. | 246 | 1,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-43 | longest | en | 0.960462 |
https://brilliant.org/problems/expected-raffle-value/ | 1,620,897,915,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990584.33/warc/CC-MAIN-20210513080742-20210513110742-00279.warc.gz | 158,366,141 | 8,328 | # Expected Raffle Value
Probability Level 4
A school is running a raffle for two prizes. $59$ tickets were sold for the raffle, numbered $1, \ldots, 59$. All the tickets are put into a hat and a teacher picks out two tickets which have numbers $i$ and $j$ from the hat. What is the expected value of $|i - j|$?
× | 88 | 315 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-21 | latest | en | 0.956557 |
https://statkat.com/stattest.php?t=18&t2=17&t3=13 | 1,713,482,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00474.warc.gz | 494,613,796 | 8,714 | # Spearman's rho - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons (max. of 3) by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Spearman's rho
Kruskal-Wallis test
Regression (OLS)
You cannot compare more than 3 methods
Variable 1Independent/grouping variableIndependent variables
One of ordinal levelOne categorical with $I$ independent groups ($I \geqslant 2$)One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables
Variable 2Dependent variableDependent variable
One of ordinal levelOne of ordinal levelOne quantitative of interval or ratio level
Null hypothesisNull hypothesisNull hypothesis
H0: $\rho_s = 0$
Here $\rho_s$ is the Spearman correlation in the population. The Spearman correlation is a measure for the strength and direction of the monotonic relationship between two variables of at least ordinal measurement level.
In words, the null hypothesis would be:
H0: there is no monotonic relationship between the two variables in the population.
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H0: the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H0: the population scores in any of the $I$ groups are not systematically higher or lower than the population scores in any of the other groups
Formulation 2:
• H0: P(an observation from population $g$ exceeds an observation from population $h$) = P(an observation from population $h$ exceeds an observation from population $g$), for each pair of groups.
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
$F$ test for the complete regression model:
• H0: $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
or equivalenty
• H0: the variance explained by all the independent variables together (the complete model) is 0 in the population, i.e. $\rho^2 = 0$
$t$ test for individual regression coefficient $\beta_k$:
• H0: $\beta_k = 0$
in the regression equation $\mu_y = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K$. Here $x_i$ represents independent variable $i$, $\beta_i$ is the regression weight for independent variable $x_i$, and $\mu_y$ represents the population mean of the dependent variable $y$ given the scores on the independent variables.
Alternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\rho_s \neq 0$
H1 right sided: $\rho_s > 0$
H1 left sided: $\rho_s < 0$
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
• H1: not all of the population medians for the $I$ groups are equal
Else:
Formulation 1:
• H1: the poplation scores in some groups are systematically higher or lower than the population scores in other groups
Formulation 2:
• H1: for at least one pair of groups:
P(an observation from population $g$ exceeds an observation from population $h$) $\neq$ P(an observation from population $h$ exceeds an observation from population $g$)
$F$ test for the complete regression model:
• H1: not all population regression coefficients are 0
or equivalenty
• H1: the variance explained by all the independent variables together (the complete model) is larger than 0 in the population, i.e. $\rho^2 > 0$
$t$ test for individual regression coefficient $\beta_k$:
• H1 two sided: $\beta_k \neq 0$
• H1 right sided: $\beta_k > 0$
• H1 left sided: $\beta_k < 0$
AssumptionsAssumptionsAssumptions
• Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Note: this assumption is only important for the significance test, not for the correlation coefficient itself. The correlation coefficient itself just measures the strength of the monotonic relationship between two variables.
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
• In the population, the residuals are normally distributed at each combination of values of the independent variables
• In the population, the standard deviation $\sigma$ of the residuals is the same for each combination of values of the independent variables (homoscedasticity)
• In the population, the relationship between the independent variables and the mean of the dependent variable $\mu_y$ is linear. If this linearity assumption holds, the mean of the residuals is 0 for each combination of values of the independent variables
• The residuals are independent of one another
Often ignored additional assumption:
• Variables are measured without error
Also pay attention to:
• Multicollinearity
• Outliers
Test statisticTest statisticTest statistic
$t = \dfrac{r_s \times \sqrt{N - 2}}{\sqrt{1 - r_s^2}}$
Here $r_s$ is the sample Spearman correlation and $N$ is the sample size. The sample Spearman correlation $r_s$ is equal to the Pearson correlation applied to the rank scores.
$H = \dfrac{12}{N (N + 1)} \sum \dfrac{R^2_i}{n_i} - 3(N + 1)$
Here $N$ is the total sample size, $R_i$ is the sum of ranks in group $i$, and $n_i$ is the sample size of group $i$. Remember that multiplication precedes addition, so first compute $\frac{12}{N (N + 1)} \times \sum \frac{R^2_i}{n_i}$ and then subtract $3(N + 1)$.
Note: if ties are present in the data, the formula for $H$ is more complicated.
$F$ test for the complete regression model:
• \begin{aligned}[t] F &= \dfrac{\sum (\hat{y}_j - \bar{y})^2 / K}{\sum (y_j - \hat{y}_j)^2 / (N - K - 1)}\\ &= \dfrac{\mbox{sum of squares model} / \mbox{degrees of freedom model}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square model}}{\mbox{mean square error}} \end{aligned}
where $\hat{y}_j$ is the predicted score on the dependent variable $y$ of subject $j$, $\bar{y}$ is the mean of $y$, $y_j$ is the score on $y$ of subject $j$, $N$ is the total sample size, and $K$ is the number of independent variables.
$t$ test for individual $\beta_k$:
• $t = \dfrac{b_k}{SE_{b_k}}$
• If only one independent variable:
$SE_{b_1} = \dfrac{\sqrt{\sum (y_j - \hat{y}_j)^2 / (N - 2)}}{\sqrt{\sum (x_j - \bar{x})^2}} = \dfrac{s}{\sqrt{\sum (x_j - \bar{x})^2}}$
with $s$ the sample standard deviation of the residuals, $x_j$ the score of subject $j$ on the independent variable $x$, and $\bar{x}$ the mean of $x$. For models with more than one independent variable, computing $SE_{b_k}$ is more complicated.
Note 1: mean square model is also known as mean square regression, and mean square error is also known as mean square residual.
Note 2: if there is only one independent variable in the model ($K = 1$), the $F$ test for the complete regression model is equivalent to the two sided $t$ test for $\beta_1.$
n.a.n.a.Sample standard deviation of the residuals $s$
--\begin{aligned} s &= \sqrt{\dfrac{\sum (y_j - \hat{y}_j)^2}{N - K - 1}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned}
Sampling distribution of $t$ if H0 were trueSampling distribution of $H$ if H0 were trueSampling distribution of $F$ and of $t$ if H0 were true
Approximately the $t$ distribution with $N - 2$ degrees of freedom
For large samples, approximately the chi-squared distribution with $I - 1$ degrees of freedom.
For small samples, the exact distribution of $H$ should be used.
Sampling distribution of $F$:
• $F$ distribution with $K$ (df model, numerator) and $N - K - 1$ (df error, denominator) degrees of freedom
Sampling distribution of $t$:
• $t$ distribution with $N - K - 1$ (df error) degrees of freedom
Significant?Significant?Significant?
Two sided:
Right sided:
Left sided:
For large samples, the table with critical $X^2$ values can be used. If we denote $X^2 = H$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
$F$ test:
• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$
$t$ Test two sided:
$t$ Test right sided:
$t$ Test left sided:
n.a.n.a.$C\%$ confidence interval for $\beta_k$ and for $\mu_y$, $C\%$ prediction interval for $y_{new}$
--Confidence interval for $\beta_k$:
• $b_k \pm t^* \times SE_{b_k}$
• If only one independent variable:
$SE_{b_1} = \dfrac{\sqrt{\sum (y_j - \hat{y}_j)^2 / (N - 2)}}{\sqrt{\sum (x_j - \bar{x})^2}} = \dfrac{s}{\sqrt{\sum (x_j - \bar{x})^2}}$
Confidence interval for $\mu_y$, the population mean of $y$ given the values on the independent variables:
• $\hat{y} \pm t^* \times SE_{\hat{y}}$
• If only one independent variable:
$SE_{\hat{y}} = s \sqrt{\dfrac{1}{N} + \dfrac{(x^* - \bar{x})^2}{\sum (x_j - \bar{x})^2}}$
Prediction interval for $y_{new}$, the score on $y$ of a future respondent:
• $\hat{y} \pm t^* \times SE_{y_{new}}$
• If only one independent variable:
$SE_{y_{new}} = s \sqrt{1 + \dfrac{1}{N} + \dfrac{(x^* - \bar{x})^2}{\sum (x_j - \bar{x})^2}}$
In all formulas, the critical value $t^*$ is the value under the $t_{N - K - 1}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20).
n.a.n.a.Effect size
--Complete model:
• Proportion variance explained $R^2$:
Proportion variance of the dependent variable $y$ explained by the sample regression equation (the independent variables):
\begin{align} R^2 &= \dfrac{\sum (\hat{y}_j - \bar{y})^2}{\sum (y_j - \bar{y})^2}\\ &= \dfrac{\mbox{sum of squares model}}{\mbox{sum of squares total}}\\ &= 1 - \dfrac{\mbox{sum of squares error}}{\mbox{sum of squares total}}\\ &= r(y, \hat{y})^2 \end{align}
$R^2$ is the proportion variance explained in the sample by the sample regression equation. It is a positively biased estimate of the proportion variance explained in the population by the population regression equation, $\rho^2$. If there is only one independent variable, $R^2 = r^2$: the correlation between the independent variable $x$ and dependent variable $y$ squared.
• Wherry's $R^2$ / shrunken $R^2$:
Corrects for the positive bias in $R^2$ and is equal to $$R^2_W = 1 - \frac{N - 1}{N - K - 1}(1 - R^2)$$
$R^2_W$ is a less biased estimate than $R^2$ of the proportion variance explained in the population by the population regression equation, $\rho^2.$
• Stein's $R^2$:
Estimates the proportion of variance in $y$ that we expect the current sample regression equation to explain in a different sample drawn from the same population. It is equal to $$R^2_S = 1 - \frac{(N - 1)(N - 2)(N + 1)}{(N - K - 1)(N - K - 2)(N)}(1 - R^2)$$
Per independent variable:
• Correlation squared $r^2_k$: the proportion of the total variance in the dependent variable $y$ that is explained by the independent variable $x_k$, not corrected for the other independent variables in the model
• Semi-partial correlation squared $sr^2_k$: the proportion of the total variance in the dependent variable $y$ that is uniquely explained by the independent variable $x_k$, beyond the part that is already explained by the other independent variables in the model
• Partial correlation squared $pr^2_k$: the proportion of the variance in the dependent variable $y$ not explained by the other independent variables, that is uniquely explained by the independent variable $x_k$
n.a.n.a.Visual representation
--Regression equations with:
n.a.n.a.ANOVA table
--
Example contextExample contextExample context
Is there a monotonic relationship between physical health and mental health?Do people from different religions tend to score differently on social economic status? Can mental health be predicted from fysical health, economic class, and gender?
SPSSSPSSSPSS
Analyze > Correlate > Bivariate...
• Put your two variables in the box below Variables
• Under Correlation Coefficients, select Spearman
Analyze > Nonparametric Tests > Legacy Dialogs > K Independent Samples...
• Put your dependent variable in the box below Test Variable List and your independent (grouping) variable in the box below Grouping Variable
• Click on the Define Range... button. If you can't click on it, first click on the grouping variable so its background turns yellow
• Fill in the smallest value you have used to indicate your groups in the box next to Minimum, and the largest value you have used to indicate your groups in the box next to Maximum
• Continue and click OK
Analyze > Regression > Linear...
• Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Independent(s)
JamoviJamoviJamovi
Regression > Correlation Matrix
• Put your two variables in the white box at the right
• Under Correlation Coefficients, select Spearman
• Under Hypothesis, select your alternative hypothesis
ANOVA > One Way ANOVA - Kruskal-Wallis
• Put your dependent variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
Regression > Linear Regression
• Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates
• If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well
• Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'
Practice questionsPractice questionsPractice questions | 3,758 | 14,212 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-18 | latest | en | 0.856859 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_TERM/TRS/TRCSR/Ex15_Luc98_Z.trs.Thm17:POLO_7172_DP:NO.html.lzma | 1,718,716,861,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861752.43/warc/CC-MAIN-20240618105506-20240618135506-00095.warc.gz | 88,066,313 | 2,995 | Term Rewriting System R:
[X, Y, X1, X2, Z]
and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(ns(X)) -> s(X)
activate(X) -> X
Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
AND(true, X) -> ACTIVATE(X)
IF(true, X, Y) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Y)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FROM(X) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(nfrom(X)) -> FROM(X)
ACTIVATE(ns(X)) -> S(X)
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polynomial Ordering`
Dependency Pairs:
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
FROM(X) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> FROM(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Y)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
Rules:
and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(ns(X)) -> s(X)
activate(X) -> X
The following dependency pair can be strictly oriented:
ACTIVATE(nfrom(X)) -> FROM(X)
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(FROM(x1)) = x1 POL(n__from(x1)) = 1 + x1 POL(0) = 0 POL(cons(x1, x2)) = x1 + x2 POL(FIRST(x1, x2)) = x1 + x2 POL(s(x1)) = x1 POL(ACTIVATE(x1)) = x1 POL(ADD(x1, x2)) = x1 + x2 POL(n__add(x1, x2)) = x1 + x2 POL(n__first(x1, x2)) = x1 + x2
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Dependency Graph`
Dependency Pairs:
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
FROM(X) -> ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Y)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
Rules:
and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(ns(X)) -> s(X)
activate(X) -> X
Using the Dependency Graph the DP problem was split into 1 DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳DGraph`
` ...`
` →DP Problem 3`
` ↳Polynomial Ordering`
Dependency Pairs:
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Y)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
Rules:
and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(ns(X)) -> s(X)
activate(X) -> X
The following dependency pair can be strictly oriented:
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(0) = 0 POL(cons(x1, x2)) = x1 + x2 POL(FIRST(x1, x2)) = x1 + x2 POL(s(x1)) = x1 POL(ACTIVATE(x1)) = x1 POL(ADD(x1, x2)) = x1 + x2 POL(n__add(x1, x2)) = x1 + x2 POL(n__first(x1, x2)) = 1 + x1 + x2
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳DGraph`
` ...`
` →DP Problem 4`
` ↳Dependency Graph`
Dependency Pairs:
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Y)
Rules:
and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(ns(X)) -> s(X)
activate(X) -> X
Using the Dependency Graph the DP problem was split into 1 DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳DGraph`
` ...`
` →DP Problem 5`
` ↳Polynomial Ordering`
Dependency Pairs:
Rules:
and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(ns(X)) -> s(X)
activate(X) -> X
The following dependency pairs can be strictly oriented:
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(0) = 0 POL(s(x1)) = 1 + x1 POL(ACTIVATE(x1)) = x1 POL(ADD(x1, x2)) = x1 + x2 POL(n__add(x1, x2)) = x1 + x2
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳DGraph`
` ...`
` →DP Problem 6`
` ↳Polynomial Ordering`
Dependency Pairs:
Rules:
and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(ns(X)) -> s(X)
activate(X) -> X
The following dependency pair can be strictly oriented:
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(0) = 0 POL(ACTIVATE(x1)) = x1 POL(ADD(x1, x2)) = x2 POL(n__add(x1, x2)) = 1 + x2
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳DGraph`
` ...`
` →DP Problem 7`
` ↳Dependency Graph`
Dependency Pair:
Rules:
and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), nfirst(activate(X), activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(ns(X)) -> s(X)
activate(X) -> X
Using the Dependency Graph resulted in no new DP problems.
Termination of R successfully shown.
Duration:
0:00 minutes | 2,740 | 8,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-26 | latest | en | 0.428147 |
http://www.varsitytutors.com/gmat_math-help/calculating-simple-interest | 1,484,628,671,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279468.17/warc/CC-MAIN-20170116095119-00187-ip-10-171-10-70.ec2.internal.warc.gz | 753,261,116 | 27,034 | # GMAT Math : Calculating simple interest
## Example Questions
← Previous 1
### Example Question #1 : Calculating Simple Interest
Grandpa Jack wants to help pay for college for his grandson, Little Jack. Little Jack is currently 8 years old. Grandpa Jack makes a one-time deposit into an account that earns simple interest every year. Grandpa Jack invests $10,000 now and in ten years, that will grow to$15,000. What rate of simple interest did Grandpa Jack receive?
3%
2%
5%
6%
4%
5%
Explanation:
To calculate simple interest, the formula is
where stands for Future Value, stands for Present Value, stands for the interest rate, and stands for the number of periods (in this case years). So plugging in,
Solving this we get
or 5%
ALTERNATE SOLUTION:
Another way of finding this is to calculate the amount of interest per year. Since this is simple interest, Grandpa Jack earns the same amount of interest per year. The total interest earned is 15,000-10,000= 5,000. $5,000 over 10 years, equates to$500 per year. $500 divided by the original$10,000 is .05, or 5%.
### Example Question #2 : Calculating Simple Interest
How much interest would an investment yield if the principal of is invested for years at a simple interest?
Explanation:
### Example Question #3 : Calculating Simple Interest
What rate does Johnny need if he would like to yield in interest from a principal of in years?
Explanation:
### Example Question #4 : Calculating Simple Interest
A bank offers a business a loan in the amount of \$13,000 with a simple annual interest rate of 9%. How much will the business owe the bank after 3 years?
Explanation:
The accrual of simple interest can be found in two steps. First, multiply the principal amount by the interest rate. Second, multiply that result by the number of years during which interest will accrue.
The question asks for the total amount that the business will owe the bank, so we must add the interest accrued to the principal amount.
### Example Question #4 : Calculating Simple Interest
John has to invest for years for return. How much simple interest will his investment yield?
Explanation:
Simple interest formula:
### Example Question #6 : Calculating Simple Interest
How many years would it take Marissa to make in interest using a simple interest investment of in principal at 5% return?
Explanation:
Simple interest formula is:
Therefore, solving for the time factor:
### Example Question #7 : Calculating Simple Interest
Using the simple interest formula, what initial investment does Terry have to make to earn in interest if there is an return for years?
Explanation:
Simple interest formula:
Therefore, solving for the principal factor in the equation:
### Example Question #5 : Calculating Simple Interest
What return does Florence have to yield in order to make in interest off of a simple investment of for years?
Explanation:
Simple interest formula:
Therefore, solving for the rate factor in the equation:
### Example Question #9 : Calculating Simple Interest
Jesse has to invest for years for return. How much simple interest will his investment yield?
Explanation:
Simple interest formula:
### Example Question #10 : Calculating Simple Interest
How many years would it take Julissa to make in interest using a simple interest investment of at a return?
Explanation:
Simple interest formula is:
Therefore, solving for the time factor:
← Previous 1 | 753 | 3,472 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2017-04 | latest | en | 0.915081 |
https://schoolworkhelper.net/copper-penny-to-silver-lab-answers/ | 1,726,559,908,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00040.warc.gz | 451,912,739 | 105,374 | Purpose: Find silver produced and compare to predicted.
Cu + 2AgNo3 → Cu(NO3)2 + 2Ag
Procedure:
-in small beaker (100-250 mL), weigh beaker
-Add 3.00 OR LESS of AgNo3, weigh AgNo3
-Weigh copper penny
-Heat and dissolve (warm, not boil)
-Put copper penny into AgNo3(aq) –observe
-Allow to sit overnight
-Wash/ decant, then dry
-weigh silver and penny
Data: beaker= 67.95 g penny= 3.14 g silver (after drying) = 2.03 g copper reacted= .58 g
Calculations: use stoichiometry to find Ag predicted
1. grams → mol Cu
(penny – copper reacted) 2.56 x 1/63.55= .0403
2. mol →mol (ratio)
.0403 x 2/1= .0806 mol Ag
3. mol → grams
.0806 mol Ag x 107.9/1= 8.70 g
William Anderson (Schoolworkhelper Editorial Team)
William completed his Bachelor of Science and Master of Arts in 2013. He current serves as a lecturer, tutor and freelance writer. In his spare time, he enjoys reading, walking his dog and parasailing. Article last reviewed: 2022 | St. Rosemary Institution © 2010-2024 | Creative Commons 4.0 | 320 | 1,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-38 | latest | en | 0.838546 |
http://wcipeg.com/problem/noi08p1 | 1,529,507,332,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863650.42/warc/CC-MAIN-20180620143814-20180620163814-00043.warc.gz | 342,925,153 | 4,426 | ## Day 1, Problem 1 - Masquerade Party
The annual masquerade ball has begun, and DongDong will be enthusiastically participating. This year's masks are specially tailored by the host. Each person going to the party can choose a mask of their liking before they enter. Each mask is labeled with a number, and the host will tell this number to the person wearing the mask.
To give the party a more mystifying atmosphere, the host has divided the masks to k (k ≥ 3) types, and using special technology, he has also marked each mask with their corresponding number. Only people wearing type i masks will be able to see the numbers of the people wearing type i + 1 masks. People wearing type k masks will be able to see the numbers of people wearing type 1 masks.
Guests of the party will not know just how many types of masks there are in the party. However, this has made DongDong very curious, so he decided to calculate for himself how many types of masks there are. Thus, he starts collecting information from the crowd of people.
DongDong's information tells him which person wearing one mask number can see which other mask numbers. For example, the person wearing mask number 2 can see the number of mask number 5. DongDong will also see some mask numbers himself, and he will also use this to make his information more complete.
Since not everybody can just remember all of the numbers they have seen, DongDong's information is not guaranteed to be perfectly complete. Now you must calculate, based on DongDong's current set of information, the maximum possible and minimum possible number of types of masks there may be. Since the host has already announced that k ≥ 3, you must take this extra information into consideration.
### Input Format
The first line of input contains two space-separated integers n and m, representing the total number of masks and the total pieces of information that DongDong has collected, respectively.
For the following m lines, each line contains two space-separated integers a and b, indicating that the person wearing mask a can see the number of mask b. Identical pairs of a, b may appear multiple times in the input data.
### Output Format
Output two space-separated integers on one line. The first integer is the maximum possible types of masks there may be, and the second integer is the minimum possible types of masks there may be. If there is no way to divide the masks into at least 3 types making the information complete, then DongDong has to believe that the information is erroneous. In this case, output -1 twice.
```6 5
1 2
2 3
3 4
4 1
3 5
```
```4 4
```
```3 3
1 2
2 1
2 3
```
```-1 -1
```
### Constraints
For 50% of the test cases, n ≤ 300, m ≤ 1000;
For 100% of the test cases, n ≤ 100000, m ≤ 1000000.
Point Value: 25 (partial)
Time Limit: 1.00s
Memory Limit: 128M | 671 | 2,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-26 | latest | en | 0.954564 |
https://dsp.stackexchange.com/questions/68215/duality-property-for-dft | 1,656,577,257,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00796.warc.gz | 273,047,454 | 67,706 | # Duality Property for DFT
I was watching a youtube video for the duality property for continuous time Fourier transforms, which shows that if Fourier transform of $$x(t)$$ is $$X(\omega)$$ then Fourier transform of $$X(t)$$ is $$2\pi x(-\omega)$$
Then how will duality look like in the case of the DFT?
Will it look like below?
If DFT of $$x[n]$$ is $$X[k]$$ then DFT of $$X[n]$$ is $$2\pi x[-k]$$
• You can check this answer : dsp.stackexchange.com/a/68103/49439 Jun 10, 2020 at 8:57
• And, if DFT of $x[n]$ is $X[k]$, then DFT of $X[n]$ will be $Nx[N-k]$. You can derive this by replacing $X[n]$ by it's analysis expression in the equation : $\displaystyle Y[k] = \sum^{N-1}_{n=0}X[n]e^{-j\frac{2\pi}{N}nk}$. Jun 10, 2020 at 9:31
• you should probably put this into an answer, Rook. maybe with the math that proves it. Jun 10, 2020 at 15:34
Duality in DFT would mean that if $$x[n]$$ has DFT coefficients as $$X[k]$$, then DFT of $$X[n]$$ would be $$Nx[(N-k) \mod N]$$
Proof:
Given, $$X[k] = \sum^{N-1}_{n=0}x[n]e^{-j\frac{2\pi}{N}nk}, k=0,1,2,3,...,(N-1)$$ If we take DFT of the sequence $$X[n]$$, then what we get is the following : $$Y[k] = \sum^{N-1}_{n=0}X[n]e^{-j\frac{2\pi}{N}nk} = N \left(\frac{1}{N}\sum^{N-1}_{n=0}X[n]e^{j\frac{2\pi}{N}n(-k)} \right)$$ Notice that the expression between "()" is the synthesis expression with $$k^{th}$$ frequency index replaced by $$(N-k \mod N)^{th}$$ index. Because, in DFT expression, $$k$$ can only take indices $$0,1,2,3,...,(N-1)$$, hence, we cannot have $$-k$$ as frequency index, but rather, $$((N-k) \mod N)$$ $$Y[k] = N \left(\frac{1}{N}\sum^{N-1}_{n=0}X[n]e^{j\frac{2\pi}{N}n(N-k)} \right) = Nx[(N-k) \mod N]$$
So, the way to interpret this is, you get a scaled and inverted sequence back when you take DFT of DFT, but $$x[0]$$ remains at $$0^{th}$$ index.
So, DFT of DFT of time-domain $$x[n]$$ gives $$\{Nx[0], Nx[N-1], Nx[N-2], ..., Nx[2], Nx[1]\}$$
Scaling by $$N$$ is the consequence of not dividing by $$\frac{1}{\sqrt{N}}$$ when taking DFT and incorporating this factor into IDFT expression.
• How do you handle/use ( Mod N) with Nx(N-k)
– LECS
Jun 12, 2020 at 6:56
• Since in math , Mod is what is left after division as remainder but i can not see any division in Nx(N-k)
– LECS
Jun 12, 2020 at 6:59
• @abtj N-k mod N means what's left as remainder when you divide N-k by N. For k=0, it's N mod N meaning 0, and hence when reversing, $x[0]$ remains at the same place whereas other k's switch places with N-k. Jun 12, 2020 at 10:43
Initial comment: duality refers to the strong similarity of mathematical expressions and properties in two different domains, here time and frequency. This is not only decorative or made to annoy learners. Duality helps a lot: one can derive results much faster, interpret classes of transformations more easily. To make this serious, you can check the Pontryagin duality with Fourier transforms.
Second comment: the classical continuous-time/continuous-frequency Fourier possesses some duality-related properties (on symmetry, shift, energy, convolution, etc.). By discretizing both time and frequency in the Discrete Fourier Transform (DFT), their developers have striven to keep, whenever possible, most of the initial properties. To me, the duality in the forward and inverse DFT is well-explained in the chapter The Discrete Fourier Transform (DFT). If you want to better observe duality between the indices, you can even modify the initial scaling factor. Thus, instead of a DFT normalized in amplitude, we can normalize it in energy. It can be useful to remark that we have two sequences $$x[n]$$ and $$X[k]$$ with the same length $$K=N$$. This is a bit artificial here, but we can rewrite the energy-normalization constant $$\sqrt{N}$$ as $$\nu_{KN}=\nu_{NK}=(KN)^{1/4}$$. Now, let $$\omega_k = 2\pi \frac{k}{N}f_s$$, then for $$n=0,1,\ldots,N-1$$, and $$k=0,1,\ldots,K-1$$ (with $$K=N$$)
$$X(\omega_k)=X[k] = \nu_{NK}\sum_{n=0}^{N-1}x[n]e^{-2\pi j \frac{n}{N}k}$$
and
$$x[n]= \nu_{KN}\sum_{k=0}^{K-1}X[k]e^{2\pi j \frac{k}{K}n}$$
where the duality in formulae is evident. The expression of a DFT of a DFT is already given by @DSP Rookie. The hidden message is that, when you have a formula or a code for the DFT, you also have one for the inverse DFT, provided you modify a bit the inputs and scaling, see for instance: Expressing the inverse DFT in terms of the DFT.
• This is not only decorative or made to annoy learners :D Sep 22, 2020 at 13:52 | 1,452 | 4,474 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 39, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-27 | longest | en | 0.786892 |
https://www.equine-psychotherapy.com/horses/how-much-can-a-horse-carry-solved.html | 1,685,588,096,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647525.11/warc/CC-MAIN-20230601010402-20230601040402-00169.warc.gz | 823,621,328 | 28,527 | # How Much Can A Horse Carry? (Solved)
The maximum weight a horse can carry is 400 pounds based on the 20% rule. Most horses can safely carry 20% of their body weight. So a large draft horse weighing 2,000 pounds can theoretically safely carry a 400-pound person.4
## Can a horse carry a 300 pound person?
Q: Can a horse carry a 300 pound person? Some horses can carry a 300 pound rider, but your balance is also important. If you don’t have a good balance then it’ll be very difficult for even the largest horses to comfortably carry the weight.
## Is there a weight limit to ride horses?
When horseback riding, the rule of thumb is that a horse can safely carry 20% of its body weight. So, if you weigh 250 pounds, you should aim to ride a horse that weighs 1,250 pounds or more. This will help ensure the horse’s safety and ability to work. Balance is also a key aspect of how much weight a horse can carry.
## Can a horse carry 100kg?
Maybe, maybe not. A “scientific study” has concluded that a horse cannot comfortably carry more than 10 percent of its own weight. According to The US Cavalry Manual of Horse Management (1941) a horse should not carry more than 20 percent of its own weight.
## Can a horse carry 30% of its body weight?
Horses carrying 25% and 30% of their body weight had higher heart and respiratory rates during exercise, and muscles that were more sore a day after exercise. So, in short, a good rule of thumb is that an average horse can carry 20% of his bodyweight (keeping in mind this is tack combined with the weight of the rider).
## Can a 500 pound person ride a horse?
The maximum weight a horse can carry is 400 pounds based on the 20% rule. Most horses can safely carry 20% of their body weight. So a large draft horse weighing 2,000 pounds can theoretically safely carry a 400-pound person. However, 20% of body weight is a safe, research-based estimate. 4
## What horse can carry 400 pounds?
The Suffolk Punch horse is usually between 16 and 17 hands tall with a weight of 2,000 – 2,200 lbs. This means the Suffolk Punch could easily carry a rider and saddle weight of 400+ pounds.
## Can a 200lb person ride a horse?
A: Laurie, the basic rule of thumb for a horse’s weight-carrying capacity is 20 percent of the horse’s weight, or, say, 200 pounds for a 1,000-pound horse. A strong, fit, well-coordinated but heavier rider can often be easier for a horse to bear than a weak, unfit, awkward but lighter rider.
## Do horses like to be ridden?
Most horses are okay with being ridden. As far as enjoying being ridden, it’s likely most horses simply tolerate it rather than liking it. However, many people argue that if horses wouldn’t want us to ride them, they could easily throw us off, which is exactly what some horses do.
## How do you know if you are too big for your horse?
If your feet are dragging on the floor or hitting poles when you are jumping, you should probably consider a larger horse… It is also true that riding a smaller or narrower horse can be more unbalancing than riding a wider or larger one and the gaits of larger horses differ from those of smaller ones.
## Can a horse be 14.5 hands?
Hands and Other Measurements A horse could never be said to be 14.5 inches, as the number after the decimal is not a fraction, but represents an entire inch. If a horse is 14.2 1/2 HH, that means he’s two and one-half inches over 14 hands.
## Is 60 too old to learn to ride a horse?
Well, the good news is that you’re never too old to ride a horse! * As long as you can manage to get in and out of the saddle, you’ll be able to embark on all the equine adventures you could wish for. Read on to discover our advice for learning to ride a horse as an adult!
## Can a horse carry a gorilla?
The rule of thumb is that a horse can safely carry 20% of its own weight, tack included. A 200 lb gorilla would be no more difficult than a 200 lb man for a horse weighing in at or over 1000 lbs, which is pretty average horse size.
## How much can a 500 kg horse carry?
As a general guideline in the UK, a rider should weigh no more than 10% of the horse’s bodyweight, but in the US, this limit is doubled to 20% of the horse’s weight. This means for a 500kg horse, the range for the maximum rider weight is large – 50kg in the UK (just less than 8st) and 100kg in the US (15st 10lb).
## How Much Weight Can a Horse Carry? (Weight Limit to Ride)
Despite the fact that horses are enormous and strong creatures, they have their limits. Any overloading can result in damage and a reduction in their capacity to function. What is the maximum amount of weight that a horse can carry? This is an important topic to ask whether you are new to horseback riding or an experienced rider who wants to learn more about the horse. Let’s have a look at this.
## How Much Weight Can a Horse Carry
The key to a safe and enjoyable ride is to keep an eye out for stiffness in the horse’s muscles. When transporting a human weighing more than 20% of the animal’s body weight, you should be aware that the animal will begin to experience substantial strain and suffering. Consequently, the maximum weight carrying capacity of the horse, including the rider and saddle, is 20 percent of its maximum carrying capacity. Keep in mind that the average western saddle weighs around 50 pounds (22.7 kg), whereas the average English saddle weighs approximately 20 pounds (9 kg).
Furthermore, stockier horses can carry more weight than the typical horse, making them more ideal for riders who weigh more than the average.
Never allow a young or aged horse to carry an excessive amount of weight in order to avoid damage.
## Proper Horses’ Sizes for Particular Riders
What is crucial for safe riding is that the horse’s size is appropriate to your own height and body weight. For example, if you are significantly larger than the horse, you will find it difficult to maintain your balance for the whole horse ride. When you are too short for the horse, on the other hand, you will have difficulty using your legs efficiently. For example, improperly wrapping the horse’s legs around the horse’s body might cause the horse discomfort. The breadth and barrel size of the horse will be acceptable for you to ride securely only if you wrap your legs over its sides in the appropriate manner.
## weight limit to ride a horse
As previously stated, 20 percent reflects the greatest amount of a person’s and equipment’s total weight that a horse is comfortable carrying. Riders should not weigh more than 15 percent of the horse’s total weight in most circumstances.
### Weight limit to ride a horse
Horse’s weight Weight carrying capacity – 15% Weight carrying capacity – 20% 700 pounds (317.5 kg) 105 pounds (47.5 kg) 140 pounds (63.5 kg) 800 pounds (363 kg) 120 pounds (54.5 kg) 160 pounds (72.5 kg) 900 pounds (408 kg) 135 pounds (61 kg) 180 pounds (81.5 kg) 1,000 pounds (453.5 kg) 150 pounds (68 kg) 200 pounds (91 kg) 1,100 pounds (499 kg) 165 pounds (75 kg) 220 pounds (99.5 kg) 1,200 pounds (544 kg) 180 pounds (81.5 kg) 240 pounds (109 kg) 1,300 pounds (590 kg) 195 pounds (88.5 kg) 260 pounds (118 kg) 1,400 pounds (635 kg) 210 pounds (95 kg) 280 pounds (127 kg) 1,500 pounds (680 kg) 225 pounds (102 kg) 300 pounds (136 kg) 1,600 pounds (726 kg) 240 pounds (109 kg) 320 pounds (145 kg) 1,700 pounds (771 kg) 255 pounds (115.5 kg) 340 pounds (154 kg) 1,800 pounds (816.5 kg) 270 pounds (122.5 kg) 360 pounds (163 kg) 1,900 pounds (862 kg) 285 pounds (129 kg) 380 pounds (172 kg) 2,000 pounds (907 kg) 300 pounds (136 kg) 400 pounds (181.5 kg)
In any case, if you want to know what horse kind will be able to appropriately transport you, you should consult one of the online calculators.
## An Ideal Horse for Riding
These are the characteristics that are most frequently taken into account when selecting a horse for riding:
### Breed
It is essential to concentrate on breeds because some of them are more slender, such as the Arabian, while others are stockier, such as the Haflinger, and so on.
You should select the one that is the most appropriate for your riding abilities and your personal preferences.
### Confirmation
Confirmation refers to the form or structure of the horse, as well as its proportions. When acquiring a horse, it is important to consider its intended use because this impacts the weight bearing capabilities of the animal.
### Training
A horse that has not been properly taught will demand a lighter rider since it is not as balanced as a horse that has been properly trained.
### Fitness
A horse that has never been used for regular labor and has not been allowed to run freely for an extended period of time is likely to be in bad condition. As a result, it will be better ideal for riders who are less in weight.
### Body condition
The amount of fat present in the horse’s body is referred to as its body condition. A horse that is underweight or overweight will always require a lighter rider since its carrying capacity will not be at its maximum level.
### Horse’s age
The carrying capacity of a horse is also determined by the horse’s age, since the very young and old animals demand a lighter rider.
### Horse breed that fits particular rider height
Rider height Horse and pony breeds Short rider, up to 65 inches (165 cm) Haflinger, Appaloosa, Fjord, Dales Pony, Highland Pony, Irish Cob, Hanoverian Average rider from 65 to 70 inches (165 – 178 cm) Irish Draught, Percheron, Fresian, Irish Cob, Haflinger, Fjord, Draft Cross, Cleveland Bay, Quarter Horse, Lusitano, Paint, Hanoverian, Knabstrupper, Holsteiner, Morgan Tall rider, over 70 inches (178 cm) Clydesdale, Irish Draught, Percheron, Draft Cross, Cleveland Bay, Hanoverian, Holsteiner, Knabstrupper
### Rider’s age
When acquiring a horse for a younger rider, selecting a horse that is taller or stockier is the most appropriate option for the situation. It will be proportionate to the rider’s potential adult height and weight. A shorter horse, on the other hand, is more ideal for seniors since it makes mounting and dismounting easier and reduces the chance of injury.
## The Best Horse Breeds for Beginners
Equine companions that are simple to teach and retain positive memories of their training are the most suitable for inexperienced riders.
### Morgan
Morgans are a kind and fearless breed that is always willing to satisfy its owners. It will put out great effort in working with any riders and will be consistent in determining their needs.
### Friesian
Because of its lively demeanor and proclivity to roll around in the mud, the Friesian is sometimes compared to a Labrador Retriever (Lab). In addition, because horses are loyal to their owners, it is the perfect option for riders who desire a long-term engagement with their mount.
### Icelandic
Many beginning riders are intimidated by the prospect of riding a large horse, therefore the Icelandic horse is a good compromise. A rider, particularly an inexperienced one, will find it more comfortable because of its height and the smooth rendition of a rapid stroll that it offers.
### Tennessee Walking Horse
Its walk is so smooth that you may comfortably have a cup of tea while riding it. Additionally, it is a fantastic answer for individuals who have saddle soreness after a lengthy riding session.
### Connemara Pony
Connemara Pony started out as a farm worker and eventually became more. Due to the fact that this horse stands around 14 hands or 56 inches (1.42 m) tall, it is ideal for accommodating shorter equestrian riders. Despite the fact that it is a pony of a smaller breed, it is an athletic animal that will become a faithful companion in future contests for you.
### Welsh Cob
Welsh Cob horses were developed via crossbreeding between the Welsh Mountain Pony and bigger breeds such as Arabians or Thoroughbreds. The fact that it may be used in a variety of disciplines makes it popular among European riding schools.
## The Best Horse Breeds for Plus-sized Riders
Despite the fact that horse size and strength can vary within a single breed, some horses are regarded to be the ideal choice for plus-size riders due to their size and strength.
These horses have good, firm hooves, thicker bones, and are often larger and stronger than other breeds.
### Appaloosa
Appaloosas belong into the group of shorter horses, making them more suited for riders who are shorter or larger in stature than other breeds. They tend to have placid demeanor, which allows them to be an excellent fit for riders of all skill levels. Traditionally, this breed has been employed in western disciplines like as cutting and reining, among others.
### Clydesdale
As a result of their height and the size of their barrels, Clydesdales are the greatest horses for tall and plus-sized riders. However, despite their outgoing personality, they are not appropriate for all riding styles. Although they are not very adept at jumping, they are an excellent choice for pleasure riding on trails, pulling, and driving duties.
### Dales Pony
Dales Ponies are a suitable match for riders who are shorter in stature and average in height and weight. They are well-known for their courageous demeanor and for their outstanding performance in driving contests. They are also good jumpers, excellent for dressage, and well-suited for leisure horseback riding as well.
### Hanoverian
Despite the fact that most of them aren’t particularly stocky, Hanoverians are suitable for riders of all sizes. Because of their exceptional athletic abilities, they make excellent sport horses. This is why jumpers for dressage and competitions are made out of them.
## The Horse’s Purpose
Another important consideration when acquiring a horse is determining whether or not the horse is fit for the discipline in which it will be utilized. Horses of a lower stature are more suited to barrel racing or gymkhanas, whilst taller animals are better suited to dressage. Furthermore, some disciplines need a greater amount of horse labor. This covers elements such as how frequently and for how long you bike, as well as how strenuous the ride is. The greater the intensity of the effort, the less weight the horse is capable of towing at a given time.
## Summary
Because the average rider is becoming heavier, it is important to ensure that the horse does not get overburdened. As a result, if you keep your animal’s weight within reasonable bounds, its performance in the duties you assign it will be improved. As a consequence, you will have a long-term partner in your endeavors in the coming years.
## How Much Weight Can a Horse Carry?
While the majority of healthy horses are capable of carrying a rider and a saddle, they do have their limitations. In recent years, experts have established a weight threshold at which a rider is too heavy for a horse to comfortably carry. According to the experts, their conclusions are based on thorough data gathered from eight horses that were ridden while carrying ranging from 15 to 30 percent of their total weight. The horses’ weights ranged from 400 to 625 kilos, depending on their size (885 to 1375 pounds).
Physical indicators altered dramatically when they were packing weights of 25 percent, and they got much more pronounced when they were packing weights of 30 percent.
Following a day of trotting and cantering with heavier weights, the horses’ muscles showed much more discomfort and tightness than the previous day.
In light of these findings, the authors of the research urge that horses should not be loaded with more than 20% of their body weight at any given time.
It’s interesting to note that this research from the Ohio State University Agricultural Technical Institute has come to the same conclusion as the US Calvary Manuals of Horse Management, which were first published in 1920.
## Guidelines for weight-carrying capacity of horses
Following the 20-percent guideline and taking into account factors such as fitness, conformation, exercise level, and equipment, you may calculate the acceptable carrying capacity of your horse. Understanding your horse’s weight-bearing capability will help you to ensure their wellbeing and a long-lasting connection for many years to come.
## Common activities where horses carry weight
Horse owners in the United States frequently employ their animals for both leisure and competitive riding purposes. According to a recent government census, the most common use of horses in the United States is for enjoyment (47 percent), which might include everything from trail riding to arena training. Farm and ranch work, which might entail sorting livestock, carrying equipment in packs, and dragging carts or timber, came in second with a quarter of the vote. We must take into consideration the welfare of our equine companions while asking them to partake in these activities with us.
Researchers have looked into the optimal weight bearing capacity of horses in order to assist in determining what this should be.
## How much weight can your horse carry?
In 2008, researchers at Ohio University examined the effects of rider and tack weight on the performance of horses. While carrying weights weighing 15, 20, 25, and 35% of their bodyweight, the horses’ heart rate, breathing rate, rectal temperature, loin muscle condition, and overall health were all measured. Using an average adult light riding horse, the researchers discovered that they could comfortably carry around 20 percent of their optimal bodyweight. This outcome is consistent with the value recommended by the Certified Horsemanship Association and the United States Cavalry Manuals of Horse Management, both of which were published in 1920.
## What affects how much your horse can carry?
| Researchers in Ohio discovered that the breadth of the loin and the size of the cannon bone are related to the animal’s weight bearing capabilities. Horses with broader loins and larger cannon bone circumferences experienced less muscular discomfort as the weight of their load was raised. As a result of this discovery, the 20-percent rule appears to be an excellent starting point. Another research looked at Arabian endurance horses who were carrying between 20 and 30 percent of their body weight at the time of the investigation.
• smaller cannon bone circumference).
• In addition, despite their small stature, Icelandic horses are sometimes seen hauling adult riders on their backs.
• They discovered that the horses did not experience muscular pain after one to two days of labor and that the majority of them were able to operate aerobically (with oxygen) until they achieved a weight load of 23 percent of their body weight.
• When oxygen is scarce, the horse must seek alternate paths, which can result in the accumulation of lactic acid and the pain of the muscles in the process.
• The reduction in stride length, on the other hand, had no effect on stride symmetry.
• In addition, as compared to a horse with long legs and a long, weak back, this horse will have a lower center of gravity.
• Weight bearing ability may also be affected by the fitness and balance of the horse and rider.
When a horse is out of shape or out of balance, he will lack the strength to properly elevate his back and support the weight of the rider while keeping his or her own balance.
In addition, an inexperienced rider can throw off the balance of a horse as they attempt to maintain proper riding posture while simultaneously combating the consequences of muscular exhaustion.
These activities should only be attempted if the horses and riders are in good enough condition to do so.
Your saddle should be properly fitted to your horse’s back so that the weight of the rider is distributed as evenly as possible without pinching or creating muscular tightness.
A balanced, smooth surface for weight bearing should be achieved via trimming the hooves.
Consult a respected equine specialist such as your farrier or veterinarian for guidance on whether to add shoes or protective boots to horses whose hooves are wearing down faster than they can regenerate or for horses with thin soles before making this decision. In 2019, the situation was reviewed.
## 8 Best Horse Breeds for Heavy Riders (Over 250 pounds)
If you weigh 250 lbs or more, you can still enjoy horseback riding as long as you ride the appropriate horse for your weight. The safety of the horse and rider should always take first, which is why it is critical to pick the correct horse breed that is capable of carrying hefty riders. Please keep in mind that we are not providing medical or veterinary advice on this website. Always consult with your veterinarian before riding to ensure that you are riding your horse in a safe manner.
## Finding the Right Horse Breed for Heavy Riders
As a general rule of thumb, a horse can comfortably carry 20 percent of its own body weight during horseback riding. As a result, if you weigh 250 pounds, you should strive to ride a horse that weighs at least 1,250 pounds. This will assist to assure the horse’s safety as well as his capacity to perform his job. The ability of a horse to carry a large amount of weight is also dependent on its ability to maintain balance. Fit and balanced riders are more likely to have an easier time supporting their horses than uncoordinated riders who might throw off the horse’s balance and induce tiredness.
## 8 Best Horse Breeds for Heavy Riders
Riding larger breeds of horses is frequently the best option for riders who weigh more than 250 pounds. A horse does not always have to be taller in order to be more durable, but they should be built with strength in mind. The ability to bear additional weight will be greater in horses with a stockier body than in polished, lean types. The circumference of the cannon bone is frequently used to determine the strength of a horse. Because the cannon bone is a weight-bearing bone, horses with bigger cannon bones are frequently able to sustain greater weight.
### Clydesdale
Clydesdales are huge horses that stand between 16 and 18 hands high and make excellent riding companions for larger riders. In fact, they are becoming increasingly popular when it comes to riding horses for show and pleasure. Because of their stocky physique and kind dispositions, they are suitable for riders of all abilities. With an average Clydesdale weighing between 1,600 and 1,800 pounds, they have a carrying capacity of around 320 pounds. That is approximately the weight of a 280-pound rider wearing a saddle weighing around 40 pounds.
Thiscold-blooded horse breedis known for being kind and loving, which makes them excellent riding mounts for beginners and experienced riders alike.
### Shire
Shire horses are one of the largest horse breeds in the world, and they are both strong and docile. They may make excellent riding partners, despite the fact that they are not often considered of as such by horse enthusiasts. With an average Shire weighing between 1,700 and 2,000 pounds, they have a carrying capacity of around 340 pounds. That is approximately the weight of a 300-pound rider wearing a saddle weighing around 40 pounds.
Despite their massive size, they are rather agile, which makes them excellent mounts for riders who want to be on the go. Shires were also included in our list of the finest horse breeds for beginning riders, which included Shires.
### Friesian
The Friesian horse is a superb riding companion because he is elegant, graceful, and strong. It is well-known for its energetic gaits and attractive carriage, as well as its long hair and feathered hooves In most cases, an AFriesian will have 15-17 hands on the table. For a Friesian to properly transport a rider weighing 250 pounds while also carrying a saddle weighing around 20 pounds, the Friesian must weigh approximately 1,360 pounds. Horses of this breed are well-known for their versatility, since they are frequently displayed in categories like saddle seat, hunt seat, western, dressage, and driving, among others.
### Irish Draught
Irish Draughts are about 15-17 hands in height and weigh between 1,300 and 1,400 pounds on average. In order for an Irish Draught to properly transport a large rider weighing over 250 pounds while also carrying a saddle weighing around 20 pounds, the Irish Draught must weigh at least 1,360 pounds. Their powerful physique, along with their athletic disposition, make them excellent partners for all sorts of riders, whether they are tiny or of bigger stature. They frequently perform well in dressage and jumping, and they make wonderful show or pleasure horses as a result.
They are well-known for their level-headed demeanor and their incredible endurance.
### Percheron
Photograph courtesy of Criadero Sumatambo The Percheron breed is renowned for its strength and dedication to its task. These gentle giants are most usually employed for driving and farmwork, but some of them may also make excellent saddle horses if they are properly trained. Percheron’s typical weight is between 1,800 and 2,000 pounds, and they have a carrying capacity of around 360 pounds. That is approximately the weight of a 320-pound rider plus the weight of a saddle that is around 40 pounds.
Despite the fact that they are not traditionally considered of as riding horses, many heavy riders have discovered that they make reliable mounts.
### Spotted Draft Horse
When fully loaded, a typical American Cream Draft can carry up to 300 pounds. It weighs between 1,500 and 1,600 pounds and stands 16-17 feet tall. This is about the weight of a 260-pound rider with tack that can weigh up to 40 pounds combined. Despite the fact that they are a rare breed, the Spotted Draft Horse has earned a reputation as a magnificent riding horse. The amazing coat patterns on these wonderful drawings make them stand out from the crowd.
Spotted Drafts are more nimble than other popular draft breeds, and they are calm and ready to please their owners. As a result, they are an excellent breed for bigger riders who desire the ability to do more than simply leisure riding.
### Cleveland Bay
The Cleveland Bay is a magnificent riding mount because it is large, strong, and graceful. They are well-known for their beauty, endurance, and reasonable temperaments, which make them excellent horses for riders of all ages. Cleveland Bay horses, which typically reach between 16 and 16 and a half hands tall and weigh between 1,400 and 1,500 pounds, can carry up to 280 pounds. That is the equivalent of a person weighing around 260 pounds with a saddle weighing up to 20 pounds. These magnificent horses must be bay, with black tips, and the only white that is authorized is a star marking on their forehead.
### American Cream Draft
Despite the fact that the American Cream Drafthorse is a rare breed, it has a large following of admirers. These lovely horses, who are distinguished by their cream coats, are the only draft breed to have originated in the United States. The American Cream Draft is around 15.1-16.3 hands in height and weighs between 1,600 and 1,800 pounds on average. They are capable of transporting riders weighing up to 320 pounds (including 40 lbs of tack). Their distinctive coats range in color from delicate cream to a deep golden, and they are well-known for having amber eyes, which are a hallmark feature of the breed.
Is it possible for a horse’s back to be damaged when riding? Riders who are well-balanced and do not weigh more than 20 percent of the horse’s body weight should not cause back problems in animals who are well tack and properly balanced. As long as these guidelines are followed, the majority of horses will not get back issues as a result of riding. What Kind of Horse Is Capable of Carrying a Large Rider? The ability to carry a hefty rider on a powerful, well-muscled horse with solid legs of greater size is common.
• Is it possible for a horse to carry 300 pounds?
• Always consult with your veterinarian before getting on a horse!
• Despite the fact that there is no established weight limit, few horses are capable of securely transporting more than 300 pounds.
• The weight restriction might range from 210 to 300 pounds, depending on the facility and the horses that are available.
• Generally speaking, your weight should not exceed 20% of the weight of a horse’s total weight.
Please keep in mind that we are not providing medical or veterinary advice on this website. Always consult with a veterinarian before attempting to ride your horse to ensure that it is safe for you to do so. You may also be interested in:
• A Horse’s Capacity for Carrying
• Horse Breeds’ Height and Weight Chart
• What Is the Average Weight of a Horse
• Differences between Shires and Clydesdales
## How Much Weight Can Your Horse Safely Carry?
Have you ever carried a typical schoolkid’s bag around with you? It was not that long ago, while some of us were in school, that we just had two or three textbooks with us at all times. Nowadays, however, because many schools have eliminated lockers for security concerns, students are frequently required to carry all of their stuff with them during the day. One survey of 3,498 middle-school pupils conducted in 2004 discovered that the average backpack weight was 10.6 pounds, with some kids carrying as much as 37 pounds.
• In other words, the bigger the weight of the backpack, the greater the possibility that the kid would complain of discomfort.
• The burdens imposed on a 1,000-pound horse would be confined to 100 to 150 pounds if the same restrictions were followed in the equestrian community.
• However, this does not imply that there is no expense.
• Our investigations focused with energetics, specifically how to measure the costs of carrying a large amount of weight, according to the study team’s leader Dr.
• The effects of weight on horse biomechanics, metabolism, and prospective soundness were among the topics covered in the research.
• Wickler adds that his results might have far-reaching ramifications, including those for recreational trail mounts and backyard horses, among other things.
• National Center for Health Statistics reports that the average height and weight of the American population has increased over the past several decades, and the number of obese individuals has increased as well, as has the number of overweight people.
The majority of the time, the answer is “It depends.” However, raising your horse’s awareness of weight concerns can go a long way toward keeping him healthy and sound for years to come in the future.
The muscles they need to run, jump, fly, and climb out of harm’s way, as well as the hoof and horn, tooth, and claw they use to wage their conflicts, must all be carried with them.
Growing and sustaining such tools, on the other hand, necessitates the use of energy, which must be obtained from readily available food supplies.
As Wickler explains, “Human engineers will overbuild in order to anticipate extremes.” “For example, an elevator with a stated capacity of eight passengers or a weight limit of 1,500 pounds may be constructed.
However, biological systems do not behave in this manner.
When a horse carries a rider, it is this “reserve capacity” that is responsible for bearing the additional weight; nevertheless, the horse must alter the way he moves and uses his muscles to accommodate the additional weight.
The amount of oxygen the horses used while trotting on a treadmill while wearing face masks was assessed by the researchers.
With the addition of weights that accounted for around 19 percent of the horses’ total weight, which is roughly similar to a 150-pound rider with equipment, the horses’ metabolic rate climbed by an average of 17.6 percent at all speeds, the researchers discovered.
With each extra pound added to the burden, the metabolic work required to move that load increases by a comparable amount–and this is on level ground.
Economy It should come as no surprise that horses who are free to pick their own speed prefer to slow down when they have additional weight placed on their backs.
This portion of the study involved the training of seven Arabian horses, including geldings and mares, to walk and trot along a level fence line in response to spoken orders.
It was estimated that the saddle and lead together weighted 85 kilograms (187 pounds), which accounted for around 19 percent of the horses’ total body weights.
According to Wickler, “Not only does their metabolic rate increase, but their preferred speed decreases as well.” He adds that the most important finding was that the horses’ preferred speed was the most economical in terms of moving a given distance while carrying the additional weight, which was the most important finding of the study.
• “When you add weight to a horse while it is standing, the force of the weight is distributed evenly among the horse’s four limbs,” Wickler explains.
• Normal (vertical) and parallel (horizontal) forces as well as each foot’s time of contact on the plate were recorded on the fore- and hind limbs; each horse was also videotaped so that stride time could be measured.
• But in fact, there are significant differences in the amount of forces borne by the front and rear legs.
• Going uphill, this pattern of distribution shifts, with 52 percent supported by the forelimbs while the hind limbs took on 48 percent.
• At higher speeds, the two feet were on the ground about the same amount of time, but at slower speeds, the hind limbs tended to spend less time on the ground–an observation that had never been made before in quadrupeds, according to Wickler.
Gait To study the biomechanical effects of loads, the Cal State researchers trotted five Arabians at a consistent speed on a treadmill under three different conditions: on the level with no load, on a 10 percent incline with no load, and on the level while carrying a saddle and weights that totaled about 19 percent of their body mass.
1. Carrying a load caused the horses to leave their feet on the ground an average of 7.7 percent longer than they did while trotting unburdened.
2. In short, explains Wickler, carrying a load causes a horse to shorten his stride, leave his feet on the ground longer and increase the distance his body travels (the “step length”) with each stride.
3. “Forces are damaging,” says Wickler, “so keeping the foot on the ground reduces peak forces and reduces that potential for injury.” Tough Road?
4. Clearly, horses the world over have been carrying riders for many centuries with little ill effect.
5. As each foot strikes the ground, whatever force is not absorbed by bone and tendon must be taken up by the muscles.
6. “A small amount of weight can make a big difference,” Wickler says.
7. “For racing performance on a short track, 10 percent is a huge amount,” Wickler says.
While carrying a single heavy rider on a one-day ride is not likely to seriously harm a horse, over the years, a consistent regimen of this sort of work could add up to chronic injury.
It’s possible that chronic overwork leads to many tiny microfractures, which can build up to a catastrophic break.
“There is a bit of normal extension and flexion during movement, and although the question has not been examined in detail, it’s likely that if you put a weight in the middle, you are going to change the way the back performs.” How Much is Too Much?
“While there seems to be some consensus, it isn’t as clear as one might think,” says Wickler.
Obviously, a horse who staggers under a pack is overloaded.
Time and terrain matter, too.
In the absence of scientific research, the next source of information on maximum weight loads for horses comes from historical sources–the result of centuries of horsemanship experience, not all of which developed with the well-being of the horse as the highest priority.
Army specifications for pack mules state that ‘American mules can carry up to 20 percent of their body weight (150 to 300 pounds) for 15 to 20 miles per day in mountains,’” Wickler says.
“Packers generally try to keep packs to 150 to 200 pounds in their animals, who must carry the dunnage on a daily basis for the entire season,” says Wickler, “so 20 percent of the animal’s body weight seems to be reasonable.
If you go faster, that means more forces on the limbs and more metabolism is needed.” Today, many dude ranches and public stables post weight limits for riders, usually around 200 pounds or less; the National Park Service, for example, does not allow riders who weigh more than 200 pounds to participate in its mule trips into the Grand Canyon.
• “Obviously, that’s not going to happen.
• Western saddles intended primarily for ranchwork or sports like as roping or cutting tend to be heavier, 40 pounds or more; those meant for trail or pleasure applications tend to be lighter, 25 to 30 pounds, but some models can range up to 40.
• English saddles vary in weight according on the discipline, but they are often 20 pounds or less, with some versions weighing as little as 10 pounds.
• Although the jury is still out on how exactly all of this weight affects individual horses, anything you can do to reduce the amount of weight your horse carries will almost certainly benefit him in the long run.
“I’d want to shed a few pounds,” Wickler confesses candidly. “It’d be better for me, and it’d be better for my horse as well,” says the author. Original publication of this essay appeared in the January 2005 edition of EQUUS magazine.
## Question of the Week: How much weight can a horse carry?
If a horse is in good form and healthy, it can normally carry more weight than a horse with a finer frame or one that is out of shape. Q:How can I estimate the safest weight for a horse to be able to transport? I’ve heard a variety of responses, ranging from 20 percent of the horse’s total weight (including equipment) to any horse over 15hh being capable of carrying 300 pounds without incident. Are there any general rules of thumb or standards that I may use to judge whether a horse is capable of transporting me safely?
1. A: In a paper published in the Journal of Equine Veterinary Science in 2008, a clear and unambiguous answer to your query was established.
2. The horses were then handled and their vital signs were monitored.
3. Horses that were carrying 25 percent and 30 percent of their body weight during exercise had greater heart and respiratory rates, as well as muscles that were more painful the next day.
4. The short answer to your query is that it is dependent on the horse in question.
5. These studies clearly demonstrate that a horse’s conformation affects his ability to bear a certain amount of weight to a certain extent.
Overall, if you have a reasonably fit horse with a large bone structure and you’re going on a relaxing trail ride consisting of walking and the occasional trot, there’s no reason why this horse can’t carry more than the 20 percent rule – in fact, this happens all the time as there are plenty of riders out there who push beyond the 240 pound limit on their 1200 pound horse, or the 200 pound limit on their 1000 pound horse, and still have a good time without worrying The 20 percent guideline and using common sense to analyze your horse’s fitness level should help you determine how much weight is “safe” for your horse to carry.
On a historical perspective, the 20 percent rule has been in use for far longer than the time period covered by this 2008 research.
More information on this subject may be found in the article “Too Heavy to Ride?” from the February 2014 edition of Horse Illustrated magazine.
## How Much Weight Can A Horse Carry?
Have you ever attempted to go on a holiday trail ride only to be required to sign a release that included your height and weight information? Have you ever pondered why they need to know this information, and why it would be essential to you throughout your ride? What is the maximum amount of weight that a horse can carry? Then this post is written specifically for you! In this post, I’ll cover the significance of weight restrictions in equestrian riding, as well as how much weight a horse is capable of securely carrying on its back.
500 pounds, perhaps?
As is always the case, the answer is that it depends!
## Why Have Weight Limit for Horse Riding
Horses are powerful, well-muscled creatures, so why do we need weight restrictions in the first place? Weight constraints are usually not something you have to deal with or worry about on a daily basis if you’re riding your own horse, in your own equipment, and doing everything the same way you do every day. Think about riding schools, scholastic and college riding programs, trail riding farms, guy ranches, and any other horse-riding businesses where you might be able to ride horses that you’ve never met before, such as a dude ranch.
• In that case, how would they determine whether they should put you on their 12 hand pony or their 18 hand draft horse?
• You must complete online disclaimers for most riding schools and trail riding stables, including your height and weight, in order to participate.
• In the Intercollegiate Horse Show Association, for example, the “height/weight” guideline is another example of this (IHSA- a competitive college riding circuit).
• Because the horses have certain weight limitations, these riders have certain horses removed from the pool of horses from which they can draw because the horses are overweight.
## What Contributes to the Weight a Horse Carries?
However, the weight of the rider is not the primary factor in determining a horse’s weight limit. Consider what else your horse has to carry when you’re out riding—the saddle, for example. Some saddles are exceptionally light in weight, while others are heavy in weight. Consider the differences between racing saddles and monoflap eventing or dressage saddles. Even certain close-contact hunt seat saddles can be incredibly light, especially if they are made of a lightweight material. In rare cases, though, western or other trail saddles can be quite hefty.
There are other occasions in which horses or mules are employed as “pack” animals, which means that they are exclusively responsible for transporting provisions and not humans. It is critical to understand how much each of these goods weighs in order to avoid overloading the horse.
## Equine Characteristics that Contribute to Weight Limit
The height of a horse is the most evident trait that will contribute to its particular weight restriction, and it is also the most important. Ponies are often shorter, smaller-boned, and weaker in general strength than a large horse of similar size. Despite the fact that this is not always the case (some “wonder” ponies have made it to the upper ranks of horse competition), horses are normally able to carry greater weight simply because they are taller than their counterparts. Another consideration is the general shape of the horse, or how it is made.
1. Horses with particular conformation that makes them highly sensitive and agile may not be ideal for hauling huge loads, as is the case with several breeds of horses.
2. The converse is true for other horses, such as draft crosses and draft horses.
3. In conclusion, it’s critical to treat each horse as a unique person.
4. Because very young horses are still growing, it is important to be as gentle as possible on their bones.
5. As a result, the weight limit of a horse will be determined by the animal’s height, build, and physical condition.
## How Much Weight Should a Horse Carry?
Despite the fact that all of this is significant, it still does not tell us how much weight a horse should be expected to carry. In the United States, the conventional rule is that a horse may carry 20% of its own body weight on its back. In addition, the typical guideline in the United Kingdom is that a horse may carry 10% of its own body weight. As a result, the average horse weighs between 1200 and 1500 pounds. In the United States, a horse of such weight would be expected to be able to carry 240-300 pounds on its back.
|
## Conclusion
Both for the safety of your horse and the safety of its riders, weight restrictions are essential for their safety. It is critical to understand each horse’s physical limitations and to make judgments in accordance with those constraints. Horses are capable of carrying enormous loads if they are not forced to do more than they are capable of.
## FAQs
Riders’ weight should not exceed 250 pounds, including tack, on average, according to the majority of those who weigh themselves. Some large breeds, such as medium build horses larger than 17.2 hands in height and draft breeds measuring 16.3 hands or more in height, are capable of carrying heavier riders. These are the breeds that are capable of carrying heavier riders. Please keep in mind that the back of larger horses may be longer than that of smaller or medium-sized horses. Despite their overall size, a huge horse with a long back may be unable to carry a significant amount of weight due to their lack of strength.
### Is 18 Stone Too Heavy To Ride A Horse?
There are 252 pounds in 18 stone when it is converted to pounds. As a result, 18 stone is too much to ride a horse. Even without equipment, the rider’s weight of 18 stone already exceeds the weight restriction for the horse. There is a maximum of 16 stone available, and that is with extremely mild tack.
### What Horse Breed Is Appropriate For A 350 lb Rider?
Whenever you are searching for a horse that can carry a rider who weighs 350 pounds, horses with a medium build that are above 17 hands in height and draft horses that are larger than 16.3 hands in height are ideal possibilities. When it comes to transporting large riders, Percherons and Percheron Crosses are an excellent option of saddle. Belgian draft horses are another kind of horse that is extremely resilient. When compared to the size of a 0 or 1 horseshoe, they wear a size five horseshoe on their feet.
### How Much Weight Can A Clydesdale Carry?
Clydesdales are more often recognized for their pulling strength than for their ability to carry a rider’s weight. A single Clydesdale can draw between 2,000 and 8,000 pounds, while a pair of Clydesdales can pull up to nearly 18,000 pounds together. Clydesdales were originally utilized as drum horses, and they were tasked with transporting the Musical Ride Officer as well as two silver drums, each weighing 123 lbs, for a total weight of 248 lbs in the drums alone. When you include in the added weight of the rider and tack, the total weight is far in excess of 300 lbs.
### Best Horse Breed For Heavy Riders
Horses with broad bones and a stocky physique are the finest breeds for heavy riders, and they include the following: Quarter horses, Irish Draught horses, Highland Ponies, Percheron horses, and Shire horses are just a few of the horses that are well-suited to heavier riders.
### Can You Be Too Heavy To Ride A Horse?
A rider’s weight can become too much for a horse to bear at some time, and this is understandable. It is often believed among horsemen that if you can get on the horse, you can ride the horse. This refers to mounting without the help of a third party or the use of a mounting block.
If you are unable to mount the horse, you should refrain from riding. In any case, this is only significant when it comes to the rider’s weight, as opposed to any potential mental or physical health issues that may make mounting problematic.
### How Much Weight Can A Pony Carry?
Ponies range in height from 9 to 14 hands and weigh between 400 and 800 pounds on average. As a result, ponies may transport riders weighing between 80 and 160 pounds, with equipment, at the most. Young children and smaller people are the most common riders on ponybacks. | 10,325 | 47,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-23 | latest | en | 0.964137 |
https://oeis.org/A335107/internal | 1,627,280,185,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152000.25/warc/CC-MAIN-20210726031942-20210726061942-00209.warc.gz | 436,278,970 | 2,858 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A335107 Least period of the length-n prefix of the Thue-Morse sequence A010060. 0
%I
%S 1,2,3,3,3,5,6,6,6,6,10,10,12,12,12,12,12,12,12,12,20,20,20,20,24,24,
%T 24,24,24,24,24,24,24,24,24,24,24,24,24,24,40,40,40,40,40,40,40,40,48,
%U 48,48,48,48,48,48,48,48,48,48,48,48,48,48,48,48,48,48,48
%N Least period of the length-n prefix of the Thue-Morse sequence A010060.
%C An integer p, 1 <= p <= n, of a length-n word x is said to be a period if x[i]=x[i+p] for 1 <= i <= n-p.
%C We have
%C a(n) = 3*2^i for 3*2^i < n <= 5*2^i, i >= 0;
%C a(n) = 5*2^i for 5*2^i < n <= 6*2^i, i >= 0.
%H D. Gabric, N. Rampersad, and J. Shallit, <a href="https://arxiv.org/abs/2005.11718">An inequality for the number of periods in a word</a>, arxiv preprint arXiv:2005.11718 [cs.DM], May 24 2020.
%e For n = 10, the first 10 symbols of the Thue-Morse sequence are 0110100110, which has period 6.
%Y Cf. A010060.
%K nonn
%O 1,2
%A _Jeffrey Shallit_, May 23 2020
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Last modified July 26 02:10 EDT 2021. Contains 346294 sequences. (Running on oeis4.) | 571 | 1,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-31 | latest | en | 0.761178 |
https://www.physicsforums.com/threads/fabry-perot-interferometer-diffraction.235418/ | 1,675,065,631,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00315.warc.gz | 939,859,790 | 15,808 | # Fabry-Perot interferometer diffraction
Math Jeans
## Homework Statement
A Fabry-Perot interferometer consists of two parallel half-silvered mirrors separated by a small distance a. Show that when light is incident on the interferometer with an angle of incidence $$\theta$$, the transmitted light will have maximum intensity when $$a = (\frac{m \cdot \lambda}{2})cos(\theta)$$
## Homework Equations
$$d \cdot sin(\theta) = m \cdot \lambda$$
## The Attempt at a Solution
I tried relating the distance a with d by creating a right triangle from the reflected rays.
That method didn't work, and my answer was the same as the one that I was trying to prove, but divided by (sin($$\theta$$))^2. I need a new way to do this.
Last edited:
Homework Helper
Hi Math Jeans,
The equation $d \sin\theta=m\lambda$ is a derived equation that applies to specific cases (e.g., two slit interference experiment).
The more fundamental relationship is:
$$\mbox{(path length difference of two rays)}=m\lambda$$
gives constructive interference. (Assuming coherent light that starts out in phase, and assuming that the path length difference effect is the only interference effect).
So when light strikes one of the mirrors inside the interferometer, some passes through the mirror and some is reflected. The maximum will occur when the extra distance that the reflected light has to travel before reaching that same mirror is equal to $m\lambda$.
What does using the right triangles indicate that the extra distance is? Once you have that you can find the given answer. What do you get?
Math Jeans
The reason that I used dsin(theta) as my phase difference, is because the light entering the interfermometer exits in such a way that there is a distance between the exiting rays that when applied to dsin(theta) will still exibit the same phase difference. It should be working, but it doesn't seem to be.
http://hyperphysics.phy-astr.gsu.edu/Hbase/phyopt/fabry.html" [Broken]
That link has info the the interfermometer, and a diagram (the second one down the page) that shows light entering and exiting it. What I was doing in my calculations was equating the distance between the exiting rays to d, and using the rays bouncing back in forth to form my right triangle. The dsin(theta) relation should work for this.
Last edited by a moderator:
Homework Helper
Math Jeans,
I think you might be focusing on the wrong distance in this problem. When we have a diffraction grating or a two slit source the light rays are in phase (at least when we use $d\sin\theta$). If that is the case, the important distance would be the distance between exit points along the mirror. But here, the phase of the light rays at the exit points are not automatically in phase; the point is we have to choose how the separation a and angle theta are related to make them be in phase when the rays leave the mirror.
So this Fabry-Perot problem is somewhat more like a thin-film type of problem than like a two-source interference problem.
When part of the light ray exits the mirror, the part that remains has to reflect back to the other mirror, then reflect again. The distance the light travels between mirrors is $d=a/\cos\theta$, and the light travels twice this distance between adjacent exit points.
I've created a quick picture at:
http://img215.imageshack.us/my.php?image=fpjl4.jpg
The extra distance that one light ray has to travel more than the one that previously exited is the distance in color. (The color was red, but seemed to change after uploading.) | 795 | 3,550 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-06 | latest | en | 0.923446 |
https://ndlib.readthedocs.io/en/latest/reference/models/epidemics/ICP.html | 1,618,803,030,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038863420.65/warc/CC-MAIN-20210419015157-20210419045157-00100.warc.gz | 505,540,189 | 9,066 | # Independent Cascades with Community Permeability¶
The Independent Cascades with Community Permeability model was introduced by Milli and Rossetti in 2019 [1].
This model is a variation of the well-known Independent Cascade (IC), and it is designed to embed community awareness into the IC model. This model exploits the idea of permeability. A community is “permeable” to a given content if it permits that content to spread from it fast (or vice-versa, if it permits the content to be easily received from nodes outside the community). Conversely, a community has a low degree of permeability if it dampens the diffusion probability across its border.
The ICP model starts with an initial set of active nodes A0; the diffusive process unfolds in discrete steps according to the following randomized rule:
• When node v becomes active in step t, it is given a single chance to activate each currently inactive neighbor u. If v and u belong to the same community, the method works as a standard IC model (it succeeds with a probability p(v,u)); instead, if the nodes are part of to different communities, the likelihood p(v,u) is dampened of a factor $$\eta$$ (the community permeability parameter).
• If u has multiple newly activated neighbors, their attempts are sequenced in an arbitrary order.
• If v succeeds, then u will become active in step t + 1; but whether or not v succeeds, it cannot make any further attempts to activate u in subsequent rounds.
• The process runs until no more activations are possible.
## Statuses¶
During the simulation a node can experience the following statuses:
Name Code
Susceptible 0
Infected 1
Removed 2
## Parameters¶
Name Type Value Type Default Mandatory Description
Edge threshold Edge float in [0, 1] 0.1 False Edge threshold
Community permeability Model float in [0, 1] 0.5 True Community Permeability
The initial infection status can be defined via:
• fraction_infected: Model Parameter, float in [0, 1]
• Infected: Status Parameter, set of nodes
The two options are mutually exclusive and the latter takes precedence over the former.
## Methods¶
The following class methods are made available to configure, describe and execute the simulation:
### Configure¶
ICPModel.__init__(graph)
Model Constructor
Parameters: graph – A networkx graph object
ICPModel.set_initial_status(self, configuration)
Set the initial model configuration
Parameters: configuration – a ndlib.models.ModelConfig.Configuration object
ICPModel.reset(self)
Reset the simulation setting the actual status to the initial configuration.
### Describe¶
ICPModel.get_info(self)
Describes the current model parameters (nodes, edges, status)
Returns: a dictionary containing for each parameter class the values specified during model configuration
ICPModel.get_status_map(self)
Specify the statuses allowed by the model and their numeric code
Returns: a dictionary (status->code)
### Execute Simulation¶
ICPModel.iteration(self)
Execute a single model iteration
Returns: Iteration_id, Incremental node status (dictionary node->status)
ICPModel.iteration_bunch(self, bunch_size)
Execute a bunch of model iterations
Parameters: bunch_size – the number of iterations to execute node_status – if the incremental node status has to be returned. progress_bar – whether to display a progress bar, default False a list containing for each iteration a dictionary {“iteration”: iteration_id, “status”: dictionary_node_to_status}
## Example¶
In the code below is shown an example of instantiation and execution of an ICP model simulation on a random graph: we set the initial set of infected nodes as 1% of the overall population, assign a threshold of 0.1 to all the edges and set the community permeability equal 0.3.
import networkx as nx
import ndlib.models.ModelConfig as mc
import ndlib.models.epidemics as ep
# Network topology
g = nx.erdos_renyi_graph(1000, 0.1)
# Model selection
model = ep.ICPModel(g)
# Model Configuration
config = mc.Configuration() | 870 | 4,002 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-17 | longest | en | 0.88664 |
http://www.theinfolist.com/html/ALL/l/square-lattice_Ising_model.html | 1,713,228,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817036.4/warc/CC-MAIN-20240416000407-20240416030407-00714.warc.gz | 59,737,543 | 8,411 | *
picture info
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2013.
# References
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CNRS
The French National Centre for Scientific Research (french: link=no, Centre national de la recherche scientifique, CNRS) is the French state research organisation and is the largest fundamental science agency in Europe. In 2016, it employed 31,637 staff, including 11,137 tenured researchers, 13,415 engineers and technical staff, and 7,085 contractual workers. It is headquartered in Paris and has administrative offices in Brussels, Beijing, Tokyo, Singapore, Washington, D.C., Bonn, Moscow, Tunis, Johannesburg, Santiago de Chile, Israel, and New Delhi. From 2009 to 2016, the CNRS was ranked No. 1 worldwide by the SCImago Institutions Rankings (SIR), an international ranking of research-focused institutions, including universities, national research centers, and companies such as Facebook or Google. The CNRS ranked No. 2 between 2017 and 2021, then No. 3 in 2022 in the same SIR, after the Chinese Academy of Sciences and before universities such as Harvard University, MIT ...
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https://www.beatthegmat.com/quadrilateral-t1193.html | 1,521,310,778,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645280.4/warc/CC-MAIN-20180317174935-20180317194935-00494.warc.gz | 766,323,674 | 34,169 | • Reach higher with Artificial Intelligence. Guaranteed
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This topic has 3 member replies
Chrystelle Junior | Next Rank: 30 Posts
Joined
07 Dec 2006
Posted:
19 messages
Fri Jan 12, 2007 6:58 am
Quadrilateral PQRS has PQ=QR=RS=5, If angle R=90 and angle Q=135 what is the area of the quadrilateral?
aim-wsc Legendary Member
Joined
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Posted:
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Followed by:
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Target GMAT Score:
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Sun Jan 14, 2007 10:10 am
Chrystelle wrote:
Quadrilateral PQRS has PQ=QR=RS=5, If angle R=90 and angle Q=135 what is the area of the quadrilateral?
take a look at the figure
split the quadrilateral into two triangles namely PQS and SRQ
both triangles are right angle triangles
that makes the problem easy.
now i want you to try this one more time...
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Please do not PM me, (not active anymore) contact Eric.
Chrystelle Junior | Next Rank: 30 Posts
Joined
07 Dec 2006
Posted:
19 messages
Sun Jan 14, 2007 9:40 am
don't hav the anwer.
thankont Senior | Next Rank: 100 Posts
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Fri Jan 12, 2007 9:27 am
If you know the answer is it --> 25*sqrt(2)/4 ?
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See More Top Beat The GMAT Experts | 746 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-13 | latest | en | 0.883046 |
http://en.wikipedia.org/wiki/Kaprekar_number | 1,432,717,781,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928923.21/warc/CC-MAIN-20150521113208-00155-ip-10-180-206-219.ec2.internal.warc.gz | 88,419,288 | 15,233 | # Kaprekar number
Not to be confused with Kaprekar's constant.
In mathematics, a Kaprekar number for a given base is a non-negative integer, the representation of whose square in that base can be split into two parts that add up to the original number again. For instance, 45 is a Kaprekar number, because 45² = 2025 and 20+25 = 45. The Kaprekar numbers are named after D. R. Kaprekar.
## Definition
Let X be a non-negative integer. X is a Kaprekar number for base b if there exist non-negative integers n, A, and positive number B satisfying:
X² = Abn + B, where 0 < B < bn
X = A + B
Note that X is also a Kaprekar number for base bn, for this specific choice of n. More narrowly, we can define the set K(N) for a given integer N as the set of integers X for which[1]
X² = AN + B, where 0 < B < N
X = A + B
Each Kaprekar number X for base b is then counted in one of the sets K(b), K(b²), K(b³),….
## Examples
297 is a Kaprekar number for base 10, because 297² = 88209, which can be split into 88 and 209, and 88 + 209 = 297. By convention, the second part may start with the digit 0, but must be positive. For example, 999 is a Kaprekar number for base 10, because 999² = 998001, which can be split into 998 and 001, and 998 + 001 = 999. But 100 is not; although 100² = 10000 and 100 + 00 = 100, the second part here is not positive.
The first few Kaprekar numbers in base 10 are:
1, 9, 45, 55, 99, 297, 703, 999, 2223, 2728, 4879, 4950, 5050, 5292, 7272, 7777, 9999, 17344, 22222, 38962, 77778, 82656, 95121, 99999, 142857, 148149, 181819, 187110, 208495, 318682, 329967, 351352, 356643, 390313, 461539, 466830, 499500, 500500, 533170, ... (sequence A006886 in OEIS)
In particular, 9, 99, 999… are all Kaprekar numbers. More generally, for any base b, there exist infinitely many Kaprekar numbers, including all numbers of the form bn - 1.
## Properties
• It was shown in 2000[1] that the Kaprekar numbers for base b are in bijection with the unitary divisors of bn − 1, in the following sense. Let Inv(a,b) denote the multiplicative inverse of a modulo b, namely the least positive integer m such that $am \equiv 1 \pmod b$. Then, a number X is in the set K(N) (defined above) if and only if X = d Inv(d, (N-1)/d) for some unitary divisor d of N-1. In particular, | 745 | 2,282 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2015-22 | latest | en | 0.833838 |
https://www.sweetstudy.com/content/math-qs-4 | 1,723,491,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641048885.76/warc/CC-MAIN-20240812190307-20240812220307-00220.warc.gz | 778,674,499 | 30,974 | # Math QS
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Could you help me with those questions? I will attach a file with the questions latter, so you can see them better. thank you
1. Using StatCrunch, calculate the chi-square statistic and degrees of freedom for the following set of data for 300 people:
Group A Group B Group C Total
Had flu shot 20 30 32 82
Didn’t have flu shot 80 70 68 218
Total 100 100 100 300
Is the value of the chi-square statistically significant at the 0.05 level?
2. Write a paragraph summarizing the results of the analysis in Exercise 1.
3. Using StatCrunch, calculate the chi-square statistic and degrees of freedom for the following set of data for 180 people undergoing a knee replacement treatment with a drug supplement:
Treatment with drug X Treatment without Drug X Total
Had > 8 wk rehab 18 32 50
Had < 8 wk rehab 70 60 130
Total 88 92 180
Is the value of the chi-square statistically significant at the 0.05 level?
4. Write a paragraph summarizing the results of the analysis in Exercise 3.
5. Given each of the following circumstances, determine whether the calculated values of chi-square are statistically significant:
a. χ2 = 3.02, df = 1, α = 0.05
b. χ2 = 8.09, df = 4, α = 0.05
c. χ2 = 10.67, df = 3, α = 0.01
d. χ2 = 9.88, df = 2, α = 0.01
6. Match each of the nonparametric tests in Column A with its parametric counterpart in Column B
A. Nonparametric Test B. Parametric Test
1. Mann-Whitney U-test a. Paired t-test
2. Friedman test b. One-way ANOVA
3. Kruskall-Wallis test c. Independent groups t-test
4. Wilcoxon signed-ranks test d. Repeated measures ANOVA
• 10 years ago | 474 | 1,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-33 | latest | en | 0.784177 |
https://forums.pocketmine.net/threads/bounds-of-an-oblique-cylinder.13140/ | 1,611,568,735,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565541.79/warc/CC-MAIN-20210125092143-20210125122143-00335.warc.gz | 341,580,297 | 14,351 | # Bounds of an oblique cylinder
Comments in 'Plugin Development' started by PEMapModder, Nov 22, 2015.
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### PEMapModderNotable MemberPlugin Developer
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I am working on cylinders in WorldEditArt, and due to my stubborn and impatient nature, I am trying to let cylinders be oblique.
So, I am defining a cylinder with these parameters:
Code:
T(x2, y2, z2)
|||
v
^
/ \
< \
\ \
\ \
\ >
\ / <= B(x1, y1, z1)
v <= distance from B = r
This is a the side view of an example of the cylinder, where B is the base point, T is the top point and r is the radius of the cylinder.
Now say, if we are having a right cuboid (a cuboid that is perpendicular/parallel to the X, Y and Z axes) to just inscribe the cylinder, like this:
Code:
________
| ^ |
| / \ |
|< \ |
| \ \ |
| \ \ |
| \ >|
| \ / |
| v |
|--------|
How can I find out the dimensions (minimum and maximum X, Y and Z coordinates) of the cuboid?
To set all blocks in the cylinder, I am trying to iterate through the integer points inside this cuboid, check distance of the point from the central vector BT (which I already found out how from Wolfram Mathworld), and if true, update the block.
Or is there another way to do this without iterating through blocks in the cuboid?
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### zombie_PowerActive Member
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Just to make sure, that I (and others) fully understand your concept, are you also trying to achieve a circular shape from a top view?
As smaller cylinders maybe not be able to achieve that circular shape. Btw the shape you provide seems like your cylinders are more of tilted, than oblique.
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### PEMapModderNotable MemberPlugin Developer
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By oblique cylinder, I meant tilted cylinder. It is still a cyliner, but it is placed at a different angle.
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### zombie_PowerActive Member
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### PEMapModderNotable MemberPlugin Developer
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### zombie_PowerActive Member
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### PEMapModderNotable MemberPlugin Developer
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oblique means tilted. I have wrote the code to check whether a point is in the cylinder, but I need code for vector-plane angle to calculate the boundaries.
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### PEMapModderNotable MemberPlugin Developer
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Let me simply the question:
How to get a line's angle with a plane in a 3D system, where the line is defined by two points of coordinates?
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### Yosshi999New MemberPlugin Developer
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Is a 'plane' always the ground?
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### Yosshi999New MemberPlugin Developer
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i think 'oblique projection vector' is useful.
Last edited: Nov 24, 2015
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### PEMapModderNotable MemberPlugin Developer
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I see what you mean. Like this?
Code:
/
/
/
/
/
/
/
/
/ ) θ
≤_________
(Calculate this point by (X2, Y1, Z2)
(Then use cosine to calculate θ)
Last edited: Nov 24, 2015
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### Yosshi999New MemberPlugin Developer
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yes
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### Yosshi999New MemberPlugin Developer
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When you project a tilted cylinder on the ground (y=0), the shadow will be one tilted rectangle + two ellipses.
The size of these ellipses is r * (r cosθ),
θ is the angle between (0,1,0) and (X2-X1,Y2-Y1,Z2-Z1).
sorry poor English
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### PEMapModderNotable MemberPlugin Developer
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### Yosshi999New MemberPlugin Developer
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Too long code for me to read!
Does it work correctly?
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### PEMapModderNotable MemberPlugin Developer
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Not tested. I used r*cos(length/projection length) to calculate the distance between lower center and lowermost point, so it should be correct. | 1,332 | 4,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-04 | longest | en | 0.90549 |
https://www.physicsforums.com/threads/balloon-pressure-homework-inside-vs-outside-bottle-balloon.740843/ | 1,723,689,873,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00275.warc.gz | 724,833,806 | 23,577 | # Balloon Pressure Homework: Inside vs Outside Bottle & Balloon
• Jimmy87
In summary: The balloon is inside the bottle,... inflated, the balloon's opening is held closed, the hole in the bottle is blocked.4. The balloon is inside the bottle,... inflated, the balloon's opening is open, the hole in the bottle is open.In summary, the balloon pressure experiment involves pushing a balloon into a plastic bottle and inflating it inside the bottle. The pressure inside the bottle is lower as air is pushed out, while the pressure outside the bottle is higher, keeping the balloon inflated. The pressure inside the balloon is equal to atmospheric pressure because it is open to the atmosphere. This can be observed by plugging the hole in the
Jimmy87
## Homework Statement
We were given homework to explain how this balloon pressure experiment works. You basically take a plastic bottle and make a small hole in it near the base. You then push a balloon in through the top and pull it over the bottle opening. You then blow it up so that it inflates inside the bottle. You then immediately plug the hole you made with your finger and it remains inflated in the bottle despite the balloon being open to the atmosphere. We have to discuss the pressure:
Inside the bottle
Outside the bottle
Inside the balloon
None
## The Attempt at a Solution
I have read up online which says that the pressure inside the bottle will be lower as air is pushed out of the bottle as you inflate it. The pressure outside the bottle is higher which keeps it inflated. but when thinking about the pressure inside the balloon I don't understand why it stays inflated and no-one seems to mention pressure inside the balloon online. The pressure inside the balloon surely must be greater than outside the balloon. In order to blow up a normal balloon apparently you need 1 psi pressure to overcome the elastic forces of the balloon (i.e. a normal balloon has a higher pressure inside it). So if the air inside the balloon is greater than outside the balloon shouldn't it deflate regardless of whether the pressure inside the bottle is lower?
Thanks for any guidance offered
Because the balloon is open into the atmosphere, the pressure inside it is equal to atmospheric. Now consider the air that is in the bottle outside the balloon. Just before you plugged the hole with your finger, what was the pressure there? What is the pressure there after the balloon is opened?
1 person
So if the air ... inside the balloon is greater than outside the balloon shouldn't it deflate regardless of whether the pressure inside the bottle is lower?
I inserted the dots; I thought about inserting the word pressure, but that is probably not what you mean, right? What do you mean there ?
I love the experiment and I like it that you did some research and came up with the 1 psi. I think that's what voko hinted at.
I'm in my study and don't have a balloon at hand. (also it's bedtime). But: as an experimental physicist I can think of an extra step in the experiment: after blowing up the balloon, plug the top bottle opening. Then the bottom hole and then unplug the top bottle opening. I bet the balloon contracts a little! Meaning the pressure difference between outside and inside of the balloon is less than 1 psi ! And before it is completely floppy it stops deflating, meaning the difference is near 1 psi again ! But the inside of the balloon is open to the outside world, so the pressure inside the balloon is equal to ..
Back to voko again!
But not without other proposals for further experimentation: can you easily re-inflate the balloon with the bottom hole still plugged? What if you did all this with the bottle in a bucket of water ?
1 person
voko said:
Because the balloon is open into the atmosphere, the pressure inside it is equal to atmospheric. Now consider the air that is in the bottle outside the balloon. Just before you plugged the hole with your finger, what was the pressure there? What is the pressure there after the balloon is opened?
Thanks for your replies. I'm still a bit confused. I kind of get that the air pressure inside the balloon must be equal to the outside pressure but I really can't see how that's possible if you need over pressure to blow up a balloon. Any inflated balloon has an air pressure greater than that outside. So how can they be equal?
To answer your questions, wouldn't the air pressure just before you plug it be the same as after you plug it which is a lower pressure than inside the balloon? What do you mean by after the balloon is opened?
BvU's questions - if you had the bottom hole plugged then you couldn't blow it up I think because the air can't move out the way to enable the balloon to inflate. Not sure about what would happen in a bucket of water?
if you had the bottom hole plugged then you couldn't blow it up I think because the air can't move out the way to enable the balloon to inflate
Air can't move out, but you can compress it (by at least 1 psi, and perhaps even 2!) That way the air volume outside the balloon inside the bottle becomes smaller. Bottle stays the same, hence balloon gets bigger ! Isn't it a good idea to trry it out ?
Jimmy87 said:
you need over pressure to blow up a balloonх
Spot on! What does that tell you about the pressure of the air that is inside the bottle but outside the balloon?
To answer your questions, wouldn't the air pressure just before you plug it be the same as after you plug it which is a lower pressure than inside the balloon? What do you mean by after the balloon is opened?
The process has the following stages.
1. The balloon is inside the bottle, not inflated.
2. The balloon is inside the bottle, inflated, the balloon's opening is held closed, the hole in the bottle is open.
3. The balloon is inside the bottle, inflated, the balloon's opening is held closed, the hole in the bottle is blocked.
4. The balloon is inside the bottle, inflated, the balloon's opening is open, the hole in the bottle is open.
It is interesting to compare pressures in the bottle (outside the balloon) at stages 2, 3, and 4, in the light of your observation about the overpressure.
1 person
voko said:
Spot on! What does that tell you about the pressure of the air that is inside the bottle but outside the balloon?
The process has the following stages.
1. The balloon is inside the bottle, not inflated.
2. The balloon is inside the bottle, inflated, the balloon's opening is held closed, the hole in the bottle is open.
3. The balloon is inside the bottle, inflated, the balloon's opening is held closed, the hole in the bottle is blocked.
4. The balloon is inside the bottle, inflated, the balloon's opening is open, the hole in the bottle is open.
It is interesting to compare pressures in the bottle (outside the balloon) at stages 2, 3, and 4, in the light of your observation about the overpressure.
Thanks for the replies. Ok so I tried it out. I blew up the balloon then quickly put my hand over the top of the bottle then (still doing this) covered the hole at the bottom. When you release your hand off the top a little bit of air comes out so would this be the over pressure? But if after this the balloon air pressure and atmospheric pressure are the same then why doesn't the elastic force of the balloon push the air out?
Also, what causes a balloon to collapse when you blow it up normally (outside a bottle) then release it? Most sources online say it is because of the higher pressure in the balloon but as demonstrated with the above experiment it is only a small over pressure. So would it be a combination of the air pressure outside pushing in on the balloon, the over pressure and the elastic forces?
It seems to me that you are confused about what "overpressure" means. Can you explain what it is?
voko said:
It seems to me that you are confused about what "overpressure" means. Can you explain what it is?
When you blow up a balloon you need a certain amount of pressure to overcome the elastic forces of the balloon which is why an inflated party balloon has a higher pressure inside it than the atmospheric pressure outside. Is that right? For that bottle experiment, I can see that the pressures inside the balloon must be equal to atmospheric pressure otherwise it would deflate but I just don't see how it does work because surely you have the elastic forces of the balloon always acting to deflate it?
Jimmy87 said:
When you blow up a balloon you need a certain amount of pressure to overcome the elastic forces of the balloon which is why an inflated party balloon has a higher pressure inside it than the atmospheric pressure outside.
So, would it be correct to state this mathematically: ##P_{\text{internal}} = P_{\text{elastic}} + P_{\text{external}} ##?
1 person
voko said:
So, would it be correct to state this mathematically: ##P_{\text{internal}} = P_{\text{elastic}} + P_{\text{external}} ##?
Thanks Voko. Yeh I suppose it's better now you have put it in an equation like that.
voko said:
So, would it be correct to state this mathematically: ##P_{\text{internal}} = P_{\text{elastic}} + P_{\text{external}} ##?
Just one last thing sorry. When the balloon is inflated inside the bottle, is the pressure inside the bottle lower like I initially said. I know that air is being pushed out of the bottle as you blow it up but the volume of the inside of the bottle is also decreasing as the balloon inflates. So is the air outside the bottle at a lower pressure or not?
I suggest you go through steps 1, 2, 3, 4 above and try to say what all those P's might be like in every case.
voko said:
I suggest you go through steps 1, 2, 3, 4 above and try to say what all those P's might be like in every case.
1. The balloon is inside the bottle, not inflated. - pressure in balloon, bottle and outside bottle are the same?
2. The balloon is inside the bottle, inflated, the balloon's opening is held closed, the hole in the bottle is open. - pressure outside balloon + inside the bottle are the same? Pressure inside the balloon is greater?
3. The balloon is inside the bottle, inflated, the balloon's opening is held closed, the hole in the bottle is blocked. Think this is the same as above?
4. The balloon is inside the bottle, inflated, the balloon's opening is open, the hole in the bottle is open. pressure inside the bottle + outside are the same. Pressure inside the balloon is higher so it deflates?
Are any of those right? Pressure is the worst part of my physics and it has always confused me so apologies for the continually slow up take. Thanks for any help
Jimmy87 said:
1. The balloon is inside the bottle, not inflated. - pressure in balloon, bottle and outside bottle are the same?
Correct. Observe that all that parts communicate freely with the atmosphere, so the pressure has to be the same everywhere.
2. The balloon is inside the bottle, inflated, the balloon's opening is held closed, the hole in the bottle is open. - pressure outside balloon + inside the bottle are the same?
Even if we assume it is the same, what is this pressure? try to reference it with the atmospheric pressure. Then think whether that little formula above is satisfied.
3. The balloon is inside the bottle, inflated, the balloon's opening is held closed, the hole in the bottle is blocked. Think this is the same as above?
Correct. By plugging the hole in the bottle we don not change the pressure anywhere.
4. The balloon is inside the bottle, inflated, the balloon's opening is open, the hole in the bottle is open. pressure inside the bottle + outside are the same. Pressure inside the balloon is higher so it deflates?
Again: what is this (and other) pressure? Try to reference them with the atmospheric pressure. Then think whether that little formula above is satisfied.
1 person
voko said:
Correct. Observe that all that parts communicate freely with the atmosphere, so the pressure has to be the same everywhere.
Even if we assume it is the same, what is this pressure? try to reference it with the atmospheric pressure. Then think whether that little formula above is satisfied.
Correct. By plugging the hole in the bottle we don not change the pressure anywhere.
Again: what is this (and other) pressure? Try to reference them with the atmospheric pressure. Then think whether that little formula above is satisfied.
2)Pressure inside bottle + outside are the same and are at atmospheric pressure. Pressure inside balloon is greater than atmospheric pressure? The pressure inside the balloon has to overcome this atmospheric pressure + the balloons elastic pressure so must be greater than atmospheric pressure?
4)Pressure inside the bottle + outside are the same (which is atmospheric pressure) and balloon has greater than atmospheric pressure. Now you release both holes, the air in the balloon will leave because the pressure is higher?
Jimmy87 said:
2)Pressure inside bottle + outside are the same and are at atmospheric pressure. Pressure inside balloon is greater than atmospheric pressure? The pressure inside the balloon has to overcome this atmospheric pressure + the balloons elastic pressure so must be greater than atmospheric pressure?
Good!
4)Pressure inside the bottle + outside are the same (which is atmospheric pressure) and balloon has greater than atmospheric pressure. Now you release both holes, the air in the balloon will leave because the pressure is higher?
Do we release both holes? That would bring us to case 1).
1 person
voko said:
Good!
Do we release both holes? That would bring us to case 1).
Thanks again for your help. So if both holes are released we go to case 1 which means the balloon must be fully deflated. So to go from inflated just before you open both holes to fully deflated air must come out of the balloon. I'm not sure whether air leaves when both holes are open because the air inside the bottle forces it to or because the elasticity of the balloon forces it to or both? The instant you open both holes the inside of the bottle is still at atmospheric pressure and so is the air above the opening so I'm thinking then it must just be the elasticity of the balloon? Could you clarify?
Case three is when the balloon is allowed to communicate with the atmosphere, but the air trapped inside the bottle is not. What can be said about the pressures in those parts? Think of the formula in #10 while you are at that.
## 1. How does the pressure inside a bottle affect a balloon?
The pressure inside a bottle affects a balloon by causing it to either inflate or deflate. If the pressure inside the bottle is greater than the pressure outside, the balloon will inflate. If the pressure inside the bottle is less than the pressure outside, the balloon will deflate.
## 2. Why does the balloon inflate when placed over the mouth of a bottle?
The balloon inflates when placed over the mouth of a bottle because the air molecules inside the bottle have a higher pressure than the air molecules outside the bottle. This causes the air to move from an area of high pressure (inside the bottle) to an area of low pressure (outside the bottle), which inflates the balloon.
## 3. What happens to the balloon when the bottle is squeezed?
When the bottle is squeezed, the pressure inside the bottle increases. This increase in pressure causes the air molecules to move from an area of high pressure (inside the bottle) to an area of low pressure (outside the bottle), which inflates the balloon.
## 4. How does the size of the bottle affect the pressure inside?
The size of the bottle does not directly affect the pressure inside. However, the volume of the bottle does affect how much air is needed to reach a certain pressure. A larger bottle will require more air to reach a certain pressure compared to a smaller bottle.
## 5. Why does the balloon deflate when the bottle is placed in hot water?
The balloon deflates when the bottle is placed in hot water because the hot water causes the air molecules inside the bottle to move faster, increasing their energy and therefore increasing the pressure inside the bottle. This causes the air to move from an area of high pressure (inside the bottle) to an area of low pressure (outside the bottle), causing the balloon to deflate.
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