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STATISTICS 101: TYPES OF DISTRIBUTIONS - Alceon Medtech Consulting
STATISTICS 101: TYPES OF DISTRIBUTIONS
In the last blog, we talked about different types of data. Data, as we know, can be discrete or continuous. Discrete data is data that can take only countable values. Continuous data is data that can
take any value, including decimal numbers.
While data in its raw form may seem unremarkable, its true value lies in its distribution. Think of a distribution as a visual storyteller, painting a picture of how the data is scattered across
different values or ranges. These patterns, when deciphered, can unlock a wealth of insights and guide us towards informed decisions.
The most used distributions are mentioned below.
A Binomial Distribution is the simplest type of distribution. It is used when an event can have only two possible values with equal probability. So, if an event can be either a pass or a fail, and we
distribute several such events, it is a binomial distribution. A researcher may want to know the success of a treatment on a set of patients. A treatment can either be a success or a failure. Using
the binomial distribution, the researcher can predict how many patients can be successfully treated with the treatment.
A Poisson Distribution gives the probability of the number of events in a given period. A Poisson Distribution can, for example, predict how many deaths are likely to occur in a day in a town or
predict how many patients might come to a doctor’s clinic in a fixed period.
For example, predict how many deaths are likely to occur in a day in a town or predict how many patients might come to a doctor’s clinic in a fixed period.
The Normal Distribution is the most used and perhaps crucial distribution for continuous data. It is a symmetric and bell-shaped distribution. It is useful because it can represent many real-life
natural phenomena in physics, biology, mathematics, finance, and economics. Another essential feature of Normal Distribution is that it can approximate other types of distributions.
Some of the commonly encountered Normal Distributions in a population are the birthweight of newborn babies, heights of males in a population and diastolic blood pressure.
Student’s t-distribution, also known as the t-distribution, is a bell-shaped, continuous probability distribution similar to the normal distribution but has heavier tails and is flatter and shorter.
It is used to analyse data sets that would otherwise be unsuitable for analysis using the normal distribution. The t-distribution is used when data are approximately normally distributed.
The t-distribution was developed in 1908 by William Sealy Gosset, an Englishman who published under the pseudonym Student. Gosset worked at the Guinness brewery in Dublin and found that existing
statistical techniques using large samples were not useful for the small sample sizes he encountered.
The Exponential Distribution is another distribution used for continuous data. It is widely used in the field of reliability. This distribution is concerned with the amount of time until some
specific event occurs. For example, the time that the battery of an electric car will last or the time taken for a metal hip implant to fail can be predicted by using Exponential Distributions.
Contact us for your data analysis and sample size calculation requirements. | {"url":"https://alceonconsulting.com/statistics-101-types-of-distributions/","timestamp":"2024-11-03T22:04:26Z","content_type":"text/html","content_length":"231475","record_id":"<urn:uuid:25321501-db81-4271-afb0-26b995f85f76>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00407.warc.gz"} |
Blackjack Using Q-Learning
Codersarts AI
Blackjack is a popular card game played in many casinos. The objective of the game is to
win money by obtaining a point total higher than the dealer’s without exceeding 21. Determining an optimal blackjack strategy proves to be a difficult challenge due to the stochastic nature of the
game. This presents an interesting opportunity for machine learning algorithms. Supervised learning techniques may provide a viable solution, but do not take advantage of the inherent reward
structure of the game.
Reinforcement learning algorithms generally perform well in stochastic environments, and could utilize blackjack's reward structure. This paper explores reinforcement learning as a means of
approximating an optimal blackjack strategy using the Q-learning algorithm.
1 Introduction
1.1 The Blackjack Problem Domain The objective of blackjack is to win money by obtaining a point total higher than the dealer’s without exceeding 21. The game works by assigning each card a point
value. Cards 2 through 10 are worth their face value, while Jacks, Queens, and Kings are worth 10 points. An ace is worth either 1 or 11 points, whichever is the most beneficial. Before the start of
every hand, each player at the table places a bet. Next, the players are dealt two cards face up. The dealer also receives two cards: One face down, and one face up. The challenge is to choose the
best action given your current hand, and the face up card of the dealer. Possible actions include hitting, standing, splitting, or doubling down.
Hitting refers to receiving another card from the dealer. Players may hit as many times as they wish, as long as they do not bust 1 If additional cards are not desirable, one may choose to stand. If
a player receives two cards of the same value, it is possible to split, and play each card as a separate hand. For example, if two 7's are dealt, one may split, and play each 7 separately. The catch
is that an additional bet of equal size to the original must be placed. At any given time a player may also choose to double down. This means doubling one's bet in the middle of a hand, on the
condition of receiving only one additional card from the dealer.
The stochastic nature of blackjack presents an interesting challenge for machine learning algorithms. A supervised learning algorithm constructs a model by learning on a labeled set of instances. One
drawback with this approach is that it does not take the inherent reward structure of the game into account. An ideal blackjack strategy will maximize financial return in the long run. Reinforcement
learning algorithms are particularly well suited for these types of problems. Also, the game can be easily modeled as a Markov Decision Process (MDP). These factors suggest that a reinforcement
learning approach is appropriate for approximating an optimal blackjack strategy. The next section provides a brief overview of reinforcement learning, which is followed by a description of the
Q-learning algorithm.
1.2 Reinforcement Learning
Reinforcement learning techniques rely on feedback from the environment in order to learn. Feedback takes the form of a numerical reward signal, and guides the agent in developing its policy. The
environment is usually modeled as an MDP, which is defined by a set of states, actions, transition probabilities, and expected rewards (Sutton & Barto, 1998). Each action has a probability of being
the action selected, as well as an associated value, which corresponds to the expected reward of taking the action. A greedy action is an action that has the greatest value. In order to learn, the
agent must balance exploration and exploitation of the environment. During exploration, the agent tries non-greedy actions in hopes of improving its estimates of their values. Value functions allow
agent to compute how “good” it is to be in a given state. Vπ(s) is called the state-value function, and allows the agent to compute the expected reward of being in state s, and following policy π
(Sutton & Barto 1998). Qπ(s, a) is called the action-value function, and allows the agent to compute the expected reward of being in state s, taking action a, and thereafter following policy π (Suton
& Barto 1998). An optimal policy consists of the actions that lead to the greatest reward over time. The optimal state-value function is
denoted by V*(s), and the optimal action-value function is denoted by Q*(s, a).
1.3 The Q-Learning Algorithm
The Q-learning algorithm is a temporal difference (TD) method that approximates Q*(s, a) directly (Sutton & Barto, 1998). TD methods such as Q-learning are incremental algorithms that update the
estimates of Q(s, a) on each time-step. The Q-learning algorithm is outlined below.
The parameter alpha is referred to as the learning rate, and determines the size of the update made on each time-step. Gama is referred to as the discount rate, which determines the value of future
rewards. The Q-learning algorithm is guaranteed to converge to Q*(s, a) with probability 1, as long as each state-action pair is continually updated (Suton & Barto, 1998). The algorithm is well
suited for on-line applications, and generally performs well in practice.
The Q-learning algorithm is an excellent method for approximating an optimal blackjack strategy because it allows learning to take place during play. This makes it a good choice for the blackjack
problem domain. Blackjack is easily formulated as an episodic task, where the terminal state of an episode corresponds to the end of a hand. The state representation consists of the agent's current
point total, the value of the dealer's face up card, whether or not the hand is soft 2 , and whether or not the agent's hand may be split. Certainly, more robust state representations could be used,
but I wanted to keep the size of the state space at a minimum. Actions include hit, stand, split, and double down. All of the possible actions where included in an attempt to stay true to the game.
The reward structure is as follows: For each action that does not result in a transition to a terminal state, a reward of 0 is given. Once a terminal state has been reached, a reward is given based
on the size of the agent's bet. For example, if the agent bets 5 dollars, and wins the hand, a reward of +5 is given. If the agent loses the hand, a reward of -5 is given. A static betting strategy
is used because the state representation does not allow predictions to be made concerning which cards may appear in the following hands. A complete list of the rules adopted for the blackjack
simulator are listed in Figure 2.
3 Experimental Evaluation
To evaluate the performance of my learning agent, two hard coded players have been implemented. The first player takes completely random actions, while the second follows a strategy known as basic
strategy. Basic strategy cuts the casinos edge down to less than one percent, and is the optimal policy for the state representation adopted (Wong 1994). My experimental approach consists of five
runs, where each run is a cycle of training hands, and test hands. During training, the agent plays with four other reinforcement learning agents, all of which share the same lookup tables. During
the test hands, the agent plays with the random player, and the basic strategy player. Each player starts with $1,000, and makes the same initial bet of $5. Within each run, the number of
training hands is continually increased to a total of 10 million, while the number of test hands remains constant at 100,000. This procedure is summarized in Figure 3.
The reinforcement learning agent's initial policy is completely random. Thus, one would expect it to perform similarly to the random player initially. As the number of training hands increases, the
learning agent's policy should converge to basic strategy because the Q- learning algorithm directly approximates Q*(s, a).
Figure 4 shows average performance results for each player over the five runs. The number of test hands remains constant at 100,000, while the number of training hands is continually increased to ten
million. Notice that all of the players lose money. This is true even of the optimal policy, which demonstrates the difficulty of actually winning money playing blackjack. A comparison between the
random player, and the learning agent reveals a significantly better performance by the learning agent. It is apparent that the agent is learning useful information during the training process. Also
notice that the learning agent's performance asymptotically approaches the performance of the basic strategy player. This is to be expected because the Q-learning algorithm directly approximates the
optimal action-value function Q*(s, a). Upon examination of the standard deviations between the performance of the basic strategy player, and the reinforcement learning player, it was revealed that
they did not overlap. This means that the learning agent did not converge to the optimal policy, but rather a near optimal policy.
While the results of the previous section indicate that the reinforcement learning agent is learning a near optimal policy, there is room for improvement. Currently, my learning algorithm uses an
epsilon-greedy exploration strategy. An algorithm with a more sophisticated exploration strategy, such as the Bayesian Q-learning algorithm described by Dearden, Friedman, and Russel (1998), may
yield better results. The Bayesian Q-learning algorithm maintains probability distributions over Q- values, in an attempt balance exploration and exploitation by maximizing information gain.
Including more information in the state representation may increase performance as well. One could image the addition of a hi-low count. This, in conjunction with a dynamic betting strategy, may
allow performance to surpass that of the basic strategy player.
Are you looking solutions of these reinforcement learning project, of any other help related to complex machine learning project then you can contact us at here: | {"url":"https://www.ai.codersarts.com/post/blackjack-using-q-learning","timestamp":"2024-11-08T15:34:12Z","content_type":"text/html","content_length":"1050041","record_id":"<urn:uuid:022e6a09-7f7c-4965-8af3-14e084c025cd>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00494.warc.gz"} |
All Implemented Interfaces:
Direct Known Subclasses:
LiteralImpl, ResourceImpl
A specialisation of Polymorphic that models an extended node in a an extended graph. An extended node wraps a normal node, and adds additional convenience access or user affordances, though the state
remains in the graph itself.
• Method Details
□ asNode
Answer the graph node that this enhanced node wraps
Specified by:
asNode in interface FrontsNode
A plain node
□ getGraph
Answer the graph containing this node
An enhanced graph
□ isAnon
public final boolean isAnon()
An enhanced node is Anon[ymous] iff its underlying node is Blank.
□ isLiteral
public final boolean isLiteral()
An enhanced node is Literal iff its underlying node is too.
□ isURIResource
public final boolean isURIResource()
An enhanced node is a URI resource iff its underlying node is too.
□ isStmtResource
public final boolean isStmtResource()
An enhanced node is a statement resource iff its underlying node is a triple term (RDF-star).
□ isResource
public final boolean isResource()
An enhanced node is a resource if it's node is a URI node, a blank node or a triple term.
□ viewAs
Answer a facet of this node, where that facet is denoted by the given type.
t - A type denoting the desired facet of the underlying node
An enhanced nodet that corresponds to t; this may be this Java object, or a different object.
□ as
allow subclasses to implement RDFNode & its subinterface
□ canAs
API-level method for polymorphic testing
□ hashCode
public final int hashCode()
The hash code of an enhanced node is defined to be the same as the underlying node.
Specified by:
hashCode in class Polymorphic<RDFNode>
The hashcode as an int
□ equals
public final boolean equals(Object o)
An enhanced node is equal to another enhanced node n iff the underlying nodes are equal. We generalise to allow the other object to be any class implementing asNode, because we allow other
implemementations of Resource than EnhNodes, at least in principle. This is deemed to be a complete and correct interpretation of enhanced node equality, which is why this method has been
marked final.
Specified by:
equals in class Polymorphic<RDFNode>
o - An object to test for equality with this node
True if o is equal to this node.
□ isValid
public boolean isValid()
answer true iff this enhanced node is still underpinned in the graph by triples appropriate to its type.
Specified by:
isValid in class Polymorphic<RDFNode> | {"url":"https://jena.apache.org/documentation/javadoc/jena/org.apache.jena.core/org/apache/jena/enhanced/EnhNode.html","timestamp":"2024-11-10T10:55:57Z","content_type":"text/html","content_length":"28071","record_id":"<urn:uuid:240ffa16-dae8-4b62-9af5-f500807d5789>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00800.warc.gz"} |
Attometers to Earth's equatorial radius Converter
Enter Attometers
Earth's equatorial radius
β Switch toEarth's equatorial radius to Attometers Converter
How to use this Attometers to Earth's equatorial radius Converter π €
Follow these steps to convert given length from the units of Attometers to the units of Earth's equatorial radius.
1. Enter the input Attometers value in the text field.
2. The calculator converts the given Attometers into Earth's equatorial radius in realtime β using the conversion formula, and displays under the Earth's equatorial radius label. You do not need
to click any button. If the input changes, Earth's equatorial radius value is re-calculated, just like that.
3. You may copy the resulting Earth's equatorial radius value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Attometers to Earth's equatorial radius?
The formula to convert given length from Attometers to Earth's equatorial radius is:
Length[(Earth's equatorial radius)] = Length[(Attometers)] / 6.378160000453972e+24
Substitute the given value of length in attometers, i.e., Length[(Attometers)] in the above formula and simplify the right-hand side value. The resulting value is the length in earth's equatorial
radius, i.e., Length[(Earth's equatorial radius)].
Calculation will be done after you enter a valid input.
Consider that the wavelength of a gamma-ray photon is around 1 attometer.
Convert this wavelength from attometers to Earth's equatorial radius.
The length in attometers is:
Length[(Attometers)] = 1
The formula to convert length from attometers to earth's equatorial radius is:
Length[(Earth's equatorial radius)] = Length[(Attometers)] / 6.378160000453972e+24
Substitute given weight Length[(Attometers)] = 1 in the above formula.
Length[(Earth's equatorial radius)] = 1 / 6.378160000453972e+24
Length[(Earth's equatorial radius)] = 0
Final Answer:
Therefore, 1 am is equal to 0 earth's equatorial radius.
The length is 0 earth's equatorial radius, in earth's equatorial radius.
Consider that the scale of nuclear interactions is on the order of 10 attometers.
Convert this scale from attometers to Earth's equatorial radius.
The length in attometers is:
Length[(Attometers)] = 10
The formula to convert length from attometers to earth's equatorial radius is:
Length[(Earth's equatorial radius)] = Length[(Attometers)] / 6.378160000453972e+24
Substitute given weight Length[(Attometers)] = 10 in the above formula.
Length[(Earth's equatorial radius)] = 10 / 6.378160000453972e+24
Length[(Earth's equatorial radius)] = 0
Final Answer:
Therefore, 10 am is equal to 0 earth's equatorial radius.
The length is 0 earth's equatorial radius, in earth's equatorial radius.
Attometers to Earth's equatorial radius Conversion Table
The following table gives some of the most used conversions from Attometers to Earth's equatorial radius.
Attometers (am) Earth's equatorial radius (earth's equatorial radius)
0 am 0 earth's equatorial radius
1 am 0 earth's equatorial radius
2 am 0 earth's equatorial radius
3 am 0 earth's equatorial radius
4 am 0 earth's equatorial radius
5 am 0 earth's equatorial radius
6 am 0 earth's equatorial radius
7 am 0 earth's equatorial radius
8 am 0 earth's equatorial radius
9 am 0 earth's equatorial radius
10 am 0 earth's equatorial radius
20 am 0 earth's equatorial radius
50 am 0 earth's equatorial radius
100 am 0 earth's equatorial radius
1000 am 0 earth's equatorial radius
10000 am 0 earth's equatorial radius
100000 am 0 earth's equatorial radius
An attometer (am) is a unit of length in the International System of Units (SI). One attometer is equivalent to 0.000000000000001 meters or 1 Γ 10^(-18) meters.
The attometer is defined as one quintillionth of a meter, making it an extremely small unit of measurement used for measuring subatomic distances.
Attometers are used in advanced scientific fields such as particle physics and quantum mechanics, where precise measurements at the atomic and subatomic scales are required.
Earth's equatorial radius
The Earth's equatorial radius is the distance from the Earth's center to the equator. One Earth's equatorial radius is approximately 6,378.1 kilometers or about 3,963.2 miles.
The equatorial radius is the longest radius of the Earth due to its equatorial bulge, caused by the planet's rotation. This bulge results in a slightly larger radius at the equator compared to the
polar radius.
The Earth's equatorial radius is used in geodesy, cartography, and satellite navigation to define the Earth's shape and for accurate measurements of distances and areas on the Earth's surface. It
provides a key parameter for understanding Earth's dimensions and its gravitational field.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Attometers to Earth's equatorial radius in Length?
The formula to convert Attometers to Earth's equatorial radius in Length is:
Attometers / 6.378160000453972e+24
2. Is this tool free or paid?
This Length conversion tool, which converts Attometers to Earth's equatorial radius, is completely free to use.
3. How do I convert Length from Attometers to Earth's equatorial radius?
To convert Length from Attometers to Earth's equatorial radius, you can use the following formula:
Attometers / 6.378160000453972e+24
For example, if you have a value in Attometers, you substitute that value in place of Attometers in the above formula, and solve the mathematical expression to get the equivalent value in Earth's
equatorial radius. | {"url":"https://convertonline.org/unit/?convert=attometers-earths_equatorial_radius","timestamp":"2024-11-09T23:05:50Z","content_type":"text/html","content_length":"92249","record_id":"<urn:uuid:36554aec-455e-4bd5-a08a-c7d186b0c55b>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00458.warc.gz"} |
Single-projection radiography for noncircular symmetries: Generalization of the Abel transform method
We present a new method which extends the use of the single projection radiographic Abel method, hitherto applicable only to objects of circular and elliptical cross sections, to objects having
general, noncircular symmetries. This is done by developing a new integral equation that is similar in applications to Abel's equation, and includes it as a special case. The use of the new equation
is discussed for objects having a smooth and convex cross-section boundary (e.g., elliptic), a piecewise smooth convex boundary (e.g., bi-parabolic), and a boundary with regions of zero curvature
(e.g., polygons). Specific examples are given for each of these three classes, and analytic inverses are calculated for these cases. Also, numerical inversion of the integral equation is given,
showing satisfactory results. We show that in contrast to Abel's equation in many cases the kernel of the integral equation is non-singular. Consequently, fairly simple inversion techniques are
sufficient. Finally, the azimuthal variation of the transmitted intensity is employed to provide a convenient and fast nondestructive evaluation test of the deviation of the radiographed object from
a prescribed symmetry.
Dive into the research topics of 'Single-projection radiography for noncircular symmetries: Generalization of the Abel transform method'. Together they form a unique fingerprint. | {"url":"https://cris.biu.ac.il/en/publications/single-projection-radiography-for-noncircular-symmetries-generali","timestamp":"2024-11-08T01:37:09Z","content_type":"text/html","content_length":"55201","record_id":"<urn:uuid:18b3d477-56e9-4893-8235-b97e2f01bc24>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00240.warc.gz"} |
More Complex Loop Practices - Building Algorithms
In this section, we will focus on building algorithms that involve multiple steps and calculations. To illustrate, let's consider a real-world example:
1# Initialize a hash with countries and their respective visit durations
2visit_durations = {
3 "France" => 5,
4 "Japan" => 7,
5 "Australia" => 10
8total_days = 0
10# Iterate through the values of the hash and calculate the total days spent by incrementing the total_days variable by each value
11visit_durations.each_value do |days|
12 total_days += days
15# Calculate the average stay by dividing the total days by the number of countries visited (size of the hash)
16average_visit = total_days / visit_durations.size.to_f # to_f is used to convert to float
18puts "Total days of travel: #{total_days}, with an average stay of #{average_visit.round(2)} days." # round(2) is used to round to 2 decimal places
In this code, we'll learn how to iterate through a hash to calculate the total days spent in different countries and then compute the average stay. This involves accumulating values and performing
calculations within loops, which are fundamental skills for any coder.
The each_value method iterates over each value in a hash. In the example, the visit_durations.each_value loop iterates through the number of days spent in each country and accumulates these values
into the total_days variable. | {"url":"https://learn.codesignal.com/preview/lessons/2152/more-complex-loop-practices--building-algorithms","timestamp":"2024-11-04T05:36:44Z","content_type":"text/html","content_length":"125337","record_id":"<urn:uuid:81872e28-5a61-4abb-a22d-829f1e85ca66>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00712.warc.gz"} |
The PiVizTool
This page provides the PiVizTool, a graphical environment for the simulation of pi-calculus-based choreographies.
The PiVizTool is a tool for the simulation and analysis of mobile systems described in the pi-calculus. PiVizTool has been developed by Anja Bog in the course of a master thesis. The tool graphically
displays the linking structure of a pi-calculus system. For the graphical representation an extended version of Robin Milner's flow graphs is utilized. The evolution of the displayed pi-calculus
system can be influenced step-by-step through a user's interaction with the graphical representation. Thereby, the interaction behavior and link passing mobility of the pi-calculus agents in the
system can be monitored. Due to the visualization of the interacting pi-calculus agents and their links among each other, an easier understanding of pi-calculus system evolution is achieved. The
pi-calculus systems as input for the tool can be described by using the same input syntax as for the Mobility Workbench and Another Bisimulation Checker. Enhancements have been made for representing
grouping processes into BPM specific pools.
The PiVizTool is written in Java and provided in two versions, either as pre-compiled .JAR file (Java 1.5) or as source code (should work from Java 1.4). Both versions require the installation of
GraphViz Dot (command line version) (GraphViz Homepage) in advance.
• The precompiled .JAR file can simply be started (e.g. using java -jar PiVizTool.jar)
• The source code requires ANT (Homepage). The following targets are provided:
□ clean: Cleans the project directory.
□ compile: Compiles the PiVizTool.
□ jar: Builds the PivVizTool.jar file.
□ run: Runs the PiVizTool.jar file.
□ Best start with "ant compile jar run".
• The dot path needs to be configured first under File/Set dot execution path... If dot is in your PATH, the default setting might work.
The PiVizTool has been presented at workshop at the EMISA 2006 as well as the BPM 2007 conferences. You can find the corresponding publications here:
• Anja Bog, Frank Puhlmann, and Mathias Weske: The PiVizTool: Simulating Choreographies with Dynamic Binding. BPM Conference 2007 Demonstration Program, CEUR Workshop Proceedings Vol. 272,
Brisbane, Australia (2007)
• Anja Bog, Frank Puhlmann: A Tool for the Simulation of Pi-Calculus Systems. Open.BPM 2006: Geschäftsprozessmanagement mit Open Source-Technologien, Hamburg (2006), Germany | {"url":"http://frapu.net/bpm/piviztool.html","timestamp":"2024-11-13T02:54:01Z","content_type":"text/html","content_length":"5290","record_id":"<urn:uuid:f7e21f3d-6a91-4bb6-8202-872e710110ca>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00022.warc.gz"} |
Digital Electronics Digital Arithmetic Operations and Circuits Online Test
Digital Electronics Digital Arithmetic Operations and Circuits Online Test
Digital Electronics Digital Arithmetic Operations and Circuits Online Test. The Digital Electronics Full online mock test paper is free for all students and Very Helpful for Exam Preparation. Digital
Electronics Question and Answers in English. Digital Electronics Online Mock test for Digital Electronics Digital Arithmetic Operations and Circuits Topic. Here we are providing Digital Electronics
Digital Arithmetic Operations and Circuits Online Test Series in English. Check Digital Electronics Mock Test Series 2019.
This paper has 30 questions.
Time allowed is 30 minutes.
The Digital Electronics online Mock Test Exam is Very helpful for all students. Now Scroll down below n click on “Start Quiz” or “Start Test” and Test yourself. | {"url":"https://gkknowledge.in/digital-electronics-digital-arithmetic-operations-circuits-online-test/","timestamp":"2024-11-08T12:55:19Z","content_type":"text/html","content_length":"169289","record_id":"<urn:uuid:a6cc38b0-b21b-4222-a3f2-e19e7ed66195>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00614.warc.gz"} |
Droste Discovered the Schwarzschild Metric
Schwarzschild metric
is famous for describing a black hole in general relativity. It was discovered in 1915 immediately after publication of the field equations by Einstein and Hilbert, although the significance of black
holes was only figured out much later.
I recently learned that this metric was independently discovered by a student of Lorentz.
From a 2002 paper:
Johannes Droste’s “Field of a single center in Einstein’s theory of gravitation, and the motion of a particle in that field.” It is a remarkable paper, arguably one of the most remarkable in the
annals of general relativity and yet, although the paper is known to historians of science, practitioners of relativity themselves have been almost universally unaware of its existence for nearly
a century, and no mention of it appears in any standard text. ...
We know little about Johannes Droste. From what we do know (see the biographical note), Einstein’s theory of gravitation was the subject of his Ph.D. thesis. As he tells us in the introduction to
the current paper, he had been working on the equations of motion in general relativity as early as 1913 after Einstein published a preliminary version of the field equations.
The birth of Newtonian gravity is considered to be the central force law, although the field equations came much later. For general relativity, the birth is considered to be the field equations, not
the central force law.
I am not sure why. The Schwarzschild-Droste metric is the analog of the central force law. It is what you want for celestial orbits. Einstein and Grossmann published an "Entwurf" theory in 1913,
saying Ricci = 0 in empty space. Einstein retracted this in subsequent papers, but it is apparently what Droste used to figure out the correct central force law.
If Einstein had never met Hilbert in 1915, and they never published their field equations, we still might have had the essence of the theory from Droste's work. Probably Lorentz contributed also.
1 comment:
1. Roger,
discussing black holes in general like playing 'whack a mole'. Any objection you logically make can be elided away from by ducking down some side alley variation. Just because you name something
a black hole does not mean in any way you actually know what one is... much less be certain of describing or modeling its unknown internals accurately.
Many of the terms used to describe black holes are only actually applicable to certain kinds of theoretical black holes, and many of the terms are entirely rubbish, much like the gem you mention
where 'Ricci = 0', (a 'gravitational field outside a body' is sheer bullshittery on such a scale I can't even measure it) which is mathematically stating that you are removing all matter and
energy from your field, so... by definition there is no source for your gravitational field despite the popular claim there is a gravitational presence still there there mathematically...well,
prey tell, just how?? Magical circular wording? In reality math can not convey actual forces, or carry mass without a defined source...outside of fantasy and in physics apparently.
Also, to add insult to injury, Ricci = 0 mathematically depends upon a little thing I like to call 'division by zero'. I can't divide by zero...and much less get infinity, and neither can you,
and more importantly, neither can Hilbert of Einstein, or even little Sean Caroll and Kaku the Magnificant. | {"url":"http://blog.darkbuzz.com/2023/05/droste-discovered-schwarzschild-metric.html","timestamp":"2024-11-02T12:10:10Z","content_type":"text/html","content_length":"108467","record_id":"<urn:uuid:4e63f9aa-b8c0-4291-a186-c6197f35c399>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00382.warc.gz"} |
Emergence of behaviour in a self-organized living matter network
What is the origin of behaviour? Although typically associated with a nervous system, simple organisms also show complex behaviours. Among them, the slime mold Physarum polycephalum, a giant single
cell, is ideally suited to study emergence of behaviour. Here, we show how locomotion and morphological adaptation behaviour emerge from self-organized patterns of rhythmic contractions of the
actomyosin lining of the tubes making up the network-shaped organism. We quantify the spatio-temporal contraction dynamics by decomposing experimentally recorded contraction patterns into spatial
contraction modes. Notably, we find a continuous spectrum of modes, as opposed to a few dominant modes. Our data suggests that the continuous spectrum of modes allows for dynamic transitions between
a plethora of specific behaviours with transitions marked by highly irregular contraction states. By mapping specific behaviours to states of active contractions, we provide the basis to understand
behaviour’s complexity as a function of biomechanical dynamics.
We have judged that the response to the referee's residual comments are sufficient to allow this paper to proceed to publication. In particular, the detailed analysis of the mode spectrum and its
relationship to behavior is novel and possibly of general use in this field. Also, the experimental data per se should be interesting to a wide spectrum of readers.
Survival in changing environments requires from organisms the ability to transition between diverse behaviours (Angilletta and Sears, 2011; Wong and Candolin, 2014). In higher organisms, a plethora
of neural dynamics enable this capacity, ranging from almost random to strongly correlated firing patterns of neurons (Mochizuki et al., 2016). Decoding the origin of behaviour from neuronal activity
has been called the ‘holy grail of neuroscience’ (Bando et al., 2019), a task especially challenging given the vastly complex networks of neurons (Berman, 2018). Significant progress has been made by
simultaneous tracking of neuronal activity and behaviour – defined as trajectories through spaces of postural dynamics – in the fruit fly Drosophila melanogaster (Honegger et al., 2020) and the
nematode Caenorhabditis elegans (Nguyen et al., 2016). Behaviours of these systems have been identified as low-dimensional (Stephens et al., 2008) and hierarchical (Berman et al., 2016).
While these discoveries have advanced our understanding of the origin of behaviour, the complexity and size of biological neural networks make the acquisition and interpretation of experimental data
especially challenging. Curiously, organisms without a nervous system may offer an ideal intermediate step towards understanding behaviour. Certain non-neural organisms readily transition between a
multitude of behaviors similar in dynamic variability to that of organisms with a nervous system (Berg and Brown, 1972; Otto and Kessin, 2001; McMains et al., 2008; Ben-Jacob et al., 1994; Ben-Jacob
et al., 2000; Wan and Goldstein, 2014; Wan, 2018) and thus provide the opportunity to study the link between the underlying biophysical process and behaviour.
A non-neural organism with an exceptionally versatile behavioural repertoire is the slime mould Physarum polycephalum - a unicellular, network-shaped organism (Sauer, 1982) of macroscopic dimensions,
typically ranging from a millimeter to tens of centimeters. P. polycephalum’s complex behaviour is most impressively demonstrated by its ability to solve spatial optimisation and decision-making
problems (Nakagaki et al., 2000; Tero et al., 2010; Nakagaki and Guy, 2007; Dussutour et al., 2010; Reid et al., 2016), exhibit habituation to temporal stimuli (Boisseau et al., 2016), and use
exploration versus exploitation strategy (Aono et al., 2014). Recently, P. polycephalum was found capable of encoding memory about food source locations in the hierarchy of its body plan (Kramar and
Alim, 2021) in a process much reminding of synaptic facilitation– the brain’s way of creating memories (Jackman and Regehr, 2017). The generation of such rich behaviour requires a mechanism allowing
not only for long-range spatial coordination but also the flexibility to enable switching between different specific behavioural states.
The behaviour generating mechanism in P. polycephalum are the active, rhythmic, cross-sectional contractions of the actomyosin cortex lining the tube walls (Yoshimoto and Kamiya, 1984; Ueda et al.,
1986; Kamiya et al., 1988). The contractions drive cytoplasmic flows throughout the organism’s network (Iima and Nakagaki, 2012; Alim et al., 2013), transporting nutrients and signalling molecules (
Alim et al., 2017). Cytoplasmic flow is responsible for mass transport across the organism and thereby contractions directly control locomotion behaviour (Rieu et al., 2015; Lewis et al., 2015; Zhang
et al., 2017; Bäuerle et al., 2020; Rodiek et al., 2015).
So far, only one type of network-spanning peristaltic contraction pattern has been described experimentally (Alim et al., 2013; Oettmeier et al., 2017). However, for small P. polycephalum plasmodial
fragments various other short-range contraction patterns have been observed (Lewis et al., 2015; Zhang et al., 2017) and predicted by theory of active contractions (Bois et al., 2011; Radszuweit et
al., 2013; Radszuweit et al., 2014; Julien and Alim, 2018; Kulawiak et al., 2019). Similarly, up to now unknown complex, large-scale contraction patterns might play a role in generating the behaviour
of large P. polycephalum networks. Furthermore, transitions between such large-scale patterns are needed to allow for switching between specific behaviours, for example taking sharp turns during
migration in the absence of stimuli (Rodiek and Hauser, 2015).
Here, we decompose experimentally recorded contractions of a large P. polycephalum network of stable morphology into a set of physically interpretable contraction modes using Principal Component
Analysis. We find a continuous spectrum of modes and high variability in the activation of modes along this spectrum. By perturbing the network with an attractive stimulus, we show that the resulting
locomotion response is coupled to a selective activation of regular contraction patterns. Guided by these observations, we design an experiment on a P. polycephalum specimen reduced in morphological
complexity to a single tube. This allows us to quantify the causal relation between locomotion behaviour, cytoplasmic flow rate and varying types of contraction patterns, thus revealing the central
role of dynamical variability in generating different behaviours.
Continuous spectrum of contraction modes reveals large variability in organism’s contraction dynamics
To characterize the contraction dynamics of a P. polycephalum network, we record contractions using bright-field microscopy (Video 1) and decompose this data into a set of modes using Principal
Component Analysis (PCA). At first, networks in bright-field images are skeletonized, with every single skeleton pixel representing the local tube intensity as a measure of the local contraction
state (Bäuerle et al., 2017). Thus, any network state at a time t[i] is represented by a list of pixels, $I→ti$, along the skeleton, see Figure 1A and ‘Data processing’ (Appendix 1). Performing PCA
on this data results in a linear decomposition of the intensity vectors $I→ti$ into a basis of modes $ϕ→i$:
(1) ${\stackrel{\to }{I}}^{{t}_{i}}=\sum _{\mu }{a}_{\mu }^{{t}_{i}}{\stackrel{\to }{\varphi }}_{\mu }.$
Figure 1
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Principal Component Analysis yields a continuous spectrum of contraction modes in the P. polycephalum network.
This video cannot be played in place because your browser does support HTML5 video. You may still download the video for offline viewing.
Raw bright field time series of a P. polycephalum network, recorded at a rate of one frame every 3 sec.
See ‘Principal Component Analysis (PCA) (Appendix 2)’ for details. The modes, $ϕ→μ$, are orthonormal eigenvectors of the covariance matrix of the data and represent linearly uncorrelated contraction
patterns of the network, and $aμti$ denotes the time-dependent coefficients of the modes.
We rank modes according to the magnitude of their eigenvalues. Contrary to the small number of large eigenvalues found in a number of biological systems (Stephens et al., 2008; Jordan et al., 2013;
Gilpin et al., 2016), here the spectrum of relative eigenvalues, see ‘Principal component analysis (PCA)’ (Appendix 2) for technical details, is continuous with no clear cutoff (Figure 1B) and as a
result the contraction dynamics is high-dimensional. Notably, this is even the case when we disregard eigenvalues which lie below the upper noise bound (black line), computed from randomised data.
Therefore, PCA does not directly lead to a dimensionality reduction of the data. Instead, we here investigate the characteristics of mode dynamics that result from a continuous spectrum and how these
shape the organism’s behaviour.
The highest-ranking modes shown in Figure 1C(i) have a smooth spatial structure that varies on the scale of network size. As we will discuss below, such large-scale modes are associated with the long
wavelength peristalsis observed in Iima and Nakagaki, 2012; Alim et al., 2013. Interestingly, we also find modes highlighting specific morphological characteristics of the network. For example, the
structure of mode $ϕ→4$, Figure 1C(i), corresponds to the thickest tubes of the network Figure 1A, which suggests a special role of these tubes in the functioning of the network. Finally, as we go to
lower ranked modes, the spatial structure of the modes becomes increasingly finer. Yet, despite lacking an obvious interpretation for their structures, like for mode $ϕ→30$, Figure 1—figure
supplement 1, it is not possible to ignore their contribution relative to high-ranking modes.
Next, we turn to the time-dependent coefficients of modes shown in Figure 1C(ii). In accordance with the known rhythmic contractions (Kamiya, 1960) the coefficient $a1$ of the highest ranked mode
$ϕ→1$ oscillates with a typical period of $T∼100sec$ . Most strikingly, amplitudes of mode coefficients vary significantly over time - even on orders of magnitude, as shown in Figure 1—figure
supplement 2.
To map out the complexity of contractions over time, we define a set of significant modes for every time point. We quantify the activity of a mode by its relative amplitude
(2) ${p}_{\mu }^{{t}_{i}}=\frac{{\stackrel{~}{{a}^{2}}}_{\mu }^{{t}_{i}}}{\sum _{u }{\stackrel{~}{{a}^{2}}}_{u }^{{t}_{i}}}\phantom{\rule{thinmathspace}{0ex}},$
where $aμ2~$ denotes the amplitude of the square of the mode’s coefficient. By definition the sum over the relative amplitudes of all modes is normalized to one at any given time, $∑μpμti=1$. For any
given time point, we order the modes by their relative amplitude from largest to smallest and take the cumulative sum of their values until a chosen cutoff percentage is reached, see Figure 2A. We
find that the percentage of modes required to reach a specified cutoff value varies considerably over time. For a 90% cumulative amplitude cutoff, we find that on average 6.06% ($≈70$ modes) of the
1500 modes are significant. As discussed in more detail in ‘Choice of the cutoff of mode coefficient amplitudes’ (Appendix 6), defining a cutoff for the cumulative sum of mode amplitudes is related
to the problem of defining a cutoff for a continuous spectrum of eigenvalues. One common method is to define the cutoff with respect to the largest eigenvalue of the spectrum computed from a
randomised version of the original data (Berman et al., 2014). In ‘Choice of the cutoff of mode coefficient amplitudes’ (Appendix 6), we find that the 90% cumulative amplitude cutoff considered above
is consistent with this definition of cutoff for eigenvalues. As an important feature, we observe that there is large variation in the number of significant modes over time, with a standard deviation
of 36.96% from the mean value. This is an indicator for the complexity of the contractions in the network.
Dynamics of network contraction pattern is subject to strong variability in the percentage of significant modes and correlations between them.
Apart from the number of significant modes, the dynamics of the network depend on the temporal correlation of modes. While the modes form a spatially uncorrelated basis, the temporal correlation of
mode activation is non-trivial. In Figure 2B, we show the distribution of temporal correlations between mode coefficients as a function of the number of significant modes, see ‘Distribution of
temporal correlations’ (Appendix 3) for technical details. For a small number of significant modes, the coefficients are strongly (anti-)correlated in time, while for a large number of significant
modes, correlations values between coefficients are more uniformly distributed. Here, correlated coefficients result in coordinated pumping behaviour/contractions, while least correlated coefficients
coincide with irregular network-wide contractions. The above analysis shows that the dynamics of network contractions covers a wide range in complexity, from superposition of few large-scale modes
strongly correlated in time, to superpositions of many modes of varying spatial scale and temporal correlations. This gives rise to strong variability in the regularity of the contraction dynamics
over time. Up to now, we investigated an ‘idle’ network not performing a specific task, so we next stimulate the network to provoke a specific behaviour and scrutinize how the continuous spectrum of
modes contributes to it.
Stimulus response behaviour is paired with activation of regular, large-scale contraction patterns interspersed by many-mode states
To probe the connection between a specific behaviour and network contraction dynamics, we next apply a food stimulus to the same network, see Figure 3A. Food acts as an attractant and causes
locomotion of the organism toward the stimulus in the long term. The stimulus immediately triggers the tubes in the network to grow in a concentric region around the stimulus site. Also, the thick
transport tubes oriented toward stimulus location increase their volume, see Figure 3A. Altogether these morphological changes are typical for the specific behaviour induced here, namely the
generation of a new locomotion front.
Figure 3
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Network growth response to an external attractive stimulus is linked to characteristic changes in the contraction dynamics.
In Figure 3B, we quantify this stimulus response behaviour by tracking the growth of the most active regions of the network, defined by the boxes shown in the 81 min in Figure 3A. The tracked regions
are located on opposing sides of the network. Starting approximately at 85 min, the part of the network next to the stimulus site grows rapidly (burgundy curve in Figure 3B), at the expense of the
fan-shaped locomotion front in the lower left corner of the network (green curve in Figure 3B). In Figure 3—figure supplement 2, we additionally show that prior to the stimulus, the network grows the
fan-like shaped locomotion front in the lower left corner. Taken together, the application of the stimulus leads to a reversal of the network’s growth direction.
To identify potential changes in the contraction dynamics due to stimulus application, we perform PCA on a 700 frames long subset of the data subsequent to the ‘idle’ data of the previous section.
First, we rediscover a continuous spectrum of modes, see Figure 3—figure supplement 1, resembling that of the ‘idle’ dynamic state. However, now the highest-ranked contraction modes, see Figure 3C,
show spatial patterns which can be directly related to the network’s growth behaviour. This includes activation of the upper region of the network close to the stimulus, as well as activation of the
thick tubes extending from top to bottom of the network. In fact, for more than 500 frames after the stimulus has been applied, the rhythmic contraction dynamics of the network are dominated by the
three highest-ranked modes, see Figure 3D and Figure 3—figure supplement 3 for the oscillatory dynamics of mode coefficients. During this period, every time a single mode is the most active one for a
duration of $§gt;30$ frames, its amplitude exceeds that of any other mode by 20–30%.
Next we link the stimulus-induced reversal in growth direction to the changes in the contraction pattern. Specifically, we observe that the time interval of growth reversal Figure 3B coincides with
the activation of the third-ranked mode $ϕ→3$, (orange curve in Figure 3D), as indicated by the pink shaded box extending across Figure 3D and B. The structure of this mode shows a clear distinction
of the growth area close to the stimulus and an activation of the two thick tubes stretching from bottom to the top of the network. This mode is followed by an activation of mode $ϕ→2$ (blue curve),
clearly marking the growth region within its spatial structure.
Finally, over time the growth of the stimulus response region tapers off and we find reactivation of mode $ϕ→1$ (red curve) which was the dominant mode before stimulus application. We note that the
spatial structure of mode $ϕ→1$ is remarkably similar to mode $ϕ→1$, the top-ranked mode that we find for PCA on the pre-stimulus ‘idle’ data Figure 3—figure supplement 2B. The reactivation of this
mode indicates that this contraction pattern is intrinsic to the network and is not simply erased by the stimulus.
Strikingly, the regular contraction dynamics shown in Figure 3D are interspersed with many-mode states where the number of significant modes increases considerably, see Figure 3E. The number of
significant modes oscillates after the stimulus. The oscillation maxima coincide with times at which the organism switches from one dominant contraction pattern to another, as indicated by the
blue-shaded boxes extending across Figure 3D and E. Our results suggest that prolonged regular dynamics dominated by a few or even a single mode are associated with specific behaviour like locomotion
and growth, while the many-mode states seem to serve as transition states between them.
While the network morphology is characteristic for P. polycephalum, reducing network complexity may help to conclude on the role of regular dynamics in driving specific behaviours, and the role of
many-mode states and the therefrom arising continuous distribution of modes.
Number of significant modes determines maximum cytoplasmic flow rate in the minimal morphological representation of the network
We next perform exactly the same course of experiments as before but on a P. polycephalum specimen reduced in complexity to a single tube with a locomotion front at either end, see inset in Figure 4
and Video 2. Strikingly, when performing PCA on this specimen of simple morphology we again find a continuous spectrum of modes (Figure 5—figure supplement 1) and large variability, including spikes
of many-mode states, in the number of significant modes (Figure 5A). This observation finally underlines that the continuous spectrum of modes and its variability in activation is intrinsic to the
organism’s behaviour, ruling out that the complexity of contraction modes only mirrors morphological complexity. Foremost, this minimal constituent of a network allows us now to directly map the
effect of variations in the contraction dynamics onto behaviour.
Number of significant modes is indicative for the volume flow rate in a cell reduced in its network complexity to a single tube.
Figure 5
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Locomotion behaviour of a single tube is determined by activation and temporal coupling of sine-and cosine-shaped contraction modes.
This video cannot be played in place because your browser does support HTML5 video. You may still download the video for offline viewing.
Raw bright field time series of a single P. polycephalum tube, recorded at a rate of one frame every 3 sec.
From the experimentally quantified tube contractions, we calculate the maximal flow rate at any point along the tube (Li and Brasseur, 1993) and over time correlate the strength of the flow rates,
driving locomotion behaviour at the tube ends, with the number of significant modes, see ‘Flow rate calculation in a P. polycephalum cell with single-tube morphology’ (Appendix 4). For both the flow
rate at the left and right end of the tube, shown in Figure 4, and Figure 4—figure supplement 1, respectively, we find that large flow rates are only achieved when the number of significant modes is
small. We had previously found that few significant modes are highly (anti-)correlated, whereas states with many significant modes are not, see Figure 2B. This observation now confirms our physical
intuition that the irregularity of states consisting of many modes goes hand in hand with reduced pumping efficiency and thus unspecific behaviour. Since a small number of significant modes not
necessarily always implies a large flow rate, we next turn to analyze their exact spatial structure and instantaneous temporal correlation to determine how cytoplasmic flow rates impact behaviour.
Instantaneous coupling and selective activation of modes determine locomotion behaviour
We now demonstrate the impact of changes in the dynamics of a small number of modes on the organism’s behaviour. For this, we quantify the locomotion behaviour of the single tube by tracking the area
of the locomotion fronts protruding from each end of the tube over time, see Figure 5A. The growth curves of the tube ends are shown in Figure 5B. While initially the right end is protruding faster
at the expense of the left end, a food stimulus applied to the left end of the tube reverses the direction of locomotion.
As for the network, we use PCA to analyse the contraction dynamics of the single tube and link it to behaviour. We apply PCA to contraction data along the tube which we parameterize by a longitudinal
coordinate. The spatial shapes of the two top-ranked modes $ϕ→1$ and $ϕ→2$ approximate Fourier modes, see Figure 5C and Figure 5—figure supplement 2. Examining the activation of modes, we find that
over long time intervals, and in particular after the stimulus, the two top-ranked modes dominate the tube’s contraction dynamics, see Figure 5D. To illustrate the connection between the nature of
tube contraction dynamics and locomotion behaviour, we pick two representative time intervals after the stimulus where either only mode $ϕ→1$, or modes $ϕ→1$ and $ϕ→2$ equally, dominate overall, see
vertical pink bars in Figure 5D. During the first interval when mode $ϕ→1$ alone is dominating, the tube is driven by a standing wave contraction pattern - yielding only a low cytoplasmic flow rate.
Correspondingly, the size of the locomotion front at either end shows no significant change in area during this interval. In contrast, during the interval when both modes $ϕ→1$ and $ϕ→2$ are equally
active, the resulting superposition is a left-traveling wave producing a large cytoplasmic flow rate in that direction. The left-traveling wave is in accordance with the growth of the left and
retraction of the right locomotion front as quantified in Figure 5B. See ‘Mode superpositions in a P. polycephalum cell with single-tube morphology’ (Appendix 4) for more details. In Figure 5E, we
highlight the most pronounced many-mode states during changes of dominant contraction dynamics.
These two examples solve the conundrum of Figure 4, which shows that a small number of significant modes does not necessarily lead to high cytoplasmic flow rates. Yet, the direct mapping of
contraction dynamics onto ensuing cytoplasmic flows confirms that a small number of significant modes is associated with specific behaviour. High cytoplasmic flow rates at the tube ends drive
locomotion, while lower flow rates likely lead to other behaviours such as mixing. Furthermore, many-mode states seem necessary for transitions in a multi-behavioural space.
Our explanation of behaviour – from contractions via flows to locomotion behaviour – in the single tube is a template for an analogous explanation in the network morphology. The analogy is justified
by the strong resemblance of the continuous mode spectrum, dynamics of significant modes, activation of regular contraction patterns and the nature of growth behaviour in both the network and single
tube. Therefore, while it is beyond the scope of this study, we expect a detailed analysis of the link between contractions and flows in the network morphology to yield qualitatively similar results
to those of the single tube, thus completing the mechanism of behaviour generation.
To uncover the origin of behaviour in P. polycephalum, we quantified the dynamics of this living matter network and linked it to its emerging behaviour. The simple build of this non-neural organism
allows us to trace contractions of the actomyosin-lined tubes, compute cytoplasmic flows from the contractions and finally link these dynamics to the emerging mass redistribution and whole-organism
locomotion behavior. Decomposing the contractions across the network into individual modes, we discover a large intrinsic variability in the number of significant modes over time along a continuous
spectrum of modes. By triggering locomotion through application of a stimulus, we identify that states with few significant modes and regular contraction patterns correspond to specific behaviors, in
this case locomotion. Yet, irregular contraction patterns consisting of a large number of significant modes are also present, particularly marking the transitions between different regular
contraction states. The use of an organism with a single-tube morphology allows us to obtain quantitative insights into the mechanism connecting contraction dynamics and locomotion behavior and in
first approximation serves as an analogue system for the large P. polycephalum with network morphology. Our findings suggest that a continuous spectrum of contraction modes allows the living matter
network P. polycephalum to quickly transition between a multitude of behaviours using the superposition of multiple contraction patterns.
Networks are ubiquitous in biology, including examples such as ecological networks (García Martín and Goldenfeld, 2006) and biomolecular interaction networks (Albert, 2005). Measurable quantities of
these networks, for instance the degree distribution of the network, typically follow continuous distributions and are oftentimes power-laws. The spectrum of eigenvalues Figure 1B that we find for
the contraction dynamics in P. polycephalum may similarly suggest a power-law. However, the presence of a power-law is generally difficult to prove and interpret. Instead, our sole focus is on the
continuous nature of the spectrum. It is important to emphasise that the continuity of the eigenvalue spectrum is not simply the result of the organism’s complex network morphology. This is
demonstrated by the fact that we find a similar spectrum also for the single-tube morphology Figure 5—figure supplement 1. Therefore, here the continuous spectrum of eigenvalues is distinctively a
property of the dynamic state of the organism.
Our observation of interlaced regular and irregular contraction patterns in P. polycephalum reminds of the strongly correlated or random firing patterns of neurons in higher organisms (Mochizuki et
al., 2016). In neural organisms, stereotyped behaviours are associated with controlled neural activity, as for example for locomotion in C. elegans (Liu et al., 2018) or the behavioural states of the
fruit fly Drosophila melanogaster (Berman et al., 2014; Berman et al., 2016). Variability in the dynamics of behaviour is also widely observed in these neuronal organisms (Grobstein, 1994; Renart and
Machens, 2014; Werkhoven et al., 2021; Honegger et al., 2020; Ahamed et al., 2020). It is thus likely that the transition role of irregular states consisting of many significant modes observed here
for P. polycephalum parallels the mechanisms of generating behaviour in the more complex forms of life.
P. polycephalum is renowned for its ability to make informed decisions and navigate a complex environment (Nakagaki et al., 2000; Tero et al., 2010; Nakagaki and Guy, 2007; Dussutour et al., 2010;
Reid et al., 2016; Boisseau et al., 2016; Aono et al., 2014; Ueda et al., 1976; Miyake et al., 1991). It would be fascinating to next follow the variability of contraction dynamics during more
complex decision-making processes. Furthermore, it would be interesting to observe ‘idle’ networks during foraging over tens of hours. It is likely that the contraction states with many significant
modes here act as noisy triggers that can spontaneously cause the organism to reorient its direction of locomotion.
In the context of P. polycephalum’s foraging behaviour, another exciting line of research opened by our results is the link between contraction modes and the organism’s metabolic changes. The
foraging networks displays a plethora of morphological patterns which are linked to the underlying metabolic states (Takamatsu et al., 2017; Lee et al., 2018). It has recently been shown that in the
neural organism Drosophila melanogaster, behaviour stemming from neural activity causes large-scale changes in metabolic activity (Mann et al., 2021). Exploring the relationship between behaviour
emergence and metabolism in P. polycephalum will bring key insight about the interplay between the mechanical and the biochemical machinery of the organism.
P. polycephalum’s body-plan as a fluid-filled living network with emerging behaviour finds its theoretical counterpart in theories for active flow networks developed recently (Woodhouse et al., 2016;
Forrow et al., 2017). Strikingly, these theories predict selective activation of thick tubes which we observe in the living network as well, prominently appearing among the top ranking modes, see
$ϕ→4$ in Figure 1C(i) or $ϕ→3$ in Figure 3C. This is a first hint that dynamics states arising from first principles in active flow networks could map onto behavioural and transition states observed
Likely our most broadly relevant finding in this work is that irregular dynamics, here arising in states with many significant modes, play an important role in switching between behaviours. This
should inspire theoretical investigations to embrace irregularities rather than focusing solely on regular dynamic states. The most powerful aspect of P. polycephalum as a model organism of behaviour
lies in the direct link between actomyosin contractions, resulting in cytoplasmic flows and emerging behaviours. The broad understanding of the theory of active contractions (Bois et al., 2011;
Radszuweit et al., 2013; Radszuweit et al., 2014; Julien and Alim, 2018; Kulawiak et al., 2019) might therefore well be the foundation to formulate the physics of behaviour not only in P.
polycephalum but also in other simple organisms. This would not only open up an new perspective on life but also guide the design of bio-inspired soft robots with a behavioural repertoire comparable
to higher organisms.
The specimen was prepared from fused microplasmodia grown in a liquid culture (Daniel et al., 1962) and plated on 1.5%-agar. The network was trimmed and imaged in the bright field setting in Zeiss
ZEN two imaging software with a Zeiss Axio Zoom V.16 microscope equipped with a Hamamatsu ORCA-Flash 4.0 digital camera and Zeiss PlanNeoFluar 1 x/0.25 objective. The acquisition frame rate was 3
sec. The stimulus was applied in a form of a heat-killed HB101 bacterial pellet in close network proximity.
The typical thickness of tubes in a P. polycephalum network is $∼50−100μm$ and the contraction amplitude about ~10% of the tube’s typical thickness (Alim et al., 2013). This change in tube thickness
can be detected from a bright-field microscopy recording. We record one bright-field frame every three seconds. Since the periodic contractions of the tubes take place on the time scale of 100 sec,
they are thus well resolved by the selected frame rate. Typically an idle network keeps a stable morphology and does not move significantly over a period of 1.5 h to 2.5 h which we use for recording
its contraction dynamics. Since no two P. polycephalum specimens ever have the same network morphology, we are naturally constrained to one biological and one technical replicate in our experiments.
Our data is a stack of bright-field images recorded from the P. polycephalum network with a rate of one frame every three seconds (Video 1; Video 2). Each bright-field frame has a time label t[i] and
the total number of frames is given by $T$. We process this data in the following steps. First, we mask the network in the bright-field images through thresholding. It is important to note that we
use the same mask for all the images in the stack. This is possible since we consider a network that does not significantly move or change its morphology. This is true even when we apply a stimulus
to the network, since we only consider the initial stages of stimulus response, before the network starts to display strong movement. From the masked regions of the bright-field frames, we extract
pixel intensity values which we convert to 8-bit format. Since we are here primarily interested in the contraction dynamics of the organism and not in the the actual base thickness of tubes or its
long-term growth dynamics, we detrend the data using a moving-average filter (rational transfer function) with a window size of two contraction periods (~ 200 sec) (Bäuerle et al., 2017). This leaves
us only with the desired information about contractions taking place in the time scale of several minutes. We store the intensity values of each frame in a vector $I→ti$ of dimension $M$ equal to the
number of pixels in the network, and $i$ indexes the frames in the range $i=1,…,T$. From the post-processed data, we define the following data matrix
(3) ${\mathbf{X}}^{t}=\left({\stackrel{\to }{I}}^{\phantom{\rule{thickmathspace}{0ex}}{t}_{1}},{\stackrel{\to }{I}}^{\phantom{\rule{thickmathspace}{0ex}}{t}_{2}},\dots ,{\stackrel{\to }{I}}^{\phantom
where $t$ denotes the matrix transpose.
Principal component analysis (PCA)
The contraction modes are computed from the covariance matrix of the data. We compute the covariance matrix from the data matrix $X$ after subtracting the mean from each column. The covariance matrix
is given by
(4) ${p}_{\mu }^{{t}_{i}}=\frac{{\stackrel{~}{{a}^{2}}}_{\mu }^{{t}_{i}}}{{\sum }_{u }{\stackrel{~}{{a}^{2}}}_{u }^{{t}_{i}}},$
The sought after contraction modes $ϕ→μ$ are the eigenvectors of the covariance matrix
(5) $\mathbf{C}{\stackrel{\to }{\varphi }}_{\mu }={\lambda }_{\mu }{\stackrel{\to }{\varphi }}_{\mu }\phantom{\rule{thinmathspace}{0ex}},$
and $λμ$ is the eigenvalue. The number of non-zero eigenvalues is equal to the rank of the covariance matrix. The eigenvalue captures the variance of the data along the direction of mode $ϕ→μ$. We
also define the relative eigenvalue as
(6) ${\stackrel{~}{\lambda }}_{\mu }=\frac{{\lambda }_{\mu }}{{\sum }_{u =1}^{T}{\lambda }_{u }}.$
The mode coefficient $aμ$ is obtained by projecting the data onto mode $ϕ→μ$.
We note that we perform PCA on data segments with at least 700 frames (=35 min). Since it is well known from the literature (Kamiya, 1960) that the period of the contraction dynamics in P.
polycephalum is on the order of 100 sec the analysed data contains on the order of 20 contraction periods at minimum. We can therefore be sure that we use enough data to resolve the characteristic
dynamical features investigated here. As a further reassuring result, we recover the typical contraction period of 100 sec in our analysis, see Figure 1C(ii).
We add a brief comment on Fourier analysis, as alternative decomposition method to PCA. First, in one dimension, PCA is equivalent to Fourier decomposition. This is indeed apparent in our PCA
analysis of the single-tube data set where the principal components shown in Figure 5C correspond precisely to half a period of a sine and cosine Fourier mode. In two dimensions, the situation is
more complicated. While we could in principle apply 2D Fourier decomposition we would need to apply Fourier analysis separately to every frame in our data set. However, this would mean that we have
no information about the temporal evolution of mode activation. The Fourier modes would be different from one frame to the next and the activation of large-scale patterns over time would be obscured.
Distribution of temporal correlations
For a given time point t[i], the significant modes are determined based on the 70% criterion curve from Figure 2A. Next, the temporal correlations among the coefficients are computed in a time
interval of ±15 frames around the time point t[i]. The correlations are then counted in bins of the appropriate row of Figure 2A. Repeating this processing for all time points and normalising each
row by the total number of correlations in that row, we obtain the final distribution shown.
Flow rate calculation in a P. polycephalum cell with single-tube morphology
To compute the flow rate of the cytoplasm in a P. polycephalum specimen with single-tube morphology we use the theory developed in Shapiro et al., 1969; Li and Brasseur, 1993. In that work the flow
of an incompressible Newtonian fluid inside an axisymmetric tube of fixed length is considered and the equations for the flow velocity field are written in the lubrication theory approximation.
Furthermore, a time-dependent thickness profile of longitudinal waves is imposed in the tube. Assuming no-slip boundary conditions, the flow field can be fully determined at every point along the
tube as a function of the time-dependent tube profile. For the case when the tube profile is a periodic train of waves, we compute the volume flow rate averaged over an oscillation period by
evaluating equation (13) of Li and Brasseur, 1993. We express the flow rate in units of volume of the entire tube divided by the oscillation period. This serves to characterize the performance in
pumping of the significantly contracting P. polycephalum cell. We determine the time period over which to average the volume flow rate directly from the flow-rate curve. Furthermore, the thickness
profile of the tube is given by the measured pixel intensity profile.
Mode superpositions in a P. polycephalum cell with single-tube morphology
We are interested in how the contraction dynamics of the cell controls the cell’s locomotion behavior. In our analysis we therefore focus on the tube segment connecting the locomotion fronts at
either end of the tube and perform Principal Component Analysis only on this part of the cell. Since the tube is effectively one-dimensional, we find that the modes we obtain closely approximate
Fourier modes. This means that superpositions of these modes afford a clear interpretation in terms of different contraction-wave patterns. Such an interpretation is even further facilitated by the
fact that we find that over large time intervals after the stimulus, the number of significant modes is very small. Indeed, over such time intervals it is sufficient to approximate the contraction
dynamics with only one or two modes, as can be seen from Figure 5D. Hence we are essentially studying a superposition of modes $ϕ→1$ and $ϕ→2$ shown in Figure 5C and Figure 5—figure supplement 2 of
the main text with their oscillating mode coefficients shown in Figure 5—figure supplement 3. To develop intuitive understanding of the nature of the superposition, we note that the modes $ϕ→1$ and
$ϕ→2$ approximate sine and cosine functions over the length of the tube. Given a sine and cosine spatial contraction profile, different types of superpositions can be formed depending on the nature
of their time-dependent coefficients. To illustrate further, let us assume the idealised case where both coefficients are sine functions that can have different phases and amplitudes. Then, if the
coefficient of one contraction profile is very small compared to the other, the resulting superposition is a standing wave. In case the coefficients have equal amplitudes, but are phase shifted by $π
/2$, the superposition is a traveling waves. Finally, if the coefficient amplitudes are not equal and the phase shift lies somewhere between zero and $π/2$, the nature of the superposition is a mix
of standing and traveling wave. Extrapolating this idealised picture allows us to infer the contraction dynamics resulting from our two-mode approximation. We see that the coefficients of the two
modes $ϕ→1$ and $ϕ→2$ shown in Figure 5—figure supplement 3 change in amplitude and phase relative to each other. It is easy to identify from this plot together with the plot of relative amplitudes
in Figure 5D time intervals which approximate one of the contraction dynamics that we have described for the idealised system. Therefore we conclude that the superposition of the two top modes
changes in its nature over time, ranging form a pure standing wave to a pure traveling wave.
Choice of the cutoff of mode coefficient amplitudes
Our analysis of contraction dynamics requires us to place a cutoff on the amplitude of mode coefficients. Our chosen cutoff of 90% is supported in two ways:
First, the problem of choosing a cutoff for the coefficients is related to the problem of choosing a cutoff for the eigenvalue spectrum, since an eigenvalue is the variance of the mode coefficient.
Given the continuous nature of the eigenvalue spectrum, there is no unique way to choose a cutoff. In (Berman et al., 2014), it is proposed to define the cutoff by the largest eigenvalue of the
spectrum of the randomised data, see the black line in Figure 1B. We tested their criterion on our data and find a 93% cutoff, equivalent to roughly 70 modes. This is consistent with our choice of a
90% cutoff for the amplitudes.
Second, our main qualitative observation - considerable variation in the number of significant modes over time - is robust to different choices of cutoff values. In Figure 2A and similarly in Figure
3E, we show the number of significant modes for two different values of the cutoff, namely 70% and 90%.
The two datasets from which Figures 1,2 and 3 and Figures 4 and 5 were generated are included as videos of raw bright-field time series in the article.
Variability in behavior and the nervous system
Encyclopedia of Human Behavior 4:447–458.
53. Book
Developmental Biology of Physarum
Cambridge, United Kingdom: Cambridge University Press.
Article and author information
Author details
Simons Foundation (400425)
IMPRS for Physics of Biological and Complex Systems
• Philipp Fleig
• Mirna Kramar
• Michael Wilczek
• Karen Alim
The funders had no role in study design, data collection and interpretation, or the decision to submit the work for publication.
© 2022, Fleig et al.
This article is distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use and redistribution provided that the original author and source are credited.
Views, downloads and citations are aggregated across all versions of this paper published by eLife.
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1. Philipp Fleig
2. Mirna Kramar
3. Michael Wilczek
4. Karen Alim
Emergence of behaviour in a self-organized living matter network
eLife 11:e62863.
Further reading
1. Computational and Systems Biology
2. Physics of Living Systems
Explaining biodiversity is a fundamental issue in ecology. A long-standing puzzle lies in the paradox of the plankton: many species of plankton feeding on a limited variety of resources coexist,
apparently flouting the competitive exclusion principle (CEP), which holds that the number of predator (consumer) species cannot exceed that of the resources at a steady state. Here, we present a
mechanistic model and demonstrate that intraspecific interference among the consumers enables a plethora of consumer species to coexist at constant population densities with only one or a handful
of resource species. This facilitated biodiversity is resistant to stochasticity, either with the stochastic simulation algorithm or individual-based modeling. Our model naturally explains the
classical experiments that invalidate the CEP, quantitatively illustrates the universal S-shaped pattern of the rank-abundance curves across a wide range of ecological communities, and can be
broadly used to resolve the mystery of biodiversity in many natural ecosystems.
1. Computational and Systems Biology
2. Physics of Living Systems
Planar cell polarity (PCP) – tissue-scale alignment of the direction of asymmetric localization of proteins at the cell-cell interface – is essential for embryonic development and physiological
functions. Abnormalities in PCP can result in developmental imperfections, including neural tube closure defects and misaligned hair follicles. Decoding the mechanisms responsible for PCP
establishment and maintenance remains a fundamental open question. While the roles of various molecules – broadly classified into “global” and “local” modules – have been well-studied, their
necessity and sufficiency in explaining PCP and connecting their perturbations to experimentally observed patterns have not been examined. Here, we develop a minimal model that captures the
proposed features of PCP establishment – a global tissue-level gradient and local asymmetric distribution of protein complexes. The proposed model suggests that while polarity can emerge without
a gradient, the gradient not only acts as a global cue but also increases the robustness of PCP against stochastic perturbations. We also recapitulated and quantified the experimentally observed
features of swirling patterns and domineering non-autonomy, using only three free model parameters - the rate of protein binding to membrane, the concentration of PCP proteins, and the gradient
steepness. We explain how self-stabilizing asymmetric protein localizations in the presence of tissue-level gradient can lead to robust PCP patterns and reveal minimal design principles for a
polarized system.
1. Computational and Systems Biology
2. Physics of Living Systems
Synthetic genetic oscillators can serve as internal clocks within engineered cells to program periodic expression. However, cell-to-cell variability introduces a dispersion in the characteristics
of these clocks that drives the population to complete desynchronization. Here, we introduce the optorepressilator, an optically controllable genetic clock that combines the repressilator, a
three-node synthetic network in E. coli, with an optogenetic module enabling to reset, delay, or advance its phase using optical inputs. We demonstrate that a population of optorepressilators can
be synchronized by transient green light exposure or entrained to oscillate indefinitely by a train of short pulses, through a mechanism reminiscent of natural circadian clocks. Furthermore, we
investigate the system’s response to detuned external stimuli observing multiple regimes of global synchronization. Integrating experiments and mathematical modeling, we show that the entrainment
mechanism is robust and can be understood quantitatively from single cell to population level. | {"url":"https://elifesciences.org/articles/62863","timestamp":"2024-11-13T12:25:26Z","content_type":"text/html","content_length":"429279","record_id":"<urn:uuid:1f3bbb7e-7afb-408b-800b-daf6444b5b8f>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00544.warc.gz"} |
Original article: https://macropolo.org/digital-projects/high-speed-rail/methodology/ https://macropolo.org/digital-projects/high-speed-rail/map/
We first explain the approach and general assumptions used in our assessment, followed by details on how we derived the cost and benefit estimate for each component.
General Approach and Assumptions
To estimate the net present value of China’s high-speed rail (HSR) network that is operational at the end of 2019, we used the discounted cash flow approach, the most commonly used valuation method.
We first forecast the annual cash flow from the entire HSR network through 2050, then converted it to the 2020 value using a 5% discount rate. All costs and benefits are expressed in 2020 values.
Although HSR is designed for a lifetime operation of 50 years, the cost/benefit analysis is for a 30-year timeframe through 2050. This is because the longer the time horizon, the less certain the
estimate, particularly if technologies like magnetic levitation displace current HSR technology in that time period.
Moreover, the benefits of HSR are future oriented. For example, in our forecast, HSR passenger traffic will not peak until 2035. So if the timeframe is set at 10 years, that would likely
underestimate the potential realized benefits of HSR.
With that general framework in mind, we made the following assumptions to forecast the annual cash flow through 2050:
HSR Traffic Density: Peaks at 30 million/km by 2035 and levels off after that point.
This is an increase from the current level of 23 million/km and is between the traffic densities of France (25 million/km) and Japan (36 million/km). This 30 million/km assumption is somewhat
conservative since it is only 40% of the current passenger density of the Beijing-Shanghai line.
With a third of the routes just completed between 2018-2019, these newer routes tend to see double-digit growth in traffic in the first few years of operation. Combined with the fact that China’s
population density is much higher than France’s, the former should easily surpass the latter’s demand for HSR.
Passenger volume: Peaks at 1.06 trillion passengers/km in 2035. This is derived by simply multiplying the traffic density with the total length of the HSR network at the end of 2019.
Shift from other modes of transportation:
• ~50% of passengers shift from conventional rail
• 20% of passengers shift from air travel
• 15% of passengers shift from bus and car
• We use the World Bank’s (WB) 2015 estimates for each mode of transportation and assume that these substitution effects last through 2025. Beyond 2025, all HSR growth represents new traffic rather
than substitution effects.
The magnitude of shifts from different modes of transport is crucial for estimating HSR’s cost and benefit. For instance, when HSR substitutes for air travel, that leads to cost savings because the
operating cost of the latter is more than twice that of the former. In contrast, shifting from conventional rail will be costlier because the operating cost of HSR is slightly higher.
Annual discount rate: 5% for both costs and benefits. We assume this discount rate because the borrowing cost for HSR investment is currently less than 4% and has been below 5% for most of the last
Calculating costs and benefits
Below we detail more specifically how we arrived at the costs and benefits of HSR over the 30-year timeframe. These are all generally accepted factors of cost and benefit for HSR, and where
appropriate, we used the WB’s estimate for our calculations.
One estimate where we differed from the WB is that of HSR’s ROI: 6.5% (our estimate) vs. 8% (WB’s estimate). This is the result of having more conservative assumptions in our estimate. First, we did
not include positive externalities such as agglomeration effects. Second, we have a lower residual value of HSR, which is equivalent to assuming a higher depreciation rate. Third, we assume that HSR
will never fully repay debt and will continue making interest payments for the entire 30-year period.
Construction Cost: Based on WB’s estimate of 130 million yuan/km per year multiplied by the estimated length of new HSR under construction each year.
Total construction cost is the sum of annual construction cost after subtracting the residual value of the HSR network in 2050. We subtract the residual value from construction cost because the HSR
network will still be in use after 2050 and has a positive value. Put differently, what is included can be interpreted as the total depreciation of the network through 2050. We assume the residual
value to be 30% of the HSR construction cost, a more conservative assumption compared to what is likely closer to 40% in reality.
Interest Payment: 4% on HSR-related debt (all HSR investments are assumed to be financed by equal portions of debt and equity).
Maintenance Cost: Based on WB’s higher-end estimate of HSR maintenance cost of 2.3 million yuan/km per year.
Rolling Stock: Based on annual spending on rolling stock, or the actual trains, before 2019 and based on an imputed annual usage cost after 2019. Usage cost is calculated by multiplying annual
traffic with usage cost per train car (each train car is assumed to have an occupancy of 60 people). The per km train car usage cost is 7.5 yuan.
Operating Cost: Based on the estimated operating cost of the Beijing-Shanghai line, the only route with detailed operating cost data, minus the rolling stock cost. Benefits
Air Travel Substitution: Estimated based on expected annual savings from the 20% passenger traffic shift to HSR. On a per passenger-km basis, travel by HSR is about 50% cheaper than by plane, which
implies that transporting the same number of passengers by HSR requires only half the upfront capital investment relative to air travel.
Lower Investment for Conventional Rail: With the shift to HSR, China can accommodate growing travel demand without investing as much in conventional rail. Since conventional rail costs less than HSR
to build, deferred conventional rail construction is assumed to be a quarter of the HSR construction cost.
Time Savings: Estimated based on the 20% passenger traffic shift from airplane to HSR, multiplied by the average hourly wage of an urban worker.
For distances no greater than 800 km, or roughly three hours of travel by HSR, it saves time to take HSR over an airplane. For instance, the “first mile” and “last mile” of air travel take much
longer than HSR because of security checks at the airport and the fact that most airports are farther away from final destinations in urban centers.
Operating Cost Savings: Estimated based on the average operating costs of various modes of transportation and the passenger traffic shift assumptions specified above (e.g. HSR’s operating cost is
higher than conventional rail but much lower than airplanes).
Generated Traffic Benefits: Estimated based on time spent traveling on HSR, multiplied by the average hourly wage of an urban worker.
In theory, the value of the trip itself must be greater than the value of time spent traveling, or else one would not bother wasting their valuable time. But to stick with our more conservative
assumptions, in this case we assume the generated traffic benefit is simply equal to the value of time spent traveling. If agglomeration effects are included, the benefit would be much greater. But
given the lack of consensus on the magnitude of agglomeration, we did not account for these positive externalities in our estimate.
• Project leads ... Damien Ma, Houze Song
• Development ... Chris Roche
• Creative leads ... Annie Cantara, Yna Mataya
• Research support ... Ya-han Cheng | {"url":"https://www.truevaluemetrics.org/DBadmin/DBtxt003.php?vv1=txt00020340","timestamp":"2024-11-01T18:53:34Z","content_type":"text/html","content_length":"15038","record_id":"<urn:uuid:882ea009-7680-4562-a20e-a38c9c8262df>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00777.warc.gz"} |
Live Online classes for kids from 1-10 | Upfunda Academy
Prime Numbers In Maths - 2023
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Did you know that prime numbers in maths are unique and special numbers?
Prime numbers in maths are whole numbers greater than one that can only be divided evenly by one and themselves. They are whole numbers greater than one that can only be divided evenly by one and
themselves. Unlike other numbers, prime numbers have no divisors apart from one and the number itself.
In this blog, we will learn more about prime numbers in maths, exploring their properties, applications, and significance in various fields of Mathematics.
What are prime numbers in maths ?
Some prime numbers in maths are 2, 3, 5, 7, and 11 and so on.
Their properties make them a very interesting subjects of study and exploration.
Twin Prime Numbers
Twin prime numbers are pairs of special numbers that are very close to each other and only have one even number between them, like 3 and 5, where there are no other prime numbers (numbers divisible
only by 1 and themselves) in between.
Co-Prime Numbers
Co-prime numbers are two numbers that don't have any common factors except for 1, which means they cannot be divided evenly by the same number other than 1. For example, 9 and 16 are co-prime because
their only common factor is 1.
Prime Number examples
There is a simple prime number formula in mathsĀ to check if a number is prime or not, you can follow the steps outlined below:
• Start with the number you want to check.
• If the number is less than 2, it is not prime.
• Check for divisibility by numbers from 2 to the square root of the given number.
• If no divisors are found up to the square root of the number, it is prime.
Here areĀ prime number formula in maths in 3 simple steps
Let's check if the number 17 is prime:
• Step - 1: Start with 17. It is greater than 1, so we proceed.
• Step - 2: Check for divisibility by numbers from 2 to the square root of 17. In this case, the square root of 17 is approximately 4.12, so we check for divisibility by 2, 3, and 4.
• Step - 3:
• No divisors are found up to the square root of 17, so we conclude that 17 is a prime number.
The Smallest Prime Number:
The smallest prime number is 2. It is the only even prime number and holds a unique place in the world of primes. Unlike other even numbers, which are divisible by 2, prime number 2 defies the common
pattern and stands alone as the only even prime.
List of Prime Numbers in maths:
Creating an exhaustive list of prime numbers is an impossible task due to their infinite nature. However, we can identify prime numbers within a given range. Here are some prime numbers up to 100:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Largest Prime Number Discovered:
The largest known prime number is called a Mersenne prime. Mersenne primes are prime numbers that can be written in the form 2^p - 1, where p is also a prime number. The current record-holder is 2^
82,589,933 - 1, a number with a staggering 24,862,048 digits. This discovery was made possible by distributed computing projects involving thousands of computers working collectively.
In conclusion, prime numbers are like special treasures in math. They have unique properties and go on forever. Mathematicians love them! They're important for many things, like keeping secrets safe
and solving problems. Prime numbers are all around us, and they make math even more exciting!
Prime Numbers Questions:
1. Which is the largest 2-digit prime number?
1. Which of the following are prime numbers?
1. Which is the smallest 3 digit prime numbers?
2. Fill in the boxes to complete the prime factorization of 99.
3. Select the correct the prime factorisation of 340 | {"url":"https://upfunda.academy/blog/ad9841dd-ee00-4ae4-881d-d5eda6bfc88a","timestamp":"2024-11-09T00:14:06Z","content_type":"text/html","content_length":"29800","record_id":"<urn:uuid:27b0284a-f68a-46e8-8586-93adb16a0b98>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00256.warc.gz"} |
Binary Tree in Python - Red And Green
Let’s create a Binary Tree in Python and understand how it works.
A binary tree is a hierarchical data structure composed of nodes, where each node has at most two children, referred to as the left child and the right child. The structure resembles a tree with a
single root node and branches extending downward, forming a hierarchical relationship between nodes.
Key characteristics of a binary tree include:
1. Root Node: The topmost node of the tree, serving as the starting point for traversing the tree.
2. Parent and Child Nodes: Each node in a binary tree (except the root) has a parent node and may have up to two children nodes. The parent node is the immediate node above the child node.
3. Left and Right Children: Each node can have at most two children, referred to as the left child and the right child. These children nodes are positioned to the left and right of the parent node,
4. Leaf Nodes: Nodes that do not have any children are called leaf nodes or terminal nodes. They are the nodes at the bottom-most level of the tree.
5. Internal Nodes: Nodes that have at least one child are called internal nodes. They are not leaf nodes and are located somewhere between the root node and the leaf nodes.
6. Depth and Height: The depth of a node is the length of the path from the root node to that node. The height of a node is the length of the longest path from that node to a leaf node. The height
of the binary tree is the height of the root node.
7. Binary Search Property: In a binary search tree (a specific type of binary tree), the values stored in the left subtree of a node are less than the value of the node, and the values stored in the
right subtree are greater than the value of the node. This property allows for efficient searching, insertion, and deletion operations.
Binary trees are used in various applications such as representing hierarchical data structures, implementing search algorithms like binary search, and serving as the foundation for more complex data
structures like binary search trees, AVL trees, and red-black trees.
Code for a Binary Tree
class Node:
A node class representing a single element in the binary tree.
def __init__(self, data):
self.data = data
self.left = None # Left child
self.right = None # Right child
def insert(root, data):
Inserts a new node with the given data into the binary tree.
if root is None:
return Node(data)
if data < root.data:
root.left = insert(root.left, data)
root.right = insert(root.right, data)
return root
def inorder_traversal(root):
Performs an in-order traversal of the binary tree, printing data.
if root is not None:
print(root.data, end=" ")
# Example usage
root = None
root = insert(root, 50)
insert(root, 30)
insert(root, 20)
insert(root, 40)
insert(root, 70)
insert(root, 60)
print("Inorder traversal: ")
/ \
/ \
/ \
/ \
/ \ /
/ \
/ \ / | {"url":"https://redandgreen.co.uk/binary-tree-in-python/python-code/","timestamp":"2024-11-07T01:23:55Z","content_type":"text/html","content_length":"104708","record_id":"<urn:uuid:a22ca0bc-2f6e-4b2f-9b40-738b07dd6a76>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00487.warc.gz"} |
Multiplication (of equal groups) | sofatutor.com
Multiplication (of equal groups)
Multiplication (of equal groups)
Basics on the topic Multiplication (of equal groups)
Equal Groups Multiplication
Imagine the following: You want to bake a cake for your mother’s birthday and you are collecting the necessary ingredients. But, the quantities of all the ingredients are labeled using multiplication
expressions on the packages, like 4 x 6 or 2 x 8. How can you determine the contents of the ingredients so you know how many ingredients are actually there? In this text you can learn how to model
multiplication using equal groups for multiplication to determine amounts.
Equal Groups Multiplication – Definition
Multiplication is an elementary math concept. Let’s learn more about multiplication as equal groups and how to use multiplication strategies with equal groups to solve equal groups multiplication
examples. Multiplication always involves joining equal groups. It is a way of finding the total number of items in equal-sized groups.What Are Equal Groups in Multiplication? When we see a
multiplication expression, like two times eight, think of 'times' as meaning 'equal groups of'. So we can also say this as two equal groups of eight. This tells us there are two equal groups with
eight mushrooms in each!
Multiplying Using Equal Groups
This is how you multiply equal groups. The first factor, or number, tells us how many groups there are.
The second factors or number, tells us how many mushrooms are in each group.
When we model this, we can easily count all sixteen mushrooms to calculate the product, or answer. Thus equal groups and multiplication go hand in hand.
Equal Groups Multiplication – Examples
Let’s practice teaching multiplication using equal groups with the example three times five. Using the idea of equal groups, we can say this is three equal groups of five. The first factor, three,
tells us to draw equal groups for multiplication. We'll draw three large boxes to represent our groups.
Then, the second factor, five, tells us how many dots to make in each group. We'll make five dots in each box.
Now we can solve this by counting up all the dots. There are fifteen dots in all, so the product of three times five is fifteen.
Multiplication Equal Groups – Summary of Steps
Remember, multiplication is a way of finding the total number of items in equal-sized groups. Using this idea of equal groups, we can restate any multiplication expression with the words 'equal
groups of'.This helps us figure out what to draw to find the answer:
Step # What to do
1 The first factor tells us how
many groups, or boxes to draw.
2 The second factor tells us
how many dots to draw in each.
3 Solve by counting up all the dots.
4 Write the product.
Have you practiced with equal groups multiplication worksheets yet? On this website, you can also find multiplication with equal groups worksheets and further equal groups multiplication activities
for third grade, such as interactive exercises for further practice.
Transcript Multiplication (of equal groups)
Chef Squeaks and Imani are doing inventory for their cooking show, “Stuff Your Cheeks With Mr. Squeaks”. They are checking to see how many ingredients are available. All the ingredients are labeled
with a multiplication expression, one number times another number, that represents how much is in the container. Let's help Chef Squeaks and Imani calculate how many ingredients there are using
"Multiplication (of equal groups)". Multiplication is a way of finding the total number of items in equal-sized groups. When we see a multiplication expression, like two times eight, think of 'times'
as meaning 'equal groups of'. So we can also say this as two equal groups of eight. This tells us there are two equal groups with eight mushrooms in each! The first factor, or number, tells us how
many groups there are. The second factor, or number, tells us how many mushrooms are in each group. When we model this, we can easily count all sixteen mushrooms to calculate the product, or answer.
For example, the jar of olives has the expression: three times five. Using the idea of equal groups, we can say this is three equal groups of five. The first factor, three, tells us to make three
groups. We'll draw three large boxes to represent our groups. Then, the second factor, FIVE, tells us how many dots to make in each group. We'll make five dots in each box. Now we can solve by
counting up all the dots. There are fifteen dots in all, so the product of three times five is fifteen. There are fifteen olives in the jar! Let's try it again with the can of tomatoes. The label
says four times six. Using this idea of equal groups, we can say this is four equal groups of six. The first factor, four, tells us to make four groups, so how many boxes should we draw? We should
draw four large boxes. Then, the second factor, six, tells us how many dots to make in each group, so how many dots should we draw in each box? We should draw six dots in each box. How do we
calculate the product? We count up all the dots! There are twenty-four dots in all, so the product of four times six is twenty-four. There are twenty-four tomatoes in the can. Before we see if Chef
Squeaks and Imani got what they needed, let's remember! Multiplication is a way of finding the total number of items in equal-sized groups. Using this idea of equal groups, we can restate any
multiplication expression with the words 'equal groups of'. This helps us figure out what to draw to find the answer. The first factor tells us how many groups, or boxes, to draw. Then, the second
factor tells us how many dots to draw in each. Solve by counting up all the dots. Finally, write the product. "Well it turns out there are enough ingredients for the next taping!" "Let's get these
down to the kitchen...." "And I'll just leave this right here."
More videos in this topic Single-Digit Multiplication | {"url":"https://us.sofatutor.com/math/videos/multiplication-of-equal-groups","timestamp":"2024-11-11T06:52:49Z","content_type":"text/html","content_length":"142500","record_id":"<urn:uuid:db966e1c-4c03-4782-ad73-10fc444c4d51>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00833.warc.gz"} |
Finding all Minimum Arborescences
There is only one thing that I need to figure out before the first coding period for GSoC starts on Monday: how to find all of the minimum arborescences of a graph. This is the set $K(\pi)$ in the
Held and Karp paper from 1970 which can be refined down to $K(\pi, d)$ or $K_{X, Y}(\pi)$ as needed. For more information as to why I need to do this, please see my last post here.
This is a place where my contributions to NetworkX to implement the Asadpour algorithm [1] for the directed traveling salesman problem will be useful to the rest of the NetworkX community (I hope).
The research paper that I am going to template this off of is this 2005 paper by Sörensen and Janssens titled An Algorithm to Generate all Spanning Trees of a Graph in Order of Increasing Cost [4].
The basic idea here is to implement their algorithm and then generate spanning trees until we find the first one with a cost that is greater than the first one generated, which we know is a minimum,
so that we have found all of the minimum spanning trees. I know what you guys are saying, “Matt, this paper discusses spanning trees, not spanning arborescences, how is this helpful?”. Well, the
heart of this algorithm is to partition the vertices into either excluded edges which cannot appear in the tree, included edges which must appear in the tree and open edges which can be but are not
required to be in the tree. Once we have a partition, we need to be able to find a minimum spanning tree or minimum spanning arborescence that respects the partitioned edges.
In NetworkX, the minimum spanning arborescences are generated using Chu-Liu/Edmonds’ Algorithm developed by Yoeng-Jin Chu and Tseng-Hong Liu in 1965 and independently by Jack Edmonds in 1967. I
believe that Edmonds’ Algorithm [2] can be modified to require an arc to be either included or excluded from the resulting spanning arborescence, thus allowing me to implement Sörensen and Janssens’
algorithm for directed graphs.
First, let’s explore whether the partition scheme discussed in the Sörensen and Janssens paper [4] will work for a directed graph. The critical ideas for creating the partitions are given on pages
221 and 222 and are as follows:
Given an MST of a partition, this partition can be split into a set of resulting partitions in such a way that the following statements hold:
□ the intersection of any two resulting partitions is the empty set,
□ the MST of the original partition is not an element of any of the resulting partitions,
□ the union of the resulting partitions is equal to the original partition, minus the MST of the original partition.
In order to achieve these conditions, they define the generation of the partitions using this definition for a minimum spanning tree
$$ s(P) = {(i_1, j_1), \dots, (i_r, j_r), (t_1, v_1), \dots, (t_{n-r-1}, v_{n-r-1}} $$
where the $(i, j)$ edges are the included edges of the original partition and the $(t, v)$ are from the open edges of the original partition. Now, to create the next set of partitions, take each of
the $(t, v)$ edges sequentially and introduce them one at a time, make that edge an excluded edge in the first partition it appears in and an included edge in all subsequent partitions. This will
produce something to the effects of
$$ \begin{array}{l} P_1 = {(i_1, j_1), \dots, (i_r, j_r), (\overline{m_1, p_1}), \dots, (\overline{m_l, p_l}), (\overline{t_1, v_1})} \\\ P_2 = {(i_1, j_1), \dots, (i_r, j_r), (t_1, v_1), (\overline
{m_1, p_1}), \dots, (\overline{m_l, p_l}), (\overline{t_2, v_2})} \\\ P_3 = {(i_1, j_1), \dots, (i_r, j_r), (t_1, v_1), (t_2, v_2), (\overline{m_1, p_1}), \dots, (\overline{m_l, p_l}), (\overline
{t_3, v_3})} \\\ \vdots \\\ \begin{multline*} P_{n-r-1} = {(i_1, j_1), \dots, (i_r, j_r), (t_1, v_1), \dots, (t_{n-r-2}, v_{n-r-2}), (\overline{m_1, p_1}), \dots, (\overline{m_l, p_l}), \\\ (\
overline{t_{n-r-1}, v_{n-r-1}})} \end{multline*} \\\ \end{array} $$
Now, if we extend this to a directed graph, our included and excluded edges become included and excluded arcs, but the definition of the spanning arborescence of a partition does not change. Let $s_a
(P)$ be the minimum spanning arborescence of a partition $P$. Then
$$ s_a(P) = {(i_1, j_1), \dots, (i_r, j_r), (t_1, v_1), \dots, (t_{n-r-1}, v_{n-r-1}} $$
$s_a(P)$ is still constructed of all of the included arcs of the partition and a subset of the open arcs of that partition. If we partition in the same manner as the Sörensen and Janssens paper [4],
then their cannot be spanning trees which both include and exclude a given edge and this conflict exists for every combination of partitions.
Clearly the original arborescence, which includes all of the $(t_1, v_1), \dots, (t_{n-r-1}, v_{n-r-1})$ cannot be an element of any of the resulting partitions.
Finally, there is the claim that the union of the resulting partitions is the original partition minus the original minimum spanning tree. Being honest here, this claim took a while for me to
understand. In fact, I had a whole paragraph talking about how this claim doesn’t make sense before all of a sudden I realized that it does. The important thing to remember here is that the union of
all of the partitions isn’t the union of the sets of included and excluded edges (which is where I went wrong the first time), it is a subset of spanning trees. The original partition contains many
spanning trees, one or more of which are minimum, but each tree in the partition is a unique subset of the edges of the original graph. Now, because each of the resulting partitions cannot include
one of the edges of the original partition’s minimum spanning tree we know that the original minimum spanning tree is not an element of the union of the resulting partitions. However, because every
other spanning tree in the original partition which was not the selected minimum one is different by at least one edge it is a member of at least one of the resulting partitions, specifically the one
where that one edge of the selected minimum spanning tree which it does not contain is the excluded edge.
So now we know that this same partition scheme which works for undirected graphs will work for directed ones. We need to modify Edmonds’ algorithm to mandate that certain arcs be included and others
excluded. To start, a review of this algorithm is in order. The original description of the algorithm is given on pages 234 and 235 of Jack Edmonds’ 1967 paper Optimum Branchings [2] and roughly
speaking it has three major steps.
1. For each vertex $v$, find the incoming arc with the smallest weight and place that arc in a bucket $E^i$ and the vertex in a bucket $D^i$. Repeat this step until either (a) $E^i$ no longer
qualifies as a branching or (b) all vertices of the graph are in $D^i$. If (a) occurs, go to step 2, otherwise go to step 3.
2. If $E^i$ no longer qualifies as a branching then it must contain a cycle. Contract all of the vertices of the cycle into one new one, say $v_1^{i + 1}$. Every edge which has one endpoint in the
cycle has that endpoint replaced with $v_1^{i + 1}$ and its cost updated. Using this new graph $G^{i + 1}$, create buckets $D^{i + 1}$ containing the nodes in both $G^{i + 1}$ and $D^i$ and $E^{i
+ 1}$ containing edges in both $G^{i + 1}$ and $E^i$ (i.e. remove the edges and vertices which are affected by the creation of $G^{i + 1}$.) Return to step 1 and apply it to graph $G^{i + 1}$.
3. Once this step is reached, we have a smaller graph for which we have found a minimum spanning arborescence. Now we need to un-contract all of the cycles to return to the original graph. To do
this, if the node $v_1^{i + 1}$ is the root of the arborescence or not.
□ $v_1^{i + 1}$ is the root: Remove the arc of maximum weight from the cycle represented by $v_1^{i + 1}$.
□ $v_1^{i + 1}$ is not the root: There is a single arc directed towards $v_1^{i + 1}$ which translates into an arc directed to one of the vertices in the cycle represented by $v_1^{i + 1}$.
Because $v_1^{i + 1}$ represents a cycle, there is another arc wholly internal to the cycle which is directed into the same vertex as the incoming edge to the cycle. Delete the internal one
to break the cycle. Repeat until the original graph has been restored.
Now that we are familiar with the minimum arborescence algorithm, we can discuss modifying it to force it to include certain edges or reject others. The changes will be primarily located in step 1.
Under the normal operation of the algorithm, the consideration which happens at each vertex might look like this.
Where the bolded arrow is chosen by the algorithm as it is the incoming arc with minimum weight. Now, if we were required to include a different edge, say the weight 6 arc, we would want this
behavior even though it is strictly speaking not optimal. In a similar case, if the arc of weight 2 was excluded we would also want to pick the arc of weight 6. Below the excluded arc is a dashed
But realistically, these are routine cases that would not be difficult to implement. A more interesting case would be if all of the arcs were excluded or if more than one are included.
Under this case, there is no spanning arborescence for the partition because the graph is not connected. The Sörensen and Janssens paper characterize these as empty partitions and they are ignored.
In this case, things start to get a bit tricky. With two (or more) included arcs leading to this vertex, it is but definition not an arborescence as according to Edmonds on page 233
A branching is a forest whose edges are directed so that each is directed toward a different node. An arborescence is a connected branching.
At first I thought that there was a case where because this case could result in the creation of a cycle that it was valid, but I realize now that in step 3 of Edmonds’ algorithm that one of those
arcs would be removed anyways. Thus, any partition with multiple included arcs leading to a single vertex is empty by definition. While there are ways in which the algorithm can handle the inclusion
of multiple arcs, one (or more) of them by definition of an arborescence will be deleted by the end of the algorithm.
I propose that these partitions are screened out before we hand off to Edmonds’ algorithm to find the arborescences. As such, Edmonds’ algorithm will need to be modified for the cases of at most one
included edge per vertex and any number of excluded edges per vertex. The critical part of altering Edmonds’ Algorithm is contained within the desired_edge function in the NetworkX implementation
starting on line 391 in algorithms.tree.branchings. The whole function is as follows.
def desired_edge(v):
Find the edge directed toward v with maximal weight.
edge = None
weight = -INF
for u, _, key, data in G.in_edges(v, data=True, keys=True):
new_weight = data[attr]
if new_weight > weight:
weight = new_weight
edge = (u, v, key, new_weight)
return edge, weight
The function would be changed to automatically return an included arc and then skip considering any excluded arcs. Because this is an inner function, we can access parameters passed to the parent
function such as something along the lines as partition=None where the value of partition is the edge attribute detailing true if the arc is included and false if it is excluded. Open edges would not
need this attribute or could use None. The creation of an enum is also possible which would unify the language if I talk to my GSoC mentors about how it would fit into the NetworkX ecosystem. A
revised version of desired_edge using the true and false scheme would then look like this:
def desired_edge(v):
Find the edge directed toward v with maximal weight.
edge = None
weight = -INF
for u, _, key, data in G.in_edges(v, data=True, keys=True):
new_weight = data[attr]
if data[partition]:
return edge, data[attr]
if new_weight > weight and not data[partition]:
weight = new_weight
edge = (u, v, key, new_weight)
return edge, weight
And a version using the enum might look like
def desired_edge(v):
Find the edge directed toward v with maximal weight.
edge = None
weight = -INF
for u, _, key, data in G.in_edges(v, data=True, keys=True):
new_weight = data[attr]
if data[partition] is Partition.INCLUDED:
return edge, data[attr]
if new_weight > weight and data[partition] is not Partition.EXCLUDED:
weight = new_weight
edge = (u, v, key, new_weight)
return edge, weight
Once Edmonds’ algorithm has been modified to be able to use partitions, the pseudocode from the Sörensen and Janssens paper would be applicable.
Input: Graph G(V, E) and weight function w
Output: Output_File (all spanning trees of G, sorted in order of increasing cost)
List = {A}
while MST ≠ ∅ do
Get partition Ps in List that contains the smallest spanning tree
Write MST of Ps to Output_File
Remove Ps from List
And the corresponding Partition function being
P1 = P2 = P
for each edge i in P do
if i not included in P and not excluded from P then
make i excluded from P1
make i include in P2
if Connected(P1) then
add P1 to List
P1 = P2
I would need to change the format of the first code block as I would like it to be a Python iterator so that a for loop would be able to iterate through all of the spanning arborescences and then
stop once the cost increases in order to limit it to only minimum spanning arborescences.
[1] A. Asadpour, M. X. Goemans, A. Mardry, S. O. Ghran, and A. Saberi, An o(log n / log log n)-approximation algorithm for the asymmetric traveling salesman problem, Operations Research, 65 (2017),
p. 1043-1061, https://homes.cs.washington.edu/~shayan/atsp.pdf.
[2] J. Edmonds, Optimum Branchings, Journal of Research of the National Bureau of Standards, 1967, Vol. 71B, p.233-240, https://archive.org/details/jresv71Bn4p233
[3] M. Held, R.M. Karp, The traveling-salesman problem and minimum spanning trees, Operations research, 1970-11-01, Vol.18 (6), p.1138-1162, https://www.jstor.org/stable/169411
[4] G.K. Janssens, K. Sörensen, An algorithm to generate all spanning trees in order of increasing cost, Pesquisa Operacional, 2005-08, Vol. 25 (2), p. 219-229, https://www.scielo.br/j/pope/a/ | {"url":"https://blog.scientific-python.org/networkx/atsp/finding-all-minimum-arborescences/","timestamp":"2024-11-02T18:08:17Z","content_type":"text/html","content_length":"42827","record_id":"<urn:uuid:aa5e4e10-9c7f-4c46-b2e1-a6b02bb4ce37>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00542.warc.gz"} |
Photometric Stereo-Based 3D Reconstruction Method for the Objective Evaluation of Fabric Pilling
Issue Wuhan Univ. J. Nat. Sci.
Volume 27, Number 6, December 2022
Page(s) 550 - 556
DOI https://doi.org/10.1051/wujns/2022276550
Published online 10 January 2023
Wuhan University Journal of Natural Sciences, 2022, Vol. 27 No.6, 550-556
CLC number: TP 399
Photometric Stereo-Based 3D Reconstruction Method for the Objective Evaluation of Fabric Pilling
School of Fashion Engineering, Shanghai University of Engineering Science, Shanghai 201620, China
^† To whom correspondence should be addressed. E-mail: xinbj@sues.edu.cn
Received: 22 September 2022
Fabric pilling evaluation has been considered as an essential element for textile quality inspection. Traditional manual method is still based on human eyes and brain, which is subjective with low
efficiency. This paper proposes an objective evaluation method based on semi-calibrated near-light Photometric Stereo (PS). Fabric images are digitalized by self-developed image acquisition system.
The 3D depth information of each point could be obtained by PS algorithm and then mapped to 2D grayscale image. After that, the non-textured image could be filtered by using the Gaussian low-pass
filter. The pilling segmentation is conducted by using global iterative threshold segmentation method, and then K-Nearest Neighbor (KNN) is finally selected as a tool for the grade classification of
fabric pilling. Our experimental results show that the proposed evaluation system could achieve excellent judging performance for the objective pilling evaluation.
Key words: photometric stereo / pilling evaluation / 3D reconstruction / image analysis / fast Fourier Transform
Biography: LUO Jian, male, Master candidate, research direction: photometric stereo-based fabric 3D reconstruction algorithm and application. E-mail: 512987427@qq.com
Supported by the National Natural Science Foundation of China (61876106)
© Wuhan University 2022
This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( https://creativecommons.org/licenses/by/4.0</ext-link>), which permits unrestricted use,
distribution, and reproduction in any medium, provided the original work is properly cited.
DOI https://doi.org/10.1051/wujns/2022276550
0 Introduction
Pilling of textile is a very unpleasant feature that results from daily washing and wearing, which is affected by fiber properties, yarn properties, and fabric structure. For the quality control of
fabric products, pilling evaluation has been always considered as an important issue. Traditional evaluation methods have some disadvantages of high cost, subjectivity, poor reliability and low
efficiency. It is necessary to develop the objective and digital technology to replace the traditional methods.
For this purpose, researchers used some objective evaluation methods based on 2D image analysis techniques in fabric pilling evaluation such as the Fourier Transform, Wavelet Transform, and
Artificial Neural Networks. Yun et al^[1] used the Fourier Transform algorithm which divided the image information into low and high frequencies where low frequencies included the deterministic
structure and high frequencies represented the noise and pills. Their experimental results showed that the method was suitable for pilling evaluation of woven fabrics. Deng et al^[2] used multi-scale
2D Dual-Tree Complex Wavelet Transform (CWT) to extract six characteristics at different scales from images of textile, indicating this evaluation system had excellent performance for knitted, woven,
and nonwoven fabrics. Xiao et al^[3] transformed the pilling image to the frequency domain using Fourier Transform, and combined it with energy algorithm, multi-dimensional Discrete Wavelet Transform
and iterative thresholding algorithm to obtain pilling segmentation images. This objective evaluation method was capable of obtaining full and accurate pilling information and deep learning
algorithms achieved 94.2% classification accuracy. Wu et al^[4] proposed a Convolutional Neural Network-based pilling evaluation system by extracting pill features and texture features. The rating
accuracy of their model reached 97.70%.
However, the analysis of pilling based on 2D images has limitations due to lighting as well as pattern variations. For more accuracy, more and more methods based on 3D information are used in fabric
pilling evaluation. Kang et al^[5] developed a noncontact 3D measurement method for reconstructing the 3D model of the fabric. A CCD camera was used to capture the image of the laser line projected
on its surface. Using a height-threshold algorithm, the 3D model was converted into a binary image, and the parameters extracted from that image were used to calculate the pilling grade. The results
of their method correlated well with the manual evaluation method. Xu et al^[6] investigated a 3D fabric surface reconstruction system that used two side-by-side images of fabric particles taken by a
pair of ordinary cameras without special illumination. To make the system resistant to fabric structures, colors, fiber contents, and other factors, robust calibration and stereo-matching algorithms
were implemented. Liu et al^[7] proposed a method based on structure from motion (SFM) and patch-based multi-view stereo (PMVS) algorithm for pilling evaluation. The pilling segmentation was achieved
by adaptive threshold segmentation and morphological analysis.
Multi-view stereo can only reconstruct the macroscopic contours of the fabric surface without significant texture details. Laser triangulation can get texture details, but this method is usually
high-cost, time-consuming and difficult in operation. In most other methods whose 3D reconstruction is based on surface features, such as stereo matching, the 3D model cannot be generated when the
surface features are blurry.
This paper designs a simple and low-cost system for pilling evaluation which can not only recover the macroscopic contour of the fabric, but also reconstruct its tissue points. This system first
reconstructs a 3D model of the fabric surface using semi-calibrated near-light Photometric Stereo (PS). In mapping the 3D model to a 2D image, a low-pass filter is used to eliminate fabric texture
for the 2D depth image. The binary image of pills is segmented by a global iterative threshold to obtain the pilling number and area. Finally, the classification is completed by K-Nearest Neighbor
(KNN). Figure 1 shows the flowchart of fabric pilling evaluation.
Fig. 1 Pilling grade evaluation system
1 System Setup
We have designed a computer image acquisition system, which consists of a data acquisition facility and image data analysis facility, as shown in Fig. 2. Imaging hardware used in the data acquisition
facility includes a box, a high-resolution digital camera (NIKKOR D7200), a macro lens (NIKKOR AF-S) and eight LED light sources (1 W). The PS is sensitive to the light information in the image, and
the inside walls of the box are painted with black matt varnish to minimize the reflection of the scattered light and the effect of stray light. To calibrate parallel light, the light sources were
concentrated with a range of 15 luminous angles. The acquired images are analyzed and reconstructed using MATLAB R2020a.
Fig. 2 Self developed PS system
2 Semi-Calibrated Near-Light PS
PS was first proposed by Woodham^[8] in 1980, which takes multiple images with different illumination using a specific LED configuration and then infers the 3D model. The traditional PS method
assumes that lighting is caused by infinitely distant point sources, and the intensity of the light source needs to be calibrated. In this paper, a near-light model is developed, and the source
intensities are assumed to be unknown. Modeling near-light sources means that low-cost lighting devices such as LEDs can be used, and the calibration procedure is simplified by assuming unknown
source intensities (semi-calibrated). This paper uses semi-calibrated near-light PS technology^[9] to complete the 3D reconstruction of fabric surfaces.
For 3D reconstruction of fabric surfaces by using semi-calibrated near-light PS, we designed the multi-light image acquisition system (Fig. 2) and conducted the calibration to obtain camera intrinsic
matrix and the light position parameters which are crucial for 3D reconstruction. The process of calibration is described in Section 2.1. The image acquisition is conducted after system calibration,
and eight light sources are controlled to irradiate and shoot the samples in turn. A variational formulation is established, which describes the relationship between the grayscale values of pixel
corresponding to the surface point of 3D model, and tackle the nonconvex variational model numerically to complete the 3D reconstruction.
2.1 System Calibration
According to the method proposed by Zhang^[10], taking a set of calibration board images, the camera is calibrated with the Matlab Camera Calibration Toolbox to obtain the camera intrinsic matrix.
Figure 3 shows one of the corner point detection results.
Fig. 3 The results of corner point detection
Triangulation is used to calibrate the location parameters of the light source^[11]. A pair of metal spheres are placed in the scene to generate visible highlights for each light source, and multiple
highlight points are extracted by thresholding. Then the Canny operator^[12] is used to detect the sphere contours, as shown in Fig.4. The light source position parameters are calculated from the
actual radius of the sphere, the focal length of the camera, and the physical pixel size of the camera sensor.
Fig. 4 Test the center and radius of the metal ball
2.2 Photometric Model
The relationship between the surface point $s$ of the 3D model and the 2D pixel point $i=(x,y)$ can be expressed as follows:
$s ( i ) = z ( i ) K - 1 [ x , y , 1 ] T$(1)
where $z(x,y)$ is the depth value of the 3D model and $K$ is the intrinsic matrix of the camera.
Considering non-parallel illumination features, the attenuation caused by distance $Flk$ can be expressed as follows:
$F l k = 1 ‖ s - x l k ‖ 2$(2)
where $k$ represents the k-th light source,$l$represents the irradiated light vector of surface point $s$, and $xlk$ is the light source position. Thus, the incident light vector $lk(s)$ at the
surface point $s$ can be expressed as:
$l k ( s ) = 1 ‖ s - x l k ‖ 2 [ x l k - s ] ‖ s - x l k ‖$(3)
The normal vector $n(s)$ is the unit-length vector proportional to $∂xs(x,y)×∂ys(x,y)$:
$n ( s ) = J ( i ) T [ ∇ l o g z ¨ ( i ) - 1 ] d ( i ; ∇ l o g z ¨ ( i ) )$(4)
$∇ l o g z ¨ ( i ) = z ( i )$(5)
$J ( i ) = [ f d X - f c o t β d X - ( x - u 0 ) 0 f d Y s i n β - ( y - v 0 ) 0 0 1 ]$(6)
$d ( i ; ∇ l o g z ¨ ( i ) ) = ‖ J ( i ) T [ ∇ l o g z ¨ ( i ) - 1 ] ‖$(7)
On the camera sensor, $f$ is the focal length, $dX$ and $dY$ are the physical lengths of pixels in the X and Y directions. In the pixel coordinate system,$u0$ and$v0$ indicate the sensor center
coordinates, $β$represents the angle between the horizontal and vertical edges of the photographic plate.
The albedo$ρ(s)$ expression was modified following the method of Quéau et al^[13]:
$ρ ( s ) = ρ ¨ ( i ) d ( i ; ∇ l o g z ¨ ( i ) )$(8)
The relationship between the grayscale values of pixel $pk(i)$ corresponding to the surface point $s$ can be expressed as:
$p k ( i ) = ρ ( s ) e k m a x { l k ( s ) · n ( s ) , 0 } , k ∈ [ 1 , M ]$(9)
where $ek$ represents the light source intensity of the $k$-th light source.
Combining the above equations into a system of nonlinear partial differential equations:
$p k ( i ) = ρ ¨ ( i ) e k m a x { [ J ( i ) l k ( i ; l o g z ( i ) ) ] T [ ∇ l o g z ( i ) - 1 ] , 0 } , k ∈ [ 1 , M ]$(10)
$j = 1 … N$ are the corresponding pixels. Then, the discrete counterpart of Eq. (10) is written as:
$p j k = ρ ¨ j e k m a x { [ J j l k ( l o g z j ) ] T [ ∇ l o g z j - 1 ] , 0 } , ∀ k ∈ [ 1 , M ] , ∀ j ∈ [ 1 , N ]$(11)
Optimize the discrete part assuming that Q consists of all rank-1 N-by-M matrices:
$m i n l o g z , θ : θ ∈ Q F ( θ , l o g z ) = ∑ j = 1 N ∑ k = 1 M λ 2 l o g ( 1 + ( ρ ¨ j e k m a x { [ J j l k ( l o g z j ) ] T [ ∇ l o g z j - 1 ] , 0 } - p j k ) 2 λ 2 )$(12)
where $λ$ is the user-defined parameter of the Cauchy estimator, and in our experiment $λ = 8$. The nonconvex model of Eq. (12) is minimized alternatively over variables $θ$ and $logz$. In each
subproblem, we solve a local quadratic model of Eq. (12) using the positive definite approximation of Hessian to achieve 3D reconstruction.
3 Experiment and Result Analysis
3.1 Preparation of Samples
In this study, 98 fabric samples were classified in pilling grades. Fabric samples were cut into squares of 30 mm×30 mm, resulting in an image of 512×512 pixels. The pilling severity is divided into
five grades. From Grade 1 to 5, the pilling degree of fabric decreases gradually until almost no pilling is seen in Grade 5^[14]. Five samples selected from the datasets and graded according to
American Society of Testing Materials (ASTM) standards were analyzed as standard samples as shown in Fig.5. The subjective evaluation results were taken based on the evaluation results of five
experts who compared each fabric with the standard samples. Those with the highest consistency were taken as the subjective evaluation results.
Fig. 5 Standard pilling images of Grade 1 to Grade 5
3.2 2D Depth Image Generation
In the feature extraction phase, the complexity of extracting 2D image features can be ignored compared with the 3D feature extraction which needs 3D point cloud processing. The 3D model can be
projected onto a 2D plane with a determined mapping relationship^[15]. The coordinates of the point $s=(x,y,z(i))$ in the depth image are converted to the pixel coordinates of the grayscale image.
Eq.(13) converts the depth value $z(i)$ into the grayscale value for the corresponding pixel.
$G ( i ) = 255 z ( i ) - z m H$(13)
where $G(i)$ is the grayscale value of the 2D depth image, $zm$ is the minimum depth value of 3D model, and $H$ is the range of the depth value.
3.3 Pilling Segmentation
To make the pilling segmentation easier in the 2D depth image, fast Fourier Transform (FFT) algorithm^[16] is used to filter the texture which is usually the patterns with a high degree of
periodicity. FFT converts the image to the different frequency domains for analysis by seperating the texture information and pilling information as the high-frequency component and the low-frequency
component, respectively. The low-frequency component is highlighted by a Gaussian low-pass filter to eliminate the textured background. For each non-textured image, a global threshold is determined
automatically using the adaptive iterative method which selects the average gray value of the whole image as the initial threshold and determines the optimal threshold by an iterative process. Figure
6 shows the processing results from sample image to binary image.
Fig. 6 Fabric pilling sample of Grade 1 to Grade 5
3.4 Pilling Evaluation
In the subjective evaluation method, the size, total area, and coverage of pills are the main factors that influence the expert evaluation^[2]. In this study, the number and area are selected as the
features for evaluating fabric pilling. In the binary image, the pilling number refers to the number of pill areas. The values of the pilling pixels are all 1, and the connected area is searched
according to the 8-connected objects. The first connected area encountered is labeled as 1, then the search is carried out successively. The pill area is the total number of pixels counted in the
detection area. It is worth noting that the total area of the pill is $Stotal$.
In this paper, the KNN classifier^[17] was used to classify pilling samples and perform a 2-fold cross-validation. The datasets are randomly divided into two subsets with equal numbers of the
samples. One subset is used as a training set, and the other as a test set. KNN classifies the test sample by comparing the distance or similarity between the training samples and the test sample,
and the K training samples closest to the test sample are found. K=3 is set in this experiment. The category with the highest frequency among the 3 points as the prediction category of the test
sample, which is the objective evaluation result of the test sample. The accuracy $PACC$ of the objective evaluation is determined by comparing the results of the subjective evaluation.
where $NM$ is the number of samples inconsistent between objective and subjective evaluation, and $NT$ is the total number of samples.
As shown in Table 1, a total of seven samples are misclassified, which gives the system a classification accuracy of 92.8%. As shown in Fig. 7, a scatter plot shows how feature parameters relate to
grade. The ordinate represents the pilling area, the abscissas represent the pilling number in samples, the point with different colors represent different pilling grades, and the misclassified
samples have also been marked with red. Figure 7 illustrates that KNN is effective in classifying samples with small intra-class spacing and large inter-class spacing.
Fig. 7 Results of different evaluation methods
Table 1
Result of objective evaluation
4 Conclusion
This paper proposed an effective way to objectively evaluate the pilling images based on PS. Self-developed image acquisition equipment is used to capture multiple images with different illumination,
and then the semi-calibrated near-light PS algorithm is used to reconstruct the fabric surface. The 3D model is then converted into a 2D depth image for texture filtering. The transformed non-texture
image is segmented into a binary image by the iterative threshold segmentation method, and the defined feature parameters of fabric pilling, including pilling number and area, were extracted.
Finally, the KNN classifier was used to identify the fabric samples. The experimental results show that the system is effective and reliable for pilling evaluation. This method performs well for
plain fabrics, but it is insufficient for patterned fabrics. The following work should focus on building non-Lambertian models.
All Tables
Table 1
Result of objective evaluation
All Figures
Fig. 1 Pilling grade evaluation system
In the text
Fig. 2 Self developed PS system
In the text
Fig. 3 The results of corner point detection
In the text
Fig. 4 Test the center and radius of the metal ball
In the text
Fig. 5 Standard pilling images of Grade 1 to Grade 5
In the text
Fig. 6 Fabric pilling sample of Grade 1 to Grade 5
In the text
Fig. 7 Results of different evaluation methods
In the text
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Estratégia Financeira
14.1 Equity Versus Debt Financing
14.1 Equity Versus Debt Financing
When corporations raise funds from outside investors, they must choose which type of security to issue.
The most common choices are financing through equity alone and financing through a combination of debt and equity.
The relative proportions of debt, equity, and other securities that a firm has outstanding constitute its capital structure.
A central question in this chapter:
If a firm decides for a specific capital structure, does it change the investment decisions?
Do capital structure decisions affect firm value?
For clarity: Throughout the chapter, we will assume the firm is a portfolio of debt and equity (and maybe other securities).
14.1 Equity Versus Debt Financing
Let’s start our discussion by analyzing again the value of a firm.
Imagine an economy with two future states; firm will receive cash flows as follows:
The cost of capital for this project is 15%. Each scenario has a 50% probability of occurrence. The NPV is:
\[NPV = -800 + \frac{\frac{1}{2} \times 1400 + \frac{1}{2} \times 900 }{1.15} = 200\]
\[PV(CF) = \frac{1150}{1.15} = 1000\]
So the firm value today is $1000. If the firm wants to finance the project all-equity, it could raise this amount selling equity. In this case, the firm is unlevered equity.
14.1 Equity Versus Debt Financing
The expected return on the unlevered equity is 15%.
Return +40% -10%
\[\frac{1}{2} \times 40\% + \frac{1}{2} \times -10\% = 15\%\]
14.1 Equity Versus Debt Financing
Let’s now analyse the case where the firm issues 500 of debt and 500 of equity. This firm is levered equity. The debt cost is 5%. Do the NPV and the firm value change?
Because the firm needs to pay debt holders no matter the state of the economy, we can say that Debt value in Date 1 is 525 (\(500 \times 1.05\)).
Because we know the firm value in each state of the economy, we can say that the difference between the firm value and the debt value is the equity value: \(V - D = E\), i.e. \(1400 - 525 = 875\) or
\(900 - 525 = 375\).
Then, we can write:
Debt (D) 500 525 525
Equity (E) ? 875 375
Firm (V) 1000 1400 900
14.1 Equity Versus Debt Financing
Wait, but if debt cost is only 5%, why the firm value still is 1.000? Shouldn’t the value of equity increase?
The answer is that the cost of equity increases for levered equity. Let’s see how.
D 500 525 525 5% 5%
E 500 875 375 75% -25%
V 1000 1400 900 40% -10%
Unlevered equity has returns of 40% or -10%. On average, 15%.
Debt has return of 5% no matter what.
Levered equity has returns of 75% (\(\frac{875}{500}-1\)) or -25% (\(\frac{375}{500}-1\)). On average, 25%.
14.1 Equity Versus Debt Financing
The takeaway:
Levered equity is riskier, so the cost of capital is higher.
Debt 5% - 5% = 0% 5% - 5% = 0%
Unlevered Equity 40% - (-10%) = 50% 15% - 5% = 10%
Levered Equity 75% - (-25%) = 100% 25% - 5% = 20%
Because the debt’s return bears no systematic risk, its risk premium is zero.
In this particular case, the levered equity has twice the systematic risk of the unlevered equity and, as a result, has twice the risk premium.
Modigliani and Miller argued that with perfect capital markets, the total value of a firm should not depend on its capital structure.
• They reasoned that the firm’s total cash flows still equal the cash flows of the project and, therefore, have the same present value.
14.1 Equity Versus Debt Financing
In summary
• In the case of perfect capital markets, if the firm is 100% equity financed, the equity holders will require a 15% expected return.
• If the firm is financed 50% with debt and 50% with equity, the debt holders will receive a return of 5%, while the levered equity holders will require an expected return of 25% (because of
increased risk).
• Leverage increases the risk of equity even when there is no risk that the firm will default.
□ Thus, while debt may be cheaper, its use raises the cost of capital for equity. Considering both sources of capital together, the firm’s average cost of capital with leverage is the same as
for the unlevered firm.
14.1 Equity Versus Debt Financing
Using the same values as before, suppose the firm borrows $700 when financing the project.
According to Modigliani and Miller, what should the value of the equity be? What is the expected return?
Because the value of the firm’s total cash flows is still 1000, if the firm borrows 700, its equity will be worth 300.
14.1 Equity Versus Debt Financing
The firm will owe 700 × 1.05 = 735 in one year to debt holders. Thus,
• if the economy is strong, equity holders will receive \(1400 − 735 = 665\), for a return of \(\frac{665}{300}-1 = 121.67\%\).
• if the economy is weak, equity holders will receive \(900 − 735 = 165\), for a return of \(\frac{165}{300}-1 = -45\%\).
Expected return is:
\[\frac{1}{2} \times 121.67\% + \frac{1}{2} \times -45\% = 38.33\%\]
Note that the equity has a return sensitivity of 121.67% − (−45.0%) = 166.67%.
Its risk premium is 38.33% − 5% = 33.33%.
14.1 Equity Versus Debt Financing
Debt 700 735 735 5% 5% 5% - 5% = 0% 5% - 5% = 0%
Equity 300 665 165 122% -45% 122% - (-45%) = 167% 38% - 5% = 33%
Firm 1000 1400 900 40% -10% 75% - (-25%) = 100% 25% - 5% = 20%
So, debt increases equity risk.
14.2 MM I
In this subsection, we will explore further the MM propositions and the theory of capital structure.
First, we will understand what Homemade Leverage means.
• Let’s say the firm selects a given capital structure. The investor likes an alternative capital structure.
• MM demonstrated that if investors would prefer an alternative capital structure to the one the firm has chosen, investors can borrow or lend on their own and achieve the same result.
14.2 MM I
Assume the firm is an all−equity firm. An investor who would prefer to hold levered equity can do so by using leverage in his own portfolio.
• An investor who would like more leverage than the firm has chosen can borrow (let’s say 500) and add leverage to his or her own portfolio (500 personal + 500 debt).
• The investor borrows 500 and buys the firm’s stock.
Unlevered equity (firm) 1000 1400 900
Margin loan (Investor borrowing) -500 -525 -525
Levered equity (Investor’s return) 500 875 375
14.2 MM I
If the cash flows of the unlevered equity serve as collateral for the margin loan (at the risk−free rate of 5%), then by using homemade leverage, the investor has replicated the payoffs to the
levered equity.
• As long as investors can borrow or lend at the same interest rate as the firm, homemade leverage is a perfect substitute for the use of leverage by the firm.
14.2 MM I
Now assume the firm uses debt, but the investor would prefer to hold unlevered equity.
The investor can replicate the payoffs of unlevered equity by buying both the debt and the equity of the firm.
Combining the cash flows of the two securities produces cash flows identical to unlevered equity, for a total cost of $1000.
Debt (Investor’s lending) 500 525 525
Levered equity (the firm) 500 875 375
Unlevered equity (Investor’s return) 1000 1400 900
14.2 MM I
Homemade leverage
In each case, the firm’s choice of capital structure does not affect the opportunities available to investors.
• Investors can alter the leverage choice of the firm to suit their personal tastes either by adding more leverage or by reducing leverage.
• With perfect capital markets, different choices of capital structure offer no benefit to investors and does not affect the value of the firm.
14.2 MM I
In summary,
• Modigliani and Miller (or simply MM) showed that leverage would not affect the total value of the firm.
• This result holds more generally under a set of conditions referred to as perfect capital markets:
□ Investors and firms can trade the same set of securities at competitive market prices equal to the present value of their future cash flows.
□ There are no taxes, transaction costs, or issuance costs associated with security trading.
□ A firm’s financing decisions do not change the cash flows generated by its investments, nor do they reveal new information about them.
Under these conditions, MM demonstrated the following result regarding the role of capital structure in determining firm value:
14.2 MM I
MM Proposition I: In a perfect capital market, the total value of a firm’s securities is equal to the market value of the total cash flows generated by its assets and is not affected by its choice of
capital structure.
In the absence of taxes or other transaction costs, the total cash flow paid out to all of a firm’s security holders is equal to the total cash flow generated by the firm’s assets.
14.3 MM II
Modigliani and Miller showed that a firm’s financing choice does not affect its value. But how can we reconcile this conclusion with the fact that the cost of capital differs for different
We will now discuss the second proposition of MM.
14.3 MM II
MM’s first proposition can be used to derive an explicit relationship between leverage and the equity cost of capital. Let’s denote:
• E = Market value of equity in a levered firm
• D = Market value of debt in a levered firm
• U = Market value of equity in an unlevered firm
• A = Market value of the firm’s assets
MMI states that:
\[E+D = U = A\]
That is, the total market value of the firm’s securities is equal to the market value of its assets, whether the firm is unlevered or levered.
14.3 MM II
Also, remember that the return of a portfolio is the weighted average of the returns.
So, we can write that the return on unlevered equity (\(R_u\)) is related to the returns of levered equity (\(R_e\)) and debt (\(R_d\)):
\[R_u = \frac{E}{E+D} \times R_e + \frac{D}{E+D} \times R_d\] Solving for \(R_e\):
\[R_e = R_u + \frac{D}{E} \times (R_u - R_d)\]
That is, the levered equity return equals the unlevered return, plus a premium due to leverage. The amount of the premium depends on the amount of leverage, measured by the firm’s market value
debt−equity ratio.
14.3 MM II
Proposition II: The cost of capital of levered equity is equal to the cost of capital of unlevered equity plus a premium that is proportional to the market value debt−equity ratio.
Using the previous example’s numbers:
• the expected return on unlevered equity is 15%
• the expected return of the debt is 5%.
• the expected return on equity for the levered firm is 25%
14.3 MM II
Suppose the entrepreneur borrows only $700 when financing the project. Recall that the expected return on unlevered equity is 15% and the risk−free rate is 5%. According to MM Proposition II, what
will be the firm’s equity cost of capital?
\[R_e = R_u + \frac{D}{E} \times (R_u - R_d) = 15\% + \frac{700}{300} \times (15\%-5\%) = 38.33\%\]
Exactly the same cost we found before.
14.3 Capital Budgeting and the WACC
We can use the insight of Modigliani and Miller to understand the effect of leverage on the firm’s cost of capital for new investments.
If a firm is unlevered, all of the free cash flows generated by its assets are paid out to its equity holders.
• The market value, risk, and cost of capital for the firm’s assets and its equity coincide and therefore
\[R_u = R_a\]
If a firm is levered, project \(R_A\) is equal to the firm’s weighted average cost of capital.
\[R_{wacc} = R_u = R_a\]
That is, with perfect capital markets, a firm’s WACC is independent of its capital structure and is equal to its equity cost of capital if it is unlevered, which matches the cost of capital of its
14.3 MM II
Assuming taxes do not exist:
14.3 MM II
The takeaway:
Although debt has a lower cost of capital than equity, leverage does not lower a firm’s WACC. As a result, the value of the firm’s free cash flow evaluated using the WACC does not change, and so the
enterprise value of the firm does not depend on its financing choices.
14.3 MM II
Honeywell International Inc. (HON) has a market debt−equity ratio of 0.5.
Assume its current debt cost of capital is 6.5%, and its equity cost of capital is 14%.
If HON issues equity and uses the proceeds to repay its debt and reduce its debt−equity ratio to 0.4, it will lower its debt cost of capital to 5.75%.
With perfect capital markets, what effect will this transaction have on HON’s equity cost of capital and WACC?
14.3 MM II
Current WACC
\[R_{wacc} = \frac{E}{E+D} \times R_e + \frac{D}{E+D} \times R_d = \frac{2}{2+1} \times 14 + \frac{1}{2+1} \times 6.5 = 11.5\%\]
New Cost of Equity:
\[R_e = R_u + \frac{D}{E} (R_u - R_a) = 11.5 +0.4 \times (11.5 - 5.75) = 13.8\%\]
“New” WACC:
\[R_{wacc_{new}} = \frac{1}{1+0.4} \times 13.8 + \frac{0.4}{1+0.4} \times 5.75 = 11.5\%\]
Multiple Securities
If the firm’s capital structure is made up of multiple securities, then the WACC is calculated by computing the weighted average cost of capital of all of the firm’s securities.
Let’s say the firm has Equity (E), Debt (D), Warrant (W) issued:
\[R_{wacc} = R_u = R_e \times \frac{E}{E+D+W} + R_d \times \frac{D}{E+D+W} + R_w \times \frac{W}{E+D+W} \]
Levered and unlevered Betas
Remember that
\[\beta_u = \frac{E}{D+E} \times \beta_e + \frac{D}{D+E} \times \beta_d\]
When a firm changes its capital structure without changing its investments, its unlevered beta will remain unaltered. However, its equity beta will change to reflect the effect of the capital
structure change on its risk.
\[\beta_e = \beta_u + \frac{D}{E} (\beta_u - \beta_d)\]
• Unlevered Beta: A measure of the risk of a firm as if it did not have leverage, which is equivalent to the beta of the firm’s assets.
Levered and unlevered Betas
In August 2018, Reenor had a market capitalization of 140 billion. It had debt of 25.4 billion as well as cash and short−term investments of 60.4 billion. Its equity beta was 1.09 and its debt beta
was approximately zero. What was Reenor’s enterprise value at time? Given a risk−free rate of 2% and a market risk premium of 5%, estimate the unlevered cost of capital of Reenor’s business.
Reenor’s net debt = 25.4 − 60.4 billion = −35.0 billion. Enterprise value 140 billion − 35billion = 105 billion.
\[\beta_u = \frac{E}{E+D} \times \beta_e + \frac{D}{E+D} \times \beta_d = \frac{140}{105} \times 1.09 + \frac{-35}{105} \times 0 = 1.45\] Unlevered cost of capital:
\[R_u = 2\% + 1.45 \times 5\% = 9.25\%\]
14.4 Capital Structure Fallacies
14.4 Capital Structure Fallacies
We will discuss now two fallacies concerning capital structure:
• Leverage increases earnings per share (EPS), thus increase firm value
• Issuing new equity will dilute existing shareholders, so debt should be issued
14.4 Capital Structure Fallacies
Leverage and EPS example
• LVI is currently an all−equity firm.
• It expects to generate earnings before interest and taxes (EBIT) of 10 million over the next year.
• Currently, LVI has 10 million shares outstanding, and its stock is trading for a price of 7.50 per share.
• LVI is considering changing its capital structure by borrowing 15 million at an interest rate of 8% and using the proceeds to repurchase 2 million shares at $7.50 per share.
Suppose LVI has no debt. Because there is no interest and no taxes, LVI’s earnings would equal its EBIT and LVI’s earnings per share without leverage would be
\[EPS=\frac{earnings}{Out. Shares}=\frac{10\;million}{10\;million}= 1\]
14.4 Capital Structure Fallacies
• If LVI recapitalizes, the new debt will obligate LVI to make interest payments each year of $1.2 million/year
□ 15 million × 8% interest/year = 1.2 million/year
• As a result, LVI will have expected earnings after interest of 8.8 million
□ Earnings = EBIT − Interest
□ Earnings = 10 million − 1.2 million = 8.8 million
\[EPS=\frac{earnings}{Out. Shares}=\frac{8.8\;million}{8\;million}= 1.1\] EPS increases. Should firm value increase?
The answer is no.
Just like equity becomes riskier when the firm issues debt, the earnings stream is also riskier when the firm issues debt.
14.4 Capital Structure Fallacies
EBIT 10 10
- Interets 0 1.2
Earnings 10 8.8
# shares (out) 10 10
EPS 1 1.1
14.4 Capital Structure Fallacies
No debt
EBIT 0 2 4 6 8 10 12 14 16 18 20
- Int. 0 0 0 0 0 0 0 0 0 0 0
Earnings 0 2 4 6 8 10 12 14 16 18 20
# shares 10 10 10 10 10 10 10 10 10 10 10
EPS 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2.0
14.4 Capital Structure Fallacies
With debt
EBIT 0 2 4 6 8 10 12 14 16 18 20
- Int. 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2
Earnings -1.2 0.8 2.8 4.8 6.8 8.8 10.8 12.8 14.8 16.8 18.8
# shares 10 10 10 10 10 10 10 10 10 10 10
EPS -0.12 0.08 0.28 0.48 0.68 0.88 1.08 1.28 1.48 1.68 1.88
14.4 Capital Structure Fallacies
Assume that LVI’s EBIT is not expected to grow in the future and that all earnings are paid as dividends. Use MM proposition I and II to show that the increase in expected EPS for LVI will not lead
to an increase in the share price.
Without leverage, expected earnings per share and therefore dividends are 1 each year, and the share price 7.50. Let r be LVI’s cost of capital without leverage. Then we can value LVI as a
\[P = 7.50 = \frac{Div}{R_u} = \frac{EPS}{R_u} = \frac{1}{R_u}\]
Therefore, current stock price implies that \(R_u = \frac{1}{7.50} = 13.33\%\)
14.4 Capital Structure Fallacies
• The market value of LVI without leverage is 7.50 × 10 million shares = 75 million.
• If LVI uses debt to repurchase 15 million worth of the firm’s equity
• then the remaining equity will be worth 75 million − 15million = 60 million.
After the transaction, \(\frac{D}{E} = \frac{1}{4}\), thus, we can write:
\[R_e= R_u +\frac{D}{E} \times (R_u - R_d) = 13.33 + 0.25 \times (13.33 - 8) = 14.66\]
Given that expected EPS is now 1.10 per share, the new value of the shares equals
\[P=\frac{1.10}{14.66} = 7.50\]
Thus, even though EPS is higher, due to the additional risk, shareholders will demand a higher return. These effects cancel out, so the price per share is unchanged.
14.4 Capital Structure Fallacies
Because the firm’s earnings per share and price-earnings ratio are affected by leverage, we cannot reliably compare these measures across firms with different capital structures.
The same is true for accounting-based performance measures such as return on equity (ROE).
Therefore, most analysts prefer to use performance measures and valuation multiples that are based on the firm’s earnings before interest has been deducted.
• For example, the ratio of enterprise value to EBIT (or EBITDA) is more useful when analyzing firms with very different capital structures than is comparing their P/E ratios.
14.4 Capital Structure Fallacies
Equity Issuances and Dilution
It is sometimes (incorrectly) argued that issuing equity will dilute existing shareholders’ ownership, so debt financing should be used instead.
• Dilution: An increase in the total of shares that will divide a fixed amount of earnings
Suppose Jet Sky Airlines (JSA) currently has no debt and 500 million shares of stock outstanding, which is currently trading at a price of $16.
Last month the firm announced that it would expand and the expansion will require the purchase of $1 billion of new planes, which will be financed by issuing new equity.
• The current (prior to the issue) value of the equity and the assets of the firm is 8 billion.
□ 500 million shares × 16 per share = 8 billion
14.4 Capital Structure Fallacies
Suppose now JSA sells 62.5 million new shares at the current price of 16 per share to raise the additional 1 billion needed to purchase the planes.
Cash 1000
Existing assets 8000 8000
Total Value 8000 9000
# shares (out) 500 562.5
Value per share 16 16
Result: share prices don’t change. Any gain or loss associated with the transaction will result from the NPV of the investments the firm makes with the funds raised. | {"url":"https://henriquemartins.net/teaching/financial_strategy/p7.html","timestamp":"2024-11-07T12:17:34Z","content_type":"text/html","content_length":"74071","record_id":"<urn:uuid:8cc19246-aa94-40de-ae4f-44832ae771c1>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00429.warc.gz"} |
SMPL Human Model Introduction
This article could be served as a bridge between the SMPL paper and a numpy-based code that synthesizes a new human mesh instance from a pre-trained SMPL model provided by the Maxplank Institute. I
wrote it as an exercise to strengthen my knowledge about the implementation of the SMPL model.
3D objects are often represented by vertices and triangles that encodes their 3D shape. The more detail an object is, the more vertices it is required. However, for objects like human, the 3D mesh
representation could be compressed down to a lower dimensional space whose axes are like their height, fatness, bust circumference, belly size, pose etc. This representation is often smaller and more
The SMPL is a statistical model that encodes the human subjects with two types of parameters:
• Shape parameter: a shape vector of 10 scalar values, each of which could be interpreted as an amount of expansion/shrink of a human subject along some direction such as taller or shorter.
• Pose parameter: a pose vector of 24x3 scalar values that keeps the relative rotations of joints with respective to their parameters. Each rotation is encoded as a arbitrary 3D vector in
axis-angle rotation representation.
As an example, the below code samples a random human subject with shape and pose parameters. The shift and multiplication are applied to bring the random values to the normal parameter range of the
SMPL model; otherwise, the synthesized mesh will look like an alien.
pose = (np.random.rand((24, 3)) - 0.5)
beta = (np.random.rand((10,)) - 0.5) * 2.5
Human synthesis pipeline
The process of synthesizing a new human instance from the SMPL model consists of 3 stages as illustrated in the below figure.
• Shape Blend Shapes: In this step, a template(or mean) mesh $\bar{T}$ is added with vertex displacements that represent how far the subject shape is from the mean shape.
• Pose Blend Shapes: After the identity mesh is constructed in the rest pose, it is further added with vertex displacements that explain for deformation correction caused by a specific pose. In
other words, the pose in the next step “Skinning” results in some amount of deformation on the rest pose at this step.
• Skinning: Each mesh vertex from the previous step is transformed by a weighted-combination of joint deformation. To put it simply, the closer to a vertex a joint is, the stronger the joint
rotates/transforms the vertex.
Shape Blend Shapes
The rest pose shape is formed by adding the mean shape with a linear combination of principal shape components (or vertex deviations), which denote the principal changes among all the meshes in the
dataset. Specifically, each principal component is a $6890\text{x}3$ matrix, which represent $(x,y,z)$ vertex displacements from the corresponding vertices of the mean mesh. To make it more clear,
below is a visualization of the first and second principal components of the SMPL model. The mesh pair for each component is constructed adding/subtract the component to/from the mean mesh an amount
of 3 standard deviations ,as showned in the below equation: $M_i = T \pm 3{\sigma}*PC_k$
From the visualization, it seems that the first component explains for the change in height and the second represents the change in weight among human meshes.
the image is taken from the Maxplank Institute
The below code creates a new mesh by linearly combining 10 principal components from the SMPL model. The more principal components we use, the less the reconstruction error is; however, the SMPL
model from the Maxplank Institute just provides us with the first 10 components.
# shapedirs: 6890x3x10: 10 principal deviations
# beta: 10x1: the shape vector of a particular human subject
# template: 6890x3: the average mesh from the dataset
# v_shape: 6890x3: the shape in vertex format corresponding to the shape vector
v_shaped = self.shapedirs.dot(self.beta) + self.v_template
Pose Blend Shapes
In the SMPL model, the human skeleton is described by a hierarchy of 24 joints as shown by the white dots in the below figure. This hierarchy is defined by a kinematic tree that keeps the parent
relation for each joint.
The 24-joints hierarchy is represented by $(23\text{x}3)$ matrix corresponding to $23$ relative rotations from parent joints. Each rotation is encoded by a axis-angle rotation representation of $3$
scalar values, which is denoted by the $\boldsymbol{\theta}$ vector in the below figure.
The image is taken from the Wikipedia
The relative rotations of 23 joints $(23\text{x}3)$ causes deformation to surrounding vertices. These deformations are captured by a matrix of (23x6890x3) which represents $23$ principal components
of vertex displacements of $(6890x3)$. Therefore, given a new pose vector of relative rotation 23x3x3 values as weights, the final deformation will be calculated as a linear combination of these
principal components.
# self.pose : 24x3 the pose parameter of the human subject
# self.R : 24x3x3 the rotation matrices calculated from the pose parameter
pose_cube = self.pose.reshape((-1, 1, 3))
self.R = self.rodrigues(pose_cube)
# I_cube : 23x3x3 the rotation matrices of the rest pose
# lrotmin : 207x1 the relative rotation values between the current pose and the rest pose
I_cube = np.broadcast_to(
np.expand_dims(np.eye(3), axis=0),
(self.R.shape[0]-1, 3, 3)
lrotmin = (self.R[1:] - I_cube).ravel()
# v_posed : 6890x3 the blended deformation calculated from the
v_posed = v_shaped + self.posedirs.dot(lrotmin)
In this step, vertex in the rest pose are transformed by a weighted combination of global joint transformations (rotation + translation). The joint rotations are already calculated from the pose
parameter of the human subject, but the joint translation part needs to be estimated from the corresponding rest-pose mesh of the subject.
Joint Locations Estimation
Thanks to the fixed mesh topology of the SMPL model, each joint location could be estimated as an average of surrounding vertices. This average is represented by a joint regression matrix learned
from the data-set that defines a sparse set of vertex weight for each joint. As shown in the below figure, the knee joint will be calculated as a linear combination of red vertices, each with a
different weight.
The image is taken from the SMPL paper
The below code shows how to regress joint locations from the rest-pose mesh.
# v_shape: 6890x3 the mesh in neutral T-pose calculated from a shape parameter of 10 scalar values.
# self.J_regressor: 24x6890 the regression matrix that maps 6890 vertex to 24 joint locations
# self.J: 24x3 24 joint (x,y,z) locations
self.J = self.J_regressor.dot(v_shaped)
Skinning deformation
The joint transformations cause the neighboring vertices transform with the same transformations but with different influence. The further from a joint a vertex is, the less it is affected by the
joint transformation. Therefore, a final vertex could be calculated as a weighted average of its versions transformed by 24 joints.
The below code first calculates the global transformation for each joint by recursively concatenating its local matrix with its parent matrix. These global transformations are then subtracted from
the corresponding transformations of the joints in the rest pose. For each vertex, its final transformation is calculated by blending the 24 global transformations with different weights. The code
for these steps are shown in the below.
# world transformation of each joint
G = np.empty((self.kintree_table.shape[1], 4, 4))
# the root transformation: rotation | the root joint location
G[0] = self.with_zeros(np.hstack((self.R[0], self.J[0, :].reshape([3, 1]))))
# recursively chain transformations
for i in range(1, self.kintree_table.shape[1]):
G[i] = G[self.parent[i]].dot(
[self.R[i],((self.J[i, :]-self.J[self.parent[i],:]).reshape([3,1]))]
# remove the transformation due to the rest pose
G = G - self.pack(
np.hstack([self.J, np.zeros([24, 1])]).reshape([24, 4, 1])
# G : (24, 4, 4) : the global joint transformations with rest transformations removed
# weights : (6890, 24) : the transformation weights for each joint
# T : (6890, 4, 4): the final transformation for each joint
T = np.tensordot(self.weights, G, axes=[[1], [0]])
# apply transformation to each vertex
rest_shape_h = np.hstack((v_posed, np.ones([v_posed.shape[0], 1])))
v = np.matmul(T, rest_shape_h.reshape([-1, 4, 1])).reshape([-1, 4])[:, :3]
# add with one global translation
verts = v + self.trans.reshape([1, 3])
In this port, we went through the steps of synthesizing a new human subject from the trained SMPL model provided by the Maxplank institute. We learned how to combine principal shape components to
reconstruct the rest-pose shape and then regress the joint locations from it. We also learned to predict the deformation correction caused by pose and apply global joint transformation to the
rest-pose mesh. For further information, please check the SMPL paper | {"url":"https://khanhha.github.io/posts/SMPL-model-introduction/","timestamp":"2024-11-08T12:10:15Z","content_type":"text/html","content_length":"36178","record_id":"<urn:uuid:507c75eb-0116-400b-96ff-a1831f8d6b45>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00383.warc.gz"} |
Trees in alley - math word problem (7195)
Trees in alley
There are four trees in the alley between which the distances are 35m, 15m, and 95m. Trees must be laid in the spaces so that the distance is equal and maximum. How many trees will they put in, and
what will be the distance between them?
Correct answer:
Did you find an error or inaccuracy? Feel free to
write us
. Thank you!
You need to know the following knowledge to solve this word math problem:
Units of physical quantities:
Grade of the word problem:
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Partial Differential Equation
In Mathematics, a partial differential equation is one of the types of differential equations, in which the equation contains unknown multi variables with their partial derivatives. It is a special
case of an ordinary differential equation. In this article, we are going to discuss what is a partial differential equation, how to represent it, its classification and types with more examples and
solved problems.
Table of Contents:
Partial Differential Equation Definition
A Partial Differential Equation commonly denoted as PDE is a differential equation containing partial derivatives of the dependent variable (one or more) with more than one independent variable. A
PDE for a function u(x[1],……x[n]) is an equation of the form
The PDE is said to be linear if f is a linear function of u and its derivatives. The simple PDE is given by;
∂u/∂x (x,y) = 0
The above relation implies that the function u(x,y) is independent of x which is the reduced form of partial differential equation formula stated above. The order of PDE is the order of the highest
derivative term of the equation.
How to Represent Partial Differential Equation?
In PDEs, we denote the partial derivatives using subscripts, such as;
In some cases, like in Physics when we learn about wave equations or sound equation, partial derivative, ∂ is also represented by ∇(del or nabla).
Partial Differential Equation Classification
Each type of PDE has certain functionalities that help to determine whether a particular finite element approach is appropriate to the problem being described by the PDE. The solution depends on the
equation and several variables contain partial derivatives with respect to the variables. There are three-types of second-order PDEs in mechanics. They are
• Elliptic PDE
• Parabolic PDE
• Hyperbolic PDE
Consider the example, au[xx]+bu[yy]+cu[yy]=0, u=u(x,y). For a given point (x,y), the equation is said to be Elliptic if b^2-ac<0 which are used to describe the equations of elasticity without
inertial terms. Hyperbolic PDEs describe the phenomena of wave propagation if it satisfies the condition b^2-ac>0. For parabolic PDEs, it should satisfy the condition b^2-ac=0. The heat conduction
equation is an example of a parabolic PDE.
Partial Differential Equation Types
The different types of partial differential equations are:
• First-order Partial Differential Equation
• Linear Partial Differential Equation
• Quasi-Linear Partial Differential Equation
• Homogeneous Partial Differential Equation
Let us discuss these types of PDEs here.
First-Order Partial Differential Equation
In Maths, when we speak about the first-order partial differential equation, then the equation has only the first derivative of the unknown function having ‘m’ variables. It is expressed in the form
F(x[1],…,x[m], u,u[x1],….,u[xm])=0
Linear Partial Differential Equation
If the dependent variable and all its partial derivatives occur linearly in any PDE then such an equation is called linear PDE otherwise a nonlinear PDE. In the above example (1) and (2) are said to
be linear equations whereas example (3) and (4) are said to be non-linear equations.
Quasi-Linear Partial Differential Equation
A PDE is said to be quasi-linear if all the terms with the highest order derivatives of dependent variables occur linearly, that is the coefficient of those terms are functions of only lower-order
derivatives of the dependent variables. However, terms with lower-order derivatives can occur in any manner. Example (3) in the above list is a Quasi-linear equation.
Homogeneous Partial Differential Equation
If all the terms of a PDE contain the dependent variable or its partial derivatives then such a PDE is called non-homogeneous partial differential equation or homogeneous otherwise. In the above four
examples, Example (4) is non-homogeneous whereas the first three equations are homogeneous.
Partial Differential Equation Examples
Some of the examples which follow second-order PDE is given as
Partial Differential Equation Solved Problem
Show that if a is a constant ,then u(x,t)=sin(at)cos(x) is a solution to \(\frac{\partial ^{2}u}{\partial t^{2}}=a^{2}\frac{\partial ^{2}u}{\partial x^{2}}\).
Since a is a constant, the partials with respect to t are
\(\frac{\partial u}{\partial t}=a\cos (at)\cos (x)\) ; \(\frac{\partial^{2} u}{\partial t^{2}}=-a^{2}\sin (at)\cos (x)\)
Moreover, u[x] = – sin (at) sin (x) and u[xx]= – sin (at)cos(x), so that
\(a^{2}\frac{\partial^{2}u }{\partial x^{2}}=-a^{2}\sin (at)\cos (x)\)
Therefore, u(x,t)=sin(at)cos(x) is a solution to \(\frac{\partial ^{2}u}{\partial t^{2}}=a^{2}\frac{\partial ^{2}u}{\partial x^{2}}\).
Hence proved.
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Question #51b13 | Socratic
Question #51b13
1 Answer
$W = - 101325 \cdot 1.248 \cdot {10}^{-} 3 = - 126.5 \text{ J}$
$\text{∆E"="∆H"-"P∆V"=-2340-126.5=-2466.5" J}$
First of all, always use SI units to get the answer out in SI units. In this case, we want work and ∆E in joules, so use volume in m^3 and pressure in Pa.
You need the volume that the N2 gas occupies:
$n \left(N a N 3\right) = \frac{2.25}{65.02} = 0.0346 \text{ mol}$
$n \left(N 2\right) = \frac{3}{2} \cdot \text{n(NaN3)"=0.0346*3/2=0.0519" mol}$
$m \left(N 2\right) = n \left(N 2\right) \cdot M \left(N 2\right) = 0.0519 \cdot 28.02 = 1.454 \text{ g}$
$V \left(N 2\right) = \frac{m}{\text{density"=1.454/1.165=1.248" L}}$
$V \left(N 2\right) = 1.248 \cdot {10}^{-} 3 {\text{ m}}^{3}$
Note that the Ideal Gas Law will give you the same volume.
As the initial volume is zero:
#∆V="Vf"-"Vi"=1.248*10^-3-0=1.248*10^-3" m"^3#
Now substitute the external pressure (convert 1 atm to Pa) and the change in volume into the expression for work done:
#W = -P∆V#
$W = - 101325 \cdot 1.248 \cdot {10}^{-} 3 = - 126.5 \text{ Pa m"^3=-126.5" J}$
By using SI units, the answer has come out neatly in J, as
$1 {\text{ J" = 1" Pa m}}^{3}$.
The heat released is the enthalpy change:
$\text{∆H"=-2340" J}$
Rearranging the formula for enthalpy allows us to solve for the internal energy change:
$\text{∆E"="∆H"-"P∆V"=-2340-126.5=-2466.5" J}$
From the system's point of view, this means that during the reaction 2340 J was transferred as heat to the surroundings and a further 126.5 J was expended as expansion work, giving a total internal
energy change/loss (∆E) of 2466.5 J.
Impact of this question
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Stephen Wolfram: The Mathematician Who Developed the Wolfram Language and the Wolfram Alpha Search Engine
Home Science Stephen Wolfram: The Mathematician Who Developed the Wolfram Language and the Wolfram...
Stephen Wolfram: The Mathematician Who Developed the Wolfram Language and the Wolfram Alpha Search Engine
Stephen Wolfram: The Mathematician Who Developed the Wolfram Language and the Wolfram Alpha Search Engine
Stephen Wolfram is a mathematician and computer scientist who has developed a revolution in computation. From a young age, he has amazed and astonished the public with his innovative ideas and
ground-breaking work. As the creator of the Wolfram Language, and the Wolfram Alpha search engine, Stephen Wolfram has changed the way we think about math, computing, and the future of technology.
Early in Stephen Wolfram’s Career
Stephen Wolfram was born in London in 1959. He went to five different schools, and attended the University of Oxford where he earned his Bachelor of Arts degree in mathematics. During his time at
Oxford, he published a paper titled “Space-Filling Curves,” in the journal Advances in Mathematics. This was one of the earliest significant results of Wolfram’s work, and it is still cited to this
He went on to study physics at the University of California at Berkeley, where he earned his Ph.D. in particle physics. After leaving Berkeley, he worked as a researcher at the Institute for Advanced
Study in Princeton. At the Institute, he made some of the earliest contributions to the field of complex systems, a field in which he remains a leader today.
The Creation of the Wolfram Language
In 1988, Stephen Wolfram created his first programming language, called Warp. Warp went on to become the basis for the Wolfram Language, a powerful programming language for scientific computing.
Since then, Wolfram Language has been used to create a vast array of computer programs, and to solve some of the world’s most difficult problems in mathematics and physics.
The Wolfram Alpha Search Engine
In 2009, Stephen Wolfram released Wolfram Alpha, a search engine that uses sophisticated algorithms to answer complex questions. Wolfram Alpha can answer questions in many areas, including
mathematics, physics, chemistry, and astronomy. Wolfram Alpha can also be used to find information related to business, finance, and economics.
Wolfram Alpha is backed by Wolfram’s vast and growing database of information, and by its sophisticated algorithms, which allow it to interpret natural language queries. Wolfram Alpha is constantly
being improved and enhanced, and new information is constantly being added.
Notable Contributions of Stephen Wolfram
Stephen Wolfram has made many notable contributions to the fields of computation, mathematics, and physics. He has made groundbreaking discoveries related to cellular automata, lattice theory, and
computational complexity. He has also developed many computational algorithms, programs, and languages, including the Wolfram Language.
In addition to his accomplishments in computation, Stephen Wolfram has also shown a great deal of interest in education. He has made numerous contributions to the fields of science and math
education, and has written several books on the subjects. He also founded Wolfram Research in 1987, which is a major provider of educational technology software.
The Impact of Stephen Wolfram
Stephen Wolfram’s work has had a massive impact on the way we think about computation and mathematics, and his contributions have been recognized with numerous awards, including the MacArthur
Fellowship and the 2017 Albert Einstein Award. His innovations have had a major impact on technology, and have helped to move science and mathematics forward in a big way.
Today, Stephen Wolfram remains actively engaged in the fields of mathematics, physics, and computation. He is the CEO of Wolfram Research, and his work continues to revolutionize the way we think
about computation, mathematics, and the future of technology.
Stephen Wolfram is a mathematician, physicist, and computer scientist who has made major contributions to the fields of mathematics, physics, and computing. He is the creator of the Wolfram Language,
and the Wolfram Alpha search engine, and his innovations have had a major impact on technology. Wolfram remains actively engaged in research, and his work continues to revolutionize the way we think
about mathematics and the future of computing. | {"url":"https://sciengist.com/stephen-wolfram-the-mathematician-who-developed-the-wolfram-language-and-the-wolfram-alpha-search-engine/","timestamp":"2024-11-05T10:20:52Z","content_type":"text/html","content_length":"178249","record_id":"<urn:uuid:03c59ec4-7c07-4633-a983-aaa857dd6700>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00026.warc.gz"} |
Margin Finder
Determine your profit, margin, and markup for selling any number of items.
Given the unit cost, determine your selling price for a desired markup or margin.
Given the unit selling price, determine your item cost for a desired margin or markup.
Make quick calculations that include tax or any other additional fees or markups.
Determine your selling price for a given margin or markup.
Determine your item cost for a given desired margin or markup.
Make quick calculations that involve any other additional fees or markups.
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Multiplication Chart 1-25 Fifth Grade 2024 - Multiplication Chart Printable
Multiplication Chart 1-25 Fifth Grade
Multiplication Chart 1-25 Fifth Grade – If you are looking for a fun way to teach your child the multiplication facts, you can get a blank Multiplication Chart. This can let your little one to fill
the information on their own. You will discover blank multiplication graphs for various product ranges, such as 1-9, 10-12, and 15 items. You can add a Game to it if you want to make your chart more
exciting. Here are several suggestions to get your little one began: Multiplication Chart 1-25 Fifth Grade.
Multiplication Charts
You can use multiplication charts as part of your child’s college student binder to assist them memorize mathematics specifics. Although many kids can remember their arithmetic specifics in a natural
way, it takes lots of others time to achieve this. Multiplication graphs are an ideal way to strengthen their boost and learning their self confidence. In addition to being educative, these graphs
might be laminated for additional toughness. The following are some helpful approaches to use multiplication graphs. You may also take a look at these web sites for valuable multiplication simple
fact solutions.
This course covers the basics from the multiplication kitchen table. In addition to studying the guidelines for multiplying, college students will comprehend the very idea of elements and patterning.
Students will be able to recall basic facts like five times four, by understanding how the factors work. They will also be able to use the home of one and zero to fix more advanced products. Students
should be able to recognize patterns in multiplication chart 1, by the end of the lesson.
As well as the standard multiplication chart, individuals should create a graph or chart with additional elements or less aspects. To create a multiplication graph with increased factors, individuals
have to create 12 tables, each with twelve lines and 3 posts. All 12 tables must match on a single page of papers. Collections must be attracted using a ruler. Graph document is perfect for this
venture. If graph paper is not an option, students can use spreadsheet programs to make their own tables.
Game suggestions
Regardless if you are training a novice multiplication lesson or working on the mastery from the multiplication table, you may develop enjoyable and interesting game ideas for Multiplication Chart 1.
Several exciting suggestions are highlighted below. This video game demands the pupils to be work and pairs about the same dilemma. Then, they are going to all endure their credit cards and talk
about the perfect solution for a min. If they get it right, they win!
When you’re teaching children about multiplication, one of the better instruments you can allow them to have can be a computer multiplication graph. These computer sheets come in a range of patterns
and may be imprinted on a single web page or several. Little ones can learn their multiplication information by copying them from the memorizing and chart them. A multiplication graph or chart may
help for a lot of factors, from helping them find out their math concepts details to training them the way you use a calculator.
Gallery of Multiplication Chart 1-25 Fifth Grade
Printable Multiplication Table 1 25 PrintableMultiplication
Free Printable Multiplication Chart 1 25 Times Table PDF
Printable Multiplication Chart 1 25 PrintableMultiplication
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Items property
The items of this form group as an array, including all fields, buttons, text boxes and named values.
Accessing all items of a form group through this property has many uses:
• Calculating the sum of all number fields of a form group.
• Enabling a button only if all fields of a form group are valid.
• Deciding which background color to use based on whether the fields of a form group are valid.
• Collecting the labels and values of all fields of a form group as a text string, for use as the body of an email.
• Only allowing a user to move forward to the next screen if they have filled out all fields of a form group.
• Only showing a form group if all switch fields of another form group have been toggled to their "on" positions.
The following sections detail how these scenarios can be realized. Skip to the examples at the bottom for the concise version.
Summing all number fields
This formula returns the sum of the values of all numeric fields, including number fields:
The formula above can also be written as follows:
However, as the SUM function is looking for an array of numbers to add together, and number fields return numbers through their Value properties, .Value,Value is inferred and does not need to be
spelled out.
In fact, you can make the formula even shorter:
In the formula above, .Items.Value,Items,Value is inferred.
Summing all fields except date and time fields
The formulas above ask all items of the form group to return numbers that the SUM function can add together. Items that cannot do that, such as text fields, are ignored.
However, number fields are not the only fields that can return numbers, that is also true for number drop-down fields and date and time fields.
Adding together the values of number fields and number drop-down fields makes sense, but mixing values of number fields and number drop-down fields with values of date and time fields only makes
sense when you want to add days to a date. (The value of a date and time field represents the number of days that have elapsed since December 31, 1899, meaning that 18,264 represents January 1,
To process only number fields, use the NumberFields property instead. This formula adds together the values of the number fields of FormGroup1, ignoring date and time fields:
To also include the values of number drop-down fields, use this formula:
Summing the other fields of a form group
Let's say that the form group FormGroup1 consists of ten fields, Field1 through Field10. If Field10 should contain the sum of the other fields, it is tempting to try to associate this formula with
the Value property of Field10:
However, that formula will not work. Instead, you'll get an error message, because you're effectively asking that the calculated value of Field10 includes the value of Field10 itself.
This is similar to trying to use the formula Field1 * 2Field1 * 2 for the Value property of Field1, effectively asking that the value of Field1 should be set to the value of Field1, multiplied by
two. This is known as a circular calculation, and results in an error message.
In order to solve this issue, the formula needs to reference the fields to include in the calculation explicitly:
Above, Field1:Field9Field1:Field9 creates an array consisting of Field1, Field9 and all items that appear between them. Notably, Field10 is not part of the array.
Enabling a button only if all fields are valid
The Enabled property of a button determines if users can interact with the button. If a button should only be enabled if all fields of FormGroup1 are considered valid, associate this formula with the
Enabled property of the button:
Above, the FormGroup1.Items.ValidFormGroup1,Items,Valid formula returns an array of logical values (TRUE or FALSE), where TRUE indicates that an item is valid and FALSE indicates that an item is
invalid. The AND function, when applied to this array, returns TRUE only if all array elements are TRUE. In effect, the button is only enabled if all fields of the form group are valid.
The items of a form group can include not only fields, but also buttons and text boxes which don't support a Valid property. The elements of the FormGroup1.Items.ValidFormGroup1,Items,Valid array
that correspond to buttons and text boxes are blank. They have no effect on the value returned from AND, as this function ignores blank values.
Making the background color red if a field is invalid
The BackgroundColor property determines the background color of a screen and all screens that follow that have no explicit background color set. That means that if the background color is set for the
first screen of an app, and no other screens have a background color set, the first screen determines the background color of the entire app.
We can make use of this knowledge to make the background of the entire app red, but only if at least one field of the form group FormGroup1 is invalid. This formula is associated with the
BackgroundColor property of the first screen:
The formula fragment FormGroup1.Items.ValidFormGroup1,Items,Valid returns a logical array, where TRUE indicates that the corresponding item is valid and FALSE indicates that the corresponding item is
invalid. Applying the NOT function to this array negates every element, meaning that TRUE indicates that an item is invalid and FALSE indicates that an item is valid. (The ! operator would have had
the same effect.)
Then, the OR function is applied to this array. It returns FALSE only if all elements of the array are FALSE. In other words, it returns TRUE if one or several elements are TRUE, meaning that it
returns TRUE if one or several items are invalid.
Finally, the IF function is used to return the color red if one or several items are invalid. Otherwise, IF returns a blank value, which has no effect on the background color. The net effect is that
the background color of the app is made red if one or several items of FormGroup1 are invalid.
Including all values of a form group in an email
The Body property of email report buttons allows the body of an email to be set through a formula. While email report buttons have built-in support for including field values, through the
IncludedFields property, building a text string manually to include in the email body allows us more flexibility.
Consider this formula, which should be associated with the Body property of an email report button:
Above, the formula fragment FormGroup1.Items.LabelFormGroup1,Items,Label returns a text array, made up of the labels of the items of FormGroup1. The formula fragment FormGroup1.Items.Value
FormGroup1,Items,Value also returns an array, this time made up of the values of the items. Using &, the labels are joined together with the values, separated by a colon.
The resulting text array, where every element consists of a label, followed by a colon and a value, is converted into a single text string using the TEXTJOIN function. Its first parameter, NEWLINE()
NEWLINE(), ensures that the array elements are separated from one another using line breaks.
Requiring all fields of a form group to be filled out
The NextScreenAvailable property of form screens determines if users are allowed to move forward to the next screen. If a user should only be allowed to move forward once all fields of a form group
have been filled out, associate this formula with the NextScreenAvailable property of the screen that the form group is part of:
The ISDEFINED function returns a logical array when its sole parameter is an array. TRUE elements in the array indicate that the fields have defined values, that is, have been filled out. Conversely,
FALSE elements indicate that the fields have not been filled out.
Finally, the AND function returns TRUE only if all elements of the array are TRUE, otherwise it returns FALSE. The net effect is that AND returns TRUE only if all fields of FormGroup1 have been
filled out, prompting the NextScreenAvailable property to only allow users to proceed once all fields have defined values.
Requiring all switch fields to be toggled "on"
The Visible property of a form group determines if it is visible to the user. If a form group should only become visible once all switch fields of FormGroup2 have been toggled to their "on"
positions, associate this formula with the Visible property of a different form group:
The AND function is looking to process logical values, a request that the switch fields of FormGroup satisfy by returning their values. In other words, this formula is equivalent:
The formula above won't work if there are items in the form group which use values which are not logical, such as number fields. To solve this issue, you can refer explicitly to switch fields using
the SwitchFields property:
When the formulas above are associated with the Visible property of another form group, the net effect is that the form group is only made visible once all switch fields have been toggled to their
"on" positions.
Ranges versus this property
If the form group FormGroup1 only consists of the fields Field1, Field2 and Field3, these formulas are equivalent:
{ Field1, Field2, Field3 }{ Field1; Field2; Field3 }
The second formula uses a range to create an array consisting of Field1, Field3 and all items that appear between them, which in this case is only Field2.
The chief advantage of the Items property, compared to a range, is that there is no need to update formulas when additional items are added to a form group. If Field4 were to be added to the form
group, the Field1:Field3Field1:Field3 range would have to be changed to Field1:Field4Field1:Field4 everywhere it is used.
By contrast, FormGroup1.ItemsFormGroup1,Items automatically includes Field4, and any other items that are added.
Filtering items
If you want to process only a subset of the items returned from this property, use the FILTER function. It can base its decision on which items to return on the property values of the items.
This formula only returns visible items:
Crucially, you can also filter on the names of the items, using standard text functions. This formula only returns items whose names include the text string "Required":
If you use a deliberate naming strategy for your items, you can use FILTER in conjunction with this property to ensure that you only process a specific subset of items.
Related properties
Use the Items property of a screen to access all items of said screen and the Items property of the app object to access all items of the entire app.
There are many other properties that return only certain items. For instance, Fields returns all fields of a form group, whereas NumberFields only returns the number fields of a form group.
Returns the sum of the values of all number fields, number drop-down fields and date and time fields that belong to FormGroup1.
Returns the sum of the values of all number fields, number drop-down fields and date and time fields that belong to FormGroup1. If .Items,Items is left out, it is inferred.
Returns the sum of the values of all number fields that belong to FormGroup1.
Returns the sum of the values of all number fields that belong to FormGroup1. If .Value,Value is left out, it is inferred.
Returns the sum of the values of all number fields and number drop-down fields that belong to FormGroup1.
Returns TRUE if all fields of FormGroup1 are valid and FALSE otherwise. FormGroup1.Items.ValidFormGroup1,Items,Valid returns a logical array, where each element reflects whether its corresponding
item is valid. Finally, the AND function returns TRUE if all elements are TRUE and FALSE otherwise.
Returns all field values of a form group as a text string, where values are separated from one another using line breaks. The formula fragment FormGroup1.Items.ValueFormGroup1,Items,Value returns an
array of values, which the TEXTJOIN function joins together with line breaks.
Returns TRUE if all fields of FormGroup1 have been filled out. When the ISDEFINED function is applied to an array of items, it returns a logical array whose elements indicate if the corresponding
item has a defined value. The AND function returns TRUE if all array elements are TRUE and FALSE otherwise.
Returns TRUE if all switch fields of FormGroup1 have been toggled to their "on" positions. FormGroup2.SwitchFields.ValueFormGroup2,SwitchFields,Value returns a logical array containing the values of
all switch fields. The AND functions returns TRUE if all array elements are TRUE and FALSE otherwise. | {"url":"https://www.calcapp.net/learn/properties/formGroup/items.html","timestamp":"2024-11-02T21:16:02Z","content_type":"text/html","content_length":"62790","record_id":"<urn:uuid:994622bb-b37a-45fa-8684-2d14983162ca>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00655.warc.gz"} |
New discovery may revolutionize our decades old understanding of integrable hierarchies
- KdV and BKP, two solitonic integrable hierarchies important in modern theoretical physics and mathematics, are surprisingly related -
Processes in nature can often be described by equations. In many non-trivial cases, it is impossible to find the exact solutions to these equations. However, some equations are much simpler to deal
with because of their extreme symmetries. An important class of such equations is given by integrable systems. Integrable systems are known to be a universal tool in theoretical physics and
mathematics. They have proven to be extremely useful in diverse areas such as statistical mechanics, gauge theories, quantum gravity, and nonlinear waves, and they are particularly important in
modern geometry.
For some mysterious reason, integrability is often closely related to solvability. Namely, when a geometry problem can be related to an integrable system, sooner or later it always can be solved
completely. There are several different types of integrable systems, and different powerful methods are constructed to solve them. Furthermore, identifying the relations between different integrable
systems allow us to apply various methods to solve these problems.
Among different families of integrable systems, integrable hierarchies of solitonic type have particularly many applications. Arguably the most important example is given by the
Kadomtsev-Petviashvili (KP) hierarchy, which is described by an infinite tower of partial differential equations. The first of them was introduced in 1970 by two physicists, Kadomtsev and
Petviashvili, for the description of the acoustic waves in plasma. The KP equation was introduced as a deformation of the Korteweg-de Vries (KdV) equation, which describes waves on shallow water
surfaces. More generally, the whole KdV hierarchy describes a reduction of the KP hierarchy.
The theory of integrable hierarchies was actively developed by Date, Jimbo, Kashiwara, Miwa, and Sato from Kyoto University in the 1980s. They found a fundamental relation between integrable
hierarchies, representation theory of the infinite-dimensional Lie algebras, and free field formalism. In particular, they described the solutions of the hierarchies in terms of tau-functions, which
are formal functions of infinitely many variables.
As for the solution of the KdV hierarchy, the Kontsevich-Witten tau function was constructed by Edward Witten and Maxim Kontsevich. It plays a special role in modern mathematical physics for its
description of two-dimensional topological gravity. Another tau-function of the KdV integrable hierarchy is the Brezin-Gross-Witten model, which has been introduced in lattice gauge theory 40 years
ago. These two tau-functions have a natural enumerative geometry interpretation; they are among of the most well-studied tau-functions of the integrable solitonic hierarchies.
Recently there appear some indications that the KdV may also have a natural relation to the B-type KP hierarchy (BKP), which is associated with the orthogonal symmetry group. Indeed, in a recent
paper by Mironov and Morozov it was noted that the Kontsevich-Witten tau-function has a simple expansion in terms of the Schur Q-functions, which are known to be closely related to the BKP hierarchy.
This result was generalized by Alexander Alexandrov to a family of KdV tau-functions related to the Brezin-Gross-Witten model. Based on these expansions, Alexander Alexandrov has conjectured (for the
Kontsevich-Witten tau-function) and proved (for the Brezin-Gross-Witten tau-function) that these KdV tau-functions also solve the BKP hierarchy.
These results have led to the question: What is the most general relationship between the KdV and BKP hierarchies? The answer to this question was recently given by Alexander Alexandrov from the
Center for Geometry and Physics within the Institute for Basic Science (IBS). Namely, he proved that any tau-function of KdV solves the BKP hierarchy.
There are several different ways to relate KdV and BKP hierarchies. In particular, already in the 80s Date, Jimbo, Kashiwara, and Miwa (DJKM) described the identification of KdV hierarchy with the
4-reduction of BKP. A new result, obtained by Alexander Alexandrov, is much simpler and elegant than any of the previously known relations and provides a new, fundamental connection between two basic
integrable hierarchies of solitonic type. As the former theory by DJKM was once a classical part of mathematical physics and was believed to be complete, the new development of this theory by
Alexandrov was unexpected by the mathematics community.
This result makes the Schur Q-functions a natural basis for an expansion of the KdV tau-functions. Such expansion can help to find new properties of the KdV tau-functions. For instance, the Schur
Q-functions expansion of the Kontsevich-Witten and Brezin-Gross-Witten tau-functions have a special form: they describe so-called hypergeometric BKP tau-functions. This class of tau-functions,
introduced by Orlov, is known to be related to an interesting class of the enumerative geometry invariants, namely, the spin Hurwitz numbers. Therefore, the identification of the KdV tau-functions
with the solutions of the BKP hierarchy leads to the new, unexpected identification between two different classes of the enumerative geometry invariants, the intersection numbers on the moduli spaces
and the spin Hurwitz numbers.
Alexander Alexandrov expects that the identification of the KdV tau-functions with the solutions of the BKP hierarchy will lead us to many new results in enumerative geometry and mathematical
Figure 1. The main theorem describes the relationship between solutions of the KdV and BKP hierarchies. The tau functions can be identified after a simple change of variables.
Figure 2. Alexander Alexandrov, the author of this study.
Notes for editors
- References
KdV solves BKP, Alexander Alexandrov, Proceedings of the National Academy of Sciences Jun 2021, 118 (25) e2101917118; DOI: 10.1073/pnas.2101917118
- Media Contact
For further information or to request media assistance, please contact Alexander Alexandrov at the Center for Geometry and Physics, Institute for Basic Science (IBS) (alex@ibs.re.kr) or William I.
Suh at the IBS Communications Team (willisuh@ibs.re.kr)
- About the Institute for Basic Science (IBS)
IBS was founded in 2011 by the government of the Republic of Korea with the sole purpose of driving forward the development of basic science in South Korea. IBS has 30 research centers as of August
2021. There are ten physics, three mathematics, six chemistry, five life science, one Earth science, and five interdisciplinary research centers. | {"url":"https://www.ibs.re.kr/cop/bbs/BBSMSTR_000000000738/selectBoardArticle.do?nttId=21055&pageIndex=1&mno=sitemap_02&searchCnd=&searchWrd=","timestamp":"2024-11-12T04:10:41Z","content_type":"text/html","content_length":"66294","record_id":"<urn:uuid:28316308-72b2-442d-bc7d-cbf2cbbba1b8>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00441.warc.gz"} |
Subtraction to 10 via inverse operation
Students learn to subtract numbers to 10 using inverse operation. They learn that each subtraction number has a partner addition problem, which they can use to solve the subtraction problem.
This strategy helps make subtraction easier and faster. It helps ground basic calculations with numbers to 10.
Start by decomposing the numbers given on the interactive whiteboard. Then discuss the number bonds for the numbers given on the interactive whiteboard. You may also encourage students to use blocks
to help represent these decompositions/number bonds. Do a few number decompositions orally with the class. For example: " I decompose the number 5 into 4 and ...". Then determine which math problems
are shown in the images on the interactive whiteboard. You see 6 fish in total, of which 1 is in the fish tank and 5 are in a fishbowl. This could represent 6 - 1 or 5 + 1. Discuss with students why
the other options are incorrect (for example 5 - 1 is incorrect, because the total amount of fish shown is 6, not 5).
Explain to students that if they know an addition problem, that they also know the subtraction problem. The numbers in the problems are simply in different places in the problem. With a subtraction
problem, you start with the total or sum of the addition problem. If you subtract or take away one of the numbers, your difference is the second number. Show this with the beads on the interactive
whiteboard. Then practice this with a new set of beads. Then show a decomposition of the number 5 and the math problems you can create with these decompositions. Practice this with students using
different decompositions. After practicing this with students using numbers that have visual support, explain that this does not change, even if you can't count the object(s). Practice using the
inverse operation with decompositions where only numbers are visible.Check that students understand subtraction to 10 via the inverse operation by asking the following questions:- You can decompose 8
into 2 and ...?- 7 + 2 = 9. What is 9 - 2?- Which number do you set as the first number when you turn an addition problem into a subtraction problem?- Which numbers can swap places?
Students are first given problems with visual support, then are given a number bond as support, and finally are asked to solve problems without any visual support. They must fill in the correct
Discuss with students that it is important to be able to use the inverse operation so that they can easily and quickly solve math problems. For example, if you know that there are 2 cats and 3 dogs
at the house, the total is 5 pets. You then also know that if you are at that house and see 2 cats, there are 3 dogs left to find. To close, ask students to solve a set of problems with visual
support, and finally a set of problems without visual support.
Students who have difficulty can be supported by making use of manipulatives like MAB blocks or a rekenrek. For example: Have them count out 7 blocks and split them into a gorup of 4 and 3. Ask them
to say what the addition problem is (3 + 4 = 7). Then ask them how many blocks are left if you take away 3 blocks. Point out to the students that they don't need to recalculate, because you already
know how many blocks are needed from the addition problem you just solved.
Gynzy is an online teaching platform for interactive whiteboards and displays in schools.
With a focus on elementary education, Gynzy’s Whiteboard, digital tools, and activities make it easy for teachers to save time building lessons, increase student engagement, and make classroom
management more efficient. | {"url":"https://www.gynzy.com/en-us/library/items/subtraction-to-10-via-inverse-operation","timestamp":"2024-11-11T06:33:08Z","content_type":"text/html","content_length":"554241","record_id":"<urn:uuid:104eab0f-7452-4fc7-abd4-fc0473657511>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00618.warc.gz"} |
Guchuan Li
李谷川   Guchuan Li
I am an Assistant Professor in the Department of Mathematics at Peking University. Previously, I was a Postdoctoral Assistant Professor in the Department of Mathematics at the University of Michigana
from 2020 to 2023 and a postdoc in the Department of Mathematical Sciences at the University of Copenhagen from 2019 t0 2020. I obtained my Ph.D. in mathematics from Northwestern University in 2019
under the supervision of Paul Goerss.
My research interest is in algebraic topology, with an emphasis on chromatic homotopy theory and equivariant homotopy theory.
My CV is here.
Email: liguchuan at math dot pku dot edu dot cn
Peking University
Department of Mathematics
I co-organize the Electronic Computational Homotopy Theory online research seminar with Jack Carlisle, Dan Isaksen, and J.D.Quigley. | {"url":"https://guchuanli.github.io/","timestamp":"2024-11-09T04:24:33Z","content_type":"text/html","content_length":"5107","record_id":"<urn:uuid:d4dae0ad-20b6-4af0-8eda-dfc2cd8a56e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00292.warc.gz"} |
How much water can gutters handle?
The most common diameter of residential gutters in Virginia is 5 inches. A 6-inch K-type foot of gutters holds 2.0 gallons of water. Gutter size is an aspect of rainwater collection that has been
studied extensively and can be calculated based on guidelines published in plumbing codes. For example, the Uniform Plumbing Code (UPC) recommends that a gutter system can withstand the runoff of the
heaviest 60-minute downpour recorded in last 100 years (e.g., sometimes referred to as storm events).
The International Plumbing Code (IPC) has a similar, but not exact, published size recommendation. Both plumbing code manuals include the size calculations needed to properly size a gutter system.
Since water weighs more than 8 pounds per gallon, it represents a significant threat to a home and its inhabitants. The weight of rainwater accumulated on a roof has caused the roofs to collapse.
The gutter system must be able to drain the roof quickly enough not to exceed the structural limits of the roof. Both plumbing codes use the highest amount of rainfall per hour recorded in the last
100 years as a way to ensure that the planned and finally installed gutter system can withstand the largest known rainfall event that has occurred in recent times. To determine system sizing
alternatives, know the size of the roof to be drained. In a typical house with two slopes, each side will have gutters and will be dimensioned separately.
Then, calculate how much rain should be drawn from the roof. You can find this number by calling your local building department or looking for the number in Appendix D, Table D1, of the UPC or in
Appendix B of the IPC manual. Copies of these manuals are usually available at the local library. For example, for Phoenix* it's 2.2 inches of rain per hour or.
So what size of gutters need to be installed to withstand this incredible volume of rain? The width of the gutters, the slope of the gutters, and the number of downspouts all come into play in
determining the correct size of the system. For example, the larger the width of the gutter (i.e., going from a 3-channel to a 5-channel), the smaller the slope required (i.e., e.g.Go from ½ to ¼) to
withstand the same amount of rain. Table 11-3-1 of the UPC indicates that a roof that needs to withstand 3450 gallons per hour would require a system designed to the specifications detailed below.
TABLE 11-3-1 Uniform Plumbing Code Any of the combinations highlighted above could be installed, depending on the UPC.
The above example assumes that there is only one downspout. Alternatively, several downspouts could be installed and the size of the gutter and the slope requirements would change. To determine these
changes, divide the number of precipitation per hour by the number of downspouts. Remember that water is extremely heavy and heavy rainfall can add a lot of weight to the roof surface.
The size of the gutters can and should be calculated and installed in accordance with the requirements of generally accepted building codes. The use of IPC or UPC manuals provides an excellent way to
correctly size a gutter system. It is important to note that the CPI and UPC tables differ considerably. The IPC table contains the square footage of the roof area and not the amount of rain that can
fall on the roof. Do not confuse the numbers with the same numbers in the table in each manual.
What's the best way to collect rain? Why do I need to collect rainwater? Do I need pumps to collect rainwater? Can I use drip irrigation or rainwater soaker hoses? What size garden can I water? How
big are rain barrels? I want more pressure, how should I increase it? In the gutter installation industry, we've done the math. There's a basic formula we use to estimate the amount of rainwater a
roof will accumulate based on its size. For every 1000 square feet of roof that receives an inch of rain, 620 gallons of water will flow through the gutters. Consequently, the gutter system must be
designed to handle up to 110 gallons per minute or 6600 gallons per hour.
You'll use this GPM calculation to select the appropriate gutter and drain size in step 4: Calculate water handling capacity for gutter and drain options. It depends on many factors, but usually the
gutter company determines the slope of the gutters once installed. The water capacity values of a gutter and downspout system are calculated based on the height (flow) of the water that a gutter can
hold, the outlet size (hole) of the drain and the number of downspouts used. To avoid this, many homeowners choose to invest in leafless gutters that trap objects at the top of the gutters so that
they cannot enter the passage and obstruct the flow of water.
There are several different factors that help determine the size of the gutters each home needs, including the average maximum amount of rainfall in the area, so it can be difficult to know what size
of gutter will be the most cost-effective for your specific home. The debris that accumulates on the top is blown away by the wind after drying, which means that these gutters never clog up like
gutters do traditional.
Leave a Comment | {"url":"https://www.wefixuglygutters.com/how-much-water-can-gutters-handle","timestamp":"2024-11-13T17:54:29Z","content_type":"text/html","content_length":"102302","record_id":"<urn:uuid:da8f66be-11c3-4437-911c-c624e540721f>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00656.warc.gz"} |
Beginning Multiplication Worksheets
Mathematics, especially multiplication, creates the keystone of numerous scholastic techniques and real-world applications. Yet, for numerous students, understanding multiplication can present an
obstacle. To address this difficulty, teachers and moms and dads have embraced an effective tool: Beginning Multiplication Worksheets.
Intro to Beginning Multiplication Worksheets
Beginning Multiplication Worksheets
Beginning Multiplication Worksheets -
Meanwhile older students prepping for a big exam will want to print out our various timed assessment and word problem multiplication worksheets Good Times Await with Multiplication Worksheets Most
children struggle with multiplication for a reason It is a really difficult skill to master And just when a kid gains a firm grasp on one
Multiplication Worksheets for Beginners Multiplication worksheets for beginners are exclusively available on this page There are various exciting exercises like picture multiplication repeated
addition missing factors comparing quantities forming the products and lots more These pdf worksheets are recommended for 2nd grade through 5th grade
Relevance of Multiplication Practice Understanding multiplication is critical, laying a strong foundation for sophisticated mathematical concepts. Beginning Multiplication Worksheets supply
structured and targeted practice, cultivating a much deeper comprehension of this essential arithmetic procedure.
Development of Beginning Multiplication Worksheets
Beginning Multiplication Worksheets
Beginning Multiplication Worksheets
Download and printout our FREE worksheets HOLIDAY WORKSHEETS Free Secret Word Puzzle Worksheets New YearsWorksheets Martin Luther King Jr Worksheets Free Multiplication Worksheets Multiplication
Here is our free generator for multiplication and division worksheets This easy to use generator will create randomly generated multiplication worksheets for you to use Each sheet comes complete with
answers if required The areas the generator covers includes Multiplying with numbers to 5x5 Multiplying with numbers to 10x10
From traditional pen-and-paper exercises to digitized interactive layouts, Beginning Multiplication Worksheets have actually progressed, dealing with varied knowing designs and choices.
Sorts Of Beginning Multiplication Worksheets
Basic Multiplication Sheets Easy workouts concentrating on multiplication tables, aiding students construct a solid arithmetic base.
Word Problem Worksheets
Real-life circumstances incorporated into problems, enhancing critical thinking and application skills.
Timed Multiplication Drills Examinations made to enhance speed and accuracy, aiding in rapid psychological math.
Benefits of Using Beginning Multiplication Worksheets
Free Printable Multiplication Worksheets 2nd Grade
Free Printable Multiplication Worksheets 2nd Grade
40 Multiplication Worksheets These multiplication worksheets extend the Spaceship Math one minute timed tests with the x10 x11 and x12 facts Even if your school isn t practicing multiplication past
single digits these are valuable multiplication facts to learn for many time and geometry problems Extended Spaceship Math
Basic Multiplication Worksheets for Kids in Grade 3 Help your 3rd grader practice multiplication using this multiplication worksheet Ask the child to count the number of objects in a row or column
and the number of rows or columns Then ask them to multiply the two numbers to find the answer Calculate the number of objects in each picture
Enhanced Mathematical Abilities
Consistent technique sharpens multiplication efficiency, boosting total math capabilities.
Boosted Problem-Solving Abilities
Word problems in worksheets create analytical reasoning and strategy application.
Self-Paced Understanding Advantages
Worksheets accommodate private knowing rates, promoting a comfy and adaptable discovering setting.
Just How to Produce Engaging Beginning Multiplication Worksheets
Including Visuals and Colors Vivid visuals and shades capture interest, making worksheets visually appealing and involving.
Including Real-Life Situations
Associating multiplication to everyday scenarios includes significance and usefulness to exercises.
Tailoring Worksheets to Various Skill Levels Customizing worksheets based upon varying effectiveness degrees guarantees comprehensive learning. Interactive and Online Multiplication Resources Digital
Multiplication Equipment and Gamings Technology-based sources provide interactive understanding experiences, making multiplication appealing and pleasurable. Interactive Websites and Applications
Online platforms give varied and available multiplication technique, supplementing typical worksheets. Personalizing Worksheets for Different Knowing Styles Aesthetic Students Visual help and
diagrams aid understanding for students inclined toward visual understanding. Auditory Learners Spoken multiplication problems or mnemonics deal with students that comprehend principles with acoustic
ways. Kinesthetic Learners Hands-on activities and manipulatives sustain kinesthetic students in understanding multiplication. Tips for Effective Implementation in Discovering Uniformity in Practice
Normal method reinforces multiplication skills, promoting retention and fluency. Balancing Repetition and Range A mix of recurring exercises and diverse trouble styles keeps passion and
comprehension. Supplying Constructive Comments Feedback help in identifying areas of improvement, encouraging continued progression. Obstacles in Multiplication Practice and Solutions Motivation and
Interaction Hurdles Tedious drills can bring about disinterest; ingenious techniques can reignite inspiration. Overcoming Concern of Math Negative understandings around math can prevent progress;
developing a favorable discovering environment is vital. Impact of Beginning Multiplication Worksheets on Academic Performance Researches and Study Searchings For Research study suggests a positive
correlation between regular worksheet usage and enhanced mathematics performance.
Final thought
Beginning Multiplication Worksheets become flexible devices, fostering mathematical efficiency in students while fitting diverse knowing styles. From basic drills to interactive online sources, these
worksheets not just improve multiplication abilities however also promote essential reasoning and problem-solving abilities.
Beginning Multiplication Worksheets
Beginning Multiplication Worksheets With Pictures Free Printable
Check more of Beginning Multiplication Worksheets below
Beginning Multiplication Worksheets Victoria McCord s 1st Grade Math Worksheets
Beginning Multiplication Worksheets
Beginning Multiplication Worksheets
Beginner Multiplication Worksheets For Grade 2 Thekidsworksheet
Simple Multiplication Worksheets Superstar Worksheets
3rd Grade Math Multiplication Arrays Worksheets Common Core Mathematics Curriculum Grade
Basic Multiplication Worksheets Math Worksheets 4 Kids
Multiplication Worksheets for Beginners Multiplication worksheets for beginners are exclusively available on this page There are various exciting exercises like picture multiplication repeated
addition missing factors comparing quantities forming the products and lots more These pdf worksheets are recommended for 2nd grade through 5th grade
Multiplication Facts Worksheets Math Drills
These multiplication worksheets include some repetition of course as there is only one thing to multiply by Once students practice a few times these facts will probably get stuck in their heads for
life Some of the later versions include a range of focus numbers In those cases each question will randomly have one of the focus numbers in
Multiplication Worksheets for Beginners Multiplication worksheets for beginners are exclusively available on this page There are various exciting exercises like picture multiplication repeated
addition missing factors comparing quantities forming the products and lots more These pdf worksheets are recommended for 2nd grade through 5th grade
These multiplication worksheets include some repetition of course as there is only one thing to multiply by Once students practice a few times these facts will probably get stuck in their heads for
life Some of the later versions include a range of focus numbers In those cases each question will randomly have one of the focus numbers in
Beginner Multiplication Worksheets For Grade 2 Thekidsworksheet
Beginning Multiplication Worksheets
Simple Multiplication Worksheets Superstar Worksheets
3rd Grade Math Multiplication Arrays Worksheets Common Core Mathematics Curriculum Grade
Beginning Multiplication Worksheets With Pictures Times Tables Worksheets
Multiplication Worksheets Grade 2 Printable Lexia s Blog
Multiplication Worksheets Grade 2 Printable Lexia s Blog
Beginning Multiplication Worksheets
Frequently Asked Questions (Frequently Asked Questions).
Are Beginning Multiplication Worksheets appropriate for any age groups?
Yes, worksheets can be customized to different age and skill degrees, making them adaptable for various students.
Exactly how typically should students exercise making use of Beginning Multiplication Worksheets?
Constant practice is essential. Regular sessions, preferably a few times a week, can produce considerable enhancement.
Can worksheets alone enhance math abilities?
Worksheets are an useful tool yet needs to be supplemented with different discovering approaches for detailed skill advancement.
Are there on the internet systems supplying cost-free Beginning Multiplication Worksheets?
Yes, many instructional web sites offer free access to a wide range of Beginning Multiplication Worksheets.
How can parents sustain their kids's multiplication technique at home?
Encouraging regular practice, offering aid, and developing a positive understanding environment are beneficial actions. | {"url":"https://crown-darts.com/en/beginning-multiplication-worksheets.html","timestamp":"2024-11-05T06:27:27Z","content_type":"text/html","content_length":"28940","record_id":"<urn:uuid:9035c0d5-6394-4e61-8b31-979cc90a6813>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00497.warc.gz"} |
Texas Go Math Grade 7 Unit 1 Study Guide Review Answer Key
Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Unit 1 Study Guide Review Answer Key.
Texas Go Math Grade 7 Unit 1 Study Guide Review Answer Key
Texas Go Math Grade 7 Unit 1 Exercise Answer Key
Write each mixed number as a whole number or decimal. Classify each number by naming the set or sets to which it belongs: rational numbers, integers, or whole numbers. (Lessons 1.1, 1.2)
Question 1.
\(\frac{3}{4}\) _____________
0.75 belongs to the set of rational numbers.
Question 2.
\(\frac{8}{2}\) ______________
4 belongs to the set of integers. the set of whole numbers and the set of rational numbers.
Texas Go Math Grade 7 Unit 1 Answer Key Question 3.
\(\frac{11}{3}\) _______________
\(3 . \overline{6}\) belongs to the set of rational numbers.
Question 4.
\(\frac{5}{2}\) _______________
2.5 belongs to the set of rational numbers.
Find each sum or difference. (Lessons 1.3, 1.4)
Question 5.
-5 + 9.5 ____________
= 9.5 – 5
= 4.5
Question 6.
\(\frac{1}{6}\) + (-\(\frac{5}{6}\)) ____________
= \(\frac{1}{6}\) – \(\frac{5}{6}\)
= –\(\frac{4}{6}\)
= –\(\frac{2}{3}\)
Question 7.
-0.5 + (-8.5) _______________
= -0.5 – 8.5
= -9
Question 8.
-3 – (-8) ___________
= -3 + 8
= 5
Question 9.
5.6 – (-3.1) _________
= 5.6 + 3.1
= 8.7
Unit 1 End of Unit Assessment Grade 7 Answer Key Question 10.
3\(\frac{1}{2}\) – 2\(\frac{1}{4}\) _____________
Write mixed fractions as proper, then find a common denominator.
= \(\frac{7}{2}\) – \(\frac{9}{4}\)
= \(\frac{14-9}{4}\)
= \(\frac{5}{4}\)
Find each product or quotient. (Lessons 1.5, 1.6)
Question 11.
-9 × (-5) __________
Product will be positive because signs are the same.
= 9 × 5
= 45
Question 12.
0 × (-7) ____________
Any number multiplied by 0 is equal to 0.
= 0 × (-7)
= 0
Question 13.
-8 × 8 _____________
The product will be negative because signs are different
= -(8 × 8)
= -64
Question 14.
– \(\frac{56}{8}\) _______________
The quotient will be negative, because signs are different
= –\(\frac{56}{8}\)
= -7
Question 15.
\(\frac{-130}{-5}\) _____________
The quotient will be positive, because signs are same.
= \(\frac{130}{5}\)
= 26
Unit 1 The Number System Answer Key 7th Grade Question 16.
\(\frac{34.5}{1.5}\) ______________
Write decimal numbers as fractions:
Write complex fractions using division:
\(\frac{345}{10}\) ÷ \(\frac{15}{10}\)
Write using multiplication:
\(\frac{345}{10}\) × \(\frac{10}{15}\) = 23
Question 17.
–\(\frac{2}{5}\) (-\(\frac{1}{2}\)) (-\(\frac{5}{6}\)) ______________
Find the product of first 2 factors. Both are negative, so the product is positive.
\(\frac{2}{5}\) (\(\frac{1}{2}\)) = \(\frac{1}{5}\)
Multiply the result by the third factor. One is negative, one is positive, so the product is negative.
–\(\frac{1}{5}\)(\(\frac{5}{6}\)) = –\(\frac{1}{6}\)
Question 18.
(\(\frac{1}{5}\)) (-\(\frac{5}{7}\)) (\(\frac{3}{4}\)) _____________
Find the product of first 2 factors. One is negative, the other positive, so the product is negative.
–\(\frac{1}{5}\) (\(\frac{5}{7}\)) = –\(\frac{1}{7}\)
Multiply the result by the third factor. One is negative, one is positive, so the product is negative.
–\(\frac{1}{7}\) (\(\frac{3}{4}\)) = –\(\frac{3}{28}\)
Question 19.
Lei withdrew $50 from her bank account every day for a week. What was the change in her account in that week?
Use negative number to represent withdrawal.
Find 7 × (-50):
7 × (-50) = -350
The change on Lei’s account that week was -$350.
Question 20.
In 5 minutes, a seal descended 24 feet. What was the average rate of change in the seal’s elevation per minute?
Use negative number to represent descent in feet.
Find \(\frac{-24}{5}\):
\(\frac{-24}{5}\) = -4.8
Sears change in elevation is -4.8 feet per minute.
Texas Go Math Grade 7 Unit 1 Performance Task Answer Key
Question 1.
CAREERS IN MATH Urban Planner Armand is an urban planner, and he has proposed a site for a new town library. The site is between City Hall and the post office on Main Street.
The distance between City Hall and the post office is 6\(\frac{1}{2}\) miles. The library site is 1\(\frac{1}{4}\) miles closer to City Hall than it is to the post office.
a. Write 6\(\frac{1}{2}\) miles and 1\(\frac{1}{4}\) miles as decimals.
First, write \(\frac{1}{2}\) and \(\frac{1}{4}\) as decimals.
Then, add 6 and 1 to the result respectively.
6\(\frac{1}{2}\) = 6 + 0.5 = 6.5 miles
1\(\frac{1}{4}\) = 1 + 0.25 = 1.25 miles
b. Let d represent the distance from City Hall to the library site. Write an expression for the distance from the library site to the post office.
d* = distance from Library site to the Post Office
d* = d + 1.25
c. Write an equation that represents the following statement: The distance from City Hall to the library site plus the distance from the library site to the post office is equal to the distance from
City Hall to the post office.
d + d* = 6.5
d. Solve your equation from part c to determine the distance from City Hall to the library site, and the distance from the post office to the library site.
d + d* = 6.5
d + d + 1.25 = 6.5
2d = 5.25
d = 2.625
d* = d + 1.25
d* = 2.625 + 1.25
d* = 3.875
Unit 1 End of Unit Assessment Grade 7 Answer Key Math Question 2.
Sumaya is reading a book with 240 pages. She has already read 90 pages. She plans to read 20 more pages each day until she finishes the book.
a. Sumaya writes the equation 330 = -20d to find the number of days she will need to finish the book. Identify the errors that Sumaya made.
First, Sumaya added 90 to 240 instead of subtracting 90 from 240. This is a mistake because if she read 90 pages, that means she has 90 pages Less to read, not more.
Second mistake is the negative sign. If the equation results in how many days more she has to read, it can not be negative.
b. Write and solve an equation to determine how many days Sumaya will need to finish the book. In your answer, count part of a day as a full day.
First, find out how many more pages she has to read:
240 – 90 = 150
Correct equation:
20d = 150
d = \(\frac{150}{20}\)
d = \(\frac{15}{2}\)
d = 7\(\frac{1}{2}\)
Sumaya will need 8 days to finish the book.
c. Estimate how many days you would need to read a book about the same length as Sumaya’s book. What information did you use to find the estimate?
Let the book have the same number of pages. I would, for example, read 10 pages a day.
10d = 330
d = 33
It would take me 33 days to read the book.
Question 3.
Jackson works as a veterinary technician and earns $12.20 per hour.
a. Jackson normally works 40 hours a week. In a normal week, what is his total pay before taxes and other deductions?
Find 40 × 12.20:
40 × 12.20 = 488
His total pay is $488.
b. Last week, Jackson was ill and missed some work. His total pay before deductions was $372.10. Write and solve an equation to find the number of hours Jackson worked.
Find 372.10 ÷ 12.20
372.10 ÷ 12.20 = 30.5
Jackson worked 30.5 hours last week.
c. Jackson records his hours each day on a time sheet. Last week when he was ill, his time sheet was incomplete. How many hours are missing? Show your work.
Find 40 – 30.5
40 – 30.5 = 9.5
Jackson missed 9.5 hours last week.
d. When Jackson works more than 40 hours in a week, he earns 1.5 times his normal hourly rate for each of the extra hours. Jackson worked 43 hours one week. What was his total pay before deductions?
Justify your answer.
His 40 hours pay is 488, as calculated iii a. He worked 43 – 40 = 3 hours overtime.
Find 3 × 12.20 × 1.5:
3 × 12.20 × 1.5 = 36.6 × 1.5
= 54.9
Now add 54.9 to his 10 hours pay.
Find 488 + 54.9:
488 + 54.9 = 5-12.9
Jackson’s pay that week was $542.9.
e. What is a reasonable range for Jackson’s expected yearly pay before deductions? Describe any assumptions you made in finding your answer.
Let’s say that Jackson will be sick couple of days in a year, and he will work overtime couple of days in a year. When it all adds up, assumption is he will work 40 hours a week on average. There
are 52 weeks in a year.
Find 488 × 52:
488 × 52 = 25376
Jackson will probably earn somewhere in between $25000 – $26000.
Texas Go Math Grade 7 Unit 1 Mixed Review Texas Test Prep Answer Key
Selected Response
Question 1.
What is -6\(\frac{9}{16}\) written as a decimal?
A. -6.625
B. -6.5625
C. -6.4375
D. -6.125
B. -6.5625
First, write \(\frac{9}{16}\) as a decimal.
Then, add 6 to the result.
6 + 0.5625 = 6.5625
Now, since the starting number was negative, this one has to be negative too.
-6\(\frac{9}{16}\) = -6.5625
7th Grade Unit 1 Performance Task Answer Key Question 2.
Working together, 6 friends pick 14\(\frac{2}{5}\) pounds of pecans at a pecan farm. They divide the pecans equally among themselves. How many pounds does each friend get?
A. 20\(\frac{2}{5}\) pounds
B. 8\(\frac{2}{5}\) pounds
C. 4\(\frac{3}{5}\) pounds
D. 2\(\frac{2}{5}\) pounds
D. 2\(\frac{2}{5}\) pounds
Start with dividing 14\(\frac{2}{5}\) by 6:
14\(\frac{2}{5}\) ÷ 6
Write mixed fraction as proper fraction:
\(\frac{72}{5}\) ÷ 6
Write using multiplication:
\(\frac{72}{5}\) × \(\frac{1}{6}\) = \(\frac{12}{5}\)
= 2\(\frac{2}{5}\)
Each friend gets 2\(\frac{2}{5}\) pounds.
Question 3.
What is the value of (-3.25)(-1.56)?
A. -5.85
B. -5.07
C. 5.07
D. 5.85
C. 5.07
The product will, be positive, because both factors are negative:
= 3.25(1.56)
= 5.07
Question 4.
Ruby ate \(\frac{1}{3}\) of a pizza, and Angie ate \(\frac{1}{5}\) of the pizza. How much of the pizza did they eat in all?
A. \(\frac{1}{15}\) of the pizza
B. \(\frac{1}{8}\) of the pizza
C. \(\frac{3}{8}\) of the pizza
D. \(\frac{8}{15}\) of the pizza
D. \(\frac{8}{15}\) of the pizza
We have to add how much Ruby ate, and how much Angie ate
\(\frac{1}{3}\) + \(\frac{1}{5}\) = \(\frac{5+3}{15}\)
= \(\frac{8}{15}\)
Ruby and Angie ate \(\frac{8}{15}\) of the pizza.
Question 5.
Jaime had $37 in his bank account on Sunday. The table shows his account activity for the next four days. What was the balance in Jaime’s account after his deposit on Thursday?
A. $57.49
B. $59.65
C. $94.49
D. $138.93
C. $94.49
Use positive numbers to represent the deposit. and negative numbers to represent withdrawal. Then, add it up to account balance before any deposits or withdrawals. 37.
37 + 17.42 – 12.60 – 9.62 + 62.29 = 54.42 – 12.60 – 9.62 + 62.29
= 41.82 – 9.62 + 62.29
= 32.2 + 62.29
= 94.49
The balance in Jaimes’s account on Friday was $94.49.
7th Grade Math Study Guide Answers Unit 1 Question 6.
A used motorcycle is on sale for $3,600. Erik makes an offer equal to of this price. How much does Erik offer for the motorcycle?
A. $4,800
B. $2,700
C. $2,400
D. $900
B. $2,700
Start by multiplying 3600 and \(\frac{3}{4}\):
3600 × \(\frac{3}{4}\) = 2700
Erik offers $2700.
Question 7.
To which set or sets does the number -18 belong?
A. integers only
B. rational numbers only
C. integers and rational numbers only
D. whole numbers, integers, and rational numbers
C. integers and rational numbers only
We can see that -18 does not belong in the set of whole numbers
Next, notice that -18 belongs to the set of integers. That implies it belongs in the set of rational numbers, since the set of integers is the subset of the set of rational numbers.
Question 8.
Mrs. Rodriguez is going to use 6\(\frac{1}{3}\) yards of material to make two dresses. The larger dress requires 3\(\frac{2}{3}\) yards of material. How much material will Mrs. Rodriguez have left to
use on the smaller dress?
A. 1\(\frac{2}{3}\) yards
B. 2\(\frac{1}{3}\) yards
C. 2\(\frac{2}{3}\) yards
D. 3\(\frac{1}{3}\) yards
C. 2\(\frac{2}{3}\) yards
Start by subtracting 3\(\frac{2}{3}\) = \(\frac{11}{3}\) from 6\(\frac{1}{3}\) = \(\frac{19}{3}\)
\(\frac{19}{3}\) – \(\frac{11}{3}\) = \(\frac{8}{3}\)
= 2\(\frac{2}{3}\)
Mrs. Rodriguez will have 2\(\frac{2}{3}\) yards of material to use on the smaller dress.
Grade 7 Unit 1 Practice Problems Answer Key Question 9.
Winslow buys 1.2 pounds of bananas. The bananas cost $1.29 per pound. To the nearest cent, how much does Winslow pay for the bananas?
A. $1.08
B. $1.20
C. $1.55
D. $2.49
C. $1.55
Start by multiplying 1.2 by 1.29:
1.2 × 1.29 = 1.548
≈ 1.55
Vins1ow pays $1.55 for the bananas.
Gridded Response
Question 10.
Roberta earns $7.65 per hour. How many hours does Roberta need to work to earn $24.48?
Given earning per hour = $7.65
Given total earning of Robert = $24.48
Hence, for the earning of $24.48, he needs to work for 3.2 hours
The table will be made as per the below instructions:
1st coLumn: mark * sign
2nd coLumn: mark 0
3rd coLumn: mark 0
4th column: mark 0
5th coLumn: mark 3
6th coLumn: mark 2
7th column: mark 0
7th Grade Math Unit 1 Test Study Guide Question 11.
What is the product of the following expression?
(-2.2)(1 .5)(-4.2)
Given expression in problem: (-2.2)(1.5)(-4.2)
(-2.2)(1.5)(-4.2) = (-3.3)( 4.2)
= 13.86
The table will be made as per below instructions:
1st column: mark * sign
2nd column : mark 0
3rd column: mark 0
4th column : mark 1
5th column : mark 3
6th column: mark 8
7th column: mark 6
Hot Tip! Correct answers in gridded problems can be positive or negative. Enter the negative skin in the first column when it is appropriate. Check your work!
Question 12.
Victor is ordering pizzas for a party. He would like to have \(\frac{1}{4}\) of a pizza for each guest. He can only order whole pizzas, not part of a pizza. If he expects 27 guests, how many pizzas
should he order?
Portion of pizza for each guest = \(\frac{1}{4}\)
Total number of guest in party = 27
Number of guest in 1 pizza = \(\frac{1}{\frac{1}{4}}\) = 4
Required pizza for party = \(\frac{27}{4}\) = 6.75
But it is mentioned in question that we can onLy order whole pizza.The required amount of pizza for 27 guest is 6.75, So Victor will order 7 pizzas.
The table will be made as per below instructions:
1st coLumn: mark – sign
2nd column : mark 0
3rd column: mark 0
4th column : mark 0
5th column : mark 7
6th column: mark 0
7th column: mark 0
Texas Go Math Grade 7 Unit 1 Vocabulary Preview Answer Key
Use the puzzle to preview key vocabulary from this unit. Unscramble the circled letters within found words to answer the riddle at the bottom of the page.
1. Any number that can be written as a ratio of two integers. (Lesson 1-1)
2. A group of items. (Lesson 1-2)
3. A set that is contained within another set. (Lesson 1-2)
4. Decimals in which one or more digits repeat infinitely. (Lesson 1-1)
5. The opposite of any number. (Lesson 13)
6. Decimals that have a finite number of digits. (Lesson 1-1)
Question 1.
Why were the two fractions able to settle their differences peacefully?
They were both ___ ___ ___ ___ ___ ___ ___ !
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Is There A Santa Claus?
As a result of an overwhelming lack of requests, and with research help from that renowned scientific journal SPY magazine (January, 1990) - I am pleased to present the annual scientific inquiry into
Santa Claus.
1. No known species of reindeer can fly. BUT there are 300,000 species of living organisms yet to be classified, and while most of these are insects and germs, this does not COMPLETELY rule out
flying reindeer which only Santa has ever seen.
2. There are 2 billion children (persons under 18) in the world. BUT since Santa doesn't (appear) to handle the Muslim, Hindu, Jewish and Buddhist children, that reduces the workload to 15% of the
total - 378 million according to Population Reference Bureau. At an average (census) rate of 3.5 children per household, that's 91.8 million homes. One presumes there's at least one good child in
3. Santa has 31 hours of Christmas to work with, thanks to the different time zones and the rotation of the earth, assuming he travels east to west (which seems logical). This works out to 822.6
visits per second. This is to say that for each Christian household with good children, Santa has 1/1000th of a second to park, hop out of the sleigh, jump down the chimney, fill the stockings,
distribute the remaining presents under the tree, eat whatever snacks have been left, get back up the chimney, get back into the sleigh and move on to the next house. Assuming that each of these
91.8 million stops are evenly distributed around the earth (which, of course we know to be false but for the purposes of our calculations we will accept), we are now talking about .78 miles per
household, a total trip of 75-1/2 million miles, not counting stops to do what most of us must do at least once every 31 hours, plus feeding and etc. This means that Santa's sleigh is moving at
650 miles per second, 3,000 times the speed of sound. For purposes of comparison, the fastest man-made vehicle on earth, the Ulysses space probe, moves at a pokey 27.4 miles per second - a
conventional reindeer can run, tops, 15 miles per hour.
4. The payload on the sleigh adds another interesting element. Assuming that each child gets nothing more than a medium-sized Lego set (2 pounds), the sleigh is carrying 321,300 tons, not counting
Santa, who is invariably described as overweight. On land, conventional reindeer can pull no more than 300 pounds. Even granting that "flying reindeer" (see point #1) could pull TEN TIMES the
normal amount, we cannot do the job with eight, or even nine. We need 214,200 reindeer. This increases the payload - not even counting the weight of the sleigh - to 353,430 tons. Again, for
comparison- this is four times the weight of the Queen Elizabeth.
5. 353,000 tons travelling at 650 miles per second creates enormous air resistance - this will heat the reindeer up in the same fashion as spacecraft re-entering the earth's atmosphere. The lead
pair of reindeer will absorb 14.3 QUINTILLION joules of energy. Per second. Each. In short, they will burst into flame almost instantaneously, exposing the reindeer behind them, and create
deafening sonic booms in their wake. The entire reindeer team will be vaporized within 4.26 thousandths of a second. Santa, meanwhile, will be subjected to centrifugal forces 17,500.06 times
greater than gravity. A 250-pound Santa (which seems ludicrously slim) would be pinned to the back of his sleigh by 4,315,015 pounds of force.
In conclusion - If Santa ever DID deliver presents on Christmas Eve, he's dead now! | {"url":"https://paganlibrary.com/humor/santa_disc.php","timestamp":"2024-11-13T14:39:29Z","content_type":"text/html","content_length":"16628","record_id":"<urn:uuid:0584a461-f7a4-49d8-97fe-523df5146173>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00773.warc.gz"} |
Advanced Encryption Standard
The more popular and widely adopted symmetric encryption algorithm likely to be encountered nowadays is the Advanced Encryption Standard (AES). It is found at least six time faster than triple DES.
A replacement for DES was needed as its key size was too small. With increasing computing power, it was considered vulnerable against exhaustive key search attack. Triple DES was designed to overcome
this drawback but it was found slow.
The features of AES are as follows −
• Symmetric key symmetric block cipher
• 128-bit data, 128/192/256-bit keys
• Stronger and faster than Triple-DES
• Provide full specification and design details
• Software implementable in C and Java
Operation of AES
AES is an iterative rather than Feistel cipher. It is based on ‘substitution–permutation network’. It comprises of a series of linked operations, some of which involve replacing inputs by specific
outputs (substitutions) and others involve shuffling bits around (permutations).
Interestingly, AES performs all its computations on bytes rather than bits. Hence, AES treats the 128 bits of a plaintext block as 16 bytes. These 16 bytes are arranged in four columns and four rows
for processing as a matrix −
Unlike DES, the number of rounds in AES is variable and depends on the length of the key. AES uses 10 rounds for 128-bit keys, 12 rounds for 192-bit keys and 14 rounds for 256-bit keys. Each of these
rounds uses a different 128-bit round key, which is calculated from the original AES key.
The schematic of AES structure is given in the following illustration −
Encryption Process
Here, we restrict to description of a typical round of AES encryption. Each round comprise of four sub-processes. The first round process is depicted below −
Byte Substitution (SubBytes)
The 16 input bytes are substituted by looking up a fixed table (S-box) given in design. The result is in a matrix of four rows and four columns.
Each of the four rows of the matrix is shifted to the left. Any entries that ‘fall off’ are re-inserted on the right side of row. Shift is carried out as follows −
• First row is not shifted.
• Second row is shifted one (byte) position to the left.
• Third row is shifted two positions to the left.
• Fourth row is shifted three positions to the left.
• The result is a new matrix consisting of the same 16 bytes but shifted with respect to each other.
Each column of four bytes is now transformed using a special mathematical function. This function takes as input the four bytes of one column and outputs four completely new bytes, which replace the
original column. The result is another new matrix consisting of 16 new bytes. It should be noted that this step is not performed in the last round.
The 16 bytes of the matrix are now considered as 128 bits and are XORed to the 128 bits of the round key. If this is the last round then the output is the ciphertext. Otherwise, the resulting 128
bits are interpreted as 16 bytes and we begin another similar round.
Decryption Process
The process of decryption of an AES ciphertext is similar to the encryption process in the reverse order. Each round consists of the four processes conducted in the reverse order −
• Add round key
• Mix columns
• Shift rows
• Byte substitution
Since sub-processes in each round are in reverse manner, unlike for a Feistel Cipher, the encryption and decryption algorithms needs to be separately implemented, although they are very closely
AES Analysis
In present day cryptography, AES is widely adopted and supported in both hardware and software. Till date, no practical cryptanalytic attacks against AES has been discovered. Additionally, AES has
built-in flexibility of key length, which allows a degree of ‘future-proofing’ against progress in the ability to perform exhaustive key searches.
However, just as for DES, the AES security is assured only if it is correctly implemented and good key management is employed. | {"url":"http://www.howcodex.com/cryptography/advanced_encryption_standard.htm","timestamp":"2024-11-07T03:05:50Z","content_type":"text/html","content_length":"43360","record_id":"<urn:uuid:1fb49a16-60ae-41fb-a541-e2e9d956ef99>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00225.warc.gz"} |
Dimensions Math® PK–5 for Homeschool
Dimensions Math® PK–5 is our flagship Singapore Math® curriculum. With its rigorous content and engaging visuals, it's easy to see why it's our most popular program.
Written by a team of Singapore math educators and experts with more than 100 years of combined classroom experience, Dimensions Math PK-5 provides a deep elementary math foundation. This is a
refined, comprehensive series that meets the needs of today’s students and educators.
Textbook lessons build on prior knowledge and develop concepts in an approachable way. Textbooks A and B for each grade correspond to the two halves of the school year.
1. Think: Stimulates interest in new concepts through a hands-on activity or problem.
2. Learn: Presents definitions and fully explains new concepts.
3. Do: Solidifies and deepens student understanding of concepts.
Workbooks offer independent practice through careful progression of exercise variation. Workbooks A and B for each grade correspond to the two halves of the school year.
4. Exercise: Provides additional problems in the workbook for students to master concepts.
5. Practice: Provides teachers with opportunities for consolidation, remediation, and assessment.
Tests help teachers systematically evaluate student progress. They align with the content of textbooks. Grades 1–5 have differentiated assessments: Test A focuses on key concepts and fundamental
problem-solving skills, while Test B focuses on the application of analytical skills and heuristics.
Home Instructor’s Guides
Dimensions Math Home Instructor’s Guides are an essential resource for teaching Dimensions Math in a home or one-on-one setting. The purpose of the Guides is to help educators understand concepts in
the Dimensions Math curriculum and teach as effectively as possible. They include extensive background notes, lesson suggestions, tips, and activities. Home Instructor’s Guides include answer keys
for corresponding Dimensions Math Textbooks and Workbooks.
Teacher's Guides
Teacher’s Guides are a comprehensive resource that help teachers understand the purpose of each lesson within the framework of the curriculum. They offer structure for thoughtfully guiding student
inquiry, and include detailed teaching notes and activities to achieve lesson objectives.
Home Instructor’s Printouts
Dimensions Math Home Instructor’s Printouts are the complete set of lesson and activity sheets needed for successful use of the Dimensions Math program at home. These printouts are referenced
throughout the Home Instructor’s Guide and used frequently in the program to demonstrate concepts and help hone skills. While they are available for free on our site, this ready-to-use resource means
you can skip printing at home and keep printouts organized in one place. The set includes multiple copies of sheets where necessary.
Dimensions Math At Home™ Video Subscription
Looking for a little more support in teaching math? This video subscription brings a professional Singapore math teacher into your home classroom and provides in-depth instruction covering all the
Dimensions Math material for an entire school year. Available for Grades 1–6.
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An Overview of Supervised Machine Learning Models
Supervised learning in machine learning is an algorithm attempting to model a target variable based on provided input data. The algorithm is provided with a collection of training data that includes
labels. The program will derive a rule from a large dataset to predict labels for fresh observations. Supervised learning algorithms are given historical data and tasked with identifying the most
effective predicting association.
Supervised learning algorithms may be categorized into two types: regression and classification algorithms. Supervised learning approaches based on regression aim to forecast outcomes using input
variables. Classification-based supervised learning approaches determine the category to which a batch of data objects belongs. Classification algorithms are based on probabilities, determining the
category with the highest chance that the dataset belongs to it. Regression algorithms predict the result of issues with a continuous collection of potential outcomes.
Academic and industrial researchers have utilized regression-based algorithms to create several asset pricing models. These models are utilized to forecast returns over different time frames and to
pinpoint important characteristics that influence asset returns. Regression-based supervised learning has several applications in portfolio management and derivatives pricing.
Classification algorithms have been utilized in several financial domains to anticipate categorical outcomes. These encompass fraud detection, default prediction, credit rating, directional asset
price movement projection, and Buy/Sell advice. Classification-based supervised learning is utilized in several applications within portfolio management and algorithmic trading.
An Overview of Supervised Learning Models
Classification predictive modeling involves predicting discrete class labels, while regression predictive modeling involves predicting continuous quantities. Both models use established factors to
predict outcomes and have a substantial amount of similarities.
Some models can serve for both classification and regression with little adjustments. The approaches mentioned are K-nearest neighbors, decision trees, support vector machines, ensemble bagging/
boosting methods, and artificial neural networks (including deep neural networks). Some models, including linear regression and logistic regression, are not suitable for both sorts of problems.
Fig: Models for regression and classification
The algorithms are designed to learn from data, and they are often used to predict the outcome of a task. The two main categories of supervised learning are:
• Regression: This is used for predicting continuous outputs, like weather forecasting or house price prediction.
• Classification: This is used for predicting discrete outputs, like spam filtering or image recognition.
The following are different categories of both
• Linear regression: This is a simple algorithm that learns a linear relationship between the input features and the output. It is a good choice for problems where the relationship between the
input and output is well-understood.
• Logistic regression: This is a variant of linear regression that is used for classification problems. It learns a linear relationship between the input features and the probability of a
particular class.
• Decision trees: These algorithms learn a tree-like structure that represents the decision process for making a prediction. They are easy to interpret and can be used for both regression and
classification problems.
• Random forests: These are ensembles of decision trees, which means that they combine the predictions of multiple decision trees to make a final prediction. They are often more accurate than
individual decision trees and are also less prone to overfitting.
• Support vector machines (SVMs): These algorithms learn a hyperplane that separates the data points into different classes. They are good for high-dimensional data and can be used for both
regression and classification problems.
• Naive Bayes: This is a simple algorithm that uses Bayes' theorem to make predictions. It is a good choice for problems where the features are independent of each other.
• K-nearest neighbors (KNN): This algorithm classifies data points based on the labels of their nearest neighbors. It is a simple and easy to implement algorithm, but it can be slow and
computationally expensive for large datasets.
• Neural networks: These are complex algorithms that are inspired by the structure of the human brain. They can learn complex relationships between the input and output data and are often used for
image recognition, natural language processing, and other challenging tasks.
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Various Second-Order Array Combinators that are operationally parallel in a way that can be exploited by the compiler.
The functions here are recognised specially by the compiler (or built on those that are). The asymptotic work and span is provided for each function, but note that this easily hides very substantial
constant factors. For example, scan is much slower than reduce, although they have the same asymptotic complexity.
Higher-order complexity
Specifying the time complexity of higher-order functions is tricky because it depends on the functional argument. We use the informal convention that W(f) denotes the largest (asymptotic) work of
function f, for the values it may be applied to. Similarly, S(f) denotes the largest span. See this Wikipedia article for a general introduction to these constructs.
Reminder on terminology
A function op is said to be associative if
(x `op` y) `op` z == x `op` (y `op` z)
for all x, y, z. Similarly, it is commutative if
x `op` y == y `op` x
The value o is a neutral element if
x `op` o == o `op` x == x
for any x.
val map 'a [n] 'x : (f: a -> x) -> (as: [n]a) -> *[n]x
val map1 'a [n] 'x : (f: a -> x) -> (as: [n]a) -> *[n]x
val map2 'a 'b [n] 'x : (f: a -> b -> x) -> (as: [n]a) -> (bs: [n]b) -> *[n]x
val map3 'a 'b 'c [n] 'x : (f: a -> b -> c -> x) -> (as: [n]a) -> (bs: [n]b) -> (cs: [n]c) -> *[n]x
val map4 'a 'b 'c 'd [n] 'x : (f: a -> b -> c -> d -> x) -> (as: [n]a) -> (bs: [n]b) -> (cs: [n]c) -> (ds: [n]d) -> *[n]x
val map5 'a 'b 'c 'd 'e [n] 'x : (f: a -> b -> c -> d -> e -> x) -> (as: [n]a) -> (bs: [n]b) -> (cs: [n]c) -> (ds: [n]d) -> (es: [n]e) -> *[n]x
val reduce [n] 'a : (op: a -> a -> a) -> (ne: a) -> (as: [n]a) -> a
val reduce_comm [n] 'a : (op: a -> a -> a) -> (ne: a) -> (as: [n]a) -> a
val hist 'a [n] : (op: a -> a -> a) -> (ne: a) -> (k: i64) -> (is: [n]i64) -> (as: [n]a) -> *[k]a
val reduce_by_index 'a [k] [n] : (dest: *[k]a) -> (f: a -> a -> a) -> (ne: a) -> (is: [n]i64) -> (as: [n]a) -> *[k]a
val reduce_by_index_2d 'a [k] [n] [m] : (dest: *[k][m]a) -> (f: a -> a -> a) -> (ne: a) -> (is: [n](i64, i64)) -> (as: [n]a) -> *[k][m]a
val reduce_by_index_3d 'a [k] [n] [m] [l] : (dest: *[k][m][l]a) -> (f: a -> a -> a) -> (ne: a) -> (is: [n](i64, i64, i64)) -> (as: [n]a) -> *[k][m][l]a
val scan [n] 'a : (op: a -> a -> a) -> (ne: a) -> (as: [n]a) -> *[n]a
val partition [n] 'a : (p: a -> bool) -> (as: [n]a) -> ?[k].([k]a, [n - k]a)
val partition2 [n] 'a : (p1: a -> bool) -> (p2: a -> bool) -> (as: [n]a) -> ?[k][l].([k]a, [l]a, [n - k - l]a)
val all [n] 'a : (f: a -> bool) -> (as: [n]a) -> bool
val any [n] 'a : (f: a -> bool) -> (as: [n]a) -> bool
val spread 't [n] : (k: i64) -> (x: t) -> (is: [n]i64) -> (vs: [n]t) -> *[k]t
val scatter 't [k] [n] : (dest: *[k]t) -> (is: [n]i64) -> (vs: [n]t) -> *[k]t
val scatter_2d 't [k] [n] [l] : (dest: *[k][n]t) -> (is: [l](i64, i64)) -> (vs: [l]t) -> *[k][n]t
val scatter_3d 't [k] [n] [o] [l] : (dest: *[k][n][o]t) -> (is: [l](i64, i64, i64)) -> (vs: [l]t) -> *[k][n][o]t
val filter [n] 'a : (p: a -> bool) -> (as: [n]a) -> *[]a
↑val map 'a [n] 'x: (f: a -> x) -> (as: [n]a) -> *[n]x
Apply the given function to each element of an array.
Work: O(n ✕ W(f))
Span: O(S(f))
↑val map1 'a [n] 'x: (f: a -> x) -> (as: [n]a) -> *[n]x
Apply the given function to each element of a single array.
Work: O(n ✕ W(f))
Span: O(S(f))
↑val map2 'a 'b [n] 'x: (f: a -> b -> x) -> (as: [n]a) -> (bs: [n]b) -> *[n]x
As map1, but with one more array.
Work: O(n ✕ W(f))
Span: O(S(f))
↑val map3 'a 'b 'c [n] 'x: (f: a -> b -> c -> x) -> (as: [n]a) -> (bs: [n]b) -> (cs: [n]c) -> *[n]x
As map2, but with one more array.
Work: O(n ✕ W(f))
Span: O(S(f))
↑val map4 'a 'b 'c 'd [n] 'x: (f: a -> b -> c -> d -> x) -> (as: [n]a) -> (bs: [n]b) -> (cs: [n]c) -> (ds: [n]d) -> *[n]x
As map3, but with one more array.
Work: O(n ✕ W(f))
Span: O(S(f))
↑val map5 'a 'b 'c 'd 'e [n] 'x: (f: a -> b -> c -> d -> e -> x) -> (as: [n]a) -> (bs: [n]b) -> (cs: [n]c) -> (ds: [n]d) -> (es: [n]e) -> *[n]x
As map3, but with one more array.
Work: O(n ✕ W(f))
Span: O(S(f))
↑val reduce [n] 'a: (op: a -> a -> a) -> (ne: a) -> (as: [n]a) -> a
Reduce the array as with op, with ne as the neutral element for op. The function op must be associative. If it is not, the return value is unspecified. If the value returned by the operator is an
array, it must have the exact same size as the neutral element, and that must again have the same size as the elements of the input array.
Work: O(n ✕ W(op))
Span: O(log(n) ✕ W(op))
Note that the complexity implies that parallelism in the combining operator will not be exploited.
↑val reduce_comm [n] 'a: (op: a -> a -> a) -> (ne: a) -> (as: [n]a) -> a
As reduce, but the operator must also be commutative. This is potentially faster than reduce. For simple built-in operators, like addition, the compiler already knows that the operator is
commutative, so plain reduce will work just as well.
Work: O(n ✕ W(op))
Span: O(log(n) ✕ W(op))
↑val hist 'a [n]: (op: a -> a -> a) -> (ne: a) -> (k: i64) -> (is: [n]i64) -> (as: [n]a) -> *[k]a
h = hist op ne k is as computes a generalised k-bin histogram h, such that h[i] is the sum of those values as[j] for which is[j]==i. The summation is done with op, which must be a commutative and
associative function with neutral element ne. If a bin has no elements, its value will be ne.
Work: O(k + n ✕ W(op))
Span: O(n ✕ W(op)) in the worst case (all updates to same position), but O(W(op)) in the best case.
In practice, linear span only occurs if k is also very large.
↑val reduce_by_index 'a [k] [n]: (dest: *[k]a) -> (f: a -> a -> a) -> (ne: a) -> (is: [n]i64) -> (as: [n]a) -> *[k]a
Like hist, but with initial contents of the histogram, and the complexity is proportional only to the number of input elements, not the total size of the histogram.
Work: O(n ✕ W(op))
Span: O(n ✕ W(op)) in the worst case (all updates to same position), but O(W(op)) in the best case.
In practice, linear span only occurs if k is also very large.
↑val reduce_by_index_2d 'a [k] [n] [m]: (dest: *[k][m]a) -> (f: a -> a -> a) -> (ne: a) -> (is: [n](i64, i64)) -> (as: [n]a) -> *[k][m]a
As reduce_by_index, but with two-dimensional indexes.
↑val reduce_by_index_3d 'a [k] [n] [m] [l]: (dest: *[k][m][l]a) -> (f: a -> a -> a) -> (ne: a) -> (is: [n](i64, i64, i64)) -> (as: [n]a) -> *[k][m][l]a
As reduce_by_index, but with three-dimensional indexes.
↑val scan [n] 'a: (op: a -> a -> a) -> (ne: a) -> (as: [n]a) -> *[n]a
Inclusive prefix scan. Has the same caveats with respect to associativity and complexity as reduce.
Work: O(n ✕ W(op))
Span: O(log(n) ✕ W(op))
↑val partition [n] 'a: (p: a -> bool) -> (as: [n]a) -> ?[k].([k]a, [n - k]a)
Split an array into those elements that satisfy the given predicate, and those that do not.
Work: O(n ✕ W(p))
Span: O(log(n) ✕ W(p))
↑val partition2 [n] 'a: (p1: a -> bool) -> (p2: a -> bool) -> (as: [n]a) -> ?[k][l].([k]a, [l]a, [n - k - l]a)
Split an array by two predicates, producing three arrays.
Work: O(n ✕ (W(p1) + W(p2)))
Span: O(log(n) ✕ (W(p1) + W(p2)))
↑val all [n] 'a: (f: a -> bool) -> (as: [n]a) -> bool
Return true if the given function returns true for all elements in the array.
Work: O(n ✕ W(f))
Span: O(log(n) + S(f))
↑val any [n] 'a: (f: a -> bool) -> (as: [n]a) -> bool
Return true if the given function returns true for any elements in the array.
Work: O(n ✕ W(f))
Span: O(log(n) + S(f))
↑val spread 't [n]: (k: i64) -> (x: t) -> (is: [n]i64) -> (vs: [n]t) -> *[k]t
r = spread k x is vs produces an array r such that r[i] = vs[j] where is[j] == i, or x if no such j exists. Intuitively, is is an array indicating where the corresponding elements of vs should be
located in the result. Out-of-bounds elements of is are ignored. In-bounds duplicates in is result in unspecified behaviour - see hist for a function that can handle this.
Work: O(k + n)
Span: O(1)
↑val scatter 't [k] [n]: (dest: *[k]t) -> (is: [n]i64) -> (vs: [n]t) -> *[k]t
Like spread, but takes an array indicating the initial values, and has different work complexity.
Work: O(n)
Span: O(1)
↑val scatter_2d 't [k] [n] [l]: (dest: *[k][n]t) -> (is: [l](i64, i64)) -> (vs: [l]t) -> *[k][n]t
scatter_2d as is vs is the equivalent of a scatter on a 2-dimensional array.
Work: O(n)
Span: O(1)
↑val scatter_3d 't [k] [n] [o] [l]: (dest: *[k][n][o]t) -> (is: [l](i64, i64, i64)) -> (vs: [l]t) -> *[k][n][o]t
scatter_3d as is vs is the equivalent of a scatter on a 3-dimensional array.
Work: O(n)
Span: O(1)
↑val filter [n] 'a: (p: a -> bool) -> (as: [n]a) -> *[]a
Remove all those elements of as that do not satisfy the predicate p.
Work: O(n ✕ W(p))
Span: O(log(n) ✕ W(p)) | {"url":"https://futhark-lang.org/docs/prelude/doc/prelude/soacs.html","timestamp":"2024-11-13T14:23:24Z","content_type":"text/html","content_length":"30840","record_id":"<urn:uuid:4cd2d261-af95-461b-8060-e56404d77edf>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00046.warc.gz"} |
Key: "S:" = Show Synset (semantic) relations, "W:" = Show Word (lexical) relations
Display options for sense: (gloss) "an example sentence"
• S: (n) mathematical process, mathematical operation, operation ((mathematics) calculation by mathematical methods) "the problems at the end of the chapter demonstrated the mathematical processes
involved in the derivation"; "they were learning the basic operations of arithmetic" | {"url":"http://wordnetweb.princeton.edu/perl/webwn/webwn?o2=&o0=1&o8=1&o1=1&o7=&o5=&o9=&o6=&o3=&o4=&s=mathematical+process&i=0&h=0","timestamp":"2024-11-04T22:20:05Z","content_type":"application/xhtml+xml","content_length":"5775","record_id":"<urn:uuid:d75a5066-829b-4878-a812-8ba00f8e44ae>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00256.warc.gz"} |
The exponential model describes population growth in an idealized, unlimited environment
Population Ecology
In animals, parental care of smaller broods may facilitate survival of offspring.
Some plants, like the dandelion, produce a large number of small seeds, ensuring that at least some of them will grow and eventually reproduce.
Other types of plants, like the coconut tree, produce a moderate number of large seeds that provide a large store of energy that will help seedlings become established.
Slide 16
Variation in the size of seed crops in plants
(a) Dandelion
(b) Coconut palm
Slide 17
The exponential model describes population growth in an idealized, unlimited environment
It is useful to study population growth in an idealized situation.
Idealized situations help us understand the capacity of species to increase and the conditions that may facilitate this growth.
Slide 18
Zero population growth occurs when the birth rate equals the death rate.
Most ecologists use differential calculus to express population growth as growth rate at a particular instant in time:
where N = population size, t = time, and r = per capita rate of increase = birth – death
Slide 19
Exponential Growth
Exponential population growth is population increase under idealized conditions.
Under these conditions, the rate of reproduction is at its maximum, called the intrinsic rate of increase.
Exponential population growth results in a J-shaped curve
Exponential Growth is not sustainable.
Slide 20
Exponential Growth Model
Number of generations
Population size (N)
Slide 21
The J-shaped curve of exponential growth characterizes some rebounding populations
Elephant population
Slide 22
The logistic model describes how a population grows more slowly as it nears its carrying capacity
Exponential growth cannot be sustained for long in any population. A more realistic population model limits growth by incorporating carrying capacity.
Carrying capacity (K) is the maximum population size the environment can support.
In the logistic population growth model, the rate of increase declines as carrying capacity is reached. | {"url":"https://www.sliderbase.com/spitem-1478-3.html","timestamp":"2024-11-12T07:26:43Z","content_type":"text/html","content_length":"19568","record_id":"<urn:uuid:6d603739-deea-40b0-9a51-5a3983bfe683>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00277.warc.gz"} |
Successive Overrelaxation Method -- from Wolfram MathWorld
The successive overrelaxation method (SOR) is a method of solving a linear system of equations Gauss-Seidel method. This extrapolation takes the form of a weighted average between the previous
iterate and the computed Gauss-Seidel iterate successively for each component,
In matrix terms, the SOR algorithm can be written as
where the matrices
If Gauss-Seidel method. A theorem due to Kahan (1958) shows that SOR fails to converge if
In general, it is not possible to compute in advance the value of | {"url":"https://mathworld.wolfram.com/SuccessiveOverrelaxationMethod.html","timestamp":"2024-11-07T07:21:11Z","content_type":"text/html","content_length":"60659","record_id":"<urn:uuid:05d31f54-5ccf-4ad5-b0f8-d9fef4855269>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00054.warc.gz"} |
Chapter 10 Designing for floods: flood hydrology | Hydraulics and Water Resources: Examples Using R
Chapter 10 Designing for floods: flood hydrology
Flood hydrology is generally the description of how frequently a flood of a certain level will be exceeded in a specified period. This was discussed briefly in the section on precipitation frequency,
Section 8.3.
The hydromisc package will need to be installed to access some of the data used below. If it is not installed, do so following the instructions on the github site for the package.
10.1 Engineering design requires probability and statistics
Before diving into peak flow analysis, it helps to refresh your background in basic probability and statistics. Some excellent resources for this using R as the primary tool are:
Rather than repeat what is in those references, a couple of short demonstrations here will show some of the skills needed for flood hydrology. The first example illustrates binomial probabilities,
which are useful for events with only two possible outcomes (e.g., a flood happens or it doesn’t), where each outcome is independent and probabilities of each are constant. R functions for
distributions use a first letter to designate what it returns: d is the density, p is the (cumulative) distribution, q is the quantile, r is a random sequence.
In R the defaults for probabilities are to define them as \(P[X~\le~x]\), or a probability of non-exceedance. Recall that a probability of exceedance is simply 1 - (probability of non-exceedance), or
\(P[X~\gt~x] ~=~ 1-P[X~\le~x]\). In R, for quantiles or probabilities (using functions beginning with q or p like pnorm or qlnorm) setting the argument lower.tail to FALSE uses a probability of
exceedance instead of non-exceedance.
Example 10.1 A temporary dam is constructed while a repair is built. It will be in place 5 years and is designed to protect against floods up to a 20-year recurrence interval (i.e., there is a \(p=\
frac{1}{20}=0.05\), or 5% chance, that it will be exceeded in any one year). What is the probability of (a) no failure in the 5 year period, and (b) at least two failures in 5 years.
# (a)
ans1 <- dbinom(0, 5, 0.05)
cat(sprintf("Probability of exactly zero occurrences in 5 years = %.4f %%",100*ans1))
#> Probability of exactly zero occurrences in 5 years = 77.3781 %
# (b)
ans2 <- 1 - pbinom(1,5,.05) # or pbinom(1,5,.05, lower.tail=FALSE)
cat(sprintf("Probability of 2 or more failures in 5 years = %.2f %%",100*ans2))
#> Probability of 2 or more failures in 5 years = 2.26 %
While the next example uses normally distributed data, most data in hydrology are better described by other distributions.
Example 10.2 Annual average streamflows in some location are normally distributed with a mean annual flow of 20 m\(^3\)/s and a standard deviation of 6 m\(^3\)/s. Find (a) the probability of
experiencing a year with less than (or equal to) 10 m\(^3\)/s, (b) greater than 32 m\(^3\)/s, and (c) the annual average flow that would be expected to be exceeded 10% of the time.
# (a)
ans1 <- pnorm(10, mean=20, sd=6)
cat(sprintf("Probability of less than 10 = %.2f %%",100*ans1))
#> Probability of less than 10 = 4.78 %
# (b)
ans2 <- pnorm(32, mean=20, sd=6, lower.tail = FALSE) #or 1 - pnorm(32, mean=20, sd=6)
cat(sprintf("Probability of greater than or equal to 30 = %.2f %%",100*ans2))
#> Probability of greater than or equal to 30 = 2.28 %
# (c)
ans3 <- qnorm(.1, mean=20, sd=6, lower.tail=FALSE)
cat(sprintf("flow exceeded 10%% of the time = %.2f m^3/s",ans3))
#> flow exceeded 10% of the time = 27.69 m^3/s
# plot to visualize answers
x <- seq(0,40,0.1)
y<- pnorm(x,mean=20,sd=6)
xlbl <- expression(paste(Flow, ",", ~ m^"3"/s))
plot(x ,y ,type="l",lwd=2, xlab = xlbl, ylab= "Prob. of non-exceedance")
abline(v=10,col="black", lwd=2, lty=2)
abline(v=32,col="blue", lwd=2, lty=2)
abline(h=0.9,col="green", lwd=2, lty=2)
legend("bottomright",legend=c("(a)","(b)","(c)"),col=c("black","blue","green"), cex=0.8, lty=2)
10.2 Estimating floods when you have peak flow observations - flood frequency analysis
For an area fortunate enough to have a long record (i.e., several decades or more) of observations, estimating flood risk is a matter of statistical data analysis. In the U.S., data, collected by the
U.S. Geological Survey (USGS), can be accessed through the National Water Dashboard. Sometimes for discontinued stations it is easier to locate data through the older USGS map interface. For any
site, data may be downloaded to a file, and the peakfq (watstore) format, designed to be imported into the PeakFQ software, is easy to work with in R.
10.2.1 Installing helpful packages
The USGS has developed many R packages, including one for retrieval of data, dataRetrieval. Since this resides on CRAN, the package can be installed with (the use of ‘!requireNamespace’ skips the
installation if it already is installed):
Other USGS packages that are very helpful for peak flow analysis are not on CRAN, but rather housed in a USGS repository. The easiest way to install packages from that archive is to follow the
installation instructions at that repository. For the exercises below, install smwrGraphs and smwrBase following the instructions at those sites. The prefix smwr refers to their use in support of the
excellent reference Statistical Methods in Water Resources.
Lastly, the lmomco package has extensive capabilities to work with many forms of probability distributions, and has functions for calculating distribution parameters (like skew) that we will use.
10.2.2 Download, manipulate, and plot the data for a site
Using the older USGS site mapper, and specifying that inactive stations should also be included, many stations in the south Bay Area in California are shown in Figure
While the data could be downloaded and saved locally through that link, it is convenient here to use the dataRetrieval command.
The data used here are also available as part of the hydromisc package, and may be obtained by typing hydromisc::Qpeak_download.
It is always helpful to look at the downloaded data frame before doing anything with it. There are many columns that are not needed or that have repeated information. There are also some rows that
have no data (‘NA’ values). It is also useful to change some column names to something more intuitive. We will need to define the water year (a water year begins October 1 and ends September 30).
Qpeak <- Qpeak_download[!is.na(Qpeak_download$peak_dt),c('peak_dt','peak_va')]
colnames(Qpeak)[colnames(Qpeak)=="peak_dt"] <- "Date"
colnames(Qpeak)[colnames(Qpeak)=="peak_va"] <- "Peak"
Qpeak$wy <- smwrBase::waterYear(Qpeak$Date)
The data have now been simplified so that can be used more easily in the subsequent flood frequency analysis. Data should always be plotted, which can be done many ways. As a demonstration of
highlighting specific years in a barplot, the strongest El Niño years (in 1930-2002) from NOAA Physical Sciences Lab can be highlighted in red.
xlbl <- "Water Year"
ylbl <- expression("Peak Flow, " ~ ft^{3}/s)
nino_years <- c(1983,1998,1992,1931,1973,1987,1941,1958,1966, 1995)
cols <- c("blue", "red")[(Qpeak$wy %in% nino_years) + 1]
barplot(Qpeak$Peak, names.arg = Qpeak$wy, xlab = xlbl, ylab=ylbl, col=cols)
10.2.3 Flood frequency analysis
The general formula used for flood frequency analysis is Equation (10.1). \[$$y=\overline{y}+Ks_y \tag{10.1}$$\] where y is the flow at the designated return period, \(\overline{y}\) is the mean of
all \(y\) values and \(s_y\) is the standard deviation. In most instances, \(y\) is a log-transformed flow; in the US a base-10 logarithm is generally used. \(K\) is a frequency factor, which is a
function of the return period, the parent distribution, and often the skew of the y values. The guidance of the USGS (as in Guidelines for Determining Flood Flow Frequency, Bulletin 17C) (England,
J.F. et al., 2019) is to use the log-Pearson Type III (LP-III) distribution for flood frequency data, though in different settings other distributions can perform comparably. For using the LP-III
distribution, we will need several statistical properties of the data: mean, standard deviation, and skew, all of the log-transformed data, calculated as follows.
mn <- mean(log10(Qpeak$Peak))
std <- sd(log10(Qpeak$Peak))
g <- lmomco::pmoms(log10(Qpeak$Peak))$skew
With those calculated, a defined return period can be chosen and the flood frequency factors, from Equation (10.1), calculated for that return period (the example here is for a 50-year return
period). The qnorm function from base R and the qpearsonIII function from the smwrBase package make this straightforward, and K values for Equation (10.1) are obtained for a lognormal, Knorm, and
LP-III, Klp3.
Now the flood frequency equation (10.1) can be applied to calculate the flows associated with the 50-year return period for each of the distributions. Remember to take the anti-log of your answer to
return to standard units.
10.2.4 Creating a flood frequency plot
Different probability distributions can produce very different results for a design flood flow. Plotting the historical observations along with the distributions, the lognormal and LP-III in this
case, can help explain why they differ, and provide indications of which fits the data better.
We cannot say exactly what the probability of seeing any observed flood exceeded would be. However, given a long record, the probability can be described using the general “plotting position”
equation from Bulletin 17C, as in Equation (10.2). \[$$p_i=\frac{i-a}{n+1-2a} \tag{10.2}$$\] where n is the total number of data points (annual peak flows in this case), \(p_i\) is the exceedance
probability of flood observation i, where flows are ranked in descending order (so the largest observed flood has \(i=1\) ad the smallest is \(i=n\)). The parameter a is between 0 and 0.5. For
simplicity, the following will use \(a=0\), so the plotting Equation (10.2) becomes the Weibull formula, Equation (10.3). \[$$p_i=\frac{i}{n+1} \tag{10.3}$$\]
While not necessary, to add probabilities to the annual flow sequence we will create a new data frame consisting of the observed peak flows, sorted in descending order.
This can be done with fewer commands, but here is an example where first a rank column is created (1=highest peak in the record of N years), followed by adding columns for the exceedance and
non-exceedence probabilities:
df_pp$rank <- as.integer(seq(1:length(df_pp$Obs_peak)))
df_pp$exc_prob <- (df_pp$rank/(1+length(df_pp$Obs_peak)))
df_pp$ne_prob <- 1-df_pp$exc_prob
For each of the non-exceedance probabilities calculated for the observed peak flows, use the flood frequency equation (10.1) to estimate the peak flow that would be predicted by a lognormal or LP-III
distribution. This is the same thing that was done above for a specified return period, but now it will be “applied” to an entire column.
df_pp$LN_peak <- mapply(function(x) {10^(mn+std*qnorm(x))}, df_pp$ne_prob)
df_pp$LP3_peak <- mapply(function(x) {10^(mn+std*smwrBase::qpearsonIII(x, skew=g))},df_pp$ne_prob)
There are many packages that create probability plots (see, for example, the versatile scales package for ggplot2). For this example the USGS smwrGraphs package is used. First, for aesthetics, create
x- and y- axis labels.
The smwrGraphs package works most easily if it writes output directly to a file, a PNG file in this case, using the setPNG command; the file name and its dimensions in inches are given as arguments,
and the PNG device is opened for writing. This is followed by commands to plot the data on a graph. Technically, the data are plotted to an object here is called prob.pl. The probPlot command plots
the observed peaks as points, where the alpha argument is the a in Equation (10.2). Additional points or lines are added with the addXY command, used here to add the LN and LP3 data as lines (one
solid, one dashed). Finally, a legend is added (the USGS refers to that as an “Explanation”), and the output PNG file is closed with the dev.off() command.
smwrGraphs::setPNG("probplot_smwr.png",6.5, 3.5)
#> width height
#> 6.5 3.5
#> [1] "Setting up markdown graphics device: probplot_smwr.png"
prob.pl <- smwrGraphs::probPlot(df_pp$Obs_peak, alpha = 0.0, Plot=list(what="points",size=0.05,name="Obs"), xtitle=xlbl, ytitle=ylbl)
prob.pl <- smwrGraphs::addXY(df_pp$ne_prob,df_pp$LN_peak,Plot=list(what="lines",name="LN"),current=prob.pl)
prob.pl <- smwrGraphs::addXY(df_pp$ne_prob,df_pp$LP3_peak,Plot=list(what="lines",type="dashed",name="LP3"),current=prob.pl)
#> png
#> 2
The output won’t be immediately visible in RStudio – navigate to the file and click on it to view it. Figure
shows the output from the above commands.
10.2.5 Other software for peak flow analysis
Much of the analysis above can be achieved using the PeakFQ software developed by the USGS. It incorporates the methods in Bulletin 17C via a graphical interface and can import data in the watstore
format as discussed above in Section 10.2.
The USGS has also produced the MGBT R package to perform many of the statistical calculations involved in the Bulletin 17C procedures.
10.3 Estimating floods from precipitation
When extensive streamflow data are not available, flood risk can be estimated from precipitation and the characteristics of the area contributing flow to a point. While not covered here (or not
yet…), there has been extensive development of hydrological modeling using R, summarized in recent papers (Astagneau et al., 2021; Slater et al., 2019).
Straightforward application of methods to estimate peak flows or hydrographs resulting from design storms can by writing code to apply the Rational Formula (included in the VFS and hydRopUrban
packages, for example) or the NRCS peak flow method.
For more sophisticated analysis of water supply and drought, continuous modeling is required. A very good introduction to hydrological modeling in R, including model calibration and assessment, is
included in the Hydroinformatics at VT reference by JP Gannon. | {"url":"https://edm44.github.io/hydr-watres-book/designing-for-floods-flood-hydrology.html","timestamp":"2024-11-05T00:46:25Z","content_type":"text/html","content_length":"83400","record_id":"<urn:uuid:190a5fca-7edc-4277-b262-61bdb4b840b3>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00231.warc.gz"} |
What is the relation between pKa and pKb?
What is the relation between pKa and pKb?
A lower pKb value indicates a stronger base. pKa and pKb are related by the simple relation: pKa + pKb = 14.
What does pKa pKb equal?
The negative log of Ka, we know that this is equal to the pKa. The negative log of Ka was our definition for our pKa, and the negative log of Kb was our definition for pKb. pKa plus pKb is equal to
finally the negative log of Kw.
What is the difference between pKb and KB?
The base dissociation constants are interpreted just like the acid dissociation constants. A large Kb value means a base has largely dissociated and indicates a strong base. A small pKb value
indicates a strong base, while a large pKb value indicates a weak base.
What is difference between KA and pKa?
What is the relationship of pKa and Ka? The smaller the value of Ka, the larger the value of pKa, the weaker the acid. If the pH of a solution of a weak acid and the pKa are known, the ratio of the
concentration of the conjugate base to the concentration of the acid may be calculated.
What is the difference between KA and KB?
The acid dissociation constant (Ka) is a quantitative measure of the strength of an acid in solution while the base dissociation constant (Kb) is a measure of basicity—the base’s general strength.
Acids are classified as either strong or weak, based on their ionization in water.
How do you convert pKa to pKb?
Steps to Convert between pKa and pKb Step 1: Write the equation that relates pKa p K a and pKb p K b . It should look like the following: pKa+pKb=14 p K a + p K b = 14 . Step 2: Insert the given pK
value provided in the question.
What is the relationship between KA and KB?
The larger the value of Kb, the stronger the base, and the larger the value of Ka, the stronger the acid. By multiplying Ka by Kb, you receive the Kw, or the dissociation constant for water, which is
1.0 x 10^-14.
What is the relation between KA and KB?
What is the relationship between Ka and Kb? The acid dissociation constant is Ka. The stronger the base, the greater the value of Kb, and the stronger the acid, the larger the value of Ka. When we
multiply Ka by Kb, we get Kw, or the water dissociation constant, which is 1.0 x 10-14.
What is pKb?
pKb is the criterion used to estimate the alkalinity of the molecule. It is used to measure basic strength. The lesser the pKb is, the more potent the base will be. It is equivalent to the negative
logarithm of base dissociation constant, Kb. pKb = – log Kb.
What is the difference between Ka and KB?
What is the relationship between Ka and KB? | {"url":"https://www.tag-challenge.com/2022/12/22/what-is-the-relation-between-pka-and-pkb/","timestamp":"2024-11-13T11:03:57Z","content_type":"text/html","content_length":"38408","record_id":"<urn:uuid:4aac91f7-b5ba-476a-bd84-7d99a5b09c25>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00602.warc.gz"} |
Publications by Author: Chekroun, Mickaël D.
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Partial Differential Equations Formula: Properties, Chemical Structure and Uses
Partial Differential Equations Formula
Partial Differential Equation Formulas
The Partial Differential Equation Formula is an equation with one or more derivatives, where a derivative is a quantity that expresses the rate at which a variable changes over time. In general, the
equation that describes the functional dependency of one variable on several variables is the Partial Differential Equation Formula solution. Usually, the Partial Differential Equation Formula
contains constant terms that were absent from the initial differential equation. Applications frequently use functions to describe physical values, derivatives to indicate the rate at which those
quantities change, and differential equations to establish the connection between the three. A function that can be used to predict the behaviour of the original system, at least under some
restrictions, is produced by the solution of the differential equation.
A differential equation is one that has a function and its derivatives. It can be referred to as either an ordinary differential equation (ODE) or a Partial Differential Equation Formula, depending
on whether partial derivatives are present or not (PDE). On the website and mobile application for Extramarks, students may download the Partial Differential Equation Formula PDF.
What are Partial Differential Equation Formulas?
A differential equation containing several unknown functions and their partial derivatives is referred to as a (PDE). Numerous phenomena, including sound, heat, diffusion, electrostatics,
electrodynamics, fluid dynamics, elasticity, gravity, and quantum mechanics are all represented by it.
The Partial Differential Equation Formula notes and solutions are also available in Hindi. All the solutions are according to the CBSE NCERT norms and regulations, making it easier to understand and
comprehend for students. The Partial Differential Equation Formula notes are curated by experts while pertaining to the NCERT book norms, which will help students to study all concepts in the book
without any hassle.
Partial Differential Equation Formulas Definition
The idea of a PDE is exciting and full of surprises, but they are also regarded as being rather challenging. So let’s dissect the idea of a PDE into smaller parts and explore each one in depth in
order to completely grasp it. First, students need to understand what an equation is. A statement in which the values of the mathematical expressions are equal is called an equation.
The solutions for the PDE are made after extensive research and while taking into consideration the Previous Years Question papers.
Experts have made sure that the framework of the solutions are such that students find it easy to comprehend the ideas and concepts contained in the solutions. The notes and solutions for the PDE are
downloadable, meaning, students can download the notes and solutions for the
PDE through the extramarks website and mobile application.
Partial Differential Equation Formulas Example
While the notes and solutions on the Partial Differential Equation Formula are students friendly, they are also extremely versatile in nature. Experts keep on improving the kind of content they
believe will help students to comprehend the ideas contained in the solutions effectively.
The notes and solutions for the Partial Differential Equation Formula can be used before the final examinations to enhance student understanding of concepts and to retain the things they have learnt
through the Partial Differential Equation Formula study materials.
Partial Differential Equation Formulas Formula
If f is a linear function of u and its derivatives, the PDE is said to be linear. The basic PDE is provided by;
∂u/∂x (x, y) = 0
The relationship mentioned above suggests that the function u(x, y), which is the reduced form of the Partial Differential Equation Formula mentioned above, is independent of x. The highest
derivative term in the equation determines the order of the PDE.
Students can download the PDE notes which are extremely helpful while studying for their examinations. The PDE notes have been curated and compiled by the top subject-matter experts at Extramarks
while keeping in mind that it pertains strictly to the CBSE norms.
Order and Degree of Partial Differential Equation Formulas
A certain finite element technique may be applicable for a given PDE-described issue, depending on the functions of each type of PDE. The equation and various variables that include partial
derivatives with regard to the variables affect the answer. In mechanics, there are three different kinds of second-order PDE. Those are
• PDE elliptic
• Inverse PDE
• PDE hyperbolic
Take the following example: aux+buy+cu=0, u=u (x, y). If b2-ac0, which is used to represent the equations of elasticity without inertial components, is true for a particular point (x, y), the
equation is said to be elliptic. If the criterion b2-ac>0 is met, hyperbolic PDE can be used to model wave propagation. It ought to be true for the parabolic PDE when b2-ac=0. One illustration of a
parabolic PDE is the equation for heat conduction.
The notes and solutions based on the PDE provided by the Extramarks experts are also available in Hindi. Students from other boards can also refer to these notes in Hindi and thus understand the
concepts better leading to a better comprehension of the subject and concepts included in it.
Order of Partial Differential Equation Formulas
PDE is frequently used in applications and simulates how quickly a physical quantity changes in relation to time and place. Solve normally only functions with PDE with two independent variables at
this level of development.
The highest derivative that appears in a PDE determines its order. A first-order PDE is an equation from earlier.
If its derivatives satisfy the specified PDE, then the function is a solution.
The above-mentioned PDE solutions can be used in making personal notes for students, which will further help them not only understand the topic but also retain everything while writing about it.
The notes based on PDE can also be used for a quick revision before the examinations, leading to improved grades.
Degree of Partial Differential Equation Formulas
The highest order derivative’s power in the PDE serves as a proxy for the differential equation’s degree.
For the degree to be specified, the differential equation must be a polynomial equation in derivatives.
The solutions on PDE provided by Extramarks experts help students by providing them with step-by-step calculations for each problem in the solutions, leading to improved learning and better grades.
Partial Differential Equation Formulas Types
Partial Differential Equation Formula can be of several sorts, including
• Partial Differential Equation Formula of First Order
• Equation for Linear Partial Differential
• Equation for a Quasi-Linear Partial Differential
• Equation for Homogeneous Partial Differential
Equation of First-Order Differential
In mathematics, the first derivative of the unknown function with variables is all that is present in the first-order Partial Differential Equation Formula. This is how it is expressed:
F(x1,…, xm, u, ux1,…., uxm) = 0
Equation for Linear Partial Differential
Any PDE is referred to as a linear PDE if the dependent variable and all of its partial derivatives occur linearly, otherwise it is referred to as a nonlinear PDE. Examples (1) and (2) in the
previous example are considered linear equations, but examples (3) and (4) are considered non-linear equations.
Equation for a Quasi-Linear Partial Differential
When all PDE terms with the highest order derivatives of the dependent variables occur linearly and the coefficients of those terms are solely functions of lower-order derivatives of the dependent
variables, the PDE is said to be quasi-linear. The occurrence of words with lower-order derivatives, however, is not restricted. In the list above, example (3) is a quasi-linear equation.
Equation for Homogeneous Partial Differential
A Partial Differential Equation Formula (PDE) is said to be homogeneous if all of its terms do not contain the dependent variable or its partial derivatives. In the four examples above, Example (4),
in contrast to the previous three homogeneous equations, is non-homogeneous.
With the help of these notes provided by Extramarks experts, students will never need any external help. The notes provided by Extramarks based on the PDE teach children all the concepts as taught in
the classroom, but with a lot of flexibility.
First-Order Partial Differential Equation Formulas
The greatest partial derivatives of the unknown function are of the first order in the first-order Partial Differential Equation Formula. Both linear and non-linear ones are possible. These
variables’ derivatives cannot be squared or multiplied.
The notes and solutions provided by Extramarks experts on PDE are extremely helpful for teachers as well. They serve as fantastic teaching aid for teachers by helping and guiding them while also
making the teaching process easier.
Second-Order Partial Differential Equation Formulas
Many concepts taught and discussed in the solutions for PDE are also topics that students might study for competitive examinations. Therefore, students must make sure that they practice these
concepts in order to score well in competitive examinations.
The highest partial derivative of any order is found in second-order PDE. They might be non-linear, semi-linear, or linear equations. The complexity of linear second-order Partial Differential
Equation Formula is significantly higher than that of non- and semi-linear second-order Partial Differential Equation Formula.
Quasi Linear Partial Differential Equation Formulas
Only linear terms in quasilinear Partial Differential Equation Formula give rise to the greatest rank of partial derivatives. In physics and engineering, first-order quasi-linear Partial Differential
Equation Formula are frequently used to address a wide range of issues.
Students can practice the solutions for PDE in order to strengthen their basics and focus on achieving the best possible results.
Homogeneous Partial Differential Equation Formulas
The formulas and concepts that require extra focus and attention from students have been marked and highlighted in the solutions for PDE. Therefore, students need not worry about looking for formulas
and definitions and other important concepts for hours. Everything is available in one study material only.
The homogeneity or non-homogeneity of a PDE depends on the type of variables used in terms. A Partial Differential Equation Formula that includes the dependent variable and its partial derivatives is
known as a non-homogeneous PDE.
Partial Differential Equation Formulas Classification
The PDE has been categorised into three different parts, namely:
• Elliptic
• Parabolic
• Hyperbolic
• Parabolic Partial Differential Equation Formula: A parabolic Partial Differential Equation Formula is produced if B2 – AC = 0. The equation for heat conduction is an illustration of a parabolic
Partial Differential Equation Formula.
• Equations with Hyperbolic Partial Differential: When B2 – AC > 0, such an equation is produced. As wave propagation may be represented by such equations, the wave equation is an illustration of a
hyperbolic Partial Differential Equation Formula.
• Elliptic Partial Differential Equation Formula: The Partial Differential Equation Formula B2 – AC 0 are elliptic. An illustration of an elliptic Partial Differential Equation Formula is the
Laplace equation.
The notes based on the PDE provided by the Extramarks experts will allow students to stop depending on other resources to study and help facilitate and encourage self-study. Once students start
practising these solutions, they will notice positive outcomes that will help them clarify doubts independently.
Solving Partial Differential Equation Formulas
The finite element method (FEM), finite volume method (FVM), and finite difference method (FDM) are the three numerical techniques that are most frequently used to solve PDEs. Additionally, there is
a class of techniques known as mesh-free methods that were developed to address issues where the aforementioned techniques have limitations.
These solutions on the other hand can also serve as a teaching guide to refer to when helping their children study and prepare for their final examinations.
Partial Differential Equation Formulas Applications
Many disciplines, including mathematics, engineering, physics, and finance, use PDEs. Here are a few of their applications:
• A Partial Differential Equation Formula with the solution uxx = ut may be used to illustrate the idea of heat waves and how they spread.
• A Partial Differential Equation Formula with the solution uxx – uyy = 0 may also be used to describe the notion of light and sound waves, as well as how they propagate.
• PDEs are also used in the fields of economics and accounting. The Black-Scholes equation, for instance, is used to build financial models.
Examples on Partial Differential Equation Formulas
1. Reduce uxx + 5uxy + 6uyy = 0 to its canonical form and solve it.
Since, b2 − 4ac = 1 > 0 for the given equation, it is hyperbolic.
Let μ(x, y)=3x − y, η(x, y)=2x − y
μx = 3, ηx = 2
μy = −1, ηy = −1
u = u(μ(x, y), η(x, y))
ux = uμμx + uηηx = 3uμ + 2uη
uy = uμμy + uηηy = −uμ − uη
uxx = (3uμ + 2uη)x = 3(uμμμx + uμηηx) + 2(uημμx + uηηηx)
=9uμμ + 12uμη + 4uηη ……(1)
uxy = (3uμ + 2uη)y = 3(uμμμy + uμηηy) + 2(uημμy + uηηηy)
= −3uμμ − 5uμη − 2uηη .…(2)
uyy = −(uμ + uη)y = −(uμμμy + uμηηy + uημμy + uηηηy)
= uμμ + 2uμη + uηη .…(3)
Thus, the canonical form is given as: uμη = 0.
The general solution is: u(x, y) = F(3x − y) + G(2x − y).
Practice Questions on Partial Differential Equation Formulas
There are a number of practice questions available based on the PDE on the Extramarks website and mobile application, there are more than enough examples that have been stated wherever they’re needed
in the solutions that serve as an aid to solve these practice questions.
FAQs (Frequently Asked Questions)
1. What are PDEs, exactly?
Differential equations with an unknown function, several dependent and independent variables, and their partial derivatives are known as Partial Differential Equation Formula.
2. Do PDEs have linear behaviour?
Partial Differential Equation Formula do not always have to have linear solutions. Partial Differential Equation Formula can also be semi- or nonlinear. One can access the study material and learning
resources concerned with differential equations from the Extramarks website or mobile application. To be able to acquire the resources provided by Extramarks, students are required to register on the
Extramarks website. Extramarks also provides online live session conducted by experts on various topics for a better understanding of students. Students may also download the needed study material in
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BCIT Physics 0312 Textbook
Chapter 2 One-Dimensional Kinematics
• Describe the effects of gravity on objects in motion.
• Describe the motion of objects that are in free fall.
• Calculate the position and velocity of objects in free fall.
Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom.
By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.
The most remarkable and unexpected fact about falling objects is that, if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the
same constant acceleration, independent of their mass. This experimentally determined fact is unexpected, because we are so accustomed to the effects of air resistance and friction that we expect
light objects to fall slower than heavy ones.
Figure 1. A hammer and a feather will fall with the same constant acceleration if air resistance is considered negligible. This is a general characteristic of gravity not unique to Earth, as
astronaut David R. Scott demonstrated on the Moon in 1971, where the acceleration due to gravity is only 1.67 m/s^2.
In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball will reach the ground after a hard baseball dropped at the same time.
(It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, while friction between objects—such as between clothes
and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them. For the ideal situations of these first few chapters, an object falling without air
resistance or friction is defined to be in free-fall.
The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called the acceleration due to gravity. The acceleration due to gravity
is constant, which means we can apply the kinematics equations to any falling object where air resistance and friction are negligible. This opens a broad class of interesting situations to us. The
acceleration due to gravity is so important that its magnitude is given its own symbol, gg size 12{g} {}. It is constant at any given location on Earth and has the average value
[latex]\boldsymbol{g\:=\:9.80\textbf{ m/s}^2.}[/latex]
Although[latex]\boldsymbol{g}[/latex]varies from[latex]\boldsymbol{9.78\textbf{ m/s}^2}[/latex]to[latex]\boldsymbol{9.83\textbf{ m/s}^2}[/latex], depending on latitude, altitude, underlying
geological formations, and local topography, the average value of[latex]\boldsymbol{9.80\textbf{ m/s}^2}[/latex]will be used in this text unless otherwise specified. The direction of the acceleration
due to gravity is downward (towards the center of Earth). In fact, its direction defines what we call vertical. Note that whether the acceleration[latex]\boldsymbol{a}[/latex]in the kinematic
equations has the value[latex]\boldsymbol{+g}[/latex]or[latex]\boldsymbol{-g}[/latex]depends on how we define our coordinate system. If we define the upward direction as positive, then[latex]\
boldsymbol{a=-g=-9.80\textbf{ m/s}^2}[/latex], and if we define the downward direction as positive, then[latex]\boldsymbol{a=g=9.80\textbf{ m/s}^2}[/latex].
One-Dimensional Motion Involving Gravity
The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So we start by considering straight up and down
motion with no air resistance or friction. These assumptions mean that the velocity (if there is any) is vertical. If the object is dropped, we know the initial velocity is zero. Once the object has
left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude[latex]\boldsymbol{g}.[/
latex]We will also represent vertical displacement with the symbol[latex]\boldsymbol{y}[/latex]and use[latex]\boldsymbol{x}[/latex]for horizontal displacement.
Example 1: Calculating Position and Velocity of a Falling Object: A Rock Thrown Upward
A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s. The rock misses the edge of the cliff as it falls back to earth. Calculate the position
and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance.
Draw a sketch.
Figure 2.
We are asked to determine the position[latex]\boldsymbol{y}[/latex]at various times. It is reasonable to take the initial position[latex]\boldsymbol{y_0}[/latex]to be zero. This problem involves
one-dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since up is positive, and the rock is thrown upward, the
initial velocity must be positive too. The acceleration due to gravity is downward, so[latex]\boldsymbol{a}[/latex]is negative. It is crucial that the initial velocity and the acceleration due to
gravity have opposite signs. Opposite signs indicate that the acceleration due to gravity opposes the initial motion and will slow and eventually reverse it.
Since we are asked for values of position and velocity at three times, we will refer to these as[latex]\boldsymbol{y_1}[/latex]and[latex]\boldsymbol{v_1}[/latex];[latex]\boldsymbol{y_2}[/latex]and
[latex]\boldsymbol{v_2}[/latex]; and[latex]\boldsymbol{y_3}[/latex]and[latex]\boldsymbol{v_3}[/latex].
Solution for Position [latex]\boldsymbol{y_1}[/latex]
1. Identify the knowns. We know that[latex]\boldsymbol{y_0=0}[/latex];[latex]\boldsymbol{v_0=13.0\textbf{ m/s}}[/latex];[latex]\boldsymbol{a=-g=-9.80\textbf{ m/s}^2}[/latex]; and[latex]\boldsymbol{t=
1.00\textbf{ s}}[/latex].
2. Identify the best equation to use. We will use[latex]\boldsymbol{y=y_0+v_0t+\frac{1}{2}at^2}[/latex]because it includes only one unknown,[latex]\boldsymbol{y}[/latex](or[latex]\boldsymbol{y_1}[/
latex], here), which is the value we want to find.
3. Plug in the known values and solve for[latex]\boldsymbol{y_1}[/latex].
[latex]\boldsymbol{y_1=0+(13.0\textbf{ m/s})(1.00\textbf{ s})+}[/latex][latex]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{(-9.80\textbf{ m/s}^2)(1.00\textbf{ s})^2=8.10\textbf{ m}}[/latex]
The rock is 8.10 m above its starting point at[latex]\boldsymbol{t=1.00}[/latex]s, since[latex]\boldsymbol{y_1>y_0}[/latex]. It could be moving up or down; the only way to tell is to calculate[latex]
\boldsymbol{v_1}[/latex]and find out if it is positive or negative.
Solution for Velocity [latex]\boldsymbol{v_1}[/latex]
1. Identify the knowns. We know that[latex]\boldsymbol{y_0=0}[/latex];[latex]\boldsymbol{v_0=13.0\textbf{ m/s}}[/latex];[latex]\boldsymbol{a=-g=-9.80\textbf{ m/s}^2}[/latex]; and[latex]\boldsymbol{t=
1.00\textbf{ s}}[/latex]. We also know from the solution above that[latex]\boldsymbol{y_1=8.10\textbf{ m}}[/latex].
2. Identify the best equation to use. The most straightforward is[latex]\boldsymbol{v=v_0-gt}[/latex](from[latex]\boldsymbol{v=v_0+at}[/latex], where[latex]\boldsymbol{a=\textbf{gravitational
3. Plug in the knowns and solve.
[latex]\boldsymbol{v_1=v_0-gt=13.0\textbf{ m/s}-(9.80\textbf{ m/s}^2)(1.00\textbf{ s})=3.20\textbf{ m/s}}[/latex]
The positive value for[latex]\boldsymbol{v_1}[/latex]means that the rock is still heading upward at[latex]\boldsymbol{t=1.00\textbf{ s}}[/latex]. However, it has slowed from its original 13.0 m/s, as
Solution for Remaining Times
The procedures for calculating the position and velocity at[latex]\boldsymbol{t=2.00\textbf{ s}}[/latex]and[latex]\boldsymbol{3.00\textbf{ s}}[/latex]are the same as those above. The results are
summarized in Table 1 and illustrated in Figure 3.
Time, t Position, y Velocity, v Acceleration, a
1.00 s 8.10 m 3.20 m/s −9.80 m/s2
2.00 s 6.40 m −6.60 m/s −9.80 m/s2
3.00 s −5.10 m −16.4 m/s −9.80 m/s2
Table 1. Results.
Graphing the data helps us understand it more clearly.
Figure 3. Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that velocity changes linearly with time and that
acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical position only. It is easy to get the impression that the graph shows some horizontal motion—the
shape of the graph looks like the path of a projectile. But this is not the case; the horizontal axis is time, not space. The actual path of the rock in space is straight up, and straight down.
The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upward, since[latex]\boldsymbol{y_1}[/latex]and[latex]\boldsymbol{v_1}[/latex]are both
positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, both[latex]\boldsymbol{y_3}[/latex]and[latex]\boldsymbol{v_3}[/
latex]are negative, meaning the rock is below its starting point and continuing to move downward. Notice that when the rock is at its highest point (at 1.5 s), its velocity is zero, but its
acceleration is still[latex]\boldsymbol{-9.80\textbf{ m/s}^2}[/latex]. Its acceleration is[latex]\boldsymbol{-9.80\textbf{ m/s}^2}[/latex]for the whole trip—while it is moving up and while it is
moving down. Note that the values for[latex]\boldsymbol{y}[/latex]are the positions (or displacements) of the rock, not the total distances traveled. Finally, note that free-fall applies to upward
motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which remains constant the entire time. Astronauts training in the famous Vomit Comet, for example,
experience free-fall while arcing up as well as down, as we will discuss in more detail later.
A simple experiment can be done to determine your reaction time. Have a friend hold a ruler between your thumb and index finger, separated by about 1 cm. Note the mark on the ruler that is right
between your fingers. Have your friend drop the ruler unexpectedly, and try to catch it between your two fingers. Note the new reading on the ruler. Assuming acceleration is that due to gravity,
calculate your reaction time. How far would you travel in a car (moving at 30 m/s) if the time it took your foot to go from the gas pedal to the brake was twice this reaction time?
Example 2: Calculating Velocity of a Falling Object: A Rock Thrown Down
What happens if the person on the cliff throws the rock straight down, instead of straight up? To explore this question, calculate the velocity of the rock when it is 5.10 m below the starting point,
and has been thrown downward with an initial speed of 13.0 m/s.
Draw a sketch.
Figure 4.
Since up is positive, the final position of the rock will be negative because it finishes below the starting point at[latex]\boldsymbol{y_0=0}[/latex]. Similarly, the initial velocity is downward and
therefore negative, as is the acceleration due to gravity. We expect the final velocity to be negative since the rock will continue to move downward.
1. Identify the knowns.[latex]\boldsymbol{y_0=0}[/latex];[latex]\boldsymbol{y_1=-5.10\textbf{ m}}[/latex];[latex]\boldsymbol{v_0=-13.0\textbf{ m/s}}[/latex];[latex]\boldsymbol{a=-g=-9.80\textbf{ m/s}
2. Choose the kinematic equation that makes it easiest to solve the problem. The equation[latex]\boldsymbol{v^2=v_0^2+2a(y-y_0)}[/latex]works well because the only unknown in it is[latex]\boldsymbol
{v}[/latex]. (We will plug[latex]\boldsymbol{y_1}[/latex]in for[latex]\boldsymbol{y}[/latex].)
3. Enter the known values
[latex]\boldsymbol{v^2=(-13.0\textbf{ m/s})^2+2(-9.80\textbf{ m/s}^2)(-5.10\textbf{ m}-0\textbf{ m})=268.96\textbf{ m}^2/\textbf{s}^2}[/latex],
where we have retained extra significant figures because this is an intermediate result.
Taking the square root, and noting that a square root can be positive or negative, gives
[latex]\boldsymbol{v=\pm16.4\textbf{ m/s}}[/latex].
The negative root is chosen to indicate that the rock is still heading down. Thus,
[latex]\boldsymbol{v=-16.4\textbf{ m/s.}}[/latex]
Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same initial speed. (See Example 1 and Figure 5(a).) This is not a coincidental
result. Because we only consider the acceleration due to gravity in this problem, the speed of a falling object depends only on its initial speed and its vertical position relative to the starting
point. For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from Example 1) when the initial velocity is 13.0 m/s straight up, a
result of[latex]\boldsymbol{\pm3.20\textbf{ m/s}}[/latex]is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value
occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction.
Figure 5. (a) A person throws a rock straight up, as explored in Example 1. The arrows are velocity vectors at 0, 1.00, 2.00, and 3.00 s. (b) A person throws a rock straight down from a cliff with
the same initial speed as before, as in Example 2. Note that at the same distance below the point of release, the rock has the same velocity in both cases.
Another way to look at it is this: In Example 1, the rock is thrown up with an initial velocity of[latex]\boldsymbol{13.0\textbf{ m/s}}[/latex]. It rises and then falls back down. When its position
is[latex]\boldsymbol{y=0}[/latex]on its way back down, its velocity is[latex]\boldsymbol{-13.0\textbf{ m/s}}[/latex]. That is, it has the same speed on its way down as on its way up. We would then
expect its velocity at a position of[latex]\boldsymbol{y=-5.10\textbf{ m}}[/latex]to be the same whether we have thrown it upwards at[latex]\boldsymbol{+13.0\textbf{ m/s}}[/latex]or thrown it
downwards at[latex]\boldsymbol{-13.0\textbf{ m/s}}.[/latex]The velocity of the rock on its way down from[latex]\boldsymbol{y=0}[/latex]is the same whether we have thrown it up or down to start with,
as long as the speed with which it was initially thrown is the same.
Example 3: Find g from Data on a Falling Object
The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface geology (whether there is dense
rock like iron ore as opposed to light rock like salt beneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physics laboratory course. An object,
usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, Figure 6. Very precise results can be produced
with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.
Figure 6. Positions and velocities of a metal ball released from rest when air resistance is negligible. Velocity is seen to increase linearly with time while displacement increases with time
squared. Acceleration is a constant and is equal to gravitational acceleration.
Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location?
Draw a sketch.
Figure 7.
We need to solve for acceleration[latex]\boldsymbol{a}[/latex]. Note that in this case, displacement is downward and therefore negative, as is acceleration.
1. Identify the knowns.[latex]\boldsymbol{y_0=0}[/latex];[latex]\boldsymbol{y=-1.0000\textbf{ m}}[/latex];[latex]\boldsymbol{t=0.45173\textbf{ s}}[/latex];[latex]\boldsymbol{v_0=0}.[/latex]
2. Choose the equation that allows you to solve for[latex]\boldsymbol{a}[/latex]using the known values.
3. Substitute 0 for[latex]\boldsymbol{v_0}[/latex]and rearrange the equation to solve for[latex]\boldsymbol{a}[/latex]. Substituting 0 for[latex]\boldsymbol{v_0}[/latex]yields
Solving for[latex]\boldsymbol{a}[/latex]gives
4. Substitute known values yields
[latex]\boldsymbol{a=}[/latex][latex]\boldsymbol{\frac{2(-1.0000\textbf{ m} - 0)}{(0.45173\textbf{ s})^2}}[/latex][latex]\boldsymbol{=-9.8010\textbf{ m/s}^2,}[/latex]
so, because[latex]\boldsymbol{a=-g}[/latex]with the directions we have chosen,
[latex]\boldsymbol{g=9.8010\textbf{ m/s}^2.}[/latex]
The negative value for[latex]\boldsymbol{a}[/latex]indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of[latex]\
boldsymbol{9.80\textbf{ m/s}^2,}[/latex] so[latex]\boldsymbol{9.8010\textbf{ m/s}^2}[/latex]makes sense. Since the data going into the calculation are relatively precise, this value for[latex]\
boldsymbol{g}[/latex]is more precise than the average value of[latex]\boldsymbol{9.80\textbf{ m/s}^2}[/latex]; it represents the local value for the acceleration due to gravity.
Check Your Understanding
1: A chunk of ice breaks off a glacier and falls 30.0 meters before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water?
Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g.[latex]\boldsymbol{y=bx}[/latex]) to see how they add to
generate the polynomial curve.
Figure 8. Equation Grapher.
Section Summary
• An object in free-fall experiences constant acceleration if air resistance is negligible.
• On Earth, all free-falling objects have an acceleration due to gravity[latex]\boldsymbol{g}[/latex], which averages
[latex]\boldsymbol{g=9.80\textbf{ m/s}^2}[/latex].
• Whether the acceleration a should be taken as[latex]\boldsymbol{+g}[/latex]or[latex]\boldsymbol{-g}[/latex]is determined by your choice of coordinate system. If you choose the upward direction as
positive,[latex]\boldsymbol{a=-g=-9.80\textbf{ m/s}^2}[/latex] is negative. In the opposite case,[latex]\boldsymbol{a=+g=9.80\textbf{ m/s}^2}[/latex]is positive. Since acceleration is constant,
the kinematic equations above can be applied with the appropriate[latex]\boldsymbol{+g}[/latex]or[latex]\boldsymbol{-g}[/latex]substituted for[latex]\boldsymbol{a}[/latex].
• For objects in free-fall, up is normally taken as positive for displacement, velocity, and acceleration.
Conceptual Questions
1: What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down?
2: An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration due to
gravity have the same sign on the way up as on the way down?
3: Suppose you throw a rock nearly straight up at a coconut in a palm tree, and the rock misses on the way up but hits the coconut on the way down. Neglecting air resistance, how does the speed of
the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down?
4: If an object is thrown straight up and air resistance is negligible, then its speed when it returns to the starting point is the same as when it was released. If air resistance were not
negligible, how would its speed upon return compare with its initial speed? How would the maximum height to which it rises be affected?
5: The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration due to gravity being the same, how many times higher could a safe fall on the Moon be than
on Earth (gravitational acceleration on the Moon is about 1/6 that of the Earth)?
6: How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (gravitational acceleration on the Moon is about 1/6 of[latex]\boldsymbol
{g}[/latex]on Earth)?
Problems & Exercises
Assume air resistance is negligible unless otherwise stated.
1: Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to
2: Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano
Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
3: A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?
4: A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it
takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air
resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.
5: A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part,
first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation,
show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.
6: A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (a) How long are her
feet in the air? (b) What is her highest point above the board? (c) What is her velocity when her feet hit the water?
7: (a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long would it take to
reach the ground if it is thrown straight down with the same speed?
8: A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of
2.20 m, and he is 1.80 m tall?
9: You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree
branch on the way back down?
10: A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?
11: Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s
later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?
12: An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground.
(c) Determine the distance traveled during the last second of motion before hitting the ground.
13: There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the
ground? (b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the
tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day.
14: A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past the window. What was the ball’s initial velocity? Hint: First
consider only the distance along the window, and solve for the ball’s velocity at the bottom of the window. Next, consider only the distance from the ground to the bottom of the window, and solve for
the initial velocity using the velocity at the bottom of the window as the final velocity.
15: Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return. (a) Neglecting the time required for sound to travel up the
well, calculate the distance to the water if the sound returns in 2.0000 s. (b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m
/s in this well.
16: A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just
after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms[latex]\boldsymbol{(8.00\times10^{-5}\textbf{ s})}.[/latex]
(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
17: A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s
after being released, and (c) the time before it hits the ground.
18: A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity
just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms[latex]\boldsymbol{(3.50\times10^{-3}\textbf{ s})}.[/
latex] (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
the state of movement that results from gravitational force only
acceleration due to gravity
acceleration of an object as a result of gravity
Check Your Understanding
1: We know that initial position[latex]\boldsymbol{y_0=0}[/latex], final position[latex]\boldsymbol{y=-30.0\textbf{ m}}[/latex], and[latex]\boldsymbol{a=-g=-9.80\textbf{ m/s}^2}[/latex]. We can then
use the equation[latex]\boldsymbol{y=y_0+v_0t+\frac{1}{2}at^2}[/latex]to solve for[latex]\boldsymbol{t}[/latex]. Inserting[latex]\boldsymbol{a=-g}[/latex], we obtain
[latex]\boldsymbol{t=\pm\sqrt{\frac{2y}{-g}}=\pm\sqrt{\frac{2(-30.0\textbf{ m}}{-9.80\textbf{ m/s}^2}}=\pm\sqrt{6.12\textbf{ s}^2}=2.47\textbf{ s}\approx2.5\textbf{ s}}[/latex]
where we take the positive value as the physically relevant answer. Thus, it takes about 2.5 seconds for the piece of ice to hit the water.
Problems & Exercises
(a)[latex]\boldsymbol{y_1=6.28\textbf{ m}};\boldsymbol{v_1=10.1\textbf{ m/s}}[/latex]
(b)[latex]\boldsymbol{y_2=10.1\textbf{ m}};\boldsymbol{v_2=5.20\textbf{ m/s}}[/latex]
(c)[latex]\boldsymbol{y_3=11.5\textbf{ m}};\boldsymbol{v_3=0.300\textbf{ m/s}}[/latex]
(d)[latex]\boldsymbol{y_4=10.4\textbf{ m}};\boldsymbol{v_4=-4.60\textbf{ m/s}}[/latex]
[latex]\boldsymbol{v_0=4.95\textbf{ m/s}}[/latex]
(a)[latex]\boldsymbol{a=-9.80\textbf{ m/s}^2};\boldsymbol{v_0=13.0\textbf{ m/s}};\boldsymbol{y_0=0\textbf{ m}}[/latex]
(b)[latex]\boldsymbol{v=0\textbf{ m/s}}.[/latex]Unknown is distance[latex]\boldsymbol{y}[/latex]to top of trajectory, where velocity is zero. Use equation[latex]\boldsymbol{v^2=v_0^2+2a(y-y_0)}[/
latex]because it contains all known values except for[latex]\boldsymbol{y},[/latex]so we can solve for[latex]\boldsymbol{y}.[/latex]Solving for[latex]\boldsymbol{y}[/latex]gives
[latex]\begin{array}{r @{{}={}} l} \boldsymbol{v^2 - v_0^2} & \boldsymbol{2a(y - y_0)} \\[1em] \boldsymbol{\frac{v^2 - v_0}{2a}} & \boldsymbol{y - y_0} \\[1em] \boldsymbol{y} & \boldsymbol{y_0 + \
frac{v^2 - v^2_0}{2a} = 0 \;\textbf{m} + \frac{(0 \;\textbf{m/s})^2 - (13.0 \;\textbf{m/s})^2}{2(-9.80 \;\textbf{m/s}^2)} = 8.62 \;\textbf{m}} \end{array}[/latex]
Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result.
(c)[latex]\boldsymbol{2.65\textbf{ s}}[/latex]
Figure 9.
(a) 8.26 m
(b) 0.717 s
1.91 s
(a) -70.0 m/s (downward)
(b) 6.10 s
(a)[latex]\boldsymbol{19.6\textbf{ m}}[/latex]
(b)[latex]\boldsymbol{18.5\textbf{ m}}[/latex]
(a) 305 m
(b) 262 m, -29.2 m/s
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Development Length FormulaDevelopment Length Formula
Development Length Formula
As a structural engineer or site engineer, we frequently work with various types of designs and details. One word, “development length,” or the denotation “Ld,” is mentioned in the drawing.
Do you understand what “development length” means? Where should development length be provided and why should we supply it?
as well as how to determine development length.
What Is the Development Length?
To achieve the appropriate bond strength between steel and concrete, the minimum length of the steel bar needs to be firmly attached to the column.
The strong link between column-to-footing and column-to-beam structural elements is caused by development length.
In buildings, we are aware of how to shift loads. The weight travels from the slab to the beam, then the beam to the column.
The column to the footing, and finally the footing to the ground. We must verify the component joints in order for this cycle to function properly.
Read More: When Do You Need Steels for Property Builds and Refurbishments?
Why Provide Development Length?
The developments length serves as a hook. As a result, there are fewer chances that the bars on a concrete beam may slip under a heavy load from the structure.
Imagine what would happen if we did not provide adequate developments length in structure. Images of the same scene with and without developments length are shown above.
Because concrete cannot withstand tensile stress, steel is added because it can. There is a danger that a beam or slab will slip from the connection if the concrete is bound without developments
length because it will appear to be externally attached to a column.
Additionally, if the structure collapses, the bars won’t be damaged; instead, they will merely separate from the column component.
We all have carried a bag at some point in our lives, for instance. What transpired, then, if the palm was open? Without a doubt, you won’t be able to carry the bag, and if you do, there’s a good
risk it’ll escape if you carry it in your open hand.
What can you then do? You can form a “U shape” with your palm-like hook. Similar to this, we create development length using steel bars.
A crucial part is played by the joint that connects the superstructure and the substructure in smoothly transferring the weight.
Therefore, joints need to be really strong and connect to one another in order to create a very strong framework.
What Did We Need to Calculate Development Length?
1. Grade of steel bars
2. Grade of concrete
3. Diameter of steel bar
4. Type of steel bar
Read More: How To Calculate Steel Quantity | Steel Quantity Formula
How to Calculate Development Length?
Once we have all of the necessary information, we can quickly determine the development length. Typically, we choose a developments length of 50D or 40D at random.
However, as it is a rule of thumb, this value is not precise. If we use a large diameter, an additional bar will be very expensive.
Note: The length of development varies for tension and compression. People typically assign the same importance to both stresses.
There are several bar types, and as there are various bar numbers, there are various calculations.
1. Calculation of development length of plain bars(compression, tension)
2. Calculation of development length of deform bars(compression, tension)
3. Calculation of development length of bundled
Calculation of development length of plain bars(compression, tension)
• L[d:] – development length
• ɸ: – nominal diameter of the bar
• σs: – stress in bars, design load
• Ï„[bd: –]design bond stress
The following equation appears in the IS code (26.2.1). Now that we have all of the answers to the equation above, we can define nominal diameter as the diameter of the steel bar that needs to be
Steel bars are valued by the stress they contain. The stress permitted in bars is Fe410, and Fe500, where 500 is the maximum acceptable stress.
Is the code on line 26.2.1.1 contains the design bond stress and other information about the design bond? From grade to grade, it differs
The ultimate developments length will be obtained once all values have been entered into the equation.
Where (Ï„[bd: – ]design bond stress) for tension in beam and slab
Grade of concrete M20 M25 M30 M35 M40 and above
Ï„[bd] 1.2 1.4 1.5 1.7 1.9
The above table is referenced in IS 456.2000, where all bd values for tension are provided. Additionally, if the component is a compression member, the value will be 25% higher than it was originally
when it was a tensile component.
Always choose 1.2 if the grade is lower than the M20 grade of concrete.
Where (Ï„[bd: – ]design bond stress) for compression in beam and slab
Grade of concrete M20 M25 M30 M35 M40 and above
Ï„[bd] 1.5 1.75 1.875 2.125 2.375
Calculation of development length of Deformed bars(compression, tension)
The equation is the same as above the difference is in Ï„[bd ]value only.
Where (Ï„[bd: – ]design bond stress) for tension in beam and slab
The value of the design bond stress in a deformed bar is 60% greater than the value of the design bond in a plain bar. for arousal.
Grade of concrete M20 M25 M30 M35 M40 and above
Ï„[bd] 1.92 2.24 2.4 2.72 3.04
The value of the design bond stress in a deformed bar is 25% greater than the value of the design bond in a plain bar. due to compression
Grade of concrete M20 M25 M30 M35 M40 and above
Ï„[bd] 2.4 2.8 3 3.4 3.8
We have one beam where the diameter of the bar is 20 mm, the steel we use is fe500 and the concrete grade is M25. So, find the development length for both compression and tension.
• Nominal diameter of bar: – 20
• stress in bars, design lode: – 500 N/mm2
• design bond stress: – 1.4(tension) , 1.75(compression)
Put the above values in the equation of development length.
T[d = ]20×500 / 4×1.4
T[d ]= 1785 mm(tension)
T[d = ]20×500 / 4×1.75
T[d ]= 1428 mm(tension)
So, both values are different.
Factors Affecting Development Length
• The compressive strength of concrete is inversely related to the development length of the steel bar. As a result of the high concrete strength, the development length is short.
• Since density is another crucial element, the length of the development will increase as we use lightweight concrete.
• Development length decreases as clear cover is increased.
• The coating of bars also impacts length of development. Because concrete and steel surfaces can be separated.
• The length of development is directly impacted by steel bar diameter. The development length is also short because of the small diameter.
Read More: Steel Fiber Reinforced Concrete
You May Also Like: | {"url":"https://civiconcepts.com/blog/development-length","timestamp":"2024-11-09T19:13:01Z","content_type":"text/html","content_length":"515010","record_id":"<urn:uuid:5f01306e-eedc-496f-a3f6-1a0cfdf87cfd>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00325.warc.gz"} |
Connection or downloading problem?!!
Hello I'm Rand, a new member to this forum.
I've been using Bitcomet for a while now and I think I got the general idea on how to use it, but my problem is that sometimes while downloading a file the program decides to stop working even though
there is an access to the network.
My question is what is the problem in this case? I'm I missing something?
click on the "READ THIS before posting" link, then better describe what you mean by "stops working".
What I meant by "stops working" is that Bitcomet can't download the file as if the connection to the internet is lost.
I'll read the FAQ's to understand my problem more. Thanks for posting.
Well, again you haven't provided the requested info, so if you want an answer without providing facts to base it on, I'll give you the most likely cause.
The torrent your trying to download has no seeders and will never complete. It's simply dead, so look for a different torrent. | {"url":"https://www.cometforums.com/topic/12796499-connection-or-downloading-problem/","timestamp":"2024-11-02T17:57:50Z","content_type":"text/html","content_length":"87891","record_id":"<urn:uuid:366c4128-514b-495c-8b58-0bab2e3e52f1>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00801.warc.gz"} |
Vertex connectivity — vertex_connectivity
Vertex connectivity
The vertex connectivity of a graph or two vertices, this is recently also called group cohesion.
vertex_connectivity(graph, source = NULL, target = NULL, checks = TRUE)
vertex_disjoint_paths(graph, source = NULL, target = NULL)
# S3 method for class 'igraph'
cohesion(x, checks = TRUE, ...)
The input graph.
The id of the source vertex, for vertex_connectivity() it can be NULL, see details below.
The id of the target vertex, for vertex_connectivity() it can be NULL, see details below.
Logical constant. Whether to check that the graph is connected and also the degree of the vertices. If the graph is not (strongly) connected then the connectivity is obviously zero. Otherwise if
the minimum degree is one then the vertex connectivity is also one. It is a good idea to perform these checks, as they can be done quickly compared to the connectivity calculation itself. They
were suggested by Peter McMahan, thanks Peter.
The vertex connectivity of two vertices (source and target) in a graph is the minimum number of vertices that must be deleted to eliminate all (directed) paths from source to target.
vertex_connectivity() calculates this quantity if both the source and target arguments are given and they're not NULL.
The vertex connectivity of a pair is the same as the number of different (i.e. node-independent) paths from source to target, assuming no direct edges between them.
The vertex connectivity of a graph is the minimum vertex connectivity of all (ordered) pairs of vertices in the graph. In other words this is the minimum number of vertices needed to remove to make
the graph not strongly connected. (If the graph is not strongly connected then this is zero.) vertex_connectivity() calculates this quantity if neither the source nor target arguments are given.
(I.e. they are both NULL.)
A set of vertex disjoint directed paths from source to vertex is a set of directed paths between them whose vertices do not contain common vertices (apart from source and target). The maximum number
of vertex disjoint paths between two vertices is the same as their vertex connectivity in most cases (if the two vertices are not connected by an edge).
The cohesion of a graph (as defined by White and Harary, see references), is the vertex connectivity of the graph. This is calculated by cohesion().
These three functions essentially calculate the same measure(s), more precisely vertex_connectivity() is the most general, the other two are included only for the ease of using more descriptive
function names.
White, Douglas R and Frank Harary 2001. The Cohesiveness of Blocks In Social Networks: Node Connectivity and Conditional Density. Sociological Methodology 31 (1) : 305-359.
g <- sample_pa(100, m = 1)
g <- delete_edges(g, E(g)[100 %--% 1])
g2 <- sample_pa(100, m = 5)
g2 <- delete_edges(g2, E(g2)[100 %--% 1])
vertex_connectivity(g, 100, 1)
#> [1] 1
vertex_connectivity(g2, 100, 1)
#> [1] 5
vertex_disjoint_paths(g2, 100, 1)
#> [1] 5
g <- sample_gnp(50, 5 / 50)
g <- as_directed(g)
g <- induced_subgraph(g, subcomponent(g, 1))
#> [1] 2 | {"url":"https://r.igraph.org/reference/vertex_connectivity.html","timestamp":"2024-11-04T15:10:59Z","content_type":"text/html","content_length":"17141","record_id":"<urn:uuid:1fe0855c-94cf-4394-a03f-e18e3b71cbfd>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00387.warc.gz"} |
This page provides a simple browsing interface for finding entities described by a property and a named value. Other available search interfaces include the page property search, and the ask query
List of results
• Model:Spbgc + (High order two dimensional simulations of turbidity currents using DNS of incompressible Navier-Stokes and transport equations.)
• Model:TransportLengthHillslopeDiffuser + (Hillslope diffusion component in the style … Hillslope diffusion component in the style of Carretier et al. (2016, ESurf), and Davy and Lague (2009).</
br></br>Works on regular raster-type grid (RasterModelGrid, dx=dy). To be coupled with FlowDirectorSteepest for the calculation of steepest slope at each timestep.lation of steepest slope at each
• Model:TaylorNonLinearDiffuser + (Hillslope evolution using a Taylor Series … Hillslope evolution using a Taylor Series expansion of the Andrews-Bucknam formulation of nonlinear hillslope flux
derived following following Ganti et al., 2012. The flux is given as:</br></br>qs = KS ( 1 + (S/Sc)**2 + (S / Sc)**4 + .. + (S / Sc)**2(n - 1) )</br></br>where K is is the diffusivity, S is the
slope, Sc is the critical slope, and n is the number of terms. The default behavior uses two terms to produce a flux law as described by Equation 6 of Ganti et al., (2012).bed by Equation 6 of
Ganti et al., (2012).)
• Model:DepthDependentTaylorDiffuser + (Hillslope sediment flux uses a Taylor Seri … Hillslope sediment flux uses a Taylor Series expansion of the Andrews-Bucknam formulation of nonlinear hillslope
flux derived following following Ganti et al., 2012 with a depth dependent component inspired Johnstone and Hilley (2014). The flux :math:`q_s` is given as:</br>q_s = DSH^* ( 1 + (S/S_c)^2 + (S/
Sc_)^4 + .. + (S/S_c)^2(n-1) ) (1.0 - exp( H / H^*)</br></br>where :math:`D` is is the diffusivity, :math:`S` is the slope, :math:`S_c` is the critical slope, :math:`n` is the number of terms,
:math:`H` is the soil depth on links, and :math:`H^*` is the soil transport decay depth. The default behavior uses two terms to produce a slope dependence as described by Equation 6 of Ganti et
al., (2012).This component will ignore soil thickness located at non-core nodes. soil thickness located at non-core nodes.)
• Model:HydroCNHS + (HydroCNHS is an open-source Python package supporting four Application Programming Interfaces (APIs) that enable users to integrate their human decision models, which can be
programmed with the agent-based modeling concept, into the HydroCNHS.)
• Model:HydroPy + (HydroPy model is a revised version of an e … HydroPy model is a revised version of an established global hydrological model (GHM), the Max Planck Institute for Meteorology's
Hydrology Model (MPI-HM). Being rewritten in Python, the HydroPy model requires much less effort in maintenance and new processes can be easily implemented.d new processes can be easily
• Model:HydroTrend + (HydroTrend v.3.0 is a climate-driven hydrological water balance and transport model that simulates water discharge and sediment load at a river outlet.)
• Model:HSPF + (Hydrological Simulation Program - FORTRAN … Hydrological Simulation Program - FORTRAN (HSPF) is a comprehensive package</br>for simulation of watershed hydrology and water quality
for both conventional</br>and toxic organic pollutants (1,2). This model can simulate the hydrologic,</br>and associated water quality, processes on pervious and impervious land</br>surfaces and
in streams and well-mixed impoundments. HSPF incorporates the</br>watershed-scale ARM and NPS models into a basin-scale analysis framework that</br>includes fate and transport in one-dimensional
stream channels. It is the</br>only comprehensive model of watershed hydrology and water quality that allows</br>the integrated simulation of land and soil contaminant runoff processes with</br>
in-stream hydraulic and sediment-chemical interactions.</br></br>The result of this simulation is a time history of the runoff flow rate,</br>sediment load, and nutrient and pesticide
concentrations, along with a time</br>history of water quantity and quality at any point in a watershed. HSPF</br>simulates three sediment types (sand, silt, and clay) in addition to a single</
br>organic chemical and transformation products of that chemical. The transfer</br>and reaction processes included are hydrolysis, oxidation, photolysis,</br>biodegradation, volatilization, and
sorption. Sorption is modeled as a</br>first-order kinetic process in which the user must specify a desorption rate</br>and an equilibrium partition coefficient for each of the three solids
types.</br></br>Resuspension and settling of silts and clays (cohesive solids) are defined in</br>terms of shear stress at the sediment water interface. The capacity of the</br>system to
transport sand at a particular flow is calculated and resuspension</br>or settling is defined by the difference between the sand in suspension and</br>the transport capacity. Calibration of the
model requires data for each of</br>the three solids types. Benthic exchange is modeled as sorption/desorption</br>and deposition/scour with surficial benthic sediments. Underlying sediment</br>
and pore water are not modeled.g sediment and pore water are not modeled.)
• Model:WACCM-EE + (I am developing a GCM based on NCAR's WACC … I am developing a GCM based on NCAR's WACCM model to studied the climate of the ancient Earth. WACCM has been linked with a
microphysical model (CARMA). Some important issues to be examined are the climate of the ancient Earth in light of the faint young Sun, reducing chemistry of the early atmosphere, and the
production and radiative forcing of Titan-like photochemical hazes that likely enshrouded the Earth at this time. likely enshrouded the Earth at this time.)
• Model:IDA + (IDA formulates the task of determing the d … IDA formulates the task of determing the drainage area, given flow directions, as a system of implicit equations. This allows the use of
iterative solvers, which have the advantages of being parallelizable on distributed memory systems and widely available through libraries such as PETSc.</br></br>Using the open source PETSc
library (which must be downloaded and installed separately), IDA permits large landscapes to be divided among processors, reducing total runtime and memory requirements per processor.</br></br>It
is possible to reduce run time with the use of an initial guess of the drainage area. This can either be provided as a file, or use a serial algorithm on each processor to correctly determine the
drainage area for the cells that do not receive flow from outside the processor's domain.</br></br>The hybrid IDA method, which is enabled with the -onlycrossborder option, uses a serial
algorithm to solve for local drainage on each processor, and then only uses the parallel iterative solver to incorporate flow between processor domains. This generally results in a significant
reduction in total runtime.</br></br>Currently only D8 flow directions are supported. Inputs and outputs are raw binary files.. Inputs and outputs are raw binary files.)
• Model:ISSM + (ISSM is the result of a collaboration betw … ISSM is the result of a collaboration between the Jet Propulsion Laboratory and University of California at Irvine. Its purpose is to
tackle the challenge of modeling the evolution of the polar ice caps in Greenland and Antarctica.</br>ISSM is open source and is funded by the NASA Cryosphere, GRACE Science Team, ICESat
Research, ICESat-2 Research, NASA Sea-Level Change Team (N-SLCT), IDS (Interdisciplinary Research in Earth Science), ESI (Earth Surface and Interior), and MAP (Modeling Analysis and Prediction)
programs, JPL R&TD (Research, Technology and Development) and the National Science Foundationvelopment) and the National Science Foundation)
• Model:IceFlow + (IceFlow simulates ice dynamics by solving … IceFlow simulates ice dynamics by solving equations for internal deformation and simplified basal sliding in glacial systems. It is
designed for computational efficiency by using the shallow ice approximation for driving stress, which it solves alongside basal sliding using a semi-implicit direct solver. IceFlow is integrated
with GRASS GIS to automatically generate input grids from a geospatial database.te input grids from a geospatial database.)
• Model:Icepack + (Icepack is a Python package for simulating … Icepack is a Python package for simulating the flow of glaciers and ice sheets, as well as for solving glaciological data
assimilation problems. The main goal for icepack is to produce a tool that researchers and students can learn to use quickly and easily, whether or not they are experts in high-performance
computing. Icepack is built on the finite element modeling library firedrake, which implements the domain-specific language UFL for the specification of PDEs.anguage UFL for the specification of
• Model:ChannelProfiler + (In order to extract channel networks, the … In order to extract channel networks, the flow connectivity across the grid must already be identified. This is typically done
with the FlowAccumulator component. However, this component does not require that the FlowAccumulator was used. Instead it expects that the following at-node grid fields will be present:<br></br>
'flow__receiver_node'<br></br>'flow__link_to_receiver_node'<br></br>The ChannelProfiler can work on grids that have used route-to-one or route-to-multiple flow directing. have used route-to-one
or route-to-multiple flow directing.)
• Model:HIM + (It is a C-grid, isopycnal coordinate, primitive equation model, simulating the ocean by numerically solving the Boussinesq primitive equations in isopycnal vertical coordinates and
general orthogonal horizontal coordinates.)
• Model:WOFOST + (It is a mechanistic model that explains cr … It is a mechanistic model that explains crop growth on the basis of the underlying processes, such as photosynthesis, respiration and
how these processes are influenced by environmental conditions. </br></br>With WOFOST, you can calculate attainable crop production, biomass, water use, etc. for a location given knowledge about
soil type, crop type, weather data and crop management factors (e.g. sowing date). WOFOST has been used by many researchers over the World and has been applied for many crops over a large range
of climatic and management conditions. WOFOST is one of the key components of the European MARS crop yield forecasting system. In the Global Yield Gap Atlas (GYGA) WOFOST is used to estimate the
untapped crop production potential on existing farmland based on current climate and available soil and water resources.te and available soil and water resources.)
• Model:FUNDY + (It solves the linearized shallow water equations forced by tidal or other barotropic boundary conditions, wind or a density gradient using linear finite elements.)
• Model:ACADIA + (It tracks any number of different depth-averaged transport variables and is usually used in conjunction with QUODDY simulations.)
• Model:LEMming + (LEMming tracks regolith and sediment fluxe … LEMming tracks regolith and sediment fluxes, including bedrock erosion by streams and rockfall from steep slopes. Initial landscape
form and stratigraphic structure are prescribed. Model grid cells with slope angles above a threshold, and which correspond to the appropriate rock type, are designated as candidate sources for
rockfall. Rockfall erosion of the cliffband is simulated by instantaneously reducing the height of a randomly chosen grid cell that is susceptible to failure to that of its nearest downhill
neighbor among the eight cells bordering it. This volume of rockfall debris is distributed across the landscape below this cell according to rules that weight the likelihood of each downhill cell
to retain rockfall debris. The weighting is based on local conditions such as slope angle, topographic curvature, and distance and direction from the rockfall source. Rockfall debris and the
bedrock types are each differentiated by the rate at which they weather to regolith and by their fluvial erodibility. Regolith is moved according to transport rules mimicking hillslope processes
(dependent on local slope angle), and bedload and suspended load transport (based on stream power). Regolith and sediment transport are limited by available material; bedrock incision occurs
(also based on stream power) where bare rock is exposed. stream power) where bare rock is exposed.)
• Model:LEMming2 + (LEMming2 is a 2D, finite-difference landsc … LEMming2 is a 2D, finite-difference landscape evolution model that simulates the retreat of hard-capped cliffs. It implements common
unit-stream-power and linear/nonlinear-diffusion erosion equations on a 2D regular grid. Arbitrary stratigraphy may be defined. Cliff retreat is facilitated by a cellular algorithm, and rockfall
debris is distributed and redistributed to the angle of repose. It is a standalone model written in Matlab with some C components.</br></br>This repo contains the code used and described by Ward
(2019) Lithosphere: "Dip, layer spacing, and incision rate controls on the formation of strike valleys, cuestas, and cliffbands in heterogeneous stratigraphy". Given the inputs in that paper it
should generate the same results.paper it should generate the same results.)
• Model:LISFLOOD + (LISFLOOD is a spatially distributed, semi- … LISFLOOD is a spatially distributed, semi-physical hydrological rainfall-runoff model that has been developed by the Joint Research
Centre (JRC) of the European Commission in late 90ies. Since then LISFLOOD has been applied to a wide range of applications such as all kind of water resources assessments looking at e.g. the
effects of climate and land-use change as well as river regulation measures. Its most prominent application is probably within the European Flood Awareness System (EFAS) operated under Copernicus
Emergency Management System (EMS).ernicus Emergency Management System (EMS).)
• Model:LOADEST + (LOAD ESTimator (LOADEST) is a FORTRAN prog … LOAD ESTimator (LOADEST) is a FORTRAN program for estimating constituent loads in streams and rivers. Given a time series of
streamflow, additional data variables, and constituent concentration, LOADEST assists the user in developing a regression model for the estimation of constituent load (calibration). Explanatory
variables within the regression model include various functions of streamflow, decimal time, and additional user-specified data variables. The formulated regression model then is used to estimate
loads over a user-specified time interval (estimation). Mean load estimates, standard errors, and 95 percent confidence intervals are developed on a monthly and(or) seasonal basis.</br></br>The
calibration and estimation procedures within LOADEST are based on three statistical estimation methods. The first two methods, Adjusted Maximum Likelihood Estimation (AMLE) and Maximum Likelihood
Estimation (MLE), are appropriate when the calibration model errors (residuals) are normally distributed. Of the two, AMLE is the method of choice when the calibration data set (time series of
streamflow, additional data variables, and concentration) contains censored data. The third method, Least Absolute Deviation (LAD), is an alternative to maximum likelihood estimation when the
residuals are not normally distributed. LOADEST output includes diagnostic tests and warnings to assist the user in determining the appropriate estimation method and in interpreting the estimated
loads.</br></br>The LOADEST software and related materials (data and documentation) are made available by the U.S. Geological Survey (USGS) to be used in the public interest and the advancement
of science. You may, without any fee or cost, use, copy, modify, or distribute this software, and any derivative works thereof, and its supporting documentation, subject to the USGS software
User's Rights Notice.to the USGS software User's Rights Notice.)
• Model:Radiation + (Landlab component that computes 1D and 2D total incident shortwave radiation. This code also computes relative incidence shortwave radiation compared to a flat surface.)
• Model:LateralEroder + (Landlab component that finds a neighbor node to laterally erode and calculates lateral erosion.)
• Model:PrecipitationDistribution + (Landlab component that generates precipita … Landlab component that generates precipitation events using the rectangular Poisson pulse model described in
Eagleson (1978, Water Resources Research).</br></br>No particular units must be used, but it was written with the storm units in hours (hr) and depth units in millimeters (mm). (hr) and depth
units in millimeters (mm).)
• Model:Flexure + (Landlab component that implements a 1 and 2D lithospheric flexure model.)
• Model:DetachmentLtdErosion + (Landlab component that simulates detachmen … Landlab component that simulates detachment limited sediment transport is more general than the stream power component.
Doesn't require the upstream node order, links to flow receiver and flow receiver fields. Instead, takes in the discharge values on NODES calculated by the OverlandFlow class and erodes the
landscape in response to the output discharge.</br>As of right now, this component relies on the OverlandFlow component for stability. There are no stability criteria implemented in this class.
To ensure model stability, use StreamPowerEroder or FastscapeEroder components instead.der or FastscapeEroder components instead.)
• Model:Vegetation + (Landlab component that simulates net primary productivity, biomass and leaf area index at each cell based on inputs of root-zone average soil moisture.)
• Model:SoilMoisture + (Landlab component that simulates root-zone … Landlab component that simulates root-zone average soil moisture at each cell using inputs of potential evapotranspiration, live
leaf area index, and vegetation cover.</br></br>This component uses a single soil moisture layer and models soil moisture loss through transpiration by plants, evaporation by bare soil, and
leakage. The solution of water balance is based on Laio et. al 2001. The component requires fields of initial soil moisture, rainfall input (if any), time to the next storm and potential
transpiration.he next storm and potential transpiration.)
• Model:Landlab + (Landlab is a Python software package for c … Landlab is a Python software package for creating, assembling, and/or running 2D numerical models. Landlab was created to facilitate
modeling in earth-surface dynamics, but it is general enough to support a wide range of applications. Landlab provides three different capabilities:</br></br>(1) A DEVELOPER'S TOOLKIT for
efficiently building 2D models from scratch. The toolkit includes a powerful GRIDDING ENGINE for creating, managing, and iterative updating data on 2D structured or unstructured grids. The
toolkit also includes helpful utilities to handle model input and output.</br></br>(2) A set of pre-built COMPONENTS, each of which models a particular process. Components can be combined
together to create coupled models.</br></br>(3) A library of pre-built MODELS that have been created by combining components together.</br></br> To learn more, please visit http://
landlab.github.ioore, please visit http://landlab.github.io)
• Model:GOLEM + (Landscape evolution model. Computes evolution of topography under the action of rainfall and tectonics.)
• Model:SpeciesEvolver + (Life evolves alongside landscapes by bioti … Life evolves alongside landscapes by biotic and abiotic processes under complex dynamics at Earth’s surface. Researchers who
wish to explore these dynamics can use this component as a tool for them to build landscape-life evolution models. Landlab components, including SpeciesEvolver are designed to work with a shared
model grid. Researchers can build novel models using plug-and-play surface process components to evolve the grid’s landscape alongside the life tracked by SpeciesEvolver. The simulated life
evolves following customizable processes. evolves following customizable processes.)
• Model:LinearDiffuser + (LinearDiffuser is a Landlab component that models soil creep using an explicit finite-volume solution to a 2D diffusion equation.)
• Model:LITHFLEX2 + (Lithospheric flexure solution for a broken … Lithospheric flexure solution for a broken plate. Load is assumed to be represented by equal width loading elements specified
distance from broken edge of plate. Inclusion of sediments as part of the restoring force effect is possible by choice of density assigned to density (2).choice of density assigned to density
• Model:LITHFLEX1 + (Lithospheric flexure solution for infinite … Lithospheric flexure solution for infinite plate. Load is assumed to be convolved with Greens function (unit load) response in
order to calculate the net effect of the load. If desired, inclusion of sediments as part of the restoring force effect can be controlled via density assigned to density (2). Each load element
can have specified density and several loadings events can be incorporated.veral loadings events can be incorporated.)
• Model:CoastMorpho2D + (Long term 2D morphodynamics of coastal are … Long term 2D morphodynamics of coastal areas, including tidal currents, wind waves, swell waves, storm surge, sand, mud, marsh
vegetation, edge erosion, marsh ponding, and stratigraphy.</br>The CoastMorpho2D model includes the MarshMorpho2D model (which was previously uploaded on CSDMS)l (which was previously uploaded on
• Model:D'Alpaos model + (Long-term ecomorphodynamic model of the initiation and development of tidal networks and of the adjacent marsh platform, accounting for vegetation influence and relative
sea level rise effects)
• Model:MARSSIM V4 + (MARSSIM is a grid based, iterative framewo … MARSSIM is a grid based, iterative framework that incorporates selectable modules, including: 1) flow routing, optionally
including event-driven flow and evaporation from lakes in depression as a function of relative aridity (Matsubara et al., 2011). Runoff can be spatially uniform or variably distributed. Stream
channel morphology (width and depth) is parameterized as a function of effective discharge; 2) bedrock weathering, following Equation 1; 3) spatially variable bedrock resistance to weathering and
fluvial erosion, including 3-D stratigraphy and surficial coherent crusts; 4) erosion of bedrock channels using either a stream power relationship (Howard, 1994) or sediment load scour (Sklar and
Dietrich, 2004; Chatanantavet and Parker, 2009); 5) sediment routing in alluvial channels including suspended/wash load and a single size of bedload. An optional sediment transport model
simulates transport of multiple grain sizes of bedload with sorting and abrasion (Howard et al., 2016); 6) geometric impact cratering modeling optionally using a database of martian fresh crater
morphology; 7) vapor sublimation from or condensation on the land surface, with options for rate control by the interaction between incident radiation, reflected light, and local topography; 8)
mass wasting utilizing either the Howard (1994) or the Roering et al. (1999, 2001a) rate law. Bedrock can be optionally weathered and mass wasted assuming a critical slope angle steeper than the
critical gradient for regolith-mantled slopes. Mass wasted debris is instantaneously routed across exposed bedrock, and the debris flux can be specified to erode the bedrock; 9) groundwater flow
using the assumption of hydrostatic pressures and shallow flow relative to cell dimensions. Both recharge and seepage to the surface are modeled. Seepage discharge can be modeled to transport
sediment (seepage erosion) or to weather exposed bedrock (groundwater sapping); 10) deep-seated mass flows using either Glen's law or Bingham rheology using a hydrostatic stress assumption; 11)
eolian deposition and erosion in which the rate is determined by local topography; 12) lava flow and deposition from one or multiple vents. These model components vary in degree to which they are
based on established theory or utilize heuristicon established theory or utilize heuristic)
• Model:MICOM + (MICOM is a primitive equation numerical model that describes the evolution of momentum, mass, heat and salt in the ocean.)
• Model:MODFLOW 6 + (MODFLOW 6 is an object-oriented program an … MODFLOW 6 is an object-oriented program and framework developed to provide a platform for supporting multiple models and multiple
types of models within the same simulation. This version of MODFLOW is labeled with a "6" because it is the sixth core version of MODFLOW to be released by the USGS (previous core versions were
released in 1984, 1988, 1996, 2000, and 2005). In the new design, any number of models can be included in a simulation. These models can be independent of one another with no interaction, they
can exchange information with one another, or they can be tightly coupled at the matrix level by adding them to the same numerical solution. Transfer of information between models is isolated to
exchange objects, which allow models to be developed and used independently of one another. Within this new framework, a regional-scale groundwater model may be coupled with multiple local-scale
groundwater models. Or, a surface-water flow model could be coupled to multiple groundwater flow models. The framework naturally allows for future extensions to include the simulation of solute
transport.nclude the simulation of solute transport.)
• Model:MODFLOW + (MODFLOW is a three-dimensional finite-diff … MODFLOW is a three-dimensional finite-difference ground-water model that was first published in 1984. It has a modular structure that
allows it to be easily modified to adapt the code for a particular application. Many new capabilities have been added to the original model. OFR 00-92 (complete reference below) documents a
general update to MODFLOW, which is called MODFLOW-2000 in order to distinguish it from earlier versions.</br></br>MODFLOW-2000 simulates steady and nonsteady flow in an irregularly shaped flow
system in which aquifer layers can be confined, unconfined, or a combination of confined and unconfined. Flow from external stresses, such as flow to wells, areal recharge, evapotranspiration,
flow to drains, and flow through river beds, can be simulated. Hydraulic conductivities or transmissivities for any layer may differ spatially and be anisotropic (restricted to having the
principal directions aligned with the grid axes), and the storage coefficient may be heterogeneous. Specified head and specified flux boundaries can be simulated as can a head dependent flux
across the model's outer boundary that allows water to be supplied to a boundary block in the modeled area at a rate proportional to the current head difference between a "source" of water
outside the modeled area and the boundary block. MODFLOW is currently the most used numerical model in the U.S. Geological Survey for ground-water flow problems.</br></br>In addition to
simulating ground-water flow, the scope of MODFLOW-2000 has been expanded to incorporate related capabilities such as solute transport and parameter estimation.solute transport and parameter
• Model:MOM6 + (MOM6 is the latest generation of the Modul … MOM6 is the latest generation of the Modular Ocean Model which is a numerical model code for simulating the ocean general circulation.
MOM6 represents a major algorithmic departure from the previous generations of MOM (up to and including MOM5). Most notably, it uses the Arbitrary-Lagrangian-Eulerian (ALE) algorithm in the
vertical direction to allow the use of any vertical coordinate system including, geo-potential coordinates (z or z*), isopycnal coordinates, terrain-following coordinates and hybrid-/user-defined
coordinates. It is also based on the horizontal C-grid stencil, rather than the B-grid used by earlier MOM versions.n the B-grid used by earlier MOM versions.)
• Model:MPeat2D + (MPeat2D incorporates realistic spatial variability on the peatland and allows for more significant insights into the interplay between these complex feedback mechanisms.)
• Model:CASCADE + (Makes use of fast Delaunay triangulation a … Makes use of fast Delaunay triangulation and Voronoi diagram calculations to represent surface processes on an irregular, dynamically
evolving mesh. Processes include fluvial erosion, transport and deposition, hillslope (diffusion) processes, flexural isostasy, orographic precipitation. Designed to model processes at the
orogenic scale. Can be easily modified for other purposes by changing process laws.r other purposes by changing process laws.)
• Model:Manningseq-bouldersforpaleohydrology + (Matlab® code for paleo-hydrological flood … Matlab® code for paleo-hydrological flood flow reconstruction in a fluvial channel: first-order magnitude
estimations of maximum average flow velocity, peak discharge, and maximum flow height from boulder size and topographic input data (channel cross-section & channel bed slope).hannel cross-section
& channel bed slope).)
• Model:Reservoir + (Measure single reservoir performance using … Measure single reservoir performance using resilience, reliability, and vulnerability metrics; compute storage-yield-reliability
relationships; determine no-fail Rippl storage with sequent peak analysis; optimize release decisions using determinisitc and stochastic dynamic programming; evaluate inflow characteristics.
gramming; evaluate inflow characteristics.)
• Model:Coastal Dune Model + (Model describing the morphodynamic evolution of vegetated coastal foredunes.)
• Model:Sun fan-delta model + (Model for fluvial fan-delta evolution, ori … Model for fluvial fan-delta evolution, originally described by Sun et al. (2002) and later adapted by Limaye et al.
(2023). The model routes water and sediment across a grid from a single inlet and via a self-formed channel network, where local divergence in sediment flux drives bed elevation change. The model
represents hydrodynamics using rules for flow routing and stress partitioning. At large scales, other heuristics determine how channels branch and avulse, distributing water and sediment. The
original model, designed for fluvial fan-deltas that debouch into standing water, is extended to allow deposition of an alluvial fan in the absence of standing water.</br></br>References: </br>
Limaye, A. B., Adler, J. B., Moodie, A. J., Whipple, K. X., & Howard, A. D. (2023). Effect of standing water on formation of fan-shaped sedimentary deposits at Hypanis Valles, Mars. Geophysical
Research Letters, 50(4), e2022GL102367. https://doi.org/10.1029/2022GL102367</br></br>Sun, T., Paola, C., Parker, G., & Meakin, P. (2002). Fluvial fan deltas: Linking channel processes with
large-scale morphodynamics. Water Resources Research, 38(8), 26-1-26–10. https://doi.org/10.1029/2001WR000284, 26-1-26–10. https://doi.org/10.1029/2001WR000284)
• Model:Avulsion + (Model stream avulsion as random walk)
• Model:GISS GCM ModelE + (ModelE is the GISS series of coupled atmos … ModelE is the GISS series of coupled atmosphere-ocean models, which provides the ability to simulate many different
configurations of Earth System Models - including interactive atmospheric chemsitry, aerosols, carbon cycle and other tracers, as well as the standard atmosphere, ocean, sea ice and land surface
components.cean, sea ice and land surface components.)
• Model:Lake-Permafrost with Subsidence + (Models temperature of 1-D lake-permafrost … Models temperature of 1-D lake-permafrost system through time, given input surface temperature and solar
radiation. Model is fully implicit control volume scheme, and cell size can vary with depth. Thermal conductivity and specific heat capacity are dependent on cell substrate (% soil and % ice) and
temperature using the apparent heat capacity scheme where freezing/thawing occurs over a finite temperature range and constants are modified to account for latent heat. Lake freezes and thaws
depending on temperature; when no ice is present lake is fully mixed and can absorb solar radiation. Upper 10 m substrate contains excess ice and, if thawed, can subside by this amount (lake then
deepens by amount of subsidence). "Cell type" controls whether cell has excess ice, only pore space ice, or is lake water.ce, only pore space ice, or is lake water.) | {"url":"https://csdms.colorado.edu/wiki/Special:SearchByProperty/:Extended-20model-20description/Landlab-20component-20that-20implements-20a-201-20and-202D-20lithospheric-20flexure-20model.","timestamp":"2024-11-07T00:02:34Z","content_type":"text/html","content_length":"103025","record_id":"<urn:uuid:e4585f0b-4314-4ea9-a403-c0e0559d5061>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00863.warc.gz"} |
Please help: Carnot cycle
I'm trying to study for my physics test and I need help.
The problem is "A carnot engine operates with a cold reservoir temperature of 194 K and has an efficiency of 74.3 %. (A) What is the temperature of the hot reservoir? (B) If the engine does 8.99*10^4
J of net work during each cycle, how much heat does it absorb form the hot reservoir during a cycle?
I took the efficiency equation e=1-Tc/ Th and got 755 K for the first part of the equation, but I cannot seem to get the second part. Any suggestions??
How do you find out how much heat is absorbed by the hot reservoir during a cycle? | {"url":"https://www.crazyengineers.com/threads/please-help-carnot-cycle.33336","timestamp":"2024-11-05T07:32:48Z","content_type":"text/html","content_length":"75075","record_id":"<urn:uuid:9ec0b197-3234-401f-bb2c-f2293671b71c>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00169.warc.gz"} |
Process Data Mining: Partitioning Variance
Manufacturing facilities can be faced with major challenges when it comes to process improvement, largely because practitioners don’t always know enough about the underlying process factors (x’s) are
that drive the improvement metric (Y).
Practitioners might have a brainstorming session to tap into the collective experience of experts involved in the process, and design experiments to first uncover and then assess the importance of
likely variables. Afterward, following rounds of experimental work that may generate thousands of pounds of off-grade material, these factors might be identified, and the process optimized around
This process may have generated a history of product performance together with associated process data – sometimes hundreds of x’s, each with thousands of data points. Yet the task is just too
cumbersome; the relationships are often too complex to search randomly for relationships between Y and various combinations of the x’s. Recursive partitioning, a data-mining strategy commonly used in
the medical field, can cut through the clutter, frequently providing the line engineer with the crucial relationship he or she is looking for in a shorter time than is needed for a traditional design
of experiments.
Optimizing a Process’ Categorical Response
Before I describe this data mining strategy, consider these two examples drawn from the chemical industry. A nylon quality factor – a pass/fail metric – is tracked in Figure 1 as it drifted through
periods of good and bad performance (in this figure, P indicates the quality factor was below the control limit, while F means the factor was above the control limit). This polymer manufacturing
process routinely generated readings for 600 x’s – things like rates, temperatures and pressures. The line engineer needed to identify those x’s responsible for driving the quality factor.
Figure 1: Nylon Quality Factor, December to January
Data from 126 production lots was collected for each x, then a recursive partitioning strategy was used to identify, in this case, the two factors most important to driving the quality factor from
the 600 variables being monitored. Run charts for these two are shown in Figure 2. The P’s and F’s over the line gate data indicate the pass/fail metric for the quality factor at that point in time.
Generally, as the polymer rate went down and/or the line gate setting moved higher, the polymer quality factor failed.
Figure 2: Two Important x’s in Process’ Quality Factor
Figure 3 contrasts these two process factors driving the nylon’s quality factor. Again, the P’s and F’s designate the pass/fail regions of Figure 1. Now, it’s obvious where this process should run
with respect to these two variables. No experiments were run; instead, a quick analysis of the process data history turned up an answer that could explain 89 percent of the quality factor’s variance
over this time period. This example demonstrates how historic process data can be used to explain a binomial pass/fail response. I call this data-mining approach “matching process variables to a
categorical Y-mask” (here, the P and F grades).
Figure 3: Nylon Quality Factor Versus Process Model
Optimizing a Process’ Continuous Response
The second example treats continuous data: The iron (Fe) concentration in a process stream that is measured monthly over a 36-month period; ten process variables were measured as well. The questions
were: Which of these variables tracked iron concentration and what was their explicit relationship?
A recursive partitioning strategy identified two process factors (x factors 1 and 2) that are influential to iron levels. Together, they explained 74 percent of the iron data’s variation. Once the
recursive partitioning strategy identified the important x’s, the next step was to code each x to its respective Z score (subtract each x data mean and divide by its standard deviation), then use
conventional multilinear least squares analysis to determine the second-order model’s parameters:
Expression 1: Y = b[0] + b[1]x[1] + b[2]x[2] + b[12]x[1]x[2] + b[11]x[1]^2 + b[22]x[2]^2
A standard regression treatment led to the following model with R^2 = 0.74 (the model explained 74 percent of the variability in the iron data):
Expression 2: Fe (predicted) = 737 + 46.3x[1] – 143.9x[1]^2 + 204.1x[2] + 188.0 x[1]x[2]
More generally, the second-order model with n x’s will require an intercept (b[o]), n main effects (b[1] to b[N]), n square effects (b[11] to b[NN]) and (n/2) x (n-1) 2-factor interactions (b[12] to
b[(N-1)bN]). Only those judged to be significant in the regression will be kept in the final expression.
Figure 4 tracks the two process factors – the actual iron concentration and the iron concentration predicted from Equation 2. Letters A, B and C denote unique group membership that was created by
partitioning the iron data twice in a recursive partitioning treatment. Over this 36-month period, iron concentration rose as either process factor increased.
Note, when building a model like Equation 1, all x data (not Y data) needs to be coded to its respective Z scores for two reasons: 1) for uncoded data, square terms are likely to be collinear
(functions of) their main effect and 2) differences in magnitudes for the various x’s can cause rounding errors in the matrix multiplication used by regression programs.
Figure 4: Iron Concentration Versus Process Variables
Figure 5 draws the relationship between actual and predicted iron levels over this 36-month period. The partitioned groups concentrated the A group at low levels, the C group at high levels and the B
group at intermediate levels. Low iron concentrations would be expected by the process operating under group A conditions (each defined by a specific range of x[1] and x[2]).
Figure 5: Modeling 36 Months of Iron Data
Generally, practitioners can use the recursive partitioning strategy to identify specific process x’s – from the many that are being collected – that drive a process Y. If the Y is categorical, as
demonstrated in the first example, a categorical Y-mask can be created against which the x’s can be matched and process charts, like Figure 3, can be drawn. When the Y data is continuous, however, as
demonstrated in the second example, those few x’s that provide the best fit to Y can be identified, and then the explicit relationship between the Y and x’s (like Expression 2) can be determined
through standard regression techniques (e.g., 36 Y’s were regressed against 36 x[1] and x[2] pairs in a full second-order model that was reduced to contain only significant parameters.
Partitioning Variance
Any group of numbers can be described by their average and variance. If nothing else is known about their underlying population, the best model for this group is just that: the average and variance.
Models, however, can be developed to explain a portion of the variance, and the degree to which a model explains a variation is quantified through an analysis of its total sum-of-squares (TSS in
Expression 3, where n equals the number of data points). The more the model explains, the better the fit.
Expression 3 results from rearranging of the standard variance expression, and R-square is the percentage of TSS explained by the model, be it a regression expression or a simple factor analysis,
where different segments of data have been grouped (e.g., 74 percent of TSS was explained by the iron concentration model described in Expression 2).
Expression 3:
If a practitioner knows something further about the set of numbers – such as the fact that different machines were involved or different shifts or different feed stock – they can factor that
variation component from the TSS as known. These known differences account for a portion of the overall TSS.
Expression 4 segregates subsets of the numbers into two or more subgroups (recursive partitioning segregates only two). The single summation breaks into double summations (partitioning n data into
their m[1] and m[2] subgroups), and adds and subtracts group means within the parenthesis for zero effect inside the parenthesis. Expression 4, thus, is identical to Expression 3; it’s simply using
more information.
Expression 4:
Expression 5 (not shown) groups terms in the expression, but again nothing’s changed. When the squared multiplication is carried out, two square terms result, (Expression 7), because the interaction
term ( Expression 6) sums to zero, as this summation is being taken across all deviations, not their squares.
Expression 6:
The right side of Expression 7 is that portion of TSS that’s been explained by grouping the Y data into two subgroups. The left-side summation is what’s left of TSS that is unexplained (the variation
within each group).
Note: if the model does not explain Y’s variation very well, the group means become nearly equivalent, and Expression 7 reduces back to Expression 3. The overall mean remains the best model for the
Expression 7:
“within” & “between” groups
Expression 7 simplifies to Expression 8, where m is the number of members in each group.
Expression 8:
TSS can now be reduced by that amount of the variance explained by the model to TSS’ as shown in Expression 9.
Expression 9:
“Explained” “Unexplained”
Recursive Partitioning Strategy
A recursive partitioning strategy systematically takes each column of x data (often as many as 600 x’s), sorts each x and Y, and then systematically partitions the sorted Y data into two subgroups,
starting with smallest Y value in subgroup 1 and the rest of the Y values in subgroup 2. The splitting operation proceeds in steps by systematically transferring the smallest value from subgroup 2 to
subgroup 1. At each step the percent of TSS explained by that partition is calculated utilizing Expression 8. The ideal split of Y for that x is the one producing the largest R^2. Thus, this x
-induced split of Y will lead to a reduced TSS (i.e., some of the original TSS will have been explained).
To this point, Y data will have been split into two subgroups based on the fit to the most important x. Each of Y’s subgroups can then be subjected to the same strategy for the string of x’s to find
the next best x and its best split. This strategy becomes clearer with the following example (Table 1), where 10 Y values have been systematically split against 10 x1 values (upper right part of
Table 1).
Table 1 tracks the partitioning possible for nine data points, where three machines constitute a categorical variable x1: Y’s average is 26.44 and its TSS is 3,100.2. The question becomes: Are we
able to explain a significant portion of Y’s TSS by partitioning the nine data points into multiple groups? That is equivalent to asking whether Y shows any dependence on x, the machine mask.
Each row of the table tracks the results of a specific split. For example, the first split of Y would have the first Y value of 10 in Group 1 and the remaining nine (12, 9, … 63) in Group 2. Each
group average, its “within sum of squares” (from Expression 7) and the number (m) of members of each group are used to calculate the percentage of variance explained by that split (%TSSEXP). In this
case, the split of six in Group 1 and three in Group 2 explained the greatest proportion of Y’s variation: 85.2 percent of the total variation. Note, appropriate formulas are depicted in the lower
right portion of the table.
Each subgroup can now be taken separately through this partitioning strategy to look for further associations – hence, the strategy’s name, recursive partitioning. The lower part of the table tracks
further partitioning of Group 1.
Note: the total variation explained in the two splits (88.26 percent of Y’s 3,100.2 variation) was equivalent to the R^2 one would obtain in a typical ANOVA (analysis of variance) calculation. Of
course, if there were hundreds of x’s to choose from, the ANOVA would be tougher to decipher.
This is a trivial example utilizing masked data. In the real manufacturing world, there might be a thousand x’s to choose from, each with a thousand data points. This adds up to 1 million data points
to sort through – the proverbial needle in a haystack.
The second example (Table 2) is a continuous one, where Y is a linear function of x. Note that “err{N(0,50)}” is an error term meaning “normally distributed with mean zero and standard deviation 50.”
Here, the greatest percentage of TSS is explained by Y partitioning based on the six smallest x1 values and the other three (split 1 on the chart). Another 77.46 percent of that group’s variation
could be explained by splitting the six into two groups of three. Thus, the partitioning strategy would have chosen x1 (from potentially hundreds of x’s not shown) to then regress Y against to
develop the final model shown at the bottom of the table.
Again, both examples are trivial and meant to show how the recursive partitioning strategy works. For a real process, there would be many more x’s, and this strategy would be used to pick out those
best fitting the Y.
In this same manner, more complicated relationships involving several x’s could be identified and then these used to build an explicit model (like that in Expression 1) through standard regression
Note: The author’s Excel program PARTITION (available on request) can be used to perform recursive partitioning and was used to generate the models developed in this work. | {"url":"https://www.isixsigma.com/sampling-data/process-data-mining-partitioning-variance/","timestamp":"2024-11-05T19:14:40Z","content_type":"text/html","content_length":"223225","record_id":"<urn:uuid:124b81fe-8f5b-4ba9-a95b-349a55671765>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00767.warc.gz"} |
Bacterial Growth Rate Calculator | Calculator Crunch
Table of Contents
In bacteriology, it is important to determine bacteria’s rate of reproduction. A calculator for the growth rate of bacteria can help with this task. This article will explain what a calculator for
the growth rate of bacteria is, why it is significant, and how to use one in simple terms. A calculator for the growth rate of bacteria is a convenient tool that enables us to observe how quickly
bacteria can multiply. You only need to input a few simple numbers and, within no time, find out the growth rate. This can be applied in various areas, depending on whether you want to study these
microorganisms at the school level or in a professional laboratory setting.
What is a Bacterial Growth Rate Calculator?
A bacterial growth rate calculator tells you how fast bacteria multiply. Over time, bacteria split into new cells, then grow and divide themselves. Understanding their multiplication speed could be
useful in various fields, such as medicine or agriculture.
Why Calculate Bacterial Growth Rate?
There are several reasons why you may want to calculate the growth rate:
• Medicine: It helps us understand how bacterial infection spreads and develop treatments accordingly;
• Farming: Managing soil nutrients so as to enhance crop productivity can benefit from knowing microbial population dynamics;
• Environment: These calculations, among others, help with the ongoing monitoring required to track changes in ecosystem health indicators like water quality parameters.
How do you find out what the Growth Rate is?
To work out the growth rate, you need two things:
1. Initial number: The total count at the beginning;
2. The final number: The amount observed after some time has elapsed.
Bacteria Growth Rate Formula:
N[t] = N[0] * ( 1 + r)^t
N[t]: The amount of time t
N[0]: The amount at time 0
r: Growth rate
t: Time passed
The formula below can be used as an example:
Growth Rate=Final Number of Bacteria −Initial Number of Bacteria / Time
[latex] Growth Rate=(Final Number of Bacteria)(Initial Number of Bacteria)/Time [/latex]
Example Calculation:
Let’s say you start with 100 bacteria; after 2 hours, you have 300. Using the formula:
1. Initial Number of Bacteria: 100
2. Final Number of Bacteria: 300
3. Time: 2 hours
Growth Rate=(300−100)2=2002=100 bacteria per hour\text{Growth Rate} = \frac{(300 – 100)}{2} = \frac{200}{2} = 100 \text{ bacteria per hour}Growth Rate=2(300−100)=2200=100 bacteria per hour
So, the bacterial growth rate is 100 bacteria per hour.
Using a Bacterial Growth Rate Calculator:
Instead of calculating it yourself, use an online calculator for the growth rate of bacteria. They’re pretty straightforward:
Enter Initial Number: Type in the number you started with;
Enter Final Number: Input current amount;
Enter Time: Specify hours or days, etc., over which change occurred;
Get Result: Click the calculate button and see the growth rate displayed on the screen.
Tips for Accurate Calculations:
Be precise: Ensure that both the initial count and final figure are accurate down to the last digit;
Choose the right time: Use exact duration to get more reliable outcomes;
Use credible tools: Whether manual or electronic (calculator), ensure accuracy using reliable ones.
Suppose you need another calculator for your daily needs. In that case, you can visit this website, or if you are not familiar with any tutorials, you can visit our YouTube channel for detailed
Frequently Asked Questions on Bacterial Growth Rate Calculator:
What is a bacterial growth rate calculator?
A bacterial growth rate calculator is a tool that helps you figure out how fast bacteria are multiplying over time. You can do this by entering the number of bacteria at the start and after a
specific interval, as well as the time elapsed, in which case it will provide you with the growth rate.
Why is it necessary to calculate bacterial growth rate?
Calculating bacterial growth rates helps us comprehend how bacteria spread and reproduce. This knowledge has applications in medicine (to devise treatments and understand infections), agriculture (to
manage soil health and crops), and environmental science (to monitor ecosystems and water quality).
How do I use a bacterial growth rate calculator?
To utilize a bacterial growth rate calculator, follow these steps:
Enter Initial Number of Bacteria: Input the number of bacteria you started with.
Enter the Final Number of Bacteria: Input the number of bacteria you have after a given time period.
Enter Time Period: Input time over which observation was made about growth.
Calculate: Click the calculate button to get the growth rate.
Is it possible for me to manually calculate bacterial growth rates?
Yes, you can manually calculate bacterial growth rates by using this formula:
[ \text{Growth Rate} = \frac{(\text{Final Number of Bacteria}) – (\text{Initial Number of Bacteria})}{\text{Time}} ]
Just ensure that your initial count, final count, and time are all correct for accurate results.
What units are used for measuring bacterial growth rates?
Bacterial growth rates are usually expressed as the number of bacteria per unit of time, such as “bacteria per hour” or “bacteria per minute,” depending on the time periods used during measurements.
What if my numbers are not accurate?
Accurate measurements are vital for obtaining reliable results. The growth rate calculation will be inaccurate if your initial or final count is wrong or the time period is incorrect. You should
strive to make your measurements as precise as possible for the best outcome.
Is it necessary to use a bacterial growth rate calculator in research?
In a research context, employing a bacterial growth rates calculator can facilitate faster, more accurate results during the analysis of such processes while saving time, especially when dealing with
exacting growth measurements coupled with data analysis.
Where can I get a bacterial growth rate calculator?
Bacterial growth rate calculators can be found online; many educational and scientific websites provide free, user-friendly ones that you can use directly from your web browser. | {"url":"https://calculatorcrunch.com/bacterial-growth-rate-calculator/","timestamp":"2024-11-07T19:12:10Z","content_type":"text/html","content_length":"210945","record_id":"<urn:uuid:e41d7844-b4fb-4d44-9e8c-a006a0235f2d>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00187.warc.gz"} |
Synthetic Control: Role of rescaling
The question without an answer
Something that seems interesting to me is that the construction of treatment effects, or estimation of optimal weights say nothing about how should data be used, nor which variation should we be
interested in.
Granted, standard approach is to just use data asis, but that seems unsatisfactory. Consider, for example, a case when the interest is on analyzing the effect on GDP of a country level policy in the
US. The US being one of the largest economies in the world, it would be difficult, if not impossible to, ex ante, find good controls.
But what if we change the measure of interest? and look into GDP percapita, or GDP relative levels, or something else. After the estimation is done, we could certainly reconstruct the original
Implementing these kind of transformations would help finding better controls, but could have important impacts when estimating the placebo tests assessing the significance of the estimated effect.
Here the question:
To what extend can we transform our explanatory variables when implementing SC Would the transformations need to be the same for all units? or panel units?
Below, I show an example of what could happen when we make these decisions:
set scheme white2
capture program drop sc_doer
program sc_doer
** Estimates the Effect for California.
tempfile sc3
synth cigsale cigsale(1971) cigsale(1975) cigsale(1980) cigsale(1985), trunit(3) trperiod(1989) keep(`sc3') replace
** And all other states
forvalues i =1/39{
if `i'!=3 {
local pool
foreach j of local stl {
if `j'!=3 & `j'!=`i' local pool `pool' `j'
tempfile sc`i'
synth cigsale cigsale(1971) cigsale(1975) cigsale(1980) cigsale(1985) , ///
trunit(`i') trperiod(1989) keep(`sc`i'') replace counit(`pool')
** Collects the Saved files to estimate the Treatment effect
** and the p-value/ratio statistic
forvalues i =1/39{
use `sc`i'' , clear
gen tef`i' = _Y_treated - _Y_synthetic
egen sef`i'a =mean( (_Y_treated - _Y_synthetic)^2) if _time<=1988
egen sef`i'b =mean( (_Y_treated - _Y_synthetic)^2) if _time>1988
gen sef`i'aa=sqrt(sef`i'a[2])
gen sef`i'bb=sqrt(sef`i'b[_N])
replace sef`i'a=sef`i'aa
replace sef`i'b=sef`i'bb
drop if _time==.
keep tef`i' sef`i'* _time
save `sc`i'', replace
use `sc1'
forvalues i = 2/39 {
merge 1:1 _time using `sc`i'', nogen
global toplot
global toplot2
** Stores which models will be saved for plotting
forvalues i = 1/39 {
global toplot $toplot (line tef`i' _time, color(gs11) )
if (sef`i'a[1])<(2*sef3a[1]) {
global toplot2 $toplot2 (line tef`i' _time, color(gs11) )
** Estimates the post/pre RMSE ratio
capture matrix drop rt
forvalues i = 1/39 {
if (sef`i'a[1])<(2*sef3a[1]) {
matrix rt=nullmat(rt)\[`i',sef`i'b[1]/sef`i'a[1]]
svmat rt
egen rnk=rank(rt2)
** and the ranking /p-value for each period
gen rnk2=0
forvalues i = 1/39 {
if (sef`i'a[1])<(2*sef3a[1]) {
local t = `t'+1
replace rnk2=rnk2+(tef`i'<=tef3)
gen pv=rnk2*100/`t' | {"url":"https://friosavila.github.io/chatgpt/synth_05_09_2023/index.html","timestamp":"2024-11-13T17:45:56Z","content_type":"application/xhtml+xml","content_length":"144624","record_id":"<urn:uuid:e6f28586-3355-4f2f-9dbc-f1406fa3c7fa>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00312.warc.gz"} |
× simplify the expression exactly. (sqrt18)(sqrt72) | Question AI
× Simplify the expression exactly. (sqrt18)(sqrt72)
Not your question?Search it
Rosalyn PrattProfessional · Tutor for 6 years
1Simplify the radicand
2Simplify the radicand
4Simplify the radicand
Step-by-step video
× Simplify the expression exactly. (sqrt18)(sqrt72)
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Circular buffers
Let's take a closer look at how the temperature samples were collected in our calibrated thermistor project.
In many scientific projects, keeping good statistics helps to understand when data are real (or significant) and when they are the result of noise (random variations in the measurement). We created
the class Circular to collect samples and do simple statistics on them so we could find out how accurate our thermometer reading were. Here is that class once again:
class Circular
double samples[ 200 ];
long int count;
double mean_value;
double standard_deviation;
double variance;
enum { COUNT = (sizeof samples / sizeof *samples) };
Circular( void ) { count = 0; }
void store( double value ) { samples[ count++ % (sizeof samples / sizeof *samples) ] = value; }
calculate_statistics( void )
double sum = 0;
int cnt = min( count, COUNT );
for( int i = 0; i < cnt; i++ )
sum += samples[ i ];
mean_value = sum / cnt;
variance = 0;
for( int i = 0; i < cnt; i++ )
double deviation = samples[ i ] - mean_value;
variance += deviation * deviation;
variance /= ( cnt - 1 );
standard_deviation = sqrt( variance );
double mean( void ) { return mean_value; }
double std_dev( void ) { return standard_deviation; }
double var( void ) { return variance; }
int num_samples( void ) { return min( count, COUNT ); }
The class gets its name from a simple concept. We want to collect up to 200 samples, and no more. But we always want to analyze the last samples that came in. We put each sample into its own slot in
an array that can hold 200 samples. But when we get the 201st sample, we copy it into the first slot, over-writing the first sample. The next sample goes into slot 2, and so on.
Let's look at how that happens. The array of samples is declared by the line
double samples[ 200 ];
The line
enum { COUNT = (sizeof samples / sizeof *samples) };
is just a clever way to set COUNT equal to 200. It says to take the number of bytes in the entire array, and divide that by the number of bytes in the first slot. We do this instead of using the
value 200 so that we can easily change the 200 to some other number if we need more samples, or we need fewer samples (to leave room for other data later). Now we just change the one number, and all
the other parts of the code that depend on the size of the array will simply work without change.
The line
void store( double value ) { samples[ count++ % (sizeof samples / sizeof *samples) ] = value; }
might be easier to read if we break it up into several lines:
store( double value )
samples[ count++ % (sizeof samples / sizeof *samples) ] = value;
There are a few tricky things going on here. First, we have the variable count, which keeps track of how many samples we have collected. The two plus signs after it mean that it gets incremented (we
add one to it) after it is used. The percent sign means to take the value of count (before it is incremented), and divide it by 200 (our sizeof trick), and use the remainder as our index into the
array. Then the variable value is stored in thesamples array at that index.
At first, count is equal to zero (we'll see how that happens in a moment).
Zero divided by 200 is 0, with a remainder of 0, so the first value goes into samples[0].
Now count is incremented, so its value is 1.
One divided by 200 is 0, with a remainder of 1. So the next value goes into slot 1.
Now lets jump ahead to when we have 200 samples. The count variable is 200. 200 divided by 200 is 1, with a 0 as the remainder.
So the next sample goes into slot 0.
It's as if we had a circle of pots to put things in, and we just keep going around the circle, putting things in the pots.
Right after the keyword public is a funny declaration:
Circular( void ) { count = 0; }
It uses the same name as the class itself, but it looks like a method declaration without a return type.
This is called a constructor. Whenever we create an instance of our class, this method will be called, just once.
It sets up things in the class. In this case, it sets out count variable to zero.
Now let's look at the method calculate_statistics().
Sometimes out count variable will be less than 200, and sometimes it will be much greater. We want to know how many samples we have collected. That number is the smaller of count or 200. That is
what min() gives us.
int cnt = min( count, COUNT );
Now we want to visit all the samples we have collected, and add them up. We use thefor() statement to do that.
for( int i = 0; i < cnt; i++ )
sum += samples[ i ];
It says to make a new variable called i and set it to zero. Then as long as i is less than cnt, do the next line, and then increment i.
The next statement says to add each sample into the variable sum.
Dividing that sum by the number of samples gives us the mean (the arithmetic average of all the samples).
The next for() statement calculates the sum of the squared differences from the mean.
for( int i = 0; i < cnt; i++ )
double deviation = samples[ i ] - mean_value;
variance += deviation * deviation;
That is the variance. Note that this for() loop has curly braces around the two statements below it. The curly braces say to treat anything between them as if it were a single line. So in this case,
the body of the loop has two statements that are executed each time through the loop.
To get the standard deviation of the sample set, we divide the variance by one less than the sample size, and take the square root. (I am being very brief here -- if you need a refresher on simple
statistics, Google for "variance", and "standard deviation").
Our circular buffer class is now a convenient package that can be copied into other programs that need to collect samples of data and do simple statistics on them. | {"url":"https://scitoys.com/circular_buffers.html","timestamp":"2024-11-14T02:17:36Z","content_type":"text/html","content_length":"33649","record_id":"<urn:uuid:906b7bf9-2206-4662-a658-107a6e21e43d>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00072.warc.gz"} |
What is the equation of linearization?
The Linearization of a function f(x,y) at (a,b) is L(x,y) = f(a,b)+(x−a)fx(a,b)+(y−b)fy(a,b). This is very similar to the familiar formula L(x)=f(a)+f′(a)(x−a) functions of one variable, only with an
extra term for the second variable.
How do you Linearize?
Mathematical form:
1. Make a new calculated column based on the mathematical form (shape) of your data.
2. Plot a new graph using your new calculated column of data on one of your axes.
3. If the new graph (using the calculated column) is straight, you have succeeded in linearizing your data.
4. Draw a best fit line USING A RULER!
Is exponential regression linear?
Observation: Since αeβ(x+1) = αeβx · eβ, we note that an increase in x of 1 unit results in y being multiplied by eβ. Observation: A model of the form ln y = βx + δ is referred to as a log-level
regression model.
How do you Linearize data?
How do you linearize a nonlinear system?
Linearization is a linear approximation of a nonlinear system that is valid in a small region around an operating point. For example, suppose that the nonlinear function is y = x 2 . Linearizing this
nonlinear function about the operating point x = 1, y = 1 results in a linear function y = 2 x − 1 .
What is Linearizing a function?
Linearizations of a function are lines—usually lines that can be used for purposes of calculation. Linearization is an effective method for approximating the output of a function at any based on the
value and slope of the function at , given that is differentiable on (or ) and that is close to .
What is meant by linearization?
Linearization. In mathematics linearization refers to finding the linear approximation to a function at a given point. In the study of dynamical systems, linearization is a method for assessing the
local stability of an equilibrium point of a system of nonlinear differential equations or discrete dynamical systems.
What is linearizing a graph?
Most relationships which are not linear, can be graphed so that the graph is a straight line. This process is called a linearization of the data. This does not change the fundamental relationship or
what it represents, but it does change how the graph looks.
Why are equations linearized?
In most cases, the equation must be modified or linearized so that the variables plotted are different than the variables measured but produce a straight line. Linearizing equations is this process
of modifying an equation to pro- duce new variables which can be plotted to produce a straight line graph.
What is the formula for exponential regression?
For the data (x,y), the exponential regression formula is y=exp(c)exp(mx). In this equation m is the slope and c is the intercept of the linear regression model best fitted to the data (x, ln(y)).
What is the difference between linear and exponential regression?
In linear regression, the function is a linear (straight-line) equation. In power or exponential regression, the function is a power (polynomial) equation of the form or an exponential function in
the form .
Why do we Linearize data?
When data sets are more or less linear, it makes it easy to identify and understand the relationship between variables. You can eyeball a line, or use some line of best fit to make the model between
Why do we Linearize equation?
Linearization can be used to give important information about how the system behaves in the neighborhood of equilibrium points. Typically we learn whether the point is stable or unstable, as well as
something about how the system approaches (or moves away from) the equilibrium point.
Is it Linearised or linearized?
verb (used with object), lin·e·ar·ized, lin·e·ar·iz·ing. to make linear; give linear form to.
Why do we use linearization? | {"url":"https://ecce216.com/reviews/what-is-the-equation-of-linearization/","timestamp":"2024-11-12T10:24:15Z","content_type":"text/html","content_length":"54196","record_id":"<urn:uuid:5c6267c9-2038-443c-8807-4850d236f0bd>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00372.warc.gz"} |
Lesson 7
Building Polygons (Part 2)
Let’s build more triangles.
7.1: Where Is Lin?
At a park, the slide is 5 meters east of the swings. Lin is standing 3 meters away from the slide.
1. Draw a diagram of the situation including a place where Lin could be.
2. How far away from the swings is Lin in your diagram?
3. Where are some other places Lin could be?
7.2: How Long Is the Third Side?
Use the applet to answer the questions.
1. Build as many different triangles as you can that have one side length of 5 inches and one of 4 inches. Record the side lengths of each triangle you build.
2. Are there any other lengths that could be used for the third side of the triangle but aren't values of the sliders?
3. Are there any lengths that are values of the sliders but could not be used as the third side of the triangle?
Assuming you had access to strips of any length, and you used the 9-inch and 5-inch strips as the first two sides, complete the sentences:
1. The third side can't be _____ inches or longer.
2. The third side can't be _____ inches or shorter.
7.3: Swinging the Sides Around
We'll explore a method for drawing a triangle that has three specific side lengths. Use the applet to answer the questions.
1. Follow these instructions to mark the possible endpoints of one side:
1. For now, ignore segment \(AC\) , the 3-inch side length on the left side
2. What shape have you drawn while moving \(BD\) around? Why? Which tool in your geometry toolkit can do something similiar?
3. Use your drawing to create two unique triangles, each with a base of length 4 inches and a side of length 3 inches. Use a different color to draw each triangle.
4. Repeat the previous instructions, letting segment \(AC\) be the 3-unit side length.
5. Using a third color, draw a point where the two traces intersect. Using this third color, draw a triangle with side lengths of 4 inches, 3 inches, and 3 inches.
If we want to build a polygon with two given side lengths that share a vertex, we can think of them as being connected by a hinge that can be opened or closed:
All of the possible positions of the endpoint of the moving side form a circle:
You may have noticed that sometimes it is not possible to build a polygon given a set of lengths. For example, if we have one really, really long segment and a bunch of short segments, we may not be
able to connect them all up. Here's what happens if you try to make a triangle with side lengths 21, 4, and 2:
The short sides don't seem like they can meet up because they are too far away from each other.
If we draw circles of radius 4 and 2 on the endpoints of the side of length 21 to represent positions for the shorter sides, we can see that there are no places for the short sides that would allow
them to meet up and form a triangle.
In general, the longest side length must be less than the sum of the other two side lengths. If not, we can’t make a triangle!
If we can make a triangle with three given side lengths, it turns out that the measures of the corresponding angles will always be the same. For example, if two triangles have side lengths 3, 4, and
5, they will have the same corresponding angle measures. | {"url":"https://im-beta.kendallhunt.com/MS/students/2/7/7/index.html","timestamp":"2024-11-07T23:05:38Z","content_type":"text/html","content_length":"900431","record_id":"<urn:uuid:6a991312-c2e2-4385-86dc-f44c81146eb3>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00159.warc.gz"} |
sum of odd numbers | Sololearn: Learn to code for FREE!
sum of odd numbers
how to print sum of first 50 odd numbers using do while loop in c
dont know c syntax, but in cs it would be like this: int num =1; int count = 0 int sum = 0; do { sum+=num; num+=2; count++; } while(count<50);
bcz ive question in my assignment...
Sergey Semendyaev
If the numbers are increased two by two, the counter must be half (count<25) =)
No, Luciano, if count<25, you'll get sum of first 25 odd numbers. Each number increases count in my logic.
Luciano, look at your own code's output and count how many numbers does it have)) (if you still didn't get what I mean the 50th odd number is 99)
Sergey Semendyaev
First, it's not my code, just add a line to the code you posted above. Basically "num" is increased by two for each interaction of the loop. In this case the amount is reached in half the time
(interactions). "Count" represents the number of interactions (50), not the first 50 numbers. I'm sorry if you do not understand well, my English is not good. =)
But each iteration adds 1 number to the sum, so if you need sum of 50 numbers there should be 49 iterations, but if you need sum of odd numbers from 1 to 50 than code should be different. So it
depends on the task and my code sums first 50 odd numbers as was asked in initial post. Peace)â
As I said, my English is not very good and I interpret odd numbers 1 to 50. See you =) | {"url":"https://www.sololearn.com/en/discuss/1217474/sum-of-odd-numbers","timestamp":"2024-11-03T05:57:28Z","content_type":"text/html","content_length":"931544","record_id":"<urn:uuid:e9382667-3a97-42c1-b55b-66766f8dcaf6>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00304.warc.gz"} |
Natural transformations
Natural transformations by Mark Seemann
Mappings between functors, with some examples in C#.
This article is part of a series of articles about functor relationships. In this one you'll learn about natural transformations, which are simply mappings between two functors. It's probably the
easiest relationship to understand. In fact, it may be so obvious that your reaction is: Is that it?
In programming, a natural transformation is just a function from one functor to another. A common example is a function that tries to extract a value from a collection. You'll see specific examples a
little later in this article.
Laws #
In this, the dreaded section on laws, I have a nice surprise for you: There aren't any (that we need worry about)!
In the broader context of category theory there are, in fact, rules that a natural transformation must follow.
"Haskell's parametric polymorphism has an unexpected consequence: any polymorphic function of the type:
alpha :: F a -> G a
"where F and G are functors, automatically satisfies the naturality condition."
While C# isn't Haskell, .NET generics are similar enough to Haskell parametric polymorphism that the result, as far as I can tell, carry over. (Again, however, we have to keep in mind that C# doesn't
distinguish between pure functions and impure actions. The knowledge that I infer translates for pure functions. For impure actions, there are no guarantees.)
The C# equivalent of the above alpha function would be a method like this:
G<T> Alpha<T>(F<T> f)
where both F and G are functors.
Safe head #
Natural transformations easily occur in normal programming. You've probably written some yourself, without being aware of it. Here are some examples.
It's common to attempt to get the first element of a collection. Collections, however, may be empty, so this is not always possible. In Haskell, you'd model that as a function that takes a list as
input and returns a Maybe as output:
Prelude Data.Maybe> :t listToMaybe
listToMaybe :: [a] -> Maybe a
Prelude Data.Maybe> listToMaybe []
Prelude Data.Maybe> listToMaybe [7]
Just 7
Prelude Data.Maybe> listToMaybe [3,9]
Just 3
Prelude Data.Maybe> listToMaybe [5,9,2,4,4]
Just 5
In many tutorials such a function is often called safeHead, because it returns the head of a list (i.e. the first item) in a safe manner. It returns Nothing if the list is empty. In F# this function
is called tryHead.
In C# you could write a similar function like this:
public static Maybe<T> TryFirst<T>(this IEnumerable<T> source)
if (source.Any())
return new Maybe<T>(source.First());
return Maybe.Empty<T>();
This extension method (which is really a pure function) is a natural transformation between two functors. The source functor is the list functor and the destination is the Maybe functor.
Here are some unit tests that demonstrate how it works:
public void TryFirstWhenEmpty()
Maybe<Guid> actual = Enumerable.Empty<Guid>().TryFirst();
Assert.Equal(Maybe.Empty<Guid>(), actual);
[InlineData(new[] { "foo" }, "foo")]
[InlineData(new[] { "bar", "baz" }, "bar")]
[InlineData(new[] { "qux", "quux", "quuz", "corge", "corge" }, "qux")]
public void TryFirstWhenNotEmpty(string[] arr, string expected)
Maybe<string> actual = arr.TryFirst();
Assert.Equal(new Maybe<string>(expected), actual);
All these tests pass.
Safe index #
The above safe head natural transformation is just one example. Even for a particular combination of functors like List to Maybe many natural transformations may exist. For this particular
combination, there are infinitely many natural transformations.
You can view the safe head example as a special case of a more general set of safe indexing. With a collection of values, you can attempt to retrieve the value at a particular index. Since a
collection can contain an arbitrary number of elements, however, there's no guarantee that there's an element at the requested index.
In order to avoid exceptions, then, you can try to retrieve the value at an index, getting a Maybe value as a result.
The F# Seq module defines a function called tryItem. This function takes an index and a sequence (IEnumerable<T>) and returns an option (F#'s name for Maybe):
> Seq.tryItem 2 [2;5;3;5];;
val it : int option = Some 3
The tryItem function itself is not a natural transformation, but because of currying, it's a function that returns a natural transformation. When you partially apply it with an index, it becomes a
natural transformation: Seq.tryItem 3 is a natural transformation seq<'a> -> 'a option, as is Seq.tryItem 4, Seq.tryItem 5, and so on ad infinitum. Thus, there are infinitely many natural
transformations from the List functor to the Maybe functor, and safe head is simply Seq.tryItem 0.
In C# you can use the various Func delegates to implement currying, but if you want something that looks a little more object-oriented, you could write code like this:
public sealed class Index
private readonly int index;
public Index(int index)
this.index = index;
public Maybe<T> TryItem<T>(IEnumerable<T> values)
var candidate = values.Skip(index).Take(1);
if (candidate.Any())
return new Maybe<T>(candidate.First());
return Maybe.Empty<T>();
This Index class captures an index value for potential use against any IEnumerable<T>. Thus, the TryItem method is a natural transformation from the List functor to the Maybe functor. Here are some
[InlineData(0, new string[0])]
[InlineData(1, new[] { "bee" })]
[InlineData(2, new[] { "nig", "fev" })]
[InlineData(4, new[] { "sta", "ali" })]
public void MissItem(int i, string[] xs)
var idx = new Index(i);
Maybe<string> actual = idx.TryItem(xs);
Assert.Equal(Maybe.Empty<string>(), actual);
[InlineData(0, new[] { "foo" }, "foo")]
[InlineData(1, new[] { "bar", "baz" }, "baz")]
[InlineData(1, new[] { "qux", "quux", "quuz" }, "quux")]
[InlineData(2, new[] { "corge", "grault", "fred", "garply" }, "fred")]
public void FindItem(int i, string[] xs, string expected)
var idx = new Index(i);
Maybe<string> actual = idx.TryItem(xs);
Assert.Equal(new Maybe<string>(expected), actual);
Since there are infinitely many integers, there are infinitely many such natural transformations. (This is strictly not true for the above code, since there's a finite number of 32-bit integers.
Exercise: Is it possible to rewrite the above Index class to instead work with BigInteger?)
The Haskell natural-transformation package offers an even more explicit way to present the same example:
import Control.Natural
tryItem :: (Eq a, Num a, Enum a) => a -> [] :~> Maybe
tryItem i = NT $ lookup i . zip [0..]
You can view this tryItem function as a function that takes a number and returns a particular natural transformation. For example you can define a value called tryThird, which is a natural
transformation from [] to Maybe:
λ tryThird = tryItem 2
λ :t tryThird
tryThird :: [] :~> Maybe
Here are some usage examples:
λ tryThird # []
λ tryThird # [1]
λ tryThird # [2,3]
λ tryThird # [4,5,6]
Just 6
λ tryThird # [7,8,9,10]
Just 9
In all three languages (F#, C#, Haskell), safe head is really just a special case of safe index: Seq.tryItem 0 in F#, new Index(0) in C#, and tryItem 0 in Haskell.
Maybe to List #
You can also move in the opposite direction: From Maybe to List. In F#, I can't find a function that translates from option 'a to seq 'a (IEnumerable<T>), but there are both Option.toArray and
Option.toList. I'll use Option.toList for a few examples:
> Option.toList (None : string option);;
val it : string list = []
> Option.toList (Some "foo");;
val it : string list = ["foo"]
Contrary to translating from List to Maybe, going the other way there aren't a lot of options: None translates to an empty list, and Some translates to a singleton list.
Using a Visitor-based Maybe in C#, you can implement the natural transformation like this:
public static IEnumerable<T> ToList<T>(this IMaybe<T> source)
return source.Accept(new ToListVisitor<T>());
private class ToListVisitor<T> : IMaybeVisitor<T, IEnumerable<T>>
public IEnumerable<T> VisitNothing
get { return Enumerable.Empty<T>(); }
public IEnumerable<T> VisitJust(T just)
return new[] { just };
Here are some examples:
public void NothingToList()
IMaybe<double> maybe = new Nothing<double>();
IEnumerable<double> actual = maybe.ToList();
[InlineData( 0)]
public void JustToList(double d)
IMaybe<double> maybe = new Just<double>(d);
IEnumerable<double> actual = maybe.ToList();
Assert.Single(actual, d);
In Haskell this natural transformation is called maybeToList - just when you think that Haskell names are always abstruse, you learn that some are very explicit and self-explanatory.
If we wanted, we could use the natural-transformation package to demonstrate that this is, indeed, a natural transformation:
λ :t NT maybeToList
NT maybeToList :: Maybe :~> []
There would be little point in doing so, since we'd need to unwrap it again to use it. Using the function directly, on the other hand, looks like this:
λ maybeToList Nothing
λ maybeToList $ Just 2
λ maybeToList $ Just "fon"
A Nothing value is always translated to the empty list, and a Just value to a singleton list, exactly as in the other languages.
Exercise: Is this the only possible natural transformation from Maybe to List?
Maybe-Either relationships #
The Maybe functor is isomorphic to Either where the left (or error) dimension is unit. Here are the two natural transformations in F#:
module Option =
// 'a option -> Result<'a,unit>
let toResult = function
| Some x -> Ok x
| None -> Error ()
// Result<'a,unit> -> 'a option
let ofResult = function
| Ok x -> Some x
| Error () -> None
In F#, Maybe is called option and Either is called Result. Be aware that the F# Result discriminated union puts the Error dimension to the right of the Ok, which is opposite of Either, where left is
usually used for errors, and right for successes (because what is correct is right).
Here are some examples:
> Some "epi" |> Option.toResult;;
val it : Result<string,unit> = Ok "epi"
> Ok "epi" |> Option.ofResult;;
val it : string option = Some "epi"
Notice that the natural transformation from Result to Option is only defined for Result values where the Error type is unit. You could also define a natural transformation from any Result to option:
// Result<'a,'b> -> 'a option
let ignoreErrorValue = function
| Ok x -> Some x
| Error _ -> None
That's still a natural transformation, but no longer part of an isomorphism due to the loss of information:
> (Error "Catastrophic failure" |> ignoreErrorValue : int option);;
val it : int option = None
Just like above, when examining the infinitely many natural transformations from List to Maybe, we can use the Haskell natural-transformation package to make this more explicit:
ignoreLeft :: Either b :~> Maybe
ignoreLeft = NT $ either (const Nothing) Just
ignoreLeft is a natural transformation from the Either b functor to the Maybe functor.
Using a Visitor-based Either implementation (refactored from Church-encoded Either), you can implement an equivalent IgnoreLeft natural transformation in C#:
public static IMaybe<R> IgnoreLeft<L, R>(this IEither<L, R> source)
return source.Accept(new IgnoreLeftVisitor<L, R>());
private class IgnoreLeftVisitor<L, R> : IEitherVisitor<L, R, IMaybe<R>>
public IMaybe<R> VisitLeft(L left)
return new Nothing<R>();
public IMaybe<R> VisitRight(R right)
return new Just<R>(right);
Here are some examples:
[InlineData("Catastrophic failure")]
[InlineData("Important information!")]
public void IgnoreLeftOfLeft(string msg)
IEither<string, int> e = new Left<string, int>(msg);
IMaybe<int> actual = e.IgnoreLeft();
Assert.Equal(new Nothing<int>(), actual);
public void IgnoreLeftOfRight(int i)
IEither<string, int> e = new Right<string, int>(i);
IMaybe<int> actual = e.IgnoreLeft();
Assert.Equal(new Just<int>(i), actual);
I'm not insisting that this natural transformation is always useful, but I've occasionally found myself in situations were it came in handy.
Natural transformations to or from Identity #
Some natural transformations are a little less obvious. If you have a NotEmptyCollection<T> class as shown in my article Semigroups accumulate, you could consider the Head property a natural
transformation. It translates a NotEmptyCollection<T> object to a T object.
This function also exists in Haskell, where it's simply called head.
The input type (NotEmptyCollection<T> in C#, NonEmpty a in Haskell) is a functor, but the return type is a 'naked' value. That doesn't look like a functor.
True, a naked value isn't a functor, but it's isomorphic to the Identity functor. In Haskell, you can make that relationship quite explicit:
headNT :: NonEmpty :~> Identity
headNT = NT $ Identity . NonEmpty.head
While not particularly useful in itself, this demonstrates that it's possible to think of the head function as a natural transformation from NonEmpty to Identity.
Can you go the other way, too?
Yes, indeed. Consider monadic return. This is a function that takes a 'naked' value and wraps it in a particular monad (which is also, always, a functor). Again, you may consider the 'naked' value as
isomorphic with the Identity functor, and thus return as a natural transformation:
returnNT :: Monad m => Identity :~> m
returnNT = NT $ return . runIdentity
We might even consider if a function a -> a (in Haskell syntax) or Func<T, T> (in C# syntax) might actually be a natural transformation from Identity to Identity... (It is, but only one such function
Not all natural transformations are useful #
Are are all functor combinations possible as natural transformations? Can you take any two functors and define one or more natural transformations? I'm not sure, but it seems clear that even if it is
so, not all natural transformations are useful.
Famously, for example, you can't get the value out of the IO functor. Thus, at first glance it seems impossible to define a natural transformation from IO to some other functor. After all, how would
you implement a natural transformation from IO to, say, the Identity functor. That seems impossible.
On the other hand, this is possible:
public static IEnumerable<T> Collapse<T>(this IO<T> source)
yield break;
That's a natural transformation from IO<T> to IEnumerable<T>. It's possible to ignore the input value and always return an empty sequence. This natural transformation collapses all values to a single
return value.
You can repeat this exercise with the Haskell natural-transformation package:
collapse :: f :~> []
collapse = NT $ const []
This one collapses any container f to a List ([]), including IO:
λ collapse # (return 10 :: IO Integer)
λ collapse # putStrLn "ploeh"
Notice that in the second example, the IO action is putStrLn "ploeh", which ought to produce the side effect of writing to the console. This is effectively prevented - instead the collapse natural
transformation simply produces the empty list as output.
You can define a similar natural transformation from any functor (including IO) to Maybe. Try it as an exercise, in either C#, Haskell, or another language. If you want a Haskell-specific exercise,
also define a natural transformation of this type: Alternative g => f :~> g.
These natural transformations are possible, but hardly useful.
Conclusion #
A natural transformation is a function that translates one functor into another. Useful examples are safe or total collection indexing, including retrieving the first element from a collection. These
natural transformations return a populated Maybe value if the element exists, and an empty Maybe value otherwise.
Other examples include translating Maybe values into Either values or Lists.
A natural transformation can easily involve loss of information. Even if you're able to retrieve the first element in a collection, the return value includes only that value, and not the rest of the
A few natural transformations may be true isomorphisms, but in general, being able to go in both directions isn't required. In degenerate cases, a natural transformation may throw away all
information and map to a general empty value like the empty List or an empty Maybe value.
Next: Functor products.
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Published: Monday, 18 July 2022 08:12:00 UTC | {"url":"https://blog.ploeh.dk/2022/07/18/natural-transformations/","timestamp":"2024-11-14T12:33:08Z","content_type":"text/html","content_length":"48167","record_id":"<urn:uuid:338c3894-4524-49b6-9186-a21ee350669c>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00430.warc.gz"} |
Review of Basic Statistical Analysis Methods for Analyzing Data - Part 3
Establishing Relationships Between Two Variables
Another important application of OLS is the comparison of two different data sets. In this case, we can think of one of the time series as constituting the independent variable x and the other
constituting the independent variable y. The methods that we discussed in the previous section for estimating trends in a time series generalize readily, except our predictor is no longer time, but
rather, some variable. Note that the correction for autocorrelation is actually somewhat more complicated in this case, and the details are beyond the scope of this course. As a general rule, even if
the residuals show substantial autocorrelation, the required correction to the statistical degrees of freedom (N' ), will be small as long as either one of the two time series being compared has low
autocorrelation. Nonetheless, any substantial structure in the residuals remains a cause for concern regarding the reliability of the regression results.
We will investigate this sort of application of OLS with an example, where our independent variable is a measure of El Niño — the so-called Niño 3.4 index — and our dependent variable is December
average temperatures in State College, PA.
The demonstration is given in three parts below:
Video: Demo - Part 1 (3:22)
Click here for a transcript
PRESENTER: Now we're going to look at a somewhat different situation where our independent variable is no longer time but it's some quantity it could be temperature it could be an index of El niño or
the North Atlantic Oscillation. Let's look at an example of that sort. We are going to look at the relationship between El niño and December temperatures in State College Pennsylvania. We can plot
out that relationship as a scatterplot. On the y-axis we have December temperature in State College, the x-axis is our independent variable the niño 3.4 index negative values indicate La niña and
positive values indicate El niños. The strength of the relationship between the two is going to be determined by the trendline. That describes how December temperatures in State College depend on El
niño and by fitting the regression, we obtain a slope of 0.7397. That means for each unit change in El niño in niño 3.4 we get a 0.74 unit change in temperature. So for a moderate El niño event where
the niño 3.4 index is in the range of plus one, that would imply that December temperatures in State College, for that year, are 0.74 degrees Fahrenheit 0.74 degrees Fahrenheit warmer than usual. And
for modestly strong La niña in weather niño 3.4 indexes on the order of minus one or so the December State College December temperatures would be about zero point seven four degrees colder than
normal. You can also see that the y-intercept here the case when the niño 3.4 index is zero we get roughly the climatological value for December temperatures 30.9. Now the correlation coefficient is
associated with that linear regression in this case 0.74. Now we have a 107 years our data set as before it goes from 1888 to 1994. If we use our table, and take n equal to 107, and R of 0.74, we
find that the one tailed value of P is 0.365 the two tailed value is 0.073. So if I threshold for significance where P of 0.05 the 95 percent significance level, then that relationship a correlation
of coefficient of 0.174 with 107 years of information would be significant for one tailed test but it would not past the 0.05, the 95% significance threshold, for a two-tailed test. So we have to ask
the question, which is more appropriate here, the one tailed test or the two tailed test. Now if you had a reason to believe that El niño events form the northeastern US, for example, you might
motivate a one tailed test since only a positive relationship would be consistent with your expectations. But if we didn't know beforehand whether El niños had a cooling influence or warming
influence on the northeastern US you might argue for a two-tailed test. So whether or not the relationship is significant at the P equals 0.05 level is going to depend on which type of hypothesis
test were able to use in this case.
Video: Demo - Part 2 (4:10)
Click here for a transcript
PRESENTER: Let's continue with this analysis. Now what I'm going to do here is plot instead the temperature as a function of the year instead of niño 3.4. That's plot number one. That's a State
College December temperatures. And now for pot number two. I'm going to plot the niño 3.4 index as a function of year. I use axis B here to put them on the same scale so here we could see the two
series. We had the State College December temperatures in blue and the niño 3.4 index in yellow. And you can see that in various years it does seem to be a little bit of a relationship between large
positive departures in the niño 3.4 index are associated with warm Decembertemperatures, and large negative departures are associated with cold temperatures. We can visually see that relationship we
also saw we plotted the two variables in a two dimensional scatterplot and looked at the slope of the line relating the two datasets. Here now we're looking at the same time looking at the time
series of the two data sets and we can see some of that positive covariance, if you will, that there does seem to be a positive relationship although we already know it's a fairly weak
relationship.So let's do a formal regression.
So I'm going to take away the niño series here. What we've got here is our State College December temperatures in blue. Now our regression model is going to use the niño 3.4 index as the independent
parameter and temperature as our dependent variable We'll run the linear regression. There is a slope 0.74 is the coefficient that describes the relationship between the niño 3.4 index of december
temperatures. It's positive, we already saw the slope was positive. There's also a constant term we're not going to worry much about here. What we're really interested in is the slope of the
regression line that describes that changes in temperature depends on changes in the niño 3.4 index. And as we've seen, 0.74 implies that for a unit increase in niño 3.4 an anomaly of +1 on the niño
3.4 scale, we'll get a temperature for December that, on average, is 0.74 degrees fahrenheit warmer than average. The r-squared value right here is 0.032 and if we take that number, take the square
root of that that's an r-value of 0.1734 and we know that's a positive correlation because the slope is positive. We already looked up the statistical significance of that number and we found that
for a one-sided hypothesis test that the relationship is significant at the 0.05 level. But if we were using a two-sided significant criterion hypothesis test, that is to say, if we didn't know a
priori whether we had a reason to believe that El niños warm or cool State College December temperatures, then the relationship would not quite be statistically significant. So we've calculated the
linear model, so now we can plot it. So now I'm going to plot year and model output on the same scale. You can change the scale up these axes by clicking on these arrows arrows. I'm gonna make them
both go from 20 to 40. This one over here. And so now, the yellow curve is showing us the component of variation in the blue curve that can be explained by El niño and we can see it's a fairly small
component. It's small compared to the overall level of variability in December state college temperatures which vary by as much as plus or minus 4 degrees or so Fahrenheit.
Video: Demo - Part 3 (3:22)
Click here for a transcript
PRESENTER: So continuing where we left off. The yellow curve is showing us the component of the variation in December State College temperatures that can be explained by El niño. In a particularly
strong El niño year, where the niño 3.4 index is say as large as +2. We get a December temperature that's about one and a half degrees Fahrenheit above average. That is to say that 0.74 degrees
Fahrenheit that we get for one unit change in niño 3.4. Particularly strong La Niña event, we get a -0.74 degrees effect that we get for the negative niño 3.4 anomaly of negative two or so. Yet, I'm
sorry, a -1.5 Fahrenheit cooling effect for negative two or so. So the influence of El niño is small compared to the overall variability of roughly four degrees Fahrenheit in the series. But it is
statistically significant. At least if we are able to motivate a one-sided hypothesis test. If we had reason to believe that the El Niño events warm State College temperatures in the winter then the
regression gives us a significant result that's significant at the 0.05 level. The standard threshold for statistical significance. Okay, so that may not be that satisfying. We're not explaining a
large amount of the variation in the data, but we do appear to be explaining a statistically significant fraction of the variability in the data. Now finally, let's look at the residuals from that
regression. So what I'll do is, I will get rid of these other graphs. Let's keep year. Let's change this to model residuals. I'm just going to plot the model residuals as a function of time, and
that's what they look like. There isn't a whole lot of obvious structure and in fact if you go back to the regression model tab, and we look at the value of the lag 1 autocorrelation coefficient,
we'll see that it's -0.09 that's slightly negative and it's quite small, close to zero. If we look up the statistical significance it's not going to be even remotely significant. So we don't have to
worry much about autocorrelation influencing our estimate of statistical significance. We also don't have much evidence here of the sort of low-frequency structure and the residuals that might cause
us to worry. So the nominal results of our regression analysis appear to be valid and again if we were to envoce a one-sided hypothesis test, we would have found a statistically significant, albiet a
weak, influence of El niño on State College December temperatures.
You can play around with the data set used in this example using this link: Explore Using the File testdata.txt | {"url":"https://www.e-education.psu.edu/meteo469/node/209","timestamp":"2024-11-04T07:44:01Z","content_type":"text/html","content_length":"56274","record_id":"<urn:uuid:52e97a9a-27e3-438f-96d0-8f41c74c882e>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00734.warc.gz"} |
A Guide to Understand Hasse Diagram | EdrawMax Online
A Guide to Understand Hasse Diagram
The hasse diagram shows the relationship between ordered sets. In the case of learning set theory, the students need to know about the hasse diagram. The students need to understand its concepts
clearly for their lessons, and for that, they should use the hasse diagrams. However, it is challenging to create a hasse diagram by hand, considering its complications. To avoid this hassle, the
students must use the EdrawMax Online tool. It can help them to create a high-quality hasse diagram for their study.
1. The Hasse Diagram
Hasse diagram is a graphical orientation of a finite partially ordered set, also known as POSETs. Dots denote the elements present in the POSETs, whereas straight lines express their relationship.
The hasse diagrams are relevant for studying the set and theories related to it and representing Boolean Algebra.
Although the initial representation of the hasse diagram involved representation of the POSETs by hand with time, the modification of the hasse diagram has allowed the students to use graphs for
their presentation.
Here are a few factors that depict the position of the line segment drawn in hasse diagrams:
1. If an element on the POSET is greater than another element, the lower among the two will be at a lower position while the higher one will also have a higher place. For example, there are two
elements, x & y in a POSET, and x>y, then x value is higher than y.
2. Draw the line segment between the elements present in the POSETs if there is a relation between x and y or y and x.
1.1 Relations in a Hasse Diagram
The hasse diagrams are named after Helmut Hasse. Although, he was not the first person to work on this. However, Helmut’s hasse diagrams initiated the graphical representation of POSETs. The hasse
diagrams may also have binary coding of a subset (1) or not (0). They are of different shapes and sizes.
An ordinary hasse diagram or the smoothest one may have the shape of a straight line. However, a complicated diagram may have a three-dimensional shape or a 4x4 matrix. There are a few rules
regarding the relationship of elements present in the hasse diagrams. These properties are:
1. Reflexivity → p ≤ p Ɐ p ꞓ B
2. Anti-symmetric → p ≤ q and q ≤ p iff p=q
3. Transitivity → if p ≤ q and q ≤ r then p ≤ r
In case of hasse diagram:
1. In the POSET, the element lesser than any other element present there is called the maximal element.
2. At the same time, the minimal element is the element that is not greater than any other elements present in the POSET.
3. The element of a POSET, which exceeds all others, is called the greatest element. Similarly, the element in the POSET that precedes the other elements is known as the least element. Both these
elements in a hasse diagram are single.
4. The maximal and minimal elements in the hasse diagram can be the same, and it depends on the set it is representing.
Source:EdrawMax Online
2. How to Draw the Hasse Diagram?
The students who want to understand the ordering of sets used in it can use a hasse diagram. It is challenging to create a hasse diagram by hand.
2.1 How to Create Hasse Diagram from Sketch
Here are a few steps which they can follow to draw a hasse diagram:
Step 1: The hasse diagram is also called the ordering diagram. Hence, to start with the hasse diagram, the students need to digraph the partial order.
Step 2: After that, they need to eliminate the self-loop present at each vertex. Then the students must also remove the elements that are present in the system due to transitivity. For example, there
are three elements, 1R2, 2R3, and 1R3. After that, they have to eliminate 1R3 for transitivity.
Step 3: For the next part, the students have to arrange the edges so that the arrows face the upward direction. Finally, they need to erase the arrowheads and represent the nodes with dots instead of
any circle or other figures.
2.2 How to Create Hasse Diagram Online
People may get confused about the nature of it and may end up with a faulty diagram. The students need to be careful about the reflexive and transitive properties while working on their hasse
diagram. To avoid this condition, the students must use the EdrawMax Online tool. The tool is user-friendly. Hence, anyone can work on it without any experience of diagramming. Its easy-to-use
interface has made it a favorite diagramming tool of many users.
You can use the tool for studying multiple subjects. It comes with more than 280 different types of diagrams. Thus, today, EdrawMax has a user base of more than 25 million people. A student can
create a hasse diagram with EdrawMax Online instead of drawing it by hand. Here are a few simple steps which they can follow to make a hasse diagram on the EdrawMax Online tool:
Step 1:To start with the hasse diagram, the students need to open the EdrawMax Online tool. It is a user-friendly tool. Therefore, the student can work on it without any difficulty. To continue with
the process, the students need to open New. Under this ‘New’ option, the students can find the Science and Education tab.
Source:EdrawMax Online
Step 2: The students can get various science and education-related videos on the tool, which they can use for their lessons. To find the hasse diagram, they need to open the Mathematics tab and
choose the hasse diagram.
Source:EdrawMax Online
Step 3: After selecting the diagram, they can edit the diagram considering the lessons they will use it. The tool has high-quality templates that require slight modification according to their need.
Source:EdrawMax Online
Step 4: Once the student completes the modification, they can save and then export the diagram for future use. The students can save the same in multiple formats for further use in lessons, projects,
and dissertation papers. They can also export the hasse diagram for future use. EdrawMax Online tool is compatible with multiple devices and operating systems. Hence, the students can smoothly work
on their hasse diagram anytime and anywhere.
Source:EdrawMax Online
3. Hasse Diagram Examples
The hasse diagram is an orientation of partially ordered sets. Here are a few examples of hasse diagrams:
• set A = {5, 6, 7, 8}. Let R be the relation ≤ on A. The hasse diagram of R.
R={{5, 5}, {5, 6}, {5, 7}, {5, 8}, {6, 6}, {6, 7}, {6, 8}, {7, 7}, {7, 8}, {8, 8}}
Removed due to reflexive property: (5, 5), (6, 6), (7, 7), (8, 8).
Removed for transitive property: (5, 7), (5, 8), (6, 8).
• The second example deals with drawing the hasse diagram to find the POSET for divisibility: (4, 6, 24, 32, 48, 72). If the set is A, then:
A = {{4, 24}, {4, 32}, {4, 48}, {4, 72}, {6, 24}, {6, 48}, {6, 72}, {24, 48}, {24, 72}}.
• Positive integer divisor of 18.
If the set is D, then D={1, 2, 3, 6, 9, 18}, then poset A = {{1, 2}{1, 3}, {1, 6}, {1, 9}, {1, 12}, {1, 18}, {2, 6}, {2, 18}, {3, 6}, {3, 9}, {3, 18}, {6, 18}, {9, 18}}.
4. Conclusion
To learn about set theory and clarify their idea about partially ordered sets, the students must use their hasse diagram. It is tough to make a hasse diagram by hand, and it may be time-consuming at
the same time. The students may fail to create the perfect hasse diagram for their lessons. Thus, they must use the EdrawMax Online tool, which is user-friendly. This tool can help them create a
high-quality hasse diagram.
In conclusion, EdrawMax Online is a quick-start diagramming tool, which is easier to make hasse diagrams and any 280 types of diagrams. Also, it contains substantial built-in templates that you can
use for free, or share your science diagrams with others in our template community. | {"url":"https://www.edrawmax.com/article/a-guide-to-understand-hasse-diagram.html","timestamp":"2024-11-13T15:07:26Z","content_type":"text/html","content_length":"141094","record_id":"<urn:uuid:5d07370d-d28f-4ce2-9b64-17b1d44a8f28>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00832.warc.gz"} |
300+ TOP Parallel Circuits MCQ Questions and Answers Quiz [2024]
300+ TOP Parallel Circuits MCQ Questions and Answers Quiz
PARALLEL CIRCUITS Multiple Choice Questions
1. An ammeter has an internal resistance of 50 Ω. The meter movement itself can handle up to 1 mA. If 10 mA is applied to the meter, the shunt resistor, RSH1, is approximately
A. 55 Ω
B. 5.5 Ω
C. 50 Ω
D. 9 Ω
Answer: B
2. The total resistance of a parallel circuit is 50 Ω. If the total current is 120 mA, the current through the 270 Ω that makes up part of the parallel circuit is approximately
A. 22 mA
B. 120 mA
C. 220 mA
D. 50 mA
Answer: A
3. The currents into a junction flow along two paths. One current is 4 A and the other is 3 A. The total current out of the junction is
A. 1 A
B. 7 A
C. unknown
D. the larger of the two
Answer: B
4. When an additional resistor is connected across an existing parallel circuit, the total resistance
A. remains the same
B. decreases by the value of the added resistor
C. increases by the value of the added resistor
D. decreases
Answer: D
5. When a 1.6 kΩ resistor and a 120Ω resistor are connected in parallel, the total resistance is
A. greater than 1.6 kΩ
B. greater than 120 but less than 1.6 kΩ
C. less than 120 but greater than 100 Ω
D. less than 100 Ω
Answer: C
6. If there are a total of 120 mA into a parallel circuit consisting of three branches, and two of the branch currents are 40 mA and 10 mA, the third branch current is
A. 50 mA
B. 70 mA
C. 120 mA
D. 40 mA
Answer: B
7. Three lights are connected in parallel across a 120 volt source. If one light burns out,
A. the remaining two will glow dimmer
B. the remaining two will glow brighter
C. the remaining two will not light
D. the remaining two will glow with the same brightness as before
Answer: D
8. Four equal-value resistors are connected in parallel. Ten volts are applied across the parallel circuit and 2 mA are measured from the source. The value of each resistor is
A. 12.5 Ω
B. 200 Ω
C. 20 KΩ
D. 50 Ω
Answer: C
9. A set of Christmas tree lights is connected in parallel across a 110 V source. The filament of each light bulb is 1.8 k. The current through each bulb is approximately
A. 610 mA
B. 18 mA
C. 110 mA
D. 61 mA
Answer: D
10. The power dissipation in each of four parallel branches is 1.2 W. The total power dissipation is
A. 1.2 W
B. 4.8 W
C. 0.3 W
D. 12 W
Answer: B
11. A 470 Ω resistor, a 220 Ω resistor, and a 100 Ω resistor are all in parallel. The total resistance is approximately
A. 790 Ω
B. 470 Ω
C. 60 Ω
D. 30 Ω
Answer: C
12. Five light bulbs are connected in parallel across 110 V. Each bulb is rated at 200 W. The current through each bulb is approximately
A. 2.2 A
B. 137 mA
C. 1.8 A
D. 9.09 A
Answer: C
13. Four resistors of equal value are connected in parallel. If the total voltage is 15 V and the total resistance is 600 Ω, the current through each parallel resistor is
A. 25 mA
B. 100 mA
C. 6.25 mA
D. 200 mA
Answer: C
14. Five 100 Ω resistors are connected in parallel. If one resistor is removed, the total resistance is
A. 25 Ω
B. 500 Ω
C. 100 Ω
D. 20 Ω
Answer: A
15. Four 8 Ω speakers are connected in parallel to the output of an audio amplifier. If the maximum voltage to the speakers is 12 V, the amplifier must be able to deliver to the speakers
A. 18 W
B. 1.5 W
C. 48 W
D. 72 W
Answer: D
16. In a certain three-branch parallel circuit, R1 has 12 mA through it, R2 has 15 mA through it, and R3 has 25 mA through it. After measuring a total of 27 mA, you can say that
A. R3 is open
B. R1 is open
C. R2 is open
D. the circuit is operating properly
Answer: A
17. A sudden increase in the total current into a parallel circuit may indicate
A. a drop in source voltage
B. an open resistor
C. an increase in source voltage
D. either a drop in source voltage or an open resistor
Answer: C
18. The following currents are measured in the same direction in a three-branch parallel circuit: 200 mA, 340 mA, and 700 mA. The value of the current into the junction of these branches is
A. 200 mA
B. 540 mA
C. 1.24 A
D. 900 mA
Answer: C
19. The following resistors are in parallel across a voltage source: 220Ω , 470Ω , and 560Ω . The resistor with the least current is
A. 220 Ω
B. 470 Ω
C. 560 Ω
D. impossible to determine without knowing the voltage
Answer: C
20. Three 47 Ω resistors are connected in parallel across a 110 volt source. The current drawn from the source is approximately
A. 2.3 A
B. 780 mA
C. 47 mA
D. 7.06 A
Answer: D
21. There is a total of 800 mA of current into four parallel resistors. The currents through three of the resistors are 40 mA, 70 mA, and 200 mA. The current through the fourth resistor is
A. 490 mA
B. 800 mA
C. 310 mA
D. 0 A
Answer: A
22. Four resistors are connected in parallel. Fifteen mA flows through resistor R. If the second resistor is 2R, the third resistor 3R, and the fourth resistor 4R, the total current in the circuit is
A. 60 mA
B. 15 mA
C. 135 mA
D. 31.25 mA
Answer: D
23. If one of the resistors in a parallel circuit is removed, the total resistance
A. decreases
B. increases
C. remains the same
D. doubles
Answer: B
24. Six resistors are in parallel. The two lowest-value resistors are both 1.2 kΩ. The total resistance
A. is less than 6 kΩ
B. is greater than 1.2 kΩ
C. is less than 1.2 kΩ
D. is less than 600 Ω
Answer: D
25. In a five-branch parallel circuit, there are 12 mA of current in each branch. If one of the branches opens, the current in each of the other four branches is
A. 48 mA
B. 12 mA
C. 0 A
D. 3 mA
Answer: B
PARALLEL CIRCUITS objective questions with answers pdf download online exam test | {"url":"https://engineeringinterviewquestions.com/parallel-circuits-questions-answers/","timestamp":"2024-11-10T21:32:00Z","content_type":"text/html","content_length":"47657","record_id":"<urn:uuid:07731545-0837-41b9-af6d-72bacc335bb3>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00524.warc.gz"} |
Symmetric vs. Asymmetric Encryption - zwilnik
Symmetric vs. Asymmetric Encryption
Previously we looked at Public Key encryption, which is also called Asymmetric Encryption because it uses two different keys for the encryption and decryption. This allows us to solve one of the
biggest problems in secure encrypted communication, which is key distribution. Because the public key can be freely distributed, you don’t need to maintain security around the process of distributing
keys. Symmetric encryption, on the other hand, relies on a shared key that is used for both encryption and decryption. An example of this is the one-time pad, where you printed up a pad of paper that
contained various keys, and each one was used only once. As long as no one can get the key, it is unbreakable, but the big weakness was key distribution. How do you get the one-time pad into the
hands of your correspondent? And you would need to do this with separate one-time pads for each person you needed to communicate with. These are the kinds of problems that made asymmetric encryption
so popular. Finally, symmetric key crypto cannot be used to reliably create a digital signature. The reason should be clear. If I have the same secret key you used to sign a message, I can alter the
message, use the shared secret key myself, and claim you sent it.
There is a downside, though, to asymmetric encryption. It requires a good deal more in computational resources to perform asymmetric encryption and decryption. Symmetric crypto, on the other hand, is
much more efficient. That is why in practice the two are actually combined. When you use GPG to encrypt a message, you use the public key of the person you are writing to. That much we have already
covered. But what are you encrypting? It turns out you are encrypting the symmetric key that was actually used to encrypt the message itself! Yes, a symmetric key is used for the efficiency, but we
solve the key distribution problem by using an asymmetric key to encrypt the symmetric key. So when someone sends you a message using your public key, you use your private key to decrypt the
symmetric key, then use the symmetric key to decrypt the message itself. This is the best combination of security and efficiency in your communication.
Symmetric Encryption Standards
Data Encryption Standard (DES)
The very first standard popularly used was developed by IBM for the U.S. government and was called the Data Encryption Standard (DES). Without going into a highly technical description. DES employed
some techniques that pop up frequently in cryptography, the Block Cipher, and XOR. In simple terms, a block cipher operates on a fixed-length block of bits to transform them in some way. And XOR
provides one of the most common transformations. XOR stands for “Exclusive Or” and in logic that means “either A is true or B is true but not both”. If used in circuit design, if either A or B is
sending a signal, the signal is output, but if both are sending a signal, nothing is output. When used in cryptography, what XOR does is to use a key that is “XORed” with the message block in such a
way that if both the message and the key have a zero in that position, the result is a zero. If the message has a one and the key has a zero, the result is a one. If the message has a zero and the
key as a one, the result is a one. And if the message has a one and the key has a one, the result is a zero. One way to think about that is that it is essentially binary addition without the “carry”
part. In binary addition, one plus one equals “10”, which is the binary form of “2”, so you write down the zero and “carry the one”. In XOR, you just throw away the carry part altogether.
Understanding how this works begins with the coding. Recall that we distinguished between codes and ciphers earlier in this series. A code is just a one-to-one transform on information from one
scheme to another. An example is Morse code. There is nothing secret about it, and the transformation usually is to render information in a way that fits the medium. In computers, everything is ones
and zeros, so there is a code that takes our letters and turns them into binary digits. In fact, there are several, but for the purpose of illustration I will use ASCII, which is the American
Standard Code for Information Interchange. In ASCII there is a numerical equivalent for every letter. I will do a very simple example, the word “cat”. I can see from the table on the Wikipedia page
that c = 1100011, a = 1100001, and t = 1110100. So the word “cat” is represented in binary as 110001111000011110100. The other thing I need is a key. This needs to be a secret in real life, but I’ll
tell you that it is “dog”, and by a similar process I can find that dog is represented in binary as 110010011011111100111. When I XOR these two, I get:
Message 1 1 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 1 0 0
Key 1 1 0 0 1 0 0 1 1 0 1 1 1 1 1 1 0 0 1 1 1
Result 0 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 0 0 1 1
And the reason this is useful as an encryption method is that if you XOR the key with the Result you get back the original message. Now, in constructing anĀ encryption algorithm, you take a number of
such methods and combine them to get something that is actually secure. In DES, the block size was selected as 64 bits. The key was also 64 bits, except that one bit from each byte was devoted to
parity checking, so the effective key length is 56 bits. In creating the DES standard, you would see that processes were repeated for multiple rounds of transformation, which is common. So the final
output could be the result of multiple XORs and other such stuff. But the reason is it is symmetric is, like we saw with the XOR process, if you have the key you can reverse all of the steps in the
algorithm and get back the original message.
DES really is the beginning of modern cryptogrpahy. Bruce Schneier said about it “”DES did more to galvanize the field of cryptanalysis than anything else. Now there was an algorithm to study.” DES
became the standard against which all other algorithms would be compared. But it was not the final word by any means. As we have seen previously, this is an arms race, and methods and technology
continually evolve. DES was found to have some weaknesses, particularly in the key length. 56 bits was just too small as computers got better. The NSA had in fact tried to limit it to 48 bits, but
resistance from the cryptographic community resulted in this slightly higher length. Nevertheless, in 1999 a DES key was cracked by brute force methods in 22 hours, and the standard has since been
Triple DES
Triple DES attempts to solve the weakness of the 56 bit key in DES by using three independent 56 bit keys which are used in a repeated process. Each block of the message is encrypted three times,
once each for the three keys. If done this way, it is generally regarded as secure, and NIST regards it as safe through 2030. On the other hand, there are theoretical attacks which, though not
considered feasible with current technology, have lead to a new standard.
Advanced Encryption Standard (AES)
The Advanced Encryption Standard was adopted by NIST in 2001. This is now considered the best available symmetric encryption. It employs a cipher known as Rijndael, which is a play on the names of
the two inventors, Vincent Rijman and Joan Daemen. This version of Rijndael used in AES has a block size of 128 bits, and key sizes of 128, 192, or 256 bits, and in general it can be referred to as
AES-128, AES-192, or AES-256 depending on the key length, and the most secure would be AES-256. As with the other symmetric ciphers, each block is subjected to repeated rounds of transformation to
get the encrypted text.
Asymmetric Encryption Standards
One thing you may have noticed in the above discussion of symmetric encryption is the lack of discussion of entropy in the process. It is not needed there, because the only thing that matters is that
the key is agreed, not that it is random. But in asymmetric key encryption entropy is essential. It is the combination of the entropy with the one-way function that makes it work.
One way functions, as you may recall from our earlier discussion, are functions that can easily be computed in one direction, but are computationally infeasible in reverse. Right now there appear to
be three types that are known and can be used in cryptography:
• Multiplying large prime numbers
• Discrete logarithm
• Elliptic curve
The mathematics of these approaches are more than I can handle. I suspect you need something like PhD in Math to make sense of this stuff, but maybe that is just me being dense. But here is the
basics of each.
Prime number factorization
This is the approach used in RSA encryption. Two large prime numbers are found and multiplied together to get a product. Multiplying them together is simple for a computer, but decomposing the
product to the two two primes you started with is computationally infeasible, so far as we know. Of course, since RSA is widely used, it has sparked intense interest in approaches to factorization,
so this may get weakened over time. And of course with improvement in computer technology what now appears to be a difficult problem may become much simpler in the future. But for now RSA appears to
be secure. The role of entropy in this case lies in finding the prime numbers. Generally they should each be in the neighborhood 1024 digits. And for security, they should not only be random, but not
“near” each other.
With the large prime numbers found and multiplied, the product is used to generate other prime numbers which help form the public key and private key. The point here is that these are just two keys
such that one key cannot decrypt anything it itself encrypted, but can decrypt anything its complementary key encrypted. So you could in theory use either one as the private key, it is not
privileged. In fact, using the public key to decrypt something encrypted with the private key is the principle behind digital signatures, which we will discuss in the following tutorial.
Discrete Logarithm
Discrete logarithm involves finding an integer that solves a logarithmic equation. It is used in Elgamal encryption and Diffie-Hellman key exchange, among many uses. Choosing the particular numbers
for the logarithmic equation is where the entropy comes in. Diffie-Hellman Key Exchange is used for perfect forward secrecy, among other uses.
Elliptic Curve
Elliptic Curve cryptography builds on the Discrete Logarithm approach. A curve with the right properties is chosen, then a point on that curve, and the problem becomes finding the discrete logarithm
of that point. Generally the curve is chosen from a small number of appropriate curves that have been agreed upon by the crypto community. NIST has recommended 15 curves as suitable. Entropy enters
when one chooses the point on the curve that will be used. Elliptic Curve Cryptography is usually faster and more efficient than RSA or general Discrete Logarithm approaches. However, it now appears
that at least one of the curves selected was chosen by NSA to have deliberate weaknesses, so that particular curve is now deprecated. The larger question of whether any NIST standard can be trusted
given the NSA’s involvement is still open.
So, this wraps up a somewhat more technical discussion of encryption methods which I hope will set us up to look at some other security problems and solutions.
Listen to the audio version of this post on Hacker Public Radio! | {"url":"https://www.zwilnik.com/security-and-privacy/hashing-passwords-and-certificates/symmetric-vs-asymmetric-encryption/","timestamp":"2024-11-05T22:38:09Z","content_type":"text/html","content_length":"121555","record_id":"<urn:uuid:6349fdd1-1ddf-44df-91d5-58542cfafb71>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00455.warc.gz"} |
Why Are So Many Students Afraid of Math?
Critical Thinking
Why Are So Many Students Afraid of Math?
Emphasizing the process of problem-solving instead of speed and right answers can help reduce students’ math anxiety.
April 8, 2017 Updated April 17, 2017
I was recently talking with a high school student about math, and she said something that really resonated with me. She had some homework problems that she did not know how to do. The teacher had
shown examples in class, but the homework problems were slightly different. She explained, “It’s like all of my math knowledge could be represented by separate chains, and each of my chains has a
broken link. I listen while the teacher describes how to do the problem, but usually there seems to be a part that they skip or that I don’t understand. Suddenly they have the answer written down,
and I don’t know how they got from what I understood last to that answer.”
This student had concluded that her struggle to understand math meant that she was just not good at math. We need to stop conveying to students that their struggle to understand ideas fully and to
process and reason through math concepts means that they are not good at math. Actually, it might mean the opposite—that they possess the critical thinking skills necessary to become an excellent
math student.
What If Our Approach to Teaching Math Is Wrong?
Studies at the Mangels Lab of Cognitive Neuroscience of Memory and Attention at Baruch College at the City University of New York found that when math is taught in a stressful and high-pressure
atmosphere where students do not feel successful, this can lead to significant math anxiety which inhibits math performance. This math anxiety tends to affect our most promising high-achieving
Often, when this happens, teachers and parents are told to address it with strategies to calm and refocus the student, but what if the real problem is the way we’re teaching math? It’s as if we’re
trying to teach students to navigate by showing them one random set of directions or by putting one starting point and destination in Google Maps and then going over each of the resulting turns.This
makes math knowledge like chains where students must remember each step or link in order to successfully navigate the problem. It also results in the student not understanding how a new piece of math
knowledge relates to what they’ve learned previously.
Instead, we should be showing them the map, and having them use it to figure out all the possible routes to get from point A to point B. As teachers, we should be emphasizing math problem-solving and
math thinking instead of quick performance and correct answers.
What Would This Look Like, and How Would It Work?
To use this method, teachers would act as facilitators. They would present the class with the problem, then in small groups or as a whole class, students would describe what they notice. The teacher
would summarize and list these observations while asking questions and providing information that would lead students in their problem-solving. Students would then work as a group to brainstorm
methods to arrive at a solution. The teacher would summarize and list each method with the group and lead the class toward deciding which method was most effective.
Students would have to describe how this method worked and why it was the easiest and most effective method.This could work at any level, and it may look different at the early elementary level from
what we see at the late elementary/middle school level.
What Are the Advantages of Teaching Math This Way?
1. Critical thinking is enhanced because students are engaged: Students discuss what they notice, explain their thinking, and are actively involved in finding solutions to the problem. This is very
different from the traditional math classroom, especially in middle and high school, where students sit quietly taking in information, writing down examples, and working on practice problems
2. It builds math confidence: This method shows students that they’re capable of solving problems on their own or by working together. Having students take an active role in their learning builds
confidence and makes them less reliant on the teacher to be the provider of solutions. It empowers them to figure out even difficult problems by doing the complex and often difficult work of thinking
through the problem themselves.
3. It teaches students a growth mindset: When students see math as a performance subject rather than a subject where learning is emphasized, they’re more likely to fear math. When teachers shift the
focus from right or wrong answer to an emphasis on mathematical thinking, they help students to understand that their math ability can grow. Instead of thinking that if they don’t understand
something right away, it means they’re bad at math, they know that if they don’t understand something, they have the ability to work on it and figure it out.
4. Math performance is improved: When students see math as a set of ideas they can explore and figure out, they have no reason to fear math, and their performance in math will improve. Speed and time
pressure block working memory, preventing students from showing what they know. However, Stanford expert Jo Boaler found that students who learn through strategies rather than simply memorizing
facts achieve superior performance because they understand the relationships between numbers.
5. This method is more applicable to how math is used in adulthood: If students get a job where they use math, they will probably need to know how to apply their knowledge of math to solve complex
and unique problems. They’ll have to use problem-solving to work through to a solution, and they’ll have the opportunity to brainstorm with others. Even students who do not grow up to pursue a career
where they use math will have to use math to solve everyday problems. These real-world problems also require that a student think critically about which math skill to apply based on the current
At first, requiring students to use critical thinking to solve problems will be frustrating to students who have always been immediately provided with a method to find the correct answer. However, if
you stick with it, in the long run it will make students into problem-solvers who are capable of asking their own questions and finding their own solutions. | {"url":"https://www.edutopia.org/discussion/why-are-so-many-students-afraid-math/","timestamp":"2024-11-03T19:18:27Z","content_type":"text/html","content_length":"98203","record_id":"<urn:uuid:20e57622-018f-4fb8-8a43-4c0580a0274d>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00567.warc.gz"} |
Build narrowband RF filters
...and learn useful RF techniques as well
LC bandpass filters have been extensively covered in the professional literature. While some of this has spilled over into the Amateur press, some confusion still remains about what type of filter to
use in a given application, and how to go about designing, building, and aligning it. In this article I hope to clear up some of this confusion and help you choose, design, build, and evaluate the
filter you need. The BASIC program (written in Commodore 64 BASIC, but translatable to others) will be presented to perform the math for you.
Fig. 1. General form of filter, showing matching capacitors and their placement.
The filters we're describing here are narrowband, capacitively-coupled designs (see fig. 1; note that the program prints out the capacitor values using these designators, so keep this figure in mind
for future reference.) By narrowband, I'm specifically referring to the results of the following, well-known relationship.(1)(2)(3)
For these filters, Q[bp] must be ≥ 10. For Q[bp] < 10 other design techniques are better. References 1 and 2 are recommended for coverage of this sort of design, based on lowpass filters, and for
converting them to bandpass.
Like most bandpass filters, these filters have a geometric symmetry about the center frequency:
Because of this, the response of the attenuation versus frequency of these filters does not follow the arithmetical symmetry one might expect. Attenuation increases more rapidly on the low side of
the passband.
Also, like most BPFs, these filters have passband versus stopband characteristics that are inevitably a compromise between passband ripple and attenuation at some out-of-band frequency. The optimally
flat Butterworth (no-ripple) designs will have the poorest attenuation out-of-band. The Chebychevs, with varying amounts of ripple, will give more attenuation at the same out-of-band frequency, with
the highest ripple designs giving the largest amount of attenuation. This attenuation also carries the price of being the most difficult of the filter types to build with real-world components.
Choosing the right filter for a job, then, becomes an exercise in compromise. In the front end of a receiver, for example, the ripple of the Chebychev might mean stations not copied (DX not worked).
You might put up with a little less rejection of an out-ofband signal to hear those. Later on in the rig, a little gain can frequently be spared - perhaps the Chebychev should go there. One of the
advantages of the program is that it allows you to compare designs before you plug in your soldering iron.
Using the program
To design a filter with this program, you need to know a few things: first, the center frequency (f[o]) and passband width. Both of these are entered in MHz. Next you need to know the required
passband characteristics (i.e., Butterworth or 0.01 dB, 0.1, 0.5, or 1.0 dB ripple Chebychev), what inductor you plan to use, and how much attenuation is required at some frequency ratio.
This last one may be unfamiliar to you. This frequency ratio is expressed as:
where BW[x] is the bandwidth at some attenuation x, and BW, is the (3dB) passband width, as defined before. Often, you won't know BW[x] directly, but will know that there is some undesired frequency
(f[un]), such as a broadcaster, or mixing product that you want to attenuate. You can solve for BW[x] by using the geometric symmetry of the filter, as follows:
You then divide this number by the bandwidth to get the ratio, f/f[c], that the progam queries for. It is entered as a single number (i.e., 3 or 4.237) rather than as a ratio.
The filter that results will inevitably have some characteristic impedance that will never be the one you're looking for - per Murphy. To overcome this, a simple matching capacitor can be calculated
for you by the program. It can match either end to another impedance, and does this by making the end sections a capacitive divider, with one capacitor in series with the input and output. It will
recalculate the changed capacitor in the body of the filter for you. The only hazard here is that a large change in impedance may make one of the capacitors go negative in value. If this happens, you
can redesign the filter with a bigger inductor, or else use another matching method of your own.
Sample filter
To illustrate the process, here's a filter that I needed for a 2-meter transverter. The bandwidth at 50 dB down was determined by a local NOAA Weatheradio signal at 162.55 MHz that I wanted to knock
down as much as possible. The 3 dB bandwidth of the filter was determined by a need to cover all 4 MHz of 2 meters in the peak of the passband. My inputs to the computer are the requirements that I
have for the filter:
• Type of filter: 0.5 dB ripple Chebychev
• Frequency Ratio f/f[c]: 4.13
• Desired Attenuation at f/f[c]: 50dB
At this point, the computer responds that an N = 4 (4 pole) filter will provide 61.08 dB (at km) and then asks for the center frequency (145), bandwidth (8), and the coil used (0.068 µH) It then
calculates its responses. Its outputs are:
Coupling Capacitors:
C12 = .6334 pF
C23 = .5327 pF
C34 = .6334 pF
Resonating Capacitors:
C1 = 17.08 pF
C2 = 16.55 pF
C3 = 16.55 pF
C4 = 17.08 pF
It concludes by telling me that the characteristic impedance of the filter is 2050 ohms and asks me if I want to match to another. I request 50 ohms, both ends, add 3.47 pF to the end sections and
change the end resonators, C1 and C4, to 13.613 pF (17.08-3.47).
Although this filter was touchy to align, it was built and aligned using sophisticated equipment, then measured on a Hewlett Packard Network Analyzer. The graphs of attenuation versus frequency for
it are in fig. 3 along with its schematic fig. 2.
Fig. 2. The two meter filter uses 68nH, Coilcraft T-113 11/2 turn coils.
Fig. 3(A) 2-meter BP filter fine response; (B) 2-meter BP filter response shows stopband attenuation.
Building the filters
Of course all this is pretty meaningless without working filters that fit your needs - and the program, like all mathematical models, is perfectly capable of generating answers that solve its
equations, but are completely absurd in the real world ... all of which requires the introduction of some sound ideas for building these filters.
To begin with, filters using conventional discrete coils and caps get progressively harder to build and align as frequency increases. The Amateur with the typical test equipment described below and
good components and assembly techniques should have no trouble getting these filters to work up through 75 to 100 MHz. Those with high-quality equipment or access to it can expect to go to beyond 2
meters. Above 200 MHz, filter construction and alignment without large amounts of equipment demands other techniques, which will not be discussed here.
When it comes to components, the rule is to get the highest Q that you can obtain. Low Q causes losses in the filter, evidenced by higher insertion loss and changes in the passband shape. Capacitors
are rarely a problem because they tend to have much higher Q[s] than coils. In an inductor:
where R is the ohmic resistance of the wire and XL is the inductive reactance. From this definition, it follows that you should use the largest practical size wire you can. Q is also affected by coil
diameter, winding pitch, and core or support material. Adding a core always degrades Q as does shielding a coil.
The tradeoff here is that airwound, large diameter coils are fine at VHF, but too large at HF. Thankfully, iron powder and ferrite toroids provide increased inductance, virtually perfect shielding,
and reasonable Q[s] from MF to the low VHF region. Amidon Associates provides data that allows you to predict Q and the number of turns to wind a given coil, on request.
While Q may not be a major concern with capacitors, stray reactances are. At RF, they contain enough stray inductance to present themselves as something far from a perfect capacitor. For example, a
0.001 µF ceramic disk capacitor with 1/4 inch leads appears as a series-resonant circuit (0 ohms) at 55 MHz, while a 0.01 is self-resonant at 15 MHz.(4) For this reason, it is imperative to keep
leads as short as possible.
As for the type of capacitor to use, ultraminiature dipped ceramics are best at and above 6 meters, although their typical 50 VDC breakdown is low for transmitters. Through the HF spectrum, silver
micas are an acceptable substitute. SMs have a higher stray inductance, but it's rarely a problem. They're available in higher voltage ratings, too.
Inductors also show stray reactance - in this case, capacitance between adjacent turns. This has the effect of making the inductor appear smaller than it is, and is especially noticeable as frequency
One last word on component Q. The filter type chosen imposes some constraints on the minimum Q you can live with. Table 1, derived from reference 2, presents the approximate minimum Q for a lowpass
filter of the given type and order. For bandpass filters, this must be multiplied by the Q[bp], derived previously. From this, you can see how high-order Chebychevs would be difficult to realize,
based on component Q alone.
Table 1.
Type of Filter Minimum Component Q
n= 2 3 4 5 6 7 8
Butterworth 2 2 3 3 4 4 5
0.10 dB Chebychev 2 3 5 7 10 14 17
0.50 dB Chebychev 3 4 7 10 14 20 30
1.0 dB Chebychev 3 5 8 12 17 26 35
This table gives an approximate minimum component Q for use in a /owpass filter of the given order. For a bandpass filter, this value must be multiplied by the passband Q,
Component layout requires some thought. While an etched PC board is not necessary, at least through 6 meters, some readers may want to etch one anyway. The optimum layout is linear, with the
components appearing as they are drawn on the schematic. If you must "bend" this line to fit into a space, try to keep input away from output and break grounds between them. A U-shaped layout isn't
asking for trouble-it's begging.
Figure 4 is a "universal" four-pole filter prototyping board I've used from 1 to 200 MHz. It's etched on double-sided G-10, with the back solid groundplane. Components are soldered directly to the
lands or the lands are jumpered out with copper foil for smaller numbers of sections. Top and bottom groundplanes are jumpered together by wires.
Fig. 4. Full size version of the "universal" 4 section filter card.
If you choose not to etch a board, or can't etch a board, take heart. Many prototypes in engineering labs are built in a style called "dead bug"; components going to ground are soldered to a piece of
unetched PC material and those not grounded are supported by their leads going to those that are, or to each other. The only precautions necessary are to keep ground connections as short as possible
and to not lay unshielded coils with their axes parallel to each other, or with turns touching. Toroids can be placed in any orientation to each other and can touch ground without trouble. As in
etched boards, the best layout is a straight line.
It should be obvious that the only differences between, say, a 1.0 dB Chebychev and a Butterworth of the same order are small component value differences. Knowing the value of the components is the
most important step in getting the filter working.
There are many ways for measuring component values; a bridge can give quite accurate results, and commercial or homebrew capacitance meters abound. Lacking one of the above, a grid dip meter will do
nicely. All you need are some capacitors and coils that will serve as your standards (known values). Using these, an unknown coil or cap can be measured by establishing it in a resonant circuit and
measuring the frequency. The ARRL Handbook, at least up through the 1982 edition, included a chart for determining LC values using standards of 100 pF and 5µH, although you can use any standard you
choose.' A frequency on the order of 10 MHz is fine for checking capacitors.
You'll find that the values the program prints out for capacitors are generally not standard and will have to be made up by paralleling one or two with a trimmer. Use the grid dipper to set the
value. Likewise, coil values will not work out to an exact number of turns (e.g., 9.39 turns). What you'll have to do, if you're not using a variable coil, is wind the nearest whole number of turns
and adjust the coil manually. This is done by setting the grid dip meter to the proper frequency and either squeezing together or spreading apart turns until the value is correct. Fix the turns in
place with coil dope. Keep your hands clear while measuring!
Fig. 5. Breaking up a filter into nodes for alignment.
Now that the parts are all the correct value, assemble the filter's nodes by shorting all of the caps around an inductor to ground (see fig. 5). All of the resulting sections are then tuned to the
geometric center frequency, f[o] of the filter. Tune only one adjustment per node! If the parts are properly measured, this should be a very small "tweak"; none may be required below 50 MHz. Connect
the filter into its final configuration.
Once the components and sections are tested and assembled, it's best to verify that the filter you have is the one you really wanted (see test setup in fig. 6). The signal source can be a cheap
generator, either purchased or home-brewed, that puts out several milliwatts of RF. If the RF voltmeter can't read frequency, use a counter. A sweep generator and receiver make an excellent
combination, although a sweeper makes some means of accurately measuring the output frequency critical.
Fig. 6. Alignment verification set-up.
If two voltmeters are available, use one to keep the output from the source at the same level. If only one is available, set the source to some level and then sweep it across the filter's passband
(still without the filter in line) while recording its level variation for a baseline. Then put the filter in line and sweep the generator across its range, say from the expected 3 dB points, or
slightly beyond, while recording levels through the filter. A power meter permits immediate comparison of dB levels; a voltmeter requires a few minutes with a calculator. Once the readings are in dB,
plot them versus frequency and see if you get what you expected. Without exceptional shielding in the generator and a more sophisticated detector, it won't be possible to determine stopband response.
It's generally valid to assume that if your passband is within your expected limits, the stopband will be also.
If the transmission characteristics of your filter are not as desired, it may be necessary to try to tweak it in. No amount of advice can help here, however, so make sure you've performed all the
preliminary steps properly before you start ripping it apart. Many of these filters - especially those built below 30 MHz - work perfectly the first time. Keep track of what you've done, and recheck
it often. If worse comes to worst, retune all nodes and start over again. (If it's any consolation, it does get easier with practice.)
If possible, build some practice filters in the HF range to familiarize yourself with the program and alignment methods. Once you've done that, you should have no problems.
Fig. 7. Narrow-band filter CAD program listing.
3 REM PROGRAMMED BY BOB LOMBARDI WB4EHS
100 PRINT CHR$(147)
105 Q=0:DIM K(7),C(8)
110 PRINT:PRINT:PRINT
120 PRINT "BANDPASS FILTER DESIGN PRO6RAM":PRINT:PRINT
122 PRINT "THIS PROGRAM COMPUTES VALUES FOR"
123 PRINT "PARALLEL RESONANT, CAPACITIVELY"
124 PRINT "COUPLED BANDPASS FILTERS":PRINT "THAT ARE (RVON)NARROW BAND(RVOF) -(Q>10)"
125 PRINT "ENTER THE NUMBER OF THE DESIRED"
126 PRINT "ROUTINE":PRINT:PRINT
130 PRINT "1 FOR BUTTERWORTH":PRINT
131 FRINT "2 FOR 0.01 DB CHEBYCHEV":PRINT
132 PRINT "3 FOR 0.1 DB CHEBYCHEV":PRINT
133 PRINT "4 FOR 0.5 DB CHEBYCHEV":PRINT
134 PRINT "S FOR 1.0 DB CHEBYCHEV":PRINT
145 INPUT O:IF O<0 AND O>5 THEN GOTO 135
146 ON O GOTO 1000,1100,1200,1300,1400
1000 REM BUTTERWORTH SECTION
1010 GOSUB 4000
1015 IF N=2 THEN Q=1.414:K(1)=0.707
1020 IF N=3 THEN Q=1.000:K(1)=0.707:K(2)=K(1)
1025 IF N=4 THEN Q=0.765:K(1)=0.841:K(2)=0.541:K(3)=K(l)
1030 IF N=5 THEN Q=0.618:K(1)=1.000:K(2)=0.556:K(3)=K(2):K(4)=K(1)
1035 IF N=6 THEN Q=0.518:K(1)=1.169:K(2)=0.605:K(3)=0.518:K(4)=K(2):K(5)=K(1)
1040 IF N=7 THEN Q=0.445:K(1)=1.342:K(2)=0.667:K(3)=0.527:K(4)=K(3)
1042 IF N=7 THEN K(5)=K(2):K(6)=K(1)
1045 IF N=8 THEN Q=0.390:K(1)=1.519:K(2)=0.736:K(3)=0.554:K(4)=0.510
1047 IF N=8 THEN K(5)=K(3):K(6)=K(2):K(7)=K(1)
1060 GOTO 2500
1100 REM 0.01 DB CHEBYCHEV SECTION
1110 R=.01:GOSUB 4500
1115 IF N=2 THEN Q=1.483:K(1)=0.708
1120 IF N=3 THEN Q=1.181:K(1)=0.682:K(2)=K(1)
1125 IF N=4 THEN Q=1.046:K(1)=0.737:K(2)=0.541:K(3)=K(1)
1130 IF N=5 THEN Q=0.977:K(1)=0.780:K(2)=0.540:K(3)=K(2):K(4)=K(1)
1135 IF N=6 THEN Q=0.937:K(1)=0.809:K(2)=0.550:K(3)=0.518:K(4)=K(2):K(5)=K(1)
1140 IF N=7 TEEN Q=0.913:K(1)=0.829:K(2)=0.560:K(3)=0.517:K(4)=K(3)
1142 IF N=7 THEN K(5)=K(2):K(6)=K(1)
1145 IF N=B THEN Q=0.997:K(1)=0.843:K(2)=0.567:K(3)=0.520:K(4)=0.510
1147 IF N=8 THEN K(5)=K(3):K(6)=K(2):K(7)=K(1)
1160 GOTO 2500
1200 REM 0,1 DB CHEBYCHEV SECTION
1210 R=0.1:GOSUB 4500
1215 IF N=2 THEN Q=1.638:K(1)=0.711
1220 IF N=3 THEN Q=1.433:K(1)=0.662:K(2)=K(1)
1225 IF N=4 THEN Q=1.345:K(1)=0.685:K(2)=0.542:K(3)=K(1)
1230 IF N=5 THEN Q=1.301:K(1)=0.703:K(2)=0.536:K(3)=K(2):K(4)=K(1)
1235 IF N=6 THEN Q=1.277:K(1)=0.715:K(2)=0.539:K(3)=0.518:K(4)=K(2):K(5)=K(1)
1240 IF N=7 THEN Q=1.262:K(1)=0.722:K121=0,542:K(3)=0.516:K(4)=K(3)
1242 IF N=7 THEN K(5)=K(2):K(6)=K(1)
1245 IF N=8 THEN Q=1.251:K(1)=0.728:K(2)=0.545:K(3)=0.516:K(4)=0.510
1247 IF N=8 THEN K(5)=K(3):K(6)=K(2):K(7)=K(1)
1260 GOTO 2500
1300 REM 0.50 DB CHEBYCHEV SECTION
1310 R=0.5:GOSUB 4500
1315 IF N=2 THEN Q=1.950:K(1)=0.123
1320 IF N=3 THEN Q=1.864:K(1)=0.647:K(2)=K(1)
1325 IF N=4 THEN Q=1.826:K(1)=0.648:K(2)=0.545:K13)=K(1)
1330 IF N=5 THEN Q=1.807:K(1)=0.652:K(2)=0.534:K(3)=K(2):K(4)=K(1)
1335 IF N=6 THEN Q=1.796:K(1)=0.655:K(2)=0.533:K(3)=0.519:K(4)=K(2):K(5)=K(1)
1340 IF N=7 THEN Q=1.790:K(1)=0.657:K(2)=0.533:K(3)=0.516:K(4)=K(3)
1342 IF N=7 THEN K(5)=K(2):K(6)=K(1)
1345 IF N=8 THEN Q=1.785:K(1)=0.658:K(2)=0.533:K(3)=0.515:K(4)=0.511
1347 IF N=8 THEN K(5)=K(3):K(6)=K(2):K(7)=K(1)
1360 GOTO 2500
1400 REM 1.0 DB CHEBYCHEV SECTION
1410 R=1.0:GOSUB 4500
1415 IF N=2 THEN Q=2.210:K(1)=0.739
1420 IF N=3 THEN Q=2.210:K(1)=0.645:K(2)=K(1)
1425 IF N=4 THEN Q=2.210:K(1)=0.638:K(2)=0.546:K(3)=K(1)
1430 IF N=5 THEN Q=2.210:K(1)=0.633:K(2)=0.535:K(3)=K(2):K(4)=K(1)
1435 IF N=6 THEN Q=2.250:K(1)=0.631:K(2)=0.531:K(3)=0.510:K(4)=K(2):K(5)=K(l
1440 IF N=7 THEN Q=2.250:K(1)=0.631:K(2)=0.530:K(3)=0.517:K(4)=K(3)
1442 IF N=7 THEN K(5)=K(2):K(6)=K(1)
1450 IF N=>8 THEN PRINT "CAN'T DO 8 POLE 1 DB CHEBUCHEV. CALCULATING 7 POLE."
1455 IF N=8 THEN N=7:GOTO 1440
1460 GOTO 2500
2500 REM CALCULATION AND DISPLAY ROUTINE
2510 PRINT "WHAT IS THE DESIRED CENTER"
2511 INPUT "FREQUENCY (IN MHZ)";F0
2515 F0=F0*1E6
2520 INPUT "WHAT IS THE 3 DB BM (IN MHZ)";BW:BW=BW*1E6
2530 INPUT "WHAT IS THE INDUCTOR (IN UH)";L:L=L*1E-6
2540 W=2*π*f0
2550 QB=F0/BW
2560 QC=QB*Q
2570 FOR I=1 TO N-1
2575 K(I)=K(I)/QB
2580 NEXT I
2590 CR=1/((W*W)*L)
2600 RE=W*L*QD
2610 FOR I=1 TO N-1
2670 K(I)=K(I)*CR
2680 NEXT I
2690 ON N-1 GOSUB 2700,2720,2740,2760,2780,2800,2820
2695 GOTO 2900
2700 C(1)=CR-K(1):C(2)=C(1):RETURN
2720 C(1)=CR-K(1):C(2)=CR-K(1)-K(2):C(3)=C(1):RETURN
2740 C(1)=CR-K(1):C(2)=CR-K(1)-K(2):C(3)=CR-K(2)-K(3):C(4)=C(1):RETURN
2760 C(1)=CR-K(1):C(2)=CR-K(1)-K(2):C(3)=CR-K(2)-K(3):C(4)=CR-K(3)-K(4)
2770 C(5)=C(1):RETURN
2780 C(1)=CR-K(1):C(2)=CR-K(1)-K(2):C(3)=CR-K(2)-K(3):C(4)=CR-K(3)-K(4)
2785 C(5)=CR-K(4)-K(5):C(6)=C(1):RETURN
2800 C(1)=CR-K(1):C(2)=CR-K(1)-K(2):C(3)=CR-K(2)-K(3):C(4)=CR-K(3)-K(4)
2805 C(5)=CR-K(4)-K(5):C(6)=CR-K(5)-K(6):C(7)=C(1):RETURN
2820 C(1)=CR-K(1):C(2)=CR-K(1)-K(2):C(3)=CR-K(2)-K(3):C(4)=CR-K(3)-K(4)
2825 C(5)=CR-K(4)-K(5):C(6)=CR-K(5)-K(6):C(7)=CR-K(6)-K(7):C(8)=C(1):RETURN
2900 PRINT:PRINT:PRINT
2910 PRINT "COUPLING CAPS ARE (IN PF)"
2915 FOR I=1 TO N-1
2920 PRINT "C";I;I+1;": ";K(I)*1E12
2930 NEXT I
2935 PRINT
2940 PRINT "RESONATING CAPS ARE (IN PF)"
2945 FOR I=1 TO N
2950 PRINT "C";I;": ";C(I)*1E12
2960 NEXT I
2970 PRINT "THIS FILTER HAS A CHARACTERISTIC Z"
2975 PRINT "OF";RE;"OHMS.":PRINT
2980 INPUT "MATCH TO ANOTHER Z";A$:IF A$<>"Y" AND A$<>"N" THEN GOTO 2980
2985 IF A$="N" GOTO 3100
2990 INPUT "NEW SOURCE AND LOAD Z (50,50):";RS,RL
2992 PRINT
2995 TM=SQR(RE*RS-RS*RS):TN=SQR(RE*RL-RL^2)
3000 CY=1/(W*TM):CZ:1/(W*TN)
3005 PRINT "ADD";CY*1E12;" PF TO THE SOURCE END"
3006 PRINT "AND ADD";CZ*1E12;" TO THE LOAD END."
3010 PRINT:PRINT "THEN CHANGE C(1) TO";(C(1)-CY)*1E12;" PF"
3020 PRINT "AND CHANGE C(";N;") TO "(C(N)-CZ)*1E12;" PF."
3100 PRINT:PRINT
3110 INPUT "ANOTHER FILTER (Y OR N)";A$
3115 IF A$<>"Y" AND A$<>"N" THEN 3110
3120 IF A$="Y" THEN 120
3200 END
4000 REM FILTER ORDER CALCULATION ROUTINE
4010 REM BUTTERWORTH SECTION
4020 PRINT "ENTER FREQUENCY RATIO F/FC AS A NUMBER, I.E. 3":INPUT WR
4025 INPUT "DESIRED ATTENUATION AT F/FC";AD
4030 FOR N=2 TO 8
4040 AC=10*(LOG(1+WR^(2*N)))/(LOG(10))
4050 IF AC=>AD GOTO 4070
4060 NEXT:GOTO 4100
4070 PRINT "N=";N;"AT A CALCULATED A=";AC;"DB":RETURN
4100 PRINT "THE REQUIRED ATTENUATION IS OUT OF RANGE FOR A BUTTERWORTH FILTER"
4110 PRINT "OF UP TO 8 POLES."
4120 PRINT "A LOW RIPPLE CHEBYCHEV MAY WORK.":GOTO 130
4130 STOP
4500 REM CHEBYCHEV CALCULATION ROUTINE
4510 DEF FNCS(X)=0.5*(EXP(X)+EXP(-X))
4520 DEF FNASH(X)=LOG(X+SQR(X^2+1))
4530 INPUT "ENTER THE FREQUENCY RATIO F/FC AS A NUMBER I.E. 3";WR
4540 INPUT "DESIRED ATTENUATION AT F/FC";AD
4560 EP=SQR(10^(R/10)-1)
4565 N=2
4570 B=(FN ASH(1/EP))/N
4580 WC=(FN CS(B))*WR
4590 CN=(2*WC^2-1)
4600 AC=10*LOG(1+EP^2*CN^2)/LOG(10)
4610 IF AC=>AD THEN PRINT "N=";N;" AT A CALCULATED A=";AC;"DB":RETURN
4620 N=3
4630 B=(FN ASH(1/EP))/N
4640 WC=(FN CS(B))*WR
4650 CN=(4*WC^3-3*WC)
4660 AC=10*LOG(1+EP^2*CN^2)/LOG(10)
4670 IF AC=>AD THEN PRINT "N=";N;" AT A CALCULATED A=";AC;"DB":RETURN
4700 N=4
4710 B=(FN ASH(1/EP))/N
4720 WC=(FN CS(B))*WR
4730 CH=(8*WC^4-8*WC^2+1)
4740 AC=10*LOG(1+EP^2*CN^2)/LOG(10)
4750 IF AC=>AD THEN PRINT "N=";N;" AT A CALCULATED A=";AC;"DB":RETURN
4800 N=5
4810 B=(FN ASH(1/EP))/N
4820 WC=(FN CS(B))*WR
4830 CN=(16*WC^5-20*WC^3+5*WC)
4840 AC=10*LOG(1+EP^2*CN^2)/LOG(10)
4850 IF AC=>AD THEN PRINT "N=";N;" AT A CALCULATED A=";AC;"DB":RETURN
4900 N=6
4910 B=(FN ASH(1/EP))/N
4920 WC=(FN CS(B))*WR
4930 CN=(32*WC^6-48*WC^4+18*WC^2-1)
4940 AC=10*LOG(1+EP^2*CN^2)/LOG(10)
4950 IF AC=>AD THEN PRINT "N=";N;" AT A CALCULATED A=";AC;"DB":RETURN
5000 N=7
5010 B=(FN ASH(1/EP))/N
5020 WC=(FN CS(B))*WR
5030 CN=(64*WC^7-112*WC^5+56*WC^3-7*WC)
5040 AC=10*LOG(1+EP^2*CN^2)/LOG(10)
3050 IF AC=>AD THEN PRINT "N=";N;" AT A CALCULATED A=";AC;"DB":RETURN
5100 N=8
5110 B=(FN ASH(1/EP))/N
5120 WC=(FN CS(B))*WR
5130 CN=(128*WC^8-2561WC^6+160*WC^4-32*WC^2+1)
5140 AC=10*LOG(1+EP^2*CN^2)/LOG(10)
5150 IF AC=>AD THEN PRINT "N=";N;" AT A CALCULATED A=";AC;"DB":RETURN
5200 PRINT "THE DESIRED ATTENUATION IS NOT WITHIN RANGE OF THIS PROGRAM."
5210 PRINT "FOR THE CHOSEN FILTER TYPE."
5220 PRINT "A HIGHER RIPPLE DESIGN MAY MEET YOUR RE-AUIREMENTS."
5230 GOTO 130
WB4EHS, Bob Lombardi. | {"url":"https://www.robkalmeijer.nl/techniek/electronica/radiotechniek/hambladen/hr/1986/03/page10/index.html","timestamp":"2024-11-03T22:45:25Z","content_type":"application/xhtml+xml","content_length":"28893","record_id":"<urn:uuid:f8d5a8be-3ad2-43ac-b103-4853d2065523>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00771.warc.gz"} |
Stelios Negrepontis Negrepontis (University of Athens): Publications - PhilPeople
• 30
Plato, most unexpectedly, in the middle of the Timaeus (48e2-49a7) declares that the sensible bodies cannot be explained solely by their participation in the intelligible, as we were led to
believe by reading the long succession of all his previous dialogues, but that it is now necessary to introduce, beside the intelligibles and the sensibles, a Third Kind, the Receptacle.We must,
however, in beginning our fresh account of the Universe make more distinctions than we did before; for whereas then…
Read more
• 27
This chapter aims to obtain a novel anthyphairetic interpretation of Knowledge as Recollection in Plato’s Meno 80d5-86c3 and 97a9-98b6, in a self-contained manner, in line with the anthyphairetic
interpretation I have developed for the whole of Plato’s work.Plato sets out to explain his philosophical notion of Knowledge in the Meno, by explaining what he means by Knowledge in the concrete
geometrical case of line a such that a2 = 2b2 for a given line b, in fact of the diameter a of a square with…
Read more
• 4
The reinterpretation of Plato’s philosophy in terms of periodic anthyphairesis, in fact of palindromically periodic anthyphairesis in the Politicus, and the reading of Book X of Euclid’s Elements
under the new light, reveal deep mathematical contributions by Theaetetus, including a proof of the general Pell equation. Fascinating similarities of Theaeteus’ reconstructed proofs with the
Hindus’ solution of the problem of Pell are noted.
• 2
In the present chapter, we provide a novel interpretation of the concept of Plato’s and Xenocrates’ indivisible line, in fact, we show that indivisibility is just another description of the
Platonic intelligible true Being. Our claim and arguments are based on our earlier interpretation of Plato’s intelligible Being as the philosophic analogue of a dyad of opposite kinds in periodic
anthyphairesis (as revealed primarily in the Meno, the second hypothesis of the One in the Parmenides, and the Sop…
Read more | {"url":"https://philpeople.org/profiles/stylianos-stelios-negrepontis/publications?order=viewings","timestamp":"2024-11-13T04:26:29Z","content_type":"text/html","content_length":"45649","record_id":"<urn:uuid:00fa8f38-127e-48ef-9c05-1776c277134c>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00111.warc.gz"} |
economics need help chap 11&12
economics need help chap 11&12
Chapter 11
2. Ajax Cleaning product is a medium sized firm operating in an industry dominated by one large firm-title King Ajax produces a multithreaded tunnel wall scrubber that is similar to a model produced
by Tile King. Ajax decides to charge the same price as Tile King to avoid the possibility of a price war. The price charged by Tile King is $20,000.
Ajax has the following short-run cost curve:
TC= 800,000-5,000Q + 100Q²
(A) Compute the marginal cost curve for Ajax
(B) Given Ajax pricing strategy. What is the marginal revenue function for Ajax
4. Unique Creations holds a monopoly position in the production and sale of mag-nometers. The cost function facing Unique is estimated to be
TC= 100,000+20Q
a. What is the marginal cost for Unique
b. If the price elasticity of demand for Unique is currently -1.5, what price should Unique charge
c. What is the marginal revenue at the price computed in part (b)
6. Wyandotte Chemical Company sells various chemical to the automobile industry. Wyandotte currently sells 30,000 gallons of polyol per year at an average price of $15 per gallon. Fixed costs of
manufacturing polyol are $90,000 per year and total variable costs equal $180,000. The operation research department has estimated that a 15 percent increase in output would not affect fixed costs
but would reduce average variable costs by 60 cent per gallon. The marketing department has estimated the arc elasticity of demand for polyol to be -2.0
a. How much would Wyandotte have to reduce the price of polyol to achieve a 15 percent increase in the quantity sold?
B. Evaluate the impact of such a price cut on (i) total revenue, (ii) total costs and (iii) total profits
Chapter 12
1. Assume that two companies (C and D) are duopolists that produce identical product. Demand for the products is given by the following liner demand function
P= 600 –Qc-Qᴰ
Where Qc and Qᴰ are the quantities sold by the respective firms and P is the selling price. Total cost functions for the two companies are
TCc= 25,000 + 100Qc
TCᴰ= 20,000 + 125 Qᴰ
Assume that the firms act independently as in the cournot model (i.e. each firm assumes that the other firms output will not change).
a. Determine the long-run Equilibrium output and selling price for each firm.
b. Determine the total profits for each firm at the equilibrium output found in part (a)
2. Assume that two companies (A and B) are duopolist who produces identical products. Determine for the products is given by the following liner demand function
P= 200 – Qᴬ-Qᴮ
Where Qᴬ and Qᴮ are the quantities sold by the respective firms and p is the selling price. Total cost functions for the two companies are
TCᴬ = 1,500+35Qᴬ+Q²ᴀ
TCᴮ= 1,200 + 20Qᴮ + 2Q²ᴃ
Assume that the firms act independently as in the Cournot model (i.e. each firm assumes that the other firms output will not change).
a. Determine Firms A, Firm B, and total industry profits at the equilibrium solution found in part (a)
5. Alchem (L) is the price leader in the polyglue market. All 10 other manufacture (follower [F] firms) sell polygue at the same price as Alchem. Alchem allows the other firms to sell as much as they
wish at the established price and supplies the remainder of the demand itself. Total demand for polygue is given by the following function (Qᴛ=Qᴌ+ Qᶠ)
P=20,000 – 4Qᴛ
Alchem’s marginal cost function for manufacturing and selling polygue is
The aggregate marginal cost function for the other manufacture of polygue is
∑MCᶠ=2,000 + 4Qᶠ
a. To maximize profits, how much polygue should Alchem produce and what price should it charge?
b. What is the total market demand for polygue at the price established by Alchem in part (a)? How much of total demand dot he following firms supply?
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2020-07-13 16:38:402020-07-13 16:38:40economics need help chap 11&12 | {"url":"https://classessay.co.uk/economics-need-help-chap-1112","timestamp":"2024-11-06T20:22:20Z","content_type":"text/html","content_length":"69044","record_id":"<urn:uuid:c5fc02c2-6313-41db-b2c3-fc9410d39a40>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00758.warc.gz"} |
Generalized nonorthogonal matrix elements. II: Extension to arbitrary excitations
Electronic structure methods that exploit nonorthogonal Slater determinants face the challenge of efficiently computing nonorthogonal matrix elements. In a recent publication [H. G. A. Burton, J.
Chem. Phys. 154, 144109 (2021)], I introduced a generalized extension to the nonorthogonal Wick’s theorem that allows matrix elements to be derived between excited configurations from a pair of
reference determinants with a singular nonorthogonal orbital overlap matrix. However, that work only provided explicit expressions for one- and two-body matrix elements between singly- or
doubly-excited configurations. Here, this framework is extended to compute generalized nonorthogonal matrix elements between higher-order excitations. Pre-computing and storing intermediate values
allows one- and two-body matrix elements to be evaluated with an $O(1)$ scaling relative to the system size, and the LIBGNME computational library is introduced to achieve this in practice. These
advances make the evaluation of all nonorthogonal matrix elements almost as easy as their orthogonal counterparts, facilitating a new phase of development in nonorthogonal electronic structure
Nonorthogonality occurs throughout modern quantum chemistry, providing chemically intuitive and compact representations of challenging electronic structures. For example, nonorthogonal configuration
interaction (NOCI) uses a linear combination of nonorthogonal Slater determinants, each with bespoke orbitals representing an important electronic configuration, to capture strong static correlation^
1–10 or provide a diabatic representation of electron transfer states.^11–13 Alternatively, nonorthogonal transition coupling elements arise for orbitally optimized excited states, which can improve
the accuracy of challenging core^14,15 or charge transfer excitations.^16–23 Beyond providing chemical intuition, nonorthogonality naturally arises in auxiliary-field quantum Monte-Carlo,^24,25 in
coupling terms between states at different molecular geometries,^26,27 and in spin-projected techniques.^10,28–30
Despite presenting many potential benefits, the development of nonorthogonal electronic structure methods is hindered by the difficulty of computing matrix elements between Slater determinants with
mutually nonorthogonal orbitals. For orthogonal orbitals, the second-quantized generalized Wick’s theorem^31 allows arbitrary matrix elements to be easily derived and evaluated using pre-computed
one- and two-electron integrals. In contrast, matrix elements for nonorthogonal determinants currently require the first-quantized generalized Slater–Condon rules,^32 with a computational scaling of
at least $O(N3)$ with respect to the number of electrons N.^4
While the generalized Slater–Condon rules can be used to couple a small number of Slater determinants,^4–7,33 their additional computational cost quickly makes larger calculations unfeasible. In
particular, many recent developments involve multiple orthogonal excited configurations from different nonorthogonal reference determinants, introducing a large number of nonorthogonal matrix
elements. This situation arises for the coupling terms between state-specific multi-determinant wave functions^20–23,34 or post-NOCI techniques for capturing dynamic correlation,^2,29,30 including
perturbative approximations such as NOCI-MP2^35–37 and NOCI-PT2.^9 Developing efficient implementations of these methods requires a theory for computing nonorthogonal matrix elements between excited
configurations that do not scale with the system size. While equations for specific applications have been independently derived many times (e.g., Refs. 5 and 35), an entirely generalized approach
has not yet been developed.
A nonorthogonal extension to Wick’s theorem has been well established for many years^38,39 and has seen limited use in quantum chemistry.^29,30,40,41 In the nonorthogonal Wick’s theorem, matrix
elements between two Slater determinants can be evaluated using contractions defined from the non-Hermitian transition density matrix.^38 Combining this framework with Lowdin’s general formula^42,43
allows nonorthogonal matrix elements to be readily evaluated between excited-configurations from different reference determinants.^24,25,44 However, the derivation of the nonorthogonal Wick’s theorem
is restricted to reference determinants with a non-singular orbital overlap matrix.^30 This restriction excludes the case where the orbitals are mutually nonorthogonal but the determinants themselves
have a zero overlap, which occurs when the orbital overlap matrix is singular but not diagonal.^45
In Ref. 46 (henceforth known as Paper I), I introduced an extension to the nonorthogonal Wick’s theorem that allows matrix elements to be computed for nonorthogonal determinants with a zero many-body
overlap. This theory allows overlap, one-body, and two-body coupling terms to be derived for excited configurations from arbitrary reference determinants, and provides a second-quantized derivation
of the generalized Slater–Condon rules. I then demonstrated that overlap and one-body coupling terms between nonorthogonal excited configurations can be evaluated with $O(1)$ scaling using a set of
pre-computed intermediates. However, these explicit derivations were difficult to extend for coupling terms beyond double excitations and the $O(1)$ scaling did not appear to be achievable for
two-electron coupling terms.
In the current work, I combine the results of Paper I with the determinantal expansion based on Löwdin’s general formula,^42,43 giving matrix elements between excitations from arbitrary Slater
determinants with any number of singular values in the reference orbital overlap matrix. Furthermore, by introducing pre-computed intermediates, I derive explicit expressions for the overlap,
one-body, and two-body coupling terms between arbitrary excitations that scale as $O(1)$ with respect to the number of electrons or basis functions. Finally, I introduce the LIBGNME computational
library to provide an open-source implementation of these expressions. These advances significantly reduce the computational cost compared with the previous state-of-the-art, creating a new paradigm
for efficient nonorthogonal matrix elements in quantum chemistry.
In Sec. II, I review the nonorthogonal Wick’s theorem and introduce the key results from Paper I, before extending these to arbitrary excitations in Sec. III. Section IV presents numerical data
illustrating the computational scaling of these generalized nonorthogonal matrix elements. The conclusions and scope for future applications are summarized in Sec. V.
In this section, I summarize Paper I’s key extensions to the nonorthogonal Wick’s theorem that are required for the derivations presented in this work. Interested readers are directed to Paper I for
a comprehensive explanation of these results and to Ref. 31 for details about the generalized Wick’s theorem.
A. Biorthogonalising the molecular orbitals
This work considers matrix elements between excited configurations constructed from a general pair of mutually nonorthogonal determinants, e.g.,
where $Ô$ is an arbitrary operator. Here, $|Φx〉$ and $|Φw〉$ are reference Slater determinants built from different molecular orbitals ${ϕpx}$ and ${ϕpw}$ that are expanded using a common n
-dimensional atomic orbital basis set {χ[μ]} as
The $C⋅pμ⋅x$ denote the orbital coefficients in the molecular orbital basis of determinant $|Φx〉$ using the nonorthogonal tensor notation of Head-Gordon et al.^47 The corresponding second-quantized
operators $bp†x$ and $bpx$ create and annihilate, respectively, an electron in the molecular orbital p for the orbital set corresponding to determinant $|Φx〉$.
To evaluate matrix elements using these nonorthogonal determinants, the molecular orbitals ${ϕpx}$ and ${ϕpw}$ must be transformed into a biorthogonal set using the Löwdin pairing approach.^48,49
First, the overlap matrix between the two sets of occupied orbitals is computed as
where g[μν] = ⟨χ[μ]|χ[ν]⟩ is the overlap matrix for the atomic orbital basis set. A singular value decomposition is then performed to diagonalize this overlap matrix, giving the modified occupied
orbital coefficients $(C̃w)i⋅⋅μ$ and $C̃i⋅⋅μx$ that satisfy
When one, or more, of the diagonal overlap terms becomes zero (i.e., $S̃ixw=0$), the many-body overlap must vanish $⟨Φx|Φw⟩=0$. The number of these biorthogonal zero-overlap orbital pairs
corresponds to the nullity of the matrix ^xwS and is denoted by the integer m. The value of m determines which instance of the generalized Slater–Condon rules is required for a matrix element between
the reference states $|Φx〉$ and $|Φw〉$, as described in Refs. 4 and 32. The product of the remaining non-zero singular values defines the reduced overlap
which also appears in the generalized Slater–Condon rules.^4
B. Extended nonorthogonal Wick’s theorem
Without loss of generality, I will represent operators in the molecular orbital basis for determinant $|Φx〉$. For example, a one-body operator $f̂$ is given by
can then be evaluated using an extension to Wick’s theorem (see Ref. 31). In particular, each term in Eq. (8) is computed as the sum of all possible ways to fully contract the second-quantized
operators, for example,
The phase of each term is given by (−1)^h where h is the number of intersecting contraction lines.
For the extended nonorthogonal Wick’s theorem outlined in Paper I, a general matrix element in this expansion (e.g., $S̃xw$ multiplied by the product of nonorthogonal contractions,
where $Xqpwx$ and $Yqpxw$ are intermediate values defined for a given pair of determinants [Eq. (11)]. If there are zero-overlap orbital pairs in the biorthogonal basis (m > 0), then a sum must be
taken over every possible way to assign the m zeros to the contractions in each product. This distribution is denoted by indices m[k] assigned to each contraction (i.e., $Xqp(mk)wx$ and $Ypq(mk)xw$
), which take values of 0 or 1 and satisfy ∑[k]m[k] = m. Accounting for these zero-overlap orbital pairs is the central result of Paper I compared with the standard nonorthogonal Wick’s theorem.^
38,45 The individual contractions, represented in the original (un-transformed) orbital basis, are then
(Warning: for convenience throughout this work, the sign of$Ypqxw$has been reversed relative to Paper I.)
Notably, the contractions in Eq. (10) are defined with respect to the original orbital coefficients such that the definition of excited configurations is not affected by the biorthogonal
transformation. The number of zero-overlap orbital pairs corresponds to the underlying biorthogonal basis and there is no relationship between the p, q and m[k] indices. Furthermore, the $Xqp(mk)wx$
and $Ypq(mk)xw$ intermediates can be evaluated and stored once for each pair of nonorthogonal reference determinants.
While the biorthogonal orbital coefficients only enter in the definition of the ^xwM and ^xwP matrices, the number of biorthogonal zero-overlap orbital pairs affects every matrix element through the
summation over the allowed combinations of m[k] values. This distribution is essential for recovering the different instances of the generalized Slater–Condon rules.^46 If m is larger than the total
number of contractions, then the corresponding matrix element is strictly zero.
In summary, a nonorthogonal matrix element between excited configurations can be evaluated through the following procedure:
1. Assemble all fully contracted combinations of the second-quantized operator and excitations, and compute the corresponding phase factors;
2. For each term, sum every possible way to distribute m zeros among the contractions {m[k]} such that ∑[k]m[k] = m;
3. For every set of {m[k]} in each term, construct the relevant contribution as a product of fundamental contractions defined in Eqs. (10) and (11);
4. Multiply the combined expression by the reduced overlap $S̃xw$.
To demonstrate this process, consider the matrix element $⟨Φiax|Φjbw⟩=⟨Φx|bi†xbaxbb†wbjw|Φw⟩$. Applying Step 1 gives two contributions,
Each term corresponds to a product of two fundamental contractions with a phase of +1. Taking a sum over the different m[1], m[2] values satisfying m[1] + m[2] = m, and multiplying by the reduced
overlap $S̃xw$, then gives
which can also be represented as the determinant,
This matrix element reduces to zero if m > 2.
A. Unification with Löwdin’s general formula
Paper I demonstrated the explicit derivation of nonorthogonal matrix elements for the overlap, one-body, and two-body operators between excited configurations from reference determinants with
zero-overlap orbital pairs.^46 However, those derivations are difficult to generalize for higher-order excitations due to the increasingly large number of fully contracted terms in the expansion of a
matrix element. A generalized approach can be achieved using Löwdin’s general formula,^42,43 which expresses the matrix element for an arbitrary product of creation and annihilation operators as a
determinant and automatically includes every fully-contracted term with the correct phase factor. Following previous nonorthogonal derivations,^25,29,30,44,50 Löwdin’s general formula can be applied
to the nonorthogonal contractions defined in Eq. (10) to give
Here, terms in the upper triangle of the determinant correspond to contractions where the annihilation operator is to the left of the creation operator and are multiplied by a factor of −1 to give
the correct sign once the determinant is expanded.
We now demonstrate how this representation of the nonorthogonal Wick’s theorem is extended to reference determinants with zero-overlap orbital pairs. First, in contrast to the conventional Wick’s
theorem, the entire determinant in Eq. (16) is multiplied by the reduced overlap $S̃xw$. Second, the distribution of m zero-overlap biorthogonal orbital pairs over products of individual contractions
is achieved by exploiting the fact that every term in the expansion of a determinant includes one element from each column of the corresponding matrix. Therefore, a summation can be taken over every
possible way to distribute the m zero-overlap orbital pairs among the columns in Eq. (16). Inserting the nonorthogonal contractions defined in Eqs. (10) and (11) then gives
where L is the total number of creation and annihilation pairs in the operator string. In Eq. (17), the lower triangle of the matrix (including the diagonal) contains “X-type” matrix elements while
the upper triangle contains “Y-type” matrix elements. Reversing the sign of the “Y-type” terms defined in Eq. (11b) relative to Paper I ensures that these elements enter the determinant in Eq. (17)
with a positive sign contribution, simplifying subsequent derivations. This modified representation of the extended nonorthogonal Wick’s theorem now provides matrix elements between excited
configurations from nonorthogonal reference determinants with an arbitrary number of zero-overlap orbital pairs, as demonstrated in Secs. III B–III D.
B. Overlap elements
The overlap between a general pair of excited configurations is given by
Using the approach presented in Sec. III A, these overlap matrix elements can be expanded as
where L is the combined number of excitations in the bra and ket states. Since fundamental contractions X and Y can be precomputed and stored once for a given pair of reference determinants, the
computational cost of evaluating Eq. (19) is only controlled by the total number of excitations L and the number of zero-overlap orbitals in the biorthogonal basis m. Specifically, the cost of
evaluating each determinant in Eq. (19) scales as $O(L3)$, while the number of ways to distribute the m zeros over the L contractions is $Lm$, giving an overall scaling of $O(L3Lm)$. In
comparison, the conventional first-quantized approach evaluates the determinant of the occupied orbital overlap matrix between a pair of excited configurations, giving an $O(N3)$ scaling.^32 Since
the number of excitations is smaller than the total number of electrons by definition, the evaluation of the extended nonorthogonal Wick’s theorem using Eq. (19) is generally much more efficient.
C. One-body coupling
Representing one-body operators in the molecular orbital basis for determinant $|Φx〉$ as
allows a general coupling term between two excitations from nonorthogonal reference determinants to be expressed as
Applying Löwdin’s general formula for the extended nonorthogonal Wick’s theorem give the expansion,
where L[x] is the number of excitations in the bra state and L is the combined number of excitations in the bra and ket states. Following Ref. 42, a Laplace expansion can then be performed along the
row corresponding to the one-body operator index q to give
where, for convenience, the element along this row corresponding to operator index p is placed first, and the dummy variables m[k] have been rearranged. On the second row of Eq. (23), and those below
it, the columns in the minor submatrices have been swapped so that the column corresponding to the one-body operator (index p) occupies the position originally held by the corresponding cofactor
(e.g., index i, j, d, or c), ensuring that these rows all have the same overall sign contribution.
While a naïve implementation of Eq. (23) scales as $O(n2)$ due to the summations over p and q, this scaling can be removed by introducing a series of intermediate quantities. The first line requires
the intermediate,
which takes different values depending on whether the contraction $Xqp(mi)xx$ is assigned to a zero-overlap orbital pair, as determined by m[i]. For the remaining lines in Eq. (23), the rules for
multiplying a determinant by a scalar can be exploited to move the summation over p and q inside the corresponding minor submatrices. Introducing the intermediate quantities
and substituting these into the corresponding column for index p gives the overall matrix element expression,
This expression comprises the term $F0(m1)xx$ multiplied by the total overlap of the excited configurations, minus terms where each column of the submatrix is replaced by the corresponding
intermediate, e.g., $Fai(m1,m2)xw$.
Computationally, the $F$ intermediates in Eq. (25) can be pre-computed with scaling $O(16n3)$ and stored with scaling $O(16n2)$, where the factor of 16 accounts for the four possible values of (m[i
], m[j]) and the four combinations xx, xw, wx, and ww. Once these intermediates have been evaluated, the total cost of computing a one-body matrix element only depends on the total number of
excitations in the bra and ket states L, which controls the size of the determinants in Eq. (26) and the number of columns that must be replaced with the intermediates, e.g., $Fai(m1,m2)xw$. The
cost of evaluating each determinant in Eq. (26) scales as $O(L3)$ and there are L + 1 determinants that must be computed. The summation over the m[i] values depends on the total number of way to
distribute m zero-overlap orbitals between the L + 1 contractions, given by $L+1m$. Consequently, the overall scaling for each one-body matrix element is $O(L3(L+1)L+1m)$, which is constant with
respect to the number of electrons or the size of the basis set. In comparison, applying the generalized Slater–Condon rules for each pair of determinants requires the biorthogonalization of the full
set of occupied orbitals, with an iterative $O(N3)$ scaling, followed by an $O(n2)$ contraction of the co-density matrices with the one-electron integrals.^4,32 Evidently, the new approach described
here is significantly more efficient when a large number of coupling elements are required.
D. Two-body coupling
Finally, two-body operators can be represented as
where the two-electron integrals $vprqsx=(pr|qs)x$ are expressed in the MO basis for reference $|Φx〉$ and are represented using Mulliken notation.^51 A general coupling term between two excitations
from nonorthogonal reference determinants is then given by
Deriving expressions for these matrix elements using the Laplace expansion approach is much more involved than the one-body operators and is much harder for the reader to follow. Instead, an
alternative approach can be used that constructs effective zero- or one-body operators by partially contracting subunits containing two or four of the indices p, r, q, s, and the one-body framework
established in Sec. III C can then be applied. The final expression is given by
$⟨Φij…ab…x|v̂|Φkl…cd…w⟩=Eq. (30) + Eq. (34) + Eq. (38),$
where each constituent equation is derived in detail below.
First, contractions containing the subunits
where the intermediate terms have been defined,
The contribution from Eq. (30) is analogous to the one-body $F0(m1)xx$ contribution in the first line in Eq. (26).
Next, the partially contracted subunits
where the intermediate term in Eq. (31a) has been employed and the factor of two arises from the equivalence of terms such as p, q, r, s. The contribution from these partially contracted subunits can
then be evaluated using the rules established for one-body operators in Sec. III C by introducing the additional intermediate terms,
Each of these intermediates can be computed with a computational scaling of $O(8n3)$, and the overall storage cost for a pair of reference determinants is 32n^2. The one-body terms that are
analogous to the first line in Eq. (26), which represent the contractions (30). Therefore, the unique contributions of these effective one-body operators to the full two-body matrix element in Eq.
(28) is
The summation within the parentheses includes the replacement of each column of the overlap determinant with the corresponding intermediate terms from Eq. (33).
The remaining terms include cases where all the indices p, q, r, s are contracted with operators that occur in the bra or ket excitation strings. An effective one-body operator can be constructed by
considering partial contractions with the form, e.g.,
where the indices p, r and p, s are contracted with bra or ket excitation operators, giving effective operators with the form, e.g.,
where the permutation of the dummy indices p, q, r, s has been exploited to incorporate the “exchange-like” terms. The operators in Eq. (36) contain a phase factor ϕ[ab] that takes the value +1 if a
and b correspond to the same excitation or are separated by an odd number of excitations, and −1 if a and b correspond to different excitations separated by an even number of excitations, as
illustrated for the example $⟨Φijabx|v̂|Φklcdw⟩$ by the checkerboard pattern,
Every effective one-body operator defined for the L^2 combinations of a creation and annihilation operator pair (excluding p, q, r, s) must be considered. The contribution to the full two-body matrix
element can then be evaluated using the one-body approach described in Sec. III C by introducing a final set of intermediates
The standard two-electron integral symmetry relation $Jab,cd(mi,mj,mk,ml)xw,yz=Jcd,ab(mk,ml,mi,mj)yz,xw$ allows intermediates for the remaining combinations of indices to be obtained. Each
intermediate can be computed with a computational scaling of $O(16n5)$, where the factor of 16 comes from the possible combinations of (m[i], m[j], m[k], m[l]). Taking into account the two-electron
symmetry relation, the maximum storage requirement is 160n^4. In practice, the storage may be reduced if not all combinations of (m[i], m[j], m[k], m[l]) are required (i.e., m < 4) or if excitations
are only considered within an active orbital space.
When applying the one-body framework (Sec. III C) to the effective operators defined in Eq. (36), the terms that arise from contracting the q and s indices are discarded as these are already taken
into account by Eq. (34). These terms correspond to the first line in Eq. (26). Therefore, the remaining unique contributions to the two-body matrix element are
Individual lines in this expression correspond to the effective one-body operators constructed from a different pair of the creation and annihilation operators selected from excitation strings [i.e.,
(ai), (aj), …, (kc)]. The sum within a line corresponds to the replacement of each column by the corresponding intermediate for the remaining excitation indices, where the excitation indices used to
build the effective one-body operators have been removed. Every contribution also includes a summation over the possible ways to distribute the m zero-overlap orbital pairs over the contractions. The
factor of 1/2 accounts for the double counting of contributions, for example, the term $Jai,bj(m1,m2,m3,m4)xx,xx$ appears for the effective one-body operator with indices (ai) and (bj) due to the
symmetry $Jai,bj(m1,m2,m3,m4)xx,xx=Jbj,ai(m1,m2,m3,m4)xx,xx$.
Once all the required intermediates have been computed and stored, the computational cost of evaluating a two-body matrix element using this approach is determined by cost of computing Eqs. (30),
(34), and (38). For each term, there are $L+2m$ ways to distribute m zero-overlap orbital pairs among the L + 2 contractions. Equation (30) involves the computation of only one L × L determinant,
giving an $O(L3L+2m)$ scaling. Equation (34) requires the computation of L determinants where each column in the overlap determinant is replaced by intermediates of the form Eq. (33), giving a
scaling of $O(L4L+2m)$. Finally, Eq. (38) involves the computation of L − 1 determinants with dimensions (L − 1) × (L − 1) for each pair of creation and annihilation operators in the excitation
strings, giving a scaling of $O(L2(L−1)4L+2m)$. The asymptotic scaling for individual two-body matrix elements using these intermediates is, therefore, $O(L6L+2m)$. Although this computational cost
increases rapidly with the number of excitations in a two-body matrix element, it has $O(1)$ scaling with respect to the number of basis functions or electrons. In contrast, applying the generalized
Slater–Condon rules for two-body operators scales as $O(n4)$ for each matrix element and rapidly makes the computation of a large number of matrix elements unfeasible.
The extended nonorthogonal Wick’s theorem for arbitrary excitations has been implemented in a developmental open-source C++ package LIBGNME, available for download at Ref. 52. The primary advantage
of the extended nonorthogonal Wick’s theorem over the generalized Slater–Condon rules is the $O(1)$ scaling with respect to the number of basis functions or electrons that can be achieved once all
the required intermediates have been pre-computed. This scaling means that the computation of nonorthogonal matrix elements for excited configurations becomes almost as straightforward as the
conventional orthogonal Slater–Condon rules or Wick’s theorem.
The acceleration relative to the generalized Slater–Condon rules can be demonstrated by comparing the average time required to compute one- and two-body matrix elements between singly excited or
doubly excited configurations with increasingly large correlation-consistent basis sets.^53 Illustrative calculations using LIBGNME have been performed for two nonorthogonal reference determinants
corresponding to the spin–flip pair of spin-symmetry-broken unrestricted Hartree–Fock solutions for the ground state of H[2]O at a bond angle of 104.5 and R(O–H) = 1.35 Å.
Average timings for each one-body matrix element are compared between the extended nonorthogonal Wick’s theorem and the generalized Slater–Condon rules in Fig. 1. This average is computed using all
the high-spin single or double excited coupling terms within a (10, 13) active space constructed from the lowest-energy molecular orbitals. This choice corresponds to 40 single and 280 double
excitations, giving a total of 1600 single–single and 78400 double–double coupling terms for each basis set. The statistical distribution is assessed using 48 replicas of each calculation. As
expected, these data demonstrate the $O(1)$ scaling of the extended nonorthogonal Wick’s theorem with respect to the basis set size, while the generalized Slater–Condon rules scale asymptotically as
$O(n2)$. For small basis sets, the scaling of the generalized Slater–Condon rules becomes constant as the computational cost is dominated by biorthogonalizing the occupied orbitals in each pair of
excited configurations. In the largest basis set considered (cc-pV5Z), the extended nonorthogonal Wick’s theorem provides a computational acceleration of nearly 3 orders of magnitude relative to the
generalized Slater–Condon rules.
Analogous average timings for two-body matrix elements between singly excited and doubly excited configurations are presented in Fig. 2. Here, the generalized Slater–Condon rules show a near-ideal $O
(n4)$ computational scaling, while the extended nonorthogonal Wick’s theorem has a constant cost for all basis sets, as predicted. The larger n^4 scaling of two-body compared with one-body matrix
elements means that the extended nonorthogonal Wick’s theorem offers an even greater advantage over the generalized Slater–Condon rules, as demonstrated by the six orders of magnitude acceleration
achieved for the $⟨Φiax|v̂|Φjbw⟩$ matrix elements in the cc-pVQZ basis set. These numerical results highlight the significant computational advance provided by the generalized nonorthogonal matrix
elements compared with the previous state-of-the-art.
This work has extended the nonorthogonal Wick’s theorem^38,45 and the results of Ref. 46 to derive coupling elements between arbitrary excited configurations from a pair of reference Slater
determinants with a singular nonorthogonal orbital overlap matrix. By pre-computing and storing various intermediate terms, subsequent one- and two-body matrix elements can be evaluated with a
computational cost that scales $O(1)$ with respect to the number of basis functions or electrons. These developments provide a significant improvement over the commonly used generalized Slater–Condon
rules, which asymptotically scale as $O(n2)$ or $O(n4)$ for one- or two-body matrix elements, respectively.
While the nonorthogonal Wick’s theorem is well established^38,45 and explicit expressions for certain nonorthogonal matrix elements have previously been reported, this work presents an entirely
general and unified framework for developing nonorthogonal techniques. The current approach shares many similarities with the nonorthogonal matrix elements derived by Mahajan and Sharma,^24,25
including the use of the determinantal expansion of Wick’s theorem. However, in contrast to their work, the derivations presented here are entirely generalized to cases where singular values occur in
the biorthogonalization of the reference orbitals, providing a unification with the various instances of the generalized Slater–Condon rules.^32 Consequently, this framework establishes the most
general and flexible version of Wick’s theorem for computing matrix elements between arbitrary Slater determinants.
In practice, the bottleneck for these generalized nonorthogonal matrix elements is the computation and storage of the intermediate terms required for two-body operators. While the evaluation of the
intermediates in Eq. (37) has the same $O(n5)$ scaling as a standard two-electron integral transformation, the cost of storing all these intermediates for a given pair of determinants scales as 160n^
4 when the two-electron symmetry is taken into account. The storage of two-electron integrals is already challenging for larger basis sets, so increasing this by a factor of 160 represents a
significant computational overhead. This cost can be reduced if there are no zero-overlap orbital pairs in the reference determinants (i.e., m = 0), such that only a subset of (m[i], m[j], m[k], m[l
]) combinations are required, or if excitations are only considered within an active orbital space. Alternatively, we can expect techniques such as the Cholesky decomposition^54 or
resolution-of-the-identity^55 to provide further reductions in the storage and computational cost of computing the two-electron intermediates. These areas for improvement are beyond the scope of the
current work but will form the focus for future computational research and development.
Until now, the development of advanced nonorthogonal techniques in electronic structure theory has been hindered by the computational cost of computing arbitrary nonorthogonal matrix elements. The
generalized extension to the nonorthogonal Wick’s theorem,^38,45 introduced in Ref. 46 and further developed in this work, now offers a universal route to overcome this challenge. Moving forwards,
the ability to rapidly compute nonorthogonal matrix elements between excited configurations will allow on-the-fly implementations of NOCI extensions such as NOCI-PT2,^9 NOCI-MP2,^36,37 and
nonorthogonal configuration interaction singles (NOCIS).^14,15 Furthermore, the generalization to arbitrary excitations will enable the evaluation of coupling terms between excited state-specific
multi-configurational wave functions^20–23,34 that are prohibitively expensive using the generalized Slater–Condon rules. Ultimately, the generality of this framework, and the availability of the
open-source LIBGNME implementation,^52 will catalyze a new phase of development in nonorthogonal electronic structure theory.
H.G.A.B. was supported by New College, Oxford through the Astor Junior Research Fellowship. The author thanks Nicholas Lee and Alex Thom for helpful feedback during the preparation of this
manuscript, and Rebecca Lloyd for careful proofreading.
Conflict of Interest
The author has no conflicts to disclose.
Author Contributions
Hugh G. A. Burton: Conceptualization (lead); Formal analysis (lead); Investigation (lead); Methodology (lead); Project administration (lead); Software (lead); Writing – original draft (lead); Writing
– review & editing (lead).
The data that support the findings of this study are available within the article.
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Theor. Chim. Acta | {"url":"https://pubs.aip.org/aip/jcp/article/157/20/204109/2842193/Generalized-nonorthogonal-matrix-elements-II","timestamp":"2024-11-08T00:10:55Z","content_type":"text/html","content_length":"597813","record_id":"<urn:uuid:866a8e44-0fae-47c5-872d-f0096c9ecb89>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00129.warc.gz"} |
Affecting A Grid With An Attractor Curve But To One Side Only
Hello All,
We are strategizing how to come up with a definition we can use to affect a grid with an attractor curve. However we only want to affect one side of the curve and not the other. The question we would
like to ask is are there any components you (the community) might suggest to help us out. We have tried googling it for tips but to no success. Any tips out there? | {"url":"https://www.grasshopper3d.com/forum/topics/affecting-a-grid-with-an-attractor-curve-but-to-one-side-only?commentId=2985220%3AComment%3A1113692","timestamp":"2024-11-08T04:26:42Z","content_type":"text/html","content_length":"75856","record_id":"<urn:uuid:036dfd7b-181f-469a-aa7d-a4864814e450>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00571.warc.gz"} |
if statement
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if statement
I am trying to make an "if statement" but I got no luck.
I don't know how to start, I tried with
=IF(sqf1 > "32", "width1*hight1*5.5", "sqf1 < "32""width1*hight1*8.5") I know its looks bad
Thank you
• hi Haithem
you're not too far off.
By using "" you are telling Smartsheet that 32 is text. You also need to consider what you want to happen if your sqft is actually 32 as this would not trigger either condition.
I've assumed you want to multiply by 5.5 if the sq ft is 32 So the final formula is
=IF(sqft1 >= 32, width1 * height1 * 5.5, IF(sqft1 < 32, width1 * height1 * 8.5))
Forgive me but I noticed you'd spelt height as hight so for the formula to work you'll need to change your column header to height
• Thanks a lot Mark! that works with me.
I don't know why I cant give your comment a Like or a reply. Thanks anyway
• what about this one?
=IF(Type37 = "feet", IF(sqft37 >= 32, width37 * height37 * Qty37 * 5.5, IF(Type37 = "feet", IF(sqft37 < 32, width37 * height37 * Qty37 * 8.5, IF(type37="inches", IF(sqft >=32,
width37*height37*qty37*5.5, IF(type37="inches" , IF(sqft37<32, width37*height37*qty37*8.5))))))))
where is the error? I am so confused
• It might be here:
=IF(Type37 = "feet", IF(sqft37 >= 32, width37 * height37 * Qty37 * 5.5, IF(Type37 = "feet", IF(sqft37 < 32, width37 * height37 * Qty37 * 8.5, IF(type37="inches", IF(sqft37 >=32,
width37*height37*qty37*5.5, IF(type37="inches" , IF(sqft37<32, width37*height37*qty37*8.5))))))))
I did not create a test case.
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[OS X TeX] Increasing font size in math mode
Michael Sharpe msharpe at ucsd.edu
Thu Dec 15 03:16:41 CET 2011
On Dec 14, 2011, at 5:19 PM, Michael Sharpe wrote:
> On Dec 14, 2011, at 4:10 PM, R Martinez wrote:
>> Hi everybody,
>> Below is a test file that describes the problem, which is as follows. I want to use the Zapf Chancery font to display the letter A in math mode. To do so, I use the command \DeclareMathAlphabet{\zcit}{T1}{pzc}{m}{it}. This produces the desired A, but the symbol is too small and I would like to increase its size to match that of the symbol obtained with \mathcal(A).
>> I have spent a fair amount of time scouring documents, books, the web, to no avail. Asking the list is my last resort.
>> Thanks in advance.
>> Happy holidays,
>> Raul Martinez
> Use it via the mathalfa package, which allows a simple mode of scaling.
> \usepackage[mathcal=zapfc,scaled=1.15]{mathalfa}
> The effect is to set mathcal to the urw clone of zapf chancery, scaled up by 15%. This bypasses a problem that can occur with pzc when used in latex+dvips+ps2pdf mode, and download base 35 was not set to True. In this case, pzc renders from the faulty font distributed by Artifex in the gs distribution. (It is supposed to be fixed in gs 9.05.)
> In my opinion, there are better options for math script---see the mathalfa documentation.
That should have been
I should have added that this provides a math alphabet with proper math metrics so that accents and subscripts are where you expect them, and spacing in formulas is math-like rather than text-like.
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Force-displacement relationships for NiTi alloy helical springs by using ANSYS: Superelasticity and shape memory effect
Shape memory alloys are smart materials which have remarkable properties that promoted their use in a large variety of innovative applications. In this work, the shape memory effect and superelastic
behavior of nickel-titanium helical spring was studied based on the finite element method. The three-dimensional constitutive model proposed by Auricchio has been used through the built-in library of
ANSYS® Workbench 2020 R2 to simulate the superelastic effect and one-way shape memory effect which are exhibited by nickel-titanium alloy. Considering the first effect, the associated
force-displacement curves were calculated as function of displacement amplitude. The influence of changing isothermal body temperature on the loading-unloading hysteretic response was studied.
Convergence of the numerical model was assessed by comparison with experimental data taken from the literature. For the second effect, force-displacement curves that are associated to a complete
one-way thermomechanical cycle were evaluated for different configurations of helical springs. Explicit correlations that can be applied for the purpose of helical spring's design were derived.
1 Introduction
Shape Memory Alloys (SMA) have attracted much attention for the last few decades due to their excellent mechanical properties as well as their essential aspects of deformation and transformation in
structural behavior [1, 2]. Recent increase in applications of these materials in a wide variety of fields, such as aerospace, medical, civil and mechanical engineering, has led to an increased focus
on modeling their thermomechanical response. SMA materials exhibit two significant macroscopic phenomena which are called the SuperElasticity (SE) and Shape Memory Effect (SME) [3]. Both these
effects are used in practice in order to design devices that enable achieving special functions. In many engineering applications, SMA helical springs are used as actuator devices. This structure is
considered in the actual work in order to derive, through numerical simulations by means of ANSYS® Workbench 2020 R2, force-displacement relationships that can be used to carry out smart design of
these devices.
Various SMA materials were discovered since they were first revealed in the thirties of the last century. They can be differentiated according to their thermomechanical characteristics, as well as
working range of temperature and cost. Among them, the nickel-titanium alloy (NiTi or Nitinol) has gained substantial interest in practice. This may be explained by its significant advantages over
the other families of SMA, such as stable transformation temperatures, effective thermal memory, high corrosion resistance, high mechanical performances and the faculty of undergoing large
deformation [1]. Stable transformation temperatures means here that transformation profiles during cooling, following thermal cycling for heat treatment purpose, tend to stabilize after performing a
sufficient number of cycles. It is then possible to reach transformation temperatures which are almost independent from the number of work cycles if the range of temperatures is consistent with heat
treatment [4]. The NiTi SMA alloy shows in practice a high quality-price ratio. It was discovered in the 1960s, at the Naval Ordnance Laboratory [5], and it is considered to be a superelastic
material with recoverable memory strains of up to 8% [6]. This SMA material is selected as the design material of helical springs considered in the present study.
To simulate SMA response, the finite element method is used under ANSYS software. Considering the SE effect, use is made of the constitutive model proposed first by Auricchio et al. [5] and improved
later by Auricchio and Taylor [7] in order to capture the asymmetrical behavior of SMA during a tension-compression test. The material option for the SME effect is based on the 3D thermomechanical
constitutive equations for solid phase transformations induced by stresses as proposed by Auricchio and Petrini [8], Auricchio [9] and Souza et al. [10]. These models have been recognized to yield
suitable results for common SMA applications. Both of these models have been successfully implemented into the finite element based commercial software ANSYS [3, 11]. It should be mentioned that to
date ANSYS code is the only software able to implement both SE and SME without needing special development of user material subroutines [12]. These two options can be accessed directly via the
Temperature Bulk (TB) – SMA command of ANSYS.
Use is made in the following of ANSYS simulation environment in order to assess in closed form force-displacement relationships for various configurations of helical springs, under both SE and SME
situations. The aim is to present an alternative way to existing analytical approaches that can be employed to ease the design procedure of arbitrary SMA applications that are based on helical
springs. This is because existing analytical models suffer in general from the shortcoming that their accuracy depends largely on the actual geometric configuration of the structure and the loading
applied to it [13, 14]. On the opposite, numerical simulations can be carried out parametrically in order to derive useful correlations that provide direct handling of the design procedure in the
framework of special helical spring's applications [15].
Considering the SE effect, the force-displacement curves are calculated as function of displacement amplitude and body isothermal temperature. Convergence of the finite element model will be assessed
through comparison of the obtained numerical predictions with experimental data taken from the literature. In contrast to the SE effect, the SME effect has been rarely studied in literature.
Considering the one-way MSE effect, the reaction force curve needed to reach a desired course of a given helical spring will be evaluated for different configurations including various values of wire
radius, coil radius and initial length.
2 Materials and methods
2.1 Numerical simulations performed on SMA NiTi alloy
SMA alloys show diverse shape-memory effects. Two common encountered effects are: superelasticity or pseudoelasticity, and one-way shape memory. The first effect SE designates the capability of
recovering the original shape after undergoing large deformations that are induced by pure mechanical loading, see Fig. 1.
Fig. 1.
Superelasticity effect: at constant high temperature the material is first loaded (ABC), it shows a nonlinear behavior, while during unloading (CDA) the reverse transformation occurs forming a
flag-shape hysteresis
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
The second effect which is termed SME indicates the ability to recover the original shape by simple heating (above austenite finish temperature) after being initially mechanically deformed at a
sufficiently low temperature. Figure 2 presents the thermomechanical trajectory associated to a one-way SME. This effect is used in many applications of SMA where self expanding of a part under the
effect of temperature is required. Deployment is performed in general at room temperature for which the part is initially stressed to deform it plastically and then unloaded to get a permanent
strained shape. Then, temperature is increased up to finish temperature of the austenitic transformation to yield countersense deformation. This enables to get back the initial shape which results to
be insensitive to temperature. In practice, the CDA branch of Fig. 2 is used by special monitoring of body temperature.
Fig. 2.
One-way shape memory effect: at the end of a mechanical loading-unloading path (ABC) performed at constant low temperature, the material presents residual deformation (AC), the residual strain may be
recovered through a thermal cycle (CDA)
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
It should be noticed that a two-way effect is also encountered in SMA, but it needs special training of the material and is discarded in the actual study.
In the current work, three-dimensional finite element analysis was used to perform simulation of helical springs undergoing static deformation when subjected to the action of a thermomechanical
loading, for both SE and SME effects scenarios. Simulation was performed by using the commercial software ANSYS® Workbench 2020 R2. This was carried out through the ANSYS CAE interface [16]. Finite
strains were automatically integrated in the analysis.
The importance of numerical simulation in the case of SMA NiTi helical springs is that it enables to predict the system response, under arbitrary situations regarding either SE or SME effect. The
deformation undergone by the particular structure of a helical spring is completely three-dimensional and cannot be rendered satisfactorily through simplified analytical modeling. Given a spring
section, partial transformation occurs between martensite and austenite phases during deformation and one-dimensional based approaches fail to capture this phenomenon. The constitutive equations used
in simulation of helical springs made of Nitinol alloy are based on the SMA Auricchio model which is implemented in the ANSYS finite element code [17, 18]. Two major features make this modeling quite
useful and appropriate [19]. First of all, the number of constitutive parameters used during the analysis is reduced to a strict minimum; so they can be accurately identified experimentally.
Secondly, it is unnecessary to utilize a USER Material Subroutine (USERMAT), since the model is implemented by default in the 2020 R2 version of ANSYS software, which we have used in this work, and
can then be directly accessed through the Temperature Bulk (TB) command.
Explicit equations of the SMA Auricchio based material model, which was chosen here to describe the coupled thermomechanical behavior of NiTi, are recalled in Appendixes A and B.
2.2 Material properties of NiTi SMA alloy used in simulations
2.2.1 Material properties for SE effect
The SMA helical springs material properties used here correspond to the experimental data given by Huang et al. [20]. The numerical simulations under ANSYS that are associated to the SE effect were
performed by using material parameters that were calculated at the reference temperature $T = 25 ° C$. They are listed in Table 1.
Table 1.
Material properties of the NiTi SMA helical springs used for the simulation of SE effect
Parameter Value
$σ s$ $96 M P a$
$σ f$ $582 M P a$
Maximum residual shear stain $ε L$ $0.032 m . m − 1$
$C A$ $11 M P a . K − 1$
$C M$ $7.6 M P a . K − 1$
Austenite Young modulus $E A$ $72 G P a$
Martensite Young modulus $E M$ $64 G P a$
Overage Young's modulus $E = ( E A + E M ) / 2$ $68 G P a$
Poisson's ratio $ν$ $0.33$
Material parameter $α$ $0.15$
Reference temperature $T$ $25 ° C$
Temperature $M s$ $268 K$
Temperature $M f$ $268 K$
Temperature $A s$ $278 K$
Temperature $A f$ $288 K$
$σ A f$ $111.6 M P a$
$σ A s$ $221.6 M P a$
$σ M s$ $325.1 M P a$
$σ M f$ $811.1 M P a$
The values of temperature parameters given in
Table 1
were calculated by using the following linearized formulas, [
], which hold for ambient temperature that verifies
$T ≥ M s$
$σ A f = C A ( T − A f )$
$σ A s = C A ( T − A s )$
$σ M s = σ s + C M ( T − M s )$
$σ M f = σ f + C M ( T − M s )$
$σ M s$
$σ M f$
are respectively the critical stresses at the beginning and end of the martensitic transformation,
$σ A s$
$σ A f$
are respectively the critical stresses at the beginning and end of the austenitic transformation, and
$σ s$
$σ f$
are respectively the critical uniaxial start and finish stresses, while the SMA material transforms from twinned martensite to detwinned martensite. The material parameters related to the martensite
and austenite phases are denoted respectively
$C M$
$C A$
$M s$
is temperature at the beginning of the martensitic transformation.
$A s$
$A f$
are respectively the temperatures at the beginning and end of the austenitic transformation.
As ANSYS admits only a single Young's modulus, an average value elastic modulus $E$ was calculated by using the austenite and martensite Young's moduli [21]. The obtained average value $E = 68 G P a$
enables to get a closer agreement with the experimental curves.
On the other hand, to use Auricchio model one needs to enter the material parameter $α$ that operates the scaling of compression response from that of tension response. Since this parameter was not
evaluated experimentally in [20], it was treated as a parameter to be identified in order to get the closest match between simulated results and experimental curves. It was found that $α = 0.15$
achieves the best fit.
2.2.2 Material properties for SME effect
To perform simulation of the SME effect by means of ANSYS software, material parameters that are needed as inputs were those given in [22]. Taking the reference temperature to be $T = 23 ° C$, the
parameters needed for simulation of SME effect for NiTi SMA material are presented in Table 2.
Table 2.
Material properties of NiTi helical springs used for the analysis of SME effect
Parameter Value
Hardening parameter $h$ $500 M P a$
Reference temperature $23 ° C$
Elastic limit $R$ $120 M P a$
Temperature scaling parameter $β$ $8.3 M P a . ° C − 1$
Lode dependency parameter $m$ 0
Maximum transformation stain $ε L$ $0.07 m . m − 1$
Martensite Young modulus $E M$ $70 G P a$
Young's modulus $E$ $60 G P a$
Poisson's ratio $ν$ $0.33$
2.3 Geometry of the SMA helical springs used in simulations
2.3.1 Geometry of the helical spring used for SE effect
Huang et al. [20] have considered four different geometries of helical springs and studied their response during a tensile test experiment. In this work, focus is on the helical spring denoted SMA
(b) in that reference and for which the geometric parameters are given in Table 3.
Table 3.
Geometric parameters of the NiTi helical spring used in SE effect simulations
Coil radius $R 0 ( m m )$ Wire radius $r 0 ( m m )$ Initial length $L 0 ( m m )$ Number of coils $N$
$6.1$ 0.4 19 7
2.3.2 Geometry of the helical springs used for SME effect
Nine different geometries of helical springs have been considered in order to assess dependency of the load-displacement curve on the actual geometry of the helical spring. All the helical springs
have the same number of coils which was fixed at
$N = 7$
, and the same wire radius which was fixed at
$r 0 = 0.4 m m$
. A full factorial design of experiment table was then constructed by choosing three levels of coil radius and initial length.
Table 4
recalls all the possible combinations of geometric parameters that are used in this work. The initial pitch angle of a spring was computed according to the following equation [
$α 0 = tan − 1 ( L 0 2 π R 0 N )$
Table 4.
Geometric parameters of NiTi helical springs used in SME effect simulations
Specimen identification Coil radius $R 0 ( m m )$ Initial length $L 0 ( m m )$ Initial pitch angle $α 0 ( ° )$
SMA (1) 5.1 17 4.334
SMA (2) 5.1 19 4.842
SMA (3) 5.1 21 5.348
SMA (4) 5.4 17 4.094
SMA (5) 5.4 19 4.574
SMA (6) 5.4 21 5.053
SMA (7) 5.7 17 3.879
SMA (8) 5.7 19 4.334
SMA (9) 5.7 21 4.788
2.4 Boundary conditions and thermomechanical loads
2.4.1 Helical spring for SE effect
The helical spring is assumed to be anchored at its left extremity, while at its right extremity it is subjected to a prescribed loading-unloading cycle of displacement which is applied in the
spring's longitudinal direction with a given amplitude. Body helical spring temperature is assumed to be uniform. The considered mechanical displacement amplitude and body isothermal temperature were
fixed by choosing three levels for each factor. Table 5 gives the resulting 9 combinations constructed on these variables. Figure 3 gives the profile of applied displacement.
Table 5.
Combinations of thermomechanical loading prescribed to the helical spring
Combination number Displacement amplitude $U ( m )$ Isothermal temperature $T ( ° C )$
1 0.06 15
2 0.06 25
3 0.06 35
4 0.08 15
5 0.08 25
6 0.08 35
7 0.1 15
8 0.1 25
9 0.1 35
Fig. 3.
Applied displacement at the right extremity of helical spring for SE analysis; $U$ is the amplitude of displacement while $u ( t )$ is the instantaneous value of displacement at chronological time
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
2.4.2 Helical springs for one-way SME effect
Helical springs considered here are assumed to be clamped at their left extremity. Their right extremity is subjected to a prescribed displacement, while body temperature, which is assumed to be
uniform, is varied. The 3D Auricchio model of the Ni-Ti helical spring was used to simulate SME in ANSYS Workbench platform, through the static structural module according to the material properties
which are given in Table 2.
The SME simulation was carried out through applying a combination of thermal and mechanical loading. The simulation was done based on five steps that are illustrated in Fig. 4. At the first step, the
material body temperature was decreased from the reference temperature of $23 ° C$ to the lower temperature of $5 ° C$. This thermal loading is applied in order to get the reverse transformation of
the material from the austenite phase zone $( A f ≤ T )$, which corresponds to the actual reference temperature of $23 ° C$, toward the twinned martensite phase $( T ≤ M f )$ for the lower
temperature of $5 ° C$. To calculate the solution during this step, ANSYS software uses the 3D thermomechanical constitutive equations that are derived from Auricchio and Petrini model [8], as given
in Appendix B.
Fig. 4.
SME analysis; applied body isothermal temperature (a) and displacement at the right extremity of helical springs (b)
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
During the second step, the temperature of the spring was fixed at $5 ° C$ and a compression displacement of − $35 m m$ was applied. This was followed in the third step by unloading the helical
spring to reach zero displacement state, at the same body temperature of $5 ° C$. In the fourth step, the body temperature of the load free spring was increased to reach $100 ° C$, so the material
was transformed to the austenite phase to fully recover the deformation. Finally, at the fifth step the material body temperature was decreased to reach again $5 ° C$. This achieves the complete
one-way SME cycle.
In order to ensure convergence of the nonlinear procedure, each step consisted of a minimum of 400 sub-steps.
2.5 Finite element mesh used in simulations
The finite element analysis (FEA) was carried out in ANSYS® Workbench 2020 R2 by generating a three-dimensional solid model for each considered geometry of SMA helical springs. For both SE and SME
simulations, mesh generation was made by using 3D structural elements of type SOLID186 and by selecting the tetrahedrons method of meshing. Mesh convergence was assessed by reducing the size of
elements until asymptotic stability of solution is reached. The total number of elements used for meshing any of the considered helical spring's domains was then fixed at 1,693, and the associated
number of nodes is 4,831.
3 Results and discussion
3.1 Simulation of SE effect in SMA helical springs
3.1.1 Convergence of the ANSYS modeling
Simulations were conducted for the SMA helical spring made of NiTi having the material properties given in Table 1 and for which the geometry parameters are those given in Table 3. The considered
thermomechanical loading conditions, consisting of uniform body temperature fixed at $25 ° C$ and prescribed cycle of displacement with positive amplitude at the right extremity of helical spring,
were fixed according to lines 2, 5 and 8 of Table 5.
Figure 5 depicts the 3D geometry of SMA helical spring elaborated under ANSYS CAE interface.
Fig. 5.
Geometry of the SMA helical spring part considered for the analysis of SE effect
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
Figure 6 gives the converged mesh used in all simulations.
Fig. 6.
Mesh for the SMA helical spring considered for the analysis de SE effect
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
Figure 7 shows the boundary conditions that are prescribed to the SMA helical spring.
Fig. 7.
Boundary conditions applied to the NiTi SMA helical spring
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
To get an interesting behavior of the SMA helical spring in terms of SE effect, finite deformation is needed. This is why the amplitude of applied displacement was chosen to be large enough. For
instance, applying the maximum displacement of 100mm represents more than five times the initial length of helical springs. Figure 8 shows the SMA helical spring in initial and deformed
Fig. 8.
SMA helical spring, initial and deformed configurations under $100 m m$ maximum displacement
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
To assess convergence of the modeling under ANSYS software, Fig. 9 shows comparison between the obtained simulation results and experimental data taken from Huang et al. [20] and corresponding to
body temperature of $25 ° C$.
Fig. 9.
Comparison of numerical force-displacement curves and experimental data [20] for a cycle loading-unloading of the SMA helical spring and three imposed displacement amplitudes U
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
Figure 9 shows that the obtained results in terms of force-displacement curves show a close agreement between numerical perditions and experimental results, revealing that the proposed model captures
the essential behavior of SMA helical springs with regards to SE effect as observed during experiment. There are, however, some variations existing between the experimental and finite element
simulation results. They are probably due to the fact that the material property $α$ was not evaluated experimentally; in fact, it was identified here to yield the least discrepancies between
simulation and experimental data. Besides, the implemented version of Auricchio model in ANSYS works with an averaged Young's modulus for both martensite and austenite domains, while these modules
are in general different as pointed out by the experimental data given in [20] and recalled in Table 1.
3.1.2 Influence of temperature on the SMA helical spring response
Simulation of the SMA helical spring response in terms of force-displacement curve was also performed by varying body temperature and amplitude of displacement according to Table 5.
Figure 10 illustrates the effect of body temperature on the helical spring reaction.
Fig. 10.
Force-displacement curves of the SMA helical spring as function of imposed amplitude of a cycle of displacement and body temperature
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
To analyze in more detail the effect of temperature on helical spring behavior, the following quantities were calculated: maximum reaction force, average (secant) stiffness at loading and total
damping energy. Table 6 summarizes the obtained response characteristics of helical spring as function of the combination number.
Table 6.
Response characteristics of helical spring as function of the combination number
Combination number Maximum reaction force $F max ( N )$ Secant stiffness at loading $K m ( N . m − 1 )$ Total damping energy $E d ( J )$
1 4.50 75.0 0.243
2 5.31 88.5 0.288
3 5.98 99.7 0.320
4 5.03 62.9 0.265
5 5.91 73.9 0.438
6 6.75 84.4 0.512
7 5.54 55.4 0.503
8 6.45 64.5 0.603
9 7.33 73.3 0.718
For all cases, the initial tangent stiffness at loading was constant and does not depend on body temperature. Its value is $K i = 114 N . m − 1$. Table 6 shows that the secant stiffness increases
with body temperature, and that the maximum reaction force and total damping energy increase with both displacement amplitude and body temperature. At lower temperatures, the stiffness is lower and
the hysteresis area is larger. On the other hand, at higher temperatures, the average stiffness is increased while the hysteresis area decreases.
To get deeper comprehension of stiffness dependency on temperature, let us recall a classical analytical relationship which links the SMA helical spring response to its body temperature, according to
linear theory of helical springs. The average stiffness can be estimated as:
$K m = G r 0 2 16 N R 0 3$
is the material shear modulus.
It is well known that for a NiTi based SMA, $G$ increases with temperature dissimilar to other metallic materials.
To be able to quantify the influence of temperature on helical spring response, explicit response surface models based on nonlinear regression of the results given in
Table 6
were derived. The obtained models for the maximum reaction load, average stiffness and total damping energy have the following quadratic polynomials forms:
$F max = 1.324 + 37.73 U + 0.06967 T + 0.3875 U T − 112.5 U 2 − 0.00035 T 2$
$K m = 1111 − 1304 U + 2.007 T − 8.5 U T + 5833 U 2 − 0.00517 T 2$
$D m = 0.4398 − 12.54 U + 0.003267 T + 0.1725 U T + 102.1 U 2 − 0.0001617 T 2$
Regressions defined by equations (7), (8) and (9) were obtained with a high value of the coefficient of determination $R 2$, which reaches respectively $99.9 %$, $100 %$ and $97.9 %$. So they can
reliably be used in sizing helical springs, as long as body temperature and applied amplitude cycle of displacement remain in their respective considered ranges here: $T ∈ [ 15,35 ] ( ° C )$ and $U ∈
[ 0.06 , 0.1 ] ( m )$.
Analysis of variance performed on the results given in Table 6 has shown that for the maximum reaction force, temperature $T$ is dominant with $66 %$, the amplitude $U$ has a participation of $33 %$
while the interaction between $U$ and $T$ is negligible. For the average stiffness, almost equal influence of temperature and amplitude is observed with respectively $45 %$ and $54 %$ for each one of
them, while their interaction remains negligible. As for the total damping energy, the amplitude of displacement is dominant with up to $73 %$ and temperature represents $22 %$ of variability source,
while again only negligible interaction between these two factors is observed.
3.2 Simulation of one-way SME in SMA helical springs
In this section, results of one-way SME simulations are presented. These were performed for helical spring having the geometries defined in Table 4, and for which the material characteristics are
those given in Table 2. Nine different 3D geometries of SMA helical springs were elaborated under ANSYS CAE interface, before calculating through static structural analysis and the option SME the
response of each spring under the thermomechanical loading given in Fig. 4.
Table 7 shows the obtained results in terms of weight of springs, strain energy and maximum force.
Table 7.
ANSYS output response characteristics of simulated helical springs under one-way SME
Case number Weight $( m k g )$ Strain energy $( μ J )$ Maximum force $( N )$
$1$ 0.1911 0.1240 0.2732
$2$ 0.1898 0.1864 0.2653
$3$ 0.1887 0.2761 0.2677
$4$ 0.1949 0.2163 0.2554
$5$ 0.1935 0.1838 0.2564
$6$ 0.1924 0.1831 0.2572
$7$ 0.2061 0.1841 0.2259
$8$ 0.2047 0.1878 0.2373
$9$ 0.2035 0.1879 0.2379
Figure 11 gives the force-displacement curves of all the considered helical springs when they are submitted to the one-way thermomechanical loading as defined in Fig. 4.
Fig. 11.
One-way SME force-displacement curves for helical springs defined in Table 4 under the action of thermomechanical loading given in Fig. 4
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
Analysis of variation of the obtained results given in Table 7 indicates that for the considered ranges of factors, helical spring radius has a dominant influence on mass with $97.3 %$ and length has
only a small influence with $2.7 %$. As to the maximum force variability, it is mainly explained by the radius $93.6 %$, which is followed by the interaction of radius with length $5.5 %$ and finally
length which participates by only $0.5 %$.
From the results given in
Table 7
, one can derive quadratic polynomial regressions of mass and maximum force as function of helical spring radius and length under the following forms:
$Mass = 1.277 − 0.4195 R 0 − 0.001283 L 0 − 8.333 × 10 − 5 R 0 L 0 + 0.04130 R 0 2 + 2.917 × 10 − 5 L 0 2$
$F max = − 0.3616 + 0.4171 R 0 − 0.03757 L 0 + 0.007292 R 0 L 0 − 0.05685 R 0 2 − 2.917 × 10 − 5 L 0 2$
The associated coefficient of determination is respectively $R 2 = 100 %$ and $R 2 = 98 %$. This states that the previous polynomial regressions can be reliability used for sizing helical springs in
the considered ranges of factors: $R 0 ∈ [ 5.1 , 5.7 ] ( m m )$ and $L 0 ∈ [ 17,21 ] ( m m )$. In particular, they can be used in the framework of a multi-objective optimization problem aiming at
minimizing both Mass and $F max$, while maximizing the helical spring course.
4 Conclusions
In this work, focus was on establishing force-displacement curves of nickel-titanium shape memory alloy based helical spring, for both superelasticity effect and one-way shape memory effect, by using
the finite element method under ANSYS software packages. These curves were derived in a systematic way which can contribute to ease the procedure of design of helical springs. At first, comparison
between numerical predictions and experimental data has been performed to assess convergence of the finite element method. Then, a parametric study was conducted to analyze the influence of
temperature on the superelastic spring response. Explicit correlations were obtained. They corroborate in particular the classical shape memory alloys behavior where the increase of temperature
causes higher stiffness and low hysteresis. Considering the one-way shape memory effect which has received small interest in the field of numerical computation, simulations were conducted and enabled
to assess dependency of the response on key geometrical parameters. The obtained results have emphasized the alternative of using a simulation based methodology in order to achieve improved design of
helical springs made of shape memory alloys.
APPENDIX A Auricchio SMA material behavior under SE deformation
To characterize superelastic behavior of a SMA, Auricchio's model considers three-phase transformations: austenite to single-variant martensite (A → S), single-variant martensite to austenite (S →A),
and reorientation of the single-variant martensite (S →S). Assuming the material to be perfectly isotropic, only two phases are considered: the austenite (A) and the single-variant martensite (S).
Two internal variables are introduced: martensite volume fraction
$ξ S$
and austenite volume fraction
$ξ A$
. These two variables satisfy the following relation:
$ξ S + ξ A = 1$
So, only one independent internal variable holds; the martensite volume fraction $ξ S$ is chosen in the following as the independent internal variable.
In order to model the pressure dependency of the phase transformation, Drucker-Prager loading function is used. This takes the following form:
$F = q + 3 α p$
$q = 3 2 s ¯ ¯ : s ¯ ¯ and p = 1 3 t r ( σ ¯ ¯ )$
is a material parameter,
$s ¯ ¯ = σ ¯ ¯ − p 1 ¯ ¯$
is the deviatoric stress tensor and
$t r$
the trace operator.
The evolution of the martensite volume fraction is expressed as follows:
$ξ ˙ S = { − H A S ( 1 − ξ S ) F ˙ F − R f A S A → S transformation − H S A ξ S F ˙ F − R f S A S → A transformation$
parameters are scalar quantities and they are defined by the following relations:
$H A S = { 1 i f R s A S < F < R f A S and F ˙ > 0 0 otherwise$
$H S A = { 1 i f R f S A < F < R s S A and F ˙ < 0 0 otherwise$
${ R s A S = σ s A S ( 1 + α ) R f A S = σ f A S ( 1 + α ) R s A S = σ s S A ( 1 + α ) R f S A = σ f S A ( 1 + α )$
$σ s A S = σ M s$
$σ f A S = σ M f$
$σ s S A = σ A s$
$σ f S A = σ A f$
are the critical stresses at the start of martensite, finish of martensite, start of austenite, and finish of austenite transformation respectively. They are shown in
Fig. A1
Fig. A1.
Idealized one-dimensional stress-strain diagram for superelastic behavior of SMA according to Auricchio model
Citation: International Review of Applied Sciences and Engineering 13, 3; 10.1556/1848.2021.00389
The parameter
characterizes the tension-compression asymmetry. In the case for which tensile and compressive behavior are symmetrical,
$α = 0$
holds. Otherwise, if the tension and compression states are uniaxial, the parameter
can be linked with the initial value of austenite to martensite phase transformation in tension
$σ t A S$
and compression
$σ c A S$
respectively as [
$α = σ c A S − σ t A S σ c A S + σ t A S$
The constitutive relation for Auricchio's model is given by the following equations:
$τ ¯ ¯ ⌢ · = D ¯ ¯ ¯ ¯ : ( ε ¯ ¯ ⌢ · − ε ¯ ¯ t r ⌢ · )$
$ε ¯ ¯ t r ⌢ · = ξ ˙ S ε ¯ L ∂ F ∂ σ ¯ ¯$
$τ ¯ ¯$
$ε ¯ ¯$
are Kirchhoff stress and left Cauchy-Green strain respectively,
$D ¯ ¯ ¯ ¯$
is the tangent elastic stiffness tensor of SMA material [
$ε ¯ ¯ t r$
is the transformation strain tensor,
$ε ¯ L$
represents the material parameter associated with maximum recoverable strain, see
Fig. A1
, and
is the Drucker-Prager like loading function which is defined in Eq. (A2).
The material parameters for the superelastic SMA model consist then of the six following constants:
1. - $σ s A S$ the starting stress value for the forward austenite-martensite phase transformation;
2. - $σ f A S$ the final stress value for the forward austenite-martensite phase transformation;
3. - $σ s S A$ the starting stress value for the reverse martensite-austenite phase transformation;
4. - $σ f A S$ the final stress value for the reverse martensite-austenite phase transformation;
5. - $ε ¯ L$ the maximum transformation shear strain;
6. - $α$ the parameter measuring the difference between material responses in tension and compression.
APPENDIX B Auricchio SMA material behavior under SME deformation
For SME simulation, a three-dimensional thermomechanical model for stress-induced solid phase transformations was developed by Auricchio and Petrini [
]. This model was recognized to be efficient and able to reproduce all of the key characteristics of SMA in a 3D setting. In this modeling, the free energy is set as follows:
$ψ ( θ , e ¯ ¯ , T , ε ¯ ¯ t r ) = 1 2 ( ε ¯ ¯ − ε ¯ ¯ t r ) : D ¯ ¯ ¯ ¯ : ( ε ¯ ¯ − ε ¯ ¯ t r ) + τ M ( T ) ‖ e ¯ ¯ t r ‖ + h 2 ‖ e ¯ ¯ t r ‖ 2 + 1 ε ¯ ¯ t r ( e ¯ ¯ t r )$
$ε ¯ ¯$
$e ¯ ¯$
$ε ¯ ¯ t r$
$e ¯ ¯ t r$
are respectively total strain tensor, total deviatoric strain tensor, transformation strain tensor and deviatoric transformation strain tensor,
is the volumetric strain,
$D ¯ ¯ ¯ ¯$
is the elastic stiffness of SMA material;
the ambient temperature,
$τ M ( T ) = 〈 β ( T − T 0 ) 〉 +$
is a positive and monotonically increasing function of the temperature in which
$〈 ⋅ 〉 +$
is the positive part of the argument,
a material parameter and
$T 0$
the temperature below which no twinned martensite is observed;
$‖ · ‖$
is the Euclidean norm;
is a material parameter related to the hardening of the material during the phase transformation. Finally,
$1 ε ¯ ¯ t r ( e ¯ ¯ t r )$
is an indicator function introduced to satisfy the constraint on the transformation strain norm and it is given by the following equation:
$1 ε ¯ ¯ t r ( e ¯ ¯ t r ) = { 0 i f 0 ≤ ‖ e ¯ ¯ t r ‖ ≤ ε ¯ L + ∞ otherwise$
Accordingly, one can write:
$σ ¯ ¯ = ∂ ψ ∂ ε ¯ ¯ = D ¯ ¯ ¯ : ( ε ¯ ¯ − ε ¯ ¯ t r )$
$X ¯ ¯ t r = − ∂ ψ ∂ e ¯ ¯ t r$
$X ¯ ¯ t r$
represents the transformation stress tensor.
Splitting the strain tensor
$ε ¯ ¯$
and the stress tensor
$σ ¯ ¯$
, one obtains:
$ε ¯ ¯ = e ¯ ¯ + 1 3 θ 1 ¯ ¯$
$σ ¯ ¯ = s ¯ ¯ + p 1 ¯ ¯$
$s ¯ ¯$
is the deviatoric stress,
is the main or hydrostatic stress and
$1 ¯ ¯$
is the unit tensor.
Equation (B4) can be rewritten as:
$X ¯ ¯ t r = s ¯ ¯ − 1 ‖ e ¯ ¯ t r ‖ [ τ M ( T ) + h ‖ e ¯ ¯ t r ‖ + γ ] e ¯ ¯ t r$
is defined as:
$γ = { 0 i f 0 ≤ ‖ e ¯ ¯ t r ‖ ≤ ε ¯ L ≥ 0 ‖ e ¯ ¯ t r ‖ = ε ¯ L$
$ε ¯ L$
represents the maximum transformation strain.
To catch the asymmetric behavior observed on SMA materials during tension–compression test, Auricchio and Petrini [
] have proposed a Prager-Lode type limit function which depends on the second
$J 2$
, and the third
$J 3$
stress deviator invariants of the transformation under the following form:
$F ( X ¯ ¯ t r ) = 2 J 2 + m J 3 J 2 − R$
is a material parameter describing Lode dependency and
is the elastic limit.
$J 2$
$J 3$
are defined in the following way:
$J 2 = 1 2 X ¯ ¯ t r 2 : 1 ¯ ¯$
$J 3 = 1 2 X ¯ ¯ t r 3 : 1 ¯ ¯$
The evolutionary equations of strain transformation are defined as:
${ ε ¯ ¯ t r ⌢ · = ξ ˙ ε ¯ L ∂ F ∂ σ ¯ ¯ ξ ˙ ≥ 0 ξ ˙ F ( X ¯ ¯ t r ) = 0$
is an internal variable, called the transformation strain multiplier. The second and third relations of Equation (B12) are the Kuhn–Tucker conditions, which reduces the problem to a constrained
optimization problem.
The SME effect option is like this described by six constants: $h$, $T 0$, $R$, $β$, $ε ¯ L$, $m$ in addition to the martensite modulus $E m$ and Poisson's ratio $ν$. These constants define the
stress-strain behavior of material in loading and unloading cycles for the uniaxial stress-state and thermal loading. | {"url":"https://akjournals.com/view/journals/1848/13/3/article-p309.xml","timestamp":"2024-11-03T19:53:57Z","content_type":"text/html","content_length":"815572","record_id":"<urn:uuid:836fbf3a-6832-4f0c-9d80-dc9b35d72951>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00596.warc.gz"} |
The Exercise Attitudes, Perceptions, and Perceived Outcomes of Older Minority Women Participating in a Fall Prevention Program
Georgia State University
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Mathematics Theses Department of Mathematics and Statistics
Noetherian Filtrations and Finite Intersection
Sara Malec
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Recommended Citation
Noetherian Filtrations and Finite Intersection Algebras
Sara Malec
Under the Direction of Dr. Florian Enescu
This paper presents the theory of Noetherian filtrations, an important concept in
com-mutative algebra. The paper describes many aspects of the theory of these objects,
presenting basic results, examples and applications. In the study of Noetherian
filtra-tions, a few other important concepts are introduced such as Rees algebras, essential
powers filtrations, and filtrations on modules. Basic results on these are presented as
well. This thesis discusses at length how Noetherian filtrations relate to important
constructions in commutative algebra, such as graded rings and modules, dimension
theory and associated primes. In addition, the paper presents an original proof of
the finiteness of the intersection algebra of principal ideals in a UFD. It concludes by
discussing possible applications of this result to other areas of commutative algebra.
INDEX WORDS: Graded Rings and Modules, Noetherian Filtrations, Rees
Sara Malec
A Thesis Submitted in Partial Fulfillment of the Requirements for the Degree of
Master of Science
in the College of Arts and Sciences
Georgia State University
Copyright by Sara Louise Childs Malec
Sara Malec
Committee Chair: Dr. Florian Enescu
Committee: Dr. Frank Hall
Dr. Yongwei Yao
Electronic Version Approved:
I am deeply grateful to all of the people who helped me create this thesis. Without
the seemingly limitless dedication and patience of my advisor, Dr. Florian Enescu,
this paper would never have happened. I would also like to thank Drs. Frank Hall
and Yongwei Yao, the other members of my committee, for their careful eye in the
editing process. Thanks are due as well to the other students in our commutative
algebra seminar, M. Brandon Meredith, Muslim Baig, and Jong Wook Kim, for their
instruction and encouragement.
Of course I also thank my parents for their unquestioning support. Finally, I’d
Chapter 1
Introduction: Graded Rings and
Noetherian filtrations are a class of mathematical objects which have certain nice
properties. In this paper we will develop the theory of filtrations, and prove the
Noetherianity of a certain class of filtrations.
All rings are assumed to be commutative with identity. We will begin the first
chapter by introducing graded rings. Then we will review several notions from
intro-ductory commutative algebra, beginning with defining Noetherian rings and modules
and presenting some related results. The rest of the first chapter contains additional
definitions and results concerning graded Noetherian rings.
Once these fundamentals have been established, Chapter 2 defines the objects in
which this thesis is primarily concerned: filtrations, Rees algebras, and associated
graded rings. We again include a number of examples. Further, we compute the
dimension of Rees algebras of an ideal in a Noetherian ring.
Chapter 3 summarizes many important results concerning a special class of
we define essential powers filtrations and discuss their relationship to Noetherian
fil-trations. Examples are given. The chapter concludes with a number of important
equivalent conditions that characterize Noetherian filtrations.
Finally, Chapter 4 presents the concept of finite intersection algebra of two ideals.
We present an original proof of the finiteness of intersection algebra of two primary
ideals in a UFD. The chapter concludes with some other related results.
We will start by giving a review of some basic facts from commutative algebra
that will be needed later in this paper. In this thesis, all rings are assumed to be
commutative with identity.
Definition 1.1. A semigroup G is a set together with a binary operation + which is closed under addition, associative, and has an identity. A semigroup is called
cancellative if for any a, b, c∈G, and a+b =a+c, thenb =c.
Definition 1.2. A graded ring over a cancellative semigroup G is a ring R that can be written as a direct sum of abelian groups R = L
i∈GRi with the additional
constraint that RiRj ⊂ Ri+j. An element r ∈ R is called homogeneous if there is
some i such that r∈Ri. Then i is called thedegree of r. A homogeneous ideal is an
ideal generated by homogeneous elements.
It should be noted that while a ring can be graded over any cancellative
semi-group, generally in this paper they are graded over [N] or [Z]. Also, in this thesis, we
will assume that [N] contains 0.
Definition 1.3. IfI is an ideal ofR, then the graded idealI∗ is defined to be the ideal generated by all of the homogeneous elements inI. An ideal is called homogeneous if
Example 1.4. The simplest example of a graded ring is the polynomial ringR=k[x]. ThenR =R0⊕R1⊕ · · · ⊕Rn⊕ · · ·, where Ri is the collection of the terms of degree
i. This is clearly a direct sum decomposition, since Ri∩
j6=iRj = ∅ for all i, and
RiRj ⊂Ri+j since xixj =xi+j.
Example 1.5. For another example, takeR=k[x, y] and use the [N]-grading induced by the total order, i.e. for any monomial xiyj ∈ R, the degree of that monomial is
i+j. Thus an example of a homogeneous ideal would be (x4y2, x3+y3, xy3+x2y2),
where the first term is in R6, the second in R3, and the third in R4.
Example 1.6. The same ring can have a different grading and produce different homogeneous elements. If we instead use the multidegree order, R=k[x, y] is graded
over [N]×[N]. The degree of any element xi[y]j [is (i, j), and therefore (x]3[y]4[, xy, x) is a]
homogeneous ideal with the first element of degree (3,4), the second of degree (1,1)
and the third of degree (1,0).
Proposition 1.7. Let R be a G-graded ring, where G is a cancellative semigroup. Then 1 is in R0.
Proof. Since 1 ∈ R, 1 = P
i∈Gxi with xi ∈ Ri. We claim x0 = 1, and thus 1 ∈
R0. Let y be homogeneous in R. Then y = y·1 = P
i∈Gyxi. We equate degrees
on both sides. Note that all of the terms of the sum are of distinct degrees, for
if deg(yxi) = deg(yxj), then deg(y)+ deg(xi) = deg(y)+ deg(xj), and since G is
cancellative, deg(xi) = deg(xj). So, as deg(x0) = 0, then y·x0 =y.
Example 1.8. (Yongwei Yao) An interesting example arises if Gis not cancellative. Let G={0, b}, where b 6= 0 is such that b+b =b. Note that this is a semigroup, as
it is closed under associative addition. LetR = 0⊕[Z]with the natural multiplication
degreeb.We claim this fits the requirements for a graded ring: R0·Rb ⊆Rb, since for
any z ∈ [Z],0·z = 0 ∈[Z] =Rb, and Rb ·Rb ⊆ Rb as Z is closed under addition. But
here, 1 is clearly in Rb and not in R0.
Proposition 1.9. Let I be an ideal of a G-graded ring R. Let I∗ be the ideal generated by the homogeneous elements of I, and I∗∗ be the ideal generated by the homogeneous components of I. Then the
following are equivalent:
1. If f ∈I and f =f1+f2+· · ·+fn with fi ∈Rgi and gi =6 gj , then fi ∈I;
2. I =I∗∗;
3. I is generated by homogeneous elements;
4. I =I∗;
Proof. 1⇒2 : First, note thatI∗ ⊆I ⊆I∗∗ always. So let f ∈I∗∗ be a generator of I∗∗. So f is a homogeneous component of an element of I by definition of I∗∗, and so by hypothesis f ∈I. Thus all of
the generators ofI∗∗ are in I, and thus I∗∗ ⊆I, therefore they are equal.
2⇒3 : SinceI∗∗is generated by homogeneous elements andI =I∗∗,Iis generated by homogeneous elements.
3 ⇒ 4 : We know already that I∗ ⊆ I, so now let f ∈ I be a in the set of homogeneous generators forI. By hypothesis, f is homogeneous, and thusf ∈I∗ by the definition of I∗. Since the generators ofI
are in I∗, I ⊂I∗ and thus I =I∗.
4⇒1 : Let f ∈I, with f =f1 +f2+· · ·+fn, and fi be homogeneous of degree
gi. Nowf ∈I∗, sof = P
rjhj where rj ∈R and thehj are homogeneous elements of
lj. Now each of the ri is a sum of homogeneous elements, so multiply out the terms
of f and identify the degrees. Then fk = P
r0[l]hl where the r0l are homogeneous, and
Before we can proceed, we need a general discussion of Noetherianity of rings and
modules and a few other items from commutative algebra. The following summary is
presented without proof, and a thorough treatment can be found in an introductory
text such as [4], [9] or [5].
Definition 1.10. A ringR is said to beNoetherian if it satisfies the ascending chain condition (A.C.C.) on ideals, i.e. for any increasing chainI1 ⊆I2 ⊆I3 ⊆ · · · of ideals
of R there exists an integer k such that In =Ik for all n≥ k. A left R-module M is
Noetherian if it satisfies the A.C.C. on submodules.
Definition 1.11. In a dual way, we can define anArtinian ringRas one that satisfies the descending chain condition, or D.C.C. That is for any decreasing chain of ideals
I1 ⊇ I2 ⊇ I3 ⊇ · · · there exists an integer k such that In =Ik for all n ≥ k. A left
R-module M is Artinian if it satisfies the D.C.C. on submodules.
Proposition 1.12. The following are equivalent: 1. R is a Noetherian ring (module);
2. Every ideal (submodule) of R (M) is finitely generated;
3. Every nonempty family of ideals (submodules) of R(M) has a maximal element
(under inclusion);
Proposition 1.13. Any homomorphic image of a Noetherian ring is Noetherian. In particular, if R is Noetherian with I an ideal of R, thenR/I is Noetherian.
Definition 1.15. Let R be a ring. The supremum of the lengths of chains of prime ideals of R is called the dimension of R, denoted dimR.
Definition 1.16. LetP be a prime ideal of a ringR. Then the height ofP, denoted ht(P), is the supremum of lengths of chains of prime idealsP0 ⊂ · · · ⊂Pn=P.
Definition 1.17. Let L/K be a field extension. The transcendence degree of the extension is the largest cardinality of an algebraically independent subset of L over
Definition 1.18. Let R be a ring. If R has a unique maximal idealm, then we say that R is a local ring, denoted (R,m).
Definition 1.19. Let R be a ring and S a subset of R with identity that is closed under multiplication. Then thelocalization ofR atS, denotedS−1[R][or][R]
S, is defined
to be {r
s|r ∈ R, s∈S}, with the additional requirement that r/s =r
0[/s]0 [if and only]
if there exists someu∈S such that u(s0r−sr0) = 0.
Definition 1.20. Let R and S be as above and let M be an R-module. Then the
localization of M at S, denoted S−1[M, is defined to be] [M] [⊗]
In the above two definitions, if S is the complement of a prime idealP inR, then
the localization of the ring or module at S is called RP or MP respectively. In this
case, S is automatically a multiplicative set.
Proposition 1.21. Let (R,m) be a Noetherian ring. Then dim(R) is finite.
Proposition 1.22. LetRbe Noetherian. ThenRis Artinian if and only if dim(R)=0. Further, if R is local with maximal ideal m, then there exists an n ∈ N such that
Definition 1.23. Let R be a ring. The collection of prime ideals of R is called the
spectrum ofRand denotedSpec(R). The collection ofminimal primes ofRis denoted
Definition 1.24. Let M be an R-module with P ∈ Spec(R). We say that P is
an associated prime if P is the annihilator of an element of M. The collection of
associated primes is denoted Ass(M).
Definition 1.25. The support of a moduleM, denotedSupp(M), is the set of prime ideals P ∈Spec(R) such that MP 6= 0.
Definition 1.26. Let I be an ideal in R. Then the radical of an ideal, denoted Rad(I) or √I, is defined to be Rad(I) ={r ∈ R|rn [∈] [I] [for some] [n][∈]
N}. Note that
for any I, I ⊂ Rad(I).
Definition 1.27. LetR be a ring andP a prime ideal. Then the nth symbolic power of P, denoted P(n), isPnRP ∩R.
The following three results are presented by Bruns and Herzog in [2] on pages
29-30. We will follow their treatment closely.
Theorem 1.28. Let R be an [N]-graded R0-algebra, and x1, . . . , xn homogeneous
ele-ments of positive degree. Then the following are equivalent:
1. x1, . . . , xn generate the ideal m= L∞
2. x1, . . . , xn generate R as an R0-algebra.
In particular, R is Noetherian if and only if R0 is Noetherian and R is a finitely
Proof. 2⇒1 : By hypothesis, for anyr∈R, there existsf(T1, . . . , Tn)∈R0[T1, . . . , Tn]
such that r =f(x1, . . . , xn). Let r ∈ m be a homogeneous element. Then we claim
that r = f(x1, . . . , xn) = P
1 · · ·xinn) ⊆ (x1, . . . , xn). Since r is
homo-geneous, so f is homogeneous of the same degree. So we can match up the degrees.
Since r ∈ m, degr ≥ 1, and so each term of f has an xi in it for some i. Hence
r0[i]·xi ∈(x1, . . . , xn). Clearly, (x1, . . . , xn)⊆m, so m= (x1, . . . , xn).
1⇒ 2 : Lety ∈R be homogeneous of degree d. We do induction on d. We want
to show thaty=y1x1+· · ·+ynxn with yi ∈R0. If degy= 0, we are done, asy ∈R0
Now assume that the homogeneous elements of R of degree less than d are
gen-erated as an R0-algebra by x1, . . . , xn. By hypothesis, we know y ∈ L
i≥1Ri =
m = (x1, . . . , xn). So y = y1x1 +· · ·+ynxn, with yi ∈ Ri. So y is homogeneous,
and the xi are homogeneous, but the yi may not be. Multiply out and combine
like terms. Then we have y = y0[1]x1 +· · · +yn0xn, where the y0i are homogeneous
of degree deg(y)−deg(xi), which is less than d. So by induction, there exists an
fi ∈ R0[T1, . . . , Tn], with y0i = fi(x1, . . . , xn). Now, non-homogeneous elements are
sums of homogeneous elements, so the statement follows.
For the last statement, if R is Noetherian, then R0 ∼=R/ L
i≥1Ri = R/m, which
implies thatR0 is Noetherian. Also, if Ris Noetherian,mis finitely generated by say
(x1, . . . , xn), and by this Theorem,R isR0 finitely generated by (x1, . . . , xn). For the
other direction, ifR0 is Noetherian, then sinceR=R0[r1, . . . , rm] =R0[T1, . . . , Th]/I,
which implies that R is Noetherian.
2. R is a Noetherian ring;
3. R0 is Noetherian, and R is a finitely generated R0-algebra;
4. R0 is Noetherian, and both S1 =
i=0Ri and S2 = L∞
i=0R−i are finitely
gen-erated R0-algebras.
Proof. The above theorems make 4⇒3⇒2⇒1 clear: assuming 4 shows that R is
a finitely generated R0-algebra, since it is a sum of S1 and S2. The previous theorem
makes 2 clear, which clearly implies 1 since every ideal of R is finitely generated.
1⇒4: Note thatR0 is a direct summand ofR as andR0-module. SoIR∩R0 =I
for any ideal I of R0. We claim that R0 is Noetherian.
Take an ascending chain of ideals I0 ⊆I1 ⊆ · · · ⊆In⊆In+1 ⊆ · · · inR0. Extend
these ideals toR. SoRI0 ⊆RI1 ⊆ · · · ⊆RIn⊆RIn+1 ⊆ · · · is a chain of ideals in R.
SinceRis Noetherian, this chain stabilizes at say thenth position. Now contract this
chain back to R0 to get RI0∩R0 ⊆RI1∩R0 ⊆ · · · ⊆RIn∩R0 =RIn+1∩R0 =· · ·.
This chain obviously stabilizes, and since IR∩R0 =I, this chain is the same as the
one we started with. A similar argument for chains of submodules shows thatRi is a
finite R0-module for every i∈Z.
Now let m=L∞
i=1Ri. We claim mis a finitely generated ideal ofS1. By
hypoth-esis, mR has a finite system of generators x1, . . . , xm, and assume each generator xi
is homogeneous of degree di. Let d = max{d1, . . . , dm}. Then y ∈ m with degy ≥ d
can be written as a linear combination of x1, . . . , xm with coefficients in S1. Thus
x1, . . . , xm together with the homogeneous generators spanningR1, . . . , Rd−1 over R0
generate m as an ideal of S1. By (1.28), S1 is a finitely generated R0-algebra and S2
follows by symmetry.
1. For every prime ideal P, the ideal P∗ [is a prime ideal.]
2. Let M be a graded R-module
(a) If P ∈Supp(M), then P∗ [∈][Supp][(M][)][.]
(b) If P ∈ Ass(M), then P is graded; furthermore P is the annihilator of a
homogeneous element.
Proof. 1. Let a, b ∈ R with ab ∈ P∗. We can write a = P
iai, with ai ∈ Ri,
jbj with bj ∈Rj. We do a proof by contradiction.
Assume a /∈ P∗ and b /∈ P∗. Then there exists a p, q ∈ Z such that ap ∈/ P∗
but ai ∈P∗ for i < p and bq ∈/ P∗, but bj ∈P∗ for j < q. Then the (p+q)th
homogeneous component of ab∈P∗ is P
i+j=p+qaibj. This sum is in P∗, since
P∗is graded. All summands of this sum are also inP∗sinceP∗is a homogeneous ideal, soapbq ∈P∗. Since P∗ ⊂P and P is prime, then ap ∈P or bp ∈P. But
ap and bq are homogeneous, so either ap or bq ∈P∗
2. (a) Assume P∗ ∈/ Supp(M). So MP∗ = 0. Let x ∈ M homogeneous. Then
there exists an a ∈ R \ P∗ such that ax = 0. Since x/1 ∈ MP∗ = 0,
there exists an a /∈ P∗ with ax = 0. It follows that aix = 0 for any ai a
homogeneous component ofa. Sincea∈R\P∗, there exists anisuch that ai ∈/ P∗. Since ai is homogeneous, ai ∈/ P. Thus x/1 = 0 in MP, which is
a contradiction.
(b) Letx∈M withP = Ann(x). Letx=xm+· · ·+xnwithxi homogeneous,
and a = ap + · · · +aq ∈ P. Since ax = 0, P
i+j=raixj = 0 for r =
m+ p, . . . , n+q. Thus apxm = 0, ap+1xm+1 = 0, etc. We claim that
Examine thep+m+ 1th [degree terms:] [a]
p·xm+1+ap+1·xm. We know this
must be 0. Thus ap(ap·xm+1 +ap+1xm = a2p·xm+1 +ap+1(ap·xm) = 0,
thus a2
p·xm+1 = 0. By induction, aipxm+i−1 = 0 for all x ≥1. So anp−m+1
annihilatesx. Since P is prime, ap ∈P, so each homogeneous component
of a is in P, and thus P is graded.
For the second part, we need to show that AnnR(x) = AnnR(xm). Now
a ∈ P = P∗, so ax = 0, and axi = 0 for all i, so P ⊆Ann(xi). Now
Ann(x) =T
i=mAnn(xi). And T
i=mAnn(xj)⊆P. SinceP is prime, there
Chapter 2
Filtrations and Rees Algebras
The fundamental objects that we will use in this paper are filtrations of ideals and the
Rees algebras generated by them. This chapter begins with some examples, and then
develops the basic ideas behind the dimension of Rees algebras of a power filtration.
Basic facts on filtrations such as [2], [5], pages 147-150, [16], pages 93-95 and [9],
pages 93-94, and a thorough treatment is given in a remarkable book by Rees. More
specific aspects, not touched on upon here, can be found in [8].
Definition 2.1. Let R be a ring. We define a filtration of ideals of R to be a chain of ideals {In}n starting with I0 = R, In ⊆ In−1 for all n ≥ 1, with the additional
requirement that the ideals satisfy In·Im ⊆In+m. LetE be an R-module. Then we
define a filtration onE, denotede={En}n, to be a descending chain ofR-submodules
En of E such that E0 =E.
Example 2.2. LetI be an ideal ofR. A typical example of a filtration is the power filtration {In[}]
n. Then I0 =R, I1 =I, I2 =I2 and so on. Clearly this satisfies both
properties of a filtration, since In [⊃][I]n+1 [and] [I]n[I]m [⊂][I]n+m[.]
using the Ii from above, that J0 =I0 =R, J1 =I4, J2 =I5, and so on. Then{Jn}is
a filtration as well, for the same reasons as above.
Example 2.4. LetP be a prime ideal in a ringR. Then for all n and m,Pn[·][P]m [⊆]
Pn+m[, so][P]n[R]
P·PmRP ⊆Pn+mRP. Hence (PnRP∩R)·(PmRP∩R)⊆Pn+mRP∩R.
In other words, P(n)[·][P](m) [⊆][P](n+m)[, and so][{][P](n)[}]
n forms a filtration.
Example 2.5. Another example is very close to a power filtration. Let I1 = (xayb),
I2 = (x2ayb),In= (xnayb) in R =k[x, y]. Again, this clearly satisfies both properties
of a filtration.
Example 2.6. A less obvious one is In = (xd
n e[) in] [k[x] with] [k] [a field. While the]
inclusion is still trivial, the other requirement requires proof.
We need to show that d√m+n e ≤ d√m e+d√n e. Obviously d√m+n e ≤
d√m+√n e. Then,√m+n ≤√m+√n≤ d√m e+d√n e. Also,d√m e+d√n e ∈
N. So by the definition of the ceiling,d
m+n e ≤ d√m e+d√n e. So the second
property of a filtration is fulfilled, and {In} is a filtration of ideals on k[x].
Definition 2.7. Letf ={In}be a filtration of ideals of a ringR. Then we can define
the graded ring associated to the filtration as
grf(R) = M
For x ∈ In and y ∈ Im, multiplication is defined to be (x + In+1)(y +Im+1) =
xy+In+m+1. If the filtration is understood to be the power filtration, we can write
the associated graded ring of an ideal I as grI(R).
If A is a ring, with I ≤A an ideal, and f = {In} is the power filtration defined
this, notice that any element in In[/I]n+1 [can be written as a linear combination of]
products of n elements of I/I2[.]
Now, using filtrations, we introduce the notion of a Rees Algebra of a filtration.
Definition 2.8. Letf ={In} be a filtration of a ring R. Then the Rees Algebra of
f is
R={F =
n X
Fktk|Fk∈Ik} ⊆R[t].
By the properties of the filtration, this is a subring of R[t]. To check this, let
F =a0+a1t+· · ·+antn and G=b0+b1t+· · ·+bmtm be in R. Then because Ii is
an ideal, ai+bi stays in Ii, so F +G is clearly still inR. Also, aiti·bjtj =aibjti+j,
and since I is a filtration, aibj ∈Ii+j.
Definition 2.9. Let u = t−1 [and] [f] [=] [{][I]
n} be a filtration. We define the extended
Rees Algebra, R0 [=][· · · ⊕][Ru]2 [⊕][Ru][⊕][R][⊕][I]
1t⊕I2t2⊕ · · · ⊆R[t, t−1].
Proposition 2.10. Let R be a ring with a filtration f and R0 [the extended Rees]
algebra as defined above. Then gf(R)∼=R0/uR0.
Proof. Let r ∈ R0[, with] [r] [=] P
n[, where] [r]
n ∈ In if n ≥ 0 and rn ∈ R when
n < 0. Construct a homomorphism ϕ : R0 [→] [g]
f(R), where ϕ(r) = P
nr¯n, where
rn ∈ In/In+1 for all n ≥ 0 and rn ∈ R for n < 0. This is clearly a surjective
homomorphism. Then, if ϕ(r0) = 0, then r0 = P nr
n+1tn, where rn0+1 ∈ In+1, which
is the same asuR0[. Thus,] [g]
Now we can compute the dimension of the Rees algebras of a power filtration.
This result is shown as Theorem 5.1.4 in [16].
Theorem 2.11. Let R be a Noetherian ring, and let I be a proper ideal of R. Then dim R is finite if and only if the dimension of either the Rees algebra or the extended
dim R[It] =
dim R+ 1, if I [*]P for some prime ideal P with dim(R/P) = dim R;
dim R otherwise.
2. dim R[It, t−1] = dim R+ 1
3. Ifmis the only maximal ideal inR, and ifI ⊆m, thenmR[It, t−1]+ItR[It, t−1]+
t−1R[It, t−1] is a maximal ideal in R[It, t−1] of height dim R+ 1.
4. dim(grI(R)) = dimR.
Proof. First, let J be an ideal of R. Then,
J ⊆J R[It]∩R ⊆J R[It, t−1]∩R⊆J R[t, t−1]∩R =J (2.1)
so the above inclusions are all equalities. So, any ideal in R is a contraction of an
ideal inR[It] and R[It, t−1[]. In addition,]
J ⊆
J R[t, t−1[]][∩][R[It]] ⊆
R[It, t−1[]]
J R[t, t−1[]][∩][R[It, t]−1[]] ⊆
R[t, t−1[]]
J R[t, t−1[]]. (2.2)
We claim that the two middle rings are isomorphic to the Rees algebra and the
extended Rees algebra, respectively, of the image of I in R/J. To see this, let ¯I =
J ⊆
J, and ¯R =R/J. Then let r ∈R[It] = r0 +r1t+· · ·, where ri ∈ I
i[. Define]
a homomorphism ϕ : R[It] → R[ ¯¯ It] by ϕ(r) = ¯r0+ ¯r1t+· · ·, where ¯ri ∈ I. Then¯
Ker(ϕ) = {r ∈ R[It]|ϕ(r) = 0} = {r ∈ R[It]|ri ∈ J for all i}, which is the same as
saying thatr ∈R[It] andr ∈J R[t]⊆J R[t, t−1[], which proves the isomorphism. The]
second is done in a similar way.
must be minimal in R[It], and P R[t, t−1[]][∩][R[It, t]−1[] must be minimal in] [R[It, t]−1[].]
Call P R[t, t−1[]][∩][R[It] = ˜][P][. To show that ˜][P] [is minimal, we need that] [R[It]]
˜ P is
Artinian, which is equivalent to having ˜P R[It][P]˜ nilpotent. So we must show that every
element in ˜P R[It][P]˜ is nilpotent. We know that ˜P ∩R =P, and thatRP is Artinian,
so P RP is nilpotent, and therefore P RP[t] is nilpotent. Let S =R\P ⊆ R[It]\P˜.
ButS−1P˜ ⊆PPRP[t], so it is nilpotent as well. Then ˜P R[It][P]˜ is a further localization
of the nilpotent ideal S−1P˜, so it is nilpotent as well.
Then, any nilpotent element ofR[It] orR[It, t−1] is certainly nilpotent inR[t, t−1], so it has to lie in the intersection of the primes ofR[t, t−1[] =]T
P∈Min(R)P R[t, t
−1[]. So]
all the minimal prime ideals of the Rees algebras are contractions of minimal primes
of R[t, t−1[] and are of the form] [P R[t, t]−1[]. So,]
dimR[It] = maxQ∈MinR[It](dim R[It]
Q ) = maxP∈MinR(dim
R[It] P R[t,t−1[]][∩][R][[][It][]])
= maxP∈MinR(dim ¯R[ ¯It]) = maxP∈MinR(dimR[P][I+[P]Pt])
Thus dimR[It] = max{dim (R[P][I+[P]Pt])|P ∈ MinR}, and similarly dimR[It, t−1] = max{dim (R[P][I+[P]Pt, t−1[])][|][P] [∈] [Min][R][}][. So, to calculate dim] [R[It], it is enough to]
show that for an integral domain R, dim R[It] = dim R if I is the zero ideal and is
dim R+ 1 otherwise. Thus we can assume thatR is a domain.
Proposition 2.12. (Dimension Inequality) Let R be a Noetherian integral domain, withS a ring extension ofR which is also a domain. LetQbe a prime ideal in S and
P =Q∩R. Then
htQ+ tr. deg[κ][(][P][)]κ(Q)≤htP + tr.deg[R]S. (2.3)
and S =R[It], that for every prime ideal Q inR[It],
htQ+ tr.deg[κ][(][Q][∩][R][)]κ(Q)≤ht(Q∩R) + tr.deg[R]R[It]. (2.4)
Clearly, tr.degRR[It] = 1, since the larger ring is simplyRwith one variable adjoined
to it. Therefore, no matter what tr.deg[κ][(][Q][∩][R][)]κ(Q) is, ht Q ≤ ht(Q∩ R) + 1 ≤
dim R + 1. So, the height of any prime in R[It] is at most one larger than the
height of any prime in R, which proves that dim R[It] ≤ dim R+ 1. Clearly dim
R[It] = dimR if I is the zero ideal, since R[(0)t] =R. So assume thatI is non-zero.
Let P0 = ItR[It]. Then P0 ∩R = (0), It ⊆ P0, htP0 > 0 (since (0) ( P0), and
R[It]/P0 ∼= R which is an integral domain, proving that P0 is prime. Since P0 is
another prime added to any chain of primes that can be made in R,
dim R[It]≥ dim R+ 1.
This proves (1).
Similarly for (2), it is enough to show that when R is a domain,
dimR[It, t−1] = dimR+ 1.
Again by the dimension inequality, dimR[It, t−1]≤ dim R+1, and the other inequal-ity follows from dim R[It, t−1[]][≥] [dim] [R[It, t]−1[]]
t−1 = dimR[t, t−1] = dim R+ 1.
Lastly, let P0 ( P1 ( · · · ( Ph = m be a saturated chain of prime ideals in R,
with h = ht m. Set Qi =PiR[t, t−1]∩R[It, t−1]. As Qi∩R =Pi, Q0 ⊆Q1 ⊆ · · · ⊆
Qh is a chain of distinct prime ideals in R. The biggest one is Qh = mR[t, t−1]∩
R[It, t−1[] =] [m][R[It, t]−1 [+][ItR[It, t]−1[], which is properly contained in the maximal]
Chapter 3
Noetherian Filtrations
Filtrations of ideals represent an important concept in commutative algebra. They
have a rich and long history and have been studied by many authors in various
contexts. Noetherian filtrations are central among filtrations of ideals and their theory
has been developed by authors such as W. Bishop, Okon, Petro, Rattliff, Rees, and
Rush among others, see [1], [10], [11], [12], [13], and [14],.
In this chapter, we define and give examples of Noetherian filtrations, and show
that they are an interesting class of filtrations with remarkable properties. Noetherian
filtrations have finiteness conditions that are similar to power filtrations. This chapter
will explain what those conditions are. Also, we will introduce and study the notion
of an e.p.f. filtration. Our presentation follows closely [1], [12] and [13].
Proposition 3.1. LetRbe a ring and I an ideal inR. Then ifR is the Rees algebra generated by the power filtration of I, R is finitely generated over R whenever I is
finitely generated. In this case, ifI = (a1, . . . , an), thenRis generated bya1t, . . . , ant.
Proof. Let I = (a1, . . . , ah). Then Ik is generated by products of k elements chosen
from I. So any element of R looks likeF =i0+i1t+i2t2+· · ·+ihth, with ij ∈Ij.
1 · · ·a jk
k , wherej1+· · ·+jh =j. Buta j1
1 · · ·a jk k t
j1+···+jk [= (a]
1t)j1· · ·(aktjk). So
every monomial inRcan be written as a sum of powers ofait, whereai is a generator
of I. So R=R[a1t, . . . , ant].
Example 3.2. LetI = (x, y)⊂R[x, y], and construct a Rees algebra with the power filtration of I. So any element of R looks like f =Pn
k=0aktk, withak∈Ik. But any
element aktk can be written as products of powers of xt and yt, so any f ∈ R can be
written as polynomial in xt and yt, so R =R[It].
Example 3.3. Now return to the example In= (xd
ne[)] [⊆][k[x]. We claim][R] [is not]
finitely generated. Assume the contrary. Then R is generated by some elements
α1 e[t]α1[, x]d
α2 e[t]α2[, . . . , x]d
αn e[t]αn[}][.]
We can write
m e
as a polynomial overR in the above generators for allm.
So we need to find a1, a2, . . . , an such that:
a1α1+a2α2+· · ·+anαn = m (3.1)
a1d √
α1 e+a2d √
α2 e+· · ·+and √
αn e = d
α1+α2+· · ·+αn e (3.2)
Assume we have theai,i= 1, . . . , n, such that equation (1) holds. Then, substituting
(1) into (2) gives:
a1d √
α1 e+a2d √
α2 e+· · ·+and √
αn e=d
We proved above that d√a+b e ≤ d√a e+d√b e, so:
a1d √
α1 e+a2d √
α2 e+· · ·+and √
αn e ≤ d √
a1α1 e+d √
a2α2 e+· · ·+d √
anαn e
≤ d√a1 ed √
α1 e+d
a2 ed √
α2 e+· · ·+d √
an ed √
αn e
Therefore, ai ≤ d √
ai e for all i = 1, . . . , n. But, since x ≥ d √
x e for any x, then
ai =d
ai e for all i. Thus, all of the inequalities above are in fact equality, and so:
d√a1α1+a2α2+· · ·+anαn e=d √
a1α1 e+d √
a2α2 e+· · ·+d √
anαn e.
Since x =d√x e only if x = 0,1 or 2, this implies that for all i, ai can be no larger
than 2. Thus the largest αi is certainly no larger than 2. So m=P[i]aiαi ≤2P[i]αi.
So m is bounded. But this is clearly impossible, so this Rees algebra is not finitely
Definition 3.4. Let R be a ring with a filtration f = {In}. Recall that grf(R) = L
n≥0 In
In+1 is the graded ring associated to the filtration. We say that f isNoetherian
if grf(R) is Noetherian.
Theorem 3.5. Let R be a Noetherian ring withf ={In[}] [the power filtration. Then]
f is Noetherian.
Proof. Examine the following isomorphism:
IR =
IR[It] 'grf(R) =
M I[n]
Thus, ifRis Noetherian, then [IR]R is as well, and thereforegrf(R) is Noetherian, which
Theorem 3.6. (P. Roberts [15]) Let R be the polynomial ring [C][x, y, z] localized at (x, y, z). Then there exists a prime ideal P in R such that L
(n) [is not]
We’ve shown a few nice properties of power filtrations. But we can generalize the
power filtration to a larger class of filtrations that behave nicely.
Definition 3.7. We say that a filtrationf ={In}of ideals of a ringRis anessentially
powers filtration (or e.p.f.) if there exists anm >0 such thatIn = Pm
i=1In−iIi for all
n≥1. If n−i <0, In−1 is assumed to be R.
Letf ={In}be a filtration on a ringR. Then we can prove a number of statements
about e.p.f.’s.
Proposition 3.8. Letf ={In} be a filtration on a ring R. Then the following are
1. f is an e.p.f.;
2. In=P(QkIjei), wherem is as given in the definition of e.p.f.’s, and the sum is
over all ei >0 such that e1+ 2e2+· · ·+kek =n;
3. There exists an m ∈ [N] with the property that f is the least filtration on R
whose first m+ 1 terms are R, I1, I2, . . . , Im;
Proof. First, what does least mean here? And how do we know a smallest filtration
exists? For the first question, for two filtrations f ={In} and g ={Jn}, we say that
f ≤ g if Ii ⊆ Ji for all i. And we know that the smallest filtration with the given
property exists, because we can simply take the intersection of all filtrations whose
first m+ 1 terms areR, I1, I2, . . . , Im.
Let In = Pk[i][=1]In−iIi for all n ≥ 1, and let In0 = P
j ). Then a ei i ∈ I
n can
be written as ai· · ·ai (ei times), which is inside Iiei. Since Iiei = Pk
j=1Iiei−jIj, any
monomial in I[n]0 can be written as a product of just two terms whose degrees sum to
n, and thus it would be in In.
In ⊆In0 by induction
(2 ⇐⇒ 3) Letg ={Jn}be any filtration onRsuch thatIi =Jifor alli= 0, . . . , k.
So by the definition of a filtration,
k Y i=1
Iei i ) =
k Y
Jei i )⊆Jn
LetKn =P(Qk[i][=1]Iiei) for all n ≥k, and Kn =In for all n < k. Then let h={Kn},
which is clearly a filtration. Since h is less than g, it is less than any filtration
that agrees with f at first on R, so it’s the smallest. So h ≤ f, but f = h, by
(1 ⇐⇒ 2)
Now assume further that R is Noetherian. Then there are a number of additional
results that we can show. This is presented as Theorem (2.7) in [12].
Theorem 3.9. Let R be a Noetherian ring with f ={In} any filtration of R. Then
the following are equivalent:
1. The extended Rees algebra R0 [of] [f][,] [· · · ⊕][Rt]−2[⊕][Rt]−1[⊕][R][⊕][I]
1t⊕I2t2⊕ · · ·, is
2. R is Noetherian;
3. R is finitely generated over R;
Proof. Notice that, from the previous sections, 1 through 3 are equivalent, since Ris
graded and Ris Noetherian. So we need only show that 4 is equivalent to the others.
(4 ⇒ 3) Since f is an e.p.f., we know that R = R[tI1, . . . , tkIk], since all terms
can be gotten from the first k terms of the filtration. (2 ⇒ 4) Let f1, . . . , fm be a
basis of N. Since N is homogeneous, we can assume the fi’s are too, since if they
are not, we can take the homogeneous components and add them to the list. So let
fi = aitei with ei > 0. Let k = max{ei|i = 1, . . . , m}, so N = (tI1, t2I2, . . . , tkIk).
Letn > k and a∈In, so x=atn∈ N. But every element of N looks like Pgifi for
some gi ∈ R, hencex=Pgifi. Assume gi =bitn−ei, andgi is homogeneous. So:
x=Xgifi =
bitn−eiaitei = X
⇒a =Xaibi ∈ n X
IeiIn−ei ⊆ k X
Thus, since a ∈ In, In = Pk[j][=1]IjIn−j for n > k, which is the definition of an
Definition 3.10. Let e = {En} be a filtration on an R-module E and f = {In} a
filtration on R. Then e is said to be compatible with f in case ImEn ⊆ Em+n for all
m and n.
Definition 3.11. Letebe as above. Theneis said to bef-good in caseeis compatible with f and there exists a positive integer m such thatEn=
i=1In−iEi for all large
n. In other words, f isf-good if and only iff is an e.p.f.
The following Proposition as well as the associated corollary are shown as (3.5)
Proposition 3.12. LetR be a Noetherian ring with an e.p.f. f ={In}and let E be
a finitely generatedR-module with anf-filtration e={En}. Theneisf-good if and
only if there exists a k≥0 such that Ek+i =IkEi for all i≥k.
Proof. Assume e is f-good. Then by definition, e is compatible with f and there
exists an m such that En = Pm
i=1In−iEi for all large n, say n > n0. Then we claim
E=PEiti is finitely generated over S =R[tI1, t2I2, . . .]. Let xn∈ En, with n > n0.
Then xn ∈ Pm[i][=0]In−iEi, so xntn ∈ Pm[i][=1]In−itn−iEiti, which is in SEiti. So if x ∈E,
then x=Pm i=0xnt
n[, which is in] Pm
i=1S(Eit i[)]
We showed before that f is an e.p.f. if and only if S =R[tI1, t2I2, . . .] is finitely
generated overR. So, there exists a g >0 such that S =R[tI1, . . . , tgIg] sincef is an
Let j be the lcm of 2,3, . . . , g. Then let mi be the positive integer such that
imi = j for all i = 1, . . . , g. Then (tiIi)mi ⊆ tjIj ⊆ A = R[tjIj]. Thus any element
of the formtix withx∈Ii is integral over A. SinceS is finitely generated over Aby
integral elements, S is integral and finitely generated over A =R[tj[I]
j]. Therefore E
is finite A-module.
Let Θ1, . . . ,Θmbe a homogeneous system of generators forEoverA, with deg(Θi) =
di and d = max di for i= 1, . . . , m. Let n > max {d, j} and letx be an element of
En. So we can write x = P
ihiΘi where hi are homogeneous elements of A. These
hi are either 0 or of degree of degree n−di. By resubscripting if necessary, assume
hi 6= 0 for i = 1, . . . , m0 ≤ m. Then n−di ≥ 1, and since all of the elements of A
have degree a multiple of j, thus for all i = 1, . . . , m0 there exists a positive integer ki such that jki =n−di. Thus:
m0 X
hiΘi ⊆ m0 X
j Edi ⊆Ij( mi X
And since Iki−1
j Edi ⊆ Ij(ki−1)Edi ⊆ Ej(ki−1)+di = En−j, we have that En ⊆ IjEn−j.
The opposite inclusion is obvious sincee is compatible withf, soEn =IjEn−j for all
n >max{d, j}.
Last, let k = jd and i ≥ k. Then by the above equation, Ek+i = Ejd+i =
IjEj(d−1)+i. Nowj(d−1) +i≥ max (d, j) + 1, so we can continue to pull outIj until
we are left with I[j]dEi, which is in IkEi. Thus, Ek ⊆IkEi.
For the converse, let there be a positive k such that Ek+i = IkEi for all i ≥ k.
Then we claim [E]=P
Eiti is generated as a module over S by E1t, . . . E2k−1t, since
the smallest iwhich is not covered in the hypothesis is i=k−1. Then according to
(2.3) in [12], if[E]is finitely generated overS, theneisf. So it remains to show that[E]
is finitely generated over S. Let [G]i be the collection of generators for Ei, i <2k−1.
This collection is finite, since by hypothesis, each Ei is finitely generated over R.
So for every e ∈ Ei, e = P
f initerjxj, where rj ∈ R and xj ∈ Ej. So to find the
generators of [E], we need only to collect all the generators from each[G]i and attach to
them the appropriate power of t, i.e. the generators of [E] over S are all of the terms
eti[, where][e][∈]
Corollary 3.13. Letf ={In} be a filtration on a Noetherian ring R. Then f is an
e.p.f. if and only if there exists a k > 0 such that Ik+i =IiIk for all i≥k.
Proof. Iff is an e.p.f. thenf is e-good, so let E =R and e=f in (3.12) to see that
there exists ak such that Ik+i =IiIk. If such ak exists, then ifn ≥2k, we can easily
In =In−kIk ⊆ 2k X
In−1Ii ⊆In,
so then f is an e.p.f. withm = 2k. If n <2k, Then In = P2k
i=1In−1Ii. Clearly, In is
is an e.p.f.
We can now show another important equivalence, but first we need a few more
Proposition 3.14. Let R be a ring with a filtration f = {In}n≥0 and let E be
an R-module with an f-filtration e = {En}n≥0 and let E be an R-module with
an f-filtration e = {En}n≥0 such that En is a finitely generated R-module for all
N ≥1. Then G+[(E, e) =] P∞
n=1En/En+1 is a finitely generated grf(R)-submodule of
G(E, e) =P∞
n=0En/En+1 if and only if there exists a positive integerg such that, for
allj ≥g,Ej+1 =IjE1+· · · ,+Ij−g+1Eg+Ej+2.
Proof. Assume thatG+[(E, e) is a finitely generated gr]
f(R)-submodule ofG(Ee).
Con-struct the following submodules: letAij =IjE1+Ij−1E2+· · ·+Ij−i+1Ei+Ej+2 and
let ¯Ai = P∞
j=0Aij/Ej+2. Then ¯Aiis a grf(R)-submodule ofG+(E, e). Also ¯Ai ⊆A¯i+1
and S∞
i=1A¯i = G+(E, e). Therefore the hypothesis implies that there exists a
pos-itive integer g such that ¯Ag = ¯Ag+t for all t ≥ 0 so it follows that Agj/A(g+t)j = 0
for all j ≥ 0 and t ≥ 0. In particular, if j ≥ g and t ≥ 1, then Ij−g−t+1Eg+t ⊆
IjE1+Ij−1E2+· · ·+Ij−g+1Eg+Ej+2 for allj ≥g, and since the opposite inclusion is
obvious when t=j−g+ 1, we obtainEj+1 =IjE1+Ij−1E2+· · ·+Ij−g+1Eg+Ej+2
for all j ≥g.
Now let g be as given in the hypothesis. Then, for every j ≥ g, Ej+1/Ej+2 =
(IjE1+· · ·+Ij−g+1Eg+Ej+2)/Ej+2 = (Ij/Ij+1)(E1/E2)+· · ·+(Ij−g+1/Ij−g+2)(Eg/Eg+1).
It follows thatG+(E, e) is generated as a gr[f](R)-submodule ofG(E, e) byEn/En+1for
n = 1, . . . , g. Therefore, since each En is finitely generated, it follows that G+(E, e)
is a finitely generated gr[f](R)-submodule of G(E, e).
Corollary 3.15. Let R be a Noetherian ring with a filtration f = {In}n≥0 and let
is a finitely generated gr[f](R)-module and for each positive integer n, there exists a
positive integer ρ(n) such that Eρ(n) ⊆(Rad(I1))nE1, thene is f-good.
Proof. Since G(E, e) is a finitely generated gr[f](R)-module, let g be as given in the
previous proposition. So, by considering consecutive values ofj, for allj ≥g, Ej+1 =
IjE1+· · ·+Ij−g+1Eg +Ej+2. Since En+1 ⊆En, it follows that for all j ≥g,
Ej+1 =IjE1+· · ·+Ij−g+1Eg +Et (3.3)
for all t ≥j + 2 by induction on t..
Assume that the ρ(n) described above exists. Since R is Noetherian, every ideal
of R contains a power of its radical, so there exists a positive integer m such that
(Rad(Ij))m) ⊆ Ij. Also, (Rad(I1))m = (Rad(Ij))m, since Ij ⊆ I1 and I1j ⊆ Ij. So
(Rad(I1))m ⊆ Ij. By assumption, or each positive integer n, there exists a positive
integer ρ(n) such that Eρ(n) ⊆ (Rad(I1))nE1. Therefore, Eρ(m) ⊆ (Rad(I1))mE1 ⊆
IjE1. Let t = ρ(m) in equation (3.3). Then Ej+1 = IjE1 +· · ·+Ij−g+1Eg for all
j ≥g. So for any m, n, we have ImEn⊆Em+n, so e is f-good.
Corollary 3.16. If f = {In}n≥0 is a filtration on a Noetherian ring R, then the
following are equivalent:
1. f is an e.p.f. ;
2. f is a Noetherian filtration and there exists a positive integer g such thatIgn ⊆
(Rad(I1))n for all large n;
3. f is a Noetherian filtration and for each positive integernthere exists a positive
integer ρ(n) such that Iρ(n)⊆(Rad(I1))
Proof. First, notice that since f is an e.p.f., then R is Noetherian. Since we showed
in Chapter 2 thatR/uR ∼= gr[f](R), and quotients of Noetherian rings are Noetherian,
then f is always a Noetherian filtration.
(1⇒2) By the previous corollary we know that there exists ak such that Ik+i =
IkIi for all i ≥ k. Therefore, Ign = Ign−1Ig for all n ≥ 1. So Ign = Ign−1Ig ⊆
(I1)n−1I1 ⊆(Rad(I1))n for all n ≥1.
(2⇒3) Clear.
(3 ⇒ 1) By 3.15, if E is a finitely generated R-module with an f-filtration e =
{En}, and if G(E, e) is a finitely generated grf(R)-module, and there exists a ρ(n)
such that Eρ(n) ⊆(Rad(I1))nE1, then e isf-good. Thus, if E =R and e=f, by (2)
Chapter 4
Finite Intersection Algebras
Now that we have established some properties of Noetherian filtrations, we can look at
one example in depth that illustrates both the concepts of graded rings and Noetherian
Definition 4.1. ([7], pages 126-127) Given a pair (I, J) of ideals of a ringR, call the
algebra B= L
r[∩][J]s[)u]r[v]s [the] [intersection algebra] [of] [I] [and] [J][. If this algebra is]
finitely generated over R, we say that I and J have finite intersection algebra.
Let R be a Noetherian ring, and let I and J be ideals of R. Then denoteBr,s =
(Ir[∩][J]s[)u]r[v]s[. Note that, because (I]r0[∩][J]s0[)][·][(I]r00[∩][J]s00[)][⊆][I]r0+r00[∩][J]s+s00 [we have that]
Br0[,s]0 · B[r]00[,s]00 ⊆ B[r]0[+][r]00[,s]0[+][s]00.
Denote Bn = L
r+s=nBr,s. With this notation, B = L
n≥0Bn, which is N-graded,
because Bn0 · Bn00 ⊆ Bn0[+][n]00. Then by (1.28), B is Noetherian if and only if B is a
finitely generated R-algebra, asB0 =R is Noetherian.
The purpose of this chapter is to prove the following theorem:
Proof. By the above definition,I andJhaving finite intersection algebra is equivalent
toB being finitely generated over R. With the notations from above, it is enough to
show the following claim.
Claim: There exists an N > 0 such that for every x ∈ Br,s there exists y∈ Br0[,s]0
and z ∈ Br00[,s]00, wherex=yz, r00+r0 =r, s00+s0 =s and 0< r0+s0 ≤N.
First, we will show that the Claim implies that B = R[Br,s|r+s ≤ N]. In our
case, Br,s are R-free submodules of B of rank 1. So B = R[Br,s|r+s ≤ N] implies
thatB is a finitely generated R-algebra, hence the Theorem. To show that the Claim
implies B = R[Br,s|r +s ≤ N], note first that R[Br,s|r + s ≤ N] ⊆ B, because
Br0[,s]0 · B[r]00[,s]00 ⊆ B[r]0[+][r]00[,s]0[+][s]00. Denote A =R[B[r,s]|r+s ≤N]. For B ⊆ A, it is enough
to show that for every r, s,Br,s ⊆A. We’ll prove this by induction on r+s.
Let x ∈ Br,s. By the Claim, there exists y ∈ Br0[,s]0 and z ∈ B[r]00[,s]00, where x =yz,
r00+r0 =r, s00+s0 =s and 0< r0+s0 ≤N. Hence, x=yz, sincer00+s00 < r+s by the induction assumption, it follows that z ∈A. In conclusion, x=yz ⊆A.
We will concentrate now on proving the Claim.
Let a, b ∈ R such that I = (a) and J = (b). R is a UFD, so a and b can be
uniquely decomposed into a product of prime elements. Thus, there existsp1, . . . , pn
primes in R, α1, . . . , αn, β1, . . . , βn ∈ N, not all zero, such that a = pα11· · ·pαnn and
1 · · ·pβnn.
To illustrate our method of proving the Claim, we will treat first the cases n= 1
and n = 2, and then move on to the general case.
First, let us examine the case where I and J are generated by one prime element.
So I = (pα[) and] [J] [= (p]β[), where where] [p] [is some prime and] [α, β] [∈]
N. Then
Ir[∩][J]s [= (p]α[)]r[∩][(p]β[)]s[= (p]αr[)][∩][(p]βs[) = (p]max(αr,βs)[).]
Examine a generic term from the algebra with its indexing dummy variables:
r0, s0 with r0 ≤r,s0 ≤s and r0 +s0 ≤N such that
vs0 =p
This equation then simplifies to
max(αr, βs)−max(αr0, βs0) = max(α(r−r0), β(s−s0)).
Letr0, s0 be such that αr0 =βs0 = [α, β]. Then, the Claim is satisfied forr0 =r0 and s0 =s0 as long as r > r0 and s > s0.
For the two prime case, let I = (pα1[q]α2[), and] [J] [= (p]β1[q]β2[). We want to find] [N]
such that for every (r, s), there exists an r0, s0 with r0 ≤ r, s0 ≤ s and r0 +s0 ≤ N such that
max((r−r0)αi,(s−s0)βi) +max(r0αi, s0βi) = max(rαi, sβi)
for i= 1,2. With an additional lemma, we can simplify these equations a bit more.
Lemma 4.3. For any a, b, c, d ∈ [N], max (a−b, c−d) + max (b, d) = max(a, c) ⇔
Proof. First we show the forward implication. Let b > d. If c−d > a−b, then we
have c−d+b≤max(a, c). Our condition implies that c−d+b > a, so since b > d,
c > a. But then the original equation can never hold. So if b > d, thena−b > c−d.
A similar calculation shows the same results for b < d.
For the converse, assume that ((a−b)−(c−d))(b−d)≥0. Then, eithera−b≥c−d
andb ≥dor vice versa. Assume that this is the case. Thena−b≥c−d⇒a≥c−d+b,
So we can rewrite these equations as
((r−r0)αi−(s−s0)βi)(r0αi−s0βi)≥0 for all i= 1,2. (4.1)
In this case, we will find two separate sets of (r0, s0) that will handle most of the
r and s.
Let r[1]0 and s0[1] such that r0[1]α1 = s01β1 = [α1, β1] and find r20 and s02 such that
2α2 =s02β2 = [α2, β2]. We will show the Claim for (r, s) as long as r≥r0i ands ≥s0i
up to a possible finite list of pairs.
Look at (4.1) with r0 =r0 1 and s
0 [=][s]0 1:
((r−r0[1])α1−(s−s01)β1)(r10α1−s01β1)≥0 (4.2)
((r−r0[1])α2−(s−s01)β2)(r10α2−s01β2)≥0 (4.3)
Note that the first equation will always hold, since r0[1]α1 = s01β1. So we look at the
second one. If it holds as well, then this r0, s0 will work. If not, then ((r−r0 1)α2 −
1)β2)(r01α2−s01β2)<0. Then repeat this process withr0 =r20 and s
0 [=][s]0 2. This
time, the second equation is automatically satisfied. If the first is as well, than this
r0 and s0 will work, and if not, ((r−r0
Since r0
iαi =s0iαi, we can rearrange these two resulting equations to give
β1 −α2
)<0 (4.4)
β2 −α1
)<0 (4.5)
Order the αi
βi, renumbering if necessary, so α1
β1 ≤
the one withr0
1 and s01. Since this equation is strictly less than 0, ( α2
β2 −
β1) can’t be
0, so it must be greater than 0. Thus ((r−r0
1)α2 −(s−s01)β2) < 0, which implies
that r < [α]1
1)β2) +r10.
Now repeat this with the other equation. Notice that now, (α2
β2 −
β1) must be less
than 0. So nowr > [α]1
2)β1) +r02.
Combining these two ranges, we get
1 α1
((s−s0[2])β1+r02 <
1 α2
((s−s0[1])β2+r10 (4.6)
⇒s(β1 α1
− β2
) < r0[1]−r[2]0− β2
s0[1]+ β1 α1
s0[2]. (4.7)
The term on the left is positive by assumption. So the equations only fail when
s < r
1 −r02− β2
0 1+
0 2 β1
α1 −
=s0[1]+s0[2] and (4.8)
r < β2 α2
(s0[1]+s0[2]−s0[1]) +r0[1] =r0[1]+r0[2]. (4.9)
This shows the Claim as long as r≥r0
i and s≥s0i, i= 1,2.
Now to the n prime case. Let a = pα1
1 p α2
2 · · ·pαnn and b = p β1
1 p β2
2 · · ·pβnn, where
pi are prime in R and αi, βi are in N. Then I = (a) and J = (b), and thus Ir ∩
Js = (pα1
1 p α2
2 · · ·pαnn)r ∩(p β1
1 p β2
2 · · ·pβnn)s = (p rα1
1 p rα2
2 · · ·prαn n) ∩(p sβ1
1 p sβ2
2 · · ·psβnn) =
1 · · ·p
n ).
Let x ∈ Br,s. Then x = c· p
1 · · ·p
n urvs, where c ∈ R. We
r0 +s0 ≤N such that
1 · · ·p
n urvs
1 · · ·p
n ur
1 · · ·pmax((r
n u
If so, then by letting
1 · · ·p
n u
z =pmax((r−r0)α1,(s−s0)β1)
1 · · ·p
n u
the Claim is proven.
What is left to be proven simplifies to
max(αir, βis)−max(αir0, βis0) = max(αi(r−r0), βi(s−s0)) for all i= 1, . . . , n
which, by Lemma (4.3), simplifies to the following:
((r−r0)αi−(s−s0)βi)(r0αi−s0βi)≥0 for all i= 1, . . . , n. (4.10)
We will produceN >0 such that for allr, sthere existsr0, s0 with 0< r0+s0 ≤N and r ≥ r0, s ≥ s0 such that (4.10) is satisfied. For clarity, we will label the ith
equation of (4.10) by Ei.
Let r0
i and s0i be such that ri0αi =s0iβi = [αi, βi], and call r0i +s0i =Ni. We will
show the Claim for all pairs (r, s) such that r ≥ r0[i], s ≥ s0[i] for all i= 0, . . . , n. The
For r0 i = r
0 [and] [s]0 i = s
0[, the equation] [E]
i is automatically satisfied. If, by some
chance, all equations are satisfied for this choice of r0 and s0, then we are done, by simply letting N =r0
i +s0i.
If, however, one equation of 4.10 is not satisfied, then there exists some ji such
that ((r−r0
i)αji −(s−s 0 i)βji)(r
iαji−s 0
iβji)<0. Further, since r 0
iαi =s0iβi, we can
simplify the system once more to:
((r−r0[i])αji −(s−s 0 i)βji)(
βji − αi
)<0. (4.11)
We will examine now this possibility:
For all i, there exists some ji such that (4.11) happens.
Order the αi
βi, renumbering if necessary, so that α1
β1 ≤
β2 ≤ · · · ≤
βn. We can
assume that all βi 6= 0 in the system (4.10) because the equation Ei becomes r ≥r0
whenever βi = 0, which is a constraint that we have to satisfy anyway.
Consider r0
1, s01. Hence, for some j1, we have
((r−r[1]0)αj1 −(s−s
0 1βj1))(
)<0. (4.12)
Due to our renumbering, we know that (αj[1] βj[1] −
β1) must be ≥ 0, and it cannot
equal zero because of (4.12). Therefore
((r−r0[1])αj1 −(s−s
1)βj1)<0 or r <
1 αj1
((s−s0[1])βj1 +r
1. (4.13)
Consider now r0 j1, s
j1. There exists j2 such that
((r−r[j]0[1])αj2 −(s−s
0 j1)βj2)(
− αj1
If (αj[2] βj2
−αj[1] βj1
)>0, then we get another upper bound on r:
r < 1 αj2
((s−s0[j][1])βj2 +r
0 j1.
If (αj2
βj2 −
βj1)<0, then
r > 1 αj2
1)βj2 +r
0 1.
If (αj[2] βj2
−αj[1] βj1
)>0, we continue on with j2 and j3, until there is a k with α[jk][+1] β[jk][+1] <
α[jk] β[jk].
There will always be such a k, since our list of αi
βi is finite. Look at the equation
relating these two terms, as well as the one previous to it, i.e. the one relating jk−1
k)βjk+1)<0 (4.15)
− αjk−1
)((r−r[k]0−1)αjk −(s−s 0
k−1)βjk)<0 (4.16)
Since (αjk+1
β[jk][+1] − α[jk]
β[jk])<0, ((r−r 0
k)αjk+1 −(s−s
k)βjk+1)>0, and so
r < 1
k)βjk+1) +r
0 jk.
Similarly, since (α[β]jk
jk − α[jk][−][1]
β[jk][−][1])<0, ((r−r 0
k−1)αjk−(s−s 0
k−1)βjk)>0, and so
r > 1 αjk
k−1)βjk) +r 0 jk−1.
Putting the two together as above, we get
− βjk
)< s0[j]
k−1 −r
Since (βjk+1
α[jk][+1] − β[jk]
α[jk]) is always positive, we always have an upper bound ons, which
induces an upper bound on r as follows:
s < s0[j]
k−s 0 jk−1
αjk−1(αjkβjk+1 −αjk+1βjk)
r < r[j]0
k −r 0 jk−1
βjk+1(αjkβjk−1 −αjk−1βjk)
βjk−1(αjkβjk+1 −αjk+1βjk)
. (4.19)
Call F the set consisting of pairs r, s satisfying all possible equations of the form
(4.18) and (4.19). Now let N0 [= max][{][r][+][s][|][(r, s)] [∈ F }][. If for all] [i,] [r] [≥] [r]0
i, s ≥ s0i,
then our Claim follows for N0 = max{N0[, N]0
1, . . . , Nn0}. This is so because either
there exists one pair (r0
i, s0i) that works as (r
0[, s]0[) or (r, s)] [∈ F][. It remains to deal]
with the case when there exists k such that for all i,r < max(r0[i]), s > max(s0[i]) (the
case r > max(r[i]0), s < max(s0[i]) is similar).
Let r0 = 0, s0 = 1 in (4.10). Then there exists an i such that
(rαi−(s−1)βi)(−βi)<0 or r >
s− βi
But r < max(r0[i]) implies that s < αi
βi max(r 0
i) + 1. Let G0 be the set of all such pairs
(r, s). G0 [is finite. Similarly the case where for all][i s <] [max(s]0
i),r > max(r0i) gives a
finite setG00[of possible pairs (r, s). Let][G] [=][G]0[∪ G]00[, and][N]00[= max][{][r][+s][|][(r, s)][∈ G}][.]
If (r, s) are such that there exists a k with r < r0
k and s > sk0 or r > rk0 and s < sk0,
then either (0,1) or (1,0) work as choices for (r0, s0) or (r, s)∈ G. In conclusion, we can let N = max{N0, N}, and the Claim follows.
Proof. Since R is a PID, I and J are principal ideals, and R is a UFD. So, by the
above theorem, I and J have finite intersection algebra.
Now let f = {In} = L
m≥nBm. This is a filtration on B, first because clearly
In+1 ⊆In. For the other part of the definition, letx∈Ik and y∈Il. Then xy⊆Ik+l
because Br0[,s]0 · B[r]00[,s]00 ⊆ B[r]0[+][r]00[,s]0[+][s]00.
Then compute grf(B).
gr[f](B) =M
We proved above thatBis Noetherian. Thus, grf(B) is Noetherian, and by definition
f is Noetherian as well. Further, it can easily be shown thatf is an e.p.f., since our
Claim (see the proof of the above theorem) shows that there exists an N > 0 such
that In=PN[i][=1]In−iIi for every n >1.
If I or J are not principal, I and J do not necessarily have finite intersection
algebra. This was shown by Fields in [7] as follows.
Example 4.5. Let P and R as in Theorem (3.6) such that the algebra R⊕P(1)[⊕]
P(2) [⊕ · · ·] [is not finitely generated. Fields has shown that there exists an] [f] [∈] [R]
such that (Pa [:][f]a[) =][P](a) [for all] [a. Then Fields shows in Lemma 5.6 in [7] that the]
algebra L
(n) [is a homomorphic image of the intersection algebra between (f)]
and P. SinceL n≥0P
(n) [is not Noetherian, (f) and] [P] [do not have finite intersection]
Although this shows that in general the intersection algebra of I, J in R is not
finite, we note that there are other known classes of ideals I and J that have finite
intersection algebra. Fields has shown in [6] that if R = k[[x1, . . . , xn]] and I and J
is a consequence of results from the theory of integer linear programming. Fields’
methods can be used to provide a proof of our theorem, 4.2. We have given a different
and original proof that also provides information on the degrees of the generators for
the finite intersection algebra. Fields’ thesis explains how intersection algebras can
be applied to the asymptotic theory of ideals. The following result illustrates more
applications of intersection algebras.
Definition 4.6. LetRbe a Noetherian ring, and I, J ideals inR withJ ⊆√I. Also assume that I is not nilpotent and T
k [= (0). Then for each positive integer] [m,]
define vI(J, m) to be the largest n such that Jm ⊆ In. Also, we can examine the
sequence {vI(J, m)}m, which here we will abbreviate tov(m).
The following Proposition appears as (3.2) in [3].
Proposition 4.7. Let I, J be ideals in a Noetherian local ringR such thatJ ⊆√I, the ideals I, J are not nilpotent, and T
k [= (0). Assume that] [J] [is principal and]
the ring B=L
m[∩][I]n [is Noetherian. Then there exists a positive integer][t] [such]
that v(m+t) = v(m) +v(t) for all m≥t.
This shows why it is of interest to establish that I and J have finite intersection
Proposition 4.8. Let R be a UFD and I and J nonzero principal ideals in R such
thatJ ⊆√I. Then there exists a positive integert such thatv(m+t) = v(m) +v(t).
Proof. Theorem (4.2) and Proposition (4.7) combined imply the result. However, we
will give a direct proof without relying on these results. Let J = (a) = (pα1
1 · · ·p αh h )
and I = (b) = (pα1
1 · · ·p αh
h ). Then v(m) is the largest n such that (am)⊆(bn), which
1, . . . , h)c =v(m). Let t be the minimum number such that tαi
βi ∈ N for all i. Then
for all m ≥t,
bmin(mαi βi
)|i= 1,· · · , h)c
+b min(tαi βi
)|i= 1, . . . , n)c
=b min((m+t)αi βi
)|i= 1, . . . , h)c
[1] W. Bishop, J. W. Petro, L. J. Ratliff, and D. E. Rush, Note on Noetherian
filtrations, Communications in Algebra 17 (1989), no. 2, 471–485.
[2] W. Bruns and J. Herzog, Cohen-Macaulay Rings, Cambridge University Press,
[3] C. Ciuperc˘a, F. Enescu, and S. Spiroff, Asymptotic growth of powers of ideals,
Illinois Journal of Mathematics 51 (2007), no. 1, 29–39. [4] D. Dummit and R. Foote,Abstract Algebra, Wiley, 2004.
[5] D. Eisenbud, Commutative Algebra, Springer, 1995.
[6] J. B. Fields, Length functions determined by killing powers of several ideals in
a local ring, Ph.D. Dissertation, University of Michigan, Ann Arbor, Michigan
[7] , Lengths of Tors determined by killing powers of ideals in a local ring,
Journal of Algebra 247 (2002), no. 1, 104–133.
[8] S. Goto and K. Nishida, The Cohen-Macaulay and Gorenstein Rees algebras
associated to filtrations, Memoirs of the American Mathematical Society 110
[9] H. Matsumura, Commutative Ring Theory, Cambridge University Press, 1989.
[10] J. S. Okon and L. J. Ratliff, Reductions of filtrations, Pacific Journal of
Mathe-matics 144 (1990), no. 1, 137–154.
[11] ,On the grade and cograde of a Noetherian filtration, Nagoya
Mathemat-ics Journal 122 (1991), 43–62.
[12] L. J. Ratliff,Notes on essentially powers filtrations, The Michigan Mathematical
Journal 26 (1979), no. 3, 313–324.
[13] L. J. Ratliff and D. E. Rush, Note on I-good filtrations and Noetherian Rees
rings, Communications in Algebra 16 (1988), no. 5, 955–975.
[14] D. Rees, Lectures on the Asymptotic Theory of Ideals, Cambridge University
Press, 1989.
[15] P. Roberts, A prime ideal in a polynomial ring whose symbolic blow-up is not
Noetherian, Proceedings of the American Mathematical Society94(1985), no. 4,
[16] I. Swanson and C. Huneke,Integral Closure of Ideals, Rings, and Modules, | {"url":"https://1library.net/document/y819vlrz-exercise-attitudes-perceptions-perceived-outcomes-minority-participating-prevention.html","timestamp":"2024-11-04T01:41:09Z","content_type":"text/html","content_length":"218141","record_id":"<urn:uuid:31711d73-2384-4125-9a00-4cbed6b3d159>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00055.warc.gz"} |
15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
Chapter 15 Thermodynamics
15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
• Define entropy and calculate the increase of entropy in a system with reversible and irreversible processes.
• Explain the expected fate of the universe in entropic terms.
• Calculate the increasing disorder of a system.
Figure 1. The ice in this drink is slowly melting. Eventually the liquid will reach thermal equilibrium, as predicted by the second law of thermodynamics. (credit: Jon Sullivan, PDPhoto.org)
There is yet another way of expressing the second law of thermodynamics. This version relates to a concept called entropy. By examining it, we shall see that the directions associated with the second
law—heat transfer from hot to cold, for example—are related to the tendency in nature for systems to become disordered and for less energy to be available for use as work. The entropy of a system can
in fact be shown to be a measure of its disorder and of the unavailability of energy to do work.
Recall that the simple definition of energy is the ability to do work. Entropy is a measure of how much energy is not available to do work. Although all forms of energy are interconvertible, and all
can be used to do work, it is not always possible, even in principle, to convert the entire available energy into work. That unavailable energy is of interest in thermodynamics, because the field of
thermodynamics arose from efforts to convert heat to work.
We can see how entropy is defined by recalling our discussion of the Carnot engine. We noted that for a Carnot cycle, and hence for any reversible processes,
for any reversible process. change in entropy
The definition of Figure 2.)
Figure 2. When a system goes from state 1 to state 2, its entropy changes by the same amount ΔS, whether a hypothetical reversible path is followed or a real irreversible path is taken.
Now let us take a look at the change in entropy of a Carnot engine and its heat reservoirs for one full cycle. The hot reservoir has a loss of entropy
Thus, since we know that
This result, which has general validity, means that the total change in entropy for a system in any reversible process is zero.
The entropy of various parts of the system may change, but the total change is zero. Furthermore, the system does not affect the entropy of its surroundings, since heat transfer between them does not
occur. Thus the reversible process changes neither the total entropy of the system nor the entropy of its surroundings. Sometimes this is stated as follows: Reversible processes do not affect the
total entropy of the universe. Real processes are not reversible, though, and they do change total entropy. We can, however, use hypothetical reversible processes to determine the value of entropy in
real, irreversible processes. The following example illustrates this point.
Example 1: Entropy Increases in an Irreversible (Real) Process
Spontaneous heat transfer from hot to cold is an irreversible process. Calculate the total change in entropy if 4000 J of heat transfer occurs from a hot reservoir at Figure 3.)
How can we calculate the change in entropy for an irreversible process when
We now calculate the two changes in entropy using
And for the cold reservoir,
Thus the total is
There is an increase in entropy for the system of two heat reservoirs undergoing this irreversible heat transfer. We will see that this means there is a loss of ability to do work with this
transferred energy. Entropy has increased, and energy has become unavailable to do work.
Figure 3. (a) Heat transfer from a hot object to a cold one is an irreversible process that produces an overall increase in entropy. (b) The same final state and, thus, the same change in entropy is
achieved for the objects if reversible heat transfer processes occur between the two objects whose temperatures are the same as the temperatures of the corresponding objects in the irreversible
It is reasonable that entropy increases for heat transfer from hot to cold. Since the change in entropy is
There is an increase in entropy for any system undergoing an irreversible process.
With respect to entropy, there are only two possibilities: entropy is constant for a reversible process, and it increases for an irreversible process. There is a fourth version of the second law of
thermodynamics stated in terms of entropy:
The total entropy of a system either increases or remains constant in any process; it never decreases.
For example, heat transfer cannot occur spontaneously from cold to hot, because entropy would decrease.
Entropy is very different from energy. Entropy is not conserved but increases in all real processes. Reversible processes (such as in Carnot engines) are the processes in which the most heat transfer
to work takes place and are also the ones that keep entropy constant. Thus we are led to make a connection between entropy and the availability of energy to do work.
Entropy and the Unavailability of Energy to Do Work
What does a change in entropy mean, and why should we be interested in it? One reason is that entropy is directly related to the fact that not all heat transfer can be converted into work. The next
example gives some indication of how an increase in entropy results in less heat transfer into work.
Example 2: Less Work is Produced by a Given Heat Transfer When Entropy Change is Greater
(a) Calculate the work output of a Carnot engine operating between temperatures of 600 K and 100 K for 4000 J of heat transfer to the engine. (b) Now suppose that the 4000 J of heat transfer occurs
first from the 600 K reservoir to a 250 K reservoir (without doing any work, and this produces the increase in entropy calculated above) before transferring into a Carnot engine operating between 250
K and 100 K. What work output is produced? (See Figure 4.)
In both parts, we must first calculate the Carnot efficiency and then the work output.
Solution (a)
The Carnot efficiency is given by
Substituting the given temperatures yields
Now the work output can be calculated using the definition of efficiency for any heat engine as given by
Solving for
Solution (b)
so that
There is 933 J less work from the same heat transfer in the second process. This result is important. The same heat transfer into two perfect engines produces different work outputs, because the
entropy change differs in the two cases. In the second case, entropy is greater and less work is produced. Entropy is associated with the unavailability of energy to do work.
Figure 4. (a) A Carnot engine working at between 600 K and 100 K has 4000 J of heat transfer and performs 3333 J of work. (b) The 4000 J of heat transfer occurs first irreversibly to a 250 K
reservoir and then goes into a Carnot engine. The increase in entropy caused by the heat transfer to a colder reservoir results in a smaller work output of 2400 J. There is a permanent loss of 933 J
of energy for the purpose of doing work.
When entropy increases, a certain amount of energy becomes permanently unavailable to do work. The energy is not lost, but its character is changed, so that some of it can never be converted to doing
work—that is, to an organized force acting through a distance. For instance, in the previous example, 933 J less work was done after an increase in entropy of 9.33 J/K occurred in the 4000 J heat
transfer from the 600 K reservoir to the 250 K reservoir. It can be shown that the amount of energy that becomes unavailable for work is
Heat Death of the Universe: An Overdose of Entropy
In the early, energetic universe, all matter and energy were easily interchangeable and identical in nature. Gravity played a vital role in the young universe. Although it may have seemed disorderly,
and therefore, superficially entropic, in fact, there was enormous potential energy available to do work—all the future energy in the universe.
As the universe matured, temperature differences arose, which created more opportunity for work. Stars are hotter than planets, for example, which are warmer than icy asteroids, which are warmer
still than the vacuum of the space between them.
Most of these are cooling down from their usually violent births, at which time they were provided with energy of their own—nuclear energy in the case of stars, volcanic energy on Earth and other
planets, and so on. Without additional energy input, however, their days are numbered.
As entropy increases, less and less energy in the universe is available to do work. On Earth, we still have great stores of energy such as fossil and nuclear fuels; large-scale temperature
differences, which can provide wind energy; geothermal energies due to differences in temperature in Earth’s layers; and tidal energies owing to our abundance of liquid water. As these are used, a
certain fraction of the energy they contain can never be converted into doing work. Eventually, all fuels will be exhausted, all temperatures will equalize, and it will be impossible for heat engines
to function, or for work to be done.
Entropy increases in a closed system, such as the universe. But in parts of the universe, for instance, in the Solar system, it is not a locally closed system. Energy flows from the Sun to the
planets, replenishing Earth’s stores of energy. The Sun will continue to supply us with energy for about another five billion years. We will enjoy direct solar energy, as well as side effects of
solar energy, such as wind power and biomass energy from photosynthetic plants. The energy from the Sun will keep our water at the liquid state, and the Moon’s gravitational pull will continue to
provide tidal energy. But Earth’s geothermal energy will slowly run down and won’t be replenished.
But in terms of the universe, and the very long-term, very large-scale picture, the entropy of the universe is increasing, and so the availability of energy to do work is constantly decreasing.
Eventually, when all stars have died, all forms of potential energy have been utilized, and all temperatures have equalized (depending on the mass of the universe, either at a very high temperature
following a universal contraction, or a very low one, just before all activity ceases) there will be no possibility of doing work.
Either way, the universe is destined for thermodynamic equilibrium—maximum entropy. This is often called the heat death of the universe, and will mean the end of all activity. However, whether the
universe contracts and heats up, or continues to expand and cools down, the end is not near. Calculations of black holes suggest that entropy can easily continue for at least
Order to Disorder
Entropy is related not only to the unavailability of energy to do work—it is also a measure of disorder. This notion was initially postulated by Ludwig Boltzmann in the 1800s. For example, melting a
block of ice means taking a highly structured and orderly system of water molecules and converting it into a disorderly liquid in which molecules have no fixed positions. (See Figure 5.) There is a
large increase in entropy in the process, as seen in the following example.
Example 3: Entropy Associated with Disorder
Find the increase in entropy of 1.00 kg of ice originally at
As before, the change in entropy can be calculated from the definition of
The change in entropy is defined as:
Now the change in entropy is positive, since heat transfer occurs into the ice to cause the phase change; thus,
This is a significant increase in entropy accompanying an increase in disorder.
Figure 5. When ice melts, it becomes more disordered and less structured. The systematic arrangement of molecules in a crystal structure is replaced by a more random and less orderly movement of
molecules without fixed locations or orientations. Its entropy increases because heat transfer occurs into it. Entropy is a measure of disorder.
In another easily imagined example, suppose we mix equal masses of water originally at two different temperatures, say
First, entropy has increased for the same reason that it did in the example above. Mixing the two bodies of water has the same effect as heat transfer from the hot one and the same heat transfer into
the cold one. The mixing decreases the entropy of the hot water but increases the entropy of the cold water by a greater amount, producing an overall increase in entropy.
Second, once the two masses of water are mixed, there is only one temperature—you cannot run a heat engine with them. The energy that could have been used to run a heat engine is now unavailable to
do work.
Third, the mixture is less orderly, or to use another term, less structured. Rather than having two masses at different temperatures and with different distributions of molecular speeds, we now have
a single mass with a uniform temperature.
These three results—entropy, unavailability of energy, and disorder—are not only related but are in fact essentially equivalent.
Life, Evolution, and the Second Law of Thermodynamics
Some people misunderstand the second law of thermodynamics, stated in terms of entropy, to say that the process of the evolution of life violates this law. Over time, complex organisms evolved from
much simpler ancestors, representing a large decrease in entropy of the Earth’s biosphere. It is a fact that living organisms have evolved to be highly structured, and much lower in entropy than the
substances from which they grow. But it is always possible for the entropy of one part of the universe to decrease, provided the total change in entropy of the universe increases. In equation form,
we can write this as
How is it possible for a system to decrease its entropy? Energy transfer is necessary. If I pick up marbles that are scattered about the room and put them into a cup, my work has decreased the
entropy of that system. If I gather iron ore from the ground and convert it into steel and build a bridge, my work has decreased the entropy of that system. Energy coming from the Sun can decrease
the entropy of local systems on Earth—that is, not violated.
Every time a plant stores some solar energy in the form of chemical potential energy, or an updraft of warm air lifts a soaring bird, the Earth can be viewed as a heat engine operating between a hot
reservoir supplied by the Sun and a cold reservoir supplied by dark outer space—a heat engine of high complexity, causing local decreases in entropy as it uses part of the heat transfer from the Sun
into deep space. There is a large total increase in entropy resulting from this massive heat transfer. A small part of this heat transfer is stored in structured systems on Earth, producing much
smaller local decreases in entropy. (See Figure 6.)
Figure 6. Earth’s entropy may decrease in the process of intercepting a small part of the heat transfer from the Sun into deep space. Entropy for the entire process increases greatly while Earth
becomes more structured with living systems and stored energy in various forms.
Watch a reaction proceed over time. How does total energy affect a reaction rate? Vary temperature, barrier height, and potential energies. Record concentrations and time in order to extract rate
coefficients. Do temperature dependent studies to extract Arrhenius parameters. This simulation is best used with teacher guidance because it presents an analogy of chemical reactions.
Figure 7. Reversible Reactions
Section Summary
• Entropy is the loss of energy available to do work.
• Another form of the second law of thermodynamics states that the total entropy of a system either increases or remains constant; it never decreases.
• Entropy is zero in a reversible process; it increases in an irreversible process.
• The ultimate fate of the universe is likely to be thermodynamic equilibrium, where the universal temperature is constant and no energy is available to do work.
• Entropy is also associated with the tendency toward disorder in a closed system.
Conceptual Questions
1: A woman shuts her summer cottage up in September and returns in June. No one has entered the cottage in the meantime. Explain what she is likely to find, in terms of the second law of
2: Consider a system with a certain energy content, from which we wish to extract as much work as possible. Should the system’s entropy be high or low? Is this orderly or disorderly? Structured or
uniform? Explain briefly.
3: Does a gas become more orderly when it liquefies? Does its entropy change? If so, does the entropy increase or decrease? Explain your answer.
4: Explain how water’s entropy can decrease when it freezes without violating the second law of thermodynamics. Specifically, explain what happens to the entropy of its surroundings.
5: Is a uniform-temperature gas more or less orderly than one with several different temperatures? Which is more structured? In which can heat transfer result in work done without heat transfer from
another system?
6: Give an example of a spontaneous process in which a system becomes less ordered and energy becomes less available to do work. What happens to the system’s entropy in this process?
7: What is the change in entropy in an adiabatic process? Does this imply that adiabatic processes are reversible? Can a process be precisely adiabatic for a macroscopic system?
8: Does the entropy of a star increase or decrease as it radiates? Does the entropy of the space into which it radiates (which has a temperature of about 3 K) increase or decrease? What does this do
to the entropy of the universe?
9: Explain why a building made of bricks has smaller entropy than the same bricks in a disorganized pile. Do this by considering the number of ways that each could be formed (the number of
microstates in each macrostate).
Problems & Exercises
1: (a) On a winter day, a certain house loses
3: A hot rock ejected from a volcano’s lava fountain cools from
5: The Sun radiates energy at the rate of
7: What is the decrease in entropy of 25.0 g of water that condenses on a bathroom mirror at a temperature of
8: Find the increase in entropy of 1.00 kg of liquid nitrogen that starts at its boiling temperature, boils, and warms to
9: A large electrical power station generates 1000 MW of electricity with an efficiency of 35.0%. (a) Calculate the heat transfer to the power station,
10: (a) How much heat transfer occurs from 20.0 kg of Chapter 15.7 Problem-Solving Strategies for Entropy. (e) Discuss how everyday processes make increasingly more energy unavailable to do work, as
implied by this problem.
a measurement of a system’s disorder and its inability to do work in a system
change in entropy
the ratio of heat transfer to temperature
second law of thermodynamics stated in terms of entropy
the total entropy of a system either increases or remains constant; it never decreases
Problems & Exercises
(b) In order to gain more energy, we must generate it from things within the house, like a heat pump, human bodies, and other appliances. As you know, we use a lot of energy to keep our houses warm
in the winter because of the loss of heat to the outside.
199 J/K | {"url":"https://pressbooks.online.ucf.edu/phy2054ehk/chapter/entropy-and-the-second-law-of-thermodynamics-disorder-and-the-unavailability-of-energy/","timestamp":"2024-11-02T01:30:27Z","content_type":"text/html","content_length":"259438","record_id":"<urn:uuid:8c722b91-423d-4dc5-81d8-daf91386b0c2>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00681.warc.gz"} |
Baccarat Chemin de Fer Practices and Strategy
October 22nd, 2024 by Jayda Leave a reply »
Baccarat Codes
Baccarat is gambled on with eight decks in a dealing shoe. Cards valued less than ten are valued at their printed value and with 10, J, Q, K are zero, and A is one. Bets are placed on the ‘bank’, the
‘player’, or for a tie (these are not really people; they just represent the 2 hands that are dealt).
Two hands of two cards are then dealt to the ‘bank’ and ‘gambler’. The score for each hand is the sum total of the cards, although the first number is ignored. For example, a hand of five and six has
a value of one (five plus 6 equals 11; ignore the initial ‘one’).
A 3rd card can be dealt depending on the rules below:
- If the gambler or bank gets a value of eight or nine, the two players stand.
- If the gambler has less than 5, she takes a card. Players otherwise stand.
- If the player holds, the bank takes a card on a total lower than five. If the gambler takes a card, a table is employed to decide if the banker stays or takes a card.
Baccarat Banque Odds
The higher of the 2 totals wins. Winning wagers on the bank payout 19 to 20 (equal cash less a 5% rake. The Rake is tracked and cleared out once you leave the table so ensure you have funds remaining
just before you depart). Winning bets on the gambler pays out at 1 to 1. Winning bets for tie normally pays out at 8 to 1 but sometimes 9 to 1. (This is a bad bet as ties occur lower than 1 in every
ten hands. Be cautious of putting money on a tie. Although odds are astonishingly greater for 9:1 vs. eight to one)
Gambled on correctly punto banco gives relatively decent odds, aside from the tie bet of course.
Punto Banco Strategy
As with all games baccarat chemin de fer has quite a few familiar myths. One of which is similar to a false impression in roulette. The past isn’t an indicator of future outcomes. Recording past
results at a table is a waste of paper and a snub to the tree that surrendered its life for our paper needs.
The most accepted and almost certainly the most accomplished strategy is the one, three, two, six plan. This tactic is employed to build up winnings and minimizing losses.
Begin by wagering one dollar. If you win, add one more to the 2 on the game table for a grand total of three dollars on the second bet. Should you win you will retain six on the game table, remove
four so you are left with 2 on the third round. Should you succeed on the third round, put down two on the four on the table for a grand total of six on the 4th bet.
If you don’t win on the initial round, you take a loss of 1. A profit on the initial bet followed by a loss on the second causes a hit of two. Success on the first 2 with a hit on the 3rd gives you
with a profit of 2. And success on the first 3 with a hit on the 4th means you experience no loss. Succeeding at all four bets gives you with twelve, a take of 10. This means you are able to give up
the second round 5 instances for every favorable run of four wagers and still experience no loss. | {"url":"http://prizediggers.com/2024/10/22/baccarat-chemin-de-fer-practices-and-strategy/","timestamp":"2024-11-11T10:54:45Z","content_type":"application/xhtml+xml","content_length":"12583","record_id":"<urn:uuid:322c2131-62e8-4e8c-adb8-e0cc1898a310>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00810.warc.gz"} |
How do you add numbers in octal systems?
Octal Addition To use this table, simply follow the directions used in this example: Add 68 and 58. Locate 6 in the A column then locate the 5 in the B column. The point in ‘sum’ area where these two
columns intersect is the ‘sum’ of two numbers. 68 + 58 = 138.
Can we add two octal numbers?
Addition of octal numbers: Addition of octal numbers is carried out by the same principle as that of decimal or binary numbers.
How is octal calculated?
In decimal to binary, we divide the number by 2, in decimal to hexadecimal we divide the number by 16. In case of decimal to octal, we divide the number by 8 and write the remainders in the reverse
order to get the equivalent octal number.
How do you do binary addition and subtraction?
Binary Addition and Subtraction
1. When we added the one’s column of the binary digit (i.e., 7+4) we get the number which is greater than the base of the decimal number (the base of the number is 10 and the sum of the digit is
2. The above sum is carried out by following step.
How do you convert from decimal to octal?
Use this method to learn the concepts. Of the two methods on this page,this method is easier to understand.
Write down the decimal number. For this example,we’ll convert the decimal number 98 into octal.
List the powers of 8.
Divide the decimal number by the largest power of eight.
Find the remainder.
Divide the remainder by the next power of 8.
How to convert from decimal to octal?
First,we divide the decimal numbers by 8.
Then we get quotient and remainder.
If the remainder is greater than 0 then again,we repeat step 1 and step 2.
Then we store the number from bottom to top and we get the octal number
How do you divide two octal numbers?
Take any two octal numbers.
Octal numbers are from 0 to 7.
Write numbers one by one.
Perform addition operation between two numbers from units place to left wards.
For example,add 1+7 is equal to 10 which means 1 as carry and 0 in the answer place.
Adding of remaining numbers which gives result less than 7 remains same.
How to divide octal numbers?
Definition. A number system which has its base as ‘eight’ is called an Octal number system.
Octal Numbers System Table. We use only 3 bits to represent Octal Numbers.
Decimal to Octal Number.
Octal to Decimal.
Binary To Octal Number.
Octal to Hexadecimal Number.
Octal Multiplication Table
Problems and Solutions. | {"url":"https://vivu.tv/how-do-you-add-numbers-in-octal-systems/","timestamp":"2024-11-12T10:53:58Z","content_type":"text/html","content_length":"121781","record_id":"<urn:uuid:2a01f613-7e96-4b34-897f-a846bb283e77>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00696.warc.gz"} |
Variants of Graph Neural Networks (GNN)
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Graphs are data structures consisting of nodes and the relationships between nodes, known as edges. Recently there have been many model architectures leveraging the power of graphs and their
characteristics. The fundamental models were called Graph Neural Networks. Once the potential was discovered of these graph-based models, the artificial intelligence community has continuously
developed the variants of graph-based models depending on various use cases.
Just as a refresher, let us look at the fundamental driving blocks of GNN. A simple GNN works based on input, i.e. node values, and the way the network propagates. There is one more parameter that
makes a particular model unique: the training methodology. In a GNN, the inputs are taken based on the propagation step, which in standard architecture is called message passing. Here it aggregates
every node value and passes it through an activation function. The networkβ s training methodology is getting the final structure/representation and goal value correct.
Now that we know what traditional architecture is, let us see how we can classify and understand different types of GNNs. The types are decided based on three categories.
1. Based on the Graph type
2. Based on the propagation step
3. Based on the training method
First based on graph types. As we know that if graphs are of many types and as the fundamental building block changes, the algorithm will change.
The types based on the graph are:
1. Directed Graph β DGP
2. Heterogeneous graph β Graph inception, HAN
3. Edge-informative graph β G2S, R-GCN
4. Dynamic graph β DCRNN, Structural-RNN
Although GNN is a powerful architecture for modeling structural data, there are several limitations, which can be removed by some tweaks. The first tweak is w.r.t the type of graph used. As mentioned
above the first is the directed graph.
An undirected edge suggests that there is a relationship between the two nodes, but a directed edge could provide more information. For example, if we have something like a class hierarchy, then we
can easily represent that type of data using a directed graph, like the head being child and tail as the parent., or vice-versa. For this, we created a DGP algorithm that uses two weight matrices. Wp
and W¬c, the weights for parents and children, respectively.
Then there are heterogeneous graphs. This kind of graph consists of various types of nodes. The computations are done by getting every node representation similar by transforming the values by
one-hot encoding. Due to the ability of different nodes, the algorithm Graph Inception was developed, which used this characteristic by grouping different neighbors and clustering them to be used as
a whole. These clusters are also called sub-graphs which can be used to have parallel computations. And keeping the same heterogeneous property in mind, the Heterogeneous Graph Attention Networks
were created.
Next are the dynamic graphs. These kinds have static graph structure and dynamic inputs. This allows for the adaptive structures or algorithms which require dynamicity in the internal structures.
When the idea of Structural-RCNN came, it seemed difficult because of two inputs, spatial as well as temporal messages at the same time. But with the dynamic graphs, it is easily possible.
Finally, the graphs with edge information. As the name suggests, the edges have some additional information like weights or the type of edge. This information can help in creating the architectures
like G2S encoders and R-GCN. Especially in R-GCN, which is the extension of GCN. The R means, relational. So, while working with relational data, it becomes easier to have the graph with edges that
can hold additional information like the relationship between those nodes.
The types based on propagation step:
1. Convolutional aggregator β Graph Convolutional Network
2. Attention aggregator β Graph Attention Network
3. Gate updater β GRU, LSTM
4. Skip connection β Highway GNN, Jump knowledge network
5. Hierarchical graph β ECC
The propagation steps do allow for versatility which is evident from the types mentioned. To understand these types of GNNs, it would be simpler to compare them to traditional neural networks. It is
important to remember that the propagation step consists of aggregating the neighboring node values. So, the main difference may also be called as aggregators.
First, convolutional aggregator. This is similar to the convolutional neural network. So, these networks work with image data. The fundamental idea remains the same. The high-level data is slowly
convoluted into lower size data. GCN works with the same core idea. With two options, one with spectral representations and second with spatial representations. The other formats of GCN in the
spectral domain are AGCN (Adaptive-GCN) and GGP, whereas in the spatial domain we have DCNN and DGCN.
In the same way, attention networks can be related to attention aggregators which are a fundamental concept of Graph Attention Networks and Gated Attention Networks. In the gate updaters, the core
blocks are like the GRU and LSTM networks. Using the GRU, we make the Gated Graph Neural Network (GGNN). With the LSTM blocks, we can build architectures like Graph LSTM, which can be further divided
into Tree LSTM, Sentence LSTM, and Graph LSTM.
With the skip connection networks, we build the architectures with core ideas parallel to residual networks. The architectures are Jump knowledge network, which can be understood by the name itself
and the Highway GNN. The hierarchical graph architectures include the Edge-conditioned convolution (ECC) networks. It uses an edge-information graph so that the information can be conditioned to
something useful. The same is then used for the computations related to propagation.
The types based on training methods:
1. Neighborhood sampling β FastGCN, GraphSAGE
2. Receptive field control β ControlVariate
3. Data augmentation β Co-training, self-training
4. Unsupervised learning β GAE, GCMC
The original GCN lacked the ability for inductive learning. To overcome this, we used neighborhood sampling architectures. The algorithm GraphSAGE is a comprehensive improvement over the original
GCN. To make the inductive learning adaptable, the algorithm replaced the full graph Laplacian with learnable aggregation functions.
With the receptive field control, the ControlVariate architectures appended the stochastic approximation algorithms for GCN, which utilizes the historic activations of the nodes as a control variate.
With data augmentation, we have two network architectures, co-training, and self-training. With the limitation of the requirement of large labeled data for training a GCN, the authors proposed these
two architectures. The co-training utilizes the power of k-means for the neighbors in the training data and self-training follows a boosting based architecture.
Unsupervised learning was proposed to take out the problem of the requirement of a large labeled dataset for training. The architectures include Graph Autoencoders (GAE) and GCMC, which follows a
Monte-Carlo based approximation. The graph autoencoders first use the GCNs to encode the nodes in the graph and then use a decoder to compute the adjacency matrix for computing the loss and taking
the training further.
So, this is how based on the limitations of the traditional graph-based algorithms, we tweak the fundamental blocks to build better architectures. We do know graph-based algorithms are always useful
in structural scenarios but now with these architectures, we can successfully train the networks for non-structural scenarios too. With the improvements over traditional networks, it is now possible
to efficiently train the models for advanced applications like semantic segmentation, sequence labeling, and event extractions.
With this article at OpenGenus, you must have the complete idea of Variants of Graph Neural Networks (GNN). Enjoy. | {"url":"https://iq.opengenus.org/types-graph-neural-network/","timestamp":"2024-11-14T12:06:08Z","content_type":"text/html","content_length":"64143","record_id":"<urn:uuid:3572c536-af4e-4af6-a507-11708f70c1bc>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00125.warc.gz"} |
Unity 3D Procedural Terrain Generation
Let us look at how we can generate procedural terrain in the unity 3d engine. So what is procedural generation useful for. Procedural generation is great technique to keep your games fresh and
interesting, by introducing randomly or algorithmic-ally generated content. Which just provides so much appeal for players who enjoy genres like open worlds and infinite scenario game play. This
technique undoubtedly is useful for making big progress in your games quickly with minimal effort.
Here is a little example of what you can do with procedural generation:
In this tutorial we will only be dealing with procedural terrain generation, but in further tutorials I will be adding we will be looking at other forms of procedural generation as well. An example
of where multiple procedural methods have been used is as per this screenshot below.
Procedural generated terrain, rocks and trees
I put this little project together to generate procedural terrain, trees and rocks. As you can see procedural generation is very powerful. You can generate entire games if done correctly. One of the
most powerful metrics in games is play time. Procedural methods can get you more play time. So now that I have shown you what it can do let’s jump into this tutorial on how we can create a terrain in
unity 3d. If you are interested in this project, I have produced a course on how you can learn to build a unity 3d rts city builder style game. Check out my course here: Build a RTS Style City
Builder from Scratch with Unity 3D – Part 1
Setting up our Unity 3D project for our terrain
Start off with a blank new unity project.
Unity hierarchy
On the unity hierarchy right click and create a empty game object. Rename it to terrain. This will be the game object we will give some geometry.
Let’s quickly cover the theory. We need to build up a mesh of triangles. Before we can get there we need to setup some vertices. For those who are beginners reading this a vertex or vertices(plural)
is a point in our 3d space which consists of a x,y,z co ordinate. Just to keep things simple we will keep our y values at 1.0f initially then will modify that with perlin noise or random noise later
on. Our vertices will be filled in with triangles which each will have 3 vertices.
To illustrate the principle. We will build up the c# script step by step so that you can actually learn how it works.
Let’s start coding, start with something basic
Our first order of business is to get a 1×1 quad set up with 4 vertices. Then we will move on to generating a grid of vertices and then finally filling in triangles.
So head over to your terrain game object and in the inspector click on add component and type in ProceduralTerrain and create and add script.
Unity 3D Procedural Terrain Script
So now open it up in Visual studio.
So to get started let’s add some variables first.
public int xWidth = 30;
public int zDepth = 30;
Vector3[] vertices;
The width and depth is going to manage the size of our grid. We adding this now just to use it later but in our basic starting script we won’t be using this as of yet. Only the Vector3 array declared
as vertices. We also now need a way to test our code so we can see what is going on. For that we will draw some gizmos. In order to do that we need to add this code to our script.
private void OnDrawGizmos()
if (vertices == null) return;
for (int i = 0; i < vertices.Length; i++)
Gizmos.DrawSphere(vertices[i], .5f);
This will allow us to draw some spheres in our scene where our vertices are supposed to be. So once this is added we just need to start off with some vertices. For this we will add a basic method. We
will just call it CreateGeometry. Before we can make this all work let’s head back to Unity so we can add a mesh filter component to our Terrain game object.
Unity 3D mesh filter for modifying our mesh
Great now jump back into visual studio. We will now need to add another variable. So add these lines to your code. We will need the mesh later.
Mesh mesh;
void Start()
While we in the start method, let’s call the CreateGeometry method. Let’s now add our code for our first block of vertices. In our CreateGeometry method we need to setup our array and setup our first
void CreateGeometry()
vertices = new Vector3[4];
vertices[0] = new Vector3(0.0f, 1.0f, 0.0f);
vertices[1] = new Vector3(1.0f, 1.0f, 0.0f);
vertices[2] = new Vector3(0.0f, 1.0f, 1.0f);
vertices[3] = new Vector3(1.0f, 1.0f, 1.0f);
So this is a very basic script. First we start by giving ourselves 4 vertices we can assign. Then we assign the 4 vertices starting at 0.0f on x and 0.0f on z all the y-axis will be 1.0f by default.
Just for learning purposes add one vertex at a time. Then check the result. Before you hit run we just need to make sure of a few things otherwise you might not see your vertices.
Set terrain transform
Start by setting the rotation and position values on your terrain to 0 like in the screenshot above. Then click on the main camera, then click on GameObject and move to view, then click GameObject
again and align to view.
Move camera to view
Align camera with view
This will ensure that the unity camera is aligned correctly to your view. Now you can go ahead and click the play button and should see something similar to this.
Unity 3D procedural terrain generation step one
Great so if you play around with the vertices by removing some you can see how this changes things. Here is an example:
Unity 3D procedural terrain changing the vertices around
Now let’s apply this in a loop so that we can expand on this. To do this we need to account for a few things, but let’s first start with just get something generated.
So in our CreateGeometry method lets add in some loops. This is what your code will look like on our 2nd iteration of the code.
void CreateGeometry()
int verticeCounter = 0;
vertices = new Vector3[xWidth * zDepth];
for (int x = 0; x < xWidth;x++) {
for(int z = 0; z < zDepth; z++) {
vertices[verticeCounter] = new Vector3(x, 1.0f, z);
So very simply put our verticeCounter just holds our number of vertices. So we just multiple our zDepth by our xWidth. Remember this is the number of vertices we want in our grid, but something to
consider is that the width and depth is not a vertex count its actually a block width. So in order to achieve this instead of multiplying xWidth by zDepth we need to add one to each. So we should
change that line to:
vertices = new Vector3[(xWidth+1) * (zDepth+1)];
Great so now moving on to the next section. The 2 for loops one will loop over the x axis and next over the z axis. We Then just add a new vector for our x and z with our default height of 1.0f on
the y-axis. Then finally since we have added a vertex we can add to our verticeCounter. Now hit the run button and you should have something like this.
Filling in our triangles
Great we have come this far now we just need to fill in our triangles and apply some noise we have our terrain. For this lets again start with a basic break down. Let’s think about filling in just
one square. So lets take our first square which will be at (0,0), (0,1), (1,0) and (1,1) all the y axis points are at 1 still so I have left them out to simplify. So to fill in two triangles between
our points we need the 6 points. Because we need a vertice at (0,0),(0,1),(1,0) then the other 3 vertices will be the other side (1,0),(0,1),(1,1). Great so let’s start with just rendering this as a
Wonderful our code ends up looking like this. So the tricky part about our triangles is we need to pass the indexes relating to our vertices. So for example we want (0,0) to be our starting point for
our triangle so the first triangle will be index 0.
int[] triangles; //We declare this at the top of our script under the class definition
//In create geometry
triangles = new int[6];
triangles[0] = 0;
triangles[1] = xWidth + 1;
triangles[2] = 1;
triangles[3] = 1; //This is 1 again because both the triangles share the point (0,1)
triangles[4] = xWidth + 1; //This also shares a point with triangles 1
triangles[5] = xWidth + 2; //This is the remaining point which is not shared
Ok so this might seem confusing but lets make sure in our unity inspector we set our xWidth zDepth = 2 so that we only see a 2×2 grid. From here will expand it. So before we can actually see our
triangles we need to do some more setup.
Add a method to update our mesh
For this we will create a new method called UpdateMesh. So let’s quickly create this code.
public void UpdateMesh()
mesh.vertices = vertices;
mesh.triangles = triangles;
So first we make sure our mesh is clear and has no Geometry. Then we assign our vertices and triangle indexes respectively. Finally we recalculate our normals to make sure our scene lighting isn’t
broken when our light source hits our mesh.
One more thing we need to get our mesh from our mesh filter component. In our start method this will be our new code.
void Start()
mesh = new Mesh();
GetComponent<MeshFilter>().mesh = mesh;
Also make sure you have a Mesh Filter component on your terrain game object like this.
Last bits of code for this tutorial on Unity 3D Procedural Terrain
Let’s now generate all our triangles and apply some perlin noise to wrap this up.
Instead our previous manual code add this code to our CreateGeometry method.
triangles = new int[xWidth * zDepth * 6];
int vertexcounter = 0;
int trianglecount = 0;
for (int z = 0; z < zDepth; z++)
for (int x = 0; x < xWidth; x++)
triangles[0 + trianglecount] = vertexcounter + 0;
triangles[1 + trianglecount] = vertexcounter + xWidth + 1;
triangles[2 + trianglecount] = vertexcounter + 1;
triangles[3 + trianglecount] = vertexcounter + 1;
triangles[4 + trianglecount] = vertexcounter + xWidth + 1;
triangles[5 + trianglecount] = vertexcounter + xWidth + 2;
trianglecount += 6;
For our perlin noise this is very simple to apply head back to this line.
vertices[verticeCounter] = new Vector3(x, 1.0f, z);
Replace it with this:
float perlinnoiseY = Mathf.PerlinNoise(x * .99f, z * .99f) * 1.1f;
vertices[verticeCounter] = new Vector3(x, perlinnoiseY, z);
We feed it x and multiply it by a seed of .99f, then we pass in z multiplied by a seed of .99f. Finally we multiply this number by a y offset of 1.1f.
Update your depth and width to something sensible like 20×20.
You can now run your unity project.
You should now have this.
Disable gizmos to see it more clearly.
You should now see some ripples in your terrain.
Unity 3D Procedural Terrain after perlin noise
You can apply different types of perlin noise to get higher peaks, smoother terrain and more experiment with the values. As a good exercise try add some on the fly updating to your script. If you
don’t want to go through the trouble you can watch my online course and learn more about this topic: Build a RTS Style City Builder from Scratch with Unity 3D – Part 1
For more tutorials and resources check out some of my other blog posts. Here is a great one about Unity 2D top down player movement for your games. | {"url":"https://generalistprogrammer.com/game-design-development/unity-3d-procedural-terrain-generation/","timestamp":"2024-11-14T01:51:27Z","content_type":"text/html","content_length":"111991","record_id":"<urn:uuid:0a2ae818-f861-4cb8-983f-2cfaa6d1fe05>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00515.warc.gz"} |
Time-Frequency Masking for Harmonic-Percussive Source Separation
Time-frequency masking is the process of applying weights to the bins of a time-frequency representation to enhance, diminish, or isolate portions of audio.
The goal of harmonic-percussive source separation (HPSS) is to decompose an audio signal into harmonic and percussive components. Applications of HPSS include audio remixing, improving the quality of
chroma features, tempo estimation, and time-scale modification [1]. Another use of HPSS is as a parallel representation when creating a late fusion deep learning system. Many of the top performing
systems of the Detection and Classification of Acoustic Scenes and Events (DCASE) 2017 and 2018 challenges used HPSS for this reason.
This example walks through the algorithm described in [1] to apply time-frequency masking to the task of harmonic-percussive source separation.
For an example of deriving time-frequency masks using deep learning, see Cocktail Party Source Separation Using Deep Learning Networks.
Create Harmonic-Percussive Mixture
Read in harmonic and percussive audio files. Both have a sample rate of 16 kHz.
[harmonicAudio,fs] = audioread("violin.wav");
percussiveAudio = audioread("drums.wav");
Listen to the harmonic signal and plot the spectrogram. Note that there is continuity along the horizontal (time) axis.
title("Harmonic Audio")
Listen to the percussive signal and plot the spectrogram. Note that there is continuity along the vertical (frequency) axis.
title("Percussive Audio")
Mix the harmonic and percussive signals. Listen to the harmonic-percussive audio and plot the spectrogram.
mix = harmonicAudio + percussiveAudio;
title("Harmonic-Percussive Audio")
The HPSS proposed by [1] creates two enhanced spectrograms: a harmonic-enhanced spectrogram and a percussive-enhanced spectrogram. The harmonic-enhanced spectrogram is created by applying median
filtering along the time axis. The percussive-enhanced spectrogram is created by applying median filtering along the frequency axis. The enhanced spectrograms are then compared to create harmonic and
percussive time-frequency masks. In the simplest form, the masks are binary.
HPSS Using Binary Mask
Convert the mixed signal to a half-sided magnitude short-time Fourier transform (STFT).
win = sqrt(hann(1024,"periodic"));
overlapLength = floor(numel(win)/2);
fftLength = 2^nextpow2(numel(win) + 1);
y = stft(mix,Window=win,OverlapLength=overlapLength,FFTLength=fftLength,FrequencyRange="onesided");
ymag = abs(y);
Apply median smoothing along the time axis to enhance the harmonic audio and diminish the percussive audio. Use a filter length of 200 ms, as suggested by [1]. Plot the power spectrum of the
harmonic-enhanced audio.
timeFilterLength = 0.2;
timeFilterLengthInSamples = timeFilterLength/((numel(win) - overlapLength)/fs);
ymagharm = movmedian(ymag,timeFilterLengthInSamples,2);
title("Harmonic Enhanced Audio")
axis tight
Apply median smoothing along the frequency axis to enhance the percussive audio and diminish the harmonic audio. Use a filter length of 500 Hz, as suggested by [1]. Plot the power spectrum of the
percussive-enhanced audio.
frequencyFilterLength = 500;
frequencyFilterLengthInSamples = frequencyFilterLength/(fs/fftLength);
ymagperc = movmedian(ymag,frequencyFilterLengthInSamples,1);
title("Percussive Enhanced Audio")
axis tight
To create a binary mask, first sum the percussive- and harmonic-enhanced spectrums to determine the total magnitude per bin.
totalMagnitudePerBin = ymagharm + ymagperc;
If the magnitude in a given harmonic-enhanced or percussive-enhanced bin accounts for more than half of the total magnitude of that bin, then assign that bin to the corresponding mask.
harmonicMask = ymagharm > (totalMagnitudePerBin*0.5);
percussiveMask = ymagperc > (totalMagnitudePerBin*0.5);
Apply the harmonic and percussive masks and then return the masked audio to the time domain.
yharm = harmonicMask.*y;
yperc = percussiveMask.*y;
Perform the inverse short-time Fourier transform to return the signals to the time domain.
h = istft(yharm,Window=win,OverlapLength=overlapLength, ...
p = istft(yperc,Window=win,OverlapLength=overlapLength, ...
Listen to the recovered harmonic audio and plot the spectrogram.
title("Recovered Harmonic Audio")
Listen to the recovered percussive audio and plot the spectrogram.
title("Recovered Percussive Audio")
Plot the combination of the recovered harmonic and percussive spectrograms.
sound(h + p,fs)
spectrogram(h + p,1024,512,1024,fs,"yaxis")
title("Recovered Harmonic + Percussive Audio")
HPSS Using Binary Mask and Residual
As suggested in [1], decomposing a signal into harmonic and percussive sounds is often impossible. They propose adding a thresholding parameter: if the bin of the spectrogram is not clearly harmonic
or percussive, categorize it as residual.
Perform the same steps described in HPSS Using Binary Mask to create harmonic-enhanced and percussive-enhanced spectrograms.
win = sqrt(hann(1024,"periodic"));
overlapLength = floor(numel(win)/2);
fftLength = 2^nextpow2(numel(win) + 1);
y = stft(mix,Window=win,OverlapLength=overlapLength, ...
ymag = abs(y);
timeFilterLength = 0.2;
timeFilterLengthInSamples = timeFilterLength/((numel(win) - overlapLength)/fs);
ymagharm = movmedian(ymag,timeFilterLengthInSamples,2);
frequencyFilterLength = 500;
frequencyFilterLengthInSamples = frequencyFilterLength/(fs/fftLength);
ymagperc = movmedian(ymag,frequencyFilterLengthInSamples,1);
totalMagnitudePerBin = ymagharm + ymagperc;
Using a threshold, create three binary masks: harmonic, percussive, and residual. Set the threshold to 0.65. This means that if the magnitude of a bin of the harmonic-enhanced spectrogram is 65% of
the total magnitude for that bin, you assign that bin to the harmonic portion. If the magnitude of a bin of the percussive-enhanced spectrogram is 65% of the total magnitude for that bin, you assign
that bin to the percussive portion. Otherwise, the bin is assigned to the residual portion. The optimal thresholding parameter depends on the harmonic-percussive mix and your application.
threshold = 0.65;
harmonicMask = ymagharm > (totalMagnitudePerBin*threshold);
percussiveMask = ymagperc > (totalMagnitudePerBin*threshold);
residualMask = ~(harmonicMask+percussiveMask);
Perform the same steps described in HPSS Using Binary Mask to return the masked signals to the time domain.
yharm = harmonicMask.*y;
yperc = percussiveMask.*y;
yresi = residualMask.*y;
h = istft(yharm,Window=win,OverlapLength=overlapLength, ...
p = istft(yperc,Window=win,OverlapLength=overlapLength, ...
r = istft(yresi,Window=win,OverlapLength=overlapLength, ...
Listen to the recovered harmonic audio and plot the spectrogram.
title("Recovered Harmonic Audio")
Listen to the recovered percussive audio and plot the spectrogram.
title("Recovered Percussive Audio")
Listen to the recovered residual audio and plot the spectrogram.
title("Recovered Residual Audio")
Listen to the combination of the harmonic, percussive, and residual signals and plot the spectrogram.
sound(h + p + r,fs)
spectrogram(h + p + r,1024,512,1024,fs,"yaxis")
title("Recovered Harmonic + Percussive + Residual Audio")
HPSS Using Soft Mask
For time-frequency masking, masks are generally either binary or soft. Soft masking separates the energy of the mixed bins into harmonic and percussive portions depending on the relative weights of
their enhanced spectrograms.
Perform the same steps described in HPSS Using Binary Mask to create harmonic-enhanced and percussive-enhanced spectrograms.
win = sqrt(hann(1024,"periodic"));
overlapLength = floor(numel(win)/2);
fftLength = 2^nextpow2(numel(win) + 1);
y = stft(mix,Window=win,OverlapLength=overlapLength, ...
ymag = abs(y);
timeFilterLength = 0.2;
timeFilterLengthInSamples = timeFilterLength/((numel(win)-overlapLength)/fs);
ymagharm = movmedian(ymag,timeFilterLengthInSamples,2);
frequencyFilterLength = 500;
frequencyFilterLengthInSamples = frequencyFilterLength/(fs/fftLength);
ymagperc = movmedian(ymag,frequencyFilterLengthInSamples,1);
totalMagnitudePerBin = ymagharm + ymagperc;
Create soft masks that separate the bin energy to the harmonic and percussive portions relative to the weights of their enhanced spectrograms.
harmonicMask = ymagharm./totalMagnitudePerBin;
percussiveMask = ymagperc./totalMagnitudePerBin;
Perform the same steps described in HPSS Using Binary Mask to return the masked signals to the time domain.
yharm = harmonicMask.*y;
yperc = percussiveMask.*y;
h = istft(yharm,Window=win,OverlapLength=overlapLength, ...
p = istft(yperc,Window=win,OverlapLength=overlapLength, ...
Listen to the recovered harmonic audio and plot the spectrogram.
title("Recovered Harmonic Audio")
Listen to the recovered percussive audio and plot the spectrogram.
title("Recovered Percussive Audio")
Example Function
The example function, HelperHPSS, provides the harmonic-percussive source separation capabilities described in this example. You can use it to quickly explore how parameters effect the algorithm
[h,p] = HelperHPSS(x,fs) separates the input signal, x, into harmonic (h)
and percussive (p) portions. If x is input as a multichannel signal, it
is converted to mono before processing.
[h,p] = HelperHPSS(...,'TimeFilterLength',TIMEFILTERLENGTH) specifies the
median filter length along the time dimension of a spectrogram, in
seconds. If unspecified, TIMEFILTERLENGTH defaults to 0.2 seconds.
[h,p] = HelperHPSS(...,'FrequencyFilterLength',FREQUENCYFILTERLENGTH)
specifies the median filter length along the frequency dimension of a
spectrogram, in Hz. If unspecified, FREQUENCYFILTERLENGTH defaults to 500
[h,p] = HelperHPSS(...,'MaskType',MASKTYPE) specifies the mask type as
'binary' or 'soft'. If unspecified, MASKTYPE defaults to 'binary'.
[h,p] = HelperHPSS(...,'Threshold',THRESHOLD) specifies the threshold of
the total energy for declaring an element as harmonic, percussive, or
residual. Specify THRESHOLD as a scalar in the range [0 1]. This
parameter is only valid if MaskType is set to 'binary'. If unspecified,
THRESHOLD defaults to 0.5.
[h,p] = HelperHPSS(...,'Window',WINDOW) specifies the analysis window
used in the STFT. If unspecified, WINDOW defaults to
[h,p] = HelperHPSS(...,'FFTLength',FFTLENGTH) specifies the number of
points in the DFT for each analysis window. If unspecified, FFTLENGTH
defaults to the number of elements in the WINDOW.
[h,p] = HelperHPSS(...,'OverlapLength',OVERLAPLENGTH) specifies the
overlap length of the analysis windows. If unspecified, OVERLAPLENGTH
defaults to 512.
[h,p,r] = HelperHPSS(...) returns the residual signal not classified as
harmonic or percussive.
% Load a sound file and listen to it.
[audio,fs] = audioread('Laughter-16-8-mono-4secs.wav');
% Call HelperHPSS to separate the audio into harmonic and percussive
% portions. Listen to the portions separately.
[h,p] = HelperHPSS(audio,fs);
HPSS Using Iterative Masking
[1] observed that a large frame size in the STFT calculation moves the energy towards the harmonic component, while a small frame size moves the energy towards the percussive component. [1] proposed
using an iterative procedure to take advantage of this insight. In the iterative procedure:
1. Perform HPSS using a large frame size to isolate the harmonic component.
2. Sum the residual and percussive portions.
3. Perform HPSS using a small frame size to isolate the percussive component.
threshold1 = 0.7;
N1 = 4096;
[h1,p1,r1] = HelperHPSS(mix,fs,Threshold=threshold1,Window=sqrt(hann(N1,"periodic")),OverlapLength=round(N1/2));
mix1 = p1 + r1;
threshold2 = 0.6;
N2 = 256;
[h2,p2,r2] = HelperHPSS(mix1,fs,Threshold=threshold2,Window=sqrt(hann(N2,"periodic")),OverlapLength=round(N2/2));
h = h1;
p = p2;
r = h2 + r2;
Listen to the recovered percussive audio and plot the spectrogram.
title("Recovered Harmonic Audio")
Listen to the recovered percussive audio and plot the spectrogram.
title("Recovered Percussive Audio")
Listen to the recovered residual audio and plot the spectrogram.
title("Recovered Residual Audio")
Listen to the combination of the harmonic, percussive, and residual signals and plot the spectrogram.
sound(h + p + r,fs)
title("Recovered Harmonic + Percussive + Residual Audio")
Enhanced Time Scale Modification Using HPSS
[2] proposes that time scale modification (TSM) can be improved by first separating a signal into harmonic and percussive portions and then applying a TSM algorithm optimal for the portion. After
TSM, the signal is reconstituted by summing the stretched audio.
To listen to a stretched audio without HPSS, apply time-scale modification using the default stretchAudio function. By default, stretchAudio uses the phase vocoder algorithm.
alpha = 1.5;
mixStretched = stretchAudio(mix,alpha);
Separate the harmonic-percussive mix into harmonic and percussive portions using HelperHPSS. As proposed in [2], use the default vocoder algorithm to stretch the harmonic portion and the WSOLA
algorithm to stretch the percussive portion. Sum the stretched portions and listen to the results.
[h,p] = HelperHPSS(mix,fs);
hStretched = stretchAudio(h,alpha);
pStretched = stretchAudio(p,alpha,Method="wsola");
mixStretched = hStretched + pStretched;
[1] Driedger, J., M. Muller, and S. Disch. "Extending harmonic-percussive separation of audio signals." Proceedings of the International Society for Music Information Retrieval Conference. Vol. 15,
[2] Driedger, J., M. Muller, and S. Ewert. "Improving Time-Scale Modification of Music Signals Using Harmonic-Percussive Separation." IEEE Signal Processing Letters. Vol. 21. Issue 1. pp. 105-109,
Related Topics | {"url":"https://uk.mathworks.com/help/audio/ug/time-frequency-masking-for-harmonic-percussive-source-separation.html","timestamp":"2024-11-11T17:00:35Z","content_type":"text/html","content_length":"102989","record_id":"<urn:uuid:a27e7088-4441-4b3c-9b66-4a7d0901eec7>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00361.warc.gz"} |
Guides resources
Activity Guide: An outdoor-inspired indoor mathematics experience
The Outdoor division at UCLan provides a team building residential Frontier Education course to many of the university's first year cohorts. It was noticed that within this course some of the skills
developed would not only foster better group cohesion, but also reflected some of the qualities desired from the mathematics undergraduates The chance to turn this idea into a project came with the
Student Internship programme offered by sigma. This booklet is aimed at lecturers without prior knowledge of coaching theory but have the desire to approach the students development from a different
direction. Developed by Andrew Burrell, Jo McCready, Zainab Munshi, Davide Penazzi.
evaluation of mathematics support centres - a review of the literature (sigma)
This sigma guide reviews published literature concerning the evaluation of mathematics support centres. There is a growing body of research studies, which have looked into a number of areas such as:
the establishment of a MSC; the usage of MSCs and mechanisms for recording usage data; feedback from students and staff and ways to collect this; effects on achievement, pass rates and retention
rates; and the prevalence of MSCs in the higher education sector. More recently researchers have begun to examine the effects of MSCs on undergraduatesâ?? mathematics learning experiences and
mathematical confidence, and to address issues concerning students who are â??at riskâ?? or underachieving and not engaging with the facilities offered by their MSC. This report reviews and
synthesises all the available published research evidence so that informed decisions can be made about the value of mathematics support activity and the targeting of future funding.
Gathering student feedback on mathematics and statistics support provision - A guide for those running mathematics support centres (sigma)
This sigma guide has been written for those who are responsible for managing mathematics support centres. It is the culmination of a project involving staff from many support centres around the UK.
Authored by Dr David Green, Mathematics Education Centre, Loughborough University, it contains a wealth of advice and information for those who want to gather student feedback, and contains examples
of forms which are currently being used.
Getting Started in Pedagogic Research within the STEM Disciplines (HE STEM)
This guide edited by Michael Grove and Tina Overton has been developed for those looking to begin pedagogic research within the science, technology, engineering and mathematics (STEM) disciplines.
Its purpose is to provide an accessible introduction to pedagogic research along with a practical guide containing hints and tips on how to get started. The guide was produced following a series of
national workshops and conferences that were started in 2011 by the National HE STEM Programme and continued in 2012 with the support of the Higher Education Academy.
Getting started with data manipulation in Microsoft Excel
This is one of four "Getting started with ..." developed by Cheryl Voake-Jones and Emma Cliffe from the Mathematics Resources Centre at the University of Bath covering equations in Microsoft Excel.
The resource includes an audio tutorial with transcript and associated files. These resources were developed with funding from sigma.
Getting started with effective entry of equations in Microsoft Word
This is one of four "Getting started with ..." developed by Cheryl Voake-Jones and Emma Cliffe from the Mathematics Resources Centre at the University of Bath covering equations in Word. The resource
includes an audio tutorial with transcript and associated files. These resources were developed with funding from sigma.
Getting started with LaTeX
This is one of four "Getting started with ..." developed by Cheryl Voake-Jones and Emma Cliffe from the Mathematics Resources Centre at the University of Bath covering LaTeX. The resource includes an
audio tutorial with transcript and associated files. These resources were developed with funding from sigma.
Good Practice in the Provision of Mathematics Support Centres (LTSN)
A second edition of the popular LTSN funded guide for those interested in the establishment and development of Mathematics Support Centres in universities and other institutes of higher education.
Authors: Lawson, D., Croft, A.C. and Halpin, M.
How to setup a mathematics and statistics support provision guide (sigma)
This sigma guide is intended for anyone who is interested in setting up or enhancing a mathematics and/or statistics support provision. Authored by Ciaran Mac an Bhaird and Duncan Lawson, the guide
covers the nature of mathematics and statistics support, staffing, resources, funding, supporting neurodiversity and provides references to literature in the field.
Learning support in mathematics and statistics in Australian universities: a guide for the university sector (ALTC)
This Guide is based on findings from a project funded by The Australian Learning and Teaching Council (ALTC). After discussion on the history, nature and roles of learning support in mathematics and
statistics in Australia, it synthesizes the findings of the project to provide information for the university sector on the need for, and the provision of, such support. The project was funded by the
ALTC's Leadership for Excellence in Learning and Teaching Program. The title of the project was Quantitative diversity: disciplinary and cross-disciplinary mathematics and statistics support in
Australian universities, and its aim was to develop national capacity and collaboration in cross-disciplinary mathematics and statistics learning support to enhance student learning and confidence.
Setting up a maths support centre (HE STEM)
The focus of this HE STEM guide is to provide mathematics support to students across all STEM disciplines to ease the transition from School/College in to University. This is a key factor influencing
drop-out from STEM degrees and a targeted provision for mathematics support is a proven way to counter this. It includes cases studies from the Universities of Coventry, Portsmouth, York. Lincoln and
Kent and articulates the experiences of the two Sigma Directors Professor Duncan Lawson, Coventry University and Professor Tony Croft, Loughborough University joint winners of the 2011 Times Higher
Award for Outstanding Support for Students
Tutoring in a Mathematics Support Centre - A Guide for Postgraduate Tutors (sigma)
This sigma Guide is written for postgraduate students who are working in, or who want to work in, mathematics support centres. It distils the wisdom of seven people, who have many years of experience
in mathematics education and in the work of support centres, into a practical resource for postgraduate students. In addition, it contains activities which can be used during training sessions to
simulate working in a mathematics support centre. The guide is edited by Tony Croft and Michael Grove and authored by A.C.Croft, J.W.Gillard, M.J.Grove, J.Kyle, A.Owen, P.C.Samuels and R.H.Wilson.
And the winner is... mathematics support
At the Times Higher Awards ceremony on 24th November 2011, it was announced that Loughborough and Coventry Universities had won the award for Outstanding Support for Students, in recognition of the
work of sigma, Centre for Excellence in University-wide mathematics and statistics support. Whilst sigma at Coventry and Loughborough Universities received the award, the real winner was mathematics
and statistics support across the country. In this booklet, we outline how sigma's work has contributed to the growing recognition of the importance of mathematics and statistics support and to the
development of a national and international community of practitioners. Authors : Ciaran Mac an Bhaird and Duncan Lawson | {"url":"https://www.mathcentre.ac.uk/topics/guides/maths-support-centres/","timestamp":"2024-11-11T19:56:29Z","content_type":"text/html","content_length":"19350","record_id":"<urn:uuid:03cb1688-9aa6-4416-b8ca-3ec76fc0ca14>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00077.warc.gz"} |
Longstaff–Schwartz Methods - codefinance.trainingLongstaff–Schwartz Methods - codefinance.training
Longstaff–Schwartz Methods
Tree returning the OAS (black vs red): the short rate is the top value; the development of the bond value shows pull-to-par clearly
A short-rate model, in the context of interest rate derivatives, is a mathematical model that describes the future evolution of interest rates by describing the future evolution of the short rate,
usually written .
The short rate
Under a short rate model, the stochastic state variable is taken to be the instantaneous spot rate.^[1] The short rate, , then, is the (continuously compounded, annualized) interest rate at which an
entity can borrow money for an infinitesimally short period of time from time . Specifying the current short rate does not specify the entire yield curve. However, no-arbitrage arguments show that,
under some fairly relaxed technical conditions, if we model the evolution of as a stochastic process under a risk-neutral measure , then the price at time of a zero-coupon bond maturing at time with
a payoff of 1 is given by
where is the natural filtration for the process. The interest rates implied by the zero coupon bonds form a yield curve, or more precisely, a zero curve. Thus, specifying a model for the short rate
specifies future bond prices. This means that instantaneous forward rates are also specified by the usual formula
Short rate models are often classified as endogenous and exogenous. Endogenous short rate models are short rate models where the term structure of interest rates, or of zero-coupon bond prices , is
an output of the model, so it is "inside the model" (endogenous) and is determined by the model parameters. Exogenous short rate models are models where such term structure is an input, as the model
involves some time dependent functions or shifts that allow for inputing a given market term structure, so that the term structure comes from outside (exogenous).^[2]
Particular short-rate models
Throughout this section represents a standard Brownian motion under a risk-neutral probability measure and its differential. Where the model is lognormal, a variable is assumed to follow an
Ornstein–Uhlenbeck process and is assumed to follow .
One-factor short-rate models
Following are the one-factor models, where a single stochastic factor – the short rate – determines the future evolution of all interest rates. Other than Rendleman–Bartter and Ho–Lee, which do not
capture the mean reversion of interest rates, these models can be thought of as specific cases of Ornstein–Uhlenbeck processes. The Vasicek, Rendleman–Bartter and CIR models are endogenous models and
have only a finite number of free parameters and so it is not possible to specify these parameter values in such a way that the model coincides with a few observed market prices ("calibration") of
zero coupon bonds or linear products such as forward rate agreements or swaps, typically, or a best fit is done to these linear products to find the endogenous short rate models parameters that are
closest to the market prices. This does not allow for fitting options like caps, floors and swaptions as the parameters have been used to fit linear instruments instead. This problem is overcome by
allowing the parameters to vary deterministically with time,^[3]^[4] or by adding a deterministic shift to the endogenous model.^[5] In this way, exogenous models such as Ho-Lee and subsequent
models, can be calibrated to market data, meaning that these can exactly return the price of bonds comprising the yield curve, and the remaining parameters can be used for options calibration. The
implementation is usually via a (binomial) short rate tree ^[6] or simulation; see Lattice model (finance) § Interest rate derivatives and Monte Carlo methods for option pricing, although some short
rate models have closed form solutions for zero coupon bonds, and even caps or floors, easing the calibration task considerably. We list the following endogenous models first.
1. Merton's model (1973) explains the short rate as : where is a one-dimensional Brownian motion under the spot martingale measure.^[7] In this approach, the short rate follows an arithmetic
Brownian motion.
2. The Vasicek model (1977) models the short rate as ; it is often written .^[8] The second form is the more common, and makes the parameters interpretation more direct, with the parameter being the
speed of mean reversion, the parameter being the long term mean, and the parameter being the instantaneous volatility. In this short rate model an Ornstein–Uhlenbeck process is used for the short
rate. This model allows for negative rates, because the probability distribution of the short rate is Gaussian. Also, this model allows for closed form solutions for the bond price and for bond
options and caps/floors, and using Jamshidian's trick, one can also get a formula for swaptions.^[2]
3. The Rendleman–Bartter model (1980)^[9] or Dothan model (1978)^[10] explains the short rate as . In this model the short rate follows a geometric Brownian motion. This model does not have closed
form formulas for options and it is not mean reverting. Moreover, it has the problem of an infinite expected bank account after a short time. The same problem will be present in all lognormal
short rate models^[2]
4. The Cox–Ingersoll–Ross model (1985) supposes , it is often written . The factor precludes (generally) the possibility of negative interest rates.^[11] The interpretation of the parameters, in the
second formulation, is the same as in the Vasicek model. The Feller condition ensures strictly positive short rates. This model follows a Feller square root process and has non-negative rates,
and it allows for closed form solutions for the bond price and for bond options and caps/floors, and using Jamshidian's trick, one can also obtain a formula for swaptions. Both this model and the
Vasicek model are called affine models, because the formula for the continuously compounded spot rate for a finite maturity T at time t is an affine function of .^[2]
We now list a number of exogenous short rate models.
1. The Ho–Lee model (1986) models the short rate as .^[12] The parameter allows for the initial term structure of interest rates or bond prices to be an input of the model. This model follows again
an arithmetic Brownian motion with time dependent deterministic drift parameter.
2. The Hull–White model (1990)—also called the extended Vasicek model—posits . In many presentations one or more of the parameters and are not time-dependent. The distribution of the short rate is
normal, and the model allows for negative rates. The model with constant and is the most commonly used and it allows for closed form solutions for bond prices, bond options, caps and floors, and
swaptions through Jamshidian's trick. This model allows for an exact calibration of the initial term structure of interest rates through the time dependent function . Lattice-based implementation
for Bermudan swaptions and for products without analytical formulas is usually trinomial.^[13]^[14]
3. The Black–Derman–Toy model (1990) has for time-dependent short rate volatility and otherwise; the model is lognormal.^[15] The model has no closed form formulas for options. Also, as all
lognormal models, it suffers from the issue of explosion of the expected bank account in finite time.
4. The Black–Karasinski model (1991), which is lognormal, has .^[16] The model may be seen as the lognormal application of Hull–White;^[17] its lattice-based implementation is similarly trinomial
(binomial requiring varying time-steps).^[6] The model has no closed form solutions, and even basic calibration to the initial term structure has to be done with numerical methods to generate the
zero coupon bond prices. This model too suffers of the issue of explosion of the expected bank account in finite time.
5. The Kalotay–Williams–Fabozzi model (1993) has the short rate as , a lognormal analogue to the Ho–Lee model, and a special case of the Black–Derman–Toy model.^[18] This approach is effectively
similar to "the original Salomon Brothers model" (1987),^[19] also a lognormal variant on Ho-Lee.^[20]
6. The CIR++ model, introduced and studied in detail by Brigo and Mercurio^[5] in 2001, and formulated also earlier by Scott (1995)^[21] used the CIR model but instead of introducing time dependent
parameters in the dynamics, it adds an external shift. The model is formulated as where is a deterministic shift. The shift can be used to absorb the market term structure and make the model
fully consistent with this. This model preserves the analytical tractability of the basic CIR model, allowing for closed form solutions for bonds and all linear products, and options such as
caps, floor and swaptions through Jamshidian's trick. The model allows for maintaining positive rates if the shift is constrained to be positive, or allows for negative rates if the shift is
allowed to go negative. It has been applied often in credit risk too, for credit default swap and swaptions, in this original version or with jumps.^[22]
The idea of a deterministic shift can be applied also to other models that have desirable properties in their endogenous form. For example, one could apply the shift to the Vasicek model, but due to
linearity of the Ornstein-Uhlenbeck process, this is equivalent to making a time dependent function, and would thus coincide with the Hull-White model.^[5]
Multi-factor short-rate models
Besides the above one-factor models, there are also multi-factor models of the short rate, among them the best known are the Longstaff and Schwartz two factor model and the Chen three factor model
(also called "stochastic mean and stochastic volatility model"). Note that for the purposes of risk management, "to create realistic interest rate simulations", these multi-factor short-rate models
are sometimes preferred over One-factor models, as they produce scenarios which are, in general, better "consistent with actual yield curve movements".^[23]
where the short rate is defined as
Other interest rate models
The other major framework for interest rate modelling is the Heath–Jarrow–Morton framework (HJM). Unlike the short rate models described above, this class of models is generally non-Markovian. This
makes general HJM models computationally intractable for most purposes. The great advantage of HJM models is that they give an analytical description of the entire yield curve, rather than just the
short rate. For some purposes (e.g., valuation of mortgage backed securities), this can be a big simplification. The Cox–Ingersoll–Ross and Hull–White models in one or more dimensions can both be
straightforwardly expressed in the HJM framework. Other short rate models do not have any simple dual HJM representation.
The HJM framework with multiple sources of randomness, including as it does the Brace–Gatarek–Musiela model and market models, is often preferred for models of higher dimension.
Models based on Fischer Black's shadow rate are used when interest rates approach the zero lower bound.
See also
Further reading
Stochastic Simulation and Applications in Finance with MATLAB Programs:
Huu Tue Huynh, Van Son Lai, Issouf Soumare -
Stochastic Simulation and Applications in Finance with MATLAB Programs explains the fundamentals of Monte Carlo simulation techniques, their use in the numerical resolution of stochastic differential
equations and their current applications in finance. | {"url":"https://codefinance.training/programming-topic/financial/longstaff-schwartz-methods/","timestamp":"2024-11-04T17:56:22Z","content_type":"text/html","content_length":"293853","record_id":"<urn:uuid:8443873d-612d-4b93-9af9-8da2c7a17ac5>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00089.warc.gz"} |
Re: QUARTILE.EXC
05-03-2020 09:29 AM
05-03-2020 15:22 PM - last edited 05-03-2020 15:25 PM
In my recent quest to create or catalog as many DAX equivalents for Excel functions, this just adds to the struggles with Excel's QUARTILE function. So, in numerous locations, such as this; not
that...God forbid, Quora is authoritative about anything, but in mulitiple places there are two documented methods of calculating 1st and 3rd quartiles and they all seem to essentially agree about
the general nature of those two methods. So I implemented the exclusive method here but the results seem to come out to more along the lines of the inclusive method, not that I trust Excel's
calculations of those either. So, who knows at this point. Here it is though.
QUARTILE.EXC =
VAR __Values = SELECTCOLUMNS('Table',"Values",[Column1])
VAR __Quart = MAX('Quartiles'[Quart])
VAR __Median = MEDIANX(__Values,[Values])
VAR __Count = COUNTROWS(__Values)
VAR __Quartile =
VAR __Median = IF(
MEDIANX(FILTER(__Values,[Values] < __Median),[Values]),
MEDIANX(FILTER(__Values,[Values] <= __Median),[Values])
RETURN __Median,
VAR __Median = IF(
MEDIANX(FILTER(__Values,[Values] > __Median),[Values]),
MEDIANX(FILTER(__Values,[Values] >= __Median),[Values])
RETURN __Median
All I can say is that apparently either everyone else in the entire world (as far as I can find) is wrong about how to calculate quartiles or...maybe I am missing something.
Something else that bugs me, all of the documentation on QUARTILE.INC, QUARTILE.EXC, PERCENTILE.INC, PERCENTILE.EXC all focus on the "inclusive/exclusive" part about the kth values from 0..1. Except
that seems like the least important part to me because there are clearly different methods going on here in terms of how these functions compute the quartiles/percentiles because you can get very
different answers, especially when dealing with even numbers of items. The fact that you can't use 0 and 1 in one of them seems like the last thing that you would want to explain but rather explain
why the calculated values are different?
And another thing with regard to the "interpolation", apparently that is why the numbers generated for the 1st and 3rd quartiles in Excel varies from the way everybody else does it so how exactly is
this interpolation happening and why is it better or worse than the way everyone else seems to do it?
05-03-2020 03:22 PM
05-03-2020 04:12 PM | {"url":"https://community.fabric.microsoft.com/t5/Quick-Measures-Gallery/QUARTILE-EXC/m-p/1064566","timestamp":"2024-11-14T09:20:56Z","content_type":"text/html","content_length":"316600","record_id":"<urn:uuid:b5825d44-076a-45b2-ade6-978d5eec551b>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00365.warc.gz"} |
What is the formula of Shannon-Hartley Theorem?
log2(1+P/N). Formula (1) is also known as the Shannon–Hartley formula, and the channel coding theorem stating that (1) is the maximum rate at which information can be transmitted reliably over a
noisy communication channel is often referred to as the Shannon–Hartley theorem (see, e.g., [4]).
Which of the following is correct formula of Shannon capacity?
Shannon’s formula C = 12log(1+P/N) is the emblematic expression for the information capacity of a communication channel.
What is Shannon channel capacity theorem?
The Shannon capacity theorem defines the maximum amount of information, or data capacity, which can be sent over any channel or medium (wireless, coax, twister pair, fiber etc.). What this says is
that higher the signal-to-noise (SNR) ratio and more the channel bandwidth, the higher the possible data rate.
What is Hartley’s Law information capacity?
The Shannon-Hartley Capacity Theorem, more commonly known as the Shannon-Hartley theorem or Shannon’s Law, relates the system capacity of a channel with the averaged received signal power, the
average noise power and the bandwidth.
What is Shannon’s theory?
In information theory, the noisy-channel coding theorem (sometimes Shannon’s theorem or Shannon’s limit), establishes that for any given degree of noise contamination of a communication channel, it
is possible to communicate discrete data (digital information) nearly error-free up to a computable maximum rate through …
What is Shannon-Hartley theorem in digital communication?
In information theory, the Shannon–Hartley theorem tells the maximum rate at which information can be transmitted over a communications channel of a specified bandwidth in the presence of noise. The
law is named after Claude Shannon and Ralph Hartley.
How is Shannon theorem different from Nyquist’s theorem?
Nyquist’s theorem specifies the maximum data rate for noiseless condition, whereas the Shannon theorem specifies the maximum data rate under a noise condition. The Nyquist theorem states that a
signal with the bandwidth B can be completely reconstructed if 2B samples per second are used.
What is the mathematical expression of Hartley’s law?
Hartley’s name is often associated with it, owing to Hartley’s rule: counting the highest possible number of distinguishable values for a given amplitude A and precision ±D yields a similar
expression C 0 = log(1 + A/D).
What is the Shannon theory?
The Shannon theorem states that given a noisy channel with channel capacity C and information transmitted at a rate R, then if. there exist codes that allow the probability of error at the receiver
to be made arbitrarily small.
What did Shannon invent?
Shannon is credited with the invention of signal-flow graphs, in 1942. He discovered the topological gain formula while investigating the functional operation of an analog computer. For two months
early in 1943, Shannon came into contact with the leading British mathematician Alan Turing.
Why is Shannon’s theorem so important in information theory?
Shannon’s theorem has wide-ranging applications in both communications and data storage. This theorem is of foundational importance to the modern field of information theory. This means that,
theoretically, it is possible to transmit information nearly without error at any rate below a limiting rate, C.
The Shannon–Hartley theorem establishes what that channel capacity is for a finite-bandwidth continuous-time channel subject to Gaussian noise. It connects Hartley’s result with Shannon’s channel
capacity theorem in a form that is equivalent to specifying the M in Hartley’s line rate formula in terms…
The Shannon–Hartley theorem establishes what that channel capacity is for a finite-bandwidth continuous-time channel subject to Gaussian noise.
What does Shannon-Hartley tell you about data rate?
Shannon-Hartley tells you that you can reduce data rate to get better range (in theory without limit). At this limit, it costs a fixed amount of power to get a bit through – so every dB of data rate
reduction buys you 1 dB of receive sensitivity.
Which is an application of the noisy channel coding theorem?
It is an application of the noisy-channel coding theorem to the archetypal case of a continuous-time analog communications channel subject to Gaussian noise. | {"url":"https://greatgreenwedding.com/what-is-the-formula-of-shannon-hartley-theorem/","timestamp":"2024-11-02T01:42:34Z","content_type":"text/html","content_length":"46143","record_id":"<urn:uuid:44bbb909-8fff-406b-85c8-cd7f8259cdb2>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00240.warc.gz"} |
Dialog Box
Find it here:
MOUS_ICO Menu: Noise > Calculation points > Dialog box
nco: Select a calculation point
MOUS_ICO Popup menu: Dialog box
This list gives an overview of all defined calculation points. For each calculation point, the following properties are shown: identity (Calculation point), number of floors (#Floors) in which noise
levels are to be calculated, relative or absolute height (Height) of the lowest calculation point, whether the calculation point is active or passive (A/P). and if the calculation point is attached
to a building (a * in Att.).
To separate relative from absolute heights for calculation points, a + will be written in front of the height value for relative heights.
In this list there may be selected several calculation points at a time for the functions Set heights, Delete, Active, Passive, and Show selection. If you want to change a calculation point, only one
point may be selected.
• New point - The dialog box New calculation point appears (see New).
• Several points - The dialog box Several new calculation points appears (see New).
• Grid - The dialog box Grid appears (see Grid).
• Change - The dialog box Change calculation point appears (see New).
• Set heights - By clicking this button, a dialog box appears, in which we may enter new heights for each floor for the selected calculation points. This function is especially relevant in the case
that we are going to run the calculation with the same heights in several points. We may also enter more floors or delete floors in this dialog box.
• Reset text - See Reset text.
• Delete - All selected calculation points are deleted.
• Active - Sets the selected calculation points active. Only active calculation points are calculated during noise calculation.
• Passive - Sets the selected calculation points passive. Passive calculation points are not calculated during noise calculation.
• Show selection < - All dialog boxes are closed, and all calculation points selected in the list will be highlighted in red on the screen. We return to the dialog box by pressing <Enter> or <Esc>.
• + The height of the window in the dialog box showing the list of calculation points is increased.
• - The height of the window in the dialog box showing the list of calculation points is decreased.
• Number - states the total number of calculation points defined in Novapoint Noise.
• In list - the number of calculation points in the filtered list.
• Selected - tells how many of the calculation points in the (filtered) list are selected.
• Select all - All calculation points are selected.
• Invert - All calculation points that are not selected are selected, and vice versa.
• Reset - All selections are canceled.
• From screen < - All dialog boxes are closed, and we may select one or more calculation points by selecting them from the drawing. We get back to the dialog box by pressing <Enter>. The
calculation points that are selected from the drawing will now be selected in the list.
AutoCAD does not allow more than 255 lines in a list to be selected at a time. If more than 255 objects are selected, a message will appear at the bottom of the screen, that the list cannot be used
directly. This means that more than 255 objects cannot be selected by clicking the list and that the functions Select All, Invert, Reset, and/or From screen have to be applied when more than 255
objects are to be selected. We cannot see from the list which objects are selected, but Selected will show the number, and the functions for the relevant objects do work!
In the box Filter, we may specify which calculation points are to be shown in the list.
• Status - Chooses which calculation points are to be shown in the list. The choices are:
□ Active and passive: All calculation points
□ Only active - Only active calculation points
□ Only passive - Only passive calculation points
□ Only calculated - Only calculation points that at least once have been calculated for the current alternative
□ Only uncalculated - Only calculation points that have never been calculated for the current alternative
• Identity - A filter based on identity, by stating certain names. The filtering by name follows AutoCAD's wildcard specifications (see Wildcard Characters in the AutoCAD help file).
• “Point*” - all entities with identity starting with “Point”, no matter what comes after
• “Point?1*” - all entities with identity having exactly one character between “Point” and “1”, e.g.: “Point01”, “PointA123”, “Point 1”, but not “Point321”.
• OK - The dialog box is closed. | {"url":"https://novapointhelp.trimble.com/Noise/Calculation-Points/Dialog-Box","timestamp":"2024-11-14T17:27:07Z","content_type":"text/html","content_length":"627427","record_id":"<urn:uuid:9a459efb-4b35-4863-89f7-9d571d640344>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00533.warc.gz"} |
Circle Misalignment | Tangram Vision Documentation
Version: 11.0
Created by: All Camera-LiDAR pairs
Circle misalignment is a metric unique to MetriCal. It's an artifact of the way MetriCal bridges the two distinct modalities of camera (primarily 2D features and projection) and LiDAR (3D point
clouds in Euclidean space).
The design of the circular target is the key. The target starts as a regular camera fiducial like a ChArUco board. That board is then sized into a circle of a known diameter and outlined with
retroreflective material, like tape. This design allows both cameras and LiDAR to pick up the fiducial via its sensing modality.
There is one circle misalignment metric group for every circular target in use.
Circle misalignment metrics contain the following fields:
Field Type Description
metadata A common metadata object The metadata associated with the point cloud this circle target was measured in.
object_space_id UUID The UUID of the object space that was being observed.
measured_circle_center An array of 3 float values The X/Y/Z coordinate of the center of the circle target, in the LiDAR coordinate frame
world_extrinsics_component_ids An array of UUIDs The camera UUIDs for each world extrinsic in world_extrinsics
world_extrinsics An array of world extrinsics objects The world pose (camera from object space) that correspond to each circle_center_misalignment
circle_center_misalignment An array of circle center coordinates The errors between the circle center location estimated between each camera and the observed LiDAR
circle_center_rmse Float The circle center misalignment RMSE over all world extrinsics.
The "Center"
Much of the circle misalignment metrics are about bridging the gap between the two modalities.
• The center of the circle in camera space is the center of the ChArUco board, given its metric dimensions
• The center of the circle in LiDAR space is the centroid of the planar 2D circle fit from the points detected on the ring of retro-reflective tape.
The measured_circle_center above is this LiDAR center; the circle_center_misalignment is the error between that LiDAR circle center and the circle center estimated from each camera. This might seem
straightforward, but there's a bit more to it than that.
Since there are no commonly observable features between cameras and LiDAR, MetriCal has to use a bit of math to make calibration work. Think of the object circle center as our origin; we'll call this
$C^O$. The LiDAR circle center is that same point, but in the LiDAR coordinate frame:
$C^L = \Gamma^{L}_{O} \cdot C^O$
...and every camera has its own estimate of the circle center w.r.t the camera board, $C^C$:
$C^C = \Gamma^{C}_{O} \cdot C^O$
We can relate these centers to one another by using the extrinsics between LiDAR and Camera, $\Gamma^{L}_{C}$:
$\hat C^L = \Gamma^{L}_{C} \cdot C^C = \Gamma^{L}_{C} \cdot \Gamma^{C}_{O} \cdot C^O$
With both $C^L$ and $\hat C^L$ in the LiDAR coordinate frame, we can calculate the error between the two and get our circle_center_misalignment:
$ccm = C^L - \hat C^L$
$\Gamma^{C}_{O}$ is what is referred to when we say world_extrinsics, and the world_extrinsics_component_ids designate what camera that extrinsic relates to. MetriCal calculates these for every pair
of synced camera-LiDAR observations. | {"url":"https://docs.tangramvision.com/metrical/11.0/core_concepts/results/residual_metrics/circle_misalignment/","timestamp":"2024-11-04T08:55:52Z","content_type":"text/html","content_length":"46355","record_id":"<urn:uuid:fa2ee2e4-c712-47f1-9918-c70013065e00>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00741.warc.gz"} |
A School Algebra
CHAPTER 1
III 20
DIVISION 48
SIMPLE EQUATIONS 59
FACTORS 84
FRACTIONS 120
FRACTIONAL EQUATIONS 144
THEORY OF EXPONENTS 217
RADICAL EXPRESSIONS 225
XVII 243
XVIII 249
XX 287
XXII 310
PROPERTIES OF SERIES 323
XII 179
INEQUALITIES 194
XIV 200
BINOMIAL THEOREM 330
LOGARITHMS 342
GENERAL REVIEW EXERCISE 356
Popular passages
The equation ad = be gives a — -£, b = — ; so that an d с extreme may be found by dividing the product of the means by the other extreme ; and a mean may be found by dividing the product of the
extremes by the other mean.
If the product of two numbers is equal to the product of two others, either two may be made the extremes of a proportion and the other two the means. For, if ad = be, then, dividing by bd, ad_ be
bd~bd' ac or j- — -
If the number is less than 1, make the characteristic of the logarithm negative, and one unit more than the number of zeros between the decimal point and the first significant figure of the given
The distance a body falls from rest varies as the square of the time it is falling.
It becomes necessary in solving an equation to bring all the terms that contain the symbol for the unknown number to one side of the equation, and all the other terms to the other side. This is
called transposing the terms. We will illustrate by examples : 1. Find the number for which x stands when...
The least common multiple of two or more numbers is the least number that is exactly divisible by each of them.
It will be seen that this third term is the square of the quotient obtained from dividing the second term by twice the square root of the first term.
To reduce a fraction to its lowest terms. A Fraction is in its lowest terms when the numerator and denominator are prime to each other. 1. Reduce - to its lowest terms.
Given that the area of a circle varies as the square of its radius...
NOTE. It is important to notice in the above examples that the terms of the quotient are all positive when the divisor is a — b, and alternately positive and negative when the divisor is a + b...
Bibliographic information | {"url":"https://books.google.com.jm/books/download/A_School_Algebra.pdf?id=9X8AAAAAMAAJ&hl=en&capid=AFLRE73_F_VRFAIEwVb_JG7cs7s4lOhic1dtOFq21AqScE1Lx6YEbCrEMMdYyhRyYLgpXx4oF6qaoYXd0exZcE6N4PpuhpMCbg&continue=https://books.google.com.jm/books/download/A_School_Algebra.pdf%3Fid%3D9X8AAAAAMAAJ%26output%3Dpdf%26hl%3Den","timestamp":"2024-11-13T08:15:49Z","content_type":"text/html","content_length":"63426","record_id":"<urn:uuid:0cd38d82-1869-4b6a-b2e0-d92ddc1faa59>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00322.warc.gz"} |
BACKGROUND: Population pharmacokinetic evaluations have been widely used in neonatal pharmacokinetic studies, while machine learning has become a popular approach to solving complex problems in the
current era of big data.
OBJECTIVE: The aim of this proof-of-concept study was to evaluate whether combining population pharmacokinetic and machine learning approaches could provide a more accurate prediction of the
clearance of renally eliminated drugs in individual neonates.
METHODS: Six drugs that are primarily eliminated by the kidneys were selected (vancomycin, latamoxef, cefepime, azlocillin, ceftazidime, and amoxicillin) as 'proof of concept' compounds. Individual
estimates of clearance obtained from population pharmacokinetic models were used as reference clearances, and diverse machine learning methods and nested cross-validation were adopted and evaluated
against these reference clearances. The predictive performance of these combined methods was compared with the performance of two other predictive methods: a covariate-based maturation model and a
postmenstrual age and body weight scaling model. Relative error was used to evaluate the different methods.
RESULTS: The extra tree regressor was selected as the best-fit machine learning method. Using the combined method, more than 95% of predictions for all six drugs had a relative error of < 50% and the
mean relative error was reduced by an average of 44.3% and 71.3% compared with the other two predictive methods.
CONCLUSION: A combined population pharmacokinetic and machine learning approach provided improved predictions of individual clearances of renally cleared drugs in neonates. For a new patient treated
in clinical practice, individual clearance can be predicted a priori using our model code combined with demographic data. | {"url":"http://mymedr.afpm.org.my/search?q=mesh_term%3A%28%22Models%2C+Biological%22%29","timestamp":"2024-11-10T18:36:02Z","content_type":"text/html","content_length":"67450","record_id":"<urn:uuid:28643213-ad39-45b8-997b-d10c3142565f>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00337.warc.gz"} |
Noncommutative Geometry
Noncommutative geometry is the geometric approach to the study of noncommutative algebra, which finds its roots in mathematical physics, representation theory of groups, and the study of singular
spaces from the world of differential geometry.
Our focus is primarily on noncommutative metric geometry, where we study quantum metric spaces, i.e. noncommutative generalizations of the algebra of Lipschitz functions over metric spaces.
Our purpose is to develop a geometric framework for the study of quantum metric spaces which arise from various fields, such as mathematical physics, dynamical systems, differential geometry and
more. A key tool in this framework is a generalization of the Gromov-Hausdorff distance to the noncommutative realm, which enables the exploration of the topology and geometry of classes of quantum
metric spaces. We thus become able to construct finite dimensional approximations for C*-algebra, establish the continuity of various families of quantum metric spaces and associated structures, and
investigate questions from mathematical physics and C*-algebra theory from a new perspective inspired by metric geometry.
Noncommutative algebra studied by noncommutative geometers typically fit within the realm of functional analysis, i.e., the analysis of infinite dimensional topological vector spaces and related
concepts. The techniques used in their study borrow from differential geometry, algebraic and differential topology, topological group theory, abstract harmonic analysis and metric geometry. | {"url":"https://science.du.edu/research/project/noncommutative-geometry","timestamp":"2024-11-02T21:09:44Z","content_type":"text/html","content_length":"56917","record_id":"<urn:uuid:03ab1fdc-aec2-467a-847c-1172573e37cd>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00264.warc.gz"} |
How to Convert From Kelvin Temperatures to Degrees Celsius | Synonym
How to Convert From Kelvin Temperatures to Degrees Celsius
How to convert from Kelvin temperatures to degrees Celsius. I'm Bon Crowder and we're talking about taking Kelvin temperatures and converting to degrees Celsius. So first thing to note is Kelvin is a
scale that measure temperature just like Celsius and Fahrenheit are but we don't use the little circle for it and we don't actually say Kelvin temperature. We just say Kelvin, they are very specific
about this in the science world. So what we note is that 0 degrees Celsius is 273.15 Kelvin. Now, you might remember that whole conversion from Celsius to Fahrenheit, it's kind of crazy, it's got
some fractions in it like 5/9ths or 9/5ths and then you add or subtract 32 and it's all crazy. Celsius and Kelvin is a whole lot easier because it's just additive, 1 degree Celsius is 274.15 Kelvin,
2 degrees Celsius is 275.15 Kelvin. So you just go up one for each degrees Celsius, you go up 1 in Kelvin which means if you want to go the other way we subtract 273.15. So here we have something
that's 300 degrees Kelvin and we're like what does that really mean? So to convert you do the 300 degrees Kelvin and subtract 273.15. Now we've got ourselves a nice little decimal subtraction
problem, 5 and 5 is 10, 1, 8, 10, 6, 7, 8, 2 so our answer is 26.85 degrees Celsius, a little warmer than room temperature. I'm Bon Crowder and that's how you convert Kelvin to degrees Celsius. | {"url":"https://classroom.synonym.com/convert-kelvin-temperatures-degrees-celsius-10126.html","timestamp":"2024-11-03T09:17:03Z","content_type":"text/html","content_length":"235127","record_id":"<urn:uuid:c710e28f-6ce2-41e5-980e-1d811661ec04>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00448.warc.gz"} |
Sun Energy Reloaded (or: Make it Look Nicer)
Fakes, Samples
Sun Energy Reloaded (or: Make it Look Nicer)
This diagram of sun radiation being absorbed and reflected when hitting earth (from Solar Energy Facts website) is a rather weak remake of the original Nasa diagram.
I find the floating powerpointish arrows kind of disturbing, and with the arrow magnitudes not to scale, would even call it misleading. Took the time to prepare two new versions of it (actually I am
beta testing the new version 2.0 of e!Sankey at the moment – so this was a nice little test case).
The first version sticks more to the original idea of the diagram shown above, but the arrow magnitudes are corrected and to scale.
The second version is closer to the original ‘Breakdown of the incoming Solar Energy’ diagram by User A1 that can be found on Wikicommons. The latter one has the flow for energy being absorbed by
atmosphere (33 PW) branching off as the first arrow horizontally.
1 Comment
1. The remake is very handsome, a definite improvement graphically. Unfortunately, I think the numbers are wrong. Chasing back to the original original (NASA), which is in terms of percentages, I
think the right reading is that every percentage is in relation to the original total solar input. So the total outgoing energy, reflected, re-emitted and all that, should total the same as the
total incoming. And that has to be happening physically, within a tiny fraction, or the Earth would be steadily heating in a dramatic way. | {"url":"https://www.sankey-diagrams.com/sun-energy-reloaded-or-make-it-look-nicer/","timestamp":"2024-11-13T13:09:31Z","content_type":"application/xhtml+xml","content_length":"38997","record_id":"<urn:uuid:c1421e60-d16d-43f4-86b1-7f62352cebda>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00017.warc.gz"} |
Knowledge Science Math Expertise
About this Course
Knowledge science programs include math—no avoiding that! This course is designed to show learners the fundamental math you’ll need with the intention to achieve success in nearly any information
science math course and was created for learners who’ve fundamental math abilities however could not have taken algebra or pre-calculus. Knowledge Science Math Expertise introduces the core math that
information science is constructed upon, with no further complexity, introducing unfamiliar concepts and math symbols one-at-a-time.
• Bayes’ Theorem
• Bayesian Likelihood
• Likelihood
• Likelihood Idea
Learner Profession Outcomes
acquired a tangible profession profit from this course
Provided by
Duke College
Duke College has about 13,000 undergraduate and graduate college students and a world-class college serving to to broaden the frontiers of data. The college has a powerful dedication to making use of
information in service to society, each close to its North Carolina campus and world wide.
Syllabus – What you’ll be taught from this course
Welcome to Knowledge Science Math Expertise
This quick module contains an outline of the course’s construction, working course of, and details about course certificates, quizzes, video lectures, and different necessary course particulars. Be
sure that to learn it immediately and refer again to it each time wanted
Constructing Blocks for Downside Fixing
This module incorporates three classes which might be construct to fundamental math vocabulary. The primary lesson, “Units and What They’re Good For,” walks you thru the fundamental notions of set
principle, together with unions, intersections, and cardinality. It additionally offers a real-world software to medical testing. The second lesson, “The Infinite World of Actual Numbers,” explains
notation we use to debate intervals on the actual quantity line. The module concludes with the third lesson, “That Jagged S Image,” the place you’ll discover ways to compactly categorical an extended
collection of additives and use this ability to outline statistical portions like imply and variance.
Capabilities and Graphs
This module builds vocabulary for graphing features within the aircraft. Within the first lesson, “Descartes Was Actually Sensible,” you’ll get to know the Cartesian Aircraft, measure distance in it,
and discover the equations of strains. The second lesson introduces the concept of a operate as an input-output machine, reveals you the right way to graph features within the Cartesian Aircraft, and
goes over necessary vocabulary.
Measuring Charges of Change
This module begins a really mild introduction to the calculus idea of the by-product. The primary lesson, “That is In regards to the By-product Stuff,” will give fundamental definitions, work just a
few examples, and present you the right way to apply these ideas to the real-world drawback of optimization. We then flip to exponents and logarithms, and clarify the foundations and notation for
these math instruments. Lastly we be taught in regards to the fee of change of steady development, and the particular fixed often known as “e” that captures this idea in a single quantity—close to
Introduction to Likelihood Idea
This module introduces the vocabulary and notation of chance principle – arithmetic for the examine of outcomes which might be unsure however have predictable charges of incidence.
We begin with the fundamental definitions and guidelines of chance, together with the chance of two or extra occasions each occurring, the sum rule and the product rule, after which proceed to Bayes’
Theorem and the way it’s utilized in sensible issues.
Begin Studying At this time
You’ll be able to share your Course Certificates within the Certifications part of your LinkedIn profile, on printed resumes, CVs, or different paperwork.
About this Course
Knowledge science programs include math—no avoiding that! This course is designed to show learners the fundamental math you’ll need with the intention to achieve success in nearly any information
science math course and was created for learners who’ve fundamental math abilities however could not have taken algebra or pre-calculus. Knowledge Science Math Expertise introduces the core math that
information science is constructed upon, with no further complexity, introducing unfamiliar concepts and math symbols one-at-a-time.
• Bayes’ Theorem
• Bayesian Likelihood
• Likelihood
• Likelihood Idea
Syllabus – What you’ll be taught from this course
Welcome to Knowledge Science Math Expertise
This quick module contains an outline of the course’s construction, working course of, and details about course certificates, quizzes, video lectures, and different necessary course particulars. Be
sure that to learn it immediately and refer again to it each time wanted
Constructing Blocks for Downside Fixing
This module incorporates three classes which might be construct to fundamental math vocabulary. The primary lesson, “Units and What They’re Good For,” walks you thru the fundamental notions of set
principle, together with unions, intersections, and cardinality. It additionally offers a real-world software to medical testing. The second lesson, “The Infinite World of Actual Numbers,” explains
notation we use to debate intervals on the actual quantity line. The module concludes with the third lesson, “That Jagged S Image,” the place you’ll discover ways to compactly categorical an extended
collection of additives and use this ability to outline statistical portions like imply and variance.
Capabilities and Graphs
This module builds vocabulary for graphing features within the aircraft. Within the first lesson, “Descartes Was Actually Sensible,” you’ll get to know the Cartesian Aircraft, measure distance in it,
and discover the equations of strains. The second lesson introduces the concept of a operate as an input-output machine, reveals you the right way to graph features within the Cartesian Aircraft, and
goes over necessary vocabulary.
Measuring Charges of Change
This module begins a really mild introduction to the calculus idea of the by-product. The primary lesson, “That is In regards to the By-product Stuff,” will give fundamental definitions, work just a
few examples, and present you the right way to apply these ideas to the real-world drawback of optimization. We then flip to exponents and logarithms, and clarify the foundations and notation for
these math instruments. Lastly we be taught in regards to the fee of change of steady development, and the particular fixed often known as “e” that captures this idea in a single quantity—close to
Introduction to Likelihood Idea
This module introduces the vocabulary and notation of chance principle – arithmetic for the examine of outcomes which might be unsure however have predictable charges of incidence.
We begin with the fundamental definitions and guidelines of chance, together with the chance of two or extra occasions each occurring, the sum rule and the product rule, after which proceed to Bayes’
Theorem and the way it’s utilized in sensible issues.
Post a Comment | {"url":"https://courses.dongthinh.co.uk/2022/07/knowledge-science-math-expertise.html","timestamp":"2024-11-03T03:34:00Z","content_type":"application/xhtml+xml","content_length":"387339","record_id":"<urn:uuid:bbaa5d3d-acb7-4be8-87d4-343e5b3beb63>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00506.warc.gz"} |
The quantum (a)muse(ment) #2: mastering random numbers generation
Photo by Susan Holt Simpson on Unsplash
Is it possible to design a quantum circuit to generate random numbers in custom intervals?
Writing code for Vera Molnar-like generative art, I wanted to generate numbers in a certain range so I found myself asking: “can we shape number generation directly in the quantum circuit, instead of
drawing numbers out of a uniform distribution and discarding those which are out-of-interval?”
Actually, this is possible by designing ad-hoc quantum circuits by applying certain operations on qubits.
Let’s say that we want to generate a random number in interval [0,3].
We can achieve this by initializing 2 qubits and inducing superposition with Hadamard:
Circuit for random numbers in interval [0,3]
Now let’s say that we want to generate numbers in [2,3]. We have to apply NOT to q1 (pi rotation on X axis):
Circuit for random numbers in interval [2,3]
It’s easier to understand what is happening here if we plot Bloch sphere for each qubit:
Leftmost qubit is always going to collapse to 1 while the other is going to collapse to 0 with probability ½ or to 1 with probability ½.
This leads to:
So we prepare one qubit with ket(0) and obtain ket(1) by applying NOT operator and make other one random by applying Hadamard.
And this quickly generalizes to higher numbers, too!
Circuit for random numbers in interval [0,7]
Amusement #1: even numbers only
Let’s take a look to even numbers “pattern”:
The figure in the rightmost position should be zero, while the other two positions should collapse half the times to 0 and the other half to 1.
So we have to “prepare” the rightmost qubit and leave it to ket(0) while applying Hadamard to the other qubits, so that they can collapse with probability ½ to 0 or 1.
Circuit for even random numbers in interval [0,7]
Amusement #2: odd numbers only
In this case, last qubit should be prepared to always collapse to 1 while the others are going to collapse to 1 or 0 with probability ½.
Circuit for odd random numbers in interval [0,7] | {"url":"https://1littleendian.medium.com/the-quantum-a-muse-ment-2-mastering-random-number-generation-36384b04ee13","timestamp":"2024-11-07T02:38:54Z","content_type":"text/html","content_length":"116012","record_id":"<urn:uuid:58292c34-8a3d-4894-bdbb-e6a5872c6fd9>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00306.warc.gz"} |
liquid velocity calculator
Liquid Velocity Calculator
In engineering and fluid dynamics, calculating liquid velocity is crucial for various applications, from designing pipelines to understanding fluid behavior in machinery. To simplify this process, we
introduce a user-friendly Liquid Velocity Calculator.
How to Use
1. Input Parameters: Enter the necessary values into the provided fields.
2. Click Calculate: Press the “Calculate” button to obtain the liquid velocity.
The formula for calculating liquid velocity is:
• v is the liquid velocity (m/s)
• g is the acceleration due to gravity (m/s²)
• h is the height of the liquid above the point of measurement (m)
• d is the density of the liquid (kg/m³)
Example Solve
Let’s say we want to calculate the velocity of water flowing through a pipe. Given:
• h=10 meters
• g=9.81 m/s² (standard gravity)
• d=1000 kg/m³ (density of water)
Plugging these values into the formula:
So, the velocity of the water is approximately 4.427m/s4.427m/s.
Q: What if I don’t know the density of the liquid?
A: If the density is unknown, you can often find it in reference materials or online databases. It’s an essential parameter for accurate calculations.
Q: Can I use this calculator for gases as well?
A: No, this calculator is specifically designed for liquids. The formula and parameters differ for gases.
Q: Is there a maximum limit for the height of the liquid h?
A: There’s no theoretical limit, but practical considerations such as pump capacity and pipe design may impose constraints.
The Liquid Velocity Calculator provides a convenient way to determine the velocity of liquids in various scenarios. By inputting the necessary parameters, users can quickly obtain accurate results,
aiding in engineering design and fluid dynamics analysis. | {"url":"https://calculatordoc.com/liquid-velocity-calculator/","timestamp":"2024-11-13T21:45:30Z","content_type":"text/html","content_length":"92681","record_id":"<urn:uuid:5fcb1dd8-8bfc-4cca-840b-8643408290ac>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00576.warc.gz"} |
Applied Math
UC Berkeley / Lawrence Berkeley Laboratory
Solving Linear Systems of Equations via Randomized Kaczmarz/Stochastic Gradient Descent
Stefan Steinerberger, University of Washington
The Randomized Kaczmarz method is a classical iterative method to solve linear systems: the solution of a system Ax = b is simply the point of intersection of several hyperplanes. The Kaczmarz method
(also known as the Projection Onto Convex Sets Method) proceeds by simply starting with a point and then iteratively projecting it on these hyperplanes. If the hyperplanes (=rows of the matrix) are
picked in random order, the algorithm was analyzed by Strohmer & Vershynin and has linear convergence. We show that the method, as a byproduct, also computes small singular vectors and, in fact, the
iterates tend to approach the true solution from the direction of the smallest singular vector in a meta-stable way. This also explains why the algorithm has such wonderful regularization properties.
The arguments are all fairly self-contained, elementary and nicely geometric. This gives a pretty clear picture – the question is: can this picture be used to improve the method? | {"url":"https://berkeleyams.lbl.gov/fall20/steinerberger.html","timestamp":"2024-11-01T19:25:20Z","content_type":"application/xhtml+xml","content_length":"4152","record_id":"<urn:uuid:7c6d033e-dc58-4013-aa50-d3abf447c105>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00129.warc.gz"} |
Somalis, A NATION WITHOUT HEROs.
Originally posted by Socod_badne:
whatever he did is between him and his lord
True now that he is dead. The Lord will punish and show him no clemency for what he got away with while here on earth. If he doesn't then the Lord would be unjust. I hope and pray he rots in hell. [/
And I thought all this time you were an athreist.....which means you don't believe in the Almight Allah. :rolleyes: Allah unjust I am very offended by this, as Allah the Almighty is just and there is
no one like him mate. Why don't you go to hell with Said Barre. As you so quick to tell others to do so. :mad:
^ Easy there, nameless-chick, no need for personal attacks.
I'm deeply suspicious of anyone labelled a national hero. History books are rife with the celebratory accounts of people hooting and hollering about a country's liberty or whatever while brutalizing
their people .
Originally posted by naden:
^ Easy there, nameless-chick, no need for personal attacks.
Mate tell him not to call the Almighty Allah unjust....you offended the Almight Allah I also get offended. :mad: :mad:
^^there a different ways of defending your beliefs, telling him to go to hell is not one of them.
Originally posted by ScarFace:
ibtisam just eat your fish & chips.......coz it seems like you havent got a single clue about somalia and its history.....how can you chat breeze when you havent even been raised in the bloody
country.......the only people that can pass judgement is the ones that were there who went through the hardship and the ones that got the pleasures......for once will you stop repeating the same ole
i know more about Somalia and its history than many Somalis that grew up there. What has being raised somewhere got to do with anything? As for people passing judgments, well at the end of the day i
know injustice when i see/ hear/read one. i don't have to be in Palestine to understand the hardship. Nor do i have to have lived in Somalia to understand their hardship.
first of all did the man kill your mother,father,brother or anyone in your family.... did he abuse your clan or your relatives....... if he did then you can hate him as much as you want........
Well according to your assessment, I am entitled to hate him as much as I want. It seems after all that I don't have to be present to have suffered under his rule.
but the funniest thing is your half breed who hasnt been raised in somalia,..........and the amount of hate thats coming out of your mouth is just making me laugh.....the man is dead f...uck....he's
dead......move on......he had his time./..... he lived his time........done things that were great in his time....done things that was bad...........
And your point is?? I am not in a position to common on Somalia because YOU consider me as half breed. So what if he is dead, I hope he rots in hell. You cannot force down my throat how great
He was, he was an asshole, and even when he is dead he is still an A*ss.
move on and stop repeatin yourself....no one is portraying the man as a saint......whatever he did is between him and his lord.........BLOODY hell YOU KEEP SAYIN YOU DONT CARE ABOUT what clan he is
but i think you do.........come correct
Everyone is aware of his crimes and it is documented. As for clan, I could not care less, I hate him as a human being, same way I hate Hitler and Sharon,
Bush (and I’ve never lived under their rule either)
It makes me sick that people are defending him and making excuses for him :rolleyes:
Mate tell him not to call the Almighty Allah unjust....you offended the Almight Allah I also get offended. :mad: :mad:
Nameless sister,
Allah Most Great can not be offended…doesn’t need our defending either….we need Allah. Even tho um happy that u sister are passionate about Islam remember this hadith:
Abû Hurayrah relates that a man said to the Prophet (peace be upon him): “Counsel me.†The Prophet (peace be upon him) said: “Do not get angry.†The man repeated his request many times, but
the Prophet (peace be upon him) kept saying: “Do not get angry.†[ Sahîh al-Bukhârî ]
Ps: I guess we have no Hero…why do I think if Somalis had a mujhid like Salahudiin ibn Ayub even….they would still curse him. The brotha asked for whose your hero (personal opinion no?)…we go
on to curse a dead man Allah forgive his sins (no I am not a distant cousin, hooyo told me
Siad Barre. Gamal Abdel-Nasser. Hafiz Al-Asad.
What do these men have in common ? They were all operating on a anti-Islamic platform. During the blessed Islamic awakening of the 60's and 70', these dictators did their utmost to combat and
persecute Muslims who were slowly coming to grips with the fact that this deen is not a cultural relic to be practiced on Friday, but that it is a complete and systematic way of life. So when they
demanded that Islam be implemented on the political arena, they were mercilessly persecuted by the state and it's Gestapo forces. Whether it was by the execution of 11 scholars, or mass jailings and
subsequent torture of the Ikhwaan, or the destruction of an entire city to the ground(Hama), they spared no effort in trying to supress the Islamic awakening(the Saxwa). Their motives doesn't matter,
nor do their beliefs and proclamations. Their actions are there for all mankind to witness and judge.
These facts must be acknowledged, and the whitewashing of public figures, just because they're dead, has gotta stop.
Originally posted by Khalaf:
quote: Mate tell him not to call the Almighty Allah unjust....you offended the Almight Allah I also get offended. :mad: :mad:
Nameless sister,
Allah Most Great can not be offended…doesn’t need our defending either….we need Allah. Even tho um happy that u sister are passionate about Islam remember this hadith:
b]Abû Hurayrah relates that a man said to the Prophet (peace be upon him): “Counsel me.†The Prophet (peace be upon him) said: “Do not get angry.†The man repeated his request many times, but
the Prophet (peace be upon him) kept saying: “Do not get angry.†[ Sahîh al-Bukhârî ]
Allah help me. Leave me alone...what I can't defend my religion now? :rolleyes: And let these low life’s Atheist people mock my religion and my believe. Why are you always so quick to correct me?
Why not these people that are always saying something bad about the religion.
I have come to conclusion now “that you are part of them and not just their supportersâ€. Someone calls Allah the almighty un-just and you try to correct me when I say that is not acceptable. Do
you see where I’m getting at?
Do we have heroes? I would say we don’t have any. The ones we have are too scared to act up and speak against the inhumanity that exists in many regions in Somalia.
I find it fascinating to see people defending the religion who only do it to look good, when they know nothing about what they’re arguing – and are quick to point the finger and raise profound
irrational and unfounded accusations when they’re participation is limited to Friday prayers.
I would suggest we consider everything that is involved, Somalis love to safeguard the deen, which is a good thing, however, we must learn all facets of the religion, and what is permitted and what
is not – even those who know much are too snobbish to actually engage people rationally. Before accusing each other, we should actually layout the facts and concrete evidence and collectively reach
a decision. I think the redundancy accusing people of being atheist is getting a bit annoying. Religion is a hot topic, but we must understand that we’re all Muslim, and we are brothers and
That being said, for the past fifteen years we have accepted the random killings of innocent people based on their clan or political associations. Even now we are seeing our lesser of evils operating
inadequately – I’m talking about the institutionalized killings of minorities by using feeble procedural methods in determining their crimes, and whether they’re legitimate claims or not based
on the Shari’a. The so-called transitional government is not making progress, and tribalism continues to spread rampantly faster than AIDS in Sub Sahara Africa, tribalism also invaded us in the
Diaspora, tell why are young people using tribal politics?
Our only heroes are clan heroes, a person who has eliminated a certain number of the opposing clan. How can we have heroes when we have hate for one another, we beginning to be divided more than
ever, by clan, by religion (fanatics, secularists, and many other sub groups) Ask me if we have a nation, before you ask me whether we have heroes or not.
Originally posted by Animal Farm:
I think the redundancy accusing people of being atheist is getting a bit annoying. Religion is a hot topic, but we must understand that we’re all Muslim, and we are brothers and sisters.
I'm guessing you are referring to me. I don't just go around accusing people of being atheist unless they show it in some form they are. Because that's just not acceptable by my nature or our diin.
The reason I called some individuals in here artiest is because not only at this post but many others. Just go to the Islamic section and see how they really are. It disputed me I tell you. And why
people attacking me for calling someone artiest when some individuals called Allah the Almighty un-just? Am I the only rational person in this whole site?
PS I don’t go to Friday prayers so I guess your judgment were wrong because I’m a girl and don’t have to.
Originally posted by nameless_chick:
And why people attacking me for calling someone artiest when some individuals called Allah the Almighty un-just? Am I the only rational person in this whole site?
Hello nameless_chick. I'm the artiest formerly known as Prince. I have some good news for you: Allah, who created the heavens and the earth, hardly needs your defence. If you can show why calling the
"Almighty unjust" is wrong, do it. If you can't, put your sword back in its sheath until you figure out how.
religist <<< what?
I was just trying to say that people on SOL who constantly jump to conclusion, who relentlessly question the devotion of other members through subliminal or suggestive sayings should stop this
Originally posted by Animal Farm:
<<< what?
I was just trying to say that people on SOL who constantly jump to conclusion, who relentlessly question the devotion of other members through subliminal or suggestive sayings should stop this
I shall not STOP my opinions or thoughts because the truth hurts. You are not the person I offended so you shall not get into others business.
PS isn't it funny how the person I called Atheist doesn't mind but these entire people think they should protect him or something. :rolleyes: The reason why this person doesn't mind would be because
they know they are.
Castro.. ummm I was not referring to you but I see there is more like you than I realised, every interested I say. And all this time I thought Somalia was 99% Muslims, now who made that figure up?
^^^who made you the judge. So what if anyone is an Atheist or not. People have to decide what they want to believe and be free to do so. There is no need for you to mock them or persist in annoying
and targeting them. if you do so and they respond by offending your religion, then you have no right to be upset and YOU will take the blame for their offence, as your provoked them.
I am sure you come across atheist people everyday, it is not strange, you don't harass them, do you? You may even consider them as your friends. There is no reason why you should treat them so well
simply because they happen to be non-Somali atheist and be horrid to Somali "non believers". IN any case no one has to justify themselves and there believes to a bunch of User names, or answer
If you had a disagreement with someone in the Islam section or in another thread. Don’t transfer that disagreement to other threads, simply because you assume something of that user name.
Now leave people's believes alone. this was not about what people believe.
p.s. I don't mean to be rude and I mean it in the nicest possible way
Holy! - is it necessary to gang up on the girl because she was offended by something someone said and called him an atheist. If that person was offended he will take it up with nameless. :eek: | {"url":"https://www.somaliaonline.com/community/topic/19300-somalis-a-nation-without-heros/page/6/","timestamp":"2024-11-04T10:32:16Z","content_type":"text/html","content_length":"302010","record_id":"<urn:uuid:1a3f8cde-5777-4c45-a743-47a787f1fb98>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00079.warc.gz"} |
Seminar in Numerical Analysis: Martin Vohralik (Inria Paris)
08 Dec 2023
11:00 - 12:00
Seminar in Numerical Analysis: Martin Vohralik (Inria Paris)
A posteriori estimates enable to certify the error committed in a numerical simulation. In particular, the equilibrated flux reconstruction technique yields a guaranteed error upper bound, where the
flux, obtained by a local postprocessing, is of independent interest since it is always locally conservative. In this talk, we tailor this methodology to model nonlinear and time-dependent problems
to obtain estimates that are robust, i.e., of quality independent of the strength of the nonlinearities and the final time. These estimates include, and build on, common iterative linearization
schemes such as Zarantonello, Picard, Newton, or M- and L-ones. We first consider steady problems and conceive two settings: we either augment the energy difference by the discretization error of the
current linearization step, or we design iteration-dependent norms that feature weights given by the current iterate. We then turn to unsteady problems. Here we first consider the linear heat
equation and finally move to the Richards one, that is doubly nonlinear and exhibits both parabolic–hyperbolic and parabolic–elliptic degeneracies. Robustness with respect to the final time and local
efficiency in both time and space are addressed here. Numerical experiments illustrate the theoretical findings all along the presentation. Details can be found in [1-4].
A. Ern, I. Smears, M. Vohralík, Guaranteed, locally space-time efficient, and polynomial-degree robust a posteriori error estimates for high-order discretizations of parabolic problems, SIAM J.
Numer. Anal.55 (2017), 2811–2834.
A. Harnist, K. Mitra, A. Rappaport, M. Vohralík, Robust energy a posteriori estimates for nonlinear elliptic problems, HAL Preprint 04033438, 2023.
K. Mitra, M. Vohralík, A posteriori error estimates for the Richards equation, Math. Comp. (2024), accepted for publication.
K. Mitra, M. Vohralík, Guaranteed, locally efficient, and robust a posteriori estimates for nonlinear elliptic problems in iteration-dependent norms. An orthogonal decomposition result based on
iterative linearization, HAL Preprint 04156711, 2023.
For further information about the seminar, please visit this webpage.
Export event as iCal | {"url":"https://dmi.unibas.ch/en/news-events/past-events/detail/seminar-in-numerical-analysis-martin-vohralik-inria-paris/","timestamp":"2024-11-13T09:51:00Z","content_type":"text/html","content_length":"23718","record_id":"<urn:uuid:eea7650c-4fc0-4376-b621-47db317b5731>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00742.warc.gz"} |
Nonlinear Continuum Mechanics
Mandatory for M.Sc. Materials Science and Engineering
(3 SWS, 5 ECTS, winter term, Modul-Nr. MW2368, Prof. M.W. Gee)
The understanding of material behavior under environmental influences is the key for the development or optimization of countless technical systems. At the same time, materials show a high variety of
different behavior patterns, which makes sophisticated modeling of materials very complex. Nonlinear Continuum Mechanics is a mathematical theory that has proven its worth in the description of
material behavior to environmental influences in case of real world engineering problems.
This course provides an introduction in the mathematical theory of continuum mechanics. Starting from the description of the motion of a continuum, measures for stretch and strain as well as the
corresponding rates are derived consistently. The concept of stress is introduced and applied in the derivation of balance laws for mass, momentum, energy and entropy. The relation between stress and
strain is established by the derivation of constitutive laws for hyperelastic, viscoelastic and plastic material behavior, so that the presented mathematical theory is self-contained and serves as a
basis for further specification or even direct application in numerical simulations.
• Introduction of the terms continuum and continuum mechanics
• Brief review of tensor calculus
• Kinematics (deformation map, deformation gradient, stretch and strain, strain rate)
• Stress (Piola-Kirchhoff stress, Cauchy stress, stress states)
• Balance laws (mass, momentum, energy, entropy)
• Constitutive laws (hyperelasticity, viscoelasticity, plasticity)
• Knowledge of the axioms of Newtonian mechanics and thermodynamics, tensor calculus in cartesian bases, basic knowledge of at least linearized continuum mechanics, basic knowledge of linear
ordinary and partial differential equations as well as linear algebra
Prof. Gee
Mi, 09:00-10:30, 8102.03.108
Office hours Prof. Gee: by appointment
Mr. Rinderer
Mi, 10:45-11:30, 8102.03.108
Office hours Mr. Rinderer: by appointment | {"url":"https://www.epc.ed.tum.de/mhpc/education/nonlinear-continuum-mechanics/","timestamp":"2024-11-09T07:37:36Z","content_type":"text/html","content_length":"33250","record_id":"<urn:uuid:e9e674bf-709f-45da-a4dd-3ea6d35d79b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00028.warc.gz"} |
Understanding How to Calibrate a Circuit to Measure Resistance
Question Video: Understanding How to Calibrate a Circuit to Measure Resistance Physics • Third Year of Secondary School
A circuit that can be used as an ohmmeter is shown. The circuit uses a galvanometer, a direct-current source with a known voltage, a fixed resistor, and a variable resistor. Which of the following
most correctly states how to calibrate the circuit to directly measure the circuit’s total resistance?
Video Transcript
A circuit that can be used as an ohmmeter is shown. The circuit uses a galvanometer, a direct-current source with a known voltage, a fixed resistor, and a variable resistor. Which of the following
most correctly states how to calibrate the circuit to directly measure the circuit’s total resistance? (A) Adjust the resistance of the variable resistor until it is equal to the mean of the
resistance of the fixed resistor and the resistance of the galvanometer. (B) Adjust the resistance of the variable resistor until it is equal to the sum of the resistance of the fixed resistor and
the resistance of the galvanometer. (C) Adjust the resistance of the variable resistor until it is equal to the difference between the resistance of the fixed resistor and the resistance of the
galvanometer. (D) Adjust the resistance of the variable resistor until the galvanometer’s arm is at a full-scale deflection position. (E) Adjust the resistance of the variable resistor until the
galvanometer’s arm is at the zero-deflection position.
In this example, our circuit consisting of a galvanometer, a device for measuring current, a fixed resistor here, a variable resistor here, and a constant voltage supply here is intended to work as
an ohmmeter, a device for measuring resistance.
The idea then is if we had a resistor of unknown resistance, we can connect it to this circuit and then solve for its resistance. For that to work though, as our problem statement tells us, this
circuit needs to first be calibrated. This is possible because the circuit involves a variable resistor. This is a resistor whose resistance can be changed. And we see that all of our answer options,
including option (A) that we can’t see on the screen right now, depend on this step of adjusting the resistance of this variable resistor.
So the question is, how do we adjust this resistor’s resistance so that the circuit does indeed function as an ohmmeter? The answer has to do with our galvanometer, the device in this circuit for
measuring current. Typically, a galvanometer reads out the measured current on a scale, like this. The scale has a minimum value of current, zero, and some maximum value that we’ve called 𝐼 sub 𝑔.
The galvanometer is capable of accurately measuring current anywhere between and including these values.
As we’ve seen though, we don’t want to use our circuit to measure current, but rather resistance. We can still do this using the current setup by making use of Ohm’s law. This law tells us that the
potential difference across a circuit is equal to the current in that circuit multiplied by the total circuit resistance.
In our circuit, we have a fixed potential difference supply due to this cell. Given that the voltage 𝑉 is fixed, we can think of the current 𝐼 and the total resistance 𝑅 in this circuit as balancing
one another out. That is, if due to a change in the resistance of the variable resistor, the overall resistance of our circuit was increased, then Ohm’s law would dictate that the current in the
circuit must decrease so that 𝐼 times 𝑅 is still equal to this fixed voltage 𝑉. The same thing holds true if the total resistance decreases; then the current 𝐼 would increase in response.
Now, if we look back at our answer options, several of them, including options (B) and (C), describe adjusting the resistance of the variable resistor based on a relationship between the resistance
of the fixed resistor and that of the galvanometer. This shows us that in our circuit, it’s not just the fixed and the variable resistors that have some resistance. The galvanometer does as well,
while for the cell, we’re treating it as an ideal voltage supply with effectively no resistance.
Speaking though of the resistances of the fixed resistor and the galvanometer, option (C) says that the resistance of the variable resistor should be adjusted until it is equal to the difference
between these two resistances. Option (B) says that the variable resistor should be adjusted until its resistance is equal to the sum of these two resistances. And if we were to import a shortened
version of answer option (A), this option says that we should adjust the resistance of the variable resistor until it is equal to the mean or the average of the resistances of a fixed resistor and
the galvanometer. However, the value to which we should adjust the resistance of our variable resistor actually has nothing to do with the resistance of our fixed resistor and rather depends only on
the current that is measured by our galvanometer.
Our circuit, recall, is intended to measure the resistance of an unknown resistor. The only way you can do this is if there’s some kind of measurement scale we can use that will indicate this
resistance. Only answer choices (D) and (E) describe adjusting the resistance of the variable resistor with reference to the galvanometer’s scale. These then are our only viable options for
developing a measurement device for resistance. We can cross out then answer choices (A), (B), and (C) and then clear them from the screen.
We’ve seen that both answer choices (D) and (E) involve adjusting the resistance of the variable resistor. Let’s look now at what makes these choices different. Answer choice (E) involves adjusting
the resistance of the variable resistor until the galvanometer’s arm is at the zero-deflection position. That’s the position that we see here in our sketch. An important thing to realize about a
zero-deflection position is that that means that effectively zero current exists in the circuit. But for zero current to exist in the circuit, that would mean that 𝑅 would need to be effectively
infinitely big. In other words, we would have to adjust the resistance of our variable resistor to a very, very high value. This may not be practical.
But actually there’s another reason why we wouldn’t want our galvanometer’s arm to be at a zero-deflection position when we use it for measuring resistance. Let’s say that we somehow had adjusted the
resistance of our variable resistor so that the arm on our galvanometer indeed indicated zero current. If we were to take our unknown resistor whose resistance we want to measure and added this
resistor to the circuit, then that addition, which would cause the overall resistance of the circuit theoretically to increase, would have no measurable impact on the current measured by our
galvanometer. That current was already zero, so it couldn’t possibly go any lower.
This would prevent us from working back from the galvanometer’s reading to solve for the resistance of our unknown resistor. By calibrating our circuit according to answer option (E) then, we
couldn’t use it as a device for measuring resistance, an ohmmeter.
But let’s say that instead we adjusted the resistance of the variable resistor in our circuit so that the arm on our galvanometer was moved to its full-scale deflection position. Note that we would
do this without the resistor whose resistance we want to measure being inserted into the circuit. So now, prior to serving as an ohmmeter, we’ve tuned the overall resistance of our circuit 𝑅 so that
the current in the circuit 𝐼 is equal to the maximum measurable current according to our galvanometer. Set up this way, if we now inserted our unknown resistor into the circuit, that would cause the
overall circuit resistance to increase since we’ve added our resistor in series, which would result in some measurable decrease in the overall current in the circuit, where that decrease is measured
by our galvanometer.
If the change in measured current in the circuit due to the addition of our unknown resistor is Δ𝐼, then we could use that measured change along with the voltage in our circuit to calculate a change
in resistance in the circuit Δ𝑅. It’s this change in resistance Δ𝑅 that is equal to the resistance of our unknown resistor. This is how our circuit can function as an ohmmeter. And so for our answer,
we choose option (D).
The question may come up though. When we tune the resistance of our variable resistor, why do we do it so that the arm on the galvanometer is at a full-scale deflection position? Why don’t we do it,
for example, so that the galvanometer’s arm is at a half-scale deflection position or some other value? The reason we set up the resistance of the variable resistor so that the galvanometer’s arm is
at the full-scale deflection position is that this allows us a maximum measurement range for the resistance of an unknown resistor inserted into the circuit. If instead our galvanometer arm were
calibrated so that it’s at half the full-scale position, then we would only have one-half of the galvanometer’s scale to work with as we measured unknown resistances.
Better than this is having the full scale of the galvanometer to work with. And this is why, as answer option (D) says, we adjust the resistance of the variable resistor until the galvanometer’s arm
is at a full-scale deflection position. | {"url":"https://www.nagwa.com/en/videos/967191320382/","timestamp":"2024-11-14T07:55:22Z","content_type":"text/html","content_length":"268759","record_id":"<urn:uuid:637ef207-8359-4816-b5ce-2a3deb69deeb>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00854.warc.gz"} |
European Alternative Fuels Observatory
Recharging and refuelling stations map
TENtec interactive mapTENtec is the European Commission’s Information System to coordinate and support the Trans-European Transport Network Policy (TEN-T). It focuses on policy-related information
by storing and managing technical, geographical and financial data for the analysis, assessment and political decision-making related to TEN-T and the underlying funding programme, the Connecting
Europe Facility (CEF).
The TENtec Geographic Information System displays the core and comprehensive TEN-T network, as well as related transport infrastructure information, on an interactive map. With the support of EAFO,
this interactive map now contains a comprehensive and up-to-date overview of the alternative fuels infrastructure deployed in the EU and neighbouring countries. It also contains a basic gap analysis,
which allows the identification of gaps in the alternative fuels infrastructure coverage on the TEN-T network. In the future, it will allow an even more sophisticated data analysis, to superimpose
alternative fuels infrastructure on different thematic layers such as traffic flows, population density, air quality etc.
Visit TENtec interactive map
1. EU alternative fuels fleet
The reference year and month for the graphs and maps in this section is the same as for road transport country reports.
1.1 Absolute numbers
This sub-section shows the total alternative fuels (AF) fleet numbers in the EU Member States in the reference year, and, where such data is available, compares these against the total vehicle fleet
numbers in those Member States.
AF passenger cars and vans
Fleet of alternative fuels (BEV, PHEV, H2, LPG, CNG, LNG) passenger cars and vans (M1+N1)
AF passenger cars and vans
Percentage of the total passenger cars and vans fleet (M1+N1) that is powered by alternative fuels (BEV, PHEV, H2, LPG, CNG, LNG)
AF buses
Fleet of alternative fuels (BEV, PHEV, H2, LPG, CNG, LNG) buses (M2 + M3)
AF trucks
Fleet of alternative fuels (BEV, PHEV, H2, LPG, CNG, LNG) trucks (N2 + N3)
BEV passenger cars and vans
Fleet of battery-electric (BEV) passenger cars and vans (M1+N1)
BEV% passenger cars and vans
Percentage of the total passenger cars and vans fleet (M1+N1) that is full battery-electric (BEV)
PHEV passenger cars and vans
Fleet of plug-in hybrid electric (PHEV) passenger cars and vans (M1+N1)
PHEV% passenger cars and vans
Percentage of the total passenger cars and vans fleet (M1+N1) that is plug-in hybrid electric (PHEV)
H2 passenger cars and vans
Fleet of hydrogen (H2) passenger cars and vans (M1+N1)
H2% passenger cars and vans
Percentage of the total passenger cars and vans fleet (M1+N1) that is powered by hydrogen (H2)
1.2 Growth rates (2021 vs. 2020)
This sub-section compares alternative fuels fleet numbers in the reference year 2021, to alternative fuels fleet numbers in the preceding year 2020, and, where such data is available, compares these
alternative fuels fleet growth rates against the overall growth rates of vehicles in the same vehicle categories.
Growth rate AF passenger cars and vans
Year-over-year growth rate of the alternative fuels (BEV, PHEV, H2, LPG, CNG, LNG) passenger cars and vans (M1+N1) fleet.
Growth rate AF buses
Year-over-year fleet growth rate of the alternative fuels (BEV, PHEV, H2, LPG, CNG, LNG) buses (M2+M3) fleet.
Growth rate AF trucks
Year-over-year growth rate of the alternative fuels (BEV, PHEV, H2, LPG, CNG, LNG) truck (N2+N3) fleet.
Growth rate BEV passenger cars and vans
Year-over-year growth rate of the BEV passenger cars and vans (M1+N1) fleet.
Growth rate BEV buses
Year-over-year growth rate of the BEV buses (M2+M3) fleet.
Growth rate BEV trucks
Year-over-year growth rate of the BEV trucks (N2+N3) fleet.
Growth rate PHEV passenger cars and vans
Year-over-year growth rate of the PHEV passenger cars and vans (M1+N1) fleet.
Growth rate PHEV buses
Year-over-year growth rate of the PHEV buses (M2+M3) fleet.
Growth rate PHEV trucks
Year-over-year growth rate of the PHEV trucks (N2+N3) fleet.
Growth rate H2 passenger cars and vans
Year-over-year growth rate of the hydrogen (H2) passenger cars and vans (M1+N1) fleet.
Growth rate H2 buses
Year-over-year growth rate of the hydrogen (H2) buses (M2+M3) fleet.
Growth rate H2 trucks
Year-over-year growth rate of the hydrogen (H2) trucks (N2+N3) fleet.
2. EU alternative fuels registrations
2.1 Absolute numbers
This sub-section shows the new registrations of alternative fuels vehicles in the Member States, and, where such data is available, compares these against the total numbers of new registrations in
the same vehicle categories in those Member States.
AF passenger cars and vans
New registrations of alternative fuels (BEV, PHEV, H2, LPG, CNG, LNG) passenger cars and vans (M1 + N1)
Market share AF passenger cars and vans
Percentage of the total new passenger cars and vans (M1 + N1) registrations that is powered by alternative fuels (BEV, PHEV, H2, LPG, CNG, LNG)
AF buses
New registrations of alternative fuels (BEV, PHEV, H2, LPG, CNG, LNG) buses (M2 + M3)
AF trucks
New registrations of alternative fuels (BEV, PHEV, H2, LPG, CNG, LNG) trucks (N2 + N3)
BEV passenger cars and vans
New registrations of battery electric passenger cars and vans (M1+N1)
Market share BEV passenger cars and vans
Percentage of the total new passenger cars and vans (M1 + N1) registrations that is full battery electric (BEV)
PHEV passenger cars and vans
New registrations of plug-in hybrid electric (PHEV) passenger cars and vans (M1+N1)
Market share PHEV passenger cars and vans
Percentage of the total new passenger cars and vans (M1 + N1) registrations that is plug-in hybrid electric (PHEV)
H2 passenger cars and vans
New registrations of hydrogen (H2) passenger cars and vans (M1+N1).
Market share H2 passenger cars and vans
Percentage of the total new passenger cars and vans (M1 + N1) registrations that is powered by hydrogen (H2)
3. EU alternative fuels infrastructure
The reference year for the graphs and maps in this section is the same as for road transport country reports. For growth rates, 2021 to 2020 figures are displayed, until data for 2022 is completely
3.1 Electricity
This sub-section provides an overview of the publicly accessible recharging points installed in the Member States, their growth rates (year-over-year, between the reference year and the preceding
year) and their total power output.
Recharging points
Total number of publicly accessible recharging points.
Growth rate of recharging points
Year-over-year (2021 vs. 2020) growth rate of publicly accessible recharging points.
AC recharging points
Total number of publicly accessible AC recharging points.
Growth rate of AC recharging points
Year-over-year (2021 vs. 2020) growth rate of publicly accessible AC recharging points.
DC recharging stations
Total number of publicly accessible DC recharging points.
Growth rate of DC recharging points
Year-over-year (2021 vs. 2020) growth rate of publicly accessible DC recharging points (EVSE)
3.2 Hydrogen (H2)
This sub-section provides an overview of the publicly accessible hydrogen refuelling points installed in the Member States, their growth rates (year-over-year, between the reference year and the
preceding year) and their relative numbers compared to the registered amount of hydrogen vehicles.
H2 refuelling points
Total number of hydrogen (H2) refuelling points
Growth rate of H2 refuelling points
Year-over-year (2021 vs. 2020) growth rate of hydrogen (H2) refuelling points.
H2 vehicles per H2 refuelling point
Number of hydrogen (H2) vehicles per hydrogen refuelling point (high and low pressure).
3.3 Natural gas
This sub-section provides an overview of the publicly accessible natural gas (CNG and LNG) refuelling points installed in the Member States, their growth rates (year-over-year, between the reference
year and the preceding year) and their relative numbers compared to the registered amount of natural gas vehicles.
CNG refuelling points
Total number of CNG refuelling points
Growth rate of CNG refuelling points
Year-over-year (2021 vs. 2020) growth rate of CNG refuelling points.
CNG vehicles per CNG refuelling point
Number of CNG vehicles per CNG refuelling point.
LNG refuelling points
Total number of LNG refuelling points
Growth rate of LNG refuelling points
Year-over-year (2021 vs. 2020) growth rate of LNG refuelling points.
LNG vehicles per LNG refuelling point
Number of LNG vehicles per LNG refuelling point
3.4 LPG
This sub-section provides an overview of the publicly accessible LPG refuelling points installed in the Member States, their growth rates (year-over-year, between the reference year and the preceding
year) and their relative numbers compared to the registered amount of LPG vehicles.
LPG refuelling points
Total number of LPG refuelling points
Growth rate of LPG refuelling points
Year-over-year (2021 vs. 2020) growth rate of LPG refuelling points.
LPG vehicles per LPG refuelling point
Number of LPG vehicles per LPG refuelling point. | {"url":"https://alternative-fuels-observatory.ec.europa.eu/interactive-map","timestamp":"2024-11-05T18:25:36Z","content_type":"text/html","content_length":"108841","record_id":"<urn:uuid:665014b5-eaa7-420c-8768-03fb77caa1f1>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00018.warc.gz"} |
Argh, argh, argh!!!!
57 thoughts on “Argh, argh, argh!!!!”
1. Ugh!!!!!!! That person teaches!
2. Yes, Bora, that’s exactly the cause of this title. It’s enough to inspire dry heaves!
3. Does it take 5 minutes to leave the board alone?
4. That’s a math WIN! (and a construction FAIL)
5. Well she practiced when she was doing the first cut, and so was able to do the second two faster.
1. Or the second board was only 75% as wide as the first board!
2. Second two? If it takes Marie two cuts to saw a board in three pieces, and she has already completed one cut, how many more cuts must she make? 🙂
1. Ah, never mind, she starts with a new board. I’m the bonehead here, not Sean 🙂
6. The extra BONUS fail is to follow the link to the FailBlog site and see a whole bunch of commenters saying, “Where’s the fail?”
(But expecting internet commenters to be intelligent is really asking too much — present company excluded, of course!)
7. Depends on the size and shape of the pieces. Notching out a teeny bit at either end of the second board would result in three pieces – two very small, one large. I bet she could do that in, say,
a couple of minutes.
Or – “if she works just as fast…” – you could take that to mean “just as little time”, meaning: first board, one cut – 10 minutes. Second board, two cuts – 10 minutes.
How long is a piece of string, anyway?
8. I can saw a board into one piece in no time. I just did it to a billion of them while you were reading this.
9. From my experience in grading and TAing, the question is actually loaded with ambiguity. First of all, it never specifies that the two boards are identical, and we can’t extrapolate well from her
cutting speed on the first board to her speed on the second board. This is what I tell students is a necessary assumption to be able to solve a multiple choice problem, but it’s surprising how
many don’t make these assumptions. So the problem really should have said “an identical board” rather than “another board.”
The second ambiguity is in the shape of the board, and how she’s cutting it into pieces. The picture implies a bit here, but we still don’t know, for instance, what angle she’s making the cuts
at. Again, if this differs between the two boards, we can’t extrapolate. Less students are likely to think of this one, though.
But here’s a fun hypothetical: What if it’s a circular board that she’s cutting into equal wedges? In this case, the teacher’s answer is in fact right, as the number of pieces will be equal to
the number of radial cuts. But the picture rules this one out, at least.
1. Funny, I immediately thought of the radial cuts option and figured that both teacher and student were, in fact, right 😀
I totally agree about the multiple ambiguities in the question.
10. @infophile
Not necessarily. The problem specifies that Marie “works just as fast,” which could be interpreted as “ten minutes per cut,” regardless of the size or composition of any subsequent board.
I agree that in a properly phrased problem, such ambiguity would have been corrected.
11. I’m guessing that the person who graded this didn’t write the test and just took a quick glance at the problem, assumed it was about division, and graded thusly without a second thought. I’m sure
they’re not that stupid.
1. I’m thinking that this is exactly the kind of stupidity we’re talking about here. Grading papers without understanding the question (or even trying to understand it)
12. I feel bad it took me a second to get that. The question isn’t ideally worded. Though that should trip up the student, not the teacher.
You might be surprised how many math teachers have just memorized the procedures and don’t actually have very good conceptual knowledge. Next time you talk to your local math teacher ask about
dividing fractions or negative numbers. See if you get a satisfactory answer. I’m not saying you won’t, but you might not.
13. Are there are some language differences in different countries? The question talks about a board, but the picture shows a baton/plank.
1. Hmm, I’m curious what you think baton means (and where you are from) since I would never have thought to describe the picture as a baton. Perhaps plank would be a more accurate term since I
suppose a board ought to be thinner than it is wide and the object in the picture appears to have a roughly square cross section. However in my own american english I would probably call the
item depicted a board.
14. If the first cut were down the length of the board–so one cut for 10 meters, then the teacher’s answer would be right if the next cut (creating the third piece) took one 10 meter board and cut 5
meters–say from an end to the edge just short of the middle (because the cut is not straight down the middle), then it could justify the teacher’s answer of 15. This would then allow for any
answer greater than 10 to be correct (up to infinity). Imagine the infinitely thin saw cutting back-and-forth in a continuous manner so that the cut of the slices are a few molecules thick.
“just as fast” is ambiguous.
15. The mistake the marker made is in assuming that the amount of work is proportional to the number of pieces rather than the number of cuts. The question itself doesn’t mention the number of cuts
and after seeing many similar problems and being in that grading trance where one thinks through the problems only superficially it’s not unlikely that a competent person would make that mistake.
16. The grader probably reasoned about time per piece rather than time per cut. So instead of thinking “ten minutes per cut,” the grader thought “five minutes per piece.” So this isn’t really a math
fail, it’s an English fail. This grader thought that piece was an activity that took time, which it isn’t. It’s not even a verb.
17. Being fair, anyone could make that mistake _once_. The real test is whether the teacher immediately accepted that 20 minutes was a better answer when the mistake was pointed out.
18. I think my daughter had this teacher for algebra.
19. Ow!
One of my colleagues made mistakes on her answer key for a final exam and is now facing a student grievance because she won’t face up to it. I wonder if she graded this test.
P.S.: Hey! There are two Zenos here!
1. How is that possible? It the mistake demonstrable? If so, how could she possibly expect to have her decision survive a grievance? That’s absurd!
1. Absurd? Yes, I quite agree. And I checked the solutions myself. When the student files a grievance the student is sure to prevail. (Even math teachers can be irrational.)
20. That is a pretty bad question; due to how it was worded I almost came to the same incorrect assumption regarding what the question asked — that is, I intially thought it read “to saw two boards”.
And then even after taking that into account, there’s the question of “just as fast”, as previously noted. I wouldn’t assume that it means that it took 10 minutes to saw the board, but if it was
about interpreting word problems…then maybe I could see that being the case.
1. But yes, I’ve read in the past about the sad state of conceptual mathematics knowledge among teachers in the US. It seems to me as though some elementary school teachers choose that field of
study even partially *because* they really don’t understand math.
Still, there are lots of schoolteachers who don’t understand multiplication and division of fractions themselves. Considering that coming to understand what a certain operation means (rather
than focusing on applying operations to model situations and do problem solving) is the brunt of grade school mathematics, it’s particularly frustrating.
21. This is another example why the system of filling in blanks is a bad one. If the pupil had been asked for a detailed explanation of how he arrived at the result, the teacher might have learned
22. This question is not designed to test mathematics skills. It is designed to catch people out, and in this it has succeeded.
The mistake is a very easy one to make, particularly if you are in the middle of a mathematics exam and not thinking about how to saw wood. To see how easily this mistake can be made, let’s
change one word.
It took Mary 10 minutes to work a board into 2 pieces. If she works just as fast, how long will it take her to work another board into 3 pieces.
Depending on the definition of work(i.e. including finishing), the answer could be either 15 minutes or 20 minutes.
There’s a more serious problem. While this question is designed to catch people out, it only accepts a final answer. The student has put down 20 here, but we have no real indication of whether
the student actually understood the question or not. They could have simply multiplied the first two numbers they saw, 10 and 2. What is this students reasoning? Are we sure they understood the
question? At least we know what the examiner’s reasoning was, even if it was flawed.
If this question was a single word problem as part of a larger exam, then it is fine. But if this is the kind of question that mathematics students are expected to answer, then there is little
wonder why they would regard mathematics as an inscrutable academic con game designed to impede their education. You can’t throw about too many questions like this in an exam.
Mathematics is not about navigating a series of artificial intellectual pitfalls. It is about understanding the world through reason and systematic methods. Perhaps this questions was designed to
test those skills, but I have my doubts.
23. This is a completely understandable off-by-one error. I have a hard time faulting the teacher for this when I make them in the regular course of programming.
Now, if the student brought it up and the teacher refused to budge, that would be “argh!” worthy.
Now, if the student brought it up and the teacher refused to budge, that would be “argh!” worthy.
Reminds me of many moons ago as a high school sophomore (age 16) when I found an error in the Plane Geometry textbook I was being taught from. I pointed it out to my teacher, and she saw the
error and agreed with me. She took it to the department head who dismissed my observation out-of-hand (i.e., with NO explanation) and insisted the textbook was right.
Of course, I did learn something from the experience …
25. Almost as annoying are the equations on the bottom. 10 = 2? I know what the grader means, but students get into the habit of thinking of ‘=’ as synonymous with ‘the answer is’ and this causes
conceptual trouble in algebra.
26. This question is a less tricky variant on “If it takes 6 seconds for a grandfather clock to strike 3 bells, how long will it take to strike 12 bells?” Although it is easy to give the wrong answer
if one is thoughtless, the correct answer is clear and not all that hard if one thinks carefully. As a teacher myself I find it inexcusable for a teacher not to carefully think through and check
all solutions on a test key.
27. I don’t see how “just as fast” is ambiguous. If I run at a rate of 6 miles an hour and my friend also runs at a rate of 6 miles an hour, then my friend runs just as fast as I do. If I run at a
rate of 6 miles per hour and my friend runs at a rate of 8 miles per hour, then my friend runs just as fast as I do, but I do not run just as fast as my friend. So, if a is “just as fast” as b,
it means that b is at least equal to and possibly more than a. The problem here, I don’t think, to be ambiguity. The problem here comes as that “just as fast” does *not* mean exactly equal.
1. and that is basicaly the definition of ambiguity. if there is more than one acceptable interpetration, then it’s ambiguous. if ‘just as fast’ does not mean exactly equal, but equal or
greater, then there is more than one answer and therefore it’s ambiguous
28. “Just as fast” could mean:
Same number of boards cut per time.
Same number of linear board-inches cut per time (same as above?).
Same number of cubic board-inches cut per time.
Same number of saw strokes per time.
29. I agree with commenters who say it’s partly an English fail. It definitely took me a second go-round to see the correct answer, because I immediately gravitated to the numbers within the
paragraph and saw the same thing the teacher saw: 10 is to 2 as x is to 3.
It would have been clear if it had said, “It took Marie 10 minutes to make 1 cut in a board. How long would it take her to make 2 cuts?”
But perhaps the point of the problem is to have the test taker work through that extra interpretive step? Still, it comes off to me like a trick question. Most people who answered as the teacher
did would see their flaw immediately, but it doesn’t prove they don’t know how to do math.
30. I do not buy that the teacher should get a pass. The teacher fails on two counts. The first is having a trick (or tricky) question on a test that appears to be a short answer answer test with a
lot of questions, especially if the student isn’t going to have much time to interpret what is being asked. The second is that the teacher is not under the time or stress pressure of the student
and should have been able to correctly answer the question.
31. I guess I fail at proof reading.
1. This. I honestly wasn’t able to find the mistake.
32. Of course there is some ambiguity in the exercise, just as there will always be some ambiguity in all ‘realistic’ word problems. Part of that ambiguity is resolved by the picture of the board &
saw to the right of the problem. In most cases its absolutely clear, even with some mild ambiguity, what the mathematical problem is that is supposed to be solved.
In this case, however, all the mild ambiguities about cutting times, shape of the board, whatever… fade to nothing compared to the fact that the teacher clearly did not see the reasoning of the
student (the correct reasoning in my opinion.) Part of being a good teacher is the ability to recognize how students think, and even if they get an incorrect answer, be able to realize by what
crooked reasoning that answer was obtained. Such an analysis enables teachers to identify trouble spots in their own teaching. This teacher clearly did not attempt that, or if he/she did, failed
The fact that there are 3 stars behind the question should also be an indication, to both student and teacher, that this requires a bit more than just standard equal-ratio reasoning.
33. Thank you for reminding me why I am not being too harsh in supporting the death penalty for innumeracy 🙂
The “bad phrasing” is on purpose. The pupil is supposed to have to work out that cutting a board into 2 pieces requires 1 cut, therefore cutting a board into 3 pieces requires 2 cuts.
When I was taking my O-level maths, calculators were not allowed and half the marks would have been awarded just for showing working — i.e., we would have been expected to write something to the
effect of “2 pieces = 1 cut = 10 mins, 3 pieces = 2 cuts = 20 mins”. Even “2 pieces = 1 cut = 10 mins, 3 pieces = 2 cuts = 30 mins” would have earned something, even though 2 * 10 != 30, for
correctly surmising how to answer the question.
The practice of showing all intermediate steps was drilled into us. Even when not required for the test, it is a useful habit to get into; because if {when} you do make an elementary mistake with
the figures, you can go back and see where you went wrong.
That’s still not a board, though. It’s not even a plank. It looks like a length of 50×50 PAR.
34. But how long does it take to saw a board into one piece?
35. If it takes 1 programmer 12 months to develop an application, then surely it would take 12 programmers 1 month to develop the same application, right?
36. Response to comments on the new real numbers:
The field axioms are listed in Royden’s Real Analysis, pp. 31 – 32. One of them is the field axioms that says: given two real numbers x, y one and only one of the following holds: x y. L. E. J.
Brouwer and myself constructed two different counterexamples to it in Benacerraf, P. and Putnam, H., (1985), Philosophy of Mathematics, Cambridge University, Press, Cambridge, 52 – 61 and
Escultura, E. E., The new real number system and discrete computation and calculus, Neural, Parallel and Scientific Computations 17 (2009), 59 – 84, respectively. The completeness axiom, a
variant of the axiom of choice, also leads to a contradiction in R^3 known as the Banach-Tarski paradox. Its contradiction comes from itself but its use of the universal or existential
quaantifier on infinite set. Nonterminating decimals are ambiguous, ill-defined, because not all its digits are known or computable. Thus, division of an integer by a prime other than 2 or 5 is
ambiguous, ill-defined because the quotient is nonterminating, e.g., 2/7. That is why it is impossible to add or multiply two nonterminating decimals; we can only approximate their quotient by a
terminating decimal. Moreover, it is improper, i.e., noesense, to apply any operation of the real number system to the dark number d* because d* is not a real number; therefore, the result will
be ill-defined. BTW, foreign writers of English sometimes write better English because they learn the language the right way.
E. E. Escultura
37. Correction: “field axioms” on line 3 of my post should read: “trichotomy axiom”
E. E. Escultura
38. Sorry, more corrections to my post:
line 4: “x y.” should read “x > y.”
line 10: “Its contradiction comes from itself but its use of the universal or existential quaantifier on infinite set.” should read “Its contradiction comes not from itself but from the use of
the universal or existential quantifier when applied to infinite set.”
E. E. Escultura
39. I just noticed that posts sometimes do not come out correctly. This error happened twice in the statement of the trichotomy axiom. It should read as follows: given real numbers x, y one and only
one of the following holds: x is less than y, x equals y, x is greater than y. Please check this out, Administrator.
Now, to the more substantive issue: I consider this blog one of the better ones along with Free Library and Larry Freeman’s False Proofs. Wikipedia is the worst. Not only does it delete posts and
block this blogger to keep unanimity, its links also post outright lies. For example, WikiPilipinas posted lies about Filipino inventors, who are long gone, and this scientist. Moreover, it
printed a fake apology purportedly coming from and blocked this blogger. Only public pressure forced deletion of both posts.
Next in the worst category are Halo Scan and Don’t Let Me Stop You. They delete posts they disagree with or don’t like.
Finally, I commend L. P. Cruz for not hiding behind username which means he is confident his post makes sense and ready to stand behind it.
E. E. Escultura
40. Any guess on the grade level for which this is intended? 1st? 2nd? 3rd?
41. The teacher is actually right. For one cut it took 10 minutes, which resulted in two pieces. Say the board was a square, when one cut has happened, you’re left with two half boards. To cut the
other half board it will take half the time it took to cut the first one. Thus it only takes 15 minutes to cut it into three pieces = two cuts = 10 minutes plus half of it.
1. Mike, you’re assuming that the board is square, that it is being cut into two perfect halves, and that one half is being cut into two squares and not lengthways. Each assumption you have made
is questionable.
43. I’ll have to admit it took me a minute to think that one through. It seems logical to think that if you saw one board into two and it takes 10 minutes, it would take 15 minutes to saw a board
into 3 pieces.
The flaw is that it is not the number of pieces that matters it is the number of times you have to saw the board to get those pieces. 2 pieces = 1 saw = 10 minutes. 3 pieces = 2 saws = 20
minutes. That is a little bit tricky at first, but easily understood if you think it through.
44. As usual I’m late, late for this blog, and late for this reply. I just hope this is not looked on as bad manners here.
I think eveyone here just failed badly.
There seems to be consensus that the teacher doing the scoring failed badly, mostly because of the ridiculous table he used as reason for the scoring.
The student also failed because he made assumptions that are just not supported by the wording of the problem.
But the one who failed the most badest was the person who worded the problem.
When I try to give an answer to the problem as worded, I can only come up with bounds. The lower bound would be 0 and the upper bound infinity. | {"url":"http://www.goodmath.org/blog/2010/10/06/argh-argh-argh/","timestamp":"2024-11-08T21:51:26Z","content_type":"text/html","content_length":"251678","record_id":"<urn:uuid:8e72d581-72ed-4a27-8622-a5f905dcb36c>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00631.warc.gz"} |