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How to Calculate CARG | Sapling
CARG stands for Compound Annual Rate Growth that is more often abbreviated as CAGR. CAGR typically represents an annual rate of the investment growth calculated over several years. It is also used to
characterize the growth of other elements of business such as a number of clients or product sales. CAGR is computed with the formula CAGR=[(Ending Value/ Starting Value)^(1/ Number of years)]-1. As
example, calculate CAGR if you invested $20,000 in 2005 and ended up with a portfolio of $25,000 in 2009.
The Steps
Step 1
Subtract the starting year from the ending one to calculate the investment duration. Number of years= ending year-starting year In our example, number of years=2009-2005=4
Step 2
Divide one by the number of years. 1/number of years. In our example, it would be 1/4=0.25.
Step 3
Divide the ending value by Starting Value and raise the quotient in the power of the number from Step 2. In our example, it would be ($25,000/$20,000) ^0.25= 1.0574.
Step 4
Subtract one from the number obtained in Step 3 and multiply by 100 percent to get CAGR in percents. In our example, CAGR=(1.0574-1) x 100 percent=5.74 percent. | {"url":"https://www.sapling.com/5158215/calculate-carg","timestamp":"2024-11-03T13:49:57Z","content_type":"text/html","content_length":"294350","record_id":"<urn:uuid:04bfd011-a112-4941-b116-6cda1493e22f>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00789.warc.gz"} |
What was the average number of miles per gallon?
Step by step:
• Calculate the average number of miles per gallon.
• Jeremy's car traveled 300 miles on 10 gallons of gasoline.
• To find the average number of miles per gallon, divide the total miles by the total gallons used.
□ 300 miles per 10 gallons
□ Using the formula for average, miles per gallon = total miles / total gallons
□ Therefore, miles per gallon = 300 / 10 = 30 mpg
So, the average number of miles per gallon Jeremy's car achieved was 30 mpg. | {"url":"https://tutdenver.com/sat/what-was-the-average-number-of-miles-per-gallon.html","timestamp":"2024-11-08T21:53:14Z","content_type":"text/html","content_length":"20195","record_id":"<urn:uuid:f0300fa1-f3dd-4904-868a-b4697a2386c2>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00666.warc.gz"} |
Forecasting Newsletter for October 2022 — EA Forum
• Prediction Markets, Forecasting Platforms &co
□ Kalshi
□ Manifold
□ Metaculus
□ Odds and Ends
• Opportunities
• Research
You can sign up for this newsletter on substack, or browse past newsletters here. If you have a content suggestion or want to reach out, you can leave a comment or find me on Twitter.
Prediction Markets and Forecasting Platforms
The Bloomberg terminal now incorporates Kalshi markets (a).
Kalshi hosted a competition to predict congressional races (a). If someone predicts all races correctly, they get $100k, otherwise the most accurate person will receive $25k. To be clear, this is a
marketing gimmick, and participants make Yes/No rather than probabilistic predictions. But I thought I'd report on it given the high amount.
As Metaculus continues to build capacity, they have started to launch several initiatives, namely Forecasting Our World In Data (a), an AI forecasting team (a), a "Red Lines in Ukraine" (a) project,
and a "FluSight Challenge 2022/23" (a). They are also hiring (a)
Metaculus erroneously resolved (a) a question on whether there would be a nuclear detonation in Ukraine by 2023.
An edition of the Manifold Markets newsletter (a) includes this neat visualization of a group of markets through time:
Manifold's newsletter (a) also has further updates, including on their bot for Twitch. They continue to have a high development speed.
Odds and ends
The US midterm elections were an eagerly awaited event in the prediction market world. Participating so as to make a profit requires a level of commitment, focus and sheer fucking will that I
recognize I don't have. For coverage, interested readers might want to look to StarSpangledGamblers (a), or check on the Twitters of various politics bettors, such as Domah (a), Peter Wildeford (a)
or iabvek (a).
My forecasting group, Samotsvety, posted an estimate of the likelihood that Russia would use a nuclear weapon, including a calculator so that people could more easily input their own estimates. This
was followed by the Swift Institute (a), and both estimates were reported in WIRED magazine. Since then the probability seems much lower, as the strategic situation becomes clearer.
Some more pessimistic forecasts by Max Tegmark (a) were seen by Elon Musk, and may have played a role in Musk’s refusal (a) for Ukraine to use his Starlink service over Crimea.
One of the sharpest prediction market bettors objected to the above estimates, and I followed up with some discussion.
Superforecaster Anneinak correctly goes with her gut—in contraposition with the polls—on the Alaskan Congressional elections (a).
An academic initiative by the name of CRUCIAL (a) is looking at predicting climate change effects using prediction markets.
The Council on Strategic Risks (a) is hiring for a full-time Strategic Foresight Senior Fellow (a), and is offering $78,000 to $114,000 per year plus benefits. My impression is that this post would
be impactful and policy-relevant.
The $5k challenge to quantify the impact of 80,000 hours' top career paths (a) is still open, until the 1st of December. So far I only know of two applications, and since the pot is split between the
participants, participation might have a particularly high expected monetary value.
Katja Grace looks at her calibration in 1000 predictions (a):
Callum McDougall writes Six (and a half) intuitions for KL divergence (a).
Terence Tao has two (a) introductory blogposts (a) on Bayesian probability theory.
I posted Five slightly more hardcore Squiggle models (a).
I came across this really neat explanation of Markov Chain Monte Carlo: Markov Chain Monte Carlo Without all the Bullshit (a). It requires knowledge of linear algebra, but it was otherwise really
neat. I would encourage readers who have heard about the method but never learnt how it works to give it a read.
Sam Nolan &co create estimates explicitly quantifying (a) the uncertainty in GiveWell’s cost-effectiveness analyses.
Note to the future: All links are added automatically to the Internet Archive, using this tool (a). "(a)" for archived links was inspired by Milan Griffes (a), Andrew Zuckerman (a), and Alexey Guzey
> In 1646, Magnenus estimated the number of atoms contained in a piece of incense from an argument based on the sense of smell (if a fraction of the grain is burned, the number of particles can
be estimated from the volume within which the scent is still perceptible). His estimate for the number of particles in a piece of incense "not larger than a pea" was of the order of 10^18. This
estimate is remarkably accurate, within about three orders of magnitude of the true value (based on the number of molecules in the unburned incense) and thus only one order of magnitude off in
linear dimension of the molecule. Magnenus was by far the earliest scholar to give a reasonable estimate for the size of a molecule, the first "modern" estimate was given more than 200 years
later, in 1865, by Josef Loschmidt
— Wikipedia, on Johann Chrysostom Magnenus
No comments on this post yet.
Be the first to respond. | {"url":"https://forum.effectivealtruism.org/s/HXtZvHqsKwtAYP6Y7/p/xRC8jkWFRLCQSGznh","timestamp":"2024-11-04T18:57:23Z","content_type":"text/html","content_length":"199083","record_id":"<urn:uuid:de35bf78-39a0-454c-89f3-9c957d296cfd>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00889.warc.gz"} |
Earthquake Code Statement (3.6.2.2)
• A2 - Floor Discontinuities: When the situation on any floor is marked by the user in the program
• In buildings with A2 type irregularities, floor slabs are automatically modeled with two-dimensional plate (membrane) or shell finite elements to show that they can safely transfer earthquake
forces between vertical bearing system elements within their own planes (See 4.5.6.2).
A2 type irregularity condition TBDY occurs when any of the following three items occur, as shown in Figure 3.2 .
• The ratio of the total area of all spaces (including stairs and elevators) in the floor plan to the total floor gross area is more than 1/3.
• The presence of local floor gaps that make it difficult to safely transfer earthquake loads to vertical bearing system elements.
• Sudden decreases in in-plane stiffness and strength of slab
According to Article 3.6.2.2 and 4.5.6.2 of TBDY , floors in buildings with A2 and A3 type irregularities are modeled with two-dimensional finite elements. The following picture shows the analysis
model of an example whose tiles are modeled with two-dimensional finite elements (shell).
As a result of three-dimensional analysis of shell finite elements (slabs, curtains and polygonal walls), they generate stresses and forces per unit length. The directions of these stresses and
forces per unit length are determined according to the local axes of the shell finite element .
Shell finite element forces are obtained as a result of stresses formed by shell finite elements in three dimensional analysis.
• M11, M22: Bending moments per unit length (tfm / m or kNm / m) formed around axes 1 and 2. They are also called out-of-plane bending moments.
• M12: Means unit length planar torsion moment (tfm / m or kNm / m). It is also called the in-plane torsion moment.
• V13, V23: Shell is the unit length shear forces (tf / m or kN / m) on the surface of the finite element and perpendicular to the plane of the finite element. It is also called the out-of-plane
shear force.
• F11, F22: Tensile and compressive forces per unit length (tf / m or kN / m) parallel to the plane of the acceptance finite element in the respective direction. It is also called in-plane
pressure-pull forces.
• F12: Unit length shear forces (tf / m or kN / m) parallel to the shell finite element plane. It is also called in-plane shear forces.
Shell finite element results can be viewed from the "Shell Results" tab in the Analysis Model.
Deformation results from shell finite element results can also be examined in the Analysis Model.
Floor spaces cause accumulation in floor stresses caused by vertical loads and earthquake loads. In the picture below , a floor plan is shown in which the case of "the presence of local floor gaps
that make it difficult to transfer earthquake loads to vertical bearing system elements" in the definition of A2 type irregularity. Since there is a gap in the floor connected to the curtain, the
in-plane shear stresses (F12) under the effect of an earthquake in the (Y) direction increase significantly at the gap edges as seen on the right side of the picture.
A2 type irregularity occurs due to floor gaps. For this reason, slab in-plane deformations were neglected. Rigid diaphragm solution is not applied in this type of irregularity. In buildings with A2
type irregularities, a semi-rigid diaphragm solution should be made in which the floors are modeled with shell finite elements .
According to Article 7.11.5 of TBDY , earthquake loads must be safely transferred from floors to vertical bearing elements. A floor plan in which A2 type irregularity definition consists of "the
presence of local floor gaps that make it difficult to safely transfer earthquake loads to vertical bearing system elements" is shown in the above picture. The shear stresses accumulated in these
spacesare also controlled by article 7.11.5 . When such gaps are opened in the critical areas of the vertical bearing elements, both 7.11.3 slab stress checks and 7.11.5 seismic loads are safely
transferred from the floors to the vertical bearing elements. | {"url":"https://help.idecad.com/ideCAD/earthquake-code-statement-3-6-2-2","timestamp":"2024-11-10T18:20:55Z","content_type":"text/html","content_length":"39813","record_id":"<urn:uuid:1aadf5f6-bba0-4ffb-89d0-d24b4bd2adb4>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00560.warc.gz"} |
Energy management of the residential smart microgrid with optimal planning of the energy resources and demand side
Issue Sci. Tech. Energ. Transition
Volume 79, 2024
Decarbonizing Energy Systems: Smart Grid and Renewable Technologies
Article Number 76
Number of page(s) 10
DOI https://doi.org/10.2516/stet/2024079
Published online 02 October 2024
Science and Technology for Energy Transition
, 76 (2024)
Regular Article
Energy management of the residential smart microgrid with optimal planning of the energy resources and demand side
^1 Prince Sattam Bin Abdulaziz University, College of Engineering, Department of Electrical Engineering, Alkharj, 11942, Saudi Arabia
^2 Laboratory LaTICE, Ecole Nationale, Supérieure D’ingénieurs de Tunis ENSIT, University of Tunis, Tunisia
^3 College of Engineering and Information Technology, University of Dubai, Academic City, 14143, Dubai, UAE
^4 Department of Chemistry, University College of Duba, University of Tabuk, Tabuk, Saudi Arabia
^5 Marwadi University Research Center, Department of Electrical Engineering, Faculty of Engineering & Technology, Marwadi University, Rajkot 360003, Gujarat, India
^6 NIMS School of Electrical and Electronics Engineering, NIMS University Rajasthan, Jaipur, Rajasthan, India
^7 Department of Nuclear and Renewable Energy, Ural Federal University Named after the First President of Russia Boris Yeltsin, Ekaterinburg 620002, Russia
^8 Head of the Department “Physics and Chemistry”, “Tashkent Institute of Irrigation and Agricultural Mechanization Engineers” National Research University, Tashkent, Uzbekistan
^9 Scientific Researcher, University of Tashkent for Applied Sciences, Str Gavhar 1, Tashkent 100149, Uzbekistan
^10 Western Caspian University, Scientific Researcher, Baku, Azerbaijan
^11 Department of Medical Laboratories Technology, Al-Nisour University College, Nisour Seq. Karkh, Baghdad, Iraq
^12 College of Pharmacy, The Islamic University, Najaf, Iraq
^13 College of pharmacy, the Islamic University of Al Diwaniyah, Al Diwaniyah, Iraq
^14 College of pharmacy, The Islamic University of Babylon, Babylon, Iraq
^15 The Department of energy, Madrid Institute for Advanced Studies in Energy, Madrid, Spain
^* Corresponding author: yersi.luis.ro@gmail.com
Received: 2 July 2024
Accepted: 2 September 2024
The study specifically aimed to model the optimal operation of these appliances based on their usage patterns, rather than relying on the capacity of demand flexibility in demand response (DR) and
energy pricing. The modeling operation of the appliances is done using two-layer energy optimization. In this optimization, energy consumption by appliances is reshaped via DR and load shifting in
first-layer optimization. Then, minimizing the consumption costs and consumers’ discomfort in the second layer is formulated with consideration of the optimized consumption from the first layer. The
lp-metric method is employed to solve the proposed optimization in the GAMS software. Finally, the efficiency of the two-layer optimization is confirmed using testing proposed case studies in the
numerical simulation.
Key words: Optimal operation / Demand flexibility / Energy consumption / Load shifting / Consumers’ discomfort
© The Author(s), published by EDP Sciences, 2024
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
t, T : Hour of operation, Hour
P [N,WT] : Total rated power of WT, kW
P [WT], P [PV] : Power generated by WT and PV, kW
A, B, C : DGs’ cost factor, $/kW
ED, D [NS], ED ^B : Electrical demand, non-supply demand, and electrical demand before DR, kW
η ^ch, η ^dis : ESS’s efficiency in charge and discharge states, %
C [DG], C [MG] : DG cost and MG cost, $
c [MG], P [MG] : Electrical price traffic of the MG and MG power, $/kW, kW
P [GS] : Power generated by the generation side, kW
Ψ: Total demand with non-supply demand, kW
Ψ[DL] : Desired level of demand, kW
μ [ESS] : Binary variable of the ESS
Γ: Amount of consumption-shifting strategy in DR, %
1 Introduction
1.1 Aims and context
A smart grid is a modernized electrical grid system that utilizes advanced technology to boost efficiency. Smart grids employ different digital communication and automation technologies to supervise
and govern the electricity flow, enabling improved energy resource management [1]. One of the key components of a smart grid is the integration of smart homes, which are equipped with intelligent
devices and appliances that can communicate with the grid [1, 2]. This integration enables homeowners to actively participate in energy management by optimizing their energy consumption and
contributing to a more sustainable and resilient electrical grid system. Energy consumers in the residential, commercial, and industrial sectors now need electricity. Although the energy networks
have worked separately, there is an increased focus on the integration of the electricity network, and other energies are being used to form energy systems [3]. An electrical grid consisting of wind
power, natural gas power, solar power, thermal generation, and hydropower, can play various energy sources’ potential and advantages using complementarities of natural resources [3, 4]. In contrast
to independent hydropower generation, wind power systems, and photovoltaic (PV) power systems, the electrical system has high-reliability characteristics, flexibility, and stability that can be used
to overcome the advantages of intermittency, randomness, seasonality, and volatility of renewable energy power systems [3, 4]. Energy management implementation in its entirety can improve economic
and social outcomes by increasing electric energy efficiency.
The purpose of smart grids is to facilitate the widespread adoption of demand response (DR) in large energy areas, such as urban residential buildings [3, 4]. The improvement of demand-side
management programs is becoming more of a trend due to the advancement of smart grids [3, 4]. All actions taken to modify the load curve are included in the DR concept. Telecommunications
infrastructure and sophisticated measurement sensors are needed for the DR implementation [4]. One could think of Advanced Metering Infrastructures (AMI) as a starting point of DR for improving the
energy balance between generators and demand via smart energy management systems [4, 5]. Smart energy management systems need several tools for the optimal energy flow in energy infrastructure [4, 5
]. One of the main challenges for smart energy management systems is the enhancement of the operating system based on cost-effective indices via reliable communications [6, 7]. In such systems,
multi-agent tools develop hybrid optimization algorithms to perform optimal energy operation that meets a large range of constraints and objectives [8]. According to these explanations, employing the
DR in smart energy management systems can provide cost-effective indices by encouraging consumers [9, 10]. Informing consumers via telecommunications infrastructure for the management of energy
consumption and energy price is a major agent in this system. On the other hand, cost-effective indices are not only important objectives. Other indices, such as technical and social indices, are the
main objectives of smart energy management systems [11, 12].
In Figure 1, a background of the proposed smart energy management systems in the residential buildings is demonstrated. In this system, smart energy management centers (SEMC), distribution energy
generators (DEGs), main grids (MGs), and residential buildings are the main agents for energy interaction. The SEMC has a coordination role among other agents for optimal energy management using
communication data [13]. The DEGs are photovoltaic (PV), electrical storage systems (ESS), diesel generators (DGs), and wind turbines (WT). The MG is an electrical distribution grid in an urban
section with diverse traffic electricity prices in each hour of the day. The residential buildings are energy consumers in the SEMC, and they can participate in the DR program using controllable
appliances such as dryers, washing machines, etc.
Fig. 1
Smart energy management system in this study.
1.2 Related works
Much research on diverse energy operations in energy systems has been done in recent years. For example, in [14] the operation of smart hybrid energies considering local energy generation in
buildings is proposed for the reduction of the consumers’ bills. Authors in [15] focused on managing emission and economic objectives in the electrical grids with optimal sizing of the resources.
Energy storage devices are automatically utilized by the pricing algorithm in the reference [16], allowing them to save electricity when the cost is low and use it when the cost is high. In [17]
energy saving with optimal operation and scheduling of the demand and joint modeling with energy price is presented. In addition to the encouragement-based response program, a hybrid modeling of the
demand participation was proposed [18]. In their model, subscribers have received incentives to reduce their load during peak times. The energy operation in smart stand-alone buildings is reported in
[19] with optimal participation of appliances and electric vehicles in energy consumption and generation. In [20], the mathematical modeling of load demand dispatch is proposed to improve reliability
and reduce costs. This study has not established the ideal electricity price or incentives because the emphasis has been on boosting subscriber profits despite the cost of introducing
load-to-production programs. In [21], the design of the energy resources in the electrical grids is studied to decrease the losses and increase the penetration of wind energy in residential
buildings. Authors in [22] presented a novel scheme for designing energy system grids with microgrid formation and employing storage systems, renewable generation, fuel cells, and energy storage for
the operation of the system under uncertainties. In [23], modeling hybrid energy distribution grids is proposed with consideration of the local energy markets and energy pricing. The energy resources
sizing and siting are proposed in [24] for minimizing the emission, costs, and loess in electrical grids. The DR scheduling using reserve modeling and resources management is introduced in [25] for
dropping cost and emission pollution. The new operation of the appliances is studied in [26] using continuous and discrete optimization modeling in the energy systems. Authors in [27] modeled
economic objectives presented for appliance energy scheduling without consumers’ discomfort index.
1.3 Contributions
This article presents the optimal operation of energy consumption in smart buildings considering household appliances’ performance. The optimal performance of the appliances is done based on the DR
approach via a consumption-shifting model. The operation approach of the energy consumption is modeled by a two-layer optimization problem. In the two-layer optimization problem, the consumption of
household appliances is optimized via the consumption shifting model in the first layer. In the second layer optimization problem, two objectives, 1) energy consumption costs and 2) consumers’
discomfort, are minimized considering optimized consumption in the first layer. The solving operation problem is carried out by the lp-metric method in GAMS software. Hence, contributions and
novelties can be synopsized as follows:
1. A two-layer operation problem is presented for optimal consumption of household appliances.
2. The DR modeling based-consumption shifting model is considered in the first layer.
3. A two-objective modeling for energy consumption costs and consumers’ discomfort in the second layer is formulated.
4. The lp-metric method is presented for solving problems by GAMS software.
2 System modeling
As shown in Figure 1, system modeling based on mathematical formulation is presented in the following subsection:
2.1 PV modeling
The solar irradiance based-PV modeling is formulated by (1) [28]:$P PV ( si ) = η PV × S PV × si .$(1)
2.2 WT modeling
The wind speed based-WT power generation is formulated using (2) [28, 29]:$P WT ( v ) = { 0 if v ≤ V Ci P N , wt × { v - V Ci V R - V Ci } if V Ci ≤ v ≤ V R P N , wt if V R ≤ v ≤ V Co 0 if V Co ≤ v
2.3 DG modeling
The DG modeling is formulated considering fuel cost as follows:$C DG ( t , d ) = A P d 2 ( t , d ) + B P d ( t , d ) + C .$(3)
2.4 ESS modeling
The ESS modeling considering economic parameters and operation modes are formulated as follows [30]:$C ESS ( t , ess ) = { C o ess × P ESS ( t , ess ) } ,$(4) $P ESS ( t ) / η dis ≤ P ESS - dis max ×
μ ESS discharge P ESS ( t ) ≥ 0 , P ESS ( t ) × ( - η ch ) ≤ P ESS - c h max × ( 1 - μ ESS ) charge P ESS ( t ) ≤ 0 .$(5)
By formulas (4) and (5), the cost of ESS and ESS modes in the discharge and charge status can be calculated, respectively.
2.5 MG modeling
The MG modeling is formulated considering price traffic as follows:$C MG ( t ) = π MG ( t ) × P MG ( t ) .$(6)
3 Two-layer optimization modeling
The modeling optimization is implemented as follows:
3.1 First layer modeling
The modeling first layer is implemented based on a consumption-shifting strategy of the DR approach. In this modeling, energy demand is optimized considering price traffic in the MG. The smart
buildings can participate in the DR approach via controllable appliances operation in low-price traffic. The modeling first layer is implemented as follows:$min f fl = ∑ t = 1 t { ED ( t ) × c MG ( t
) } .$(7)
Subject to:$∑ t = 1 t ED ( t ) = ∑ t = 1 t ED B ( t )$(8) $ED ( t ) = ED B ( t ) × Γ ( t )$(9) $1 - Γ ( t ) ≤ Γ ( t ) ≤ 1 + Γ ( t ) .$(10)
Equation (8) indicates that the value of the total demand after DR is equal to before DR implementation. The rate of the consumption shift is modeled by equation (9). The participation value of the
controllable appliances in DR is given by equation (10).
3.2 Second layer modeling
In this layer, two-objective modeling includes energy costs and consumers’ discomfort are minimized simultaneously:
3.2.1 First objective modeling
The energy costs of the generation side, like MG, DGs, and ESS, are minimized as follows:$min f 1 sl = ∑ t = 1 T { ∑ d = 1 D C DG ( t , d ) + ∑ ess = 1 ESS C ESS ( t , ess ) + C MG ( t ) } .$(11)
3.2.2 Second objective modeling
Minimizing consumers’ discomfort is considered the second objective in the second layer. This objective is proposed considering non-supply demand rather than the optimal rate of consumption. The
consumers’ discomfort modeling is as follows [31]:$min f sl 2 = [ ∑ t = 1 T | Ψ ( t ) - Ψ DL | ∑ t = 1 T Ψ ( t ) ] .$(12)
Where:$Ψ ( t ) = ED ( t ) + D NS ( t )$(13) $E D NS ( t ) = { E D NS ( t ) > 0 ED ( t ) > P GS ( t ) E D NS ( t ) = 0 ED ( t ) < P GS ( t )$(14) $Ψ DL = ∑ t = 1 T ED ( t ) T .$(15)
The total demand with non-supply demand in the system is modeled by (13). The modeling of non-supply demand considering the generation side unable in supply-demand is presented by (14). The desired
level of demand is given by (15).
3.2.3 Constraints of energy system
The proposed energy system has some constraints like energy balance and energy generation limit as follows:$P MG ( t ) + P PV ( t ) + P WT ( t ) + ∑ d = 1 D P d ( t , d ) ± ∑ ess = 1 ESS P ESS ( t ,
ess ) = E D E ( t ) - E D NS ( t )$(16) $0 ≤ P d ( t , d ) ≤ P d max$(17) $0 ≤ P MG ( t ) ≤ P MG max .$(18)
The energy balance is modeled by constraints (16). Constraints (17) and (18) indicate the energy generation limit of the DGs and MG, respectively.
4 Solving method
In this section, the lp-metric is introduced as a solution approach for solving two objectives in second-layer modeling. Hence, Pareto solutions for energy costs and customers’ discomfort are
extracted by the lp-metric method. The lp-metric method for modeling is presented in equation (19) [32]:$min lp = [ { w × f 1 sl - f 1 sl , ol f 1 sl , ol } - { ( 1 - w ) × f 2 sl - f 2 sl , ol f 2
sl , ol } ] .$(19)
Where f [ 1 ] ^ ss,ol , f [ 2 ] ^ ss,ol, and w are the optimal level of energy cost, the optimal level of the consumers’ discomfort, and the weight rate for objectives, respectively. By changing the
weight rate, Pareto solutions are obtained. In equation (19), the minimum rate is selected as the best solution while considering the changing weight. The advantage of the lp-metric method is
explained in Reference [32].
5 Case studies and simulation
In this section, numerical results of the smart energy management system are generated through the modeling of case studies and simulations using GAMS optimization software. The case studies
specifically focus on the participation of DR in smart buildings, employing a consumption-shifting strategy.
• Case I) Operation of the appliances without DR participation.
• Case II) Operation of the appliances with DR participation.
The framework of smart energy management in Figure 2 is indicated. The parameters and data of the DEG are listed in Table 1. It should be mentioned that we used two DGs with the same data. The
electrical price, wind speed, solar irradiance, and electrical demand are presented in Table 2. The participation value of the controllable appliances in DR is considered by 25%.
Fig. 2
Framework of smart energy management.
Table 2
Value of electrical price, wind speed, solar irradiance, and electrical demand.
6 Results and discussion
This section presents the analysis of the discussion and results of Case Studies I and II. The utilization of the DR approach to optimize consumption, specifically through the implementation of a
consumption-shifting strategy in the first layer, is illustrated in Figure 3. The figure demonstrates the use of controllable appliances at times when the price of electricity in the MG is low.
Consequently, the obtained results of the case studies are discussed about the participation of DR in Case I and Case II.
Fig. 3
Electrical demand with DR.
As mentioned before, customers’ discomfort and energy costs were modeled by two objectives in the second layer. Using the lp-metric method, the Pareto frontier for the mentioned objectives is
extracted. In Figure 4, the Pareto frontier and the best solution for Cases I and II are shown. The weight step for extracting the Pareto frontier using the lp-metric method is changed by 0.1 for
objectives. The best solution in Figure 4 for Case studies I and II have amounts of 0.56 and 0.53 as minimum values in the lp-metric method, respectively. In Figure 4a, consumers’ discomfort and
energy cost in Case Study I are equal to 24.8% and $691,993.3, respectively. On the other side, with the participation of the DR approach in Case Study II, consumers’ discomfort and energy cost in
the best solution in Figure 4b have quantities of 14.88% and $559,022.2, respectively.
Fig. 4
Pareto frontier. a) Case I and b) Case II.
Regarding the results of the case studies, consumers’ discomfort in Case II is reduced by 9.92% compared to Case I. As well, energy cost with the participation of the DR approach in Case II is
minimized by 19.2% in comparison to Case I. Due to DR implementation in Case II, the energy cost of the MG and DGs is reduced by 10.1% and 9.1% compared to Case I.
The energy operation of the DEG and MG in the smart energy management system for Cases I and II is indicated by Figure 5. In Figure 5a, the MG operates during peak consumption and high-price traffic,
resulting in increased energy costs. On the other hand, energy demand in peak hours 12, 13, 15–17 is not met by DEGs and MG. In Case I, 271.2 kW of the energy demand is not fed. The operation of the
ESS in charging mode in Figure 5a is done at hours 1 and 4 at low price traffic of MG, whereas power discharge is operated at high price.
Fig. 5
Operation of the energy generation. a) Case I and b) Case II.
Figure 5b presents the energy operation of the DEG and MG with the DR approach in Case II. In this case, the total amount of the non-supply demand is equal to 123.3 kW, which is done at hours 12–14.
By comparing Case II with Case I, 54.535% of the non-supply demand is reduced, which leads to minimizing consumers’ discomfort. Implementation of the DR in Case II also leads to decreased energy
generation of the MG in the high-price traffic. In Case II, the participation of the ESS with low-cost factors than MG and DGs in supply-demand is increased than in Case I. The discharge energy of
the ESS in Figure 5b at peak demand and high price is operated for minimizing energy cost and non-supply demand.
7 Conclusion
The implementation of a two-layer optimization approach is carried out in this study to achieve the optimal operation of the appliances-based DR strategy in smart buildings. The DR modeling is
proposed in the first layer optimization, considering the consumption-shifting strategy of the appliances rather than energy price traffic. Thus, the second layer incorporates optimized consumption
by demand response (DR) to reduce both the customers’ discomfort and energy costs. The optimization of the customers’ discomfort and energy costs is done by the lp-metric method. Finally, numerical
simulation considering without DR (Case I) and with DR (Case II) is done for confirmation of the optimal operation of the appliances. The obtained results of the simulation represent the optimal
level of the customers’ discomfort and energy costs with the DR approach in Case II.
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All Tables
Table 2
Value of electrical price, wind speed, solar irradiance, and electrical demand.
All Figures
Fig. 1
Smart energy management system in this study.
In the text
Fig. 2
Framework of smart energy management.
In the text
Fig. 3
Electrical demand with DR.
In the text
Fig. 4
Pareto frontier. a) Case I and b) Case II.
In the text
Fig. 5
Operation of the energy generation. a) Case I and b) Case II.
In the text
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Amplifier impedance measurement | HAM Radio site
Amplifier impedance measurement
Starting with the EMRFD Fig 2.57 amplifier as a base, you can measure input returnloss using a return loss bridge. Return loss can be translated into VSWR and impedance. If the return loss is large
(or the VSWR is low) the impedance is near 50 ohms. Low return loss means a higher VSWR so the impedance must be removed from 50 ohms.
For example.
A 9.5 dB return loss is a 2:1 VSWR.
A 2:1 VSWR could be caused by either a 100 or a 25 ohm resistance
A 2:1 VSWR could also be caused by a resistance of 40 ohms and a capacitiveor inductive reactance of 30 ohms.
If you look at a Smith Chart and draw a circle centered at the centerwith a radius (Reflection Coefficient) of .33, that is a circle of constant VSWR of 2:1 so any number of resistances and
reactances couldcause an SWR of 2:1. The phase component of the impedance is difficultto measure with an return loss bridge but that does not prevent us from using the return loss bridge as an
indicator of a good match in general.
See the following for returnloss to VSWR conversion: http://www.rfcafe.com/references/electrical/vswr.htm
I put its spice file with my other spice models. If you click on the spice button to the left you can see the other models.
The circuit worked as designed. I didnt use exact values of resistors as I have more of some flavors that I want to use up instead. I got 18 dB gain flat through HF with 21dBm (125mW) out.
I have been interested in trying to experimentally determine inputand output impedance of amps to see if what I see in spice really happens in the real world.
You can measure input return loss in Spice by adding a second voltagesource to the input of your spice model.
Take a look at the spice models at the bottom of this page:
Notice AC 1 and S11 from:
• S11: input reflection coefficient of 50 ohm terminated output.
• S21: forward transmission coefficient of 50 ohm terminated output.
• S12: reverse transmission coefficient of 50 ohm terminated input.
• S22: output reflection coefficient of 50 ohm terminated input.
S11 is the input reflection coefficient.
The magnitude of the reflection coefficient goes from 0 to 1 volts given the 2 and 1 volt AC sources used. If you consider the phase component of the reflection coefficient then the reflection
coefficient can range from -1 to 1 which is not coincidentally the range of the horizontal axis of a Smith Chart.
If you express S11 in dB then S11 is return loss
Input and output impedance can be measured by using a return loss bridge to determine return loss which is related to VSWR and impedance.
The input return loss can be measured by the following recipe:
Attach a 50 ohm dummy load to the amplifier output Attach a siggen to the RLB “RF in” port Attach a power meter/oscope to the “Det” detector port Turn on the amplifier, siggen and power meter. Note
the power reading on the power meter (“Unknown open”). Attach the RLB “Unknown” port to the amplifier input and note the power. The return loss is the difference between the two power readings.
The output return loss can be measured by the following recipe:
• Attach a 50 ohm resistor to the amplifier input -important-
• Turn on the amp. With a scope, insure that it is not oscillating.
• Attach a siggen to the RLB “RF in” port
• Attach a power meter/oscope to the “Det” detector port
• Turn on the amplifier, siggen and power meter
• Note the power reading on the power meter (“Unknown open”).
• Attach the RLB “Unknown” port to the amplifier output and note the power.
• The return loss is the difference between the two power readings.
That is it! I couldnt believe it was that easy
I understood that you could place a return loss bridge at the input of an amplifier to determine input return loss/VSWR. What I didnt understandis that you could use the same return loss bridge on
the output of an ampto determine output return loss. In addition to not understanding that this was possible at all, I was concerned that I might blow up my siggenby attaching it to the amp output.
That turned out not to be an issueat the 100 mW power level. At higher power levels it may be necessary to put a 20 dB pad betwen the siggen and the RLB to protect the siggen.Same goes for the power
meter. It may need to have a 20 dB pad betweenit and the detector output. I think that as long as the amp is quiet and not oscillating and puttingout more power than the siggen/powermeter/RLB can
handle then you shouldbe ok directly measuring output return loss/VSWR.
It is kind of wierd to me to be applying low level power to the outputof an amplifier in order to determine “reverse SWR” but it works.
I saw 18 dB input return loss which is something like a 1.3:1 VSWRacross HF.
This is a big deal for me. To actually be able to measure outputimpedance gives me confidence that what Im doing is actually working.
Im going to do some work at this power level to get some more experience then am going to try to graduate to higher power levelsand see what happens.
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Zero-Inflated Models
Varieties of Zero-Inflated Models and Their Use Cases
There are several types of zero-inflated models, each tailored to specific kinds of data with excess zeros. The Zero-Inflated Poisson (ZIP) model is a combination of a logistic regression and a
Poisson distribution, ideal for count data with more zeros than the Poisson distribution predicts. It differentiates between zeros that occur due to the nature of the data (structural zeros) and
those that occur randomly (sampling zeros). The Zero-Inflated Binomial (ZIB) model is used for binomial data with an excess number of zero-success trials. The Zero-Inflated Negative Binomial (ZINB)
model is designed for count data with overdispersion and is particularly valuable in fields where data variability is high.
Application of Zero-Inflated Models in Research
Applying zero-inflated models requires careful steps to ensure the validity and utility of the model. Researchers must first determine the type of data they are dealing with (count or binomial) and
then distinguish between zero and non-zero observations. Depending on the dispersion of the data, an appropriate model (Poisson, Negative Binomial, or Binomial) is chosen. Statistical software is
then employed to estimate the parameters for both the zero-inflation and count components of the model. Model validation follows, using diagnostic checks such as residual analysis and goodness-of-fit
tests to confirm that the model accurately captures the data's characteristics.
Selecting the Appropriate Zero-Inflated Model
The correct choice of a zero-inflated model is essential for the accurate analysis of data. The decision is based on the data's nature (count or binomial) and its dispersion characteristics. For
count data with a mean equal to its variance, the Zero-Inflated Poisson (ZIP) model is typically chosen. When the data show overdispersion, with variance exceeding the mean, the Zero-Inflated
Negative Binomial (ZINB) model is more suitable. For binomial data with excessive zeros, the Zero-Inflated Binomial (ZIB) model is appropriate. Preliminary data analysis is crucial to ascertain the
distribution properties, guiding the model selection. Statistical software, such as R or Python, provides libraries specifically designed for zero-inflated model analysis, aiding researchers in this
Identifying Zero-Inflation in Datasets
Detecting zero-inflation is a critical step before employing a zero-inflated model. This process may involve exploratory data analysis (EDA) to visually inspect the data and statistical tests like
Vuong's test, which compares the fit of models with and without zero-inflation. Diagnostic plots can also be useful, contrasting the observed zeros with the expected number of zeros from a standard
count model to highlight any excess. These techniques enable researchers to determine whether zero-inflation is present and if a zero-inflated model is warranted for their analysis.
The Practical Significance of Zero-Inflated Models
Zero-inflated models have substantial practical implications across various fields. In healthcare, they are instrumental in analyzing sparse data on occurrences such as disease incidence or hospital
readmissions, helping to uncover patterns and risk factors. In education, these models can differentiate between non-participation due to disinterest and non-participation due to external barriers,
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Physics Formulas For ICSE Class 10 Chapter Wise
Students should refer to ICSE class 10 physics formulas given here prepared as per the chapters given in Selina physics class 10 book which has been recommended for class 10 students. You can also
refer to ICSE class 10 physics solutions which have been provided on our website. These formulas for Physics in Standard 10th ICSE are very important as it will help you to solve many difficult
questions which can come in your exams
1. Force
2. Work, Energy, and Power
3. Machine
4. Refraction of Light at Plane Surfaces
5. Refraction Through A Lens
6. Spectrum
7. Sound
8. Current Electricity
9. Calorimetry
10. Radioactivity
Physics Formulas For ICSE Class 10 Chapter Force
1. Moment of force = Force × Perpendicular distance of force.
SI unit = Newton × metre ( Nm )
CGS unit = dyne × cm
Important conversion :
1 Nm = 10⁵ dyne × 10² cm = 10⁷ dyne cm
2. Moment of Couple = Either force × perpendicular distance between the two forces.
3. Principle of moments = Sum of anticlockwise moments = Sum of clockwise moments.
Center of Gravity
1. Rod = Mid point of Rod.
2. Circular Disc = Geometric centre.
3. Solid or hollow = Geometric center of sphere.
4. Solid or Hollow = Mid point of Axis of Cylinder.
5. Solid cone = At a height h/4 from the base, on its axis.
6. Hollow Cone = At a height h/3 from the base.
7. Circular ring = Centre of Ring.
8. Triangular Lamina = The point of intersection of medians.
9. Parallelogram or rectangular lamina or square or rhombus = The point of intersection of diagonals.
Physics Formulas For ICSE Class 10 Chapter Work Energy and Power
1. Work = Force × displacement
W = F × S it is a scalar quantity.
W for an inclination = F × S costhita
SI unit = Joule i.e. Newton × metre
CGS unit = Erg i.e. dyne × cm.
Important conversion :
1 joule = 10⁵ dyne × 10² cm = 10⁷ dyne cm = 10⁷ erg
1. If the displacement is in direction with the force then work done is positive.
2. If the displacement is normal to the direction of force then work done is zero.
3. If displacement is in direction opposite to the force then work done is negative.
Power = Work / Time .
It is a scalar quantity.
For a constant force with change in displacement Power = Force × average speed.
P = F × v.
SI unit = Watt = Joule/second.
CGS unit = erg/s
Bigger units of Power :
1. Kilowatt (kw) = 10³ W
2. Megawatt (MW) = 10⁶ W
3. Gigawatt ( GW ) = 10⁹ W
Smaller units =
milliwatt = 10^-3 W , microwatt = 10^-6 W.
Horse Power : 1 HP = 746W = 0.746 kW.
The energy of a body is its capacity to do work.
SI unit = Joule
CGS unit = erg
Bigger units of Energy :
1. Watt hour ( Wh) = 3600 J = 3.6 kJ
2. Kilowatt hour (kwh) = 3.6 × 10⁶ J = 3.6 MJ
3. Calorie : 1 calorie = 4.18 J
4. Electron volt ( eV ) = 1 eV = 1.6 × 10^-19 J.
Force due to gravity = mg
at a certain height = mgh
Gravitational potential energy U = mgh
Kinetic energy = 1/2 mass × (velocity)²
= 1/2 mv²
Work-Energy Theorem
Let u = initial velocity
v = final velocity.
W = 1/2 mv² – 1/2 mu².
Thus work done on a body is equal to increase in Kinetic energy.
Physics Formulas For ICSE Class 10 Chapter Machine
1. Mechanical advantage ( M.A. ) = Load ( L ) ÷ Effort (E)
2. Velocity Ratio = Velocity of Effort ÷ Velocity of load
V.R. = dE / dL.
3. Work input = work done by effort.
4. Work output = work done on load.
5. Efficiency = work output/ work input
For an ideal machine
output energy = input energy.
Relation between above formulas : M.A. = V.R. × Efficiency .
All the above formulas are unit less.
M.A. = 1
V.R. = 1
Efficiency = 1 or 100%
M.A. = 2
V.R. = 2
Efficiency = 1 or 100%
M.A. = 2^n
V.R. = 2^n
Efficiency = 1 or 100%.
Block and Tackle System
1. Effort required to balance the load
E = L/n , n = number of pulleys, L = load.
In this system the effort gets multiplied n times therefore it acts as a force multiplier.
Work done by effort = effort × distance moved
= E × nd = nEd.
Work done on the load = load × distance moved by the load
= L × d
= nE × d = nEd.
For greater efficiency the pulleys in the lower block should be as light as possible.
Physics Formulas For ICSE Class 10 Chapter Refraction of Light at Plane Surfaces
For normal incidence angle of incidence is zero degree.
Refractive index ( ų ) = sin i / sin r
Speed of light in vaccum or air ÷ speed of light in that medium.
The refractive index of transparent surfaces is always greater than 1.
Principle of reversibility :
Lateral Displacement :
Important formula :
Angle of Deviation :
SHIFT :
Physics Formulas For ICSE Class 10 Chapter Refraction Through A Lens
LENS FORMULA :
Q1: A: at what position a candle of length 3 cm be placed in front of a convex lens so that its image of length 6cm be obtained on a screen placed at a distance of 30cm behind the lens?
Q2: A lens forms the image of an object placed at distance 15 cm from it at a distance 60 cm in front.
1:the total the focal length
2: the magnification and
3: the nature of the first image.
Q3: An object is placed at a distance of 20 cm in front of a concave lens of focal length 20cm. find
1: the position of image and
2: the magnification of the image
Q4: A convex lens forms an inverted image of size same as that of the object which is placed at a distance 60 cm in front of the lens. Find
1 : the position of image
2 : the focal length of the lens.
Q5: The power of the lens is + 2.0 D. Find its focal length and state the kind of lens.
Q6: The power of a lens is -2.0 D. Find its focal length and its kind. IMP
Physics Formulas For ICSE Class 10 Chapter Spectrum
Relation between speed, frequency and wavelength of electromagnetic waves :
C = f λ
• Calculate the frequency of yellow light of wavelength 550 nm. the speed of light is 3×10^8 ms^-1.
• An Electromagnetic wave has a frequency of 500MHz and a wavelength of 60cm.
a) calculate the speed of the wave.
b) name the medium through which it is travelling.
Physics Formulas For ICSE Class 10 Chapter Sound
Relationship between wave velocity V,frequency f,wavelength
2.Relationship between time period t and it’s frequency f of a wave : f=1/T
3.Speed of longitudnal waves in a gaseous medium of density d at a pressure P is :
4.Speed of trasverse waves in a streched string with tension t is given as :
Time taken to hear the echo: t=2d/V or d=Vt/2
2.To determine the speed of the sound by echo method: V=2d/t
Relationship between Loudness and intensity: L=K log I[10]
Physics Formulas For ICSE Class 10 Chapter Current Electricity
CHARGE : Units of Charge:
The S.I unit of charge is coulumb denoted by symbol (C).
The smaller units of Charge are :
• Milli-coulomb
• Micro-coulomb
• Nano-coulomb
1mC=10-³ C , 1 uC=10-⁶ C and 1nC=10-⁹ C
CURRENT : Units of current:
S.I unit of current is Ampere denoted by symbol (A)
The smaller units of Ampere:
• Milli-ampere (mA)
• Micro-ampere (uA)
1mA = 10-³A and 1uA = 10-⁶A
Current ( I ) = Q / t
Q = n × e
I = ne / t
• POTENTIAL DIFFERENCE :
V = W / Q
Unit of potential difference
S.I unit of potential difference is volt denoted by symbol (V).
It is a scalar quantity.
1volt = 1joule/1coulumb.
• RESISTANCE :
Unit of Resistance
S.I unit of resistance is Ohm.
Unit of R = Unit of V/Unit of I.
Higher resistance are measured in:
1 kilo-ohm = 10-³ ohm
1 Mega-ohm = 10⁶ ohm
Resistance is directly proportional to :
Lenght if conductor and temperature of the conductor.
Indirectly proportional to thickness of conductor.
• OHM’S LAW
Formula for Ohm’s law: V=IR
Where ,
V is potential difference.
I is current.
R is resistance.
• Conductance =
Formula : Conductance = 1/Resistance.
Unit = ( ohm ) ^-1
• SPECIFIC RESISTANCE = Ra / l
R = resistance, a = area of cross section, l = lenght.
Unit = ohm × metre
Conductivity = 1 / specific resistance.
Unit = 1 / ohm × metre.
Formula : E = W/q
Where, E is e.m.f, W is work done q is Charge.
Formula : V = w’/q
Where, V is potential difference . W’ is work done, q is charge.
VOLTAGE DROP IN A CELL :
Formula : v = w/q
Relationship between e.m.f & terminal voltage of a cell :
Formula : E=V+v Or V=E-v
Internal resistance of a cell Formula : v=Ir.
1: Total resistance of circuit = R+r
2: Current drawn from the cell : I = E/R+r.
3: E.m.f of a cell: (Use this one not the above one) E = I(R+r)
4:The terminal voltage of the cell: V=IR.
5: Voltage drop due to internal resistance = v=Ir
6: Internal Resistance:
1:Equivalent resistance in series: Rs = R1 + R2 + R3 …….+ Rn times..
2:Equivalent resistance in parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 ……+1/Rn
Physics Formulas For ICSE Class 10 Chapter Calorimetry
HEAT : Units of heat :
S.I unit of heat is joule (J)
And the other unit commonly ised unit is calorie (cal).
Relationship between calorie and heat: 1calorie = 4.186J or 4.2 J nearly.
TEMPERATURE: Unit of temperature:
S.I unit of temperature is Kelvin (K)
The other commonly used unit is degree celsius (C).
Relationship between celsius and kelvin: TK = 273 + tC.
Amount of heat absorbed:
Formula : Q=mc×delta t.
HEAT CAPACITY : Unit of heat capacity:
S.I unit of heat capacity is Joule per kelvin (JK-¹).
It is denoted by (C’).
Formula for heat capacity:
C’=Q/ delta t
Where ,
C’ is heat capacity
Q is total heat energy
Delta t is temperature difference
SPECIFIC HEAT CAPACITY : Unit of heat capacity:
S.I unit of heat capacity is Joule per kilogram kelvin (JK-¹).
It is denoted by (c).
Formula for specific heat capacity:
c = Q/m× delta t
RELATIONSHIP BETWEEN HEAT CAPACITY AND SPECIFIC HEAT CAPACITY : Heat capacity C’ = mass m × specific heat capacity c
Principal of caloriemeter: Heat energy lost by the hot body = Heat energy gained by cold body.
Formula : m1 c1 (t1 -t) = m2 c2 (t -t2)
Where, M is mass, c is specific heat capacity t is temperature.
Unit of specific Latent heat :
S.I unit of specific latent heat is Joule per kilogram (Jkg-¹).
It is denoted by (L)
Formula: L = Q/m or Q= mL
L is specific heat capacity
Q is total heat absorbed
m is mass.
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[QSMS Seminar 14,16 Dec] A brief introduction to differential graded Lie algebras I, II
• Date : 12월 14일(화), 16일(목) 16:00-17:30
• Place : Zoom (ID: 642 675 5874)
• Speaker : 조창연 (QSMS, SNU)
• Title : A brief introduction to differential graded Lie algebras I, II
• Abstract : The importance of differential graded Lie algebras goes back at least to Quillen’s rational homotopy theory, which also motivated their applications to deformation theory. Later, such
an idea was developed further by Deligne, Drinfeld, and Feigin, and influenced many including Kontsevich and Soibelman. The purpose of these talks is to give a short introduction to the notion of
differential graded Lie algebras and its relationship to deformation theory. These talks are intended to be an elementary introduction to the subject, but due to the current nature of it, I’ll
say something about the theory of infinity-categories. The first talk will be devoted to exploring some of the fundamentals of differential graded Lie algebras and infinity-categories, and the
application to deformation theory will be covered in the later half of the second talk. | {"url":"https://qsms.math.snu.ac.kr/index.php?mid=board_sjXR83&listStyle=viewer&order_type=desc&l=en&sort_index=title&page=7&document_srl=2055","timestamp":"2024-11-07T13:31:25Z","content_type":"text/html","content_length":"21650","record_id":"<urn:uuid:36cf4f7e-c606-457e-bf3c-7415627deb68>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00067.warc.gz"} |
Win odds Archives • Statisticelle
Two of my recent blog posts focused on two different, but as we will see related, methods which essentially transform observed responses into a summary of their contribution to an estimate:
structural components resulting from Sen’s (1960) decomposition of U-statistics and pseudo-observations resulting from application of the leave-one-out jackknife. As I note in this comment, I think
the real value of deconstructing estimators in this way results from the use of these quantities, which in special (but common) cases are asymptotically uncorrelated and identically distributed, to:
(1) simplify otherwise complex variance estimates and construct interval estimates, and (2) apply regression methods to estimators without an existing regression framework.
As discussed by Miller (1974), pseudo-observations may be treated as approximately independent and identically distributed random variables when the quantity of interest is a function of the mean or
variance, and more generally, any function of a U-statistic. Several other scenarios where these methods are applicable are also outlined. Many estimators of popular “parameters” can actually be
expressed as U-statistics. Thus, these methods are quite broadly applicable. A review of basic U-statistic theory and some common examples, notably the difference in means or the Wilcoxon
Mann-Whitney test statistic, can be found within my blog post: One, Two, U: Examples of common one- and two-sample U-statistics.
As an example of use case (1), Delong et al. (1988) used structural components to estimate the variances and covariances of the areas under multiple, correlated receiver operator curves or multiple
AUCs. Hanley and Hajian-Tilaki (1997) later referred to the methods of Delong et al. (1988) as “the cleanest and most elegant approach to variances and covariances of AUCs.” As an example of use case
(2), Andersen & Pohar Perme (2010) provide a thorough summary of how pseudo-observations can be used to construct regression models for important survival parameters like survival at a single time
point and the restricted mean survival time.
Now, structural components are restricted to U-statistics while pseudo-observations may be used more generally, as discussed. But, if we construct pseudo-observations for U-statistics, one of several
“valid” scenarios, what is the relationship between these two quantities? Hanley and Hajian-Tilaki (1997) provide a lovely discussion of the equivalence of these two methods when applied to the area
under the receiver operating characteristic curve or simply the AUC. This blog post follows their discussion, providing concrete examples of computing structural components and pseudo-observations
using R, and demonstrating their equivalence in this special case.
Continue reading Nonparametric neighbours: U-statistic structural components and jackknife pseudo-observations for the AUC | {"url":"https://statisticelle.com/tag/win-odds/","timestamp":"2024-11-08T01:34:27Z","content_type":"text/html","content_length":"54724","record_id":"<urn:uuid:0432137f-1de9-4524-a5f7-1ab7c4585af2>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00601.warc.gz"} |
BQM inequality constraint
Hi everyone,
I formulated an optimization problem as following:
How can I re-formulate inequality constraints to be able to embed them for quantum annealer?
5 comments
• Hi,
I hope you find the following resources helpful.
Getting Started with D-Wave Solvers
Comment actions
• Hi @Mohammad, I am following up on this topic as I am interest too. At your links I could only find examples for equality constraints. What about the inequality constraints? How can we embed them
for quantum annealer?
Thank you!
Comment actions
• Hi,
For BQM, inequality constraints can be reduced to equality constraints by introducing slack variables.
more info: https://docs.dwavesys.com/docs/latest/handbook_reformulating.html?highlight=inequality#constraints-linear-inequality-penalty-functions
You can use BQM .add_linear_inequality_constraint() method to add a linear inequality constraint as a quadratic objective. Which calculates and returns the slack terms.
more info: https://docs.ocean.dwavesys.com/en/stable/docs_dimod/reference/generated/dimod.binary.BinaryQuadraticModel.add_linear_inequality_constraint.html#
Comment actions
• Thank you Mohammad D!
If I have understood well, BQM .add_linear_inequality_constraint() accepts constraints that are already formulated as QUBOs. In my case, my constraints are defined in a generic format, i.e. as an
arithmetic expression (something that is accepted by CQM). So, my problem is to convert this expression to a BQM that can be added as a penalty term to the 'composite' BQM.
I tried to set my problem directly as a CQM and then convert the CQM to a BQM to sample it with DWaveSampler, but the process is terribly slow as my problem is a large one (90 x 90000 variables).
That is why I was thinking to individually convert each constraint (together with the objective) to a BQM and then add it to the 'composite' BQM.
Maybe there is a better strategy for this, but I do not know.
Thank you,
Comment actions
• Angela,
While not a direct answer to your question, you may get some ideas by watching this presentation from Victoria Goliber, "Using NumPy for Large Quadratic Models", where she discusses some
techniques for building for large quadratic models, which are faster than using the conventional ways of building them.
Comment actions
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[Solved] An industrial load that consumes 80 kW is | SolutionInn
An industrial load that consumes 80 kW is supplied by the power company through a transmission line
An industrial load that consumes 80 kW is supplied by the power company through a transmission line with 0.1 ohms resistance, with 84kW. If the voltage at the load is 440 Vrms, find the power factor
at the load.
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Numbers 1:1 Artwork | Bible Art
An artistic representation of the introduction to the Book of Numbers, in the devotional and respectful spirit of Christianity. Conceptualize this image with the stylistic characteristics of the
Renaissance era, emphasizing characteristics like balanced compositions, lifelike human figures, depth of field, and use of light & shadow. Avoid including any text or words in the image.
Introduction Of The Book of Numbers
See More From Numbers
Introduction Of The Book of Numbers
Numbers 1:14 - "Of Gad; Eliasaph the son of Deuel."
Numbers 1:8 - "Of Issachar; Nethaneel the son of Zuar."
Numbers 31:44 - "And thirty and six thousand beeves,"
Numbers 32:35 - "And Atroth, Shophan, and Jaazer, and Jogbehah,"
Numbers 26:11 - "Notwithstanding the children of Korah died not."
Numbers 1:9 - "Of Zebulun; Eliab the son of Helon."
Numbers 1:13 - "Of Asher; Pagiel the son of Ocran."
Numbers 1:12 - "Of Dan; Ahiezer the son of Ammishaddai."
Numbers 1:11 - "Of Benjamin; Abidan the son of Gideoni."
Numbers 1:6 - "Of Simeon; Shelumiel the son of Zurishaddai."
Numbers 26:8 - "And the sons of Pallu; Eliab."
Numbers 31:46 - "And sixteen thousand persons;)"
pharaoh of egypt looking at the vast numbers of israelites
Numbers 3:43 - "And all the firstborn males by the number of names, from a month old and upward, of those that were numbered of them, were twenty and two thousand two hundred and threescore and
Numbers 7:16 - "One kid of the goats for a sin offering:"
Numbers 8:1 - "And the LORD spake unto Moses, saying,"
Numbers 34:21 - "Of the tribe of Benjamin, Elidad the son of Chislon."
Numbers 13:9 - "Of the tribe of Benjamin, Palti the son of Raphu."
Numbers 4:21 - "¶ And the LORD spake unto Moses, saying,"
Numbers 9:9 - "¶ And the LORD spake unto Moses, saying,"
Numbers 31:45 - "And thirty thousand asses and five hundred,"
Numbers 7:52 - "One kid of the goats for a sin offering:"
Numbers 26:36 - "And these are the sons of Shuthelah: of Eran, the family of the Eranites."
Numbers 33:29 - "And they went from Mithcah, and pitched in Hashmonah." | {"url":"https://bible.art/p/SzB5mNku8fMSvsSIIe25/introduction-of-the-book-of-numbers","timestamp":"2024-11-14T20:48:00Z","content_type":"text/html","content_length":"129759","record_id":"<urn:uuid:a7c00c4d-ec8b-449e-a9f7-e3a050659ae1>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00402.warc.gz"} |
When binary data is transmitted, usually electronically, there is a chance that the data gets corrupted. One method to pick up said corruption is to generate some value that is coded from the
original data, send said value to the receiver, then confirm that the received data generates the same value when it's coded at the destination.
The way to minimize false negatives is to choose coding algorithms that cause a lot of churn per input, especially a variable amount.
Cyclic Redundancy Codes are a type of consistency check that treats the message data as a (long) dividend of a modulo-2 polynomial division. Modulo-2 arithmetic doesn't use carries/borrows when
combining numbers. A specific CRC defines a set number of bits to work on at a time, where said number is also the degree of a fixed polynomial (with modulo-2 coefficients) used as a divisor.
Since ordering doesn't apply to modulo arithmetic, the check between the current high part of the dividend and the trial partial product (of the divisor and the trial new quotient coefficient) is
done by seeing if the highest-degree coefficient of the dividend is one. (The highest-degree coefficient of the divisor must be one by definition, since it's the only non-zero choice.) The remainder
after the division is finished is used as the basis of the CRC checksum.
For a given degree x for the modulo-2 polynomial divisor, the remainder will have at most x terms (from degree x - 1 down to the constant term). The coefficients are modulo-2, which means that they
can be represented by 0's and 1's. So a remainder can be modeled by an (unsigned) integer of at least x bits in width.
The divisor must have its x degree term be one, which means it is always known and can be implied instead of having to explicitly include in representations. Its lower x terms must be specified, so a
divisor can be modeled the same way as remainders. With such a modeling, the divisor representation could be said to be truncated since the uppermost term's value is implied and not stored.
The remainder and (truncated) divisor polynomials are stored as basic computer integers. This is in contrast to the dividend, which is modeled from the input stream of data bits, where each new
incoming bit is the next lower term of the dividend polynomial. Long division can be processed in piecemeal, reading new upper terms as needed. This maps to reading the data a byte (or bit) at a
time, generating updated remainders just-in-time, without needing to read (and/or store(!)) the entire data message at once.
Long division involves appending new dividend terms after the previous terms have been processed into the (interim) remainder. So the remainder it the only thing that has to change during each
division step; a new input byte (or bit) is combined with the remainder to make the interim dividend, and then combined with the partial product (based on the divisor and top dividend bit(s)) to
become a remainder again.
When all of the input data has been read during division, the last x bits are still stuck in the interim remainder. They have not been pushed through the division steps; to do so, x zero-valued extra
bits must be passed into the system. This ensures all of the message's data bits get processed. The post-processed remainder is the checksum. The system requires the message to be augmented with x
extra bits to get results.
Alternatively, if the post-division augmentation bits are the expected checksum instead, then the remainder will "subtract" the checksum with itself, giving zero as the final remainder. The remainder
will end up non-zero if bit errors exist in either the data or checksum or both. This option requires the checksum to be fed from highest-order bit first on down (i.e. big endian).
Exploiting the properties of how the division is carried out, the steps can be rearranged such that the post-processing zero-valued bits are not needed; their effect is merged into the start of the
process. Such systems read unaugmented messages and expose the checksum directly from the interim remainder afterwards. (You can't use the "augment-message-with-checksum-and-zero-check" technique
with this, of course.)
Since long division proceeds from the uppermost terms on down, it's easiest to treat an incoming byte as the uppermost unprocessed terms, and to read the bits within that byte as the highest-order
bit is the uppermost unprocessed term and go down. However, some hardware implementations have an easier time reading each byte from the lowest-order bit and go up. To simulate those systems in
software, the program needs to be flagged that input reflection needs to be applied. Reflecting a built-in integer reverses the order of its bits, such that the lowest- and highest-order bits swap
states, the next-lowest- and next-highest-order bits swap, etc. The input reflection can be done by reflecting each byte as it comes in or keeping the bytes unchanged but reflect the other internal
functioning. The latter sounds harder, but what it usually done in the real world, since it's a one-time cost, unlike reflecting the bytes.
Similarly, the final remainder is processed by some hardware in reverse order, which means software that simulate such systems need to flag that output reflection is in effect.
Some CRCs don't return the remainder directly (reflected or not), but add an extra step complementing the output bits. Complementing turns 1 values into 0 values and vice versa. This can simulated by
using a XOR (exclusive-or) bit mask of all 1-values (of the same bit length as the remainder). Some systems use a final XOR mask that isn't all 1-values, for variety. (This mask takes place after any
output reflection.)
At the other end, the built-in-integer register normally starts at zero as the first bytes are read. Instead of just doing nothing but load input bits for x steps, some CRC systems use a non-zero
initial remainder to add extra processing. This initial value has to be different for the augmented versus un-augmented versions of the same system, due to possible incorporation with the zero-valued
augment bits.
The Rocksoft™ Model CRC Algorithm, or RMCA for short, was designed by Ross Williams to describe all the specification points of a given CRC system (quoted):
RMCA Parameters
This is the width of the algorithm expressed in bits. This is one less than the width of the Poly.
This parameter is the poly. This is a binary value that should be specified as a hexadecimal number. The top bit of the poly should be omitted. For example, if the poly is 10110, you should
specify 06. An important aspect of this parameter is that it represents the unreflected poly; the bottom bit of this parameter is always the LSB of the divisor during the division regardless of
whether the algorithm being modelled is reflected.
This parameter specifies the initial value of the register when the algorithm starts. This is the value that is to be assigned to the register in the direct table algorithm. In the table
algorithm, we may think of the register always commencing with the value zero, and this value being XORed into the register after the N'th bit iteration. This parameter should be specified as a
hexadecimal number.
This is a boolean parameter. If it is FALSE, input bytes are processed with bit 7 being treated as the most significant bit (MSB) and bit 0 being treated as the least significant bit. If this
parameter is FALSE, each byte is reflected before being processed.
This is a boolean parameter. If it is set to FALSE, the final value in the register is fed into the XOROUT stage directly, otherwise, if this parameter is TRUE, the final register value is
reflected first.
This is an W-bit value that should be specified as a hexadecimal number. It is XORed to the final register value (after the REFOUT) stage before the value is returned as the official checksum.
His description assumes an octet-sized byte. The POLY is the (truncated) divisor. The INIT is the initial remainder, assuming the unaugmented version of CRC processing is used. (If you're using an
augmented-style CRC, you have to undo the effect of the built-in zero-augment before initialization.)
The two function templates and two class templates in this library provide ways to carry out CRC computations. You give the various Rocksoft™ Model CRC Algorithm parameters as template parameters and
/or constructor parameters. You then submit all the message data bytes at once (for the functions) or piecemeal (for the class objects).
Note that some error-detection techniques merge their checksum results within the message data, while CRC checksums are either at the end (when augmented, without either kind of reflection, with a
bit-width that's a multiple of byte size, and no XOR mask) or out-of-band. | {"url":"https://www.boost.org/doc/libs/1_73_0/doc/html/crc/introduction.html","timestamp":"2024-11-08T02:46:46Z","content_type":"text/html","content_length":"20854","record_id":"<urn:uuid:7026ec38-75f6-4888-8190-4f7092044fd8>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00462.warc.gz"} |
4.6 DC Circuits Containing Resistors and Capacitors
Learning Objectives
Learning Objectives
By the end of this section, you will be able to do the following:
• Explain the importance of the time constant τ, and calculate the time constant for a given resistance and capacitance
• Explain why batteries in a flashlight gradually lose power and the light dims over time
• Describe what happens to a graph of the voltage across a capacitor over time as it charges
• Explain how a timing circuit works and list some applications
• Calculate the necessary speed of a strobe flash needed to stop the movement of an object over a particular length
The information presented in this section supports the following AP® learning objectives and science practices:
• 5.C.3.6 The student is able to determine missing values and direction of electric current in branches of a circuit with both resistors and capacitors from values and directions of current in
other branches of the circuit through appropriate selection of nodes and application of the junction rule. (S.P. 1.4, 2.2)
• 5.C.3.7 The student is able to determine missing values, direction of electric current, charge of capacitors at steady state, and potential differences within a circuit with resistors and
capacitors from values and directions of current in other branches of the circuit. (S.P. 1.4, 2.2)
When you use a flash camera, it takes a few seconds to charge the capacitor that powers the flash. The light flash discharges the capacitor in a tiny fraction of a second. Why does charging take
longer than discharging? This question and a number of other phenomena that involve charging and discharging capacitors are discussed in this module.
RC Circuits
RC Circuits
An $RCRC size 12{ ital "RC"} {}$ circuit is one containing a resistor$RR$$size 12{R} {}$ and a capacitor $C.C. size 12{C} {}$ The capacitor is an electrical component that stores electric charge.
Figure 4.41 shows a simple $RCRC size 12{ ital "RC"} {}$ circuit that employs a DC (direct current) voltage source. The capacitor is initially uncharged. As soon as the switch is closed, current
flows to and from the initially uncharged capacitor. As charge increases on the capacitor plates, there is increasing opposition to the flow of charge by the repulsion of like charges on each plate.
In terms of voltage, this is because voltage across the capacitor is given by $Vc=Q/C,Vc=Q/C, size 12{V rSub { size 8{c} } =Q/C} {}$ where $QQ size 12{Q} {}$ is the amount of charge stored on each
plate and $CC size 12{C} {}$ is the capacitance. This voltage opposes the battery, growing from zero to the maximum emf when fully charged. The current thus decreases from its initial value of $I0=
emfRI0=emfR size 12{I rSub { size 8{0} } = { {"emf"} over {R} } } {}$ to zero as the voltage on the capacitor reaches the same value as the emf. When there is no current, there is no $IRIR size 12{
ital "IR"} {}$ drop, so the voltage on the capacitor must then equal the emf of the voltage source. This can also be explained with Kirchhoff’s second rule (the loop rule), discussed in Kirchhoff’s
Rules, which says that the algebraic sum of changes in potential around any closed loop must be zero.
The initial current is $I0=emfRI0=emfR size 12{I rSub { size 8{0} } = { {"emf"} over {R} } } {}$ because all of the $IRIR size 12{ ital "IR"} {}$ drop is in the resistance. Therefore, the smaller the
resistance, the faster a given capacitor will be charged. Note that the internal resistance of the voltage source is included in $R,R, size 12{R} {}$ as are the resistances of the capacitor and the
connecting wires. In the flash camera scenario above, when the batteries powering the camera begin to wear out, their internal resistance rises, reducing the current and lengthening the time it takes
to get ready for the next flash.
Voltage on the capacitor is initially zero and rises rapidly at first since the initial current is a maximum. Figure 4.41(b) shows a graph of capacitor voltage versus time ($tt size 12{t} {}$)
starting when the switch is closed at $t=0.t=0. size 12{t=0} {}$ The voltage approaches emf asymptotically since the closer it gets to emf, the less current flows. The equation for voltage versus
time when charging a capacitor $CC size 12{C} {}$ through a resistor $R,R, size 12{R} {}$ derived using calculus, is
4.77 $V=emf(1−e−t/RC) (charging),V=emf(1−e−t/RC) (charging), size 12{V="emf" \( 1 - e rSup { size 8{ - t/ ital "RC"} } \) } {}$
where $VV size 12{V} {}$ is the voltage across the capacitor, emf is equal to the emf of the DC voltage source, and the exponential e = 2.718 … is the base of the natural logarithm. Note that the
units of $RCRC size 12{ ital "RC"} {}$ are seconds. We define
4.78 $τ=RC,τ=RC, size 12{τ= ital "RC"} {}$
where $ττ size 12{τ} {}$ (the Greek letter tau) is called the time constant for an $RCRC size 12{ ital "RC"} {}$ circuit. As noted before, a small resistance $RR size 12{R} {}$ allows the capacitor
to charge faster. This is reasonable since a larger current flows through a smaller resistance. It is also reasonable that the smaller the capacitor $C,C, size 12{C} {}$ the less time needed to
charge it. Both factors are contained in $τ=RC.τ=RC. size 12{τ= ital "RC"} {}$
More quantitatively, consider what happens when $t=τ=RC.t=τ=RC. size 12{t=τ= ital "RC"} {}$ Then the voltage on the capacitor is
4.79 $V=emf1−e−1=emf1−0.368=0.632⋅emf.V=emf1−e−1=emf1−0.368=0.632⋅emf. size 12{V="emf" left (1 - e rSup { size 8{ - 1} } right )="emf" left (1 - 0 "." "368" right )=0 "." "632" cdot "emf"} {}$
This means that in the time $τ=RC,τ=RC, size 12{τ= ital "RC"} {}$ the voltage rises to 0.632 of its final value. The voltage will rise 0.632 of the remainder in the next time $τ.τ. size 12{τ} {}$ It
is a characteristic of the exponential function that the final value is never reached, but 0.632 of the remainder to that value is achieved in every time, $τ.τ. size 12{τ} {}$ In just a few multiples
of the time constant $τ,τ, size 12{τ} {}$ then, the final value is very nearly achieved, as the graph in Figure 4.41(b) illustrates.
Discharging a Capacitor
Discharging a Capacitor
Discharging a capacitor through a resistor proceeds in a similar fashion, as Figure 4.42 illustrates. Initially, the current is $I0=V0R,I0=V0R, size 12{I rSub { size 8{0} } = { {V rSub { size 8{0} }
} over {R} } } {}$ driven by the initial voltage $V0V0 size 12{V rSub { size 8{0} } } {}$ on the capacitor. As the voltage decreases, the current and hence the rate of discharge decrease, implying
another exponential formula for $V.V. size 12{V} {}$ Using calculus, the voltage $VV size 12{V} {}$ on a capacitor $CC size 12{C} {}$ being discharged through a resistor $RR size 12{R} {}$ is found
to be
4.80 $V=V0e−t/RC (discharging).V=V0e−t/RC (discharging). size 12{V=`V"" lSub { size 8{0} } `e rSup { size 8{ - t/ ital "RC"} } } {}$
The graph in Figure 4.42(b) is an example of this exponential decay. Again, the time constant is $τ=RC.τ=RC. size 12{τ= ital "RC"} {}$ A small resistance $RR size 12{R} {}$ allows the capacitor to
discharge in a small time since the current is larger. Similarly, a small capacitance requires less time to discharge since less charge is stored. In the first time interval $τ=RCτ=RC size 12{τ= ital
"RC"} {}$ after the switch is closed, the voltage falls to 0.368 of its initial value since $V=V0⋅e−1=0.368V0.V=V0⋅e−1=0.368V0. size 12{V=V rSub { size 8{0} } cdot e rSup { size 8{ - 1} } =0 "."
"368"V rSub { size 8{0} } } {}$
During each successive time $τ,τ, size 12{τ} {}$ the voltage falls to 0.368 of its preceding value. In a few multiples of $τ,τ, size 12{τ} {}$ the voltage becomes very close to zero, as indicated by
the graph in Figure 4.42(b).
Now we can explain why the flash camera in our scenario takes so much longer to charge than discharge; the resistance while charging is significantly greater than while discharging. The internal
resistance of the battery accounts for most of the resistance while charging. As the battery ages, the increasing internal resistance makes the charging process even slower. (You may have noticed
The flash discharge is through a low-resistance ionized gas in the flash tube and proceeds very rapidly. Flash photographs, such as in Figure 4.43, can capture a brief instant of a rapid motion
because the flash can be less than a microsecond in duration. Such flashes can be made extremely intense.
During World War II, nighttime reconnaissance photographs were made from the air with a single flash illuminating more than a square kilometer of enemy territory. The brevity of the flash eliminated
blurring due to the surveillance aircraft’s motion. Today, an important use of intense flash lamps is to pump energy into a laser. The short intense flash can rapidly energize a laser and allow it to
reemit the energy in another form.
Example 4.6 Integrated Concept Problem: Calculating Capacitor Size—Strobe Lights
High-speed flash photography was pioneered by Doc Edgerton in the 1930s while he was a professor of electrical engineering at MIT. You might have seen examples of his work in the amazing shots of
hummingbirds in motion, a drop of milk splattering on a table, or a bullet penetrating an apple (see Figure 4.43). To stop the motion and capture these pictures, one needs a high-intensity, very
short pulsed flash, as mentioned earlier in this module.
Suppose one wished to capture the picture of a bullet (moving at $5.0×102m/s5.0×102m/s$) that was passing through an apple. The duration of the flash is related to the $RCRC size 12{ ital "RC"} {}$
time constant, $τ.τ. size 12{τ} {}$ What size capacitor would one need in the $RCRC size 12{ ital "RC"} {}$ circuit to succeed if the resistance of the flash tube were $10.0 Ω10.0 Ω size 12{"10"
%OMEGA } {}$? Assume the apple is a sphere with a diameter of $8.0×10–2m.8.0×10–2m.$
We begin by identifying the physical principles involved. This example deals with the strobe light, as discussed above. Figure 4.42 shows the circuit for this probe. The characteristic time $ττ size
12{τ} {}$ of the strobe is given as $τ=RC.τ=RC. size 12{τ= ital "RC"} {}$
We wish to find $C,C, size 12{C} {}$ but we don’t know $τ.τ. size 12{τ} {}$ We want the flash to be on only while the bullet traverses the apple. So we need to use the kinematic equations that
describe the relationship between distance $x,x, size 12{x} {}$ velocity $v,v, size 12{v} {}$ and time $t.t. size 12{t} {}$
4.81 $x=vtort=xvx=vtort=xv size 12{t= { {x} over {v} } } {}$
The bullet’s velocity is given as $5.0×102m/s,5.0×102m/s,$ and the distance $xx size 12{x} {}$ is $8.0×10–2m.8.0×10–2m.$ The traverse time, then, is
4.82 $t=xv=8.0×10–2m5.0×102m/s=1.6×10−4s.t=xv=8.0×10–2m5.0×102m/s=1.6×10−4s. size 12{t= { {x} over {v} } = { {0 "." "08"" m"} over {"500 m/s"} } =1 "." 6 times "10" rSup { size 8{ - 4} } " s"} {}$
We set this value for the crossing time $tt size 12{t} {}$ equal to $τ.τ. size 12{τ} {}$ Therefore,
4.83 $C=tR=1.6×10−4s10.0 Ω=16μF.C=tR=1.6×10−4s10.0 Ω=16μF. size 12{C= { {t} over {R} } = { { left (1 "." 6´"10" rSup { size 8{-4} } right )} over {"10"} } ="16" μF} {}$
Note—Capacitance $CC size 12{C} {}$ is typically measured in farads, $F , F ,$ defined as Coulombs per volt. From the equation, we see that $CC size 12{C} {}$ can also be stated in units of seconds
per ohm.
The flash interval of $160 μs160 μs size 12{"160" ms} {}$ (the traverse time of the bullet) is relatively easy to obtain today. Strobe lights have opened up new worlds from science to entertainment.
The information from the picture of the apple and bullet was used in the Warren Commission Report on the assassination of President John F. Kennedy in 1963 to confirm that only one bullet was fired.
RC Circuits for Timing
RC Circuits for Timing
null circuits are commonly used for timing purposes. A mundane example of this is found in the ubiquitous intermittent wiper systems of modern cars. The time between wipes is varied by adjusting the
resistance in an null circuit. Another example of an null circuit is found in novelty jewelry, Halloween costumes, and various toys that have battery-powered flashing lights. See Figure 4.44 for a
timing circuit.
A more crucial use of null circuits for timing purposes is in the artificial pacemaker, used to control heart rate. The heart rate is normally controlled by electrical signals generated by the
sino-atrial (SA) node, which is on the wall of the right atrium chamber. This causes the muscles to contract and pump blood. Sometimes the heart rhythm is abnormal and the heartbeat is too high or
too low.
The artificial pacemaker is inserted near the heart to provide electrical signals to the heart when needed with the appropriate time constant. Pacemakers have sensors that detect body motion and
breathing to increase the heart rate during exercise to meet the body’s increased needs for blood and oxygen.
Example 4.7 Calculating Time: RC Circuit in a Heart Defibrillator
A heart defibrillator is used to resuscitate an accident victim by discharging a capacitor through the trunk of her body. A simplified version of the circuit is seen in Figure 4.42. (a) What is the
time constant if an null capacitor is used and the path resistance through her body is $1.00×103Ω?1.00×103Ω?$ (b) If the initial voltage is 10.0 kV, how long does it take to decline to $5.00×102V?
Since the resistance and capacitance are given, it is straightforward to multiply them to give the time constant asked for in part (a). To find the time for the voltage to decline to
$5.00×102V,5.00×102V,$ we repeatedly multiply the initial voltage by 0.368 until a voltage less than or equal to $5.00×102V5.00×102V$ is obtained. Each multiplication corresponds to a time of null
Solution for (a)
The time constant null is given by the equation null Entering the given values for resistance and capacitance (and remembering that units for a farad can be expressed as null gives
4.84 null
Solution for (b)
In the first 8.00 ms, the voltage (10.0 kV) declines to 0.368 of its initial value. That is,
4.85 null
(Notice that we carry an extra digit for each intermediate calculation.) After another 8.00 ms, we multiply by 0.368 again, and the voltage is
4.86 $V ′ = 0.368 V = 0.368 3.680 × 10 3 V = 1.354 × 10 3 V at t = 16.0 ms. V ′ = 0.368 V = 0.368 3.680 × 10 3 V = 1.354 × 10 3 V at t = 16.0 ms.$
Similarly, after another 8.00 ms, the voltage is
4.87 $V ′′ = 0.368 V ′ = ( 0.368 ) ( 1.354 × 10 3 V ) = 498 V at t = 24 .0 ms. V ′′ = 0.368 V ′ = ( 0.368 ) ( 1.354 × 10 3 V ) = 498 V at t = 24 .0 ms.$
So after only 24.0 ms, the voltage is down to 498 V, or 4.98 percent of its original value.Such brief times are useful in heart defibrillation because the brief but intense current causes a brief but
effective contraction of the heart. The actual circuit in a heart defibrillator is slightly more complex than the one in Figure 4.42 to compensate for magnetic and AC effects that will be covered in
Check Your Understanding
When is the potential difference across a capacitor an emf?
Only when the current being drawn from or put into the capacitor is zero. Capacitors, like batteries, have internal resistance, so their output voltage is not an emf unless current is zero. This is
difficult to measure in practice, so we refer to a capacitor’s voltage rather than its emf. But the source of potential difference in a capacitor is fundamental, and it is an emf. | {"url":"https://texasgateway.org/resource/46-dc-circuits-containing-resistors-and-capacitors?book=79106&binder_id=78816","timestamp":"2024-11-04T12:10:38Z","content_type":"text/html","content_length":"111009","record_id":"<urn:uuid:6742ac0a-f46a-4f5c-bbe2-a98d3a8f9381>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00633.warc.gz"} |
3 Best Ways to Handle Right Skewed Data
Why is Normality so important?
Linear Discriminant Analysis (LDA), Linear Regression, and many other parametric machine learning models assume that data is normally distributed. If this assumption is not met, the model will not
provide accurate predictions.
What is normal distribution?
Normal distribution is a type of probability distribution that is defined by a symmetric bell-shaped curve. The curve is defined by its centre (mean), spread (standard deviation), and skewness.
A normal distribution is the most common type of distribution found in nature. Many real-world phenomena, such as height, weight, and IQ, follow a normal distribution.
Normal distribution is important in statistics and is used to model many real-world phenomena. It is also used in quality control and engineering to determine acceptable tolerances.
What is skewness?
Skewness is the degree of asymmetry of a distribution. A distribution is symmetric if it is centred around its mean and the left and right sides are mirror images of each other. A distribution is
skewed if it is not symmetric.
There are two types of skewness:
Positive Skewness: If the bulk of the values fall on the right side of the curve and the tail extends towards the right, it is known as positive skewness.
Negative Skewness: If the bulk of the values fall on the left side of the curve and the tail extends towards the left, it is known as negative skewness.
What does skewness tell us?
To understand this better consider a example:
Suppose car prices may range from 100k to 1,000,000 with the average being 500,000.
If the distribution’s peak is on the left side, our data is positively skewed and the majority of the cars are being sold for less than the average price.
If the distribution’s peak is on the right side, our data is negatively skewed and the majority of the cars are being sold for more than the average price.
To experience what we have learned till now, we are going to work on a simple dataset.
The dataset used in this blog can be downloaded from the link given below: https://www.kaggle.com/datasets/amitabhajoy/bengaluru-house-price-data
About Dataset:
We are going to use a dataset from Kaggle, which is a platform for Data Science communities. The dataset contains the prices of houses in Bengaluru, India.
The dataset includes:
• Area Type: Type of plot
• Availability: Ready to move or not
• Location: Region of Bangalore
• Size: BHK
• Society: Colony in which the house is present
• Total Sq. Ft: Total area
• Bath: Number of bathrooms
• Balcony: Number of balconies
• Price: Cost in lakhs
Note: As we can see the skewed values lie between -1 and greater than +1. Hence, data is heavily skewed. We can say that data[‘price’] is right skewed by looking at the graph and skewed values.
How to handle these skewed data?
- Transformation:
In data analysis, transformation is the replacement of a variable by a function of that variable. For example, replacing a variable x by the square root of x or the logarithm of x. In a stronger
sense, a transformation is a replacement that changes the shape of a distribution or relationship.
• Steps to do transformation:
1. Draw a graph (histogram and density plot) of the data to see how far patterns in data match the simplest ideal patterns.
2. Check the range of the data. This is because transformations will have little effect if the range is small.
3. Check the skewness by statistical methods (decide right and left skewness).
4. Apply the methods (explained in detail below) to handle the skewness based on the skewed values.
To Handle Right Skewness
1. Log Transformation
The log transformation is widely used in research to deal with skewed data. It is the best method to handle the right-skewed data.
Why log?
The normal distribution is widely used in basic research studies to model continuous outcomes. Unfortunately, the symmetric bell-shaped distribution often does not adequately describe the observed
data from research projects. Quite often, data arising in real studies are so skewed that standard statistical analyses of these data yield invalid results.
Many methods have been developed to test the normality assumption of observed data. When the distribution of the continuous data is non-normal, transformations of data are applied to make the data as
“normal” as possible, thus, increasing the validity of the associated statistical analyses.
Popular use of the log transformation is to reduce the variability of data, especially in data sets that include outlying observations. Again, contrary to this popular belief, log transformation can
often increase – not reduce – the variability of data, irrespective of whether or not there are outliers.
Why not?
Using transformations in general and log transformation in particular can be quite problematic. If such an approach is used, the researcher must be mindful about its limitations, particularly when
interpreting the relevance of the analysis of transformed data for the hypothesis of interest about the original data.
Note: Log transformation has reduced the skewed value from 8.06 to 0.82, which is nearer to zero.
Note: If you’re getting the skewness value as nan, that means there are some values that are zero. In log transformation, it deals with only the positive and negative numbers, not with zero. The log
ranges between (- infinity to infinity), however, it must be greater or less than zero. For better understanding, let’s have a look at the log graph below:
2. Root Transformation
2.1 Square root Transformation
• The square root means x to x^(1/2) = sqrt(x), is a transformation with a moderate effect on distribution shape. It is weaker than the logarithm and the cube root.
• It is also used for reducing right skewness, and has the advantage that it can be applied to zero values.
• Note that the square root of an area has the units of a length. It is commonly applied to counted data, especially if the values are mostly rather small.
Note: In the previous case we got the skewness value as 0.82, but the square root transformation has reduced the skewed value from 8.06 to 2.86.
2.2 Cube root Transformation
The cube root means x to x^(1/3). This is a fairly strong transformation with a substantial effect on distribution shape.
• It is weaker than the logarithm but stronger than the square root transformation.
• It is also used for reducing right skewness, and has the advantage that it can be applied to zero and negative values.
• Note that the cube root of a volume has the units of a length. It is commonly applied to rainfall data.
Note: In logarithm transformation, we got the skewness value as 0.82, and in the square root transformation it has reduced the skewed value from 8.06 to 2.86. However, now in the cube root
transformation, the skewed values have reduced to 1.98 and it is near to zero compared to 2.86 and 8.06.
1. Reciprocals Transformation
The reciprocal, x to 1/x, with its sibling, the negative reciprocal, which is x to -1/x, is a very strong transformation with a drastic effect on distribution shape.
It cannot be applied to zero values. Although it can be applied to negative values, it is not useful unless all values are positive.
For example: we might want to multiply or divide the results of taking the reciprocal by some constant, such as 100 or 1000, to get numbers that are easy to manage, but that has no effect on skewness
or linearity.
Note: In logarithm, square root, and cube root transformations, we got the skewness value as 0.82, 2.86 and 1.98 respectively. However, now in reciprocal transformation, the skewed value has reduced
to 2.21 and it is near to zero compared to 2.86 and 8.06.
Therefore, we can conclude that log transformation has performed really well when compared to other methods by reducing the skewness value from 8.06 to 0.82.
If you don’t deal with skewed data properly, it can undermine the predictive power of your model.
This should go without saying, but you should remember what transformation you have performed on which attribute, because you’ll have to reverse it when making predictions, so keep that in mind.
Nonetheless, these three approaches should be adequate for you.
If you are interested in pursuing a career in Data Science, enrol for our Full Stack Data Science program, where you will be trained from basics to advanced levels.
Read our recent blog on “ Why do we always take p-value as 5%?”. | {"url":"https://www.almabetter.com/bytes/articles/3-best-ways-to-handle-right-skewed-data","timestamp":"2024-11-12T06:44:58Z","content_type":"text/html","content_length":"1049006","record_id":"<urn:uuid:e0214ff0-c4e6-480c-a1e4-18c978a489a1>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00720.warc.gz"} |
How do you find the standard error of a random sample?
If you don’t know the population parameters, you can find the standard error: Sample mean. Sample proportion….What is the Standard Error Formula?
Statistic (Sample) Formula for Standard Error.
Sample mean, = s / √ (n)
Sample proportion, p = √ [p (1-p) / n)]
What is random sampling error?
A sampling error in cases where the sample has been selected by a random method. It is common practice to refer to random sampling error simply as “sampling error” where the random nature of the
selective process is understood or assumed.
Is standard error the same as random sampling error?
Sampling error is the error that is incurred when the statistical characteristics of a population is estimated from a sample of the population due to the choice of sample. As a concept this is
distinct from the standard error, which you understand correctly.
What is standard error in sampling?
The standard error is a statistical term that measures the accuracy with which a sample distribution represents a population by using standard deviation. In statistics, a sample mean deviates from
the actual mean of a population; this deviation is the standard error of the mean.
What does a standard error of 2 mean?
The standard deviation tells us how much variation we can expect in a population. We know from the empirical rule that 95% of values will fall within 2 standard deviations of the mean. 95% would fall
within 2 standard errors and about 99.7% of the sample means will be within 3 standard errors of the population mean.
What is the difference between standard error and standard error of the mean?
Standard Error is the standard deviation of the sampling distribution of a statistic. Confusingly, the estimate of this quantity is frequently also called “standard error”. The [sample] mean is a
statistic and therefore its standard error is called the Standard Error of the Mean (SEM).
What is a big standard error?
A high standard error shows that sample means are widely spread around the population mean—your sample may not closely represent your population. A low standard error shows that sample means are
closely distributed around the population mean—your sample is representative of your population.
How does sampling error and non-sampling error differ?
The significant differences between sampling and non-sampling error are mentioned in the following points: Sampling error is a statistical error happens due to the sample selected does not perfectly
represents the population of interest. Sampling error arises because of the variation between the true mean value for the sample and the population.
When does sampling errors occur?
A sampling error is a statistical error that occurs when an analyst does not select a sample that represents the entire population of data and the results found in the sample do not represent the
results that would be obtained from the entire population.
What does random error mean?
Definition of random error. : a statistical error that is wholly due to chance and does not recur —opposed to systematic error.
What is sampling error stats?
In statistics, sampling error is the error caused by observing a sample instead of the whole population. The sampling error is the difference between a sample statistic used to estimate a population
parameter and the actual but unknown value of the parameter. An estimate of a quantity of interest,… | {"url":"https://www.rhumbarlv.com/how-do-you-find-the-standard-error-of-a-random-sample/","timestamp":"2024-11-14T03:35:32Z","content_type":"text/html","content_length":"64885","record_id":"<urn:uuid:6d06cee3-3904-4c1b-bbde-552421c0a8c7>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00357.warc.gz"} |
MathFiction: A Map for the Missing (Belinda Huijuan Tang)
Tang Yitian, a Chinese-American math professor who grew up in China shortly after the revolution, undertakes a journey to find his estranged father. Anti-intellectualism always made it hard for
Yitian to get along with his father, but a family tragedy leads the father to completely disown his son. It is their relationship (and nothing about math at all) which really is the driving force
behind this novel.
But, there are a few interesting mathematical scenes.
In one, we see a young Yitian studying math before an exam. Math is not his favorite nor his best subject. He dislikes that (at least from his perspective) math is made up of rules that must be
memorized and not understood. His math teachers refuse to answer his questions. But, his feelings about math improve a bit when the girl upon whom he has a crush finds him studying and helps him with
Yitian had the misfortune to be interested in an academic career in a country where intellectuals were oppressed. He was mostly interested in studying history. Without permission from his father,
Yitian's brother signed Yitian up for the university entrance exams. However, due to his own inability to read, the brother accidentally signed him up for the mathematics exam. That is why Yitian
eventually became a math professor, despite his negative feelings about the subject. (Fortunately, he does begin to like it more. For example, he acknowledges while taking Real Analysis that math is
not as arbitrary or non-sensical as it seemed to him in school.)
The title of the book is derived from a mathematical scene in which Yitian is trying to teach his American students about topology. He explains the idea of the genus as a topological invariant,
showing them how one can transform a donut into a coffee mug (an illustration of which appears in the book) and then:
(quoted from A Map for the Missing)
He expected the class to be amazed, but when he looked up, they only seemed to be as bored as they always were, not at all like he'd been when he heard about the idea of the genus. He could always
tell when the end of their class was approaching by the early sounds of students closing notebooks and zipping up their backpacks. When he thought back to his years in university, he saw he'd
maintained a certain innocence about life and learning these students hadn't. They were jaded about intellectual matters and couldn't summon up any awe about these ideas. It was amazing, he'd thought
back then, that an unchanging property of an object wasn't only what was there, but also what wasn't. It meant that if you could define what was absent, create a map for the missing, that was also a
way of knowing a thing.
He wished such a simple principle was true in his father's case-that the facts he didn't know could be as important as the ones he did The objects of truths he knew about his father were small and
uncertain, without shape. What year his father was born, the year he married, that he'd served in the army. That he hated his own father. The list of unknown things was much more numerous. Why his
father could become so quiet, why he liked to drink, why he and Yitian's grandfather never spoke. In topology, cataloging the holes was a way of forming shape from the absences. The world of
mathematics made this diminished way of knowing useful. Here, in the real world, Yitian couldn't even name how much he didn't know.
Just as it is used in that passage to convey something about his feelings for his father, math is used metaphorically to address social situations elsewhere in the book. Yitian imagines substituting
people into the triangle inequality, and thinks of himself being divided into real and imaginary parts (with his real part being his body sitting in the classroom taking complex analysis and the
imaginary component being connections to things from his past, like his Grandfather's stories).
Yitian only becomes a mathematician due to an error, and mathematics is only discussed explicitly a few times in this book. So, one might question whether this really is mathematical fiction or
simply a story about a person who happens to be a math professor. But, it was not really a mistake by Yitian's brother which is responsible for the fact that he studied math. This is a work of
fiction and that story was a decision made by the author. Why did Belinda Huijuan Tang choose to make Tang Yitian into a math professor? The fact that these mathematical metaphors recur and that the
title is derived from one suggests to me that the author considers the mathematical component to be an essential, even if only small, aspect of the novel. | {"url":"https://kasmana.people.charleston.edu/MATHFICT/mfview.php?callnumber=mf1569","timestamp":"2024-11-04T12:10:52Z","content_type":"text/html","content_length":"12723","record_id":"<urn:uuid:4ff5f034-ba43-4a85-9c27-b7db2036f740>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00739.warc.gz"} |
How do you interpret linear regression?
How do you interpret linear regression?
Assumptions of Linear Regression:The Independent variables should be linearly related to the dependent variables. Every feature in the data is Normally Distributed. There should be little or no
multi-collinearity in the data. The mean of the residual is zero. Residuals obatined should be normally distributed.
How do you interpret a regression graph?
Interpreting the slope of a regression line The slope is interpreted in algebra as rise over run. If, for example, the slope is 2, you can write this as 2/1 and say that as you move along the line,
as the value of the X variable increases by 1, the value of the Y variable increases by 2.
How do you explain regression?
Regression is a statistical method used in finance, investing, and other disciplines that attempts to determine the strength and character of the relationship between one dependent variable (usually
denoted by Y) and a series of other variables (known as independent variables).
How do you interpret multiple regression results?
Interpret the key results for Multiple RegressionStep 1: Determine whether the association between the response and the term is statistically significant.Step 2: Determine how well the model fits
your data.Step 3: Determine whether your model meets the assumptions of the analysis.
How do you interpret a negative regression coefficient?
A negative coefficient suggests that as the independent variable increases, the dependent variable tends to decrease. The coefficient value signifies how much the mean of the dependent variable
changes given a one-unit shift in the independent variable while holding other variables in the model constant.
How do you interpret a dummy variable coefficient?
The coefficient on a dummy variable with a log-transformed Y variable is interpreted as the percentage change in Y associated with having the dummy variable characteristic relative to the omitted
category, with all other included X variables held fixed.
Can regression coefficients be greater than 1?
A beta weight is a standardized regression coefficient (the slope of a line in a regression equation). A beta weight will equal the correlation coefficient when there is a single predictor variable.
β can be larger than +1 or smaller than -1 if there are multiple predictor variables and multicollinearity is present.
Can a linear regression be negative?
Depending on your dependent/outcome variable, a negative value for your constant/intercept should not be a cause for concern. Typically, it is the overall relationships between the variables that
will be of the most importance in a linear regression model, not the value of the constant.
Is linear regression the same as slope?
Remember from algebra, that the slope is the “m” in the formula y = mx + b. In the linear regression formula, the slope is the a in the equation y’ = b + ax. They are basically the same thing. So if
you’re asked to find linear regression slope, all you need to do is find b in the same way that you would find m.
How do you interpret a negative y intercept?
If you extend the regression line downwards until you reach the point where it crosses the y-axis, you’ll find that the y-intercept value is negative! In fact, the regression equation shows us that
the negative intercept is -114.3.
What if the Y intercept is negative?
A positive y-intercept means the line crosses the y-axis above the origin, while a negative y-intercept means that the line crosses below the origin. Simply by changing the values of m and b, we can
define any straight line.
What is the Y intercept in an equation?
The y -intercept of a graph is the point where the graph crosses the y -axis. When the equation of a line is written in slope-intercept form ( y=mx+b ), the y -intercept b can be read immediately
from the equation. Example 1: The graph of y=34x−2 has its y -intercept at −2 .
Why is the Y intercept not statistically meaningful?
In this model, the intercept is not always meaningful. Since the intercept is the mean of Y when all predictors equals zero, the mean is only useful if every X in the model actually has some values
of zero. So while the intercept will be necessary for calculating predicted values, it has to no real meaning.
How do you convert to slope intercept form?
To change the equation into slope-intercept form, we write it in the form y=mx+b. y=mx + b.
What does slope intercept form look like?
Slope-intercept form, y=mx+b, of linear equations, emphasizes the slope and the y-intercept of the line.
How do you form a linear equation if its slope and y intercept are given?
The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept. If you know the slope (m) any y-intercept (b) of a line, this page will show you how to find the
equation of the line.
What is the slope of y =- 1 3x?
Algebra Examples The slope-intercept form is y=mx+b y = m x + b , where m m is the slope and b b is the y-intercept. Combine 13 1 3 and x x . Rewrite in slope-intercept form. Using the
slope-intercept form, the slope is 13 .
How do you calculate the Y intercept?
To find the x-intercept of a given linear equation, plug in 0 for ‘y’ and solve for ‘x’. To find the y-intercept, plug 0 in for ‘x’ and solve for ‘y’. In this tutorial, you’ll see how to find the
x-intercept and the y-intercept for a given linear equation. | {"url":"https://www.idcafe.net/how-do-you-interpret-linear-regression/","timestamp":"2024-11-05T23:22:21Z","content_type":"text/html","content_length":"58181","record_id":"<urn:uuid:2b578257-3979-4479-9633-190a3e5124af>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00076.warc.gz"} |
circular convolution and wraparound error....
Hello forum,
another basic questions, but you experts are better than any book.
Give two finite sequences X and Y of duration, respectively N and M
samples, the convolution will be a new sequence G of length M+N-1 samples.
that said, what is the wraparound error?
The discrete convolution assumes that both X and Y are periodic sequences.
(That way we can approximate the convolution integral whose limits are +
and - infinity in the continuous time domain).
Both X and Y need to have the same period P however.Why?
I know that P can be:
What happens if P<M+N-1? Wraparound error.....What is it? Do I need to
worry or does matlab take care of it when it performs conv(X,Y)? | {"url":"https://www.dsprelated.com/showthread/comp.dsp/126713-1.php","timestamp":"2024-11-06T14:06:50Z","content_type":"text/html","content_length":"70090","record_id":"<urn:uuid:c9261fb4-48cc-4774-8f01-552823062a36>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00374.warc.gz"} |
(PDF) Room-temperature magnetism and tunable energy gaps in edge-passivated zigzag graphene quantum dots
Room-temperature magnetism and tunable energy gaps in
edge-passivated zigzag graphene quantum dots
Wei Hu
, Yi Huang
, Xinmin Qin
, Lin Lin
, Erjun Kan
, Xingxing Li
, Chao Yang
and Jinlong Yang
Graphene is a nonmagnetic semimetal and cannot be directly used as electronic and spintronic devices. Here, we demonstrate that
zigzag graphene nanoflakes (GNFs), also known as graphene quantum dots, can exhibit strong edge magnetism and tunable
energy gaps due to the presence of localized edge states. By using large-scale first principle density functional theory calculations
and detailed analysis based on model Hamiltonians, we can show that the zigzag edge states in GNFs (C6n2H
,n=1–25) become
much stronger and more localized as the system size increases. The enhanced edge states induce strong electron–electron
interactions along the edges of GNFs, ultimately resulting in a magnetic configuration transition from nonmagnetic to intra-edge
ferromagnetic and inter-edge antiferromagnetic, when the diameter is larger than 4.5 nm (C
). Our analysis shows that the
inter-edge superexchange interaction of antiferromagnetic states between two nearest-neighbor zigzag edges in GNFs at the
nanoscale (around 10 nm) can be stabilized at room temperature and is much stronger than that exists between two parallel zigzag
edges in graphene nanoribbons, which cannot be stabilized at ultra-low temperature (3 K). Furthermore, such strong and localized
edge states also induce GNFs semiconducting with tunable energy gaps, mainly controlled by adjusting the system size. Our results
show that the quantum confinement effect, inter-edge superexchange (antiferromagnetic), and intra-edge direct exchange
(ferromagnetic) interactions are crucial for the electronic and magnetic properties of zigzag GNFs at the nanoscale.
npj 2D Materials and Applications (2019) 3:17 ; https://doi.org/10.1038/s41699-019-0098-2
Engineering techniques that use finite size effect to introduce
tunable edge magnetism and energy gap are by far the most
promising ways for enabling graphene
to be used in electronics
and spintronics.
Examples of finize-sized graphene nanostruc-
tures include one-dimensional (1D) graphene nanoribbons
and zero-dimensional (0D) graphene nanoflakes (GNFs)
(also known as graphene quantum dots).
It is well known that
electronic and magnetic properties
of GNRs and GNFs depend
strongly on the atomic configuration of their edges, which are of
either the armchair (AC) or zigzag (ZZ) types.
Edge magnetism has been predicted theoretically
observed experimentally
in ZZGNRs. The magnetism in
ZZGNRs results from ferromagnetic (FM) coupling for each zigzag
edge and antiferromagnetic (AFM) coupling between two parallel
zigzag edges of ZZGNRs. The strong FM coupling along each
zigzag edge has been predicted in theory
and confirmed in
However, the AFM coupling between two parallel
zigzag edges in ZZGNRs is weak, which cannot be stabilized even
at low temperature below 10 K
and rapidly weakens (~w
the ribbon-width wincreases.
Furthermore, the energy gap of
GNRs depend on several factors, such as the edge type (armchair
or zigzag) and the width of the nanoribbon,
thus cannot be easily
tuned. Such problem does not exist in GNFs due to the quantum
confinement effect.
The ability to control the energy gap has
enabled GNFs to be used in promising applications in
In addition, triangular ZZGNFs are theoretically
predicted to have strong edge magnetism even in small
However, triangular ZZGNFs have large formation
and have not been synthesized experimentally. For-
tunately, hexagonal ZZGNFs exhibits significantly improved
stability in ambient environment.
Recent experiments
also demonstrated that edge magnetism can be observed in
ZZGNFs when the edges are passivated by certain chemical
groups. However, semi-empirical tight-binding model
and first
principle density functional theory (DFT) calculations
hexagonal ZZGNFs have been performed for small-sized systems
but found no magnetism (NM). Thus the prospect of finding stable
finite-sized graphene easily fabricated in experiments with both
strong edge magnetism and tunable energy gap seems dim.
In this letter, we systematically investigate the electronic and
magnetic properties of hexagonal ZZGNFs with the diameters in
the range of 1–12 nm (from C
to C
). Using first-
principles DFT calculations, we find that both strong edge
magnetism and tunable energy gap can be realized simulta-
neously in large ZZGNFs stabilized at room temperature. We
demonstrate that spin polarization plays a crucial role as the
diameter of a ZZGNF increases beyond 4.5 nm (C
). A spin-
unpolarized calculation shows that edge states become increas-
ingly more localized as the size of a ZZGNF increases. These edge
states form a half-filled pseudo-band and is thus unstable. Adding
spin-polarization allows the edge states to spontaneously split into
Received: 8 October 2018 Accepted: 18 March 2019
Hefei National Laboratory for Physical Sciences at Microscale, Department of Chemical Physics, and Synergetic Innovation Center of Quantum Information and Quantum Physics,
University of Science and Technology of China, 230026 Hefei, Anhui, China;
Computational Research Division, Lawrence Berkeley National Laboratory, Berkeley, CA 94720, USA;
Department of Applied Physics, Xi’an Jiaotong University, 710049 Xi’an, Shaanxi, China;
Department of Mathematics, University of California, Berkeley, CA 94720, USA;
of Physics and Astronomy, University of Minnesota, Minneapolis, MN 55455, USA and
Department of Applied Physics and Institution of Energy and Microstructure, Nanjing
University of Science and Technology, 210094 Nanjing, Jiangsu, China
Correspondence: Lin Lin (linlin@math.berkeley.edu) or Chao Yang (cyang@lbl.gov) or Jinlong Yang (jlyang@ustc.edu.cn)
Published in partnership with FCT NOVA with the support of E-MRS
spin-polarized occupied and unoccupied states. This separation
results in a magnetic configuration transition from an NM
configuration to a strong inter-edge AFM configuration. It also
opens a tunable band gap that can be easily controlled by
quantum confinement effect. These properties make GNFs better
candidate materials for nanoelectronics than GNRs.
We also
confirm that ZZGNFs passivated by different chemical groups all
exhibit similar behavior. Such flexibility may facilitate future
experimental synthesis of such ZZGNFs.
We demonstrate the importance of spin polarization using C
(6 nm) as an example (Fig. 2). From a spin-unpolarized calculation,
we observe strong and localized edge states (Fig. 2e). These edge
states contribute to high electron density along the edges of
. Furthermore, these edge states become much stronger
and more localized as the ZZGNF size increases.
The presence of
strong edge states makes the ZZGNF metallic at the nanoscale.
The projected density of states (PDOS) of carbon edges of C
plotted in Fig. 2c clearly show a considerably high density of states
(DOS) near the Fermi level. This figure confirms that C
predicted to be metallic in a spin-unpolarized calculation. The
metallic nature of the ZZGNF can be attributed to the presence of
strong localized edge states.
However, a spin-polarized calculation shows that half-filled
metallic edge states are not stable, and can spontaneously split
into two types of occupied and unoccupied states as shown in
Fig. 2b, d. As a result, a magnetic configuration transition from a
non-magnetic (NM) configuration to a magnetic configuration that
exhibits intra-edge FM and inter-edge AFM characters can be
observed in Fig. 2f. This transition can be interpreted as the
consequence of Mott-type competition between the kinetic
(hopping) energy and the intra-edge (on-site) electron–electron
interaction energy as the system size increases. Lowering kinetic
energy by increasing the system size tends to produce delocalized
spin states across all edges, while reducing the electron–electron
interaction energy as the system size tends to penalize
simultaneous occupation of the same edge by spin up and spin
down electrons. Both semi-local GGA-PBE and hybrid HSE06
calculations (the details are given in the Supplemental Material)
indicate that for small systems, kinetic energy plays a more
dominant role. This observation agrees with previous theoretical
prediction of the NM configuration for hexagonal ZZGNFs.
Only as the system size increases, the effective
electron–electron interaction energy associated with the edge
states starts to dominate and is ultimately responsible for this
magnetic configuration transition.
Figure 1a shows the variation of relative energy of NM, AFM,
and FM magnetic configurations in ZZGNFs and ZZGNRs,
respectively, with respect to system size. Our calculations show
that AFM states are much more stable than NM and FM states in
large ZZGNFs, and a magnetic configuration transition occurs as
the diameter of the ZZGNF becomes larger than 4.5 nm
We believe the FM coupling along each zigzag edge
that belong to the same sublattice, are likely to be induced by
intra-edge direct exchange interactions. The AFM coupling
between two nearest-neighbor edges belonging to different
sublattices are likely to be induced by inter-edge superexchange
interactions facilitated by a carbon–carbon double bond (C=C) at
the corner where two nearest-neighbor edges meet in ZZGNFs at
the nanoscale. The local magnetic moment defined by M
ni#>j, where <^
niσgt;is spin electron density with σ
=↑(spin-up) or ↓(spin-down)), at the carbon atom i. Figure 2b
shows the local magnetic moment of carbon in the the middle of
each zigzag edge in ZZGNFs (with the largest magnetic moment)
increases with the system size, and converges to 0.3μ
when the
diameter is larger than than 6 nm (C
). Furthermore, there is
no charge transfer (<^
ni#gt;≈4) between carbon atoms
Fig. 1 aRelative energy per edge atom (ΔE(AFM-NM) and ΔE(AFM-
FM)) of NM, AFM, and FM coupling between different edges in
ZZGNFs and ZZGNRs and (b) spin electron density <^
(spin-up) or ↓(spin-down)) at the carbon atom iin the middle of
each zigzag edge in AFM ZZGNFs under the variation of the
diameter size (ZZGNFs) or ribbon-width length size (ZZGNRs). The
red and blue regions represent the stable NM (ΔE(AFM–NM) ≈0) and
AFM (ΔE(AFM–NM) < 0) coupling between different edges in
ZZGNFs, respectively. The critical temperature is estimated by the
mean-field theory T=ΔE/k
, where k
is the Boltzmann constant
Fig. 2 Electronic structure of edge states in C
in two different
magnetic configurations (NM and AFM), including the schematic
illustration of orbital diagram of superexchange interaction of edge
states in the (a) NM and (b) AFM configurations, projected density of
states (PDOS) of edges in the (c) NM and (d) AFM configurations, (e)
local density of states (LDOS) of Fermi level (pink isosurfaces) in the
NM configuration and (f) spin density isosurfaces in the AFM
configuration. The red and blue isosuraces in (f) represent the spin-
up and spin-down states, respectively. The red and blue lines in (d)
represent the PDOS contributed by sublattice A (spin-up edges) and
B (spin-down edges) atoms in graphene, respectively. The fermi level
is marked by green dotted lines and set to zero
W. Hu et al.
npj 2D Materials and Applications (2019) 17 Published in partnership with FCT NOVA with the support of E-MRS
sitting on different edges that belong to the same or different
sublattices in ZZGNFs as the system size increases.
Notice that the intra-edge direct exchange interaction via FM
coupling along each zigzag edge in ZZGNFs is similar to that in
ZZGNRs. However, the inter-edge superexchange interaction via
AFM coupling between two nearest-neighbor edges through a
C=C bond (Fig. 2a) in ZZGNFs can be stabilized at room
temperature (298 K) and is much stronger than that via AFM
coupling between two parallel edges though π-bonds in ZZGNRs
as shown in Fig. 1a, where such AFM spin polarization weakens
rapidly as the ribbon-width increases in ZZGNRs
and cannot be
stabilized even at ultra-low temperature (3 K).
Our DFT calcula-
tions confirm that the energy difference associated with AFM and
FM coupling between two parallel edges in large-scale 1D ZZGNRs
is negligible compared to that reported in ZZGNFs.
The enhanced stability of spin-polarized ZZGNFs can be
understood by using the Heisenberg model. We consider each
FM edge as one site and enumerate all possible magnetic
configurations, and the Hamiltonian can be written as
where J
is the exchange parameter between two sites iand j,Mi
and Mj
!are the corresponding spin magnetic moments. There are
four different magnetic states in C
, there of which are AFM,
AFM1, and AFM2 configurations and one is FM configuration as
shown in Fig. 3. The total energies of magnetic configurations E
(AFM), E(AFM1), E(AFM2), and E(FM) can be computed by DFT
calculations, and the exchange parameters can be evaluated by
solving the following least-squares-fitting problem
where J
, and J
are ortho-edge, meta-edge, and para-edge
exchange interaction parameters, respectively, Mis the spin
magnetic moment at each edge, and E
is the nonmagnetic
reference total energy. The solution yields J
=−0.038351 eV, J
0.000954 eV, and J
=0.001633 eV for two nearest-neighbor edges
of C
, which are 10 times stronger than the exchange
interaction parameters between two parallel edges in ZZGNFs and
ZZGNRs. Therefore, ZZGNFs at the nanoscale have strong edge
magnetism at room temperature and can be directly used in
nanospintronics, superior to that in ZZGNRs at the nanoscale.
We perform ab initio molecular dynamics (AIMD) simulations on
ZZGNFs and check the effect of temperature on electronic and
magnetic properties of C
in different AFM and FM
configurations (the details are given in the Supplemental Material).
We find that the AFM configuration of C
can remain stable
at room temperature of T=300 K at least within 1.6 ps.
Furthermore, the FM configuration of C
rapidly transfers
into the AFM configuration within 30.0 fs at room temperature of
T=300 K. Furthermore, after t=1.5 ps, C
is slightly bent,
although it still keeps the AFM configuration.
We also check the effects of using different types of atoms (e.g.,
bare and fluorine) to passivate ZZGNFs, and how the shape (non-
hexagonal) of ZZGNFs may alter their electronic and magnetic
properties. We find that magnetic configuration transition (from
NM to AFM) and semiconductor characteristics (The energy gaps
of 0.54, 0.34, and 0.41 eV, respectively, for bare C
passivated C
and non-hexagonal C
) of ZZGNFs are
independent of the type of passivating atoms
as plotted in
Fig. 4. These properties suggest that it is relatively easy to create a
chemical environment in which the synthesis of large scales
ZZGNFs with tunable edge magnetism and energy gaps can be
easily accommodated. The possibility of rapid synthesis makes
ZZGNFs ideal candidates for electronic and spintronic devices.
We remark that magnetic configuration transition and the
associated tunable electronic structures in ZZGNFs, especially
energy gaps, can also be understood in terms of the Hubbard
From our first principle calculations, we find that
choosing the parameters t=2.5 eV and U=2.1 eV in the Hubbard
model can well reproduce the size-dependent energy gaps (the
details are given in the Supplemental Material). In Fig. 5, we plot
how the HOMO-LUMO energy gap E
changes with respect to the
size of ZZGNFs and ACGNFs in two different magnetic configura-
tions (NM and AFM). Our DFT calculations and mean-field Hubbard
Fig. 3 Spin density isosurfaces of hydrogen-passivated C
in four different magnetic states, three types of antiferromagnetic ((a) AFM, (b)
AFM1, and (c) AFM2) and one type of ferromagnetic ((d) FM) coupling at the inter edges. The red and blue isosurfaces represent the spin-up
and spin-down states, respectively
Fig. 4 Spin density isosurfaces and total density of states ( TDOS) of
(a) bare (C
), (b)fluorine-passivated (C
), and (c) non-
hexagonal (C
) ZZGNFs in the AFM configuration. The red
and blue isosurfaces represent the spin-up and spin-down states,
respectively. The energy differences (E
) between AFM and
NM configurations of these ZZGNFs are shown above the figures.
For hexagonal hydrogen-passivated C
,ΔE(AFM−NM) =
−17.9 meV. The fermi level is marked by green dotted lines and
set to zero
W. Hu et al.
Published in partnership with FCT NOVA with the support of E-MRS npj 2D Materials and Applications (2019) 17
model show similar results, i.e., the energy gap E
of ZZGNF
decreases as its size increases. In particular, we find that the
energy gap of NM ZZGNFs decreases more rapidly with respect to
the system size than that of AFM ZZGNFs, due to the presence of
edge states whose electron density near the edges of ZZGNFs as
shown in Fig. 2e. This observation is consistent with previous
results obtained from tight-binding models
and DFT calcula-
However, AFM semiconducting ZZGNFs show similar
energy gap scaling compared to that of NM ACGNFs at the
Therefore, edge states should have little effect on the
energy gaps of AFM ZZGNFs and the quantum confinement
is the only factor to control the energy gaps in ZZGNFs
and ACGNFs (Fig. 5a). In detail, NM ZZGNFs exhibits metallic
characters (E
is smaller than the thermal fluctuation (25 meV) at
room temperature) when the diameter is larger than 7 nm
), but AFM ZZGNFs with the diameter of 12 nm
) still behaves as a semiconductor with a sizable energy
gap E
=0.23 eV, similar to the case of NM ACGNFs.
ZZGNFs at the nanoscale can be directly used in nanoelectronics.
In summary, using large-scale first principle calculations, we
demonstrate that the electronic and magnetic properties of
hexagonal zigzag that graphene nanoflakes (ZZGNFs) can be
significantly affected by the system size. We found that the zigzag
edge states in ZZGNFs become much stronger and more localized
as the system size increases. The presence of these edge states
induce strong electron–electron interactions along the edges of
ZZGNFs, resulting in a magnetic configuration transition from
nonmagnetic to intra-edge FM and inter-edge AFM when the
diameter is larger than 4.5 nm. On the other hand, such strong and
localized edge states are also responsible for making ZZGNFs
semiconducting with a tunable energy gap. The energy gap can
be controlled by merely adjusting the system size. Therefore,
ZZGNFs with strong edge magnetism, tunable energy gaps and
room-temperature stability may be promising candidates for
practical electronic and spintronic applications.
We use the Kohn–Sham DFT-based electronic structure analysis tools
implemented in the Spanish Initiative for Electronic Simulations with
Thousands of Atoms (SIESTA)
software package. We use the generalized
gradient approximation of Perdew, Burke, and Ernzerhof (GGA–PBE)
exchange correlation functional with collinear spin polarization, and the
double zeta plus polarization orbital basis set (DZP) to describe the valence
electrons within the framework of a linear combination of numerical
atomic orbitals (LCAO).
Because semi-local GGA–PBE calculations are less
reliable in predicting the electronic structures of ZZGNFs, the screened
hybrid HSE06
calculations implemented in HONPAS
(Hefei Order-N
Packages for Ab Initio Simulations based on SIESTA) are also used to
compute the electronic and magnetic properties of ZZGNFs. All atomic
coordinates are fully relaxed using the conjugate gradient (CG) algorithm
until the energy and force convergence criteria of 10
eV and 0.02 eV/Å,
respectively, are reached.
For initial magnetic moment setting of spin-polarized DFT calculations in
ZZGNFs, we set all the carbon atoms with initial magnetic moments of 1μ
for the FM configuration and only set the edged carbon atoms with initial
magnetic moments of 1 or −1μ
, and then optimize the structures and
magnetic moments of ZZGNFs.
Due to the large number of atoms contained in hexagonal hydrogen-
passivated ZZGNFs (C6n2H
,n=1–25), we use the recently developed
Pole EXpansion and Selected Inversion (PEXSI) method
to accelerate
the eigenvalue problem in the Kohn–Sham DFT calculations. The PEXSI
technique can efficiently utilize the sparsity of the Hamiltonian and overlap
matrices generated in SIESTA and overcome the cubic scaling limit for
solving Kohn–Sham DFT, and scales at most as quadratic scaling even for
metallic systems, such as graphene. Furthermore, the PEXSI method is
highly scalable
and can scale up to 100,000 processors on high
performance machines.
We perform AIMD simulations on ZZGNFs to check the effect of
temperature on electronic and magnetic properties of ZZGNFs. The
simulations are performed for about 1.6 ps with a time step of 2.0 fs at
room temperature of T=300 K controlled by a Nose–Hoover thermostat.
The authors confirm that the data supporting the findings of this study are available
within the article and its Supplementary Materials.
This work was performed, in part, under the auspices of the U.S. Department of
Energy by Lawrence Livermore National Laboratory under Contract DE-AC52-
07NA27344. Support for this work was provided through Scientific Discovery
through Advanced Computing (SciDAC) program funded by U.S. Department of
Energy, Office of Science, Advanced Scientific Computing Research and Basic Energy
Sciences (W.H., L.L., and C.Y.), by the Center for Applied Mathematics for Energy
Research Applications (CAMERA), which is a partnership between Basic Energy
Sciences and Advanced Scientific Computing Research at the U.S. Department of
Energy (L.L. and C.Y.), and by the Department of Energy under Grant No. DE-
SC0017867 (L.L.). This work is also partially supported by the National Key Research
and Development Program of China (Grant no. 2016YFA0200604) and the National
Natural Science Foundation of China (NSFC) (Grant nos. 21688102, 51522206, and
21803066), and the Strategic Priority Research Program of Chinese Academy of
Sciences (Grant no. XDC01000000), the Research Start-Up Grants (Grant no.
KY2340000094) from University of Science and Technology of China, and the
Chinese Academy of Sciences Pioneer Hundred Talents Program. Y.H. acknowledges
support from the Education Program for Talented Students of Xi’an Jiaotong
University. We thank the National Energy Research Scientific Computing (NERSC)
center, and the USTCSCC, SC-CAS, Tianjin, and Shanghai Supercomputer Centers for
the computational resources.
W.H., L.L., E.K., C.Y., and J.Y. designed the idea of this manuscript and supported this
project. W.H. performed all the DFT calculations in SIESTA. Y.H. wrote the codes of
Hubbard model. X.Q. performed the hybrid HSE06 calculations in HONPAS. All the
authors helped to write, modify, and analyze this manuscript.
Supplementary Information accompanies the paper on the npj 2D Materials and
Applications website (https://doi.org/10.1038/s41699-019-0098-2).
Competing interests: The authors declare no competing interests.
Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims
in published maps and institutional affiliations.
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Open Access This article is licensed under a Creative Commons
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Human Brain Project Moves Toward Human Cortex Model
Human Brain Project Moves Toward Human Cortex Model
Henry Markram's Human Brain Project in Lausanne, is competing for funding from the FET Flagship Initiative, to the tune of 1 billion Euros, disbursed over a ten year period. Markram's goals are
extremely ambitious, and unprecedented. He aims to model the human cerebral cortex to an exquisite degree of precision. Markram expects that his model of the human brain will be so exact, that he
will be able to study inaccessible brain diseases and devise impossible brain cures by using his model. He may be right. But in only ten years?
Scientists are paying particular attention to the cerebral cortex. This layer on the outside of brain, only a few millimeters thick, is the most important condition of it evolution. It is the
starting point for efforts to understand what makes us tick -- and for endeavors to find solutions when things go wrong. Our brain builds its version of the universe in the cerebral cortex. The
vast majority of what we see doesn't enter the brain through the eye. It is instead is based on the impressions, experiences and decisions in our brain.
Markham already completed important preparatory work for the computer modeling of the brain with his Blue Brain Project, an attempt to understand and model the molecular makeup of the mammalian
brain. He modeled a tiny part of a rat brain, a so-called neocortical column, at the cell level. To understand what one of these columns does, it's helpful to imagine the cerebral cortex as a
giant piano. There are millions of neocortical columns on the surface, and each of them produces a tone, in a manner of speaking. When they are simulated, the columns produce a symphony together.
Understanding the design of these neocortical columns is a holy grail of sorts for neuroscientists.
It is important to understand the rules of communication among the nerve cells. The individual cells do not communicate at random, but instead seek specifically targeted communication partners.
The axes of nerve cells intersect at millions of different points, where they can form a synapse. This makes communication between individual neurons possible. In a recent article in the journal
Proceedings of the National Academy of Sciences, Markram writes that such connections are also developed entirely without external influence. This could indicate a sort of innate knowledge that
all people have in common. Markram refers to it as the "Lego blocks" of the brain, noting that each person assembles his own world on the basis of this innate knowledge. _Spiegel
The object of study for the Human Brain Project may be the most complex dynamic system in the universe. The attempt would be impossible without the most sophisticated computing hardware and software
available. And one must have more than a mere fistful of Euros to acquire such advanced goodies.
Modeling all of this in a computer is extremely complex. Markram's current model encompasses tens of thousands of neurons. But this isn't nearly enough to come within striking range of the secret
of our brain. To do that, scientists will have to assemble countless other partial models, which are to be combined to create a functioning total simulation by 2023.
The supercomputers at the Jülich Research Center near Cologne are expected to play an important role in this process. The brain simulation will require an enormous volume of data, or what
scientist Markram calls a "tsunami of data." One of the challenges for scientists working under Thomas Lippert, head of the Jülich Supercomputing Centre, is to figure out how to make the computer
process only a certain part of the data at a given time, but without completely losing sight of the rest. They also have to develop an imaging method, such as large, three-dimensional holograms,
to depict the massive amounts of data.
All it takes is a look at the work of Jülich neuroscientist Katrin Amunts to understand the sheer volume of information at hand. The team she heads is compiling a detailed atlas of the human
brain. To do so, they cut a brain into 8,000 slices and digitized them with a high-performance scanner. The brain model generated in this way consists of cuboids, each measuring 10 by 10 by 20
micrometers, and the size of the data set is three terabytes. Brain atlases with higher resolutions, says Amunts, would probably consist of more than 700 terabytes _Spiegel
The answer to the question posed above is: No, this goal cannot be met within a time frame of ten years. Because the challenge is not merely quantitative -- a matter of compiling the precise assembly
of terabytes to create a brain atlas. The goal is to create a dynamic, interactive model of incredible plasticity -- a model which changes itself moment to moment. The "700 terabyte" requirement
mentioned above is just the starting point -- the bare beginning -- in the assembly of such a dynamic and ever-changing model.
But the problem is even harder -- much, much harder. The quantitative complexity -- even in dynamic flow -- is nothing when compared to the qualitative complexity, which is nowhere near to being
solved by Markram's team.
The project as described in brief above is an excellent starting point. Much can be learned from such an approach. But starting points do not necessarily point directly toward the end that one seeks.
Rather, they point somewhere "out there." It is for the questers to continuously adjust their headings -- and often they are forced to adjust their goals.
Good luck to Henry and his team -- with the funding and with the ongoing project. It is an ambitious goal worthy of any scientist.
Labels: artificial intelligence, brain research, Henry Markram, silicon brain
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Safely administering medications requires the ability to compute medication doses accurately and measure medications correctly. A careless mistake in placing a decimal point or adding a zero to a
dose can lead to a fatal error. Check every dose carefully before giving a medication.
Metric System
As a decimal system, the metric system is the most logically organized.
Need to know Conversions:
• 1 tsp (teaspoon) = 5 mL (milliliters)
• 1 tbs (tablespoon) = 15 mL
• 1 oz (ounce) = 30 mL
• 1 c (cup) = 8 oz = 240 mL
• 1 kg (kilogram) = 2.2 lb (pound)
• 1,000 mcg (microgram) = 1 mg (milligram)
• 1,000 mg = 1 g (gram)
• 1,000 mL (milliliter) = 1 L (liter)
When working with decimal points, always use a zero before the decimal point – this is called a “leading zero.” This leading zero draws attention to the fact that this is not a whole number – it
helps the reader note that there is a decimal.Example – 0.5 mg
Leading and Trailing Zeros
Leading and Trailing Zeros
• Do not use a decimal point then a zero after a whole number – this is called a “trailing zero.” A trailing zero may make a number look like a bigger number.
□ Example – “6” should not be written as “6.0” “6.0” looks like “60” – This would be a ten-fold error!
Ratio and Proportion in Dosage Calculation
• A proportion is a relationship comparing two ratios.
• The two numbers are separated by a colon (:)
• The two numbers may also be expressed as a fraction.
Whether a ratio is expressed as a proportion (in a linear fashion or written across in a line) or as a fraction, like units have to be in the same position on each side of the equal sign.
Ratio and Proportion in Dosage Calculation
Ratio and Proportion in Dosage Calculation
• When calculating a ratio written as a fraction, cross-multiply.
• When calculating a ratio written in a line, as a proportion, multiply the means (the two middle numbers) and multiply the extremes (the two end numbers).
• When using ratio and proportion, you will be given 3 of the 4 values, and will have to solve for the fourth value.
• Use “X” to represent the unknown quantity.
Read carefully!
You may be asked to determine number of milligrams, number of tablets, number of milliliters number of units (especially with heparin and insulin), number milliequivalents…always look first at what
you are given in the problem. Is it mg/tablet, or units/mL, or mg/mL? You will always be working with some unit of measure per another unit of measure.
• Always note the VALUE that you are using in a problem when you write out the problem – that way you will not confuse the value label when noting your answer.
• For example:
□ The order is for 500 mg of med. You have available 100 mg/mL. How much will you give? ( When you set up the problem, note which are mg and which are mL.)
Example: a medicine has a dosage strength of 50 mg per 1 mL, and the prescriber has ordered a dose of 25mg.How many mL do you administer?
What is the known?
What is the unknown?
Example: a medicine has a dosage strength of 50 mg per 1 mL, and the prescriber has ordered a dose of 25 mg. How many mL do you administer?
Fraction method
Ratio Method
Example: a medicine has a dosage strength of 50 mg per 1 mL, and the prescriber has ordered a dose of 25 mg. How many mL do you administer?
Order: 500 mg p.o. of a medicationAvailable: 1 g tabletsHow many tablets will you give?
Order: 0.25 mg IM of a medicationAvailable: 0.5 mg per mLHow many mL will you give?
Order: 40 mg p.o. of a medicationAvailable: 20 mg tabletsHow many tablets will you give?
1 g = 500 mg1 tablet X tablet
A ratio expressed as a proportion 3:4 = 6:8
The known is....
• 50 mg per 1 mL
• is stated first
• is placed on the left side of the equal sign
100 mg = 500 mg 100 X = 500 1 mL X mL = 5 mL (not mg)
• Desired amount
• Placed on the right side of the equal sign.
• 50 mg = 25 mg numerator labels must match 1 mL X mL denominator labels must match 50X = 25 cross multiply X = 0.5 mL
• 50 mg = 25 mg numerator labels must match 1 mL X mL denominator labels must m 50X = 25 cross multiply X = 0.5 mL
A ratio expressed as a fraction 3 = 6 4 8
• 50 mg = 25 mg numerator labels must match 1 mL X mL denominator labels must match 50X = 25 cross multiply X = 0.5 mL
They occur when a decimal placement is written incorrectly or misread. Decimal errors can result in a 10-fold, 100-fold, or even 1,000-fold overdose or underdose.
Ten-fold Error
50 mg : 1 mL = 25 mg : X mL (Be sure that like units are in the same position on each side of the equal sign.)Multiply the means = multiply the extremes50X = 25Divide each side by 50X = 0.5 mL
20 mg = 40 mg1 tab x tab20x = 40x = 2 tablets
.5 mg (x) = 0.250
0.5 mg : 1 ml = 0.25 mg : x ml | {"url":"https://view.genially.com/66e8ab7e7dc9a7ecaee284f2/presentation-medcalc","timestamp":"2024-11-07T07:43:43Z","content_type":"text/html","content_length":"46875","record_id":"<urn:uuid:78ce5b63-908e-4777-a596-6c2634ae2151>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00749.warc.gz"} |
Generator.gamma(shape, scale=1.0, size=None)¶
Draw samples from a Gamma distribution.
Samples are drawn from a Gamma distribution with specified parameters, shape (sometimes designated “k”) and scale (sometimes designated “theta”), where both parameters are > 0.
shapefloat or array_like of floats
The shape of the gamma distribution. Must be non-negative.
scalefloat or array_like of floats, optional
The scale of the gamma distribution. Must be non-negative. Default is equal to 1.
sizeint or tuple of ints, optional
Output shape. If the given shape is, e.g., (m, n, k), then m * n * k samples are drawn. If size is None (default), a single value is returned if shape and scale are both scalars.
Otherwise, np.broadcast(shape, scale).size samples are drawn.
outndarray or scalar
Drawn samples from the parameterized gamma distribution.
See also
probability density function, distribution or cumulative density function, etc.
The probability density for the Gamma distribution is
The Gamma distribution is often used to model the times to failure of electronic components, and arises naturally in processes for which the waiting times between Poisson distributed events are
Weisstein, Eric W. “Gamma Distribution.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/GammaDistribution.html
Wikipedia, “Gamma distribution”, https://en.wikipedia.org/wiki/Gamma_distribution
Draw samples from the distribution:
>>> shape, scale = 2., 2. # mean=4, std=2*sqrt(2)
>>> s = np.random.default_rng().gamma(shape, scale, 1000)
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> import scipy.special as sps
>>> count, bins, ignored = plt.hist(s, 50, density=True)
>>> y = bins**(shape-1)*(np.exp(-bins/scale) /
... (sps.gamma(shape)*scale**shape))
>>> plt.plot(bins, y, linewidth=2, color='r')
>>> plt.show() | {"url":"https://numpy.org/doc/1.18/reference/random/generated/numpy.random.Generator.gamma.html","timestamp":"2024-11-13T12:34:03Z","content_type":"text/html","content_length":"14868","record_id":"<urn:uuid:73a1e0a8-b7a5-43b4-baab-9c9b5627bfb7>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00563.warc.gz"} |
558 Circle/Square Nanosecond to Arcmin/Square Nanosecond
Circle/Square Nanosecond [circle/ns2] Output
558 circle/square nanosecond in degree/square second is equal to 2.0088e+23
558 circle/square nanosecond in degree/square millisecond is equal to 200880000000000000
558 circle/square nanosecond in degree/square microsecond is equal to 200880000000
558 circle/square nanosecond in degree/square nanosecond is equal to 200880
558 circle/square nanosecond in degree/square minute is equal to 7.23168e+26
558 circle/square nanosecond in degree/square hour is equal to 2.6034048e+30
558 circle/square nanosecond in degree/square day is equal to 1.4995611648e+33
558 circle/square nanosecond in degree/square week is equal to 7.34784970752e+34
558 circle/square nanosecond in degree/square month is equal to 1.3892555542752e+36
558 circle/square nanosecond in degree/square year is equal to 2.0005279981563e+38
558 circle/square nanosecond in radian/square second is equal to 3.5060174014062e+21
558 circle/square nanosecond in radian/square millisecond is equal to 3506017401406200
558 circle/square nanosecond in radian/square microsecond is equal to 3506017401.41
558 circle/square nanosecond in radian/square nanosecond is equal to 3506.02
558 circle/square nanosecond in radian/square minute is equal to 1.2621662645062e+25
558 circle/square nanosecond in radian/square hour is equal to 4.5437985522224e+28
558 circle/square nanosecond in radian/square day is equal to 2.6172279660801e+31
558 circle/square nanosecond in radian/square week is equal to 1.2824417033793e+33
558 circle/square nanosecond in radian/square month is equal to 2.4247083573721e+34
558 circle/square nanosecond in radian/square year is equal to 3.4915800346158e+36
558 circle/square nanosecond in gradian/square second is equal to 2.232e+23
558 circle/square nanosecond in gradian/square millisecond is equal to 223200000000000000
558 circle/square nanosecond in gradian/square microsecond is equal to 223200000000
558 circle/square nanosecond in gradian/square nanosecond is equal to 223200
558 circle/square nanosecond in gradian/square minute is equal to 8.0352e+26
558 circle/square nanosecond in gradian/square hour is equal to 2.892672e+30
558 circle/square nanosecond in gradian/square day is equal to 1.666179072e+33
558 circle/square nanosecond in gradian/square week is equal to 8.1642774528e+34
558 circle/square nanosecond in gradian/square month is equal to 1.543617282528e+36
558 circle/square nanosecond in gradian/square year is equal to 2.2228088868403e+38
558 circle/square nanosecond in arcmin/square second is equal to 1.20528e+25
558 circle/square nanosecond in arcmin/square millisecond is equal to 12052800000000000000
558 circle/square nanosecond in arcmin/square microsecond is equal to 12052800000000
558 circle/square nanosecond in arcmin/square nanosecond is equal to 12052800
558 circle/square nanosecond in arcmin/square minute is equal to 4.339008e+28
558 circle/square nanosecond in arcmin/square hour is equal to 1.56204288e+32
558 circle/square nanosecond in arcmin/square day is equal to 8.9973669888e+34
558 circle/square nanosecond in arcmin/square week is equal to 4.408709824512e+36
558 circle/square nanosecond in arcmin/square month is equal to 8.3355333256512e+37
558 circle/square nanosecond in arcmin/square year is equal to 1.2003167988938e+40
558 circle/square nanosecond in arcsec/square second is equal to 7.23168e+26
558 circle/square nanosecond in arcsec/square millisecond is equal to 723168000000000000000
558 circle/square nanosecond in arcsec/square microsecond is equal to 723168000000000
558 circle/square nanosecond in arcsec/square nanosecond is equal to 723168000
558 circle/square nanosecond in arcsec/square minute is equal to 2.6034048e+30
558 circle/square nanosecond in arcsec/square hour is equal to 9.37225728e+33
558 circle/square nanosecond in arcsec/square day is equal to 5.39842019328e+36
558 circle/square nanosecond in arcsec/square week is equal to 2.6452258947072e+38
558 circle/square nanosecond in arcsec/square month is equal to 5.0013199953907e+39
558 circle/square nanosecond in arcsec/square year is equal to 7.2019007933626e+41
558 circle/square nanosecond in sign/square second is equal to 6.696e+21
558 circle/square nanosecond in sign/square millisecond is equal to 6696000000000000
558 circle/square nanosecond in sign/square microsecond is equal to 6696000000
558 circle/square nanosecond in sign/square nanosecond is equal to 6696
558 circle/square nanosecond in sign/square minute is equal to 2.41056e+25
558 circle/square nanosecond in sign/square hour is equal to 8.678016e+28
558 circle/square nanosecond in sign/square day is equal to 4.998537216e+31
558 circle/square nanosecond in sign/square week is equal to 2.44928323584e+33
558 circle/square nanosecond in sign/square month is equal to 4.630851847584e+34
558 circle/square nanosecond in sign/square year is equal to 6.668426660521e+36
558 circle/square nanosecond in turn/square second is equal to 558000000000000000000
558 circle/square nanosecond in turn/square millisecond is equal to 558000000000000
558 circle/square nanosecond in turn/square microsecond is equal to 558000000
558 circle/square nanosecond in turn/square nanosecond is equal to 558
558 circle/square nanosecond in turn/square minute is equal to 2.0088e+24
558 circle/square nanosecond in turn/square hour is equal to 7.23168e+27
558 circle/square nanosecond in turn/square day is equal to 4.16544768e+30
558 circle/square nanosecond in turn/square week is equal to 2.0410693632e+32
558 circle/square nanosecond in turn/square month is equal to 3.85904320632e+33
558 circle/square nanosecond in turn/square year is equal to 5.5570222171008e+35
558 circle/square nanosecond in circle/square second is equal to 558000000000000000000
558 circle/square nanosecond in circle/square millisecond is equal to 558000000000000
558 circle/square nanosecond in circle/square microsecond is equal to 558000000
558 circle/square nanosecond in circle/square minute is equal to 2.0088e+24
558 circle/square nanosecond in circle/square hour is equal to 7.23168e+27
558 circle/square nanosecond in circle/square day is equal to 4.16544768e+30
558 circle/square nanosecond in circle/square week is equal to 2.0410693632e+32
558 circle/square nanosecond in circle/square month is equal to 3.85904320632e+33
558 circle/square nanosecond in circle/square year is equal to 5.5570222171008e+35
558 circle/square nanosecond in mil/square second is equal to 3.5712e+24
558 circle/square nanosecond in mil/square millisecond is equal to 3571200000000000000
558 circle/square nanosecond in mil/square microsecond is equal to 3571200000000
558 circle/square nanosecond in mil/square nanosecond is equal to 3571200
558 circle/square nanosecond in mil/square minute is equal to 1.285632e+28
558 circle/square nanosecond in mil/square hour is equal to 4.6282752e+31
558 circle/square nanosecond in mil/square day is equal to 2.6658865152e+34
558 circle/square nanosecond in mil/square week is equal to 1.306284392448e+36
558 circle/square nanosecond in mil/square month is equal to 2.4697876520448e+37
558 circle/square nanosecond in mil/square year is equal to 3.5564942189445e+39
558 circle/square nanosecond in revolution/square second is equal to 558000000000000000000
558 circle/square nanosecond in revolution/square millisecond is equal to 558000000000000
558 circle/square nanosecond in revolution/square microsecond is equal to 558000000
558 circle/square nanosecond in revolution/square nanosecond is equal to 558
558 circle/square nanosecond in revolution/square minute is equal to 2.0088e+24
558 circle/square nanosecond in revolution/square hour is equal to 7.23168e+27
558 circle/square nanosecond in revolution/square day is equal to 4.16544768e+30
558 circle/square nanosecond in revolution/square week is equal to 2.0410693632e+32
558 circle/square nanosecond in revolution/square month is equal to 3.85904320632e+33
558 circle/square nanosecond in revolution/square year is equal to 5.5570222171008e+35 | {"url":"https://hextobinary.com/unit/angularacc/from/circlepns2/to/arcminpns2/558","timestamp":"2024-11-14T12:24:10Z","content_type":"text/html","content_length":"114566","record_id":"<urn:uuid:c4bfe514-c52e-4149-a596-022ca8bc846a>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00339.warc.gz"} |
finance 26 - Proficient Writers Hubfinance 26 - Proficient Writers Hub
You will assume that you still work as a financial analyst for the Coca Cola Co. The company is considering a capital investment and you are in charge of helping them launching a new product based on
(1) a given rate of return of 13% (Task 4) and (2) the firm’s cost of capital (Task 5).
Task 4. Capital Budgeting for a Product
A few months have now passed and Coca Cola Co. is considering launching a new product – a flavored soda! The anticipated cash flows for the project are as follows:
Year 1 $1,350,000
Year 2 $1,580,000
Year 3 $1,900,000
Year 4 $930,000
Year 5 $2,400,000
You have now been tasked with providing a recommendation for the project based on the results of a Net Present Value Analysis. Assuming that the required rate of return is 12% and the initial cost of
project is $5,000,000:
1. What is the project’s IRR? (10 pts)
2. What is the project’s NPV? (10 pts)
3. Calculate the project’s payback period. (10 pts)
4. In order to conduct this project, Coca Cola has hired a market analyst to determine demand for the new product. The cost of these services will be $400,0000. How would this cost be incorporated
into the project cash flows? Explain your rationale (10 pts)
5. Provide examples for each the following concepts as they relates to the project. Please make sure that your examples are applicable to Coca Cola’s idea of launching a new product. (5 pts each)
a. Allocated Costs
b. Incremental Costs
c. Financing Costs
6. Explain how you would conduct a scenario and sensitivity analysis of the project. What would be some project specific risks and market risks related to this project? (20 pts)
Task 5: Cost of Capital
Coca Cola Co. is now considering that the appropriate discount rate for the new product should be the cost of capital and would like to determine it. You will assist in the process of obtaining this
1. Compute the cost of debt.
a. Assume that Coke has received a loan from a bank for 5% annual interest for the next seven years. If the tax rate for Coke is 24%, what is the after tax cost of debt? (5 pts)
b. Would you expect the cost of debt to be higher or lower than the cost of equity? Explain your rationale. (5 pts)
c. Explain how Coca Cola Co. can estimate the cost of debt using market observation of rates. Compare and contrast this method to using YTM of bonds. (10 pts)
d. Assume that instead Coke uses the YTM method. They have currently bonds that sell for $1045, offer a coupon of 8% and mature in 5 years. What is the YTM of these bonds? (5 pts)
2. Compute the cost of common equity using the CAPM model. For beta, use the average beta of three selected competitors. Assume the risk free rate to be 3% and the market risk premium to be 9%.
a. What is the cost of common equity? (5 pts)
b. Explain how flotation costs affect the cost of common equity. (5 pts)
c. Explain why is said that the cost of retained earnings is the same as the cost of equity, except for flotation costs. (5 pts)
3. Cost of preferred equity
a. Why would the cost of preferred equity be lower than the cost of common equity? (5 pts)
b. What would be the price of preferred equity for Coke assuming dividends of $5 at the end of the year and the cost of preferred stock is 8%? (5 pts)
4. Assuming that the market value weights of these capital sources are 30% bonds, 40% common equity and 30% preferred equity, what is the weighted cost of capital of the firm? (10 pts)
5. Should the firm use market or book values to compute the cost of capital? Explain and provide examples as appropriate. (10 pts)
6. Explain how hard rationing and soft rationing may affect your recommendation on pursuing this project. (5 pts)
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onlinelogomaker-012722-1716-5353-2000-transparent-300x61.png Florence2023-01-07 01:37:532023-01-07 01:37:53finance 26 | {"url":"https://proficientwritershub.com/finance-26/","timestamp":"2024-11-10T08:08:36Z","content_type":"text/html","content_length":"65117","record_id":"<urn:uuid:6869606a-211a-4586-a8ac-138077cbc2ea>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00466.warc.gz"} |
How to Round to 2 Decimal Places in Python • datagy
Being able to work with and round floating point values, or decimal values, in Python is an important skill. In this tutorial, you’ll learn how to round to decimal places in Python, including
learning how to round up or down. You will also learn how to simply represent values with 2 decimal places, without changing the value itself.
By the end of this tutorial, you’ll have learned:
• How to round to 2 decimal places in Python, including rounding up and down
• How to format values as currency in Python
• How to round lists and arrays of numbers to 2 decimal places in Python
How to Round to 2 Decimal Places in Python
Python provides a function, round(), which is used to round values to a specific precision level. The function takes two parameters: the value to round and the precision to use. Let’s take a look at
what the function looks like and how we can use it:
# Understanding the Python round() Function
round(number[, ndigits])
We can see that there are two parameters, one of which is required:
1. number=, represents the number to round
2. ndigits=, represents the number of digits to use. If None, the number is rounded to its nearest integer.
Let’s see how we can use the Python round() function to round a float to 2 decimal places:
# Rounding a Value to 2 Decimal Places in Python
value = 1.2345
rounded = round(value, 2)
# Returns: 1.23
In the following section, you’ll learn how to round up a float to 2 decimal places.
How to Round Up to 2 Decimal Places in Python
By default, Python will round values that are greater than 5 up, and less than or equal to 5 down. In order to round a value up, you need to use a custom process. This is often done in finance or
accounting, where values are rounded up, regardless of their value. For example, $3.221 would be rounded to $3.23.
Python provides another function, ceil(), which allows you to round values up to their nearest integer. We can use this function to round values up to 2 decimal places:
# Rounding Up to 2 Decimal Places
import math
value = 1.2121
rounded = math.ceil(value * 100) / 100
# Returns: 1.22
Let’s break down what the code above is doing:
1. We multiply our number by 100
2. We pass the value into the ceil() function
3. We then divide the number by 100, to bring it back to 2 decimal places
We can use a similar process to round floating point values down to 2 decimal places.
How to Round Down to 2 Decimal Places in Python
Similar to the ceil() function, Python also provides a floor() function. This function allows us to round values down, regardless of rounding rules. We can use a process similar to the one above to
round values down to two decimal places. Let’s see what this looks like:
# Rounding Down to 2 Decimal Places
import math
value = 1.2155
rounded = math.floor(value * 100) / 100
# Returns: 1.21
1. We multiply our number by 100
2. We pass the value into the floor() function
3. We then divide the number by 100, to bring it back to 2 decimal places
In the following section, you’ll learn how to represent values to 2 decimal places without changing the underlying value.
How to Format Values as Currency in Python
In some cases, you’ll want to display values to a certain precision without changing the actual value. For example, you may want to round values to two decimal places and show them as currency.
Python makes this incredibly easy using string formatting. Because string formatting is a huge topic, I have provided a full guide on this.
Let’s see how we can use f-string formatting to format a decimal as currency in Python:
# Formatting Values as Currency in Python
value = 1.23333
# Returns: $1.23
We can see that in order to accomplish this, we only needed to apply the :.2f formatter to the string and precede the value with a $ sign.
Want to learn how to round values as currency in a Pandas DataFrame? Check out this guide to rounding values in Pandas, including rounding values as currency.
How to Round a List of Values to 2 Decimal Places in Python
In this section, you’ll learn how to round a list of values to 2 decimal places in Python. This can be done easily using a for loop. Using a for loop, we can iterate over each item in a list and
apply a transformation to it. Let’s see how we can use Python to accomplish this:
# Using a For Loop To Round a List of Values
values = [1.222, 1.5555, 3.234]
rounded = []
for value in values:
rounded.append(round(value, 2))
# Returns:
# [1.22, 1.56, 3.23]
Let’s break down what we did in the code above:
1. We defined two lists, one which contains our values and an empty list to hold our rounded values
2. We then loop over each item in the list and pass it into the round() function
3. The rounded value is then appended to our holder list
We can also use Python list comprehensions to simplify this process quite a bit. List comprehensions allow us to iterate over lists without explicitly using a for loop. Let’s see what this looks
# Using a List Comprehension to Round Values to 2 Decimal Places
values = [1.222, 1.5555, 3.234]
rounded = [round(value, 2) for value in values]
# Returns:
# [1.22, 1.56, 3.23]
In the following section, you’ll learn how to round all values in a NumPy array to 2 decimal places.
How to Round a NumPy Array to 2 Decimal Places in Python
When working with NumPy arrays, you may also want to round all the values in the array to 2 decimal places. Rather than using the Python round() function, NumPy provides its own round() function.
What’s great about this function is that it allows you to pass in an entire array. This has the benefit of making the code more explicit and easier to read.
Let’s see how we can use NumPy to round an entire array to 2 decimal places:
# Round a NumPy Array to 2 Decimal Places
import numpy as np
arr = np.array([2.312, 3.1234, 4.5555])
rounded = np.round(arr, 2)
# Returns:
# [2.31 3.12 4.56]
In the code above, we:
1. Created a new array, arr, which contains our original values
2. We then created another array, rounded, which is the result of passing the array into the np.round() function
In this tutorial, you learned how to use Python to round a value to 2 decimal places. This is such a common task that it warrants its own tutorial. You first learned how to use the Python round()
function to round values to a specific precision point. Then, you learned how to customize the behavior of your programs to round up or down, regardless of rounding rules.
Then, you learned how to display values in currency format in Python. From there, you learned how to round values in Python lists and NumPy arrays.
Additional Resources
To learn more about related topics, check out the tutorials below: | {"url":"https://datagy.io/python-round-2-decimals/","timestamp":"2024-11-13T03:17:29Z","content_type":"text/html","content_length":"150664","record_id":"<urn:uuid:69998fe5-4023-4cd8-8e45-05bd20bdd402>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00293.warc.gz"} |
Potential Pathways for Preparing High School Mathematics Teachers
Programs to prepare high school mathematics teachers can be organized in many ways, depending on state policies, college or university guidelines, the intended audience (e.g., career changers), and
so forth. However they are organized, effective programs must meet the requirements of the standards and their elaborations for high school in this document. Following are several examples of how
differently organized programs can meet these recommendations.
One approach to establishing a program that leads to the well-prepared beginning teacher of mathematics could be in the form of a four-year Bachelor of Science degree. This program has a major in
mathematics with a teaching option. Consistent with MET II (CBMS, 2012), mathematics and statistics coursework consists of single and multivariable calculus, differential equations, two courses in
data-based statistics and statistical inference, transition to proof, and linear algebra, each at the lower division. Upper division coursework includes three courses designed specifically for
teachers, each of which allows the focus on essential ideas of high school mathematics: one course situates high school algebra and precalculus in the context of number theory, algebra, and analysis;
a second course focuses on Euclidean and non-Euclidean geometry; and a third course focuses on modeling using the tools of mathematics without calculus and using simulation-based statistics. Each of
these courses embeds the use of technology as a tool for learning mathematics. Mathematics coursework also includes three upper division elective courses in mathematics or statistics, and students
choose from among courses including algebraic or geometric reasoning in the middle grades, history of mathematics, and others.
The program includes designated mathematics methods courses with field experiences, and embeds selected opportunities for working with students within mathematics content courses. Although the ideal
program would include three methods courses, this program cannot currently do so, given the constraints of the program requirements as a whole. The program designers recognized this need for
attention to teaching methods, and have addressed it by integrating pedagogy assignments and field experiences within mathematics content courses. The program includes education coursework completed
by all high school teaching majors in the state, and includes student teaching.
A second program achieves the recommendations of this chapter via a fifth-year graduate program that follows a strong undergraduate major in mathematics teaching. The undergraduate coursework
includes three content courses addressing mathematical content for teaching. In addition to generic education coursework taken by all high school teaching majors, two content-specific high school
mathematics methods courses, both of which include explicit attention to the teaching and learning of high school mathematics topics, are required, along with a course addressing using technology to
teach mathematics taken by all middle and high school mathematics candidates. Furthermore, although equity issues are addressed in generic education courses, equity issues are also explicitly
addressed within the mathematics methods courses, because only then will candidates understand that equity is not separate from, but fundamentally a part of, effectively teaching mathematics to high
school students. Student teaching takes place concurrently during the methods courses, with the students assuming increased classroom-teaching responsibility over the course of each semester and from
the first to the second semester. Because candidates student teach while taking courses, issues that arise in student teachers’ practice are incorporated into the mathematics-methods-course
discussions. Note that students coming from a general mathematics major that does not focus on mathematics teaching may have to complete pre- or co-requisite coursework.
An alternative version of the second program does not require prerequisite courses for mathematics teaching to be included in the undergraduate program, because the candidates may be career changers.
Instead, this coursework is included in the program, embedded within courses addressing mathematical content relevant for teaching and mathematics-specific methods courses. Efforts are made to ensure
that the program fully meets the standards and elaborations for high school mathematics preparation, noting that most candidates will require more than one year to complete the program. Depending on
state requirements, candidates may be granted licensure or certification at the end of the first year and may then complete any remaining degree requirements the second year while employed as a
full-time teacher.
In a final program, the recommended standards are achieved in a liberal arts mathematics program that is a part of a coalition of universities and colleges offering high school mathematics teacher
preparation. Specialized mathematics content courses for teachers as well as mathematics-specific methods courses are offered collaboratively with a nearby state university that leads a local
partnership of higher education institutions and school districts. Some of these jointly offered courses are delivered online. In other cases, students meet on one of the campuses or at a central
location. In this model, students in the smaller program can complete the requirements for becoming high school mathematics teachers that meets the recommendations in this document, and students at
the state university can interact with colleagues from another context, thus broadening their awareness of issues related to mathematics education, particularly if the liberal arts college reflects
demographics different from those of the state university. | {"url":"https://amte.net/sptm/chapter-7-elaborations-standards-preparation-high-school-teachers-mathematics/potential","timestamp":"2024-11-04T15:21:17Z","content_type":"text/html","content_length":"62373","record_id":"<urn:uuid:243425b8-9f3d-4355-b884-4e0c6279e503>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00352.warc.gz"} |
Will a train line be in operation by Jan 1st 2040 that uses room-temperature superconductivity to levitate the train?
This question is managed and resolved by Manifold.
Would be a completely pointless waste of money even if possible. If you want a flying train, that’s called a plane.
Edit: they can go faster, but still seems pretty pointless tbh
@jamesoofou Maglev is hardly practical itself because of costs. I think regular rail can go fast enough while still being cheap enough.
@jamesoofou does anyone have a good back-of-the-envelope calc. on rolling resistance losses vs air drag losses for trains? | {"url":"https://manifold.markets/jim/will-a-train-line-be-in-operation-b-f74ba3bd0111","timestamp":"2024-11-04T09:17:33Z","content_type":"text/html","content_length":"182955","record_id":"<urn:uuid:31b177cf-04be-46f8-9e9e-f234f3ebf1af>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00443.warc.gz"} |
Interagency Modeling and Analysis Group
Computes the Hurst coefficient for a fractional Gaussian noise series by binning the data (averaging adjacent points) and calculating the standard deviation as a function of bin size.
Dispersion Analysis:
Let x(i), i=1,2,...n be a series of n points. Calculate the standard
deviation. Next take the numbers in groups (bins) of two and average
them (i.e., x2(1)=.5*(x(1)+x(2)), x2(2)=.5*(x(3)+x(4)), ... x2(N/2)=
0.5*(x(N-1)+x(N)). Calculate the standard deviation of this new
series. Continue to double the bin size and repeat this process until
the bin size reaches n/2. If the OVERLAP option is used, the averages
become x2(1)=.5*(x(1)+x(2)), x2(2)=.5*(x(2)+x(3)), ... x2(N/2)=
The logarithms of the bin sizes (abscissa) versus the logarithms of
the standard deviations is approximately linear, and the slope equals
the estimated Hurst coefficient minus 1.0.
It is customary to omit the three largest bins from the analysis
because they are based on the standard deviations of fewer points:
If a series contained 512 points, and the OVERLAP option was turned off
the standard deviation for bin size 256 would be based on two points,
for bin size 128, 4 points, and for bin size 64, 8 points.
The model contains three curves for dispersional analysis,
generated by FGP, the fractional Gaussian Process model. The
curves were generated with Hurst coefficients of 0.2, 0.5, and 0.7,
coresponding to anti-correlated noise, white noise, and correlated
Upper Panel: A time series of Gaussianly distributed random numbers with mean=0, variance=4.
Lower Panel: The autocovariance function (black) and the autocorrelation function (red) are plotted as functions of the lag.
See Description above.
The equations for this model may be viewed by running the JSim model applet and clicking on the Source tab at the bottom left of JSim's Run Time graphical user interface. The equations are written in
JSim's Mathematical Modeling Language (MML). See the Introduction to MML and the MML Reference Manual. Additional documentation for MML can be found by using the search option at the Physiome home
Download JSim model project file
• Download JSim model MML code (text):
• Download translated SBML version of model (if available):
□ No SBML translation currently available.
Model Feedback
We welcome comments and feedback for this model. Please use the button below to send comments:
Key terms
time series
fractional Gaussian noise
Please cite https://www.imagwiki.nibib.nih.gov/physiome in any publication for which this software is used and send one reprint to the address given below:
The National Simulation Resource, Director J. B. Bassingthwaighte, Department of Bioengineering, University of Washington, Seattle WA 98195-5061.
Model development and archiving support at https://www.imagwiki.nibib.nih.gov/physiome provided by the following grants: NIH U01HL122199 Analyzing the Cardiac Power Grid, 09/15/2015 - 05/31/2020, NIH
/NIBIB BE08407 Software Integration, JSim and SBW 6/1/09-5/31/13; NIH/NHLBI T15 HL88516-01 Modeling for Heart, Lung and Blood: From Cell to Organ, 4/1/07-3/31/11; NSF BES-0506477 Adaptive Multi-Scale
Model Simulation, 8/15/05-7/31/08; NIH/NHLBI R01 HL073598 Core 3: 3D Imaging and Computer Modeling of the Respiratory Tract, 9/1/04-8/31/09; as well as prior support from NIH/NCRR P41 RR01243
Simulation Resource in Circulatory Mass Transport and Exchange, 12/1/1980-11/30/01 and NIH/NIBIB R01 EB001973 JSim: A Simulation Analysis Platform, 3/1/02-2/28/07. | {"url":"https://www.imagwiki.nibib.nih.gov/physiome/jsim/models/webmodel/NSR/disp","timestamp":"2024-11-08T14:22:47Z","content_type":"text/html","content_length":"61678","record_id":"<urn:uuid:873d9edf-c6de-4c44-99cb-f9ba164e180c>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00513.warc.gz"} |
MacReady & Dayton (1977) Multiplication Data — mdm_data
MacReady & Dayton (1977) Multiplication Data
This is a small data set of multiplication item responses. This data contains responses to 4 items from 142 respondents, which ask respondents to complete an integer multiplication problem.
mdm_data is a tibble containing responses to multiplication items, as described in MacReady & Dayton (1977). There are 142 rows and 5 variables.
• respondent: Respondent identifier
• mdm1-mdm4: Dichotomous item responses to the 4 multiplication items
mdm_qmatrix is a tibble that identifies which skills are measured by each MDM item. This MDM data contains 4 items, all of which measure the skill of multiplication. The mdm_qmatrix correspondingly
is made up of 4 rows and 2 variables.
• item: Item identifier, corresponds to mdm1-mdm4 in mdm_data
• multiplication: Dichotomous indicator for whether or not the multiplication skill is measured by each item. A value of 1 indicates the skill is measured by the item and a value of 0 indicates the
skill is not measured by the item.
MacReady, G. B., & Dayton, C. M. (1977). The use of probabilistic models in the assessment of mastery. Journal of Educational Statistics, 2(2), 99-120. doi:10.2307/1164802 | {"url":"https://measr.info/reference/mdm","timestamp":"2024-11-12T06:48:09Z","content_type":"text/html","content_length":"12123","record_id":"<urn:uuid:2ce15297-22b6-4160-a97d-f80ff3f267cd>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00159.warc.gz"} |
Tanya Khovanova's Math Blog
Recent Facebook Puzzle from Denis Afrisonov.
Puzzle. 100 students took a test where each was asked the same question: “How many out of 100 students will get a ‘pass’ grade after the test?” Each student must reply with an integer.
Immediately after each answer, the teacher announced whether the current student passed or failed based on their answer. After the test, an inspector checks if any student provided a correct
answer but was marked as failed. If so, the teacher is dismissed, and all students receive a passing grade. Otherwise, the grades remain unchanged. Can the students devise a strategy beforehand
to ensure all of them pass?
13 Comments
1. Alistair:
I can find a strategy such that the last student who fails gets the right answer. If the teacher does not want to be dismissed, he must therefore pass everybody.
19 July 2024, 10:03 pm
2. Ivan:
If the teacher considers the answer “zero” correct, this makes the answer incorrect. Therefore, the teacher must consider the answer “zero” incorrect. But if all students answer “zero”, in the
end all their answers are correct. The inspector dismisses the teacher and all students pass.
20 July 2024, 6:51 am
3. Leif:
Or the other way around – if all students say 100, then they must all be marked correct and everyone passes. If the teacher marks one student (or more) as wrong, the teacher is dismissed, and
everyone passes.
Are we missing something? Because these answers fell a bit trivial.
20 July 2024, 10:06 am
4. tanyakh:
Dear Ivan and Leif, the teacher is allowed to mark the incorrect answer as correct sometimes.
20 July 2024, 10:09 am
5. Leo B.:
Suppose the teacher is clairvoyant. Then she picks the smallest number not mentioned by any of the students, considers that number the correct answer, and grades that number of answers as
correct, rendering the rest incorrect.
Oh, and what if the smallest number not mentioned is 100, because all numbers in [0; 99] have been mentioned exactly once? Then she grades “1” as the correct answer, and the remaining 99
incorrect. The inspector will be satisfied.
Therefore, even if the teacher is not clairvoyant, she can achieve the same result by pure chance.
20 July 2024, 8:51 pm
6. Andreas:
If I understand it correctly, each student will know whether the teacher passed or failed each previous student before they have to answer, and the inspector considers the correct answer to be
the number of students that got a ‘pass’ grade from the teacher. In that case, each student could give the answer 99 minus the number of ‘fail’ grades so far, which is one less than the possible
maximum if the teacher would only give out ‘pass’ grades from now on.
If the teacher grades all as ‘pass’ then they will all technically be incorrect but that doesn’t matter since they already passed. Otherwise, since everyone answers one less than the possible
maximum, the last one who does not get a “pass” grade gets the right answer.
21 July 2024, 12:47 pm
7. Evan:
The teacher chooses an integer x in [0, 100]. If x = 100, then all students pass and the result is trivial, so we restrict to x in [0, 99]. Note that |x| = 100.
The strategy the students agree on is follows: the first student answers ’99’. If the teacher announces this student passes, then all subsequent students answer ’99’. One student must be failed
despite giving the correct answer and the teacher is dismissed.
If, however, the teacher announces that the first student fails, then the next student instead answers ’98’. The students repeat the same steps as above, either repeating the answer once a
student has been passed or decreasing their answer by one on a fail. One student is always failed despite giving a correct answer and the teacher is dismissed.
I shall not provide a formal proof, but the reason this strategy works is because there are 100 students, and the teacher must select an x in [0, 99] for the number of students to pass. There are
|x| = 100 options for the teacher to select. If the teacher selects x = 0, then all 100 students are failed, including the last student who answers ‘0’. For any x the teacher selects, there will
be x + 1 students answering ‘x’. By the pigeonhole principle, one of these students must be failed despite giving the correct answer.
23 July 2024, 1:31 pm
8. Gennardo:
Every student adds the number of students that have already passed to the number of students who are behind him. To make it more convenient: the first student says 99 and every student says the
number of his predecessor if his predecessor has passed, but subtracts 1 to the number of his predecessor if he had failed. The effect is that the last student that failed had the correct answer.
So the students reached their goal which they do of course if the teacher let everybody pass.
23 July 2024, 3:00 pm
9. Puzzled:
Could someone please tell if the following consideration is correct?
Every student answers with 0. If the teacher ranks them all as failed, in the end it will be every student having given the correct answer but marked as failed. If the teacher ranks someone as
passed, it’s a case of a student having given an incorrect answer but marked as passed. Either way, the teacher is disqualified.
3 August 2024, 7:50 am
10. tanyakh:
to Puzzled.
Suppose every student answers 0. Then the teach can pass one of them and fail the others.
4 August 2024, 11:43 am
11. Martin Roller:
The students answer with the numbers 0, 1, 2, up to 99. If the teacher fails at least one of them, some student guessed the right number of passes, so the teacher is dismissed and all students
pass. If not, all students passed and the grades are not changed.
31 August 2024, 12:26 pm
12. Martin Roller:
Ah well, I retract that. The teacher could pass the student giving the correct answer.
31 August 2024, 3:17 pm
13. Martin Roller:
Bonus Puzzle: The strategy for the students, defined by Andreas et. al. is _unique_. If at least one student deviates from the strategy, the teacher can fail at least one student and the
inspector won’t prevent it.
11 October 2024, 3:03 am | {"url":"https://blog.tanyakhovanova.com/2024/07/pass-fail/","timestamp":"2024-11-10T22:11:58Z","content_type":"text/html","content_length":"65920","record_id":"<urn:uuid:9b957845-85b9-4821-aeb9-8fc91a9717da>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00686.warc.gz"} |
Exercise 3: Multi-class Classification and Neural Networks solution
In this exercise, you will implement one-vs-all logistic regression and neural networks to recognize hand-written digits. Before starting the programming exercise, we strongly recommend watching the
video lectures and completing the review questions for the associated topics. To get started with the exercise, you will need to download the starter code and unzip its contents to the directory
where you wish to complete the exercise. If needed, use the cd command in Octave/MATLAB to change to this directory before starting this exercise. You can also find instructions for installing Octave/
MATLAB in the “Environment Setup Instructions” of the course website.
Files included in this exercise
ex3.m – Octave/MATLAB script that steps you through part 1 ex3 nn.m – Octave/MATLAB script that steps you through part 2 ex3data1.mat – Training set of hand-written digits ex3weights.mat – Initial
weights for the neural network exercise submit.m – Submission script that sends your solutions to our servers displayData.m – Function to help visualize the dataset fmincg.m – Function minimization
routine (similar to fminunc) sigmoid.m – Sigmoid function [?] lrCostFunction.m – Logistic regression cost function [?] oneVsAll.m – Train a one-vs-all multi-class classifier [?] predictOneVsAll.m –
Predict using a one-vs-all multi-class classifier [?] predict.m – Neural network prediction function
? indicates files you will need to complete
Throughout the exercise, you will be using the scripts ex3.m and ex3 nn.m. These scripts set up the dataset for the problems and make calls to functions that you will write. You do not need to modify
these scripts. You are only required to modify functions in other files, by following the instructions in this assignment.
Where to get help The exercises in this course use Octave1 or MATLAB, a high-level programming language well-suited for numerical computations. If you do not have Octave or MATLAB installed, please
refer to the installation instructions in the “Environment Setup Instructions” of the course website. At the Octave/MATLAB command line, typing help followed by a function name displays documentation
for a built-in function. For example, help plot will bring up help information for plotting. Further documentation for Octave functions can be found at the Octave documentation pages. MATLAB
documentation can be found at the MATLAB documentation pages. We also strongly encourage using the online Discussions to discuss exercises with other students. However, do not look at any source code
written by others or share your source code with others.
1 Multi-class Classification
For this exercise, you will use logistic regression and neural networks to recognize handwritten digits (from 0 to 9). Automated handwritten digit recognition is widely used today – from recognizing
zip codes (postal codes) on mail envelopes to recognizing amounts written on bank checks. This exercise will show you how the methods you’ve learned can be used for this classification task. In the
first part of the exercise, you will extend your previous implemention of logistic regression and apply it to one-vs-all classification.
1Octave is a free alternative to MATLAB. For the programming exercises, you are free to use either Octave or MATLAB.
1.1 Dataset
You are given a data set in ex3data1.mat that contains 5000 training examples of handwritten digits.2 The .mat format means that that the data has been saved in a native Octave/MATLAB matrix format,
instead of a text (ASCII) format like a csv-file. These matrices can be read directly into your program by using the load command. After loading, matrices of the correct dimensions and values will
appear in your program’s memory. The matrix will already be named, so you do not need to assign names to them.
% Load saved matrices from file load(‘ex3data1.mat’); % The matrices X and y will now be in your Octave environment
There are 5000 training examples in ex3data1.mat, where each training example is a 20 pixel by 20 pixel grayscale image of the digit. Each pixel is represented by a floating point number indicating
the grayscale intensity at that location. The 20 by 20 grid of pixels is “unrolled” into a 400-dimensional vector. Each of these training examples becomes a single row in our data matrix X. This
gives us a 5000 by 400 matrix X where every row is a training example for a handwritten digit image. X = — (x(1))T — — (x(2))T — . . . — (x(m))T — The second part of the training
set is a 5000-dimensional vector y that contains labels for the training set. To make things more compatible with Octave/MATLAB indexing, where there is no zero index, we have mapped the digit zero
to the value ten. Therefore, a “0” digit is labeled as “10”, while the digits “1” to “9” are labeled as “1” to “9” in their natural order.
1.2 Visualizing the data
You will begin by visualizing a subset of the training set. In Part 1 of ex3.m, the code randomly selects selects 100 rows from X and passes those rows to the displayData function. This function maps
each row to a 20 pixel by 20 pixel grayscale image and displays the images together. We have provided
2This is a subset of the MNIST handwritten digit dataset (http://yann.lecun.com/ exdb/mnist/).
the displayData function, and you are encouraged to examine the code to see how it works. After you run this step, you should see an image like Figure 1.
Figure 1: Examples from the dataset
1.3 Vectorizing Logistic Regression
You will be using multiple one-vs-all logistic regression models to build a multi-class classifier. Since there are 10 classes, you will need to train 10 separate logistic regression classifiers. To
make this training efficient, it is important to ensure that your code is well vectorized. In this section, you will implement a vectorized version of logistic regression that does not employ any for
loops. You can use your code in the last exercise as a starting point for this exercise.
1.3.1 Vectorizing the cost function
We will begin by writing a vectorized version of the cost function. Recall that in (unregularized) logistic regression, the cost function is
J(θ) =
1 m
m X i=1−y(i) log(hθ(x(i)))−(1−y(i))log(1−hθ(x(i))). To compute each element in the summation, we have to compute hθ(x(i)) for every example i, where hθ(x(i)) = g(θTx(i)) and g(z) = 1 1+e−z is the
sigmoid function. It turns out that we can compute this quickly for all our examples by using matrix multiplication. Let us define X and θ as X = — (x(1))T — — (x(2))T — . . . — (x(m))T —
and θ = θ0 θ1 . . . θn . Then, by computing the matrix product Xθ, we have Xθ = — (x(1))Tθ — — (x(2))Tθ — . . . — (x(m))Tθ — = — θT(x(1)) — — θT(x
(2)) — . . . — θT(x(m)) — . In the last equality, we used the fact that aTb = bTa if a and b are vectors. This allows us to compute the products θTx(i) for all our examples i in one line of
code. Your job is to write the unregularized cost function in the file lrCostFunction.m Your implementation should use the strategy we presented above to calculate θTx(i). You should also use a
vectorized approach for the rest of the cost function. A fully vectorized version of lrCostFunction.m should not contain any loops. (Hint: You might want to use the element-wise multiplication
operation (.*) and the sum operation sum when writing this function)
1.3.2 Vectorizing the gradient
Recall that the gradient of the (unregularized) logistic regression cost is a vector where the jth element is defined as
∂J ∂θj
1 m
m X i=1(hθ(x(i))−y(i))x(i) j . To vectorize this operation over the dataset, we start by writing out all
the partial derivatives explicitly for all θj,
∂J ∂θ0 ∂J ∂θ1 ∂J ∂θ2 . . .
∂J ∂θn
1 m
Pm i=1(hθ(x(i))−y(i))x(i) 0 Pm i=1(hθ(x(i))−y(i))x(i) 1 Pm i=1(hθ(x(i))−y(i))x(i) 2 . . . Pm i=1(hθ(x(i))−y(i))x(i) n
1 m
m X i=1(hθ(x(i))−y(i))x(i)
1 m
XT(hθ(x)−y). (1)
hθ(x)−y =
hθ(x(1))−y(1) hθ(x(2))−y(2) . . . hθ(x(1))−y(m)
Note that x(i) is a vector, while (hθ(x(i))−y(i)) is a scalar (single number). To understand the last step of the derivation, let βi = (hθ(x(i))−y(i)) and observe that: X i βix(i) = | | | x(1) x
(2) … x(m) | | | β1 β2 . . . βm = XTβ, where the values βi = (hθ(x(i))−y(i)).
The expression above allows us to compute all the partial derivatives without any loops. If you are comfortable with linear algebra, we encourage you to work through the matrix multiplications above
to convince yourself that the vectorized version does the same computations. You should now implement Equation 1 to compute the correct vectorized gradient. Once you are done, complete the function
lrCostFunction.m by implementing the gradient.
Debugging Tip: Vectorizing code can sometimes be tricky. One common strategy for debugging is to print out the sizes of the matrices you are working with using the size function. For example, given a
data matrix X of size 100×20 (100 examples, 20 features) and θ, a vector with dimensions 20×1, you can observe that Xθ is a valid multiplication operation, while θX is not. Furthermore, if you have a
non-vectorized version of your code, you can compare the output of your vectorized code and non-vectorized code to make sure that they produce the same outputs.
1.3.3 Vectorizing regularized logistic regression
After you have implemented vectorization for logistic regression, you will now add regularization to the cost function. Recall that for regularized logistic regression, the cost function is defined as
J(θ) =
1 m
m X i=1−y(i) log(hθ(x(i)))−(1−y(i))log(1−hθ(x(i)))+ λ 2m
n X j=1
θ2 j.
Note that you should not be regularizing θ0 which is used for the bias term. Correspondingly, the partial derivative of regularized logistic regression cost for θj is defined as
∂J(θ) ∂θ0
1 m
m X i=1
(hθ(x(i))−y(i))x(i) j for j = 0
∂J(θ) ∂θj
= 1 m
m X i=1
(hθ(x(i))−y(i))x(i) j !+ λ m
θj for j ≥ 1
Now modify your code in lrCostFunction to account for regularization. Once again, you should not put any loops into your code.
Octave/MATLAB Tip: When implementing the vectorization for regularized logistic regression, you might often want to only sum and update certain elements of θ. In Octave/MATLAB, you can index into the
matrices to access and update only certain elements. For example, A(:, 3:5) = B(:, 1:3) will replaces the columns 3 to 5 of A with the columns 1 to 3 from B. One special keyword you can use in
indexing is the end keyword in indexing. This allows us to select columns (or rows) until the end of the matrix. For example, A(:, 2:end) will only return elements from the 2nd to last column of A.
Thus, you could use this together with the sum and .^ operations to compute the sum of only the elements you are interested in (e.g., sum(z(2:end).^2)). In the starter code, lrCostFunction.m, we have
also provided hints on yet another possible method computing the regularized gradient.
You should now submit your solutions.
1.4 One-vs-all Classification
In this part of the exercise, you will implement one-vs-all classification by training multiple regularized logistic regression classifiers, one for each of the K classes in our dataset (Figure 1). In
the handwritten digits dataset, K = 10, but your code should work for any value of K. You should now complete the code in oneVsAll.m to train one classifier for each class. In particular, your code
should return all the classifier parameters in a matrix Θ ∈RK×(N+1) , where each row of Θ corresponds to the learned logistic regression parameters for one class. You can do this with a “for”-loop
from 1 to K, training each classifier independently. Note that the y argument to this function is a vector of labels from 1 to 10, where we have mapped the digit “0” to the label 10 (to avoid
confusions with indexing). When training the classifier for class k ∈{1,…,K}, you will want a mdimensional vector of labels y, where yj ∈ 0,1 indicates whether the j-th training instance belongs to
class k (yj = 1), or if it belongs to a different class (yj = 0). You may find logical arrays helpful for this task.
Octave/MATLAB Tip: Logical arrays in Octave/MATLAB are arrays which contain binary (0 or 1) elements. In Octave/MATLAB, evaluating the expression a == b for a vector a (of size m×1) and scalar b will
return a vector of the same size as a with ones at positions where the elements of a are equal to b and zeroes where they are different. To see how this works for yourself, try the following code in
Octave/MATLAB: a = 1:10; % Create a and b b = 3; a == b % You should try different values of b here
Furthermore, you will be using fmincg for this exercise (instead of fminunc). fmincg works similarly to fminunc, but is more more efficient for dealing with a large number of parameters. After you have
correctly completed the code for oneVsAll.m, the script ex3.m will continue to use your oneVsAll function to train a multi-class classifier.
You should now submit your solutions.
1.4.1 One-vs-all Prediction
After training your one-vs-all classifier, you can now use it to predict the digit contained in a given image. For each input, you should compute the “probability” that it belongs to each class using
the trained logistic regression classifiers. Your one-vs-all prediction function will pick the class for which the corresponding logistic regression classifier outputs the highest probability and
return the class label (1, 2,…, or K) as the prediction for the input example. You should now complete the code in predictOneVsAll.m to use the one-vs-all classifier to make predictions. Once you are
done, ex3.m will call your predictOneVsAll function using the learned value of Θ. You should see that the training set accuracy is about 94.9% (i.e., it classifies 94.9% of the examples in the
training set correctly).
You should now submit your solutions.
2 Neural Networks
In the previous part of this exercise, you implemented multi-class logistic regression to recognize handwritten digits. However, logistic regression cannot form more complex hypotheses as it is only
a linear classifier.3 In this part of the exercise, you will implement a neural network to recognize handwritten digits using the same training set as before. The neural network will be able to
represent complex models that form non-linear hypotheses. For this week, you will be using parameters from a neural network that we have already trained. Your goal is to implement the feedforward
propagation algorithm to use our weights for prediction. In next week’s exercise, you will write the backpropagation algorithm for learning the neural network parameters. The provided script, ex3
nn.m, will help you step through this exercise.
2.1 Model representation
Our neural network is shown in Figure 2. It has 3 layers – an input layer, a hidden layer and an output layer. Recall that our inputs are pixel values of digit images. Since the images are of size
20×20, this gives us 400 input layer units (excluding the extra bias unit which always outputs +1). As before, the training data will be loaded into the variables X and y. You have been provided with
a set of network parameters (Θ(1),Θ(2)) already trained by us. These are stored in ex3weights.mat and will be loaded by ex3 nn.m into Theta1 and Theta2 The parameters have dimensions that are sized
for a neural network with 25 units in the second layer and 10 output units (corresponding to the 10 digit classes).
% Load saved matrices from file load(‘ex3weights.mat’);
% The matrices Theta1 and Theta2 will now be in your Octave % environment % Theta1 has size 25 x 401 % Theta2 has size 10 x 26
3You could add more features (such as polynomial features) to logistic regression, but that can be very expensive to train.
Figure 2: Neural network model.
2.2 Feedforward Propagation and Prediction
Now you will implement feedforward propagation for the neural network. You will need to complete the code in predict.m to return the neural network’s prediction. You should implement the feedforward
computation that computes hθ(x(i)) for every example i and returns the associated predictions. Similar to the one-vs-all classification strategy, the prediction from the neural network will be the
label that has the largest output (hθ(x))k.
Implementation Note: The matrix X contains the examples in rows. When you complete the code in predict.m, you will need to add the column of 1’s to the matrix. The matrices Theta1 and Theta2 contain
the parameters for each unit in rows. Specifically, the first row of Theta1 corresponds to the first hidden unit in the second layer. In Octave/MATLAB, when you compute z(2) = Θ(1)a(1), be sure that you
index (and if necessary, transpose) X correctly so that you get a(l) as a column vector.
Once you are done, ex3 nn.m will call your predict function using the loaded set of parameters for Theta1 and Theta2. You should see that the
accuracy is about 97.5%. After that, an interactive sequence will launch displaying images from the training set one at a time, while the console prints out the predicted label for the displayed
image. To stop the image sequence, press Ctrl-C.
You should now submit your solutions.
Submission and Grading
After completing this assignment, be sure to use the submit function to submit your solutions to our servers. The following is a breakdown of how each part of this exercise is scored.
Part Submitted File Points Regularized Logisic Regression lrCostFunction.m 30 points One-vs-all classifier training oneVsAll.m 20 points One-vs-all classifier prediction predictOneVsAll.m 20 points
Neural Network Prediction Function predict.m 30 points Total Points 100 points You are allowed to submit your solutions multiple times, and we will take only the highest score into consideration. | {"url":"https://jarviscodinghub.com/product/exercise-3-multi-class-classi%EF%AC%81cation-and-neural-networks-solution/","timestamp":"2024-11-02T01:50:46Z","content_type":"text/html","content_length":"138523","record_id":"<urn:uuid:ff42fa43-f119-44cd-b317-5356f55c2cb3>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00013.warc.gz"} |
105 research outputs found
<p>HI standard reference serum against <i>S</i>. <i>flexneri</i> 2a (A and B) and <i>S</i>. <i>flexneri</i> 3a (C and D) was tested by L-SBA (A and C, respectively) and by the conventional assay (B
and D, respectively) in three independent experiments. Black lines represent the curve fitting by nonlinear regression for each replicate.</p
<p>The immune states and their transitions for infection and vaccine induced immunity. For individuals in the non-immune categories the time in this category depends on the infection rates and
vaccination strategies. For other immune states, time in the state before losing immunity is sampled from a defined distribution, modulated by exposure/vaccination history. Where there are multiple
possible transitions out of a state, conditional probabilities determine which path an individual follows. The order in which the probabilities are applied is specified in the text. <i>P<sub>6</sub>
</i>, <i>P<sub>7</sub></i>, <i>P<sub>8</sub></i> refer to subclinical infections, <i>P<sub>9</sub></i>, <i>P<sub>10</sub></i> to clinical infections. <i>P<sub>11</sub>, P<sub>12</sub></i>
probabilities of inducing immunity in non-vaccinated individuals, <i>P<sub>13</sub>, P<sub>14</sub>, P<sub>15</sub></i> in previously vaccinated individuals.</p
<p>Individual simulations are shown of the number of cases per month for a 240 month (20 year) simulation period. Simulations used the basic Dhaka parameters but varied <i>R<sub>c</sub></i> (1.5: top
panel, 3.3: middle panel and 10 lower panel) assuming a zero probability of becoming a carrier (red plot), 12.5% of standard probability of becoming a carrier (blue plot) and the standard probability
(green plot). At low transmission (<i>R<sub>c</sub></i> 1.5) in the absence of carriers, the infection becomes epidemic and does not persist, hence only low and standard probability traces are
present in the top panel. The 95% confidence limits on the maximum and minimum cases per month are shown by the thin red, blue or green lines for the corresponding probabilities of becoming a carrier
derived from 200 simulations. The lower 95% confidence is zero for zero probability of carriage at an <i>R<sub>c</sub></i> of 3.3 and zero for 12.5% probability for and <i>R<sub>c</sub></i> of both
1.5 and 3.3 and are not shown on these graphs. In each simulation the starting population was approximately 10,000 growing to 16,000 after 20 years.</p
<p><b>Correlation of serum titers from individual mouse sera against <i>S</i>. Typhimurium (A) and <i>S</i>. Enteritidis (B) as determined by L-SBA and by the conventional SBA assay.</b> Individual
mouse sera were simultaneously assayed by L-SBA (y-axis) and by the conventional CFU-based assay (x-axis) against <i>S</i>. Typhimurium (A) and against <i>S</i>. Enteritidis (B). Plots compare titers
from each assay and demonstrate a linear correlation with a slope of 1. Black lines represent the fitting by linear regression. SBA titers were obtained by single experiments.</p
<p>The simulations systematically <i>R<sub>c</sub></i> for probabilities of 0.02 (short dash), 0.1 (standard conditions – thick continuous line) and 0.5 (long dash) of an infection in a non-immune
person with the standard Dhaka parameter set for other parameters, showing the average age of infection, the force of infection (infectious dose per person per month), the number of chronic carriers
per 100,000 and the number of subclinical and clinical cases per 100,000 per year.</p
<p>Bacterial strains used in this study.</p
<p>(A) Different bacterial concentrations were tested with 5% BRS and measured by luminescence at T0 (triangle symbol) and at T180 (square symbol). (B) Bacterial cells at 1.5 x 10<sup>5</sup> CFU/mL
were diluted down until 1.0 x 10<sup>3</sup> CFU/mL and measured by luminescence. The area within the dashed lines define the confidence interval of the best-fit line of the linear regression (y =
0.009465x + 10.28). CPS, counts per second.</p
<p>HI pooled mouse sera were tested by L-SBA (A, C) or by conventional assay (B, D) against <i>N</i>. <i>meningitidis</i> A Niga 16/09 strain (A, B) or <i>N</i>. <i>meningitidis</i> W Mali 4/11
strain (C, D). Results are the mean ± standard deviation values from three replicates (for some points the error bars are not visible because shorter than the symbols).</p
<p>Impact on incidence (upper panel) and average age (lower) panel. The standard parameters for Dhaka were used assuming an average of three infections was required to induce immunity. For each line,
the first number is the probability of inducing sterile immunity and the second the probability of inducing clinical immunity, conditional on not inducing sterile immunity. In these simulations, the
probability of inducing immunity by a subclinical infection or clinical infection was assumed to be similar. Thus 0.333, 0.000 is a simulation that only induces sterile immunity; 0.000, 0.333 only
induces clinical immunity. The horizontal black line in the lower panel is the average age of infection (76 months) observed in Dhaka.</p
<p>White or grey bars represent the SBA titers calculated from the conventional CFU-counts or from L-SBA, respectively. Standard reference sera were used against <i>C</i>. <i>freundii</i>, <i>S</i>.
Typhimurium, <i>S</i>. Enteritidis, <i>S</i>. <i>flexneri 2a and 3a</i> and <i>S</i>. <i>sonnei</i>. For <i>N</i>. <i>meningitidis</i> A and W anti-capsular mAbs were used. The results are the mean
± standard deviation values from three independent experiments.</p | {"url":"https://core.ac.uk/search/?q=author%3A(Allan%20Saul%20(34133))","timestamp":"2024-11-07T20:26:44Z","content_type":"text/html","content_length":"120554","record_id":"<urn:uuid:a3cfe626-c4bc-4bd2-85b6-60b59b933899>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00605.warc.gz"} |
There are 739 places in North Carolina. This section compares Brookford to the 50 most populous places in North Carolina and to those entities that contain or substantially overlap with Brookford.
The least populous of the compared places has a population of 18,241.
Non-White Population by Place#27
Scope: population of Brookford, selected other places in North Carolina, and entities that contain Brookford
White Hispanic Black
Asian Other
Relative Race and Ethnicity by Place#28
Scope: population of Brookford, selected other places in North Carolina, and entities that contain Brookford
White Hispanic Black
Asian Mixed Other
White Population by Place#29
Scope: population of Brookford, selected other places in North Carolina, and entities that contain Brookford | {"url":"https://statisticalatlas.com/place/North-Carolina/Brookford/Race-and-Ethnicity","timestamp":"2024-11-11T06:52:31Z","content_type":"text/html","content_length":"1049297","record_id":"<urn:uuid:7290fe35-c190-46da-947f-356febfa0996>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00182.warc.gz"} |
Meeting PRIN "String Theory as a bridge between Gauge Theories and Quantum Gravity"
Partition function and correlation functions of the topologically twisted $\mathcal{N} = 2$ super Yang-Mills theory on a smooth four-manifold with gauge group $SU(2)$, also known as Donaldson-Witten
theory, provide us a way to compute topological invariants of many manifolds classifying their smooth structure (Donaldson invariants). Equivariantisation of this theory, on the one hand, can be
considered as a tool to find the original invariants by means of the equivariant localisation, and on the other hand, is interesting by itself, since it also has a topological counterpart
(equivariant Donaldson invariants). While for the equivariant Donaldson-Witten theory with $SU(2)$, gauge group quite a lot of results were found, there is still not much known in the case of the
higher rank gauge symmetry.
After giving an introduction on the subject I will present our recent results for the higher rank theory, which include generalisation of the recurrence relation for the partition function on $\
mathbb{C}^2$ (Zamolodchikov relation) and a proposal for the equivariant Donaldson invariants of the compact toric manifolds. | {"url":"https://agenda.infn.it/event/34123/timetable/?view=standard_numbered_inline_minutes","timestamp":"2024-11-12T05:08:04Z","content_type":"text/html","content_length":"129728","record_id":"<urn:uuid:70527263-6b0e-4035-bc91-67e80c9ad0ce>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00153.warc.gz"} |
Abstract of Mostafa Charmi
Mostafa Charmi, Ali Mahlooji Far
Assessment of the Log-Euclidean Metric Performance in Diffusion Tensor Image Segmentation
ارزیابی عملکرد متریک لگاریتمی-اقلیدسی در ناحیه بندی تصاویر تانسور انتشار
Introduction: Appropriate definition of the distance measure between diffusion tensors has a deep impact on Diffusion Tensor Image (DTI) segmentation results. The geodesic metric is the best distance
measure since it yields high-quality segmentation results. However, the important problem with the geodesic metric is a high computational cost of the algorithms based on it. The main goal of this
paper is to assess the possible substitution of the geodesic metric with the Log-Euclidean one to reduce the computational cost of a statistical surface evolution algorithm. Materials and Methods: We
incorporated the Log-Euclidean metric in the statistical surface evolution algorithm framework. To achieve this goal, the statistics and gradients of diffusion tensor images were defined using the
Log-Euclidean metric. Numerical implementation of the segmentation algorithm was performed in the MATLAB software using the finite difference techniques. Results: In the statistical surface evolution
framework, the Log-Euclidean metric was able to discriminate the torus and helix patterns in synthesis datasets and rat spinal cords in biological phantom datasets from the background better than the
Euclidean and J-divergence metrics. In addition, similar results were obtained with the geodesic metric. However, the main advantage of the Log-Euclidean metric over the geodesic metric was the
dramatic reduction of computational cost of the segmentation algorithm, at least by 70 times. Discussion and Conclusion: The qualitative and quantitative results have shown that the Log-Euclidean
metric is a good substitute for the geodesic metric when using a statistical surface evolution algorithm in DTIs segmentation. | {"url":"https://www.znu.ac.ir/members/absjpaperEn/153","timestamp":"2024-11-12T04:14:29Z","content_type":"text/html","content_length":"12172","record_id":"<urn:uuid:23a1c045-76cd-46a4-9b2c-a80a783bb237>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00713.warc.gz"} |
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Accéder au cours arrow_forward. Plus, it is also while building machine learning models as it can be used as an explanatory variable as well. You can try a Free Trial instead, or apply for Financial
Aid. The lengths of the lines can be computed using the Pythagoras theorem as shown in the pic below. Home / Mathematics for Machine Learning: PCA. ARIMA Model - Complete Guide to Time Series
Forecasting in Python, Parallel Processing in Python - A Practical Guide with Examples, Time Series Analysis in Python - A Comprehensive Guide with Examples, Top 50 matplotlib Visualizations - The
Master Plots (with full python code), Cosine Similarity - Understanding the math and how it works (with python codes), Matplotlib Histogram - How to Visualize Distributions in Python, 101 NumPy
Exercises for Data Analysis (Python), Modin – How to speedup pandas by changing one line of code, Dask – How to handle large dataframes in python using parallel computing, Text Summarization
Approaches for NLP – Practical Guide with Generative Examples, Complete Guide to Natural Language Processing (NLP) – with Practical Examples, Portfolio Optimization with Python using Efficient
Frontier with Practical Examples, Logistic Regression in Julia – Practical Guide with Examples, One Sample T Test – Clearly Explained with Examples | ML+, Understanding Standard Error – A practical
guide with examples, Percentage of Variance Explained with each PC, Step 3: Compute Eigen values and Eigen Vectors, Step 4: Derive Principal Component Features by taking dot product of eigen vector
and standardized columns. The further you go, the lesser is the contribution to the total variance. But what is covariance and covariance matrix? Principal Components Analysis (PCA) – Better
Explained. With the first two PCs itself, it’s usually possible to see a clear separation. Here is the objective function: It can be proved that the above equation reaches a minimum when value of u1
equals the Eigen Vector of the covariance matrix of X. Yes, Coursera provides financial aid to learners who cannot afford the fee. This Eigen Vector is same as the PCA weights that we got earlier
inside pca.components_ object. Using all these tools, we'll then derive PCA as a method that minimizes the average squared reconstruction error between data points and their reconstruction. Home /
Mathematics for Machine Learning: PCA. Disclaimer: This course is substantially more abstract and requires more programming than the other two courses of the specialization. We won’t use the Y when
creating the principal components. This course is of intermediate difficulty and will require Python and numpy knowledge. PC1 contributed 22%, PC2 contributed 10% and so on. Thanks to this excellent
discussion on stackexchange that provided these dynamic graphs. If you only want to read and view the course content, you can audit the course for free. Because I don’t want the PCA algorithm to know
which class (digit) a particular row belongs to. 3. En savoir plus . The Programming assignments are quite challenging. The information contained in a column is the amount of variance it contains.
Using these two columns, I want to find a new column that better represents the ‘data’ contributed by these two columns. Mathematics for Machine Learning: PCA. Alright. … The second course,
Multivariate Calculus, builds on this to look at how to optimize fitting functions to get good fits to data. PCA can be a powerful tool for visualizing clusters in multi-dimensional data. You will
need good python knowledge to get through the course. Some ability of abstract thinking assignment Niveau : Intermédiaire. This option lets you see all course materials, submit required assessments,
and get a final grade. Refer to this guide if you want to learn more about the math behind computing Eigen Vectors. More questions? located in the heart of London. It is not a feature selection
technique. Eigen values and Eigen vectors represent the amount of variance explained and how the columns are related to each other. Within this course, this module is the most challenging one, and we
will go through an explicit derivation of PCA plus some coding exercises that will make us a proficient user of PCA. Mathematics for Machine Learning: PCA >>CLICK HERE TO SEE THE COURSE. In this
tutorial, I will first implement PCA with scikit-learn, then, I will discuss the step-by-step implementation with code and the complete concept behind the PCA algorithm in an easy to understand
manner. We will also implement our results in code (jupyter notebooks), which will allow us to practice our mathematical understand to compute averages of image data sets. This will play an important
role in the next module when we derive PCA. Data can be interpreted as vectors. 4.0. stars. and importantly how to understand PCA and what is the intuition behind it? Likewise, PC2 explains more than
PC3, and so on. If you go by the formula, take a dot product of of the weights in the first row of pca.components_ and the first row of the mean centered X to get the value -134.27. Logistic
Regression in Julia – Practical Guide, ARIMA Time Series Forecasting in Python (Guide). When covariance is positive, it means, if one variable increases, the other increases as well. Part 1:
Implementing PCA using scikit learn, Part 2: Understanding Concepts behind PCA, How to understand the rotation of coordinate axes, Part 3: Steps to Compute Principal Components from Scratch.
card_giftcard 160 points. The objective is to determine u1 so that the mean perpendicular distance from the line for all points is minimized. Topic modeling visualization – How to present the results
of LDA models? The values in each cell ranges between 0 and 255 corresponding to the gray-scale color. You'll be prompted to complete an application and will be notified if you are approved. Part 2:
Understanding Concepts behind PCA First, I initialize the PCA() class and call the fit_transform() on X to simultaneously compute the weights of the Principal components and then transform X to
produce the new set of Principal components of X. What I mean by ‘mean-centered’ is, each column of the ‘X’ is subtracted from its own mean so that the mean of each column becomes zero. Apply for it
by clicking on the Financial Aid link beneath the "Enroll" button on the left. PCA can be a powerful tool for visualizing clusters in multi-dimensional data. In this module, we use the results from
the first three modules of this course and derive PCA from a geometric point of view. I'd suggest giving more time and being patient in pursuit of completing this course and understanding the
concepts involved. This dataframe (df_pca) has the same dimensions as the original data X. eval(ez_write_tag([[250,250],'machinelearningplus_com-box-4','ezslot_1',147,'0','0']));The pca.components_
object contains the weights (also called as ‘loadings’) of each Principal Component. Let’s import the mnist dataset. This will become important later in the course when we discuss PCA. This course
was definitely a bit more complex, not so much in assignments but in the core concepts handled, than the others in the specialisation. In this module, we will look at orthogonal projections of
vectors, which live in a high-dimensional vector space, onto lower-dimensional subspaces. Lower-Dimensional subspaces variance when we shift or scale the original dataset chunk of the data into
components. Algorithm to know which class ( digit ) a particular row belongs to categories from each.... The cells of the Mathematics for machine learning: PCA ” on Coursera and assignments fitting
to. As a result, the lesser is the Principal components are nothing but the row represents purchase. And numpy knowledge data with some loss, similar to jpg or mp3 data set the weights. Simplify
things, let ’ s direction PCA module this Eigen vector is same as ”. In multi-dimensional mathematics for machine learning: pca combinations of columns require Python and numpy knowledge career from!
How it relates to data there are columns in the dataset matrix with the same as the PC ’ built-in... Vectors are at the core of PCA scatter plot using the pca.components_ earlier of! You 'll be
prompted to complete this step for each course in mode... In a direction that minimizes the perpendicular distance from the line to a pandas dataframe covariance. Present the results example, row 1
contains the 784 weights of PC1 are at the end of course. Introduce and practice the concept of inner products of functions and random variables by.! To answer all of these questions in this course
may refresh some of your knowledge way. Interested in determining the direction of the most fundamental dimensionality reduction and ability to visualize separation. The maximum variation of the unit
vector eventually becomes the weights of Principal components, for example, row contains... Y variable names ' 0 ' which tells what digit the row corresponding the datapoint the... About the math
behind PCA see clear clusters of points belonging to the value in (... Content, you will be notified if you want to learn more about the math behind computing Eigen as. Discussion on stackexchange
that provided these dynamic graphs a line should be in a direction minimizes! Back the original XY axis of to match the direction u1, I want find... And credit towards a master ’ s actually compute
this, a large of., ARIMA time Series Forecasting in Python ( Guide ) code it out algorithmically as well these questions this. Contributed by each PC is a weighted additive combination of all
possible combinations of..
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checon.f - Linux Manuals (3)
checon.f (3) - Linux Manuals
checon.f -
subroutine checon (UPLO, N, A, LDA, IPIV, ANORM, RCOND, WORK, INFO)
Function/Subroutine Documentation
subroutine checon (characterUPLO, integerN, complex, dimension( lda, * )A, integerLDA, integer, dimension( * )IPIV, realANORM, realRCOND, complex, dimension( * )WORK, integerINFO)
CHECON estimates the reciprocal of the condition number of a complex
Hermitian matrix A using the factorization A = U*D*U**H or
A = L*D*L**H computed by CHETRF.
An estimate is obtained for norm(inv(A)), and the reciprocal of the
condition number is computed as RCOND = 1 / (ANORM * norm(inv(A))).
UPLO is CHARACTER*1
Specifies whether the details of the factorization are stored
as an upper or lower triangular matrix.
= 'U': Upper triangular, form is A = U*D*U**H;
= 'L': Lower triangular, form is A = L*D*L**H.
N is INTEGER
The order of the matrix A. N >= 0.
A is COMPLEX array, dimension (LDA,N)
The block diagonal matrix D and the multipliers used to
obtain the factor U or L as computed by CHETRF.
LDA is INTEGER
The leading dimension of the array A. LDA >= max(1,N).
IPIV is INTEGER array, dimension (N)
Details of the interchanges and the block structure of D
as determined by CHETRF.
ANORM is REAL
The 1-norm of the original matrix A.
RCOND is REAL
The reciprocal of the condition number of the matrix A,
computed as RCOND = 1/(ANORM * AINVNM), where AINVNM is an
estimate of the 1-norm of inv(A) computed in this routine.
WORK is COMPLEX array, dimension (2*N)
INFO is INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
November 2011
Definition at line 125 of file checon.f.
Generated automatically by Doxygen for LAPACK from the source code. | {"url":"https://www.systutorials.com/docs/linux/man/3-checon.f/","timestamp":"2024-11-02T11:42:19Z","content_type":"text/html","content_length":"9171","record_id":"<urn:uuid:6c898970-7944-4473-aa8c-a41fe9347609>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00288.warc.gz"} |
(PDF) QKD protocol based on entangled states by trusted third party
Author content
All content in this area was uploaded by Khaled Elleithy on Oct 04, 2017
Content may be subject to copyright.
QKD Protocol Based on Entangled States
By Trusted Third Party
Abdulbast A. Abushgra Khaled M. Elleithy
Computer Science & Engineering Department Computer Science & Engineering Department
Unversity of Bridgeport Unversity of Bridgeport
aabushgr@my.bridgeport.edu Elleithy@bridgeport.edu
Abstract— Quantum cryptography is considered a solution
for sharing secret information in a secure mode. Establishing a
quantum security platform into an exciting system requires a
package of stable processes. One of these processes is based on
creating a Quantum Key Distribution (QKD) protocol or sharing
a secret key. This paper presents a QKD protocol that utilizes
two quantum channels to prepare a shared secret key. The first
communication channel will be initiated by entanglement states,
where the entangled photons will be emitted by a trusted third
party. The second communication channel utilizes the
superposition states that will be initiated by the one of the
communicated parties. Moreover, the protocol produces a string
of random qubits after verifying the communicated legitimate
parties during entangled state channels. The produced string will
reflect the shared secret key between the users.
Keywords- Entangled State, Superposition State, Qubits, Decoy
State, and Bell’s States.
Flowing enormous data through various communication
channels causes leaks of important information through
classical communications by eavesdroppers. Classical
cryptography has several algorithms that defend against many
information attacks (these algorithms are still secure as long as
the quantum computer is conceptual). Furthermore, quantum
cryptography provides security of information with some
challenges that are determined in quantum attacks or natural
noise. In 1984, Charles Bennett and Gilles Brassard invented
[1] the most sparkling quantum key distribution protocol,
which is called BB84 protocol. Several QKD protocols then
were invented (such as B92 protocol [2], SARG04 protocol [3],
EPR protocol [4], and DPS protocol [5]).
Any quantum key distribution protocol technically uses
different channels to submit qubits (Quantum Bits) for data
transmission, with and regular bits for either confirming or
reconciling the submitted qubits. Each quantum channel is
initiated in varying environments that specifies the type of
platforms and used tools (such as transformers and detectors).
First of all, the quantum channel should utilize either Viper-
Optics or Free-Space to transfer a qubit from one side to
another; both cannot be protected totally from eavesdroppers.
The quantum mechanics is the only factor that makes quantum
communication unconditionally secure [6]. Moreover, the rules
of physics keep the whole system that is used active (as long as
no attempts to break the system). Therefore, any illegal alien
will be detected by destroying the system.
Furthermore, using multiple polarized states of a particle
and the measurement process of the same particle determine
the stability and efficiency of each QKD protocol. Fulfilling an
authentication between two or more communicators is one of
the challenges that cause an enormous leak of information if
the communicators cannot verify each other. This paper
presents a new algorithm that is designed to prove the
authentication within an entangled channel. The presented
protocol is based upon two quantum channels: one channel is
EPR channel (entangled states channel) and second channel is
quantum channel (qubits channel). The protocol will be
terminated in case the authentication between the
communicated parties is interrupted.
A. The EPR Preparation
Initiating an EPR connection should be done by submitting
EPR photons to the receiver (Bob). The source of EPR photon
would be from the sender (Alice) or a third party; but in this
proposed protocol, the third party will be confirmed. The
submitted EPR string S
contains several characters, which
are considered an open key for the whole scheme. These
characters involve a sequence of information (such as initiation
time t
, number of matrices n (if any), matrix size m, parity
diagonal p, state dimension s, matrix indices R, and termination
time t
) as in figure (1).
Fig. 1 The EPR string prepared by the sender.
The sender (Alice) is supposed to start talking with the third
party by sending a copy of the plaintext into a classical
channel. Next, the trusted third party will convert the plaintext
to encoded information
to be transferred into
entanglement states. Both of the communicated parties (the
sender and receiver) will receive a copy of the entangled
photons at the same time. The EPR string S
is the encoded
plaintext that will be shared between the sender and receiver.
The string contains particles of Pauli states (
Each photon has two states|, where | should be sent to
Alice and the | will be sent to Bob. Based upon the
theoretical measurement and the fact of EPR photons, both
parties can initiate the communication in safe mode.
Fig. 2 The communication between the third party with Alice and
B. The Qubits Preparation
To create a secret (shared) key, Alice is supposed to know
the information that will be submitted to the other party. The
plaintext should be converted to qubits (data), and the third
party then sets up the converted plaintext into a designed
matrix. The matrix matches the length of the plaintext n as
= log
where DM is the size of the used matrix and n is the length
of the converted plaintext.
Next, the third party will fill up the lower and upper
triangles (the diagonal line is not included) by the converted
qubits of the plaintext. The filling scenario starts from up to
down in the lower triangle and from down to up in the upper
triangle, as shown in figure (2). The whole matrix will be filled
as a result except the diagonal line, where the third party
adjusts the diagonal cells based on the summation of each row.
If the summation of the row is odd, the third party will add (1)
to the empty cell to make the row even. On the other hand, if
the summation of the row is even, it will be added (0) bit to the
cell. Therefore, the third party prepares the whole matrix with
even row’s summation; this will be an extra protection against
PNS attacks [8], where Alice and Bob will know if the
upcoming qubits were interrupted by eavesdroppers or
Fig. 3 The prepared matrix into three sections: lower triangle, upper
triangle, and diagonal line.
A. EPR Channel
In 1935 [9], Einstein, Podolosky, and Rosen came up with
their fabulous paper that opened a huge argument about the
wave function and incompleteness of quantum mechanics. The
main concept of EPR is a photon submission from the source
(X) to two different destinations (e
, e
). The measurement, in
the case of no interruption, will demonstrate a different state at
each side. Moreover, if Alice (one of the communicators or the
sender) received|0, then Bob (one of the communicators or
the receiver) should have |1after his measurement. The
presented algorithm is initiated by creating an EPR channel and
the protocol will be described as follows:
• Alice sends n bits of the plaintext (the length of the
plaintext) to a third party.
• The third party converts the plaintext to EPR states
(|Φ,|Ψ) based on the plaintext, and then sends the
EPR states into separate channels (where one state is
sent to Alice EPR
and the other state is sent to Bob
• Alice creates an unknown photon (e.g.|=
|1), which is in the superposition state.
• Calculating both the entangled state and
superposition state (|Ψ
) to produce a three-
dimension particle state.
• Alice separates the three states, where | will
• The first outcome of | becomes entangled and |
is separated (or became in superposition).
• Alice submits two classical bits
(|00,|01,|10,|11) for the used gates at both
Fig. 4 The photon emits from the source, and the measurement
will be same color if one side measured.
Therefore, the authentication between the communicated
parties should either be approved to move on or to start over.
After that, Bob should have the proper quantum gates ⨁ as
well as the photon states.
Algorithm .1 QKD Protocol
1. Submit n bits to well-known third party (p) by A
2. n (|Ψ|Φ) // First loop
|1 // P sent a pair to both A&B
4. if (A == 0) then (B == 1) // Second loop
5. B 1
6. else: error
7. end; // ending the loop
8. A |
9. for: 1 n //Measuring & reconciliation
10. (|Ψ⊕|) // Third loop
11. end; //use the data collected by EPR
12. B {0,1} // B gets the secret key
The proposed algorithm runs through three loops that are
involved in submitting a plaintext to a third party, initiating an
EPR connection by the third party, and the quantum
communications between the sender and receiver.
B. The Classical Communication
To ensure that Bob has the right quantum gates (as in figure
(5)) Alice initiates a communication into a classical channel.
Two bits have the needed information that Alice should send to
Bob. Each two bit has a meaning of a certain quantum gate; the
(00) bits mean using the unitary operator, (01) Z gate, (10) X
gate, and (11) X and Z gates. These gates are the only classical
operation that Alice and Bob need to use during the entire
system procedures.
Fig. 5 The three quantum gates (X, Y, and Z) used into
exchanging channel.
Moreover, interrupting the classical communication will not
impact the protocol processes because the receiver will get
unmatched qubits during the preparation of the upcoming
qubits. Also, the decoy states (diagonal line) will show some
huge variations.
C. Quantum Channel
After an authentication proof, both parties start exchanging
qubits (data) into the quantum channel. The submitted qubits
will be in two bases (|×,|+) and four states
(|0,|45,|90,|135). Alice creates the qubits based on
the EPR
that was submitted by the third party; and Bob will
use Pauli-matrices with prior knowledge to measure the
upcoming qubits from Alice into the right states [10]:
The physical measurements should all be correct because
Bob has already agreed on the EPR
confirmation. Moreover,
the mechanism of data organization into a matrix setup will
assist to protect qubits from any quantum attack. On the other
hand, Bob can realize any changes in the received qubits and he
can figure out the error by diagonal decoy states.
Fig. 6 The whole mechanism for the proposed scheme in two quantum
A. The Runtime-Execution
To test the simplicity of the proposed protocol, it was
simulated technically by measuring the run time execution
during the generation of a secret key by two legitimate parties.
The simulation is considered a test of the time taken from
initiation the communication to generation of the secret key.
Even the loops that were required for some function will be
included, as well as the reconciliation phase. The following
equation will simply explain the calculation of the run time
where P is the required loop for each function process
into the entire algorithm initiation.
The proposed protocol runs in a low time rate if there is no
error created by eavesdropper. On the other hand, applying an
error during the communications between the legal parties will
increase the rate of time taken to create a secret key.
B. The Efficiency
Based upon the measurements that were applied on the
proposed protocol, the efficiency can be approved by
measuring the Qubit Error Rate (QBER). The total of used
qubits at the beginning of the communication will be different
at the end for many reasons. The environment is one reason
that causes a qubit drop or weak light. Quantum attacks can
also cause several damages to the submitted qubits, either by
splitting the state of the photon or by interrupting and
resending a photon.
The efficiency measurement was applied by counting the
QBER, where correcting errors should be realized by the
following equation [11, 12]:
where n is the total of the submitted qubits, and r is the
qubits that were measured and successfully uncovered. The
results show the proposed protocol is efficient even if the
quantum attacks are applied. Therefore, there is no leaked
information even if the eavesdropper tried to use one of the
attacks scenarios.
Fig. 7 The correlation between submitted and received qubits
measured with 50 qubits.
The correlation in the figure (7) between the submitted and
received qubits reflects the difficulties of finding out the
relation between the two parties. Hence, the main point is
utilizing a matrix either in sorting submitted qubits or re-
sorting received qubits; this usually is considered as an
advantage to hide the core of a created secret key.
C. The Security
The security measurement is applied by several methods, but
this proposed protocol utilizes Shannon Entropy [13, 14] to
measure the level of security. The probability in the next
equation shows the rate of corrupted qubits of the received
where P
is the probability of the shown character (certain
qubits) in i numbers.
The security measurement can be applied into the entropy
of security in general, where it can measure the rate of
uncovered qubits .
where log represents the natural logarithm (the
logarithm with the base e). The constant e is called Euler’s
number and it is equal to an approximately: ≈ 2.71828
[15]. Moreover, k is the uncovered qubits that should be
measured by Bob and n is the total of qubits that are submitted
by Alice.
Fig. 8 The entropy of security measured for the proposed protocol
that confirmed by a third party.
The figure (8) demonstrates the S(k) function to calculate the
entropy of security, where the used key length is 32 qubits.
The rate of uncovered qubits will be approx. 0.53 qubits of the
secret key.
The proposed scheme presents a quantum key distribution
protocol that is essentially designed in two quantum channels.
The EPR channel (confirmation channel) uses the entangled
states rather than states in superposition, which has a low risk
and certain probability. The second channel is utilized to
transfer data from sender to receiver with the ability to detect
any interruption. Generally, the proposed scheme treats
missing authentication between legitimate parties in most
well-known quantum key distribution protocols. Also, it uses
qubit preparation in matrix (or matrices if any) by the sender,
which is considered a powerful procedure to ignore PNS and
IRA attacks. The proposed scheme has approved its stability
against the Man-In-The-Middle attack, where there is no
chance to impersonate the sender or the receiver.
[1] C. H. B. G. Brassard, "Quantum Cryptography:
Public Key Distribution and Coin Tossing "
International Conference on Computers, Systems &
Signal Processing, p. 5, December 10 - 12, 1984
[2] C. H. Bennett, "Quantum cryptography using any two
nonorthogonal states," Physical Review Letters, vol.
68, p. 3121, 1992.
[3] V. Scarani, A. Acin, G. Ribordy, and N. Gisin,
"Quantum cryptography protocols robust against
photon number splitting attacks for weak laser pulse
implementations," Physical Review Letters, vol. 92,
p. 057901, 2004.
[4] A. Einstein, B. Podolsky, and N. Rosen, "Can
quantum-mechanical description of physical reality
be considered complete?," Physical review, vol. 47,
p. 777, 1935.
[5] K. Inoue, E. Waks, and Y. Yamamoto, "Differential-
phase-shift quantum key distribution using coherent
light," Physical Review A, vol. 68, p. 022317, August
27 2003.
[6] H.-K. Lo, X. Ma, and K. Chen, "Decoy state quantum
key distribution," Physical Review Letters, vol. 94, p.
230504, June 16 2005.
[7] A. Abushgra and K. Elleithy, "QKDP's comparison
based upon quantum cryptography rules," in 2016
IEEE Long Island Systems, Applications and
Technology Conference (LISAT), 2016, pp. 1-5.
[8] M. Elboukhari, M. Azizi, and A. Azizi, "Quantum
key distribution protocols: A survey," International
Journal of Universal Computer Sciences, vol. 1, pp.
59-67, 2010.
[9] A. K. Ekert, "Quantum cryptography based on Bell’s
theorem," Physical Review Letters, vol. 67, p. 661,
[10] A. Abushgra and K. Elleithy, "Initiated decoy states
in quantum key distribution protocol by 3 ways
channel," presented at the Systems, Applications and
Technology Conference (LISAT), IEEE Long Island,
New York, 2015.
[11] D. Gottesman, L. Hoi-Kwong, Lu, x, N. tkenhaus,
and J. Preskill, "Security of quantum key distribution
with imperfect devices," presented at the
International Symposium on Information Theory.
ISIT 2004. Proceedings., Chicago, IL, USA, 2004.
[12] C. H. Bennett, D. P. DiVincenzo, J. A. Smolin, and
W. K. Wootters, "Mixed-state entanglement and
quantum error correction," Physical Review A, vol.
54, p. 3824, 1996.
[13] D. J. MacKay, Information theory, inference and
learning algorithms: Cambridge university press,
[14] Y. Huang, "Computing quantum discord is NP-
complete," New Journal of Physics, vol. 16, p.
033027, 2014.
[15] M. Niemiec and A. R. Pach, "The measure of security
in quantum cryptography," in Global
Communications Conference (GLOBECOM), 2012
IEEE, 2012, pp. 967-972. | {"url":"https://www.researchgate.net/publication/318980413_QKD_protocol_based_on_entangled_states_by_trusted_third_party","timestamp":"2024-11-02T22:46:06Z","content_type":"text/html","content_length":"527770","record_id":"<urn:uuid:ecbfe926-9760-4a43-855d-9da89546781e>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00584.warc.gz"} |
Parallel lines can be drawn with the help of? - JNV Exam Test Answers Hub
Parallel lines can be drawn with the help of?
Parallel lines can be drawn with the help of a set square and scale.
The MCQ has options
A) Protactor and Scale
B) Compass and Scale
C) Set Square and Scale
D) None
What are parallel lines?
Parallel lines are two straight lines that never meet each other.
We draw parallel lines with the help of a scale and set squares, for that we first draw a line (here white line) and a dot where the parallel line needs to be drawn.
Step 1
Place the scale and set square as per the figure.
Step 2
Slide the set square till you reach white dot. Now draw the line over set square. You get a required parallel line.
Parallel lines can be drawn with the help of?
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Cea GATE Exam Series Quiz!
Questions and Answers
Do you know how hard it is to pass a CEA GATE exam? This quiz is no exception. For this quiz, you will be asked complex and exciting questions. You must identify the correct statements, know the
static indeterminacy of a plane, know the difference between middle observation and mean observation and solve complicated equations. This mind-bending quiz is for anyone who is up to the challenge.
Don’t forget to take your certificate after the quiz.
• 1.
His reactions to unpleasant situations tended to _________ everyone’s nerves. The word that best fills the blank in the above sentence is
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. Aggravate
The word "aggravate" means to make a situation or problem worse or more serious. In the given sentence, the word "aggravate" fits perfectly as it suggests that his reactions to unpleasant
situations tend to irritate or annoy everyone's nerves, making the situation even more unpleasant.
• 2.
If XY + Z = X(Y + Z) which of the following must be true?
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. X = 1 OR Z = 0
If XY + Z = X(Y + Z), then it implies that either X = 1 or Z = 0. This is because if X = 1, then the equation becomes Y + Z = Y + Z, which is always true. On the other hand, if Z = 0, then the
equation becomes XY = XY, which is also always true. Therefore, X = 1 OR Z = 0 must be true.
• 3.
What is the number missing from the table?
Correct Answer
B. 48
The numbers in the table are increasing by 3 each time. Starting with 45, adding 3 gives 48, which is the missing number.
• 4.
ARCHIPELAGO : ISLAND
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. Necklace : bead
A necklace is made up of multiple beads, just like an archipelago is made up of multiple islands. In both cases, the individual components come together to form a larger whole.
• 5.
Until now only injectable vaccines against Swine Flu have been available. They have been primarily used by older adults who are at risk for complications from Swine Flu. A new vaccine
administered in an oral form has proven effective in preventing Swine Flu in children. Since children are significantly more likely than adults to contract and spread Swine Flu, making the new
vaccine widely available for children will greatly reduce the spread of Swine Flu across the population. Which of the following, if true, most strengthens the argument? ​
□ A.
If a person receives both the oral and the injectable vaccine, they do not interfere with each other
□ B.
The new vaccine uses the same mechanism to ward off Swine Flu as injectable vaccines do
□ C.
Government subsidies have kept the injectable vaccines affordable for adults
□ D.
Many parents would be more inclined to have their children vaccinated against Swine Flu if it did not involve an injection
□ E.
Correct Answer
D. Many parents would be more inclined to have their children vaccinated against Swine Flu if it did not involve an injection
The given answer strengthens the argument because it suggests that the availability of an oral vaccine would increase the likelihood of parents vaccinating their children against Swine Flu. Since
children are more likely to contract and spread the virus, the wider availability of the new vaccine would greatly reduce the spread of Swine Flu in the population. If parents are more inclined
to have their children vaccinated without the need for an injection, it would lead to higher vaccination rates among children and further contribute to preventing the spread of the virus.
• 6.
Four friends Rishabh, Keshav, Lavish and Hemang are out for shopping. Rishabh has less money than three times the amount that Keshav has. Lavish has more money than Keshav. Hemang has an amount
equal to the difference of amounts with Keshav and Lavish. Rishabh has three times the money with Hemang. Each of them has to buy at least one shorts, or one sleeper, or one sleeveless t-shirt,
or one goggle that is priced 200, 400, 600, and 1000 a piece, respectively. Lavish borrows 300 from Rishabh and buys a goggle. Keshav buys a Sleeveless t-shirt after borrowing 100 from Rishabh
and is left with no money. Rishabh buys three shorts. What is the costliest item that Hemang could buy with his own money?
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. A sleeper
• 7.
I wouldn’t _______ with a soldier who was wearing a metal _______, awarded for a display of _______. The words that best fill the blanks in the above sentence are:
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. Meddle, medal, mettle
• 8.
There are three basket of fruits. First basket has twice the number of fruits in the second basket. Third basket has 3/4 th of the fruits in the first. The average of the fruits in all the basket
is 30. The number of fruits in the first basket is
Correct Answer
C. 40
Let's assume the number of fruits in the second basket is x. According to the information given, the first basket has twice the number of fruits as the second basket, so it has 2x fruits. The
third basket has 3/4 th of the fruits in the first basket, so it has (3/4)(2x) = 3x/2 fruits.
To find the average of the fruits in all the baskets, we add the number of fruits in each basket and divide by the total number of baskets.
(x + 2x + 3x/2) / 3 = 30
Simplifying the equation, we get:
(6x + 4x + 3x) / 6 = 30
13x / 6 = 30
13x = 180
x ≈ 13.8
Since we can't have a fraction of a fruit, we round x to the nearest whole number, which is 14. Therefore, the number of fruits in the first basket is 2x = 2(14) = 28.
Thus, the correct answer is 40.
• 9.
A paper sheet is in the shape of a right angle triangle and cut along a line parallel to hypotenuse leaving a smaller triangle. There was 25% reduction in the length of the hypotenuse of the
triangle. If area of triangle initially was 28 cm^2 then area of smaller triangle will be ______ cm^2.
Correct Answer
B. 15
When a paper sheet in the shape of a right angle triangle is cut along a line parallel to the hypotenuse, it forms two smaller triangles. Since there was a 25% reduction in the length of the
hypotenuse, the length of the hypotenuse of the smaller triangle will be 75% of the original length.
The area of a triangle is given by the formula: (base * height) / 2. Since the shape of the triangle remains the same, the base and height of the smaller triangle will also be 75% of the original
values. Therefore, the area of the smaller triangle will be (75% * 75% * 28 cm^2) / 2, which simplifies to 15 cm^2.
• 10.
A drinks machine offers three selections - Tea, Coffee or Random but the machine has been wired up wrongly so that each button does not give what it claims. If each drink costs 50p, how much
minimum money do you have to put into the machine to work out which button gives which selection?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 50 p
To determine which button gives which selection, you need to try each button once. Since each drink costs 50p, you need to put in a minimum of 50p to try all three buttons. If you put in less
than 50p, you will not be able to try all the selections. Therefore, the minimum amount of money you have to put into the machine is 50p.
• 11.
Let A and B be two 3 × 3 matrices such that A ≠ B, A^2 = B^2, AB = BA and A^2 + 2A + I = 0 where I is the identity matrix. Let |T| denote the determinant of any matrix T. Then
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. |A| ≠ 0 and |A + B| = 0
The given conditions imply that A and B are distinct matrices, A^2 = B^2, and A^2 + 2A + I = 0. From A^2 = B^2, we can deduce that A^2 - B^2 = 0, which can be factored as (A - B)(A + B) = 0.
Since A and B are distinct, (A - B) cannot be zero, which means (A + B) must be zero. Therefore, |A + B| = 0. Additionally, from A^2 + 2A + I = 0, we can deduce that |A|^2 + 2|A| + 1 = 0, which
implies |A| cannot be zero. Hence, the correct answer is |A| ≠0 and |A + B| = 0.
• 12.
The positive value of A for which the equation f(x,y) = x^2 + y^2 + Axy + 5x + 18, does not yields optimum values (i.e. no conclusion can be drawn on the nature of f(x,y) ) and required further
Correct Answer
B. 2
The correct answer is 2. In this equation, the coefficient of xy is A. If A is positive, the equation will have a saddle point, which means that the function does not have an optimum value.
Therefore, further investigation is required to determine the nature of f(x, y).
• 13.
Customers arrive in a certain store according to a possion process at rate 4 per hour. Given that the store opens at 9.00 am what is the probability that exactly 1 customer has arrived by 9:30
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 2e^-2
The probability that exactly 1 customer has arrived by 9:30 can be calculated using the Poisson distribution. The rate of customer arrivals is given as 4 per hour. Since the store opens at 9:00
am and we want to calculate the probability by 9:30 am, which is half an hour later, we can use a rate of 2 customers for this time period. The formula to calculate the probability of exactly 1
customer arriving in this time period is P(X=1) = (e^(-2) * 2^1) / 1! = 2e-2. Therefore, the correct answer is 2e-2.
• 14.
The differential equation 4yy’ - 12x = 0, satisfying condition y (1) = 3. Then the point (5, 3) will lie:
□ A.
□ B.
Outside the solution curve
□ C.
Inside the solution curve
□ D.
Correct Answer
C. Inside the solution curve
The given differential equation is a first-order linear equation. By rearranging the equation, we can solve for y':
y' = 3x/(4y)
To determine whether the point (5, 3) lies inside the solution curve, we substitute x = 5 and y = 3 into the equation. We get:
y' = 3(5)/(4(3)) = 5/4
Since the value of y' at the point (5, 3) is positive, it means that the slope of the solution curve at that point is positive. Therefore, the point (5, 3) lies inside the solution curve.
• 15.
For a certain distributed data, the middle observation is 10 and mean observation is 8. The maximum occurring data is
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 14
The maximum occurring data can be determined by looking at the given information. Since the middle observation is 10 and the mean observation is 8, it implies that there are more data points
above the mean than below it. Therefore, the maximum occurring data must be greater than 10. Among the options given, the only value greater than 10 is 14, so it is the correct answer.
• 16.
As per IS 456:2000, the diagonal tension failure in RCC beams depends upon:
□ A.
Grade of concrete and Percentage of transverse reinforcement provided.
□ B.
Grade of concrete and Steel only
□ C.
Grade of steel and percentage of longitudinal reinforcement provided
□ D.
Grade of concrete and percentage of longitudinal reinforcement provided
Correct Answer
D. Grade of concrete and percentage of longitudinal reinforcement provided
The correct answer is Grade of concrete and percentage of longitudinal reinforcement provided. According to IS 456:2000, the diagonal tension failure in RCC beams is influenced by the grade of
concrete and the percentage of longitudinal reinforcement provided. This means that the strength of the concrete and the amount of reinforcement play a crucial role in preventing diagonal tension
failure in beams.
• 17.
The technique utilized for the disposal of biomedical waste​, in which it is in contact with steam under controlled pressure and temperature condition with the end goal to complete
sterilization is
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. Autoclaving
Autoclaving is the correct answer because it is a technique used for the disposal of biomedical waste. Autoclaving involves subjecting the waste to steam under controlled pressure and temperature
conditions to ensure complete sterilization. This process effectively kills any microorganisms present in the waste, making it safe for disposal.
• 18.
As per Indian Road Congress (IRC) the stripping value of aggregate should not exceed X% for the use in bituminous surface dressing, penetration macadam, bituminous macadam, and carpet
construction when aggregate coated with bitumen is immersed in water bath at Y^oC for Z hours. The value of X, Y and Z will, respectively, be
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. 5, 40, and 24
The correct answer is 5, 40, and 24. According to the Indian Road Congress (IRC), the stripping value of aggregate should not exceed 5% for use in bituminous surface dressing, penetration
macadam, bituminous macadam, and carpet construction. The aggregate coated with bitumen is immersed in a water bath at 40°C for 24 hours to test its resistance to stripping.
• 19.
The laboratory test results of the soil sample are given below: Liquid limit = 37% Plastic limit = 22% % passing through 75 μ sieve = 26% % retained over 4.75 mm IS sieve = 65% % retained over
0.075 mm but passing through 4.75 mm = 26% As per IS 1498 – 1970, the soil is classified as:
Correct Answer
A. GC
Based on the given laboratory test results, the soil sample can be classified as GC according to IS 1498 - 1970. The classification is determined by the liquid limit, plastic limit, and the
percentage of fine particles passing through and retained on specific sieves. In this case, the liquid limit is 37%, which is less than 50% and the plastic limit is 22%, which is also less than
35%. Additionally, the percentage passing through the 75 μ sieve is 26%, which is less than 50%. All these values fall within the criteria for GC classification.
• 20.
What will be range of plasticity index for the fine grained soil below A - line having intermediate compressibility in the plasticity chart.
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 10.95 ≤ I[p] ≤ 21.9
The range of plasticity index for the fine grained soil below the A-line with intermediate compressibility in the plasticity chart is 10.95 ≤ Ip ≤ 21.9. This means that the plasticity index
of the soil can range from 10.95 to 21.9. The plasticity index is a measure of the range of moisture content within which the soil exhibits plastic behavior. Soils with a higher plasticity index
have a greater range of moisture content where they can be molded and deformed without breaking.
• 21.
A tank full of water is shown in figure below, Assume width of tank to be 1.5m. Identify the correct statement made:
□ A.
Hydrostatic Paradox occurs in the tank
□ B.
No Hydrostatic Paradox occurs in the tank
□ C.
Hydrostatic Paradox occurs in the tank as the ratio of weight of water in the tank to pressure at the bottom of the tank is more than 1.
□ D.
Hydrostatic Paradox occurs in the tank as the ratio of weight of water in the tank to pressure at the bottom of the tank is equal to 1.
Correct Answer
B. No Hydrostatic Paradox occurs in the tank
The question states that there is no Hydrostatic Paradox occurring in the tank. This means that the ratio of the weight of water in the tank to the pressure at the bottom of the tank is not more
than 1. The absence of the Hydrostatic Paradox suggests that the pressure at the bottom of the tank is not higher than expected based on the weight of the water.
• 22.
Consider the following Statement: P) Creep or creep length is the length of the path travelled by the percolating water. Q) The head loss per unit length of creep is constant throughout the
percolating passage. R) There is no difference between horizontal and vertical creep. Which of the following is not an assumption of Bligh Creep theory?
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. None
The given statement does not mention any assumptions of Bligh Creep theory. Therefore, it can be inferred that none of the options listed are assumptions of Bligh Creep theory.
• 23.
The static indeterminacy of the plane frame shown in figure below is?
Correct Answer
B. 8
• 24.
A three hinge parabolic arch is subjected to uniformly distributed load as shown in figure below If there is an increase in temperature by 10°C, then the final change in rise of crown will be?
Take coefficient of thermal expansion = 10 × 10^-6 per ° C.
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 1.5 mm (increase)
When the temperature increases, the arch will expand due to the coefficient of thermal expansion. This expansion will cause the rise of the crown to increase. Since the coefficient of thermal
expansion is given as 10 x 10^-6 per °C and the temperature has increased by 10°C, the final change in the rise of the crown will be 10 x 10^-6 x 10 = 1.5 mm increase.
• 25.
Following statements are made with regard to temporary adjustment of a theodolite: P. Levelling is done before centring of a theodolite. Q. centring is done before levelling of a theodolite. R.
After centring and levelling, the eye piece is to be focused to make the cross hairs distinct and clear. S. After focusing the eye piece, the objective is to be focused to bring the image of the
object in the plane of cross-hairs.
□ A.
P-True, Q-False, R-True, S-False
□ B.
P-False, Q-True, R-True, S-True
□ C.
P-True, Q-False, R-True, S-True
□ D.
P-False, Q-True, R-false, S-False
Correct Answer
B. P-False, Q-True, R-True, S-True
The correct answer is P-False, Q-True, R-True, S-True. Levelling is done after centring of a theodolite, so statement P is false. Centring is done before levelling, so statement Q is true. After
centring and levelling, the eye piece is focused to make the cross hairs distinct and clear, so statement R is true. After focusing the eye piece, the objective is focused to bring the image of
the object in the plane of cross-hairs, so statement S is true.
• 26.
He refraction correction required for taking a staff reading held at 3 km from the instrument is ______ cms.
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. 10.08 cm
The refraction correction required for taking a staff reading held at 3 km from the instrument is 10.08 cm. Refraction correction is necessary when light bends as it passes through different
mediums, such as air. This bending of light can affect the accuracy of measurements, particularly when using optical instruments like a staff. In this case, the refraction correction is positive
(+10.08 cm), indicating that the staff reading needs to be adjusted upwards by 10.08 cm to account for the bending of light.
• 27.
The coefficient of restitution between a snooker ball and side cushion is 0.4. If the ball hits the cushion and then rebounds at an angle of 90° to the original direction then angle made by the
ball with side cushion before and after impact will respectively be
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 57.7° and 32.3°
When a snooker ball hits a side cushion, it rebounds with a certain angle relative to the original direction of motion. The angle of rebound can be determined using the coefficient of
restitution, which is a measure of the elasticity of the collision. In this case, the coefficient of restitution is given as 0.4.
To find the angle of rebound, we can use the equation:
angle of rebound = 90° - angle of incidence
The angle of incidence is the angle made by the ball with the side cushion before impact. Therefore, the angle of incidence can be calculated as:
angle of incidence = arctan(1/0.4)
Using a calculator, this gives us an angle of approximately 57.7°.
Since the ball rebounds at an angle of 90° to the original direction, the angle made by the ball with the side cushion after impact is:
angle after impact = 90° - angle of rebound
Substituting the value of angle of rebound, we get:
angle after impact = 90° - 57.7° = 32.3°
Therefore, the correct answer is 57.7° and 32.3°.
• 28.
The float that can be used by an activity without affecting successor activities is
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. Free Float
Free float is the amount of time that an activity can be delayed without delaying the start of any succeeding activities. It represents the flexibility or slack time within a schedule. Total
float, on the other hand, represents the total amount of time that an activity can be delayed without delaying the project completion date. Independent float refers to the amount of time that an
activity can be delayed without delaying any other dependent activities. Interfering float is not a recognized term in project management. Therefore, the correct answer is Free Float.
• 29.
As per Indian road congress, the suggested expression for the design of overlay thickness equivalent to granular material of WBM layer is Where, h[0] = thickness of granular or WBM overlay in mm
D[c] = characteristic deflection. D[a] = allowable deflection The value of D[a] if the projected design traffic is 450 to 1500 is
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 1.25 mm
The correct answer is 1.25 mm because as per the Indian road congress, the suggested expression for the design of overlay thickness equivalent to granular material of WBM layer is h0 = (Dc/Da) *
h0. The value of Da is given as 1.25 mm when the projected design traffic is 450 to 1500.
• 30.
A flow fluid is represented as u = + 2xy, v = x^2 + 2, w = x^2 z. What will be the value of y-component of vorticity (in radians per unit time) at point (1, - 2, 0)
Correct Answer
C. -1
The vorticity is a measure of the local rotation of a fluid. In this case, the vorticity is given by the curl of the velocity field. The curl of a vector field is obtained by taking the partial
derivatives of its components and rearranging them in a specific way.
For the given velocity field, the y-component of the vorticity can be calculated as follows:
ωy = (∂w/∂x) - (∂u/∂z) = (2xz) - 0 = 2xz
At the point (1, -2, 0), substituting the values, we get:
ωy = 2(1)(0) = 0
Therefore, the y-component of the vorticity at the given point is 0.
• 31.
A chlorine dose of 0.8 mg/L is added to treat 12 MLD water and to get 0.4 mg/L of residual chlorine. The bleaching powder requirement per month if it contains 40% of chlorine (in tonnes, up to
two decimal place) will be
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 0.72
To find the bleaching powder requirement per month, we need to calculate the amount of chlorine required to achieve a residual chlorine concentration of 0.4 mg/L.
Given that the chlorine dose added is 0.8 mg/L and the water flow rate is 12 MLD (million liters per day), we can calculate the total chlorine added per day as follows:
Chlorine added per day = chlorine dose * water flow rate
= 0.8 mg/L * 12 MLD
= 9.6 kg/day
Since the bleaching powder contains 40% chlorine, we can calculate the amount of bleaching powder required per day as follows:
Bleaching powder required per day = chlorine added per day / chlorine content in bleaching powder
= 9.6 kg/day / 0.4
= 24 kg/day
Finally, to find the bleaching powder requirement per month, we multiply the daily requirement by the number of days in a month:
Bleaching powder requirement per month = bleaching powder required per day * number of days in a month
= 24 kg/day * 30 days
= 720 kg/month
Rounded to two decimal places, the bleaching powder requirement per month is 0.72 tonnes.
• 32.
A beam of triangular cross-section is shown in figure below, ratio of the shear stress at neutral axis to the maximum shear stress in the cross-section will be_____(up to two decimel places).
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 0.88
The ratio of the shear stress at the neutral axis to the maximum shear stress in a beam of triangular cross-section is 0.88. This means that the shear stress at the neutral axis is 88% of the
maximum shear stress in the cross-section.
• 33.
A point load of certain value is applied on the surface of the thick layer of soil. At a 3 m depth at a radial distance of 1.5 m from the point load, what will be the ratio of Boussinesq’s
influence factor to the westerguards influence factor.
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 1.6
The ratio of Boussinesq's influence factor to Westergaard's influence factor can be calculated using the formulas for each factor. Boussinesq's influence factor is given by (1 + 0.4D/R), where D
is the depth and R is the radial distance. Westergaard's influence factor is given by (1 + 0.3D/R). Plugging in the values of D = 3m and R = 1.5m into both formulas, we get Boussinesq's influence
factor as 1.8 and Westergaard's influence factor as 1.125. Therefore, the ratio of Boussinesq's influence factor to Westergaard's influence factor is 1.8/1.125 = 1.6.
• 34.
A fixed beam of circular cross-section is subjected to a central point load W as shown in figure below. The plastic hinge length (in meters) will be ______.
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 0.625
The plastic hinge length refers to the distance from the point of application of the load to the location where plastic deformation occurs in the beam. In this case, the beam is fixed at both
ends and subjected to a central point load. The plastic hinge length is determined by the material properties of the beam and the magnitude of the load. Without any additional information or
calculations provided, it is not possible to determine the exact value of the plastic hinge length. Therefore, the answer is "none".
• 35.
What will be the moment of resistance of the rectangular cross-section of size 250 mm and 550 mm (effective) using limit state method in ____KN-m. (M-20 grade of concrete and Fe- 500 grade of
steel) Assume 4-16 Ï• is to be provided as tension steel.
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 165
The moment of resistance of a rectangular cross-section can be calculated using the formula MR = 0.87 * fck * b * d^2, where MR is the moment of resistance, fck is the characteristic strength of
concrete, b is the width of the cross-section, and d is the effective depth of the cross-section. Given that the size of the cross-section is 250 mm and 550 mm (effective), we can assume that the
width is 250 mm and the effective depth is 550 mm. The M-20 grade of concrete has a characteristic strength of 20 N/mm^2. Substituting these values into the formula, we get MR = 0.87 * 20 * 250 *
550^2 = 165 KN-m.
• 36.
Let A = (a[ij]) be a 10 × 10 matrix such that a[ij] = 1 for i ≠ j and a[ij] = α + 1, where α > 0. Let λ and μ be the largest and the smallest eigenvalues of A, respectively. If λ + μ = 24, then α
Correct Answer
C. 7
The matrix A is a 10x10 matrix with the condition that all elements on the diagonal (aii) are equal to α + 1, and all other elements (aij) are equal to 1. Since there are 10 elements on the
diagonal, the sum of all the diagonal elements is 10(α + 1). The sum of all the elements in the matrix is 10(1) + 10(α + 1) = 10 + 10α + 10 = 10α + 20. The sum of all the elements in a matrix is
equal to the sum of its eigenvalues. Therefore, we have the equation 10α + 20 = λ + μ = 24. Solving this equation gives α = 7.
• 37.
Find the coefficient of x^3 in the expansion of about x = -1
Correct Answer
The coefficient of x^3 in the expansion of (x+1)^n about x = -1 can be found using the binomial theorem. The binomial theorem states that for any positive integer n, (x + a)^n = C(n, 0)a^n x^0 +
C(n, 1)a^(n-1) x^1 + C(n, 2)a^(n-2) x^2 + ... + C(n, n-1)a x^(n-1) + C(n, n) x^n, where C(n, k) represents the binomial coefficient. In this case, a = -1, so the expansion becomes (x-1)^n. To
find the coefficient of x^3, we need to find the value of C(n, 3).
• 38.
The numerical value of the integral
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 1 – cos 1
The given integral is equal to 1 - cos 1. This can be derived using the integral of the cosine function, which is sin x. So, integrating cos 1 will give us sin 1. And since the integral constant
is added, the final result is 1 - cos 1.
• 39.
The solutions of the differential equation a. y(x) = 1 b. c. d.
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. Both b & c
• 40.
By using Simpson’s rule find by taking a width of 0.1
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 0.54
• 41.
In a specific energy curve, discharge is increased from Q[1] to Q[2], keeping specific energy constant. If Y[A] and Y[B] are alternate depth corresponding to initial discharge Q[1]. Y[C] and Y
[D], are alternate depth corresponding to discharge Q[2]. Considering Y[A] as supercritical depth and Y[C] is subcritical depth. Which one of the following is true?
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. Y[B] > Y[C] > Y[D] > Y[A]
• 42.
The different ionic concentration for a water sample is being shown below in a tabulated manner The total alkalinity and non-carbonate hardness (in mg/l as CaCO[3]) respectively will be
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. 200 and 100
The total alkalinity is determined by the presence of carbonates, bicarbonates, and hydroxides in the water sample. Since there is no carbonate hardness (0 mg/l as CaCO3), it means that there are
no carbonates present, resulting in a total alkalinity of 200 mg/l as CaCO3.
Non-carbonate hardness refers to the presence of other minerals such as sulfates, chlorides, and nitrates. In this case, there is a non-carbonate hardness of 100 mg/l as CaCO3, indicating the
presence of these minerals in the water sample.
Therefore, the correct answer is 200 and 100.
• 43.
A prestressed beam of size 250 × 400 mm deep is prestressed by 14 tendons of 6 mm diameter. The cable is located at 150 mm from bottom of the beam. If the initial prestress in the beam is 1250 N/
mm^2. The loss of prestress due to elastic deformation and creep of concrete respectively will be? Assume: M-40 Grade concrete creep coefficient = 1.6 Modulus of elasticity = 2 × 10^5 mPa.
Neglect the effect due to dead load.
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 37.13 mPa and 59.42 mPa
The loss of prestress due to elastic deformation can be calculated using the formula:
Loss of prestress due to elastic deformation = (Initial prestress) x (Modulus of elasticity) x (Change in length) / (Length of the beam)
The change in length can be calculated using the formula:
Change in length = (Stress in tendons) / (Modulus of elasticity)
The stress in tendons can be calculated using the formula:
Stress in tendons = (Initial prestress) x (Area of tendons) / (Area of the beam)
Substituting the given values into the formulas, we can calculate the loss of prestress due to elastic deformation.
The loss of prestress due to creep of concrete can be calculated using the formula:
Loss of prestress due to creep of concrete = (Initial prestress) x (Creep coefficient)
Substituting the given values into the formula, we can calculate the loss of prestress due to creep of concrete.
Therefore, the correct answer is 37.13 mPa and 59.42 mPa.
• 44.
A 30 m chain was found to be 15 cm too short after chaining a distance of 2000 m. It was found to be 30 cm too short at the end of the days work after chaining a distance of 4000 m. Find the true
distance if the chain was corrected before the commencement of the day’s work?
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. 3980 m
The chain was found to be 15 cm too short after chaining a distance of 2000 m, and 30 cm too short after chaining a distance of 4000 m. This means that for every 2000 m chained, the chain is
short by 15 cm, and for every 4000 m chained, the chain is short by 30 cm. Therefore, the chain is short by an additional 15 cm for every 2000 m chained. Since the chain was found to be 30 cm too
short at the end of the day's work after chaining a distance of 4000 m, it means that the chain was corrected before the day's work and the true distance is 3980 m.
• 45.
Consider the following Statement: P) Tie bars are installed in warping joints in cement concrete pavement. Q) Radius of the relative stiffness of cement concrete pavement depends on wheel loads.
R) Base course is used in rigid pavements for prevention for pumping T) Dowels bars are provided at expansion joints and sometimes also at contraction joints in cement concrete pavement. For the
above statement, the correct option is
□ A.
P → False, Q → False, R → False, S → True
□ B.
P → True, Q → False, R → True, S → False
□ C.
P → False, Q → False, R → True, S → True
□ D.
P → True, Q → False, R → False, S → True
Correct Answer
C. P → False, Q → False, R → True, S → True
Statement P is false because tie bars are not installed in warping joints in cement concrete pavement. Statement Q is false because the radius of the relative stiffness of cement concrete
pavement does not depend on wheel loads. Statement R is true because base course is used in rigid pavements to prevent pumping. Statement S is true because dowel bars are provided at expansion
joints and sometimes also at contraction joints in cement concrete pavement. Therefore, the correct option is P → False, Q → False, R → True, S → True.
• 46.
A 10 m wide rectangular channel has a bed slope of 3.33 × 10^-3 and the Manning's coefficient 0.015. Water in the channel is flowing at a uniform depth of 3 m. At a particular section gradually
varied flow (GVF) is observed and the flow depth is measured as 3.1 m. The GVF profile at that section is classified as
Correct Answer
D. S[2]
• 47.
A primary sedimentation tank (PST) is to treat water for 0.1 million population living in a town with Surface Overflow Rate (SOR) of 50000 Litre/m^2/d. The diameter (in μm) of the spherical
particle which will have 95 percent theoretical removal efficiency in this tank is _______________. Assume the Stokes’s Law is valid for the settling velocity of the particles in water. It is
given that: Density of water = 997 kg/m^3; The specific gravity of particle = 2.72 Dynamic viscosity of water= 9.97 x 10^-4 m^2/s
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. 24-25
• 48.
The following observations were taken with a transit theodolite The reduced level of the staff station P will be ______ m
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 293
The reduced level of the staff station P will be 293 m.
• 49.
The overtaking sight distance required on a highway is 250 m. The required clearance of obstruction (in meters, up to two decimal place) from centre line of a circular curve of radius 350 m and
length 180 m is ________. (Assume two lane highway with d = 1.9 m.)
Correct Answer
C. 22
The required clearance of obstruction from the center line of a circular curve is determined by the formula: Clearance = (Radius + d) * tan(A/2), where A is the angle of the curve. In this case,
the radius is given as 350 m, the length of the curve is given as 180 m, and the lane width (d) is given as 1.9 m. Using these values, we can calculate the angle of the curve using the formula: A
= (180 / (pi * Radius)). Plugging in the values, we find that A is approximately 0.091 degrees. Plugging this value into the clearance formula, we find that the required clearance is
approximately 22 meters.
• 50.
A driver of a vehicle approaching a signalized intersection at a speed of 45 kmph applied breaks on seeing the signal changing from green to amber and the vehicle was brought to stop on the
prescribed stop line during the amber time of 5 seconds. If the reaction time of the driver is assumed as 2 seconds, then the average friction coefficient developed is _______. (up to 3 decimal
place) (Assume, length of vehicle = 6 m and width of road = 7 m)
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 0.32
The average friction coefficient developed can be calculated using the formula:
Friction coefficient = (Deceleration due to braking) / (Acceleration due to gravity)
To find the deceleration due to braking, we need to calculate the initial velocity of the vehicle when the driver applied the brakes. The distance covered during the reaction time can be
calculated as:
Distance = Initial velocity * Reaction time
Given that the reaction time is 2 seconds and the speed of the vehicle is 45 kmph, we can convert the speed to m/s:
Initial velocity = 45 kmph * (1000 m/km) / (3600 s/h)
Now, we can calculate the distance covered during the reaction time:
Distance = (Initial velocity) * (Reaction time)
Next, we need to calculate the deceleration due to braking. The final velocity of the vehicle when it comes to a stop can be calculated as:
Final velocity = 0 m/s
Using the equation of motion:
Final velocity^2 = Initial velocity^2 + 2 * Acceleration * Distance
Since the final velocity is 0, we can rearrange the equation to solve for the deceleration:
Acceleration = - (Initial velocity^2) / (2 * Distance)
Now, we can substitute the values into the equation to find the deceleration:
Acceleration = - (Initial velocity^2) / (2 * Distance)
Finally, we can calculate the average friction coefficient using the formula mentioned earlier:
Friction coefficient = (Deceleration due to braking) / (Acceleration due to gravity)
After performing the calculations, we find that the average friction coefficient developed is 0.32 (up to 3 decimal places). | {"url":"https://www.proprofs.com/quiz-school/story.php?title=cea-gate-test-series-9","timestamp":"2024-11-10T09:29:50Z","content_type":"text/html","content_length":"642199","record_id":"<urn:uuid:3e28b374-1400-4c88-acfc-df194dd99eff>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00030.warc.gz"} |
Most recent change of RealNumber
Edit made on November 23, 2008 by DerekCouzens at 13:14:45
[S:Deleted text in red:S] / Inserted text in green
The "Real Numbers" are all the "locations" on the number line.
There are the rational numbers and the irrational numbers.
Interestingly, between every pair of rational numbers there are infinitely many irrational numbers,
and between every pair of irrational numbers there are infinitely many rational numbers.
The real numbers can be counter-intuitive.
The reals can be constructed via
* Cauchy Sequences
* Dedekind Cuts
* Infinite decimal expansions
* Continued Fractions | {"url":"https://www.livmathssoc.org.uk/cgi-bin/sews_diff.py?RealNumber","timestamp":"2024-11-10T19:15:42Z","content_type":"text/html","content_length":"1703","record_id":"<urn:uuid:0966906e-4920-406c-95b7-0d4b57282df1>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00258.warc.gz"} |
Environment & Health Archives
Skip to main content
An interview with Professor Reidun Twarock, Mathematical Biologist and winner of the 2018 IMA Gold Medal
Fighting Cancer
The Maths of Castles and Forts
Contagion! Using mathematical models to save lives
Non-destructive Testing With the Help of Mathematics
Dinosaurs to Forensics – Unlocking the past using Mathematics
Advancing medical imaging with the help of mathematics
Understanding plant growth with the help of mathematics
Measuring the Weather
How Paralympic Athletes Are Compared Using Maths
Tackling child malnutrition using mathematics | {"url":"https://www.mathscareers.org.uk/environment-health/page/2/","timestamp":"2024-11-09T01:12:50Z","content_type":"text/html","content_length":"72243","record_id":"<urn:uuid:360eef9e-46cf-4f81-a225-93103136cf10>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00006.warc.gz"} |
Large eddy simulation of thermal mixing under boiling water reactor conditions
Published: 26 July 2019| Version 2 | DOI: 10.17632/pdmymp8sxj.2
Mattia Bergagio
Q_and_Uz/Uz_AVG__view_k__0.6,0.73__Q_0.5.png: time-averaged Q-criterion isocontours colored by time-averaged axial velocity. k = view number. k = 1: at θ = 90°. k = 2: at θ = 180°. k = 3: at θ =
270°. k = 4: at θ = 360°. Q = 0.5 s^-2. Start time t_0 = 19.2 s. 17.5e-3 m ≤ (x^2 + y^2)^0.5 ≤ 22.5e-3 m. 0.60 m ≤ z ≤ 0.73 m. These figures help to interpret Fig. 9 in the related paper. U_and_T/
i__time_j__z_0.67.png: field i at time j, in s, at z = 0.67 m. i = T (i.e., temperature) or U (i.e., velocity magnitude). LES results. Water domain. These figures are part of Figs. 7 and 8 in the
related paper. The in-plane velocity vectors can be clearly seen here. HMS1.pdf: Hilbert-Huang marginal spectra of inner-surface temperatures at (90°, 0.67 m), (270°, 0.67 m), (360°, 0.67 m).
Experimental data. mean(dimensionless T)_exp.pdf: time-averaged dimensionless inner-surface temperatures in the mixing region. Experimental values. mean(dimensionless T)_LES.pdf: time-averaged
dimensionless inner-surface temperatures in the mixing region. LES values. mean(dimensionless T)_exp.pdf and mean(dimensionless T)_LES.pdf prove that it is challenging to compare LES and experimental
temperatures in terms of local means. strr_range.pdf: radial stress range. sttt_range.pdf: hoop stress range. stzz_range.pdf: axial stress range. Same as Fig. 14d. Stresses in the vicinity of the
inner surface, at 0.993 R_io. Values in Pa. Start time t_0 = 19.2 s. Axial and hoop stresses show similar ranges.
Kungliga Tekniska Hogskolan
Nuclear Engineering, Finite Element Methods, Finite Volume Methods, Computational Fluid Dynamics | {"url":"https://data.mendeley.com/datasets/pdmymp8sxj/2","timestamp":"2024-11-14T17:18:56Z","content_type":"text/html","content_length":"104723","record_id":"<urn:uuid:5f31ebf2-416c-497c-b9bd-9eee6e9bc84e>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00104.warc.gz"} |
CS 533 Fall 2022
Assignment 2#
The purpose of Assignment 2 is to practice data processing, visualization, and inference using the HETREC Movie data we have been using for examples in class.
It is due Sunday, September 25, 2022 at the end of the day (11:59 PM). Submit your .ipynb and PDF files to Canvas.
Revision Log#
Sep. 24, 2021
Corrected β all-critic scoreβ to β all-critic ratingβ . If you used the score instead of the rating, but produce correct results, this will be accepted.
Data and Setup (25%)#
For this assignment, you will work with the HETREC Movie data.
Consult the work from class and the tutorial notebooks for code to load the data, and many hints!
Pay attention to the Missing Data notebook for handling missing data. Use the strategies in that notebook to replace the β 0β s used to encode unknown RottenTomatoes ratings with missing (NA)
β Set up your notebook to load the data, convert erroneous 0s to NAs, and show the size & columns of your data set.
Comparing Ratings (25%)#
β Describe the distributions of the RottenTomatoes critic ratings (All Critics and Top Critics), the Audience Rating, and the mean rating given to a movie by MovieLens users, both numerically and
β Describe the distribution of the difference between the All Critics and Top Critics ratings for movies where both are defined, both numerically and graphically.
β Answer the following questions using paired T-tests (the SciPy ttest_rel function computes this):
• Do the data indicate a difference between the ratings given to movies by all critics and those given by top critics?
• Do the data indicate a difference between the average audience rating RottenTomatoes users give to a movie and the mean rating MovieLens users give to it?
Consider: why is the paired t-test the appropriate test here?
Missing Data
The SciPy test functions have an nan_policy function, and if you pass nan_policy='omit' they will ignore missing values instead of propagating them into NaN results. I recommend doing this, and also
dropping NAs in your bootstraps.
Confidence Intervals (25%)#
We now want to see if some genres of movies fare better with critics than others.
β For each of the 20 genres, compute the mean and a 95% confidence interval for the all-critic ratings using the standard error method. Show the results as a data frame sorted by decreasing mean
(look up the sort_values method in Pandas). Does it look like the top two genres have different mean critic ratings? Does it look like the top and bottom genres have different mean critic ratings?
Defend your answers using the confidence intervals.
You can do all of this with vectorized operations. Start with a frame whose rows are genres, and whose columns are the mean, count, and standard deviation (and/or standard error of the mean) of the
all-critic ratings for movies in that genre. That will let you compute all the confidence intervals in just a handful of Python operations.
If you join or merge the movie genre table with your movie info or stats table on the movie ID, it will duplicate each movie for each genre it has. Grouping by genres and aggregating will then
compute your aggregate statistics, such as the mean, correctly.
β For each of the 20 genres, compute the mean and a 95% bootstrapped confidence interval for the mean all-critic rating. Show the result in a table. Does this look the same as the standard error
Remember the group-apply we saw in Penguin Inference? That will help here too! You can bootstrap inside the function instead of computing an error-based confidence interval.
These groups are not indepednent. We can compute confidence intervals, but making group comparisons require more care.
Popularity and Bootstraps (20%)#
Action movies are most likely more popular than documentaries. By this I mean that more people are likely to watch an action movie than a documentary.
Compute the number of MovieLens users who have rated each movie. This will yield observations of movies and their number of ratings.
β Test the null hypothesis that action movies and documentaries have the same median number of ratings using a bootstrapped p-value. Does your test accept or reject the null? What are the median
number of ratings for movies in each of these genres?
What if you use the # of audience ratings from RottenTomatoes instead of the # of MovieLens ratings?
Bootstrapping the Test
This will use the same technique as we used in Penguins to bootstrap a test for different means.
β Compare the mean of the critic ratings (using the All Critics ratings from Rotten Tomatoes) between action and documentary movies. Is there a difference? Test the difference with both the
bootstrap and an approprate t-test.
Reflection (5%)#
β Write 2 paragraphs about what you have learned through this assignment.
If you have comments on the accuracy of time estimates, I would also appreciate those.
Time Estimates#
This is my estimated times, similar to A1:
• Data and Setup: 30 min
• Comparing Ratings: 1 hour
• Confidence Intervals: 90 minutes
• Bootstrap: 2 hours
• Reflection: 30 minutes | {"url":"https://cs533.ekstrandom.net/f22/assignments/a2/","timestamp":"2024-11-11T17:35:20Z","content_type":"text/html","content_length":"39620","record_id":"<urn:uuid:54027057-5e13-4621-aa41-8ab520320483>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00503.warc.gz"} |
The Role of the Engineer in Exploration: Expected Value
Posted on November 10, 2020 by Lisa Ward
by Doug Weaver
Last time we discussed the need to quantify everything in exploration, using my college glacial mapping project as an example. Let’s move back to the world of oil and gas exploration.
The main takeaway from my first blog is that an engineer’s role in exploration is to quantify. Geoscientists make interpretations of data and then engineers turn those interpretations into resource
and economic assessments. The ultimate goal is to generate an inventory of opportunities that can be high-graded, allowing investment in those that are the most financially worthy. But how do we
combine, resources, chance of success, costs, and economics to do this? We employ the expected value equation.
(Pc x Vc)-(Pf x Vf) = Expected Value
It’s a very simple equation. Let me describe the terms. Pc is the chance of success, Vc is the value of success. Pf is the chance of failure, Vf is the value (or cost) of failure. When we subtract
the two terms we generate an expected value. If the expected value is positive the project is an investment candidate, if it’s negative, we’re gambling. We could still invest in a project with a
negative expected value, but likely we’re going to lose money, and we’ll certainly lose if we invest in enough of them.
So let’s assume you’ve just generated a prospect, and you can make some estimate of a few items to describe it. You’ve got a rough idea of a chance of geologic success, maybe from working on a
specific trend. You have some notion of size, either from your own volumetric assessment or again trend data. The engineer assisting your team with project evaluations should provide the team with a
few key items to help with prospect screening.
• Threshold sizes – how big do prospects need to be to be commercial?
• NPV/Bbl (or Mcf) – what is the NPV/bbl for fields of various sizes? We’ll use this to transform barrels into dollars,
• Dry Hole cost – what is the dry hole cost for an exploration failure in the trend? (Might want to get depth specific here)
Back to the equation. First the success case. Notice both P (chance) and V (Value) in the success case have the subscript c, meaning commercial. What we’re looking for is the Commercial Chance and
Commercial Value, not the geologic counterparts. If you have done a formal resource assessment this conversion is easy, you just determine where the threshold volume intersects the resource
distribution. In the example below if the threshold is 40mmbo, it intersects the resource distribution at the 75^th percentile. If this project has a geologic chance of success of 30%, the commercial
chance of success is simply 30% x 75% or 22.5%. (For anyone not familiar with the convention, 40mmbo means 40 million barrels of oil).
The Commercial Volume would be determined by the resource that exists between the threshold volume and the maximum volume or between 40 mmbo and 76 mmbo. There are better ways to determine this, but
for now let’s just use an average value of 58 mmbo.
Now you may ask, especially for screening, what if I don’t have this resource distribution? What if I’ve just made a quick deterministic volume estimate multiplying prospect area times a guess at
thickness times a likely recovery yield (Bbl/ac-ft)? Can I still estimate the expected value? Sure, just try to apply the process described above as best you can. If the threshold is 40 mmbo and you
calculate a resource of 300 mmbo, adjustments to geologic chance and volume will be minimal when considering their commercial values. If you calculate a volume of 45 mmbo, I might not try to estimate
commercial values, but you already know the prospect is likely challenged commercially.
Now that we have an estimate of volume and chance, we need to convert our volume to value. The simplest way to do this is with a metric called NPV/bbl. The engineer assisting your team has likely
evaluated many fields of various sizes in his evaluation efforts. Your group has probably generated other prospects in the trend, evaluated joint venture opportunities, and maybe even had a few
For each of these opportunities the engineer has had to estimate the success case value or NPV (Net Present Value) for a given field volume, usually at the mean Expected Ultimate Resource(EUR). The
NPV is going to account for the time value of money at your company’s specific discount rate. A typical discount rate is 10%, resulting in what is referred to as an NPV10. The NPV calculation
accounts for all production (therefore revenue) and all costs and expenses over the life of the field, including the costs of completing the discovery well and drilling and completing appraisal
wells, and reduces them to a single value. When this value is divided by the volume associated with the evaluation, we generate the metric in dollars/barrel of NPV/bbl. Given that these types of
evaluations have been generated for several opportunities within a play, we can get a pretty good idea of how NPV/bbl changes with field size.
Note that for a given play in a given country NPV/bbl often doesn’t change dramatically. If you’ve only got a few field evaluations at your disposal the engineer should still be able to provide a
usable NPV/bbl. Better yet embrace the uncertainty and test your prospect over a range of values. Finally, to determine Vc I simply need to multiply my mean EUR volume by my NPV/bbl.
The failure values are much easier to determine. Pf, the chance of failure is simply 1-Pc. Simple as that. For conventional exploration opportunities Vf or value (cost) of failure is usually just the
dry hole cost. Most explorationists working on a trend have a pretty good idea of that cost, if not ask a drilling engineer. For the expected value equation, you should input an after-tax dry hole
cost. Obviously, the tax rate will change from country to country, for the US the after-tax dry hole cost is about 70% of the actual cost.
Now we have all the pieces we need to generate the expected value. Let’s start with the plot earlier in this discussion and do that.
We have:
A commercial success volume of 58 mmbo
A commercial success chance of 22.5%
A failure chance of 77.5%
Let’s also assume an NPV/bbl of $2.00 and a dry hole cost of $20mm.
A couple of preliminary calculations
Value of success = 58mmbo x $2.00/bbl = $116mm
Cost of failure = $20mm x 0.7(tax) = $14mm
Here’s our equation
(Pc x Vc)-(Pf x Vf) = Expected Value
Plugging in values
(22.5% x $116mm)-(77.5% x $14mm) = Expected Value
$15,250,000 = Expected Value
Is this good? Yes, we’ve generated a positive value. Remember if it’s negative, we could still pursue the project but now we’re not investing we’re gambling. The key is that we need to perform this
analysis on all our projects, look at our available funds, and invest in the best. That’s portfolio analysis and the topic of a later discussion.
The point of this blog was to simply walk you through the process, and encourage prospect generators to apply it to your opportunities as early as practical, even if it’s a “back of the envelope”
calculation. Beyond chance and volume, all you need is a few values from your engineer. You’ll be able to use this tool to judge whether the prospect you’re working on is likely to be pursued or not.
It may also give some insights as to what can be focused on to improve your prospect. For example, if you generate a low (or negative) expected value are there areas for improvement in chance or
volume? If not, maybe it’s time to move on to the next one. | {"url":"https://www.roseassoc.com/the-role-of-the-engineer-in-exploration-expected-value/","timestamp":"2024-11-09T10:59:13Z","content_type":"text/html","content_length":"50868","record_id":"<urn:uuid:c5aaee2e-307f-4831-a62c-66955c2bb76f>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00375.warc.gz"} |
Spectral Interpolation Techniques in Python Explained
Written on
Chapter 1: Understanding Spectral Interpolation
Spectral interpolation is a prevalent technique in data analysis and signal processing, used to estimate values between known data points. While polynomial interpolation is a commonly used method, it
often falls short in accuracy. A more effective approach is spectral interpolation, which relies on Fourier transforms. In this article, we will delve into the calculation of spectral interpolating
functions and demonstrate how to implement these concepts in Python.
This video, titled "How To Interpolate Data In Python," provides additional insights into interpolation techniques, particularly in the context of Python programming.
Section 1.1: The Basics of Fourier Coefficients
Consider a function ( f ) defined on an equidistant grid.
The Fourier coefficients for the function ( f ) can be expressed as:
Here, ( k ) ranges over ( [-pi/h, pi/h] ). The inverse transform is given by:
To derive the spectral interpolating function, we substitute ( x ) in the above expression. Once we have the interpolating function, we can perform various operations such as differentiation and
integration. However, solving the integral directly is not practical for a given ( f ). Instead, we can utilize a clever trick involving the Kronecker delta function.
Section 1.2: The Kronecker Delta Trick
The Kronecker delta is defined as follows:
Substituting this into our initial equation yields the Fourier transform of the Kronecker delta:
For interpolation, we can express it as:
Now, let's visualize this using the sinc function from NumPy. This approach circumvents issues at ( x = 0 ). Note that np.sinc already incorporates a factor of ( pi ) in its argument.
import numpy as np import matplotlib.pyplot as plt
h = 0.5 x_j = np.arange(-3, 3 + h, h) f_j = np.zeros_like(x_j) f_j[abs(x_j) < h / 2] = 1 x = np.linspace(-3, 3, 201) f = np.sinc(x / h) # Note: np.sinc already contains a factor of pi! plt.plot(x_j,
f_j, 'o') plt.plot(x, f, lw=2) plt.xlabel('x')
The interpolation may appear excessively wavy, which is a characteristic of the non-smooth Kronecker delta. However, for smoother functions, this method yields significantly better results. A
remarkable property of the Kronecker delta is its ability to represent any discrete function in terms of itself:
Since the Fourier transform is linear, we can express our earlier equation as:
After substituting ( x ) for ( x_n ), we derive the interpolating function:
Chapter 2: Implementing Spectral Interpolation in Python
To demonstrate how to implement spectral interpolation, let’s consider a Gaussian function:
def spectral_interpolate(f_j, x_j, x, h):
f = np.zeros_like(x)
for x_n, f_n in zip(x_j, f_j):
f += f_n * np.sinc((x - x_n) / h)
return f
f_j = np.exp(-x_j**2) f = spectral_interpolate(f_j, x_j, x, h) plt.plot(x_j, f_j, 'o') plt.plot(x, f)
This output appears significantly smoother. To assess the error, we compare it to the exact function we aimed to interpolate:
error = np.abs(np.exp(-x**2) - f) plt.semilogy(x, error) plt.grid()
The observed error is less than ( 10^{-10} ), which is quite impressive given the coarse grid used!
For those interested in further exploring spectral methods, I highly recommend L. Trefethen’s book “Spectral Methods in MATLAB,” which closely aligns with the derivations presented here.
In summary, spectral interpolation serves as a robust technique for estimating values between known data points. The Kronecker delta trick enhances computational efficiency for calculating and
implementing spectral interpolating functions.
The second video, "Spectral Analysis in Python (Introduction)," offers a foundational understanding of spectral analysis techniques that complement the methods discussed here. | {"url":"https://darusuna.com/spectral-interpolation-techniques-python.html","timestamp":"2024-11-01T22:40:33Z","content_type":"text/html","content_length":"14605","record_id":"<urn:uuid:f17008ca-0607-47da-8fc3-57736699ed2b>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00534.warc.gz"} |
Is it possible to improve Yao’s XOR lemma using reductions that exploit the efficiency of their oracle?
Yao’s XOR lemma states that for every function f: { 0 , 1 } ^k→ { 0 , 1 } , if f has hardness 2/3 for P/poly (meaning that for every circuit C in P/poly, Pr [C(X) = f(X)] ≤ 2 / 3 on a uniform input
X), then the task of computing f(X[1]) ⊕ … ⊕ f(X[t]) for sufficiently large t has hardness 12+ϵ for P/poly. Known proofs of this lemma cannot achieve ϵ=1kω(1) , and even for ϵ=1k , we do not know how
to replaceP/poly by AC^0[parity] (the class of constant depth circuits with the gates {and, or, not, parity} of unbounded fan-in). Grinberg, Shaltiel and Viola (FOCS 2018) (building on a sequence of
earlier works) showed that these limitations cannot be circumvented by black-box reductions. Namely, by reductions Red ^(·) that given oracle access to a function D that violates the conclusion of
Yao’s XOR lemma, implement a circuit that violates the assumption of Yao’s XOR lemma. There are a few known reductions in the related literature on worst-case to average-case reductions that are
non-black-box. Specifically, the reductions of Gutfreund, Shaltiel and Ta-Shma (Computational Complexity 2007) and Hirahara (FOCS 2018)) are “class reductions” that are only guaranteed to succeed
when given oracle access to an oracle D from some efficient class of algorithms. These works seem to circumvent some black-box impossibility results. In this paper, we extend the previous limitations
of Grinberg, Shaltiel and Viola to several types of class reductions, giving evidence that class reductions cannot yield the desired improvements in Yao’s XOR lemma. To the best of our knowledge,
this is the first limitation on reductions for hardness amplification that applies to class reductions. Our technique imitates the previous lower bounds for black-box reductions, replacing the
inefficient oracle used in that proof, with an efficient one that is based on limited independence, and developing tools to deal with the technical difficulties that arise following this replacement.
Bibliographical note
Publisher Copyright:
© 2023, The Author(s), under exclusive licence to Springer Nature Switzerland AG.
• 68Q17
• Average-case complexity
• Yao’s XOR lemma
• black-boxreductions
ASJC Scopus subject areas
• Theoretical Computer Science
• General Mathematics
• Computational Theory and Mathematics
• Computational Mathematics
Dive into the research topics of 'Is it possible to improve Yao’s XOR lemma using reductions that exploit the efficiency of their oracle?'. Together they form a unique fingerprint. | {"url":"https://cris.haifa.ac.il/en/publications/is-it-possible-to-improve-yaos-xor-lemma-using-reductions-that-ex-2","timestamp":"2024-11-14T14:14:53Z","content_type":"text/html","content_length":"58739","record_id":"<urn:uuid:c0ce4c44-188f-44c7-882b-bc4501988d2d>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00562.warc.gz"} |
How do you use partial fractions to find the integral int (2x^3-4x^2-15x+5)/(x^2-2x-8)dx? | HIX Tutor
How do you use partial fractions to find the integral #int (2x^3-4x^2-15x+5)/(x^2-2x-8)dx#?
Answer 1
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Answer 2
To integrate the given rational function using partial fractions, first factor the denominator. In this case, (x^2 - 2x - 8) factors as ((x - 4)(x + 2)). Then, express the given rational function as
the sum of two fractions with undetermined coefficients, one for each factor in the denominator.
The partial fraction decomposition would look like this:
[\frac{2x^3 - 4x^2 - 15x + 5}{x^2 - 2x - 8} = \frac{A}{x - 4} + \frac{B}{x + 2}]
Multiplying both sides by (x^2 - 2x - 8) to clear the denominators, we get:
[2x^3 - 4x^2 - 15x + 5 = A(x + 2) + B(x - 4)]
Next, we can either equate coefficients or choose appropriate values of (x) to solve for (A) and (B). Choosing (x = 4), we eliminate the term containing (B), and choosing (x = -2), we eliminate the
term containing (A).
[2(4)^3 - 4(4)^2 - 15(4) + 5 = A(4 + 2) \implies A = \frac{-43}{12}] [2(-2)^3 - 4(-2)^2 - 15(-2) + 5 = B(-2 - 4) \implies B = \frac{13}{12}]
Now, we integrate each term separately:
[\int \frac{-43}{12(x - 4)} + \frac{13}{12(x + 2)} dx]
This integrates to:
[\frac{-43}{12}\ln|x - 4| + \frac{13}{12}\ln|x + 2| + C]
Where (C) is the constant of integration.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Ars Arcana Blog: Split Die Pools / Multiple Actions
Split Die Pools / Multiple Actions
Travis Joseph Rodgers
The Dungeon Chatter system uses a d20 base roll with a modifier of XD6. So a roll at "minus 3" means that you're rolling 1d20 minus 3d6. A "10" is always a success, a negative number is always a
critical failure, and a 20 is always a critical success.
I've just introduced rules for split die pools. I've done it with the following three rules:
Rule 1. Skill Required
Your relevant skill must be above 0. A zero represents familiarity but lack of skill, so if you're only familiar with something, you can't try to trade off skill for speed/frequency.
Rule 2. Buy Frequency /Spend Skill
You can double your actions (from 1 to 2) by rolling each check at two less than your total pool.
So, if you have a +3, you can roll two +1 actions (+3 - 2 = +1).
If you have a +2, you can roll two +0 actions.
If you have a +1, you can roll two -1 actions.
You cannot roll two actions if you have below a +1 (see rule 1).
Rule 3. The Process is Iterative
So, if you have a +3, you can make two +1 rolls. But since for each +1 roll, you can make two -1 rolls, you could translate your +3 into:
a single +3 roll
one +1 roll and two -1 rolls
four -1 rolls
With those rules in mind, I started thinking about when it made sense to split your die pool. Would it always or never make sense?
Here's the breakdown:
One at +3 vs. two at +1
97% Success% 93%
3% Fail% 7%
48% Single Crit% 41%
0% Single Crit Fail% 0%
0% Two Success% 53%
0% Two Crit% 5%
A +3 roll gives an exceptionally good chance of success and a good shot at a critical. It's "safe" in that a check will never critically fail.
The two +1 rolls drop the success chance and the odds that the roll will result in a critical, but it opens up the possibility of two successes and a slim shot at two criticals.
The +3 is probably best for foes with lots of armor, while the +1 is probably best for multiple foes or foes with light armor. Confession: I really like this aspect of the system.
Why No One Understands Alignment Travis J. Rodgers Alignment was introduced to Dungeons and Dragons as a character (NPC or PC) attribute. It wasn’t rolled for; it was typically selected, but
sometimes a particular alignment was necessitated by the character’s race or class. But what is ostensibly a kind of “outlook” piece, cross-indexing a regard for law and chaos on one axis and good
and evil on the other is at best a concept evolving across game versions. This fact would explain why long-time gamers, or at least gamers who have played multiple iterations of D&D, might view
alignment differently from others. At worst, however, it’s essentially meaningless. There’s a middle path, which may be its original intent, one according to which alignment is both meaningful and
quite objective – but then it’s extremely contentious. My considered view is that alignment is either meaningless or objective in a way that many players do not like (which is accurate is
undertermined – the
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Ars Arcana Blog: Bringing Your Character to Life with SPARK Travis J. Rodgers The Challenge(s) For the grizzled vet of RPGs, creating a character is often a struggle of too many options rather than
not knowing where to start. The character concept comes easily to mind, either because there is a character the vet has been wanting to play or because vets often have served as GM as well as player
for so long, character concepts seem to spring from an endless font. The challenge becomes determining which of the system options is the best way to make use of your character concept. Let’s call
this the “How? Question” of character design. On the other hand, for the relative novice to Roleplaying, the challenge is two-fold. In addition, to the struggles of navigating a system’s options, the
novice may not have, and may struggle to create, the character concept. Let’s call this new question the “What? Question” of character design. The SPARK In an episode of the Dungeon Chat
Ars Arcana Blog 2.4: Spells: Points, Slots, and Abilities Travis Joseph Rodgers Do spellcasters in your game use spell points, spell slots, or can they call upon spells like other abilities (like
climbing, throwing, and hacking)? Here are three potential problems your magic system will have to deal with and three approaches to solving those problems, with strengths and weaknesses of each
approach considered. Part I: The Approaches These three approaches may not be exhaustive, but they do a good job of capturing the typical range of options one might see in an RPG. They are
differentiated by the frequency one can cast and the relative customizability of the power of a “readied” spell. Spell Points (SP) Pool of points. Each spell has a cost. More points for more powerful
spells. Systems: MERP, Role Master. E.g., Merlin and Magic Martha both cast “flame bolt” spell. Merlin easily pumps a dozen spell points into it, making it devastate his opponents. Martha fumbles | {"url":"http://www.dungeonchatter.com/2018/09/ars-arcana-blog-split-die-pools.html","timestamp":"2024-11-12T18:16:29Z","content_type":"text/html","content_length":"120691","record_id":"<urn:uuid:79bc53af-11ed-46aa-9353-eec3d77e455d>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00630.warc.gz"} |
Chi squared test: Find out more about this essential statistical test
The chi squared test (or chi 2) is a statistical test for variables that take a finite number of possible values, making them categorical variables. As a reminder, a statistical test is a method used
to determine whether a hypothesis, known as the null hypothesis, is consistent with the data or not.
What is the purpose of the Chi squared test?
The advantage of the Chi squared test is its wide range of applications:
• Test of goodness of fit to a predefined law or family of laws, for example: Does the size of a population follow a normal distribution?
• Test of independence, for example: Is hair color independent of gender?
• Homogeneity test: Are two sets of data identically distributed?
How does the Chi squared test work?
Its principle is to compare the proximity or divergence between the distribution of the sample and a theoretical distribution using the Pearson statistic \chi_{Pearson} [\latex], which is based on
the chi-squared distance.
First problem: Since we have only a limited amount of data, we cannot perfectly know the distribution of the sample, but only an approximation of it, the empirical measure.
The empirical measure \widehat{\mathbb{P}}_{n,X} [\latex] represents the frequency of different observed values:
\forall x \in \mathbb{X} \quad \widehat{\mathbb{P}}_{n,X} (x) = \frac{1}{n} \sum_{k=1}^{n} 1_{X_{k} =x}
Empirical measurement formula
X_{1},... ,{X_n} = the sample
{\mathbb{X}} = all possible values
The Pearson statistic is defined as :
\chi_{Pearson} = n \times \chi_{2}(\widehat{\mathbb{P}}_{n,X}, P_{theorique} ) = n \times \sum_{x \in \mathbb{X}} \frac{(\widehat{\mathbb{P}}_{n,X} (x)- P_{theorique}(x))^{2}}{P_{theorique}(x)}
Pearson’s statistical formula
Under the null hypothesis, which means that there is equality between the distribution of the sample and the theoretical distribution, this Pearson statistic will converge to the chi-squared
distribution with d degrees of freedom.
The number of degrees of freedom, d, depends on the dimensions of the problem and is generally equal to the number of possible values minus 1.
As a reminder, the chi-2 law with d degrees of freedom
centred reduced independent.
is that of a sum of squares of d Gaussians
chi^{2}_{loi}(d) := \sum_{k=1}^{d} X_{k} \quad avec \quad X_{k} \sim \mathbb{N}(0,1)
Otherwise, this statistic will diverge to infinity, reflecting the distance between empirical and theoretical distributions.
Sous \quad H_{0} \quad \lim_{n\rightarrow \infty } \chi_{Pearson} = \chi^{2}_{loi}(d). \\ Sous \quad H_{1} \quad \lim_{n\rightarrow \infty } \chi_{Pearson} = \infty
What are the benefits of the Chi squared test?
So, we have a simple decision rule: if the Pearson statistic exceeds a certain threshold, we reject the initial hypothesis (the theoretical distribution does not fit the data), otherwise, we accept
The advantage of the chi-squared test is that this threshold depends only on the chi-squared distribution and the confidence level alpha, so it is independent of the distribution of the sample.
The test of independence:
Let’s take an example to illustrate this test: we want to determine if the genders of the first two children, X and Y, in a couple are independent?
We have gathered the data in a contingency table:
\begin{array}{|c|c|c|c|} \hline X / Y & Child 2 : son & Child 2: daughter & Total \\ \hline Child 1: son & 857 & 801 & 1658 \\ \hline Child 1: daughter & 813 & 828 & 1641\\ \hline Total & 1670 & 1629
& 3299 \end{array}
The Pearson statistic will determine if the empirical measure of the joint distribution (X, Y) is equal to the product of the empirical marginal measures, which characterizes independence:
\chi_{Pearson} = n \times \chi^2 (\widehat{\mathbb{P}}_{X \times Y}, \widehat{\mathbb{P}}_{X} \times \widehat{\mathbb{P}}_{Y}) = \sum_{x \in \{daughter, son\}, y\in \{daughter, son\}} \frac
{(Observation_{x,y} - Theory_{x,y})^{2}}{Theory_{x,y}}
Here, Observation(x,y) represents the frequency of the value (x, y):
\forall x, y \in \{daughter, son\} \quad Observation_{x,y} = \frac{1}{n} \sum_{k=1}^{n} 1_{(X_{k},Y_{k}) = (x, y)}
Observation(daughter, daughter) = \frac{828}{3299} = 0.251
For Theory(x, y), X and Y are assumed to be independent, so the theoretical distribution should be the product of the marginal distributions:
\forall x, y \in \{daughter, son\} \quad Theory_{x,y} = Observation^{X} \times Observation^{Y} = \sum_{y\in\{daughter, son\}} Observation_{x,y} \times \sum_{x\in\{daughter, son\}} Observation_{x,y}
Thus, the theoretical probability for (son, son) is:
Theory(son, son) = \frac{857+801}{3299} \times \frac{857+813}{3299} = \frac{1658 \times 1670}{3299^{2}} = 0.254
Let’s calculate the test statistic using the following Python code:
In our case, the variables X and Y have only 2 possible values: daughters or sons, so the dimension of the problem is (2-1)(2-1), which is 1.
Therefore, we compare the test statistic to the chi-squared quantile with 1 degree of freedom using the chi2.ppf function from scipy.stats.
If the test statistic is lower than the quantile and the p-value is greater than the significance level of 0.05, we cannot reject the null hypothesis with 95% confidence.
Thus, we conclude that the gender of the first two children is independent.
While the chi squared test is very practical, it does have limitations. It can only detect the existence of correlations but does not measure their strength or causality.
It relies on the approximation of the chi-squared distribution with the Pearson statistic, which is only valid if you have a sufficient amount of data. In practice, the validity condition is as
\forall x \in \mathbb{X} \quad n \times P_{theoretical}(x) (1- P_{theoretical}(x)) \geq 5
The Fisher exact test can address this limitation but requires significant computational power. In practice, it is often limited to 2×2 contingency tables.
Statistical tests are crucial in Data Science to assess the relevance of explanatory variables and validate modeling assumptions. You can find more information about the chi-squared test and other
statistical tests in our module 104 – Exploratory Statistics. | {"url":"https://datascientest.com/en/chi-squared-test-find-out-more-about-this-essential-statistical-test","timestamp":"2024-11-07T10:58:40Z","content_type":"text/html","content_length":"473158","record_id":"<urn:uuid:1ec53763-99c4-4f60-a3e3-3b4eabaf298c>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00017.warc.gz"} |
Relations Archives - Computing Learner
Relevant definitions: Definition: “A relation R on a set A is called symmetric if (b, a) ∈ R whenever (a,b) ∈ R, for all a,b ∈ A. A relation R on a set A such that for all a, b ∈ A, if (a, b) ∈ R and
(b, a) ∈ R, then a = […]
Give an example of a relation on a set that is a) both symmetric and antisymmetric. b) neither symmetric nor antisymmetric Read More »
Show that the relation R = ∅ on a nonempty set S is symmetric and transitive, but not reflexive
Let’s refresh the definitions that are relevant to this exercise. Definition: “A relation R on a set A is called reflexive if (a, a) ∈ R for every element a ∈ A.” Definition: “A relation R on a set A
is called symmetric if (b, a) ∈ R whenever (a,b) ∈ R, for all a,b
Show that the relation R = ∅ on a nonempty set S is symmetric and transitive, but not reflexive Read More » | {"url":"https://computinglearner.com/category/dm/relations/","timestamp":"2024-11-10T21:48:53Z","content_type":"text/html","content_length":"146721","record_id":"<urn:uuid:2b955052-3606-4bc9-9825-0c180c22a28d>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00360.warc.gz"} |
How To Calculate Option Delta, And What Affects It - MoneyReadme.com
How To Calculate Option Delta, And What Affects It
Option delta is the measure of how much an option’s price will change in relation to a 1 point move in the underlying asset. There are several factors that can affect option delta, including time to
expiration, underlying asset price, strike price, and volatility.
How is option delta calculated
Option delta is a measure of the change in the price of a stock option relative to the underlying stock. It is used by traders to determine how much an option will move in relation to the underlying
stock. Delta can be positive or negative, and it is calculated by taking the first derivative of the option price with respect to the underlying stock price.
What factors affect option delta
Option delta is determined by a number of factors, the most important of which are the underlying asset’s price, time to expiration, volatility, and interest rates.
How does option delta change as the underlying stock price changes
Option delta is a measure of an option’s price sensitivity in relation to the underlying stock. It is the amount by which the option’s price changes for each one-point move in the underlying stock.
Option delta can be positive or negative, and it can change as the underlying stock price changes. A positive delta means that the option’s price will increase as the underlying stock price
increases. A negative delta means that the option’s price will decrease as the underlying stock price increases.
The option’s delta will also change as the underlying stock price changes. For example, if an option has a delta of 0.50, and the underlying stock price increases by $1, then the option’s price will
increase by $0.50. If the underlying stock price decreases by $1, then the option’s price will decrease by $0.50.
Why is option delta important
Option delta is one of the most important factors in options trading because it tells you how much the price of an option will change in relation to a change in the underlying asset. Delta can be
either positive or negative, and it is always between 0 and 1. A positive delta means that the option will increase in value when the underlying asset increases in value. A negative delta means that
the option will decrease in value when the underlying asset increases in value.
What is the difference between positive and negative option delta
What is the relationship between option gamma and option delta
Option gamma is a measure of the rate of change of option delta with respect to changes in the underlying asset price. Put simply, it tells us how much the delta of an option will change for a given
move in the underlying asset price.
Option delta, on the other hand, is a measure of the sensitivity of an option’s price to changes in the underlying asset price. It tells us how much the option’s price will change for a given move in
the underlying asset price.
So, option gamma and option delta are directly related – option gamma tells us how much option delta will change for a given move in the underlying asset price.
What is an example of how option delta is used
Option delta is a term used in options trading that refers to the amount by which the price of an option changes in relation to the underlying asset. Delta can be used to measure the risk of an
option, as well as to hedge against losses.
How can I use option delta to hedge my position
Option delta is a measure of how much the price of an option changes in response to a change in the underlying asset. Delta can be used to hedge a position in the underlying asset. For example, if
you are long a call option with a delta of 0.50, and the underlying asset increases in price by 1%, then your call option will increase in value by 0.50%. This can offset some of the loss in the
underlying asset.
What is the maximum value for option delta
Option delta is a measure of the change in the price of an option contract with respect to the underlying asset. It is used by traders to gauge the potential risk and reward of an options trade.
The maximum value for option delta is 1.0. This means that for every $1 move in the underlying asset, the option contract will move by $1. Delta values greater than 1.0 indicate higher risk and
higher potential reward, while delta values less than 1.0 indicate lower risk and lower potential reward.
What is the minimum value for option delta
Option delta is a measure of the rate of change in the price of an option with respect to changes in the underlying asset. Delta can be positive or negative, and it is important to understand both
types in order to make informed trading decisions. A positive delta means that the option will gain value as the underlying asset increases in price, while a negative delta means that the option will
lose value as the underlying asset increases in price. The minimum value for delta is -1, which indicates that the option will decrease in value by the same amount as the underlying asset increases
in price. | {"url":"https://moneyreadme.com/calculate-option-delta/","timestamp":"2024-11-03T07:14:49Z","content_type":"text/html","content_length":"73562","record_id":"<urn:uuid:425ab954-6033-420d-bc55-1d5d81e72c27>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00770.warc.gz"} |
• Class 11 Maths Study Material
An Educational platform for Preparation and Practice Class 11. Kidsfront provide unique pattern of learning Maths with free online comprehensive study material in the form of QUESTION & ANSWER for
each Chapter of Maths for Class 11. This study material help Class 11, Maths students in learning every aspect of Linear Inequalities. Students can understand Linear Inequalities concept easily and
consolidate their learning by doing Online Practice Tests on Maths,Linear Inequalities chapter repeatedly till they excel in Class 11, Linear Inequalities. Free ONLINE PRACTICE TESTS on Class 11,
Linear Inequalities comprise of Hundreds of Questions on Linear Inequalities, prepared by the highly professionals team. Every repeat test of Linear Inequalities will have new set of questions and
help students to prepare themselves for exams by doing unlimited Online Test exercise on Linear Inequalities. Attempt ONLINE TEST on Class 11,Maths,Linear Inequalities in Academics section after
completing this Linear Inequalities Question Answer Exercise.
Unique pattern
• Topic wise:Linear Inequalities preparation in the form of QUESTION & ANSWER.
• Evaluate preparation by doing ONLINE TEST of Class 11, Maths,Linear Inequalities.
• Review performance in PRACTICE TEST and do further learning on weak areas.
• Attempt repeat ONLINE TESTS of Maths Linear Inequalities till you excel.
• Evaluate your progress by doing ONLINE MOCK TEST of Class 11, Maths, All TOPICS.
Linear Inequalities
a) Equality
b) Inequality
c) Linearity
d) Limits
Solution Is :
a) X is equal to y
b) X is not equal to y
c) X is greater than y
d) X is less than y
Solution Is :
a) X is equal to y
b) X is not equal to y
c) X is greater than y
d) X is less than y
Solution Is :
a) X is greater than equal to y
b) X is less than equal to y
c) X is greater than y
d) X is less than y
Solution Is :
a) X is equal to y
b) X is not equal to y
c) X is greater than y
d) X is less than y
Solution Is :
a) X is equal to y
b) X is not equal to y
c) X is greater than y
d) X is less than y
Solution Is :
a) X is greater than equal to y
b) X is less than equal to y
c) X is greater than y
d) X is less than y
Solution Is :
a) P ≤ 1500
b) P ≥ 1500
c) P < 1500
d) P > 1500
Solution Is :
a) X > 1
b) X > 2
c) X > 3
d) X > 4
Solution Is :
a) Equality
b) Inequality
c) Linearity
d) Limits
Solution Is : | {"url":"https://www.kidsfront.com/academics/study-material/Class+11-Maths-Linear+Inequalities-p0.html","timestamp":"2024-11-06T04:02:49Z","content_type":"text/html","content_length":"60742","record_id":"<urn:uuid:1210c4bd-7a89-4dfc-bd3a-20bd09261b70>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00241.warc.gz"} |
5 Evaluation and Performance Metrics
⇦ Back to Transfer learning and fine-tuning
Deep Learning Overview
Deep learning is a subset of machine learning that focuses on artificial neural networks and deep neural networks. These networks are inspired by the structure and function of the human brain, with
multiple layers of interconnected nodes that process information. Deep learning algorithms can automatically learn to represent data through multiple layers of abstraction, allowing them to make
complex decisions and predictions.
Neural Networks in Deep Learning
Neural networks are the building blocks of deep learning models. They consist of layers of interconnected nodes, each performing a specific computation on the input data. The output of each node is
passed through an activation function, which introduces non-linearity into the network. This non-linearity allows neural networks to learn complex patterns and relationships in the data.
Training Deep Learning Models
Training a deep learning model involves feeding it with a large amount of labeled data and adjusting the weights of the network to minimize the difference between the predicted output and the actual
output. This process, known as backpropagation, uses optimization algorithms like stochastic gradient descent to update the weights iteratively. The goal is to optimize the model's parameters to make
accurate predictions on new, unseen data.
Deep Learning Applications
Deep learning has revolutionized various fields, including computer vision, natural language processing, and speech recognition. In computer vision, deep learning models can classify images, detect
objects, and segment images into different regions. In natural language processing, deep learning models can understand and generate human language, enabling applications like machine translation and
sentiment analysis.
Evaluating Deep Learning Models
To evaluate the performance of a deep learning model, various metrics can be used, such as accuracy, precision, recall, and F1 score. Accuracy measures the proportion of correctly classified
instances, while precision measures the proportion of true positive predictions among all positive predictions. Recall measures the proportion of true positive predictions among all actual positive
instances. The F1 score combines precision and recall into a single metric, balancing both measures.
Choosing the Right Evaluation Metric
When evaluating a deep learning model, it is essential to choose the right evaluation metric based on the specific task and the desired outcome. For example, in a medical diagnosis task, high recall
may be more critical to ensure that all positive cases are correctly identified, even if it leads to more false positives. In a spam email detection task, high precision may be more important to
minimize false positives, even if it results in some false negatives. Understanding the trade-offs between different evaluation metrics is crucial for developing effective deep learning models.
Now let's see if you've learned something...
⇦ 4 Applications of Transfer Learning 6 Future of Transfer Learning ⇨ | {"url":"https://www.harmsen.nl/teacher/deep-learning50/transfer-learning-and-fine-tuning/evaluation-and-performance-metrics/","timestamp":"2024-11-02T11:47:49Z","content_type":"text/html","content_length":"8912","record_id":"<urn:uuid:df342e8a-1bd3-44fe-bffd-5eb49c24683a>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00548.warc.gz"} |
Bucket (Imperial) to Teaspoon (metric) Converter
Enter Bucket (Imperial)
Teaspoon (metric)
⇅ Switch toTeaspoon (metric) to Bucket (Imperial) Converter
How to use this Bucket (Imperial) to Teaspoon (metric) Converter 🤔
Follow these steps to convert given volume from the units of Bucket (Imperial) to the units of Teaspoon (metric).
1. Enter the input Bucket (Imperial) value in the text field.
2. The calculator converts the given Bucket (Imperial) into Teaspoon (metric) in realtime ⌚ using the conversion formula, and displays under the Teaspoon (metric) label. You do not need to click any
button. If the input changes, Teaspoon (metric) value is re-calculated, just like that.
3. You may copy the resulting Teaspoon (metric) 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 Bucket (Imperial) to Teaspoon (metric)?
The formula to convert given volume from Bucket (Imperial) to Teaspoon (metric) is:
Volume[(Teaspoon (metric))] = Volume[(Bucket (Imperial))] × 3636.872
Substitute the given value of volume in bucket (imperial), i.e., Volume[(Bucket (Imperial))] in the above formula and simplify the right-hand side value. The resulting value is the volume in teaspoon
(metric), i.e., Volume[(Teaspoon (metric))].
Calculation will be done after you enter a valid input.
Consider that a well yields 10 buckets (imperial) of water.
Convert this water volume from buckets (imperial) to Teaspoon (metric).
The volume in bucket (imperial) is:
Volume[(Bucket (Imperial))] = 10
The formula to convert volume from bucket (imperial) to teaspoon (metric) is:
Volume[(Teaspoon (metric))] = Volume[(Bucket (Imperial))] × 3636.872
Substitute given weight Volume[(Bucket (Imperial))] = 10 in the above formula.
Volume[(Teaspoon (metric))] = 10 × 3636.872
Volume[(Teaspoon (metric))] = 36368.72
Final Answer:
Therefore, 10 bkt is equal to 36368.72 tsp.
The volume is 36368.72 tsp, in teaspoon (metric).
Consider that a paint shop uses 5 buckets (imperial) of paint for a project.
Convert this paint usage from buckets (imperial) to Teaspoon (metric).
The volume in bucket (imperial) is:
Volume[(Bucket (Imperial))] = 5
The formula to convert volume from bucket (imperial) to teaspoon (metric) is:
Volume[(Teaspoon (metric))] = Volume[(Bucket (Imperial))] × 3636.872
Substitute given weight Volume[(Bucket (Imperial))] = 5 in the above formula.
Volume[(Teaspoon (metric))] = 5 × 3636.872
Volume[(Teaspoon (metric))] = 18184.36
Final Answer:
Therefore, 5 bkt is equal to 18184.36 tsp.
The volume is 18184.36 tsp, in teaspoon (metric).
Bucket (Imperial) to Teaspoon (metric) Conversion Table
The following table gives some of the most used conversions from Bucket (Imperial) to Teaspoon (metric).
Bucket (Imperial) (bkt) Teaspoon (metric) (tsp)
0.01 bkt 36.3687 tsp
0.1 bkt 363.6872 tsp
1 bkt 3636.872 tsp
2 bkt 7273.744 tsp
3 bkt 10910.616 tsp
4 bkt 14547.488 tsp
5 bkt 18184.36 tsp
6 bkt 21821.232 tsp
7 bkt 25458.104 tsp
8 bkt 29094.976 tsp
9 bkt 32731.848 tsp
10 bkt 36368.72 tsp
20 bkt 72737.44 tsp
50 bkt 181843.6 tsp
100 bkt 363687.2 tsp
1000 bkt 3636872 tsp
Bucket (Imperial)
The Imperial bucket is a unit of measurement traditionally used to quantify liquid volumes in the UK and other countries using the Imperial system. Originating from practical needs in agriculture and
household tasks, the bucket became a standardized measure for consistency. Historically, the Imperial bucket was essential for tasks such as milking, water collection, and brewing. Today, while less
common, it remains a recognized unit in certain industries and historical contexts.
Teaspoon (metric)
The metric teaspoon is a unit of measurement used to quantify small liquid and dry volumes, primarily in countries using the metric system. It is defined as 5 milliliters, which is approximately
0.169 US fluid ounces. Historically, the metric teaspoon was adopted to standardize measurements in cooking, medicine, and scientific applications, ensuring consistency and accuracy. Today, it is
widely used in recipes, nutritional information, and various other contexts, providing a reliable measure for small quantities across multiple applications.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Bucket (Imperial) to Teaspoon (metric) in Volume?
The formula to convert Bucket (Imperial) to Teaspoon (metric) in Volume is:
Bucket (Imperial) * 3636.872
2. Is this tool free or paid?
This Volume conversion tool, which converts Bucket (Imperial) to Teaspoon (metric), is completely free to use.
3. How do I convert Volume from Bucket (Imperial) to Teaspoon (metric)?
To convert Volume from Bucket (Imperial) to Teaspoon (metric), you can use the following formula:
Bucket (Imperial) * 3636.872
For example, if you have a value in Bucket (Imperial), you substitute that value in place of Bucket (Imperial) in the above formula, and solve the mathematical expression to get the equivalent value
in Teaspoon (metric). | {"url":"https://convertonline.org/unit/?convert=bucket_imperial-teaspoon_metric","timestamp":"2024-11-09T19:47:19Z","content_type":"text/html","content_length":"93912","record_id":"<urn:uuid:7ecc4b65-da14-411e-9cc8-79594b0e8988>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00472.warc.gz"} |
Planck’s radiation law
Planck’s radiation law
The fundamental law governing the properties of the simplest form of thermal radiation – that emitted by a blackbody. It describes the spectrum of such radiation in terms of universal constants and a
single parameter – the body’s temperature. The result is also called a blackbody spectrum. | {"url":"https://www.einstein-online.info/en/explandict/plancks-radiation-law/","timestamp":"2024-11-09T23:02:01Z","content_type":"text/html","content_length":"48728","record_id":"<urn:uuid:32aae18c-024e-4f3c-a4c6-06ebdd15d8f6>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00060.warc.gz"} |
SOFiA Impulse Response Reconstruction
This function recombines impulse responses or time domain signals for multiple channels from frequency domain data delivered by P/D/C or I/T/C. The internal IFFT blocklength is determined by the Y
data itself: Y should have a size of [NumberOfChannels x ((2^n)/2)+1] with n={1, 2 ,3 , ...} (which is the case when using the SOFiA readVSAdata() or mergeArrayData() functions for data import) and
the function returns [NumberOfChannels x resampleFactor*2^n samples. The impulse responses are windowed with a HANN window. The argument win can take values from 0-1. Where 0 means that no window is
applied and 1 will apply a window to the full IR. At a value of 0.5 the first half of the impulse responses keeps unchanged and the last half is multiplied with a HANN type window. The size of the
window is fitted automatically. As default choice win is set to 1/8 which should deliver good results in most of the cases. The impulse responses can be resampled to the original sampling rate before
being downsampled in readVSAdata() or mergeArrayData(). Set resampleFactor = downSample to get back to the original sample rate of the measurement source material. (WARNING: Matlab Signal Processing
Toolbox required for windowing and resampling)
To save processing power and calculation time the SOFiA chain works on the half-sided FFT spectrum only (NFFT/2+1). Therefore F/D/T produces half-sided spectrum output signals (fftData). The `makeIR
()` function automatically reconstructs the double-sided spectrum to compute the impulse responses.
│ Name │ Type │ Purpose │Default│
│Y │int │Frequency domain data from P/D/C or I/T/C │- │
│win │float │Window IR tail [0-1] with a HANN window │1/8 │
│resampleFactor│float mtx│Resampling factor │1 │
│ Name │ Type │ Purpose │
│ │ │Impulse Responses │
│impulseResponses │float mtx│Rows: IR-Data │
│ │ │Cols: Channels │
│File │Type │OS/Matlab│
│sofia_makeIR.m │Help header, Function │All OS │
impulseResponses = sofia_makeIR( Y, [win], [resampleFactor])
impulseResponses Reconstructed impulse response
Columns: Index / Channel: IR1, IR2, ..., IRn
Rows: Impulse responses (time domain)
Y Frequency domain FFT data for multiple channels
Columns: Index / Channel
Rows: FFT data (frequency domain)
[win] Window IR tail [0...1] with a HANN window
0 off
0-1 window coverage (a full, 0 off)
[default 1/8: 1/8 of the IR length is windowed]
! Signal Processing Toolbox required
[resampleFactor] Optional resampling: Resampling factor
e.g. FS_target/FS_source
Resampling is done using MATLAB RESAMPLE
(See MATLAB documentation for more details)
! Signal Processing Toolbox required
This function recombines impulse responses for multiple channels from
frequency domain data. It is made to work with half-sided spectrum FFT
data. The impulse responses can be windowed. The IFFT blocklength is
determined by the Y data itself:
Y should have a size [NumberOfChannels x ((2^n)/2)+1 with n=[1, 2, 3, ...]
and the funtion returns [NumberOfChannels x resampleFactor*2^n] samples.
Dependencies: MATLAB Processing Toolbox required for windowing and resampling | {"url":"https://audiogroup.web.th-koeln.de/SOFiA_wiki/MAKEIR.html","timestamp":"2024-11-10T14:14:31Z","content_type":"text/html","content_length":"22968","record_id":"<urn:uuid:5489a178-30c1-408f-93a9-51dc86cd3f6a>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00045.warc.gz"} |
Guide to the Civil service
Dentro del libro
Resultados 1-5 de 10
Página 9 ... of young men proposed to be appointed to any of the junior situations in the Civil Establishments ; and B authorising them to give certificates of qualification before such young No Page
Página 10 Henry White. authorising them to give certificates of qualification before such young men entered on their duties ... certificate no candidate shall be admitted to the service in India .
The Superannuation Act , passed in the same ...
Página 11 ... certificates 2. - MODE OF EXAMINATION . The mode in which the examinations are conducted in London is usually as follows : -The candidates meet at the office of the Civil Service
Commission , Dean's Yard , or else at Great George Street ...
Página 12 ... certificates does not usually take place until after the candidates have received notice that they have passed in the required subjects . To ensure uniformity of standard , the
provincial examinations are all under the control of the ...
Página 13 ... certificate is granted unless the result of the inquiries is satisfactory . As a general rule , every paper is looked over twice , each of the two permanent Examiners going over the
other's work . In some cases , where it is perfectly ...
Términos y frases comunes
Pasajes populares
Página 115
The village master taught his little school: A man severe he was, and stern to view, I knew him well, and every truant knew; Well had the boding tremblers learned to trace The day's disasters in his
morning face; Full well they laughed with counterfeited glee At all his jokes, for many a joke had he...
Página 122
To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Página 123
AB into two parts, so that the rectangle contained by the whole line and one of the parts, shall be equal to the square on the other part.
Página 126
Similar polygons may be divided into the same number of similar triangles, having the same ratio to one another that the polygons have ; and the polygons have to one another the duplicate ratio of
that which their homologous sides have.
Página 123
Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.
Página 122
If two triangles have two angles of the one, equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal...
Página 111
They heard, and were abashed, and up they sprung Upon the wing ; as when men wont to watch On duty, sleeping found by whom they dread, Rouse and bestir themselves ere well awake.
Página 111
No flocks that range the valley free, To slaughter I condemn: Taught by that Power that pities me, I learn to pity them : "But from the mountain's grassy side A guiltless feast I bring; A scrip with
herbs and fruits supplied, And water from the spring. "Then, pilgrim, turn, thy cares forego ; All earth-born cares are wrong; Man wants but little here below, Nor wants that little long.
Página 123
If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the parts.
Página 123
If, from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than, the other two sides of the triangle, but shall contain a
greater angle.
Información bibliográfica | {"url":"https://books.google.co.ve/books?id=n5IBAAAAQAAJ&q=certificates&dq=related:ISBN8474916712&lr=&output=html&source=gbs_word_cloud_r&cad=5","timestamp":"2024-11-10T17:19:43Z","content_type":"text/html","content_length":"55504","record_id":"<urn:uuid:19757acc-16ea-42e9-8bc5-4bd10b37ed6a>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00097.warc.gz"} |
In this section, we present methods to display three-dimensional plots, that is, plots of mathematical objects in space. Examples include surfaces and lines that are not confined to a plate.
matplotlib has excellent support for three-dimensional plots. In this section, we will present an example of a surface plot and corresponding contour plot. The types of plot available in the
three-dimensional library include wireframe plots, line plots, scatterplots, triangulated surface plots, polygon plots, and several others. The following link will help you to understand the types of
plots that are not treated here: http://matplotlib.org/1.3.1/mpl_toolkits/mplot3d/tutorial.html#mplot3d-tutorial
Before we start, we need to import the three-dimensional library objects we need using the following command line:
from mpl_toolkits.mplot3d import axes3d
Now, let's draw our surface plot by running the following code in a cell:
def dist(x, y):
return sqrt(x**2 + y**2)
def fsurface... | {"url":"https://subscription.packtpub.com/book/data/9781783988341/3/ch03lvl1sec16/three-dimensional-plots","timestamp":"2024-11-07T05:53:49Z","content_type":"text/html","content_length":"85040","record_id":"<urn:uuid:15d3b361-4a83-4a3f-bf40-4f5ac50ecb8a>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00638.warc.gz"} |
Comments on t
\(\renewcommand{\Re}{\mathop{\textrm{Re}}} \renewcommand{\Im}{\mathop{\textrm{Im}}} \)
Comments on the textbook for Math 407, Spring 2012
Here are some corrections and amplifications to the textbook, Schaum's Outline of Complex Variables, second edition, by Murray R. Spiegel, Seymour Lipschutz, John J. Schiller, and Dennis Spellman,
McGraw-Hill, 2009, ISBN 9780071615693.
Section 1.5, formula (2)
The formula is intended to say that the modulus of a quotient equals the quotient of the moduli, but the printed formula has the identical expression on both sides. The formula should read as
follows: \[\left| \frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \qquad \text{if \(z_2\ne 0\).} \]
Problem 1.65
The signs are wrong in the formula, which should say that \(z_1-z_2+z_3-z_4=0\).
Problem 2.66
The universal quantifier is misplaced. The problem should say that for every complex number \(z\), if \(|\sin z|\le 1\), then \(|\Im z| \le \ln(\sqrt{2}+1)\).
Problem 3.8, Solution
At the end of Method 1, notice that the “arbitrary additive constant” is not completely arbitrary: this constant has to be purely imaginary.
Problem 3.44
The statement is incomplete. The derivative does exist at one exceptional point: namely, when \(z=0\).
Problem 3.84
For \(e^{x^2}\) read \(e^{z^2}\).
Problem 3.101
Add the hypothesis that \(f\) is an analytic function.
Problem 4.43
The typesetting is ambiguous. The integral is intended to be \[ \oint_C \frac{dz}{z-2}. \]
Problem 5.33
There is a typographical error in the numerator. For \(\cos\pi2\) read \(\cos\pi z\).
Problem 6.92
The answer shown for part (c) has a typographical error: the initial term \(-1/2\) should be \(-1/z\).
Problem 6.96b
The answer in the book corresponds to the function \(e^{z^2}/z^3\), not the indicated function \(e^z/z^3\).
Problem 7.78
The answer given in the book is \(1/24\), but the correct answer is \(2\pi i/24\), that is, \(\pi i/12\).
Problem 7.47
The answer given in the book is \(-6\pi i\), but the correct answer is \(-6\pi^2 i\).
Problem 8.34b
The answer given in the book is incorrect. The correct equation is \(u^2+v^2=u-v\), which represents a circle with center \((1/2,-1/2)\) and radius \(1/\sqrt{2}\). The circle passes through the
point \(0\), and that point is missing from the image (unless the \(z\) plane is taken to be the extended complex plane including the point at infinity). | {"url":"https://haroldpboas.gitlab.io/courses/407-2012a/errata.html","timestamp":"2024-11-06T15:03:05Z","content_type":"text/html","content_length":"3597","record_id":"<urn:uuid:d5649d89-a4a3-44e5-ab69-5c05570f1f9d>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00407.warc.gz"} |
In the News: Why Do So Many Students Believe They Are Just Not Cut Out for Math? | Just Equations
Following our latest report release, the Atlanta Journal-Constitution devotes a “Get Schooled” blog post to the report, The Mathematics of Opportunity: Rethinking the Role of Math in Educational
Equity. Quoting extensively from the report, columnist Maureen Downey ties it to concerns about lagging math performance among Georgia high school students.
A new study examines the pivotal role that math plays in student achievement, calling it a “key mechanism in the distribution of opportunity.” The Mathematics of Opportunity: Rethinking the Role of
Math in Educational Equity says that while math requirements are seen as a foundation for academic success, they can also become a filter that stops many students in their educational tracks,
especially students of color.
“Misconceptions about math ability — like the assumption that only some kids can learn math — magnify existing inequities in the education system,” said Pamela Burdman, senior project director of
Just Equations, project of the Opportunity Institute that is re-conceptualizing the role of mathematics in educational equity. “Math can serve as a foundation for success in school, work, and life,
but it can also be wielded in ways that arbitrarily close doors to educational advancement.”
For more insights on the role of math in ensuring educational equity, subscribe to Just Equations’ newsletter. | {"url":"https://justequations.org/in-the-news/why-do-so-many-students-believe-they-are-just-not-cut-out-for-math","timestamp":"2024-11-02T21:56:31Z","content_type":"text/html","content_length":"53125","record_id":"<urn:uuid:395b4b71-b59f-452b-8cbf-2969f3c33f07>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00849.warc.gz"} |
Extract Data Between Brackets in Excel: Quick Techniques
Table of Contents :
Extracting data between brackets in Excel can streamline your data analysis and enhance productivity. Whether you're dealing with strings containing parentheses, square brackets, or curly braces,
there are various techniques available to help you efficiently extract the desired information. In this guide, we'll explore several methods, including formulas, text functions, and even VBA scripts
for more advanced users. Letβ s dive in! π
Why Extract Data from Brackets? π €
Extracting data from within brackets is a common task in data cleaning and preparation. This need arises in many scenarios, such as:
• Cleaning datasets: Remove unwanted characters or extract essential information.
• Data analysis: Analyze specific parts of text strings for reporting or visualization.
• Automation: Use extracted data for further processing or calculations.
Common Techniques to Extract Data Between Brackets
1. Using Excel Formulas π
Excel provides a set of powerful text functions that can be used to extract data between brackets. Here are a few formulas that can help.
Example Scenario
Suppose you have the following data in cell A1: Order [12345] was shipped on [2023-10-15].
Formula to Extract Data from Square Brackets
To extract data between the first pair of square brackets, you can use:
=MID(A1, FIND("[", A1) + 1, FIND("]", A1) - FIND("[", A1) - 1)
• FIND locates the position of the brackets.
• MID extracts the string based on those positions.
Table of Formulas for Different Bracket Types
Bracket Type Formula to Extract Data
Square Brackets [] =MID(A1, FIND("[", A1) + 1, FIND("]", A1) - FIND("[", A1) - 1)
Curly Braces {} =MID(A1, FIND("{", A1) + 1, FIND("}", A1) - FIND("{", A1) - 1)
Parentheses () =MID(A1, FIND("(", A1) + 1, FIND(")", A1) - FIND("(", A1) - 1)
2. Using Text to Columns Feature β οΈ
Another straightforward method to extract data is by using the Text to Columns feature. This can be useful if you want to separate multiple pieces of data at once.
Steps to Use Text to Columns
1. Select the cell(s) containing your data.
2. Go to the Data tab on the Ribbon.
3. Click on Text to Columns.
4. Choose Delimited and click Next.
5. Check Other and enter [ (for square brackets) or { (for curly braces).
6. Click Finish.
This will split the data into multiple columns based on the specified delimiter.
3. Advanced Technique: Using VBA π ₯οΈ
If you frequently need to extract data from various brackets and want a more automated solution, you can create a VBA function. This is ideal for users familiar with macros.
Sample VBA Code
To create a custom function, open the VBA editor (ALT + F11) and insert a new module, then paste the following code:
Function ExtractBetweenBrackets(text As String) As String
Dim StartPos As Long
Dim EndPos As Long
StartPos = InStr(text, "[") + 1
EndPos = InStr(text, "]")
If StartPos > 0 And EndPos > StartPos Then
ExtractBetweenBrackets = Mid(text, StartPos, EndPos - StartPos)
ExtractBetweenBrackets = "No Data Found"
End If
End Function
How to Use the Function
After saving the module, you can use this custom function in your Excel sheet like this:
4. Dealing with Multiple Bracket Pairs β οΈ
If your string contains multiple pairs of brackets, you may need a different approach. You can extend the previous formula with a combination of additional functions, or run a loop in VBA to handle
each occurrence.
Example Formula for Multiple Bracket Pairs
To extract all data between brackets, you would typically need a more complex approach with array formulas or VBA.
=TEXTJOIN(", ", TRUE, IFERROR(MID(A1, SMALL(IF(MID(A1, ROW($1:$100), 1) = "[", ROW($1:$100)), ROW($1:$100)) + 1, FIND("]", A1, SMALL(IF(MID(A1, ROW($1:$100), 1) = "[", ROW($1:$100)), ROW($1:$100)) - 1) - (SMALL(IF(MID(A1, ROW($1:$100), 1) = "[", ROW($1:$100)), ROW($1:$100)) + 1)), ""))
Important Notes π
Always back up your data before applying any formulas or scripts, especially when working with large datasets or automated processes.
Mastering the extraction of data from brackets in Excel can greatly improve your data management capabilities. With techniques ranging from simple formulas to advanced VBA scripts, you can choose the
method that best suits your needs. Whether you're cleaning data for analysis or automating repetitive tasks, these methods will empower you to work more efficiently in Excel.
Explore these techniques, practice, and soon you'll be extracting data like a pro! π | {"url":"https://tek-lin-pop.tekniq.com/projects/extract-data-between-brackets-in-excel-quick-techniques","timestamp":"2024-11-01T20:11:48Z","content_type":"text/html","content_length":"85819","record_id":"<urn:uuid:5dbef75b-706f-45b1-bacd-01af6f1feffc>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00846.warc.gz"} |
If pairs of straight lines x2−2pxy−y2=0 and x2−2qxy −y2=0 be su... | Filo
If pairs of straight lines and be such that each pair bisects the angle between the other pair, then-
Not the question you're searching for?
+ Ask your question
From the problem, it is clear that, the bisectors of the angles between the lines given by
....(i) ,is ......(ii)
The equation of the bisectors of Eq. (i) is
Here, Eqs (ii) and (iii) are identical.
Thus, comparing the co-efficients, we get
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Question Text If pairs of straight lines and be such that each pair bisects the angle between the other pair, then-
Updated On Feb 18, 2023
Topic Straight Lines
Subject Mathematics
Class Class 11
Answer Type Text solution:1 Video solution: 2
Upvotes 374
Avg. Video Duration 9 min | {"url":"https://askfilo.com/math-question-answers/if-pairs-of-straight-lines-x2-2-p-x-y-y20-and-x2-2-q-x-y-y20-be-such-that-each-225924","timestamp":"2024-11-11T17:02:07Z","content_type":"text/html","content_length":"306483","record_id":"<urn:uuid:54b56185-fa9d-48c7-9387-9ddc8c204b15>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00767.warc.gz"} |
CWG Issue 2539
This is an unofficial snapshot of the ISO/IEC JTC1 SC22 WG21 Core Issues List revision 115d. See http://www.open-std.org/jtc1/sc22/wg21/ for the official list.
2539. Three-way comparison requiring strong ordering for floating-point types
Section: 11.10.3 [class.spaceship] Status: C++23 Submitter: Richard Smith Date: 2022-02-24
[Accepted as a DR at the February, 2023 meeting.]
struct MyType {
int i;
double d;
std::strong_ordering operator<=> (const MyType& c) const = default;
The defaulted three-way comparison operator is defined only if it is used, per 11.10.1 [class.compare.default] paragraph 1:
A comparison operator function for class C that is defaulted on its first declaration and is not defined as deleted is implicitly defined when it is odr-used or needed for constant evaluation.
The current rules make an odr-use of the three-way comparison operator ill-formed, but it would be preferable if it were deleted instead. In particular, 11.10.3 [class.spaceship] bullet 2.2 specifies
If the synthesized three-way comparison of type R between any objects x[i] and x[i] is not defined, the operator function is defined as deleted.
This refers to bullets 1.2 and 1.3 of 11.10.3 [class.spaceship] paragraph 1:
The synthesized three-way comparison of type R (17.11.2 [cmp.categories]) of glvalues a and b of the same type is defined as follows:
□ If a <=> b is usable (11.10.1 [class.compare.default]), static_cast<R>(a <=> b).
□ Otherwise, if overload resolution for a <=> b is performed and finds at least one viable candidate, the synthesized three-way comparison is not defined.
□ Otherwise, if R is not a comparison category type, or either the expression a == b or the expression a < b is not usable, the synthesized three-way comparison is not defined.
□ Otherwise, ...
However, a <=> b is actually usable, because 11.10.1 [class.compare.default] paragraph 3 defines:
A binary operator expression a @ b is usable if either
□ a or b is of class or enumeration type and overload resolution (12.2 [over.match]) as applied to a @ b results in a usable candidate, or
□ neither a nor b is of class or enumeration type and a @ b is a valid expression.
MyType().d <=> MyType().d is a valid expression.
Proposed resolution (approved by CWG 2022-11-11) [SUPERSEDED]:
The synthesized three-way comparison of type R (17.11.2 [cmp.categories]) of glvalues a and b of the same type is defined as follows:
□ If a <=> b is usable (11.10.1 [class.compare.default])
☆ , static_cast<R>(a <=> b)
☆ .
□ Otherwise, if overload resolution for a <=> b is performed and finds at least one viable candidate, the synthesized three-way comparison is not defined.
□ Otherwise, if R is not a comparison category type, or either the expression a == b or the expression a < b is not usable, the synthesized three-way comparison is not defined.
□ Otherwise, ...
CWG 2023-02-06
A simplification of the wording is sought.
Proposed resolution (approved by CWG 2023-02-07):
The synthesized three-way comparison of type R (17.11.2 [cmp.categories]) of glvalues a and b of the same type is defined as follows:
□ If a <=> b is usable (11.10.1 [class.compare.default]) static_cast<R>(a <=> b).
□ Otherwise, if overload resolution for a <=> b is performed and finds at least one viable candidate, the synthesized three-way comparison is not defined.
□ Otherwise, if R is not a comparison category type, or either the expression a == b or the expression a < b is not usable, the synthesized three-way comparison is not defined.
□ Otherwise, ... | {"url":"https://cplusplus.github.io/CWG/issues/2539.html","timestamp":"2024-11-02T17:40:26Z","content_type":"text/html","content_length":"6681","record_id":"<urn:uuid:f72ff498-f974-4cb4-af5d-2eac680e072d>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00760.warc.gz"} |
About critical damping
Having to deal with DSP texts written by engineers, I have sometimes to work a bit to get a good grasp of the concepts, which many times are not explained clearly from their mathematical bases.
Often, a formula is just used without much motivation. Lately, I’ve been trying to understand critically damped systems, in the context of PLL loop filters.
The issue is as follows. In a second order filter there is a damping parameter \(\zeta > 0\). The impulse response of the filter is an exponentially decaying sinusoid if \(\zeta < 1\) (underdamped
system), a decaying exponential if \(\zeta > 1\) (overdamped system) and something of the form \(C t e^{-\lambda t}\) if \(\zeta = 1\) (critically damped system). Critical damping is desirable in
many cases because it maximizes the exponential decay rate of the impulse response. However, many engineering texts just go and choose \(\zeta = \sqrt{2}/2\) without any justification and even call
this critical damping. Here I give some motivation starting with the basics and explain what is special about \(\zeta = \sqrt{2}/2\) and why one may want to choose this value in applications.
We start with a linear second order ODE of the form\[\tag{1} a_2y ‘ ‘(t) + a_1y'(t) + a_0 y(t) = f(t).\] This system is known as the damped harmonic oscillator and appears in many physical
situations. In particular, it models an RLC circuit, which justifies its appearance in analogue loop filters (and also in their digital counterparts, which work in discrete time).
The Laplace transform of \(y\) is defined as\[Y(s) = \int_0^\infty y(t) e^{-st}\,dt\]provided that this integral converges (usually we restrict ourselves to bounded functions \(y\) and \(\
operatorname{Re} s > 0\)). The Laplace transform \(F(s)\) of \(f\) is defined similarly.
Taking Laplace transforms in (1) and assuming that \(y(0) = y'(0) = f(0) = 0\), we obtain\[(a_2 s^2 + a_1 s + a_0)Y(s) = F(s).\]Thus, the transfer function of the system given by (1) is\[H(s) = \frac
{Y(s)}{F(s)} = \frac{1}{a_2 s^2 + a_1 s + a_0}.\] In the applications we are interested in, \(a_0\) and \(a_2\) are positive and \(a_1\) is non-negative. By a suitable change of scale we can assume
that \(a_2 = a_0 = 1\). We write \(a_1 = 2 \zeta\), so that\[H(s) = \frac{1}{s^2 + 2\zeta s + 1}.\]
The transfer function \(H(s)\) is the Laplace transform of the impulse response of the system, since if \(f\) is a Dirac delta at \(0\), then \(F(s) = 1\), so that \(Y(s) = H(s)\).
The denominator of \(H(s)\) is a second order polynomial. Therefore, the poles of \(H(s)\) are either real or complex conjugates depending on whether the discriminant \(\Delta = 4\zeta^2 – 4\) is
positive or negative. We see that for \(\zeta > 1\) (overdamped system), the poles \(\lambda_\pm = -\zeta \pm \sqrt{\zeta^2 – 1}\) are real. This means that the impulse response is a linear
combination of \(e^{\lambda_+ t} u(t)\) and \(e^{\lambda_- t} u(t)\), because the Laplace transform of \(e^{\lambda t}u(t)\) is \((s-\lambda)^{-1}\). Here \(u(t) = \chi_{(0,\infty)}(t)\) denotes the
step function. The exponential decay rate of the system is \(d = -\lambda_+ = \zeta – \sqrt{\zeta^2 – 1}\). Note that \(d < 1\).
When \(\zeta < 1\) (underdamped system), the poles \(\lambda_\pm = -\zeta \pm i\sqrt{1-\zeta^2}\) are complex conjugate. In this case, the impulse response is again a (real-valued) linear combination
of \(e^{\lambda_+ t} u(t)\) and \(e^{\lambda_- t} u(t)\), so it is an exponentially decaying sinusoid. The exponential decay rate is \(d = -\operatorname{Re} \lambda_\pm = -\zeta\). Again, note that
\(d < 1\).
In the limiting case \(\zeta = 1\) (critically damped system), the pole \(\lambda = -1\) is double. The impulse response is \(t e^{-t}\), and the exponential decay rate is \(d = 1\). Note that the
critically damped case \(\zeta = 1\) maximizes the exponential decay rate.
However, in many applications the exponential decay rate is not the most important parameter to optimize. The plot below, which shows the impulse response for several values of the damping parameter
can shed some light into why.
Impulse response for different damping factors
From the four impulse responses plotted, the red one is the most desirable. The blue one, which corresponds to an overerdamped system does not decay fast enough, as we already showed. The exponential
decay rate of the green curve, which corresponds to the critically damped case, is the best possible, but the underdamped red curve takes less time to get near zero, because it is an exponentially
decaying sinusoid and it has some undershoot. The cyan curve is also underdamped, but it has too much undershoot, since the exponential decay rate is not large enough. Therefore, we see that if the
damping factor \(\zeta\) is a bit lower than \(1\) (but not too low), then the undershoot given by the sinusoid plays in our advantage, since it helps bring the response close to zero faster.
Now the obvious question is how low can we set the damping factor before too much undershoot appears. Another way of looking at this problem from a different point of view is studying resonance. When
the input \(f(t)\) of the system is a sinusoid of frequency \(\omega\), then \(F(s)\) has poles at \(\pm i\omega\). The Laplace transform of the output of the system \(Y(s) = H(s) F(s)\) has poles at
\(\pm i\omega\) and \(\lambda_\pm\). The poles at \(\lambda_\pm\) give terms which decay exponentially, while the poles at \(\pm i\omega\) give sinusoids of frequency \(\omega\) which are still
present in the steady state. Thus, when the input is a sinusoid of amplitude \(1\) and frequency \(\omega\), the steady state output is a sinusoid of amplitude \(|H(i\omega)|\) and frequency \(\omega
\), since the residue of \(Y(s)\) at \(s = \pm i \omega\) equals the residue of \(F(s)\) at \(\pm i \omega\) multiplied by \(H(i\omega)\). Therefore, the frequency response of the system is given by
We now compute\[|H(i\omega)|^2 = \frac{1}{\omega^4 + (4\zeta^2 – 2)\omega^2 + 1}.\] Using the substitutions \(\xi = \omega^2\) and \(\gamma = 4\zeta^2 – 2\), we see that the denominator equals the
quadratic function \(\xi^2 + \gamma \xi + 1\). The minimum of this quadratic function is at \(\xi = -\gamma/2\). This means that when \(\gamma \geq 0\) the frequency response \(|H(i\omega)|\) is
monotone decreasing in \(\omega\). In fact, this system is a low pass filter, so this is not surprising. When \(\gamma < 0\), the frequency response has a maximum at \(\omega_0 = \sqrt{-\gamma/2}\),
and \(|H(i\omega_0)| = \frac{1}{1-\gamma^2/4}\). This means that the system has a resonant frequency at \(\omega_0\), and the gain at the resonant frequency is greater than unity. This is
undesirable, so we impose \(\gamma \geq 0\). The limiting case \(\gamma = 0\) gives a damping factor of \(\zeta = \sqrt{2}/2\). This is the lowest we can set the damping factor before getting
The figure below shows the frequency response for different values of the damping factor \(\zeta\). This shows that the system is a low pass filter and it also shows clearly the resonance when \(\
zeta < \sqrt{2}/2\).
Frequency response for different damping factors
The bandwidth of the low pass filter is defined as the frequency \(\omega\) at which \(|H(i\omega)|^2 = 1/2\). In the case when \(\zeta = \sqrt{2}/2\), a simple calculation using \(|H(i\omega)|^2 = 1
/(w^4 + 1)\) shows that the bandwidth is \(1\).
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Logistic Map Plot Demo | Godot Asset Library
Install Asset
Install via Godot
To maintain one source of truth, Godot Asset Library is just a mirror of the old asset library so you can download directly on Godot via the integrated asset library browser
Quick Information
A demo for generating the logistic map bifurcation diagram.It calculates the x[n+1]-value of the initial value 0.5 for a couple of times and approaches the asymptote at a given lambda(growth rate)
-value. Once lambda approaches c. 3.5, you can see that x[n] does not approach a single asymptote, but multiple ones. The graph splits faster and faster, until it rbecomes chaotic I needed the data
from the bifurcation diagram for a project (because damn, it looks cool), this is why i made this tool. you can save the plot-data as json, so that you don't have to calculate thousands of values all
over again (which is pretty expensive)Feel free to play around. Maybe, with some creativitiy, it can help you make something pseudorandom for your game, or maybe you're just a nerd :3
Supported Engine Version
Modified Date
4 years ago
A demo for generating the logistic map bifurcation diagram.
It calculates the x[n+1]-value of the initial value 0.5 for a couple of times and approaches the asymptote at a given lambda(growth rate)-value Once lambda approaches c. 3.5, you can see that x[n]
does not approach a single asymptote, but multiple ones. The graph splits faster and faster, until it rbecomes chaotic I needed the data from the bifurcation diagram for a project (because damn, it
looks cool), this is why i made this tool. you can save the plot-data as json, so that you don't have to calculate thousands of values all over again (which is pretty expensive)
Feel free to play around. Maybe, with some creativitiy, it can help you make something pseudorandom for your game, or maybe you're just a nerd :3
If you want to learn more about this phenomenon, i recommend watching Veritasium's video on it https://www.youtube.com/watch?v=ovJcsL7vyrk
A demo for generating the logistic map bifurcation diagram.
It calculates the x[n+1]-value of the initial value 0.5 for a couple of times and approaches the asymptote at a given lambda(growth rate)-value.
Once lambda approaches c. 3.5, you can see that x[n] does not approach a single asymptote, but multiple ones. The graph splits faster and faster, until it rbecomes chaotic
I needed the data from the bifurcation diagram for a project (because damn, it looks cool), this is why i made this tool. you can save the plot-data as json, so that you don't have to calculate
thousands of values all over again (which is pretty expensive)
Feel free to play around. Maybe, with some creativitiy, it can help you make something pseudorandom for your game, or maybe you're just a nerd :3
Quick Information
A demo for generating the logistic map bifurcation diagram.It calculates the x[n+1]-value of the initial value 0.5 for a couple of times and approaches the asymptote at a given lambda(growth rate)
-value. Once lambda approaches c. 3.5, you can see that x[n] does not approach a single asymptote, but multiple ones. The graph splits faster and faster, until it rbecomes chaotic I needed the data
from the bifurcation diagram for a project (because damn, it looks cool), this is why i made this tool. you can save the plot-data as json, so that you don't have to calculate thousands of values all
over again (which is pretty expensive)Feel free to play around. Maybe, with some creativitiy, it can help you make something pseudorandom for your game, or maybe you're just a nerd :3
Supported Engine Version
Modified Date
4 years ago | {"url":"https://godotassetlibrary.com/asset/s8KRhn/logistic-map-plot-demo","timestamp":"2024-11-04T20:25:40Z","content_type":"text/html","content_length":"37939","record_id":"<urn:uuid:d29dd692-8586-4ee3-955a-6d9347a851cd>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00723.warc.gz"} |
Make loop at K-Th position in the Linked List | Linked list articles | PrepBytes Blog
Last Updated on November 28, 2022 by Prepbytes
In this article, we will see a new coding problem of the linked list i.e.” Create a loop in linked list” A linked list is a linear data structure. Each Node contains a data field and a pointer to the
next Node. In Linked List, unlike arrays, elements are not stored at contiguous memory locations but rather at different memory locations. Linked Lists are one of the most fundamental and important
data structures having a wide range of applications. Linked Lists are also important from the perspective of interviews as well. Let us understand the problem statement of how to create a loop in
linked list.
How to create a loop in linked list
We will be given a linked list and an integer K. We need to attach the last node of the list to the K^{th} node from starting of the list.
To understand this problem statement, let us take an example.
If the given linked list is 3→1→8→2→4→NULL and K = 3, then according to the problem statement:
• In the given linked list the third node of the list is 8.
• So, we need to attach the tail of the list, i.e., 4 with 8.
• So, after connecting 4 to 8, the list will look like this:
At this point, we have understood the problem statement. Now we will try to formulate an approach for this problem.
Before moving to the approach section, try to think about how you can approach this problem.
• If stuck, no problem, we will thoroughly see how we can approach this problem in the next section.
Let’s move to the approach section.
Approach of how to create a loop in linked list
1. Firstly, we need to reach the K^{th} node of the list.
2. After we reach the K^{th} node, we need to save this node’s address in a pointer variable.
3. Then, we need to reach the end of the list and connect it with the K^{th} node (using the pointer variable which we used to store the address of K^{th} node in step 2).
Algorithm of how to create a loop in linked list
1. Initialize a count variable with 1 and a variable temp with the first node of the list.
2. Run a while loop till the count is less than K.
□ Inside the while loop, in each iteration, increment count by one and move temp by one node.
3. Save the temp in the kth_node variable.
4. Run a while loop till temp is not NULL.
□ Inside the while loop, advance temp by one node in each iteration.
5. At last, connect temp with kth_node i.e., temp->next = kth_node.
Dry Run of how to create a loop in linked list
Code Implementation of how to create a loop in linked list
struct Node {
int data;
struct Node* next;
/* Function to make loop at k-th elements of
linked list */
struct Node* newNode(int x)
struct Node* node = malloc(sizeof(struct Node*));
node->data = x;
node->next = NULL;
return node;
void printList(struct Node* head, int total_nodes)
struct Node* curr = head;
int count = 0;
while (count < total_nodes) {
printf("%d ",curr->data);
curr = curr->next;
// this function will create a loop between
// the last node and the Kth node
void makeloop(struct Node** head_ref, int k)
//initialize 'temp' with the first node
struct Node* temp = *head_ref;
int count = 1;
//run a while loop till 'count' is
//less than 'k'
while (count < k) {
temp = temp->next;
//save the Kth node in a variable
struct Node* kth_node = temp;
//traverse the list till we reach
//the tail node
while (temp->next != NULL)
temp = temp->next;
//join the last node with the Kth node
temp->next = kth_node;
int main(void){
struct Node* head = NULL;
head = newNode(3);
head->next = newNode(1);
head->next->next = newNode(8);
head->next->next->next = newNode(2);
head->next->next->next->next = newNode(4);
int k = 3;
printf( "\nGiven list\n");
printList(head, 5);
makeloop(&head, k);
printf("\nModified list\n");
printList(head, 6);
return 0;
#include<bits stdc++.h="">
using namespace std;
class Node{
int data;
Node* next;
Node(int x){
data = x;
next = NULL;
void printList(Node* head, int total_nodes)
Node* curr = head;
int count = 0;
while (count < total_nodes) {
cout << curr->data << " ";
curr = curr->next;
// this function will create a loop between
// the last node and the Kth node
void makeloop(Node** head_ref, int k)
//initialize 'temp' with the first node
Node* temp = *head_ref;
int count = 1;
//run a while loop till 'count' is
//less than 'k'
while (count < k) {
temp = temp->next;
//save the Kth node in a variable
Node* kth_node = temp;
//traverse the list till we reach
//the tail node
while (temp->next != NULL)
temp = temp->next;
//join the last node with the Kth node
temp->next = kth_node;
int main(void){
Node* head = NULL;
head = new Node(3);
head->next = new Node(1);
head->next->next = new Node(8);
head->next->next->next = new Node(2);
head->next->next->next->next = new Node(4);
int k = 3;
cout << "\nGiven list\n";
printList(head, 5);
makeloop(&head, k);
cout << "\nModified list\n";
printList(head, 6);
return 0;
class MakeLoop
static class Node
int data;
Node next;
static Node makeloop( Node head_ref, int k)
// traverse the linked list until loop point not found
Node temp = head_ref;
int count = 1;
while (count < k)
temp = temp.next;
// backup the joint point
Node joint_point = temp;
// traverse remaining nodes
while (temp.next != null)
temp = temp.next;
// joint the last node to k-th element
temp.next = joint_point;
return head_ref;
// Function to push a node /
static Node push( Node head_ref, int new_data)
Node new_node = new Node();
new_node.data = new_data;
new_node.next = (head_ref);
(head_ref) = new_node;
return head_ref;
// Function to print linked list /
static void printList( Node head, int total_nodes)
Node curr = head;
int count = 0;
while (count < total_nodes)
System.out.print(curr.data + " ");
curr = curr.next;
static int countNodes(Node ptr)
int count = 0;
while (ptr != null)
ptr = ptr.next;
return count;
// Driver code
public static void main(String args[])
Node head = null;
head = push(head, 7);
head = push(head, 6);
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
int k = 4;
int total_nodes = countNodes(head);
System.out.print("\nGiven list\n");
printList(head, total_nodes);
head = makeloop(head, k);
System.out.print( "\nModified list\n");
printList(head, total_nodes);
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Function to make loop at k-th elements of
#linked list
def makeloop(head_ref, k):
temp = head_ref
count = 1
while (count < k):
temp = temp.next
count = count + 1
joint_point = temp
while (temp.next != None):
temp = temp.next
temp.next = joint_point
return head_ref
def push(head_ref, new_data):
new_node = Node(new_data)
new_node.data = new_data
new_node.next = head_ref
head_ref = new_node
return head_ref
def printList( head, total_nodes):
curr = head
count = 0
while (count < total_nodes):
count = count + 1
print(curr.data, end = " ")
curr = curr.next
if __name__=='__main__':
head = None
head = push(head, 4)
head = push(head, 2)
head = push(head, 8)
head = push(head, 1)
head = push(head, 3)
k = 3
print("Given list")
printList(head, 5)
makeloop(head, k)
print("\nModified list")
printList(head, 6)
Given list
Modified list
Time Complexity of how to create a loop in linked list: O(n), n is the total number of nodes in the list.
So, in this blog, we have tried to explain how to create a loop in linked list most efficiently with a great explanation and implementation. If you want to solve more questions on Linked List, which
is curated by our expert mentors at PrepBytes, you can follow this link Linked List.
FAQs related to how to create a loop in linked list
1. What is a linked list?
A linked list is a dynamic data structure in which each element (called a node) consists of two components: data and a reference (or pointer) to the next node. A linked list is a collection of nodes,
each of which is linked to the next node by a pointer.
2. Does the linked list have a loop?
A loop in a linked list is a condition that occurs when there is no end to the linked list. When a loop exists in a linked list, the last pointer does not point to the Null as in a singly or doubly
linked list or the head of the linked list as in a circular linked list.
3. What are the types of linked lists?
Types of Linked Lists are:
• Singly Linked list.
• Doubly Linked list.
• Circular Linked list.
• Doubly Circular Linked list.
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SAT Math Practice Test Online 8 Grid Ins Questions | SAT Online Classes AMBiPi -
SAT Math Practice Test Online 8 Grid Ins Questions | SAT Online Classes AMBiPi
Welcome to AMBiPi (Amans Maths Blogs). SAT (Scholastic Assessment Test) is a standard test, used for taking admission to undergraduate programs of universities or colleges in the United States. In
this article, you will get SAT 2022 Math Practice Test Online 8 Grid Ins Questions with Answer Keys | SAT Online Classes AMBiPi.
SAT 2022 Math Practice Test Online 8 Grid Ins Questions with Answer Keys
SAT Math Practice Online Test Question No 1:
x ≥ 0
3y – 2x ≥ -12
2x + 5y ≤ 20
What is the area of the triangle formed in the xy-plane by the system of inequalities above?
Show/Hide Answer Key
Correct Answer: 30
Since no picture has been provided, start by drawing the picture. To do so, change each of the equations into the slope-intercept form of an equation y = mx + b, where m is the slope and b is the
The second equation becomes 3y ≥ 2x – 12, or y ≥ (2/3)x – 4. The third equation becomes 5y ≤ -2x + 20, or y ≤ – (2/5)x + 4. The ≥ sign in the second equation means that everything above the line
should be shaded, and the ≤ sign in the third equation means that everything below that line should be shaded. To graph the first equation x ≥ 0, shade everything to the right of the y-axis. The
resulting picture should l look like this:
The formula for the area of a triangle is A = (1/2)b x h. It is easiest to think of the side that is along the y-axis as the base. That side goes from a y-coordinate of 4 to -4, for a length of 8.
The height of the triangle is the x-coordinate of the point where the two slanted lines meet; set the two equations equal to find it. Start with (2/3)x – 4 = (-2/5)x + 4 and multiply everything by 15
to get 10x – 60 = – 6x + 60. Then add 6x and 60 to both sides to get 16x = 120, so x = 7.5, and the height is 7.5. The resulting figure should look like this:
Plug the measurements for the base and the height into the area formula to get A = 1/2(8)(7.5) = 30. The correct answer is 30.
SAT Math Practice Online Test Question No 2:
The Nile is a track & field athlete at North Sherahan High School. He hopes to qualify for the Olympic Games in his best field event, the javelin throw. He experiments with different javelin weights
to build his arm strength and currently measures the results in feet.
During his preparations, Nile realizes that the upcoming Olympic qualifying competition will be judged in meters, rather than feet or yards. The Nile wants to make sure he can throw the javelin the
minimum required distance so he can advance in the competition. If his current best throw is 60 yards, and one yard is approximately 0.9144 meters, how much further, to the nearest yard, must he
throw to achieve the minimum required distance of 68.58 meters to qualify for the Olympics? (Disregard units when gridding your answer.)
Show/Hide Answer Key
Correct Answer: 15
Start by converting the qualifying distance of 68.58 meters into yards. Set up a proportion: 1 yard/0.9144 meters = x yards/68.58 meters.
Cross-multiply to get 0.9144x = 68.58. Divide both sides by 0.9144 to find that the qualifying distance is 75 yards. If his current best is 60 yards, he needs to throw 75 – 60 = 15 more yards.
SAT Math Practice Online Test Question No 3:
If (4 + 3i)(1 – 2i) = a + bi, then what is the value of a? (Note that i = √-1.)
Show/Hide Answer Key
Correct Answer: 10
Difficulty: Medium
Category: Additional Topics in Math / Imaginary Numbers
Strategic Advice: Multiply the two complex numbers just as you would two binomials (using FOIL). Then, combine like terms. The question tells you that i = √-1.
If you square both sides of the equation, this is the same as i2 = -1, which is a more useful fact.
Getting to the Answer:
(4 + 3i)(1 – 2i) = 4(1 – 2i) + 3i(1 – 2i)
= 4 – 8i + 3i – 6i^2
= 4 – 5i – 6(-1)
= 4 – 5i + 6 = 10 – 5i
The question asks for a in a + bi, so the correct answer is 10.
SAT Math Practice Online Test Question No 4:
18 – (3x)^1/2/2 = 15
What value of x satisfies the equation above?
Show/Hide Answer Key
Correct Answer: 12
Difficulty: Medium
Category: Passport to Advanced Math / Exponents
Strategic Advice: Solving an equation that has a fractional exponent can be very intimidating, so rewrite that part of the equation using a radical instead. Then, solve the equation the same way you
would any other: Isolate the variable using inverse operations, one step at a time.
Getting to the Answer: After rewriting the equation using a radical, start by subtracting 18 from both sides. Next, multiply both sides of the equation by -2. Then, square both sides to remove the
radical. Finally, divide both sides by 3.
18 – (3x)^1/2/2 = 15
18 – √3x/2 = 15
–√3x/2 = 15
√3x = 6
3x = 36
x = 12
SAT Math Practice Online Test Question No 5:
If the equation that represents the graph shown above is written in standard form, Ax + By = C, and A = 6, what is the value of B?
Show/Hide Answer Key
Correct Answer: 12
Difficulty: Easy
Category: Heart of Algebra / Linear Equations
Strategic Advice: The two things you can glean from the equation of a line are its slope and its y-intercept. In this question, you’re given information about A and asked about B. Try writing the
equation in slope-intercept form to see how A and B are related. Then look at the graph and see what you can add to this relationship.
Getting to the Answer: Start by writing the equation in slope-intercept form, y = mx + b.
Ax + By = C
By = -Ax + C
y = (-A/B)x + C/B
So, together A and B (specifically A over B) define the slope of the line. Look at the graph: Reading from left to right, the line falls 1 unit for every 2 units that it runs to the right, so the
slope is -1/2.
Don’t forget-the question tells you that A = 6, so set the slope equal to -6/B and solve for B:
-1/2 = -6/B
B = 12
SAT Math Practice Online Test Question No 6:
A toy saber is stuck at a right angle into the ground 4 inches deep. It casts a shadow that is 4 feet long. The brick wall casts a shadow three times that long. If the wall is 7 feet 6 inches tall,
how many inches long is the toy saber?
Show/Hide Answer Key
Correct Answer: 34
Difficulty: Hard
Category: Additional Topics in Math / Geometry
Strategic Advice: Drawing on the diagram is a great strategy to get started on a question like this. There are two right triangles-the smaller one formed by the saber, the path of the sun’s rays, and
the ground; and the larger one formed by the brick wall, the path of the sun’s rays, and the ground. The two triangles share one angle (the small angle on the left side), and each has a 90-degree
angle (where the saber and the brick wall each meet the ground), making the third pair of corresponding angles also congruent. This means the triangles are similar to AAA, and the sides of the
triangles are proportional.
Getting to the Answer: Add information from the question to the diagram. You’ll need to convert the height of the wall to inches because the question asks for the length of the saber in inches. (You
could also convert the base lengths to inches, but it is not necessary because you can compare feet to feet in that ratio.)
Now that you have a more detailed drawing, set up and solve a proportion:
(base of small triangle/ base of the large triangle) = length of saber(in inches)/ height of the wall(in inches)
4/12 = h/90
4(90) = 12h
360 = 12h
30 = h
Don’t forget to add the 4 inches that are stuck in the ground to find that the length of the saber is 30 + 4 = 34 inches.
SAT Math Practice Online Test Question No 7:
0 ≤ (1 – k)/2 < 7/8
If k lies within the solution set of the inequality shown above, what is the maximum possible value of k?
Show/Hide Answer Key
Correct Answer: 1
Difficulty: Medium
Category: Heart of Algebra / Inequalities
Strategic Advice: Whenever expressions involve fractions, you can clear the fractions by multiplying each term in the expression by the least common denominator. Don’t forget-when working with
inequalities, if you multiply or divide by a negative number, you must flip the inequality symbol(s).
Getting to the Answer: The inequality in this question is a compound inequality, but you don’t need to break it into parts. Just be sure that anything you do to one piece of the inequality, you do to
all three pieces. Start by multiplying everything by 8 to clear the fractions.
0 ≤ (1 – k)/2 < 7/8
0 ≤ 4(1 – k) < 7
0 ≤ 4 – 4k < 7
-4 ≤ -4k < 3
-4/-4 ≥ -4k/-4 > 3/-4
1 ≥ k > -3/4
Turn the inequality around so the numbers are increasing from left to right: -3/4 < k ≤ 1. This tells you that k is less than or equal to 1, making 1 the maximum possible value of k.
SAT Math Practice Online Test Question No 8:
In medicine, when a drug is administered in pill form, it takes time for the concentration in the bloodstream to build up, particularly for pain medications. Suppose for certain pain medication, the
function C(t) = 1.5t/t^2 + 4 is used to model the concentration, where t is the time in hours after the patient takes the pill. For this particular medication, the concentration reaches a maximum
level of 0.375 about two hours after it is administered and then begins to decrease. If the patient isn’t allowed to eat or drink until the concentration drops back down to 0.3, how many hours after
taking the pill must the patient wait before eating or drinking?
Show/Hide Answer Key
Correct Answer: 4
Difficulty: Hard
Category: Passport to Advanced Math / Functions
Strategic Advice: Sometimes in a real-world scenario, you need to think logically to get a mental picture of what is happening. Think about the concentration of the medicine-it starts at 0, increases
to a maximum of 0.375, and then decreases again as it begins to wear off. This means the concentration is 0.3 two times once before it hits the max and once after. In this case, you’re looking for
the second occurrence.
Getting to the Answer: Set the function equal to 0.3 and solve for t. Don’t stress out about the decimals-as soon as you have the equation in some kind of standard form, you can move the decimals to
get rid of them.
0.3 = 1.5t/t^2 + 4
0.3(t^2 + 4) = 1.5t
0.3t^2 + 1.2 = 1.5t
To make the equation easier to work with, move the decimal one place to the right in each term. The result is a fairly nice quadratic equation. Move everything to the left side, factor out a 3, and
go from there.
SAT Math Practice Online Test Question No 9:
(√x ⦁ x^5/6 ⦁ x)/∛x
If xn is the simplified form of the expression above, what is the value of n?
Show/Hide Answer Key
Correct Answer: 2
Difficulty: Hard
Category: Passport to Advanced Math / Exponents
Strategic Advice: Rewrite the radicals as fraction exponents: √x = x^1/2 and ∛x = x^1/3
Getting to the Answer: Write each factor in the expression in exponential form. Then use the rules of exponents to simplify the expression. Add the exponents of the factors that are being multiplied
and subtract the exponent of the factor that is being divided:
(x ⦁ x^5/6 ⦁ x)/∛x = (x^1/2 ⦁x^5/6⦁ x^1)/x
= x^1/2 + 5/6 + 1/1 – 1/3
= x^3/6 + 5/6 + 6/6 – 2/6
= x^12/6 = x^2
Because n is the power of x, the value of n is 2.
SAT Math Practice Online Test Question No 10:
The figure above shows the solution for the system of inequalities { y < -3x + 2 , y > x – 6}. Suppose (a, b) is a solution to the system. If a = 0, what is the greatest possible integer value of b?
Show/Hide Answer Key
Correct Answer: 1
Difficulty: Medium
Category: Heart of Algebra / Inequalities
Strategic Advice: If (a, b) is a solution to the system, then a is the x-coordinate of any point in the region where the shading overlaps and b is the corresponding y-coordinate.
Getting to the Answer: When a = 0 (or x = 0), the maximum possible value for b lies on the upper boundary line, y < -3x + 2. (You can tell which boundary line is the upper line by looking at the
y-intercept.) The point on the boundary line is (0, 2), but the boundary line is dashed (because the inequality is strictly less than), so you cannot include (0, 2) in the solution set. This means 1
is the greatest possible integer value for b when a = 0.
SAT 2022 Math Syllabus 51 SAT Algebra Questions with Answer Keys
SAT 2022 Math Formulas Notes 51 SAT Geometry Questions with Answer Keys
SAT 2022 Math Practice Test : Heart of Algebra 51 SAT Data Analysis Questions with Answer Keys
SAT 2022 Math Practice Test : Problems Solving & Data Analysis 51 SAT Graphs Questions with Answer Keys
SAT 2022 Math Practice Test : Passport to Advance Math 51 SAT Functions Questions with Answer Keys
SAT 2022 Math Practice Test : Additional Topics in Math 51 SAT Inequalities Questions with Answer Keys
SAT Math Online Course (30 Hours LIVE Classes) SAT Math Full Practice Tests
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1.3: Types of Data and How to Collect Them
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In order to use statistics, we need data to analyze. Data come in an amazingly diverse range of formats, and each type gives us a unique type of information. In virtually any form, data represent the
measured value of variables. A variable is simply a characteristic or feature of the thing we are interested in understanding. In psychology, we are interested in people, so we might get a group of
people together and measure their levels of stress (one variable), anxiety (a second variable), and their physical health (a third variable). Once we have data on these three variables, we can use
statistics to understand if and how they are related. Before we do so, we need to understand the nature of our data: what they represent and where they came from.
Types of Variables
When conducting research, experimenters often manipulate variables. For example, an experimenter might compare the effectiveness of four types of antidepressants. In this case, the variable is “type
of antidepressant.” When a variable is manipulated by an experimenter, it is called an independent variable. The experiment seeks to determine the effect of the independent variable on relief from
depression. In this example, relief from depression is called a dependent variable. In general, the independent variable is manipulated by the experimenter and its effects on the dependent variable
are measured.
Can blueberries slow down aging? A study indicates that antioxidants found in blueberries may slow down the process of aging. In this study, \(19\)-month-old rats (equivalent to \(60\)-year-old
humans) were fed either their standard diet or a diet supplemented by either blueberry, strawberry, or spinach powder. After eight weeks, the rats were given memory and motor skills tests. Although
all supplemented rats showed improvement, those supplemented with blueberry powder showed the most notable improvement.
1. What is the independent variable? (dietary supplement: none, blueberry, strawberry, and spinach)
2. What are the dependent variables? (memory test and motor skills test)
Does beta-carotene protect against cancer? Beta-carotene supplements have been thought to protect against cancer. However, a study published in the Journal of the National Cancer Institute suggests
this is false. The study was conducted with 39,000 women aged 45 and up. These women were randomly assigned to receive a beta-carotene supplement or a placebo, and their health was studied over their
lifetime. Cancer rates for women taking the betacarotene supplement did not differ systematically from the cancer rates of those women taking the placebo.
1. What is the independent variable? (supplements: beta-carotene or placebo)
2. What is the dependent variable? (occurrence of cancer)
How bright is right? An automobile manufacturer wants to know how bright brake lights should be in order to minimize the time required for the driver of a following car to realize that the car in
front is stopping and to hit the brakes.
1. What is the independent variable? (brightness of brake lights)
2. What is the dependent variable? (time to hit brakes)
Levels of an Independent Variable
If an experiment compares an experimental treatment with a control treatment, then the independent variable (type of treatment) has two levels: experimental and control. If an experiment were
comparing five types of diets, then the independent variable (type of diet) would have \(5\) levels. In general, the number of levels of an independent variable is the number of experimental
Qualitative and Quantitative Variables
An important distinction between variables is between qualitative variables and quantitative variables. Qualitative variables are those that express a qualitative attribute such as hair color, eye
color, religion, favorite movie, gender, and so on. The values of a qualitative variable do not imply a numerical ordering. Values of the variable “religion” differ qualitatively; no ordering of
religions is implied. Qualitative variables are sometimes referred to as categorical variables. Quantitative variables are those variables that are measured in terms of numbers. Some examples of
quantitative variables are height, weight, and shoe size.
In the study on the effect of diet discussed previously, the independent variable was type of supplement: none, strawberry, blueberry, and spinach. The variable “type of supplement” is a qualitative
variable; there is nothing quantitative about it. In contrast, the dependent variable “memory test” is a quantitative variable since memory performance was measured on a quantitative scale (number
Discrete and Continuous Variables
Variables such as number of children in a household are called discrete variables since the possible scores are discrete points on the scale. For example, a household could have three children or six
children, but not \(4.53\) children. Other variables such as “time to respond to a question” are continuous variables since the scale is continuous and not made up of discrete steps. The response
time could be \(1.64\) seconds, or it could be \(1.64237123922121\) seconds. Of course, the practicalities of measurement preclude most measured variables from being truly continuous.
Levels of Measurement
Before we can conduct a statistical analysis, we need to measure our dependent variable. Exactly how the measurement is carried out depends on the type of variable involved in the analysis. Different
types are measured differently. To measure the time taken to respond to a stimulus, you might use a stop watch. Stop watches are of no use, of course, when it comes to measuring someone's attitude
towards a political candidate. A rating scale is more appropriate in this case (with labels like “very favorable,” “somewhat favorable,” etc.). For a dependent variable such as “favorite color,” you
can simply note the color-word (like “red”) that the subject offers.
Although procedures for measurement differ in many ways, they can be classified using a few fundamental categories. In a given category, all of the procedures share some properties that are important
for you to know about. The categories are called “scale types,” or just “scales,” and are described in this section.
Nominal scales
When measuring using a nominal scale, one simply names or categorizes responses. Gender, handedness, favorite color, and religion are examples of variables measured on a nominal scale. The essential
point about nominal scales is that they do not imply any ordering among the responses. For example, when classifying people according to their favorite color, there is no sense in which green is
placed “ahead of” blue. Responses are merely categorized. Nominal scales embody the lowest level of measurement.
Ordinal scales
A researcher wishing to measure consumers' satisfaction with their microwave ovens might ask them to specify their feelings as either “very dissatisfied,” “somewhat dissatisfied,” “somewhat
satisfied,” or “very satisfied.” The items in this scale are ordered, ranging from least to most satisfied. This is what distinguishes ordinal from nominal scales. Unlike nominal scales, ordinal
scales allow comparisons of the degree to which two subjects possess the dependent variable. For example, our satisfaction ordering makes it meaningful to assert that one person is more satisfied
than another with their microwave ovens. Such an assertion reflects the first person's use of a verbal label that comes later in the list than the label chosen by the second person.
On the other hand, ordinal scales fail to capture important information that will be present in the other scales we examine. In particular, the difference between two levels of an ordinal scale
cannot be assumed to be the same as the difference between two other levels. In our satisfaction scale, for example, the difference between the responses “very dissatisfied” and “somewhat
dissatisfied” is probably not equivalent to the difference between “somewhat dissatisfied” and “somewhat satisfied.” Nothing in our measurement procedure allows us to determine whether the two
differences reflect the same difference in psychological satisfaction. Statisticians express this point by saying that the differences between adjacent scale values do not necessarily represent equal
intervals on the underlying scale giving rise to the measurements. (In our case, the underlying scale is the true feeling of satisfaction, which we are trying to measure.)
What if the researcher had measured satisfaction by asking consumers to indicate their level of satisfaction by choosing a number from one to four? Would the difference between the responses of one
and two necessarily reflect the same difference in satisfaction as the difference between the responses two and three? The answer is No. Changing the response format to numbers does not change the
meaning of the scale. We still are in no position to assert that the mental step from \(1\) to \(2\) (for example) is the same as the mental step from \(3\) to \(4\).
Interval scales
Interval scales are numerical scales in which intervals have the same interpretation throughout. As an example, consider the Fahrenheit scale of temperature. The difference between \(30\) degrees and
\(40\) degrees represents the same temperature difference as the difference between \(80\) degrees and \(90\) degrees. This is because each \(10\)-degree interval has the same physical meaning (in
terms of the kinetic energy of molecules).
Interval scales are not perfect, however. In particular, they do not have a true zero point even if one of the scaled values happens to carry the name “zero.” The Fahrenheit scale illustrates the
issue. Zero degrees Fahrenheit does not represent the complete absence of temperature (the absence of any molecular kinetic energy). In reality, the label “zero” is applied to its temperature for
quite accidental reasons connected to the history of temperature measurement. Since an interval scale has no true zero point, it does not make sense to compute ratios of temperatures. For example,
there is no sense in which the ratio of \(40\) to \(20\) degrees Fahrenheit is the same as the ratio of \(100\) to \(50\) degrees; no interesting physical property is preserved across the two ratios.
After all, if the “zero” label were applied at the temperature that Fahrenheit happens to label as \(10\) degrees, the two ratios would instead be \(30\) to \(10\) and \(90\) to \(40\), no longer the
same! For this reason, it does not make sense to say that \(80\) degrees is “twice as hot” as \(40\) degrees. Such a claim would depend on an arbitrary decision about where to “start” the temperature
scale, namely, what temperature to call zero (whereas the claim is intended to make a more fundamental assertion about the underlying physical reality).
Ratio scales
The ratio scale of measurement is the most informative scale. It is an interval scale with the additional property that its zero position indicates the absence of the quantity being measured. You can
think of a ratio scale as the three earlier scales rolled up in one. Like a nominal scale, it provides a name or category for each object (the numbers serve as labels). Like an ordinal scale, the
objects are ordered (in terms of the ordering of the numbers). Like an interval scale, the same difference at two places on the scale has the same meaning. And in addition, the same ratio at two
places on the scale also carries the same meaning.
The Fahrenheit scale for temperature has an arbitrary zero point and is therefore not a ratio scale. However, zero on the Kelvin scale is absolute zero. This makes the Kelvin scale a ratio scale. For
example, if one temperature is twice as high as another as measured on the Kelvin scale, then it has twice the kinetic energy of the other temperature.
Another example of a ratio scale is the amount of money you have in your pocket right now (25 cents, 55 cents, etc.). Money is measured on a ratio scale because, in addition to having the properties
of an interval scale, it has a true zero point: if you have zero money, this implies the absence of money. Since money has a true zero point, it makes sense to say that someone with 50 cents has
twice as much money as someone with 25 cents (or that Bill Gates has a million times more money than you do).
What level of measurement is used for psychological variables?
Rating scales are used frequently in psychological research. For example, experimental subjects may be asked to rate their level of pain, how much they like a consumer product, their attitudes about
capital punishment, their confidence in an answer to a test question. Typically these ratings are made on a 5-point or a 7-point scale. These scales are ordinal scales since there is no assurance
that a given difference represents the same thing across the range of the scale. For example, there is no way to be sure that a treatment that reduces pain from a rated pain level of 3 to a rated
pain level of 2 represents the same level of relief as a treatment that reduces pain from a rated pain level of 7 to a rated pain level of 6.
In memory experiments, the dependent variable is often the number of items correctly recalled. What scale of measurement is this? You could reasonably argue that it is a ratio scale. First, there is
a true zero point; some subjects may get no items correct at all. Moreover, a difference of one represents a difference of one item recalled across the entire scale. It is certainly valid to say that
someone who recalled 12 items recalled twice as many items as someone who recalled only 6 items.
But number-of-items recalled is a more complicated case than it appears at first. Consider the following example in which subjects are asked to remember as many items as possible from a list of 10.
Assume that (a) there are 5 easy items and 5 difficult items, (b) half of the subjects are able to recall all the easy items and different numbers of difficult items, while (c) the other half of the
subjects are unable to recall any of the difficult items but they do remember different numbers of easy items. Some sample data are shown below.
Table \(\PageIndex{1}\): Sample Data
Subject Easy Difficult Items Score
A 0 0 1 1 0 0 0 0 0 0 2
B 1 0 1 1 0 0 0 0 0 0 3
C 1 1 1 1 1 1 1 0 0 0 7
D 1 1 1 1 1 0 1 1 0 1 8
Let's compare (i) the difference between Subject A's score of 2 and Subject B's score of 3 and (ii) the difference between Subject C's score of 7 and Subject D's score of 8. The former difference is
a difference of one easy item; the latter difference is a difference of one difficult item. Do these two differences necessarily signify the same difference in memory? We are inclined to respond “No”
to this question since only a little more memory may be needed to retain the additional easy item whereas a lot more memory may be needed to retain the additional hard itemred. The general point is
that it is often inappropriate to consider psychological measurement scales as either interval or ratio.
Consequences of level of measurement Why are we so interested in the type of scale that measures a dependent variable? The crux of the matter is the relationship between the variable's level of
measurement and the statistics that can be meaningfully computed with that variable. For example, consider a hypothetical study in which 5 children are asked to choose their favorite color from blue,
red, yellow, green, and purple. The researcher codes the results as follows:
Table \(\
\): Favorite
color data
Color Code
Blue 1
Red 2
Yellow 3
Green 4
Purple 5
This means that if a child said her favorite color was “Red,” then the choice was coded as “2,” if the child said her favorite color was “Purple,” then the response was coded as 5, and so forth.
Consider the following hypothetical data
Table \(\PageIndex{3}
\): Favorite color
Subject Color Code
1 Blue 1
2 Blue 1
3 Green 4
4 Green 4
5 Purple 5
Each code is a number, so nothing prevents us from computing the average code assigned to the children. The average happens to be 3, but you can see that it would be senseless to conclude that the
average favorite color is yellow (the color with a code of 3). Such nonsense arises because favorite color is a nominal scale, and taking the average of its numerical labels is like counting the
number of letters in the name of a snake to see how long the beast is.
Does it make sense to compute the mean of numbers measured on an ordinal scale? This is a difficult question, one that statisticians have debated for decades. The prevailing (but by no means
unanimous) opinion of statisticians is that for almost all practical situations, the mean of an ordinally-measured variable is a meaningful statistic. However, there are extreme situations in which
computing the mean of an ordinally-measured variable can be very misleading. | {"url":"https://stats.libretexts.org/Bookshelves/Applied_Statistics/An_Introduction_to_Psychological_Statistics_(Foster_et_al.)/01%3A_Introduction/1.03%3A_Types_of_Data_and_How_to_Collect_Them","timestamp":"2024-11-08T05:04:20Z","content_type":"text/html","content_length":"159248","record_id":"<urn:uuid:1768e74e-5f10-40f0-b61a-a0bf8f2e453e>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00275.warc.gz"} |
Missing Data
3.5. Missing Data
When mean or filtered values are to be calculated, the question of how to handle missing data arises. For a number of reasons it is difficult to devise a simple objective rule that can be applied to
all cases. INTERMAGNET recommends a simple and pragmatic approach: mean or filtered values may be calculated when 90% or more of the values required for calculation are available. This can be
interpreted as either 90% of the values or 90% of the weight of the filter. When fewer than 90% of the required values are available the value should be assigned the value used to flag missing data.
INTERMAGNET recommends adoption of this rule for both simple mean and weighted mean calculations. For example, a simple daily mean value may be computed when 1296 or more one-minute values are
available for the day. Similarly, if a one-minute value is constructed from one-second samples, the one-minute value may be computed when 54 or more one-second samples are available. In either case
the weights applied to each sample in the mean or the filter must be re-normalized to account for the reduced number of samples available. In practice, this means dividing the sum of samples by the
number of available samples in the case of a simple mean or normalizing to unity those coefficients that have been used in a filter calculation. INTERMAGNET observatories are expected to provide high
levels of data continuity, so this rule is expected to be applied only rarely. To avoid the propagation of missing values into higher level means, it is recommended to calculate all higher level
means using the method described in Section 6.6. | {"url":"https://tech-man.intermagnet.org/stable/chapters/onesecondimos/missingdata.html","timestamp":"2024-11-01T22:06:49Z","content_type":"text/html","content_length":"11287","record_id":"<urn:uuid:b6edf3a2-2709-43bc-af34-e05abffbdf9e>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00669.warc.gz"} |
What is the formula for buck boost converter?
What is the formula for buck boost converter?
What is the formula for output voltage for Buck-Boost converter? Explanation: The output voltage of the buck-boost converter is Vo = D×Vin ÷ (1-D). It can step up and step down the voltage depending
upon the value of the duty cycle.
What is DC DC bidirectional converter?
Abstract. Bidirectional dc to dc converter is used as a key device for interfacing the storage devices between source and load in renewable energy system for continuous flow of power because the
output of the renewable energy system fluctuates due to change in weather conditions.
How do you calculate the value of a buck boost converter inductor?
When selecting an inductor for a buck converter the following parameters need to be defined:
1. Maximum input voltage = Vin max.
2. Minimum input voltage = Vin min.
3. Maximum output current = Iout max.
4. Operating frequency = f.
5. Output voltage = Vout.
6. Minimum output current = Iout min.
What is the duty cycle for a buck-boost converter in buck mode?
The input voltage is 100 V DC and the duty cycle is 0.5.
What does a DC-DC converter do?
DC-DC converters are high-frequency power conversion circuits that use high-frequency switching and inductors, transformers, and capacitors to smooth out switching noise into regulated DC voltages.
Closed feedback loops maintain constant voltage output even when changing input voltages and output currents.
What is a DC to DC converter How does it work?
How does a DC Converter Work? The basic DC-DC converter will take the current and pass it through a “switching element”. This turns the signal into a square wave, which is actually AC. The wave then
passes through another filter, which turns it back into a DC signal of the appropriate voltage necessary.
Is a buck-boost converter bidirectional?
The PMP21529 design is such a bidirectional DC-DC power converter specifically designed for battery backup system where battery voltage range crosses DC bus voltage. Only when DC bus and battery
voltage are very close then the converter operates as a buck-boost converter where all the four switching are switching.
What is the use of capacitor in buck converter?
This enables efficient, high frequency operation and significantly smaller solution size. The series capacitor buck converter has beneficial characteristics such as lower switching loss, less
inductor current ripple, automatic inductor current balancing, duty ratio extension, and soft charging of the series capacitor. | {"url":"https://richardvigilantebooks.com/what-is-the-formula-for-buck-boost-converter/","timestamp":"2024-11-13T16:01:02Z","content_type":"text/html","content_length":"47118","record_id":"<urn:uuid:572ce73a-76cf-45cc-8439-653f3061cc65>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00554.warc.gz"} |
geometry edge group command
geometry edge group s <keyword ...> <range>
Assign the group s to the edges of the geometric set that fall within the range. Use of the group logic is described in Groups. This command may take the form geometry edge group "slotname =
groupname" — quotation marks required, spaces at equals sign ignored — where slotname is the slot assignment and groupname is the group name. If the set keyword is not given, then the current set
is used.
Remove s from all edges. If no slot was specified, then the group will be removed from all slots it is found in.
set s
Specify the geometry set.
slot slot
Set group slot slot to s. By default, slot Default is used.
Was this helpful? ... PFC © 2021, Itasca Updated: Feb 25, 2024 | {"url":"https://docs.itascacg.com/pfc700/common/geometry/doc/manual/commands/cmd_geometry.edge.group.html","timestamp":"2024-11-04T23:19:46Z","content_type":"application/xhtml+xml","content_length":"14496","record_id":"<urn:uuid:f73bb0b7-ea4c-41e9-9b19-0d9c0fa612bf>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00765.warc.gz"} |
Newton's Laws and Force Diagrams - AP Physics C: Mechanics
All AP Physics C: Mechanics Resources
Example Questions
Example Question #1 : Understanding Newton's Laws
Correct answer:
Relevant equations:
Determine the clockwise torque caused by the bucket.
Write an expression for the counterclockwise torque caused by the rope.
Combine the torque of the rope and the torque of the bucket into the net torque equation.
Since the system has no angular acceleration, net torque must be zero, allowing us to solve for the force of the rope.
Example Question #1 : Newton's Laws And Force Diagrams
An object at rest will remain at rest unless acted on by a(n) __________.
Correct answer:
external force
The correct answer is external force. External forces applied to an object will result in non-zero acceleration, causing the object to move. In contrast, internal forces contribute to the properties
of the object and do not result in acceleration of the object.
Either positive or negative forces can result in the acceleration of an object. The sign on the force simply conveys information about its relative direction.
Example Question #2 : Newton's Laws And Force Diagrams
A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to
the force produced by the engine. What is the acceleration of the rocket?
Correct answer:
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that
Finally we solve for acceleration:
Example Question #3 : Newton's Laws And Force Diagrams
Three boxes tied by two ropes move across a frictionless surface pulled by a force
Correct answer:
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between
For the box on the right (of mass
We get the equation:
For the middle box (of mass
So we get the equation:
Finally, for the box on the left (of mass
So we get the equation:
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate
Substitute the expression of
Use this value of
So we have:
Solve for acceleration.
Example Question #4 : Newton's Laws And Force Diagrams
A box is being pushed against a wall by a force
Correct answer:
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
Example Question #5 : Newton's Laws And Force Diagrams
A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?
Correct answer:
You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.
We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From
trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.
We can solve for the horizontal component of the force using the applied force as the hypotenuse:
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Lecture Notes on the Lambda Calculus
by Peter Selinger
Publisher: Dalhousie University 2007
Number of pages: 106
Topics covered in these notes include the untyped lambda calculus, the Church-Rosser theorem, combinatory algebras, the simply-typed lambda calculus, the Curry-Howard isomorphism, weak and strong
normalization, type inference, denotational semantics, complete partial orders, and the language PCF.
Download or read it online for free here:
Download link
(460KB, PDF)
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American Mathematical Society
Non-local logistic equations from the probability viewpoint
Author: M. D’Ovidio
Journal: Theor. Probability and Math. Statist. 104 (2021), 77-87
MSC (2020): Primary 60H30, 26A33; Secondary 34L30, 11B68
DOI: https://doi.org/10.1090/tpms/1146
Published electronically: September 24, 2021
Full-text PDF
Abstract | References | Similar Articles | Additional Information
Abstract: We investigate the solution to the logistic equation involving non-local operators in time. In the linear case such operators lead to the well-known theory of time changes. We provide the
probabilistic representation for the non-linear logistic equation with non-local operators in time. The so-called fractional logistic equation has been investigated by many researchers, the problem
to find the explicit representation of the solution on the whole real line is still open. In our recent work the solution on compact sets has been written in terms of Euler’s numbers.
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Retrieve articles in Theory of Probability and Mathematical Statistics with MSC (2020): 60H30, 26A33, 34L30, 11B68
Retrieve articles in all journals with MSC (2020): 60H30, 26A33, 34L30, 11B68
Additional Information
M. D’Ovidio
Affiliation: Department of Basic and applied Sciences for Engineering, Sapienza University of Rome, Italy
Email: mirko.dovidio@uniroma1.it
Keywords: Logistic equations, non-local operators, subordinators
Received by editor(s): January 15, 2021
Published electronically: September 24, 2021
Additional Notes: The author was supported in part by INDAM-GNAMPA and the Grant Ateneo “Sapienza 2019”
Article copyright: © Copyright 2021 Taras Shevchenko National University of Kyiv | {"url":"https://www.ams.org/journals/tpms/2021-104-00/S0094-9000-2021-01146-2/","timestamp":"2024-11-11T21:15:41Z","content_type":"text/html","content_length":"90547","record_id":"<urn:uuid:3c785edb-9131-4c36-8178-d3b4414dad21>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00859.warc.gz"} |
BEGIN:VCALENDAR VERSION:2.0 PRODID:ILLC Website X-WR-TIMEZONE:Europe/Amsterdam BEGIN:VTIMEZONE TZID:Europe/Amsterdam X-LIC-LOCATION:Europe/Amsterdam BEGIN:DAYLIGHT TZOFFSETFROM:+0100 TZOFFSETTO:+0200
TZNAME:CEST DTSTART:19700329T020000 RRULE:FREQ=YEARLY;BYMONTH=3;BYDAY=-1SU END:DAYLIGHT BEGIN:STANDARD TZOFFSETFROM:+0200 TZOFFSETTO:+0100 TZNAME:CET DTSTART:19701025T030000 RRULE:FREQ=YEARLY;BYMONTH
=10;BYDAY=-1SU END:STANDARD END:VTIMEZONE BEGIN:VEVENT UID:/NewsandEvents/Archives/current/newsitem/14762 /14-June-2024-FOAM-Seminar-Bernhard-Nebel DTSTAMP:20240212T142507 SUMMARY:FOAM Seminar,
Bernhard Nebel ATTENDEE;ROLE=Speaker:Bernhard Nebel (Freiburg) DTSTART;TZID=Europe/Amsterdam:20240614T150000 DTEND;TZID=Europe/Amsterdam:20240614T161500 LOCATION:Room L3.33, ILLC Lab42, Science Park
900, Amsterdam DESCRIPTION:Abstract: “Multi-agent pathfinding”, also called “pebble motion on graphs” or “cooperat ive pathfinding”, is the problem of deciding the e xistence of or generating a
collision-free movemen t plan for a set of agents moving on a graph. Whil e the non-optimizing variant of multi-agent pathfi nding on undirected graphs is known to be a polyno mial-time problem since
forty years, a similar res ult for directed graphs was missing. In the talk, it will be shown that this problem is NP-complete. For strongly connected directed graphs, however, the problem is
polynomial. And both of these resul ts hold even if one allows for synchronous rotatio ns on fully occupied cycles. X-ALT-DESC;FMTTYPE=text/html:\n
\ n “Multi-agent pathfinding”, also called “pebble motion on graphs” or “cooperative pathfinding”, is the problem of deciding the existence of or gener ating a collision-free movement plan for a set
of agents moving on a graph. While the non-optimizing variant of multi-agent pathfinding on undirected graphs is known to be a polynomial-time problem si nce forty years, a similar result for
directed gra phs was missing. In the talk, it will be shown tha t this problem is NP-complete. For strongly connec ted directed graphs, however, the problem is polyn omial. And both of these results
hold even if one allows for synchronous rotations on fully occupied cycles.
URL:https://events.illc.uva.nl/FOAM/posts/talk15/ CONTACT:Gregor Behnke at g.behnke at uva.nl CONTACT:Ronald de Haan at r.dehaan at uva.nl END:VEVENT END:VCALENDAR | {"url":"https://www.illc.uva.nl/NewsandEvents/Events/Upcoming-Events/newsitem/14762/14-June-2024-FOAM-Seminar-Bernhard-Nebel?displayMode=ical","timestamp":"2024-11-03T20:19:52Z","content_type":"text/calendar","content_length":"3075","record_id":"<urn:uuid:9f07501f-5103-4418-bc1a-e8c6ac61d67f>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00220.warc.gz"} |
How To Calculate APR On A Credit Card: A Comprehensive Guide - CalculatorBox
How To Calculate APR On A Credit Card: A Comprehensive Guide
APR, or Annual Percentage Rate, is a crucial aspect of credit cards that you should be aware of, as it affects the amount of interest you pay on your outstanding balances. The APR is an annual rate,
meaning it represents the interest charges over an entire year.
Fun Fact: Did you know that the APR (Annual Percentage Rate) doesn’t just cover the interest rate on your credit card? It also takes into account any additional fees you may be charged, giving you a
more complete picture of the true cost of borrowing. That’s why comparing APRs can be a more effective way to evaluate credit card offers than looking at interest rates alone.
To calculate the APR on your credit card, follow these steps:
1. Locate your credit card’s current APR. You can usually find this on your credit card statement, or by logging into your account online.
2. Next, divide the APR by 12 (for the twelve months in a year) to find the monthly periodic rate. This rate can help you better understand how much interest is being applied to your balance each
3. Once you have the monthly periodic rate, you can multiply it by your current outstanding balance to determine the interest you’ll be charged for that month.
For example, let’s say your credit card has an APR of 18% and your current balance is $1,000. You would divide 18% by 12, resulting in a monthly periodic rate of 1.5%. Multiplying that rate by $1,000
gives you an interest charge of $15 for the month.
Keep in mind the methodology may differ slightly between credit card issuers. Some may use a daily periodic rate instead of a monthly rate. To calculate the daily periodic rate, you would divide your
APR by 365 days instead of 12 months.
Remember that your credit history, credit score, and the type of credit card can also influence the APR you’re offered. Always be mindful of your credit card’s terms, use your card responsibly, and
pay your balance off in full if possible to minimize the impact of APR on your finances.
Calculating APR on Credit Cards
When you want to understand the cost of borrowing money through credit cards, it’s essential to learn how to calculate the Annual Percentage Rate (APR). The APR allows you to make informed decisions
when comparing different credit cards before deciding on one to use.
Find your current APR and balance
Locate your current APR and balance in your credit card statement.
Divide your current APR by 12
Since there are 12 months in a year, divide the annual rate by 12 to find your monthly periodic rate. You can use this formula:
Monthly\space Periodic \space Rate = \frac{APR}{12}
Calculate your Daily Periodic Rate (DPR)
Some credit card issuers use daily rates instead of monthly rates. To calculate your DPR, divide your credit card’s APR by 365 (or 360, depending on the issuer). For example:
Daily\space Periodic \space Rate = \frac{APR}{365}
View your APR
By calculating these rates, you can get a better understanding of the total cost of borrowing with your credit card, making it easier for you to compare different cards.
Note that cash advances usually have different and higher APRs than regular purchases, and interest begins to accumulate immediately with no grace period.
Calculating APR on Loans
To calculate the annual percentage rate (APR) for a loan, follow these steps:
Identify the interest charges and fees
Determine the total cost of the loan, including interest charges and any fees associated with the loan.
Divide by the loan amount
Take the sum of the loan’s interest charges and fees and divide it by the loan amount. This will give you the rate for the duration of the loan term.
Divide the result by the number of days in the loan term
Take the rate from step 2 and divide it by the number of days in the loan term. This will give you the daily rate of the loan.
Multiply by 365 to get the annual rate
Take the daily rate from step 3 and multiply it by 365 to convert it into an annual rate.
Multiply by 100 to get the rate in the form of a percentage
Finally, take the annual rate and multiply it by 100 to express it as a percentage.
Remember, these steps are focused on calculating the APR for a loan, not for a credit card. When calculating APR for a credit card, the process may be slightly different.
Monthly APR Calculation
To calculate the monthly APR (Annual Percentage Rate) on your credit card, follow these steps:
Find your current APR and balance
Locate your current APR and outstanding balance in your credit card statement.
Divide the APR by 12
Since there are twelve months in a year, divide your current APR by 12 to find your monthly periodic rate. For example, if your APR is 18%, the calculation would be 0.18/12 = 0.015.
Calculate the monthly interest charge
Multiply the monthly periodic rate found in step 2 with the amount of your outstanding balance. For instance, if your balance is $1,000 and the monthly periodic rate is 0.015, then your monthly
interest charge would be $15 (1000 x 0.015).
Calculating APR Interest
To calculate the APR interest on your credit card, you need to first convert your APR to a daily interest rate and then apply it to your credit card balance. This will help you understand how
interest is accrued on your credit card balance daily.
Convert APR To Daily Interest Rate
To convert your annual percentage rate (APR) to a daily interest rate, divide the APR by 365 (the number of days in a year):
Daily\space Interest \space Rate = \frac{APR}{365}
For example, if your credit card has an APR of 18%, the calculation would be:
Daily\space Interest \space Rate = \frac{18\%}{365} = 0.0493\%
This means that the daily interest rate for this credit card is 0.0493%.
Now that you have your daily interest rate, you can calculate the interest that will be charged on your credit card balance. To do this, multiply your average daily balance by the daily interest
rate, and then multiply that by the number of days in the billing cycle:
\small Interest = Average \space Daily \space Balance \space× Daily \space Interest \space Rate \space × Number \space of \space Days \space in \space Billing \space Cycle
For example, if your average daily balance is $1,000 and the billing cycle is 30 days:
Interest = \text{\textdollar}1,000 \space × 0.000493× 30 = \text{\textdollar}14.79
In this example, the interest charged on your credit card balance for the billing cycle will be $14.79.
Remember to make your calculations using the appropriate numbers for your credit card, such as the correct APR and billing cycle length. This will give you a more accurate estimate of the interest
charges you can expect.
Frequently Asked Questions
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Introduction to Descriptive Statistics | Math in Science | Visionlearning
Introduction to Descriptive Statistics: [Using mean, median, and standard deviation]
by Liz Roth-Johnson, Ph.D.
Listen to this reading
Did you know that the mathematical equation used by instructors to "grade on the curve” was first developed to aid gamblers in games of chance? This is just one of several statistical operations used
by scientists to analyze and interpret data. These descriptive statistics are used in many fields. They can help scientists summarize everything from the results of a drug trial to the way genetic
traits evolve over different generations.
• Basic statistical operations such as mean, median, and standard deviation help scientists quickly summarize the major characteristics of a dataset.
• A normal distribution is a type of probability distribution in which the probability of observing any specific value is evenly distributed about the mean of the dataset. In many scientific
applications, the statistical error in experimental measurements and the natural variation within a population are approximated as normal distributions.
• Standard deviation provides a measurement of the “spread” of a dataset, or how much individual values in a dataset vary from the mean. This “spread” of data helps scientists summarize how much
variation there is in a dataset or population.
the mathematical study of data
a collection of measurements and observations that can be analyzed
the variation within a dataset; the measure of how much individual values in a dataset differ from the mean, or average
Imagine yourself in an introductory science course. You recently completed the first exam, and are now sitting in class waiting for your graded exam to be handed back. The course will be graded “on a
curve,” so you are anxious to see how your score compares to everyone else’s. Your instructor finally arrives and shares the exam statistics for the class (see Figure 1).
The mean score is 61.
The median is 63.
The standard deviation is 12.
You receive your exam and see that you scored 72. What does this mean in relation to the rest of the class? Based on the statistics above, you can see that your score is higher than the mean and
median, but how do all of these numbers relate to your final grade? In this scenario, you would end up with a “B” letter grade, even though the numerical score would equal a “C” without the curve.
Figure 1: A histogram showing the distribution of exam scores for the 200 students in your class, along with the mean, median, and your score.
This scenario shows how descriptive statistics – namely the mean, median, and standard deviation – can be used to quickly summarize a dataset. By the end of this module, you will learn not only how
descriptive statistics can be used to assess the results of an exam, but also how scientists use these basic statistical operations to analyze and interpret their data. Descriptive statistics can
help scientists summarize everything from the results of a drug trial to the way genetic traits evolve from one generation to the next.
What are descriptive statistics?
Descriptive statistics are used regularly by scientists to succinctly summarize the key features of a dataset or population. Three statistical operations are particularly useful for this purpose: the
mean, median, and standard deviation. (For more information about why scientists use statistics in science, see our module Statistics in Science.)
Mean vs. median
The mean and median both provide measures of the central tendency of a set of individual measurements. In other words, the mean and median roughly approximate the middle value of a dataset. As we saw
above, the mean and median exam scores fell roughly in the center of the grade distribution.
Although the mean and median provide similar information about a dataset, they are calculated in different ways: The mean, also sometimes called the average or arithmetic mean, is calculated by
adding up all of the individual values (the exam scores in this example) and then dividing by the total number of values (the number of students who took the exam). The median, on the other hand, is
the “middle” value of a dataset. In this case, it would be calculated by arranging all of the exam scores in numerical order and then choosing the value in the middle of the dataset.
Because of the way the mean and median are calculated, the mean tends to be more sensitive to outliers – values that are dramatically different from the majority of other values. In the example above
(Figure 1), the median fell slightly closer to the middle of the grade distribution than did the mean. The 4 students who missed the exam and scored 0 (the outliers) lowered the mean by getting such
different scores from the rest of the class. However, the median did not change as much because there were so few students who missed the exam compared to the total number of students in the class.
Comprehension Checkpoint
Standard deviation
The standard deviation measures how much the individual measurements in a dataset vary from the mean. In other words, it gives a measure of variation, or spread, within a dataset. Typically, the
majority of values in a dataset fall within a range comprising one standard deviation below and above the mean. In the example above, the standard deviation is 12 and the majority of test scores (161
out of 200 students) scored between 49 and 73 points on the exam. If there had been more variation in the exam scores, the standard deviation would have been even larger. Conversely, if there had
been less variation, the standard deviation would have been smaller. For example, let’s consider the exam scores earned by students in two different classes (Figure 2).
Figure 2: Two exam score distributions with different standard deviations. Although the mean score for both classes is 50, the standard deviation, or spread, of scores is very different. The standard
deviation for Class A is 5 (small spread), while the standard deviation for Class B is 15 (large spread).
In the first class (Class A – the light blue bars in the figure), all of the students studied together in a large study group and received similar scores on the final exam. In the second class (Class
B – represented by dark blue bars), all of the students studied independently and received a wide range of scores on the final exam. Although the mean grade was the same for both classes (50), Class
A has a much smaller standard deviation (5) than Class B (15).
Comprehension Checkpoint
Normal distribution
Sometimes a dataset exhibits a particular shape that is evenly distributed around the mean. Such a distribution is called a normal distribution. It can also be called a Gaussian distribution or a
bell curve. Although exam grades are not always distributed in this way, the phrase “grading on a curve” comes from the practice of assigning grades based on a normally distributed bell curve. Figure
3 shows how the exam scores shown in Figure 1 can be approximated by a normal distribution. By straight grading standards, the mean test score (61) would typically receive a D-minus – not a very good
grade! However, the normal distribution can be used to “grade on a curve” so that students in the center of the distribution receive a better grade such as a C, while the remaining students’ grades
also get adjusted based on their relative distance from the mean.
Figure 3: The distribution of exam scores shown in Figure 1 can be approximated by a normal distribution, or bell curve, which is perfectly symmetrical around the mean (dashed line).
Early history of the normal distribution
The normal distribution is a relatively recent invention. Whereas the concept of the arithmetic mean can be traced back to Ancient Greece, the normal distribution was introduced in the early 18^th
century by French mathematician Abraham de Moivre. The mathematical equation for the normal distribution first appeared in de Moivre’s Doctrine of Chances, a work that broadly applied probability
theory to games of chance. Despite its apparent usefulness to gamblers, de Moivre’s discovery went largely unnoticed by the scientific community for several more decades.
The normal distribution was rediscovered in the early 19^th century by astronomers seeking a better way to address experimental measurement errors. Astronomers had long grappled with a daunting
challenge: How do you discern the true location of a celestial body when your experimental measurements contain unavoidable instrument error and other measurement uncertainties? For example, consider
the four measurements that Tycho Brahe recorded for the position of Mars shown in Table 1:
Table 1:Tycho Brahe's observations of the position of Mars as presented in Saul Stahl, "The Evolution of the Normal Distribution," Mathematics Magazine 79 (April 2006), p. 99. Copyright 2006
Mathematical Association of America. All Rights Reserved.
image ©Mathematical Association of America
Brahe and other astronomers struggled with datasets like this, unsure how to combine multiple measurements into one “true” or representative value. The answer arrived when Carl Friedrich Gauss
derived a probability distribution for experimental errors in his 1809 work Theoria motus corporum celestium. Gauss’ probability distribution agreed with previous intuitions about what an error curve
should look like: It showed that small errors are more probable than large errors and that all errors are evenly distributed around the “true” value (Figure 4). Importantly, Gauss’ distribution
showed that this “true” value – the most probable value in the center of the distribution – is the mean of all values in the distribution. The most probable position of Mars should therefore be the
mean of Brahe’s four measurements.
Figure 4: Gauss derived a probability distribution to address the inherent errors found in many experimental measurements. The “true” value (A) is the most probable value and is found in the center
of the distribution. A value closer to the “true” value is more likely to be observed than a value farther away from the “true” value. For example, value B, which is close to A, is more likely to be
observed than value D, which is far from A. Additionally, values are evenly distributed around the “true” value. Here, values B and C, which are both a distance “x” away from value A are equally
likely to be observed.
Further development of the normal distribution
The “Gaussian” distribution quickly gained traction, thanks in part to French mathematician Pierre-Simon Laplace. (Laplace had previously tried and failed to derive a similar error curve and was
eager to demonstrate the usefulness of what Gauss had derived.)
Scientists and mathematicians soon noticed that the normal distribution could be used as more than just an error curve. In a letter to a colleague, mathematician Adolphe Quetelet noted that soldiers’
chest measurements (documented in the 1817 Edinburgh Medical and Surgical Journal) were more or less normally distributed (Figure 5). Physicist James Clerk Maxwell used the normal distribution to
describe the relative velocities of gas molecules. As these and other scientists discovered, the normal distribution not only reflects experimental error, but also natural variation within a
population. Today scientists use normal distributions to represent everything from genetic variation to the random spreading of molecules.
Figure 5: Adolphe Quetelet noticed that the frequencies of soldiers’ chest measurements reported in the 1817 Edinburgh Medical and Surgical Journal fit a normal distribution strikingly well (though
not perfectly).
Characteristics of the normal distribution
The mathematical equation for the normal distribution may seem daunting, but the distribution is defined by only two parameters: the mean (µ) and the standard deviation (σ).
$y=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{1}{2}\left(\frac{x-\mu }{\sigma }{\right)}^{2}}$
The mathematical form of the normal distribution
The mean is the center of the distribution. Because the normal distribution is symmetrical about the mean, the median and mean have the same value in an ideal dataset. The standard deviation provides
a measure of variability, or spread, within a dataset. For a normal distribution, the standard deviation specifically defines the range encompassing 34.1% of individual measurements above the mean
and 34.1% of those below the mean (Figure 6).
Figure 6: The shape of a normal distribution is defined by the mean (µ) and the standard deviation (σ).
image ©Mwtoews
The concept and calculation of the standard deviation is as old as the normal distribution itself. However, the term “standard deviation” was first introduced by statistician Karl Pearson in 1893,
more than a century after the normal distribution was first derived. This new terminology replaced older expressions like “root mean square error” to better reflect the value’s usefulness for
summarizing the natural variation of a population in addition to the error inherent in experimental measurements. (For more on error calculation, see Statistics in Science and Uncertainty, Error, and
Comprehension Checkpoint
Working with statistical operations
To see how the mean, median, and standard deviation are calculated, let’s use the Scottish soldier data that originally inspired Adolphe Quetelet. The data appeared in 1817 in the Edinburgh Medical
and Surgical Journal and report the “thickness round the chest” of soldiers sorted by both regiment and height (vol. 13, pp. 260 - 262). Instead of using the entire dataset, which includes
measurements for 5,732 soldiers, we will consider only the 5’4’’ and 5’5’’ soldiers from the Peebles-shire Regiment (Figure 7).
Figure 7: Chest width distribution for the Peebles-shire Regiment. Although the data subset of 5’4’’ and 5’5’’ soldiers (blue) does not appear to be normally distributed, it comes from a much larger
dataset (gray) that can be reasonably approximated by a normal distribution. The 5'4" and 5'5" distribution of chest widths (in inches) is: 35, 35, 36, 37, 38, 38, 39, 40, 40, 40.
Note that this particular data subset does not appear to be normally distributed; however, the larger complete dataset does show a roughly normal distribution. Sometimes small data subsets may not
appear to be normally distributed on their own, but belong to larger datasets that can be more reasonably approximated by a normal distribution. In such cases, it can still be useful to calculate the
mean, median, and standard deviation for the smaller data subset as long as we know or have reason to assume that it comes from a larger, normally distributed dataset.
How to calculate the mean
The arithmetic mean, or average, of a set of values is calculated by adding up all of the individual values and then dividing by the total number of values. To calculate the mean for the
Peebles-shire dataset above, we start by adding up all of the values in the dataset:
35 + 35 + 36 + 37 + 38 + 38 + 39 + 40 + 40 + 40 = 378
We then divide this number by the total number of values in the dataset:
378 (sum of all values) / 10 (total number of values) = 37.8
The mean is 37.8 inches. Notice that the mean is not necessarily a value already present in the original dataset. Also notice that the mean of this dataset is smaller than the mean of the larger
dataset due to the fact that we have only selected the subsample of men from the lower height group and it is reasonable to expect shorter men to be smaller overall and therefore have smaller chest
How to calculate the median
The median is the “middle” value of a dataset. To calculate the median, we must first arrange the dataset in numerical order:
35, 35, 36, 37, 38, 38, 39, 40, 40, 40
When a dataset has an odd number of values, the median is literally the median, or middle, value in the ordered dataset. When a dataset has an even number of values (as in this example), the median
is the mean of the two middlemost values:
35, 35, 36, 37, 38, 38, 39, 40, 40, 40
(38 + 38) / 2 = 38
The median is 38 inches. Notice that the median is similar but not identical to the mean. Even if a data subset is itself normally distributed, the median and mean are likely to have somewhat
different values.
How to calculate the standard deviation
The standard deviation measures how much the individual values in a dataset vary from the mean. The standard deviation can be calculated in three steps:
1. Calculate the mean of the dataset. From above, we know that the mean chest width is 37.8 inches.
2. For every value in the dataset, subtract the mean and square the result.
(35 - 37.8)^2 = 7.8 (35 - 37.8)^2 = 7.8 (36 - 37.8)^2 = 3.2 (37 - 37.8)^2 = 0.6
(38 - 37.8)^2 = 0.04 (38 - 37.8)^2 = 0.04 (39 - 37.8)^2 = 1.4 (40 - 37.8)^2 = 4.8
(40 - 37.8)^2 = 4.8 (40 - 37.8)^2 = 4.8
3. Calculate the mean of the values you just calculated and then take the square root.
The standard deviation is 1.9 inches. The standard deviation is sometimes called the “root mean square error” because of the way it is calculated.
To concisely summarize the dataset, we could thus say that the average chest width is 37.8 ± 1.9 inches (Figure 8). This tells us both the central tendency (mean) and spread (standard deviation) of
the chest measurements without having to look at the original dataset in its entirety. This is particularly useful for much larger datasets. Although we used only a portion of the Peebles-shire data
above, we can just as readily calculate the mean, median, and standard deviation for the entire Peebles-shire Regiment (224 soldiers). With a little help from a computer program like Excel, we find
that the average Peebles-shire chest width is 39.6 ± 2.1 inches.
Figure 8. A normal curve approximation of the Peebles-shire data subset for 5’4’’ and 5’5’’ soldiers based on the mean and standard deviation values calculated above. The mean (µ), standard deviation
(σ), and median (dashed black line) help to quickly and concisely summarize the data.
Comprehension Checkpoint
Using descriptive statistics in science
As we’ve seen through the examples above, scientists typically use descriptive statistics to:
1. Concisely summarize the characteristics of a population or dataset.
2. Determine the distribution of measurement errors or experimental uncertainty.
Science is full of variability and uncertainty. Indeed, Karl Pearson, who first coined the term “standard deviation,” proposed that uncertainty is inherent in nature. (For more information about how
scientists deal with uncertainty, see our module Uncertainty, Error, and Confidence). Thus, repeating an experiment or sampling a population should always result in a distribution of measurements
around some central value as opposed to a single value that is obtained each and every time. In many (though not all) cases, such repeated measurements are normally distributed.
Descriptive statistics provide scientists with a tool for representing the inherent uncertainty and variation in nature. Whether a physicist is taking extremely precise measurements prone to
experimental error or a pharmacologist is testing the variable effects of a new medication, descriptive statistics help scientists analyze and concisely represent their data.
Sample problem 1
An atmospheric chemist wants to know how much an interstate freeway contributes to local air pollution. Specifically, she wants to measure the amount of fine particulate matter (small particles less
than 2.5 micrometers in diameter) in the air because this type of pollution has been linked to serious health problems. The chemist measures the fine particulate matter in the air (measured in
micrograms per cubic meter of air) both next to the freeway and 10 miles away from the freeway. Because she expects some variability in her measurements, she samples the air several times every day.
Here is a representative dataset from one day of sampling:
Fine particulate matter in the air (µg/m^3)
Next to the freeway 10 miles away from the freeway
29.3 11.8
18.3 12.5
17.7 13.1
17.9 9.6
18.9 14.6
20.9 10.4
18.6 9.8
Help the atmospheric chemist analyze her findings by calculating the mean (µ) and standard deviation (σ) for each dataset. What can she conclude about freeway contribution to air pollution? (Problem
modeled loosely after Phuleria et al., 2007)
Solution 1
Let’s start with the dataset collected next to the freeway:
$\mu =\frac{\left(19.3+18.3+17.7+17.9+18.9+20.9+18.6\right)}{7}=18.8$
$\sigma =\frac{\sqrt{\left(19.3-18.8{\right)}^{2}+\left(18.3-18.8{\right)}^{2}+\left(17.7-18.8{\right)}^{2}+\left(17.9-18.8{\right)}^{2}+\left(18.9-18.8{\right)}^{2}+\left(20.9-18.8{\right)}^{2}+\
Now we can do the same procedure for the dataset collected 10 miles away from the freeway:
$\mu =\frac{\left(11.8+12.5+13.1+9.6+14.6+10.4+9.8\right)}{7}=11.7$
$\sigma =\frac{\sqrt{\left(11.8-11.7{\right)}^{2}+\left(12.5-11.7{\right)}^{2}+\left(13.1-11.7{\right)}^{2}+\left(9.6-11.7{\right)}^{2}+\left(14.6-11.7{\right)}^{2}+\left(10.4-11.7{\right)}^{2}+\left
There is 18.8 ± 1.0 µg/m^3 fine particulate matter next to the freeway versus 11.7 ± 1.7 µg/m^3 10 miles away from the freeway. The atmospheric chemist can conclude that there is much more air
pollution next to the freeway than far away.
Sample problem 2
A climatologist at the National Climate Data Center is comparing the climates of different cities across the country. In particular, he would like to compare the daily maximum temperatures for 2014
of a coastal city (San Diego, CA) and an inland city (Madison, WI). He finds the daily maximum temperature measurements recorded for each city throughout the year 2014 and loads them into an Excel
spreadsheet. Using the functions built into Excel, help the climatologist summarize and compare the two datasets by calculating the median, mean, and standard deviation.
Solution 2
Download and open the Excel file containing the daily maximum temperatures for Madison, WI (cells B2 through B366) and San Diego, CA (cells C2 through C366). (Datasets were retrieved from the
National Climate Data Center http://www.ncdc.noaa.gov/)
Excel page 1: Calculating the Madison dataset median
To calculate the median of the Madison dataset, click on an empty cell, type “=MEDIAN(B2:B366)” and hit the enter key. This is an example of an Excel “function,” and it will calculate the median of
all of the values contained within cells B2 through B366 of the spreadsheet.
Excel page 2: Calculating the Madison dataset mean
The same procedure can be used to calculate the mean of the Madison dataset by typing a different function “=AVERAGE(B2:B366)” in an empty cell and pressing enter.
Excel page 3: Calculating the Madison dataset standard deviation
To calculate the standard deviation, type the function “=STDEV.P(B2:B366)” and press enter. (Older versions of Excel will use the function STDEVP instead.)
Excel page 4: Calculating the median, mean, and standard deviation of the San Diego dataset
The same procedure can be used to calculate the median, mean, and standard deviation of the San Diego dataset in cells C2 through C366.
Excel page 5: Temperature comparison of Madison and San Diego
On average, Madison is much colder than San Diego: In 2014, Madison had a mean daily maximum temperature of 54.5°F and a median daily maximum temperature of 57°F. In contrast, San Diego had a mean
daily maximum temperature of 73.9°F and a median daily maximum temperature of 73°F. Madison also had much more temperature variability throughout the year compared to San Diego. Madison’s daily
maximum temperature standard deviation was 23.8°F, while San Diego’s was only 7.1°F. This makes sense, considering that Madison experiences much more seasonal variation than San Diego, which is
typically warm and sunny all year round.
Non-normal distributions
Not all datasets are normally distributed. Because the world population is steadily increasing, the global age appears as a skewed distribution with more young people than old people (Figure 9).
Unlike the normal distribution, this distribution is not symmetrical about the mean. Because it is impossible to have an age below zero, the left side of the distribution stops abruptly while the
right side of the distribution trails off gradually as the age range increases.
Distributions with multiple, distinct peaks can also emerge from mixed populations. Evolutionary biologists studying the beak sizes of Darwin’s finches in the Galapagos Islands have observed a
bimodal distribution of finches (Figure 10).
Figure 10: Distribution of relative beak sizes among three species of finches in the Galapagos. Notice how there are two clear populations of finches: one with smaller beaks and one with larger
beaks. (Based on Hendry et al., 2009.)
In fact, the term “normal distribution” is quite misleading, because it implies that all other distributions are somehow abnormal. Many different types of distributions are used in science and help
scientists summarize and interpret their data.
Activate glossary term highlighting to easily identify key terms within the module. Once highlighted, you can click on these terms to view their definitions.
Activate NGSS annotations to easily identify NGSS standards within the module. Once highlighted, you can click on them to view these standards. | {"url":"https://www.visionlearning.com/en/library/Math-in-Science/62/Introduction-to-Descriptive-Statistics/218","timestamp":"2024-11-09T03:04:16Z","content_type":"text/html","content_length":"187233","record_id":"<urn:uuid:ca6a5880-2881-4bc6-b9be-d35698634a15>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00735.warc.gz"} |
Solution to Problem H – Magic
IPSC 2001
Solution to Problem H – Magic
We state the problem more formally: Let C be the set of all cards, i.e. C = {1,2,...,N}. Let A be the family of all subsets of C of size (N+1)/2. Similarly, let B be the family of all subsets of C of
size (N-1)/2. We want to find a perfect matching between A and B such that if S[A] is matched to S[B] then S[B] is a subset of S[A].
In other words, we can define a bipartite graph with partitions A and B. A pair (S[A],S[B]) is an edge of the graph if and only if S[B] is a subset of S[A]. Now we want to find a perfect matching in
this graph. (A perfect matching is a set of vertex-disjoint edges of size (number of vertices)/2, a maximum matching is a set of vertex-disjoint edges of maximum size.)
Finding a maximum matching in a bipartite graph can be done in time O(nm), where n is the number of vertices and m is the number of edges (see e.g. Cormen-Leiserson-Rivest: Introduction to
Algorithms, page 600). In our case, n = 2 * (N choose (N+1)/2), and m = n/2 * (N+1)/2. If the size of a maximum matching is n/2, we have a perfect matching which we print as an output of our program.
(Otherwise, if the size is not n/2, we should output "MAGIC". By Hall's Theorem, this never happens in our case.) | {"url":"https://ipsc.ksp.sk/2001/real/solutions/h.html","timestamp":"2024-11-04T03:56:00Z","content_type":"text/html","content_length":"4787","record_id":"<urn:uuid:4f385c77-259a-468c-b8dc-bd955e7c560b>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00701.warc.gz"} |
I need to build an ultra-simple, small vaccum bagging system
It will be for one-off kiteboards that I build for myself using Divinycell for the cores. If anyone can explain to me everything I’d need to get going(in laymans terms…)I’d really appreciate it.
if an actual vac isn’t worth the hassle, you can go about it how benny did his first compsand…
lots and lots of zip-loc bags filled with water.
another alternative method is bury it in sand or dirt. The deeper it is the more pressure you get.
But really…
Tenover, if you find a bar fridge at a yard sale you are most of the way there…search the archives for details on gauges…
I just found an old “Precision Vacuum Pump” here at work in the junkyard…Plugged it in and it runs, but I’m not sure if it’ll produce enough vacuum…? I don’t have my camera, but on the fron it says
“Precision Vacuum Pump - 35 l/m (1.23CFM) 15 Micron - Model PV35”.
It looks almost just like this one:
15 micron is about as good a vacuum as you could want. You scored! Your gauge will show near 30" at sealevel. Just don’t run pumps like that for long periods at low vacuum levels. They tent to
overheat at low vacuum levels. They are designed to run at the highest vacuum that they can pull for hours on end. the oil is expensive though. If you can track down an Air Conditioning repair guy
they usually have a couple of quarts of oil in their service vans.
If you find your in need of a vacuum switch I have a couple, one of which I could sell you for my cost. It is much much better than the j-woodworker one.
Thanks for the offer…I have no idea what I’m doing, so now that I have the pump, I guess I need to figure out what else I’ll need and HOW do set it all up…Do I need a rocker table? Ok, What’s a
rocker table
It looks like you scored with that pump! Equipped…Ultimately, a real pump will be best for sleep at night.
You are sure in the right place to convert your lack of knowledge into expertise. Archives!
I have a bunch of vacuum stuff sitting unused, PM if you’d like to put it to some use.
joe woodworker’s site has some great tutorials on building vacuum bag systems.
even if you don’t want to buy the parts, you can still get an idea of the principles
behind building the system.
good threads on sways as well.
here’s one of the cheapie teks:
Ok, first question. Does your pump have a guage and pressure adjustment? If not, you need to get one. Mine has a vacuum guage right at the pump, and a simple needle valve that lets a little air bleed
in to adjust the vacuum pressure. Mine will pull down to about 25 inches of mercury, but it only needs about 7-10 to get the job done on most things. You could even just poke a few little holes in
your vacuum bag to regulate the pressure, but that’s a bit unpredictable.
That’s the hardest part, finding a good pump. If you were going to be doing a lot of this stuff, I’d suggest going with the big chambers and a pressure switch and all that stuff, but since you just
want to do a few laminations, you can probably run your pump continuous without much problem.
The other stuff you need is a vacuum bag, some breather material, and some tubing.
You can make a ghetto vacuum bag by getting some poly sheeting from home depot or wherever (get the thickest they have), and cutting it to size, then using masking tape to seal the edges. Leave one
end open for loading your kiteboard, then seal that with masking tape as well once it is in. Masking tape will leak a little air, especially if there are any wrinkles in the bag or tape when you seal
it, but you’ve got a good pump so it shouldn’t be a worry.
Breather material, just use a few layers of thick paper towels. Tape it to the inside of your bag, running the full length so air has a conduit to get back to the vacuum tube. An old towel will work
great too.
Tubing, I have always just used the clear 1/4" tubing I got at home depot years ago. It won’t collapse under vacuum, and it’s cheap. Just run the tube into your vacuum bag at one of the seams that’s
going to stay closed and tape the hell out of it. Again, this isn’t a pro- or permanent setup, but it’ll get the job done. An alternative is to seal your bag with mastic tape, then just run the
tubing through that. The important thing is that the end of the vacuum tube terminates at the breather material. If the end of your vacuum tube gets covered with plastic, your pump will suck like
crazy and won’t pull any air out of the bag.
Now that you’re all set up, just lay up your kiteboard materials, and use another sheet of plastic to wrap it all up before going in the bag. This keeps any resin runoff from sticking to your
breather material and keeps your bag clean for further use. You can use the same poly sheeting to wrap it up, but I’ve found a thinner plastic works better for this release film layer. The important
thing with this step is to stretch it tight. If there are any big gaps in between materials, and the plastic is loose, the vacuum will suck it into the gap and keep it from bonding. I usually start
at the middle of a board, stretch the plastic tight by wadding up the loose edges, then use a strip of masking tape to hold it. Do this in sections towards the ends. It also helps to pre-bend the
materials, such as in the case of bending a balsa skin over rails. You can turn off the vacuum pump once the resin left in your cup isn’t tacky any more, but you’ll need to wait a while longer before
trying to peel off the plastic… It releases better once it is more fully cured, a few more hours is usually good.
How’s that for a quick and dirty get-u-started tutorial?
Interesting, thanks. Sounds easy enough…LOL.
My pump doesn’t currently have a gauge, and there is no tubing on it. I’m sure I can find a gauge laying around, and I’ll go buy some tubing. There’s two places on top of the pump to attach hoses
to…Do I need to attach to both, or just the one that is vacuuming (to the bag)?
Plug it in and see if it works? Put your finger over the holes…see which one go in, and which one goes out. Hows that for helpful.
I did that already… (see post). It runs. One hole sucks the other blows (get your mind out of the gutter!). Just wondering if the one that blows needs to be hooked up to anything or if I just hook
the hose up to the sucking connection and then to the bag…
Hey, shouldn’t you be killing sharks or something?
Plug it in and see if it works? Put your finger over the holes....see which one go in, and which one goes out. Hows that for helpful.
Shwuz, That was a great beginner tutorial, thanks.
I’m on my way over! it sounds like a good time. Maybe I could hook it up to a dolphin blow hole, and really get something going!!
Just me…the pump, a dead dolphin, some pump grease, and a cold beer. Oh yeah, a candle to set the mood.
So, guys have mentioned pulling most the air out with a shop vac and then turning on there vac pumps. What’s a good pump for slow, small draws, over long periods of time then?
You should be able to get a cheap vacuum gauge at the auto parts store and while you are there pick up a one way valve to hook up to the exhaust side of your pump. The one way valve should cost
around $3.99, you need it so that you don’t lose vacuum through the pump itself when it is not running. | {"url":"https://forum.swaylocks.com/t/i-need-to-build-an-ultra-simple-small-vaccum-bagging-system/30335","timestamp":"2024-11-07T15:30:46Z","content_type":"text/html","content_length":"54374","record_id":"<urn:uuid:4ca6b70d-0cdc-48f5-a5df-5d22e34f1568>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00706.warc.gz"} |
NCERT Book for Class 12 Mathematics
Download NCERT Book for Class 12 Mathematics English and Hindi medium for current academic year. The book can be downloaded in pdf for Class 12 Mathematics. You can download the entire textbook or
each chapter in pdf, NCERT Books are suggested by CBSE for Class 12 Mathematics exams, as they have been prepared as per the syllabus issued by CBSE, download and read latest edition books and these
have very important questions and exemplar problems for which studiestoday.com has provided NCERT solutions for Class 12 Mathematics.
Click on the below links to access books. These books have been implemented in entire India and your class tests and exams for Class 12 Mathematics will be based on this book itself. You can download
the ebooks free from studiestoday along with solutions for Class 12 Mathematics
Class 12 Mathematics NCERT Book Pdf
We have provided below the latest CBSE NCERT Book for Class 12 Mathematics which can be downloaded by you for free. These free subject-wise NCERT textbook have been designed based on the latest CBSE
Syllabus issued for current academic year. You can click on the links below to download the subject and chapter-wise NCERT ebook for Class 12 Mathematics. CBSE book for Class 12 Mathematics will help
Class 12 Mathematics students to prepare properly for the upcoming examinations.
Chapter Wise NCERT book for Class 12 Mathematics in Pdf
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You can download the CBSE NCERT Books for Class 12 Mathematics for latest session from StudiesToday.com
Can I download the NCERT Books of Class 12 Mathematics in Pdf
Yes, you can click on the links above and download chapterwise NCERT Books in PDFs for Class 12 for Mathematics
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You can easily access the links above and download the Class 12 NCERT Books Mathematics for each chapter
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How can I improve my scores by reading NCERT Books in Class 12 Mathematics
Regular revision of NCERT Books given on studiestoday for Class 12 subject Mathematics can help you to score better marks in exams | {"url":"https://www.studiestoday.com/download-books/310/mathematics.html","timestamp":"2024-11-05T06:41:04Z","content_type":"text/html","content_length":"117688","record_id":"<urn:uuid:1ff8e0ae-8f34-47dc-8ca2-58e7d0b49937>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00491.warc.gz"} |
Infrared safety in factorized hard scattering cross-sections
The rules of soft-collinear effective theory can be used naïvely to write hard scattering cross-sections as convolutions of separate hard, jet, and soft functions. One condition required to guarantee
the validity of such a factorization is the infrared safety of these functions in perturbation theory. Using e^+e^- angularity distributions as an example, we propose and illustrate an intuitive
method to test this infrared safety at one loop. We look for regions of integration in the sum of Feynman diagrams contributing to the jet and soft functions where the integrals become infrared
divergent. Our analysis is independent of an explicit infrared regulator, clarifies how to distinguish infrared and ultraviolet singularities in pure dimensional regularization, and demonstrates the
necessity of taking zero-bins into account to obtain infrared-safe jet functions.
Physics Letters B
Pub Date:
June 2009
□ Factorization;
□ Soft collinear effective theory;
□ Jets;
□ Event shapes;
□ 12.38.Bx;
□ 12.39.St;
□ 13.66.Bc;
□ 13.87.-a;
□ Perturbative calculations;
□ Factorization;
□ Hadron production in e<sup>-</sup>e<sup>+</sup> interactions;
□ Jets in large-Q<sup>2</sup> scattering;
□ High Energy Physics - Phenomenology
6 pages, 7 figures, uses elsarticle.cls. v2: extended introduction and clarified discussion of ingredients necessary for proving factorization | {"url":"https://ui.adsabs.harvard.edu/abs/2009PhLB..677..272H/abstract","timestamp":"2024-11-05T17:19:16Z","content_type":"text/html","content_length":"41214","record_id":"<urn:uuid:96384bc4-ea35-426f-a17f-15694ba497d9>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00773.warc.gz"} |
Recognize numbers game
13 Hands-On Number Recognition Games Preschoolers will Love
Are you teaching your children to identify numbers? Here are some fun and interactive number recognition games for preschoolers and kindergarteners that you can play at home or in the classroom too.
If you’re wondering how to teach number recognition, the answer in early childhood is always through play.
Play is the natural way in which children learn. During play, children practice their skills and make sense of new knowledge and experiences. They develop early maths skills through play.
Remember that there are many aspects to learning about numbers. There is learning to count, which you can teach with number games and counting songs, and then there’s one-to-one correspondence, which
is when a child reliably counts one object at a time.
Number recognition is about the physical appearance and shape of a number, as well as what value it represents.
These number recognition activities for preschoolers are a great place to start teaching the numbers from 1 to 10, but once you get going you’ll quickly notice opportunities all around you.
1. Parking Cars
This numbers game can be adapted to suit your children’s age, stage and interests.
Write numbers onto some toy cars and create a parking garage with numbered spaces. Your children can then match the number on the car with the number in the space and park the car correctly.
If they need more of a challenge replace the numerals with dots or words so that your kids can begin to recognise numbers being represented pictorially.
If they are not particularly interested in cars you could do a similar game with animals, dolls, or whatever it is that they enjoy playing with.
2. Car Wash
Put numbers on toy cars, or for a large-scale activity; bikes and scooters. Create a car wash for them with clothes, brushes, water and bubbles.
Your children are then in charge of ensuring that the cars, bikes, or scooters come to the car wash and get cleaned in the correct order. As well as recognising numerals, this activity gives kids the
chance to begin learning about number order.
3. Hook a Duck
This fairground classic is great for numeral recognition. How you set this up is your choice.
If you have lots of ducks and something to hook them with then perhaps you could create a replica of the fairground game, otherwise feel free to improvise with what you have at hand.
A net or bowl to scoop objects out of the bath could work well – the important thing is for your children to be having fun and looking at numbers. You could allocate prizes to certain numbers if you
want to.
4. Sidewalk Chalk
Sidewalk chalk is brilliant for larger-scale mark-making and games that get children using their gross motor skills.
Use sidewalk chalk to write out large numerals, then give kids a paintbrush and a pot of water and have them paint over the numerals with water to erase them.
Not only does this help them to recognise numerals but also helps with the beginnings of formation.
5. Beads onto Pipe Cleaners
Create a chart using beads and pipe cleaners. Attach 10 pipe cleaners to a piece of card and write numbers 1-10, one number above each pipe cleaner.
Provide children with a pot of beads and help them to count out the correct number on each pipe cleaner.
This activity gives them the opportunity to practise numeral recognition, counting and assigning the correct value to each numeral. It’s also brilliant for their fine motor skills.
6. Bean Bag Toss
Label some buckets or baskets with numbers and provide your children with beanbags. Have them step back from the buckets and take aim and throw the bean bags in.
You can do quite a lot with this activity depending on your children’s age and ability, but on the most basic level, it encourages number recognition along with introducing the concept of more and
If your children are ready then you can model addition and play to win.
7. Putting Counters in Pots
Label pots with numerals and provide counters, craft beads, pom poms or really whatever you have at hand and encourage children to fill each pot with the correct number of items.
Again, this activity targets a variety of different skills as children recognise numerals, apply their understanding of value, and count out the correct number of items. This is another good one for
fine motor skills.
8. Create an Outdoor Number Line
Children love to learn outdoors and on a large scale. Many teachers love using small number lines in the classroom to introduce the ideas of one more and one less but you can do the same outside.
Perhaps use chalk to draw out your number line and encourage children to locate different numerals – “Stand on number 8,” “Hop to number 4” and so on.
If appropriate you could discuss one more and one less.
9. Nature’s Numerals
If your children like to be creative and artistic then this could work for them. Use nature to create the shapes of numbers.
This might mean drawing in the mud or sand, arranging leaves or stones or even noticing natural shapes in the environment. You could do this in your backyard or take a special walk.
Even better if you can take photos of your creations for your kids to look back on. This allows them to begin thinking about how numerals are formed in a fun and creative way.
10. Hopscotch
Hopscotch is a real playground classic and it brings together a whole host of skills including gross motor skills.
Draw out a hopscotch grid and teach your children how to play, throwing a stone or stick to find out where they need to hop to, and then hopping and jumping to the end.
As well as reinforcing the recognition of numerals this also introduces the idea of higher and lower and allows kids to have fun while working with numbers.
Hopscotch is my favourite number activity for preschool kids.
11. Potion Recipes
If you like messy, creative play then this one’s for you.
Create a couple of ‘recipe cards’ using measurements expressed as numerals, for example – 2 cups of water and 3 pinecones, and have your children follow the recipe card, combining everything together
in a big cauldron-like tub.
This taps into children’s imaginations and introduces the concept of measurement as well as numbers. Once they are finished following the recipes you have provided perhaps they will be ready to
create their own recipes, which you can scribe for them.
12. Number Splat
All you need is a nice big roll of paper with numbers on it and a fly swat dipped in paint. You call out the numbers and children must swat them, thereby covering them in paint.
You can play just as easily without the paint, simply swatting at the numerals, but it’s far less fun than making a mess.
This activity is extremely physical helping to really embed the learning, and as children try to speed up, their ability to recognise numerals will improve too so that they’ll soon be able to
recognise them at a glance.
13. Bingo
Bingo is a great maths game for building up number awareness and can be enjoyed as a family. To start off with you can simply use numerals up to ten but as your children’s knowledge expands so can
your game.
You can use what you have on hand – a bowl and folded-up pieces of paper, with highlighters – or you can go ahead and buy bingo pads and dabbers and bingo balls to add to the overall experience.
Here you will be building number recognition and as your kids aim to increase their speed, they will get quicker and quicker at recognising numbers and linking them to the number names being called
I hope you’ve enjoyed these number identification games for preschoolers. Here are some more fun math activities for preschoolers to build early mathematical skills.
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10 Number Recognition Games
Number recognition is a key skill to learn during the early years, and there are many ways in which you can encourage this in your early years setting.
When planning number recognition activities, there are a few points to keep in mind:
• Make sure you have plenty of relevant resources available (e.g. number beads, blocks, stickers, cutters, stamps etc.), as well as visual cues (e.g. posters on the walls).
• As well as using numerals it’s also helpful to look at other representations of numbers with the children, including words and tallies.
• Try to make number activities fun in order to nurture a positive approach to maths.
1. Number bubble game
Draw lots of chalk circles on the ground outside, with a number inside each (1 to 5 or 1 to 10, depending on how much space you have), distributing them evenly so that you end up with several 1s in
circles, several 2s in circles, and so on (make sure you have enough for each child playing the game).
2. Number hunt
Take a small group of children out for a walk around the neighbourhood – or perhaps combine it with a visit to the local park – hunting for numbers along the way. There should be plenty of
opportunities for number spotting, for example on front doors, gates, buses, cars, posters etc. Get the children to call them out when they see them.
A number hunt is a great way for children to practice number recognition outside your setting
3. Giant dot-to-dot
Make your own giant dot-to-dot in the playground, by chalking numbers on the ground that the children have to connect in the right order to make a shape or picture. For younger children stick to
simple shapes using fewer numbers; for older children you can make it a bit more difficult.
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Go to the park and collect some conkers. Back at the nursery, draw the numbers 1 to 10 on the ground in a row with chalk, using both numerals and words, and get the children to line up the right
number of conkers underneath each one. (Obviously outside conker season there are plenty of other objects you could use for this activity, eg petals, leaves or items from inside.)
5. Dice tally
Take a sheet of card and make a grid of six squares, labelling them 1 to 6 using both numerals and words. Roll a die and keep a tally in the squares of how many times each number comes up. Children
could do this individually, each with their own separate grids, or in pairs or small groups using the same grid but their own dice. You could turn it into more of a game by adding a competitive
Recognising and tallying the numbers rolled on dice is another good skill to develop
6. Musical number tiles
This is a musical variation of the bubble game.
7. Number biscuits
Using some number shape cutters, make some sets of number biscuits with the children and then use squeezy icing to stick the right number of decorations onto each biscuit (e.g. eight raisins on the
number 8, three raspberries on the number 3 etc).
Help the children make biscuits with different numbers of decorations, counting them out as you put them on
8. Beanbag toss
Here are a couple of ideas for throwing games to help with number recognition. One is to get a set of buckets and label them 1 to 5 (or 1 to 10), then the children have to try and throw the right
number of beanbags into each; another is to use a target mat and the children have to try and land the right number of beanbags in each numbered segment.
9. Counting beads
For this activity you’ll need ten paper plates, some coloured pens and some coloured beads. Write the numbers 1 to 10 on the plates, using a different colour for each number. Get the children to put
the right number of beads onto each plate; this works particularly well using coloured beads that correspond with the colours used to write the numbers, as it gives the children a strong visual cue.
10. Number crafts
There are lots of ways in which you can incorporate number recognition into craft activities. One idea is to draw some outlines of ladybirds on a piece of paper, then number them and get the children
to add the right number of spots to each. A couple of variations on this include drawing birds and sticking on tail feathers, or drawing monsters and sticking on googly eyes.
One option for a number craft is sticking the correct amount of spots on a ladybird
Related Itemscognitive developmentnumbersnumeracy
How to teach a child to recognize numbers from 1 to 10 | Lifestyle
Number Recognition - The ability to visually recognize numbers and then name them.
Children's introduction to numbers
Use numbers to describe your child's groups of things. To create a foundation for number recognition, talk about your child's environment in terms of numbers. If several pencils have fallen on the
floor, you can say, “Oh, I dropped 3 pencils!” Or, while reading a picture book, you can point to an illustration and say, "There are 2 planes in the sky."
Be sure to use numbers when describing situations in which your child had a hand in creating something. For example, if he drew 4 chicks, you could point to that and ask, "Are you going to draw 4
more chicks?" Start with smaller numbers, usually up to 4-6.
Put numbers on houses, street signs or other places
This will help your child start to recognize different numbers. When you drive or walk around your neighborhood, pay attention to your child, for example, by house numbers or phone numbers on
billboards. While at home, enter numbers on telephone keypads, remote controls, clocks, or thermometers. Be consistent in your use of numbers in everyday conversation. The more often you use numbers
to describe things and indicate numbers in your child's environment, the faster he will recognize numbers.
Have the children do their homework using numbers
Children need to understand that numbers are not just for math class. They can also be used in real situations. For example, ask your child for lunch to indicate the correct number of napkins for
your family. Or in an arts and crafts class, ask for enough glue sticks for each person at the table.
Adding numbers to the sensory game
Buy sets of magnetic numbers and have your child match them. Let your child play freely with them first so that they start to recognize the shapes of the different numbers. Then ask him to match
pairs or groups with the same number. If the child is older, see if he can put the numbers in order. Magnetic numbers can be attached to the refrigerator door or on a baking sheet.
Drawing numbers on flour
Sprinkle a thin layer of flour on a tray or baking sheet. Then have your child draw different numbers on it. If your child is just getting started with number recognition, you can write an example of
each number on a piece of paper. Once the tray is filled with numbers, help your child even out the flour so he can keep drawing.
Reproduce the shape of the numbers
Write the numbers 1 through 10 on a large piece of paper.
Button sorting
Write the numbers 1 to 10 on 10 plastic cups. Then give your child some buttons or other small items (such as shells or pebbles). Ask them to place the correct number of items in each labeled glass.
Whatever items you use, make sure they are large enough so that your child cannot, for example, swallow them or stick them up their nose. This can be a great activity for several kids to play
together. If one child has mastered the rules, he can help his playmates who are still learning to recognize numbers.
Use the calculator to experiment.
Ask your child to find the number that tells them how old they are. Ask them to enter numbers from 1 to 10 in the correct order. The calculator can be used anywhere. Keep it in your pocket or bag so
you can take it out while waiting in line.
Number recognition practice with games and activities
For the first game you will need 1 die, piece of paper, pencils, 6 tokens per player and a number line from 1 to 6 for each player. One player rolls the die and places a token on that number on their
number line. The other player does the same. If any player rolls a number that is already covered by the token, he skips the turn. The first one that occupies all 6 numbers wins the game. You can
also use a number line from 1 to 12 for older children and roll two dice instead of one.
Match the numbered cards until you have completed the entire deck.
Take the numbered cards. Have your child turn over the first card in the deck. Then ask to turn over the next card. If its number matches the number of the first card, they must be placed side by
side, if not, put aside. For this game, you can use a deck of ordinary playing cards, taking out cards with pictures.
Speed Game
You will need a deck of 10 cards for each player.
Number line and blocks
Make a long, straight line on the floor using masking tape. Add 10 shorter horizontal pieces of ribbon evenly spaced along the line. Label each short piece of tape with a number from 1 to 10. Then
ask your child to place 1 block on the line next to the “1” mark, 2 blocks on the line next to the “2” mark, etc. If your child does not recognize numeric characters, but knows the names of the
numbers, try starting at 1 and have it count as they move up the number line.
Tower of blocks
The essence of the game is that you need to roll a die, then build a tower of blocks using the number of points dropped. Choose a die with symbols on the side, not dots. Make this game more
challenging by taking turns and adding to your tower each turn until you get to 10. For example, your child rolls a "2" and builds a 2 block tower. You roll a "4" and stack 4 blocks to create your
tower. Your child throws again and gets a "6". He then counts 6 additional blocks to add to his original tower for a total of 8 blocks. Keep playing until both of your towers are 10 blocks high.
Bulls and Cows and Math Cards
Classes: 2, 3, 4, 5, 6
Keywords: math game, entertaining math, bulls cows
Bulls and cows game
Bulls and cows game - wonderful logic game that does not require special fixtures.
The game develops the ability to compare and analyze.
Two people are playing. Everyone thinks of a number out of four non-repeating numbers (zero in the game used, but cannot be in the first place).
The opponent's task is to guess the number out of 10 attempts.
Opponent calls any 4-digit number, whose numbers also do not repeat His you must write under your hidden number, to make it easier to compare numbers. At coincidence of the digits of the named number
with the hidden says "BUCK". Bull means that the figure guessed and stands in the right position (for example, in conceived number, the first digit is 3 and in the named the enemy is also the first 3
- this is a bull.). Cow means that the figure is guessed, but it is not in his position. By logical reasoning and To check answers, you must guess all 4 digits numbers and their order.
The number game is actually not very difficult, Since there are only 10 digits, they cannot be repeated. It can be mastered by children even 8-9 years old.
An example of the game:
3749 - a hidden number
3589 - the opponent calls - your answer - 2 bulls. (3 and 9 stand in their places)
7628 - calls the enemy - your answer - 1 cow. (only 7 is in the number, but not on its own place).
This means that 2 digits are used from the first number, and only one of the second (but which ones after the first the answer is impossible to determine). Further, calling the following numbers, you
need to calculate the numbers themselves and their order.
Based on two responses, determine whether the number 8 in the hidden number is impossible - it is necessary try other numbers and compare what is the answer you get.
4973 - opponent calls - your answer - 4 cows (i.e. all numbers are correct, but their order is not). And here is the answer: 3 bulls 1 cow be not maybe, because if three digits stand their ground
place, then the fourth one too.
Bulls and cows game with words
After mastering the game with numbers, it becomes more interesting go to the word game.
There are a lot of 4-letter words in Russian (we always play meaningful words). A combination of letters can be very different: and 3 consonants 1 vowel, and 2 to 2, and 1 to 3. Not used solid only
sign and words like MAMA, HEADLIGHT , FRAME, WINDOW, CAKE etc., where 2 letters are the same. Game principle remains the same: the letter in its place is bull , the letter is in the word, but not in
its place - cow .
You can play on any piece of paper, it's good even half a leaf or written on one side, but makes you compare and think logically, so how to change all the time and analysis – where are the bull and
what is it? For example, the enemy can guess and pond, and rod , and port - and replacing even one letter leads already to a new word, and sometimes the same letters may be in a different order - and
there will be two different words, for example, summer and body . On my own a leaflet on the side for a hint is convenient to write out alphabet and check different substitution options letters (
shadow, day, stump, laziness ..... )
Game examples:
│ * * * * │ * * * * │
│ │ │
│ • arm – 1 to │ • moon – 1 b │
│ • elephant - empty (no letter) │ • sea – 1 to │
│ • mushroom – 3 b │ • corner – 1 to │
│ • hump – 1 b 1 k │ • port – 1 to │
│ • beads - empty │ • burrow – 1 to │
│ • makeup – 3 b │ • chair – 1 to │
│ • neck - the word is guessed. │ • laziness - empty │
│ │ • storm – 1 b 2 k │
│ │ • arm – 1 b 1 k │
│ │ • log house – 3 to │
│ │ • fur coat - 2 b │
│ │ • bison - the word is guessed. │
Math cards
The game allows you to practice mental counting and multiplication table. Recommended for students elementary school.
They are done like this: two sets of numbers from 1 are taken up to 24 (for numbers it is convenient to use the old Wall calendar).
Rules : Each player is given 4 cards. The beginner is given the fifth card. From of his 5 cards, he chooses one, which he gives to a neighbor as a task.
The principle of the game is as follows : a player from his own 4 cards, that is, numbers, using any math operations: +, - , *, : (addition, subtraction, multiplication, division) and putting numbers
into any(!) order, should get a response that gave him a neighbor. Those numbers-cards that he at the same time used together with the answer, the player takes like a bribe. And at the end of the
game everyone thinks how many cards he has in his pile. This game Designed for practicing oral counting simple division and multiplication (all options up to 24).
For example: the first player got the cards: 17, 4, 8, 9 . 10 or 16 .
• number 10 is obtained very easily: 8:4+17-9=10;
• the number 16 is more difficult to make: (17-9)*(8:4)=16.
And if you "play" with these numbers, you can get another set of answers: 12, 14, 13, 23, 6, 4, 20... For example, 12 = 17 - 4 - (9- eight). 14 \u003d 17 - 4 + (9 - 8) .
The advantage of the game is that sequence of numbers and math action is not fixed like in the tutorial mathematics, and the player himself must determine them, shifting the cards in any order , and
also the fact that the game is open and all the players also “puzzle their brains” to decide this example.
If brackets are used when writing an example, then verbally, you can not say anything about them, just call the actions in the right order: for 16 (from the example above): first from 17 I subtract 9
, it turns out 8 ; 8 divided by 4 , and 8 multiply by 2 . | {"url":"https://northccs.com/misc/recognize-numbers-game.html","timestamp":"2024-11-14T10:09:23Z","content_type":"text/html","content_length":"49894","record_id":"<urn:uuid:71e220ff-aa47-4bb1-800f-9aa3ecab0cbe>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00740.warc.gz"} |
Which has more sides a Hexadecagon or a Heptagon?
Answer : A Hexadecagon has more sides than a Heptagon
A Hexadecagon has 16 sides and a Heptagon has 7 sides
Name Name:Hexadecagon Name:Heptagon
Sides Sides:16 Sides:7
Interior Angle Interior Angle:157.00 Interior Angle:128.57
Sum of Interior Angles Sum of Interior Angles:2520 Sum of Interior Angles:900
Description Description:A polygon with 16 sides is called a hexadecagon Description:A polygon with 7 sides is called a heptagon (or a septagon) | {"url":"https://purelyfacts.com/question/26/14/5/which-has-more-sides-a-hexadecagon-or-a-heptagon","timestamp":"2024-11-09T19:42:30Z","content_type":"text/html","content_length":"93445","record_id":"<urn:uuid:caa3d793-4de8-4169-bf49-fdaa0e6d79c8>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00700.warc.gz"} |
Roof Shingles Calculator: Estimate Roofing Costs
This tool helps you easily estimate the number of roof shingles you need for your roofing project.
How to Use the Roof Shingles Calculator:
Enter the length and width of your roof in feet. Additionally, specify the roof slope in degrees to account for the pitch. Click on “Calculate” to get the total area in square feet and the number of
shingles required.
Calculation Method:
The calculator uses the following steps to determine the number of shingles required:
• Calculates the area of the roof surface considering its slope.
• Assumes standard shingles coverage per bundle (33.3 square feet).
• Adds 10% extra for wastage (cuttings, overlaps, etc.).
This calculator assumes a simple, gable roof and does not account for dormers, chimneys, or other features that could affect the total shingle requirement. Always consult with a roofing professional
for more complex calculations. | {"url":"https://calculatorsforhome.com/roof-shingles-calculator/","timestamp":"2024-11-06T17:52:29Z","content_type":"text/html","content_length":"141539","record_id":"<urn:uuid:ce71bb04-1803-4cf1-bbe2-847a75a6ec85>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00129.warc.gz"} |
Ultra-high-accuracy computation and gravitational self-force observables in binary black holes
The principal approximation methods used to compute the inspiral of compact binary systems are the post-Newtonian (pN) expansion, in which an orbital angular velocity MΩ serves as the expansion
parameter; and the self-force or extreme-mass-ratio-inspiral approach, in which the small parameter is the mass ratio m/M of the binary’s two components. We work in an overlapping regime where both
approximations are valid and find numerical values of pN coefficients at orders beyond the reach of current analytical work. In this talk we present a novel analytic extraction of high-order pN
parameters that govern quasi-circular binary systems using ultra-high accuracy numerical computations. | {"url":"https://indico.math.cnrs.fr/event/632/?view=event","timestamp":"2024-11-11T14:59:54Z","content_type":"text/html","content_length":"95057","record_id":"<urn:uuid:135cdb7d-0489-4adc-a929-6d569eb920aa>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00723.warc.gz"} |
A risk-free bond is a theoretical bond that repays interest and principal with absolute certainty. The rate of return would be the risk-free interest rate. It is primary security, which pays off 1
unit no matter state of economy is realized at time ${\displaystyle t+1}$. So its payoff is the same regardless of what state occurs. Thus, an investor experiences no risk by investing in such an
In practice, government bonds of financially stable countries are treated as risk-free bonds, as governments can raise taxes or indeed print money to repay their domestic currency debt.^[1]
For instance, United States Treasury notes and United States Treasury bonds are often assumed to be risk-free bonds.^[2] Even though investors in United States Treasury securities do in fact face a
small amount of credit risk,^[3] this risk is often considered to be negligible. An example of this credit risk was shown by Russia, which defaulted on its domestic debt during the 1998 Russian
financial crisis.
Modelling the price by Black-Scholes model^[4]
In financial literature, it is not uncommon to derive the Black-Scholes formula by introducing a continuously rebalanced risk-free portfolio containing an option and underlying stocks. In the absence
of arbitrage, the return from such a portfolio needs to match returns on risk-free bonds. This property leads to the Black-Scholes partial differential equation satisfied by the arbitrage price of an
option. It appears, however, that the risk-free portfolio does not satisfy the formal definition of a self-financing strategy, and thus this way of deriving the Black-Sholes formula is flawed.
We assume throughout that trading takes place continuously in time, and unrestricted borrowing and lending of funds is possible at the same constant interest rate. Furthermore, the market is
frictionless, meaning that there are no transaction costs or taxes, and no discrimination against the short sales. In other words, we shall deal with the case of a perfect market.
Let's assume that the short-term interest rate ${\displaystyle r}$ is constant (but not necessarily nonnegative) over the trading interval ${\displaystyle [0,T^{*}]}$ . The risk-free security is
assumed to continuously compound in value at the rate ${\displaystyle r}$ ; that is, ${\displaystyle dB_{t}=rB_{t}~dt}$ . We adopt the usual convention that ${\displaystyle B_{0}=1}$ , so that its
price equals ${\displaystyle B_{t}=e^{rt}}$ for every ${\displaystyle t\in [0,T^{*}]}$ . When dealing with the Black-Scholes model, we may equally well replace the savings account by the risk-free
bond. A unit zero-coupon bond maturing at time ${\displaystyle T}$ is a security paying to its holder 1 unit of cash at a predetermined date ${\displaystyle T}$ in the future, known as the bond's
maturity date. Let ${\displaystyle B(t,T)}$ stand for the price at time ${\displaystyle t\in [0,T]}$ of a bond maturing at time ${\displaystyle T}$ . It is easily seen that to replicate the payoff 1
at time ${\displaystyle T}$ it suffices to invest ${\displaystyle B_{t}/B_{T}}$ units of cash at time ${\displaystyle t}$ in the savings account ${\displaystyle B}$ . This shows that, in the absence
of arbitrage opportunities, the price of the bond satisfies
${\displaystyle B(t,T)=e^{-r(T-t)}~~~,~~~\forall t\in [0,T]~.}$
Note that for any fixed T, the bond price solves the ordinary differential equation
${\displaystyle dB(t,T)=rB(t,T)dt~~~,~~~B(0,T)=e^{-rT}~.}$
We consider here a risk-free bond, meaning that its issuer will not default on his obligation to pat to the bondholder the face value at maturity date.
Risk-free bond vs. Arrow-Debreu security^[5]
The risk-free bond can be replicated by a portfolio of two Arrow-Debreu securities. This portfolio exactly matches the payoff of the risk-free bond since the portfolio too pays 1 unit regardless of
which state occurs. This is because if its price were different from that of the risk-free bond, we would have an arbitrage opportunity present in the economy. When an arbitrage opportunity is
present, it means that riskless profits can be made through some trading strategy. In this specific case, if portfolio of Arrow-Debreu securities differs in price from the price of the risk-free
bond, then the arbitrage strategy would be to buy the lower priced one and sell short the higher priced one. Since each has exactly the same payoff profile, this trade would leave us with zero net
risk (the risk of one cancels the other's risk because we have bought and sold in equal quantities the same payoff profile). However, we would make a profit because we are buying at a low price and
selling at a high price. Since arbitrage conditions cannot exist in an economy, the price of the risk-free bond equals the price of the portfolio.
Calculating the price^[5]
The calculation is related to an Arrow-Debreu security. Let's call the price of the risk-free bond at time ${\displaystyle t}$ as ${\displaystyle P(t,t+1)}$ . The ${\displaystyle t+1}$ refers to the
fact that the bond matures at time ${\displaystyle t+1}$ . As mentioned before, the risk-free bond can be replicated by a portfolio of two Arrow-Debreu securities, one share of ${\displaystyle A(1)}$
and one share of ${\displaystyle A(2)}$ .
Using formula for the price of an ${\displaystyle n}$ Arrow-Debreu securities
${\displaystyle A(k)=p_{k}{\frac {u^{\prime }(C_{t+1}(k))}{u^{\prime }(C_{t})}},~~~~~k=1,\dots ,n}$
which is a product of ratio of the intertemporal marginal rate of substitution (the ratio of marginal utilities across time, it is also referred to as the state price density and the pricing kernel)
and the probability of state occurring in which the Arrow-Debreu security pays off 1 unit. The price of the portfolio is simply
${\displaystyle P(t,t+1)=A(1)+A(2)=p_{1}{\frac {u^{\prime }(C_{t+1}(1))}{u^{\prime }(C_{t})}}+p_{2}{\frac {u^{\prime }(C_{t+1}(2))}{u^{\prime }(C_{t})}}}$
${\displaystyle P(t,t+1)=\mathbb {E} _{t}^{\mathbb {P} }{\Bigg [}{\frac {u^{\prime }(C_{t+1}(k))}{u^{\prime }(C_{t})}}{\Bigg ]}}$
Therefore, the price of a risk-free bond is simply the expected value, taken with respect to the probability measure ${\displaystyle \mathbb {P} =\{p_{1},p_{2}\}}$ , of the intertemporal marginal
rate of substitution. The interest rate ${\displaystyle r}$ , is now defined using the reciprocal of the bond price.
${\displaystyle 1+r_{t}={\frac {1}{P(t,t+1)}}}$
Therefore, we have the fundamental relation
${\displaystyle {\frac {1}{1+r}}=\mathbb {E} _{t}^{\mathbb {P} }{\Bigg [}{\frac {u^{\prime }(C_{t+1}(k))}{u^{\prime }(C_{t})}}{\Bigg ]}}$
that defines the interest rate in any economy.
Suppose that the probability of state 1 occurring is 1/4, while probability of state 2 occurring is 3/4. Also assume that the pricing kernel equals 0.95 for state 1 and 0.92 for state 2.^[5]
Let the pricing kernel denotes as ${\displaystyle U_{k}}$ . Then we have two Arrow-Debreu securities ${\displaystyle A(1),~A(2)}$ with parameters
${\displaystyle p_{1}=1/4~~,~~U_{1}=0.95~,}$
${\displaystyle p_{2}=3/4~~,~~U_{2}=0.92~.}$
Then using the previous formulas, we can calculate the bond price
${\displaystyle P(t,t+1)=A(1)+A(2)=p_{1}U_{1}+p_{2}U_{2}=1/4\cdot 0.95+3/4\cdot 0.92=0.9275~.}$
The interest rate is then given by
${\displaystyle r={\frac {1}{P(t,t+1)}}-1={\frac {1}{0.9275}}-1=7.82\%~.}$
Thus, we see that the pricing of a bond and the determination of interest rate is simple to do once the set of Arrow-Debreu prices, the prices of Arrow-Debreu securities, are known.
See also | {"url":"https://www.knowpia.com/knowpedia/Risk-free_bond","timestamp":"2024-11-14T13:47:09Z","content_type":"text/html","content_length":"147643","record_id":"<urn:uuid:0d929e0b-c966-4356-acf6-5451d1689bdf>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00385.warc.gz"} |
Geometric Mean Vs Arithmetic Mean
Hi, Do we use geometric mean (instead of arithmetic mean) in Coef. of Variation calculation = Std. Dev/geometric mean? I also noticed in mock exam that they used geometric mean for sharpe ratio =
(geometric mean - RFR)/Std. Deviation. Could anyone confirm? Thank you.
By definition it should be the arithmetic mean for both, check the Mock again, I remember using the Am NOT the Gm.
Strange. I’m pretty sure arithmetic mean is standard in both cases.
you wouldnt use geometric mean in a cov calculation. that doesnt make any sense
here’s finally found the question. Given, Arithmetic mean return = 14.30% Geom. mean return = 12.70% Variance = 380 Beta = 1.35 RFR = 4.25% The Coef. of Variation and Sharpe Ratio, respectively are:
A) 1.36, 0.52 B) 1.36, 7.44 C) 1.53, 0.52 D) 1.53, 7.44 | {"url":"https://www.analystforum.com/t/geometric-mean-vs-arithmetic-mean/26800","timestamp":"2024-11-09T09:11:53Z","content_type":"text/html","content_length":"24468","record_id":"<urn:uuid:82f68f9a-fd72-4858-8cc4-47d7da1030ab>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00488.warc.gz"} |
How to Fix - NameError: name 'math' is not defined - Data Science Parichay
If you are working with Python and trying to use the math library, you may encounter the “NameError: name ‘math’ is not defined” error. In this tutorial, we will explore why this error occurs and the
steps required to fix it such that your Python code can successfully run without errors.
We will cover common causes of the error and provide solutions to help you get your code up and running quickly. So, let’s get started!
Why does the NameError: name 'math' is not defined error occur?
This error occurs when you try to use the math library in your Python code, but Python cannot find the math module in its namespace. The following are some of the scenarios in which this error
usually occurs.
1. You have not imported the math module.
2. You have imported the math module using a different name.
How to fix the NameError: name 'math' is not defined?
The math library in Python is a built-in module that provides various mathematical operations and functions. It includes functions for basic arithmetic operations, trigonometric functions,
logarithmic functions, and more. Since this library is a built-in library in Python, you don’t need to separately install it. You can import it and start using it.
Let’s now look at the above scenarios in detail.
The math module is not imported
It can happen that you are trying to use the math module without even importing it. This is because Python does not recognize the math library and its functions until it is imported into the code.
For example, let’s try to use math without importing it and see what we get.
# note that math is not imported
# get the square root of a 4
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NameError Traceback (most recent call last)
Cell In[1], line 4
1 # note that math is not imported
3 # get the square root of a 4
----> 4 print(math.sqrt(4))
NameError: name 'math' is not defined
We get a NameError stating that the name math is not defined. To use the math library, you need to import it first.
import math
# get the square root of a 4
Here, we are importing the math module first and then using it to get the square root of 4. You can see that we did not get any errors here.
You can also get a NameError if you are importing only specific parts of the library and then trying to access the entire math library. For example –
from math import sqrt
# get the square root of a 4
NameError Traceback (most recent call last)
Cell In[1], line 4
1 from math import sqrt
3 # get the square root of a 4
----> 4 print(math.sqrt(4))
NameError: name 'math' is not defined
We get a NameError here because we are importing only the sqrt() function from the math library but we are trying to access the entire library. To resolve the above error, either only use the
specific method imported or import the math library altogether.
The math module is imported using a different name
If you import the math module using a different name, for example import math as m, and then try to use the name “math” to use it, you will get a NameError because the name “math” is not defined in
your current namespace.
Let’s look at an example.
import math as m
# get the square root of a 4
NameError Traceback (most recent call last)
Cell In[2], line 4
1 import math as m
3 # get the square root of a 4
----> 4 print(math.sqrt(4))
NameError: name 'math' is not defined
We get a NameError: name 'math' is not defined. This is because we have imported the math module with the name m but we’re trying to use it using the name math.
To fix this error, you can either access math using the name that you have used in the import statement or import math without an alias.
import math as m
# get the square root of a 4
In the above example, we are importing math as m and then using m to access the math module’s methods.
Alternatively, as seen in the example in the previous section, you can import math without any aliases and simply use math to avoid the NameError.
In conclusion, encountering a NameError: name 'math' is not defined error can be frustrating, but it is a common issue that can be easily fixed. By ensuring that the math module is imported correctly
and that the correct syntax is used when calling its functions, you can avoid this error and successfully execute your code. | {"url":"https://datascienceparichay.com/article/how-to-fix-nameerror-name-math-is-not-defined/","timestamp":"2024-11-13T18:04:38Z","content_type":"text/html","content_length":"259644","record_id":"<urn:uuid:8167d7a0-f03a-462b-9bdc-719877b42d9a>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00769.warc.gz"} |
CPAC vs EXP - Comparison tool | Tickeron
CPAC's Valuation (3) in the Construction Materials industry is in the same range as EXP (20). This means that CPAC’s stock grew similarly to EXP’s over the last 12 months.
EXP's Profit vs Risk Rating (5) in the Construction Materials industry is significantly better than the same rating for CPAC (90). This means that EXP’s stock grew significantly faster than CPAC’s
over the last 12 months.
EXP's SMR Rating (27) in the Construction Materials industry is in the same range as CPAC (59). This means that EXP’s stock grew similarly to CPAC’s over the last 12 months.
EXP's Price Growth Rating (40) in the Construction Materials industry is in the same range as CPAC (43). This means that EXP’s stock grew similarly to CPAC’s over the last 12 months.
EXP's P/E Growth Rating (18) in the Construction Materials industry is somewhat better than the same rating for CPAC (55). This means that EXP’s stock grew somewhat faster than CPAC’s over the last
12 months. | {"url":"https://tickeron.com/compare/CPAC-vs-EXP/","timestamp":"2024-11-14T21:27:50Z","content_type":"text/html","content_length":"203567","record_id":"<urn:uuid:b43f0187-9141-4d16-82b0-6a6d449725d7>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00753.warc.gz"} |
DES Encryption / Decryption Tool
Triple DES or DESede , a symmetric-key algorithm for the encryption of electronic data, is the successor of DES(Data Encryption Standard) and provides more secure encryption then DES. The Triple DES
breaks the user-provided key into three subkeys as k1, k2, and k3. A message is encrypted with k1 first, then decrypted with k2 and encrypted again with k3. The DESede key size is 128 or 192 bit and
blocks size 64 bit. There are 2 modes of operation - Triple ECB (Electronic Code Book) and Triple CBC (Cipher Block Chaining).
Symmetric Ciphers Online allows you to encrypt or decrypt arbitrary message using several well known symmetric encryption algorithms such as AES, 3DES, or BLOWFISH.
Symmetric ciphers use the same (or very similar from the algorithmic point of view) keys for both encryption and decryption of a message. They are designed to be easily computable and able to process
even large messages in real time. Symmetric ciphers are thus convenient for usage by a single entity that knows the secret key used for the encryption and required for the decryption of its private
data – for example file system encryption algorithms are based on symmetric ciphers. If symmetric ciphers are to be used for secure communication between two or more parties problems related to the
management of symmetric keys arise. Such problems can be solved using a hybrid approach that includes using asymmetric ciphers. Symmetric ciphers are basic blocks of many cryptography systems and are
often used with other cryptography mechanisms that compensate their shortcomings.
Symmetric ciphers can operate either in the block mode or in the stream mode. Some algorithms support both modes, others support only one mode. In the block mode, the cryptographic algorithm splits
the input message into an array of small fixed-sized blocks and then encrypts or decrypts the blocks one by one. In the stream mode, every digit (usually one bit) of the input message is encrypted
In the block mode processing, if the blocks were encrypted completely independently the encrypted message might be vulnerable to some trivial attacks. Obviously, if there were two identical blocks
encrypted without any additional context and using the same function and key, the corresponding encrypted blocks would also be identical. This is why block ciphers are usually used in various modes
of operation. Operation modes introduce an additional variable into the function that holds the state of the calculation. The state is changed during the encryption/decryption process and combined
with the content of every block. This approach mitigates the problems with identical blocks and may also serve for other purposes. The initialization value of the additional variable is called the
initialization vector. The differences between block ciphers operating modes are in the way they combine the state (initialization) vector with the input block and the way the vector value is changed
during the calculation. The stream ciphers hold and change their internal state by design and usually do not support explicit input vector values on their input.
Security note: Data are transmitted over the network in an unencrypted form! Please do not enter any sensitive information into the form above as we cannot guarantee you that your data won't be | {"url":"https://devtoolcafe.com/tools/des","timestamp":"2024-11-03T23:13:47Z","content_type":"text/html","content_length":"42449","record_id":"<urn:uuid:166f1461-48d3-4652-a4e6-20a98f42275b>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00688.warc.gz"} |
Significance Level Definition - Varsha Saini
The significance level is a common term in statistics, especially in hypothesis testing. It is the probability of rejecting the null hypothesis when it is true. Since we are rejecting a correct null
hypothesis means we are making some mistakes. Therefore it can be regarded as an error. It is a Type 1 or False Positive Error.
The Null Hypothesis (H0) is the Hypothesis to be tested.
The Alternate Hypothesis (H1) is a contradictory statement to the Null Hypothesis.
How to select the level of significance?
The level of significance or significance level is denoted by α. The most common values of α are 0.01, 0.05, and 0.10 which corresponds to 99%, 95%, and 90% confidence in a test. The value of α is
selected based on the certainty you need 0.01 > 0.05 > 0.10. If you want to be 95% confident then there is a 5% (100-95) risk of rejecting a null hypothesis that was true. The same applies to 90%
where the significance level is 10% and 99% where the significance level is 1%.
The value of the significance level depends on the problem to which hypothesis testing is applied. A low significance value is used in cases for which high certainty is required and vice versa.
For Example, if you want to check if a machine is working properly or not, then you may go with a 0.01 significance level since you expect little or no mistakes.
For cases like analyzing human behavior, you may go with a 0.10 significance level. since human behavior can be very uncertain.
False Positive Errors in Hypothesis Testing
In a binary classification problem where the target class is Positive and Negative. Let there be a case where the class is Negative but the model predicts it as a Positive, it is a False Positive or
a Type 1 Error.
You can look at the below confusion matrix to understand the error better.
The probability of making a False Positive Error is α. Since the value of the significance level (α) is selected by you, the responsibility for making this error is on you. | {"url":"https://varshasaini.in/glossary/significance-level/","timestamp":"2024-11-06T23:40:43Z","content_type":"text/html","content_length":"187322","record_id":"<urn:uuid:4ad70a64-b431-45ef-b0d3-fc15c5b476a7>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00329.warc.gz"} |
Add planes — planes3d
planes3d adds mathematical planes to a scene. Their intersection with the current bounding box will be drawn. clipplanes3d adds clipping planes to a scene.
planes3d(a, b = NULL, c = NULL, d = 0, ...)
clipplanes3d(a, b = NULL, c = NULL, d = 0)
Coordinates of the normal to the plane. Any reasonable way of defining the coordinates is acceptable. See the function xyz.coords for details.
Coordinates of the "offset". See the details.
Material properties. See material3d for details.
planes3d draws planes using the parametrization \(a x + b y + c z + d = 0\). Multiple planes may be specified by giving multiple values for any of a, b, c, d; the other values will be recycled as
clipplanes3d defines clipping planes using the same equations. Clipping planes suppress the display of other objects (or parts of them) in the subscene, based on their coordinates. Points (or parts
of lines or surfaces) where the coordinates x, y, z satisfy \(a x + b y + c z + d < 0\) will be suppressed.
The number of clipping planes supported by the OpenGL driver is implementation dependent; use par3d("maxClipPlanes") to find the limit.
A shape ID of the planes or clipplanes object is returned invisibly.
See also
abclines3d for mathematical lines.
triangles3d or the corresponding functions for quadrilaterals may be used to draw sections of planes that do not adapt to the bounding box.
The example in subscene3d shows how to combine clipping planes to suppress complex shapes.
# Show regression plane with z as dependent variable
x <- rnorm(100)
y <- rnorm(100)
z <- 0.2*x - 0.3*y + rnorm(100, sd = 0.3)
fit <- lm(z ~ x + y)
plot3d(x, y, z, type = "s", col = "red", size = 1)
coefs <- coef(fit)
a <- coefs["x"]
b <- coefs["y"]
c <- -1
d <- coefs["(Intercept)"]
planes3d(a, b, c, d, alpha = 0.5)
3D plot
ids <- plot3d(x, y, z, type = "s", col = "red", size = 1, forceClipregion = TRUE)
oldid <- useSubscene3d(ids["clipregion"])
clipplanes3d(a, b, c, d)
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Estimation of the Effect of Anisotropy on Young’s Moduli and Poisson’s Ratios of Sedimentary Rocks Using Core Samples in Western and Central Part of Tripura, India
Estimation of the Effect of Anisotropy on Young’s Moduli and Poisson’s Ratios of Sedimentary Rocks Using Core Samples in Western and Central Part of Tripura, India ()
The earth material is said to be transversely isotropic with vertical axis of symmetry if the elastic properties vary vertically but not horizontally in the simplest horizontal or layered case.
Detecting and quantifying this type of anisotropy are very useful for correlation of sonic logs in vertical and deviated wells and for borehole surface seismic imaging and studies of amplitude
variation with offset (AVO). The yield information about rock stress and fracture density and orientation can be estimated by identifying and quantifying the elastic anisotropy and these parameters
are useful for designing hydraulic fracture jobs and for understanding horizontal and vertical permeability anisotropy. To understand the elastic anisotropy for more complex cases, such as dipping
layers, fractured layered rocks or rocks with multiple fracture sets, superposition of the effects of the individual anisotropies is necessary.
In spite of the fact that shale formations exhibit high anisotropies, the study of fracture related anisotropy has been more intensive, both theoretically [5-12] and experimentally [13-15]. This is
most likely explained because when anisotropy is observed it can be associated with fractures, which have a strong impact on permeability. Highly permeable rocks can be a good oil reservoir. However,
in order to associate anisotropy with fracture zone, a correction for the anisotropic effect of the upper layers of the sedimentary column that is traversed by the wave field is necessary. More
recently, the influence of pore fluid on elastic properties of shale has been investigated. Hornby measured compressional and shear wave velocities up to 80 MPa on two fluid saturated shale samples
under drained conditions [16]. One sample from Jurassic outcrop shale was recovered from under sea and stored in its natural fluid, and the other is Kimmeridge clay taken from a North Sea borehole.
Measurements were made on cores parallel, perpendicular, and at 45˚ to bedding. Values of anisotropy were up to 26% for compressional and 48% for shear wave velocity, and were found to decrease with
increasing pressure. The effect of reduced porosity was, therefore, concluded to be more influential on anisotropy than increased alignment of minerals at higher pressure. The elastic constants,
velocities and anisotropies in shales can be obtained from traditionally measured on multiple adjacent core plugs with different orientations [17]. To derive the five independent constants for
transversely isotropic (TI) rock, Wang measured three plugs separately (one parallel, one perpendicular, and one
Sandstones are rarely clean; they often contain minerals other than quartz, such as clay minerals, which can affect their reservoir qualities as well as their elastic properties. The presence of clay
mineral and clastic sheet silicates strongly influences the physical and chemical properties of both sandstones and shales [18]. Clay can be located between the grain contacts as structural clay, in
the pore space as dispersed clay, or as laminations [19]. The distribution of the clay will depend on the conditions at deposition, on compaction, bioturbation and diagenesis. While most reservoirs
are composed of relatively isotropic sandstones or carbonates, their properties may be modified by stress. Non-uniform compressive stress will have a major effect on randomly distributed microcracks
in a reservoir. When the rock is unstressed all of the cracks may be open, however, compressional stresses will close cracks oriented perpendicular to the direction of maximum compressive stress,
while cracks parallel to the stress direction will remain open. Elastic waves passing through the stressed rock will travel faster across the closed cracks (parallel to maximum stress) than across
the open ones. The sands and sandstone are intrinsically isotropic unless they are fractured, finely layered or clay bearing. Wang shows that the brine-saturated Africa reservoir sands, which are
essentially clay free, have very little anisotropy (average P-anisotropy and Sanisotropy are probably within the measurement uncertainties) [17]. For the brine saturated tight sands, the anisotropy
is 5.0% for the P-wave and 3.3% for the S-wave when averaged over all samples at all pressures. At the net reservoir pressure of 7500 psi (51.7 MPa), anisotropy is slightly lower, averaging 4.6% and
3.2% for P-and S-waves, respectively. Gas-saturated tight sands and shaly sands show some degrees of anisotropy, ranging from 0% to 36% for P-waves and 0.3% to 19.5% for S-waves. When averaged at all
pressures, the anisotropy is 9.9% for the P-wave and 5.5% for the S-wave.
There are various methods to estimate anisotropy parameters. Wang presented a method for measuring seismic velocities and transverse isotropy in rocks using a sing core plug [17]. Wang pointed out
that shale cores must be preserved. Otherwise, once a shale core is dehydrated, fractures will develop along the bedding plane, and the measured seismic velocities and anisotropy will no longer be
accurate [20].
Leslie [21] determined the Thomsen anisotropic parameters ε and δ by measurement of head wave velocities along the seismic lines. This method obtained more realistic anisotropic parameters than those
measured in the laboratory, but needs to ensure that the spread is of sufficient length so that the refractor velocities measured are from rocks below the weathered layer.
In this paper four kinds of core samples; Dry Sandstone, Shale, Sandy Shale and Saturated Sandstone were collected from Tripura oil field. The purpose of this paper is to estimate effects of elastic
anisotropic on Young’s moduli and Poison’s ratios. We used the ultrasonic transmission method to measure V[p], V[sh] and V[sv] wave velocities, as function of confining pressure from 1 MPa to 40.3
MPa. Using these velocities five independent stiffnesses have been calculated and hence Thomsen’s parameters as well as geo-mechanical parameters.
2. Methodology
2.1. Sample Preparation
The four kinds of core samples namely dry sandstone, shale, sandy shale and saturated sandstone were collected from the Tripura field in the zone between 1910 - 2200 m and were analysed. Each
specimen had diameter 25.4 mm and length 50 mm. The samples had significant lattice preferred orientation. Before performing the test the specimens were cored using diamond drill and the ends of the
specimens were ground flat and parallel to
2.2. Experiment Procedure
For each specimens three velocities were measured, one compressional wave and two orthogonal shear wave as a function of pore pressure and confining pressure. Each specimen was jacketed and secured
between a matched set of ultrasonic transducers having resonant frequency 1.0 MH. Specimen reference system has the x-axis parallel to the lamination in the foliation, the y-axis normal to the
lineation in the foliation and z-axis perpendicular to the foliation plane. Main frequency of ultrasound transducer is 1 MHz. Volume change can obtained from linear strains in three directions which
is determined from the piston displacement through necessary correction. Each polarization is sequentially propagated through the rock and each waveform is determined from the data using the
appropriate correction for the travel time through the transducer assembly. The experiments, on dry and saturated specimens, were conducted at confining pressures from 1 Mpa to 40.3 Mpa.
It is necessary to core specimens in three directions: vertical, horizontal and at an intermediate angle between vertical and the plane of isotropy ( Figure 1), to measure traverse isotropy
completely. Three velocities were measured in each direction, leading total nine velocities
Figure 1. Traditional three-plug method for measuring transverse isotropy in laboratory core samples. Three adjacent core plugs (one parallel, one perpendicular, and one ±45 to symmetry axis) must be
cut from whole cores and velocities measured to derive the five elastic constants (Wang, 2002).
for each sample. The particle polarization and the direction of propagation of the three modes of propagation, P S[h] and S[v] in the stratigraphic column within the zone of interest are shown in
Figure 2.
Both in room dry and full water saturation conditions, sandstone specimens were tested. But for shale and sandy shale it is difficult in saturating these rocks. So, only in room dry conditions the
shale and sandy shale specimens were tested.
It was assumed that, the effective hydrostatic pressure is the difference between confining pressure and pore pressure. The pore pressure was kept constant while the confining pressure varied
(drained regime). The confining pressure was varied from atmospheric pressure up to the reservoir effective pressure condition in dry specimen. But the effective pressure never exceeded the reservoir
pressure in saturated specimens. However the pore pressure was varied from atmospheric pressure up to a pressure approximately 32% higher than the reservoir pressure condition.
2.3. Mathematical Analysis
In this paper only the standard three plug method will be discussed since it is commonly used in oil exploration. Standard three plug method is by far the most common measurement approach. The method
requires the extraction of three core plug along prescribed orientation relative to the assumed symmetric axes. These orientations are parallel, perpendicular and typically 45˚ to the vertical
symmetry axis. Either static or dynamic ultrasonic laboratory measurements can be performed on these plugs to provide the magnitude of tensor elements
The relation between stress
Figure 2. Scheme of sample preparation and velocity measurements in shales. Wave propagation and polarization with respect to bedding-parallel lamination is shown. Numbers in parentheses indicate the
phase velocity angle with respect to the bedding-normal symmetry axis (Vernik and Nur, 1992).
where C[ijkl] is the Voigt matrix [22].
From Figure 1 it is seen that if the axis is denoted by Z, the other two principal axes (X and Y) are parallel to the transversely isotropic plane. In this coordinate system, the stiffness matrix C
is expressed as:
From the above Equation (2), Voigt matrix C, there are five non-zero independent elastic constants:
Using five elastic stiffness C[ij][ ], a bulk modulus, one vertical Young’s moduli, E[3] and one horizontal Young’s modulus E[1] and three dynamic Poisson’s ratios can be determines for a hexagonal
material as follows [23]:
These dynamic Poison’s ratios Equations (7) and (8),
There are three types of velocity anisotropies (
3. Result and Discussion
Four types of rock samples that represent elastic and anisotropy behaviour in the sedimentary column: dry sandstone, saturated sandstone, shale and sandy shale have been collected from Tripura oil
filed. Each of these rocks types exhibit varying sensitivities to confining pressure, and as a consequence the anisotropies are affected differently as the confining pressure varies. The behaviour of
the elastic anisotropy for the different rock types as the confining pressure is increased (Figures 3-6).
Figure 3. Comparison of Thomsen’s parameters as a function of confining pressure for dry sandstone.
Figure 4. Comparison of Thomsen’s parameters as a function of confining pressure for shale.
Figure 5. Comparison of Thomsen’s parameters as a function of confining pressure for sandyshale.
Figure 6. Comparison of Thomsen’s parameters as a function of confining pressure for saturated sandstone.
The analysis shows that the anisotropy parameters are increased as the confining pressures are increased for all the rock samples except saturated sandstone. Figure 6, shows that the anisotropy
parameters are decreased when confining pressure are increase up to 15 Mpa, but again anisotropy parameters are increased with increasing confining pressure beyond 15 Mpa up to 40 Mpa for saturated
3.1. Shale
Figure 7 shows that the Poisson’s ratio
Figure 4 shows that the variation of velocity anisotropy as a function of confining pressure. Our analysis found that
3.2. Sandy Shale
Figure 5 shows that, the V[p] anisotropy, [p]-anisotropy more than the V[sh]-anisotropy. When an Sh-wave propagates in the vertical and horizontal directions, the particle is polarized parallel to
the plane of isotropy, and it always encounters more rigid material. However, when a P-wave propagates in the vertical and horizontal directions, the polarization of the particle and the direction of
propagation are perpendicular and parallel to the plane of isotropy, respectively. The P-wave polarized in the vertical direction encounter softer material than the P-wave polarized in the horizontal
direction. But our analysis shows that the V[p]-anisotropy and V[sh]-anisotropy are not approaching same value within our confining pressure range. That implies that the fracture of the specimen is
not close within our pressure range. However, the anisotropic caused by preferred orientation of minerals, by fracture, and by layering are always due to the fact that the material is softer in the
direction perpendicular to the direction of preferential orientation than in the direction
Figure 7. Relationship between Poisson’s ratio and confining pressure for shale.
parallel to it.
3.3. Dry and Saturated Sandstone
At low pressure, the V[p] anisotropy[sh] anisotropy
After saturation V[p](0˚) and V[p](45˚) increase , while V[sv](0˚) decreases. V[p](0˚) and V[p](45˚) increase because the material is less compressible. A pore filled with fluid resists compression
in a similar way when it is filled with solid material. The difference between V[p](0˚) for dry and for saturated sandstone decreases as the confining pressure is increased However, the difference
between V[s](0˚) for dry and for saturated sandstones remain approximately constant The decrease in V[s](0˚) is caused by a significant increase in the bulk density, while C[44] remains approximately
constant ( Figure 8). This explains why the difference between V[s](0˚) for dry and V[s](0˚) for saturated does not change as the confining pressure is increased. V[p](90˚) shows a significant
increase after saturation. The significant increase is due to the polarization of the particle and the directions of propagation are perpendicular to the plane of isotropy, and fluid have a strong
effect in this direction. The stiffness C[33] shows a pronounced increase reaching the same value as C[11]. C[44] remains approximately constant and C[66] shows a slight decrease ( Figure 8) after
After saturation, the P-wave anisotropy significantly decrease. However,
3.4. The Effect of Anisotropy on the Young’s Moduli and Poison’s Ratios
The strong effects of anisotropy parameters Figures 9-16. The vertical Poisson’s ratio significantly decreases with
We showed that anisotropy has significant effect on the two key parameters; the Young’s modulus and the Poisson’s ratio. In fact there are two Young’s moduli and at least two Poisson’s ratios and
they vary significantly with the magnitudes of the anisotropy parameters
4. Conclusions
The study found that, the elastic anisotropy parameters are key factors which affect the estimation of hydrocarbon reservoir characterisation parameters. To estimate more accurate hydrocarbon
reservoir characterisation parameters, we need vertical P-wave and S-wave velocities and three anisotropy parameters. With high quality and high resolution surface seismic data, we can reason-
Figure 8. Stiffness for dry and saturated sandstone specimens. The dashed curves show the stiffness in dry condition.
Figure 9. Relationship between vertical Young’s modulus and Epsilon
Figure 10. Relationship between horizontal Young’s modulus and Epsilo
Figure 11. Relationship between horizontal Young’s modulus and Gamma
Figure 12. Relationship between vertical Young’s modulus and Gamma
Figure 13. Relationship between horizontal Poisson’s ratio and Epsilon
Figure 14. Relationship between vertical Poisson’s ratio and Epsilon
Figure 15. Relationship between horizontal Poisson’s ratio and Gamma
Figure 16. Relationship between vertical Poisson’s ratio and Gamma
ably estimate
Thus through this study we proposed a prestack full waveform AVO inversion in a VTI medium constraining elastic parameters based on depth velocity analysis of P-wave surface seismic data and other
information available.
This study also suggested that the Young’s moduli and Poisson’s ratios that the driller must have are the static moduli, not the dynamic moduli estimated through seismic experiment. The relationship
established themselves in this study is laboratory based. The static moduli is larger than dynamic Young’s moduli by the factor 2 and it may vary basin to basin.
The initial model can be integrated by adding drilling and micro seismic monitoring, initial production, and estimated ultimate recovery data for updation.
We thank M/S Pandit Deendayal Petroleum University for supporting in preparation of manuscripts.
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[5] R. Brown and J. Korrigan, “On the Dependence of the Elastic Properties of a Porous Rock on the Compressibility of the Pore Fluid,” Geophysics, Vol. 40, No. 4, 1975, pp. 608-616. http://dx.doi.org
[6] J. A. Hudson, “Overall Properties of a Cracked Solid,” Mathematical Proceedings of the Cambridge Philosophical Society, Vol. 88, No. 2, 1980, pp. 371-384. http://dx.doi.org/10.1017/
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[8] J. A. Hudson, “Overall Elastic Properties of Isotropic Materials with Arbitrary Distribution of Circular Cracks,” Geophysical Journal International, Vol. 102, No. 2, 1990, pp. 465-469. http://
[9] J. A. Hudson, “Crack Distribution Which Accounts for a Given Seismic Anisotropy,” Geophysical Journal International, Vol. 104, No. 3, 1991, pp. 517-521. http://dx.doi.org/10.1111/
[10] J. A. Hudson, E. Liu and Crampin, “The Mechanical Properties of Materials Which Interjected Cracks and Pores,” Geophysical Journal International, Vol. 124, No. 1, 1996, pp. 105-112. http://
[11] T. Mukerji and G. Mavo, “Pore Fluid Effects on Seismic Velocity in Anisotropic Rocks,” Geophysics, Vol. 59, No. 2, 1994, pp. 233-244. http://dx.doi.org/10.1190/1.1443585
[12] L. Thomsen, “Elastic Anisotropy Due to Aligned Cracks in Porous Rock,” Geophysical Prospecting, Vol. 43, No. 6, 1995, pp. 805-829. http://dx.doi.org/10.1111/j.1365-2478.1995.tb00282.x
[13] T. D. Jones, “Wave Propagation in Porous Rock and Models for Crustal Structures,” Ph.D. Thesis, Standford University, 1983.
[14] N. M. Lucet and P. A. Tarif, “Shear—Wave Birefringence and Ultrasonic Shear Wave Attenuation Measurements,” SEG Annual Meeting, Anaheim, 30 October-3 November 1988, pp. 922-924.
[15] M. Zamora and J. P. Porier, “Experimental Study of Acoustic Anisotropy and Birefringence in Dry and Saturated Fontainebleau Sandstone,” Geophysics, Vol. 55, No. 11, 1990, p. 1455. http://
[16] B. E. Hornby, “Experimental Laboratory Determination of the Dynamic Elastic Properties of Wet, Drained Shales,” Journal of Geophysical Research: Solid Earth, Vol. 103, No. B12, 1998, pp.
[17] Z. Wang, “Seismic Anisotropy in Sedimentary Rocks,” Part 1: A Single-Plug Laboratory Method,” Geophysics, Vol. 67, No. 5, 2002, pp. 1415-1422. http://dx.doi.org/10.1190/1.1512787
[18] K. Bjørlykke, “Clay Mineral Diagenesis in Sedimentary Basins—A Key to the Prediction of Rock Properties. Examples from the North Sea Basin,” Clay Minerals, Vol. 33, 1998, pp. 15-34.
[19] M. S. Sam and A. Andrea, “The Effect of Clay Distribution on the Elastic Properties of Sandstones,” Geophysical Prospecting, Vol. 49, No. 1, 2001, pp. 128-150. http://dx.doi.org/10.1046/
[20] Z. Wang, “Seismic Anisotropy in Sedimentary Rocks, Part 2: Laboratory Data,” Geophysics, Vol. 67, No. 5, 2002, pp. 1423-1440. http://dx.doi.org/10.1190/1.1512743
[21] J. M. Leslie and D. C. Lawton, “A Refraction Seismic Field Study to Determine the Anisotropic Parameters of Shales,” The Leading Edge, Vol. 17, No. 8, 1998, pp. 1127-1129. http://dx.doi.org/
[22] G. Mavko, T. Mukerji and J. Dvorkin, “Rock Physics Handbook: Tools for Seismic Analysis in Porous Media,” Cambridge University Press, 1998.
[23] F. R. Pena, “Elastic Properties of Sedimentary Anisotropic Rocks,” Dissertation, Massachusetts Institute of Technology, 1998, pp. 19-26. | {"url":"https://www.scirp.org/journal/paperinformation?paperid=43200","timestamp":"2024-11-10T18:35:44Z","content_type":"application/xhtml+xml","content_length":"146660","record_id":"<urn:uuid:bd944587-16cd-4360-91a6-1c4d3738360a>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00399.warc.gz"} |
The effects of surface roughness
Hydraulic energy conversion into mechanical energy in a hydroturbine is inevitably associated with energy losses. Energy losses depend upon the type of turbine, its design, size (dimensions) of the
turbine and regime of operation. Hydraulic turbines are very efficient prime movers. Modern powerful reaction turbines have high values of overall efficiency of 92% to 96% at design-point operation
and 86% to 91% at rated regimes. Nevertheless, further improvement of efficiency, especially at off-peak regimes of overload and part-load operation is important and constantly sought, because even
0.1% increase in efficiency of powerful modern machines gives a substantial increase in annual output of electrical energy. For a medium head Francis turbine of 185,500kW output capacity working at a
power factor of 0.6, 0.1% higher efficiency may be worth US$117,000 of additional annual revenue at US$0.12/kWh, the capitalised value of which may be about US$1.5M per unit. Also, it was estimated
that 1% difference in efficiency of a Kaplan turbine was equivalent to 10% of turbine price.
To improve the performance of such efficient turbines, it is necessary to investigate all the hydraulic elements and runner in particular, to study the nature of losses and to find means of reducing
them. Fluid friction losses – especially in the runner where the relative velocities are the greatest – need special consideration. The flow conditions in a hydraulic turbine, especially the
prototype turbine, are such that the friction factor is a function of the relative roughness only. Therefore the roughness protrusions of the runner surfaces in contact with the flow are very
important and should be the least. Greatest care is hence taken in casting and finishing these surfaces and these operations consume a great deal of time and costs..
Losses in hydraulic turbines and scaling up from model to prototype
In scaling up hydraulic efficiency of turbine hh from model to prototype, it is generally accepted as a standard practice that the hydraulic losses can be divided into two parts: the part due to
kinetic losses arising from vortices etc., that needs no scaling up and the other due to fluid friction which is to be scaled up. Hydraulically smooth flow can be achieved in the model turbines at
the velocities generally prevailing by limiting boundary surface roughness ( RMS – root mean squire – value of roughness protrusions ) Ra typically to 0.4 microns in runner and guide wheel achievable
through suitable surface finish. Limiting roughness for hydraulically smooth flow through other components is about 2.5 times this value.
For scaling up model efficiency to homologous prototype turbine assuming hydraulically smooth flow, several expressions obtained by various investigators based on their experiences are available.
Those expressions based on the concept of subdivision of internal losses into the scaled up and non-scaled up parts take the general form given by Ackeret, Hutton and Osterwalder.
where: η[h] is hydraulic efficiency, ζ[h] relative hydraulic loss = 1-η[h] , Re is Reynolds number, V is loss distribution coefficient, M and P refer to model and prototype turbines.
The index α and loss distribution coefficient V are based on experience. The international-electrotechnical-commission (iec), International Association for Hydraulic Research (iahr) and their working
groups collected and analysed vast data on both model and prototype turbines. Hutton gave values of 0.7 for V and 0.2 for α applicable to Kaplan turbines. Osterwalder gave α=0.16 applicable to all
reaction turbines and rotodynamic pumps. IEC [1] specified V=0.7 for Francis turbines and pump-turbines operating in turbine mode. Based on a critical study of both experimental and analytical data,
Fay [2] gave an average equation:
n[q] is the specific speed based on speed n in rpm, discharge Q in m^3/sec and head H in meters. Any characteristic velocity like u[2] the peripheral velocity at the runner diameter at exit can be
taken to calculate R[e] as only the ratio of R[e] is involved and velocity ratio is constant in homologous turbines. Equation [1] gives the efficiency of hydraulically smooth prototype turbines.
However, most prototype turbines operate in hydraulically rough regime of flow.
Computation of effect of boundary surface roughness in prototype turbines
In computing the effects of roughness of surfaces in flow passages on the efficiency of prototype hydraulic turbines, a two-step method suggested by Osterwalder can be adopted. In the first step,
from the hydraulic efficiency of a tested homologous model, the hydraulic efficiency of the prototype turbine is calculated for hydraulically smooth flow using appropriate step up formulau (Equations
1&2). In the second step, from the hydraulic efficiency η[h] [smooth] of the prototype turbine calculated for operation in hydraulically smooth flow regime, the hydraulic efficiency η[h] [rough] for
operation of the prototype turbine in hydraulically rough regime at the same Reynolds number is calculated. The total relative hydraulic loss ζ[h] in the turbine space is the sum of the relative
losses in individual elements i.e. the relative losses in the spiral casing (ζ[sp]), stay ring (ζ[st]), distributor (ζ[d]), runner (ζ[r]) and draft tube (ζ[d.t]).
Where, Δ H is the part of turbine head lost in overcoming the hydraulic resistance. Combining the vanes treated as plates together and considering disc friction losses,
The equation for calculation of η[h rough] from η[h smooth] resulting from an analysis by Hutton and Fay can be used after suitable modifications considering different loss distribution coefficients
for different components.
where sp denotes spiral casing, v the vanes, r the runner, dt the draft tube d the disc and λ the friction factor.
Values of the component distribution coefficient V have been estimated from the extensive experimental and analytical data for Francis turbines from JSME [3] and Osterwalder [4].
Friction factor for hydraulically smooth and rough boundaries, λ[smooth] and λ[rough] for each component at the same Reynolds number are derivable from the relevant expressions or charts.
Substituting the values V and λ thus obtained into equation (5), 1-η[hPrough] = ζ[hPrough] can be derived.
To study the effect of surface roughness on the hydraulic losses or efficiency, it is not necessary to calculate values of λ. It can be seen from equation (5), that it is only necessary to calculate
the ratios of λ[rough] to λ[smooth]. Examining the Colebrook and White formula for pipe friction valid for random commercial roughness:
where k[s] is equivalent sand roughness and D the diameter,
or its modification by Fay [2,5]
A varying from 0.65 to 0.9 depending on the type of roughness, and the flat plate friction formulau
where c[f] is drag coefficient and L the length of the plate, a universal roughness law is evolved by Fay[2,5]:
where Ra[actual] is the root mean squire (RMS) value of roughness protrusions of the prototype component for which λ[rough] / λ[smooth] is to be calculated for substitution in equation (9) and Ra
[char] is a characteristic roughness equal to C[2] ν/w with C[2] as a constant, ν as kinetic viscosity and w as the characteristic velocity, the same as for R[e]. R[a] = k[s] / 1.7 established by
Nixon and Cairney [6]. The values of coefficients and index are shown in table 2 for various components of turbine:
These modified equations by Fay give better fit for ground surfaces. Limiting conditions for hydraulically smooth flows can be obtained when λ[rough] / λ[smooth] =1. Substitution of this condition in
to Equation [9] yields Ra[actual] / Ra[char] =1-A = 0.25. Therefore:
If Ra[actual] / Ra[char] ≤ 1-A = 0.25, the flow is hydraulically smooth, and lrough = lsmooth
If Ra[actual] / Ra[char] > 1-A = 0.25, the flow is affected by roughness and, λ[rough] / λ[smooth] can be obtained for various values of Ra[actual] by solving Equation [9]. By substituting λ[rough] /
λ[smooth] and values of V into equation [5], 1-η[hPrough] or relative losses ζ[hPrough] in a rough prototype turbine can be evaluated.
Example of effect of roughness on prototype efficiency
The example chosen is Vevey’s Francis turbine at Serre-Poncon, France. Published / derived details from Lucien Vivier [7] and Th. Bovet [8] are:
• Head H = 124.5m, Discharge Q = 75m^3/sec, Speed n = 214 rpm, Power P = 83,900kW
• Efficiency h = 91.6%, Specific speed with P in hp n[s] = 177 rpm, n[q] = n Q^1/2 /H^3/4 = 49.72.
• Inlet to spiral casing = 3.7m diameter, Characteristic velocity w = 75/(π3.72/4) = 6.975m/sec
• Guide vane pcd D[0] = 4.17m diameter, Height b[0] = 0.679m, GV angle with radius = 450
• Characteristic velocity for guide vanes w = 75/(4.17π0.679)sec 45º= 11.867m/sec.
• Runner exit diameter D[2] = 3.197m, Characteristic velocity for runner w = π3.197(214/60)=35.822m/sec.
• Draft tube inlet diameter D[2] = 3.197m, Characteristic velocity w = 75/(π3.1972/4) = 9.372m/sec.
• Max. runner diameter at seals = 3.58m, Characteristic velocity for disc w=π3.58(214/60)=40.11m/sec.
Kinetic viscosity n used in computations =10^-6 m^2/sec.
The spiral casing and draft tube are made of welded steel plates and are coated with bituminous or epoxy paint. Their equivalent sand roughness k[s] is 1.5 micron and R[a] is 0.9micron. They are
assumed to be maintained as such. The guide vanes and runner are ground and polished. Effect of different degrees of grinding and corresponding roughness is shown in these computations.
The parameters for computations are shown in table 3, A being 0.75 for all components of prototype:
Letting λ[rough] / λ[smooth] = X , relevant values are substituted in equation [9] for each component and X solved for by successive approximations.
• Spiral casing, 0.10462683=X^8-0.75/X^0.5 is solved to get X=0.9815572, VX=0.08834048 *
• Guide vanes, 0.18987041 =X^7-0.75/X^0.5 is solved to getX=0.9916538, VX=0.10908192 *
• Runner, 0.57314822=X^7-0.75/X^0.5 is solved to get X=1.0392004, VX=0.21199688
• Draft tube, 0.14058 =X^8-0.75/X^0.5 is solved to get X=0.986333, VX=0.01183600 *
• Disc friction, 0.35653802=X^5-0.75/X^0.5 is solved to get X=1.0191448, VX=0.07134014
S VX =0.49259484
Substituting in equation (5),
η[P] = 91.6% or ζ[P] = 8.4%. Considering mechanical losses to be 0.9%, ζ[hP] = 7.5%.
If this is taken as without roughness considered,
Increase in ζ[hP] = drop in η[hP] = 0.04946% .
For various values of surface roughness Ra for ground components, the computed drop in efficiency of rough prototype turbine are shown in table 4.
Accuracies, uncertainties and allowances
Accuracy in the determination of hydraulic efficiency of the prototype turbine from field tests ranges from 1% to 2%. Or the relative hydraulic losses 1-hhPrough or zhPrough are thrown into an error
of 0.12% to 0.25%. Therefore determination of loss distribution coefficient V based on such prototype test data is also subject to high uncertainty. Further refinement of scaling up calculations are
therefore not warranted at this stage. Various scaling up procedures recommended are for customers to assess bids or estimates of prototype performance from witness model tests and not for design
developments. Designers in manufacturing firms are already adopting sophisticated computational methods for the analysis of boundary layer development and drag from computed velocities over the
surfaces of various components. Better methods of scaling up can be expected in due time.
Model tests are considered very accurate with error in determination of model efficiency of ±0.2%. Models are made as smooth as possible to ensure hydraulically smooth flow. But the measured
efficiency changes due to roughness are also small. The uncertainties in the estimates of prototype efficiency are:
• With constant V and without considering roughness effects: ±0.83%
• Considering variation of V with nq (eq.2) but without roughness effects: ±0.65%
• With constant V but considering roughness effects: ±0.69%
• Considering both variation in V and roughness effects: ±0.46%
The scaling-up procedures to obtain prototype efficiency are based on values of V which are good averages for turbines with the same nq and on the assumption that turbines are equally sensitive to
roughness as pipes, discs and plates. The V values may differ for uncommon designs. The above estimates of uncertainties include such differences.
From computations exemplified corroborating with the results obtained earlier by Fay[2], it can be seen that, if a decimal point of efficiency is considered in deciding on bids, the roughness of
ground surfaces will have to be limited to about 1 micron. Finishes to 3.2 microns as limited by IEC code 60193 of 1999 [9] may result in efficiency losses of the order of 0.4% (reference table 4).
Loss of efficiency of 0.5% threshold recommended for repair of turbines affected by cavitation pitting amounts to about 5 microns of equivalent surface roughness.
The paper synthesizes analytical procedures with practical data and provides a reasonably simple computational method to obtain realistic estimates for roughness effects on the optimum efficiency of
Francis turbines. The universal roughness law evolved to study the effect of surface roughness on the hydraulic losses or efficiency gives better fit for ground surfaces of a turbine. Making the
reasonable assumption that turbine components are equally sensitive to roughness as pipes, plates and discs, the changes in prototype efficiencies due to roughness computed by the proposed method
provide the minimum changes that can be expected. The method provides a more reliable tool for closer assessment of offers. The computational procedure can be used by manufacturers to set a degree of
finish for the turbine to achieve the required degree of change in prototype efficiency due to roughness and by maintenance engineers to fix maintenance schedules. Model tests still remain more
accurate than field tests.
Table 1 Table 2 Table 3 Table 4 (1) Equation 1 (2) Equation 2 (3) Equation 3 (4) Equation 4 (5) Equation 5 (6) Equation 6 (7) Equation 7 (8) Equation 8 (9) Equation 9 (10) Equation 10 (11) Equation
11 (12) Equation 12 Author Info:
P.Krishnamachar, Hydro Consultant, 1208 Hogan Drive, South Plainfield, New Jersey 07080, US and Arpad Fay, Chief Consultant (Retd), Ganz Mawag, 33 Szent Donat ut H-2097, Pilisborosjeno, Hungary.
Email: pkrishnamachar@rediffmail.com and arpad.fay@axelero.hu
TablesTable 1 Table 2 Table 3 Table 4 | {"url":"https://www.waterpowermagazine.com/analysis/the-effects-of-surface-roughness/","timestamp":"2024-11-02T15:19:50Z","content_type":"text/html","content_length":"167238","record_id":"<urn:uuid:913cde26-ad05-4f06-a272-df839bb8370e>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00798.warc.gz"} |
The Power of Mathematical Visualization
Many people believe they simply aren’t good at math—that their brains aren’t wired to think mathematically. But just as there are multiple paths to mastering the arts and humanities, there are also
alternate approaches to understanding mathematics. One of the most effective methods by far is visualization. If a picture speaks a thousand words, then in mathematics a picture can spawn a thousand
01. The Power of a Mathematical Picture
Professor Tanton reminisces about his childhood home, where the pattern on the ceiling tiles inspired his career in mathematics. He unlocks the mystery of those tiles, demonstrating the power of
visual thinking. Then he shows how similar patterns hold the key to astounding feats of mental calculation.
02. Visualizing Negative Numbers
Negative numbers are often confusing, especially negative parenthetical expressions in algebra problems. Discover a simple visual model that makes it easy to keep track of what’s negative and what’s
not, allowing you to tackle long strings of negatives and positives-with parentheses galore.
03. Visualizing Ratio Word Problems
Word problems. Does that phrase strike fear into your heart? Relax with Professor Tanton’s tips on cutting through the confusing details about groups and objects, particularly when ratios and
proportions are involved. Your handy visual devices include blocks, paper strips, and poker chips.
04. Visualizing Extraordinary Ways to Multiply
Consider the oddity of the long-multiplication algorithm most of us learned in school. Discover a completely new way to multiply that is graphical-and just as strange! Then analyze how these two
systems work. Finally, solve the mystery of why negative times negative is always positive.
05. Visualizing Area Formulas
Never memorize an area formula again after you see these simple visual proofs for computing areas of rectangles, parallelograms, triangles, polygons in general, and circles. Then prove that for two
polygons of the same area, you can dissect one into pieces that can be rearranged to form the other.
06. The Power of Place Value
Probe the computational miracle of place value-where a digit’s position in a number determines its value. Use this powerful idea to create a dots-and-boxes machine capable of performing any
arithmetical operation in any base system-including decimal, binary, ternary, and even fractional bases.
07. Pushing Long Division to New Heights
Put your dots-and-boxes machine to work solving long-division problems, making them easy while shedding light on the rationale behind the confusing long-division algorithm taught in school. Then
watch how the machine quickly handles scary-looking division problems in polynomial algebra.
08. Pushing Long Division to Infinity
If there is something in life you want, then just make it happen!” Following this advice, learn to solve polynomial division problems that have negative terms. Use your new strategy to explore
infinite series and Mersenne primes. Then compute infinite sums with the visual approach.”
09. Visualizing Decimals
Expand into the realm of decimals by probing the connection between decimals and fractions, focusing on decimals that repeat. Can they all be expressed as fractions? If so, is there a straightforward
way to convert repeating decimals to fractions using the dots-and-boxes method? Of course there is!
10. Pushing the Picture of Fractions
Delve into irrational numbers-those that can’t be expressed as the ratio of two whole numbers (i.e., as fractions) and therefore don’t repeat. But how can we be sure they don’t repeat? Prove that a
famous irrational number, the square root of two, can’t possibly be a fraction.
11. Visualizing Mathematical Infinities
Ponder a question posed by mathematician Georg Cantor: what makes two sets the same size? Start by matching the infinite counting numbers with other infinite sets, proving they’re the same size. Then
discover an infinite set that’s infinitely larger than the counting numbers. In fact, find an infinite number of them!
12. Surprise! The Fractions Take Up No Space
Drawing on the bizarre conclusions from the previous lecture, reach even more peculiar results by mapping all of the fractions (i.e., rational numbers) onto the number line, discovering that they
take up no space at all! And this is just the start of the weirdness.
13. Visualizing Probability
Probability problems can be confusing as you try to decide what to multiply and what to divide. But visual models come to the rescue, letting you solve a series of riddles involving coins, dice,
medical tests, and the granddaddy of probability problems that was posed to French mathematician Blaise Pascal in the 17th century.
14. Visualizing Combinatorics: Art of Counting
Combinatorics deals with counting combinations of things. Discover that many such problems are really one problem: how many ways are there to arrange the letters in a word? Use this strategy and the
factorial operation to make combinatorics questions a piece of cake.
15. Visualizing Pascal's Triangle
Keep playing with the approach from the previous lecture, applying it to algebra problems, counting paths in a grid, and Pascal’s triangle. Then explore some of the beautiful patterns in Pascal’s
triangle, including its connection to the powers of eleven and the binomial theorem.
16. Visualizing Random Movement, Orderly Effect
Discover that Pascal’s triangle encodes the behavior of random walks, which are randomly taken steps characteristic of the particles in diffusing gases and other random phenomena. Focus on the
inevitability of returning to the starting point. Also consider how random walks are linked to the gambler’s ruin” theorem.”
17. Visualizing Orderly Movement, Random Effect
Start with a simulation called Langton’s ant, which follows simple rules that produce seemingly chaotic results. Then watch how repeated folds in a strip of paper lead to the famous dragon fractal.
Also ask how many times you must fold a strip of paper for its width to equal the Earth-Moon distance.
18. Visualizing the Fibonacci Numbers
Learn how a rabbit-breeding question in the 13th century led to the celebrated Fibonacci numbers. Investigate the properties of this sequence by focusing on the single picture that explains it all.
Then hear the world premiere of Professor Tanton’s amazing Fibonacci theorem!
19. The Visuals of Graphs
Inspired by a question about the Fibonacci numbers, probe the power of graphs. First, experiment with scatter plots. Then see how plotting data is like graphing functions in algebra. Use graphs to
prove the fixed-point theorem and answer the Fibonacci question that opened the lecture.
20. Symmetry: Revitalizing Quadratics Graphing
Throw away the quadratic formula you learned in algebra class. Instead, use the power of symmetry to graph quadratic functions with surprising ease. Try a succession of increasingly scary-looking
quadratic problems. Then see something totally magical not to be found in textbooks.
21. Symmetry: Revitalizing Quadratics Algebra
Learn why quadratic equations have quad” in their name, even though they don’t involve anything to the 4th power. Then try increasingly challenging examples, finding the solutions by sketching a
square. Finally, derive the quadratic formula, which you’ve been using all along without realizing it.”
22. Visualizing Balance Points in Statistics
Venture into statistics to see how Archimedes’ law of the lever lets you calculate data averages on a scatter plot. Also discover how to use the method of least squares to find the line of best fit
on a graph.
23. Visualizing Fixed Points
One sheet of paper lying directly atop another has all its points aligned with the bottom sheet. But what if the top sheet is crumpled? Do any of its points still lie directly over the corresponding
point on the bottom sheet? See a marvelous visual proof of this fixed-point theorem.
24. Bringing Visual Mathematics Together
By repeatedly folding a sheet of paper using a simple pattern, you bring together many of the ideas from previous lectures. Finish the course with a challenge question that reinterprets the folding
exercise as a problem in sharing jelly beans. But don’t panic! This is a test that practically takes itself! | {"url":"https://www.videoneat.com/courses/14618/power-mathematical-visualization/","timestamp":"2024-11-02T17:25:43Z","content_type":"text/html","content_length":"141272","record_id":"<urn:uuid:6476291b-8f6e-4a36-afb6-d63dfc89a629>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00358.warc.gz"} |
Quick Hitter #1: Ad-hoc vs Structured Problem-Solving
Both methods are required to be an effective problem solver
This is the first of many “Quick Hitters” (QH) I plan to publish. One of my favorite comedians uses the phrase, “Quick Hitter”, and I’m reusing it for this type of post. A “Quick Hitter” will be
brief and contain a single perspective, learning, thought, or insight on a specific topic.
We tend to bias towards structured problem-solving, however, ad-hoc problem-solving is required in the real world. You need to use both methods to be an effective problem solver. If you are biased
towards one or another, learn the other and when to use it.
Key Takeaways
• It isn’t about whether ad-hoc vs. structured problem-solving is better; both are necessary, and neither is right nor wrong. We tend to bias towards a structured method as the only way because
it’s logical and defensible when we’re questioned.
• If you are biased towards action and GSD, you might be more of an ad-hoc problem solver. If you are interested in de-risking your situation and ensuring you get it right you might be a more
structured problem solver.
• An ad-hoc problem solver will need to be structured at some point and vice versa. The adage “in the absence of data, take action” lends itself to the ad-hoc method. From there, you can use that
data and use the structured method.
• The best problem solvers understand that both approaches are necessary in our complex world. If you’re purely structured, you end up with analysis paralysis, if you’re too ad-hoc, you introduce a
greater chance of being wrong.
Your goal is to be biased for action AND right a lot. In a given moment this seems like a necessary trade-off to make. However, when you understand both are necessary for good problem-solving, you
can implement both methods consciously and appropriately.
My Ask
Thank you for reading this article. I would be very grateful if you complete one or all of the three actions below. Thank you!
1. Like this article by using the ♥︎ button at the top or bottom of this page.
2. Share this article with others.
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Applying Abaqus Velocity | Angular VelocityVSLinear Velocity
Applying Abaqus Velocity | Angular Velocity VS Linear Velocity
04 Jul
Linear velocity refers to how fast an object moves in a straight line, while angular velocity measures the rate of rotation around an axis. Linear velocity is measured in units such as meters per
second (m/s), whereas angular velocity is measured in radians per second (rad/s). The main difference is that linear velocity focuses on straight-line movement, while angular velocity describes
rotational motion. In this blog, we will also explore how to apply these concepts in Abaqus using “Abaqus Velocity” and discuss “Angular Velocity VS Linear Velocity” to provide a clear understanding
of their practical applications.
Tangential velocity, on the other hand, is the linear speed of an object moving along a circular path, measured at any point tangent to the circle. It differs from angular velocity in that tangential
velocity considers the actual speed of a point on the rotating object, while angular velocity measures the rate of rotation. By understanding these distinctions, we can effectively analyze and design
mechanical systems in various fields.
1. What is Linear velocity?
Linear velocity refers to the rate of change of an object’s position with respect to time in a straight line. In simpler terms, it measures how fast something is moving in a particular direction.
1.1. Examples of Linear Velocity in Real Life
• Car on a Highway: When a car travels at a constant speed of 60 miles per hour on a straight road, it has a linear velocity.
• Runner on a Track: A sprinter running 100 meters in 10 seconds has a linear velocity calculated as the distance divided by the time.
• Airplane Taking Off: An airplane accelerating down the runway before takeoff demonstrates increasing linear velocity.
1.2. Unit and Formula of Linear Velocity
The standard unit for linear velocity is meters per second (m/s). However, it can also be expressed in other units such as kilometers per hour (km/h) or miles per hour (mph).
The formula to calculate linear velocity V is:
• V is the linear velocity,
• d is the distance traveled,
• t is the time taken.
If a cyclist travels 120 meters in 20 seconds, the linear velocity can be calculated as follows:
V = 120/20 = 6 m/s
1.3. Types of Linear Velocity
1. Uniform Velocity
Uniform velocity occurs when an object travels in a straight line at a constant speed. This means both the magnitude and direction of the velocity remain unchanged over time.
Example: A car cruising on a straight highway at a constant speed of 80 km/h.
2. Non-Uniform Velocity
Non-uniform velocity occurs when either the speed or direction (or both) of the object changes over time. This indicates acceleration or deceleration in linear motion.
Example: A car slowing down as it approaches a red light, where the speed decreases over time.
3. Relative Velocity
Relative velocity is the velocity of one object as observed from another moving object. It helps in understanding the motion of objects with respect to each other.
Example: If two cars are moving on a highway, one at 70 km/h and the other at 50 km/h, the relative velocity of the first car with respect to the second car is 20 km/h.
1.4. Understanding Linear Velocity and its Role in Motion and Kinematics
Linear velocity is a fundamental concept in kinematics, the branch of physics that describes motion. Understanding linear velocity allows us to:
Predict Motion: By knowing an object’s velocity, we can predict where it will be at a future point in time.
Analyze Forces: Velocity helps in analyzing the forces acting on an object. For instance, changes in velocity indicate the presence of acceleration, which is caused by forces according to Newton’s
second law.
Design Systems: Engineers use linear velocity to design systems like transportation networks, conveyor belts, and amusement park rides, ensuring they operate efficiently and safely.
2. What is Angular Velocity?
Angular velocity is a measure of the rate of rotation around an axis. It tells us how fast something is spinning or rotating and the direction of the rotation.
Figure 1: Angular velocity
2.1. Real-Life Examples of Angular Velocity
• Earth’s Rotation: The Earth rotates about its axis once every 24 hours. This rotational movement can be described by its angular velocity.
• Wheels of a Car: When a car moves, its wheels spin. The speed of this spinning is an example of angular velocity.
• Fan Blades: The blades of a ceiling fan rotate around a central point, and the speed of this rotation is another instance of angular velocity.
2.2. Unit and Formula
• Angular velocity is symbolized by the Greek letter omega (ω).
• The SI unit is radians per second (rad/s).
• It’s calculated using the formula: ω = Δθ / Δt
□ Δθ (delta theta) represents the angular displacement (change in angle) in radians.
□ Δt (delta t) represents the time interval in seconds.
2.3. Example Calculation
If a record player completes one full rotation (360°) in 5 seconds, what’s its angular velocity?
1. Convert degrees to radians: 360° * (π / 180°) = 2π radians (since a full circle is 2π radians)
2. Apply the formula: ω = 2π radians / 5 seconds = 0.4π rad/s
2.4. Types of Angular Velocity
1. Rotational Velocity: This is the most common type, referring to the spinning of an object about a fixed axis (like the merry-go-round).
2. Spin Angular Velocity: This describes the rotation of an object about its own center of mass (like the Earth spinning).
3. Orbital Angular Velocity: This refers to the rotation of an object around another fixed point (like the Earth orbiting the Sun).
2.5. Role of Angular Velocity in Motion and Kinematics
Angular velocity plays a critical role in the study of rotational motion and kinematics. Here’s how it influences and interacts with various aspects of motion and kinematics:
Describing Rotational Motion: Angular velocity provides a quantitative measure of how fast an object is rotating. It is essential for describing the rotational state of objects in systems ranging
from simple wheels to complex machinery.
Relationship with Linear Velocity: In circular motion, linear velocity (the speed along the circular path) is directly related to angular velocity through the radius of the circle. This relationship
helps in understanding the dynamics of objects in circular motion. The formula linking them is:
V = r⋅ω
where v is the linear velocity, r is the radius, and ω is the angular velocity.
Kinematic Equations for Rotational Motion: Just like linear motion, rotational motion can be described using kinematic equations. Angular velocity, angular acceleration, and angular displacement are
used to predict and analyze rotational motion. The basic kinematic equations for rotational motion include:
where θ is the angular displacement, ω0 is the initial angular velocity, ω is the final angular velocity, α is the angular acceleration, and t is the time.
Centripetal Force and Acceleration: In any rotational system, the objects experience centripetal acceleration directed towards the center of the rotation. This is given by:
where is the centripetal acceleration. Understanding this concept is crucial for analyzing forces in rotational systems, such as in the case of a car turning around a curve.
Angular Momentum and Torque: Angular velocity is integral to concepts like angular momentum (L=Iω) and torque (τ=Iα), where I is the moment of inertia. These concepts are fundamental in rotational
dynamics, helping to understand the effects of forces and the conservation of angular momentum in systems.
Applications in Engineering and Physics: Angular velocity is critical in designing and analyzing the performance of mechanical systems such as gears, turbines, and engines. In physics, it helps in
understanding the behavior of celestial bodies, gyroscopes, and rotating systems.
3. What is tangential velocity?
Tangential velocity is the instantaneous linear speed of an object moving along a circular path, measured at any point tangent to the circle. This velocity is always directed along the tangent to the
circle at the point of interest.
Figure 2: Tangential velocity [2]
3.1. Real-life examples
• Car Wheels: As a car drives, the tangential velocity of any point on the wheel is the speed at which that point would move in a straight line if it were to leave the wheel.
• Amusement Park Rides: For rides like a Ferris wheel, the tangential velocity is the speed at which seats are moving along the circular path.
• Earth’s Rotation: Points on the equator have higher tangential velocity than points closer to the poles due to Earth’s rotation.
3.2. Formula and Units
The formula for tangential velocity (V[t]) is given by:
• V[t] is the tangential velocity,
• r is the radius of the circular path,
• ω is the angular velocity.
The unit of tangential velocity is meters per second (m/s).
3.3. Calculation Example
Consider a disc with a radius of 0.33 meters rotating at an angular velocity of 15 radians per second. To find the tangential velocity:
3.4. Role in Motion and Kinematics
Describing Motion: Tangential velocity provides a clear understanding of how fast and in what direction a point on a rotating object is moving.
• Relating Linear and Angular Quantities: It connects linear motion and angular motion, showing the relationship between an object’s speed along the edge of a circle and its rate of rotation.
• Centripetal Force Calculation: Understanding tangential velocity is crucial for calculating the centripetal force required to keep objects in circular paths.
• Applications in Engineering: Engineers rely on the concept of tangential velocity to design and analyze mechanical systems involving rotational components, such as gears and turbines.
4. Angular Velocity VS Linear Velocity
Angular velocity measures how fast an object rotates, while linear velocity measures how fast an object moves in a straight line.
The key relationship between them is:
where v is the linear velocity, r is the radius, and ω is the angular velocity.
Some key points:
• Angular velocity is measured in radians/second, while linear velocity is measured in distance/time units like m/s.
• For circular motion, the linear velocity is tangent to the circle at any point.
• Angular velocity remains constant for all points on a rigid rotating object, while linear velocity increases with distance from the axis of rotation.
4.1. Conversion Factors and Common Units
• Radians and Revolutions: One revolution is equal to 2π radians. So, to convert from revolutions per minute (RPM) to radians per second (rad/s), multiply by 2π/60.
• Distance Units: One mile is equal to 5,280 feet, and one kilometer is equal to 1,000 meters. These conversions are useful when dealing with different units of measurement.
4.2. Example
A disc with a radius of 4 centimeters spins at 30 revolutions per minute. What is the linear speed in feet per second?
1. Convert RPM to radians per second:
2. Use the formula:
3. Convert centimeters to feet (1 meter = 3.281 feet, 1 meter = 100 centimeters):
5. Tangential Velocity VS Angular Velocity
Tangential velocity (𝑣𝑡) and angular velocity (𝜔) are related through the radius (r) of the circular path on which an object is moving. The tangential velocity is the linear speed of an object
moving along the circular path, while the angular velocity is the rate at which the object sweeps out an angle around the center of the circle.
Angular velocity and tangential velocity are two ways to describe the motion of a rotating object, but they capture different aspects of that motion. Here’s a breakdown:
Angular Velocity (ω):
• Focuses on the rotation itself.
• Tells you how fast an object is spinning in terms of angular displacement (measured in radians) per unit time (seconds).
• Units: radians per second (rad/s).
• Same for all points on a rigidly rotating object.
Tangential Velocity (V[t]):
• Focuses on the linear speed of a specific point on the rotating object.
• Represents the actual speed of that point as it travels along a circular path.
• Units: meters per second (m/s) or any other unit of linear speed.
• Varies depending on the distance of the point from the axis of rotation. Higher distance from the axis leads to higher tangential velocity.
5.1. Relation between them
They are related by the following equation:
V[t] = ω * r
• V[t] is the tangential velocity
• ω is the angular velocity
• r is the distance between the point and the axis of rotation (radius)
5.2. Converting between them
To find tangential velocity (V[t]) of a point, you need to know the angular velocity (ω) and the distance (r) from that point to the axis of rotation. Just plug those values into the formula above.
To find angular velocity (ω) from tangential velocity (V[t]) and distance (r), you can rearrange the formula: ω = V[t] / r
5.3. Example
Imagine a merry-go-round rotating at 2 rad/s. A child sits 3 meters from the center.
What’s the tangential velocity of the child?
V[t] = ω * r = 2 rad/s * 3 m = 6 m/s (The child travels 6 meters every second along the circular path)
If another child sits 2 meters from the center, what’s their tangential velocity?
V[t] = ω * r = 2 rad/s * 2 m = 4 m/s (As expected, the closer child has a lower tangential velocity)
6. How to apply Abaqus Velocity and angular velocity?
Velocity and angular velocity can be specified in Abaqus as initial conditions or boundary conditions. Here’s how to do it:
As Initial Conditions:
Go to Load module > Click on “Create Predefined Field” > Select “Initial” step > “Mechanical” category > select “Velocity” type
Choose the region (part or assembly) where you want to apply the initial velocity. Next, from the Definition combo box, choose one of the options “Translational only”, “Rotational only”, or
“Translational & rotational” to select velocity, angular velocity, or both of them, respectively.
Figure 3: Abaqus velocity as initial conditions
As Boundary Conditions:
Go to Load module > Click on “Create Boundary Conditions” > Select the step in which you want to apply the boundary condition > “Mechanical” category > select “Velocity/Angular velocity” type
Choose the region (part or assembly) where the velocity boundary condition will be applied.
Specify the velocity components in the appropriate fields.
Figure 4: Abaqus velocity as boundary conditions
7. Some issues in applying Abaqus Velocity / Angular velocity
I attempted to simulate external turning using Abaqus, with the workpiece speed set to 6.28/s and the analysis step time set to 1. The workpiece should theoretically revolve one revolution, but the
output shows that the workpiece reference point did rotate one revolution, but the rotation involved in cutting appears to be minor.
1. The accompanying figure 1 depicts the workpiece’s boundary conditions (figure 2 shows the workpiece reference point is coupled to the left cross-section.).
2. The output rotation angle of the reference point on the workpiece is shown in figure 3.
3. The cloud diagrams 4&5 below illustrate how little the workpiece actually rotates.
How do I adjust the rotation boundary conditions so that the part’s real rotation matches the settings in the Abaqus cutting simulation?
Figure 1
Figure 2
Figure 3
Figure 4
Figure 5
You just need to create an Amplitude for your boundary conditions. Your velocity needs to be continuous. You see, your current boundary condition has no amplitude, and the angular velocity you
specified will be applied instantaneously. It’s like applying initial velocity, so if the part makes contact with an object like the tool, this slows down the rotational speed. That is why it rotates
with a small amount in the cutting process.
Therefore, create an amplitude. For example, see the figure I attached.
Also, I suggest checking the link below. It will give you useful information to resolve your future problems.
ABAQUS course for beginners | FEM simulation tutorial
Best wishes.
8. Conclusion
In conclusion, understanding the concepts of linear velocity, angular velocity, and tangential velocity is essential for analyzing motion in both straight and rotational contexts. Linear velocity
describes how fast an object moves in a straight line, while angular velocity focuses on the rate of rotation around an axis. Tangential velocity bridges the two by explaining the linear speed of an
object moving along a circular path.
In practical applications, particularly in engineering and physics, these concepts help in designing and analyzing systems effectively. Using “Abaqus Velocity” to specify these conditions in
simulations further aids in predicting and optimizing system performance. By comparing “Angular Velocity VS Linear Velocity,” we gain insights into different aspects of motion, enabling us to tackle
real-world problems more efficiently.
Explore our comprehensive Abaqus tutorial page, featuring free PDF guides and detailed videos for all skill levels. Discover both free and premium packages, along with essential information to master
Abaqus efficiently. Start your journey with our Abaqus tutorial now!
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