text:
User talk:Rhalah

 Hello Rhalah, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Adambro 19:58, 15 March 2010 (UTC)

title: User talk:Rhalah


text:
MediaWiki:Licenses/own

 * [I hereby release my own work (works by you):]
 * self|GFDL|cc-by-sa-3.0,2.5,2.0,1.0|GFDL and Creative Commons ShareAlike 3.0 and earlier
 * self|GFDL|cc-by-sa-3.0|GFDL and Creative Commons ShareAlike 3.0
 * self|GFDL|FAL|GFDL and Free Art License
 * self|GFDL|cc-by-3.0|GFDL and Creative Commons Attributions 3.0
 * self|GFDL|GNU Free Documentation License (GFDL)
 * subst:nld|For non-commercial use only
 * PD-self|Public domain with all rights released
 * subst:nld|I don't know what license to use

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: MediaWiki:Licenses/own


text:
Wikiversity:Upload

Wikimedia Commons acts as a common repository of freely-licensed educational media content (images, sound and video clips) for the various projects of the Wikimedia Foundation. Any content uploaded to Commons will be immediately available for use on Wikiversity and its sister projects.

Whilst Commons only accepts freely-licensed content, limited non-free content can be uploaded locally for use on Wikiversity. Please select the source of the content you wish to upload and you will be directed to the appropriate upload form.

Where is the work from? (Click on the appropriate link)
 * It is [ entirely my own work]
 * It is [ someone else's work from Flickr] (more information on uploading Flickr images)
 * It is a derivative work of a file from Commons
 * It is [ from another Wikimedia project] (Wikipedia, Wikibooks, Wikinews, etc.)
 * It is [ from a US federal government source]
 * It is [ a free software screenshot]
 * It is [ from somewhere else]
 * It is [ a non-free fair use image] (Non-free logo, cover, screenshot, artwork etc.)
 * [ I don't know who the author is, or I don't know what license applies]
 * I need help figuring out what the license is

title: Wikiversity:Upload


text:
User:Stanistani

My interests are writing, space, computer science, history, and books. I am a wikipedia editor under the same name, making mostly gnome-like edits. I have a writing blog and forum at: Often Inspired.

I hope to use this site to explore the connection between writer and reader in detail.

title: User:Stanistani


text:
Human anatomy quizzes/Deep muscles of the pectoral region

Deep dissection of pectoral region

{ Label the picture  A: { Pectoralis Major (i) } B: { Deltoids (i)|Deltoid (i)| } C: { Subscapularis (i) } D: { Tendon of long head of bicep (i)|bicep tendon (i) } E: { Tendon of pectoralis major(i)|Pectoralis major tendon (i) } F: { Terez major (i) } G: { Latissimus Dorsi (i) } H: { Coracobrachialis (i) }

I: { Long head of bicep (i)|long head of biceps (i)|long head of biceps brachii (i) }

J: { Short head of bicep (i)|Short head of biceps (i)|Short head of biceps brachii (i) }

K: { Serratus anterior (i) }

L: { Pectoralis minor (i) }

M: { Pectoralis major (i) }

 * type="{}"}


text:
Human anatomy quizzes/Superficial muscles of the anterior compartment

The superficial anterior compartment of the forearm

{ Label the picture  A: { Biceps (i)|biceps brachii (i) } B: { Supinator (i)| } C: { Pronator teres (i) } D: { Flexor Carpi Radialis (i) } E: { Palmaris longus (i) } F: { Flexor carpi ulnaris (i) } G: { Flexor digitorum superficialis (i) } H: { Brachioradialis (i) }</li>
 * type="{}"}

</ol>


text:
Template:JacobFrank

Hello , you can call me Jacob Frank. I'd like to be your mascot and welcome you to Wikiversity. Even though I was a plain and untutored man back in 18th-century Europe, that still didn't stop me from contributing my ideas to the world. Neither should it stop you. As a religious leader who developed the doctrine of "purification through transgression," my advice to you is to be bold and to ignore all rules—yes, these are two of Wikiversity's official policies! Happy editing.

title: Template:JacobFrank


text:
User talk:Sundayakanbi

 Hello Sundayakanbi, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --AFriedman (talk) 02:46, 16 March 2010 (UTC)

 Hi! My name is WikiversityJack. I am a pumpkin and a Wikiversity mascot. My aunt told me that fun + learning = Wikiversity. Let me know if I can help you out.

title: User talk:Sundayakanbi


text:
Careers of Health Education Specialist

Health Education is a broad field where you can experience your knowledge on a variety of jobs. Since there are lot of opportunities for the qualified and certified candidates, Health Educators are becoming valuable in every field. These are some of the places where Health Educators are working at present.

Community Health
Job Titles:
 * Health Educator I, II, III
 * Hospital Health Education and outreach coord
 * Community Health Educator
 * Behavioral Scientist
 * State Environmentalist
 * Health Education Specialist
 * Health Coach
 * School Health Coordinator
 * Health Communication specialist
 * Reproductive Health Education Specialist

Postions:
 * US Centers for Disease control and Prevention
 * US and state Depts of Health and Human services
 * National Institute of Health
 * Substance Abuse and Mental Health services
 * Food and Drug Administration
 * World Health Organisation
 * Maternal and Child Health
 * Minority Health
 * Community Programs and Food Inspection
 * Coordinated School Health Programs
 * Local/County Health Departments
 * Voluntary Health Agencies
 * American Red Cross
 * American Heart Association

Worksite Health Promotion
Job Titles:
 * Worksite Health Promotion specialist
 * Worksite Health Director
 * Occupational Safety and Health Coordinator

Positions:
 * Office of Disease Prevention and Health Promotion
 * Managed Care Organizations
 * State and Fed Agencies w/worksite Hlth Prog
 * National Center for Chronic disease prev. and Hlth promotion
 * All US fortune 500 Companies
 * Wellness Councils of America
 * Occupational safety and Health

Public Health- Epidemiology
Job Titles:
 * Epidemiologist
 * State Health Laboratory Director
 * Health Surveillance Specialist
 * Bioterrorism specialist
 * Sanitarian I,II
 * Public Health Educator
 * Research Epidemiologist
 * State Epidemiologist
 * Public Health Preparedness Specialist
 * Pharmacoepidemiologist

Positions:
 * Public Health/Statistics
 * Chronic Disease Epidemiologist
 * Environmental Health and Protection
 * Agency for Toxic substances and Disease Registry
 * HIV/STD/TB prevention and control
 * Infectious Disease Epidemiology
 * Bioterrorism Investigation/Prevention
 * State and local Health Departments
 * Center for Injury Prevention and Control

Health Care Administration
Job Titles:
 * Personal Management Specialist
 * Health Care Administrator
 * Regional Coordinator of Voluntary Agency
 * Management Policy Advisor
 * Hospital/Health Department Administrator
 * Health Services Administrator
 * Nursing Home Director
 * Health Policy Advisor

Postions:
 * Hospitals
 * Long-Term Care Facilities
 * Health Resources and Services Administration
 * Managed Care and Quality Assurance
 * Nursing Homes
 * Agency for Care Research and Quality
 * Health Care Financing Administration
 * Medical Clinics

title: Careers of Health Education Specialist


text:
Facts about Health Educators

According to the U.S. Bureau of Labor and Statistics (BLS) Health educators held about 62,000 jobs in 2006, that number is expected to grow to 78,000 by 2016. More than 250 colleges and universities provide baccalaureate, masters and/or doctoral degrees in health education. Entry-level positions for Health Educators generally require a bachelor’s degree in health education, but many employers prefer to hire workers with a master’s degree. Many health educators have achieved the designation of certified health education specialists (CHES) by the National Commission for Health Education Credentialing, Inc. A list of the areas of responsibility, competencies and sub-competencies that guide the professional preparation and employment of health.

Where do health educators work?

 * A small percent of health educators work in grant-making services and social advocacy organizations.
 * 5 out of 10 health educators work in health care and social assistance and an additional 2 out of 10 work in State and local government.

What do Health Educators Do?
Among the many skills health educators have are: You can expect a trained health educator to be able to:
 * Assessing individual and community needs
 * Planning health education programs
 * Developing health education programs
 * Coordinating health education programs
 * Implementing health education programs
 * Managing health education programs
 * Coalition building
 * Identifying resources & making referrals
 * Developing social marketing & media campaigns
 * Organize / mobilize communities
 * Handle controversial health issues / content
 * Advocate for health-related issues
 * Encourage healthy behavior
 * Use a variety of educational / training methods
 * Develop audio, visual, print and electronic resource materials
 * Conduct research
 * Write scholarly articles

title: Facts about Health Educators


text:
Talk:Facts about Health Educators

 Hello Facts about Health Educators, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --AFriedman (talk) 05:55, 16 March 2010 (UTC)

 Hi! My name is WikiversityJack. I am a pumpkin and a Wikiversity mascot. My aunt told me that fun + learning = Wikiversity. Let me know if I can help you out.

title: Talk:Facts about Health Educators


text:
User talk:Adrianjneubauer

 Hello Adrianjneubauer, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --AFriedman (talk) 05:55, 16 March 2010 (UTC)

 Woof! My name is Jack Russell. I am a dog and a Wikiversity mascot. I am pretty new around Wikiversity. Perhaps we can learn together! Tail wag...

to main namespace
Hi! Please note that Categories are not a good place for educational resources such as this. Move them into the main namespace where people can find them. --138.102.140.115 08:29, 16 March 2010 (UTC)

title: User talk:Adrianjneubauer


text:
User talk:Lrg805

 Hello Lrg805, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --AFriedman (talk) 05:56, 16 March 2010 (UTC)

 Hello , we are the Baah and Bahh twins and we'd both like to welcome you to Wikiversity. While you sit back and explore Wikiversity, enjoy some of our fresh goat milk. You can help us learn by creating new learning resources or help us by improving the guided tours and other exploration tools. We look forward to seeing you around. Baaaahbye for now.

title: User talk:Lrg805


text:
User talk:Mcirish3

 Hello Mcirish3, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --AFriedman (talk) 05:56, 16 March 2010 (UTC)

 Hello , you can call me Jacob Frank. I'd like to be your mascot and welcome you to Wikiversity. I live here because even though I was a plain and untutored man back in 18th-century Europe, that still didn't stop me from contributing my ideas to the world. Neither should it stop you. As a religious leader who developed the doctrine of "purification through transgression," my advice to you is to be bold and to ignore all rules—yes, these are two of Wikiversity's official policies! Happy editing.

Hi!
Very interesting learning resource you created. Hope to see you around. I just created the Mascot template of Jacob Frank and I'm excited that you're the first User to have it on your Userpage. What do you think of the site? --AFriedman (talk) 05:58, 16 March 2010 (UTC)

title: User talk:Mcirish3


text:
User talk:Vin~enwikiversity

 Hello Vin, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 07:18, 16 March 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:18, 16 March 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Vin. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Vin~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 00:12, 18 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:24, 19 April 2015 (UTC)

title: User talk:Vin~enwikiversity


text:
User talk:MattStarsky

 Hello MattStarsky, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 07:18, 16 March 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:18, 16 March 2010 (UTC)

title: User talk:MattStarsky


text:
User talk:Josephlyons

 Hello Josephlyons, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 07:18, 16 March 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:18, 16 March 2010 (UTC)

title: User talk:Josephlyons


text:
User talk:RobDzi

 Hello RobDzi, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 07:18, 16 March 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:18, 16 March 2010 (UTC)

title: User talk:RobDzi


text:
User:Florian Schimpf~enwikiversity

= E-Learning and E-Teaching =

This is a shared project between Ludwigsburg University of Education and Šiauliai University [].

Web 2.0
The Web 2.0 offers a lot of tools that can improve the way how we learn and work. Using Web 2.0 applications students can express their own thoughts in blogs and wikis and discuss them with others via the internet. It is easy to create knowledge collaboratively with an online mind map or to cooperate in a shared project. Before and during my stay for the Erasmus Staff Mobility Event 2010, I would like to share and create knowledge by using a Wiki with both of you: students in Germany and students in Lithuania

On both universities we have a group of students who are freshman in using E-Learning and E-Teaching technologies. Using a Wiki like this one is pretty easy: There is nothing what can go wrong. At least you always can go back to former versions. To get familiar to work in an Wiki, please watch this Video: After this please open this link: []and go to "edit this page". Please write down some information about you and your intrests.

If you have any questions feel free to contact me by mail:[floschimpf@googlemail.com]

Participants
Florian:Teacher and researcher in the field of educational informatic, 43 years old, beekeeping as hobby

Mantgirdas:

Aurimas i am in education as a student i want to reviw your in the coutry concemend with raising a child with something innovaltive, we do not

Karolina: I am learning in faculty of Education, and I am 1 course student. So, these things is new for me. :)

Brigita: I am a student from the Siauliai University, Faculty of Education, 20 years and always with a smile:)

Laura: Hello I am a student at Siauliai University, Laura 1 year of education specialty:)

Ingrida: I am lecturer at the University, faculty of Education. I like innovation; I hope an interesting co-operation.

Margarita: I am doctoral student and lecturer at the university, faculty of Education. I am not so old :)

Rytis: scientists, project manager and lecturer in the field of teacher training and science education, 45 years old, eco-tourism as hobby.

Sabrina: study teaching post (elementary school) in PH-Ludwigsburg, 23 years old, different sports as hobby (Rhönrad)

Benjamin: student at the PH-Ludwigsburg, 23 years old, different sports (soccer, ice-/inline-hockey) as hobby

Lisa:I am a student and 20 years old. I like to read and visit the world =)

Melanie: Hello, I am a student in my 6. and last semester. ;) I am 23 years old and my hobbys are ice hockey und football.

Ann Kathrin: Hi, I`m a student at PH Ludwigsburg. I`m 22 years old and I like playing handball.

Florian U.: I am 24 years old and I am a student at PH Ludwigsburg. My hobbys are playing handball and guitar.

Pelagia:I study at the PH - Ludwigsburg. I`m 24 years old and I like playing basketball and dancing (especially hip hop:) )

Jan: I am a student at the PH in Ludwigsburg. I am 27 years old. After this Semester I'm going to make the exam. My passion is VW Cars.

Johannes: I'm studying in Ludwigsburg in the fifth semester. One year ago I've taken part in a media-didactics course but still feel like a beginner in this field. I like gardening and biking.

Ilaria: Hello, I'm 22 years old and I study at the PH Ludwigsburg. I like to read and go out with my dog.

Yeliz:Hi. I am 23 years old and I am a student at PH Ludwigsburg. I like holidays ;)

Evelyn: Hello, I'm a student at PH Ludwigsburg. I'm 22 years old and I like swimming.

Diana: Hi, I`m 22 years old and a student at PH Ludwigsburg. I like travelling.

Isabel:Hey I am 21 years old and also a student at the PH Ludwigsbrug. I love travelling and I am hooked on chokolate.

E-Learning Pro (+) and Con (-)
pro:

learning is everywhere possible, e.g from home

reducing cost of learning

a lot of useful information

learning is everytime possible (at my favorit time) and we can learning in our temp

con:

no social contact with others

a lot of not useful information

the information is not always reliable

we need a new (some time very expensive) technology

title: User:Florian Schimpf~enwikiversity


text:
User talk:JohnLee

 Hello JohnLee, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --138.102.140.115 08:30, 16 March 2010 (UTC)

 Woof! My name is Jack Russell. I am a dog and a Wikiversity mascot. I am pretty new around Wikiversity. Perhaps we can learn together! Tail wag...

to main namespace
Hi! Please note that Categories are not a good place for educational resources such as this. Move them into the main namespace where people can find them. --138.102.140.115 08:31, 16 March 2010 (UTC)

title: User talk:JohnLee


text:
User talk:Florian Schimpf~enwikiversity

 Hello Florian Schimpf, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
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title: User talk:Florian Schimpf~enwikiversity


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User talk:Assistant2010

4 AM  2010

title: User talk:Assistant2010


text:
Portal talk:Time

Vernal Equinox   Mar 20 2010   1:32 PM EDT - Equinoxes and Solstices 2000-2010

countdown
10 ... CQ (discuss • contribs) 20:47, 2 December 2018 (UTC)

Santa's list of commands:

 * Please don't fall for me.
 * Thanks in advance.
 * Stay on time.
 * Work hard.
 * Play fair.
 * Be good.

I'm watching ... CQ (discuss • contribs) 20:51, 2 December 2018 (UTC)

title: Portal talk:Time


text: File:SeducedByTheDarkSideScript(Animation).pdf


text:
User:Egm6341.s10.team3.sa/HW5

Problem Set 5 =Trapezoidal Error - Derivation: Evaluation of the constants$$c_5 \,and\, c_6$$ in the polynomial $$p_{5}(t) $$= Ref: Lecture Notes [[media:Egm6341.s10.mtg26.djvu|p.26-3]]

Problem Statement
Given that:
 * {| style="width:100%" border="0" align="left"

p_{5}(t) = -\frac{t^5}{120} + \frac{t^3}{36} + c_{5} t + c_{6} $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * {| style="width:100%" border="0" align="left"

 * with,
 * }
 * {| style="width:100%" border="0" align="left"

p_{5}(0) =0 $$ $$
 * $$ \displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"

p_{5}(\pm 1) = 0 $$ $$
 * $$ \displaystyle
 * $$\displaystyle (Eq. 3)
 * }

Solution

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle p_{5}(0) = -\frac{0}{120} + \frac{0}{36} + c_{5} (0) + c_{6} = 0 $$
 * Substitute Eqn(2) in Eqn(1),
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle c_{6} = 0 $$
 * This yields,
 * }

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle p_{5}(1) = -\frac{1}{120} + \frac{1}{36} + c_{5}  = 0 $$
 * Substitute Eqn(3) in Eqn(1),
 * }

 * {| style="width:100%" border="0" align="left"

$$ \displaystyle c_{5} = -\frac{7}{360} $$
 * This yields,
 * }

 * {| style="width:100%" border="0" align="left"

 * Substituting $$\displaystyle p_{5}(-1) = 0$$ also will give the same value of $$ \displaystyle c_5 $$
 * }
 * }
 * }

Hence,
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle c_6 = 0 \, and\, c_5 = -\frac{7}{360} $$ Subramanian Annamalai 01:13, 24 March 2010 (UTC)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }

= Proof of Trapezoidal error - Evaluating the polynomials $$p_{6}(t)$$ and $$p_{7}(t)$$ =

Problem Statement
$$ \displaystyle i)\, Do\, steps\, 4a\, and\, 4b\, in\, order\, to\, determine\, P_{6}(t)\, and\, P_{7}(t)\, $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

$$ \displaystyle ii)\, Steps\, 3a\, and\, 3b\, are\, given\, by\, Lecture\, 26\,$$

$$ E=[P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}P_5(t)g^{(5)}(t)dt}_{Call\, D\,} $$

$$ \begin{align} P_4(t)&=-\frac{t^4}{4!}+\frac{t^2}{12}+\alpha \\ &=c_{1}\frac{t^4}{4!}+c_{3}\frac{t^2}{2!}+c_{5}\, (c_{5}=\alpha) \\ &=-\frac{t^4}{24}+\frac{t^2}{12}-\frac{7}{360} \\\end{align}\ $$

$$ \begin{align} P_5(t)&=-\frac{t^5}{120}+\frac{t^3}{36}+\alpha t + \beta \\ &=c_{1}\frac{t^5}{5!}+c_{3}\frac{t^3}{3!}+c_{5}t+c_{6}\, (c_{5}=\alpha\, and\, c_{6}=\beta) \\ &=-\frac{t^5}{120}+\frac{t^3}{36}-\frac{7t}{360} \\\end{align}\ $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg26.djvu|p.26-2&3]]

Solution
$$ \displaystyle i)\, Step\, 4a\, :\, Recall\, D\, and\, integrate\, by\, parts\, $$

$$ D=:\int_{-1}^{+1}\underbrace{P_5(t)}_{v'}\underbrace{g^{(5)}(t)}_{u}dt = [P_6(t)g^{(5)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}{P_6(t)g^{(6)}(t)}\, dt}_{Call\, E} $$

$$P_6(t) = c_1(\frac{t^6}{6!})+c_3(\frac{t^4}{4!})+c_5(\frac{t^2}{2!})+c_7 = -\frac{t^6}{720}+\frac{t^4}{144}-\frac{7t^2}{720}+\alpha $$

$$ \displaystyle ii)\, Step\, 4b\, :\, Recall\, E\, and\, integrate\, by\, parts\, $$

$$ E=:\int_{-1}^{+1}\underbrace{P_6(t)}_{v'}\underbrace{g^{(6)}(t)}_{u}dt = [P_7(t)g^{(6)}(t)]_{-1}^{+1}-\underbrace{\int_{-1}^{+1}{P_7(t)g^{(7)}(t)}\, dt}_{Call\, F} $$

$$P_7(t) = c_1(\frac{t^7}{7!})+c_3(\frac{t^5}{5!})+c_5(\frac{t^3}{3!})+c_7(t)+c_8 = -\frac{t^7}{5040}+\frac{t^5}{720}-\frac{7t^3}{2160}+\alpha t+\beta $$

$$ \displaystyle iii)\, Select\, points\, such\, that\, $$

$$ \displaystyle P_{7}(\pm 1)=0 $$

$$ \displaystyle P_{7}(0)=0 $$

$$\displaystyle Recall\, P_7(t)\, and\, plug\, in\, t=0\, and\, t=-1\,$$ $$P_7(t=0) = c_1(\frac{0}{7!})+c_3(\frac{0}{5!})+c_5(\frac{0}{3!})+c_7(0)+c_8 = -\frac{0}{5040}+\frac{0}{720}-\frac{0}{2160}+\alpha \times 0+\beta $$ $$\displaystyle Hence\, c_8\, =\, \beta\, =\, 0\,$$ $$P_7(t=-1) = c_1(\frac{(-1)}{7!})+c_3(\frac{(-1)}{5!})+c_5(\frac{(-1)}{3!})+c_7(-1) = -\frac{(-1)}{5040}+\frac{(-1)}{720}-\frac{(-1)}{2160}+\alpha \times (-1) $$ $$\displaystyle Hence\, c_7\, =\, \alpha\, =\, \frac{31}{15120}\,$$

$$\displaystyle Therefore\, $$

$$ P_6(t)=-\frac{t^6}{600}+\frac{t^4}{108}-\frac{7t^2}{720}+\frac{31}{15120} $$

$$ P_7(t)=-\frac{t^7}{4200}+\frac{t^5}{540}-\frac{7t^3}{2160}+\frac{31}{15120} t $$

$$ E= [P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)+P_6(t)g^{(5)}(t)]_{-1}^{+1}-F $$

$$ \displaystyle where\, F=:\int_{-1}^{+1}{P_7(t)g^{(7)}(t)}\, dt $$

--Heejun Chung 13:06, 26 March 2010 (UTC)

= Find $$ {t}_{k}(x) $$ = Ref: Lecture Notes [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

Problem Statement
Find $$ {t}_{k}(x) $$ by the reversion of $$ x ({t}_{k}) $$

where
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle x(t_k)=\frac{1}{2}(x_k+x_{k+1})+t_k \frac{h}{2}
 * $$\displaystyle x(t_k)=\frac{1}{2}(x_k+x_{k+1})+t_k \frac{h}{2}
 * $$\displaystyle (Eq. 1)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle h=x_{k+1}-x_k
 * $$\displaystyle h=x_{k+1}-x_k
 * }
 * }

Solution
Revise Eq.1 and express $$t(x_k)$$by x:

 * {| style="width:100%" border="0" align="left"

 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$\displaystyle \Rightarrow t_k(x)=\frac{2}{h}(x- \frac{x_k+x_{k+1}}{2}) $$ $$
 * $$\displaystyle (Eq. 2)
 * style = |
 * }

When $$ x = x_k $$
 * {| style="width:100%" border="0" align="left"

$$ When $$ x = x_{k+1} $$
 * $$\displaystyle t_k(x_k)=\frac{2}{h}(x_k- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_k-x_{k+1}}{2}=-1
 * $$\displaystyle t_k(x_k)=\frac{2}{h}(x_k- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_k-x_{k+1}}{2}=-1
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle t_k(x_{k+1})=\frac{2}{h}(x_{k+1}- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_{k+1}-x_k}{2}=1
 * $$\displaystyle t_k(x_{k+1})=\frac{2}{h}(x_{k+1}- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_{k+1}-x_k}{2}=1

-- By Min Zhong 12:00, 23 March 2010 (UTC)

= Find coefficients of error terms of Trapezoidal rule $$\bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ from polynomials $$P_{2}, P_{4}, P_{6}$$ = Ref: Lecture notes [[media:Egm6341.s10.mtg27.djvu|p.27-2]]

Problem Statement
Find coefficients $$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ using the following relations.

 * {| style="width:100%" border="0" align="left"

d_{r} = \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Solution
We already obtained the polynomials $$\displaystyle P_{2}(t), \; P_{4}(t)$$ in class meeting 26. (see lecture notes [[media:Egm6341.s10.mtg26.djvu|p.26-1]] and [[media:Egm6341.s10.mtg26.djvu|p.26-3]].)

 * {| style="width:100%" border="0" align="left"

P_{2}(t) = -\frac{t^2}{2!} + \frac{1}{6} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

P_{4}(t) = -\frac{t^4}{24} + \frac{t^2}{12} - \frac{7}{360} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Also, at the problem 2 in the HW5 (Ref:[[media:Egm6341.s10.mtg27.djvu|p.27-1]]), $$\displaystyle P_{6}(t) $$ have been obtained.

 * {| style="width:100%" border="0" align="left"

P_{6}(t) = -\frac{t^6}{720} + \frac{t^4}{144} - \frac{7t^2}{720} + \frac{31}{15120} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

From the (Eq.1) and (Eq.2),

 * {| style="width:100%" border="0" align="left"

$$\displaystyle d_{1} = \bar{d_{2}} = \frac{P_{2}(1)}{2^2}=\frac{1}{4}\times\frac{-3+1}{6}=-\frac{1}{12} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

From the (Eq.1) and (Eq.3),

 * {| style="width:100%" border="0" align="left"

$$\displaystyle d_{2} = \bar{d_{4}} = \frac{P_{4}(1)}{2^4}=\frac{1}{16}\times\frac{-15+30-7}{360}=\frac{1}{720} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

From the (Eq.1) and (Eq.4),

 * {| style="width:100%" border="0" align="left"

$$\displaystyle d_{3} = \bar{d_{6}} = \frac{P_{6}(1)}{2^6}=\frac{1}{64}\times\frac{-21+105-147+31}{15120}=-\frac{1}{30240} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Yong Nam Ahn 14:46, 24 March 2010 (UTC)

= Derive Trapezoidal Error expression = Ref: Lecture Notes [[media:Egm6341.s10.mtg28.djvu|p.28-1]]

Problem Statement
Derive $$ E_n^1 = \sum_{r=1}^{l} \bar{d}_{2r}\big[ f^{2r-1}(b) - f^{2r-1}(a) \big] - \frac{h^{2l}}{2l} \sum_{k=0}^{n-1} \int_{x_k}^{x_{k+1}} p_{2l}(t_k(x))f^{(2l)}(x) \,dx$$ Ref:Lecture Notes [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

Solution
We have

$$ E = \sum_{r=1}^l P_{(2r)}(+1) [g^{(2r-1)}- g^{(2r-1)}(-1)] - \int\limits_{-1}^{+1} P_{(2l)}(t) g^{(2l)}(t) dt]$$

and $$ g_k^i(t) = (\frac{h}{2})^i f^i(x(t)) $$

$$ g^{(2r-1)}(t) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (x(t)) $$

$$ g^{(2r-1)}(+1) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (x(1)) $$

and $$ x(1) = b $$

$$ g^{(2r-1)}(+1) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (b) $$

Similarly

$$ g^{(2r-1)}(-1) = (\frac{h}{2})^{(2r-1)} f^{(2r-1)} (a) $$

and $$ g^{(2l)} = (\frac{h}{2})^{(2l)} f^{(2l)} (x) $$

using above equations into first equation we obtain

$$ E = \sum_{r=1}^l P_{(2r)}(+1) \frac{h}{2} (\frac{h}{2})^{(2r-1)} [f^{(2r-1)}(b)- f^{(2r-1)}(a)] - \sum_{k=1}^{n-1} \int\limits_{-1}^{+1} P_{(2l)}(t_k) (\frac{h}{2})^{2l} f^{(2l)}(x) dx]$$

$$ = \sum_{r=1}^l h^{2r} \frac{P_{(2r)}(+1)}{2^{2r}}[f^{(2r-1)}(b)- f^{(2r-1)}(a)] - (\frac{h}{2})^{2l} \sum_{k=1}^{n-1} \int\limits_{x_k}^{x_{k+1}} P_{(2l)}(t_k(x))  f^{(2l)}(x) dx]$$

but $$ \frac{P_{(2r)}(+1)}{2^{2r}} = \overline{d}_{2r}$$

So finally we obtain

$$ = \sum_{r=1}^l h^{2r} \overline{d}_{2r} [f^{(2r-1)}(b)- f^{(2r-1)}(a)] - (\frac{h}{2})^{2l} \sum_{k=1}^{n-1} \int\limits_{x_k}^{x_{k+1}} P_{(2l)}(t_k(x))  f^{(2l)}(x) dx]$$

Abhishekksingh 06:18, 24 March 2010 (UTC)

= Find coefficients of error terms of Trapezoidal rule $$\bar{d_2}, \; \bar{d_4}, \; \bar{d_6}$$ from Bernoulli numbers = Ref: Lecture notes [[media:Egm6341.s10.mtg28.djvu|p.28-2]]

Problem Statement
Using Bernoulli numbers obtained by the Taylor series expansion of $$\displaystyle x\,coth(x)$$,

 * {| style="width:100%" border="0" align="left"

x\,coth(x) = \sum_{r=0}^{\infty}\underbrace{\frac{B_{2r}}{(2r)!}}_{-\bar{d_{2r}}}x^{2r} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

i) Verify $$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ by comparing the results from $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$. ii) Compute $$\displaystyle \bar{d_{8}}, \; \bar{d_{10}}$$

Solution
i)

Let's explicitly express first several terms of the Taylor series expansion of $$\displaystyle x\,coth(x)$$,

 * {| style="width:100%" border="0" align="left"

x\,coth(x) = B_{0} + \frac{B_2}{2!}x^{2} + \frac{B_4}{4!}x^{4} + \frac{B_6}{6!}x^{6} + \frac{B_8}{8!}x^{8} + \frac{B_{10}}{10!}x^{10} + \cdot\cdot\cdot. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Note that $$\displaystyle B_{0}=1, \, B_{2}=\frac{1}{6}, \, B_{4}=-\frac{1}{30}, \, B_{6}=\frac{1}{42}$$.

Then, based on the following relation between $$\displaystyle \bar{d_{2r}} $$ and Bernoulli numbers,

 * {| style="width:100%" border="0" align="left"

\frac{B_{2r}}{(2r)!} = -\bar{d_{2r}}, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

we can obtain $$\displaystyle \bar{d_{2}}, \,\bar{d_{4}}, \,\bar{d_{6}}$$ as follows.

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \bar{d_{2}}=-\frac{B_{2}}{2!}=-\frac{1}{2}\times\frac{1}{6}=-\frac{1}{12} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \bar{d_{4}}=-\frac{B_{4}}{4!}=\frac{1}{24}\times\frac{1}{30}=\frac{1}{720} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \bar{d_{6}}=-\frac{B_{6}}{6!}=-\frac{1}{720}\times\frac{1}{42}=-\frac{1}{30240} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Above results are exactly same with the results from $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$ as we obtained at problem 4 in HW5.

ii)

Since $$\displaystyle B_{8}=-\frac{1}{30}, \, B_{10}=\frac{5}{66}$$,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \bar{d_{8}}=-\frac{B_{8}}{8!}=\frac{1}{40320}\times\frac{1}{30}=\frac{1}{1209600} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \bar{d_{10}}=-\frac{B_{10}}{10!}=-\frac{1}{3628800}\times\frac{5}{66}=-\frac{1}{47900160} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Yong Nam Ahn 14:46, 24 March 2010 (UTC)

= Cancel odd order of derivative of g in the proof of Trap error  = Ref: Lecture Notes [[media:Egm6341.s10.mtg28.djvu|p.28-2]]

Problem Statement
Redo steps in proof of Trap. error by trying to cancel terms with odd order of derivative of g. From Lecture Notes [[media:Egm6341.s10.mtg21.djvu|p.21-1,p.21-2,p.21-3]], the composite trap error is given by:
 * {| style="width:100%" border="0" align="left"

E_n^1=\frac{h}{2} \sum_{k=0}^{n-1} \left[ \underbrace{\int_{-1}^{1}g_k(t)dt-(g_k(-1)+g_k(+1))}_{E}\right] $$ Integrate E by part we have:
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{align} E &=\int_{-1}^{1}p_{1}(t)g_k^{(1)}(t)dt \\ &= \left[ P_2(t)g^{(1)}(t) \right]_{-1}^{+1} - \int_{-1}^{1}P_2(t)g^{(2)}(t)dt \\ &= \left[ P_2(t)g^{(1)}(t)\right]_{-1}^{+1}-\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \int_{-1}^{+1} P_3(t)g^{(3)}(t)dt \\ \end{align} $$ $$ Where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle P_1(t)=-t
 * $$\displaystyle P_1(t)=-t

$$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \alpha
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \alpha
 * $$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta
 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta

$$
 * }
 * }

$$ \alpha $$ and $$\beta$$ are the integration constants

Solution
We start the proof from Equation 1. i) Step 1 we try to cancel the term contains first order of derivative of g in Eq. 1 by having

 * {| style="width:100%" border="0" align="left"

$$ Then
 * $$\displaystyle \left[ P_2(t)g^{(1)}(t)\right]_{-1}^{+1}=0
 * $$\displaystyle \left[ P_2(t)g^{(1)}(t)\right]_{-1}^{+1}=0
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ From Equation 2.
 * $$\displaystyle P_2(+1)= P_2(-1)=0
 * $$\displaystyle P_2(+1)= P_2(-1)=0
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ Summary:
 * $$\displaystyle P_2(+1)=-\frac{1^2}{2}+ \alpha \Rightarrow \alpha=\frac{1}{2}
 * $$\displaystyle P_2(+1)=-\frac{1^2}{2}+ \alpha \Rightarrow \alpha=\frac{1}{2}
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \frac{1}{2}
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \frac{1}{2}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \underbrace {\int_{-1}^{+1} P_3(t)g^{(3)}(t)dt}_{A}
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \underbrace {\int_{-1}^{+1} P_3(t)g^{(3)}(t)dt}_{A}
 * }
 * }

From Lecture Notes [[media:Egm6341.s10.mtg26.djvu|p.26-2]], The result of integration by part of A is

 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle A= \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt
 * $$\displaystyle A= \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt

 * }
 * }

Then we have
 * {| style="width:100%" border="0" align="left"

$$ $$ Where
 * $$\displaystyle E= \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt
 * $$\displaystyle E= \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt
 * $$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta = -\frac{t^3}{6}+ \frac{1}{2}t+ \beta
 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta = -\frac{t^3}{6}+ \frac{1}{2}t+ \beta
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ $$ where $$\omega$$ is the integration constant
 * $$\displaystyle P_4(t)= \int P_3= -\frac{t^4}{24}+ \frac{1}{4}t^2+ \beta t + \omega
 * $$\displaystyle P_4(t)= \int P_3= -\frac{t^4}{24}+ \frac{1}{4}t^2+ \beta t + \omega
 * $$\displaystyle (Eq. 4)
 * }
 * }

ii) Step 2

we try to cancel the term contains third order of derivative of g in Eq. 3

Let
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}=0
 * $$\displaystyle \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}=0
 * }
 * }

Then
 * {| style="width:100%" border="0" align="left"

$$ From Equation 4, then
 * $$\displaystyle P_4(+1)= P_4(-1)=0
 * $$\displaystyle P_4(+1)= P_4(-1)=0
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_4(+1)= -\frac{1}{24}+ \frac{1}{4}+ \beta  + \omega =0
 * $$\displaystyle P_4(+1)= -\frac{1}{24}+ \frac{1}{4}+ \beta  + \omega =0
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ Solve the above two equations, we can find
 * $$\displaystyle P_4(-1)= -\frac{1}{24}+ \frac{1}{4}- \beta  + \omega =0
 * $$\displaystyle P_4(-1)= -\frac{1}{24}+ \frac{1}{4}- \beta  + \omega =0
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle  \beta =0
 * $$\displaystyle  \beta =0
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle  \omega = -\frac{5}{24}
 * $$\displaystyle  \omega = -\frac{5}{24}
 * }
 * }

Summary:
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_3(t)= -\frac{t^3}{6}+ \frac{1}{2}t
 * $$\displaystyle P_3(t)= -\frac{t^3}{6}+ \frac{1}{2}t
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_4(t)=-\frac{t^4}{24}+ \frac{1}{4}t^2 -\frac{5}{24}
 * $$\displaystyle P_4(t)=-\frac{t^4}{24}+ \frac{1}{4}t^2 -\frac{5}{24}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ $$ From Lecture Notes [[media:Egm6341.s10.mtg26.djvu|p.26-2]], The result of integration by part of B is
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}- \underbrace{\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt}_{B}
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}- \underbrace{\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt}_{B}
 * $$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle B= \left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt
 * $$\displaystyle B= \left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt

$$
 * }
 * }

Then
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}-\left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}+\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}-\left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}+\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt

$$
 * }
 * }

iii) Step 3

We can continually cancel the items with odd order derivative of g and find the general expression of E as

 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle E= -\left[ \underbrace{P_3(t)g^{(2)}(t)}_{C}+P_5(t)g^{(4)}(t)+ P_7(t)g^{(6)}(t)+...+P_{2l+1}(t)g^{(2l)}(t)\right]_{-1}^{+1} - \int_{-1}^{+1} P_{2l+1}(t)g^{(2l+1)}(t)dt
 * $$\displaystyle E= -\left[ \underbrace{P_3(t)g^{(2)}(t)}_{C}+P_5(t)g^{(4)}(t)+ P_7(t)g^{(6)}(t)+...+P_{2l+1}(t)g^{(2l)}(t)\right]_{-1}^{+1} - \int_{-1}^{+1} P_{2l+1}(t)g^{(2l+1)}(t)dt
 * $$\displaystyle (Eq. 6)
 * }
 * }

Let
 * {| style="width:100%" border="0" align="left"

$$ Since $$P_3(t) and P_{3i}(t) $$ are odd function, Then
 * $$\displaystyle C = \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}=P_3(+1)g^{(2)}(+1)-P_3(-1)g^{(2)}(-1)
 * $$\displaystyle C = \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}=P_3(+1)g^{(2)}(+1)-P_3(-1)g^{(2)}(-1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_3(+1)=-P_3(-1)
 * $$\displaystyle P_3(+1)=-P_3(-1)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle C = - P_3(-1)\left[(g^{(2)}(+1)+g^{(2)}(-1))\right]
 * $$\displaystyle C = - P_3(-1)\left[(g^{(2)}(+1)+g^{(2)}(-1))\right]
 * }
 * }

Thus Equation 6 can be simplified as follows:

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Rightarrow  E= \sum_{r=1}^{l}P_{2r+1}(-1) \left[g^{2r}(+1)+g^{2r}(-1) \right] - \int_{-1}^{+1} P_{2l+1}(t)g^{(2l+1)}(t)dt $$ $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 7)
 * style = |
 * }
 * }

iv) Step 4 From the equation (5) in lecture note [[media:Egm6341.s10.mtg21.djvu|p.21-2]], We find the composite trap error $$E_n^1$$ through Equation 7

 * {| style="width:100%" border="0" align="left"

$$\displaystyle E_n^1 = \frac{h}{2} \sum_{r=1}^{l} \sum_{k=0}^{n-1}P_{2r+1}(-1) \left[g_k^{2r}(+1)+g_k^{2r}(-1) \right]-\frac{h}{2} \sum_{k=0}^{n-1} \int_{-1}^{1}P_{2l+1}(t)g_k^{2l+1}(t)dt $$ $$ From the lecture note [[media:Egm6341.s10.mtg21.djvu|p.21-2]], $$\displaystyle g_k(t)$$can be changed to $$\displaystyle f_(x_k)$$ and below was obtained.
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{align} &g_k^{(2r)}(+1)=\left(\frac{h}{2}\right)^{2r}f^{(2r)}(x_{k+1}),\\ &g_k^{(2r)}(-1)=\left(\frac{h}{2}\right)^{2r}f^{(2r)}(x_{k}),\\ &g_k^{(2l+1)}(t)=\left(\frac{h}{2}\right)^{2l+1}f^{(2l+1)}(x_{t}) \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Plug the above equations into Equation 8. We have
 * {| style="width:100%" border="0" align="left"

$$\displaystyle E_n^1 = \sum_{r=1}^{l} \sum_{k}^{n-1}\left(\frac{h}{2}\right)^{2r+1}P_{2r+1}(-1) \left[f^{2r}(x_{k+1})+f^{2r}(x_{k})\right] - \left(\frac{h}{2}\right)^{2r+2}\sum_{k=0}^{n-1} \int_{x_k}^{x_{k+1}}P_{2l+1}(t)f_{2l+1}(x_t)dx $$ $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * style = |
 * }
 * }

When we derive the composite trap error, we use the simple trap error which contains additive term $$ g^{2r}(+1)+g^{2r}(-1) $$. Because of the additive sign, no term could be canceled in the summation of $$E_n^1$$. We have to add all points $$\sum_{k}^{n-1}\left[f^{2r}(x_{k+1})+f^{2r}(x_{k})\right]$$ when evaluate the error. It is very complicated compared with that derived by canceling items with even oder derivative of g. In that case, we can get simple trap with item $$ g^{2r-1}(+1)-g^{2r-1}(-1) $$ which can simplify the expression of composite trap error by cancel same term when summation and result in only initial and end point $$[f^{(2r-1)}(x_{n})-f^{(2r-1)}(x_0)$$. Thus Trap error proofed by canceling terms with the even order derivative of g is more effective. -- By Min Zhong 12:00, 23 March 2010 (UTC)

= Obtain expressions for $$(P_2,P_3),(P_4,P_5),(P_6,P_7)$$ Using recurrence formula  =

Ref: Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]]

Problem Statement
Obtain expressions for $$ (P_2,P_3),(P_4,P_5),(P_6,P_7) $$ using $$ \frac{c_1}{(2i+1)!} + \frac{c_3}{(2i-1)!} +\frac{c_5}{(2i-3)!} + ............ + \frac{c_{2i-1}}{3!} + c_{2i+1}$$ Ref:Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-2]]

Solution
$$ (P_2,P_3)$$

$$ P_{2i}(t) = \sum_{j=0}^i C_{2j+1} \frac{t^{2(i-j)}}{(2(i-j))!} $$

$$ P_{2}(t) = \sum_{j=0}^i C_{2j+1} \frac{t^{2(1-j)}}{(2(1-j))!} $$

$$ P_{2}(t) = C_1 \frac{t^{2)}}{2!} + C_3 $$

Using eq (6) from Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]], we obtain

$$ \frac {C_1}{3!}+ C_3 = 0 $$

$$ C_3 = - \frac {C_1}{3!} $$

$$    = - \frac{-1}{3!} = \frac{1}{6} $$

so $$ P_{2}(t) = - \frac{t^2}{2} + \frac{1}{6} $$

$$ P_3 = \sum_{j=0}^i C_{2j+1} \frac{t^{2(i-1)}}{(2(i-j))!} + C_{2i+2} $$

where $$ C_{2i+2} = 0  $$

$$ P_3 = C_1 \frac{t}{3!} + C_3 \frac{t}{1!}$$

where $$ C_1 = -1 and C_3 = \frac{1}{6} $$ from previous result

$$ P_3 = - \frac{t}{3!} + \frac{t}{6} = 0 $$

$$ P_4, P_5 $$

$$ P_4 = C_1 \frac{t^4}{4!} + C_3 \frac{t^2}{2!}+ C_5 $$

Using eq (6) from Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]], we obtain

$$ \frac{C_1}{5!} + \frac {C_3}{3!}+\frac {C_5}{1!} = 0 $$

$$ C_5 = \frac {1}{5!} - \frac{1}{36} = \frac {-7}{360}$$

$$ P_4 = - \frac{t^4}{24} + \frac{t^2}{12} - \frac{7}{360} $$

$$ P_5 = C_1 \frac{t^3}{5!} + C_3 \frac{t^3}{3!}+ C_5 \frac{t^3}{1!} $$

$$ P_5 = (\frac {C_1}{5!} + \frac {C_3}{3!}+ \frac {C_5}{1!})t^3 $$

$$ P_5 = 0 $$

$$ P_6, P_7$$

$$ P_6 = C_1 \frac{t^6}{12!} + C_3 \frac{t^4}{10!} + C_5 \frac{t^2}{8!} + C_7 \frac{1}{6!} $$

By using eq 6 from Lecture Notes [[media:Egm6341.s10.mtg29.djvu|p.29-3]], we obtain

$$ \frac{C_1}{7!} + \frac {C_3}{5!}+\frac {C_5}{3!} + C_7 = 0 $$

$$ C_7 = \frac{1}{7!} - \frac{1}{6!} - \frac{7}{2160}$$

$$   = \frac{11} {15120} $$

$$ P_6 = - \frac{t^6}{12!} + \frac{t^4}{6*10!}- \frac{7 t^2}{360 * 8!} + \frac {11}{15120} $$

$$ P_7 = C_1 \frac {t^5}{7!} + C_3 \frac {t^5}{5!} + C_5 \frac{t^5}{3!}+C_7 \frac{t^5}{1!} $$

$$ P_7 = (\frac {C_1}{7!} + \frac {C_3}{5!} + \frac {C_5}{3!}+ \frac {C_7}{1!}) t^5 $$

$$ P_7 = 0 * t^5 $$

$$ P_7 = 0 $$

Abhishekksingh 06:18, 24 March 2010 (UTC)

=Kessler's MATLAB code: Line by line explanation= Ref: Lecture Notes [[media:Egm6341.s10.mtg30.djvu|p.30-2]]

Problem Statement

 * {| style="width:100%" border="0" align="left"

Provide an explanation to the code used by the author to find the error in the trapezoidal rule.
 * Refer to the paper by P.Kessler on Trapezoidal rule error, Math 128a, Berkeley, 2006.
 * Refer to the paper by P.Kessler on Trapezoidal rule error, Math 128a, Berkeley, 2006.
 * }
 * }

Solution

 * {| style="width:100%" border="0" align="left"

$$ c_1 $$ = -1 $$ f=1 \,\,\, g=2$$ $$ f=f.*g.*(g+1) $$ yields f=6. $$ fracsum(-1*cn,cd.*f) $$ yields fracsum(-1, 6) Now looking into the fracsum function,

div=gcd(round(n),round(d))

will yield div = 1. because, round(-1) = -1 round(6) = 6 gcd(-1,6) = 1.

n=round(n./div); will give n=1

d=round(d./div); will give d=6

Next, for k=1:length(d) dsum=lcm(dsum,d(k)); end

will yield dsum = 6.

nsum=dsum*sum(n./d); is, nsum = 6*sum(1/6) = 6*(1/6) =1.

div=gcd(round(nsum),round(dsum)); gives div = 1.

So. nsum = 1 and  dsum =6 Hence $$ c_3 = \frac{1}{6} $$

Now going back, [newcn newcd] = [1 6] cn=[cn;newcn]; cd=[cd;newcd]

cn = [-1;1]; cd=[1;6]

f=[f;1]; So, f=[6;1]

and

g=[2+g(1);g]; which means, g = [2+2, 2] = [4;2]

[newpn,newpd]=fracsum((g-1).*cn,f.*cd); i.e, fracsum(-3,6) and fracsum(-1,1)

Again repeating the exercise for fracsum as specified above will give, would yield, $$-\frac{1}{2} \,\,\, and  \frac{1}{6} $$ respectively. Hence,

$$p_2(1) = -\frac{1}{2} + \frac{1}{6} = -\frac{1}{3} $$

So this explains the way to calculate $$ c_3 \,\,\, and p_2(1) $$.

The other constants and the values of polynomial at t=1 can be similarly calculated. It is to be noted that once one constant is found the succeeding constants can be calculated from the previous ones.

Subramanian Annamalai 01:13, 24 March 2010 (UTC)
 * }

=  Compare the result of the given circumference formular with the complete elliptic integral of the second kind method =

Problem Statement
$$ \displaystyle Compare\, the\, results\, i)\, and\, ii)\, $$

$$ \displaystyle i)\, Evaluate\, the\, given\, circumference\, formular\, with\, Comp. Trap,\, Romberg\, and\, Clencurt\, $$

$$ \int_{\theta=0}^{\theta=2\pi} dl = C $$

$$ \displaystyle Where\,\,\,\, \begin{align} dl &=d\theta [r^2+(\frac{dr}{d\theta})^2]^(\frac{1}{2})\\ r(\theta) &= \frac{1-e^2}{1-ecos(\theta)} \\ e &= sin(\frac{\pi}{12}) \\\end{align}\ $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg25.djvu|p.25-2]] and [[media:Egm6341.s10.mtg30.djvu|p.30-3]]

$$ \displaystyle ii)\, The\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

$$ \displaystyle C = 4a E (e) $$

$$ \displaystyle E(e) = \int_{0}^{\pi /2} [1-e^2sin^2(\theta)]^{(\frac{1}{2})} d\theta $$

Ref: Team 4 in HW 4 [|problem_11] and Wikipedia[|Elliptic_integral]

Solution
$$ \displaystyle i)\, Quad\, function\, $$

$$ \displaystyle a)\, Define\, function\,\, dl\, $$

$$ \displaystyle b)\, Integrate\,\, dl\,\, with\, Quad\, $$

$$ \displaystyle Therefore\,\,\, \int_{\theta=0}^{\theta=2\pi} dl = 6.176601938988738 $$

$$ \displaystyle ii)\, Comp.\, Trap.\, $$

$$ \displaystyle a)\, Define\, function\,\, dl\, $$

$$ \displaystyle b)\, Integrate\,\, dl\,\, with\, Comp.\, Trap.\, $$

$$ \displaystyle Therefore\,\,\, dl = 6.176601987658693 $$

$$ \displaystyle iii)\, Romberg\, $$

$$ \displaystyle a)\, Define\, function\,\, dl\, $$

$$ \displaystyle b)\, Integrate\,\, dl\,\, with\, Romberg\, $$

$$ \displaystyle Therefore\,\,\, dl = 6.176601987700596 $$

$$ \displaystyle iiii)\, Clencurt\, $$

$$ \displaystyle iiiii)\, The\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

$$ \displaystyle Therefore\,\,\, C = 4aE(e) = 6.176601933080789\,\,\, when\, a=1\, $$

$$ \displaystyle Therefore\, $$

--Heejun Chung 13:11, 26 March 2010 (UTC)

= Derivation of the formula to obtain arc length of ellipse from cosine raw = Ref: Lecture notes [[media:Egm6341.s10.mtg31.djvu|p.31-1]]

Problem Statement
Derive the following formula for arc length of ellipse from cosine raw.

 * {| style="width:100%" border="0" align="left"

Arclength(PQ) = \int_{\theta_{P}}^{\theta_{Q}}d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^{2}\right]^{\frac{1}{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Solution

Recall the cosine raw:

 * {| style="width:100%" border="0" align="left"

\overline{AB}^2=\overline{OA}^2+\overline{OB}^2-2\overline{OA}\,\overline{OB}\cos d\theta $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }

To simply the derivation, let's use the following notations.

$$\displaystyle \overline{AB} = dl $$

$$\displaystyle \overline{OA} = r(\theta) = r $$

$$\displaystyle \overline{OB} = r(\theta+d\theta) \approx r + dr $$

Also, from the Taylor series expansion of $$\displaystyle \cos d\theta $$,
 * {| style="width:100%" border="0" align="left"

\cos d\theta = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}d\theta^{2n} = 1 - \frac{d\theta^2}{2!} + \frac{d\theta^4}{4!} - \cdots, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

we can use the following approximated expression since the angle $$\displaystyle d\theta $$ is extremely small,
 * {| style="width:100%" border="0" align="left"

\cos d\theta \approx 1 - \frac{d\theta^2}{2} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

As a consequence, (Eq.1) can be expressed as below,

 * {| style="width:100%" border="0" align="left"

\begin{align} dl^2 &= r(\theta)^2+r(\theta+d\theta)^2-2r(\theta)r(\theta+d\theta)\cos d\theta\\ &\approx r^2 + (r+dr)^2 -2r(r+dr)\left(1-\frac{1}{2}d\theta^2\right)\\ &=2r^2+2rdr+dr^2-(2r^2+2rdr)+(r^2+rdr)d\theta^2\\ &=dr^2+r^2d\theta^2+\underbrace{rdrd\theta^2}_{\approx 0}\\ &\approx d\theta^2\left[r^2+\left(\frac{dr}{d\theta}\right)^2\right] \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Therefore,
 * {| style="width:100%" border="0" align="left"

dl = d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Finally, the arc length PQ can be obtained by

 * {| style="width:100%" border="0" align="left"

$$\displaystyle Arclength(PQ)=\int_{\theta_{P}}^{\theta_{Q}}dl = \int_{\theta_{P}}^{\theta_{Q}}d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Above result is exactly same with (2) [[media:Egm6341.s10.mtg30.djvu|p.30-3]].

Yong Nam Ahn 14:46, 24 March 2010 (UTC)

= Clarification for HW4 problem 8 =

The I_exact is obtained from the online integration calculator. There are several online sites and one of them is on line integral

Using such online sites is not a good way to get the integration value because we don't know what kind of algorithm is behind such calculators. However, we needed the integration value which comes from the different method with our investigated methods in HW4 as a criterion in order to compare the integration values from our investigated methods with it. So, this online calculator is used.

There is a better way:

In HW4, we compared the integration value with the criterion which is obtained from online calculator and increased the number of nodes if the difference b/w these two values is larger than O(10^-10).

Instead of comparing the obtained value with the criterion from online, we can compare the obtained value with the value of the previous step. For example, compare the value from 4 nodes with the value from the 2 nodes (which is the values of the previous step). If the difference b/w them is greater than O(10^-10), obtain the value from 8 nodes and compare it with the values from 4 nodes (which is the values of the previous step) and so on... By doing this, we can get integration value with the error less than O(10^-10) without using a separate criterion.

-- Uploaded by Min Zhong 12:00, 23 March 2010 (UTC)

= Contributing Team Members = 1.Subramanian Annamalai 01:09, 24 March 2010 (UTC) Authored: 1 and 9 Proof-read: 4,6 and 11

2. Abhishekksingh 06:18, 24 March 2010 (UTC) Authored: 5 and 8 Proof-read: 3 and 7

3. Yong Nam Ahn 14:46, 24 March 2010 (UTC) Authored: 4, 6, and 11 Proof-read: 2 and 10

4. Min Zhong 16:46, 24 March 2010 (UTC) Authored: 3, 7, Proof-read: 5 and 8

5. Heejun Chung 13:08, 26 March 2010 (UTC) Authored: 2 and 10 Proof-read: 1 and 9

=References= 1. Introduction to Numerical Methods,2nd Edition by Kendall E Atkinson Second Edition 2. Numerical Methods for Engineers,5th Edition by Steven C Chapra and Raymond P Canale

title: User:Egm6341.s10.team3.sa/HW5


text:
User talk:Noj Cire Snave

This is yet another test- I learn by doing!

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Jet engine design point performance

Design point performance calculation
In fixed-wing aircraft driven by one or more jet engines, the performance of the jet engine is important to the operation of the aircraft. Strictly speaking, the performance of the jet engine includes among other parameters the measurement/estimation of thrust, fuel consumption, noise and engine emissions. This article addresses mainly the calculation of thrust.

Suggested reading is the topic Jet Engine Off-Design and Transient Performance.

TS diagram
A reciprocating engine, such as the petrol or diesel unit in a vehicle, has a Constant Volume (during combustion) Cycle. Similarly a Jet Engine (and gas turbine engines in general) has a Constant Pressure (during combustion) Cycle. Also, the Jet Engine has a continuos flow process, whereas the reciprocating engine cycle is intermittent. Temperature vs. entropy (TS) diagrams (see example RHS) are usually used to illustrate the cycle of gas turbine engines. Entropy represents the degree of disorder of the molecules in the fluid. It tends to increase as energy is converted between different forms, i.e. chemical and mechanical.

The TS diagram shown on the RHS is for a single spool turbojet, where a single drive shaft connects the turbine unit with the compressor unit. Apart from stations 0 and 8s, stagnation pressure and stagnation temperature are used. Station 0 is ambient. Stagnation quantities are frequently used in gas turbine cycle studies, because no knowledge of the flow velocity is required.

The processes depicted are:

 * Freestream (stations 0 to 1)
 * In the example, the aircraft is stationary, so stations 0 and 1 are coincident. Station 1 is not depicted on the diagram.

 * Intake (stations 1 to 2)
 * In the example, a 100% intake pressure recovery is assumed, so stations 1 and 2 are coincident.

 * Compression (stations 2 to 3)
 * The ideal process would appear vertical on a TS diagram. In the real process there is friction, turbulence and, possibly, shock losses, making the exit temperature, for a given pressure ratio, higher than ideal. The shallower the positive slope on the TS diagram, the less efficient the compression process.

 * Combustion (stations 3 to 4)
 * Heat (usually by burning fuel) is added, raising the temperature of the fluid. There is an associated pressure loss, some of which is unavoidable

 * Turbine (stations 4 to 5)
 * The temperature rise in the compressor dictates that there will be an associated temperature drop across the turbine. Ideally the process would be vertical on a TS diagram. However, in the real process, friction and turbulence cause the pressure drop to be greater than ideal. The shallower the negative slope on the TS diagram, the less efficient the expansion process.

 * Jetpipe (stations 5 to 8)
 * In the example the jetpipe is very short, so there is no pressure loss. Consequently, stations 5 and 8 are coincident on the TS diagram.

 * Nozzle (stations 8 to 8s)
 * These two stations are both at the throat of the (convergent) nozzle. Station 8s represents static conditions. Not shown on the example TS diagram is the expansion process, external to the nozzle, down to ambient pressure.

Design point performance equations
In theory, any combination of flight condition/throttle setting can be nominated as the engine performance Design Point. Usually, however, the Design Point corresponds to the highest corrected flow at inlet to the compression system (e.g. Top-of-Climb, Mach 0.85, 35,000 ft, ISA).

The design point net thrust of any jet engine can be estimated by working through the engine cycle, step by step. Below are the equations for a single spool turbojet.

Freestream
The stagnation (or total) temperature in the freestream approaching the engine can be estimated using the following equation, derived from the Steady Flow Energy Equation:

$$T_1 = t_0 \cdot (1 + ({\gamma}_c-1)\cdot M^2/2)$$

The corresponding freestream stagnation (or total) pressure is:

$$P_1 = p_0 \cdot (T_1/t_0)^ {{\gamma}_c/({\gamma}_c-1)}$$

Intake
Since there is no work or heat loss in the intake under steady state conditions:

$$T_2=T_1 \,$$

However, friction and shock losses in the intake system must be accounted for:

$$P_2=P_1 \cdot \mathrm{prf}$$

Compressor
The actual discharge temperature of the compressor, assuming a polytropic efficiency is given by:

$$T_3 = T_2 \cdot (P_3/P_2) ^ {{(\gamma}_c-1)/({\gamma}_c \cdot {\eta}pc)}$$

Normally a compressor pressure ratio is assumed, so:

$$P_3 = P_2 \cdot (P_3/P_2)$$

Combustor
A turbine rotor inlet temperature is usually assumed:

$$T_4 = \mathrm{RIT} \,$$

The pressure loss in the combustor reduces the pressure at turbine entry:

$$P_4 = P_3 \cdot (P_4/P_3)$$

Turbine
Equating the turbine and compressor powers and ignoring any power offtake (e.g. to drive an alternator, pump, etc), we have:

$$w_4 \cdot C_{\mathrm{pt}}(T_4-T_5) = w_2 \cdot C_{\mathrm{pc}}(T_3-T_2)$$

A simplyfying assumption sometimes made is for the addition of fuel flow to be exactly offset by an overboard compressor bleed, so mass flow remains constant throughout the cycle.

The pressure ratio across the turbine can be calculated, assuming a turbine polytropic efficiency:

$$P4/P5 = (T4/T5)^{{\gamma}_t/(({\gamma}_t-1).{\eta}_\mathrm{pt})}$$

Obviously:

$$P_5 = P_4 / (P_4/P_5) \,$$

Jetpipe
Since, under Steady State conditions, there is no work or heat loss in the jetpipe:

$$T_8 = T_5 \,$$

However, the jetpipe pressure loss must be accounted for:

$$P_8 = P_5 \cdot (P_8/P_5) \,$$

Nozzle
Is the nozzle choked? The nozzle is choked when the throat Mach number = 1.0. This occurs when the nozzle pressure ratio reaches or exceeds a critical level:

$$(P_8/p_{\mathrm{8s}})crit = (({\gamma}_t+1)/2)^{{\gamma}_t/({\gamma}_t-1)} \,$$

If $$(P_8/p_0) >= (P_8/p_{\mathrm{8s}})crit \,$$ then the nozzle is CHOKED.

If $$(P_8/p_0) < (P_8/p_{\mathrm{8s}})crit \,$$ then the nozzle is UNCHOKED.

Choked Nozzle
The following calculation method is only suitable for choked nozzles.

Assuming the nozzle is choked, the nozzle static temperature is calculated as follows:

$$t_{\mathrm{8s}} = T_8/(({\gamma}_t+1)/2) \,$$

Similarly for the nozzle static pressure:

$$p_{\mathrm{8s}} = P_8/(T_8/t_{\mathrm{8s}})^{{\gamma}_t/({\gamma}_t-1)}$$

The nozzle throat velocity (squared) is calculated using the Steady Flow Energy Equation:

$$V_8^2 = 2gJC_{pt}(T_8 - t_{\mathrm{8s}})$$

The density of the gases at the nozzle throat is given by:

$${\rho}_{\mathrm{8s}} = p_{\mathrm{8s}}/(R \cdot t_{\mathrm{8s}})$$

Nozzle throat effective area is estimated as follows:

$$A_8 = w_8/({\rho}_{\mathrm{8s}} \cdot V_8)$$

Gross thrust
There are two terms in the nozzle gross thrust equation; ideal momentum thrust and ideal pressure thrust. The latter term is only non-zero if the nozzle is choked:

$$F_g = C_\mathrm{x}((w_8 \cdot V_8/g) + A_8(p_{\mathrm{8s}} - p_0)) \,$$

Unchoked nozzle
The following special calculation is required, if the nozzle happens to be unchoked.

Once unchoked, the nozzle static pressure is equal to ambient pressure:

$$p_{\mathrm{8s}} = p_0 \,$$

The nozzle static temperature is calculated from the nozzle total/static pressure ratio:

$$t_{\mathrm{8s}} = T_8/(P_8/p_{\mathrm{8s}})^{{(\gamma}_t-1)/{\gamma}_t}$$

The nozzle throat velocity (squared) is calculated, as before, using the steady flow energy equation:

$$V_8^2 = 2gJC_{pt}(T_8 - t_{\mathrm{8s}})$$

Gross thrust
The nozzle pressure thrust term is zero if the nozzle is unchoked, so only the Momentum Thrust needs to be calculated:

$$F_g = C_\mathrm{x}((w_8 \cdot V_8/g) \,$$

Ram drag
In general, there is a ram drag penalty for taking air onboard via the intake:

$$F_r = w_0 \cdot V_0/g$$

Net thrust
The ram drag must be deducted from the nozzle gross thrust:

$$F_n = F_g - F_r \,$$

The calculation of the combustor fuel flow is beyond the scope of this text, but is basically proportional to the combustor entry airflow and a function of the combustor temperature rise.

Note that mass flow is the sizing parameter: doubling the airflow, doubles the thrust and the fuel flow. However, the specific fuel consumption (fuel flow/net thrust) is unaffected, assuming scale effects are neglected.

Similar design point calculations can be done for other types of jet engine e.g. turbofan, turboprop, ramjet, etc.

The method of calculation shown above is fairly crude, but is useful for gaining a basic understanding of aeroengine performance. Most engine manufacturers use a more exact method, known as True Specific Heat. High pressures and temperatures at elevated levels of supersonic speeds would invoke the use of even more exotic calculations: i.e. Frozen Chemistry and Equilibrium Chemistry.

Worked example
Question

Calculate the net thrust of the following single spool turbojet cycle at Sea Level Static, ISA, using Imperial units for illustration purposes:

Key design parameters:

Intake air mass flow, $$w_2 = 100 \ \mathrm{lb/s} \,$$ (use 45.359 kg/s if working in SI units)

Assume the gasflow is constant throughout the engine.

Overall pressure ratio, $$P_3/P_2 = 10.0 \,$$

Turbine rotor inlet temperature, $$T_4 = \mathrm{RIT}=1400 \ \mathrm{K} \,$$

(factor-up by 1.8, if working with degrees Rankine) Design component performance assumptions:

Intake pressure recovery factor, $$\mathrm{prf} = 1.0 \,$$

Compressor polytropic efficiency, $${\eta}pc = 0.89 \ (i.e. 89\%) \,$$

Turbine polytropic efficiency, $${\eta}pt = 0.90 \ (i.e. 90\%) \,$$

Combustor pressure loss 5%, so the combustor pressure ratio $$P_4/P_3 =0.95 \,$$

Jetpipe pressure loss 1%, so the jetpipe pressure ratio $$P_8/P_5 = 0.99 \,$$

Nozzle thrust coefficient, $$C_\mathrm{x} = 0.995 \,$$

Constants:

Ratio of specific heats for air, $${\gamma}_c = 1.4 \,$$

Ratio of specific heats for combustion products, $${\gamma}_t = 1.333 \,$$

Specific heat at constant pressure for air, $$C_{\mathrm{pc}} = 0.6111 \ \frac{\mathrm{hp} \cdot \mathrm {s}}{\mathrm{lb} \cdot \mathrm{K}} \,$$

(use 1.004646 kW·s/(kg·K) when working with SI units and use 0.3395 hp·s/(lb·°R) if working with American units)

Specific heat at constant pressure for combustion products, $$C_{\mathrm{pt}} = 0.697255 \ \frac{\mathrm{hp} \cdot \mathrm {s}}{\mathrm{lb} \cdot \mathrm{K}} \,$$ (use 1.1462 kW·s/(kg·K) when working with SI units and use 0.387363889 hp·s/(lb·°R) if working with American units) Acceleration of gravity, $$g = 32.174 \ \mathrm{ft}/\mathrm{s}^{2} \,$$ (use 1000 when working with SI units) Mechanical equivalent of heat, $$J = 550 \ \mathrm{ft} \cdot \mathrm{lb}/(\mathrm{s} \cdot \mathrm{hp}) \,$$ (use 1 when working with SI units)

Gas constant, $$R = 96.034 \ \mathrm{ft} \cdot \mathrm{lbf}/(\mathrm{lb} \cdot \mathrm{K}) \,$$ (use 0.287052 kN·m/(kg·K) when working with SI units and use 53.3522222 ft·lbf/(lb·°R) if working with American units including degrees Rankine)

Answer

Ambient conditions

A sea level pressure altitude implies the following:

Ambient pressure, $$p_0 =14.696 \ \mathrm{psia} \,$$ (assume 101.325 kN/m² if working in SI units)

Sea level, ISA conditions (i.e. Standard Day) imply the following:

Ambient temperature, $$t_0 =288.15 \ \mathrm{K} \,$$

(Note: this is an absolute temperature i.e. $$15 \ ^\circ \mathrm{C} + 273.15 \ ^\circ \mathrm{C} \,$$)

(Use 518.67 °R, if working with American units) Freestream

Since the engine is static, both the flight velocity, $$V_0 \,$$ and the flight Mach number, $$M \,$$are zero

So:

$$T_1 = t_0 = 288.15 \ \mathrm{K} \,$$

$$P_1 = p_0 = 14.696 \ \mathrm{psia} \,$$

Intake

$$T_2 = T_1 = 288.15 \ \mathrm{K} \,$$

$$P_2 = P_1 \cdot \mathrm{prf} \,$$

$$P_2 = 14.696 * 1.0 = 14.696 \ \mathrm{psia} \,$$

Compressor

$$T_3 = T_2 \cdot ((P_3/P_2) ^ {({\gamma}_c-1)/({\gamma}_c \cdot {\eta}pc)} = 288.15 * 10 ^ {(1.4-1)/(1.4 * 0.89)} = 603.456 \ \mathrm{K}$$

$$P_3 = P_2 \cdot (P_3/P_2) \,$$

$$P_3 = 14.696 * 10 = 146.96 \ \mathrm{psia} \,$$

Combustor

$$T_4 = \mathrm{RIT}=1400 \ \mathrm{K} \,$$

$$P_4 = P_3 \cdot (P_4/P_3) = 146.96 * 0.95 = 139.612 \ \mathrm{psia} \,$$

Turbine

$$w_4 \cdot C_{\mathrm{pt}}(T_4-T_5) = w_2 \cdot C_{\mathrm{pc}}(T_3-T_2) \,$$

$$100 * 0.697255 * (1400 - T_5)= 100 * 0.6111 * (603.456 - 288.15) \,$$

$$T_5 = 1123.65419 \ \mathrm{K} \,$$

$$P4/P5 = (T4/T5)^{{\gamma}_t/(({\gamma}_t-1).{\eta}_\mathrm{pt})} \,$$

$$P4/P5 = (1400/1123.65419)^ {1.333/((1.333-1) * 0.9)} \,$$

$$P4/P5 = 2.65914769 \,$$

Jetpipe

$$T_8 = T_5 = 1123.65419 \ \mathrm{K} \,$$

$$P_5 = P_4 /(P_4/P_5) \,$$

$$P_5 = 139.612/2.65914769 = 52.502537 \ \mathrm{psia}\,$$

$$P_8 = P_5 \cdot (P_8/P_5) \,$$

$$P_8 = 52.502537 * 0.99 = 51.9775116 \ \mathrm{psia} \,$$

Nozzle

$$P_8/p_0 = 51.9775116/14.696 = 3.53684755 \,$$

$$(P_8/p_{\mathrm{8s}})crit = (({\gamma}_t+1)/2))^{{\gamma}_t/(({\gamma}_t-1)} \,$$

$$(P_8/p_{\mathrm{8s}})crit  = ((1.333+1)/2)^{1.333/(1.333-1)} = 1.85242156 \,$$

Since $$P_8/p_0 > P_8/p_{\mathrm{8s}} \,$$, the nozzle is CHOKED

Choked Nozzle

$$t_{\mathrm{8s}} = T_8/(({\gamma}_t+1)/2) \,$$

$$t_{\mathrm{8s}} = 1123.65419/((1.333+1)/2) \,$$

$$t_{\mathrm{8s}} = 963.269773 \ \mathrm{K} \,$$

$$p_{\mathrm{8s}} = P_8/((T_8/t_{\mathrm{8s}})^{{\gamma}_t/({\gamma}_t-1)})$$

$$p_{\mathrm{8s}} = 51.9775116/ (1123.65419/963.269773)) ^ {1.333/((1.333-1))} \,$$ $$p_{\mathrm{8s}} = 28.059224 \ \mathrm{psia} \,$$ $$V_8^{2} = 2gJC_{pt}(T_8 - t_{\mathrm{8s}}) \,$$

$$V_8^{2} = 2 * 32.174 * 550 * 0.697255 * (1123.65419-963.269773) = 3957779.09 \,$$

$$V_8 = 3957779.09 ^ {0.5} = 1989.41677 \ \mathrm{ft}/\mathrm{s} \,$$

$${\rho}_{\mathrm{8s}} = p_{\mathrm{8s}}/(R \cdot t_{\mathrm{8s}}) \,$$

$${\rho}_{\mathrm{8s}} = (28.059224 * 144)/ (96.034 * 963.269773) = 0.0436782467 \ \mathrm{lb}/ \mathrm{ft}^3 \,$$

NOTE: inclusion of 144 in²/ft² to obtain density in lb/ft³.

$$A_8 = w_8/({\rho}_{\mathrm{8s}} \cdot V_8) \,$$

$$A_8 = (100 * 144)/ (0.0436782467 * 1989.41677) = 165.718701in^2 \,$$

NOTE: inclusion of 144 in²/ft² to obtain area in in².

Gross Thrust

$$F_g = C_\mathrm{x}((w_8 \cdot V_8/g) + A_8(p_{\mathrm{8s}} - p_0)) \,$$

$$F_g = 0.995 (((100 * 1989.41677)/32.174) + (165.718701 * (28.059224 - 14.696))) \,$$

$$F_g = 6152.38915 + 2203.46344 \,$$

The first term is the momentum thrust which contributes most of the nozzle gross thrust. Because the nozzle is choked (which is the norm on a turbojet), the second term, the pressure thrust, is non-zero.

$$F_g = 8355.85259 \ \mathrm{lbf} \,$$

Ram Drag

$$F_r = w_0 \cdot V_0/g \,$$

$$F_r = (100 * 0)/32.174 = 0 \,$$

The ram drag in this particular example is zero, because the engine is stationary and the flight velocity is therefore zero.

Net thrust

$$F_n = F_g - F_r \,$$

$$F_n = 8355.85259 - 0 = 8356 \ \mathrm{lbf} \,$$

To retain accuracy, only the final answer should be rounded-off.

Cooling Bleeds
The above calculations assume that the fuel flow added in the combustor completely offsets the bleed air extracted at compressor delivery to cool the turbine system. This is pessimistic, since the bleed air is assumed to be dumped directly overboard (thereby bypassing the propulsion nozzle) and unable to contribute to the thrust of the engine.

In a more sophisticated performance model, the cooling air for the first row of (static) turbine nozzle guide vanes (immeditely downstream of the combustor) can be safely disregarded, since for a given (HP) rotor inlet temperature it has no effect upon either the combustor fuel flow or the net thrust of the engine. However, the turbine rotor cooling air must be included in such a model. The rotor cooling bleed air is extracted from compressor delivery and passes along narrow passage ways before being injected into the base of the rotating blades. The bleed air negotiates a complex set of passageways within the aerofoil extracting heat before being dumped into the gas stream adjacent to the blade surface. In a sophisticated model, the turbine rotor cooling air is assumed to quench the main gas stream emerging from turbine, reducing its temperature, but also increasing its mass flow:

i.e.

$$w_{\mathrm{rotorexit}} \cdot C_{\mathrm{pt}}\cdot T_{\mathrm{rotorexit}} = w_{\mathrm{rotorbleed}} \cdot C_{\mathrm{pc}}\cdot T_{\mathrm{rotorbleed}} + w_{\mathrm{rotorentry}} \cdot C_{\mathrm{pt}}\cdot T_{\mathrm{rotorentry}} \, $$

$$w_{\mathrm{rotorexit}} = w_{\mathrm{rotorbleed}} + w_{\mathrm{rotorentry}} \, $$

The bleed air cooling the turbine discs is treated in a similar manner. The usual assumption is that the low energy disc cooling air cannot contribute to the engine cycle until it has passed through one row of blades or vanes.

Naturally any bleed air returned to the cycle (or dumped overboard) must also be deducted from the main air flow at the point it is bled from the compressor. If the some of the cooling air is bled from part way along the compressor (i.e. interstage), the power absorbed by the unit must be adjusted accordingly.

Cycle improvements
Increasing the design overall pressure ratio of the compression system raises the combustor entry temperature. Therefore, at a fixed fuel flow and airflow, there is an increase in turbine inlet temperature. Although the higher temperature rise across the compression system implies a larger temperature drop over the turbine system, the nozzle temperature is unaffected, because the same amount of heat is being added to the total system. There is, however, a rise in nozzle pressure, because turbine expansion ratio increases more slowly than the overall pressure ratio (which is inferred by the divergence of the constant pressure lines on the TS diagram). Consequently, net thrust increases, implying a specific fuel consumption (fuel flow/net thrust) decrease.

So turbojets can be made more fuel efficient by raising overall pressure ratio and turbine inlet temperature in unison.

However, better turbine materials and/or improved vane/blade cooling are required to cope with increases in both turbine inlet temperature and compressor delivery temperature. Increasing the latter may also require better compressor materials. Also, higher combustion temperatures can potentially lead to greater emissions of nitrogen oxides, associated with acid rain.

Adding a rear stage to the compressor, to raise overall pressure ratio, does not require a shaft speed increase, but it reduces core size and requires a smaller flow size turbine, which is expensive to change.

Alternatively, adding a zero (i.e. front) stage to the compressor, to increase overall pressure ratio, will require an increase in shaft speed (to maintain the same blade tip Mach number on each of the original compressor stages, since the delivery temperature of each of these stages will be higher than datum). The increase in shaft speed raises the centrifugal stresses in both the turbine blade and disc. This together with increases in the hot gas and cooling air(from the compressor) temperatures implies a decrease in component lives and/or an upgrade in component materials. Adding a zero stage also induces more airflow into the engine, thereby increasing net thrust.

If the increase overall pressure ratio is obtained aerodynamically (i.e. without adding stage/s), an increase in shaft speed will still probably be required, which has an impact on blade/disc stresses and component lives/material.

Other gas turbine engine types
Design point calculations for other gas turbine engine types are similar in format to that given above for a single spool turbojet.

The design point calculation for a two spool turbojet, has two compression calculations; one for the Low Pressure (LP) Compressor, the other for the High Pressure (HP) Compressor. There is also two turbine calculations; one for the HP Turbine, the other for the LP Turbine.

In a two spool unmixed turbofan, the LP Compressor calculation is usually replaced by Fan Inner (i.e. hub) and Fan Outer (i.e. tip) compression calculations. The power absorbed by these two "components" is taken as the load on the LP turbine. After the Fan Outer compression calculation, there is a Bypass Duct pressure loss/Bypass Nozzle expansion calculation. Net thrust is obtained by deducting the intake ram drag from the sum of the Core Nozzle and Bypass Nozzle gross thrusts.

A two spool mixed turbofan design point calculation is very similar to that for an unmixed engine, except the Bypass Nozzle calculation is replaced by a Mixer calculation (where the static pressures of the core and bypass streams at the mixing plane are usually assumed to be equal) followed by a Final (Mixed) Nozzle calculation.

Nomenclature

 * $$A\, $$  flow area
 * $$A_{\mathrm{8calc}}\, $$ calculated nozzle effective throat area
 * $$A_{\mathrm{8 des pt}}\, $$ design point nozzle effective throat area
 * $$A_{\mathrm{8 geometricdesign}}\, $$ nozzle geometric throat area
 * $${\alpha}\, $$ shaft angular acceleration
 * $${\beta}\, $$ arbitrary lines which dissect the corrected speed lines on a compressor characteristic
 * $$C_{\mathrm{pc}}\, $$ specific heat at constant pressure for air
 * $$C_{\mathrm{pt}}\,$$ specific heat at constant pressure for combustion products
 * $$C_{\mathrm{dcalc}}\, $$ calculated nozzle discharge coefficient
 * $$C_x\, $$ thrust coefficient
 * $${\delta}\, $$ ambient pressure/Sea Level ambient pressure
 * $$({\delta}H/T)_{\mathrm{turb}} \, $$ turbine enthalpy drop/inlet temperature
 * $${\delta}N\, $$ change in mechanical shaft speed
 * $${\delta}P_w\, $$ excess shaft power
 * $${\delta}\,{\tau}\, $$ excess shaft torque
 * $${\eta}_{\mathrm{pc}}\, $$ compressor polytropic efficiency
 * $${\eta}_{\mathrm{pt}}\, $$ turbine polytropic efficiency
 * $$g\, $$  acceleration of gravity
 * $$F_g\, $$ gross thrust
 * $$F_n\, $$ net thrust
 * $$F_r\, $$ ram drag
 * $${\gamma}_{\mathrm{c}}\, $$ ratio of specific heats for air
 * $${\gamma}_{\mathrm{t}}\, $$ ratio of specific heats for combustion products
 * $$I\, $$  spool inertia
 * $$J\, $$  mechanical equivalent of heat
 * $$K\, $$  constant
 * $$K_1\, $$  constant
 * $$K_2\, $$  constant
 * $$M\, $$  flight Mach number
 * $$N\, $$ compressor mechanical shaft speed
 * $$N_{\mathrm{cor}}\, $$ compressor corrected shaft speed
 * $$N_{\mathrm{turb cor}}\, $$ turbine corrected shaft speed
 * $$p\, $$  static pressure
 * $$P\, $$  stagnation (or total) pressure
 * $$P_3/P_2\, $$ compressor pressure ratio
 * $$prf\, $$ intake pressure recovery factor
 * $$R\, $$  gas constant
 * $${\rho}\, $$ density
 * $$SFC\, $$ specific fuel consumption
 * $$RIT\, $$ (turbine) rotor inlet temperature
 * $$t\, $$  static temperature or time
 * $$T\, $$ stagnation (or total) temperature
 * $$T_1\, $$ intake stagnation temperature
 * $$T_3\, $$ compressor delivery total temperature
 * $${\theta}\, $$ ambient temperature/Sea Level, Standard Day, ambient temperature
 * $${\theta}_T\, $$ total temperature/Sea Level, Standard Day, ambient temperature
 * $$V\, $$ velocity
 * $$w\, $$ mass flow
 * $$w_{\mathrm{4cor calc}}\, $$ calculated turbine entry corrected flow
 * $$w_{\mathrm{2cor}}\, $$ compressor corrected inlet flow
 * $$w_{\mathrm{4cor des pt}}\, $$ design point turbine entry corrected flow
 * $$w_{\mathrm{4cor turb char}}\, $$ corrected entry flow from turbine characteristic (or map)
 * $$w_{\mathrm{fe}}\, $$ combustor fuel flow

title: Jet engine design point performance