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Why You Rarely Hear About Hayden Panettiere Anymore
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By Jane Harkness/Updated: Jan. 31, 2022 11:52 am EST
Back in the early 2000s, it seemed like Hayden Panettiere was on track to become a household name. She was, for a while, totally inescapable: A starring role on the popular series Heroes and plum parts in high school rom-coms like I Love You, Beth Cooper made her a regular feature of the entertainment landscape. As a child star who learned the Hollywood ropes with roles on soap operas like One Life to Live and Guiding Light, she had always lived life in front of the camera. She was, in every sense, a star on the rise, poised to become a true A-lister.
For all that exposure, however, Panettiere has largely faded from public view in the years since. Today, you're unlikely to find her appearing in blockbuster films or on hotly anticipated television shows. Her attempted music career never really got off the ground either, to the point that this might very well be the first you've heard of it. But Panettiere hasn't been resting on her laurels. She came back to the spotlight with a role on the drama series Nashville, and has popped up on a number of other productions. Notably, she's also thrown herself into campaigning for worthy causes — including some that are incredibly close to her heart.
What has Hayden Panettiere been up to? How have the challenges in her personal life affected her career? What are her plans for the future? We're here to answer those questions — and more.
Heroes' mixed ratings
Panettiere rose to mainstream prominence by starring as Claire Bennet on Heroes. For a while, this show was enormously popular and uniquely headline-making, something of a predecessor to widely-discussed genre TV behemoths like Game of Thrones. Those who were around for its burst of success will likely recall its infamous tagline, "Save the cheerleader, save the world." Claire Bennet is this very cheerleader, and was featured prominently in the show's promotional campaigns. Claire has the power of cellular regeneration, which allows her to heal quickly from injuries and become practically immortal. When she initially discovers her powers, she is both curious and frightened, and eventually realizes that her own fate might be tied to that of the entire world.
Heroes didn't just earn the public's acclaim — critics initially responded to it with effusive praise as well. The first season inspired widespread excitement and outright adulation, scoring multiple wins and nominations from a wide variety of industry awards, including the Golden Globes and the Emmys. But in later seasons, Heroes' ratings took a tumble. It began to feel as though the writers had no real plan for giving the show a satisfying conclusion, and the narrative became increasingly ludicrous and convoluted. Unfortunately for Panettiere, Heroes has become remembered for its stellar first season and not much else. In the end, the show was canceled on a low note, which very well might have impacted Panettiere's career.
She was typecast after Heroes
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When Panettiere started her career as a child actor, she worked on soap operas, television movies, and the occasional miniseries. Disney came calling in the late 1990s, leading to a few prominent voice roles in movies including A Bug's Life, in which Panettiere played the adorable Dot, and Dinosaur, where she played Suri. As she got older and made a name for herself on Heroes, Panettiere found swift and dazzling success — but it came with a price. She quickly found that she was being typecast. She had already played a talented ice skater and high school student in Ice Princess, and now she was playing a cheerleader on Heroes. She also starred as a cheerleader in Bring It On: All or Nothing. A theme was developing, and she wanted out.
After playing a cheerleader again in I Love You, Beth Cooper, Panettiere expressed that she was ready for something new. "If I don't get out of this category pretty quickly, it will just get more and more difficult to be seen as anything besides a cheerleader," Panettiere told Contactmusic in 2009. "Like ... you get in a corset, it's hard to get out of a corset again." Unfortunately, it doesn't seem like Panettiere was able to entirely escape being typecast until several years later, when she landed the role of country singer Juliette Barnes on Nashville.
Panettiere's music career
As fans of Nashville know, Panettiere is one heck of a talented singer who knows how to light up a stage. But somehow, her real-world music career never really got off the ground. Panettiere made her way down the early sections of the Disney actress-to-pop star path, recording songs for several Disney movies including Ice Princess and Cinderella III: A Twist in Time. She even started working on her own album, releasing a single entitled "Wake Up Call" in 2008. But she wasn't happy with the music that her record label was pushing her to sing, and has since revealed her dissatisfaction with this release. Panettiere did not feel like she could express herself authentically, and eventually, the project was shelved.
"At that point I was a puppet, basically. It wasn't me," Panettiere said in an interview with Vulture. However, these unpleasant experiences in the music industry have proved useful to her as an actress. While playing Juliette, an aspiring singer and one of the main roles on Nashville, Panettiere took her experiences into account as her character was encouraged to record music she was not passionate about. "When I read this script and it became that for Juliette, I was like, I understand. I know what that is," the actress remarked in 2013. Though things ended positively in this regard, she doesn't think she'll try to record another album in the future, as she feels that people will conflate her music with her character.
She's struggled with depression
After having her daughter Kaya in December 2014, Panettiere found herself struggling with postpartum depression. She sought treatment multiple times, most recently checking into a Utah treatment center for support in May 2016. Panettiere has repeatedly chosen to share her struggles, despite how difficult they are, as she believes greater openness about the condition can help women who fear they might be alone in dealing with such a fraught illness.
"It takes you a while and you feel off, you don't feel like yourself," Hayden said in an interview on Good Morning America. "Women are so resilient and that's the incredible thing about them ... I think I'm all the stronger for it. I think I'm a better mom because of it because you never take that connection for granted." Panettiere's character on Nashville, Juliette Barnes, also ends up battling the condition. "It doesn't make you a bad person," Panettiere further remarked, "doesn't make you a bad mother. It makes you a very strong, resilient woman. You've just got to let it make you stronger."
Her daughter lives in Ukraine
Today, Hayden's daughter Kaya lives with her father Wladimir Klitschko in Ukraine. Naturally, Panettiere still wants to spend as much time with her daughter as possible, so she travels to Ukraine whenever she gets the chance. When she can't get on a plane and travel to spend hands-on time with her daughter, she picks up the phone for a FaceTime call. It's easy to imagine this might be a big part of why Panettiere doesn't always have the free time to commit to new roles — she has to balance co-parenting on another continent with her acting career.
Although they have been separated for several years now, Panettiere and Klitschko enjoy a genuinely amicable bond, and seek to set a positive example for their daughter. "We're still respectful of each other, and we still have that friendship with each other," Panettierre told Us Magazine. "It's not a situation that you find yourself in all that often, so it's a new one. But I think we've done a really good job." Though the situation undoubtedly involves unique struggles, Panettiere seems more than up to the challenge.
She values privacy
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It may seem contradictory to say that a celebrity like Panettiere is actually a very private person, but that is, in fact, the truth. Unlike so many highly visible stars, Panettiere makes a point of not making every detail of her life public. While she does share her struggles with fans when she's ready, like her battle with postpartum depression, she rarely posts about her life on social media — even when she's busy working or attending star-studded events. Panettiere's Instagram is a modest affair, with few photos posted at all (although you've got a smile at her proudly posting her nearly completed 1,000-piece puzzle). And while Panettiere has a Twitter account as well, she often goes weeks without posting any Tweets. It's a far cry from the ultra-connected world of many stars, who have multiple feeds updating on an hourly basis, keeping fans informed of what they're working on, where they're going, what they're eating, and who they're doing it all with.
It seems that Panettiere chooses to stay out of the public eye to the fullest extent she can these days — and it's not necessarily because of a lack of opportunities. While plenty of celebrities enjoy sharing their day-to-day lives on social media, Panettiere has strong boundaries when it comes to privacy.
She's appeared in controversial films
As Panettiere tried to move forward in her career and break away from her "cheerleader" image, she took on a majorly controversial role. In 2011, she starred as Amanda Knox, the college student who was convicted of the murder of Meredith Kercher, in Amanda Knox: Murder on Trial in Italy. In the real world, Knox was eventually exonerated when another man was identified as the murderer, and has since spoken openly about her ordeal. Kercher's family was unhappy with the contents of the film, especially because it depicts a fictionalized version of Kercher's murder.
"It's so awful what these film people have done. Your imagination runs riot as it is about what happened, but to actually see it like this is very different and very distressing," said Meredith Kercher's father, John Kercher, via Time. "I was told the original brief and synopsis of the film was to take it up to the point of the killing but not actually show the killing itself. It's obvious from these pictures they are showing the killing." While Panettiere may have had good intentions when she accepted the role, it's clear that this film was deeply upsetting for those who lived through these events.
She felt judged in Hollywood
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After she began working on Nashville, Panettiere decided to pack up and move to the city for which the show is named. She purchased a home in Nashville, and subsequently realized that she was much happier living in the South than New York City or Los Angeles. She felt like she could be more laid-back, and thus more true to herself. Moreover, she appreciated that her new neighbors seemed to value the little things in life, allowing her to slow down and stop focusing on the nonstop hustle of life in the entertainment industry. When her glitzy job took her away from Hollywood, she finally felt like she fit in.
"The people here are amazing. Living in Nashville gives me a sense of normalcy that I wasn't sure I was ever going to get back in my life," Panettiere said in a 2014 interview with Southern Living. "When I first got here, all my defenses were still up, and I was waiting for that world to suddenly creep back in. When it didn't, I realized that I can live a normal life here with people who don't judge you based on what you do or who you are in the public eye." That said, the humidity of the region did prove to be something of a challenge — without careful hair styling, she claims to look like "a Pomeranian."
Rebuilding her life
Recently, Panettiere has opened up about her struggle to leave an abusive relationship. In the summer of 2020, her ex-boyfriend, Brian Hickerson, was arrested on several charges of domestic abuse. Now, Panettiere is speaking out about what happened, and why she hopes her story can help other people who are in similar situations break free.
"I am coming forward with the truth about what happened to me with the hope that my story will empower others in abusive relationships to get the help they need and deserve," Panettiere said in a 2020 statement. "I am prepared to do my part to make sure this man never hurts anyone again." Now, she is focusing on herself and rebuilding her life. For Panettiere, stepping away from the camera for a while has been necessary: After going through such a traumatic experience, she wants to put her health and wellbeing first. Healing has to take priority. Happily, she has found the courage to speak out and has surrounded herself with supportive people.
She might question her career choices
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Although Panettiere has taken time off from acting in the past, she has never publicly stated that she regrets becoming an actress. But based on certain statements in recent interviews, it seems like she wouldn't necessarily encourage her daughter to go down the same path she has. This is understandable: Panettiere has spent almost her entire life in Hollywood, and life in the spotlight brings some unique challenges with it.
During an appearance on The Late Show With Stephen Colbert, Colbert talked to Panettiere about her daughter, Kaya. When Colbert asked Panettiere if her daughter had any interest in becoming an actress, she sighed and said, "Please, God, no. I hope not, I hope not." She continued, "I will support whatever she wants to do ... let's go doctor, veterinarian ... I will live vicariously through her!" There's a chance that Kaya will feel inspired by her mother's career and choose to become an actress, but it looks as though Panettiere herself hopes she'll pursue goals that don't involve living in the public eye.
She is passionate about activism
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Panettiere is well-known for her activism, particularly in regards to animal welfare. After appearing on Heroes, she realized that she could use her new platform for good. "The show put me in a place to speak for the things that I'm passionate about," she told Teen Vogue in 2008, a notion she's made good on.
Panettiere joined up with the Whaleman Foundation in 2007 to protest dolphin hunting. She was personally involved in a confrontation between a group of fishermen and several surfers in Japan, who had arrived to protest the annual dolphin hunt in Taiji. This incident is explored in the 2009 documentary The Cove, which analyzes this annual dolphin hunt and offers viewers a call to action for the welfare of these animals.
Today, Panettiere still serves as a spokesperson for the Whaleman Foundation. "I'm an activist who's aware of what's realistic to achieve and what's not," Panettiere said in a 2011 interview with Conde Nast Traveler. She continued, "[the Whaleman Foundation's goals] are achievable, and they make sense."
She is currently on hiatus
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In 2018, the final episode of Nashville aired, bringing the engaging series to a conclusion. Panettiere's time playing Juliette Barnes thus ended after a six-year run. According to her IMDb page as of 2022, Panettiere hasn't appeared in any more TV shows or movies, and doesn't have any upcoming projects in the works. (Her most recent screen appearance: In the 2022 Scream revival, her Scream 4 character, Kirby Reed, pops up in a YouTube video viewed by someone else.) At this point, it looks like Panettiere is purposefully enjoying a well-earned hiatus, and that she does not have any immediate plans to get back in front of the camera.
Considering everything Panetierre has gone through over the course of the past decade, it's easy to see why she feels the need to take some time off from acting. She's welcomed a daughter, faced postpartum depression, and spoken out about abuse — that's enough work for anyone to be going on with. But she has not made any statements about retiring, meaning she'll likely be back onscreen in the future. In the meantime, fans of her work will be waiting to see what her next move is. There's no denying her talent: When Panettiere does decide to act, sing, or campaign again, she'll certainly enjoy a warm welcome back.


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Home inspirational Top 10 Largest Religions In The World Today 2018
Top 10 Largest Religions In The World Today 2018
Top Largest Religions In The World
We all have a specific religion that we follow. In short, a 'Religion' term may be defined as that it is a set of beliefs that is held by a group of people or as a cultural system of designated behaviours and practices, worldviews, texts that relate humanity with the supernatural, spiritual or devotional. It is the identification component for human-beings to give more importance. Globally, there are hundreds of religions in the world. Here, in this article, we are discussing the top 10 largest religions in the world.
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Christianity is the oldest and prominent type of religion and one of the largest religions in the world. Christianity is also an Abrahamic monotheistic religion. It was founded around 27 AD. This religion is followed by the people called Christians, who think that Jesus was the son of God. This Christianity religion generally relies on what Jesus Christ has taught and also depends on all lessons taught by Jesus Christ. 'Bible' is the holy book of Christians. This Christianity religion is on 1st position on this list. There are 2.3 Billion followers and 32 percent of total population of Christianity religion.
In 610 AD, Islam was founded by Prophet Muhammad. This religion is also one of the largest religions in the world. This is the oldest and fast growing religion all over the world. The people who follow Islamic religion are known as Muslims and they believe in Muhammad, Allah and Quran. 'Quran' is the Islamic holy book which is mostly read by the Islamic or Muslim people. This Islam religion is in 2nd position on this list. Muslim community includes two groups: Shia and Sunni. There are 1.7 Billion followers and 23 percent of total population of Islam Community.
Hinduism is presently the third biggest religion in the world. This is an Indian religion and dharma was found around 1500 BC. Some practitioners refer to it as Sanatana Dharma. It is also one of the oldest religions on this earth. The most vital factor of this particular religion is that it has also some other religions in it. There is no particular single coordinator of this specific religion. The people who follow Hinduism religion are known as Hindus. Diwali is the crucial festival of Hindus. The 'Bhagavad Gita' is the holy book of Hinduism and it is read by most of the Hindus. This Hindu religion is on the 3rd position on this list. There are 1 Billion followers and 15 percent of total population of Hinduism religion.
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Buddhism is the religion of unanimity. This is also world's largest religion. In 600 BC, Buddhism religion is found by the Siddhartha Gautama who better known as "The Buddha". Just like Hinduism, Buddhism is also a Dharmic religion. Buddhism is followed by the people who are called Buddhists. 'The Tripitaka' is the holy book of Buddhism. Three Baskets is the actual meaning of 'The Tripitaka'. This holy book was written in Pali language. The religion has different denominations such as Mahayana, Vajrayana and Hinayana. This Buddhist religion is in 4th position on this list. There are 400 Millon followers and 7 percent of total population of Buddhism religion.
Sikhism is the world popular religion which is generally found in the states of Punjab in India. Means the religion is originated from the Punjab region. This Sikhism religion was founded by Guru Nanak in the 15th century or during 1469-1539 AD. It is the largest Dharmic religion and people who follow this Sikhism religion are known as Sikhs. Their holy book is called "Guru Granth Sahib" and it is read by most of the Sikhs. This Guru Granth Sahib holy book was initially written in 'Gurmukhi' script. This Sikhism religion is on 5th rank on this list. There are 30 Millon followers and 0.4 percent of total population of Sikhism.
Judaism is the monotheistic, Abrahamic religion or philosophy of Jewish people. It is also one of the largest religions on the globe. The founders of Judaism are Abraham, Jacob, Issac. They found this religion in the 1300 BC. It is one of the ancient religions in the world. Judaism includes a huge collection of texts, practices, theological positions, and forms of organization. It is the 6th largest religion throughout the world. Jews are those who follow the Judaism religion. There are 20 Millon followers and 0.3 percent of total population of Judaism.
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Bahaism is also an Abrahamic religion as like the previous religion that is Judaism. In 1900 AD, Bahaism was founded by Mirza Husayn Ali who is also known as Bahaullah. It was originally started in Iran in the 19th century. This Bahaism is the unity and equality of all people. The people who follow this religion are known as Bahai and Babis. 'Kitab al-Aqdas' is the holy book of Bahaism. This religion secures the 7th position on the list of top 10 largest religions in the world. There are 8 Million followers and 0.15 percent of total population of Judaism.
Confucianism is also known as 'Ruism'. It is a part of the famous Chinese philosopher Confucius where it teaches how to deal with the life in the better way. Confucianism is founded by Kong Qiu in 600 BC. The central core of Confucianism is humanistic. Confucianism is a Taoist religion just like Shintoism. Confucianism is the 8th biggest religion in the world. The followers of this religion are known as Confucius and Confucians. The Holy books of Confucianism include four major books and five classic books. It has over 7 million followers all across the globe and 0.3 percent of total population of Confucianism.
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Jainism is conventionally is also known as 'Jain Dharma'. Jainism is also one of the largest religion founded "The Great Hero" Mahavira in 600 BC. It is the oldest religion from ancient times. The followers of this religion are known as 'Jains'. It is an old Indian religion and vital standards of Jainism are ahimsa means non-violence and non-attachment. 'Jain Agamas' is the holy book of Jainism. There are more than forty Jain Agamas which are written in Prakrit language. In this religion, Shvetambaras and The Digambaras are two categories of followers. Jainism is the 9th biggest religion in the world. There are 4.5 Millon followers and 0.06 percent of total population of Jainism.
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Shintoism is a traditional religion of Japan was also known as 'Shinto' and 'Kami-no-Michi'. The word Shinto was derived, initially as 'Jindo' or 'Shindo'. Shintoism religion was founded in 300 BC. It is a form of Toaic religion. There are two holy books of Shintoism are Nihon Shoki and The Kojiki which are written and completed in 712 AD. Generally, this Shintoism religion focuses on ritual practices. It is the 10th largest religion throughout the world. It has over 4 million followers all across the globe and 0.01 percent of total population of Shintoism.
Number 10: Baha'i
Founded in 1900AD by Mirza Husayn Ali (Bahaullah), the followers of the religion are known as Bahai and Babis. It was initially started in Iran in the 19th Century where Bahaullah preached the religion. The religion follows the Abrahamic teachings with their holy book called Kitab al-Aqdas. The followers of the religion are currently spread across the Middle Eastern countries and it is 10th largest religion in the world.
Number 9: Judaism
Judaism is an Abrahamic religion and one of the oldest religions in the world. Judaism is the 9th largest religion with over 14 million followers or 0.20% of the global population known as Jews.
Number 8: Spiritism 0.21%
Spiritism is the 8th largest religion in the world with over 15 million followers worldwide.
Number 7: Sikhism 0.32%
Sikhism is one of the youngest of the major world religions and it was originated in the Punjab, India about the end of the 15th century. It is based on the Spiritual teachings of Guru Nanak and the Tenth Guru Gobind Singh named the Sikh scripture Guru Granth Sahib. Sikhism is the 7th largest religion in the world with over 30 million followers or 0.32% of the global population known as Sikh.
Number 6: African tradition religion 1.40%
African tradition is the 6th largest religion in the world with over 100 million followers worldwide. There are many traditional religions practiced by African people. However, they share the same general concepts. Generally these traditions are oral rather than scriptural.
Top 10 biggest religion In The World Today 2018
Top 10 Largest Religions In The World
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redpajama_set_name: RedPajamaCommonCrawl


SimpliVity, a software startup that helps companies manage their hardware, is reportedly planning to IPO in the first quarter of 2016. CEO Doron Kempel said in an interview that the six-year-old company is leaning toward Morgan Stanley as an underwriter. SimpliVity was last valued at over $1bn in March and has raised $276m in total capital. Investors in the company include Accel Partners, Charles River Ventures, DFJ Growth, Kleiner Perkins Caufield & Byers Growth, Waypoint Capital and Meritech Capital Partners.
SimpliVity is a member of the Triton Research IPO Watchlist, a continuously updating database of companies we expect to either IPO or be acquired. The current IPO Watchlist is comprised of 108 companies organized by investment theme.


What does the rise of pan-African banks mean for Kenya and other African countries? What do new research findings on these indigenous banks mean for bank managers, private sector practitioners and financial sector regulators?
Pan-African banks are on the rise. The rapid expansion of these indigenous banks is so phenomenal that Pan-African banks have succeeded to dislodge some well-entrenched European and American banks, which had been very active in the continent for more than five decades.
A team of 20 CGF researchers from Africa, Europe and North America have been looking into the rise of pan-Africa banks and related issues, as part of a research project funded by DFID and ESRC under the DFID-ESRC Growth Research Programme (DEGRP), ES/N013344/2. Their findings provide the material for this symposium.
"Being customer-centric is the way to have a sustainable business model", commented Dr Njoroge.
The Deputy Governor concluded that the CBK looks forward to continuing an active partnership with Professor Murinde and SOAS University of London.
Following the DG's comment on future collaboration, Mr Raphael Otieno, the Director of Research Department at the CBK, reinforced the need of continuing this joint event and embracing a wider scope of participants across Africa.
The symposium also brought together other key stakeholders who have important roles to play in shaping the developments of the financial system, industry, commerce, especially the private in Kenya, in light of the rise of Pan-African banks. These stakeholders include bank managers, banking sector regulators, private sector practitioners and researchers.
In addition, symposium participants also used the opportunity to shape current and future research on Pan-African banks, from the perspective of Kenya. For example, does the rise of Pan-African banks translate into an increase in firms' access to finance – if not, what should be done, in terms of policy and practice? How do different types of private capital flows affect productivity across the different recipient sectors (agriculture, trade, infrastructure, services, extractives, construction, manufacturing and tourism) in Kenya or in Africa? How do fixed and mobile telecommunications and mobile money contribute to economic growth, human development and financial inclusion in African countries, especially in Kenya? Does microcredit increase hope, aspirations and well-being?


Sheezter.Pink "Live From LAX" EP | Rosecrans Ave.
Sheezter.Pink is a name you must get familiar with immediately, he's already on the runway headed to his next destination, stardom. His fly new video for "Frequent Flyer" was just premiered by No Jumper, and his song "Don't Get Mad" is already on your favorite Instagram model's playlist, trust me I checked. The EP "Live From LAX" is in reference to his recent trip to LA where Sheezter was not only mentally inspired by the women, weed and weather in Los Angeles, but also physically inspired by the airport itself. The Pinkzter was actually stuck in the Los Angeles International Airport (LAX) due to a flight delay for way more hours than he had predicted and ended up having enough time to record a freestyle at his terminal which ended up being the title track, and write a few others, which completed the three song EP. I can't help, but thank whatever air traffic controller who chose to delay Sheezter's flight from LAX back home to Connecticut, because we wouldn't have gotten this amazing body of work. Let's hope all of Sheez's flights from LA get delayed in the future and he has enough time to mix, master and shoot videos for a full length "Stuck at LAX" LP.


The rows of pinot grapevines at Williams Seleym Vineyards in the Russian River Valley were the subject for this painting. See this painting in the "Art on the Green" show at Viewpoints Gallery in August.
I was painting in Santa Cruz the other day, on West Cliff Drive near the Santa Cruz Surfing Museum and lighthouse. I chose another view but took a photo of Steamer Lane, a famous surfing spot in Santa Cruz. Steamer Lane has consistent surf and two distinct beautiful waves, add to that a warm, sunny day and there were plenty of surfers in the water.
I chose this composition because of the cast shadows on the cliffs created by the angle of the sun. It is a long drop down from the cliffs to the ocean where the rocky bottom contrasts with the shallow greenish water.
About ten of my Southern California paintings are in a show called "Staycation" through August 13 at the Red Brick Gallery, 315 E Main St, Ventura.
I've painted many townscapes and bungalows in Ventura and Chrisman Avenue is one of my favorite streets. The small Spanish style bungalows in Southern California and the unique landscape plantings of palms, cycads, and cactus, always capture my attention. In this particular painting I like the shadow of the palm tree on the house and the dark underside of the queen palm tree canopy.
Stop by Viewpoints Gallery in Los Altos to see my latest work. The photo above shows the exhibit of my artwork for the month of July.
You can see cityscapes of San Francisco, surfers, sailboats and even a landscape of Ojai. Viewpoints Gallery is located at 315 State St, Los Altos and is open seven days a week, Mon-Sat 11-5 and Sun 11-3.


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\section*{Introduction}

By definition, the mobility of a mechanical linkage is the dimension of its configuration space.
We say that a mechanical linkage moves paradoxically if the mobility is positive, but
one does not expect this by counting parameters and constraints. We are especially interested
in the case when the expected mobility is zero, but the linkage is still mobile.
Examples are closed linkages
with 6 revolute joints: for a generic choice of parameters, the closure equations have 16
complex solutions. But there are many families of special cases with mobility~1, such as
Hooke's linkage~\cite{Hooke}, Bricard's line symmetric linkage\cite{bricard},
or Wohlhart's partially symmetric linkage~\cite{partsym}.

The theory of bonds was introduced in \cite{hss2} as a tool for systematically explaining
and analyzing paradoxical mobility of closed loops with only revolute joints. That paper
contains a simplified proof of Karger's classification of mobile closed 5R linkages (Karger's
original proof \cite{karger} uses computer algebra). In \cite{hlss}, the theory is used
to prove that the genus of the configuration curve of a mobile 6R linkage is at most 5, and
to classify all cases where the maximum is attained. In \cite{ls}, the theory is used to
obtain equations in the Denavit/Hartenberg parameters of a 6R linkage that are necessary
for mobility. The paper \cite{als} introduces bonds for prismatic joints and the paper
\cite{nawratil} introduces bonds for Stewart platforms. The theory can also be used
if the mobility is bigger than one; however, in this paper we will focus on mobility one linkages.

The main purpose of this paper is to make the theory more accessible, because we think that there
is still potential to derive new results on paradoxically moving linkages. The paper is
therefore a survey on bond theory with a tutorial ambition.

In Section~1, we recall
a well-known isomorphism between the Euclidean group $\SE$ of direct isometries from $\R^3$ into itself
and the quotient group of dual quaternions with nonzero real norm by the subgroup of nonzero
real scalars. We use the language of dual quaternions to formulate configuration spaces
of linkages and the closure equations. In Section~3, we recall a well-known description of
all paradoxically moving closed $n$R-loops with revolute joints for $n=3,4,5$;
the classification of paradoxically moving closed 6R-loops is an open problem which will be
the main question addressed in the subsequent sections. Bonds for R- and P-joints are introduced in Section~3;
this section also contains properties that translate into geometric conditions on the
Denavit/Hartenberg parameters of the linkage under consideration and a proposition illustrating
the use of bond theory for showing non-trivial (but known)
geometric conditions for paradoxically moving loops of type PRRRR. Section~4 introduces
the bond diagram of a linkage, which is useful for ``reading off'' the degree of various
coupler motions. The last section summarizes the current knowledge on paradoxically moving
6R-loops and their bond diagrams and points out open questions. It also contains the single
new result of this paper (Example~\ref{ex:333}): by specializing a line symmetric linkage,
one may obtain a linkage with three additional rotations in its configuration set.

This article has been accepted for publication in the IMA Journal of Mathematical
Control and Information published by Oxford University Press.

\section{Dual Quaternions and the Closure Equations}

The algebra $\D\H$ of dual quaternions is defined as the 8-dimensional vector space over $\R$
with basis $(1,\qi,\qj,\qk,\eps,\eps\qi,\eps\qj,\eps\qk)$. The multiplication is defined in the
usual way for quaternions; the symbol $\eps$ is supposed to commute with all quaternions and
to fulfill the equation $\eps^2=0$.
Every dual quaternion $h$ can be written as $h=p+\eps q$ with quaternions $p,q$ called
the primal and dual part of $h$. Alternatively, we may write $h$ as an expression
$h=a_0+a_1\qi+a_2\qj+a_3\qk$ with coefficients $a_0,a_1,a_2,a_3\in\D:=\R\oplus\eps\R$
in the ring of dual numbers. The conjugate of $h$ is defined as $\overline{h}:=a_0-a_1\qi-a_2\qj-a_3\qk$;
conjugation is an anti-automorphism of $\D\H$.

The norm of the dual quaternion is defined as $N(h):=h\overline{h}$. Norm is a
homomorphism of the semigroup $(\D\H,\cdot)$ to the semigroup $(\D,\cdot)$.
Let ${\mathbb S}:=\{ h\in\D\H \mid N(h)\in\R \}$ and ${\mathbb S}^\ast:=\{h\in\D\H \mid N(h)\in\R^\ast\}$,
where $\R^\ast:=\R \setminus\{0\}$. Then ${\mathbb S}^\ast$ is a group and $\R^\ast$ is
a normal subgroup. The quotient group ${\mathbb S}^\ast/\R^\ast$ is isomorphic to the
group $\SE$, see \cite[Section~9.3]{selig05}; we may consider it as a locally closed subset $G$ in $\P^7$. Its closure
is the Study quadric $S$, represented by all dual quaternions in ${\mathbb S}$.
We also introduce the \emph{null cone $Y$} defined by the quadratic
form $p+\eps q \mapsto p\overline{p}$.
For instance, if $t\in\R$, then $[t-\qi]$ corresponds to a rotation around the first axis by
an angle $2\mathrm{arccot}(t)$, and $[1-\eps\qi t]$ corresponds to a translation by
a distance $2t$ in the direction of the first axis.

\begin{Remark} \label{rem:inf}
In order to parametrize the full rotation group around a fixed axis, we choose the parameter $t$ in
$\P^1_\R = \R\cup \{\infty\}$. For any dual quaternion $h$, the element $(\infty-h)$ is not finite but
we still can consistently say that the class $[\infty-h]$ is equal to $[1]$. The corresponding group element
is the identity.
\end{Remark}

A {\em linkage} is a collection of rigid bodies, called {\em links}, where two links may be
connected by a {\em joint}. A joint restricts the relative position of the joined links.
We consider two types of joints:
\begin{enumerate}
\item (R) revolute joints: allow rotations around a fixed axis;
\item (P) prismatic joints: allow translations in a fixed direction;
\end{enumerate}

The {\em link graph} of a linkage is defined by putting a vertex for each link, and an edge
whenever two links are joined by a joint. In order to specify the linkage completely, it suffices
to specify the allowed subset of $\SE$ for each joint.

Relative positions can be composed by the group operation: the relative position of link~1 with
respect to link~2 times the relative position of link~2 with respect to link~3 is equal to the
relative position of link~1 with respect to link~3. By multiplying relative positions in
a cycle in the link graph, we get the {\em closure equations}. The solution set of the
closure equations is the {\em configuration set} of the linkage.

\begin{Example} \label{ex:6rhom}
Let $n\ge 3$ be an integer.
The link graph of a closed $n$R linkage is an $n$-cycle. For $r=1,\dots,n$, the set of allowed relative
position can be written as $\{ [t_r-h_r] \mid t_r\in\R\cup\{\infty\} \} = \{ [t_r-h_r] \mid t_r\in\R \} \cup \{[1]\}$, where $h_r$ is a dual
quaternion such that $h_r^2=-1$ specifying the rotation axis in an initial position.
This gives the closure equation
\begin{equation}\label{eq:ch}
 [(t_1-h_1)(t_2-h_2)\dots(t_n-h_n) ] = [1] .
\end{equation}
The class on the left hand side is $[1]$ if and only if 7 of the 8 coordinates of the product
are zero, hence we have 7 polynomial equations in $t_1,\dots,t_n$. Actually, one of the 7 equations
is redundant, because it is clear that the product is contained in the Study quadric $S$; and
if $h=a_0+a_1\qi+a_2\qj+a_3\qk$, $a_0,\dots,a_3\in\D$, is a dual quaternion with norm in $\R^\ast$
such that $a_1=a_2=a_3=0$, then it follows that $h\in\R^\ast$.

Assume $n=6$.
For generic choice of $h_1,\dots,h_6\in\D\H$ such that $h_r^2=-1$ for $r=1,\dots,6$, one gets
at most 16 real solutions for these 6 equations in $t_1,\dots,t_6$, including the solution
$t_r=\infty$ for $r=1,\dots,6$. The number of complex solutions
is infinite, but if one excludes solutions contained in the null cone $Y$,
then one generically gets 16 complex solutions \cite[pp.~262--264]{selig05}.
\end{Example}

\begin{Example} \label{ex:6rinhom}
In the case of a closed $n$R linkage (or more general simply closed linkages), it is possible
to use a more invariant specification which does not depend on the choice of an initial position.
In essence, this is the method of Denavit/Hartenberg~\cite{Denavit}.
For $r=1,\dots,n$, let $\phi_r$ be the angle between the $r$-th and $(r+1)$-th rotation axis,
with indices modulo $n$; let $d_r$ be the normal distance between these axes; let $s_r$ be the
signed distance of the intersections of the common normals of neighboring axes on the $r$-th
rotation axis. For each linkage, we introduce an internal frame of reference in which the first
joint is the first coordinate axes, the intersection with the common normal with the previous
axes is the origin, and the second axis lies in the first coordinate plane. When we express relative
positions in this frame of reference, then the allowed positions are composed by a translation
in direction of the first axes by $s_r$, rotation around first axis by $\phi_r$, translation
in direction of the second axis by $d_r$, and rotation around the second axis by an arbitrary angle.
So the closure equation is
\begin{equation}\label{eq:cinh}
  [(t_1-\qi)g_1(t_2-\qi)g_2\cdots(t_n-\qi)g_n] = [1],
\end{equation}
where
\begin{equation}\label{gi:1}
g_r=\left(1-\frac{s_r}{2}\eps\qi\right)\left(w_r-\qk\right)\left(1-\frac{d_r}{2}\eps\qk\right)
\end{equation}
and $w_r=\cot(\frac{\phi_r}{2})$ for $r=1,\dots,n$. If $\phi_r$ is a multiple of $\pi$ for some $r$,
then we set $w_r=\infty$ and
$g_r=\left(1-\frac{s_r}{2}\eps\qi\right)\left(1-\frac{d_r}{2}\eps\qk\right)$.

Similar as in the previous example, we get 6 equations in $n$ parameters, and if $n=6$,
then for generic choice
of the parameters $s_1,w_1,d_1,\dots,s_6,w_6,d_6$ we get 16 isolated complex solutions.
(But now there is no trivial solution at infinity.)
\end{Example}

\begin{Remark}
The invariant parameters for a closed $n$R linkage above do depend on a choice of orientation
of the rotation axis. Generically, there are $2^n$ choices leading to different parameters.
If we change the orientation of the $k$-th axis, then $w_k$ gets replaced by $-1/w_k$ and $s_k$
gets replaced by $-s_k$, and all other parameters stay the same. Also, the parameter $d_k$ can be
replaced by $-d_k$ without changing the linkage, one just needs to reparametrize the
sets of allowed positions (replacing $t_k$ by $-1/t_k$).

If $w_k=0$ for some $k$, then one can add a constant to $s_k$ and $s_{k-1}$ without changing
the linkage, because then the axes are parallel and the common normal is not unique. Similarly,
if $w_k=\infty$, then one can add a constant to $s_k$ and subtract the constant from $s_{k-1}$.
\end{Remark}

\begin{figure}
  \centering
  \includegraphics{bennett}
  \caption{Bennett linkage}
  \label{fig:bennett}
\end{figure}

\begin{Example} \label{ex:bennett}
With the notation from Example~\ref{ex:6rinhom}, we set $n=4$ and
\[ (w_1,w_2,w_3,w_4) = (1,2,1,2) , \]
\[ (d_1,d_2,d_3,d_4) = (4,5,4,5) , \]
\[ (s_1,s_2,s_3,s_4) = (0,0,0,0) . \]
The closure equation~\eqref{eq:cinh} for $(t_1,t_2,t_3,t_4)$ can be simplified using computer algebra (we used Maple).
The simplified system is
\[ 3t_1t_2+1 = t_1+t_3 = t_2+t_4 = 0 , t_1^2+1\ne 0, t_2^2+1\ne 0. \]
Its solution set is a curve that can be easily parametrized; it is
\[ (t_1,t_2,t_3,t_4) = \left( t,\frac{-1}{3t}, -t, \frac{1}{3t} \right) . \]
This linkage is therefore mobile. It is an example of a Bennett linkage \cite{bennett14} (Figure~\ref{fig:bennett}).
The general description of Bennett linkages is given by the equations
\begin{equation}
  \label{eq:1}
  \begin{gathered}
s_1=s_2=s_3=s_4=0,\ w_1=w_3\ne 0,\infty,\ w_2=w_4\ne 0,\infty\\
d_1=d_3\ne 0,\ d_2=d_4\ne 0,\
 \frac{2d_1w_1}{w_1^2+1}=\frac{2d_2w_2}{w_2^2+1} ,
  \end{gathered}
\end{equation}
which lead to a similar one-dimensional solution set.
\end{Example}

\section{Mobile Closed 4R and 5R Linkages}

For $n=3,4,5$, and for generic choice of parameters, the closure equation~\eqref{eq:ch} has
only the trivial solution, and the closure equation~\eqref{eq:cinh} has no solution at all
(disregarding the complex solutions with factors in the null cone $Y$). Nevertheless, there are
special cases of mobile linkages of these types. Our main question is what are the implications
of the assumption of mobility for the structure of a linkage. The hope is to find enough conditions
that allow a classification of mobile loops with 4,5 (and later 6) links.

\begin{Example} \label{ex:ca}
If $h_1=h_s$ for some $s, 2\le s\le n$, then the closure equation~\eqref{eq:ch} has the solution
\[ t_s=-t_1, t_r=\infty \mbox{ for } r\ne 1,s. \]
\end{Example}
These are configurations where two of the axes coincide, and one part of the linkage just
rotates about the coinciding axes.

It is easy to show that every mobile 3R linkage is of this type. In case all 3 axes coincide,
the mobility is 2.

\begin{Example} \label{ex:sph}
Assume that $n=4$. Assume that the dual parts of $h_1,\dots,h_4$ are all zero. Then the
dual part of the left hand side of equation~\eqref{eq:ch} is automatically zero and the
closure equation boils down to three polynomial equations in $t_1,t_2,t_3,t_4$. In general,
the solution has complex dimension~1.

Geometrically, the vanishing of the dual part means that all four axes pass through the origin.
This type of linkage is known as {\em spherical 4R linkage}. It can also be
characterized by the conditions
\[ s_1=s_2=s_3=s_4=d_1=d_2=d_3=d_4=0 . \]

A similar case (which may actually be considered as limiting case of a spherical linkage) is the
{\em planar 4R linkage}, where all 4 axes are parallel.
\end{Example}

It is well-known \cite{delassus} that every mobile 4R linkage either has two coinciding axes
or is spherical, planar, or Bennett.

\begin{Example} \label{ex:frozen}
Let $n=4$. Let $h_1,h_2,h_3,h_4$ be the dual quaternions defining the rotation axes of
a mobile 4R linkage (e.g. spherical). Choose an arbitrary dual quaternion $h_5$ such that
$h_5^2=-1$. Then any configuration of the mobile 4R linkage can be extended to a configuration
of the 5R linkage by setting $t_5:=\infty$. Hence the 5R linkage is again movable.

The 5th joint in this linkage remains frozen during this particular motion.
\end{Example}

\begin{figure}[ht]
\begin{center}
\includegraphics{./Goldberg0}
\end{center}
\caption{The link graph of a triply closed linkage with 6 links and 8 joints (three joints with
coinciding axes) of mobility 1. If the link in the center is removed,
	one obtains a Goldberg 5R linkage.}
\label{fig:gblg}
\end{figure}

\begin{Example} \label{ex:goldberg}
Consider the triply closed linkage with 6 links that are connected according to the link graph
in Figure~\ref{fig:gblg}. Its mobility is one: the motion of the joint corresponding
to the vertical edge in the middle determines the motion of the two Bennett 4R linkages, and then
the motion of the 3R linkage below is also determined. If we remove the link in the center,
then we get a mobile closed 5R linkage. This construction is due to Goldberg \cite{goldberg}.
\end{Example}

It is well-known that every mobile closed 5R linkage either has coinciding axes, or a frozen joint,
or is planar, spherical, or a Goldberg linkage. The original proof \cite{karger} uses
computer algebra; a simpler proof without computers is based on bond theory \cite{hss2}.

Let ${\cal L}_n:=\R^{2n}\times (\P^1)^n$ denote the parameter space of closed $n$R-linkages.
Let $D_n:={\cal L}_n\times (\P^1_\C)^n$ be the Zariski closure of the set of all solutions
of equation~\eqref{eq:cinh}. The projection $D_n\to{\cal L}_n$ is a proper morphism,
hence the dimension of its fiber is upper semicontinuous in the Zariski topology
as a function in ${\cal L}_n$. The subset ${\cal M}_n\subset {\cal L}_n$ of all parameters
of mobile linkages is then also Zariski closed, i.e., it is a subset defined by algebraic
equations in the parameters $s_1,w_1,d_1,\dots,s_n,w_n,d_n$. The linkages with parameters in ${\cal M}_n$
have an infinite solution set over the complex numbers. It is possible that none of these
solutions are real, for instance for a planar linkage where $d_1>d_2+d_3+d_4$.

For $n=3,4,5$, these equations are known (for $n=3,4$, we essentially gave them above; for $n=5$, see \cite{dietmaier}).
The dimension of ${\cal M}_n$
is determined by the largest components, corresponding to linkages with coinciding/parallel axes
($\dim({\cal M}_3)=3$, $\dim({\cal M}_4)=7$, $\dim({\cal M}_5)=10$); but the more interesting components have
smaller dimension (Bennett linkages form a 3-dimensional component of ${\cal M}_4$, and
Goldberg linkages form a 5-dimensional component of ${\cal M}_5$). For ${\cal M}_6$,
we do know the dimension: it is 14, again because of components with coinciding axes.
Several other components are known and will be discussed in the following. Even more
algebraic subsets of ${\cal M}_6$ are known which are not contained in any known component,
but for which it is not clear whether they form a component or they are properly contained
in some yet unknown component. It is an open problem
to determine all components and to give equations for them. A broad discussion of partial
results can be found in \cite{li}.

\section{Definition and First Properties of Bonds}

Assume that we have a linkage with $e$ joints. If the $k$-th joint is of type R, then the
set of allowed motions can be parametrized by $t_k\mapsto m_k:=(t_k-h_k)g_k$ for some $h_k,g_k\in\D\H$
of norm 1 with $h_k^2=-1$, with $t_k\in\R$. If we pass to classes, we may even allow $t_k\in\P^1_\R$,
see Remark~\ref{rem:inf}.
The parametric dual quaternion $m_k$ also appears as a factor in closure
equations. Note that $N(m_k)=t_k^2+1$.

If the $k$-th joint is of type P, then the set of allowed motions can be parametrized by
$t_k\mapsto m_k:=(t_k-\eps p_k)g_k$ for some purely vectorial $p_k\in\H$ (that is, $p_k + \overline{p_k} = 0$) and $g_k\in \D\H$, both of norm 1, with
$t_k\in\R\setminus \{0\}$. Here, we have $N(m_k)=t_k^2$.

Recall that the configuration set $K$ is the set of all $(t_1,\dots,t_e)$ such that
$m_{i_1}\dots m_{i_r}\in\R^\ast$ for all loops with edges $i_1,\dots,i_r$. In order to define
bonds, we have to allow also complex parameters $t_1,\dots,t_e$. Recall that the Zariski
closure of any set $X$ is defined as the set of all points, maybe with complex coordinates,
which satisfy all polynomial equations that are satisfied by all points in $X$.
The Zariski closure of $K$ in the product of complex projective lines
is denoted by $\overline{K}$. The set $B$ of {\em bonds} is defined as the set of
all elements $(t_1,\dots,t_e)\in\overline{K}$ such that at least one of the $m_k(t_k)$ has
norm zero.
For a fixed bond, the subset of joints $k$ such that $N(m_k(t_k))=0$ are called the joints
attached to the bond. The bond also induces a partition of all links into the connected components
of the subgraph which is obtained by deleting all attached joints.

The following proposition guarantees that the set of bonds of a mobile linkage
is non-empty.

\begin{Proposition} \label{thm:frozen}
A joint is frozen in a linkage if and only if it is not attached to any bond.
\end{Proposition}

\begin{proof}
If the $k$-th joint is frozen, then $t_k=c$ for some constant $c$ for all configurations
in $K$. Hence we also have $t_k=c$ for all points in $\overline{K}$, which includes all bonds.
Hence $N(m_k)\ne 0$.

Conversely, if the $k$-th joint is not attached to any bond, then the projection from $\overline{K}$
to the coordinate $t_k$ is not surjective. On the other hand, the image is a closed subvariety.
It follows that the image is finite and therefore the joint is frozen.
\end{proof}

\begin{Example} \label{ex:bondbenett}
The Zariski closure of the configuration set of the Bennett linkage in Example~\ref{ex:bennett} is
\[ \{ (t_1,t_2,t_3,t_4) \mid 3t_1t_2+1 = t_1+t_3 = t_2+t_4 = 0 \}. \]
It has 4 bonds:
\[ \beta_1 := (\ci,-\ci/3,-\ci,\ci/3),\ \beta_2 := (-\ci,\ci/3,\ci,-\ci/3), \]
\[ \beta_3 := (\ci/3,-\ci,-\ci/3,\ci),\ \beta_4 := (-\ci/3,\ci,\ci/3,-\ci). \]
The joints 1 and 3 are attached to $\beta_1$ and $\beta_2$.
The joints 2 and 4 are attached to $\beta_3$ and $\beta_4$.
\end{Example}

\begin{Proposition}
Let $i_1,\dots,i_k$ be a set of joints forming a path in the link graph such that starting point
and ending point are in the same subset of the partition induced by the bond $\beta = (t_1,\ldots,t_e)$. If at least one
of the joints is attached to the bond $\beta$, then $m_{i_1}(t_{i_1})\cdots m_{i_k}(t_{i_k})=0$.
\end{Proposition}

\begin{proof}
Since the starting and ending point are in the same subset of the partition, there exists a path $(j_1,\dots,j_l)$ of joints
not attached to the bond $\beta$, with the same starting and ending point. The closure equation
\[ [m_{i_1}\cdots m_{i_k}] = [m_{j_1}\cdots m_{j_l}]  \]
is valid for all configurations, but it is not valid for $\beta$ because the left side has norm zero and the right
side has norm different from zero. Since $\beta$ is in the closure of the configuration set, there is only one possibility:
the left side is not defined, because the product is zero.
\end{proof}

\begin{Example} \label{ex:bennettbc}
In our running example of the Bennett linkage (Example~\ref{ex:bennett} and Example~\ref{ex:bondbenett}),
the following equations
and their conjugate counterparts obtained by replacing $\ci$ by $-\ci$ are valid:
\[ (\ci-\qi)g_1(-\ci/3-\qi)g_2(-\ci-\qi)g_3 = 0 , \]
\[ (\ci-\qi)g_2(\ci/3-\qi)g_3(-\ci-\qi)g_4 = 0 , \]
\[ (\ci-\qi)g_3(-\ci/3-\qi)g_4(-\ci-\qi)g_1 = 0 , \]
\[ (\ci-\qi)g_4(\ci/3-\qi)g_1(-\ci-\qi)g_2 = 0. \]
\end{Example}

Assume that we have a minimal chain $i_1,\dots,i_k$ such that $m_{i_1}(t_{i_1})\cdots m_{i_k}(t_{i_k})=0$
for some fixed bond $(t_1,\dots,t_e)$. Then $N(m_{i_1}(t_{i_1}))=N(m_{i_k}(t_{i_k}))=0$ - otherwise we could
multiply by $\overline{m}_{i_1}(t_{i_1})$ from the left or by $\overline{m}_{i_k}(t_{i_k})$ from the right and make the chain
shorter. The condition
\begin{equation} \label{eq:bond}
 m_{i_1}(t_{i_1})\cdots m_{i_k}(t_{i_k})=0,\ N(m_{i_1}(t_{i_1}))=N(m_{i_k}(t_{i_k}))=0
\end{equation}
is called {\em bond condition}. The validity of a bond condition for some chain in the link graph
has some interesting geometric consequences on the geometric parameters of the linkage.

\begin{Lemma} \label{lem:bcon1}
\begin{enumerate}
\item If joint 1 is of type R and joint 2 is of type R or P, then the bond condition 1-2 is never valid.
\item If joint 1 is of type P and joints 2 and 3 of type R, and the bond condition 1-2-3 is valid,
	then the axes of joints 2 and 3 are parallel.
\end{enumerate}
\end{Lemma}

\begin{proof}
(1): Assume that joints 1 and 2 are of type R, and the axis have distance $d$ and twist angle $2\mathrm{arccot}(w)$.
Assume, without loss of generality, that the bond coordinates at joints 1 and 2 are both $\ci$ and not $-\ci$
(this can always be achieved by a change of orientation of the axes). With
\[ g= \left(w-\qk\right)\left(1-\frac{d}{2}\eps\qk\right), \]
the bond condition reduces to the equation
\[ (\ci-\qi)g(\ci-\qi)= (2w\ci-\eps d\ci)(\ci-\qi) = 0 ,\]
hence $w=d=0$ and the axes are equal.

Assume now that joint 2 is of type P. Then the bond coordinate at the second joint is $0$, and the bond condition
has the form $(t_{1}-\qi)\eps p=0$ for some quaternion $p$ specifying the direction of the P-joint. Since $p$ is invertible,
it follows that $t_1-\qi=0$, which is impossible.

The statement (2) reduces to a similar short and straightforward calculations.
\end{proof}

For any chain $c:=(i_1,\dots,i_k)$, the {\em coupling space} $L_c$ is defined as the linear subspace
of $\D\H$ generated by all products $m_{i_1}(t_{i_1})\cdots m_{i_k}(t_{i_k})$, where $t_{i_1},\dots,t_{i_k}$
range over the full parameter space. If all joints are of type R or P, then $L_c$ has a
generating set of cardinality $2^k$ which can be obtained by expanding the product and taking
all coefficients with respect to $t_{i_1},\dots,t_{i_k}$. The projectivization of the coupling
spaces contains the {\em coupling varieties}, consisting of all relative positions of the two links
that are connected by the chain.

\begin{Lemma} \label{lem:bcon2}
Let $c:=(i_1,\dots,i_k)$ be a chain of joints.
\begin{enumerate}
\item If the joint $i_1$ or the joint $i_k$ is of type R, then $\dim(L_c)$ is even.
\item If $\dim(L_c)=2$, then all joints are of the same type and have the same axis (for R-joints)
	resp. directions (for P-joints).
\item If all joints are of type $R$ and $\dim(L_c)=4$, then all axes are parallel or pass through a
	common point.
\item If $k=3$ and there is a bond such that the bond condition for $c$ is valid, then $\dim(L_c)<8$.
\end{enumerate}
\end{Lemma}

\begin{proof}
This is \cite[Theorem 1]{hss2}. The proofs of (1) and (4) do give some insight, so we include them here.

(1): assume that $i_1$ is an R-joint. Then $L_c$ is closed under multiplication by $h_1$ from the left.
Since left multiplication by $h_1$ is a linear map whose square is negative identity, it follows that $L_c$
may be considered as a vector space over $\C$. Its real dimension is two times its dimension over $\C$,
which proves the claim.

(4): Expanding the bond condition, we obtain a nontrivial linear equation between the products generating
$L_c$. Hence these products cannot be linearly independent.
\end{proof}

Assume that we have a chain 1-2-3 of three joints of type R. Then we say that the chain satisfies the
{\em Bennett condition} if $\dim(L_c)=6$. The following proposition expresses the condition
in terms of the Denavit/Hartenberg parameters. The proof is straightforward.

\begin{Lemma} \label{lem:bcon3}
Let $d_1,d_2,w_1,w_2,s_2$ be the distances, angles, and offset of a 3-chain of R-joints.
Then the Bennett condition is equivalent to
\[ s_2=0,\ \frac{2d_1w_1}{w_1^2+1}=\frac{2d_2w_2}{w_2^2+1} \]
(compare with Equation~\eqref{eq:1}).
\end{Lemma}

The lemmas ~\ref{lem:bcon1}, \ref{lem:bcon2},and \ref{lem:bcon3} above give necessary conditions on the geometric
parameters of a linkage for the existence of bonds.
The following proposition, which is taken from \cite[Theorem 6]{als}, demonstrates how these lemmas
are applied to classify linkages with a given link diagram and types of joints.

\begin{Proposition} \label{ex:prrrr}
Consider a mobile closed PRRRR linkage. Then one of the following conditions must be satisfied.
\begin{enumerate}
\item The P-joint is frozen.
\item Two of the axes of the rotational joints coincide.
\item Three of the rotational axes are parallel, and the fourth axes is frozen.
\item The axes of joints 2 and 3 are parallel, and the axes of joints 4 and 5 are parallel.
\end{enumerate}
\end{Proposition}

\begin{proof}
By Lemma~\ref{lem:bcon1}, we get that either joints 2 and 3 are parallel {\em or} joints 4 and 5 are parallel.
In order to show that actually both are necessary, we consider the closure equation modulo $\eps$.
If, say, joints 2 and 3 are parallel, and joints 4 and 5 are not, then there are three different rotation axes, since parallel
axes only differ in their dual part. Modulo $\eps$, we get a closure equation of a 3R loop with three different
axes, but such a link is never movable.
\end{proof}

For chains of four R-joints, the bond condition does not imply geometric conditions on the four axes.
Indeed, given four generic dual quaternions $h_1,h_2,h_3,h_4$, there are two solutions of the equation
\[ (\ci-h_1)(t_2-h_2)(t_3-h_3)(\ci-h_4) = 0 \]
in the unknowns $t_2,t_3$. This gives room for two bonds with $t_1=t_4=\ci$. Varying the signs of $t_1,t_4$,
one has up to 8 choices for the coordinates $(t_1,t_2,t_3,t_4)$ of a bond in such a chain.

In a 6R loop with joints 1-2-3-4-5-6-1, one can get a geometric condition by comparing the possible solutions of bond coordinates
in the chains 1-2-3-4 and 4-5-6-1. The idea is to take into account the equality
\[ [(t_1-h_1)(t_2-h_2)(t_3-h_3)] = [(t_6+h_6)(t_5+h_5)(t_4+h_4)] , \]
which holds for all points in the configuration set, and therefore also for the bonds in case both sides are defined.
The lengthy calculations have been done in \cite{ls} for 6R linkages given in terms of their Denavit/Hartenberg parameters
$d_1,w_1,s_1,\dots.d_6,w_6,s_6$. For $i=1,\dots,6$, we define $c_i:=\frac{w_i^2-1}{w_i^2+1}$ and $b_i:=\frac{2d_iw_i}{w_i^2+1}$.
Then we define the {\em quad polynomial} as the quadratic polynomial in a variable $x$
\[ Q_1^+(x) = \left(x+\frac{b_{3}c_3-b_{1}c_1}{2}-\frac{s_{1}}{2}\ci\right)^2 + \]
 \[ \frac{\ci}{2}\left(b_1 s_{2}+b_{3} s_{3}+s_{2} b_{3} c_{2}+s_{3} b_1 c_{2}\right) -\]
 \[ \frac{b_1 b_{3} c_{2}-s_{2} s_{3} c_{2}}{2}
   + \frac{s_{2}^2+s_{3}^2-b_1^2+b_{2}^2-b_{3}^2-b_{2}^2 c_{2}^2}{4}. \]
For $i=2,\dots,6$, we define the quad polynomial $Q_i^+(x)$ by a cyclic shift of indices that
shifts $1$ to $i$. Finally, we define $Q_i^-(x)$ by replacing the parameters $c_1,\dots,c_6,b_1,\dots,b_6$
and $s_2,s_4,s_6$ by their negatives, and leaving $s_1,s_3,s_5$ as they are. For instance,
\[ Q_1^-(x) = \left(x+\frac{b_{3}c_3-b_{1}c_1}{2}-\frac{s_{1}}{2}\ci\right)^2 + \]
 \[ \frac{\ci}{2}\left(b_1 s_{2}-b_{3} s_{3}-s_{2} b_{3} c_{2}+s_{3} b_1 c_{2}\right) -\]
 \[ \frac{-b_1 b_{3} c_{2}-s_{2} s_{3} c_{2}}{2}
   + \frac{s_{2}^2+s_{3}^2-b_1^2+b_{2}^2-b_{3}^2-b_{2}^2 c_{2}^2}{4}. \]

\begin{Theorem} \label{thm:q}
Let $k$ be the number of bond connections of $1$ and $4$. Then
\[ k \le \deg(\gcd(Q_1^+,Q_4^+)) + \deg(\gcd(Q_1^-,Q_4^-)) . \]
\end{Theorem}

One can use this to derive a necessary condition for the existence of a bond connecting 1 and 4, since two polynomials
have a nontrivial $\gcd$ if and only if their resultant is zero. The equations get simpler if one assumes that some of the $\gcd$'s
have degree~2, because then the two quad polynomials need to be equal.

Similar conditions should be possible for 6-loops with P-joints, but the equations have not yet been derived, and so
its consequences are not yet known.

\section{Bond Diagrams}

The combinatorial structure of the bonds -- how many bonds are attached to which joints -- can be visualized
in a diagram. This diagram can be used to read off the degree of coupling curves.

Consider a linkage of mobility 1 with $e$ joints. Let $K$ be its configuration set and let
$\beta=(\beta_1,\dots,\beta_e)\in\overline{K}$ be a bond ($\overline{K}$ denotes the Zariski closure of $K$).
For any two links $i,j$, define the {\em coupling map} $f_{i,j}:\overline{K}\to\SE$ as the map that computes
the relative position of link $j$ with respect to the link $i$.
We define the {\em local distance} $d_\beta(i,j)$ as the order of the Taylor expansion
of the analytic function $N\circ f_{i,j}\circ \lambda$ divided by~2, where $\lambda$ is a local parametrization of $\overline{K}$
around~$\beta$.

\begin{Example} \label{ex:ldist1}
In the Bennett linkage of Example~\ref{ex:bennett} whose bonds are given in Example~\ref{ex:bondbenett}, we consider the bond
$\beta_1=(\ci,-\ci/3,-\ci,\ci/3)$. A local parametrization $\lambda$ of $\overline{K}$ is
$(t-\ci,(-t+\ci)/3,-t+\ci,(t-\ci)/3)$.
We label the links cyclically, the link attached to joints 1 and 2 is indexed by 2.
Then $f_{4,2}\circ \lambda$ is $(t-\ci-\qi)g_1((-t+\ci)/3\qi)g_2$, and its norm has Taylor expansion
\[ N((t-\ci-\qi)N(g_1)N((-t+\ci)/3-\qi)N(g_2)= -160/9\ci t+40/9 t^2 + {\cal O}(t^3), \]
which is of order 1. Hence $d_{\beta_1}(4,2)=1/2$.

For the other links, one gets $d_{\beta_1}(4,1)=d_{\beta_1}(1,4)=d_{\beta_1}(2,3)=d_{\beta_1}(3,2)=1/2$,
and all other values of $d_{\beta_1}$ are zero.
\end{Example}

\begin{Lemma} \label{lem:ldist1}
The local distance is a pseudo-metric on the set of links, i.e. for any three links $i,j,k$ we have
\[ d_\beta(i,i)=0, \ d_\beta(i,j) = d_\beta(j,i), \]
\[ d_\beta(i,k) \le d_\beta(i,j) + d_\beta(j,k) . \]
Moreover, the perimeter $d_\beta(i,j)+d_\beta(j,k)+d_\beta(k,i)$ of the triangle $(i,j,k)$ is even.
\end{Lemma}

\begin{Lemma} \label{lem:ldist2}
Assume that the link graph contains a chain from $i$ to $k$ passing $j$, for which the bond condition imposed by $\beta$
is not valid. Then
\[ d_\beta(i,k) = d_\beta(i,j) + d_\beta(j,k) . \]
\end{Lemma}

\begin{proof}
Both Lemmas are consequences of \cite[Theorem 3]{hss2}. The necessary adaptions to include prismatic joints are easy.
\end{proof}

The vast majority of bonds we studied so far have a very simple local distance: the link graph is partitioned into two
subsets; $d_\beta(i,j)=\frac{1}{2}$ if $i$ and $j$ lie in different subsets, and $d_\beta(i,j)=0$ otherwise
(see e.g. Example~\ref{ex:ldist1} above). We visualize linkages with these bonds by adding to the link diagram
additional lines connecting the edges, one for each conjugated pair of bonds,
that separate the vertices in the same way. Figure~\ref{fig:bbbg}
shows the bond diagrams of the Bennett 4R linkage and
of the Goldberg 5R linkage.

\begin{figure}
\begin{center}
\includegraphics{./Bennett}\quad\quad
\includegraphics{./Goldberg}
\end{center}
\caption{The bond diagrams of the Bennett 4R linkage and of the Goldberg 5R linkage.
	As explained in Exercise~\ref{ex:gbb}, one can ``read off'' the degree of its various
	coupler curves, be counting the number of lines that cross a line connecting to vertices corresponding
	to links.
	For instance, the coupling curves $C_{14}$ in the Goldberg linkage is a cubic.}
\label{fig:bbbg}
\end{figure}

For any two links $i,j$, the algebraic degree of the coupling curve $C_{i,j}$ is defined as the number of all points
$p\in\overline{K}$ such that $f_{i,j}(p)$ lies in a fixed generic hyperplane of $\P^7$. If $f_{i,j}:\overline{K}\to C_{i,j}$
is birational, then this is simply the degree of $C_{i,j}$ as a curve in $\P^7$. In general, it is equal to the degree
of $C_{i,j}$ multiplied with the mapping degree of $f_{i,j}$, i.e. the number of preimages of a generic point of $C_{i,j}$.

\begin{Theorem} \label{thm:adeg}
For any two links, the algebraic degree of $C_{i,j}$ is equal to the sum of all local distances $d_\beta(i,j)$ over
all bonds $\beta$.
\end{Theorem}

\begin{proof}
  For R joints, this is \cite[Theorem 4]{hss2}. The adaptions to make it work for P joints are easy to make. The idea of the proof is the following:
  instead of computing the algebraic degree
by intersection with a hyperplane, we can also intersect with a quadric not containing any coupler curve. We take
the null cone $Y$. Then we do the counting taking multiplicities into account, and finally divide by 2.
\end{proof}

In the bond diagram, the local distance of a bond visualized by a single line separating the link diagram into
two is $0$ for any two vertices in the same subset and it is $\frac{1}{2}$ for any two vertices in different subsets. Summing up over all bonds, and taking into account
that conjugate bonds separate the same links, we obtain that the algebraic degree of a coupling curve is equal to
the number of lines that must be crossed when one draws a line between the two vertices.

\begin{Example} \label{ex:gbb}
Figure~\ref{fig:bbbg} shows the bond diagram of Goldberg's 5R linkage. The coupling curves $C_{13}$, $C_{35}$,
and $C_{24}$ are conics (the two first actually appear in the Bennett linkage in the construction). The coupling curves
$C_{14}$ and $C_{25}$ are cubics. The remaining 5 coupling curves
are lines, parametrizing rotations around the joint axes. The coupling map $f_{15}$ is 2:1, all other coupling
maps are birational.
\end{Example}

\begin{figure}
\begin{center}
\includegraphics{./bDiet} \quad\quad
\includegraphics{./bWGG}
\end{center}
\caption{The bond diagrams of Dietmaier's 6R linkage and of Wohlhart's double Goldberg linkage. The degree of the coupler curve $C_{25}$ is 8
  in Dietmaier's linkage and 4 in Wohlhart's linkage
  (see also Example~\ref{ex:dietb} and Example~\ref{ex:wgg}).}
\label{fig:bdiet}
\end{figure}

\begin{Example} \label{ex:dietb}
The mobile 6R linkage found by Dietmaier~\cite{dietmaier} can be characterized by the following condition: the coupling
spaces of two disjoint chains of length 3 both have dimension~6, and intersect in a space of dimension~5. Then the coupling curve
with respect to the two links connected by the two chains
is contained in the projectivization of the intersection, which is a $\P^4$. It is defined by three quadratic
equations and therefore it has degree~8.

The bond diagram is shown in Figure~\ref{fig:bdiet}. One can see the algebraic degree of $C_{25}$. Also, the Bennett
conditions need to hold for two triples of axes because of the bond connections 1-5 and 2-4.
\end{Example}

\begin{Example} \label{ex:wgg}
In \cite{wgg}, Wohlhart constructed a movable 6R linkage by combing four Bennett linkages (see Example~\ref{ex:compb}
for the construction). Its bond diagram is resembling
the bond diagram of Dietmair's linkage. The dimensions of coupling spaces are the same, and the
number of bonds attached to any pair of joints is exactly half as in Example~\ref{ex:dietb}. Consequently, the degree
of each coupler curve is exactly half of the degree of the corresponding coupler curve in Dietmaier's linkage.
\end{Example}

\begin{figure}
\begin{center}
\includegraphics{./compboth1}
\includegraphics{./compboth2}
\includegraphics{./compboth3}
\includegraphics{./compboth4}
\includegraphics{./compboth5}
\includegraphics{./compboth6}
\end{center}
\caption{Bond diagrams of multiply closed linkages consisting of Bennett 4R linkages and 3R linkages with the same axis (``s.a.'')
	in all three joints, including bond diagrams. If one leaves away the links in the interior, then one obtains
	Waldron's double Bennett linkage, the Goldberg L-form 6R linkage, Goldberg's serial 6R linkage,
	Goldberg's second serial 6R linkage, the cube linkage, and Wohlhart's double Goldberg linkage.}
\label{fig:compb}
\end{figure}

\begin{Example} \label{ex:compb}
By combining Bennett linkages and 3R linkages such that all 3 axes coincide, one can construct various multiply closed
linkages. Leaving away the linkages with vertices drawn in the interior of the planar representation of the link graph,
we obtain the following 6R linkages: Waldron's double Bennett linkage~\cite{waldron}, the Goldberg L-form 6R linkage,
Goldberg's serial 6R linkage, Goldberg's second serial 6R linkage (all in \cite{goldberg,baker}), the cube linkage~\cite{hss1},
and Wohlhart's double Goldberg linkage~\cite{wgg}. Figure~\ref{fig:compb} shows the bond diagrams
of the multiply closed linkages.

A complete list of all linkages with three or four conjugate pairs of bonds can be found in \cite{li}.
\end{Example}

\section{Open questions for 6R linkages}

Bond theory has helped to discover new families of 6R linkages (see \cite{hss2,sharp,ls}), sometimes containing
known families (see \cite{hlss}). The possible list of bond diagrams is finite. For some of these diagrams,
we know all linkages, for other diagrams we do have examples but no proof of completeness, and for other possible
diagrams we do not if they appear as the diagrams of any linkage. In this section, we give a summary of
the open cases.

In this overview we exclude degenerate cases where one of the joints is frozen or where two consecutive
axes coincide. Also, we exclude cases where three consecutive axes are incident to a single
point; then the three joints can be replaced by a spherical joint, and the mobile SRRR linkages are
well understood. We also exclude cases with three consecutive axes being parallel, since they can be seen
as limit cases of three consecutive axes that are incident to a single point. By Lemma~\ref{lem:bcon2},
the dimension of the coupling spaces $L_{i,i+1,i+2}$ is either 6 or 8. We label the links and joints cyclically
modulo 6, the link attached to joint 1 and 2 has index 1.

We start by restricting the possible candidates of bond diagrams of such 6R linkages.
We may distinguish ``long bond connections'' connecting joints 1-4, 2-5, or 3-6, and ``short bond connections''
connecting $i$-$i+2$ for $i=1,\dots 6$ modulo $6$. By Lemma~\ref{lem:bcon1}, consecutive joints are not connected.

\begin{Theorem} \label{thm:bdia}
For any mobile 6R linkage, the following conditions on the bonds are known.
\begin{enumerate}
\item Any joint is attached to at most 4 conjugate pairs of bonds (counted with multiplicities).
\item If $\dim(L_{i,i+1,i+2})=\dim(L_{i+1,i+2,i+3})=6$, then there is no bond connecting joints $i$ and $i+3$.
\item If $\dim(L_{i,i+1,i+2})=\dim(L_{i+3,i+4,i+5})=6$, and the algebraic degree of $C_{i+2,i+5}$ is bigger
	than 4, then the linkage is a Dietmaier linkage (see Example~\ref{ex:dietb}).
\item If $\dim(L_{i,i+1,i+2})=6$ and $\dim(L_{i+3,i+4,i+5})=8$, and the algebraic degree of $C_{i+2,i+5}$ is bigger
	than 6, then it is 8.
\end{enumerate}
\end{Theorem}

\begin{proof}
(1): by Theorem~\ref{thm:adeg}, the number $k$ of pairs of bonds attached to joint $i$ is equal to the algebraic
degree of $C_{i-1,i}$. Since $C_{i-1,i}$ is a line parametrizing rotations around a fixed axis, $k$ is the
number of configurations with fixed joint parameter $t_i$. Since the maximal number of configurations of a
non-mobile 5R linkage with no three consecutive axes incident to a single point is 4, the claim follows.

For (2), we refer to \cite[Lemma~5.6]{li}. For (3) and (4), we refer to \cite{hlss}.
\end{proof}

When the dimensions of all coupling spaces $L_{i,i+1,i+2}$ is 8, then there are no short connections. Let us assume
that the number of bonds connecting joints 1 and 4 is $k_{14}$, the number of bonds connecting joints 2 and 5 is $k_{25}$,
and the number of bonds connecting joints 3 and 6 is $k_{36}$. Without loss of generality, we may assume that
$k_{14}\le k_{25}\le k_{36}$; also, we have $k_{15}\ge 1$ by Proposition~\ref{thm:frozen} and $k_{36}\le 4$ by Theorem~\ref{thm:bdia}.
Here is a summary about what is known for these cases.

\begin{itemize}
\item All linkages with $k_{14}=k_{25}=1$ are known: there is one family \cite{hss1} with $k_{36}=1$ and another family \cite{sharp}
	with $k_{36}=2$.
	Both families are maximal, i.e. they are irreducible components of the variety ${\cal M}_6$ of movable 6R loops.
\item All linkages with $k_{25}=k_{36}=4$ are known: there is one family with $k_{14}=2$ \cite{ls} and two families
	with $k_{14}=4$ \cite{hlss}. All three families are maximal.
\item The family of line symmetric linkages has $k_{14}=k_{25}=k_{36}=2$. It is maximal. Another family with the same bond
	diagram can be found in \cite{par}. But we do not know if this family is maximal, or if there
	are other families with $k_{14}=k_{25}=k_{36}=2$.
\item There are examples with linkages with $k_{14}=k_{25}=k_{36}=3$ with reducible configuration space (see below). But we do not
	know if there are other examples.
\item For any triple $(k_{14},k_{25},k_{36})$ not covered by the above cases, we do not know if there are any linkages.
\end{itemize}

\begin{Example} \label{ex:333}
Let $h_1,h_2,h_3$ be three dual quaternions corresponding to three random lines. Set
\[ h_4:=h_1,\ h_5:=h_2,\ h_6:=h_3 . \]
The configuration space of the closed linkage specified by the axes corresponding to $h_1,\dots,h_6$ is a reducible curve with
four components: three lines parametrizing rotations around a coincident axis, and one curve parametrizing the motion of a line
symmetric linkage. The first three lines intersect in a common point, corresponding to the initial configuration with three
coincident lines. The fourth component does not meet the other three.
\end{Example}

The question whether a known family of linkages is maximal is more difficult when short connections are present. In this case,
maximality is known only for just one family, namely Dietmaier's linkage \cite{dietmaier}. The situation is more complicated because
one has to exclude that the family under consideration is a specialization of another family yet unknown for which the dimension
of the coupling spaces is 8. In the case of Dietmaier's linkage, it is possible to prove maximality by
semicontinuity of the genus of the configuration curve (see \cite{hlss}).

Let us consider only bond diagrams with at least one short connection. The thesis \cite{li} contains
a complete list of linkages with three or four conjugated pair of bonds (i.e. the bond diagram contains three or four lines; some
of these diagrams are depicted in Figure~\ref{fig:compb}). For most diagrams compatible with Theorem~\ref{thm:bdia} and
Lemmas~\ref{lem:bcon1}, \ref{lem:bcon2}, we do not know if they are bond
diagrams of movable linkages, and for many others we have examples but we do not know if the known examples are all linkages
with the bond diagram under consideration. There are just two diagrams with more than four lines and at least one short connection
for which all linkages are known, namely the diagram of Wohlhart's partially symmetric linkage~\cite{partsym} and its third isomerization
shown in Figure~\ref{fig:bwbw}. Isomerization is a technique introduced in \cite{iso}
that allows to construct new families from known ones by interchanging two links; it is possible only if their affected joints
satisfy the Bennett condition.

\begin{figure}
\begin{center}
\includegraphics{./bWps}\quad\quad
\includegraphics{./b666}
\end{center}
\caption{The bond diagrams of the Wohlhart's partially symmetric linkage and its third isomerization.
	For a generic partially symmetric linkage, three consecutive triples of axes satisfy the Bennett condition and three do not,
	which one can be seen from the short connections in the diagram. Isomerization works only if all six triples satisfy
	the Bennett condition.}
\label{fig:bwbw}
\end{figure}

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


text:
Diary of a Wimpy Kid: Rewriting attitudes towards journaling and male mental health?
Nov 08, 2017 by Joel Sodzi	in Literature
"First off let me get something straight: this is a journal not a diary. I know what it says on the cover, but when Mom went out to buy this thing I specifically told her to get one that didn't say "diary" on it"
It would terrify Greg Heffley if he knew that his first "private" diary had been read by over 150 million young adults, according to the Independent in 2015. Though perhaps he might find some comfort in realising that his story – the Diary of a Wimpy Kid – has changed how a generation views an activity which, though trivial, has great benefits for mental health.
Epistolary writing is nothing new to the literary scene, being a form of storytelling through a series of documents, it has evolved over a long period of time. It is common in many of the world's most famous novels from Bram Stoker's Dracula to Stephen King's Carrie. Both of these use diaries as a way of telling the story. Over time, diaries have become more prominent in fiction, proving to be a different and interesting way of presenting a narrative and so these kinds of epistolary novels have become very popular, as was seen by one of the earliest of it's kind, "Diary of a Nobody" written in 1892 by the Grossmith brothers.
Weedon (left) and George (right) Grossmith, pictured in 1897
Despite this early novel having a male protagonist and it being written by two men, the diary, over the course of the 20th century, became more synonymous with femininity. Likely a knock-on effectt from the success of The Diary of a Young Girl by Anne Frank and, later, the publication of Virginia Woolf's diaries, the appeal of journaling seemed to dissipate in the male view. This had a knock-on effect on the fictional journaling genre also, as a simple Google search for books about fictional diaries reveals. The picks Google turns up first are Diary, Bridget Jones's Diary and Go Ask Alice which are all good books, but linked In having female protagonists, with suggestions containing men being far less common.
The view that diaries are purely a girly activity might be changed with the recent emergence of more modern epistolary novels, featuring fictional male diary writers. One of the most famous is Diary of a Wimpy Kid by Jeff Kinney. While there is no evidence suggesting a relationship between the success of the series and a rise in diary writing in young males, the outstanding success of the books could pave the road for greater appreciation of the pastime in males. Greg Heffley has the potential to be a titan in the fight to improve men's mental health. His stories, so full of humour and childish innocence, are appealing to the young male audience, and perhaps they might be influencing more to take up the pen and record their emotions.
The basis of this theory stems from what we see in the books themselves. Greg is undoubtedly flawed; when picturing his character, the words selfish, jealous and conceited come to mind, however, he encapsulates well many of emotions that modern tweenage and teenage boys deal with daily. While each book continuously puts him in more humorous and ludicrous scenarios, they are all grounded in some relatability. Fundamentally, if boys reading these books see themselves in Greg, then they might not be so reluctant to write down their experiences and recognise the importance of allowing their own emotions to flourish. In fact, there is evidence that journaling can be linked to positive effects – the prominent mental health website Psych Central wrote in their article The Health Benefits of Journaling:
"Writing removes mental blocks and allows you to use all your brain power to better understand yourself, others and the world around you."
This idea is supported by the research of James Pennebaker, a psychologist and researcher at the University of Texas, whom they mention in their article. He believes that regular journaling can strengthen immune cells called T-lymphocytes as well as generally reducing stress, as writing about them can help come to terms with their problems, and so reduce the impact of stress on the individual. Pennebaker is not isolated in his views; other researchers Karen Baikie and Kay Wilhelm concluded that physical and psychological health can be vastly improved by writing about stressful, traumatic or emotional events for as little as 15 minutes on three separate occasions. There is then, no doubt that journaling has some positive effects, and yet it seems that the activity is still almost exclusively enjoyed by women when it could be highly beneficial to male mental health.
With the number of recorded suicide rates in 2016 exceeding 5000, 75% of which were men, mental health is an issue which we must be wary of,
especially in men who have been encouraged for too long to be steadfast in all situations. Journaling is a good place to start for tackling mental health issues, and Jeff Kinney's book has not only entertained many readers worldwide but also inspired many more books with protagonists like Greg. The Private Blog of Joe Cowley by Ben Davis and Middle School: The Worst Years of My Life by James Patterson are some prominent ones in a similar style. They have all branched off into series of their own, and have paved the way for a more unisex outlook on journaling to return after it has been dominated by female writers for so long.
Gary Dunaier, CC BY-SA 2.0
Header image: Gary Dunaier, CC BY-SA 2.0
Tagged with: Diary, Diary of a Wimpy Kid, Jeff Kinney, Literature, Mental health
About the author Joel Sodzi
Writer at Berkeley Squares for articles focused on literature and language. Giving a new spin on an old medium. Oh, and let's throw some classics in there too
Culture Mental health: a conversation about services
Music Ghost: Welcome to the Spectacle
Business & Economics The storm forming around China's tech sector


Meghan Markle and Prince Harry's hopes for working outside of the palace's watchful eye has reportedly been royally squashed.
It was previously announced that the duke and duchess of Sussex would be getting their own court apart from the one that also represents Prince William and Kate Middleton, which is based in Kensington Palace, their shared residence. And now, Britain's The Sunday Times reported that Meghan and Harry had suggested that this new court be independent from the royal family.
But grandmom Queen Elizabeth and dad Prince Charles apparently nixed the idea.
The couple reportedly had plans to create a worldwide "Sussex brand" of philanthropy and humanitarianism. And Meghan also wanted to continue growing her role as an activist.
Instead, Meghan and Harry's official team of new aides and staff members will be based inside Buckingham Palace, the official residence of the Queen.
Meghan and Harry have been separating themselves for some time after tying the knot. It was announced in November that they would move out of the Kensington Palace complex to Frogmore Cottage at Windsor Castle in the spring, when Meghan is due to give birth.
Meghan and Harry will be getting their new court this spring to reportedly "reflect their increasingly different responsibilities."


text:
dir=${0%/*}
if [ -d "$dir" ]; then
  cd "$dir"
fi

# set up your app name, architecture, and background image file name
APP_PACKAGE=$1
APP_VERSION=$2
APP_NAME=$3
APP_FULLNAME=$4

rm entitlements-child.plist
ln -s ../resources/bitcoin.com/mac/pkg/entitlements-child.plist entitlements-child.plist

rm entitlements-parent.plist
ln -s ../resources/bitcoin.com/mac/pkg/entitlements-parent.plist entitlements-parent.plist

rm build.cfg
ln -s ../resources/bitcoin.com/mac/pkg/build.cfg build.cfg

rm build_mas.py
ln -s ../resources/bitcoin.com/mac/pkg/build_mas.py build_mas.py

echo "Signing ${APP_FULLNAME}"
export CURRENT_PATH=`pwd`
export APP_PATH="pkg/${APP_FULLNAME}/osx64/${APP_FULLNAME}"
export TMP_PATH="tmp"
export DIST_PATH="dist"

rm -rf $TMP_PATH
mkdir $TMP_PATH

if [ ! -d $DIST_PATH ]; then
  mkdir $DIST_PATH
fi

cd "${APP_PATH}.app/Contents/Versions"
ln -s "55.0.2883.87" "Current"

cd $CURRENT_PATH
chmod -vR 777 "${APP_PATH}.app/Contents"

python build_mas.py -C build.cfg -O "${TMP_PATH}/${APP_FULLNAME}.app" -I "${APP_PATH}.app" -P "$DIST_PATH/${APP_PACKAGE}-wallet-${APP_VERSION}-osx.pkg"

echo "Signing Done"

echo "Done."

exit

redpajama_set_name: RedPajamaGithub


text:
EHUD BARAK MAY TOPPLE ISRAELI PRIME MINISTER BIBI NETANYAHU
Categories Hamas, Historic Figures, Internal Politics, Palestinian Issues | Author David EssingPosted on | June 26, 2019 | 2:32 pm
Former Israeli PM Ehud Barak
It's still early, but Bibi Netanyahu's days as Prime Minister may soon be over. Netanyahu's nemesis, Ehud Barak, has just announced that he is entering Israel's current election campaign intending to defeat Netanyahu, who has dominated Israeli politics as long as most people can remember. At a Tel Aviv news conference.
Barak declared, "The time has come to topple Netanyahu, not to save him from criminal prosecution"! This Israeli Prime Minister is now facing three possible indictments for bribery and breach of trust. And Barak went on to say that this was Bibi's last moment to leave office and retire. "Do not try and drag the state through chaos in order to save yourself."
It has long been rumored that the 77-year-old Barak was contemplating a comeback in light of the surprise election on September 17th. Several decades ago, Barak served as Bibi's commander in the vaunted Sayeret Matkal special forces. I believe it's fair to say that during both of their political careers, Barak has always held a psychological whip-hand over Bibi, even when Bibi was his political boss as Prime Minister and Barak served under him as Defense Minister. Barak's dramatic entrance into the election race comes at a critical crossroads. Israeli politics is in a state of chaos.
In an apparent attempt to forestall the legal charges against him, Netanyahu called an early election in April. But he outsmarted himself. He believed that he would win the recent election hands down, form a new government, and then declare that the people had reelected him knowing full and well about the criminal allegations. However, the ballot ended in a dead heat, and one of Bibi's former political chums surprised the country by refusing to join Bibi's right-wing coalition because Lieberman charged Netanyahu with selling out to the ultra-orthodox parties. But, never at a loss for political tricks, Bibi forced an unprecedented new election rather than let Benny Gantz of the Blue & White party try his luck at forming a new government.
"We are not blind, and understand that you are simply trying to derail the indictments against you. Netanyahu, you have reached the end of the road."
But that's not all. After recently seeing polls indicating that he would not win the next election, the Prime Minister has been trying to overrule the Knesset decision to hold another election in September. So much for Bibi's shenanigans. Barak blasted the Prime Minister by declaring, "We are not blind, and understand that you are simply trying to derail the indictments against you. Netanyahu, you have reached the end of the road." Barak went on to say that the PM's closest advisors are aware of this, as are his palls at the Knesset and the Cabinet.
However, Netanyahu rules Likud with an iron fist, and his colleagues are both paralyzed and fearful. Then Barak referred to their army days together, some 50 years ago. "As your former commander, I tell you that you must not continue to try and steer the ship of state. For your own good, for the good of the country, and for all you have contributed throughout your life, your persistence in trying to hold on may wind up badly." For his part, Barak promised to get Israel back on track by forming a new political framework with the goal of ending Netanyahu's regime.
Barak, Clinton, and Arafat during peace negotiations. Photo credit: PD-USGOV
Of all the leaders in Israel, Ehud Barak is the most qualified to take on Netanyahu. But, Barak's own image with many Israeli voters took a nosedive after he tried to reach a wide-ranging peace agreement with the former Palestinian leader Yasser Arafat at Camp David 2000, hosted by then-President Bill Clinton. In his autobiography, "My country, my life," Barak detailed how he pulled out every stop to finalize a peace accord with Arafat.
Low and behold, he offered far-reaching compromises that highlighted with a Palestinian state in the West Bank and Gaza. This was the two-state solution some 19 years ago. The borders of this Palestinian state would involve Palestinian territorial compromises to maintain Israel's security. But in return, Israel would compensate with territory in less critical security areas. And get this, Barak, in effect, agreed to the partition of Jerusalem, with the city also serving as the Palestinian capital.
Moreover, a limited number of Palestinian refugees would be allowed to return to the Jewish state. But, Arafat (as testified by Clinton) refused even to reply to Barak's offer. The Palestinian leader simply packed his bags and flew back to the West Bank to launch a second Intifada, which has, in effect, continued to this very day.
"Judaism is a religion, and practitioners of religions are not entitled to a state."
What was the hitch? Arafat's aids told US negotiator, Dennis Ross, that Barak's proposal would have granted Palestinian recognition of Israel as the Jewish state. And if the Palestinian leader would have done so, "Palestinians would have killed him when he returned home!" This stalemate has existed until this very day – even the moderate president Mahmud Abbas categorically denies the right of Jewish self-determination and their right to an independent state. Doctor Abbas, and other Palestinian academics such as doctor Saeed Erekat, contend that "Judaism is a religion, and practitioners of religions are not entitled to a state." This should be born in mind by both the J Street crowd and the BDS anti-Semites.
In any case, Arafat's refusal to accept the PM's generous proposal put paid to Barak's political career. He had offered everything, including the kitchen sink, but came home empty-handed to be hauled over the coals by Israel's right-wing that accused him of selling out the country. And consider this – Barak's own Labor party blamed him for astounding Arafat by offering too much at one go. Since then, the Labor party, that founded and developed the modern state of Israel, has never recovered.
Barak's own Labor party blamed him for astounding Arafat by offering too much at one go. Since then, the Labor party, that founded and developed the modern state of Israel, has never recovered.
So what impact is Barak's comeback likely to have on Israel's fragmented political seam? Barak, who knows Bibi inside-out, is the only political rival who can take on Netanyahu, who wows the right-wing with his rhetoric.
And this just in: A snap poll after Barak's announcement indicates that he is now a viable player in the election campaign. Barak's party could take six seats in the September ballot. Moreover, the poll shows that the left-wing bloc could win 61 seats, the required majority in the 120-member Knesset if elections were held today. And Barak is just warming up. Overall, Netanyahu's Likud party would slip to 32, the same as the Blue & White party led by Benny Gantz. Incidentally, Barak's old Labor party would take a bare five seats – just over the four required to enter the Knesset. This would leave open the possibility for Barak teaming up with his old Labor palls. More bad news for Bibi – Avigdor Lieberman of the "Russian Immigrants' party, has climbed to seven seats. Remember, it was Lieberman who barred Bibi from forming a new coalition government – the country is still reeling politically, and more surprises can be expected.
Author David Essing
Senior Correspondent & News Analyst
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I don't often jump into the world of politics with my blog. The times I have are when crime was mixed with politics. I am making an exception here though. I'm making this exception because for the first time in recent memory I see a person who is perfect for a job. I just noticed that Ray Goldenchild is running for Park Board Commissioner in Vancouver. This man has been involved at almost every level of sports for virtually his entire life. And much of that involvement has been with kids in sports. I can assure you that Ray Goldenchild puts families, kids and sports first.
If you want to find Ray Goldenchild on the weekend, look for kids and a soccer ball. Ray will be there pacing the sidelines and encouraging our children. He'll be there with his ever present upbeat attitude and his smile. Ray wants these kids to win, but winning to Ray Goldenchild means that the kids had a good time. It means that they better their skills, and most importantly, that they take something from the field that is rare these days, honour. You see, Ray Goldenchild keeps score, not by the number of goals, but by the number of lives changed.
If Vancouver's parks are not about families, kids and sports, what are they about? If they are not about laughter and challenge, what are they about? These things don't just occur though because there happens to be an open field and a bathroom.
No, it takes more than that. It takes organization. It takes dedication, and it takes understanding.
You see, when some people think "Park Board Commissioner", they think a political position from where to start their climb. But when Ray sees "Park Board Commissioner", Ray sees an opportunity to have a louder voice for families, for kids and for sports. Ray Goldenchild sees an opportunity to make even a bigger difference in the quality of lives of people.
True beauty comes from the seamless manner in which the parks integrate with the city.
It comes from proper use of the parks.
It comes from the skill to manage this valuable resource.
Everybody who wants use of a park should find time and space available.
Ray Goldenchild will accomplish all of this.
Ray will manage your parks with same level of skill and caring that have been trademarks of his success. With Ray, he's "all in" all the time. The commitment is there whether he is mentoring athletes who go onto to win college scholarships or whether working with the African Canadian Soccer and Cultural Association. Typical of his work ethic Ray won their annual award for "Outstanding Contribution to the Community". That is what you get when you get Ray. You get OUTSTANDING.


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"Ours" - Taylor Swift (Cover by Jameson Bass, Brad Kirsch and Tristin Hagen)
Get this track for FREE! Go to http://www.facebook.com/songswithfriends and click on "Band Profile" to download the song :)
Hey everyone! Here is our new cover of Taylor Swift's new single "Ours"! I heard this track and knew that I had to bring in Tristin and Brad to make it a duet. Hope everyone enjoys our Taylor Swift cover!
Vocals performed by Tristin Hagen and Brad Kirsch
All instruments performed and recorded by Jameson Bass
Filmed by Tristin Hagen and Jameson Bass
Edited by Jameson Bass
If you like it, hit the Like button, Favorite, Comment, or even share on your Facebook or twitter profiles if you feel so inclined! Thank you so much :)
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Entertainment/recreation, Creative pastimes in Saint-Laurent-du-Pont
From January 12, 2023
until June 29, 2023
Centre Social des Pays du Guiers, 1 rue Charles Hérold
38380 Saint-Laurent-du-Pont
Updated on December 26, 2022 at 11:21 AM
by Office de Tourisme Coeur de Chartreuse


text:
namespace laconomie {
    namespace tests {
        class EventPosting : public testing::Test {
            const MediaRuntime runtime;
        };

        TEST_F(EventPosting, NoOpPostsNoEvents)
        {
            post_appevent(AppEvent::NoOp);

            SDL_Event expectedEvent;
            ASSERT_EQ(0, SDL_PollEvent(&expectedEvent));
        }

        TEST_F(EventPosting, QuitPostsSDLQuitEvent)
        {
            post_appevent(AppEvent::Quit);

            SDL_Event expectedEvent{};
            ASSERT_EQ(1, SDL_PollEvent(&expectedEvent));
            ASSERT_EQ(SDL_QUIT, expectedEvent.type);
            ASSERT_LT(0u, expectedEvent.quit.timestamp);
        }
    }
}

void laconomie::tests::details::register_eventposting_tests() noexcept {}