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Giuditta Pasta

Giuditta Angiola Maria Costanza Pasta (26 October 1797 – 1 April 1865) was an Italian opera singer. A soprano, she has been compared to the 20th-century soprano Maria Callas.

Early career
Pasta was born Giuditta Angiola Maria Costanza Negri in Saronno, near Milan, on 26 October 1797. She was born of the Negri family, who came from Lomazzo, where the family practiced medical art. Her father, Carlo Antonio Negri or Schwarz, was Jewish and a soldier in the Napoleonic Army. She studied in Milan with Giuseppe Scappa and Davide Banderali, and later with Girolamo Crescentini and Ferdinando Paer among others. In 1816, she married fellow singer Giuseppe Pasta and took his surname as her own. She made her professional opera début in the world première of Scappa's Le tre Eleonore in Milan that same year. Later that year she performed at the Théâtre Italien in Paris as Donna Elvira in Don Giovanni, Giulietta in Niccolò Antonio Zingarelli's Giulietta e Romeo, and in two operas by Paer.

Pasta's first appearance in London in 1817 was a failure. Further studies with Scappa were followed by a successful debut in Venice in 1819. She caused a sensation in Paris in 1821–22, in the role of Desdemona in Gioachino Rossini's opera Otello.

Roles written specifically for Pasta
She sang regularly in London, Paris, Milan and Naples between 1824 and 1837. In Milan she created three roles which were written for her voice. They were the title role of Donizetti's Anna Bolena given at the Teatro Carcano in 1830 (and which was that composer's greatest success to date), the Amina in Bellini's La sonnambula and the protagonist's part of his Norma (both in 1831), which became three of her major successes. Stendhal had argued persuasively in 1824 for the necessity of a score composed expressly for Pasta.

Later career
Pasta retired from the stage in 1835 and performed only infrequently after that date (including performances in London in 1837 and in Germany and Russia in 1840–1841.)

Pasta later taught singing in Italy. Among her notable pupils were contralto Emma Albertazzi and soprano Marianna Barbieri-Nini and the English soprano Adelaide Kemble. Another pupil was Carolina Ferni, herself a noted Norma, who in her turn taught the soprano Eugenia Burzio whose recordings are known for their passionate expression.

Pasta died in Blevio, a town in the province of Como on 1 April 1865, at the age of 67.

Pasta's voice

Giuditta Pasta's voice was described by a New Monthly Magazine reviewer in 1824 as follows:

 * It is a mezzo-soprano, somewhat similar to that of Madame Vestris, but clearer, more powerful, and of greater compass. She commands two octaves, but two or three of the highest notes of this range are forced, and not agreeable. Her middle tones are fine and full-bodied; but, occasionally, notes escape in the lowest half octave, which are husky and harsh. In point of cultivation and science, she possesses, first of all, the rare merit of a pure intonation. We have not heard her once out of tune.

Her voice type was what could be called a soprano sfogato. It was described by Stendhal as follows:
 * She can achieve perfect resonance on a note as low as bottom A, and can rise as high as C♯, or even to a slightly sharpened D; and she possesses the rare ability to be able to sing contralto as easily as she can sing soprano. I would suggest ... that the true designation of her voice is mezzo-soprano, and any composer who writes for her should use the mezzo-soprano range for the thematic material of his music, while still exploiting, as it were incidentally and from time to time, notes which lie within the more peripheral areas of this remarkably rich voice. Many notes of this last category are not only extremely fine in themselves, but have the ability to produce a kind of resonant and magnetic vibration, which, through some still unexplained combination of physical phenomena, exercises an instantaneous and hypnotic effect upon the soul of the spectator.

 * This leads to the consideration of one of the most uncommon features of Madame Pasta's voice: it is not all moulded from the same metallo, as it is said in Italy (which is to say that it possesses more than one timbre); and this fundamental variety of tone produced by a single voice affords one of the richest veins of musical expression which the artistry of a great cantatrice is able to exploit.

In 1829 named cantante delle passioni by Carlo Ritorni, one of the most erudite critics of the period, he described her as such because her voice was directed "towards expressing the most intense passions, accompanying it with expressions of physical action, unknown before her in the lyric theatre".

In modern times Susan Rutherford has made a specific comparison with Callas:
 * For the impact of corporeality on vocal timbre and delivery, and in the absence of Pasta's own explanations of its effect, we might turn to another distinctive attrice cantante (and one who sang much of Pasta's repertory) from a quite different period, Maria Callas. She also argued that gesture and facial expression must precede word in order to create the appropriate vehicle.

 * It isn't merely fame that makes Pasta interesting:... Pasta's singularity is measured rather by the tone and extent of the debates her celebrity provoked, by her influence on the operatic stage, and by the timing of her career at the transition from Rossinian opera to the works of Bellini and Donizetti (with all the stylistic ramifications this implied). No other singer during that period attracted as much intellectual discussion, or was regarded as of such significance in the articulation of theories around operatic practices. For such reasons alone, Pasta is deserving of critical attention.

title: Giuditta Pasta


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Talk:Sergius Paulus

Untitled
If my oral family history is correct, Sergius was exiled from Rome to Germania for his faith. This might explain why Paul does not salute him in the Letter to the Romans. This is only an oral history that has been passed down for quite a few generations. I do not swear that it actually occurred, but evidence of the Church being in the region of Germany to which he was supposedly exiled prior to AD 100 does tend to support it though.
 * Oral history handed down almost 2 millennia; that is impressive. -- llywrch (talk) 07:54, 7 September 2020 (UTC)

So, is it Paulus or Paullus?
It seems like many sources give the name as Paullus, a spelling that also appears in the painting by Raphael and in the article itself. Aviad2001 (talk) 09:26, 19 March 2012 (UTC)

Lucius or Quintus?
Mitford's article in Annual of the British School at Athens has an interesting passage: "The problem is further complicated by an inscription from Kythrea of the Julio-Claudian period, which raises the possibility that a Sergius Paulus, with however the praenomen Quintus, was proconsul of Cyprus." The inscription is IGR iii.935. Acts 13:6-12 does not provide a praenomen for the Sergius Paullus who was interested in Paul's preaching; there is no good reason for calling him Lucius, except that's what the older scholars told us it was.Maybe the older scholars got his name wrong all along. -- llywrch (talk) 08:11, 7 September 2020 (UTC)
 * I stuck a reference to that information in a footnote. I'd be happy for you to improve it. John Foxe (talk) 21:37, 7 September 2020 (UTC)
 * Actually, I was thinking this might require a re-write of the whole article. The author of Acts implies that Paul converted the proconsul Sergius Paulus (I say "imply" because had a Roman Senator embraced Christianity, it would have been a notorious scandal that might have destroyed his family -- who yet appear afterwards as consuls), which would be one article. Another article, perhaps, to discuss the family of the Sergii Pauli, who thrived for at least another century; it is possible that the famous Anicii of the 4th century are their distant descendants. I'm just not confident I know enough to untangle what is in this article & reweave the information I have found in the relevant secondary literature about the Sergii Pauli. -- llywrch (talk) 16:48, 8 September 2020 (UTC)
 * I can imagine scenarios in which Paul's Sergius Paulus was converted but without subsequent scandal—as for instance, if he had died shortly thereafter. If the man had had the faintest of reputation thereafter, I'd expect some early Christian stories about him. I mean, if Longinus can be a saint then why not Sergius Paulus? Also, a strange coincidence I never noticed before is that Acts 13 begins by calling the protagonist "Saul," he's called "Paul" for the first time while in the presence of Sergius Paulus, and he remains "Paul" ever after.John Foxe (talk) 21:11, 8 September 2020 (UTC)

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

title: Talk:Sergius Paulus


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User talk:Thizrark

Please stop vandalizing articles. -- Decumanus 01:25, 17 Mar 2004 (UTC)

Please stop or you will be blocked from making further contributions. -- Decumanus 01:28, 17 Mar 2004 (UTC)

title: User talk:Thizrark


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Vanuatu rain forests

The Vanuatu rain forests are tropical and subtropical moist broadleaf forests ecoregion which includes the islands of Vanuatu, as well as the Santa Cruz Islands group of the neighboring Solomon Islands. It is part of the Australasian realm, which includes neighboring New Caledonia and the Solomon Islands, as well as Australia, New Guinea, and New Zealand.

Geography
The islands were created by the subduction of the northward-moving Australian Plate beneath the Pacific Plate. The surface geology of Vanuatu consists mostly of Pliocene-Pleistocene volcanic rocks and uplifted coral limestone. The Santa Cruz Islands have areas of both uplifted limestone and volcanic ash over limestone. The oldest rocks in Vanuatu are 38 million years old. The Santa Cruz islands are younger, with the oldest rocks less than 5 million years old.

Most of the islands are low-lying. The largest island is Espiritu Santo (3,955.5 km2). The highest peak is Mount Tabwemasana on Espiritu Santo (1,879 m).

Nendö, also known as Santa Cruz, is the largest of the Santa Cruz Islands with an area of 519 km2. Vanikoro is 190 km2, and Utupua is 69 km2. The highest peak in the Santa Cruz islands (924 m) is on Vanikoro. Nendö reaches over 500 meters elevation, and Utupua 350 m. They are made mostly of basaltic volcanic rocks of Pliocene origin, less than 5 million years old. The southeastern lowlands of Nendö are composed of uplifted Pleistocene reef limestone.

Smaller islands in the group include Tinakula (8 km2 and 800 meters elevation), a conical active stratovolcano 30 km north of Nendö, and the Reef Islands northeast of Nendö, composed of uplifted Pleistocene reef limestone and rising only 5 meters above sea level. The Duff Islands are a small chain with four main islands about 130 km northeast of Nendö, with a combined area of 14 km2 and which rise up to 300 m elevation. Tikopia (5 km2), Anuta (1 km2), and Fataka (5 km2) are small isolated islands east and southeast of Vanikoro.

Climate
The ecoregion has a tropical wet climate. The windward southeastern sides of the islands receive more rainfall. The leeward northwestern slopes of islands have a distinct dry season between April and October. Tropical cyclones occur regularly.

Flora
The natural plant communities on the islands include lowland rain forest, montane rain forest, seasonal forest and scrub, coastal strand, mangroves, vegetation on recent volcanic rocks, and secondary vegetation.

Lowland rain forest occurs on the southeastern, or windward, sides of Vanuatu's islands. There are several lowland rain forest types. Complex forest scrub densely covered with lianas is the most widespread forest type on the larger northern islands. Other types include high- and medium-stature forests, alluvial and floodplain forests, and mixed-species forests without conifers. Agathis-Calophyllum lowland forest is found on the southern islands of Erromango and Aneityum.

Lowland rainforest is the predominant plant community on the Santa Cruz Islands, and has some differences from the lowland rain forests on the islands further south. Typical species include Campnosperma brevipetiolatum, Calophyllum vitiense, Gmelina salomonensis, Maranthes corymbosa, Falcataria falcata, Pterocarpus indicus, and Endospermum medullosum. There is no well-developed montane forest, but the trees Metrosideros vitiensis, Syzygium branderhorstii, Syzygium buettnerianum, Syzygium effusum, Syzygium myriadenum, and Syzygium onesimum, and the conifers Agathis macrophylla and Retrophyllum vitiense, which are typically montane species elsewhere, grow in the islands' lowland forests. Agathis macrophylla grows only on Nendö and Vanikoro, and is found on ridges and slopes as an emergent tree in mixed forests or in monotypic stands.

In Vanuatu, montane rain forests extent from as low as 500 meters elevation up to patches of stunted cloud forest on the islands' highest peaks. They include the conifers Agathis macrophylla and Dacrycarpus imbricatus, together with broadleaf evergreen trees Metrosideros vitiensis, Syzygium spp., Pterophylla spp., Geissois spp., Quintinia spp., and Ascarina spp. The tree ferns Cyathea and Dickinsonia are common, and the endemic palm Clinostigma harlandii is found on the islands Ambrym, Aneityum, and Erromango. Podocarpus vanuatuensis is a Vanuatu endemic, native to Aneityum and Erromango. Agathis silbae is endemic to the Cumberland Peninsula and Mount Tabwemasana, also known as Santo Peak, on the west coast of Espiritu Santo, from 450 to 760 meters elevation. It is a large emergent tree in lower montain rainforest on the wetter western and northwestern slopes of Espiritu Santo's central mountain range, with an average annual rainfall of about 4,500 mm. Associated trees include Calophyllum neo-ebudicum, Cryptocarya turbinata, Didymocheton sp., Myristica sp., and Podocarpus sp. Metrosideros tabwemasanaensis is a tree endemic to the montane forests of Mount Tabwemasana.

Seasonal forest, scrub, and grassland grow on the leeward sides of the islands. Semideciduous Kleinhovia hospita-Castanospermum australe forests are a transition between rain forest and dry forest, and include some rain forest species. Forest of gaiac (Acacia spirorbis) is found in drier areas, with a canopy up to 15 meters high. Thickets and savannas of the introduced tree Leucaena leucocephala and grasslands are also found on the leeward sides of the islands.

Littoral forests include Casuarina equisetifolia, Pandanus spp., Barringtonia asiatica, Terminalia catappa, Hernandia spp., and Thespesia populnea.

Coastal mangrove forests are found on some islands, and contain species of Rhizophora, Avicennia, Sonneratia, Xylocarpus, and Ceriops.

Fauna
Bats are the only native mammals in the ecoregion. There are twelve species – four megabats and eight microbats – five of which are endemic. The four megabats – Vanuatu flying fox (Pteropus anetianus), Temotu flying fox (Pteropus nitendiensis), Vanikoro flying fox (Pteropus tuberculatus), and Banks flying fox (Pteropus fundatus) – are endemic. Native microchiroptera include the Fijian blossom bat (Notopteris macdonaldi), Fijian mastiff bat (Chaerephon bregullae), Pacific sheath-tailed bat (Emballonura semicaudata), large-footed bat (Myotis adversus), little bent-wing bat (Miniopterus australis), great bent-winged bat (Miniopterus tristis), Temminck's trident bat (Aselliscus tricuspidatus), and fawn leaf-nosed bat (Hipposideros cervinus). The endemic Nendo tube-nosed fruit bat (Nyctimene sanctacrucis) is presumed extinct.

The Pacific boa (Candoia bibroni), also known as Bibron’s bevel-headed boa, the Solomon Islands boa, or the Pacific ground boa (among several other names), is native to the island and surrounding region. It is unique among Boidae snakes for its “bevel” or “spade”-shaped snout, used for digging; perhaps the closest comparable species would be the Kenyan sand boa, which spends much of its time burrowing, where it will lie in wait to ambush its passing prey above.

There are 79 native bird species in Vanuatu. Fifteen species are endemic – Vanuatu scrubfowl (Megapodius layardi), Santa Cruz ground-dove (Gallicolumba sanctaecrucis), Tanna ground-dove (Gallicolumba ferruginea), Tanna fruit-dove (Ptilinopus tannensis), Baker's imperial pigeon (Ducula bakeri), palm lorikeet (Charmosyna palmarum), chestnut-bellied kingfisher (Todirhamphus farquhari), Vanikoro monarch (Mayrornis schistaceus), buff-bellied monarch (Neolalage banksiana), black-throated shrikebill (Clytorhynchus nigrogularis), Vanikoro flycatcher (Myiagra vanikorensis), Santa Cruz white-eye (Zosterops santaecrucis), yellow-fronted white-eye (Zosterops flavifrons), Sanford's white-eye (Woodfordia lacertosa), New Hebrides honeyeater (Phylidonyris notabilis), royal parrotfinch (Erythrura regia), Polynesian starling (Aplonis tabuensis), rusty-winged starling (Aplonis zelandica), and Mountain starling (Aplonis santovestris). The ecoregion corresponds to the Vanuatu and Temotu endemic bird area.

Conservation
4.3%, or approximately 515 km2, of the ecoregion is in protected areas. Protected areas include:
 * Ambrym Megapode Reserve
 * Central Efate (Teouma) Forest Conservation Area
 * Erromango Kauri Forest Conservation Area
 * Lasenuwi Forest Conservation Area
 * Loru Protected Area
 * Pankumo Protected Area
 * Vatthe Forest Conservation Area
 * Western Peninsular Forest Conservation Area
 * Wiawi Conservation Area

title: Vanuatu rain forests


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Ray-Ban

Ray-Ban is a brand of luxury sunglasses and eyeglasses created in 1936 by Bausch & Lomb. The brand is best known for its Wayfarer and Aviator lines of sunglasses. In 1999, Bausch & Lomb sold the brand to Italian eyewear conglomerate Luxottica Group for a reported $640 million.

History
In 1929, US Army Air Corps Colonel John A. Macready worked with Bausch & Lomb, a Rochester, New York–based medical equipment manufacturer, to create aviation sunglasses that would reduce the distraction for pilots caused by the intense blue and white hues of the sky.

Specifically, MacCready was concerned that pilots' goggles would fog up, greatly reducing visibility at high altitude. The prototype, created in 1936 and known as "Anti-Glare", had plastic frames and green lenses that could cut out the glare without obscuring vision. The name "Ray-Ban" was hence derived from the ability of these glasses to limit the ingress of either ultra-violet or infra-red rays of light. Impact-resistant lenses were added in 1938. The sunglasses were redesigned with a metal frame the following year and patented as the Ray-Ban Aviator. According to the BBC, the glasses used "Kalichrome lenses designed to sharpen details and minimise haze by filtering out blue light, making them ideal for misty conditions."

Product lines
Ray-Ban's most popular sunglasses are the Wayfarer, Erika, and Aviator models. During the 1950s, Ray-Ban released the Echelon (Caravan), which had a squarer frame. In 1965, the Olympian I and II were introduced; they became popular when Peter Fonda wore them in the 1969 film Easy Rider. The company has also produced special edition lines, such as The General in 1987, bearing similarity to the original aviators worn by General Douglas MacArthur during the Second World War. In the 1980s the Ray-Ban Clubmaster was added to the model line. The Clubmaster has a browline frame and went on to become the third best selling sunglasses style of the 1980s, behind the Wayfarer and Aviator.

In 2007, Luxottica Group launched Ray-Ban Youth, a collection of prescription eyewear aimed at children ages eight through twelve. Modeled after popular adult Ray-Bans styles, these hypoallergenic titanium frames featured both neutral and bold colorways as well as sturdy flex hinges.

In 2009, Luxottica released the Ray-Ban Tech collection, which included the Carbon Fibre line. This line of sunglasses were factory stress-tested and designed to be durable and lightweight, and featured polarized lenses of polycarbonate or crystal.

On the brand's website, Ray-Ban also offers the option to customize and purchase individual pairs of many of their popular styles, giving buyers express choice between frames, lenses, and engraving, among other options.

In 2021, Ray-Ban commercialized a model of smart glasses that they developed with Facebook Reality Labs called Ray-Ban Stories. Building on recent trends of wearable technology, the glasses feature a built-in camera and Bluetooth earphones. Like other wearable, camera-equipped tech, Ray-Ban Stories have come under scrutiny for their susceptibility to privacy issues and potential for misuse.

Counterfeits
Like other luxury brands, Ray-Ban has been a notable target for retail counterfeiters. The advertisement of Ray-Bans for unrealistically low prices has been consistently linked to fraudulent websites peddling counterfeit products. Studies have shown that one of every four ads for discounted luxury products on Facebook links users to such sites. Ray-Bans are often involved in international seizures of counterfeit designer goods. For one example, in 2016, law enforcement in Thailand seized an alleged import of hundreds of thousands of counterfeit sunglasses including Ray-Bans and Oakleys (also owned by Luxottica Group). Counterfeiters were said to have applied tags and stickers to the fakes to deceive consumers about their authenticity.

Luxottica has taken measures to thwart the trade of counterfeit products, such as converting the India Ray-Ban website from a reference site to functional e-commerce platform, and pursuing legal action against online retailers that market fake Ray-Ban products. In 2016, Luxottica introduced a "Minimum Advertised Price (MAP)" policy for Ray-Ban in contracts with their wholesale customers. While the policy does not name a specific minimum price point, it forbids the advertisement of Ray-Ban products at extreme discounts, as well as any advertising that could otherwise devalue the products.

title: Ray-Ban


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Daley Thompson

Francis Morgan Ayodélé Thompson, (born 30 July 1958 ), better known as Daley Thompson, is an English former decathlete. He won the decathlon gold medal at the Olympic Games in 1980 and 1984, and broke the world record for the event four times. He was unbeaten in competition for nine years.

With four world records, two Olympic gold medals, three Commonwealth titles and wins in the World and European Championships, Thompson is considered by many to be one of the greatest decathletes of all time. Robert Chalmers described him as "the greatest all-round athlete this country has ever produced."

Early life and education
Thompson was born in Notting Hill, London, the second son of a British Nigerian father, Frank Thompson, who ran a minicab firm, and Scottish mother, Lydia, from Dundee. When Thompson was six, his father left home. At seven years old, Lydia sent Thompson to Farney Close Boarding School, Bolney, Sussex, which he described as "a place for troubled children". When Thompson was eleven or twelve, his father was shot dead in Streatham by the husband of a woman whom the father and a friend had dropped off. Thompson's forename is a contraction of Ayodelé, a Yoruba word meaning "joy comes home". Thompson's first ambition was to become a professional footballer, but he later switched his interests to athletics.

Early career
Initially, he was a member of Haywards Heath Harriers, but when he returned to London in 1975 he joined the Newham and Essex Beagles Athletics club, training as a sprinter. He began to be coached by Bob Mortimer, who suggested he try for decathlon. He competed in his first decathlon later that year in Cwmbran, Wales, which he won along with his next competition. In 1976, he won the AAA title and was 18th at the Montréal Olympic Games. The following year, he won the European Junior title and in 1978 came the first of his three Commonwealth titles. In 1979, he failed to finish in his only decathlon of that year, but won the long jump at the UK Championships.

1980–1986: Breaking records
Thompson opened the 1980 Olympic season with a world decathlon record of 8,648 points at Götzis, Austria, in May, and followed this by winning gold at the Moscow Olympics. After a quiet 1981 season, he was in top form in 1982; back at Götzis in May, he raised the world record to 8,730 points and then in September, at the European Championships in Athens, he took the record up to 8,774 points. The following month in Brisbane, Thompson took his second Commonwealth title.

In 1983, Thompson won the inaugural World Championships and became the first decathlete to hold a continental title, in his case the European title and the World and Olympic titles simultaneously. He also became by virtue of his World title, the first athlete in any athletics event to hold Olympic, World, continental and Commonwealth Games titles in a single event simultaneously.

Thompson spent much of the summer of 1984 in California preparing for the defence of his Olympic title, with Jürgen Hingsen, the West German who had succeeded Thompson as the world record holder, expected to be a major threat. Thompson took the lead in the first event, a lead he never relinquished throughout the competition. It seemed that, by easing off in the 1,500 metres, he had missed tying the world record by just one point. When the photo-finish pictures were examined, however, it was found that Thompson should have been credited with one more point in the 110 metres hurdles so he had in fact, equalled Hingsen's record.

When the new scoring tables were introduced, Thompson became the sole record holder once more with a recalculated score of 8,847 points – a world record that stood until 1992, when it was surpassed by the American athlete Dan O'Brien with a score of 8,891. Thompson's two victories in the Olympic decathlon are a feat shared only with the Americans Bob Mathias and Ashton Eaton. Thompson's 1984 performance is still the UK record. In 2002, Thompson's successful defence of his Olympic title was ranked number 34 on Channel 4's poll of the 100 Greatest Sporting Moments.

1987–1990: Defeat and injury
Thompson won his third Commonwealth title in 1986 and also won another European Championship but after that he never quite recaptured the excellent form of earlier years. In 1987, he suffered his first decathlon defeat for nine years when he finished ninth in the World Championships, and at his fourth Olympics in Seoul in 1988 he finished fourth. He made the Commonwealth Games team for the fourth time in 1990 but was forced to withdraw because of injury.

Rivalry with Jürgen Hingsen
Thompson's rivalry with West German athlete Jürgen Hingsen was legendary in the sport throughout the 1980s. The pair constantly traded world records, but Thompson always had the upper hand in the major events, remaining undefeated in all competitions for nine years between 1978 and 1987.

Promotional activities
Thompson was also well known for his appearances in commercials for the drink Lucozade in the 1980s. Thompson's name was used for three officially licensed home computer games by Ocean Software in the 1980s: Daley Thompson's Decathlon, Daley Thompson's Supertest and Daley Thompson's Olympic Challenge. He is also a brand ambassador for a travel agency network "Not Just Travel".

Personal bests
Information from World Athletics profile unless otherwise noted.

After athletics
Thompson was forced to retire from athletics in 1992, due to a persistent hamstring injury. In the 1990s, he briefly played reserve team football for Mansfield Town and played one first-team game for Non-League side Stevenage F.C. before having a short spell at Non-League side Ilkeston F.C. He also worked as fitness coach for Wimbledon F.C. and Luton Town football clubs. He also took part in motorsport, entering the Ford Credit Fiesta Challenge Championship in 1994. Thompson worked as a fitness trainer and motivational speaker, as well as appearing at corporate events. In 1994 Thompson trained with Reading Football Club and scored in a friendly against Leatherhead.

Thompson was an ambassador for the London 2012 Summer Olympics, focusing during the bid stage on highlighting the benefits that hosting the Olympics would bring to education and sport in schools. He also took part in the 2011 TV series Jamie's Dream School.

In 2015, Thompson opened his own gym, Daley Fitness, located on Upper Richmond Road in London. In 2018, he joined Masterchef Gary Barnshaw and co founded DT10 Sports, creating and selling a range of low-sugar protein shakes and sports bars.

Public persona
Thompson was a natural showman who endeared himself to the British public with his irreverent personality and anti-establishment attitude. He won the BBC Sports Personality of the Year award in 1982, was appointed an MBE in the 1983 New Year Honours and promoted to the CBE in the 2000 New Year Honours.

At the 1984 Los Angeles Olympics, Thompson, who had just won his second Olympic decathlon gold medal, wore a T-shirt that read "IS THE WORLD'S SECOND GREATEST ATHLETE GAY?". United States sprinter Carl Lewis had been dealing with rumours of homosexuality and the shirt was regarded as being cruelly directed at Lewis. "The second athlete could be anybody, Carl Lewis, anybody" Thompson stated.

In 2012, Thompson was accused of anti-Irish sentiment after commenting on live BBC TV that the creator of a misspelled tattoo "must have been Irish". The tattoo artist was actually American.

In August 2014, Thompson was one of 200 public figures who were signatories to a letter to The Guardian expressing their hope that Scotland would vote to remain part of the United Kingdom in September's referendum on that issue.

In 2015, Thompson co-presented the mid morning show on talkSPORT once a week alongside Colin Murray until Murray quit the station the following year.

Personal life
Thompson has two children with his girlfriend Lisa and three with ex-wife Tish. Thompson has sporting sons Elliot and Alex. Elliot Thompson became the national decathlon champion at the 2022 British Athletics Championships. The title was won 46 years after Daley claimed his first national title. Alex plays rugby for the University of Bath and England Sevens.

title: Daley Thompson


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You're a Good Man, Charlie Brown

You're a Good Man, Charlie Brown is a 1967 musical with music and lyrics by Clark Gesner and (in a 1999 revision) Andrew Lippa. It is based on the characters created by cartoonist Charles M. Schulz in his comic strip Peanuts. The musical has been a popular choice for amateur theatre productions because of its small cast and simple staging.

Background
John Gordon was credited with the book of the show, but according to Gesner's foreword in the published script, "John Gordon" is a collective pseudonym that covers Gesner, the cast members, and the production staff, all of whom worked together to assemble the script. The Guide to Musical Theatre notes that "John Gordon is a pseudonym for the staff and cast of the show. The original cast included Bob Balaban, Gary Burghoff, Bill Hinnant, Skip Hinnant, Karen Johnson and Reva Rose."

History
During the early 1960s, Gesner had begun writing songs based on Charles Schulz's  Peanuts characters, but was unable to get permission from United Feature Syndicate to use the characters in his songs. Eventually Gesner sent Schulz a demo recording of some of the songs and Gesner soon had permission to properly record them, which he did in 1966. Orson Bean sang the role of Charlie Brown, Clark Gesner sang Linus, Barbara Minkus sang Lucy, and Bill Hinnant sang Snoopy (he reprised his role in the Off-Broadway production).

At the time, Gesner had no plans for a musical based on this pre-production "concept album". However, producer Arthur Whitelaw, who would later go on to write another musical based on Peanuts entitled: Snoopy! The Musical, encouraged Gesner to turn the album into a musical.

The stage adaptation of the concept album, titled You're a Good Man, Charlie Brown, went into rehearsal in New York City on February 10, 1967. Prior to its opening, the musical had no actual libretto; it was several vignettes with a musical number for each one.

Original New York productions and U.S. tours
On March 7, 1967, the musical premiered off-Broadway at Theatre 80 in the East Village, featuring Gary Burghoff as Charlie Brown, Skip Hinnant as Schroeder, Reva Rose as Lucy, Bob Balaban as Linus, Karen Johnson as Patty (an early Peanuts character not to be confused with Peppermint Patty), and Bill Hinnant as Snoopy. Joseph Hardy directed and choreographer Patricia Birch was billed as "Assistant to the Director". Joe Raposo, later of Sesame Street fame, was billed as "Music Director" and composer of incidental music for the show. This production of You're A Good Man, Charlie Brown lasted 1,597 performances, closing on February 14, 1971.

The off-Broadway cast recording, originally released on MGM Records, was later remastered by Decca Broadway/Universal Classics and re-released on September 31, 2000. A 1970 U.S. tour lasted 202 performances on the road.

A Broadway production opened at the John Golden Theatre on June 1, 1971, and closed on June 27, 1971, after 32 performances and 15 previews. Directed by Joseph Hardy and with choreography by Patricia Birch, the new cast consisted of Carter Cole as Schroeder, Grant Cowan as Snoopy, Stephen Fenning as Linus, Liz O'Neal as Lucy, Dean Stolber as Charlie Brown, and Lee Wilson as Patty.

1968 West End premiere
The musical opened in the West End in London on February 1, 1968, produced by Harold Fielding and Bernard Delfont, and directed by original 1967 director Joseph Hardy. It played at the Fortune Theatre for 116 performances, until closing on May 11, 1968. The whole original Toronto cast reprised their roles for West End. Don Potter, who portrayed Snoopy, reprised his role in the original San Francisco cast of Snoopy! The Musical.

1998 U.S. tour and 1999 Broadway revival
A U.S. tour began on November 18, 1998, in Skokie, Illinois. The tour was expected to become a full-scale revival to open at the Longacre Theatre on Broadway, but was moved to the Ambassador Theatre after Bring in 'Da Noise, Bring in 'Da Funk's closing. After the tour ended on January 17, 1999, the revival opened on February 4, 1999, and closed on June 13, 1999, having played 14 previews and 149 performances. It featured new dialogue by Michael Mayer, who also directed, and additional songs and orchestration written by Andrew Lippa; choreography was by Jerry Mitchell and sets by David Gallo, Mayer's frequent collaborator.

In this revival, the character of Patty was replaced with Sally Brown, inspired by the same change Schulz made in the animated TV adaptation. The cast featured Anthony Rapp as Charlie Brown, B.D. Wong as Linus, Ilana Levine as Lucy, and Stanley Wayne Mathis as Schroeder. Also featured were Kristin Chenoweth and Roger Bart as Sally and Snoopy, with each winning the Tony award in the respective category. Kirsten Wyatt was the standby for Lucy and Sally.

The original Broadway revival recording was released by RCA Victor/BMG on March 9, 1999.

The 1999 cast appeared on The Rosie O'Donnell Show to promote the show, singing "Beethoven Day." O'Donnell joined in with the cast to close her show by performing the finale as the credits rolled.

2008 Manhattan benefit concert
On December 15, 2008, a one-night-only benefit performance of Charlie Brown was staged at the Gerald W. Lynch Theater at John Jay College in Manhattan for the Make-A-Wish Foundation, directed by David Lefkowich. The cast featured Morgan Karr as Charlie Brown, David Larsen as Schroeder, Tom Deckman as Snoopy, Matt Crowle as Linus, Carmen Ruby Floyd as Lucy, and Kenita R. Miller as Sally.

2016 Off-Broadway revival
The musical was revived at the Off-Broadway York Theatre Company. The revival used some young actors from Broadway productions. The six-member “Peanuts” gang featured Joshua Colley as Charlie Brown, Gregory Diaz as Schroeder, Aidan Gemme as Snoopy, Milly Shapiro as Sally, Mavis Simpson-Ernst as Lucy, and Jeremy T. Villas as Linus. Graydon Peter Yosowitz played the role of Charlie Brown from June 1–7. The production ran from May 24 – June 26, 2016.

Original
Charlie Brown and Linus are together, as his friends give their various opinions of him ("Opening"). Today everyone is calling him a "good man". They state he could be king. Lucy, however says he could be king if he wasn't so wishy-washy ("You're A Good Man, Charlie Brown"). Lucy expresses her deep infatuation with Schroeder and asks him what he thinks of the idea of marriage. Schroeder is aware of her feelings, but remains aloof as he plays his piano. Lucy then exclaims: "My Aunt Marion was right. Never try to discuss marriage with a musician" ("Schroeder").
 * Act I

Schroeder is walking around in the sun, happy with all the peace and quiet. However, that is soon interrupted by Lucy screaming for a ball. Snoopy is lying on top of his doghouse, relaxing vacantly and peacefully. He daydreams that all the children are adoring him, and saying he's “the best dog in the world”. Birds start to poke his stomach, and then he begins to daydream about being a wild jungle beast. In a few minutes, however, he is back to his peaceful state ("Snoopy").

Linus enters, holding his blanket and sucking his thumb. Lucy and Patty show up and mock him for this habit. Lucy tries to pull away the blanket, but fails. Linus decides to abandon his blanket and move on, only to come running back to it in desperation. After the girls leave, Linus decides to relax with his blanket until he can outgrow it ("My Blanket and Me"). Charlie Brown appears, trying to get his unusually stubborn kite to soar in the air. Eventually, he succeeds in doing this, and he enjoys a few minutes of triumph before the kite plummets to the ground ("The Kite").

After this trauma, Charlie Brown tries to find the right way to give The Little Red-Headed Girl her Valentine's Day card, but he ends up saying "Merry Christmas", making a fool out of himself. He finds Patty handing out valentines, and she drops a valentine with the initials C.B. Charlie Brown gives it back to Patty, but soon learns it was Craig Bowerman's valentine. He goes to see Lucy, who is at her psychiatrist booth. He tells her all the things he thinks of himself. Lucy then clears it up by saying that Charlie Brown is unique the way he is, then asks for the five cent price ("The Doctor Is In").

Later, Lucy comes up to Schroeder again and talks about if they got married, and they were so poor that Schroeder had to sell his piano for saucepans, and, again, Schroeder cannot stand it. At noon, Linus, Lucy, Schroeder, and Charlie Brown are working on their Peter Rabbit book reports, each in his or her own way. Lucy is simply babbling to fit the 100-word requirement, Schroeder is doing a "comparison" between the book and Robin Hood, Linus is doing an overcomplicated psychological analysis, and Charlie Brown hasn't even started out of worry ("The Book Report").

Snoopy, in his World War I flying ace uniform, climbs atop his doghouse. He goes through a scene, with him being a pilot searching for the Red Baron. In his imagination, he is defeated by the Red Baron and returns to the aerodrome in France ("The Red Baron"). Meanwhile, Charlie Brown returns, and, with his friends, plays the Little League Baseball Championship. After some mishaps, the team finally manages to make some progress. Charlie Brown steps up to the plate, and despite his valiant efforts, strikes out and loses the game. We learn that this was a flashback, and Charlie Brown expresses his deep sorrow to his pen pal ("T-E-A-M (The Baseball Game)").
 * Act II

Later on, Linus and Lucy arrive home. Linus starts to watch TV, but Lucy tells him to switch channels. Lucy tells Linus that what she intends to do when she grows up, she will become a queen. Linus denies, and Lucy decides to cultivate her life for her "natural beauty" ("Queen Lucy"). The next day, at lunchtime, Charlie Brown talks about his bad days. Then he notices the Little Red-Haired Girl approaching and puts a paper bag over his head. It turns out it is Lucy and Patty, chatting together.

At Schroeder's Glee Club, Patty is the only one to arrive early. When everyone comes, it's chaos. Unfortunately, a fight ensues between Lucy and Linus over a pencil. Lucy threatens to tell Patty that Linus called her an enigma. The fight spreads, and Charlie Brown decides to leave with his angry friends, leaving Schroeder and Snoopy the only ones singing ("Glee Club Rehearsal"). Later, Charlie Brown comes across Lucy teaching Linus about nature the way she views it, with "facts" such as bugs pulling the grass to make it grow or snow coming out of the ground in winter. Charlie Brown tries to correct her, but she retaliates with a false explanation, and Charlie Brown bangs his head against a tree in frustration ("Little Known Facts").

The next morning, Snoopy is puzzled why he has his supper in the red dish, and the water in the blue dish. Meanwhile, Linus and Schroeder are walking to school. Schroeder asks if Linus filled out the form that Ms. Othmar gave them. Linus puts down “Dr. Seuss“. Snoopy talks about how he hates cats, but he is also scared of them ("Peanuts Potpourri"). That evening, Snoopy complains that he hasn't been fed yet, and begins to overly complicate and dramatize the matter until Charlie Brown shows up with his dinner. Snoopy bursts into song about his craving for supper until Charlie Brown firmly tells him to eat his meal ("Suppertime").

That night, everyone is looking at the stars in wonder. Charlie Brown soon discovers a pencil which has been dropped by the Little Red-Haired Girl (his perennial crush). As he examines it, he discovers that "there are teeth-marks all over it . . . she nibbles her pencil . . . she's HUMAN!" With that realization, he concludes that today hasn't been so bad, after all, and he's done a lot of things that make him happy. As Charlie Brown expresses what makes him happy, everyone, touched by his love of life, begin to express what makes them happy as well ("Happiness"). Right then, he realizes being a "good man" means trying your best and making the most of the things you've been given in life. As his other friends leave the stage, Lucy turns to him then tells him, "You're a good man, Charlie Brown."

Revised
Charlie Brown stands alone as his friends give their various opinions of him ("Opening"). Today everyone is calling him a "good man". Charlie Brown is happy and hopeful as usual, but he nevertheless wonders if he really is what they say. He decides to find out how he can really become a good person ("You're a Good Man, Charlie Brown").
 * Act I

Alone one day, during lunch, Charlie Brown talks about his bad days. Then he notices the Little Red-Haired Girl and decides to go sit with her. However, he cannot find the courage to do so, and puts a paper bag over his head. It turns out it is Lucy and Sally, chatting together. They draw a dress on the paper bag, while Charlie Brown feels like he has no confidence ("You're a Good Man, Charlie Brown" (reprise))

Lucy expresses her deep infatuation with Schroeder and asks him what he thinks of the idea of marriage. Schroeder is aware of her feelings, but remains aloof as he plays his piano. Lucy then exclaims: "My Aunt Marion was right. Never try to discuss marriage with a musician" ("Schroeder"). Sally is sad because her jump rope tangled up.

Snoopy is lying on top of his doghouse, relaxing vacantly and peacefully. He begins to daydream about being a wild jungle beast. In a few minutes, however, he is back to his peaceful state ("Snoopy"). Linus enters, holding his blanket and sucking his thumb. Lucy and Sally show up and mock him for this habit. Linus decides to abandon his blanket and move on, only to come running back to it in desperation. After the girls leave, Linus daydreams of a blanket fantasy where everyone can relax with their blankets ("My Blanket and Me"). Lucy later tells him that she would someday like to be a queen ("Queen Lucy"). However, Linus tells her that she can't and she threatens to punch him. Sally gets a C for her pathetic coat-hanger sculpture.

Charlie Brown appears, trying to get his unusually stubborn kite to soar in the air. Eventually, he succeeds in doing this, and he enjoys a few minutes of triumph before the notorious Kite-Eating Tree eats it up ("The Kite"). After this trauma, Charlie Brown tries to find the right way to give Lucy a Valentine's Day card, but he ends up saying "Merry Christmas", making a fool out of himself. He later learns that no one sent him a card, whereas everyone else, including Snoopy, got several. He goes to see Lucy, who is at her psychiatrist booth. He tells her all the things he thinks of himself. Lucy then clears it up by saying that Charlie Brown is unique the way he is, then asks for the five cent price ("The Doctor Is In"). Later, Charlie Brown sees a happy Schroeder spreading the word of Beethoven's birthday and pulling together a celebration. He and company join Schroeder in the song of jubilation ("Beethoven Day").

The next morning, Sally wakes up Snoopy to go rabbit chasing, and they go into strange places, like the Sahara. At noon, Linus, Lucy, Schroeder, and Charlie Brown are working on their Peter Rabbit book reports, each in their own way. Lucy is simply babbling to fit the 100-word requirement, Schroeder is doing a "comparison" between the book and Robin Hood, Linus is doing an overcomplicated psychological analysis, and Charlie Brown hasn't even started out of worry, while Sally and Snoopy continue to chase rabbits ("The Book Report").

Snoopy, in his World War I flying ace uniform climbs atop his doghouse. He goes through a scene, with him being a pilot searching for the Red Baron. In his imagination, he is defeated by the Red Baron and returns to the aerodrome in France ("The Red Baron").
 * Act II

Sally is clearly cross about a D her teacher gave her on her homework assignment. In response, she says, "Oh, yeah? That's what you think!" Schroeder hears and asks why Sally is telling him that. It quickly becomes Sally's new "philosophy", and she bursts into song about her philosophies. Schroeder, after failing to explain to her how philosophies work, leaves in bafflement while Sally continues ("My New Philosophy").

Charlie Brown returns, and, with his friends, plays the Little League Baseball Championship. After some mishaps, the team finally manages to make some progress. Charlie Brown steps up to the plate, and despite his valiant efforts, strikes out and loses the game. We learn that this was a flashback, and Charlie Brown expresses his deep sorrow to his pen pal ("T-E-A-M (The Baseball Game)"). After Schroeder tells her she is a very crabby person, Lucy takes a crabbiness survey of all her friends to determine her crabbiness: Charlie Brown gives a waffling answer, while Sally is more definitive. Upon asking Linus, he refuses to answer until she promises not to punch him; when he gives her a score of ninety-five, though, she punches him anyway. After tallying the score, she realizes that she, in reality, is really very crabby and becomes depressed. Linus cheers her up by reminding her she has a little brother who loves her, remarking as she sobs in his shoulder, "Every now and then I say the right thing!"

Determined not to let what happened at the championship bother him, Charlie Brown decides to join Schroeder's Glee Club and cheer up by singing "Home on the Range" with his friends. Unfortunately, a fight ensues between Lucy and Linus over a pencil. The fight spreads, and Charlie Brown decides to leave with his angry friends, leaving Schroeder and Snoopy the only ones singing ("Glee Club Rehearsal").

Later, Charlie Brown comes across Lucy teaching Linus about nature the way she views it, with "facts" such as bugs pulling the grass to make it grow or snow growing out of the ground in winter. Charlie Brown tries to correct her, but she retaliates with a false explanation, and Charlie Brown bangs his head against a tree in frustration ("Little Known Facts"). That evening, Snoopy complains that he hasn't been fed yet, and begins to overly complicate and dramatize the matter until Charlie Brown shows up with his dinner. Snoopy bursts into song about his craving for supper until Charlie Brown firmly tells him to eat his meal ("Suppertime").

That night, Charlie Brown is still sad that he has not discovered what it means to be a "good man", then he discovers a pencil which has been dropped by the Little Red-Haired Girl (his perennial crush). As he examines it, he discovers that "there are teeth-marks all over it...she nibbles her pencil...she's HUMAN!" With that realization, he concludes that today hasn't been so bad, after all, and he's done a lot of things that make him happy. As Charlie Brown expresses what makes him happy, everyone, touched by his love of life, begin to express what makes them happy as well ("Happiness"). Right then, he realizes being a "good man" means trying your best and making the most of the things you've been given in life. As his other friends leave the stage, Lucy turns to him and puts out her hand, making him shrink back. As he reaches out, she shakes his hand firmly, then tells him, "You're a good man, Charlie Brown."

A medley of "Happiness" and "You're a Good Man, Charlie Brown" is performed as the cast comes out for a final curtain call.

Musical numbers

 * Original

 * Act I
 * "Opening" – Company
 * "You're a Good Man, Charlie Brown" – Company
 * "Schroeder" – Lucy (sung over Beethoven's "Moonlight Sonata")
 * "Snoopy" – Snoopy
 * "My Blanket and Me" – Linus
 * "The Kite" – Charlie Brown
 * "The Doctor is In" (Dr. Lucy) – Lucy and Charlie Brown
 * "The Book Report" – Company

 * Act II
 * "Opening, Pt. 2" - Company
 * "T-E-A-M (The Baseball Game)" – Charlie Brown and Company
 * "The Red Baron" – Snoopy
 * "Queen Lucy" – Lucy and Linus
 * "Glee Club Rehearsal" – Company (sung over "Home on the Range")
 * "Little Known Facts" – Lucy with Linus and Charlie Brown
 * "Peanuts Potpourri" – Snoopy, Schroeder, and Linus
 * "Suppertime" – Snoopy
 * "Happiness" – Company
 * Bows – Company (includes a reprise of "You're a Good Man, Charlie Brown")

Note: "Opening, Pt. 2", "Glee Club Rehearsal", and "Bows" was not included in the original cast recording.
 * Revised

 * Act I
 * "Opening" – Company
 * "You're a Good Man, Charlie Brown" – Company
 * "You're a Good Man, Charlie Brown" (reprise) – Charlie Brown, Linus, and Sally
 * "Schroeder" – Lucy (sung over Beethoven's "Moonlight Sonata")
 * "Snoopy" – Snoopy
 * "My Blanket and Me" – Linus and Company
 * "Queen Lucy" – Lucy and Linus
 * "The Kite" – Charlie Brown (optional company background)
 * "The Doctor is In" (Dr. Lucy) – Lucy and Charlie Brown
 * "Peanuts Potpourri" – Sally, Linus, Snoopy, Lucy, and Schroeder
 * "Beethoven Day" – Schroeder and Company
 * "Rabbit Chasing" - Sally and Snoopy
 * "The Book Report" – Company

 * Act II
 * "The Red Baron" – Snoopy
 * "My New Philosophy" – Sally and Schroeder
 * "T-E-A-M (The Baseball Game)" – Charlie Brown and Company
 * "Glee Club Rehearsal" – Company (sung over "Home on the Range")
 * "Little Known Facts" – Lucy with Linus and Charlie Brown
 * "Suppertime" – Snoopy
 * "Happiness" – Company
 * Bows – Company (includes a partial reprise of "Happiness," and "You're a Good Man, Charlie Brown")

''Note: The three new songs added to the show ("Beethoven Day", "Rabbit Chasing", and "My New Philosophy") are by Andrew Lippa. "You're a Good Man, Charlie Brown Reprise", "Queen Lucy", "Peanuts Potpourri", "Rabbit Chasing", and "The Red Baron" are not included in the 1999 revival Broadway cast recording.''

Instrumentation
The instrumentation varies greatly and three kinds exist.

In the original Off-Broadway production, the instrumentation was simply a piano (doubling on toy piano and melodica), a bass, and percussion. It can be heard on the original cast recording.

When Tams-Witmark acquired the rights to Charlie Brown, the orchestration was rewritten from the original version. The complete orchestration contained a piano, bass, guitar, percussion, five woodwind parts, two trumpets, horn, trombone, and strings. The piano player can also be doubled on celeste, toy piano, and melodica; the first woodwind plays flute and piccolo; the second is the second flute part; the third and fourth are the first and second clarinet parts respectively; the fifth on bass clarinet and tenor sax. Any guitar, horn, and string parts (excluding bass) were all optional.

When Charlie Brown was brought back to Broadway in 1999, the orchestration was deeply revised, containing a five-piece orchestra that consisted of a piano, bass, percussion, a woodwind player, and a violinist. The piano player can double on keyboard synthesizer and kazoo; the bass player doubles on electric and acoustic bass, tenor recorder, and kazoo (in the original Broadway pit the bass player also doubled on acoustic and electric guitar); the woodwind part doubles on piccolo, flute, clarinet, soprano and alto sax, soprano recorder, and kazoo; the violin part also doubles on viola, alto recorder, kazoo, and tambourine. The percussionist primarily plays drum set but doubles on vibraphone, bells, triangle, timpani, and xylophone, with the parts intended to be played with a synthesizer. This version is also available through Tams-Witmark.

Casts
''Notes: The character of "Sally" was added in the 1999 revival, replacing "Patty" from the original version. Sally was then used for the 2016 revival as well. The characters of Patty and Schroeder do not appear on the concept album.''

Articles about the 1999 revision while it was in previews noted that the one difference between the original production and the 1999 version was that the latter reflected the increased ethnic diversity of casting over the decades that had passed, with Schroeder being played by an African American actor (Mathis) and Linus by an Asian American (Wong).

Response
The off-Broadway production was well received, with The Village Voice praising the simplistic set and "strikingly talented" cast. Walter Kerr in The New York Times called the show "a miracle", saying, "Almost everything works, because almost everything is effortless."

In reviewing the 1999 revival, Playbill's Steven Suskin found it "overblown and underwhelming. The scenic and musical enhancements were especially harmful, it seemed to me; the unassuming, child-size characters were overwhelmed… Which is not to say that the 1999 music department did a bad job; it's simply that the concept of a big, new 'You're a Good Man, Charlie Brown' worked against the inherent qualities of the material." In The New York Times, Ben Brantley wrote a lukewarm review: "The real problem is a matter of scale… there's an uncomfortable feeling of dead air that the cast must work much too hard to fill… Songs that were created as droll, low-key character portraits have been reconceived as showstoppers, and the frail, winsome little bodies of these numbers just aren't up to the job. When Linus sings a duet with his famous security blanket, which has been wired to dance on its own, the sequence has a flailing, improvised quality that is the stuff of actors' nightmares."

Brantley did praise some of the cast, saying, "Kristin Chenoweth's performance as Sally will be the part that should seal her reputation. This glow cast by a star-in-the-making gives a real Broadway magic to a show that otherwise feels sadly shrunken… And Roger Bart, in the plum role of Snoopy, the charismatic beagle, incorporates some delightful doglike mannerisms."

Adaptations
In 1973, the show was adapted for television in a Hallmark Hall of Fame TV special, broadcast on NBC. Actors featured in the adaptation included original 1967 cast member Bill Hinnant as Snoopy. Hinnant was the only member of the original off-Broadway cast to reprise their role in the special.

CBS aired a new prime-time animated TV special in 1985, based on the original musical. This version was the first animated depiction of Snoopy with comprehensible dialogue, voiced by Robert Towers, who previously portrayed the role in the 1967 Los Angeles production alongside Burghoff as Charlie Brown and Judy Kaye as Lucy.

Original cast albums have been released for all three versions of the stage show, however the 1973 Hallmark Hall of Fame recording on Atlantic Records is no longer in print.

title: You're a Good Man, Charlie Brown


text:
Descartes' theorem

In geometry, Descartes' theorem states that for every four kissing, or mutually tangent, circles, the radii of the circles satisfy a certain quadratic equation. By solving this equation, one can construct a fourth circle tangent to three given, mutually tangent circles. The theorem is named after René Descartes, who stated it in 1643.

Frederick Soddy's 1936 poem The Kiss Precise summarizes the theorem in terms of the bends (signed inverse radii) of the four circles: "The sum of the squares of all four bends Is half the square of their sum"

Special cases of the theorem apply when one or two of the circles is replaced by a straight line (with zero bend) or when the bends are integers or square numbers. A version of the theorem using complex numbers allows the centers of the circles, and not just their radii, to be calculated. With an appropriate definition of curvature, the theorem also applies in spherical geometry and hyperbolic geometry. In higher dimensions, an analogous quadratic equation applies to systems of pairwise tangent spheres or hyperspheres.

History
Geometrical problems involving tangent circles have been pondered for millennia. In ancient Greece of the third century BC, Apollonius of Perga devoted an entire book to the topic, Ἐπαφαί [Tangencies]. It has been lost, and is known largely through a description of its contents by Pappus of Alexandria and through fragmentary references to it in medieval Islamic mathematics. However, Greek geometry was largely focused on straightedge and compass construction. For instance, the problem of Apollonius, closely related to Descartes' theorem, asks for the construction of a circle tangent to three given circles which need not themselves be tangent. Instead, Descartes' theorem is formulated using algebraic relations between numbers describing geometric forms. This is characteristic of analytic geometry, a field pioneered by René Descartes and Pierre de Fermat in the first half of the 17th century.

Descartes discussed the tangent circle problem briefly in 1643, in two letters to Princess Elisabeth of the Palatinate. Descartes initially posed to the princess the problem of Apollonius. After Elisabeth's partial results revealed that solving the full problem analytically would be too tedious, he simplified the problem to the case in which the three given circles are mutually tangent, and in solving this simplified problem he came up with the equation describing the relation between the radii, or curvatures, of four pairwise tangent circles. This result became known as Descartes' theorem. Descartes did not provide the reasoning through which he found this relation.

Japanese mathematics frequently concerned problems involving circles and their tangencies, and Japanese mathematician Yamaji Nushizumi stated a form of Descartes' circle theorem in 1751. Like Descartes, he expressed it as a polynomial equation on the radii rather than their curvatures. The special case of this theorem for one straight line and three circles was recorded on a Japanese sangaku tablet from 1824.

Descartes' theorem was rediscovered in 1826 by Jakob Steiner, in 1842 by Philip Beecroft, and in 1936 by Frederick Soddy. Soddy chose to format his version of the theorem as a poem, The Kiss Precise, and published it in Nature. The kissing circles in this problem are sometimes known as Soddy circles. Soddy also extended the theorem to spheres, and in another poem described the chain of six spheres each tangent to its neighbors and to three given mutually tangent spheres, a configuration now called Soddy's hexlet. Thorold Gosset and several others extended the theorem and the poem to arbitrary dimensions; Gosset's version was published the following year. The generalization is sometimes called the Soddy–Gosset theorem, although both the hexlet and the three-dimensional version were known earlier, in sangaku and in the 1886 work of Robert Lachlan.

Multiple proofs of the theorem have been published. Steiner's proof uses Pappus chains and Viviani's theorem. Proofs by Philip Beecroft and by H. S. M. Coxeter involve four more circles, passing through triples of tangencies of the original three circles; Coxeter also provided a proof using inversive geometry. Additional proofs involve arguments based on symmetry, calculations in exterior algebra, or algebraic manipulation of Heron's formula (for which see ). The result also follows from the observation that the Cayley–Menger determinant of the four coplanar circle centers is zero.

Statement
Descartes' theorem is most easily stated in terms of the circles' curvatures. The signed curvature (or bend) of a circle is defined as $k=\pm1/r$, where $$r$$ is its radius. The larger a circle, the smaller is the magnitude of its curvature, and vice versa. The sign in $$k=\pm1/r$$ (represented by the $$\pm$$ symbol) is positive for a circle that is externally tangent to the other circles. For an internally tangent circle that circumscribes the other circles, the sign is negative. If a straight line is considered a degenerate circle with zero curvature (and thus infinite radius), Descartes' theorem also applies to a line and three circles that are all three mutually tangent (see Generalized circle).

For four circles that are tangent to each other at six distinct points, with curvatures $$k_i$$ for $i=1,\dots, 4$, Descartes' theorem says:

If one of the four curvatures is considered to be a variable, and the rest to be constants, this is a quadratic equation. To find the radius of a fourth circle tangent to three given kissing circles, the quadratic equation can be solved as

The $$\pm$$ symbol indicates that in general this equation has two solutions, and any triple of tangent circles has two tangent circles (or degenerate straight lines). Problem-specific criteria may favor one of these two solutions over the other in any given problem.

The theorem does not apply to systems of circles with more than two circles tangent to each other at the same point. It requires that the points of tangency be distinct. When more than two circles are tangent at a single point, there can be infinitely many such circles, with arbitrary curvatures; see pencil of circles.

Locating the circle centers
To determine a circle completely, not only its radius (or curvature), but also its center must be known. The relevant equation is expressed most clearly if the Cartesian coordinates $$(x,y)$$ are interpreted as a complex number $$z=x+iy$$. The equation then looks similar to Descartes' theorem and is therefore called the complex Descartes theorem. Given four circles with curvatures $$k_i$$ and centers $$z_i$$ for $i\in\{1,2,3,4\}$, the following equality holds in addition to equation (1):

Once $$k_4$$ has been found using equation (2), one may proceed to calculate $$z_4$$ by solving equation (3) as a quadratic equation, leading to a form similar to equation (2):

$$z_4 = \frac{z_1 k_1 + z_2 k_2 + z_3 k_3 \pm 2 \sqrt{k_1 k_2 z_1 z_2 + k_2 k_3 z_2 z_3 + k_1 k_3 z_1 z_3} }{k_4}.$$

Again, in general there are two solutions for $z_4$ corresponding to the two solutions for $k_4$. The plus/minus sign in the above formula for $z_4$ does not necessarily correspond to the plus/minus sign in the formula for $k_4$.

Three congruent circles
When three of the four circles are congruent, their centers form an equilateral triangle, as do their points of tangency. The two possibilities for a fourth circle tangent to all three are concentric, and equation (2) reduces to

$$k_4 = (3 \pm2 \sqrt{3})k_1.$$

One or more straight lines
If one of the three circles is replaced by a straight line tangent to the remaining circles, then its curvature is zero and drops out of equation (1). For instance, if $k_3=0$, then equation (1) can be factorized as

$$\begin{align} & \bigl(\sqrt{k_1} + \sqrt{k_2} + \sqrt{k_4}\bigr) \bigl({\sqrt{k_2} + \sqrt{k_4} - \sqrt{k_1}}\bigr) \\[3mu] & \quad {} \cdot \bigl(\sqrt{k_1} + \sqrt{k_4} - \sqrt{k_2}\bigr) \bigl(\sqrt{k_1} + \sqrt{k_2} - \sqrt{k_4}\bigr) = 0, \end{align}$$

and equation (2) simplifies to

$$k_4=k_1+k_2\pm2\sqrt{k_1k_2}.$$

Taking the square root of both sides leads to another alternative formulation of this case (with $k_1\ge k_2$),

$$\sqrt{k_4}=\sqrt{k_1}\pm \sqrt{k_2},$$

which has been described as "a sort of demented version of the Pythagorean theorem".

If two circles are replaced by lines, the tangency between the two replaced circles becomes a parallelism between their two replacement lines. In this case, with $k_2=k_3=0$, equation (2) is reduced to the trivial

$$\displaystyle k_4=k_1.$$

This corresponds to the observation that, for all four curves to remain mutually tangent, the other two circles must be congruent.

Integer curvatures
When four tangent circles described by equation (2) all have integer curvatures, the alternative fourth circle described by the second solution to the equation must also have an integer curvature. This is because both solutions differ from an integer by the square root of an integer, and so either solution can only be an integer if this square root, and hence the other solution, is also an integer. Every four integers that satisfy the equation in Descartes' theorem form the curvatures of four tangent circles. Integer quadruples of this type are also closely related to Heronian triangles, triangles with integer sides and area.

Starting with any four mutually tangent circles, and repeatedly replacing one of the four with its alternative solution (Vieta jumping), in all possible ways, leads to a system of infinitely many tangent circles called an Apollonian gasket. When the initial four circles have integer curvatures, so does each replacement, and therefore all of the circles in the gasket have integer curvatures. Any four tangent circles with integer curvatures belong to exactly one such gasket, uniquely described by its root quadruple of the largest four largest circles and four smallest curvatures. This quadruple can be found, starting from any other quadruple from the same gasket, by repeatedly replacing the smallest circle by a larger one that solves the same Descartes equation, until no such reduction is possible.

A root quadruple is said to be primitive if it has no nontrivial common divisor. Every primitive root quadruple can be found from a factorization of a sum of two squares, $n^2+m^2=de$, as the quadruple $(-n,\,d+n,\,e+n,\,d+e+n-2m)$. To be primitive, it must satisfy the additional conditions $\gcd(n,d,e)=1$, and $-n\le 0\le 2m\le d\le e$. Factorizations of sums of two squares can be obtained using the sum of two squares theorem. Any other integer Apollonian gasket can be formed by multiplying a primitive root quadruple by an arbitrary integer, and any quadruple in one of these gaskets (that is, any integer solution to the Descartes equation) can be formed by reversing the replacement process used to find the root quadruple. For instance, the gasket with root quadruple $(-10,18,23,27)$, shown in the figure, is generated in this way from the factorized sum of two squares $10^2+2^2=8\cdot 13$.

Ford circles

The special cases of one straight line and integer curvatures combine in the Ford circles. These are an infinite family of circles tangent to the $x$-axis of the Cartesian coordinate system at its rational points. Each fraction $$p/q$$ (in lowest terms) has a circle tangent to the line at the point $$(p/q,0)$$ with curvature $$2q^2$$. Three of these curvatures, together with the zero curvature of the axis, meet the conditions of Descartes' theorem whenever the denominators of two of the corresponding fractions sum to the denominator of the third. The two Ford circles for fractions $$p/q$$ and $$r/s$$ (both in lowest terms) are tangent when $ When they are tangent, they form a quadruple of tangent circles with the $x$-axis and with the circle for their mediant $$(p+r)/(q+s)$$.

The Ford circles belong to a special Apollonian gasket with root quadruple $(0,0,1,1)$, bounded between two parallel lines, which may be taken as the $x$-axis and the line $y=1$. This is the only Apollonian gasket containing a straight line, and not bounded within a negative-curvature circle. The Ford circles are the circles in this gasket that are tangent to the $x$-axis.

Geometric progression
When the four radii of the circles in Descartes' theorem are assumed to be in a geometric progression with ratio $\rho$, the curvatures are also in the same progression (in reverse). Plugging this ratio into the theorem gives the equation

$$2(1+\rho^2+\rho^4+\rho^6)=(1+\rho+\rho^2+\rho^3)^2,$$

which has only one real solution greater than one, the ratio

$$\rho=\varphi + \sqrt{\varphi} \approx 2.89005 \ ,$$

where $$\varphi$$ is the golden ratio. If the same progression is continued in both directions, each consecutive four numbers describe circles obeying Descartes' theorem. The resulting double-ended geometric progression of circles can be arranged into a single spiral pattern of tangent circles, called Coxeter's loxodromic sequence of tangent circles. It was first described, together with analogous constructions in higher dimensions, by H. S. M. Coxeter in 1968.

Soddy circles of a triangle
Any triangle in the plane has three externally tangent circles centered at its vertices. Letting $$A, B, C$$ be the three points, $$a, b, c$$ be the lengths of the opposite sides, and $s = \tfrac12(a + b + c)$ be the semiperimeter, these three circles have radii $s-a, s-b, s-c$. By Descartes' theorem, two more circles, sometimes called Soddy circles, are tangent to these three circles. They are separated by the incircle, one interior to it and one exterior. Descartes' theorem can be used to show that the inner Soddy circle's curvature is $(4R + r + 2s) / \Delta$, where $$\Delta$$ is the triangle's area, $$R$$ is its circumradius, and $$r$$ is its inradius. The outer Soddy circle has curvature $(4R + r - 2s) / \Delta$. The inner curvature is always positive, but the outer curvature can be positive, negative, or zero. Triangles whose outer circle degenerates to a straight line with curvature zero have been called "Soddyian triangles".

One of the many proofs of Descartes' theorem is based on this connection to triangle geometry and on Heron's formula for the area of a triangle as a function of its side lengths. If three circles are externally tangent, with radii $$r_1, r_2, r_3,$$ then their centers $$P_1, P_2, P_3$$ form the vertices of a triangle with side lengths $$r_1+r_2,$$ $$r_1+r_3,$$ and $$r_2+r_3,$$ and semiperimeter $$r_1+r_2+r_3.$$ By Heron's formula, this triangle $$\triangle P_1P_2P_3$$ has area

$$\sqrt{r_1r_2r_3(r_1+r_2+r_3)}.$$

Now consider the inner Soddy circle with radius $$r_4,$$ centered at point $$P_4$$ inside the triangle. Triangle $$\triangle P_1P_2P_3$$ can be broken into three smaller triangles $$\triangle P_1P_2P_4,$$ $$\triangle P_4P_2P_3,$$ and $$\triangle P_1P_4P_3,$$ whose areas can be obtained by substituting $$r_4$$ for one of the other radii in the area formula above. The area of the first triangle equals the sum of these three areas:

$$\begin{align} \sqrt{r_1r_2r_3(r_1+r_2+r_3)} = {} & \sqrt{r_1r_2r_4(r_1+r_2+r_4)}+{}\\ &\sqrt{r_1r_3r_4(r_1+r_3+r_4)}+{}\\ &\sqrt{r_2r_3r_4(r_2+r_3+r_4)}. \end{align}$$

Careful algebraic manipulation shows that this formula is equivalent to equation (1), Descartes' theorem.

This analysis covers all cases in which four circles are externally tangent; one is always the inner Soddy circle of the other three. The cases in which one of the circles is internally tangent to the other three and forms their outer Soddy circle are similar. Again the four centers $$P_1, P_2, P_3, P_4$$ form four triangles, but (letting $$P_4$$ be the center of the outer Soddy circle) the triangle sides incident to $$P_4$$ have lengths that are differences of radii, $$r_4 - r_1,$$ $$r_4 - r_1,$$ and $$r_4 - r_3,$$ rather than sums. $$P_4$$ may lie inside or outside the triangle formed by the other three centers; when it is inside, this triangle's area equals the sum of the other three triangle areas, as above. When it is outside, the quadrilateral formed by the four centers can be subdivided by a diagonal into two triangles, in two different ways, giving an equality between the sum of two triangle areas and the sum of the other two triangle areas. In every case, the area equation reduces to Descartes' theorem. This method does not apply directly to the cases in which one of the circles degenerates to a line, but those can be handled as a limiting case of circles.

Arbitrary four-circle configurations
Descartes' theorem can be expressed as a matrix equation and then generalized to other configurations of four oriented circles by changing the matrix. Let $$\mathbf{k}$$ be a column vector of the four circle curvatures and let $$\mathbf{Q}$$ be a symmetric matrix whose coefficients $$q_{i,j}$$ represent the relative orientation between the $$th and $$th oriented circles at their intersection point: $$ \mathbf{Q} = \begin{bmatrix} \phantom{-}1 & -1 & -1 & -1 \\ -1 & \phantom{-}1 & -1 & -1 \\ -1 & -1 & \phantom{-}1 & -1 \\ -1 & -1 & -1 & \phantom{-}1 \\ \end{bmatrix}, \qquad \mathbf{Q}^{-1} = \frac14 \begin{bmatrix} \phantom{-}1 & -1 & -1 & -1 \\ -1 & \phantom{-}1 & -1 & -1 \\ -1 & -1 & \phantom{-}1 & -1 \\ -1 & -1 & -1 & \phantom{-}1 \\ \end{bmatrix}. $$

Then equation (1) can be rewritten as the matrix equation

$$\mathbf{k}^\mathsf{T}\mathbf{Q}^{-1}\mathbf{k} = 0.$$

As a generalization of Descartes' theorem, a modified symmetric matrix $$\mathbf{Q}$$ can represent any desired configuration of four circles by replacing each coefficient with the inclination $$q_{i,j}$$ between two circles, defined as

$$ q_{i,j} = \frac{r_i^2 + r_j^2 - d_{i,j}^2}{2 r_i r_j}, $$

where $$r_i, r_j$$ are the respective radii of the circles, and $$d_{i,j}$$ is the Euclidean distance between their centers. When the circles intersect, $q_{i,j} = \cos(\theta_{i,j})$, the cosine of the intersection angle between the circles. The inclination, sometimes called inversive distance, is $$1$$ when the circles are tangent and oriented the same way at their point of tangency, $$-1$$ when the two circles are tangent and oriented oppositely at the point of tangency, $$0$$ for orthogonal circles, outside the interval $$[-1, 1]$$ for non-intersecting circles, and $$\infty$$ in the limit as one circle degenerates to a point.

The equation $$\mathbf{k}^\mathsf{T}\mathbf{Q}^{-1}\mathbf{k} = 0$$ is satisfied for any arbitrary configuration of four circles in the plane, provided $$\mathbf{Q}$$ is the appropriate matrix of pairwise inclinations.

Spherical and hyperbolic geometry

Descartes' theorem generalizes to mutually tangent great or small circles in spherical geometry if the curvature of the $$j$$th circle is defined as $k_j = \cot \rho_j,$ the geodesic curvature of the circle relative to the sphere, which equals the cotangent of the oriented intrinsic radius $$\rho_j.$$ Then:

$$(k_1 + k_2 + k_3+k_4)^2 = 2(k_1^2+k_2^2+k_3^2+k_4^2) + 4.$$

Solving for one of the curvatures in terms of the other three,

$$ k_4 = k_1 + k_2 + k_3 \pm2 \sqrt{k_1 k_2 + k_2 k_3 + k_3 k_1 - 1}. $$

As a matrix equation,

$$\mathbf{k}^\mathsf{T}\mathbf{Q}^{-1}\mathbf{k} = -1.$$

The quantity $$1/k_j = \tan \rho_j$$ is the "stereographic diameter" of a small circle. This is the Euclidean length of the diameter in the stereographically projected plane when some point on the circle is projected to the origin. For a great circle, such a stereographic projection is a straight line through the origin, so $$k_j = 0$$.

Likewise, the theorem generalizes to mutually tangent circles in hyperbolic geometry if the curvature of the $$j$$th cycle is defined as $k_j = \coth \rho_j,$ the geodesic curvature of the circle relative to the hyperbolic plane, the hyperbolic cotangent of the oriented intrinsic radius $$\rho_j.$$ Then:

$$(k_1 + k_2 + k_3+k_4)^2 = 2(k_1^2+k_2^2+k_3^2+k_4^2) - 4.$$

Solving for one of the curvatures in terms of the other three,

$$ k_4 = k_1 + k_2 + k_3 \pm2 \sqrt{k_1 k_2 + k_2 k_3 + k_3 k_1 + 1}. $$

As a matrix equation,

$$\mathbf{k}^\mathsf{T}\mathbf{Q}^{-1}\mathbf{k} = 1.$$

This formula also holds for mutually tangent configurations in hyperbolic geometry including hypercycles and horocycles, if $$k_j$$ is the geodesic curvature of the cycle relative to the hyperbolic plane, the reciprocal of the stereographic diameter of the cycle. This is the diameter under stereographic projection (the Poincaré disk model) when one endpoint of the diameter is projected to the origin. Hypercycles do not have a well-defined center or intrinsic radius and horocycles have an ideal point for a center and infinite intrinsic radius, but $$|k_j| > 1$$ for a hyperbolic circle, $$|k_j| = 1$$ for a horocycle, $$|k_j| < 1$$ for a hypercycle, and $$k_j = 0$$ for a geodesic.

Higher dimensions
In $$n$$-dimensional Euclidean space, the maximum number of mutually tangent hyperspheres is $$n+2$$. For example, in 3-dimensional space, five spheres can be mutually tangent. The curvatures of the hyperspheres satisfy

$$\biggl(\sum_{i=1}^{n+2} k_i\biggr)^{\!2} = n\,\sum_{i=1}^{n+2} k_i^2$$

with the case $$k_i=0$$ corresponding to a flat hyperplane, generalizing the 2-dimensional version of the theorem. Although there is no 3-dimensional analogue of the complex numbers, the relationship between the positions of the centers can be re-expressed as a matrix equation, which also generalizes to $$n$$ dimensions.

In three dimensions, suppose that three mutually tangent spheres are fixed, and a fourth sphere $$S_1$$ is given, tangent to the three fixed spheres. The three-dimensional version of Descartes' theorem can be applied to find a sphere $$S_2$$ tangent to $$S_1$$ and the fixed spheres, then applied again to find a new sphere $$S_3$$ tangent to $$S_2$$ and the fixed spheres, and so on. The result is a cyclic sequence of six spheres each tangent to its neighbors in the sequence and to the three fixed spheres, a configuration called Soddy's hexlet, after Soddy's discovery and publication of it in the form of another poem in 1936.

Higher-dimensional configurations of mutually tangent hyperspheres in spherical or hyperbolic geometry, with curvatures defined as above, satisfy

$$\biggl(\sum_{i=1}^{n+2} k_i\biggr)^{\!2} = nC + n\,\sum_{i=1}^{n+2} k_i^2,$$

where $$C = 2$$ in spherical geometry and $$C = -2$$ in hyperbolic geometry.

title: Descartes' theorem


text:
Combat Action Ribbon

The Combat Action Ribbon (CAR) is a United States Navy, United States Coast Guard, and United States Marine Corps military decoration awarded to United States sea service members "who have actively participated in ground or surface combat." Coast Guardsmen, Navy sailors, and Marines active in clandestine, stealth or special operations are deemed eligible for consideration of the award. The ribbon is awarded to members of the Navy and Marine Corps with a rank no higher than captain and colonel, respectively.

The U.S. Navy first authorized the Combat Action Ribbon on 17 February 1969. The Navy ribbon was originally retroactive to March 1961; in 1999 it was made retroactive to 7 December 1941.

The Coast Guard Combat Action Ribbon was authorized on 16 July 2008, and may be awarded to members of the Coast Guard in the rank of captain and below, "who have actively participated in ground or maritime combat." The Coast Guard ribbon is retroactive to 1 May 1975 (during the Vietnam war Coast Guard members were awarded the Navy Combat Action ribbon). Coast Guardsmen in Vietnam riverine warfare operations were deemed eligible for award of the Navy Combat Action Ribbon.

Air combat does not meet the criteria for the Combat Action Ribbon; naval aviators, naval flight officers and enlisted naval air crewmen, while in the performance of aerial flight, are instead eligible for consideration for the Air Medal.

Eligibility criteria
For a military member to be awarded a Combat Action Ribbon evidence must establish the member engaged the enemy, was under hostile fire, or was physically attacked by the enemy. The service member must have demonstrated satisfactory performance under enemy fire while actively participating in a ground or surface engagement. The Combat Action Ribbon will not be awarded to personnel for aerial combat, since the Strike/flight Air Medal provides recognition for aerial combat exposure; however, a pilot, flight officer/navigator, or other crew member forced to escape or evade, after being forced down, may be eligible for the award. Direct exposure to the detonation of an Improvised Explosive Device (IED) used by an enemy, with or without the immediate presence of enemy forces, constitutes active participation in a ground or surface engagement. Eligibility under this criterion is retroactive to 7 October 2001.

The Combat Action Ribbon is awarded to individuals only. The CAR is not awarded to a military unit, station, or group, although multiple individual service members (e.g., those stationed on a boat or warship) may be nominated for the award arising from combat action in which all participated.

A military member does not automatically qualify for the Combat Action Ribbon. The CAR is awarded after the service member's command takes into consideration the evidence and specified criteria. The evidence must show a service member was engaged in direct combat, not indirect or in an area where combat is occurring; mere presence in a designated combat zone does not qualify a service member for the award.

Only one Combat Action Ribbon is awarded to a service member per theater of war. The CAR is a positive assessment of a service member's battle effectiveness during a combat engagement within a specific theater of war and is not awarded for additional combat interactions therein. Colloquially speaking, it's one CAR per war.

A military member who serves in special operations, who by the nature of their mission, are restricted in their ability to return fire, and who are operating in conditions where the risk of enemy fire was great and expected to be encountered, may be eligible for the Combat Action Ribbon.

The Combat Action Ribbon is a ribbon-only decoration in contrast to military branches that award a badge or medal. Compare: the U.S. Army awards the Combat Infantryman Badge, Combat Action Badge, or Combat Medical Badge. U.S. Air Force and U.S. Space Force combat participant is awarded the Air Force Combat Action Medal (AFCAM), which includes both full and miniature size suspension medals and a ribbon. The Combat Action Ribbon is worn in order of precedence on a member's "ribbon rack" displayed on the left (or right) side of a service member's uniform. The outermost blue stripe is always to the wearer's center, with the central colors in order of red, white, and blue (left to right when worn on left side of uniform; right to left when worn on right side of uniform); only one Combat Action Ribbon is authorized for wear, with each additional CAR award signified with a $5/16$ inch gold or silver star device attached to the center of the ribbon. For example, two CAR awards are signified with the CAR and the addition of one gold star device. Three awards are signified with the ribbon and two gold star devices.

In full dress uniform medals are worn on the member's left side. The CAR is therefore worn on the right side of the uniform. The CAR is the highest rated ribbon-only award in the U.S. military decorations order of precedence and is superseded only by medals of meritorious achievement and combat valor.

An individual whose eligibility has been established in combat in any of the following listed operations may be authorized award of the Combat Action Ribbon. Only one award per operation is authorized. This listing is not all-inclusive, as the Combat Action Ribbon has been awarded in minor operations, as well as for specific actions.  Southeast Asia: 1 Mar 61 to 15 Aug 73.

Dominican Republic: 28 Apr 65 to 21 Sep 66.

USS LIBERTY (AGTR 5): 8 Jun 67 to 9 Jun 67.

USS PUEBLO (AGTR 2): 23 Jan 68.

Operation FREQUENT WIND (Saigon evacuation): 29 to 30 Apr 75. (No ships qualified. )

Operation MAYAGUEZ: 15 May 75. (No ships qualified.)

Grenada: 240ct83 to 2 Nov 83. (No ships qualified.)

Lebanon: 20Aug82 to 1 Aug 84. (No ships qualified.)

Persian Gulf
COMNAVSPECWAR Task Unit Tango: 22 Sep 87.

USS SAMUEL B. ROBERTS (FFG 58) : 14 Apr 88.

Operation PRAYING MANTIS: 18 Apr 88.

USS ELMER MONTGOMERY (FF 1082) and USS VINCENNES (CG 49) : 3 Jul 88.

Persian Gulf MCM Operations: Specific units during the periods 19 Nov 87 to 1 Apr 88; 14 to 20 Apr 88; 20 to 23 Apr 88; and 2 Aug 90 to 10 Sep 91.

HM-14 Persian Gulf MCM Operations: 2 Aug 90 to 10 Sep 91.

Operation JUST CAUSE (Panama) : 20 Dec 89 to 31 Jan 90. (No ships qualified. )

Operation SHARP EDGE: 5 to 24 Aug 90. (No ships qualified.)

Operation DESERT STORM: 17 Jan 91 to 28 Feb 91.
The Secretary of the Navy approved the CAR as an exception to policy for the following ships that operated north of 28.30N.and west of 49.303 from 17 Jan 91 to 28 Feb 91 USS ADROIT (MSO 509)

USS BEAUFORT (ATS 2)

USS CARON (DD 970)

USS DURHAM (LKA 114)

USS FORD (FFG 54)

USS PAUL F. FOSTER (DD964)

USNS HASSAYAMPA (T-AO 145)

USS IMPERVIOUS (MSO 449)

USS AVENGER (MCM 1)

USS BUNKER HILL (CG 52)

USS CURTS (FFG 38)

USS FIFE (DD 991)

USS FORT MCHENRY (LSD 43)

USS HAWES (FFG 53)

USS HORNE (CG 30)

USS JARRETT (FFG 33)

USS KIDD (DDG 993)

USS LEADER (MSO 490)

USS MACDONOUGH (DDG 39)

USS MISSOURI (BB 63)

USS NASSAU (LHA 4)

USS NICHOLAS (FFG 47)

USS OLDENDORF (DD 972)

USS PORTLAND (LSD 37)

USS RALEIGH (LPD 1)

USS VREELAND (FF 1068)

USS WORDEN (CG 18)

USS LASALLE (AGF 3)

USS LEFTWICH (DD 984)

USS MCINERNEY (FFG 8)

USS MOBILE BAY (CG 53)

USS NIAGARA FALLS (AFS 3)

USS OKINAWA (LPH 3)

USNS PASSUMPSIC (T-A0 107)

USS PRINCETON (CG 59) (hit by mine)

USS TRIPOLI (LPH 10) (hit by mine)

HM-14 Vanguard Detachment

USS WISCONSIN (BB 64)

The following ships were approved for the dates indicated USNS COMFORT (T-AH 20) 26 Feb 91

USS GUAM (LPH 9) 25 to 26 Feb 91

USS IWO JIMA (LPH 2) 26 Feb 91

USS OGDEN (LPD 5) 25 to 26 Feb 91

USS MISSOURI (BB 63) 12 Feb 91 and 25 Feb 91

VC-6 Detachment

EODMU Detachments

USS RICHMOND K. TURNER (CG 20) 19 to 24 Feb 91

USS VALLEY FORGE (CG 50) 16 to 28 Feb 91

USS LEADER (MSO 490) 23 Mar 91

El Salvador: 1 Jan 81 to 1 Feb 92.

Operation RESTORE HOPE (Somalia): 5 Dec 92 to 31 Mar 95. (No ships qualified. )

Cambodia: 1 Jun 92 to 15 Nov 93.

Operation ASSURED RESPONSE (Monrovia, Liberia) : 7 to 18 Apr 96. (No ships qualified.)

Kosovo Campaign: 24 Mar 99 to 27 Jan 00.

Operation ENDURING FREEDOM: 11 Sep 01 to 28 Dec 14.

Operation IRAQI FREEDOM: 19 Mar 03 to 18 Dec 11.

Israel–Hamas war: October 2023 - April 2024

 * USS Mason (DDG-87)
 * USS Carney (DDG-64)
 * USS Gravely (DDG-107)
 * USS Thomas Hudner (DDG-116)
 * USS Laboon (DDG-58)
 * USS Philippine Sea (CG-58)
 * USS Dwight D. Eisenhower (CVN-69)

Combat Action Ribbon versions
The Combat Action Ribbon is currently authorized with a U.S. Navy/Marine Corps design, and as of 2008, a U.S. Coast Guard version.

The Navy CAR covers the Navy and Marine Corps since the establishment of the CAR in 1969. Additionally, the award was made retroactive to 7 December 1941.

Prior to 2008, eligible U.S. Coast Guardsmen were awarded the Navy Combat Action Ribbon (because in most conflicts and wars Coast Guard members usually operated with or under the U.S. Navy). From 2009 forward, Coast Guard members who engage in combat are now awarded the Coast Guard Combat Action Ribbon.

Award inquiries
Navy, Marine Corps, and Coast Guard veterans are advised to submit requests about eligibility or other matters to their respective branches to inquire about the Combat Action Ribbon or other awards of this nature.

Navy CAR
The Navy's Combat Action Ribbon ("CAR") was established during the Vietnam War by a Secretary of the Navy Notice, dated February 17, 1969, with retroactive award to 1 March 1961. The Navy CAR is awarded to members of the Navy and Marine Corps (and Coast Guard, when operating under the control of the Navy during a war or national emergency), with the grade of captain/colonel and below, who have actively participated in ground or surface combat.

Marines, Navy sailors, or Coast Guardsmen who are awarded the Combat Infantryman Badge, Combat Medical Badge, Combat Action Badge, or Combat Action Medal while assigned to United States Army, United States Air Force, and United States Space Force units, or who earned such for prior service in the U.S. Army, Air Force, or Space Force, may be authorized to wear the Combat Action Ribbon upon application to the Department of the Navy (or Department of Homeland Security, as applicable).

After the destroyer USS Cole (DDG-67) was attacked by suicide bombers in 2000, the entire crew of the ship was awarded the Combat Action Ribbon a year later.

In January 2013, the awarding criteria were expanded to include dangerous exposure to IEDs, mines, and scatterable munitions, be it the detonation of such or direct action taken to disable, render safe, or destroy such; servicemembers may be deemed eligible if the IEDs are detonated or specifically emplaced by the enemy. Previous eligibility applied only to exposure to IEDs actually detonated by the enemy. Eligibility under this criterion is retroactive only to 7 October 2001.

Blanket lists of units and operations whose members or participants are deemed to be "in-combat", and thus potentially eligible, can be found in OPNAVNOTE 1650 (for specific units and ships) and in chapter 2, appendix E, of recent SECNAVINSTs (for specific operations and ships) though, "Neither service in a combat area nor being awarded the Purple Heart Medal automatically makes a service member eligible", and specific sailors and marines may receive the ribbon in recognition of individual actions or various minor operations.

In 2017, sailors from USS Nitze, USS Mason, USS Ponce, and USS San Antonio were awarded the Combat Action Ribbon after their ships were fired upon by rebels off the coast of Yemen in 2016, one of the few instances in the 21st century when U.S. Navy sailors aboard a commissioned warship were awarded the ribbon for combat occurring at sea. In 2005, the crew of the USS Kearsarge were awarded the CAR as well.

In 2024, sailors from the destroyers USS Carney, USS Mason, USS Gravely, USS Laboon, USS Thomas Hudner, Cruiser USS Philippine Sea and Carrier USS Dwight D. Eisenhower were awarded the Combat Action Ribbon after their ships were fired upon by Houthi rebels during the 2024 missile strikes in Yemen and shooting down drones in the Red Sea for the time period of October 2023 to April 2024.

World War II and Korean War
In October 1999, World War II and Korean War veterans became retroactively eligible for the Navy Combat Action Ribbon by Public Law 106-65 on 5 October 1999, which permitted the Secretary of the Navy (SECNAV) to award the Combat Action Ribbon to a member of the Navy or Marine Corps for participation in ground or surface combat during any period on or after 7 December 1941, and before 1 March 1961, if the Secretary determines that the member has not been previously recognized in an appropriate manner for such participation. Two specific blocks of time were later designated by then SECNAV Danzig: Dec. 7, 1941 – Apr. 14, 1946 (World War II), and June 25, 1950 – July 27, 1954 (Korean War)

Coast Guard CAR
In 2009, the U.S. Coast Guard began awarding a Coast Guard Combat Action Ribbon (CGCAR), that is stylized similarly in design to the Navy Combat Action Ribbon.

Prior to 2008 Coast Guard members awarded the Combat Action Ribbon received the U.S. Navy CAR because in times of conflict and war the Coast Guard in combat areas typically operated with or under the U.S. Navy. For example, in the Vietnam War's Operation Market Time the U.S. Coast Guard had at any one time approximately 1,200 Coast Guard members participating in brown water navy riverine warfare operations. Those Coast Guard members were awarded the U.S. Navy Combat Action Ribbon by the Commander U.S. Naval Forces Vietnam.

The U.S. Coast Guard Combat Action Ribbon (CGCAR) was established by the approval of the Secretary of Homeland Security on 16 July 2008, in ALCOAST 361/08. The CGCAR is awarded to members who have actively participated in ground or maritime combat. Satisfactory performance under fire with the enemy is required.

From 2009 forward U.S. Coast Guard members will receive the CGCAR regardless if a member is operating in conjunction with or under the control of the U.S. Navy. Those USCG members awarded the U.S. Navy CAR prior to 2009 are authorized to continue wearing the Navy award, or at the member's option can choose to wear the CGCAR.

Eligibility for the CGCAR also includes personnel having direct exposure to the detonation of an enemy's improvised explosive device, and to personnel who served in clandestine/special operations and were restricted in their ability to return fire, where the risk of enemy fire was great.

The Coast Guard Combat Action Ribbon is authorized for:

 * Operation Allied Force (Kosovo)
 * Operation Desert Storm
 * Operation Enduring Freedom
 * Operation Iraqi Freedom

Other minor operations and specific actions may allow the award, as determined by the Commandant of the Coast Guard. Only one award per operation is authorized.

Initially, all other similar military awards from other services were required to be converted to the Coast Guard Combat Action Ribbon in order to be worn on the Coast Guard uniform, but the policy was modified effective January 1, 2009 to allow wearing the Navy Combat Action Ribbon, Air Force Combat Action Medal, and Army Combat Infantry/Medical/Action Badge for service prior to May 1, 1975, and which cannot be converted to the Coast Guard Combat Action Ribbon.

Additional awards of the Coast Guard Combat Action Ribbon are denoted by $5/16$ inch gold stars on the ribbon.

Notable Recipients: DC3 Nathan Bruckenthal, CWO Joseph Ruggiero, CAPT Holly Harrison, LT Michael Brooks, IS1 Daniel Duffin

title: Combat Action Ribbon