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# William Slim, 1st Viscount Slim

The Viscount Slim
William Slim in 1950
13th Governor-General of Australia
In office
8 May 1953 – 2 February 1960
MonarchElizabeth II
Prime MinisterRobert Menzies
Preceded bySir William McKell
Succeeded byThe Viscount Dunrossil
Personal details
Born
William Joseph Slim

6 August 1891
Bishopston, Bristol, England
Died14 December 1970 (aged 79)
London, England
Resting placeMemorial plaque in St Paul's Cathedral
Spouse(s)
Aileen Robertson
(m. 1926)
Children2nd Viscount Slim
Una Mary Slim
Alma materStaff College, Quetta
Nickname(s)"Uncle Bill"
Military service
AllegianceUnited Kingdom
Branch/serviceBritish Army
British Indian Army
Years of service1914–1952
RankField Marshal
Unit
Commands
Battles/wars
Awards

Field Marshal William Joseph Slim, 1st Viscount Slim, KG, GCB, GCMG, GCVO, GBE, DSO, MC, KStJ (6 August 1891 – 14 December 1970), usually known as Bill Slim, was a British military commander and the 13th Governor-General of Australia.

Slim saw active service in both the First and Second World Wars and was wounded in action three times. During the Second World War he led the Fourteenth Army, the so-called "forgotten army" in the Burma campaign. After the war he became the first British officer who had served in the Indian Army to be appointed Chief of the Imperial General Staff. From 1953 to 1959 he was Governor-General of Australia.

In the early 1930s, Slim also wrote novels, short stories, and other publications under the pen name Anthony Mills.

## Early years

William Slim was born at 72 Belmont Road, St Andrews, Bristol, the son of John Slim by his marriage to Charlotte Tucker, and was baptised there at St Bonaventure's Roman Catholic church, Bishopston. He was brought up first in Bristol, attending St Bonaventure's Primary School, then St Brendan's College, before moving to Birmingham in his teens. In Birmingham, he attended St Philip's Grammar School, Edgbaston and King Edward's School, Birmingham.

After leaving school, his father's failure in business as a wholesale ironmonger meant that the family could afford to send only one son, Slim's older brother, to the University of Birmingham, so between 1910 and 1914 Slim taught in a primary school and worked as a clerk in Stewarts & Lloyds, a metal-tube maker.

## First World War

Despite having no other connection to the university, in 1912 Slim joined the Birmingham University Officers' Training Corps, and he was thus able to be commissioned as a temporary second lieutenant into the Royal Warwickshire Regiment on 22 August 1914, on the outbreak of the First World War; in later life, as a result of his modest social origins and his unpretentious manner, he was sometimes wrongly supposed to have risen from the ranks. He was badly wounded at Gallipoli. On return to England, he was granted a regular commission as a second lieutenant in the West India Regiment.

In October 1916, he rejoined the Royal Warwickshire Regiment in Mesopotamia. On 4 March 1917, he was promoted to lieutenant (with seniority back-dated to October 1915). He was wounded a second time in 1917. Having been previously given the temporary rank of captain, he was awarded the Military Cross on 7 February 1918 for actions in Mesopotamia.

Evacuated to India, he was given the temporary rank of major in the 6th Gurkha Rifles on 2 November 1918. He was formally promoted to captain and transferred to the Indian Army on 22 May 1919.

## Interwar career

Slim became battalion adjutant with the 6th Gurkha Rifles in 1921.

On 1 January 1926, he married Aileen Robertson, daughter of Rev John Anderson Robertson (d.1941) minister of Cramond near Edinburgh. They had one son and one daughter. Later that year Slim was sent to the Staff College, Quetta. On 5 June 1929, he was appointed a General Staff Officer, Second Grade.

On 1 January 1930, he was given the brevet rank of major, with formal promotion to this rank made on 19 May 1933. His performance at Staff College resulted in his appointment first to Army Headquarters India in Delhi and then to Staff College, Camberley, in England (as a General Staff Officer, Second Grade), where he taught from 1934 to 1937. During this period, he also wrote novels, short stories, and other publications under the pen name of Anthony Mills, in order to further his literary interests, as well as to supplement his then modest army salary.

He attended the Imperial Defence College in 1937. The following year he was promoted to lieutenant colonel and given command of the 2nd Battalion, 7th Gurkha Rifles. In 1939 he was briefly given the temporary rank of brigadier as commander of his battalion. On 8 June 1939, he was promoted to colonel (again with temporary rank of brigadier) and appointed head of the Senior Officers' School, Belgaum in India.

## Second World War

### East African Campaign

On the outbreak of the Second World War, Slim was given command of the 10th Indian Infantry Brigade of the 5th Indian Infantry Division and was sent to Sudan. He took part in the East African campaign to liberate Ethiopia from the Italians. Slim was wounded again during the fighting in Eritrea. On 21 January 1941, Slim was hit when his position was strafed during the advance on Agordat.

### Middle East

Recovering from his wounds but still unfit for active service, Slim was temporarily employed on the General Staff at GHQ in Delhi. He was involved in the planning for potential operations in Iraq where trouble was expected. By early May 1941 Slim had been appointed Brigadier General Staff (chief staff officer) to Edward Quinan the commander designate for operations in Iraq, arriving in Basra on 7 May. Not long afterwards, Major-General Fraser, commanding Indian 10th Infantry Division, fell ill and was relieved of his command, and Slim was promoted to take his place on 15 May 1941 with the acting rank of major-general. He led the Indian 10th Infantry Division as part of Iraqforce during the Anglo-Iraqi War, the Syria-Lebanon Campaign (where the division advanced up the river Euphrates to capture Deir ez-Zor), and the invasion of Persia. He was twice mentioned in despatches during 1941.

### Burma campaign

Field Marshal Sir William Slim, General Officer Commanding Fourteenth Army in Burma, 5 March 1945.
Portrait of General Slim as commander of the Fourteenth Army, commissioned by the Ministry of Information.

In March 1942, Slim was given command of Burma Corps, also known as BurCorps, consisting of the 17th Indian Infantry Division and 1st Burma Division. Slim was made acting lieutenant general on 8 May 1942. The corps was under attack in Burma by the Japanese and, heavily outclassed by the more mobile and flexible Japanese, was soon forced to withdraw to India. On 28 October 1942, Slim was appointed a Commander of the Order of the British Empire (CBE).

Slim then took over XV Corps under the command of the Eastern Army. His command covered the coastal approaches from Burma to India, east of Chittagong. He had a series of disputes with Noel Irwin, commander of Eastern Army and, as a result, Irwin (although an army commander) took personal control of the initial advance by XV Corps into the Arakan Peninsula. The operations ended in disaster, during which Slim was restored to command of XV Corps, albeit too late to salvage the situation. Generals Irwin and Slim blamed each other for the result, but in the end Irwin was removed from his command, and Slim was promoted to command the new Fourteenth Army—formed from IV Corps (United Kingdom) (Imphal), XV Corps (Arakan) and XXXIII Corps (reserve) – later joined by XXXIV Corps. On 14 January 1943, Slim was awarded the Distinguished Service Order (DSO) for his actions in the Middle East during 1941. The American historians Alan Millet and Williamson Murray described Slim as:

A hardened field soldier who had learned his trade on the Western Front and in the Indian Army, Slim combined troop-leading and training skills with personal and moral courage as well as charm, a sound grasp of soldiering, and a solid appreciation of Asian warfare and the excellence of the Japanese Army. He had experienced the catastrophe of the 1942 retreat from Burma and the abortive attack on the Arakan. His honesty and character made him the obvious choice to reshape the Fourteenth Army, a force built on the Indian Army but including the ever-dependable Gurkha Rifles of Nepal, unproven infantry battalions from East and West Africa, and infantry battalions and supporting arms from the British Army.

Slim quickly got on with the task of training his new army to take the fight to the enemy. His basic premise was that off-road mobility was paramount: much heavy equipment was exchanged for mule- or air-transported equipment, and motor transport was kept to a minimum and restricted to vehicles that could cope with some of the worst combat terrain on Earth. The new doctrine dictated that if the Japanese had cut the lines of communication, then they too were surrounded. All units were to form defensive 'boxes', to be resupplied by air and assisted by integrated close air support and armour. The boxes were designed as an effective response to the tactics of infiltration practised by the Japanese in the war. Slim also supported increased offensive patrolling and night training, to encourage his soldiers to lose both their fear of the jungle and their belief that Japanese soldiers were better jungle fighters. He also instructed them to hold their positions once outflanked.

The Chin Hills formed a natural defensive barrier into Burma, which Slim would have preferred to outflank by an amphibious operation by landing further down the coast of Burma, but demands of the war in Europe meant the necessary landing craft were not available, forcing Slim to devise plans for advancing into Burma overland through the Chin hills. At the same time, the Japanese 15th Army, which formed the main striking force of the Burma Area Army, had grown from four divisions at the beginning of 1943 to eight divisions by the end of 1943 as the Japanese made preparations for invading India, which increased the difficulties of an overland advance into Burma. By 1943, the Burma Railway, which cost the lives of thousands of prisoners of war who built it, was finished; this allowed the Japanese to reinforce the Burma Area Army, and made invading India barely possible.

As Slim went about training his men for the rigours of jungle warfare, he clashed with Brigadier Orde Wingate, who took away some of Slim's best Gurkha, British and African units for his Chindit raiding group. Slim argued against the loss of his better units to Wingate, and maintained that though Wingate had a successful career in Palestine and Ethiopia he would discover that the Japanese were a considerably tougher foe than the Palestinians and the Italians that Wingate had hitherto been fighting. However, Slim did approve of Wingate's plans for aid to the hill tribes of Burma. The various hill peoples of Burma such as the Kachins, Karens, Chin, Nagas and the Shan collectively amounted to about 7 million of Burma's 17 million people, and unlike the Bamars, who had welcomed the Japanese as liberators, had stayed loyal to the British when the Japanese invaded. The hill peoples of Burma had suffered under Japanese rule, and were more than willing to wage guerrilla warfare against them. Slim approved of the plans of the SOE and OSS to provide arms and training to the hill tribes as a way to tie down Japanese forces that would otherwise be deployed against him.

At the start of 1944, Slim held the official rank of colonel with a war-time rank of major-general and the temporary rank of lieutenant-general. In January 1944, when the Second Arakan Offensive was met by a Japanese counter-offensive, the Indian 7th Infantry Division was quickly surrounded along with parts of the Indian 5th Infantry Division and the 81st (West Africa) Division. The 7th Indian Division's defence was based largely on the "Admin Box" formed initially from drivers, cooks and suppliers. They were supplied by air, thus negating the importance of their lost supply lines. The Japanese forces were able to halt the offensive into Arakan but were unable to decisively defeat the allied forces or advance beyond the surrounded formations.

In early 1944, the Japanese Prime Minister General Hideki Tojo approved of plans for victory in Asia, calling for two operations, Operation U-Go as the invasion of India was code-named and Operation Ichi-Go which was intended to defeat China once and for all. The two operations in India and China were closely linked given that American supplies for China were flown over "the Hump" of the Himalayas and the Japanese wanted to take the Indian province of Assam in part to close the American air bases in India that sustained China at the same time that they were launching Operation Ichi-Go, the biggest Japanese offensive of all time, involving 2 million men. The Japanese knew that they lacked the logistics to invade India, and the plans for U-Go were based on the assumption that the British Fourteenth Army would just collapse, allowing the Japanese 15th Army to capture enough food to prevent its men from starving to death. Following the Japanese 15th Army into India was the Indian National Army commanded by Subhas Chandra Bose, an ardent nationalist. The Japanese believed that the mere presence of Bose in India would inspire the men of the Indian Army to mutiny and murder their British officers, and set off an anti-British revolution that would allow the Japanese 15th Army to take all of India.

Unlike the Japanese, who killed their wounded, Slim went out of his way to ensure good medical care for his wounded and to evacuate his wounded by air to hospitals in India. Slim knew his men would fight better if they knew that they would receive the best possible medical care under the conditions if they were wounded. On 8 August 1944, Slim was promoted to lieutenant general, and, on 28 September 1944, he was appointed a Knight Commander of the Order of the Bath (KCB). In December 1944, during a ceremony at Imphal in front of the Scottish, Gurkha and Punjabi regiments, Slim and three of his corps commanders (Christison, Scoones and Stopford) were knighted by the viceroy Lord Wavell and invested with honours. Slim was presented with his insignia as KCB, and the others with their KBEs. Slim was also mentioned in despatches. By the end of 1944, the majority of the men serving in the British Fourteenth Army were in fact not British as of the 12 divisions that made up the British Fourteenth Army, 2 were British, 7 were Indian and 3 came from Britain's African colonies. In addition, there were six Chinese divisions, two regiments from the U.S. Army and various tribal militias made up of Shan, Chin, Naga, Kachin and Karen peoples raised by the OSS and the SOE fighting on the Allied side in Burma, requiring Slim to play the role of the diplomat as much as a general to hold these disparate forces made up of so many different peoples together.

Lieutenant General Sir William Slim being knighted by the Viceroy of India, Field Marshal the Viscount Wavell, near Imphal, December 1944.

Lieutenant General Sir William Slim (GOC-in-Chief Fourteenth Army, left), Air Vice Marshal Stanley Vincent (AOC 221 Group South East Asia Air Forces, centre) and Major General Henry Chambers (GOC 26th Indian Division, right) at Government House, Rangoon, 8 May 1945.

The Japanese garrison in Mandalay chose not to surrender, using the forts built by the British and the maze of pagodas in downtown Mandalay to fight to the death in an urban battle that destroyed much of the city, which finally fell to the British Fourteenth Army on 20 March 1945. Slim's plan was a masterpiece of operational art, and the capture of Meiktila left most of Japan's troops stranded in Burma without supplies. The Allies had reached the open plains of central Burma, sallying out and breaking Japanese attacking forces in isolation, maintaining the initiative at all times, backed up by air-land cooperation, including resupply by air and close air support, performed by both RAF and USAAF units. Slim followed up this victory by ordering his coastal corps to seize the mouth of the Irrawaddy where it flowed into the Bay of Bengal.

In combination with these attacks, Force 136 helped initiate a countrywide uprising of the Bamar people against the Japanese. In addition to fighting the allied advance south, the Japanese were faced with heavy attacks from behind their own lines. As he advanced into Burma, Slim discovered gruesome evidence of the nature of Japanese rule in Burma, finding in village after village, Burmese peasants who had been tied to trees and bayoneted to death as the Japanese preferred bayonet practice with people rather than sandbags as normally is the case. Toward the end of the campaign, the army raced south to capture Rangoon before the start of the monsoon. It was considered necessary to capture the port because of the length of the supply lines overland from India and the impossibility of supply by air or land during the monsoon. Rangoon was eventually taken by a combined attack from the land (Slim's army), the air (parachute operations south of the city) and a seaborne invasion. Also assisting in the capture of Rangoon was the Anti-Fascist People's Freedom League led by Thakin Soe, with Aung San (the future Prime Minister of Burma and father of Aung San Suu Kyi) as one of its military commanders.

As the Burma campaign came to an end, Slim was informed in May by Oliver Leese, the commander of Allied Land Forces South-East Asia (ALFSEA) that he would not be commanding Fourteenth Army in the forthcoming invasion planned for Malaya but would take command of the new Twelfth Army being formed to mop up in Burma. Slim refused the appointment, saying he would prefer to retire. As the news spread, Fourteenth Army fell into turmoil and Alan Brooke, the Chief of the Imperial General Staff, furious at not having been consulted by Leese, and Claude Auchinleck, the C-in-C India who was at the time in London, brought pressure to bear. The Supreme Allied Commander of the Southeast Asia Theatre, Louis Mountbatten was obliged to order Leese to undo the damage. On 1 July 1945, Slim was promoted to general and was informed that he was to succeed Leese as C-in-C ALFSEA. However, by the time he took up the post, having taken some leave, the war was at an end.

### Relations with troops

Slim had an excellent relationship with his troops – the "Forgotten Army", as they called themselves. In his book, Defeat into Victory, he tells of the malaria rates among his units being 70%, largely due to noncompliance by his soldiers with the foul-tasting mepacrine they refused to take. Slim did not blame his medics for this problem, but placed the responsibility on his officers. "Good doctors are no use without good discipline. More than half the battle against disease is fought not by the doctors, but by the regimental officers". After Slim dismissed a few officers for high unit malaria rates, the others realised he was serious and malaria treatment was enforced, dropping the rate to less than five percent. The combat effectiveness of his army was thus greatly enhanced. This physical and mental turnaround in the army under him was a contributing factor to the eventual defeat of the Japanese in Burma.

George MacDonald Fraser, later author of The Flashman Papers series of novels, then a nineteen-year-old lance corporal, recalled:

But the biggest boost to morale was the burly man who came to talk to the assembled battalion … it was unforgettable. Slim was like that: the only man I've ever seen who had a force that came out of him... British soldiers don't love their commanders much less worship them; Fourteenth Army trusted Slim and thought of him as one of themselves, and perhaps his real secret was that the feeling was mutual.

and:

...I see him clear, with that robber-baron face under that Gurkha hat, and his carbine slung, looking like a rather scruffy private with a general's tabs, which of course is what he was.

## Post-war career

### Initial retirement from the army

At the end of 1945 Slim returned to the UK. On 1 January 1946, he was made a Knight Grand Cross of the Order of the British Empire (GBE). and took the post of Commandant of the Imperial Defence College for its first course since 1939. On 7 February 1947 he was made an Aide-de-camp (ADC) to the King. At the end of his two-year appointment at the Imperial Defence College Slim retired as ADC and from the army on 11 May 1948. He had been approached by both India and Pakistan to become C-in-C of their respective armies post independence but refused and instead became Deputy Chairman of the Railway Executive.

However, in November 1948 the British Prime Minister Clement Attlee rejected the proposal by Viscount Montgomery that he should be succeeded as Chief of the Imperial General Staff (CIGS) by John Crocker and instead brought back Slim from retirement in the rank of field marshal in January 1949 with formal appointment to the Army Council from 1 January 1949. Slim thus became the first Indian Army officer to become CIGS.

On 2 January 1950, he was promoted to Knight Grand Cross of the Order of the Bath (GCB) and later that year was made a Chief Commander of the Legion of Merit by the United States having previously, in 1948, been awarded the lower ranking Commander of the Legion of Merit. On 1 November 1952, he relinquished the position of Chief of the Imperial General Staff.

### Governor-General of Australia

On 10 December 1952 Slim was made a Knight Grand Cross of the Order of St Michael and St George (GCMG) on his appointment as Governor-General of Australia which post he took up on 8 May 1953. On 2 January 1953, he was appointed a Knight of the Order of St. John (KStJ). Slim was a popular choice for Governor-General since he was an authentic war hero who had fought alongside Australians at Gallipoli and in the Middle East. In 1954 he was able to welcome Queen Elizabeth II on the first visit by a reigning monarch to Australia. For his services to the Queen during the tour, he was appointed a Knight Grand Cross of the Royal Victorian Order (GCVO) on 27 April 1954. Liberal leader Robert Menzies held office throughout Slim's time in Australia. His Official Secretary throughout his term was Murray Tyrrell.

### Retirement

In 1959, Slim retired and returned to Britain, where he published his memoirs, Unofficial History. He had already published his personal narrative of the Burma Campaign, Defeat into Victory, in 1956, which has never been out of print, and in which he candidly talked about his mistakes and the lessons he learned. On 24 April 1959, he was appointed a Knight Companion of the Order of the Garter (KG). On 15 July 1960, he was created "Viscount Slim of Yarralumla in the Capital Territory of Australia and of Bishopston in the City and County of Bristol".

After a successful further career on the boards of major UK companies, he was appointed Constable and Governor of Windsor Castle on 18 June 1964. He died in London on 14 December 1970, aged 79 and was given a full military funeral at St. George's Chapel, Windsor and was afterwards cremated.

## Allegations

During his tenure as Governor-General of Australia, Slim was patron of the Fairbridge Farm school child migration to Australia. Forty-seven years later, in 2007, three former child migrants alleged Slim sexually assaulted them during visits to Fairbridge Farm. The allegations were dismissed at the time both by those who had served under Slim in the army and by his son. The allegations were aired again on ABC television in the programme The Long Journey Home, broadcast on 17 November 2009. On 27 February 2017, the ABC again aired allegations against Slim and Fairbridge Farm.

Somehow or another, I was sat on his [Slim's] knee and, ah, um, these silky white hands were right up, because I was wearing shorts, right up my trousers and yeah, it was not, not very nice.

— Robert Stephens' claims of a drive in Governor-General Slim's car, 2017

## Historical assessment

Statue of General Slim on Whitehall.

Lieutenant General Sir John Kiszely has recommended Slim's memoirs (Defeat into Victory) (1956) in which he (Slim) candidly talks about his own mistakes in the Burma campaign. Kiszely has described Slim as "perhaps the Greatest Commander of the 20th Century" and has commenting on Slim's "self-deprecating style" Military historian Max Hastings stated:

In contrast to almost every other outstanding commander of the war, Slim was a disarmingly normal human being, possessed of notable self-knowledge. He was without pretension, devoted to his wife, Aileen, their family and the Indian Army. His calm, robust style of leadership and concern for the interests of his men won the admiration of all who served under him ... His blunt honesty, lack of bombast and unwillingness to play courtier did him few favours in the corridors of power. Only his soldiers never wavered in their devotion.

The spirit of comradeship Slim created within Fourteenth Army lived on after the war in the Burma Star Association, of which Slim was a co-founder and first President.

A statue of Slim on Whitehall, outside the Ministry of Defence, was unveiled by Queen Elizabeth II in 1990. Designed by Ivor Roberts-Jones, the statue is one of three British Second World War military leaders (the others being Alan Brooke and Bernard Montgomery).

Slim's papers were collected by his biographer, Ronald Lewin, and given to the Churchill Archives Centre by Slim's wife, Aileen, Viscountess Slim, and son, John Slim, 2nd Viscount Slim, and other donors, 1977–2001. Lewin's biography, Slim: The Standardbearer, was awarded the 1977 WH Smith Literary Award.

## List of honours

Arms as a Knight of the Garter
 Knight of the Order of the Garter (KG) 24 April 1959 Knight Grand Cross of the Order of the Bath (GCB) 2 January 1950 Knight Commander of the Order of the Bath (KCB) 28 September 1944 Companion of the Order of the Bath (CB) 1 January 1944 Knight Grand Cross of the Order of St Michael and St George (GCMG) 10 December 1952 Knight Grand Cross of the Royal Victorian Order (GCVO) 16 February 1954 Knight Grand Cross of the Order of the British Empire (GBE) 1 January 1946 Commander of the Order of the British Empire (CBE) 28 October 1942 Knight of the Order of St John (KStJ) 2 January 1953 Companion of the Distinguished Service Order (DSO) 14 January 1943 Military Cross (MC) 7 February 1918 Chief Commander of the Legion of Merit (United States)

## Bibliography

• Defeat into Victory by Field Marshal Sir William Slim; Originally published 1956. More recent editions are NY: Buccaneer Books ISBN 1-56849-077-1, Cooper Square Press ISBN 0-8154-1022-0; London: Cassell ISBN 0-304-29114-5, Pan ISBN 0-330-39066-X.
• Other publications include Courage and Other Broadcasts (1957); and Unofficial History (1959).

## Footnotes

1. ^ "No one called him William" (Hastings 2008, p. 68).
2. Heathcote 1999, p. 259.
3. ^ "The Papers of Field Marshal Lord Slim". Churchill Archives Centre. Retrieved 30 June 2020.{{cite web}}: CS1 maint: url-status (link)
4. ^ "No. 30183". The London Gazette (Supplement). 14 July 1917. p. 7076.
5. ^ a b "No. 30514". The London Gazette (Supplement). 7 February 1918. pp. 1800–1802.
6. ^ "No. 31993". The London Gazette (Supplement). 26 July 1920. p. 7835.
7. ^ "No. 31574". The London Gazette. 30 September 1919. p. 12035.
8. ^ a b Heathcote 1999, p. 260.
9. ^ "The succession of ministers in the Church of Scotland from the reformation". Retrieved 6 March 2018.
10. ^ "No. 33544". The London Gazette. 18 October 1929. p. 6620.
11. ^ "No. 33576". The London Gazette. 4 February 1930. pp. 726–727.
12. ^ "No. 33952". The London Gazette. 23 June 1933. p. 4205.
13. ^ "No. 34017". The London Gazette. 23 January 1934. p. 541.
14. ^ Lewin 1976, p. 50.
15. ^ Smart, p. 288
16. ^ "No. 34527". The London Gazette. 1 July 1938. p. 4249.
"No. 34547". The London Gazette. 2 September 1938. p. 5610.
17. ^ "No. 34625". The London Gazette. 12 May 1939. p. 3194.
"No. 34647". The London Gazette. 21 July 1939. p. 5036.
18. ^ "No. 34656". The London Gazette. 18 August 1939. p. 5676.
19. ^ "No. 34651". The London Gazette. 4 August 1939. pp. 5408–5409.
20. ^ a b Heathcote 1999, p. 261.
21. ^ Mead 2007, p. 418.
22. ^ Mead 2007, p. 419.
23. ^ "No. 35183". The London Gazette. 6 June 1941. p. 3243.
24. ^ "No. 35120". The London Gazette (Supplement). 1 April 1941. p. 1871.
"No. 35396". The London Gazette. 30 December 1941. p. 7353.
25. ^ "No. 35550". The London Gazette (Supplement). 5 May 1942. p. 2021.
26. ^ a b "No. 35763". The London Gazette (Supplement). 28 October 1942. p. 4687.
27. ^ a b "No. 35862". The London Gazette (Supplement). 14 January 1943. p. 321.
28. ^ Murray & Millet 2000, pp. 227–228.
29. ^ Murray & Millet 2000, pp. 227–229.
30. ^ Murray & Millet 2000, p. 228.
31. Weinberg 2005, p. 641.
32. ^ a b Murray & Millet 2000, p. 229.
33. ^ a b c d Murray & Millet 2000, p. 230.
34. ^ "No. 36331". The London Gazette. 14 January 1944. p. 319.
35. ^ Heathcote 1999, p. 262.
36. ^ a b Weinberg 2005, pp. 640–641.
37. ^ a b "No. 36309". The London Gazette (1st supplement). 31 December 1943. p. 4.
38. Murray & Millet 2000, p. 350.
39. ^ a b c d Weinberg 2005, p. 642.
40. ^ Murray & Millet 2000, p. 351.
41. ^ "No. 36642". The London Gazette (Supplement). 8 August 1944. p. 3659.
42. ^ a b "No. 36720". The London Gazette. 28 September 1944. p. 4473.
43. ^ "No. 36753". The London Gazette (Supplement). 17 October 1944. p. 4794.
44. Murray & Millet 2000, p. 489.
45. ^ Murray & Millet 2000, pp. 489–491.
46. ^ a b c Murray & Millet 2000, p. 492.
47. ^ Dower 1986, p. 53.
48. Murray & Millet 2000, p. 491.
49. ^ Croke 2015, pp. 255–276.
50. ^ Ford 2005, p. 470.
51. ^ McLynn 2011, p. 379.
52. ^ Dower 1986, p. 44.
53. ^ "The Burma Campaign 1941 – 1945". BBC. Retrieved 12 September 2015.
54. ^ a b Mead 2007, p. 425.
55. ^ Alanbrook in his diary entry of 17 May wrote "...(met) with Auk about appointment of Slim to Burma Command. Leese is going quite wild and doing mad things, prepared a fair rap on the knuckles for him!"
56. ^ "No. 37239". The London Gazette (Supplement). 28 August 1945. p. 4319.
57. ^ Slim 1961, p. 153.
58. ^ Fraser 1995, pp. 35–36.
59. ^ Fraser 1995, p. 37.
60. ^ Fraser 1995, pp. xiv.
61. ^ a b "No. 37407". The London Gazette. 28 December 1945. p. 17.
62. ^ "No. 37875". The London Gazette. 4 February 1947. p. 662.
63. ^ "No. 38289". The London Gazette. 14 May 1948. p. 2935.
"No. 38321". The London Gazette. 11 June 1948. p. 3470.
64. ^ a b Mead 2007, p. 426.
65. ^ "No. 38500". The London Gazette (Supplement). 31 December 1948. p. 49.
66. ^ "No. 38553". The London Gazette. 4 March 1949. p. 1127.
67. ^ a b "No. 38797". The London Gazette (Supplement). 2 January 1950. p. 2.
68. ^ a b "No. 38973". The London Gazette (Supplement). 21 July 1950. p. 3741.
69. ^ "No. 38359". The London Gazette. 20 July 1948. p. 4189.
70. ^ "No. 39706". The London Gazette (Supplement). 28 November 1952. p. 6331.
71. ^ a b "No. 39716". The London Gazette. 12 December 1952. p. 6544.
72. ^ "No. 39743". The London Gazette. 2 January 1953. p. 94.
73. ^ a b "No. 40159". The London Gazette. 27 April 1954. p. 2500.
74. ^ "National Library of Australia". Nla.gov.au. Retrieved 12 September 2015.
75. ^ a b "No. 41691". The London Gazette. 24 May 1959. p. 2687.
76. ^ "No. 42094". The London Gazette. 15 July 1960. p. 4925.
77. ^ "No. 43360". The London Gazette. 19 June 1964. p. 5337.
78. ^ Hill 2007, p. [page needed].
79. ^
80. ^
81. ^ "Inquiry to expose sexual abuse of British child migrants – 27/02/2017". Abc.net.au. 27 February 2017. Retrieved 29 March 2017.
82. ^ Lowrey, Tom (16 June 2019). "How renaming Canberra's William Slim Drive could trigger a rethink of history". ABC News. Australian Broadcasting Corporation. Retrieved 12 December 2021.
83. ^ Groch, Sherryn (6 June 2019). "William Slim Drive in Canberra to be renamed in wake of sexual abuse claims". Canberra Times. Retrieved 7 June 2019.
84. ^ Hayne, Jordan; Scott, Elise (6 June 2019). "William Slim Drive to be renamed over sexual abuse allegations against former governor-general". ABC News. Australian Broadcasting Corporation. Retrieved 6 June 2019.
85. ^ Bladen, Lucy (16 February 2021). "William Slim Drive renamed ahead of duplication works". The Canberra Times. Retrieved 17 December 2021.
86. ^ "Defence Management Journal, Issue 32". Archived from the original on 29 July 2012.
87. ^ Plaque in Slim's honour unveiled at Bristol cenotaph on 7 September 2008
88. ^ "Salvation Army". Where it is. Retrieved 12 September 2015.
89. ^ "A student's guide to Shrivenham" (PDF). Cranfield University. Archived from the original (PDF) on 21 October 2014. Retrieved 21 September 2015.
90. ^ "Slim School". Slim School. Retrieved 21 February 2016.
91. ^ "The Defence Academy: The Director's Reading List". Archived from the original on 23 January 2014.
92. ^ Hastings 2008, p. 69.
93. ^ Burma Star Association history Archived 11 May 2015 at the Wayback Machine
94. ^ "UK attractions: Statue of Viscount Slim". Archived from the original on 11 January 2008. Retrieved 26 January 2008.
95. ^ "Churchill Archives Centre in Cambridge". Archived from the original on 9 July 2012.
96. ^ "WH Smith Literary Award Winners". Good Reads. Retrieved 12 September 2015.
97. ^ "Governors-General". Archived from the original on 13 March 2014. Retrieved 13 March 2014.

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text:
wren ng thornton wren at freegeek.org
Mon May 25 22:04:01 EDT 2009

voigt at tcs.inf.tu-dresden.de wrote:
> > 2009/5/24 Petr Pudlak <deb at pudlak.name>:
> > If all Haskell had would be HM, it would be System F.
>
> That cannot be quite right, can it? System F has more powerful
> polymorphism than HM.

As I recall HM is along the edge to \lambda^2.

Haskell 98 is typically described in terms of System F_\omega. I've
heard people say GADTs are a subset of (or equivalent to?) System F_c,
though I'm not familiar with F_c myself. Like most interesting
languages, Haskell as a whole doesn't belong on one of the vertices.
I've yet to see a nice diagram that presents non-vertex languages like
HM and what arcs can cause one to (often unexpectedly) degenerate into
another.

--
Live well,
~wren

format: plain


text:
# The number of partitions between two fixed partitions

Given two partitions M and N, with $M_i \leq N_i$ for all $1\leq i\leq \max\{l(M),l(N)\}$. Is there a formula for the generating function: $$\sum_{\lambda: M_i\leq \lambda_i\leq N_i} q^{|\lambda|}$$

I remember seeing a determinant formula before, the elements of the determinant being q-binomial coefficients. But now I can't find it.

• Maybe this is only a minor technicality, but the upper bound on $i$ is throwing me off. Can you clarify? For example, between (2,1,1) and (2,2) are we allowing longer partitions such as (2,1,1,1) and/or larger partitions such as (2,2,1)? Or are there zero partitions in between these (presumably because of padding by zeros)? I would be happier if $l(M) \le l(N)$, but the way the upper bound on $i$ is written goes against this. Mar 24, 2014 at 8:44
• @PeterDukes Well, let's restrict our attention to what's called 'containment order', i.e. the containment of corresponding Ferrers diagram. (2,1,1) and (2,2) would not be comparable, under this order. Mar 26, 2014 at 5:00
• OK good. I am outside my comfort level here, but I agree this should be well known. Do you vaguely recall if the determinant was of some Hessenberg matrix? Mar 26, 2014 at 7:50
• Well i have found it now, see my answer below @PeterDukes Apr 4, 2014 at 4:27


text:
$$\require{cancel}$$

# 13.9.3: The Wien Bridge

• Page ID
5857
• Figure $$\text{XIII.11}$$

This bridge can be used for measuring frequency.

The reader will, I think, be able to show that

$\dfrac{R_4}{R_3} + \dfrac{C_3}{C_4} = \dfrac{R_2}{R_1} \tag{13.9.8}\label{13.9.8}$

and

$\omega^2 =\dfrac{1}{R_3R_4C_3C_4} \tag{13.8.10}\label{13.9.10}$


text:
# Probability Question

1. Feb 5, 2013

### GreenPrint

1. The problem statement, all variables and given/known data

Suppose
P(A|B) = P(A|B^c)
(P(B) > 0 and P(B^c) > 0 both understood). Show that A and B are independent.

2. Relevant equations

3. The attempt at a solution

I don't know where to go from here. Thanks for any help

$\frac{P(A \bigcap B)}{P(B)} = \frac{P(A \bigcap B^{c})}{P(B^{c})}$

Last edited: Feb 5, 2013
2. Feb 5, 2013

### HallsofIvy

Please restate (in fact, go back are reread) the problem. As you have it now it says "Suppose A and B are independent. Show that A and B are independent"!

3. Feb 5, 2013

### GreenPrint

Ok sorry about that I fixed that. Thanks for helping me.

4. Feb 5, 2013

### HallsofIvy

Okay, good. Now what is your definition of "independent events"?

5. Feb 5, 2013

### GreenPrint

P(A intersection B) = P(A)P(B)

not sure what to do from here because it's conditional probabilities.

6. Feb 5, 2013

### HallsofIvy

Good! So you are given that $P(A|B)= P(A|B^c)$ and want to prove that $P(A\cap B)= P(A)P(B)$.

You should know, then, that $P(A)= P(A|B)P(B)+ P(A|B^c)B^c$, for any A and B. Further, $P(B^c)= 1- P(B)$.

(The reason I asked about the definition was that some texts use "P(A)= P(A|B)" as the definition of "A and B are independent. Of course it is easy to show that is equivalent to your "$P(A\cap B)= P(A)P(B)$".)


text:
# Equation problem...Lost on it..

• Jan 31st 2010, 04:31 PM
victorfk06
Equation problem...Lost on it..
Solve the equation and express each solution in the form of A + Bi
X^3 – A =0
• Jan 31st 2010, 04:36 PM
VonNemo19
Quote:

Originally Posted by victorfk06
Solve the equation and express each solution in the form of A + Bi
X^3 – A =0

$x=\sqrt[3]a+0i$?
• Feb 1st 2010, 04:41 AM
HallsofIvy
That's one solution!

Now, victorfk06, divide $x^3- A$ by $x- \sqrt[3]{A}$ to get the remaining, quadratic, equation. Use the quadratic formula (or complete the square) to solve for the other two roots.


text:
# Projectile Motion

## What is called a Projectile Motion?

When a particle is thrown in space near the earth’s surface then it moves along a curved path under the influence of gravity, then its motion is called a projectile motion.

In a projectile motion –

1. Particle thus thrown is called a projectile.
2. The curved path traced by the projectile motion is called its trajectory.
3. The trajectory of a projectile motion is a parabolic curve shown by Galileo.
4. If the particle is thrown vertically upward, it will return back moving on same path. Then its trajectory will be a vertical straight line.
5. In the absence of air resistance, the motion of a projectile is considered as the combination of two independent motions – (1) motion along horizontal direction with constant velocity and (2) motion in vertical direction with constant acceleration of ‘g’. This is called “principle of independence of motions”.

EXAMPLES OF PROJECTILE MOTION –

1. A javelin thrown by a player.
2. A jet of water coming out from the side hole of a vessel.
3. A stone thrown horizontally from the top of a building.

## Principle of Independence of Motions

In the absence of air resistance, the motion of a projectile is considered as the combination of following two independent motions –

1. Motion along horizontal direction with uniform velocity.
2. Motion along vertical direction with uniform acceleration equal to acceleration due to gravity, ( g ) .

The above two motions ( i.e. motion in horizontal and vertical directions ) of a projectile are independent of each other. This is called the principle of physical independence of motions.

EXAMPLE –

The principle of physical independence of motion can be understood with the following example –

1. If a ball A is dropped downward from the roof of a building and at the same time from the same place another ball B is thrown in a horizontal direction, then both the balls will reach the ground at different places but at the same time.
2. It proves that the vertical motion of ball B is not affected by its horizontal motion.

## Angular projection from ground

Consider about a body which is projected with initial velocity ( u ) from a point O on the ground level at an angle ( \theta ) with the horizontal as shown in figure.

The initial velocity ( u ) of body has two rectangular components

1. Horizontal component ( u \cos \theta ) .
2. Vertical component ( u \sin \theta ) .

### Equation of Trajectory

The equation of the path traversed by a projectile is called equation of trajectory.

Suppose, the body reaches the point P after time ( t ) .

Horizontal motion has no acceleration. Thus, using kinematic equation, horizontal distance covered will be –

x = u \cos \theta t

Or, \quad t = ( \frac { x }{ u \cos \theta } )

Vertical motion has constant acceleration ( g ) . Thus, distance covered will be –

y = ( u \sin \theta )  t - \left ( \frac {1}{2} \right ) g t^2

= ( u \sin \theta ) \left ( \frac {x}{u \cos \theta} \right ) - \left ( \frac {1}{2} \right ) g \left ( \frac {x}{u \cos \theta} \right )^2

= \left ( \tan \theta \right ) x - \left ( \frac {g}{2 u^2 \cos^2 \theta} \right ) x^2

In this equation, ( \theta, \ u \ \text {and} \ g ) are constants. Thus,

1. Term \left ( \tan \theta \right ) is a constant, let it is ( p )
2. Term \left [ \left ( \frac {g}{2 u^2 \cos^2 \theta} \right ) \right ] is also a constant, let it is ( q )

So, \quad y = p x - q x^2

Therefore, ( y \propto x^2 ) , which is a required condition of a parabola.

Therefore, the trajectory of the projectile is a parabola.

### Time of Maximum height

As the body is projected it goes up. Vertical component of velocity ( u \sin \theta ) gradually diminishes and becomes zero at the maximum height of flight. After that, body starts moving downwards.

Let, ( t_m ) is the time to reach at maximum height ( h_m ) of flight.

Therefore, from kinematic equation, we have –

0 = u \sin \theta - g t_m

Or,  \quad t_m = \left ( \frac {u \sin \theta}{g} \right )

### Time of Flight

Total time taken by the projectile between the instant it is projected and till it reaches at a point in the horizontal plane of its projection is called Time of flight.

Let, the body reaches at point B on ground after time ( T_f ) of projection. Then –

Net vertical displacement covered during the time of flight is zero. Using kinematic equation of motion, we get –

0 = ( u \sin \theta ) T_f - \left ( \frac {1}{2} \right ) g \ ( T_f )^2

Or, \quad T_f = \left ( \frac {2 u \sin \theta}{g} \right ) = 2 \left ( \frac {u \sin \theta}{g} \right )

= 2 t_m

Thus, \quad \text {Total time of flight} = \text {Time of ascent} + \text {Time of descent}

= 2 \times \text {Time of maximum height.}

### Maximum height of Flight

It is the maximum height reached by a projectile. It is denoted by ( h_m )

At the highest point of flight, the vertical component of velocity becomes zero.

From kinematic equation of motion, we have –

v^2 = u^2 + 2 a s

Therefore, \quad 0^2 - ( u \sin \theta )^2 = 2 ( - g ) h_m

Or, \quad h_m = \left ( \frac {u^2 \sin^2 \theta}{2 g} \right )

### Horizontal Range

Horizontal distance traveled by a projectile during its total time of flight is called Horizontal Range.

\text {Horizontal range (R)} = \text {Horizontal velocity} \times \text {Time of flight}

So, \quad R = ( u \cos \theta ) \times 2 \left ( \frac {u \sin \theta}{g} \right ) = \left ( \frac {u^2}{g} \right ) ( 2 \sin \theta \cos \theta )

= \left ( \frac {u^2 \sin 2 \theta}{g} \right )

See numerical problems based on this article.

## Horizontal Projection from Height

Consider about a body projected horizontally with initial velocity ( u ) from a point O at a height ( h ) above the ground level as shown in figure.

The body is under the influence of two independent motions –

1. Uniform horizontal velocity ( u ) .
2. Vertical downward accelerated motion with constant acceleration ( g ) .

Under the combined effect of these two motions, the body moves along a curved path OPA .

### Trajectory of Projectile

After time ( t ) , suppose the body reaches at point P .

Acceleration in horizontal direction is zero and initial velocity is ( u ) . Using kinematic equation, the horizontal distance covered by the body will be –

x = u t

Or, \quad t = \left ( \frac {x}{u} \right )

Acceleration in vertical direction is ( g ) but initial velocity is zero. Using kinematic equation, vertical distance traveled will be –

y = 0 \times t + \left ( \frac {1}{2} \right ) g t^2

Putting the value of time \left ( t = \frac {x}{u} \right ) \quad , we get –

y = \left ( \frac {1}{2} \right ) g \left ( \frac {x}{u} \right )^2 = \left ( \frac {g}{2u^2} \right ) x^2

But, \left ( \frac {g}{2u^2} \right ) is constant because u \ \text {and} \ g are constants.

So, \quad y = k x^2

Therefore, ( y \propto x^2 ) , which is a required condition of a parabola.

Therefore, the trajectory of the projectile is a parabola.

### Time of Flight

Total time for which a projectile remains in space is called its time of flight.

Let, ( T ) is the time of flight.

Consider vertical motion of body. Using kinematic equation, we will get –

h = 0 \times T + \left ( \frac {1}{2} \right ) g T^2

Therefore, \quad T = \sqrt { \left ( \frac {2h}{g} \right ) }

### Horizontal Range

Total horizontal distance covered by the projectile during time of flight is called horizontal range. It is equal to OA = R .

\text {Horizontal range (R)} = \text {Horizontal velocity} \times \text {Time of flight}

Or, \quad R = u \times T

Putting the value of T = \left [ \sqrt { \left ( \frac {2h}{g} \right ) } \right ] , we will get –

R = u \sqrt { \left ( \frac {2h}{g} \right ) }

### Velocity at an instant

At any instant ( t ) , let the body is at point P and its velocity is ( v ) .

This velocity has two rectangular components

1. Horizontal component ( v_x = u )
2. Vertical component ( v_y = 0 + gt = gt )

Therefore, resultant velocity at point P is –

v = \sqrt { ( v_x )^2 + ( v_y )^2 } = \sqrt { ( u^2 + g^2 t^2 ) }

If the velocity ( v ) makes an angle ( \beta ) with the horizontal, then –

\tan \beta = \left ( \frac {v_y}{v_x} \right ) = \left ( \frac {g t}{u} \right )

Or, \quad \beta = \tan^{-1} \left ( \frac {g t}{u} \right )

See numerical problems based on this article.


text:
# A geometric realization of a 4-polytope with 7 vertices

I am looking for a geometric realization of a 4-polytope with 7 vertices. A list can be found in "An Enumeration of Simplicial 4-Polytopes with 8 vertices" by Grünbaum and V. P. Sreedharan. For an example we have a 4 polytope called $$P_1^7$$ with a list of facets: 1256, 1245, 1234, 1237, 1345, 1356, 1267, 1367, 2367, 2345, 2356, where a digit represents a vertex of $$P_1^7$$. My question would be: how should I choose 7 arbitrary, but indexed (what I mean is that I know which one is 1 for example) points on the sphere, so that the 4-polytope given by the convex hull of these points have the same combinatorial type as listed above? Is there any intelligent way to do this?

• It might be easier to consider the dual objects? Objects with larger numbers of vertices can constructed form smaller ones by a process of "slicing off" vertices (or edges, sometimes). – Donald Splutterwit Apr 21 at 14:00
• Are you asking in general, or for this specific example? – M. Winter Apr 22 at 20:49
• Thanks for the answers, I tried it with dual objects, but unfortunately I failed. For the second question, this specific example would be enough, but if there is a general method that would be more than enough. – Csaba Medgyes Apr 23 at 21:14

For a general approach, you can use the technique of Gale diagrams to transform a $$d$$-dimensional polytope with $$n$$ vertices into an $$(n-d-2)$$-dimensional arrangement of (signed) points with $$n$$ vertices. In your case, your have $$n-d-2=1$$, so the arrangement is of a very low dimension and can be handled with geoemtric intuition. The process also works the other way around too, and so to construct your $$4$$-dimensional polytope with seven vertices we can start from a 1-dimensional Gale diagram.

This is a rather technical approach, but I will lay out the general approach (the details can be found in Chapter 6 of Ziegler's "Lectures on Polytopes"). You will need to find a 1-dimensional arrangement of seven signed points:

• indexed by $$\{1,...,7\}$$,
• where signed means that each point is either colored black or white,
• and so that if $$\bar F\subseteq \{1,...,7\}$$ is the complement of a facet of your polytope (which in your case is always of size three) then the corresponding poins are two black points surrounding a white point, or vice versa.

The complements of your facets are

$$145,\;147,\;167,\;245,\;247,\;267,\;345,\;347,\;367,\;456,\;567.$$

You can check that the following diagram has the desired properties:

Technically, the diagram has to satisfy some further properties, and this one does, so I will not talk about them.

Now, there is a sequence of steps we have to perform to obtain the vertices of the polytope. I will not explain them but just perform them here:

1. We have to choose some actual coordinates for the shown points. Since the arrangement is 1-dimensional, each position is described by a single number. In the order of the indexing we can choose the coordinates $$6,7,1,2,3,4,5$$.

2. Now we lift the arrangement by one dimension and get rid of the signs: a black point at position $$x$$ is replaced by a vector $$(x,1)\in\Bbb R^2$$, and a white point is replaced by $$(-x,-1)\in\Bbb R^2$$. If we put all these vectors as columns into a matrix, we get

$$M:=\begin{bmatrix} 6 & 7 & -1 & 2 & -3 & 4 & -5 \\ 1 & 1 & -1 & 1 & -1 & 1 & -1 \end{bmatrix}\in\Bbb R^{2\times 7}.$$

1. Let $$U:=\mathrm{span}(M^\top)\subseteq\Bbb R^7$$ be the column span of the transpose $$M^\top\in\Bbb R^{7\times 2}$$. This is a $$2$$-dimensional subspace of $$\smash{\Bbb R^7}$$. Thus, if $$U^\bot$$ denotes its orthogonal complement, then $$U^\bot$$ is a $$5$$-dimensional subspace of $$\Bbb R^7$$ (this is the crucial "duality step"; we transitioned from an arrangement of dimension $$2=1+1$$ to an arrangement of dimension $$5=4+1$$).

2. Determine a basis of $$U^\bot$$. For example \begin{align} u_1 &= [-5,\phantom+6,1,0,0,0,0],\\ u_2 &= [\phantom+4,-5,0,1,0,0,0],\\ u_3 &= [-3,\phantom+4,0,0,1,0,0],\\ u_4 &= [\phantom+2,-3,0,0,0,1,0],\\ u_5 &= [-1,\phantom+2,0,0,0,0,1]. \end{align} We put these as rows into a matrix: $$\bar M = \begin{bmatrix} -5 & \phantom+6&1&0&0&0&0\\ \phantom+4&-5&0&1&0&0&0\\ -3&\phantom+4&0&0&1&0&0\\ \phantom+2&-3&0&0&0&1&0\\ -1&\phantom+2&0&0&0&0&1 \end{bmatrix}\in\Bbb R^{5\times 7}.$$

3. Read out the columns of that matrix, that is $$v_1=[-5,4,-3,2,-1],\;v_2=[6,-5,4,-3,2]\;\text{and}\;v_i=e_{j-2}\;\text{for 3\le i\le 7}.$$

4. This set of vectors is "acyclic", that is, there is a vector $$c\in\Bbb R^5$$ with $$\langle c,v_i\rangle >0$$ for all $$i\in\{1,...,7\}$$. This follows from a property of the Gale diagram that I have not mentioned (namely, being totally cyclic). One choice of such a vector is

$$c=(1,7,1,1,17)\in\Bbb R^5.$$

1. Finally, the vertices of your polytopes are the points $$p_i$$, where $$p_i$$ is the intersection of the ray $$\Bbb R v_i$$ with the hyperplane $$\langle c,\cdot\rangle=1$$ (which is 4-dimensional). Choosing an appropriate basis in this hyperplane gave me the following result:

$$p_1=\begin{bmatrix} -112\\-14\\0\\-56 \end{bmatrix},\; p_2=\begin{bmatrix} -184\\-16\\-4\\-86 \end{bmatrix},\; p_3=\begin{bmatrix} 7 \\ 0\\ 0 \\ 0 \end{bmatrix},\; p_4=\begin{bmatrix} 0\\7\\0\\0 \end{bmatrix},$$ $$p_5=\begin{bmatrix} 0 \\ 0\\7\\0 \end{bmatrix},\; p_6=\begin{bmatrix} 0\\0\\0\\7 \end{bmatrix},\; p_7=\begin{bmatrix} -119\\-7\\-7\\-49 \end{bmatrix}.$$

It is likely that you can find nicer coordinates if you make a better choice in any of the steps above (other coordinates for the diagram or a different basis for $$U^\bot$$). But at least all of the coordinates here are integers.

Here is Mathematica code that automates the Gale transformation:

vec = Transpose@NullSpace[{
{-3,-2, 1, 0,-1, 2,-3},
{ 1, 1,-1, 1,-1, 1,-1}
}]

sol = FindInstance[
Dot[#, {c1,c2,c3,c4,c5}]>0 &/@ vec,
{c1,c2,c3,c4,c5}
]
null = NullSpace[
{{c1,c2,c3,c4,c5}}] /. sol[[1]
]

pnts = Table[
Dot[v,n]/Dot[v,{1,1,1,1,1}],
{v,vec2}, {n,null}
]


text: New user? . Réponse préférée 1 ⁄ 0 = infinity = ∞ ... it is NOT undefined.... so infinity is obviously too big a value for any fixed display. should be the solution x of the equation It does not, however, make sense to ask for a "value" of this distribution at x = 0; a sophisticated answer refers to the singular support of the distribution. 2 x→0+limx1=+∞. 1 0 Again, any number multiplied by 0 is 0 and so this time every number solves the equation instead of there being a single number that can be taken as the value of 0/0. or The statement is true \color{#3D99F6}{\textbf{true}}true. If you're seeing this message, it means we're having trouble loading external resources on … 2 For any positive a, the limit from the right is. {\displaystyle 1/0=\infty } You can divide 1 by 0.091 to check that we got the right answer. 11 Answers. 1 divided by 0.1= 10 1 divided by 0.01=100 1 divided by 0.001=1000. ), if b ≠ 0 then the equation a/b = c is equivalent to a = b × c. Assuming that a/0 is a number c, then it must be that a = 0 × c = 0. Everybody told you that's undefined, but nobod y showed you WHY IS IT UNDEFINED: let's suppose the result of 1/0 is x; 1/0 = x . What is 1 divided by 0? In the hyperreal numbers and the surreal numbers, division by zero is still impossible, but division by non-zero infinitesimals is possible. End of long division (Remainder is 0 and next digit after decimal is 0). So for example, you take 0.1 divided by 0.1. = Similarly, to support division of any integer by any other, the realm of numbers must expand to the rational numbers. 1 So we say that division by zero is undefined, for it is not consistent with division by other numbers. The operation that you lears as 15 divided by 5 is really the multiplication : 5 * ? 1 divided by 0=infinity. The problem with this question is the "when". [8], The concept that explains division in algebra is that it is the inverse of multiplication. In two's complement arithmetic, attempts to divide the smallest signed integer by −1 are attended by similar problems, and are handled with the same range of solutions, from explicit error conditions to undefined behavior. The meaning of the expression − Well that's gonna be one. What number should be divided by (0.81)1/2 to give the result as 81? Considering the 10/0 example above, setting x = 10/0, if x equals ten divided by zero, then x times zero equals ten, but there is no x that, when multiplied by zero, gives ten (or any number other than zero). The Brāhmasphuṭasiddhānta of Brahmagupta (c. 598–668) is the earliest text to treat zero as a number in its own right and to define operations involving zero. a This set is analogous to the projectively extended real line, except that it is based on the field of complex numbers. ∞ The IEEE floating-point standard, supported by almost all modern floating-point units, specifies that every floating point arithmetic operation, including division by zero, has a well-defined result. 1.62 divided by 0.8 16.2 divided by 8 0.0162 divided by 0.008 0.162 divided by 0.08 There are actually two different ways to complete the expressions above with the given numbers so that each expression has the same value. What is 1.0 divided by 8? In matrix algebra (or linear algebra in general), one can define a pseudo-division, by setting a/b = ab+, in which b+ represents the pseudoinverse of b. So 10/0, at least in elementary arithmetic, is said to be either meaningless, or undefined. The set ∞ {\displaystyle \textstyle {\frac {a}{b}}} . 1 month ago; RT @ArcadeDaydream: If you remember using Silicon Graphics’ Irix Unix OS fondly, check out MaXX Desktop for multiple Linux distributions. This infinity can be either positive, negative, or unsigned, depending on context. It can be proven that if b−1 exists, then b+ = b−1. In the zero ring, division by zero is possible, which shows that the other field axioms are not sufficient to exclude division by zero in a field. Well that's gonna be one. Any number system that forms a commutative ring—for instance, the integers, the real numbers, and the complex numbers—can be extended to a wheel in which division by zero is always possible; however, in such a case, "division" has a slightly different meaning. For example,[9], since 2 is the value for which the unknown quantity in, requires a value to be found for the unknown quantity in. Log in to reply to the answers Post; Steve . ∞ {\displaystyle -\pi /2} Log in here. 1 Answer sente Mar 16, 2016 5. math. / is only shorthand for the formal expression ab−1, where b−1 is the multiplicative inverse of b. Nevertheless, a (non-rigorous) justification can be given in this setting. Some calculators, the online Desmos calculator is one example, allow arctangent(1/0). b So there are situations where 10\frac1001 is defined, but they are defined in a tightly controlled way. The mathematical justification is that the limit as x goes to zero of arctangent 1/x is This equation has two distinct solutions, x = 1 and x = 4, so the expression If b equals 0, then b+ = 0. answers something/0:. Zero divided by zero is zero. If instead of x = 10/0, x = 0/0, then every x satisfies the question 'what number x, multiplied by zero, gives zero?'. Let's get super close to zero: 0.000001 divided by 0.000001. Here's why: Remember that a b \frac{a}{b} b a means … The above explanation may be too abstract and technical for many purposes, but if one assumes the existence and properties of the rational numbers, as is commonly done in elementary mathematics, the "reason" that division by zero is not allowed is hidden from view. During this gradual expansion of the number system, care is taken to ensure that the "extended operations", when applied to the older numbers, do not produce different results. Although division by zero cannot be sensibly defined with real numbers and integers, it is possible to consistently define it, or similar operations, in other mathematical structures. {\mathbb C} \cup \{\infty\}.C∪{∞}. _\square There are some common responses to this logic, but they all have various flaws. Mettre à jour: I tried it on calculator and it said ERROR. However, the resulting algebraic structure is not a field, and should not be expected to behave like one. → First, the natural numbers (including zero) are established on an axiomatic basis such as Peano's axiom system and then this is expanded to the ring of integers. Arrggh! 0 divided by 0 is not defined, although one could define it … See the consequences of assuming that 10\frac{1}{0}01 is defined for yourself in the following problem: What is wrong with the following "proof"? So if 1 divided by zero is infinite. Certain words can be pinpointed in the question to highlight the problem. } How do you divide rational numbers? 1 divided by 0 is not 0, nor 0.1/0 or 0.01/0 etc. A logically rigorous (as opposed to formal) computation would assert only that, Since the one-sided limits are different, the two-sided limit does not exist in the standard framework of the real numbers. [4] Similarly, when the realm of numbers expands to include the rational numbers, division is replaced by multiplication by certain rational numbers. More p… 1 month ago 0 The infinity signs change when dividing by −0 instead. Some programs (especially those that use fixed-point arithmetic where no dedicated floating-point hardware is available) will use behavior similar to the IEEE standard, using large positive and negative numbers to approximate infinities. Because there's just no sensible way to define it. There are two zeroes: +0 (positive zero) and −0 (negative zero) and this removes any ambiguity when dividing. Historically, one of the earliest recorded references to the mathematical impossibility of assigning a value to a/0 is contained in George Berkeley's criticism of infinitesimal calculus in 1734 in The Analyst ("ghosts of departed quantities").[1]. Solve the inequality W > Y plus H all divided by P for H. W divided by P – Y > H W times P divided by Y > H WP – Y > H W + P – Y > H . ∞ 2 205 ÷ 2 = 102.5 Lv 7. is undefined (the limit is also undefined for negative a). is undefined in this extension of the real line. Well, that also equals one. Math and Arithmetic. Pertinence. \lim\limits_{x \to 0^-} \frac{1}{x} = - \infty. / Well, that also equals one. For instance, in the realm of integers, subtraction is no longer considered a basic operation since it can be replaced by addition of signed numbers. This impossibility was first noted in philosopher George Berkeley's [4] … Therefore as the denominator becomes smaller, the result of the equation becomes greater. The problem with 5 cookies and 0 people, on the other hand, cannot be solved in any way that preserves the meaning of "divides". Most calculators will either return an error or state that 1/0 is undefined; however, some TI and HP graphing calculators will evaluate (1/0)2 to ∞. R SUBSCRIBE!!! / For instance, suppose a,b,c,da,b,c,da,b,c,d are complex numbers such that ad−bc≠0. ad-bc\ne 0.ad−bc=0. Integer division by zero is usually handled differently from floating point since there is no integer representation for the result. a firnd made a calculator in his programing class and forgot to put in safty catches, so when he divided by zero the pc crashed! - Dr. Robert. 0 It is even better if the kids can make sense out of it! {\displaystyle \textstyle {\frac {2}{2}}} {\displaystyle a/\infty =0} It is still the case that 10\frac1001 can never be a real (or complex) number, so—strictly speaking—it is undefined. The disguised division by zero occurs since x − 1 = 0 when x = 1. We are assuming that we can divide by zero, so 0/0 should work the same as 5/5, which is 1). When you divide by 1 the answer stays the same. Why some people say it's 0: Zero divided by any number is 0. Log in. Since the field axioms only guarantee the existence of such inverses for nonzero elements, this expression has no meaning when b is zero. b It is the natural way to view the range of the tangent function and cotangent functions of trigonometry: tan(x) approaches the single point at infinity as x approaches either Any number divided by itself equals 1. ex: 24 / 24 = 1 and 2,154,378,549,215,044.32158 / 2,154,378,549,215,044.32158 = 1. Divided By What Equals Calculator Please enter another problem for us to solve below: In keeping with this change of viewpoint, the question, "Why can't we divide by zero? In computing, a program error may result from an attempt to divide by zero. The reason 0/0 is undefined is because it's an Indeterminate form, not because of our inability to calculate it. Example: Let's get even closer to zero: 0.001 divided by 0.001. So, for dividing by zero, what is the number of cookies that each person receives when 10 cookies are evenly distributed amongst 0 people at a table? Lv 5. Answering this revised question precisely requires close examination of the definition of rational numbers. Dividing by 1, 10 or 100. {\displaystyle {\tfrac {\pi }{2}}} This relation is shown to be an equivalence relation and its equivalence classes are then defined to be the rational numbers. {\displaystyle \textstyle {\frac {1}{x}}} The set 0 × ( 1 / 0) = 0. Why some people say it's 1: A number divided by itself is 1. Understand the mathematics of continuous change. This set has the geometric structure of a sphere, called the Riemann sphere. Why some people say it's false: 10=∞.\frac10 = \infty.01=∞. is an unsigned infinity – or, as it is often called in this context, the point at infinity. https://www.youtube.com/HaxHatcherFollow me on twitter! Geronimo. . The justification for this definition is to preserve the sign of the result in case of arithmetic underflow. { to a distribution on the whole space of real numbers (in effect by using Cauchy principal values). You might be wondering after seeing these answers. ∞ Sign up to read all wikis and quizzes in math, science, and engineering topics. \lim\limits_{x\to 0}\frac{1}{x}.x→0limx1. Maple and SageMath return an error message for 1/0, and infinity for 1/0.0 (0.0 tells these systems to use floating point arithmetic instead of algebraic arithmetic). It is true that, in some situations, the indeterminate form 10\frac1001 can be interpreted as ∞: \infty:∞: for instance, when taking limits of a quotient of functions. 1 divided by infinity: In this case, if we divide a small number with a large number, the result gets very close to zero. If there are, say, 5 cookies and 2 people, the problem is in "evenly distribute". Zero divided by a negative or positive number is either zero or is expressed as a fraction with zero as numerator and the finite quantity as denominator. Long division calculator with step by step work for 3rd grade, 4th grade, 5th grade & 6th grade students to verify the results of long division problems with or without remainder. {\displaystyle \infty } For instance, to make it possible to subtract any whole number from another, the realm of numbers must be expanded to the entire set of integers in order to incorporate the negative integers. Explanation: #1/0.2 = (0.2+0.2+0.2+0.2+0.2)/0.2# #= (5*0.2)/0.2# #= 5*0.2/0.2# #= 5*1# #=5# Answer link. ∪ Thus, the answer to "1 divided by what equals 11?" Some modern calculators allow division by zero in special cases, where it will be useful to students and, presumably, understood in context by mathematicians. There are 10mm in 1cm, so 124 divided by 10 will give you your answer of 12.4cm The limit. One, you could start taking numbers closer and closer to zero and dividing them by themselves. Such a division can be formally expressed as .mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;white-space:nowrap}a/0 where a is the dividend (numerator). Write the remainder after subtracting the bottom number from the top number. In some programming languages, an attempt to divide by zero results in undefined behavior. 2 First, infinity is not a real number. This is part of a series on common misconceptions. Microsoft Math and Mathematica return ComplexInfinity for 1/0. One, you could start taking numbers closer and closer to zero and dividing them by themselves. Let's get even closer to zero: 0.001 divided by 0.001. Conclusion: By substituting in a=b=1, a = b = 1,a=b=1, we have 1+1=1  ⟹  2=1.1+1 = 1 \implies 2 = 1.1+1=1⟹2=1. Thus, the answer to "1 divided by what equals 4?" Each person would receive 10/5 = 2 cookies. Home Science Math History Literature Technology Health Law Business All Topics Random. we know, 0.81 = 0.9 × 0.9 = (0.9)² . Only one of these explanations is valid, and choosing the other explanations can lead to serious contradictions. ∪ Technically 1 divided by infinite would be zero. When division is explained at the elementary arithmetic level, it is often considered as splitting a set of objects into equal parts. 21 ÷ 1 = 21; When you divide by 10, move all the digits one place to the right. Note that our answers are rounded to the nearest thousandth if necessary. ∞ For example, formally: As with any formal calculation, invalid results may be obtained. Already have an account? are undefined. 0 The graphical programming language Scratch 2.0 and 3.0 used in many schools returns Infinity or −Infinity depending on the sign of the dividend. 7 years ago. 1.24 divided by 0.04 is 31. 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# Tellegen Theorem

This theorem has been introduced in the year of 1952 by Dutch Electrical Engineer Bernard D.H. Tellegen. This is a very useful theorem in network analysis. According to Tellegen theorem, the summation of instantaneous powers for the n number of branches in an electrical network is zero. Are you confused? Let’s explain. Suppose n number of branches in an electrical network have i1, i2, i3, …………. in respective instantaneous currents through them. These currents satisfy Kirchhoff’s Current Law.

Again, suppose these branches have instantaneous voltages across them are v1, v2, v3, ……….. vn respectively. If these voltages across these elements satisfy Kirchhoff Voltage Law then,

vk is the instantaneous voltage across the kth branch and ik is the instantaneous current flowing through this branch. Tellegen theorem is applicable to lumped networks that consist of linear, non-linear, time variant, time-invariant, and active and passive elements.

This theorem can easily be explained by the following example.

In the network shown, arbitrary reference directions have been selected for all of the branch currents, and the corresponding branch voltages have been indicated, with positive reference direction at the tail of the current arrow.
For this network, we will assume a set of branch voltages satisfy the Kirchhoff voltage law and a set of branch current satisfy Kirchhoff current law at each node.

We will then show that these arbitrary assumed voltages and currents satisfy the equation.

And it is the condition of Tellegen theorem.
In the network shown in the figure, let v1, v2 and v3 be 7, 2 and 3 volts respectively. Applying Kirchhoff Voltage Law around loop ABCDEA. We see that v4 = 2 volt is required. Around loop CDFC, v5 is required to be 3 volt and around loop DFED, v6 is required to be 2. We next apply Kirchhoff’s Current Law successively to nodes B, C and D.
At node B let ii = 5 A, then it is required that i2 = – 5 A. At node C let i3 = 3 A and then i5 is required to be – 8. At node D ssume i4 to be 4 then i6 is required to be – 9. Carrying out the operation of equation,

We get,
Hence Tellegen theorem is verified.


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# Hoffman-Wielandt Theorem Proof

Exercise 3.3 of Izenman's Modern Multivariate Statistical Techniques: let $\mathbf{A}$, $\mathbf{B}$ be symmetric $J \times J$ matrices, with eigenvalues $\{\lambda_j(\mathbf{A})\}$ and $\{\lambda_j(\mathbf{B})\}$ respectively, arranged in descending order with respect to $j$ (so $\lambda_1$ is largest, $\lambda_J$ is the smallest for both matrices). Prove that $$\sum_{j=1}^{J}\left[\lambda_j(\mathbf{A}) - \lambda_j(\mathbf{B})\right]^2 \leq \text{tr}\{(\mathbf{A}-\mathbf{B})(\mathbf{A}-\mathbf{B})^{T}\}\text{.}$$ The hint says to use spectral decomposition, so \begin{align*} \mathbf{A} &= \sum_{j=1}^{J}\lambda_j(\mathbf{A})\mathbf{v}_j(\mathbf{A})\mathbf{v}^T_j(\mathbf{A}) \\ \mathbf{B} &= \sum_{j=1}^{J}\lambda_j(\mathbf{B})\mathbf{v}_j(\mathbf{B})\mathbf{v}^T_j(\mathbf{B}) \end{align*} where the $\mathbf{v}_j(\cdot)$ denote the eigenvectors of the matrix $\cdot$ corresponding to $\lambda_j$. Then it says to express $$\text{tr}\{(\mathbf{A}-\mathbf{B})(\mathbf{A}-\mathbf{B})^{T}\}$$ in terms of the decomposition. I have $$\mathbf{A}-\mathbf{B} = \sum_{j=1}^{J}[\lambda_j(\mathbf{A})\mathbf{v}_j(\mathbf{A})\mathbf{v}^T_j(\mathbf{A})-\lambda_j(\mathbf{B})\mathbf{v}_j(\mathbf{B})\mathbf{v}^T_j(\mathbf{B})]$$ and $$(\mathbf{A}-\mathbf{B})^{T} = \mathbf{A}^{T}-\mathbf{B}^{T} = \sum_{j=1}^{J}[\lambda_j(\mathbf{A})\mathbf{v}^T_j(\mathbf{A})\mathbf{v}_j(\mathbf{A})-\lambda_j(\mathbf{B})\mathbf{v}^T_j(\mathbf{B})\mathbf{v}_j(\mathbf{B})]\tag{1}\text{.}$$ I suppose we could assume the vectors are normalized, so we get $\mathbf{v}^T_j(\mathbf{A})\mathbf{v}_j(\mathbf{A}) = \mathbf{v}^T_j(\mathbf{B})\mathbf{v}_j(\mathbf{B}) = 1$. But I'm not sure what else to do. Direct multiplication looks like a very messy approach (which would possibly involve induction on $J$), but I thought I'd ask here for suggestions.

• The statement is ambiguous (if not wrong) if the order of the eigenvalues are not specified. E.g. if $A=B=\operatorname{diag}(1,2)$ but the eigenvalues of $A$ and $B$ are arranged in opposite order, then the inequality would imply that $2\le 0$. Things also get complicated if the eigenvalues do not have the same signs. Mar 24, 2016 at 15:41
• @user1551 Thanks for catching that. It so happens that $\lambda_1$ is used as the largest eigenvalue and $\lambda_J$ is used as the smallest eigenvalue. (This wasn't directly in the statement of the theorem in this text, but was somewhere else.) Mar 24, 2016 at 15:43
• @user1551 The post has been edited. Mar 24, 2016 at 15:44

Symmetric matrices are orthogonally diagonalisable. Let $A=U\Lambda U^T$ and $B=V\Sigma V^T$, where $U,V$ are real orthogonal and the eigenvalues in the two diagonal matrices $\Lambda,\Sigma$ are arranged in descending order. Let also $W=U^TV$. Then the inequality in question is equivalent to $$\operatorname{tr}\left((\Lambda-\Sigma)^2\right) \le \operatorname{tr}\left((U\Lambda U^T-V\Sigma V^T)^2\right).\tag{1}$$ Expand the square terms on both sides, we may in turn rewrite $(1)$ as $$\operatorname{tr}(\Lambda W\Sigma W^T) \le \operatorname{tr}(\Lambda\Sigma).\tag{2}$$ It is well-known that the LHS of the above inequality is maximised when $W=I$ (and therefore the inequality is true). To see this, let $S$ be the entrywise square of the real orthogonal matrix $W$. Then $S$ is a doubly stochastic matrix and the LHS is equal to $\sum_{i,j}\lambda_i\sigma_js_{ij}$, which is a linear function in the entries of $S$. By Birkhoff-von Neumann theorem, the set of all doubly stochastic matrices is the convex hull of all permutation matrices. Therefore the LHS of $(2)$ is maximised when $W$ is a permutation matrix. It is easy to see that among all permutation matrices, $W=I$ gives the global maximum.


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# Combining reinforcement learning with rule-based controllers for transparent and general decision-making in autonomous driving

Amarildo Likmeta, Alberto Maria Metelli, Andrea Tirinzoni, Riccardo Giol, Marcello Restelli, and Danilo Romano

Robotics and Autonomous Systems, 2020.

Abstract
The design of high-level decision-making systems is a topical problem in the field of autonomous driving. In this paper, we combine traditional rule-based strategies and reinforcement learning (RL) with the goal of achieving transparency and robustness. On the one hand, the use of handcrafted rule-based controllers allows for transparency, i.e., it is always possible to determine why a given decision was made, but they struggle to scale to complex driving scenarios, in which several objectives need to be considered. On the other hand, black-box RL approaches enable us to deal with more complex scenarios, but they are usually hardly interpretable. In this paper, we combine the best properties of these two worlds by designing parametric rule-based controllers, in which interpretable rules can be provided by domain experts and their parameters are learned via RL. After illustrating how to apply parameter-based RL methods (PGPE) to this setting, we present extensive numerical simulations in the highway and in two urban scenarios: intersection and roundabout. For each scenario, we show the formalization as an RL problem and we discuss the results of our approach in comparison with handcrafted rule-based controllers and black-box RL techniques.

[Paper] [BibTeX]

 @article{likmeta2020combining,
author = "Likmeta, Amarildo and Metelli, Alberto Maria and Tirinzoni, Andrea and Giol, Riccardo and Restelli, Marcello and Romano, Danilo",
title = "Combining reinforcement learning with rule-based controllers for transparent and general decision-making in autonomous driving",
journal = "Robotics and Autonomous Systems",
volume = "131",
pages = "103568",
year = "2020",
issn = "0921-8890",
doi = "https://doi.org/10.1016/j.robot.2020.103568",
url = "http://www.sciencedirect.com/science/article/pii/S0921889020304085",
keywords = "Autonomous driving, Decision making, Interpretability, Reinforcement learning, Parameter-based exploration"
}


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# Food as Energy

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“ Coley, D., Goodliffe, E. and Macdiarmid, D. (1998) The Embodied Energy of Food: The Role of Diet, Energy Policy, 6, 455–459.[\note]: Baked beans $r \approx$ 10 Sausages $r \approx$ 10 Yoghurt $r \approx$ 9 Beef $r \approx$ 8 Pork $r \approx$ 6 Eggs $r \approx$ 6 Pasta $r \approx$ 4 Rice $r \app…” Basal metabolism We need a constant input of food energy to keep our internal organs functioning and to allow us to move – this is called our basal metabolism. For mammals, Mb ≈ 50 (A/m2) W; for humans, surface area A ≈ 1.7 m2, i.e. ~ 85 W for sleeping Fast walk or cycle (20 km/h): M ~ 700 W Overall mean M ~ 100 – 150 W (2000 – 3000 kcal/d) Where does this energy come from? Most of the energy required for agriculture comes from sunlight, right? Wrong. Most of the energy comes from burning oil. The usable energy content of beef and bread is about 2500 kcal/kg or 10 MJ/kg (+/- 30%). For chicken and fish the number is about half that [1] Beef calories and nutrients, https://www.freedieting.com/calories-in-meat [2019-10-04]. [2] Peter Rez, Simple Physics of Energy Use, Oxford UP 2017. Overall for a typical Western diet, the ratio of embodied energy of the food to the dietary energy in the food$r$is about 6, i.e. for every joule (or calorie) of dietary energy, six joules (or calories) of oil have been burned. The ratio for vegetables is generally more favourable; that for meat (especially beef) is much worse. However, the picture is complicated, as can be seen from a British study by David Coley et al.[3] Coley, D., Goodliffe, E. and Macdiarmid, D. (1998) The Embodied Energy of Food: The Role of Diet, Energy Policy, 6, 455–459.[\note]: • Baked beans$r \approx$10 • Sausages$r \approx$10 • Yoghurt$r \approx$9 • Beef$r \approx$8 • Pork$r \approx$6 • Eggs$r \approx$6 • Pasta$r \approx$4 • Rice$r \approx$3 • Milk$r \approx$3 • Potatoes$r \approx$2 • Bread$r \approx$1 Note however, that these figures are highly variable, depending on circumstances. What about Animals? The metabolic rate,$\Gamma$, for animals is related to the mass of the animal by Eqn.1:$\Gamma = 4 M^{3/4} \tag{1}$Such a relation where a physical property of an animal scales with mass to some power that is not unity is called an allometric relation[4] Ahlborn, Zoological Physics, p.12-16 , Springer, 2004. It is obvious that an elephant which is large and consumes vast amounts of food will have a larger metabolic rate than a mouse that is much less massive and consumes much less. Considering, however, the metabolic rate per unit mass of these animals (Eqn.2):$\dfrac{\Gamma}{M} = 4 M^{-1/4} \tag{2}$we find to sustain itself an elephant needs less energy per unit mass than a mouse. This explains why even though elephants eat a larger quantity of food compared to mice, mice actually need to eat more compared to their body weight than elephants. Our goal is to understand how heat loss contributes to smaller animals having a higher metabolic rate per unit mass than larger animals. As mentioned some of the energy generated inside an animal goes to heat which is radiated out through the surface area of the animal. The rate of heat loss is proportional to the surface area, while the mass of an animal is proportional to its volume. The rate at which heat is radiated per unit mass then is proportional to surface area per unit volume. For simplicity, if we take an animal to be spherical with radius,$r$, then we can relate the heat loss per unit mass to size (Eqn.3):$\frac{4 \pi r^{2}}{\frac{4}{3} \pi r^{3}} \propto \dfrac{1}{r} \tag{3}$As animals increase in size the rate of heat loss per unit mass decreases and less energy is needed to sustain the animal per unit mass. It might appear an obvious advantage then to be a large animal. Large animals, however, still consume more energy on the whole than smaller animals so that they need more food. If food is not abundant larger animals cannot sustain themselves[5] Ahlborn, Zoological Physics, p.12-16 , Springer, 2004. In conclusion, larger animals have a smaller heat loss per unit mass and require less food per unit mass even though they eat larger amounts than smaller animals. Bicycling and Calories How long do I have to ride my bicycle to burn off a doughnut? This example looks at the relationship between physical exercise and calorie consumption. In magazine articles and on the web, you often find numbers relating exercise and calorie consumption. In this example, we will explore where the numbers come from. For example, you find that a 68 kg person cycling at 15 km/h for one hour burns approximately 400 Calories[note]Many web-based resources. See for example: www.dietandfitnesstoday.com or www.nutristrategy.com,[6]Knight, Jones, Field, “College Physics: a strategic approach”, 1st edition, p. 339, Pearson Addison-Wesley (2007) (corresponding to approximately 1.5 – 2.0 donuts[7]Tim Hortons Nutritional Information, https://www.timhortons.com/ca/en/menu/nutrition-and-wellness.php [2019-10-16].). Let’s check. Our calculation is based on the principle that the energy in food is like other forms of energy, so it can be transformed into mechanical energy. However, like other real-life engines, our body cannot transfer 100% of the chemical energy in food into mechanical energy (motion). We need two additional bits of information: the amount of energy that the body consumes during sedentary periods (typically 100 W; this is mostly needed to keep the internal organs functioning) and the efficiency to convert chemical energy into mechanical energy (approximately 25%). The two numbers given above vary somewhat with the person and the activity but can conveniently be used to get a good estimate. So why does bicycling consume energy, even on a flat road and a day without wind? You have to overcome rolling friction and air drag, since the relative wind speed is 15 km/h. Assumptions: • level road, no wind (relative wind speed = 15 km/h) • 68 kg person plus 10 kg bike • rolling friction: rubber on concrete: μ r = 0.02 (table 5.1[fn value=2][/fn]). Calculations: • Force due to air drag: FD = ¼ ρ A v 2, with density of air ρ = 1.28 kg/m 3 (The simplified equation above follows from the more general formula for air drag FD = ½ ρ CD A v 2 by using CD = 0.5, which is typical for everyday moving objects.) We need to estimate the frontal area A of person (+ bike) A = (1.5 m) · (0.6 m) = 0.9 m 2. v = 4.2 m/s, so the distance traveled is d = 4.2 m each second. So F = 5.1 N and the work done is W = F d = 21 J, so P = 21 W • Rolling friction: f r = μ r m g = (0.02) · (68 kg+10 kg) · (9.8 m/s2) = 15.3 N, so the work done by rolling friction is W = (15.3) · (4.2 m) = 64 J and the corresponding power is P = 64 W. • So we need a total of 85 W of mechanical power to maintain our speed. • Now we want to relate this to the total metabolic power. We need to multiply our result by 4 and add 100 W to account for our efficiency and the energy consumed by our internal organs. Result = 440 W. • Finally, we have to perform a unit conversion to relate our result to food calories: 1 Cal = 1 kcal = 4200 J. • 440 W = 440 J/s correspond to (440 J/s) · (3600 s/h) = 1.58 MJ/h or 377 Calories burned in one hour. • You can replenish your energy by eating almost two doughnuts (62 g each) or six timbits (17 g each). Interpretation: Our result is close to the published value showing that the calorie consumption in bicycling is mainly due to overcoming air drag and rolling friction. Finally, how long will someone have to bike to burn off eating a donut? A chocolate glazed donut from Tim Horton’s has 260 Calories and is 70 g. If you eat one it will take you (260 Cal)(1 h/377 Cal)(60 min/h) = 41 min of biking at 15 km/h to burn it off. Updated (CEW) 2019-10-16 Footnotes  ↑1 Beef calories and nutrients, https://www.freedieting.com/calories-in-meat [2019-10-04]. ↑2 Peter Rez, Simple Physics of Energy Use, Oxford UP 2017. ↑3 Coley, D., Goodliffe, E. and Macdiarmid, D. (1998) The Embodied Energy of Food: The Role of Diet, Energy Policy, 6, 455–459.[\note]: Baked beans$r \approx$10 Sausages$r \approx$10 Yoghurt$r \approx$9 Beef$r \approx$8 Pork$r \approx$6 Eggs$r \approx$6 Pasta$r \approx$4 Rice$r \approx$3 Milk$r \approx$3 Potatoes$r \approx$2 Bread$r \approx$1 Note however, that these figures are highly variable, depending on circumstances. What about Animals? The metabolic rate,$\Gamma$, for animals is related to the mass of the animal by Eqn.1:$\Gamma = 4 M^{3/4} \tag{1}$Such a relation where a physical property of an animal scales with mass to some power that is not unity is called an allometric relation[note] Ahlborn, Zoological Physics, p.12-16 , Springer, 2004 ↑4 Ahlborn, Zoological Physics, p.12-16 , Springer, 2004. It is obvious that an elephant which is large and consumes vast amounts of food will have a larger metabolic rate than a mouse that is much less massive and consumes much less. Considering, however, the metabolic rate per unit mass of these animals (Eqn.2):$\dfrac{\Gamma}{M} = 4 M^{-1/4} \tag{2}$we find to sustain itself an elephant needs less energy per unit mass than a mouse. This explains why even though elephants eat a larger quantity of food compared to mice, mice actually need to eat more compared to their body weight than elephants. Our goal is to understand how heat loss contributes to smaller animals having a higher metabolic rate per unit mass than larger animals. As mentioned some of the energy generated inside an animal goes to heat which is radiated out through the surface area of the animal. The rate of heat loss is proportional to the surface area, while the mass of an animal is proportional to its volume. The rate at which heat is radiated per unit mass then is proportional to surface area per unit volume. For simplicity, if we take an animal to be spherical with radius,$r$, then we can relate the heat loss per unit mass to size (Eqn.3):$\frac{4 \pi r^{2}}{\frac{4}{3} \pi r^{3}} \propto \dfrac{1}{r} \tag{3}\$ As animals increase in size the rate of heat loss per unit mass decreases and less energy is needed to sustain the animal per unit mass.  It might appear an obvious advantage then to be a large animal. Large animals, however, still consume more energy on the whole than smaller animals so that they need more food. If food is not abundant larger animals cannot sustain themselves[note] Ahlborn, Zoological Physics, p.12-16 , Springer, 2004 ↑5 Ahlborn, Zoological Physics, p.12-16 , Springer, 2004. In conclusion, larger animals have a smaller heat loss per unit mass and require less food per unit mass even though they eat larger amounts than smaller animals.   Bicycling and Calories How long do I have to ride my bicycle to burn off a doughnut? This example looks at the relationship between physical exercise and calorie consumption. In magazine articles and on the web, you often find numbers relating exercise and calorie consumption. In this example, we will explore where the numbers come from. For example, you find that a 68 kg person cycling at 15 km/h for one hour burns approximately 400 Calories[note]Many web-based resources. See for example: www.dietandfitnesstoday.com or www.nutristrategy.com ↑6 Knight, Jones, Field, “College Physics: a strategic approach”, 1st edition, p. 339, Pearson Addison-Wesley (2007) ↑7 Tim Hortons Nutritional Information, https://www.timhortons.com/ca/en/menu/nutrition-and-wellness.php [2019-10-16].