Verse 16 · The Rules for finding the true places of the planets. · Surya Siddhanta · siva.sh

siva.sh/surya-siddhanta/2/16

आद्येनैवम् क्रमात् पिण्डान् भक्त्वा *लब्धोनसम्युताः ॥(C लब्धोनितैर् युतैः) *खण्डकाः स्युश् चतुर्विम्शज्यार्धपिण्डाः क्रमाद् अमी ॥(C खण्डकैस्)

sanskrit

In the same manner, divide successively the sines (found) by the first sine; subtract (the sum of) the quotients from the divisor and add the remainder to the sine last found and the sum will be the next sine.[1] Thus you will get twenty-four sines (in a quadrant of a circle whose radius is 3438). These are as follows. [1. This method is proved thus. Let sin . A − sin . 0 = d 1 {\displaystyle \sin .{\phantom {2}}A-\sin .{\phantom {2}}0=d_{1}}; sin ⁡ .2 A − sin . A = d 2 {\displaystyle \sin .2A-\sin .{\phantom {2}}A=d_{2}}; sin ⁡ .3 A − sin ⁡ .2 A = d 2 {\displaystyle \sin .3A-\sin .2A=d_{2}}; & c . = & c . {\displaystyle \mathrm {\&c.} =\mathrm {\&c.} } sin . n A − sin . ( n − 1 ) A = d n {\displaystyle \sin .nA-\sin .(n-1)A=d_{n}}; sin . ( n + 1 ) A − sin . n A = d n + 1 {\displaystyle \sin .(n+1)A-\sin .nA=d_{n+1}}. Then since d 1 − d 2 = 2 vers ⁡ A . sin ⁡ A ÷ R {\displaystyle d_{1}-d_{2}=2\operatorname {vers} A.\sin {\phantom {2}}A\div R}; d 2 − d 3 = 2 vers ⁡ A . sin ⁡ 2 A ÷ R {\displaystyle d_{2}-d_{3}=2\operatorname {vers} A.\sin 2A\div R} d 3 − d 4 = 2 vers ⁡ A . sin ⁡ 3 A ÷ R {\displaystyle d_{3}-d_{4}=2\operatorname {vers} A.\sin 3A\div R} & c . = & c . {\displaystyle \mathrm {\&c.} =\mathrm {\&c.} } d n − d n + 1 = 2 vers ⁡ A . sin ⁡ n A ÷ R {\displaystyle d_{n}-d_{n+1}=2\operatorname {vers} A.\sin nA\div R}; we have by addition d 1 − d n + 1 = 2 vers ⁡ A R ( sin . A + sin ⁡ .2 A + … + sin . n A ) {\displaystyle d_{1}-d_{n+1}={\frac {2\operatorname {vers} A}{R}}(\sin .A+\sin .2A+\ldots +\sin .nA)} or, sin . A − sin . n A − sin . ( n + 1 ) A = 2 vers ⁡ A R ( sin . A + sin ⁡ .2 A + … + sin . n A ) {\displaystyle \sin .A-\sin .nA-\sin .(n+1)A={\frac {2\operatorname {vers} A}{R}}(\sin .A+\sin .2A+\ldots +\sin .nA)} ∴ sin . ( n + 1 ) A = sin . n A + sin . A {\displaystyle \therefore \sin .(n+1)A=\sin .nA+\sin .A} 2 vers ⁡ A R ( sin . A + sin ⁡ .2 A + … + sin . n A . ) {\displaystyle {\frac {2\operatorname {vers} A}{R}}(\sin .A+\sin .2A+\ldots +\sin .nA.)} Here, A = 3 ∘ 45 ′ {\displaystyle A=3^{\circ }45'}, ∴ 2 vers ⁡ A r = .0042822 = 1 233.5 {\displaystyle \therefore {\frac {2\operatorname {vers} A}{r}}=.0042822={\frac {1}{233.5}}}, which is roughly given in the text = 1 225 {\displaystyle ={\frac {1}{225}}}. ]

english translation

Adyenaivam kramAt piNDAn bhaktvA *labdhonasamyutAH ॥(C labdhonitair yutaiH) *khaNDakAH syuz caturvimzajyArdhapiNDAH kramAd amI ॥(C khaNDakais)

hk transliteration by Sanscript

siva.sh/surya-siddhanta/2/16

आद्येनैवम् क्रमात् पिण्डान् भक्त्वा *लब्धोनसम्युताः ॥(C लब्धोनितैर् युतैः) *खण्डकाः स्युश् चतुर्विम्शज्यार्धपिण्डाः क्रमाद् अमी ॥(C खण्डकैस्)

sanskrit

In the same manner, divide successively the sines (found) by the first sine; subtract (the sum of) the quotients from the divisor and add the remainder to the sine last found and the sum will be the next sine.[1] Thus you will get twenty-four sines (in a quadrant of a circle whose radius is 3438). These are as follows. [1. This method is proved thus. Let sin . A − sin . 0 = d 1 {\displaystyle \sin .{\phantom {2}}A-\sin .{\phantom {2}}0=d_{1}}; sin ⁡ .2 A − sin . A = d 2 {\displaystyle \sin .2A-\sin .{\phantom {2}}A=d_{2}}; sin ⁡ .3 A − sin ⁡ .2 A = d 2 {\displaystyle \sin .3A-\sin .2A=d_{2}}; & c . = & c . {\displaystyle \mathrm {\&c.} =\mathrm {\&c.} } sin . n A − sin . ( n − 1 ) A = d n {\displaystyle \sin .nA-\sin .(n-1)A=d_{n}}; sin . ( n + 1 ) A − sin . n A = d n + 1 {\displaystyle \sin .(n+1)A-\sin .nA=d_{n+1}}. Then since d 1 − d 2 = 2 vers ⁡ A . sin ⁡ A ÷ R {\displaystyle d_{1}-d_{2}=2\operatorname {vers} A.\sin {\phantom {2}}A\div R}; d 2 − d 3 = 2 vers ⁡ A . sin ⁡ 2 A ÷ R {\displaystyle d_{2}-d_{3}=2\operatorname {vers} A.\sin 2A\div R} d 3 − d 4 = 2 vers ⁡ A . sin ⁡ 3 A ÷ R {\displaystyle d_{3}-d_{4}=2\operatorname {vers} A.\sin 3A\div R} & c . = & c . {\displaystyle \mathrm {\&c.} =\mathrm {\&c.} } d n − d n + 1 = 2 vers ⁡ A . sin ⁡ n A ÷ R {\displaystyle d_{n}-d_{n+1}=2\operatorname {vers} A.\sin nA\div R}; we have by addition d 1 − d n + 1 = 2 vers ⁡ A R ( sin . A + sin ⁡ .2 A + … + sin . n A ) {\displaystyle d_{1}-d_{n+1}={\frac {2\operatorname {vers} A}{R}}(\sin .A+\sin .2A+\ldots +\sin .nA)} or, sin . A − sin . n A − sin . ( n + 1 ) A = 2 vers ⁡ A R ( sin . A + sin ⁡ .2 A + … + sin . n A ) {\displaystyle \sin .A-\sin .nA-\sin .(n+1)A={\frac {2\operatorname {vers} A}{R}}(\sin .A+\sin .2A+\ldots +\sin .nA)} ∴ sin . ( n + 1 ) A = sin . n A + sin . A {\displaystyle \therefore \sin .(n+1)A=\sin .nA+\sin .A} 2 vers ⁡ A R ( sin . A + sin ⁡ .2 A + … + sin . n A . ) {\displaystyle {\frac {2\operatorname {vers} A}{R}}(\sin .A+\sin .2A+\ldots +\sin .nA.)} Here, A = 3 ∘ 45 ′ {\displaystyle A=3^{\circ }45'}, ∴ 2 vers ⁡ A r = .0042822 = 1 233.5 {\displaystyle \therefore {\frac {2\operatorname {vers} A}{r}}=.0042822={\frac {1}{233.5}}}, which is roughly given in the text = 1 225 {\displaystyle ={\frac {1}{225}}}. ]

english translation

Adyenaivam kramAt piNDAn bhaktvA *labdhonasamyutAH ॥(C labdhonitair yutaiH) *khaNDakAH syuz caturvimzajyArdhapiNDAH kramAd amI ॥(C khaNDakais)

hk transliteration by Sanscript