Journal der indischen Mathematical Society, v. 1913, S. 132

Es sei PQR ein Kreis mit dem Mittelpunkt O, von dem PR der Durchmesser ist. Halbiere PO in H, und T sei der Punkt aus der Dreiteilung von OR nahe R. Zeichne TQ senkrecht zu PR und setze die Sehne RS = TQ.

Verbinde P mit S und zeichne OM und TN parallel zu RS. Setze eine Sehne PK = PM und zeichne die Tangente PL = MN. Verbinde R mit L, R mit K und K mit L. Abschnitt RC = RH. Zeichne CD parallel zu KL, [CD] trifft auf RL in D.

Dann ist das Quadrat über RD annähernd gleich dem Kreis PQR.

Denn ${\displaystyle RS^{2}={\frac {5}{36}}d^{2},}$

worin ${\displaystyle d}$ der Durchmesser des Kreises ist.

Somit ${\displaystyle PS^{2}={\frac {31}{36}}d^{2}.}$

Aber ${\displaystyle PL}$ und ${\displaystyle PK}$ sind gleich ${\displaystyle MN}$ bzw. ${\displaystyle PM.}$

Somit ${\displaystyle PK^{2}={\frac {31}{144}}d^{2},}$ und ${\displaystyle PL^{2}={\frac {31}{324}}d^{2}.}$

Folglich ${\displaystyle RK^{2}=PR^{2}-PK^{2}={\frac {113}{144}}d^{2},}$

und ${\displaystyle RL^{2}=PR^{2}+PL^{2}={\frac {355}{324}}d^{2}.}$

Aber ${\displaystyle {\frac {RK}{RL}}={\frac {RC}{RD}}={\frac {3}{2}}{\sqrt {\frac {113}{355}}},}$

und ${\displaystyle RC={\frac {3}{4}}d.}$

Darum ${\displaystyle RD={\frac {d}{2}}{\sqrt {\frac {355}{113}}}=r{\sqrt {\pi }},}$ nahezu gleich.

Anmerkung: Wenn die Fläche des Kreises 140.000 Quadratmeilen ist, dann ist RD um etwa einen Zoll größer als die wahre Länge.

Deshalb ${\displaystyle \textstyle RD={\frac {d}{2}}\cdot {\sqrt {\frac {355}{113}}}}$ dies ist ${\displaystyle r\cdot {\sqrt {\pi }}}$ sehr nahe. Ergänzung: Abweichung = 7,52...E-8 [LE]

Bemerkung

Wenn die Fläche des Kreises 140.000 Quadratmeilen ist, dann ist RD etwa einen Zoll größer als die wahre Länge.

Ergänzung zur Bemerkung

1 Meile [mile] = 1.609,344 m

${\displaystyle \textstyle {\sqrt {140.000}}}$ mile² = 374,165738... mile

RD = 1,772453926158302 m

${\displaystyle \textstyle 374,165738...\cdot {\frac {1.609,344}{1,772453926158302...}}=339.733{,}1668...\;[o.D.]}$

Abweichung${\displaystyle \textstyle =7{,}52...E-8\;m\cdot 339.733{,}1668...\cdot 1000\approx 25{,}55\;mm>1\;Zoll=25{,}4\;mm}$

Ergänzung

RD² = 3,14159292035398 (sechs Nachkommastellen sind gleich denen von ${\displaystyle \pi }$)

(-) ${\displaystyle \pi }$ = 3,14159265358979...

Abweichnung zu ${\displaystyle \pi }$ ≈ 2,667E-7 mm

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