Steps / Historical AI

Leonhard Euler · 1744 · Internet Archive

Methodus inveniendi lineas curvas

All 336 leaves · leaf 268

Transcription

ELASTICIS. 261 requisita finitae magnitudinis & quidem $= \frac{Ekk}{ff} \times \frac{\pi\pi}{4}$. Scilicet si extremitates A & B colligentur filo AB, hoc filum contrahi debebit vi $= \frac{Ekk}{ff} \times \frac{\pi\pi}{4}$. 26. Secundam speciem constituat casus, quo $c > 0$, attamen $c < a$; scilicet si $c$ contineatur intra limites 0 & $a$. His enim casibus angulus DAM recto erit minor; est namque anguli PAM sinus, seu anguli DAM cosinus $= \frac{aa-cc}{aa}$. Hoc ergo casu, forma lineae curvae talis fere erit qualem Figura 6, repraesentat. Quia igitur est $c < a$ erit $\frac{cc}{2aa} < \frac{1}{2}$; cum vero fit $\frac{cc}{2aa} > 0$, erit utique $AC = f > \frac{\pi a}{2\sqrt{2}}$ unde $aa < \frac{8ff}{\pi\pi}$; quare vis, qua extremitates laminae A & B ope fili AB ad se invicem attrahuntur, major erit quam casu praecedente, nempe $> \frac{Ekk}{ff} \times \frac{\pi\pi}{4}$. 27. In tertia specie unicum complector casum, quo $c = a$, quia hoc casu axis AP curvam in puncto A tangit: haecque species singulare nomen curvae Elasticae rectangulae obtinuit. Erit ergo $dy = \frac{xxdx}{\sqrt{(a^4-x^4)}}$, & $ds = \frac{aadx}{\sqrt{(a^4-x^4)}}$; hoc igitur casu AD & AC ita se habebunt ut fit: $AC = f = \frac{\pi a}{2\sqrt{2}} (1 + \frac{1^2}{2^2} \times \frac{1}{2} + \frac{1^2 \cdot 3^2}{2^2 \cdot 4^2} \times \frac{1}{4} + \frac{1^2 \cdot 3^2 \cdot 5^2}{2^2 \cdot 4^2 \cdot 6^2} \times \frac{1}{8} + \&c.)$ $AD = b = \frac{\pi a}{2\sqrt{2}} (1 - \frac{1^2}{2^2} \times \frac{3}{1 \cdot 2} - \frac{1^2 \cdot 3^2}{2^2 \cdot 4^2} \times \frac{5}{3 \cdot 4} - \frac{1^2 \cdot 3^2 \cdot 5^2}{2^2 \cdot 4^2 \cdot 6^2} \times \frac{7}{5 \cdot 8} - \&c.)$. Quanquam autem hinc, neque $b$, neque $f$ per $a$ accurate assignari potest; tamen alibi insignem relationem inter has quantitates locum habere demonstravi. Scilicet ostendi esse $4bf = \pi aa$, seu rectangulum ex AD & AC formatum erit aequale areae Circuli cujus diameter est $= AE$. Reperietur autem, calculum subducendo, proxime $f = \frac{5a}{6} \times \frac{\pi}{2}$, ita ut fit $a = \frac{12f}{5\pi}$; hinc vis: Kk 3 qua
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