De linea isochrona, in qua grave sine acceleratione descendit, et de controversia cum Dn. Abbate De Conti

On the Isochronous Curve, on which a Heavy Body Descends without Acceleration, and on the Controversy with the Abbé De Conti

Gottfried Wilhelm Leibniz · 1689 · Acta Eruditorum · skimmed

Source: original scan (archive.org:leibnizensmathe07leibgoog#leaves n250–n253 (printed pp. 234–237))
"De linea isochrona…", Acta Eruditorum, Leipzig (April 1689). Transcribed here from C. I. Gerhardt (ed.), Leibnizens mathematische Schriften, printed pp. 234–237.

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III. De linea isochrona, in qua grave sine acceleratione descendit, et de controversia cum Dn. Abbate De Conti

Cum a me in his Actis Martio 1686 editis publicata esset demonstratio contra Cartesianos, qua vera virium aestimatio traditur, ostenditurque non quantitatem motus, sed potentiae, a quantitate motus differentem servari, Vir quidam doctus in Gallia, Dn. Abbas De Conti, pro Cartesianis respondit, sed, ut post apparuit, vi mei argumenti non satis perspecta. Credidit enim, recepta quaedam alia principia a me impugnari, quae in Novellis Reip. Litterar. mens. Jun. 1687 p. 579 enumerat, et negat p. 519 seq. se agnoscere contradictionem, quam ego in illis invenire mihi videar: cum tamen nunquam mihi de illis dubitare in mentem venerit, quemadmodum ipsum admonui Novell. Reip. Lit. Septemb. 1687. Idem ut eluderet objectionem meam, conjecerat se in diverticulum temporis, quod eo modo, quo conceptus a me erat status controversiae, plane est accidentale. Eadem enim manente altitudine, eadem vis acquiritur aut impenditur a gravibus quocunque tempore indulto, quod pro inclinatione descensus majore minoreve augetur aut minuitur. Ea occasione, quo magis appareret, tempus atque adeo distinctionem inter potentias isochronas vel anisochronas hoc loco nihil ad rem facere, et ut ex disputatione nostra aliquid incrementi scientia caperet, problema tale, a me inter scribendum solutum, et, ut videtur, non inelegans, ipsi proposui in dictis Novellis Septembr. 1687: “Invenire lineam isochronam, in qua grave descendat uniformiter, sive aequalibus temporibus aequaliter accedat ad horizontem, atque adeo sine acceleratione et aequali semper velocitate deorsum feratur.” Sed Dn. Abbas De Conti nihil ultra reposuit, sive quod problema attingere nollet, sive quod agnita tandem mente mea, satisfactum sibi judicaret. Sed ejus loco problema hoc sua opera dignum judicavit Vir celeberrimus Christianus Hugenius, cujus solutio mea prorsus consona extat in Novellis Reip. Lit. Octbr. 1687, sed suppressa demonstratione et explicatione

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discriminis inter diversas lineas ejusdem, ut ait, generis, quas satisfacere notat. Haec igitur ego supplere hoc loco volui, facturus citius, nisi aliquid hic a Dn. Abbatis industria exspectavissem.

Problema. Invenire lineam planam, in qua grave sine acceleratione descendit.

Solutio. Sit (fig. 116)Editorial note: The plate containing fig. 116 could not be located in the scans available for this work (the Gerhardt edition scan has no plates section); it is not reproduced here. linea paraboloeides quadrato-cubica quaecunque $\beta Ne$ (nempe ubi solidum sub quadrato basis $NM$ et parametro $aP$ aequale est cubo altitudinis $\beta M$) ita sita, ut verticis $\beta$ tangens $\beta M$ sit perpendicularis horizonti, in cujus lineae puncto quocunque $N$ si ponatur grave ea descendendi ulterius celeritate praeditum, quam potuit acquirere descendendo ex horizonte $Aa$, cujus elevatio $a\beta$ supra verticem $\beta$ sit $\frac{4}{9}$ parametri curvae, tunc idem grave descendet porro uniformiter per lineam $Ne$, utcunque continuatam, ut desiderabatur.

Demonstratio. Recta $NT$ curvam $\beta Ne$ tangat in $N$ et ipsi $\beta M$ occurrat in $T$. Utique (ex nota proprietate tangentium hujus curvae) erit $TM$ ad $NM$ in subduplicata ratione $a\beta$ ad $\beta M$. Ergo $TM$ ad $TN$ erit in subduplicata ratione $a\beta$ ad $a\beta + \beta M$ seu ad $aM$. Jam ratio $TM$ ad $TN$ eadem est, quae velocitatis per curvam descendendi (seu horizonti porro in curva appropinquandi), quam grave habet positum in $N$ ad velocitatem, qua idem ex $N$ porro, non per curvam, sed libere descenderet, si posset (ut constat ex natura motus inclinati). Sed velocitas haec libera porro est ad constantem quandam in subduplicata ratione $aM$ ad $a\beta$; sunt enim (ut ex motu gravium constat) velocitates liberae in altitudinum

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(unde descendendo quaesitae sunt) $aM$ subduplicata ratione: ergo velocitas descendendi per curvam $Ne$, quam grave habet in quocunque curvae puncto $N$ positum, est ad velocitatem constantem in composita subduplicata ratione $a\beta$ ad $aM$ et $aM$ ad $a\beta$, quae est ratio aequalitatis. Ipsamet igitur velocitas illa per curvam descendendi est constans, seu ubique in curva $Ne$ eadem. Quod praestandum erat.

Consectaria: 1) Grave celeritatem habens tamquam lapsum ab altitudine aliqua $Aa$, descendere potest ex eodem puncto $N$ per curvas isochronas infinitas, sed ejusdem speciei, seu sola magnitudine parametri differentes, ut $Ne$, $N(e)$, $NE$, quae omnes sunt paraboloeides quadrato-cubicae, adeoque similes inter se. Imo quaelibet hujusmodi paraboloeidum hic inservit, modo ita collocetur, ut $a\beta$ vel $(a)(\beta)$ distantia verticis ab horizontali $a(a)$, unde descendere incepit grave, sit $\frac{4}{9}$ parametri curvae $\beta e$ vel $(\beta)(e)$: nec refert, an grave isochrone descensurum in curva $N(e)$ pervenerit ad $N$ ex $a(a)$ per viam $(a)(\beta)N$, an per aliquam aliam, aut sine descensu ullo ob aliam causam eandem celeritatem atque directionem acquisiverit. Ex infinitis tamen istis lineis isochronis, in quibus grave ex $N$ porro sine acceleratione descendere potest, ea celerrimum ipsi descensum praebet, cujus vertex est ipsum punctum $N$, qualis est $NE$, quam recta $AN$ horizonti perpendicularis tangit.

2) Tempus descensus per rectam $a\beta$ est ad tempus descensus per curvam $\beta N$, ut dimidia altitudo $\beta M$ ad ipsam $a\beta$: ac proinde si $\beta M$ sit dupla $a\beta$, aequalia erunt tempora descensuum per $a\beta$ et per $\beta N$. Quorum ratio manifesta est: nam tempora descensus uniformis sunt inter se ut altitudines, et ex demonstratis a Galilaeo, tempus quo mobile percurrit altitudinem $a\beta$ motu accelerato, est duplum ejus, quo percurrit aequalem altitudinem $\beta M$ (ut hoc loco fit, licet per curvam $\beta N$) motu uniformi, qui celeritatem habet aequalem ultimae per accelerationem acquisitae in $\beta$.

Hoc autem problema fateor me non Geometris primariis proposuisse, qui interiorem quandam Analysin callent, sed his potius, qui cum erudito illo Gallo sentiunt, quem mea de Cartesianis plerisque hodiernis (Magistri paraphrastis potius, quam aemulatoribus) querela suboffendisse videbatur. Tales enim cum alias receptis inter Cartesianos dogmatibus, tum etiam analysi inter ipsos pervulgatae nimium tribuunt, adeo ut se ipsius ope quidvis in Mathesi

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(si modo velint scilicet calculandi laborem sumere) praestare posse arbitrentur, non sine detrimento scientiarum, quae falsa jam inventorum fiducia negligentius excoluntur. His materiam exercendae suae Analyseos praebere volueram in hoc problemate, quod non prolixo calculo, sed arte indiget.

Si quis tamen praereptam sibi jam solutionem queratur, poterit aliam isochronam huic vicinam quaerere, in qua non, ut hactenus, grave uniformiter recedat ab horizontali (vel ad eam accedat), sed a certo puncto. Unde problema erit tale, invenire lineam, in qua descendens grave recedat uniformiter a puncto dato, vel ad ipsam accedat.

Talis foret linea $NQR$, si ejus esset naturae, ut ex puncto dato seu fixo $A$, ductis rectis quibuscunque ad curvam ut $AN$, $AQ$, $AR$, esset excessus $AR$ super $AQ$, ad excessum $AQ$ super $AN$, ut tempus quo descenditur per arcum $QR$, ad tempus quo descenditur per arcum $NQ$.

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III. On the isochronous curve, on which a heavy body descends without acceleration, and on the controversy with the Abbé De Conti

When, in these Acta published in March 1686, I had brought out a demonstration against the Cartesians, in which the true estimation of forces is set forth and it is shown that it is not the quantity of motion, but potentia, differing from the quantity of motion, that is conserved, a certain learned man in France, the Abbé De Conti, replied on behalf of the Cartesians, but — as afterwards appeared — without having sufficiently grasped the force of my argument. For he believed that certain other received principles were being attacked by me, which he lists in the Nouvelles de la République des Lettres for June 1687, p. 579, and he denies (p. 519 ff.) that he recognises the contradiction which I seem to myself to find in them: although it never came into my mind to doubt those principles, as I reminded him in the Nouvelles de la République des Lettres for September 1687. In order to evade my objection, he had betaken himself to a side-issue of time, which — in the way in which the state of the controversy had been conceived by me — is plainly accidental. For, the height remaining the same, the same force is acquired or expended by heavy bodies whatever the time allowed, which is increased or diminished according as the inclination of the descent is greater or less. On that occasion, so that it might appear more clearly that time — and hence the distinction between isochronous and non-isochronous powers — is here nothing to the purpose, and that our dispute might yield some increase of knowledge, I proposed to him, in the said Nouvelles for September 1687, the following problem, solved by me in the course of writing and, as it seems, not inelegant: “To find an isochronous curve on which a heavy body descends uniformly — that is, in equal times draws equally near to the horizon — and so is carried downward without acceleration and always with equal velocity.” But the Abbé De Conti made no further reply, whether because he was unwilling to touch the problem, or because, having at last understood my meaning, he judged himself satisfied. But in his stead the most celebrated Christiaan Huygens judged this problem worthy of his effort; his solution, wholly consonant with mine, appears in the Nouvelles de la République des Lettres for October 1687, though with the demonstration suppressed and the explanation

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of the distinction between the various lines of the same kind, as he says, which he notes as satisfying it. These things, therefore, I wished to supply in this place, and would have done so sooner, had I not been awaiting something here from the Abbé's industry.

Problem. To find a plane curve on which a heavy body descends without acceleration.

Solution. Let there be (fig. 116)Editorial note: The plate containing fig. 116 could not be located in the scans available for this work (the Gerhardt edition scan has no plates section); it is not reproduced here. any semicubical paraboloid curve $\beta Ne$ (namely, one in which the solid under the square of the base $NM$ and the parameter $aP$ is equal to the cube of the height $\beta M$), so placed that the tangent $\beta M$ at the vertex $\beta$ is perpendicular to the horizon. If at any point $N$ whatever of this curve a heavy body be placed, endowed with that further speed of descent which it could have acquired by descending from the horizontal $Aa$ — whose elevation $a\beta$ above the vertex $\beta$ is $\frac{4}{9}$ of the parameter of the curve — then the same body will descend further uniformly along the curve $Ne$, however far continued, as was desired.

Demonstration. Let the straight line $NT$ touch the curve $\beta Ne$ at $N$ and meet $\beta M$ at $T$. Then certainly (from the known property of the tangents of this curve) $TM$ will be to $NM$ in the subduplicate ratio of $a\beta$ to $\beta M$. Therefore $TM$ to $TN$ will be in the subduplicate ratio of $a\beta$ to $a\beta + \beta M$, that is, to $aM$. Now the ratio of $TM$ to $TN$ is the same as that of the velocity of descending along the curve (that is, of drawing nearer to the horizon along the curve) which the heavy body has when placed at $N$, to the velocity with which the same body would descend from $N$ onward, not along the curve, but freely, if it could (as is clear from the nature of inclined motion). But this free velocity is further to a certain constant one in the subduplicate ratio of $aM$ to $a\beta$; for (as is clear from the motion of heavy bodies) free velocities are in the subduplicate ratio of the heights

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(from which, by descending, they are sought), namely $aM$: therefore the velocity of descending along the curve $Ne$, which the body has when placed at any point $N$ of the curve, is to the constant velocity in the ratio compounded of the subduplicate ratio of $a\beta$ to $aM$ and of $aM$ to $a\beta$ — which is a ratio of equality. That very velocity of descending along the curve is therefore constant, that is, everywhere the same on the curve $Ne$. Which was to be shown.

Corollaries: 1) A heavy body having a speed as though fallen from some height $Aa$ can descend from the same point $N$ along infinitely many isochronous curves, but of the same species — differing only in the magnitude of the parameter — such as $Ne$, $N(e)$, $NE$, which are all semicubical paraboloids, and therefore similar to one another. Indeed any paraboloid of this kind serves here, provided it be so placed that $a\beta$ (or $(a)(\beta)$), the distance of the vertex from the horizontal $a(a)$ from which the body began to descend, is $\frac{4}{9}$ of the parameter of the curve $\beta e$ (or $(\beta)(e)$): nor does it matter whether the body, about to descend isochronously along the curve $N(e)$, has reached $N$ from $a(a)$ by the path $(a)(\beta)N$, or by some other, or has acquired the same speed and direction without any descent at all, from some other cause. Yet of those infinitely many isochronous curves, along which the body can descend from $N$ onward without acceleration, that one affords it the swiftest descent whose vertex is the point $N$ itself — such as $NE$, which the straight line $AN$, perpendicular to the horizon, touches.

2) The time of descent along the straight line $a\beta$ is to the time of descent along the curve $\beta N$ as half the height $\beta M$ is to $a\beta$ itself: and accordingly, if $\beta M$ be double $a\beta$, the times of the descents along $a\beta$ and along $\beta N$ will be equal. The reason for this is manifest: for the times of uniform descent are to one another as the heights; and, from what has been demonstrated by Galileo, the time in which a moving body traverses the height $a\beta$ by accelerated motion is double that in which it traverses an equal height $\beta M$ (as happens here, though along the curve $\beta N$) by uniform motion having a speed equal to the last one acquired by acceleration at $\beta$.

I confess, however, that I proposed this problem not to the foremost geometers, who are versed in a certain deeper Analysis, but rather to those who side with that learned Gaul, whom my complaint about most present-day Cartesians (paraphrasts of the Master, rather than his rivals) seemed to have somewhat offended. For such men attribute too much both to the dogmas received among the Cartesians and to the analysis widely current among them, so much so that they suppose themselves able, by its aid, to accomplish anything whatever in Mathematics

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(provided only they are willing to take on the labour of calculation), not without detriment to the sciences, which are now cultivated too negligently out of a false confidence in what has been discovered. To these I had wished to offer, in this problem, matter for exercising their Analysis — a problem that needs not lengthy calculation, but art.

If, however, anyone should complain that the solution has already been snatched from him, he may seek another isochrone neighbouring on this one, in which the heavy body recedes uniformly not, as hitherto, from the horizontal (or approaches it), but from a certain point. Whence the problem will be this: to find a curve on which a descending heavy body recedes uniformly from a given point, or approaches it.

Such would be the curve $NQR$, if it were of such a nature that, from a given or fixed point $A$, straight lines being drawn to the curve at will, such as $AN$, $AQ$, $AR$, the excess of $AR$ over $AQ$ were to the excess of $AQ$ over $AN$ as the time in which the arc $QR$ is descended is to the time in which the arc $NQ$ is descended.

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@article{Leibniz1689De,
  author = {Leibniz, Gottfried Wilhelm},
  title = {De linea isochrona, in qua grave sine acceleratione descendit, et de controversia cum Dn. Abbate De Conti},
  journal = {Acta Eruditorum},
  year = {1689},
  url = {https://archive.org/details/leibnizensmathe07leibgoog/page/n250/mode/1up}
}

@misc{Leibniz1689De-rtm,
  author = {Leibniz, Gottfried Wilhelm},
  title = {De linea isochrona, in qua grave sine acceleratione descendit, et de controversia cum Dn. Abbate De Conti --- LaTeX transcription},
  howpublished = {ReadTheMasters, \url{https://readthemasters.pages.dev/works/leibniz-1689-isochrona/}},
  note = {AI transcription (claude-opus-4-8); status: skimmed. Original: Acta Eruditorum 1689},
  year = {2026}
}

@misc{Leibniz1689De-en,
  author = {Leibniz, Gottfried Wilhelm},
  title = {On the Isochronous Curve, on which a Heavy Body Descends without Acceleration, and on the Controversy with the Abbé De Conti --- English translation},
  howpublished = {ReadTheMasters, \url{https://readthemasters.pages.dev/works/leibniz-1689-isochrona/}},
  note = {English translation (AI, claude-opus-4-8); status: ai-draft},
  language = {English},
  year = {2026}
}