Some time ago, when I was still a
boy, the most learned Mersenne proposed to me and to many others
the investigation of centers of oscillation or agitation, a very
famous problem among mathematicians of that time, as far as I can
gather from the letters he sent me and from the recently edited
letters of Descartes containing his response to Mersenne on these
matters.^{[1]}
Mersenne proposed further that I should find these
centers for sectors of circles, hung both from the vertex and
from the middle of the arc, and moved from side to side, as well
as for segments of circles and for triangles hung first from the
vertex and then from the midpoint of the base. The problem comes
down to finding a simple pendulum, i.e. a weight hung on a
string, of such a length that it makes oscillations of the same
time as those aforesaid suspended figures. At the same time,
indeed, he promised me a rather large and coveted reward for my
labors, if perchance I should solve the questions. But no one at
that time obtained what he desired. And, as for me, since I found
nothing that made clear even a first approach to this
contemplation, repulsed from the start, I abstained then from a
lengthier investigation. Indeed, those distinguished men who
hoped to accomplish the thing, Descartes, Honoré Fabri, and
others, in no way hit the mark, except in some few simpler cases,
and even of the latter, it seems to me, they adduced no proper
demonstration. This will, I hope, be manifest by comparison with
the things I set forth here, if perchance someone should compare
what has been set forth by them with our material, which I judge
to be demonstrated by more certain principles and which I have
found to agree precisely with experiments. Indeed, an occasion to
test these things again has been offered by the need to adjust
the timing of the pendulums of our machine [e*x pendulorum
automati nostri temperandorum ratione*], if I apply the
movable weight to them, as was explained in the description of
the clock, in addition to what is below. Hence, having begun the
matter from the very beginning and under better auspices, I
overcame at last all difficulties and found not only the solution
of Mersenne's problems, but also other results more difficult
than these, and, finally, a way which permits the investigation
of this center in lines, surfaces, and solid bodies by a sure
method [*certa ratione*]. Whence, moreover, besides the
pleasure of discovering what was much sought after by others and
of knowing the laws and decrees of nature in these matters, I
have also gained the usefulness for the sake of which I first
applied my mind to these things, having found an easy and
expeditious method for calibrating clocks. Added to this is what
I think will be of great value to many: I can set forth from
these results a most absolute definition of a sure measure which
will endure through all the centuries; it is that which will be
found added on at the end of these results.

I. One calls a

pendulumany figure endowed with weight [gravitas],whether it be a line, or a surface, or a solid, so suspended that it can continue, by the force of its weight, reciprocal motion about some point, or rather axis, which is understood to be parallel to the plane of the horizon.II. This axis parallel to the plane of the horizon, about which the motion of the pendulum is imagined to take place, is called the

axis of oscillation.III. One calls a

simple pendulumthat which is imagined to consist of a string or inflexible line devoid of weight and bearing a weight [pondus] affixed to its lowest end; the gravity of the weight is to be thought of as gathered together at one point.IV. A

composite pendulumis that which consists of several weights which maintain constant distances, both between each other and from the axis of oscillation. Whence, any suspended figure endowed with weight [gravitas] can be called a composite pendulum, since it is divisible mentally into any number of parts.V. Pendulums are called

isochronous,the oscillations of which along similar arcs are completed in equal times.VI. One calls the

plane of oscillationthe plane that is imagined to be drawn through the center of gravity of the suspended figure and at right angles to the axis of oscillation.VII. The

line of centeris the straight line drawn through the center of gravity of the figure perpendicular to the axis of oscillation,VIII. The

line of the perpendicularis the straight line in the plane of oscillation drawn from the axis of oscillation perpendicular to the plane of the horizon.IX. One calls the

center of oscillation(oragitation) of any figure the point in tile line of center as distant from the axis of oscillation as the length of a simple pendulum that is isochronous with the figure.X. An

axis of gravityis any straight line passing through the center of gravity of the figure.XI. A plane figure, or a line lying in a plane, is said to be

agitated in a planewhen the axis of oscillation is in the same plane as the figure or line.XII. The same things are said to be

agitated from side to side[in latus] when the axis of oscillation is perpendicular to the plane of the figure or line.XIII. Whenever weights

[pondera] are said to be multiplied by straight lines, it is to be understood as if numbers or lines expressing the quantities of the weights and their mutual ratio are so multiplied.

I. If any number of weights begin to move by the force of their gravity, the center of gravity composed of them cannot ascend higher than where it was located when the motion began.

Height is measured [*consideratur*]
in these propositions by the distance from a horizontal plane,
and heavy bodies are assumed to tend to fall to this plane along
straight lines perpendicular to it. The same thing is either
expressly assumed by all who have dealt with the center of
gravity or is to be supplied by the readers, since without it
there can be no consideration of the center of gravity.

But, lest this hypothesis of ours cause doubt, we will make clear that it intends nothing other than what no one has ever denied, to wit, that heavy bodies do not move upward. For, first, if we suppose any heavy body, it is beyond doubt that it cannot ascend higher by the force of its gravity. And it is understood to ascend when its center of gravity ascends. But one must also concede the same for any number of weights, joined to each other by inflexible lines, since nothing prevents their being considered as some one body. And thus neither will their common center of gravity be able to ascend further.

But now, if any number of weights are not assumed to be connected with one another, we know that they also have some common center of gravity, the height of which will be so much. I say that the height of the weight composed of all of them should be thought of as that much, inasmuch as all can be brought to this same height of the center of gravity without summoning any other power than is in the weights themselves, but only by joining them at will by inflexible lines and moving them about the center of gravity, for which no determinate force nor power is necessary. Whence, just as it cannot happen that weights placed in the same horizontal plane all rise above that plane by the force of their gravity, so neither can the center of gravity of any number of weights, however disposed, reach a greater height than it has. And our statement that any bodies can be brought to the horizontal plane passing through their center of gravity, with no added force, is shown thus.

Let A, B, C be weights given in
position, of which the common center of gravity is D. Suppose a
horizontal plane, of which EF is a right section, to have been
drawn through D. Now let DA, DB, DC be inflexible lines which
rigidly [*invariabiliter*] connect the weights with each
other. Next, let the weights be moved until A is in plane EF at
E. All rods having been moved through equal angles, B will now be
at G and C at H.

Now imagine further that B and C are connected by rod HG, which cuts plane EF at F, where necessarily the center of gravity of these two connected weights will also be, since the center of gravity of the three placed at E, B, H is D and the center of gravity of that which is at E is also in the plane EDF. Therefore, weights H, G are again moved without any force around point F as axis, and both are brought to plane EF at the same time, so that now the three, which were formerly at A, B, C, have manifestly been transferred to the height D of their center of gravity with their equilibrium maintained. Q.E.D. The demonstration is the same for any number of other bodies.

This hypothesis of ours also applies to liquid bodies; by it can be demonstrated not only all that Archimedes has on floating bodies, but also many other theorems of mechanics. Indeed, if the builders of new machines, who strive with fruitless effort after perpetual motion, will learn to use this same hypothesis, they will easily perceive their own errors and will understand that the thing is in no way possible for mechanical reasons.

II. Air and any other manifest impediment having been removed (just as we want it understood in the following demonstrations), the center of gravity of an agitated pendulum traverses equal arcs in descending and ascending.

This has been demonstrated for a
simple pendulum in Proposition 9 of "On the fall of heavy
bodies" [= Part 2 of the *Horologium*]. Experience
declares that the same should also be held for composite
pendulums, inasmuch as, whatever be the shape of the pendulum, it
is equally fitted to continue its motion, unless to some greater
or less extent it is impeded by the air striking it.

If, from each of the centers of gravity of any number of weights located on the same side of a plane, perpendiculars are drawn to that plane, these perpendiculars, each multiplied by its weight, together make as much as the perpendicular falling from the center of gravity of all the weights onto the same plane, multiplied by all the weights.

Let A, B, C be weights located on the same side of a plane of which the right section is DF, and to it let perpendiculars AD, BE, CF be drawn from each of the weights respectively. And let the center of gravity of all the weights A, B, C be point G, from which perpendicular GH is drawn to the same plane. I say that the sum of the products that result from the individual weights times their perpendiculars is equal to the product of the straight line GH times all the weights A, B, C.

For imagine the perpendiculars
drawn from each weight to be extended [*continuari*__]__
on the other side of plane DF, and let each of DK, EL, FM be
equal to HG; all the lines may be thought of [*referant*] as
inflexible rods. And suppose the weights at K, L, M to be such
that each is in equilibrium with the one opposite it, A, B, C,
with respect to the intersection DEF of the plane. Therefore, all
of K, L, M will balance all of A, B, C. But, as length AD is to
DK, so weight K will be to weight A, and consequently DA times
the magnitude of A will be equal to DK, or GH, times K.
Similarly, EB times B will be equal to EL, or GH, times L, and FC
times C will be equal to FM, or GH, times M. Therefore the sum of
the products of AD times A, BE times B, CF times C will be equal
to the sum of the products of GH times each of K, L, M. But,
since K, L, M balance these A, B, C, they will also balance the
same A, B, C suspended from their center of gravity G. Whence,
since distance GH is equal to each of DK, EL, FM, it is necessary
that the magnitudes A, B, C taken together be equal to these K,
L, M. Thus also, the sum of the products of GH times each of A,
B, C will be equal to the products of DA times A, EB times B, and
FC times C. Q.E.D.

Indeed, even if in the demonstration the straight lines AD, GH, CF were placed parallel to the horizon, and the plane perpendicular to the horizon, it is clear that, if all were transposed together to some other position, the same equality of products would remain, since all the straight lines are the same as before. Wherefore the proposition holds.

Under the same assumptions as before, if all the weights A, B, C are equal, I say that the sum of all the perpendiculars AD, BE, CF is equal to the perpendicular GH drawn from the center of gravity, multiplied by the number of weights.

For, since the sum of the products of the individual weights times their perpendiculars is equal to the product of GH times all the weights and here, because of the equality of the weights, that sum of products is equal to the product of one weight times the sum of all the perpendiculars, and also the product of GH times all the weights is the same as the product of one weight times GH multiplied by the number of weights, it is clear that the sum of the perpendiculars is then necessarily equal to this GH multiplied by the number of weights. Q.E.D.

If any magnitudes all descend or ascend, albeit through unequal intervals, the heights of descent or ascent of each, multiplied by the magnitude itself, yield a sum of products equal to that which results from the multiplication of the height of descent or ascent of the center of gravity of all the magnitudes times all the magnitudes.

Let A, B, C be magnitudes which descend from A, B, C to D, E, F or ascend from D, E, F to A, B, C. And let the center of gravity of all of them, while they are at A, B, C, be at the same height as point G, and, when they are at D, E, F, at the same height as point H. I say that the sum of the products of height AD times A, BE times B, CF times C is equal to the product of GH times all of A, B, C.

For imagine a horizontal plane, the right section of which is MP, and let AD, BE, CF, and GH produced fall on it at M, N, O, P.

Then, because the sum of the products of AM times A, BN times B, CO times C is equal to the product of GP times all of A, B, C, and, similarly, the sum of the products of DM times A, EN times B, FO times C is equal to the product of HP times all of A, B, C, it follows that the excess of the former products over the latter is equal to the product of GH times all the magnitudes A, B, C. Indeed, the said excess is manifestly equal to the products of AD times A, BE times B, CF times C. Therefore, these together are also equal to the product of GH times all of A, B, C. Q.E.D.

If a pendulum composed of several weights and released from rest were to traverse some part of its complete oscillation, and then its individual weights were imagined, under release from constraint [

relicto communi vinculo], to convert their acquired speeds upward and to ascend as far as they can, when this has occurred, the center of gravity composed of all of them will have returned to the same height that it had before the oscillation began.

Let a pendulum be composed of any number of weights A, B, C lying on a rod or surface devoid of weight. And let it be suspended from an axis drawn through point D, which is imagined to be perpendicular to the plane that is here viewed. Let the center of gravity E of the weights A, B, C be placed in this same plane, and let the line of center DE be inclined to the line of the perpendicular DF at angle EDF, the pendulum, of course, having been drawn up this far. From here, assume now that it is released and that it traverses some part of its oscillation, so that the weights A, B, C arrive at G, H, K, from where the individual weights, released from constraint, are imagined to convert their acquired speeds upward (which can happen by striking against some inclined planes, such as QQ) and to ascend as far as they can, say to L, M, N. When they reach there, let point P be the center of gravity of all of them. I say that this is at the same height as point E.

Now first, it holds by the first
of the assumed hypotheses that P cannot be higher than E. But
that it will also not be lower is shown thus. For, if it is
possible, let P be lower than E, and imagine the weights to fall
again from the same heights to which they ascended, which are LG,
MH, NK. Whence it holds that they will acquire the same speeds
that they had to ascend to these heights, that is, those same
speeds that they acquired from the motion of the pendulum from
CBAD to KHGD. Whence, if they are now brought back at the said
speeds to the rod or surface to which they were connected and
adhere to it at the same time, and continue their motion along
the arcs already begun (which will happen if, before they reach
the rod, they are imagined to bounce off inclined planes QQ
again), the pendulum, restored in this manner, will complete the
remaining part of the oscillation, in the same way [*aequè*]
as if the motion had continued without any interruption. So that
the center of gravity of the pendulum, E, traverses equal arcs
EF, FR in descending and ascending, and consequently is found at
the same height at R as at E. But it was assumed that E was
higher than the center of gravity P of the weights placed at L,
M, N. Therefore R will also be higher than P, and thus the center
of gravity of the weights which have fallen from L, M, N has
ascended higher than from where it descended, which is absurd.
Therefore, the center of gravity P is not lower than E. But
neither was it higher. Therefore it must necessarily be equally
high. Q.E.D.

Given a pendulum composed of any number of weights, if the individual weights are multiplied by the squares of their distances from the axis of oscillation, and the sum of the products is divided by the product of the sum of the weights times the distance of the center of gravity of all of them from the same axis of oscillation, the result is the length of a simple pendulum isochronous with the composite pendulum, or the distance between the axis and the center of oscillation of the composite pendulum itself.

Let the weights composing the
pendulum be A, B, C, (of which neither the shape nor the
magnitude, but only the weight is considered), and let them be
suspended from an axis through point D, which is imagined to be
perpendicular to the plane which is viewed. Also, let their
common center of gravity in that plane be E; it does not matter
if the weights are in different planes. Let the distance of point
E from the axis, that is, straight line ED, be called *d*,
the distance of weight A be *e*, of BD be *f*, of CD be
*g*. Thus, if the individual weights are multiplied by the
squares of their distances, the sum of the products will be *aee
*+ *bff *+ *cgg*. Furthermore, if the sum of the
weights is multiplied by the distance of the center of gravity of
all of them, the product will be *ad *+ *bd *+ *cd*.
Whence, dividing the former product by the latter, one will have

For, assume both the pendulum FG and the line of center DE to be moved away from the line of the perpendicular through equal angles, the former from FH, the latter from DK, and then, when released, to swing; and in line DE take DL equal to FG. Thus, weight G of pendulum FG will, in a full oscillation, traverse arc GM, which is bisected by the line of the perpendicular FH, and point L will traverse an arc LN similar and equal to GM, which is bisected by DK. The center of gravity E will also traverse a similar arc EI. Now, if, when one takes in arcs GM, NL any points, say O, P, which divide them similarly, the speed of weight G at O is shown to be the same as the speed of point L at P, it will then hold that both arcs are traversed in equal times, and consequently that pendulum FG is isochronous with the pendulum composed of A, B, C. And the antecedent is shown in the following way.

First, if possible, let the speed
of point L when it arrives at P be greater than that of weight G
at O. It holds that, while point L traverses arc LP, at the same
time the center of gravity E traverses a similar arc EQ. From
points Q, P, O draw perpendiculars upward, which meet the chords
of the arcs EI, LN, GM at R, S, Y. And let SP be called *y*.
Whence, since LD (or *x*) is to ED (or *d*) as SP (or *y*)
is to RQ, RQ will be equal to *dy*/*x*. Now, because
weight G has at O the speed sufficient to ascend to the same
height from which it descended, i.e. along arc OM or the
perpendicular OY equal to this PS, therefore point L, when it
arrives at P, will have a greater speed there than that by which
it would ascend to height PS. And, while L passes by P, at the
same time weights A, B, C traverse arcs similar to this LP, that
is, AT, BV, CX. And the speed of point L at P is to the speed of
weight A at T as the distance DL to DA, since they are contained
in the same chord. But as the square of the speed of point L at P
is to the square of the speed of point A at T, so the height
which can be ascended at the former speed is to the height which
can be ascended at the latter speed. Also, as the square of
distance DL, which is *xx*, is to the square of distance DA,
which is *ee*, so the height which is ascended at the speed
of point L, when it is at P (which height has been said to be
greater than PS, or y), is to the height which is ascended at the
speed of weight A at T; that is, if, after it has arrived at T
and freed from the pendulum, it separately converts its motion
upward. Consequently, the height will be greater than *eey*/*xx*.

By the same argument, the height
to which weight B would ascend at the speed acquired along arc BV
will be greater than *ffy*/*xx*. And the height to
which weight C would ascend at the speed acquired through arc CX
will be greater than *ggy*/*xx*. Whence, if these
individual heights are multiplied by their weights, the sum of
the products will be greater than

can go. Indeed, the quantity *dy*/*x*, as was found above)
times the same sum of the weights, *a *+ *b* + *c*.
Hence, since the first product above was shown to be greater than
this last product, it follows that, if the individual weights A,
B, C, freed from the pendulum, convert their acquired speeds
upward when they arrive at T, V, X, the ascent of their center of
gravity will be greater than the descent of the same center of
gravity, when from A, B, C they are moved to T, V, X. Which is
absurd, since the said ascent must be equal to the descent, by
the preceding proposition.

In the same way, if it is said that the speed of point L when it arrives at P is less than the speed of weight G when it arrives at 0, we can show that the possible ascent of the center of gravity of the weights A, B, C is less than the descent, which contradicts the same antecedent proposition. Whence it remains that the speed of point L passing P is the same as that of weight G at O. Whence, as has been said above, it follows that the simple pendulum FG is isochronous with the pendulum composed of A, B, C.

Given a pendulum composed of any number of equal weights, if the sum of the squares of the distances of each of the weights from the axis of oscillation is applied to the distance of the common center of gravity from the same axis of oscillation, multiplied by the number of these weights, the result will be the length of a simple pendulum isochronous with the composite pendulum.

[The proof, which follows *mutatis
mutandis* directly from the preceding, is here omitted.]

If any figure and a straight line externally tangent to it lie in the same plane, and another straight line perpendicular to that plane is moved around the perimeter of the figure and describes some surface, which is then cut by a plane drawn through the said tangent and inclined to the plane of the said figure, the solid contained by these two planes and the portion of the surface intercepted between them is called a

truncated wedge[cuneus abscissus] on that figure as base.

In the attached drawing ABEC is the given figure, MD the straight line tangent to it, and EF the straight line which is moved along its perimeter. And the wedge is the solid figure contained by planes ABEC, MFG, and the portion of the surface described by the straight line EF.

The distance between the straight line through which the wedge is cut off and the point of the base on which falls the perpendicular from the center of gravity of the wedge is called the

subcentricof the wedge. That is, in the same figure, if K is the center of gravity of the wedge, and the straight line KI is the perpendicular drawn to its base ABEC, and further IM is perpendicular to AD, IM will be what we call the subcentric.

If the [cutting] plane is inclined at a semiright angle, the truncated wedge on any plane figure is equal to the solid that results from multiplying the same figure by a height equal to the distance of the center of gravity of the figure from the straight line through which the wedge is cut off.

On the plane figure ACB, let ABD be a wedge cut off by a plane inclined at a semiright angle and passing through the straight line EE tangent to figure ACB and lying in its plane. And let F be the center of gravity of the figure, from which let the perpendicular FA be drawn to straight line EE. I say that the wedge ABD is equal to the solid that results from multiplying figure ACB by a height equal to this FA.

For imagine figure ACB to be
divided into very small equal particles [*particulas minimas
aequales*], of which one is G. Thus it holds that, if each of
these is multiplied by its distance from straight line EE, the
sum of the products will be equal to the product of straight line
AF times all the particles, i.e. to the product of the figure ACB
itself times a height equal to AF. But the individual particles
such as G, multiplied by their distances GTI, are equal to very
small parallelepipeds, or prisms, erected on them and bounded at
the oblique surface AD, as in the case of GK. The reason is that
the heights of the parallelepipeds are equal to these distances
GH due to the semiright angle of inclination of planes AD and
ACB. And it is clear that the whole wedge ABD is composed of
these parallelepipeds. Therefore, the wedge will also be equal to
the solid on base ACB having an altitude equal to line FA. Q.E.D.

If a straight line is tangent to a plane figure, and the figure is imagined to be divided into very small equal particles, and also perpendiculars are drawn from each of the particles to this straight line, the squares of all these perpendiculars, taken together, will be equal to some rectangle multiplied by the number of these particles, which rectangle is the product of the distance of the center of gravity of the figure from the same straight line and of the subcentric of the wedge on the figure, which is cut off through that line.

For, having assumed all the other things assumed in the preceding construction, let LA be the subcentric of wedge ABD to straight line EE. It is therefore necessary to show that the sum of the squares of the distances of all the particles of figure ACB is equal to the rectangle formed by FA, LA multiplied by the number of particles.

It holds from the preceding
demonstration that the heights of the individual parallelepipeds,
such as GK, are equal to the distances of the particles that are
their bases, such as G, from the straight line AE. Wherefore, if
we now multiply parallelepiped GK by distance GH, it is just as
if particle G were multiplied by the square of distance GH. The
situation is the same for all the others. But all the products of
the parallelepipeds times their distances from straight line AE
together equal the product of wedge ABD times distance LA,
because the wedge gravitates [*gravitat*] on point L. Hence
also, the sum of the products of the individual particles G times
the squares of their distances from straight line AE will be
equal to the product of wedge ABD times straight line LA, i.e.
the product of figure ACB times rectangle FA, LA. For the wedge
ABD is equal to the product of figure ACB times straight line FA.
Furthermore, because figure ACB is equal to the product of one
particle G times the number of these particles, it follows that
the said product of the figure ACB times the rectangle formed
from FA, LA is equal to the product of particle G times the
rectangle formed from FA, LA multiplied by the number of
particles G. To which, consequently, the said sum of products of
the individual particles G times the squares of their distances
from straight line AE, or of one particle times the sum of all
these squares, will also be equal. Wherefore, if the
multiplication by particle G is omitted on both sides, it is
necessary that the same sum of squares be equal to the rectangle
formed from FA, LA multiplied by the number of particles into
which figure ACB is imagined to be divided. Q.E.D.

Given a plane figure and a straight line in the same plane, which either cuts the figure or not and to which perpendiculars fall from individual very small and equal particles, into which the figure is imagined to be divided, to find the sum of the squares of all these perpendiculars, or a plane which, multiplied by the number of particles, is equal to the said sum of squares.

Let ABC be a given plane figure and ED a straight line in the same plane; having divided the figure mentally into very small equal parts, imagine perpendiculars to be drawn from each of them to straight line ED, as FK is drawn from particle F. One must find the sum of the squares of all these perpendiculars.

Let the straight line AL, which is tangent to the figure and is supposed totally outside of it, be parallel to the given ED. It can be tangent to the figure either on the same side as ED or on the opposite side. And let the distance of the center of gravity of the figure from straight line AL be straight line GA, which cuts ED at E; let HA be the subcentric of a wedge on the figure cut off by a plane through line AL. I say that the sum of the squares sought for is equal to rectangle AGH plus the square of EG, [both] multiplied by the number of particles into which the figure is imagined to be divided.

For let FK, produced if necessary, meet tangent AL at point L. First, then, in the case where straight line ED lies off the figure and tangent AL is drawn on the same side of the figure, the proposition is shown thus. The sum of all the squares of FK is equal to the same number of squares of KL plus twice the same number of rectangles KLF plus the same number of squares of LF. Then the squares of KL are equal to the same number of squares of EA. And the rectangles KLF are equal to the same number of rectangles EAG, because all the FL are equal to the same number of GA. And finally, the squares of LF are equal to the same number of rectangles HAG, i.e. to the same number of squares of AG plus the same number of rectangles AGH. Hence, all the squares of FK will be equal to the same number of squares of EG plus the same number of rectangles EAG doubled plus also the same number of squares of AG plus the same number of rectangles AGH. But these three, that is, the square of EA plus twice rectangle EAG plus the square of AG, form the square of EG. Hence, it is clear that all the squares of FK are equal to the same number of squares of EG plus the same number of rectangles AGH. Q.E.D.

Further, in all remaining cases, all the squares of FK are equal to the same number of squares of KL minus twice the same number of rectangles KLF plus the same number of squares of L.F, i.e. to the same number of squares of EA minus the same number of rectangles EAG doubled plus the same number of squares of AG plus the same number of rectangles AGH. But, in all these cases, the square of EA plus the square of AG minus twice the rectangle EAG is equal to the square of EG. Hence again all the squares of FK will be equal to the same number of squares of EG plus the same number of rectangles AGH. Wherefore the proposition holds.

From this it follows that rectangle AGH is of the same magnitude, whether AH is the subcentric of either wedge, i.e. [the one] cut off by either the former or the latter of the parallel tangents AL. And thus as AG of the one case is to AG of the other, so is H of the former to H of the latter. But, as the straight lines AG are to each other, so in both cases are the wedges cut off through AL, as is gathered from Proposition VII. Hence, so too reciprocally is GH to GH.

It is clear also that, given the center of gravity G of the plane figure and the subcentric of the wedge cut off through either of the parallel tangents AL, the subcentric of the wedge cut off through the other tangent AL is also given.

On the assumptions of the preceding proposition, if the given straight line ED passes through G, the center of gravity of the figure ABC, the sum of the squares of the distances of the particles into which the figure is imagined to be divided from straight line ED will be equal to rectangle AGH alone multiplied by the number of these particles.

For this is manifest, since then there is no square of EG.

Again on the assumptions of the penultimate proposition, if DE is the axis of the plane figure ABC, dividing it into two equal and similar portions, and in addition VG is the distance of the center of gravity of the half of the figure, DAD, from straight line ED, and GX the subcentric of the wedge on this half cut off through that ED, the rectangle XGV will be equal to the rectangle AGH.

For rectangle XGV, multiplied by the number of particles of figure DAD, is equal to all the squares of the perpendiculars from the particles of the same half-figure falling on straight line ED. And, consequently, the same rectangle XIV, multiplied by the number of particles of the whole figure ABC, will be equal to the squares of the perpendiculars let fall on line ED from all the particles of this figure, i.e. to rectangle AGH multiplied by the same number of particles, as holds from the preceding proposition. Whence it follows that rectangles XIV, AGH are equal to one another. Q.E.D.

Given any number of points in a plane, if a circle is described about their center of gravity, and straight lines are drawn from all of the given points to some point on the circumference of this circle, the sum of the squares of all of them will always be equal to the same plane [area].

[The proof of this proposition is
very long and filled with diffuse algebraic calculations. The
theorem is not original; it may be found as Proposition V of Book
II of Apollonius' *Plane Loci.* Fermat's proof of the
theorem, contained in his restoration of that Apollonian text,
circulated widely after it was sent to Roberval in 1637. Though
this proof differs from the one given here, Huygens knew of it,
as he himself attests in a letter to Fermat. Huygens' teacher,
Franz van Schooten, also provided a demonstration of the theorem
as part of his own restoration of the *Plane Loci*. Owing to
its length and its purely mathematical content, the demonstration
is here omitted, even though the theorem itself is crucial for
what follows.]

If a plane figure, or a line lying in a plane, is suspended in different ways from points which, taken in the same plane, are equally distant from its center of gravity, and it is agitated in a side-to-side motion, it is isochronous with itself.

Let ABC be a plane figure, or a line lying in a plane, of which the center of gravity is D. About that center describe the circumference ECF of a circle in the same plane. I say that, if the figure, suspended from any point on the circumference, such as E, C, or G, is a agitated from side to side, it is isochronous with itself, or with the same simple pendulum.

Let the suspension first be from point E, and, when it is outside the figure, as here, line EH, from which the figure hangs, should be thought of as rigid and unmovably fixed to the figure.

Imagine figure ABC to be divided into very small equal particles, from the centers of gravity of all of which straight lines are drawn to point E. It is manifest that, since the figure is moved in a side-to-side motion, these lines are perpendicular to the axis of agitation. Therefore, the squares of all of these perpendiculars, divided by straight line ED multiplied by the number of particles into which the figure is divided, yield the length of a simple pendulum, which is KL, isochronous with the figure. And if the figure is suspended from point G, the length of an isochronous simple pendulum is again found by dividing the squares of all the lines which are drawn to point F from the particles of the figure by the straight line GD multiplied by the number of the same particles. Since therefore points G and E are in the circumference described about center D, which is the center of gravity of figure ABC, or the center of gravity of all the points which are centers of the equal particles of the figure, consequently the sum of the squares of the lines which are drawn from the said particles to point G will be equal to the sum of the squares of the lines which are drawn from the same particles to point E. In both suspensions, these sums of squares are applied to equal magnitudes, that is, in suspension from E, to straight line ED multiplied by the number of all the particles, and in suspension from G, to straight line DG multiplied by the number of the same particles. Hence it is clear that, from this last application, namely when suspension is from G, the same length of an isochronous pendulum results as from the former application, i.e. the same as this KL.

If the figure is suspended from C or any other point of the circumference ECF, it will be proved in the same way to be isochronous with the same pendulum KL. And thus the proposition holds.

[Propositions XIV and XV, which extend the result of Proposition IX to solid bodies, are here omitted. They are not required for the propositions immediately following, and their proofs are quite long and complex.]

If any figure, be it a line, a surface, or a solid, is suspended in different ways and agitated about axes which are parallel to one another and which are equally distant from the center of gravity of the figure, it is isochronous with itself.

Suppose any magnitude, of which the center of gravity is point E, and let it first be suspended from an axis which is imagined to pass through point F at right angles to the plane of this page. Thus, the same plane will also be the plane of oscillation. If in that plane a circumference FHG is described about center E with radius EF, and any point is taken in that circumference, such as H, the magnitude is then secondly imagined to be suspended from an axis fixed at this point and to be agitated while remaining in the same plane of oscillation. I say that it will be isochronous with itself agitated about the axis at F.

For imagine the proposed magnitude to be divided into very small equal particles. Thus, because in either suspension it remains in the same plane of oscillation with respect to the parts of the magnitude, it is manifest that, if from all of the particles into which the magnitude is divided, perpendiculars are imagined to fall onto the said plane of oscillation, they will meet it at the same points in either suspension. Let them be those points that appear in the area ABCD.

Since therefore E is the center
of gravity of the proposed magnitude and this magnitude, whatever
its position, maintains equilibrium about the axis which is
perpendicular to the plane of ABCD through point E, it is easily
seen that, if equal weight is assigned to all of the aforesaid
points, which are designated in area ABCD, the center of gravity
of all of them will be point E. Indeed, if, as can happen,
several perpendiculars coincide at some points, these points
should be imagined to be multiplied [*geminata*__]__ that
many times, and that many multiple weights are to be taken. And
it is clear that the center of gravity of the points so
considered is again point E.

Further, it is clear that the sum of the squares of the straight lines drawn from all the said points to point F is the same as the sum of the squares of those straight lines that are drawn from the individual particles of the proposed magnitude perpendicular to the axis of oscillation passing through F, since indeed those lines, whose squares are imagined, have the same length in both cases. And it is similarly clear that, when the suspension is from the axis through H, the sum of the squares of the straight lines which are drawn to point H from all the points designated in area ABCD is the same as the sum of the squares of those which are drawn from all of the particles of the proposed magnitude perpendicular to the axis of oscillation passing through 13. Hence, in both cases, if the sum of the squares of the straight lines drawn from all the aforesaid points to points F or 11 is divided by straight lines EF or EH multiplied by the number of particles into which the proposed magnitude is imagined to be divided, there will result from this application the length of a simple pendulum which is isochronous with the magnitude suspended from F or H. For, in either case, the sum of the squares is equal, and straight lines EF, EH are also equal to each other, and the number of particles is the same. Hence, since both the applied quantities and those to which they are applied are equal in both cases, the lengths resulting from the application will also be equal, i.e. the lengths of pendulums isochronous with the proposed magnitude suspended from F or from H. Wherefore the proposition holds.

Given a plane which, multiplied by the number of particles into which a suspended figure is imagined to be divided, is equal to the squares of all the distances from the axis of oscillation, if that plane is applied to a straight line equal to the distance between the axis of oscillation and the center of gravity of the suspended magnitude, there will result the length of a simple pendulum isochronous with the magnitude.

Let ABC be a figure, with center of gravity E, suspended from an axis which is perpendicular to the plane which is viewed through point F. On the assumption that the figure has been divided into very small equal particles, from all of which perpendiculars are imagined to fall to the said axis, let, by what has been shown above, the plane H be found which, multiplied by the number of the said particles, is equal to the squares of all the said perpendiculars. Let length FG be the result of the application of plane H to straight line FE. I say that this length is the length of a simple pendulum having isochronous oscillations with the magnitude ABC agitated about the axis through F.

For, because the sum of the squares of the distances from axis FE, applied to the distance FE multiplied by the number of particles, yields the length of a simple isochronous pendulum, and plane H multiplied by the same number of particles is supposed equal to this sum of squares, therefore, if plane H multiplied by the same number of particles (or, omitting the common multiple, if plane H is applied to distance FE), there will also result the length of a simple isochronous pendulum. Consequently it holds that it will be the length FG. Q.E.D.

If a plane area multiplied by the number of particles of a suspended magnitude is equal to the squares of the distances from the axis of gravitation parallel to the axis of oscillation, if, I say, this area is applied to a straight line equal to the distance between the said axes, there will result a straight line equal to the interval by which the center of oscillation is lower than the center of gravity of the same magnitude.

Let ABCD be a magnitude of which the center of gravity is E and which, suspended from an axis imagined to pass through point F perpendicular to the plane of this page, has center of oscillation G. Next, imagine another axis passing through the center of gravity E and parallel to the axis through F. Having divided the magnitude mentally into very small equal particles, let the squares of the distances from the said axis through E be equal to plane I multiplied by the number of the said particles. If plane I is applied to distance FE, some straight line results. I say that it is equal to the interval EG by which the center of oscillation is lower than the center of gravity of the magnitude ABCD.

For, in order to comprehend what is proposed in a universal demonstration, imagine a plane figure OQP analogous to magnitude ABCD and placed alongside it. That is, if the plane figure is cut by the same horizontal planes as magnitude ABCD, its segments intercepted between any two planes are in the same ratio to one another as the segments of the said magnitude which correspond to them. And let the individual segments of figure OQP be divided into as many equal particles as are contained by the segments corresponding to them in figure ABCD. It is, moreover, possible to imagine this to be done, whatever the magnitude ABCD might be, a line, a surface, or a solid. Indeed, the center of gravity of figure OQP, which is at T, is manifestly at the same height as the center of gravity of magnitude ABCD; and thus, if a horizontal plane drawn through F cuts the line of center of figure OQP, say here at S, the distances ST, FE are equal.

Next, it also holds that the squares of the distances from the axis of oscillation F, applied to distance FE multiplied by the number of particles, yields the length of an isochronous pendulum, which length was assumed to be FG. Indeed, it is quite clear that the sum of those squares is equal to the squares of the distances from the horizontal plane through F plus the squares of the distances from the vertical plane FE drawn through the axis F and the center of gravity E. But the squares of the distances of magnitude ~BAD from the horizontal plane through F are equal to the squares of the distances of figure OQP from straight line SF. Which squares (if O is the highest point of figure OQP and H the center of gravity of the wedge on it cut off by a plane through straight line OV and parallel to SF) are equal to rectangle OTH and the square of ST multiplied by the number of particles of the said figure or of the magnitude ABCD. Indeed, the squares of the distances of magnitude ABCD from plane FE, whatever be the distance of the axis of oscillation F from the center of gravity E, are always the same. Consequently, let us think of them as equal to area Z multiplied by the number of particles of magnitude ABCD.

Thus, since the squares of the distances of magnitude ABCD from the axis of oscillation F are equal to the square of ST, rectangle OTH and plane Z, all multiplied by the number of particles of the same magnitude, if all of these are applied to the distance FE, or ST, there will result the length FG of a pendulum isochronous with magnitude ABCD. But from the application of the square of ST to its side ST, there will result ST, or FE. Therefore, the remainder EG is that which results from the application of rectangle OTH and plane Z to the same ST, or FE.

Wherefore it remains for us to demonstrate that rectangle OTH plus plane Z is equal to plane I. For then it will hold that plane I applied to distance FE yields a length equal to this EG. But this is demonstrated as follows. Rectangle OTM multiplied by the number of particles of figure OQP, or of magnitude ABCD, is equal to the squares of the distances of the figure from straight line XT, which is drawn through the center of gravity T parallel to this ST. And, consequently, the rectangle is also equal to the squares of the distances of magnitude ABCD from the horizontal plane KK drawn through the center of gravity E, since the distances are the same in either case. But plane Z similarly multiplied has been assumed equal to the squares of the distances of magnitude ABCD from the vertical plane FE. And, moreover, it is clear that these squares of the distances from plane FE, together with the said squares of the distances from the horizontal plane through E, are equal to the squares of the distances from the axis of gravity through E, which is parallel to axis F. Thus, rectangle OTH plus plane Z, both multiplied by the number of particles of magnitude ABCD, will be equal to the squares of the distances of the same magnitude from the said axis through E. But plane I multiplied by the same number of particles has also been assumed to be equal to the same squares of the distances. Therefore, plane I is equal to rectangle OTH and plane Z taken together. Which remained to be shown.

From this is further manifest what has been demonstrated in Proposition XVI; to wit, if any magnitude is suspended and agitated in different ways from parallel axes equidistant from its center of gravity, it is isochronous with itself.

For, whether magnitude ABCD is suspended from axis F or from axis L parallel to it, it is clear that in either case the squares of the distances from the axis through E, which is parallel to the axes F or L, are the same. Whence also plane I, which when multiplied by the number of particles is equal to the sum of the squares, will be the same in either case. Indeed, this plane, applied to the distance of the center of gravity from the axis of oscillation, which distance is assumed to be the same in both cases, yields the distance by which the center of oscillation is lower than the center of gravity. Hence also this distance will be the same in both cases. If, for example, under suspension from L, the said distance is EY, this will be equal to EG, and the whole YL equal to GF. And thus, in either suspension, the same simple pendulum isochronous with magnitude ABCD will result.

If the same magnitude suspended at different lengths is agitated, the distances of the axes of oscillation from the center of gravity will be to each other in the inverse ratio of the distances of the centers of oscillation from the same center of gravity.

Let a magnitude, of which the center of gravity is A, be suspended and agitated first from an axis at B and then from an axis at C. In the first suspension let the center of oscillation be D, and in the second let the center of oscillation be E. I say that BA is to CA as EA to DA.

For, since in the suspension from B the distance AD, by which the center of oscillation is lower than the center of gravity, results from applying to distance AB some area which, multiplied by the number of very small equal particle, into which the magnitude is imagined to be divided, is equal to the squares of the distances from the axis through A and parallel to the axis at B, consequently rectangle BAD will be equal to the said area. Also, in suspension from C, since the distance AE results from applying to distance CA the same said area, rectangle CAE will also be equal to the same area. Thus, the rectangles BAD, CAE are equal to each other, and consequently the ratio of BA to CA is the same as that of AE to AD. Q.E.D.

From this it is clear that, given a simple pendulum that is isochronous with the suspended magnitude in one suspension, and given its center of gravity, then in any other suspension, longer or shorter, provided it maintains the same plane of oscillation, the length of the isochronous pendulum is also given.

The center of oscillation and the point of suspension are convertible with one another.

In the above figure, because, assuming suspension from B, the center of oscillation is D, then, inverting everything and supposing suspension from D, the center of oscillation will be B. For this is manifest by the preceding proposition.

How to find the centers of oscillation of plane figures.

If one understands what has been demonstrated up to now, it will then be easy to determine the centers of oscillation in many figures which one is wont to consider in geometry. To speak first of plane figures, we have defined above two sorts of motion of oscillation in them, to wit, either about an axis lying in the same plane as the figure or about one which is perpendicular to the plane of the figure. We called the former agitation in the plane, the latter agitation from side to side.

If it is agitated in the former manner, i.e. about an axis lying in the same plane, such as figure BAD about axis EF, here, if one imagines a wedge on the figure cut off by a plane which so cuts the plane of the figure that the intersection, which here is DD, is parallel to the axis of oscillation, and if the distance of the center of gravity of the figure from this intersection is given, as here AD, and if also the subcentric of the said wedge with respect to the same intersection is given, which is here DH, then one will have the center of oscillation K of the figure BDC by applying rectangle DAH to distance FA, since from this application results the distance AK, by which the center of oscillation is lower than the center of gravity. For rectangle DAH multiplied by the number of particles of figure BAD is equal to the squares of the distances from straight line BAC, which is drawn through the center of gravity A parallel to the axis of oscillation EF. Wherefore, the application of the same rectangle to distance FA yields the distance AK, by which the center of oscillation is lower than the center of gravity A.

From this it is manifest that, if the axis of oscillation is DD, the center of oscillation turns out to be point H. Hence the length DH of a simple pendulum isochronous with figure BAD is then the subcentric of the wedge cut off by the plane through DD and with respect to this DD. This is one result revealed before, but not demonstrated.

But it is not our intention to pursue how one finds the centers of gravity of wedges on plane figures; they are already known for many wedges. For example, if figure BAD is a circle, DH will be equal to 5/8 of the diameter. If it is a rectangle, DH = 2/3 of the diameter. Whence also the reason is clear why a rod, or a line endowed with weight, suspended from either end, is isochronous with a pendulum of subsesquialterate length; that is, by a consideration of the line just as if it were a rectangle of minimum width.

If the figure is a triangle with the vertex upward, DH is 3/4 of the diameter. If the vertex hangs downward, DH is 1/2 the diameter.

But one should know that what was demonstrated in Proposition XVI pertains to this sort of motion of plane figures in the following way: if we give different positions to figure BAD, by inverting it about axis BAC such that it either lies parallel to the horizon or is inclined obliquely, while maintaining the same axis of agitation FE, the length of the isochronous pendulum FK will remain the same. For this is manifest by that proposition.

Next, when a plane figure is
agitated about an axis perpendicular to the plane of the figure,
which we called agitation from side to side, for example, if
figure BAD is moved about an axis which is imagined to pass
through point F perpendicular to plane DCB, here, in addition to
the wedge on the figure which is cut off by the plane drawn
through DD tangent to the figure at its highest point, one must
consider another wedge, which is cut off by a plane through BD
tangent to the figure on the side, which is at right angles to
tangent DD. And one must be given besides the center of gravity A
of the figure and the subcentric HD of the first wedge, also the
subcentric LB of the second wedge. For thus rectangles DAH, BAL
are known, which taken together yield the area to be applied;
this area will henceforth be called the *rectangle *of
oscillation. Applied to distance FA, it will give distance AK, by
which the center of oscillation K is lower than the center of
gravity A.

And if FA is the axis of figure BAD, one can use, in place of the wedge cut off through BD on the whole figure, the wedge on half the figure DBM cut off by a plane through DM. For, if the subcentric of this wedge on DM is OA, and the distance of the center of gravity of the plane figure DBM from the same DM is NA, it holds that rectangle OAN is equal to rectangle BAL. And thus rectangle OAN, added to rectangle DAH, will also constitute the plane to be applied to distance FA, such that distance AK results.

The demonstration of these things
is manifest from the preceding propositions, inasmuch as
rectangles DAH, BAL, or DAH, OAN, multiplied by the number of
particles of the figure, are equal to the squares of the
distances from the center of gravity A, or (what is here the same
thing) from the axis of gravity parallel to the axis of
oscillation. And consequently the said rectangles, applied to
distance FA, yield the length of the interval AK* [* *marginal
note* Prop. 18 of this [part]].

[Huygens then goes on to carry out determinations of the centers of oscillation of the following figures: circle, rectangle, isosceles triangle, parabola, sector of a circle, circumference of a circle, regular polygons; adding on at the end a substantial algebraic treatment of "plane and solid loci," i.e. the complete family of conic sections.]

How to find the centers of oscillation of solid figures.

Furthermore, one may also easily find, by what has been demonstrated previously, the center of oscillation in solid figures. For, if ABC is a solid suspended from an axis which is imagined to be at right angles to this page through point E, and the center of gravity is F, then, having drawn through F planes EFD, GFH, of which the latter is parallel to the horizon and the former passes through axis E, and having found by Proposition XIV the sums of the squares of the distances of the particles of solid ABC from plane GFH and again from plane EFD, that is, having found both rectangles which, multiplied by the number of the said particles, are equal to the said sums of squares, these rectangles, applied to distance EF, by which the axis of suspension is distant from the center of gravity, will give the interval FK, by which the center of agitation K is lower than the center of gravity F. For this is clear from Proposition XVIII. But we will also give some examples.

[Huygens treats: pyramid, cone, sphere, cylinder, paraboloid, hyperboloid, semicone.]

[Omitted are Propositions XXIII through XXV.]