In current notation, it is the curve given parametrically by the equation (x(q),y(q)) = a(q - sin(q),1 - cos(q)), where a is the radius of the circle and q the angle of rotation of the radius to the point in question. It is the resultant of two motions: a uniform translation from 0 to 2 p a and a uniform rotation of the circle which takes the same time. Because of the beauty of the curve and the controversy it created, the cycloid was nicknamed the Helen of Geometry in the 17th Century.

One of the first mathematicians to study the cycloid was Galileo (1564--1643), who tried to find the area under one arch by cutting it out of sheet metal and weighing it, concluding it is about 3 times the area of the generating circle. Many others tried their methods of quadrature on this curve, including Galileo's student Cavalieri (1598--1647).

Gilles Personne de Roberval (1602--1675) was the only professional mathematician in the Mersenne group. He was Professor of Mathematics at the College Royale, a chair contested every three years by competitive examination--set by the incumbent! For this reason, Roberval kept his techniques of integration secret until his death.

Suppose a circle radius a initially at point A on an axis rolls uniformly till the point A meets the axis again at C. Using Roberval's diagram, denote the left hand semicircle by AGB. Divide the axis AC and the semicircle AGB into infinitely many equal parts. Around the semicircle these parts are AE = EF = FG =... Along the axis they are AE = AM, EF = MN, FG = NO.

So rotating the circle A goes to E while sliding along the axis A goes to M , so the resultant A goes to 8, a point on the cycloid. Similarly, 9, 10,.. are points on the cycloid corresponding to N, O , etc.

We suppose that the diameter AB of the circle AEFGB moves parallel to itself, as carried by some other body, until it has arrived at CD and turned through a semi-circle. While it travelled, the point A at the extremity of the said diameter moves round the circumference of this circle AEFGB, and moves as far as the diameter, in such a way that when the diameter is at CD, the point A has reached B, and the line AC is equal to the circumference AGHB. Now, the path of the diameter is divided into infinitely many parts equal amongst themselves and to each part of the circumference AGB which is also divided into infinitely many parts all equal to themselves and to the parts of AC run through by the diameter, as was said.

And then I consider the path of the said point A carried by the two movements: the one of the diameter forwards, the other its proper motion along the circumference. To find the said path, I see that when it has reached E it has risen above its initial position which it has left; the height is marked by drawing the sine El through the point E to the diameter AB, and the versed sine A1 is the height of the said A when it has reached E. Likewise when it has reached F, I draw the sine F2 through the point F to AB, and A2 will be the height of A when it has reached G; and doing this at all the places on the circumference that A runs through, I find all the heights and elevations above the end of the diameter A, which are Al, A2, A3, A4, A5, A6, A7; therefore to find the places which the said point A passes through, and to know the curve which is drawn by the two movements, I carry each of the heights on each of the diameters M, N, 0, P, Q, R, S, T and I find that Ml, N2, 03, P4, Q5, R6, S7 are the same as those taken on AB.

Then I take the same sines El, F2, G3 etc. and I carry them at the height found on this diameter and draw them towards the circle, and the ends of these sines form two curves of which one is A, 8, 9, 10, 11, 12, 13, 14, D and the other A, 1, 2, 3, 4, 5, 6, 7, D. I know how the curve A89D is drawn, but to know which movements produce the other I say that while AB has run along the line AC the point A has risen up the line AB and marked all the points 1, 2, 3, 4, 5, 6, 7 -- the first space while AB reached M, the second while AB reached N, and so always the one space is equal to the other until the diameter has arrived at CD, when the point A has risen to B. That is how the curve A, 1, 2, 3, D is drawn.

Now, these two curves enclose a space, being separated one from the other by all the sines and joining together again at the two ends A, D. Now, each part contained between these two curves is equal to each part of the area of the circle AEB contained in that circumference, for the ones and the others are made of equal lines, i.e. of height Al, A2 etc. and of sines El, F2 etc., which are the same as those of the diameters M, N, 0 etc., and thus the figure A, 4, D, 12 is equal to the semi-circle AHB.

Now, the line A123D divides the parallelogram ABCD in two equally, because the lines of one half are equal to the lines of the other half, and the line AC to the line BD, and consequently, according to Archimedes, the half is equal to the circle, to which on adding the semi-circle, i.e. the space between the two curves, one will have a circle and a half for the space A89DC; and doing the same for the other half, the whole figure of the cycloid will make three times the circle.

To find the tangent to the figure at a given point, I draw a tangent to the circle which passes through the said point, for each point of the circle moves along the tangent to the circle. I then consider the movement which we have given to our point by carrying it along the diameter moving parallel to itself. Drawing through the same point the curve of this movement, if I obtain a parallelogram (which will always have its four sides equal when the path of the point A on the circumference is equal to the path of the diameter AB on the line AC) and if at the same point I draw the diagonal, I have the tangent to the figure which has these two movements as its components, i.e. the circular and the direct. That is how one proceeds when one supposes the movements are equal. If one had supposed that, instead of being equal, the movements had been in some other ratio, the parallelogram would have been constructed with its sides in that ratio.

To find the tangent at any point on the cycloid, Roberval takes the resultant of the tangent to the circle and the longitudinal velocity: as we would say,