Page:Newton's Principia (1846).djvu/315

rectly and the same particle of time inversely. Therefore, since the resistance is, by the supposition, as the square of the velocity, the increment of the resistance will (by Lem. II) be as the velocity and the increment of the velocity conjunctly, that is, as the moment of the space and V - R conjunctly; and, therefore, if the moment of the space be given, as V - R; that is, if for the force V we put its exponent PIGR, and the resistance R be expressed by any other area Z, as PIGR - Z.

Therefore the area PIGR uniformly decreasing by the subduction of given moments, the area Y increases in proportion of PIGR - Y, and the area Z in proportion of PIGR - Z. And therefore if the areas Y and Z begin together, and at the beginning are equal, these, by the addition of equal moments, will continue to be equal and in like manner decreasing by equal moments, will vanish together. And, vice versa, if they together begin and vanish, they will have equal moments and be always equal; and that, because if the resistance Z be augmented, the velocity together with the arc Ca, described in the ascent of the body, will be diminished; and the point in which all the motion together with the resistance ceases coming nearer to the point C, the resistance vanishes sooner than the area Y. And the contrary will happen when the resistance is diminished.

Now the area Z begins and ends where the resistance is nothing, that is, at the beginning of the motion where the arc CD is equal to the arc CB,

and the right line RG falls upon the right line QE; and at the end of the motion where the arc CD is equal to the arc Ca, and RG falls upon the right line ST. And the area Y or $\scriptstyle {\frac {OR}{OQ}}$ IEF - IGH begins and ends also where the resistance is nothing, and therefore where $\scriptstyle {\frac {OR}{OQ}}$ IEF and IGH are equal; that is (by the construction), where the right line RG falls successively upon the right lines QE and ST. Therefore those areas begin and vanish together, and are therefore always equal. Therefore the area $\scriptstyle {\frac {OR}{OQ}}$ IEF - IGH is equal to the area Z, by which the resistance is expressed, and therefore is to the area PINM, by which the gravity is expressed, as the resistance to the gravity.   Q.E.D.