Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

SOLƲTION.

Make the side of the given cube = a, the excess of the ••gth above the breadth NO=b, and of the breadth above 〈◊〉〈◊〉 depth PQ=c, make the depth = x, the breadth = x+c, ••d the length = x+b+c. Multiplying therefore these ••ree dimensions together.

〈 math 〉〈 math 〉

•• according to the forms of Baker and Cartes 〈 math 〉〈 math 〉

Wherefore the Central Rule contracted by the supposition ••hich will hereafter follow will be this, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH. •• e. by virtue of the supposition just now mentioned (which ••kes LM viz. a for unity and also for Lat. Rectum)

〈 math 〉〈 math 〉 = AD. And 〈 math 〉〈 math 〉 = DH.

Or more short; 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = AD; and 〈 math 〉〈 math 〉 (∽ 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = DH.

Geometrical Construction. If LM or a be taken for unity or Latus Rectum of the Parabola to be described, that being described (Fig. 50. n. 3.) you are first of all to determine two quantities AD and DH; which may be done two ways; either by Baker's form of his Central Rule, or by ours immediately divided by the quantities of the last Equation.

1. For AD by our form, 〈 math 〉〈 math 〉 = AD, you must make (Fig. 50. n. 1.) as a to b so b to a third (AB to AC so BE to CF) which will be bb, and D{powerof2} the eighth part of this CF must be added to A{powerof2} the half of AB. And by Baker's form you must make, 1. as AB (=a, n. 2.) to AC (=½p i. e. ½b+c) so BE (=¼p i. e. ¼b+½c) to a fourth CF (which will be = 〈 math 〉〈 math 〉▪) 2. Make moreover as AB to AG (a to b) so BH to GI (c to bc) and, as AB to AK (a to c) so BH to KL (c to cc) and the two quantities found GI and KL (bc and cc) being added into one sum will give the quantity q=MN, the half whereof MO will express the quantity 〈 math 〉〈 math 〉 in the

Rule, and to be substracted from the former 〈 math 〉〈 math 〉. Actually therefore to determine the quantity AD not on the Ax, but on another Diameter of the described parabola n. 3. (because the quantity p is in the Equation) having made a perpendi∣cular to the Ax aE=〈 math 〉〈 math 〉 i. e. to the line BE n. 2. and from E having drawn EA parallel to the ax, according to our form AD n. 1. transfer it only on the diameter of the parabola n. 3. from A to D, either by parts 〈 math 〉〈 math 〉 i. e. ½LM from A to c, and 〈 math 〉〈 math 〉 i. e. ⅛CF n. 1. from c to D: But according to Baker's form, first you must put 〈 math 〉〈 math 〉=½LM n. 1. from A to 1. Se∣condly you must put from 1 to 2 the quantity 〈 math 〉〈 math 〉=CF n. 2. Thirdly you must put from 2 to 3 backwards the quantity to be substracted 〈 math 〉〈 math 〉=MO n. 2. which being done, the point D will be determined.

[It is evident by comparing these two ways of Construction, that we may join our forms not incommodiously to Baker's; because by ours the quantity AD was obtained more compendiously than by Baker's, which will also often happen hereafter. And where this Compendium cannot be had, there is another not inconside∣rable one, that, if the quantities AD and DH determined ac∣cording to both ways shall coincide, (which happens in the pre∣sent case) we may be so much the more sure of the truth]

2. As for DH by our form, you must put it from D to e on a perpendicular erected from D on the left hand, the quantity 〈 math 〉〈 math 〉 = BE n. 2. falling here upon the Ax. Then for the quantity 〈 math 〉〈 math 〉 make n. 4. as a to ½bb (LM to LN) so ⅛b to 〈 math 〉〈 math 〉 (MO to NP) and this quantity must be put from e to f in a perpendicular to the Diam. Thirdly, for the quantity 〈 math 〉〈 math 〉

you must farther make n. 4. as a to ½bb (DM to LN) so ¼c to a fourth (MQ to NR) which must be put from f to g. Lastly the quantity 〈 math 〉〈 math 〉 (which is = MO n. 2.) must be put backwards (because to be substracted) from g to H, which is the centre sought. In like manner by Baker's form, first 〈 math 〉〈 math 〉=BE is put from D to 1 even to the Ax. Secondly, for the quantity 〈 math 〉〈 math 〉 make, n. 4. as a to 〈 math 〉〈 math 〉 (LM to LS=CF n. 2.) so 〈 math 〉〈 math 〉 to a fourth (MT=AC n. 2. to SV) and this SV is further put (n. 3.) from 1 to 2. Thirdly, make in the same Fig. n. 4. for the quantity 〈 math 〉〈 math 〉, as a to 〈 math 〉〈 math 〉 (LM to MT) so 〈 math 〉〈 math 〉 to a fourth (LX to XZ) and this XZ is put back∣wads (n. 3.) from 2 to 3, which precisely coincides with the point g. Lastly the remaining quantity 〈 math 〉〈 math 〉 (=MO n. 2. and so by what we have said above, precisely coinciding with the interval g H) is put backwards from g to H the Centre sought.

[Hence it appears again that Baker's form is more laborious than ours; tho' both accurately agree, and hereafter, for the most part, we shall use them both together, tho in' the work it self, rather in Figures, than in that tedious process of words, which we have here for once made use of, that it might be as an Example for the following Constructions.]

Having therefore found by one or both ways the Center H, and thence described a circle thro' the vertex of the parabola A, the intersection N will give the perpendicular to the dia∣meter NO, the value of the quantity x sought.