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

PROBLEM VIII.

THere is given AB the capital line of a horn work (which we represent (tho' rudely) n. 1. Fig. 56) and the Gorge AD, also part of the line of defence EF, to find the face BE, the flank DE, the curtain (or the chord) DF, also the angle of the Bastion ABE, &c. and so the whole delineation of the horn work. It is evident if you have the flank DE or the curtain DF, the rest will be had also. Suppose therefore, the capital line AB and the gorge AD, and part of the line of defence EF to be of the magnitudes denoted by the Letters a, b, c, on the right hand.

SOLƲTION.

Make AB=a, AD=b, and EF=c, and DF=x; then will AF=x+b, and by reason of the similitude of the ▵▵BAF and EDE and ECB, as FA to AB so FD to DE 〈 math 〉〈 math 〉

But now □ □ DF+DE are = = □EF i. e. 〈 math 〉〈 math 〉; or giving the same denomination to all the quan∣tities on the left hand, 〈 math 〉〈 math 〉

and multiplying both sides by 〈 math 〉〈 math 〉 and translating all the quantities on the right hand by the contrary signs to the left, 〈 math 〉〈 math 〉

Wherefore (putting a for 1 and L. Rectum) the Central Rule will be, 〈 math 〉〈 math 〉 (because the quantity q is negative in the E∣quation, for cc is greater than aa+bb)=AD; and 〈 math 〉〈 math 〉 = DH. or according to our Redu∣ction, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = DH.

Wherefore the Geometrical Construction requires no other preparatory determination by our form than of the quantities 〈 math 〉〈 math 〉 for AD, 〈 math 〉〈 math 〉 for DH at the center, and bbcc to determine the radius of the circle; which are exhibited by n. 2. Fig. 56. viz NP is = cc, RS=bcc, RV=bbcc, which are found by means of LM=a, LR=b, LN and MO=c, MQ=NP and MT=RS. Having therefore described a para∣bola, n. 3. and drawn its diameter, transfer AD=½NP upon it (because the quantity p is in the Equation) and also ½RS from D upon H perpendicularly, and on the right hand, (because DH=−〈 math 〉〈 math 〉;) and so you'l have the center H;

thro' which having drawn KAI so that AK shall be = to the quantity bbcc or S, i. e. RV, &c. a circle described at the interval HL will cut the parabola in M and N, and applying the magnitude NO it will be that of the Curtain sought; up∣on which, n. 4. having laid down the circumference of the horned work by help of the given lines AB and AD, you'l have the line EF, of the magnitude which was above supposed. Now if any one has a mind to do the same thing by Baker's way; by laying down first the interval AB =〈 math 〉〈 math 〉 and then making bc=〈 math 〉〈 math 〉, and lastly, putting cd for the quantity 〈 math 〉〈 math 〉; he will fall upon the same point D, and in like manner may express the other parts of the Central Rule by the intervals De, ef, fg, and setting back the last gh, he will fall upon the same center H: But this is done with a great deal more trou∣ble and labour to determine so many quantities, and also is in more danger of erring by cutting off so many parts separately, as experience will shew; and thus we have by a new argument shewn the advantage of our Reduction.

Another Solution of the same Problem.

Things remaining as before (only assuming the given lines AB, AD and EF, n. 1. Fig. 56. one half less, that the Scheme may take up less room) make BE=x, as the first or chief unknown quantity; then will BF=x+c, and its □xx+2cx +cc: And since as BE to BC=AD so BF to AF a fourth, which will be 〈 math 〉〈 math 〉 and its square = 〈 math 〉〈 math 〉. Where∣fore if this square be substracted from the square of BF, there will remain the square of BA, i. e.

〈 math 〉〈 math 〉; i. e. all being reduced to the same denomination, 〈 math 〉〈 math 〉;

or according to the forms of Cartes and Baker, 〈 math 〉〈 math 〉

Therefore the Central Rule (putting again a for 1 and the L. R.) will be 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH; or by our Reduction, 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = AD; and 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉=DH.

Geometrical Construction. Having therefore described a para∣bola (Fig. 56. n. 5.) and drawn the diameter Ay, make A1=a and 1, 2=〈 math 〉〈 math 〉, so you'l have the point D; make moreover D{powerof3} or 2, 3=c, and backwards 3, 4=〈 math 〉〈 math 〉 (we here omit to express the Geometrical determination of these quantities 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉 as being very easy) and you'l have the point H, &c. and there will come out the quantity sought NO; which since it is equal to half BE n. 4. the business will be done; which Baker's form will also give, exactly the same, but after a more tedious process.