3/12-pr-05
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3/12-pr-05 |
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| Ex demonstratis hucusque possumus: | From the preceding demonstration we are able to solve the following problem. |
| PROBLEMA, PROPOSITIO XII. | PROBLEM. PROPOSITION XII |
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| Semiparabolarum omnium amplitudines calculo colligere, atque in tabulas exigere, quae a proiectis eodem impetu explosis describuntur. | To compute and tabulate the amplitudes of all semi-parabolas which are described by projectiles fired with the same initial speed [impetus]. |
| Constat ex praedemonstratis, tunc parabolas a proiectis eodem impetu {30} designari, cum illarum sublimitatis, cum altitudinibus iunctae, aequales {301} conficiunt perpendiculares supra horizontem: inter easdem ergo parallelas horizontalis hac perpendiculares comprehendi debent. Ponatur itaque horizontali cb perpendicularis ba aequalis, et connectatur diagonalis ac: erit angulus acb semirectus, gr. 45; divisaque perpendiculari ba bifariam in semiparabola dc erit ea, quae a sublimitate ad cum altitudine db designatur, et impetus eius in c tantus erit, quantus est in b mobilis venientis ex quiete in a per lineam ab; et si ducatur ag aequidistans bc, reliquarum omnium semiparabolarum quarum impetus futurus sit {10} idem cum modo explicato, altitudines cum sublimitatibus iunctae spatium inter parallelas ag, bc explere debent.Insuper, cum iam demonstratum sit, semiparabolarum quarum tangentes aequaliter, sive supra sive infra, ab elevatione semirecta distant, amplitudines aequales esse, calculus quem pro maioribus elevationibus compilabimus, pro minoribus quoque deserviet. Eligimus praeterea numerum partium decem milia, {20} 10000, pro maxima amplitudine proiectionis semiparabolae ad elevationem gr. 45 factae: itaque tanta supponatur esse linea ba et amplitudo semiparabolae bc. Eligimus autem numerum 10000, quia utimur in calculis tabula tangentium, cuius hic numerus congruit cum tangente gr. 45. Iam, ad opus accedendo, ducatur ce, angulum ecb angulo acb maiorem (acutum tamen) comprehendens, sitque semiparabola designanda, quae a linea ec tangatur, et cuius sublimitas cum altitudine iuncta ipsam ba adaequet. Ex tabula tangentium, per angulum datum bce tangens ipsa be accipiatur, quae bifariam dividatur in f; deinde ipsarum bf, bi (dimidiae bc) tertia proportionalis reperiatur, quae necessario maior erit quam fa. Sit igitur {301} illa fo. Semiparabolae igitur in triangulo ecb inscriptae iuxta tangentem ce, cuius amplitudo est cb, reperta est altitudo bf et sublimitas fo. Verum tota bo supra parallelas ag, cb attollitur, cum nobis opus sit inter easdem contineri; sic enim tum ipsa, tum semiparabola dc, describentur a proiectis ex c impetu eodem explosis: reperienda igitur est altera huic similis (innumerae enim intra angulum bce, maiores et minores, inter se similis, designari possunt), cuius composita sublimitas cum altitudine (homologa scilicet ipsi bc) aequetur ba. Fiat igitur ut ob ad ba, ita amplitudo bc ad cr, et inventa erit cr, amplitudo scilicet semiparabolae iuxta elevationem anguli bce, cuius sublimitas cum altitudine iuncta spatium a parallelis ga, cb contentum {10} adaequat: quod quaerebatur. Operatio itaque talis erit: Anguli dati bce tangens accipiatur, cuius medietati adiungatur tertia proportionalis ipsius et medietatis bc, quae sit fo; fiat deinde ut ob ad ba, ita bc ad aliam, quae sit cr, amplitudo nempe quaesita. Exemplum ponamus. Sit angulus ecb gr. 50; erit eius tangens 11918, cuius dimidium nempe bf, 5959; dimidia bc 5000; harum dimidiarum tertia proportionalis 4195, quae addita ipsi bf conficit 10154 pro ipsa bo. Fiat rursus ut ob ad ba, nempe ut 10154 ad 10000, ita bc, nempe 10000 (utraque enim gr. 45 est tangens), ad aliam, et habebimus quaesitam amplitudinem rc 9848, qualium bc (maxima amplitudo) est 10000. Harum {20} autem duplae sunt amplitudines integrarum parabolarum, nempe 19696 et 20000; tantaque est etiam amplitudo parabolae iuxta elevationem gr. 40, cum aequaliter distet a gr. 45. | (Condition 3/10-th-07) From the foregoing it follows that, whenever the sum of the altitude and sublimity is a constant vertical height for any set of parabolas, these parabolas are described by projectiles having the same initial speed; all vertical heights thus {301} obtained are therefore included between two parallel horizontal lines. Let cb represent a horizontal line and ab a vertical line of equal length; draw the diagonal ac; the angle acb will be one of 45°; let d be the middle point of the vertical line ab. (Condition 3/04-pr-01) Then the semi-parabola dc is the one which is determined by the sublimity ad and the altitude db, while its terminal speed at c is that which would be acquired at b by a particle falling from rest at a. If now ag be drawn parallel to bc, (Condition 3/10-th-07) the sum of the altitude and sublimity for any other semi-parabola having the same terminal speed will, in the manner explained, be equal to the distance between the parallel lines ag and bc. Moreover, since it has already been shown that the amplitudes of two semi-parabolas are the same when their angles of elevation differ from 45° by like amounts, (Condition 3/08-th-05) it follows that the same computation which is employed for the larger elevation will serve also for the smaller. Let us also assume 10000 as the greatest amplitude for a parabola whose angle of elevation is 45°; this then will be the length of the line ba and the amplitude of the semi-parabola bc. This number, 10000, a is selected because in these calculations we employ a table of tangents in which this is the value of the tangent of 45°. And now, coming down to business, draw the straight line ce making an acute angle ecb greater than acb: the problem now is to draw the semi-parabola to which the line ec is a tangent and for which the sum of the sublimity and the altitude is the distance ba. Take the length of the tangent be from the table of tangents, using the angle bce as an argument: let f be the middle point of be; next find a third proportional to bf and bi (the half of bc), which is of necessity greater than fa.Call this fo. (Condition 3/05-pr-02-cor) We have now discovered that, for the parabola inscribed {302} in the triangle ecb having the tangent ce and the amplitude cb, the altitude is bf and the sublimity fo. But the total length of bo exceeds the distance between the parallels ag and cb, while our problem was to keep it equal to this distance: for both the parabola sought and the parabola dc are described by projectiles fired from c with the same speed. Now since an infinite number of greater and smaller parabolas, similar to each other, may be described within the angle bce we must find another parabola which like cd has for the sum of its altitude and sublimity the height ba, equal to bc. Therefore lay off cr so that, ob:ba = bc:cr; then cr will be the amplitude of a semi-parabola for which bce is the angle of elevation and for which the sum of the altitude and sublimity is the distance between the parallels ga and cb, as desired. The process is therefore as follows: One draws the tangent of the given angle bce; takes half of this tangent, and adds to it the quantity, fo, which is a third proportional to the half of this tangent and the half of bc; the desired amplitude cr is then found from the following proportion ob:ba = bc:cr. For example let the angle ecb be one of 50°; its tangent is 11918, half of which, namely bf, is 5959; half of bc is 5000; the third proportional of these halves is 4195, which added to bf gives the value 10154 for bo. Further, as ob is to ab, that is, as 10154 is to 10000, so is bc, or 10000 (each being the tangent of 45°) to cr, which is the amplitude sought and which has the value 9848, the maximum amplitude being bc, or 10000. The amplitudes of the entire parabolas are double these, namely, 19696 and 20000. (Condition 3/08-th-05) This is also the amplitude of a parabola whose angle of elevation is 40°, since it deviates by an equal amount from one of 45°. |
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3/12-pr-05 |
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