Fitzwilliam Notebook

1665

March 25 1666.

1666.

I went into the Countrey December 4^th 1667.

I returned to Cambridg Feb 12. 1667.

I went to London on Wednesday Aug 5^t & returned to Cambridge on Munday Sept 28, 1668.

Aprill 1669.

Friedadolff Lewys Langerhanss.

Carolus Gottlob de Theler

Georgius Bernhardus de Theler Equites ex Superiore Lusatia

Iohannes Christophorus Ritter Wurcenâ-Misnicus.

Sep. 25 1727
Not fit to be printed

T Pellet

Nova
Cubi Hæbræi Tabella

Of right angled triangles.

h = hypotenusa.
b = basis.
c = Cathetus.
p = perpendicular.
hdc = diff: hypot & Cath
bdc = diff: basis & cathet:
bdh = difference basis & hyp{ot}
dsh = diff: seg: hypoten:
sh = segment: hypoten:
bh = greater seg hyp:
ch = lesse seg: hypot:

I. Any two leggs given to find the other

1. bq + cq = hq.

2 r: hq - bq: = c.

3 r: hq - cq = b

Eucl. lib 1. pr: 47.

II the b. c. & h given to find p.

1. $\frac{\mathrm{b\; x\; c}}{\mathrm{h}}$ = p Euclid 6 .8.

III c. h. p. given to find dsh.

1. H - 2r: bq - pq: = dsh.

IIII. b. p. h given to find dsh.

1. 2r: bq - pq: h = dsh.

V. b. c. h given to find dsh.

1. H - 2r: cq - Q: $\frac{\mathrm{b\; x\; c}}{\mathrm{h}}$: = dsh.

2 2r: bq - Q: $\frac{\mathrm{b\; x\; c}}{\mathrm{h}}$: + h = dsh.

VI b.c or b. h or h. c given to find p:

1 $\frac{\mathrm{b\; x\; c}}{\mathrm{r:\; bq\; +\; cq:}}$ = p

2 $\frac{\mathrm{b\; x\; r:\; hq\; -\; bq:}}{\mathrm{h}}$ = p.

3 $\frac{\mathrm{c\; x\; r:\; hq\; -\; cq:}}{\mathrm{h}}$ = p

VII b. h. or c. h. or b. c given to find dsh.

Theorem 1

As the difference twixt the base & cath (in rectang: triang:) is to the greater side:: so is the difference of the segment of the base; to the greater segment of the base & perpendicular.

Theorem 2.

As the difference twixt the base & cathetus to the less side:: so the diff of the segments of the base to the lesse segment of the base & perpendicular

Theorem 3^d.

base – Cathetus: hypotenusa:: :: greater segment: base - less seg base : base + Cathetus.

Theor. 4.

If within a circle be described an Ellipsis touching the Circle in 2 opposite points if the Diameter cut it at right angle in any points except the touch point yⁿ a line drawn fm either touch point perpendicular to the former diameter will bisect it & being produced will cut the circle in the other touch point & all the lines drawne twixt the circle & that line <4v> parallell to that diameter shall be divided by the Ellipsis so as one segment shall bee to the other as the segments of the semidiameter are to one another they being divided by the same Ellip: let ab bee equall to 10 pts. eb = 157979 = Periph: & priph - Rad: Rad:: Rad: db. db = 175, 1938394. de = 18,1142067

To describe an ellipsis

Let fe & gc be two lines ef make right angles with gc. let a point be taken in bd as at a & let that point move along the line gc. & d the one end of the line db move on the line ef & the other end b shall describe the Ellipsis gbc. f.

Let c & a be two fixed points about which let a loose cord be put haveing both ends tyed together. as is signified by the 3 lines cb. ba. ac. Strech it out with another point as b. & keeping it so streched out draw the point b about & it shall describe the Ellipsis bd. Chartesij Dioptr

Let the line ae be infinitely extended in it take the point o about the line oc shall turne at the point c in oc let the point c in the line ab be fastened & let a the end of the line ab move on the line ae & oc turning round, each point of the line ab betwixt ac will describe an Ellipsis whose transvers axis is equall to oc & parallell to ae but each point on the other side c describes Ellipsis whose right axis <5v> is equall to oc & parallell to ae

Extend de both ways take the lines ca & ab equall to one another fasten together at one end as at a. set the other end of ca at the point c in db. & let the other end of ab slide on db. yⁿ take a point in ab as o & turne ac about & it shall describe the ellipsis dgoe Shooten in lib. 2^d Cartesij Geometria:

Cut the cone abc so that the diam of the section ed produced cute the base of the triangle ac produced without the cone as at r & makes right angles with gh the base of the section

If eg be moved twixt the lines ed & gd. a point in it as (θ) shall describe an ellipsis whose semi-axis ad is equall to bd & semiaxis dc = eb

If dc revolve abute the center d. & to the other end b be fastend a triangle bca & db = ba = bc & the angle a moves on the line ad the other end c will describe the streight line cd & the angle cba = 2cda & a point in the line (ca) as (e) shall describe an Ellipsis ehg whose diam 2dh =²dg = ²ec & the other diameter conjugated to it is od & od = $\sqrt{\mathrm{4db }x \mathrm{db }- \mathrm{ec }x \mathrm{ec }- {\mathrm{ }}^{\mathrm{2 }}\mathrm{ec }x \mathrm{ea }}$ for op = ec. oq = ea. dp = 2db.

& if in the line bc be taken a point as s, it shall describe an ellipsis the one diam: being ²ab + ²bs, the other diam = 2cs.

If o & a be the foci & cp = oa & ca = op = it theire section in s shall describe an ellipsis

If ab = bc = ci = ai = if or greater yⁿ (if) & bh = fp & ac bisects the angles bai. bci. yⁿ if bh turne round the intersections of bh & ac shall describ{e} an Ellipsis. & hi & i are the foci.

To describe a Parabola

Let bc fall perpendicular on ad & let c the one end there of move uppon ad a given line & if bc x k a given line be equall to ac x cd yⁿ shall b the other end of bc describe the Parabola afd.

Draw ah perpendicu{lar} to ap. & ab from ah parallell to ap divid{e} bh into equall parts as bcdefgh. & divide ap into parts equall to the former as iklmnop. draw lines cros to each part of the lines ah & ap as cb. kc. ld. me. nf. &c with half of each line descri{bing} a circle as brc with $\frac{1}{2}$ cb. from bu in the poi{nt} cut by the diameters of the circle draw lines perpendicular to the diameter <7v> untill they reach the circle from whose diameter they are drawne as the lines pw, qx, ry, sz, t&, u+. Erect those lines perpendicular to the line bu as p♉, q♈, r♊, s♋, t♌, u♍. & by the end of those lines draw a line & it shall be a parabola . as b♉♈♊♋♌

If abc be a cone: de (the diameter of the Section fgd) parallell to ac: & fg (the base thereof) cutting bc at right angles yⁿ is the section dfg a Parab

Make db perpendicular to ef on the center b let the right angled figure pbgh turne. Let gh move perpendicularly on ef ever intersecting ef & bh in one point yⁿ pbgh moveing rownd the intersections made twixt pg gh describe the parabola qbg.

If ab = bd = do = ao is greater then ac & ac = cs the corner (a) fasten{ed} to the focus (a) . & the line de fastened to the corner d & moveing perpendicularly o{r} on sd & the line boe crossing the corners b & o. yⁿ the line boe & de at theire intersections shall describe a Parab & the line boe always toucheth the Parabola in (e) &c

If (d) be the focus od = oe the ruler fc = to the thred fad & thred fastened to the ruler at f & to the focus d & the ruler move perpendicular to ce & parallell to de. yⁿ the parting of the thred from the ruler as at (a) shall describe a Parabola

To describe an Hyperbole

Let fa fall on ag suppose at right angles let one end of the line lg move up & downe in the line fa & towards the other end let it cut the line ga in g. let mp keepe parallel to df haveing one end p moveing in the line fa but yet keeping an equall distance from l the end of gl. that is let the triangle npl be immutable. let yⁿ the lines mp & gl thus move to & fro & theire intersections shall describe an Hyperbola. & the rectangle de x ea = ic x cb = qo x op. Cartes Geom:

ffasten a pegg as at a & another as at b upon which let the line de be turned at the pin a fasten one end of a cord & the other at e the end of the line de. yⁿ streching the cord from a & e with the pin c turne de about & the pin c will slip towards e & describe $\frac{1}{2}$ the Hyper: oce

{If} the rectangle twixt ad & db is equall to the rectangle twixt ae & ec {so} that each point c in the Hyperb: bc is found by makeing ec = $\frac{\mathrm{ad\; x\; db}}{\mathrm{ae}}$ or ae = $\frac{\mathrm{ad\; x\; db}}{\mathrm{ec}}$. also be x ce = be x da - db x ec

Cut the cone abc so that the diameter of the section er produced cuteth one side of the Cone bc produced as at d. the base thereof gh cutteth ac the base of the triang: abc at right angles.

If (of) touch the Hyperb: & (as) be its transverse diam: & (gb) keepe parallel to (eo) & (cag) aways pass through (a). the vertex of the Hyperb. & (bc) be always in the line (fh) fastend to (gb) & equall to fd = de = $\frac{\mathrm{fh}}{4}$. yⁿ the lines (agc) & (gb) moveing by theire intersection shall describe an Hyperbola whose asymtotes are oea, fe; eb, eb, & wx is a right line conjugate to the transverse diameter (as.) viz: it is the right diameter

If dk = er be (latus transversum) & de = kr, be latus rectum yⁿ is sd = sr = se = sk = sa = sx. at (a) & (x) fasten 2 pins on which let the (acbp, xobq) revolve, & if ac = ox = zi = dk = er, & co = ax yⁿ the intersection of the lines cabp, & qbox (when they move) shall describe a Hyperb whose focus is a, & the opposite Hyperbola (whose focus is x is described by the same lines after qbox, esk & cabp are parallell

If de = dc = ex = cx is not lesse yⁿ ix = az & 2 of theire ends loose pind together at (e) & 2 at (c) on which 2 corners lyes the line (coe) two of theire ends are loosely pinnd on the focus (x) the last two are pind on the line (adp) at (d) soe that the ruler adp being pinnd to the focus (a), ad = zi yⁿ the intersections of the lines (adp, coe) describe the Hyperbola oiq. & after they are parallell they shall describe the opposite Hyperbola hzk.

The Asymptotes aq, an, & (m) point in the Hyperbola draw mq || an. & mn || aq. Then draw en at a venture & make er = mc || er & r shall bee a point in the Hyperbola

If the position of the Asymptotes (ad) (ab) bee given & any point as (c) in the Hyperbola. then draw ucbf || ad. ud || ab || fg making bf = bu = 4bc. Then at a venter draw bewh, through the point b. & make ak = fh = uw Or dw = bk & from the point k draw ke, which shall touch the Hyperbola. in n, if kn = ne.

The foci (a, d) & (c) a point in one Hyerbo{la} given to describe them.

Draw ac, cd, from the given point c to the foci, yⁿ upon the center c with any radius ce describe the circle erf. soe that ec = ef. yⁿ with the Rad ae & df upon the centers a & d describe the circles hep fhp their points of intersection p, h, shall bee in the hyperbola. The intermediate distance twixt divers points thus found may bee completed by the helpe of tangent lines or circles or a steady hand.

The properties of the Parabola

ab = a. bc = b. ac = c. eb = d. ei = x. fi = y. b : c :: x : (ik)$\frac{\mathrm{cx}}{\mathrm{b}}$. a : c :: d : (es, or il)$\frac{\mathrm{cd}}{\mathrm{a}}$ whence yy = $\frac{\mathrm{ccd}}{\mathrm{ab}}$x. ab : cc :: d : (en)$\frac{\mathrm{ccd}}{\mathrm{ab}}$. $\frac{\mathrm{ccd}}{\mathrm{ab}}$ = r. rx = yy. that is ne a given line multipling ei = if square. Or breifly a : c :: d : (es or il)$\frac{\mathrm{cd}}{\mathrm{a}}$. b : c :: $\frac{\mathrm{cd}}{\mathrm{a}}$ : (en)$\frac{\mathrm{ccd}}{\mathrm{ab}}$ = r Ne is called latus rectum of Apollon & Parameter by Mydorgius. gh is its base ed its Diameter.

ang pbh = phg. kg parallell to ac tangent no parallell to the tangent ac. yⁿ nm = mo. (2). db x bk = kg x kg.

kg x kg : nm x nm :: db x bk : db x bm :: bk : bm

a = foco. ac = $\frac{1}{4}$ lateris recti. ac = oc. ah = do. sit (sh) Parallela ad. (dr) & (rh) contingat Parab: in h. & (dh) perpend: ad (dr) erit ang : ahr = rhs.

If cs = sb & su parallell to ab yⁿ the triang cea : cab :: l : 4. & so it may be saide infinitely.

If ab & cd, are ordinately applyed the Parabola ceadb is to the triangle cda as Eight to six. & rf x rf = rs x re. or, re : rf :: rf : rs.

If rs, is parallell to gx yⁿ are the 2 segments of Parabolas gproxa = gcsqxa) equall & po = cq. & if ga = ax then the diameters ar as cut the line rs in its touch points.

The properties of the Hyperbola

rx + $\frac{\mathrm{acxx}}{\mathrm{bb}}$ = yy.

☞ rx + $\frac{\mathrm{r}}{\mathrm{q}}$ xx = yy. for

$\frac{\mathrm{acq}}{\mathrm{bb}}$ = r. & $\frac{\mathrm{ac}}{\mathrm{bb}}$ = $\frac{\mathrm{r}}{\mathrm{q}}$.

{am} = a. mb = b. mc = c. de = q. ei = x. di = q + x fi = y. b : c :: q + x : (il) $\frac{\mathrm{cq\; +\; cx}}{\mathrm{b}}$ b : a :: x : (ik)$\frac{\mathrm{ax}}{\mathrm{b}}$. il x ik = yy = $\frac{\mathrm{cqax\; +\; caxx}}{\mathrm{bb}}$ bb : ac :: q : (en)$\frac{\mathrm{acq}}{\mathrm{bb}}$(r). bb : ac :: q : $\frac{\mathrm{acq}}{\mathrm{bb}}$ :: :: x : (qpcron) or $\frac{\mathrm{acx}}{\mathrm{bb}}$. whenc $\frac{\mathrm{acqx\; +\; acx}}{\mathrm{bb}}$ = rx + $\frac{\mathrm{acxx}}{\mathrm{bb}}$ = pi x ie = yy

More breifly thus.

b : c :: q :: (es)$\frac{\mathrm{cq}}{\mathrm{b}}$ : b : a :: $\frac{\mathrm{cq}}{\mathrm{b}}$ : $\frac{\mathrm{acq}}{\mathrm{bb}}$ (= r)

de is called latus transversum & en latus rectum by Appolonius. but Parameter by Mydordgius.

mn = pd = bq = q. fg = db = pq = p nu = x. au = y. ha = ck = b. st = r (1) q : r :: qx + xx : yy. & yy = rx + $\frac{\mathrm{rxx}}{\mathrm{q}}$ = yy. (2) 2by + bb :: = $\frac{1}{4}$ pp. (3) q : p :: p ; r. (4) q : r :: qq : pp. (5) yy : qx + xx :: qq : pp.

pq = fg = db = axi secundo, & recto & diam rectæ

pd = mn = qb = axi primo, transverso & lateri sive diametro transversæ.

st = r = Lateri recto.

If xt = p. sr = q : r = Parameter & iry = a. eno = b. in = y en = z. rn = x.

Then if p = q = r as in (a) : (a) is the simplest of all Hyperbola's, & yⁿ, yy = xx + qx. & if (q) is the same in both (a & b) & (xt = p) is propper to (b) then yy : zz :: qq : pp. & therefore Hyperbolas are to one another as theire rigt axis are supposeing theire transverse axes equall. viz iryeon : eron :: in : en :: p : p. therefore if (rs) is parallell to ao, & ae = co. yⁿ (arextc = csoext.) & if at = te = cx = xo tr & xs (cutting rs in the touch points) are ordinately applyed to the Diameters & bisect the Hyperbolas.

If (o & a) are the foci & (u) a point in one of the Hyperb: s. then au + ei = ou & if as = ei = or. yⁿ us = uo. & rs = oa & (iu) bisecting the angle (ria.) it shall touch the Hyperb in u.

The Properties of the Ellipsis

rx - $\frac{\mathrm{acxx}}{\mathrm{bb}}$ = yy. that is

rx - $\frac{\mathrm{r}}{\mathrm{q}}$xx = yy for

$\frac{\mathrm{acq}}{\mathrm{bb}}$ = r & $\frac{\mathrm{ac}}{\mathrm{bb}}$ = $\frac{\mathrm{r}}{\mathrm{q}}$

am = a. bm = b. cm = c. ed = q. ei = x. id = q - x fi = y. en = r

b : c :: q - x : (ib)$\frac{\mathrm{cq\; -\; cx}}{\mathrm{b}}$. b : a :: x : (ik)$\frac{\mathrm{ax}}{\mathrm{b}}$ ki x il = $\frac{\mathrm{cqax\; -\; acxx}}{\mathrm{bb}}$ = fi x fi = yy

bb : ac :: q : en = $\frac{\mathrm{aqc}}{\mathrm{bb}}$ = r. bb : ac :: x : on = $\frac{\mathrm{acx}}{\mathrm{bb}}$ wherefore rx - (onx) = $\frac{\mathrm{acxx}}{\mathrm{bb}}$. = yy.

Af = q = axi transo: sive primo : ch = p. fg = r = lateri recto. ad = x df = q - x. dh = y.

(1) q : p :: p : r. (2) yy : xq - xx :: r : q. therefore 3 yy = rx - $\frac{\mathrm{rxx}}{\mathrm{q}}$ as before.

af is the first & transverse axis or side

ch is the seacond & right axis

fg is the Parameter or right side

sit p = qn. nc = no. erit segmentum oeth ad segmentum cbd, ut cbd ad gbhcd :: fh : ab :: (afbhcd) elipsis {illeg} : (ahbg) circulum.

If the lines (pq, rs) are parallell & co the common axis of both the Ellipses yⁿ are the 2 Ellipses equall to one another, for ax = be. the conjugated diam: cut the touch points of pq, rs & parallells to these are also conjugated.

If ab touch an Ellipsis & (o) & (x) be the foci yⁿ the angle aco = bcx. & if (ocx) be bisected by (cr) yⁿ acr = bcr = right angle

If xu = ot = ys. & uo bisected in a then uac = oac = to a right angle.

If also ut = ox & ut & xo be produced till they meete in h. the angle uho shall be bisected by the line acb.