Bd, Enlargement of the oesophagus, armed with chitinous teeth..
1, Bd) are very intimately fused together to form what is called the "lower lip" or labium, a firm transverse plate representing the fused basal portions of the maxillae, which may carry a small median "ligula," representing apparently the fused inner maxillary lobes, a pair of paraglossae (outer maxillary lobes), and a pair of palps.
(1901), p. 547; Staub, Semon's Forschungsreisen (5 Bd., 1900); C. B.
From B the curve of equilibrium (BD) between rhombic and liquid sulphur proceeds; and from C (along CE) the curve of equilibrium between liquid sulphur and sulphur vapour.
Of especial interest is the 0 curve BD; along this line liquid and rhombic sulphur are in equilibrium, which means that at above 131° and 400 atmospheres the rhombic (and not the monoclinic) variety would separate from liquid sulphur.
Cantor's Vorlesungen fiber Geschichte der Mathematik (Leipzig, 1st Bd., 1880; 2nd Bd., 1892; 3rd Bd., 1898; 4th Bd., 1908; 1st Bd., von den eiltesten Zeiten bis zum Jahre 1200, n.
Chr.; 2nd Bd., von 1206-1668; 3rd Bd., von 1668-1758; 4th Bd., von 1 795 bis 1 799); W.
We may in any relation substitute for any pair of quantities any other cogredient pair so that writing -}-d 2, -d l for x 1 and x 2, and noting that gx then becomes (gd), the above-written identity bceomes (ad)(bc)+(bd)(ca)+(cd)(ab) = 0.
(IV.) and herein writing d 2, -d 1 for x l, x2, 2 (ac) (bc) (ad) (bd) = (bc) 2 (ad) 2 +(ac) 2 (bd) 2 - (ab) 2 (cd) 2.
From (ac) 2 (bd) 2 (ad)(bc) we obtain (bd) 2 (bc) cyd x +(ac) 2 (ad) c xdx - (bd) 2 (ad)axb x - (ac)2(bc)axbx =4(bd) 2 (bc)c 2.
In the case of oblique reflection at an angle cb, the error of retardation due to an elevation BD (fig.
5) is QQ'- QS = BD sec 4:,(I - cos SQQ') = BD sec cI)(1 +cos 20) = 2BD cos 4); from which it follows that an error of given magnitude in the figure of a surface is less important in oblique than in perpendicular reflection.
Cl., 15, Bd., iii.
19) represent a gun at height BD above water-level DC, elevated to such an angle that a shot would strike the water at C. Draw EB parallel to DC. It is clear that under these conditions, if a tangent sight AF be raised to a height F representing the elevation due to the range BC, the object C will be on the line of sight.
Then MA'B'N is a right trapezium, whose area is equal to that of Cabd; and it is related to the latter in such a way that, if any two lines parallel to AC and BD meet AB, CD, MN, A'B', in E, G, P, E', and F, H, Q, F', respectively, the area of the piece PE'F'Q of the right trapezium 'B.
BC, BD, AD, in P, Q, R, S (fig.
By drawing Ac and Ad parallel to BC and BD, so as to meet the plane through CD in c and d, and producing QP and RS to meet Ac and Ad in q and r, we see that the area of Pqrs is (x/h - x 2 /h 2) X area of cCDd; this also is a quadratic function of x.
Patricius, § BD) he decided to throw in his lot with the cause of Rome.
If these are first drawn it is easy, for any position of the loads, to draw the lines B'C, B'D, B'E, and to find the sum of the intercepts which is the total bending moment under a load.
Let a frame consist of the five members AB, BD, DC, CA, CB (fig.
Arbogast's rule of the last and the last but one; in fact, taking the value of a to be unity, and, understanding this letter in each term, the rule gives b; c, b2; d, bc, b; e, bd, c, b c, b, &c., which, if b, c, d, e, &c., denote I, 2, 3, 4, &c., respectively, are the partitions of 1, 2, 3, 4, &c., respectively.
This quantity may readily be expressed in terms of the refractive indices for the three colours, for if A is the angle of the prism (supposed small) bc=(/1c - I)A, bD =(/ AD - OA, F - I)A, where µc, A n, µ F are the respective indices of refraction.
8), joined A and B with 0 the centre, bisected the angle AOB by OD, so that BD became the semi-side of a circumscribed regular 12-gon; then as AB :BO: OA :: I: d 3: 2 he sought an approximation to X 1 3 and found that AB: BO >153:265.
Next he applied his theorem 4 BO+OA: AB:: OB: BD to calculate BD; from this in turn he calculated the semi-sides of the circumscribed regular 24-gon, 48-gon and 96-gon, and so finally established for the circumscribed regular 96-gon that perimeter: diameter 3+V :I.
9): ACB being a semicircle whose centre is 0, and AC the arc to be rectified, he produced AB to D, making BD equal to the radius, joined DC, 1 Vieta, Opera math.
Io) so that BD was equal to r, he produced it E only so far that, when the extremity D' was joined with circle was equal to r; in C, the part D'F outside the A o FIG.
Rod as in Foucault's construction, and BD is another rod which can be set to the direction in which the rays are to be reflected.
He had now three distinct space-units, i, j, k; and the following conditions regulated their combination by multiplication: - I T = 12 '=' 2 = _ 1, ij= - ji=k, jk= - kj=i, ki= - ik =j.3 And now the product of two quaternions could be at once expressed as a third quaternion, thus (a+ib+jc+kd) (a'+ib'+jc'+kd') = A+iB+jC+kD, where A=aa' - bb' - cc' - dd', B = ab'+ba'+cd' - dc', C = ac'+ca'+db' - bd', D =ad' +da'+bc' - cb'.
Be the middle points of AC BD...; then BH, CK...
Of the vectors AB and AC (or BD), and of their sum AD, on a line perpendicular to AO, this is obvious.
As a simple example, take the case of a light frame, whose bars form the slides of a rhombus ABCD with the diagonal BD, suspended from A and carrying a weight W at C; and let it be required to find the stress in BD.
If we remove the bar BD, and apply two equal and opposite forces S at B and D, the equation is W.I(2lcosO) + 2S .1(1 sin 8)=o, A where 2 is the length of a side of the rhombus, and 8 its inclination to the vertical.
Set off ab = ac = 1/2p. Draw radii bd, Ce; draw fb, cg, making angles of e 753/4 with those radii.
B 0 c relatively to the fixed, ~ ___a 4 - link a, a fact indicated o o ~ by the suffix ac placed bd ti after the letter 0.