Encyclopaedia Br
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共 86 题- ENCYCLOPAEDIA BRITANNICA
- ELEVENTH EDITION
- VOLUME VIII, SLICE IV
- Diameter to Dinarchus
- ARTICLES IN THIS SLICE:
- PLATE I.
- PLATE II.
- From Gardner Williams's _Diamond Mines of South Africa_. FIG. 8.]
- The following are some of the most famous diamonds of the world:--
- FOOTNOTE:
- FOOTNOTE:
- C6H5NH2.HNO3 + HNO2 = C6H5N2.NO3 + 2H2O,
- FOOTNOTES:
- DICTIONARY.
- AUSTRALIA AND POLYNESIA.
- EUROPE
- TEUTONIC: (1) _Scandinavian._
- ASIA
- AFRICA
- AUSTRALIA AND POLYNESIA
- AMERICA
- FOOTNOTES:
- FOOTNOTES:
- TABLE I.
- Substitutes for fresh meat when the latter is not available:--
- TABLE III.
- TABLE IV.--_Weekly, per Statute Adult._
- The Russian conscript is allowed weekly:--
- TABLE I.--_Percentage Composition of some Common Food Materials._
- FOOTNOTES:
- 5. As an example, take the arithmetical series
- This gives the relation between [Delta] and D. Also we have
- Adding a series of similar expressions, we find
- For practical calculations this may more conveniently be written
- Another simple form is
- A third simple and important form is that denoted by
- An extended form of Clairaut's equation expressed by
- M dy/dx = N,
- A linear differential equation of which the left side has the form
- The equation
- More generally the equation
- The equation
- Another method of solution of the equations
- Equations occur to be integrated of the form
- Now compare with the given equation another equation
- AdF/dx + BdF/dy = dF/dt
- We have above considered the integration of an equation
- Z - z = p(X - x) + q(Y - y)
- U0 = dz0/du - p0dx0/du - q0dy0/du
- Hence this equation involves dz - [psi]dx - qdy = 0, or we have
- Putting x = 1/t, it becomes
- FIG. 1.
- If dS = 2[pi]xdx, we have for the whole effect
- To calculate the roots of (5) we may assume
- Writing for brevity
- Now by definition
- The value of C for an annular aperture of radius r and width dr is thus
- The roots of J0(z) after the first may be found from
- The maxima of C occur when
- Now by (17), (18)
- When z is great, the descending series (10) gives
- Thus, if [lambda] = 1/4000, R = 1/10, we find
- FIG. 13.--xy. FIG. 14.--x^2y. FIG. 15.--y^3.]
- Now as far as [omega]^4
- 4 sin^2 1/2[omega] = sin^2[omega] + 1/4sin^4[omega],
- I = u^2 + v^2 + 2uv cos[alpha],
- By differentiation with respect to g1 it may be proved that
- The integrated intensity, I', or
- I = [pi]2h1 + (2h1 - [sqrt] g1^2) cos[rho]' (12).
- With the best thickness
- Now
- Thus, if we take
- C = 1/2 - G cos u + H sin u, S = 1/2 - G sin u - H cos u (14).
- From the series for G and H just obtained it is easy to verify that
- If a be very great in relation to b, the condition becomes
- In particular, if
- The first of equations (9) gives
- In like manner we find
- FOOTNOTES:
- These involve
- From (2) (3) we find by elementary algebra
- Hence equations (1) (2) gives
- FOOTNOTES:
- TABLE I.
- Physical abnormalities
- TABLE II.