**Bt**, Basi-temporal.

**Bt**, Basi-temporal.

**Bt**, Basi-temporal.

The circular space on each side of the basi-temporal (**bt**.) is the opening of the anterior tympanic recess.

**Bt**, Basi-temporal.

**Bt**, Basi-temporal.

The third equation gives 2 d dt2 = b dx (3), of which the solution is = f (**bt** (4), where f is an arbitrary function.

" Let E = o,7 7 = o, =f (**bt** - x) be the displacements corresponding to the incident light; let O l be any point in the plane P (of the wave-front), dS an element of that plane adjacent to 01; and consider the disturbance due to that portion only of the incident disturbance which passes continually across dS.

Then the displacement at 0 will take place in a direction perpendicular to 0 1 0, and lying in the plane Z0 1 0; and, if 1' be the displacement at 0, reckoned positive in the direction nearest to that in which the incident vibrations are reckoned positive, = 4?y (1 +cos 0) sin 4 f' (**bt** - r).

F(**bt** - x) =c sin 2 i n: (**bt** - x).

= (**bt** - r).

Draw the tangents at A and B, meeting at T; draw TV parallel to the axis of the parabola, meeting the arc in C and the chord in V; and M draw the tangent at C, meeting AT and **BT** in a and b.

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To find the angular acceleration A, draw kI, **bt** respectively parallel to and at right angles to the link KB.

To find the angular acceleration A, draw kI, **bt** respectively parallel to and at right angles to the link KB.

The following formulae are some of those employed for this purpose by different observers E t = **bt** +ct 2..

E t = **bt** +c log T/273, (c = Ts.).

Let APB be a semicircle, **BT** the tangent at B, and APT a line cutting the circle in and **BT** at T; take a point Q on AT so that AQ always equals PT; then the locus of Q is the cissoid.

The curve is symmetrical about the axis of x, and consists of two infinite branches asymptotic to the line **BT** and forming a cusp at the origin.

X = cq - **bt**, y = at -cp; if we take aq-bp= - I, we have x=**bt-cq**, y=cp-at.