PENUMBRA

, in Astronomy, a faint or partial shade, in an eclipse, observed between the perfect shadow, and the full light.

The Penumbra arises from the magnitude of the sun's body: were he only a luminous point, the shadow would be all perfect; but by reason of the diameter of the sun it happens, that a place which is not illuminated by the whole body of the sun, does yet receive rays from some part of it.

Thus, suppose S the sun, and T the moon, and the shadow of the latter projected on a plane, as GH (Plate xix, fig. 3). The true proper shadow of T, viz GH, will be encompassed with an imperfect shadow, or Penumbra, HL and GE, each portion of which is illuminated by an entire hemisphere of the sun.

The degree of light or shade of the Penumbra, will be more or less in different parts, as those parts lie open to the rays of a greater or less part of the sun's body; thus from L to H, and from E to G, the light continually diminishes; and in the consines of G and H, the Penumbra is darkest, and becomes lost and confounded with the total shade: as near E and L it is thin and confounded with the total light.

A Penumbra must be found in all eclipses, whether of the sun, the moon, or the other planets, primary or secondary; but it is most considerable with us in eclipses of the sun; which is the case here referred to.

The Penumbra extends infinitely in length, and grows still wider and wider; two rays drawn from the two extremities of the earth's diameter, and which proceed always diverging, form its two edges; all that infinite diverging space, included between lines passing through E and L, is the Penumbra, except the cone o<*> the shadow in the middle of it.

To determine how much of the surface of the earth can be involved in the Penumbra, let the apparent semidiameter of the sun be supposed the greatest, or about 16′ 20″, which is when the earth is in her perihelion; also let the moon be in her apogee, and therefore at her greatest distance from the earth, or about 64 of the earth's semidiameters. Let KNC be the earth, T the moon, and MKN the Penumbra, involving the part of the earth from K to N, which it is required to find. Here then are given the angle KMC = 16′ 20″, TC = 64, KC = 1, and OT = 11/40 of KC. Hence, in the right-angled triangle OTM, as fin. OMT: radius : : OT : TM = 210 1/2OT = 58KC nearly. Therefore semidiameters of the earth. Then, in the triangle KMC, there are given| KC = <*>, and MC = 122, also the angle KMC = 16′ 20″, to sind the angle C; thus, as KC:MC::sin. [angle] KMC:sin. [angle] MKP = 35° 25′ 35″; from this take the [angle] KMC - - 0 16 20, leaves the [angle] C - - - 35 9 11, the double of which is the arc KN 70 18 22, or nearly a space of 4866 miles in diameter.