*Intermediate Physics for Medicine and Biology*.

A plane angle measures the diverging of two lines in two dimensions. Solid angles measure the diverging of a cone of lines in three dimensions. Figure A.3 shows a series of rays diverging from a point and forming a cone. The solid angleIt is useful to have an intuitive idea of how big a steradian is. Viewed from the center of the earth, Asia subtends about one steradian, and Switzerland subtends about one millisteradian. From its center a sphere subtends 4π steradians, so one steradian is 1/4π = 0.08, or 8% of the sphere area. Suppose we use spherical coordinates to determine the area, centered at the north pole (Ωis measured by constructing a sphere of radiusrcentered at the vertex and taking the ratio of the surface areaSon the sphere enclosed by the cone tor^{2}:

Ω=S/r^{2}.

…The unit of solid angle is thesteradian(sr). A complete sphere subtends a solid angle of 4π steradians, since the surface area of a sphere is 4πr^{2}.

*θ*= 0), that subtends one steradian. It is the area subtended by the “cap” of the sphere having an angle

*θ*= cos

^{-1}(1-1/2π) = cos

^{-1}(0.84) = 32.8 degrees, or 0.57 radians.

One square degree is (π/180)

^{2}= 0.000305 sr = 305 μsr. In other words, there are 3283 square degrees per steradian. Put in yet another way, a steradian is one square radian. The moon has a radius of 1737 km, and the distance between the earth and the moon is 384,400 km. The solid angle subtended by the moon in the night sky is therefore π 1737

^{2}/384400

^{2}= 0.000064 sr, or 64 μsr. Interestingly, the sun, with a radius of 696,000 km and an earth-sun distance of 149,600,000 km, subtends almost the same solid angle, which makes solar eclipses so interesting. Viewed from earth, Mars at its closest approach subtends about 12 nanosteradians.

At the Battle of Bunker Hill, the order went out to “don't fire until you see the whites of their eyes.” This may be a figure of speech, but let’s take it literally. You can see the whites of a person’s eyes at a distance of about 10 meters (I would definitely be shooting before the enemy got that close). The area of the “whites of the eye” is difficult to estimate accurately, but let’s approximate it as one square centimeter. What solid angle is subtended by the whites of the eye at a distance of ten meters? It would be about (0.01 m)

^{2}/(10 m)

^{2}, or one microsteradian. This is not bad for an estimate of our visual acuity. We may sometimes do a little better than this, but probably not during battle.

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