T 448 (Andrews Catalogue) O is the center of circles 1,2,3, and the center of equilateral triangle ADE. ABC is also equilateral with height AD. The moon has center D, radius AD. But OB is also a height of triangle ABC, therefore circle 3 with radius OB is the same size as the moon. Circle 2 is tangent to the moon on OD produced, and circle 1 is the exterior circle of the hexagon tangent to circle 2. This construction fits the crop formation to within the limits of measurement, and we can find the areas of the circles exactly. They give diatonic ratios. From 1 to 2 we get a ratio of 4/3, and from 1 to 3 we get closely a ratio of 10/3. This geometry is repeated three times by rotating 60º and 120º. The terminator or shadow-line of the moon is an arc of radius CB centered on C. Point B is exactly at the middle of the terminator, and exactly where circle 3 intersects the terminator. The circle of the disc of the moon also passes through E, that is why it touches circle 2 on OD produced. This makes the tip of the moon in the crop formation curve-in slightly from the outer circle. Is this all an artistic accident, or is it clever design? Are we suppoed to discover where the triangles are,and the exact sizes of the three circles, 1,2, and 3? Is it confirmation of our work that when we get the answer the circles give diatonic ratios? The six outer loops are embellishments giving a hint of the hexagon. The formation gives the rotational geometry, accurate to a few inches on the ground. The music notes are F and A in the second octave.