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If all three circles are great circles, the triangle is called a spherical triangle. Obviously, spherical triangles are a subset of small circle triangles.

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Our goal is to partition a particular spherical triangle, an icosahedron face , into equal area sub-triangles. The edges of the sub-triangles could be either small or great circles. We will consider two ways of dividing a triangle: four-fold and nine-fold subdivision. In the four-fold subdivision method, the three edges of a triangle are bisected and the bisection points are connected with small circles, producing four sub-triangles within the triangle.

In the nine-fold subdivision method, the three edges of a triangle are trisected , and the trisection points as well as an inner center point are connected with one another by small circles, producing nine sub-triangles within the triangle. We define the initial spherical triangle e. If a recursion level 0 triangle is partitioned, we define the process as a recursion level 1 process and the resulting sub-triangles as recursion level 1 triangles.

If the recursion level 1 triangles are partitioned, we define the process as a recursion level 2 process and the resulting sub-triangles as recursion level 2 triangles, and so on. In the four-fold subdivision method, the three edges of a triangle are bisected and the bisection points are connected with small circles, producing four sub-triangles within the parent triangle. If the bisection points were connected with great circles, the four resulting sub-triangles would not have equal areas because only one great circle passing through two points on a sphere is available, that is, the curvature of the great circle is fixed.

For example, if we partition an icosahedron face, the great circle segment between two bisection points always bends outward from the center of the parent triangle, resulting in a larger area of the middle sub-triangle Figure 4. If the bisection points are connected with small circles, we could obtain four sub-triangles equal in area, because there exist infinite numbers of small circles between two points on a sphere and we can arbitrarily change the radius and axis or curvature of a small circle by moving its pole up or down on the sphere.

In Figure 4 , if we move the pole of a small circle toward the center sub-triangle, the radius of the small circle will become smaller and the arc of the small circle segment will shift toward the center of the parent triangle, resulting in a smaller area for the center sub-triangle and a larger area for the outer sub-triangle associated with the small circle segment. The area of the outer sub-triangle is determined only by its associated small circle segment, independent of the other two small circle segments.

If the pole of the small circle is moved to a position on the sphere such that the area of the outer sub-triangle is just one-fourth of the area of its parent triangle, the small circle segment is determined. The other two small circle segments can be determined in the same way. Because each of the four sub-triangles has an area one-fourth of the parent triangle, they obviously are equal in area. Figure 4.

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Four-fold small circle and great circle subdivision geometry for an icosahedron face. We now derive mathematical equations and determine numerical methods to implement the subdivision process discussed above, so that the following essential information about sub-triangles at each recursion level will be obtained: 1 the latitudes and longitudes of all vertices; 2 the relative co-latitudes of all edges; 3 the geodetic latitudes and longitudes of poles for all edges; 4 the arc lengths of all edges; and 5 the area of all sub-triangles.

For simplicity, we represent the earth as a unit sphere, so that the arc length radian measure of a great circle segment between two points on the sphere is the same as the angle it subtends. For a sphere with radius r, linear measurements are scaled by a factor of r and area measurements by r 2. Figure 5. Four-fold small circle subdivision for an arbitrary triangle. Consider an arbitrary initial triangle ABC on a unit sphere in Figure 5 , with its three edges made of either small or great circles. We assume that the following information about the triangle is known a priori:.

Our problem is to find three small circles connecting each pair of the midpoints D, E and F, so that four resulting sub-triangles will have equal areas. We first find the coordinates of midpoints D, E and F. Denote the spherical coordinates for the vertices A and B as and , respectively. Construct a Cartesian coordinate system with the origin located at the center of the earth, the x-y plane on the equator and the z-axis pointing to the north pole.

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Standard formulas for transforming from spherical coordinates to Cartesian coordinates of a point on the unit sphere are:. Figure 6. Geometry of a small circle midpoint calculation.

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So we get:. The vector which points from the earth center O to the midpoint D of the small circle segment AB is given by:. Transforming equation 5 from vector expression to scalar expressions, we get the Cartesian coordinates of the midpoint D:. The spherical coordinates of the midpoint D are calculated using the following equations:. The other two midpoints, E and F, can be found in the same manner. We have shown that the curvature of a small circle depends both on the position of its pole and on its relative co-latitude. But for a small circle crossing two fixed points on a sphere i.

To find the relative latitude of the small circle, we connect two midpoints, D and F, with a great circle Figure 7. We call it a semi-lune in spherical trigonometry a lune is a region on a sphere bounded by two great circle segments. The area of triangle ADGF i. For convenience, we assume that the area of a semi-lune has a positive value if its small circle edge is outside its great circle edge,. Figure 7. A semi-lune confined by a great circle and a small circle. So the area of the semi-lune SL can be written as:. Our problem simplifies to finding the relative co-latitude of the small circle DSF. Figure 8. Geometry of area representation of the unknown semi-lune SL by areas of two known semi-lunes and one spherical triangle. Heavy lines represent small circles and light lines represent great circles. The following steps are needed to calculate the area of the semi-lune, S SL :. The arc lengths of great circle segments AGD, DGF and AGF denoted by f, a and d, respectively, can be obtained by calculating the spherical distances between pairs of vertices.

The arc length f of great circle segment AGD, for example, is:. The law of cosines for sides of a spherical triangle gives:. The area of the spherical triangle ADF is the spherical excess sum of the interior angles minus p , or:. Compute the areas and of two known semi-lunes SL 1 and SL 2. First consider semi-lune SL 1 Figure 9. The area.

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Figure 9. Geometry of area calculation for the known semi-lune SL 1. The arc length, p AD , of the great circle segment AGD is the spherical distance from point A to point D, which can be calculated using equation 9. The law of cosines for angles of a spherical triangle gives:. A manipulation of equation 20 gives the vertex angle A:. Substituting equations 19 and 21 into 16 , the area of semi-lune SL 1 is computed. Similarly, we can compute the area of semi-lune SL 2.

Next, determine the small circle DSF in Figure 7. As mentioned above, if we find the relative co-latitude of the small circle, the small circle will be determined. Similar to the area calculation above for known semi-lune SL 1 , the area of the semi-lune SL can be expressed by a DF as follows:. It is very difficult to get an analytical solution from these equations because the set of equations is nonlinear.

So we turn to numerical methods to get an approximate solution with prescribed precision. We assume an initial value of a DF and find the area of the semi-lune by solving equation Then we adjust the value of a DF up or down by comparing the computed area with the desired amount obtained above. If the computed area is larger or smaller than the desired amount, the value of a DF is increased or decreased, respectively.

If the difference between the computed area and the desired amount is less than a prescribed precision, we stop computation and select the value of a DF as the final relative co-latitude of. Figure To speed up the convergence, we will apply the binary-tree method. We first find the range of the relative co-latitude, and then determine the relative co-latitude iteratively using bisection beginning with the middle point of the range. This consideration is reasonable because we require that sub-triangles have similar shapes. If one of the small circle edges of a sub-triangle has an arc length more than a semi-circle, the shape of the small circle triangle will be unacceptably strange.

This requirement sets a limitation on subdivision precision. In some cases where a triangle has an unusual shape and the prescribed precision is high, a small circle segment larger than a semi-circle may be needed to subdivide the triangle. In practice, this situation has never been encountered when partitioning an icosahedron face up to recursion level 9, but was encountered once when partitioning an octahedron face at recursion level 8.

After the relative co-latitude of the small circle DSF has been found, we next determine the spherical coordinates of the pole of the small circle, , in Figure Denote the Cartesian coordinate of the pole P DF by. The pole P DF is on the sphere:. The Cartesian coordinates of the endpoints, D and F, denoted by and , respectively, can be obtained from equation 1. The angle between the vectors and or is the relative co-latitude a DF of the small circle DF Figure 11 , so the inner products give:. Equations 23 and 24 are solved together for , and.

There are two possible solutions here. If the small circle segment DSF is outside the great circle segment.

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• Geometry of pole determination of the small circle DF. DGF, we choose one solution close to the vertex A. Otherwise, we choose the solution far from the vertex A. Then we transform the Cartesian coordinates back to spherical coordinates:. The expression for the small circle a rc length is:. We determined the small circle DF in Figure 5. The other two small circles DE and EF can be obtained in the same way as above. Each of the four sub-triangles has an area equal to one-fourth the area of their parent triangle, so the triangle ABC is divided in to four equal-area sub-triangles.

In the nine-fold method, all three edges of a triangle are trisected and the trisection points, as well as a central inner point, are connected by small circles, producing nine sub-triangles within the parent triangle.

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The basic problem is still to determine the small circles so that the resulting nine sub-triangles have equal areas. The method of small circle determination is similar to the four-fold case, with a few exceptions. First, the central inner point of a triangle needs to be found and to be connected with six trisection points of three sides with small circles so as to construct nine sub-triangles Figure If six trisection points are directly connected with one another by small circles, three inner small circles, each dividing the hexagon DEFGHI into two equal-area quadrilaterals, may not intersect at one point.

The position of the inner point will affect the compactness of the subdivision method. We have tried several ways to choose the inner point, and will illustrate these in detail later. Secondly, each of six inner sub-triangles has two unknown sides. This is different from the two-frequency method where each sub-triangle has only one unknown side.

Thus, to determine six inner small. Small circles directly connect six trisecting points with one another. Three inner small circles heavy lines , each of them dividing the hexagon DEFGHI into two equal-area quadrilaterals, do not intersect at one point and so an inner point is required. The others will be derived one by one from the initial small circles. Consider the small circle triangle ABC in Figure First, we need to find the trisection points for the three sides of the triangle. Nine-fold small circle subdivision for an arbitrary triangle ABC.

Assume that D and E are two trisection points of side AB. Because D and E are on the unit sphere, their Cartesian coordinates and satisfy:. We know that the angle between the vector and each of the vectors and is the relative co-latitude, a AB , of the small circle AB, so the inner product gives:. Denote the angle between vectors and as d , so the inner product gives:. The angle between vectors and is also d , so we get:. We need to compute the angles d. Applying the plane law of cosines to the plane triangle connecting A, D and the earth center O Figure 14 , the chord length between A and D is:.

So the plane law of cosines gives:. Solving the equations 33 and 34 for cos d :. Substituting into equations 31 and 32 , we get:. We must solve equations 27 , 29 and 36 for , and equations 28 , 30 and 37 for. Each has two solutions. We only select the solution close to the point B as the Cartesian coordinates for point D, and the solution close to the point A as the Cartesian coordinates for point E. The trisection points of other two sides can be computed in the same way. The method to find the three small circles has been introduced previously in the four-fold subdivision method.

Next, we need to determine the inner point of triangle ABC. As mentioned previously, the determination of the inner point will affect the compactness of the subdivision method. The ideal way is to select the center point of the hexagon as the inner point. But the calculation of the center point is complex because it involves integration over the unit sphere bounded by six small circles. These three small circles intersect at three points and define a small triangle see Figure We arbitrarily choose a point inside the small triangle as an inner point e. However, a difficulty will be encountered later when we determine six inner small circles using the inner point.

As we explain later, at least one initial small circle should be known a priori to find other small circles. But there is no proper way to obtain such a small circle if we use the inner point. To simplify our calculation, we use two other methods to find the inner point. Second, find the small circles DG and IF in the same way as above, and then select the intersection point of the two small circles as an inner point Figure The Cartesian coordinates of the intersection point T, can be calculated by the following steps:.

Nine-fold small circle subdivision with the inner point being the midpoint of the small circle EH which divides the small circle hexagon DEFGHI into two equal-area quadrilaterals. The angle between the vector and the vector is a DG and the angle between vector and is a IF. So the inner product gives:. Inner point determination for nine-fold small circle subdivision. Solving equations 38 , 39 and 40 , we get two solutions for the Cartesian coordinates of the intersection point T.

We selected the solution close to vertex A. There are six variables for the six relative co-latitudes of these small circles. If we apply the same method as in the four-fold subdivision case to the six inner sub-triangles so that each of them has an area one-ninth of the triangle ABC, six equations can be obtained. But only five of the equations are independent, so one more equation is needed to solve the six relative co-latitudes. So many instances within my small circle of friends is remarkable View in context. We need be careful how we deal with those about us, when every death carries to some small circle of survivors, thoughts of so much omitted, and so little done--of so many things forgotten, and so many more which might have been repaired View in context.

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