Circular Spiral Envelope Intersection

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{{Image Description |ImageName=Circular Spiral Envelope Intersection |Image=Screen shot 2013-05-02 at 8.47.07 AM.png |ImageIntro=The interesting shape to the right is really just a number of overlaid circles whose intersections result in beautiful and, as we will see, mathematically predictable patterns. |ImageDescElem=The main image is really just a bunch of circles created according to a certain pattern. The result is several "envelopes." An [[#Envelope|envelope] is a family of related curves whose tangent lines seem to create other shapes. Here we do not exactly have an envelope in a strict sense, since many of the circles overlap, but the patterns related to where the various circles intersect in effect create other circles that "aren't really there." We'll explore how that happens on this page. First we will need to understand how the main image was generated.

I started with an awesome mathematical program called Geometer's Sketchpad.

Figure 1
Figure 1

I created a circle with 24 evenly spaced points along its circumference. We will call this first circle the original circle and the points on its circumference will henceforth be referred to as nodes. In Figure 1, the original circle has center B. C, D, E, F, and G are all nodes. These are the only nodes we have labeled, but we can see that the original circle in fact has 19 more. This allows us to refer to node distance, or how many nodes one must shift to get from one to another. For example, in Figure 1, D is one node away from E, E is two nodes away from G, and C is four nodes away from G.

Figure 2
Figure 2

The rest of the figure is constructed by creating what we will call a set of circles around each node; a set of circles about, say, node D consists of all circles centered on D and passing through two nodes equidistant from D. In this case, we would call D the central node of the particular set in question. Figure 2 shows an example such a set of circles, centered on D. Note that there is a caveat to the rule we've established: since there is an even number of nodes, there is one node which is opposite of the central node. Since there is no other node equidistant to the node opposite of D, for instance, the circle passing through that node only passes through one node, rather than two.

Figure 3
Figure 3

As another example, consider the set of circles around E, which in Figure 3 is added to the overall image we are developing. Each circle in the set is centered on E. One circle passes through D and F (which are both one node away from E). Another circle passes through C and G (which are both two nodes away from E). And so on for every pair of points that are equidistant from E, the point which is the center of every circle in the set.


Figure 4
Figure 4

The radius of any circle in a set is the length of the chord from the set's central node to either of the other two nodes that the circle passes through. For example, the radius of the circle centered at E and passing through G is simply the length of the line \overline{EG}. This fact will help us later in analyzing the circular envelope.

In Figure 5, we repeat this process of constructing a set of circles once again, this time around the node F. Each set, centered around its own node, is represented in a different color. The intersections of these set of circles form circular envelopes within and outside of the original circle.

Figure 5
Figure 5

The goal of this page is to examine the mathematical relations between the intersections and circular envelope formed by these circles.


Below is another image of a single set of circles around the node E. In this example, again, we have 24 equally spaced nodes, so the central angle is 15° (obtained by dividing the 360° in the circle by 24 nodes).

Figure 6
Figure 6

The above example shows that the measure of the central angle depends on the how many equally spaced nodes are constructed on the circle's circumference. It seems, therefore, that we can change the measure of the central angle by choosing to construct a different amount of nodes. Below is an example in which there are 10 nodes, rather than 24, on the original circle, so the central angle 36° rather than 15°.

Figure 7
Figure 7

Another way of thinking of this is as nodes being spaced by a certain, fixed central angle measure. In this case, the amount of nodes we have follows from our chosen central angle. The central angle measurement must, then, divide evenly into 360°.

The number of nodes we construct likewise affects the number of circles in each set as well. How so? Circles pass through every node except the central node. Each circle in a set is centered on around a single node and passes through two other nodes (or passes through one node, in the exceptional case of the node opposite the central node). As such, in our example with 24 nodes, there are 12 circles in each set: 11 which pass through two nodes and one which passes through one node. In our example with 10 nodes, there are 5 circles in each set: four which pass through two nodes and one which passes through one node. In general, when we have an even number 2n of nodes, each set of circles consists of n circles.

There is also the case when we have an odd number, say 3 or 5, of nodes. (Respectively, this would mean that the central angle subtending the arc between any two nodes would be 120° or 72°.) Then how many circles are there in each set? In odd cases, there is no node opposite the central node. In the case of three nodes, there is one circle in each set which passes through the two other nodes. In the case of five nodes, there are two circles in each set. With seven, there would be 3 circles in each set. So in general, when there are 2n+1 nodes, there will be n circles in each set. In other words, to get the number of circles in each set for an odd number of nodes, we must subtract 1 from the number of nodes and divide by 2.

|ImageDesc===Chord/Radius Distance== As mentioned before, the chord lengths from one node to another are the radii of circles in a set. In general, the length of a chord is given by the formula:

2r \sin \frac{\theta}{2}

where r is the radius of the circle and \theta is the central angle.

We can find the radius, then, for a circle in a given set by knowing:

  • the node distance from the center of the set to the node through which the circle passes, and
  • the total number of nodes in the circle.

We can know the angle subtended by any arc between two nodes based on this information. The central angle between two adjacent nodes is 360° divided by the total number of nodes. If the two nodes are n nodes apart, then the central angle between them is equivalent to the central angle between two adjacent nodes times n.

Instead of 2r \sin \frac{\theta}{2}, we will modify the formula for our purposes:

2 \sin \left( \frac{15n}{2} \right)^{\circ}.
Figure 8
Figure 8

Here, 15° is the central angle measurement for two adjacent nodes (from here on we will pursue the case with 24 nodes, although as discussed this angle measurement could be anything that divides evenly into 360°). n, again, represents the node distance from a set's central node to a node on the particular circle in question. We have also done away with r; we will assume that the original circle is the unit circle.

For example, say A is the central node of a set of circles, and traveling clockwise around the circle we have the points p_1, p_2, p_3, and so on. The subscript of the previous points denotes n. Then we know the following:

  • The circle centered around A and passing through p_1 has radius 2 \sin \left( \frac{15 \cdot 1}{2} \right)^{\circ} = 2 \sin \left( \frac{15}{2} \right)^{\circ}.
  • The circle centered around A and passing through p_2 has radius 2 \sin \left( \frac{15 \cdot 2}{2} \right)^{\circ} = 2 \sin 15^{\circ}.
  • The circle centered around A and passing through p_3 has radius 2 \sin \left( \frac{15 \cdot 3}{2} \right)^{\circ} = 2 \sin \left( \frac{45}{2} \right)^{\circ}.
  • And so on.

Again, it is important to note that this formula changes when the total number of nodes is different. In the prior example with 10 nodes, the formula would be:

2 \sin \left( \frac{36n}{2} \right)^{\circ}

Points of Intersections and Perpendicular Distances

Consider two adjacent sets of circles; that is, the sets of circles around two adjacent nodes. As shown below, the various intersections of the circles in these sets form an envelope. Furthermore, the intersections of circles in each set with equal radii fall onto a straight line that passes through a diameter of the circle.

Figure 9
Figure 9
Figure 10
Figure 10

What is the relevance of this relationship? Since this relationship can be repeated, in principle, for any pair of adjacent sets of circles, it seems likely that these intersections contribute to the appearance of envelopes in the main image. In other words, since these intersections all seem to occur at consistent distances from the center B of the original circle, and since those intersections occur any diameter passing midway between two adjacent nodes, the intersections are likely candidates for the circular envelopes formed (highlighted in the image below). We find confirmation of this in the images below, in which the intersections are shown and the resulting envelopes are highlighted in orange.

If we want to find the radii of these circular envelopes, then, we must locate these points of intersection. We will do this by finding the perpendicular distances from the chord between the two central nodes to a certain point of intersection. We will try to find this in terms of n; that is, the node distance that determines the radius of any given pair of circles. We will use the Pythagorean Theorem and the previously established formula for the radii by chord length.

We will refer to the chord from one node to its adjacent node as the main chord. Its length, as established previously, is

2 \sin \frac{15}{2}.

As noted, the intersections of the two sets of circles lie on a line. By symmetry this line bisects the main chord. We construct a triangle with vertices at the two adjacent nodes and the point of intersection. By symmetry, this triangle is isosceles (its edges are the main chord and two segments drawn from the point of intersection to each of the nodes). We let the line perpendicular to the main chord (on which all of the points of intersection lie) divide the isosceles triangle into two right triangles, on which we can of course use the Pythagorean Theorem. Figure 12 shows how these triangles are constructed.

Figure 11
Figure 11

The points of intersections in the sets of circles in the original circle is the main points of intersections for the circular envelopes. If we take a closer look, the circular envelopes forms a circle shape within the inner and outer area of the original circle. These circular shapes could be formed with the original circle's center point and an inter section.

The images above shows two screenshot of the circular envelope. The first image is a close up view of the circular envelope without the points of intersections. In this image, you can see the circular envelope develop and form a circular shape (a circle shape). The next image is the close up view of the circular envelope with some points of intersection plotted. As you can see the circles formed with the original circle's center and the points of intersections matches the circular shapes that the circular envelopes create.

Figure 12
Figure 12

Figuring these factors allows us to use the Pythagorean Theorem, a^2+b^2=c^2. Using reverse calculation c^2-a^2=b^2. In this situation 'c' is the hypotenuse,'a' is the short leg, and 'b'is the long leg, also the perpendicular distance that we are trying to find out. We now can use what we found out already to form a function to summarize the work.

RIGHT TRIANGLE

short leg = (1/2)(2rsin(15/c))

hypotenuse = 2rsin(15n/c)

long leg = √(2rsin(15n/c))^2 - (1/2)(2rsin(15/c))^2

To make this easier (by substitution):

x=(1/2)(2rsin(15/c))

y= 2rsin(15n/c)

perpendicular distance= √(y^2)-(x^2)

This function could be used to find the perpendicular distances from the main chord to the intersections of circles for any pair sets of adjacent circles.

Figure Y
Figure Y

More Relations with Circular Envelope Distances

Differences in perpendicular distances between two points of intersection coud be found using the following formula:

(√(2rsin(15n/c))(^2) - (1/2)(2rsin(15/c)))(^2)- (√(2rsin(15(n-1)/c)))(^2) - (1/2)(2rsin(15/c))(^2)

Figure V

Image:Ttt.PNG

To find the difference, we use the function or formula to find the perpendicular distance of a certain intersection and then subtract the distance before the first intersection point we found. This is why I included the (n-1). As mentioned before, 'n' replaces continuing and extending points on the original circle. |other=Basic Triganometry |AuthorName=Geometry Sketchpad |Field=Algebra |Field2=Geometry |InProgress=No }}

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