Educational toy and teaching tool using three-dimensional shapes and a balance scale.

Abstract

To provide an educational toy-cum-teaching tool that, by making use of solid bodies that are a circular column, sphere, and a cone, and a volume ratio thereof being 3:2:1, serves as means for allowing for concretely recognizing spatial graphics and also as means for allowing for understanding what the volume is, and further provides an opportunity to deepen the recognition and the understanding while enhancing the purpose and making the opportunity enjoyable.SOLUTION: A circular column including a sphere, a cone, and cones each with height being a half of height of said cone is divided, at borders between overlapping portions and non-overlapping portions, into 8 solids, with respective volumes thereof capable of being represented by a simple ratio. Then, 8 solids are added with 4 pieces of supplementary solids conforming to the simple ratio and dimensions of the circular columns. The respective solid have the simple ratio, a scale can be balanced by simple additions and various combinations. Through visual perception and the weights, A level of applications can be enhanced, and recognition of spatial graphics and understanding of the volume can be deepened in an enjoyable manner.SELECTED DRAWING: Figure 1

Description

Technical Field

[0001] The present invention relates to a stereoscopic educational toy and teaching tool for spatial figure learning.

Background Art

[0002] A balance-type educational building block is proposed in Japanese Patent Laid-Open No. 2005-073729, which aims to clarify the concept of addition through weight and length by placing a cube (only one in quantity) on a balance with the weight quantity marked on the surface of a rectangular parallelepiped in proportion to the length of the rectangular parallelepiped. It mainly focuses on the understanding of addition and does not focus on the understanding of spatial figures.

[0003] Two-dimensional plane figures are relatively easy to understand even on the same two-dimensional paper surface. On the other hand, it is difficult to understand three-dimensional spatial figures from the solid figures on the two-dimensional paper surface.

[0004] The volume of a sphere is obtained by the following formula.

[0005]

Number

[0006] There is no interest in memorizing only numbers and symbols without understanding the logic of this formula (1). If even one character is forgotten, it is impossible to reconstruct the formula based on the logic in one's own mind, and it will immediately be lost. Many people must have experienced that in all matters related to arithmetic and mathematics, there is no logic and rote memorization leads to forgetting.

Patent Document

[0009] A sphere, a cone, and a cylinder containing a cone half its height were subdivided at the boundary between overlapping and non-overlapping parts. These subdivided solids can be represented by a simple, uniform volume ratio. To increase the number of combinations for each of the subdivided cylinders, four supplementary solids, which are subdivided semicylinders, were added. When these four supplementary solids are combined, they form a semicylinder with half the height and half the volume of the cylinder, and are formed according to the simple, uniform volume ratio of each of the subdivided cylinder solids. If the materials are the same, the weight ratios of each of the subdivided cylinder solids and each of the subdivided semicylinder solids are also simple, uniform ratios. Within the relationship of solids that are subdivided from the same cylinder, the weight ratio values ​​can be used directly to balance a scale by making the values ​​on the left pan and the right pan equal. Similarly, the four supplementary solids, which correspond to the size-to-weight ratio of the cylinder, can be balanced using the weight ratio values. Because of the simple weight ratios, the scales can be balanced by simple addition and various combinations. It incorporates elements of simple puzzles and building blocks to add playfulness. The addition of addition adds depth to this simplicity. [Effects of the Invention]

[0010] This invention allows users to recognize difficult-to-perceive spatial figures in an enjoyable way, using both visual and weight-based methods with actual three-dimensional objects and a balance scale. Furthermore, it allows users to deepen their understanding of the volume of three-dimensional objects in an enjoyable way, similarly using three-dimensional objects and a balance scale, both visually and through weight. [Brief explanation of the drawing]

[0011] [Figure 1] An exploded perspective view of a subdivided cylinder according to an embodiment and example of the present invention. [Figure 2]Exploded perspective view of a semi-cylindrical column subdivided according to embodiments and examples of the present invention [Figure 3] Regarding embodiments and examples of the present invention, (a) is a perspective view of a cylinder, and (b) is a cross-sectional view taken along line A-A in (a). [Figure 4] Regarding embodiments and examples of the present invention, (a) is a perspective view of a semi-cylindrical column, and (b) is a cross-sectional view taken along line A-A in (a). [Figure 5] Coordinate diagram according to an embodiment of the present invention [Figure 6] Front view according to an embodiment and the first example of the present invention [Figure 7] Front view according to an embodiment and the second example of the present invention [Figure 8] Front view showing an example according to the third example of the present invention [Figure 9] Front view showing an example according to the third example of the present invention [Figure 10] Front view showing an example according to the third example of the present invention [Figure 11] Perspective view showing an example according to the third example of the present invention [Figure 12] Front view according to an embodiment and the fourth example of the present invention [Figure 13] Front view according to the fifth example of the present invention

Modes for Carrying Out the Invention

[0012] FIG. 1 is an exploded perspective view in which a sphere, a cone, and a cylinder containing a cone with half the height are subdivided with the overlapping and non-overlapping parts as the boundary. FIG. 2 is an exploded perspective view in which a semi-cylindrical column is subdivided into four supplementary solids to increase the combination. The volume of the semi-cylindrical column in FIG. 2 is half the height of the cylinder in FIG. 1 and is equal to half the volume. Each supplementary solid is formed according to a uniform and simple ratio. Each solid with that volume ratio will have the same ratio of weight if the materials are the same. The actual size and weight of each solid vary depending on the size at the time of manufacture, but this uniform and simple ratio is regarded as the weight of each solid. Each solid with a uniform and simple ratio is placed on a balance in various combinations to balance them.

[0013] If they are made of the same material, the weight ratio is equal to the volume ratio. However, when the interior is made hollow for reasons such as cost reduction during manufacturing, intentionally adjusting the weight to match the ratio is also considered the same as the uniform ratio.

[0014] When the ratio value of the cylinder 13 is 24, the values of each solid are as follows: Solid A1 = 1, Solid B2 = 3, Solid C3 = 4, Solid D4 = 4, Solid E5 = 4.76, Solid F6 = 3.24, Solid G7 = 2.24, Solid H8 = 1.76, Solid I9 = 1, Solid J10 = 2, Solid K11 = 4, Solid L12 = 5. The proof that each solid has these values is shown below.

[0015] The volume ratio of a cylinder, a sphere, and a cone is 3:2:1, which is well-known. This ratio can also be regarded as the volume as it is. For example, when the volume of the cylinder is 3 cm 3 or 3 m 3 , the sphere is 2 cm 3 or 2 m 3 , and the cone is 1 cm 3 or 1 m 3 . In the proof of the volume ratio of each solid of the present invention, the volume ratio is regarded as the volume for proof, and since it holds even without units, units such as cm 3 or m 3 will not be attached.

[0016] The cone 16 shown in FIG. 6 is a solid formed by combining Solid A1, Solid E5, and Solid G7, and is the cone in terms of the volume ratio, cylinder (3): sphere (2): cone (1). The sphere 17 shown in FIG. 7 is a solid formed by combining Solid A1, Solid B2, Solid C3, Solid E5, and Solid F6, and is the sphere in terms of the volume ratio, cylinder (3): sphere (2): cone (1).

[0017] We will prove the ratio of each solid of cylinder 13. First, to simplify the calculation, we will calculate the volume of cylinder 13 as a cylinder with a base radius of 1 and a height of 2. In this case, the volume of cylinder 13 is 2π. The solids whose values ​​can be determined from this calculation are solid A1 and solid B2. It is obvious that solid A1 is a similar solid to cone 16 with half the height, and from this we can see that the length of the radius of the base is halved (1 / 2). When solid A1 and solid B2 are combined, they form a cone with a base radius of 1 and a height of 1. The volume is given by the following formula.

[0018]

number

number

[0019] Since the volume of cylinder 13 is 2π, we need to adjust it so that its volume becomes 24. That is, we multiply the volume of 2π by 12 and divide by π. The values ​​taken within the volume ratio of each solid are the same number, and by taking the least common multiple and canceling them out, the value changes but the volume ratio does not change, so we similarly multiply solid A1 and solid A1 + solid B2 by 12 and divide by π. Solid A1 has a volume of 1, and solid A1 + solid B2 has a volume of 4. Since solid A1 has a volume of 1, solid B2 has a volume of 3.

[0020] The volumes that can be determined by comparison are those of solids C3 and D4. The volume ratio of cylinders, spheres, and cones is 3:2:1, so if the volume of cylinder 13 is 24, then the volume of sphere 17 is 16, and the volume of cone 16 is 8. When solids A1, B2, and C3 are combined, they form a hemisphere, which is half the volume of sphere 17, so the volume is 8. Since the volume of solids A1 + B2 is known to be 4, the volume of solid C is hemisphere - (solids A1 + solids B2) = 4. When solids A1, B2, C3, and D4 are combined, they form a semicylinder. The volume of the semicylinder is 12. Since the volume of solids A1 + B2 + C3 is known to be 8, the volume of solid D4 is semicylinder - hemisphere = 4.

[0021] The solids whose volume difference can be determined by comparison are solids E5, F6, G7, and H8. The volume of solid A1 + solid E5 + solid G7 (cone 16) is 8. Solids E5 and F6, when combined, form a hemisphere, so their volume is 8. Solid E5 has the same volume in its overlapping portion, so the non-overlapping portion also has the same volume. That is, solid A1 + solid G7 = solid F6. Since the volume of solid A1 is known to be 1, we can see that solid F6 is 1 larger than solid G7. It is obvious that the solid formed by combining solid G7 and solid H8 is congruent to solid D4. The volume of solid D4 is 4, so solid D4 + solid G7 + solid H8 = volume 8. Solids A1 + solid E5 + solid G7 (cone 16) have a volume of 8, and since solid G7 overlaps, the non-overlapping portion, solid D4 + solid H8 = solid A1 + solid E5, has the same volume. Since we know that solid A1 has a volume of 1 and solid D4 has a volume of 4, we can deduce that solid E5 has a volume 3 greater than solid H8.

[0022] The volume of solid G7 is calculated using the Baumkuchen integral. To simplify the calculation, cylinder 13 is assumed to have a base radius of 1 and a height of 2. Figure 5 is a diagram showing the equation of a circle 21 and a line 22 drawn on a coordinate system, and also a diagram showing the cross-section of solid G7 in Figure 3(b) divided into the cross-section of solid Ga 23 and the cross-section of solid Gb 24 by drawing auxiliary lines. The equations for the circle (21) and the line (22) are shown below.

[0023]

number

number

[0024] Solve the equations of the circle (21) and the line (22) simultaneously to find the solution for x. The solutions for x are 0 and 4 / 5. When x is 4 / 5, y is 2 / 5. Draw an auxiliary line perpendicular to the x-axis from the intersection point (4 / 5, 2 / 5), and divide the cross section of solid G7 in Figure 3(b) into the cross section of solid Gb24 formed by the cross section of solid Ga23 shown in Figure 5. Solids Ga23 and Gb24 are also solids, and when combined they are equal to solid G7.

[0025] We will find the volume of the solid of revolution Ga23. First, we transform the equation of the circle 21 as follows.

[0026]

number

[0027] Equation 6 is divided into Equation 7 and Equation 8, and the volume is found using the Baumkuchen integral.

[0028]

number

number

[0029] Add the answers to equation 7 and equation 8 together as shown in the following equation.

number

[0030] The volume of the solid of revolution Gb24 can be found using the Baumkuchen integral as follows:

[0031]

number

[0032] The sum of the answers from equations 9 and 10 gives the volume of solid G7. This is shown in the following equation.

[0033]

number

[0034] To find the volume of cylinder 13 when its volume is 2π, we take the answer from equation 11 and convert it to the value when the volume of cylinder 13 is 24, that is, multiply by 12 and divide by π. The value is 840 / 375. Converting this to a decimal, it is 2.24. Comparing these, we can see that the volume differences are E5 = 4.76, F6 = 3.24, and H8 = 1.76.

[0035] The values ​​of each subdivided supplementary solid of semicylinder 14 were arbitrarily set by the creator to be similar to cylinder 13, so there is no particular proof, but semicylinder 14 is set to be half the height and half the volume of cylinder 13. Solid I9 is ​​congruent to solid A1 and has a volume of 1. Solid I9 + solid J10 becomes cylinder 18 as shown in Figure 12. Since it is half the height of cylinder 13 and the radius of the base is half, the volume is 1 / 8 of cylinder 13, which is 3. Since solid I9 has a volume of 1, solid J10 has a volume of 2. Solid L12 is congruent to the solid F6 + solid H8 and has a volume of 5. The volume of semicylinder 14 is 12, and solid I9 + solid J10 + solid L12 = volume 8, so semicylinder 14 - (solid I9 + solid J10 + solid L12) = solid K11, which has a volume of 4. [Examples]

[0036] In the educational toy and teaching aid of the first embodiment and the educational toy and teaching aid of the second to fifth embodiments, each solid is set as follows. The values ​​taken in the ratio of each solid will change depending on whether it is a common multiple or a common decrease, but the easiest values ​​to understand are when the ratio of the weights of cylinder 13 is 600 and when it is 24. In this embodiment, the ratio of the weights of each solid is set when the ratio of the weights of cylinder 13 is 24. Although solids E5, F6, G7, and H8 are finite decimals, they become integers when combined. These are described below. Solid E5 + Solid F6 = 8, Solid E5 + Solid G7 = 7, Solid F6 + Solid H8 = 5, Solid G7 + Solid H8 = 4. As shown in Figure 3(b) Cross-section of cylinder 13 and Figure 4(b) Cross-section of semi-cylinder 14, both cylinder 13 and semi-cylinder 14 of the present invention are cut in half vertically, and the cross-sections can be joined by methods such as magnets or by creating protrusions and indentations. The cross-sections are marked with the symbol of each solid and the ratio of its weight. Solids E5, F6, G7, and H8 are marked with the symbol of the solid and the ratio of its weight, as shown in Figure 3(b): Solid E5 is E+F=8, E+G=7; Solid F6 is F+E=8, F+H=5; Solid G7 is G+E=7, G+H=4; Solid H8 is H+F=5, H+G=4. One pair each of cylinders 13 and semicylinders 14 are used.

[0037] In this embodiment 1 of the educational toy and teaching aid, Figure 6 shows a balance scale 20 with a solid 15 (solid B2 + solid C3 + solid D4 + solid F6 + solid H8) placed on the left pan and two cones 16 (solid A1 + solid E5 + solid G7) placed on the right pan. The cone 16 is removed from the cylinder 13. The removed cone 16 has a weight of 8. The remaining solid 15 has a weight of 16. The solid 15 is placed on the left pan of the balance scale 20, the removed cone 16 is placed on the right pan, another cone 16 is removed from the other cylinder 13 and placed on the right pan as well. The two cones 16 add up to a weight of 16. Since both pans have a weight of 16, they can balance. From this, it can be understood that the weight (volume) of cylinder 13 is equivalent to three cones 16. Even elementary school students in the lower grades who do not understand volume calculations can grasp the sense of size. While methods for calculating the volume of cylinders and cones are taught in the upper grades of elementary school, this invention allows for a logical understanding of these concepts. [Examples]

[0038] In this second embodiment of the educational toy and teaching aid, Figure 7 shows a solid 15 placed on the left pan of a balance scale 20, and a sphere 17 (solid A1 + solid B2 + solid C3 + solid E5 + solid F6) placed on the right pan. The cone portion 16 is removed from the cylinder 13, and the remaining solid 15 has a weight of 16, which is placed on the left pan of the balance scale 20. The sphere 17 portion is removed from the other cylinder 13. Its weight is 16, and it is placed on the right pan. Since both pans have a weight of 16, they can be balanced. In the first embodiment, it is known that the solid 15 on the left pan is equivalent to two cones 16, so the volume of the sphere 17 is equivalent to two cones. If we replace it with cylinder 13, it is found to be two-thirds of the volume of cylinder 13. The method for calculating the volume of a cylinder is taught in the upper grades of elementary school, but if you can calculate the volume of a cylinder, you can naturally derive the formula for calculating the volume of a sphere. The formula for calculating the volume of cylinder 13 in this invention is as follows.

[0039]

number

[0040] The cylinder 13 of this invention has a height of twice its radius r (2r). That is, the area of ​​the base circle is multiplied by the height 2r. Since the volume of the sphere 17 is two-thirds of the volume of the cylinder 13 in equation 12, the formula for the sphere in equation 1 can be derived naturally. [Examples]

[0041] In this third embodiment of the educational toy and teaching aid, Figure 8 shows a balance scale 20 with solid B2 placed on the left pan and two solids A1 and one solid I9, for a total of three, placed on the right pan. Solid B2 is taken out of cylinder 13. Its weight is 3 and it is placed on the left pan. Solid A1 is taken out from each of the pair of cylinders 13. Each weighs 1 and is placed on the right pan. Solid I9 is ​​taken out from semicylinder 14. Its weight is 1 and it is placed on the right pan. The two solids A1 and the one solid I9 add up to a weight of 3, so they can be balanced.

[0042] Figure 9 shows a balance scale 20 with solid C3 on the left pan and two solids A1 and two solids I9 on the right pan. Solid C3 is taken from cylinder 13. Its weight is 4 and it is placed on the left pan. Solid A1 is taken from each of the pair of cylinders 13. Each weighs 1 and is placed on the right pan. Solid I9 is ​​taken from each of the pair of semi-cylinders 14. Each weighs 1 and is placed on the right pan. The two solids A1 and the two solids I9 add up to a weight of 4, so the scale is balanced.

[0043] Figure 10 shows a balance scale 20 with a cone 16 placed on the left pan and solids A1, B2, and C3 on the right pan. The cone 16 has a weight of 8 and is placed on the left pan. Solids A1, B2, and C3 add up to a weight of 8 and are placed on the right pan. Both pans have a weight of 8 and are therefore balanced. This third embodiment shows that the cone 16 is equivalent to eight solids A1, which are also cones. As shown in Figure 11, one solid A1 and one solid I9 are placed side by side with their bases aligned and matched with the base of the cone 16. It can be seen that the length of the base of the cone 16 is twice the length of solid A1 (solid A1 is half the length of the base of the cone 16). It is obvious from the beginning that the height of solid A1 is half the height of the cone 16. From this third embodiment, it can be seen that a similar solid A1, which has half the length and height of the base of the cone 16, has one-eighth the weight (volume). Also, since solids A1, B2, and C3 are assembled to form a hemisphere, it can be seen that the cone 16 has half the weight (volume) of a sphere. [Examples]

[0044] In this fourth embodiment of the educational toy and teaching aid, Figure 12 shows a cylinder 13 placed on the left pan of a balance scale 20, and cylinders 18 (solids I9 + J10) and solids 19 (solids A1 + B2 + C3 + D4 + K11 + L12) placed on the right pan. Cylinder 13 has a weight of 24 and is placed on the left pan. Cylinder 18 is taken out of the semi-cylinder 14. Its weight is 3 and it is placed on the right pan. Solids A1 + B2 + C3 + D4 are taken out of the other cylinder 13 and combined with the remaining solids K11 + L12 from the semi-cylinder 14 (solid 19). Its weight is 21 and it is placed on the right pan. Since both pans now have a weight of 24, they can be balanced. If solid 19 is removed and another combination that results in a weight of 21 is placed on top, the balance scale 20 will balance. If any combination that does not result in a weight of 21 is used, the balance scale 20 will lose its balance. The same thing happens if cylinder 18 is removed and other combinations are tried. From this, it can be confirmed that cylinder 18 has a weight of 3. It can also be seen that cylinder 18 is 1 / 8 the size of cylinder 13 on the left pan. If the bases of the two cylinders 18 are placed side by side and matched with the base of cylinder 13, it can be seen that the length of the base of cylinder 13 is twice the length of cylinder 18 (cylinder 18 is half the length of the base of cylinder 13). It is obvious from the beginning that the height of cylinder 18 is half the height of cylinder 13. It can be seen that similar solids with half the length of their bases and half the height will have 1 / 8 the weight (volume). [Examples]

[0045] In this fifth embodiment of the educational toy and teaching aid, Figure 12 shows solids A1 and G7 on the left plate and solid F6 on the right plate. The fascinating phenomenon of purely equal weight (volume) within the relationship of subdivided solids of cylinder 13 can be experienced; many other combinations with equal weight exist. By embodying this invention, recognition and understanding of each solid will increase, and new perspectives on this relationship may be discovered in the future. [Explanation of symbols]

[0024] 1 Solid A 2 Solid B 3 Solid C 4 3D D 5 3D E 6 3D F 7 3D G 8 Solid H 9 Solid I 10 3D J 11 3D K 12 3D L 13 cylinders 14 semi-cylindrical 15 3D 16 cones 17 balls 18 cylinders 19 3D 20 balance 21. The equation of a circle 22 straight line 23. Solid of revolution Ga 24 Rotating body Gb

Claims

1. The radius of a sphere is equal to the radius of the base of a cylinder and a cone, and the diameter of a sphere is equal to the height of a cylinder and a cone. The volume ratio of a cylinder, sphere, and cone is known to be 3:2:

1. If we further add a cone that shares a vertex with a cone, has the same base radius, and is half the height, and enclose the cylinder, then subdivide the cylinder along the overlapping and non-overlapping parts, each solid can be represented by a uniform volume ratio. This educational toy and teaching tool consists of a three-dimensional shape, a cylinder, a semi-cylinder with an equal base and half its height, a frustum of a cone with an equal height and an upper base radius half the radius of the lower base, a cylinder sharing the center point with the semi-cylinder's base and half its radius, and an equal height with the semi-cylinder, and a cone sharing the upper base and base of the frustum of a cone and sharing the center point and vertex of the lower base, and is subdivided into four solid shapes separated by overlapping and non-overlapping parts.

2. In claim 1, Each solid obtained by subdividing the cylinder and each supplementary solid obtained by subdividing the semicylinder can be represented by a uniform volume ratio, and if the material is the same, the weight will also be the same ratio. When each solid with that weight ratio is placed on the right and left pans of a balance scale, it will balance if the ratio of the weights on the right pan is equal to the ratio of the weights on the left pan. Similarly, when multiple solids are combined or added together, they will balance if the ratio of their weights is equal. This educational toy and teaching tool is characterized by its ability to allow children to experience and learn, while having fun, that solids of different shapes are equal in various combinations through weight and visual perception.

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