SE523921C2 - Ruler for dividing angle into three equal segments, has length direction lines provided on either side of central line, and two spaced apart cross direction lines - Google Patents

Abstract

The transparent ruler has a black line (0.2 mm thick) running down the middle between its two ends, forming a central line. Thinner lines (0.1 mm thick) spaced apart by intervals of 5 mm extend the full length of the ruler, parallel to the central line, between the central line and one length edge of the ruler. The other half of the ruler is marked in the same way, except that the four lines measure only 25 mm in length from one end of the ruler. Two cross-direction lines (0.2 mm thick) extending at 90 deg. to the central line are located at distances of 5 and 25 mm from one end of the ruler.

Description

From this it can be concluded that the distance c-e (the two catheters) is equal to the double distance b-c (catheter) and that the catheter a-d constitutes the center point to the distance c-e. At the same time, the same midpoint normal a-d constitutes a bisector in the angle cge = Zß.

Results of the study: When dividing an arbitrary angle less than 270 ”into three equal sub-angles, three congruent right-angled triangles are formed by the three equal angles, where the opposite catheters to the equally large sub-angles are arbitrary. In order for the adjacent triangles to be congruent, one of the catheters must form a right angle to one of the angular legs of the original angle (a).

In addition, the corresponding catheters of the other two triangles form a straight line (the sum of two catheters) whose midpoint normal (the common hypotenuse) intersects through the tip (u) of the original angle.

The midpoint normal is allowed to be parallel or coincide with one of the angle legs only in the extreme cases 0 ° and 270 ”.

To carry out the angular division, the three congruent triangles that meet the conditions according to the study done above are first formed.

Method theory for dividing angle less than 270 ”into three equal parts.

Given: The angle u and the arbitrary distance k Step 1.

First, an arbitrary distance k is chosen to form the three equally large catheters in the congruent right-angled triangles. These catheters must be in the opposite position to the sub-angles (3 x ß) of the angle a. To achieve this, a guide line is drawn parallel to the angle leg ab on its inside with the arbitrary distance k. The position of the catheters ki in relation to the angle leg ab is determined later in the implementation of the angle division. (Appendix 2 and 4).

Step 2.

The next step is to search for the position of the distance c-e, which must therefore be 2 x k. Its center point normal must intersect through the vertex a to form the bisector of the angle cge.

(Appendix 3 fi g. 5.) Step. 3.

When the exact position of the section c-e has been formed, so that its midpoint normal constitutes the bisector of the angle cge, the exact position of the catheter b-c is also determined. In this way, three congruent triangles are formed according to the congruence cases. The angle a is thus divided into three equal sub-angles 3 x ß = a.

Observation: At angles between 180 ° and 270 ”, the distance c-e (2 x k) can be placed perpendicular to the parallel guide line and then its midpoint normal coincides with an angle leg. That position is an incorrect variant. Therefore, a position of the distance c-e must be sought so that its midpoint needle does not coincide with the angle leg. Only at 270 ° does the midpoint normal coincide with the angle leg correctly. 523 921 B. Angles greater than 270 ”.

Given: Arbitrary angle a (Greater than 270 °).

Searched for: Basic theory for practically being able to perform a division of the arbitrary angle u into three equal sub-angles ß without the aid of a protractor. (Appendix 4 ñg. 6 and 7) The relationship between the three equal segments is first studied when division has been carried out.

Each sub-angle ß is divided into 90 ° + ö. In this way the straight line e-f is formed which passes through the intersection point a. (Appendix 5 fi g. 8) On the line e-f (Appendix 5 fig. 9) the angle leg a-g is a normal. An arbitrary distance (k) is plotted on the line e-f in both directions towards its end points measured from the point a or the normal a-g. The endpoints of the sections (k) are marked i and m. (Ñg. 9) These endpoints are connected with the corresponding angle legs b and h including the angle ö.

With a line perpendicular to the distance (k) in the points i and m., Two congruent right-angled triangles are formed, (Appendix 5 fig. 9) namely a-i-j and a-m-n. These triangles contain two equal angles and a distance, which means that they are congruent according to the third congruence case.

(The distance (k) and the angles 90 ° and ö).

The point (triangle tip a-j-i) j is connected perpendicular to the normal to the angle leg a-c at the point of intersection a and thus forms a third triangle a-j-d. This triangle has the same hypotenuse a-j as the triangle a-j-i and has the same angles as that triangle, namely 90 ° and ö. They are thus congruent according to the congruence rules.

The catheter a-d which is at the same time normal to the angle leg a-c is thus equal to the distance (k) in its length because the triangles are congruent. The catheters a-d, a-i and a-m are equal in size i.e. equal to the distance (k).

At the same time, it appears that the catheter j-d is parallel to the angle leg a-c (Angle leg of the original angle) because the catheter a-d in the triangle a-j-d is normal to the same angle leg a-c.

The catheter j-d and the angle leg a-c are parallel according to the first parallel set. (Appendix 5 fi g. 9) "If a straight line intersects two other straight lines so that two alternate angles are equal, then the two lines are parallel" If the point j (Appendix 6 fi g. 10) is connected to the point n, the distance jn is formed which is parallel to the distance im and forms with the distances ij and mn a right-angled rectangle. This means that the distance j-n = 2 x (k). The extension of the angle leg a-g towards the section j-n becomes the center point normal of this section.

Thus, the midpoint normal to the distance j-n (2 x (k)) passes through the point of intersection a and its extension forms the angle leg a-g. With the angle leg ab, the angle leg ag forms one third ß of the angle u. (Appendix 5 fi g. 8) Results of the study: (Appendix 6 fl g. 10) If when dividing an arbitrary angle u greater than 270 ° into three equal sub-angles ß , lets a straight line (ef) intersect through the intersection of the angle a in a perpendicular relationship to one of the internal angular legs of the resulting sub-angles (For example ag), then it forms the line with each of the other two adjacent angle legs (In this case ab and ah) pointed angles ö which are equal in size and which are located directly opposite each other at the point of intersection a. 525 921 4 If on that line (ef) an arbitrary distance (k) is deposited on both sides of the angle leg (ag) which forms the nonnal to the straight line (ef), and at the same time connecting the end points (i and m) of these two deposited distances (k) on the straight line with the adjacent angle legs (ah and ab) at right angles to the straight line (ef), then two right-angled congruent triangles (aij o ch a-m-n) which are placed opposite each other at the point of intersection a.

The hypotenuses of these two right-angled triangles (a-j and a-n) coincide with the two adjacent angle legs (a-h and a-b). One of these coincides with the angle legs (a-b) of the original angle (a). The second coincides with an angle leg (a-h) formed during the division which is located closest to the inside of the second original angle leg (a-c).

When the outer end points (j and n) of these two hypotenuses (j and n) are connected by a straight line, a distance is formed which is twice as large as the deposited distance (k) and is parallel to the straight line (ef), and intersects the extended intermediate angle leg (ag) at right angles.

The elongated inner newly formed angle leg (a-g) forms the midpoint normal with the distance (j-n), when the distant angular tips (9028), (j and n) of the two congruent right-angled triangles are connected to each other.

Through the distant angular tip (90 ° - kong) of the second conjugate right-angled triangle (a-i-j) (i-j-a) a parallel line is drawn with the nearest angle leg (a-c) of the original angle (a). The perpendicular distance between the point of intersection (a) and the straight parallel line forms the catheter (a-d) in a further right-angled congruent triangle. Its length then becomes equal to the initially arbitrarily deposited distance (k) on the straight line (e-f) passing through the point of intersection (a).

Thus, when a parallel line is drawn at an arbitrary distance to one of the angle legs at an angle greater than 270 °, and at the same time seeks a point on the drawn parallel line connected to a point on the other angle leg, so that the distance between them both sought points become twice the arbitrary distance to the parallel line, and at the same time the midpoint normal of the distance passes through the point of intersection of the original vertex, the result is that the midpoint normal forms with one of the angle legs exactly one third of the original angle.

Method theory for dividing an angle greater than 270 ”into equal parts.

Given: The angle a and the arbitrary distance k.

Step l: First, a guide line is drawn parallel to the angle leg a-c (Appendix 7 fl gll) at an arbitrary distance (k). On that line, a point is sought that is to be combined with another sought point on the angle leg a-b. Its two searched points join together and form the distance 2 x (k).

Position is sought so that the center point standard on the section 2 x (k) intersects through the angle point b-g-c.

Step 2: The elongation of the midpoint normal ar (Appendix 7 fi g. 12) constitutes the angle bone ag with the angle bone ab, according to the study done above, one third ß of the angle a. of the angle a. The remaining third angle ß logically constitutes the third of a. 523 921 Observation: At angles between 270 ° and 360 ° the distance jn (Appendix 7 fi g. 12) which is 2 x (k), can be placed perpendicular to it parallel guide and then its midpoint normal coincides with an angle leg. That position is an incorrect variant. Therefore, a position of the distance j-n is sought, so that the midpoint nonnel does not coincide with any angle leg. Only at 270 ° does the midpoint normal coincide with an angle leg correctly. 525 921 C. Total summary: All angles can be divided into three equal sub-angles by applying the following method: l. Parallel to one of the angle legs in the arbitrary angle, a guide line is drawn at an arbitrary distance k from the angle leg on its inside. (Appendix 8 fi g. 13 y 14) 2. Then measure the double arbitrary distance k (The distance between the parallel guide line and the angle leg). That distance is placed between the second angle leg and the parallel line in such a way that its midpoint normal intersects through the tip of the angle. 4. Then three equally large partial angles arise: a) The angle between the center point normal and the outer angle leg a-c. b) The angle between the center point nonnel and a line a-p drawn from the vertex through the intersection point p. (Intersection point between the section 2 x k and the parallel guideline according to Appendix 8 fi g. l3 and 14) c) The angle between the angle leg a-b and the line a-p.

According to previous basic theory, these three angles are exactly equal.

According to the study carried out above, the following geometric theorem can be formed as a basis for dividing any angle into exactly three equal parts.

'If at any angle one angle leg is joined by a straight line parallel to the other angle leg on its inside about the angle, so that the connecting distance formed is twice as great as the distance between the parallel line and the adjacent angle leg.

If at the same time the center point norm of the section intersects through the tip of the angle, a third of the angle is formed between the center point norm and the adjacent angle leg. ”In practice, the dividing operation can be performed easily with a ruler, compass and pen.

The origin and use of the patent-pending special ruler is based on this geometric theorem. Angle division can thus be carried out only with the help of a pen and a single special ruler (The patent-pending). 523 921 IL SPECIAL PART - PRACTICAL USE OF THE SPECIAL RULER.

A. Technical description of the sgggiallinjalen.

The special ruler is designed and graded (calibrated) as follows: l. The ruler is made of transparent hard plastic. Dimensions and markings appear from the drawing attached to Patent Application 0000150-3. 2. The ruler has a center line (2) (Appendix 9 fi g. 15) that is parallel to its long sides (4 and 7).

In use, the center line (2) is placed over one of the legs of the arbitrary angle to be divided. 3. The ruler has two groups of thinner parallel lines (1 and 3) (Appendix 9 fi g. 15) which in turn are parallel to the center line ( 2). The distances between the thin lines (1 and 3), which are grouped in [vra and fi / ra on each side of the center line (2), are exactly equal.

These thin lines, groups (I and 3), come into use when the physical space does not allow to use the full width of the ruler in the angle division operation. 4. The ruler is marked with two transverse lines (5 and 6) (Appendix 9 ñg. 15) which are perpendicular to the center line (2) and the long sides of the ruler (4 and 7). In the splitting operation, one of these lines (5 and 6) (either of them is selected depending on the physical space), are catheters in the two congruent triangles that are formed and whose other two catheters coincide with the center line (2) of the ruler which simultaneously intersects the tip of the arbitrary angle.

The new thing about the ruler is that it does not consist of several moving parts assembled into a kind of tool or apparatus, as previously patented rulers have been presented.

The new thing is simplicity. (No moving parts ~ Only a single unit.) In addition, the ruler can be used to split angles up to 360 degrees. Previously known tools lack that capacity.

The new thing is the ability to divide arbitrary angles up to 360 degrees.

The ruler also does not need to be maintained or serviced.

The ruler can also be combined with other drawing tools, such as angle hooks, standard rulers, geometric drawing tools, drawing machines, protractors and even pens.

Thus, the patent-pending SPECIAL RULER differs in several respects from previously known angle dividers. 525 921 s B. Use of the ruler for dividing angles less than 270 °.

Practical approach: Aids: Pen and special ruler. (Appendix 9, p. 15).

Given: Angle a. (Less than 270 °) (Appendix 9 fi g. 16) Step 1: The special ruler is placed so that its center line coincides with one of the angle legs of the arbitrary angle, point 2 of the appendix (Appendix 10 fi g. 17). With the pen, a line that runs parallel to the angle leg is marked at the long side of the ruler. Point 1 of the Annex.

Step 2: Then place the center line of the special ruler so that it intersects the tip of the arbitrary angle and that the end points of one of the transverse lines of the special ruler touch the parallel guide line and the other angle leg of the arbitrary angle. (Appendix 10 tig. 18).

The pen marks the points "p". One point p is marked at the tangent point between the transverse line of the special ruler and the drawn guide line. The second point p is marked at the end point of the center line of the special ruler. (Appendix 10 fi g. 18) Step 3: With the help of the special ruler, the two inner new angle legs that cut through the tip of the arbitrary angle and the two marked points “p” are drawn. (Appendix 11 ñg. 19) The angle a is thus divided into exactly three equal sub-angles ß. (Appendix 1 1 fi g. 19). 523 921 9 C. The use of the ruler for dividing angles greater than 270 °.

Practical approach: Aids: Pen and special ruler. (Appendix 9, p. 15).

Given: Angle a (Greater than 270 °) (Appendix ll fi g. 20) Step l: The special ruler is placed so that its center line coincides with one of the angle legs of the arbitrary angle. (Appendix 11 fi g. 20) The pen marks a line that runs parallel to the angle leg at the long side of the ruler. Point 1 of the Annex.

Step 2: Then place the center line of the special ruler so that it intersects the tip of the arbitrary angle and that the end points of one of the transverse lines of the special ruler touch the parallel guide line and the other angle leg of the arbitrary angle (Appendix 12 ñg. 21) Use the pen to mark the points "p". One point p is marked at the tangent point between the transverse line of the special ruler and the drawn guide line. The second point on is marked by the center line endpoint of the special ruler. (Appendix 12 fi g. 21) Step 3: with the help of the special ruler, draw two inner new angle legs that cut through the tip of the arbitrary angle and the two marked points “p”. (Appendix 12 and 22).

The angle a is thus divided into exactly three equal angles ß. (Appendix 12 fig. 22).

Stockholm, May 27, 2003 ¿i rtil Holgerss

Claims (1)

1. 523 921 PATENT CLAIMS. Special ruler for dividing an angle into three parts, characterized in that it has a center line parallel to its long sides, that it has two groups of thinner parallel lines on either side of and parallel to the center line, the distances between the thin lines being equal, and that it is marked with two transverse lines which are perpendicular to the center line and the long sides of the liujal. Stockholm, September 29, 2003 QaáD / ß / Bertil Holge son