Connection of wooden parts. Measurements and construction of angles during various works. golden egyptian triangle them at a certain angle with

In geometry, an angle is a figure that is formed by two rays that come out of the same point (it is called the vertex of the angle). In most cases, the unit of measure for an angle is degrees (°) - remember that a full angle or one revolution equals 360°. You can find the value of the angle of a polygon by its type and the values of other angles, and if a right triangle is given, the angle can be calculated from two sides. Moreover, the angle can be measured with a protractor or calculated with a graphing calculator.

Steps

How to find the interior angles of a polygon

Count the number of sides of the polygon. To calculate the interior angles of a polygon, you first need to determine how many sides the polygon has. Note that the number of sides of a polygon is equal to the number of its angles.

For example, a triangle has 3 sides and 3 interior angles, while a square has 4 sides and 4 interior angles.

Calculate the sum of all interior angles of the polygon. To do this, use the following formula: (n - 2) x 180. In this formula, n is the number of sides of the polygon. The following are sums of angles of commonly occurring polygons:
- The sum of the angles of a triangle (polygon with 3 sides) is 180°.
- The sum of the angles of a quadrilateral (polygon with 4 sides) is 360°.
- The sum of the angles of a pentagon (polygon with 5 sides) is 540°.
- The sum of the angles of a hexagon (polygon with 6 sides) is 720°.
- The sum of the angles of an octagon (polygon with 8 sides) is 1080°.
Divide the sum of all angles of a regular polygon by the number of angles. A regular polygon is a polygon with equal sides and equal angles. For example, each angle of an equilateral triangle is calculated as follows: 180 ÷ 3 = 60°, and each angle of a square is calculated as follows: 360 ÷ 4 = 90°.
- An equilateral triangle and a square are regular polygons. And the Pentagon building (Washington, USA) and the Stop road sign have the shape of a regular octagon.
Subtract the sum of all known angles from the total angle sum of the irregular polygon. If the sides of the polygon are not equal to each other, and its angles are also not equal to each other, first add up the known angles of the polygon. Now subtract the resulting value from the sum of all the angles of the polygon - this is how you find the unknown angle.
- For example, given that the 4 angles of a pentagon are 80°, 100°, 120° and 140°, add these numbers: 80 + 100 + 120 + 140 = 440. Now subtract this value from the sum of all the angles of the pentagon; this sum is equal to 540°: 540 - 440 = 100°. Thus, the unknown angle is 100°.
Advice: the unknown angle of some polygons can be calculated if you know the properties of the figure. For example, in an isosceles triangle, two sides are equal and two angles are equal; in a parallelogram (it's a quadrilateral) opposite sides are equal and opposite angles are equal.

Measure the length of two sides of the triangle. The longest side of a right triangle is called the hypotenuse. The adjacent side is the side that is near the unknown corner. The opposite side is the side that is opposite the unknown angle. Measure two sides to calculate the unknown angles of a triangle.

Advice: use the graphical calculator to solve the equations, or find an online table with the values of sines, cosines and tangents.

Calculate the sine of an angle if you know the opposite side and the hypotenuse. To do this, plug the values into the equation: sin(x) = opposite side ÷ hypotenuse. For example, the opposite side is 5 cm and the hypotenuse is 10 cm. Divide 5/10 = 0.5. So sin(x) = 0.5, i.e. x = sin -1 (0.5).

Let AB be a segment lying on the line, point M is an arbitrary point that does not belong to the line (Fig. 284). The angle a at the vertex M of the triangle AMB is called the angle at which the segment AB is visible from the point M. Let us find the locus of points from which this segment is visible at the same constant angle a. To do this, we describe a circle around the triangle AMB and consider its arc AMB containing the point M. According to the previous one, from any point of the constructed arc, the segment AB will be visible at the same angle measured by half of the arc ASB (in Fig. 284 it is shown by a dotted line). In addition, a segment from and from will be visible at the same angle. points of an arc located symmetrically with AMB relative to the straight line AB. From no other point of the plane that does not lie on one of the found arcs, the segment can be seen at the same angle a.

Indeed, from a point P lying inside the figure bounded by the arcs AMB, the segment will be visible at an angle ARB greater than a, since the angle ARB will be measured by half the sum of the arc ASB and some other arc, i.e., it will certainly be greater than the angle a. It is also seen that for the corner with vertex Q outside this figure we will have . Therefore, points of arcs AMB and AMB and only they have the required property: The locus of points from which a given segment is visible at a constant angle consists of two arcs of circles symmetrically located relative to this segment.

Task 1. A segment AB and an angle a are given. Construct a segment that contains the given angle a and rests on the segment AB. Here, a segment containing a given angle is understood to mean a segment bounded by a given segment and any of two circular arcs from the points of which the segment is visible at an angle a.

Solution. Let's draw a perpendicular to the segment AB in its middle (Fig. 285). The center of the circle, the segment of which you want to build, will be placed on this perpendicular. From end B of segment AB we draw a ray that forms an angle with it; it will intersect the perpendicular at the center of the desired arc O (prove it!).

Task 2. Construct a triangle by angle A, side and median.

Solution. On an arbitrary straight line, we set aside the segment BC, equal to the side a of the triangle (Fig. 286). The vertex of the triangle must be placed on the arc of the segment, from the points of which this segment is visible at an angle a (the construction process is not shown in Fig. 286). Then, from the middle M of the side BC, as from the center, we draw a circle with a radius equal to m. The points of its intersection with the arc of the segment will give the possible positions of the vertex A of the desired triangle. Explore the number of solutions!

Problem 3. Tangents to a circle are drawn from an external point. The points of contact divide the circle into parts, the ratio of which is equal to

Find the angle between the tangents.

These are simple text problems from the Unified State Examination in Mathematics 2012. However, some of them are not so simple. For a change, some problems will be solved using the Vieta theorem (see the lesson " Vieta Theorem"), others - in the standard way, through the discriminant.

Of course, B12 problems will not always be reduced to a quadratic equation. Where a simple linear equation arises in a problem, no discriminants and Vieta's theorems are required.

Task. For one of the monopoly enterprises, the dependence of the volume of demand for products q (units per month) on its price p (thousand rubles) is given by the formula: q = 150 − 10p . Determine the maximum price level p (in thousand rubles), at which the value of the company's monthly revenue r = q · p will be at least 440 thousand rubles.

This is the simplest word problem. Substitute the demand formula q = 150 − 10p into the revenue formula r = q · p . We get: r = (150 − 10p ) p .

According to the condition, the company's revenue should be at least 440 thousand rubles. Let's make and solve the equation:

(150 − 10p ) p = 440 is a quadratic equation;
150p - 10p 2 \u003d 440 - opened the brackets;
150p - 10p 2 - 440 = 0 - collected everything in one direction;
p 2 − 15p + 44 = 0 - divided everything by the coefficient a = −10.

The result is a quadratic equation. According to Vieta's theorem:
p 1 + p 2 = −(−15) = 15;
p 1 p 2 \u003d 44.

Obviously, the roots: p 1 = 11; p2 = 4.

So, we have two candidates for the answer: the numbers 11 and 4. We return to the condition of the problem and look at the question. It is required to find the maximum price level, i.e. from the numbers 11 and 4, you need to choose 11. Of course, this problem could also be solved through the discriminant - the answer will be exactly the same.

Task. For one of the monopoly enterprises, the dependence of the volume of demand for products q (units per month) on its price p (thousand rubles) is given by the formula: q = 75 − 5p . Determine the maximum price level p (in thousand rubles), at which the value of the company's monthly revenue r = q · p will be at least 270 thousand rubles.

The problem is solved similarly to the previous one. We are interested in revenue equal to 270. Since the company's revenue is calculated by the formula r \u003d q p, and demand - by the formula q \u003d 75 - 5p, we will compose and solve the equation:

(75 − 5p ) p = 270;
75p - 5p 2 = 270;
−5p 2 + 75p − 270 = 0;
p2 − 15p + 54 = 0.

The problem is reduced to the given quadratic equation. According to Vieta's theorem:
p 1 + p 2 = −(−15) = 15;
p 1 p 2 \u003d 54.

Obviously, the roots are the numbers 6 and 9. So, at a price of 6 or 9 thousand rubles, the revenue will be the required 270 thousand rubles. The problem asks you to specify the maximum price, i.e. 9 thousand rubles.

Task. The stone-throwing machine model shoots stones at a certain angle to the horizon with a fixed initial speed. Its design is such that the flight path of the stone is described by the formula y = ax 2 + bx, where a = −1/5000 (1/m), b = 1/10 are constant parameters. At what greatest distance (in meters) from the fortress wall 8 meters high should the car be located so that the stones fly over it?

So, the height is given by the equation y = ax 2 + bx. In order for the stones to fly over the fortress wall, the height must be greater than or, in extreme cases, equal to the height of this wall. Thus, in the indicated equation, the number y \u003d 8 is known - this is the height of the wall. The remaining numbers are indicated directly in the condition, so we make up the equation:

8 = (−1/5000) x 2 + (1/10) x - rather strong coefficients;
40,000 = −x 2 + 500x is already a perfectly sane equation;
x 2 − 500x + 40,000 = 0 - moved all terms to one side.

We got the given quadratic equation. According to Vieta's theorem:
x 1 + x 2 \u003d - (-500) \u003d 500 \u003d 100 + 400;
x 1 x 2 = 40,000 = 100 400.

Roots: 100 and 400. We are interested in the largest distance, so we choose the second root.

Task. The stone-throwing machine model shoots stones at a certain angle to the horizon with a fixed initial speed. Its design is such that the flight path of the stone is described by the formula y = ax 2 + bx, where a = −1/8000 (1/m), b = 1/10 are constant parameters. At what maximum distance (in meters) from a fortress wall 15 meters high should a car be placed so that the stones fly over it?

The task is completely similar to the previous one - only the numbers are different. We have:

15 \u003d (−1/8000) x 2 + (1/10) x;
120,000 = −x 2 + 800x - multiply both sides by 8000;
x 2 − 800x + 120,000 = 0 - collected all the elements on one side.

This is a reduced quadratic equation. According to Vieta's theorem:
x 1 + x 2 \u003d - (-800) \u003d 800 \u003d 200 + 600;
x 1 x 2 = 120,000 = 200 600.

Hence the roots: 200 and 600. The largest root: 600.

Task. The stone-throwing machine model shoots stones at a certain angle to the horizon with a fixed initial speed. Its design is such that the flight path of the stone is described by the formula y = ax 2 + bx, where a = −1/22 500 (1/m), b = 1/25 are constant parameters. At what greatest distance (in meters) from the fortress wall 8 meters high should the car be located so that the stones fly over it?

Another problem with crazy odds. Height - 8 meters. This time we will try to solve through the discriminant. We have:

8 \u003d (−1/22 500) x 2 + (1/25) x;
180,000 = −x 2 + 900x - multiply all numbers by 22,500;
x 2 − 900x + 180,000 = 0 - collected everything in one side.

Discriminant: D = 900 2 − 4 1 180,000 = 90,000; Root of the discriminant: 300. Equation roots:
x 1 \u003d (900 - 300) : 2 \u003d 300;
x 2 \u003d (900 + 300) : 2 \u003d 600.

Largest root: 600.

Task. The stone-throwing machine model shoots stones at a certain angle to the horizon with a fixed initial speed. Its design is such that the flight path of the stone is described by the formula y \u003d ax 2 + bx, where a \u003d -1/20,000 (1/m), b \u003d 1/20 are constant parameters. At what greatest distance (in meters) from the fortress wall 8 meters high should the car be located so that the stones fly over it?

A similar task. The height is again 8 meters. Let's make and solve the equation:

8 \u003d (−1/20 000) x 2 + (1/20) x;
160,000 = −x 2 + 1000x - multiply both sides by 20,000;
x 2 − 1000x + 160,000 = 0 - collected everything on one side.

Discriminant: D = 1000 2 − 4 1 160,000 = 360,000. Root of the discriminant: 600. Equation roots:
x 1 \u003d (1000 - 600) : 2 \u003d 200;
x 2 \u003d (1000 + 600) : 2 \u003d 800.

Largest root: 800.

Task. The stone-throwing machine model shoots stones at a certain angle to the horizon with a fixed initial speed. Its design is such that the flight path of the stone is described by the formula y \u003d ax 2 + bx, where a \u003d -1/22 500 (1 / m), b \u003d 1/15 are constant parameters. At what maximum distance (in meters) from a fortress wall 24 meters high should a car be placed so that the stones fly over it?

Another task is a clone. Required height: 24 meters. We make an equation:

24 = (−1/22 500) x 2 + (1/15) x;
540,000 = −x 2 + 1500x - multiply everything by 22,500;
x 2 − 1500x + 540,000 = 0 - collected everything in one side.

We got the given quadratic equation. We solve by Vieta's theorem:
x 1 + x 2 = −(−1500) = 1500 = 600 + 900;
x 1 x 2 = 540,000 = 600 900.

It can be seen from the decomposition that the roots are: 600 and 900. We choose the largest: 900.

Task. A crane is fixed in the side wall of the cylindrical tank near the bottom. After it is opened, water begins to flow out of the tank, while the height of the water column in it changes according to the law H (t) \u003d 5 - 1.6t + 0.128t 2, where t is time in minutes. How long will water flow out of the tank?

Water will flow out of the tank as long as the height of the liquid column is greater than zero. Thus, we need to find out when H (t) \u003d 0. We compose and solve the equation:

5 − 1.6t + 0.128t 2 = 0;
625 - 200t + 16t 2 = 0 - multiply everything by 125;
16t 2 − 200t + 625 = 0 - put the terms in the normal order.

Discriminant: D = 200 2 − 4 16 625 = 0. Hence, there will be only one root. Let's find it:

x 1 \u003d (200 + 0) : (2 16) \u003d 6.25. So, after 6.25 minutes the water level will drop to zero. This will be the moment until which the water will flow out.

Since ancient times, after mastering the tools of labor, a person began to build a dwelling made of wood. Having gone through evolution, a person continues to improve the construction of his home for thousands of years. Certainly modern technologies simplified construction, gave a wide opportunity for imagination, but basic knowledge about the properties wooden structures pass from generation to generation. Consider ways to connect wooden parts.

Consider the ways of connecting wooden parts that beginner craftsmen face. These are mainly carpentry joints passed down from generation to generation, these skills have been used for more than one century. Before joining wood, we assume that the wood has already been processed and is ready for use.

The first basic rule that should be followed when joining wooden parts is that a thin part is attached to a thicker one.

The most common ways of joining wood, which will be needed in the construction of household buildings, are of several types.

End connection

This is one of the most simple ways connections (rallying). With this method, it is necessary to fit the surfaces of the two elements to be joined as closely as possible. The parts are pressed tightly against each other and fastened with nails or screws.

The method is simple, but to obtain the quality of the product, several conditions must be met:

The length of the nails should be such that, having passed through the entire thickness of the first workpiece, they would enter with their sharp end into the base of another part to a depth equal to at least ⅓ of the length of the nail;

Nails should not be located on the same line, and their number should be at least two. That is, one of the nails is displaced from the center line upwards, and the second, on the contrary, downwards;

The thickness of the nails should be such that when they are hammered into the wood, a crack does not appear. Pre-drilling holes will help to avoid cracks in the wood, and the diameter of the drill should be equal to 0.7 of the diameter of the nails;

For getting best quality pre-lubricate the joints, the surfaces to be joined with glue, and it is better to use a moisture-resistant glue, such as epoxy.

Invoice connection

With this method, two parts are superimposed one on top of the other and fastened with nails, screws or bolts. Wooden blanks, with this method of connection, can be placed in one line or shifted at a certain angle relative to each other. In order for the angle of connection of the workpieces to be rigid, it is necessary to fasten the parts with at least four nails or screws in two rows of two pieces in a row.

If you fasten with only two nails, screws or bolts, then they should be placed diagonally. If the nails will have a through exit through both parts, followed by bending of the protruding ends - this connection method will significantly increase strength. The connection to the invoice does not require a high qualification of the master.

Half tree connection

This method is more complex, it requires already certain skills and a more scrupulous approach to work. For such a connection, in both wooden blanks, wood is sampled to a depth equal to half their thickness, and a width equal to the width of the parts to be joined.

You can connect parts in half a tree at different angles.

It is important to observe the following rule:

So that the sampling angle on both parts is equal, and the width of both samples strictly corresponds to the width of the part. Under these conditions, the parts fit snugly against each other, and their edges will be placed in the same plane. The connection is fastened with nails, screws or bolts, and glue is still used to enhance strength. If necessary, such a connection may be partial. That is, the end of one of the workpieces is cut at a certain angle, and the corresponding sample is made in the other part. Such a connection is used for angular rallying. Both spikes (samples) in this case are cut at an angle of 45 degrees, and the joint between them is located diagonally.

Splicing to length

Such splicing of bars and beams along the length has its own characteristics.

Note for vertical supports splicing is simple.

But it’s a completely different matter when a beam or beam at the splicing point is subject to bending or torsion loads, in which case you can’t get by with simple fastening with nails or screws.

The parts to be joined are cut at an angle (into an oblique overlay) and compressed with bolts. The number of bolts depends on the applied loads, but there must be at least two.

Sometimes additional overlays are installed, for example, metal plates, it is better on both sides, top and bottom, for strength, you can additionally fasten with wire.

Cleat

Such a connection is used when laying the floor or for sheathing boards. To do this, a spike is made in the face of one board, and a groove in the other.

With this splicing, gaps between the boards are excluded, and the sheathing itself acquires beautiful view. Appropriately processed lumber enters the distribution network, where they can be purchased ready-made.

Examples of such materials are batten or lining.

Connector "socket-thorn"

This is one of the most common joints of wooden parts.

Such a connection will provide a strong, rigid and neat rallying.

It goes without saying that it requires certain skills and accuracy in work from the performer.

When making this connection, you need to remember that a poor-quality spiked connection will not add reliability and will not have a beautiful appearance.

A spike connection consists of a groove hollowed out or drilled in one of the wooden parts, as well as a spike made at the end of another attached element.

The parts must have the same thickness, but if the thickness is different, then the socket is made in the thicker part, and the spike is made in the second, thinner part. The connection is carried out on glue with additional fastening with nails, screws. When driving a screw, remember that pre-drilling will facilitate this process. It is better to hide the head of the screw, and the pilot hole should be ⅔ of the diameter of the screw and be 6 mm less than its length.

One of the very important conditions is the same humidity of the parts to be joined. If the elements to be joined have different moisture content, then when dried, the spike will decrease in size, which will lead to the destruction of the entire connection. That is why the parts to be joined must have the same humidity, close to the operating conditions. For outdoor structures, humidity should be in the range of 30-25%.

The use of wood to decorate buildings.

Choice of wood.

In carving, to perform large crafts with large elements, they often use coniferous wood as the main one. They are available, and the striped texture can be used in ornaments.

As a background for overhead and slotted threads, it is used fir.

The valuable material is cedar, its soft, with a beautiful texture and a pleasant yellow-pink or light pink color of the wood core. The wood is easy to cut, cracks little during shrinkage and is resistant to decay.

Wood pears used for highly artistic carving details, as it is durable and warps little from atmospheric influences.

Poplar, the wood is very soft and light - it is used to make a carved decorative column or background shields for attaching false threads.

It is good to use wood to make chains from round rings. apple trees. This wood is used in small crafts, in applied carvings. In this case, the springy properties of the apple tree are used.

Wood is also used lindens. Very light, well planed, well drilled and polished.

carving from oak difficult to manufacture due to its hardness.

But oak is not afraid of moisture, it does not warp. Products made of natural wood are very beautiful, but they can afford it. Veneering is used to reduce the cost of the product. For example, veneered doors are made, by order of the client, "under the oak". We get beautiful doors, outwardly similar to natural ones, but at a much lower price.

From a certain angle

Sub certa specie

Latin-Russian and Russian-Latin dictionary of winged words and expressions. - M.: Russian Language. N.T. Babichev, Ya.M. Borovskoy. 1982 .

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