§1. C center of gravity of a homogeneous body.

Consider a rigid body weighing P and volume V in the coordinate system Oxyz, where are the axes x And y connected to the surface of the earth, and the axis z aimed at the zenith.

If we break the body into elementary parts with a volume∆ V i , then the force of attraction will act on each part of it∆ P i, directed towards the center of the Earth. Let us assume that the dimensions of the body are significantly smaller than the dimensions of the Earth, then the system of forces applied to the elementary parts of the body can be considered not converging, but parallel (Fig. 1), and all the conclusions of the previous chapter are applicable to it.

Fig.1. Parallel force system

Center of gravity of a solid body is called the center of parallel forces of gravity of the elementary parts of this body.

Several theorems are useful in determining the center of gravity.

1) If a homogeneous body has a plane of symmetry, then its center of gravity is in this

plane.

2) If a homogeneous body has an axis of symmetry, then the center of gravity of the body is on this axis.

3) If a homogeneous body has a center of symmetry, then the center of gravity of the body is at this point.

§2. Methods for determining the coordinates of the center of gravity.

1. Symmetry. If a homogeneous body has a plane, axis or center of symmetry (Fig. 2), then its center of gravity lies, respectively, in the plane of symmetry, axis of symmetry or in the center of symmetry.

Fig.2. Center of gravity of bodies having an axis of symmetry

2. Partitioning. The body is divided into a finite number of parts (Fig. 3), for each of which the position of the center of gravity and area are known.

Fig.3. Center of gravity solid

complex geometric figure

Center of gravity and area of ​​the first figure;

Coordinate of the center of gravity of a solid complex geometric figure along the axis x;

Coordinate of the center of gravity of a solid complex geometric figure along the axisy;

3. Negative area method. A special case of the partitioning method (Fig. 4). It applies to bodies that have cutouts if the centers of gravity of the body without the cutout and the cutout part are known. A body in the form of a plate with a cutout is represented by a combination of a solid plate (without a cutout) with an area S 1 and the area of ​​the cut part S2.

Fig.4. Center of gravity complex geometric figure,

Having a hole

- center of gravity and area of ​​the first figure;

- center of gravity and area of ​​the second figure;

x;

Coordinate of the center of gravity of a complex geometric figure along the axisy;

§3.Coordinates of the center of gravity of some simple figures.

1. Center of gravity of the triangle. Cthe center of gravity of a triangle lies at the point of intersection of its medians(Fig.5). TO the coordinates of the center of gravity of the triangle are the arithmetic mean of the coordinates of its vertices:x c =1/3 (x 1 +x 2 +x 3) ; y c =1/3 (y 1 +y 2 +y 3).

Fig.5. Triangle center of gravity

2. Center of gravity of the rectangle. CThe center of gravity of a rectangle lies at the point of intersection of its diagonals(Fig. 6). TO The coordinates of the center of gravity of the rectangle are calculated using the formulas:x c =b/2 ; y c =h/2.

Rice. 6. Triangle center of gravity

3. Center of gravity of the semicircle. Cthe center of gravity of a semicircle lies on the axis of symmetry(Fig. 7). TO The coordinates of the center of gravity of the semicircle are calculated using the formulas:x c =D/2 ; y c =4R/3π.

Rice. 7.Center of gravity of a semicircle

4. Center of gravity of the circle. Cthe center of gravity of a circle lies in the center(Fig. 8). TO The coordinates of the center of gravity of the circle are calculated using the formulas:x c =R ; y c =R.

Rice. 8.Circle center of gravity

Self-test questions:

What is the center of parallel forces called?

What is the center of gravity of a body?

Why can the gravitational forces of the Earth acting on a point on a body be taken as a system of parallel forces?

Write down the formula for determining the position of the center of gravity of inhomogeneous and homogeneous bodies, the formula for determining the position of the center of gravity of flat sections?

Write down the formula for determining the position of the center of gravity of simple geometric shapes: rectangle, square, trapezoid and half a circle?

How are the properties of symmetry used in determining the centers of gravity of bodies?

What is the essence of the negative area method?

What graphical construction can be used to find the center of gravity of a triangle?

Write down the formula that determines the center of gravity of a triangle.

Draw a diagram of the system and mark the center of gravity on it. If the found center of gravity is outside the object system, you received an incorrect answer. You may have measured distances from different reference points. Repeat the measurements.

• For example, if children are sitting on a swing, the center of gravity will be somewhere between the children, and not to the right or left of the swing. Also, the center of gravity will never coincide with the point where the child is sitting.
• These arguments are valid in two-dimensional space. Draw a square that will contain all the objects of the system. The center of gravity should be inside this square.

Check your math calculations if you get a small result. If the reference point is at one end of the system, a small result places the center of gravity near the end of the system. This may be the correct answer, but in the vast majority of cases this result indicates an error. When you calculated the moments, did you multiply the corresponding weights and distances? If instead of multiplying you added the weights and distances, you would get a much smaller result.

Correct the error if you found multiple centers of gravity. Each system has only one center of gravity. If you found multiple centers of gravity, you most likely did not add up all the moments. The center of gravity is equal to the ratio of the “total” moment to the “total” weight. There is no need to divide “every” moment by “every” weight: this way you will find the position of each object.

Check the reference point if the answer differs by some integer value. In our example, the answer is 3.4 m. Let's say you got the answer 0.4 m or 1.4 m, or another number ending in ".4". This is because you did not choose the left end of the board as your starting point, but a point that is located a whole amount to the right. In fact, your answer is correct no matter which reference point you choose! Just remember: the reference point is always at position x = 0. Here's an example:

• In our example, the reference point was at the left end of the board and we found that the center of gravity was 3.4 m from this reference point.
• If you choose as a reference point a point that is located 1 m to the right from the left end of the board, you will get the answer 2.4 m. That is, the center of gravity is 2.4 m from the new reference point, which, in turn, is located 1 m from the left end of the board. Thus, the center of gravity is at a distance of 2.4 + 1 = 3.4 m from the left end of the board. It turned out to be an old answer!
• Note: when measuring distances, remember that the distances to the “left” reference point are negative, and to the “right” reference point are positive.

Measure distances in straight lines. Suppose there are two children on a swing, but one child is much taller than the other, or one child is hanging under the board rather than sitting on it. Ignore this difference and measure the distances along the straight line of the board. Measuring distances at angles will give close but not entirely accurate results.

• For the see-saw board problem, remember that the center of gravity is between the right and left ends of the board. Later, you will learn to calculate the center of gravity of more complex two-dimensional systems.

Note. The center of gravity of a symmetrical figure is on the axis of symmetry.

The center of gravity of the rod is at the middle of the height. The following methods are used to solve problems:

1. symmetry method: the center of gravity of symmetrical figures is on the axis of symmetry;

2. separation method: complex sections are divided into several simple parts, the position of the centers of gravity of which is easy to determine;

3. negative area method: cavities (holes) are considered as part of a section with a negative area.

Examples of problem solving

Example 1. Determine the position of the center of gravity of the figure shown in Fig. 8.4.

Solution

We divide the figure into three parts:

Defined similarly at C = 4.5 cm.

Example 2. Find the position of the center of gravity of a symmetrical bar truss ADBE(Fig. 116), the dimensions of which are as follows: AB = 6m, DE = 3 m and EF = 1m.

Solution

Since the truss is symmetrical, its center of gravity lies on the axis of symmetry D.F. With the selected (Fig. 116) coordinate axes system, the abscissa of the center of gravity of the truss

Consequently, only the ordinate is unknown at C center of gravity of the farm. To determine it, we divide the truss into separate parts (rods). Their lengths are determined from the corresponding triangles.

From ΔAEF we have

From ΔADF we have

The center of gravity of each rod lies in its middle; the coordinates of these centers are easily determined from the drawing (Fig. 116).

The found lengths and ordinates of the centers of gravity of individual parts of the truss are entered into the table and according to the formula

determine the ordinate y s the center of gravity of a given flat truss.

Therefore, the center of gravity WITH the entire truss lies on the axis DF symmetry of the truss at a distance of 1.59 m from the point F.

Example 3. Determine the coordinates of the center of gravity of the composite section. The section consists of a sheet and rolled profiles (Fig. 8.5).

Note. Often frames are welded from different profiles to create the required structure. Thus, metal consumption is reduced and a high-strength structure is formed.

For standard rolled profiles, their own geometric characteristics are known. They are given in the relevant standards.

Solution

1. Let’s designate the figures by numbers and write out the necessary data from the tables:

1 - channel No. 10 (GOST 8240-89); height h = 100 mm; shelf width b= 46 mm; cross-sectional area A 1= 10.9 cm 2;

2 - I-beam No. 16 (GOST 8239-89); height 160 mm; shelf width 81 mm; cross-sectional area A 2 - 20.2 cm 2;

3 - sheet 5x100; thickness 5 mm; width 100mm; cross-sectional area A 3 = 0.5 10 = 5 cm 2.

2. The coordinates of the centers of gravity of each figure can be determined from the drawing.

The composite section is symmetrical, so the center of gravity is on the axis of symmetry and the coordinate X C = 0.

3. Determination of the center of gravity of a composite section:

Example 4. Determine the coordinates of the center of gravity of the section shown in Fig. 8, A. The section consists of two angles 56x4 and channel No. 18. Check the correctness of determining the position of the center of gravity. Indicate its position on the section.

Solution

1. : two corners 56 x 4 and channel No. 18. Let's denote them 1, 2, 3 (see Fig. 8, A).

2. We indicate the centers of gravity each profile, using table 1 and 4 adj. I, and denote them C 1, C 2, C 3.

3. Select a system of coordinate axes. Axis at compatible with the axis of symmetry, and the axis X draw through the centers of gravity of the corners.

4. Determine the coordinates of the center of gravity of the entire section. Since the axis at coincides with the axis of symmetry, then it passes through the center of gravity of the section, therefore x s= 0. Coordinate y s we will determine by the formula

Using the tables in the appendix, we determine the areas of each profile and the coordinates of the centers of gravity:

Coordinates at 1 And at 2 are equal to zero, since the axis X passes through the centers of gravity of the corners. Let's substitute the obtained values ​​into the formula to determine y s:

5. Let us indicate the center of gravity of the section in Fig. 8, a and denote it by the letter C. Let us show the distance y C = 2.43 cm from the axis X to point C.

Since the corners are symmetrically located and have the same area and coordinates, then A 1 = A 2, y 1 = y 2. Therefore, the formula for determining at C can be simplified:

6. Let's check. For this purpose the axis X Let's draw along the lower edge of the corner shelf (Fig. 8, b). Axis at Let's leave it as in the first solution. Formulas for determining x C And at C do not change:

The areas of the profiles will remain the same, but the coordinates of the centers of gravity of the angles and channels will change. Let's write them down:

Find the coordinate of the center of gravity:

According to the found coordinates x s And y s draw point C on the drawing. The position of the center of gravity found in two ways is at the same point. Let's check it out. Difference between coordinates y s, found in the first and second solutions is: 6.51 - 2.43 = 4.08 cm.

This is equal to the distance between the x-axis in the first and second solution: 5.6 - 1.52 = 4.08 cm.

Answer: s= 2.43 cm if the x axis passes through the centers of gravity of the corners, or y c = 6.51 cm if the x-axis runs along the bottom edge of the corner flange.

Example 5. Determine the coordinates of the center of gravity of the section shown in Fig. 9, A. The section consists of I-beam No. 24 and channel No. 24a. Show the position of the center of gravity on the section.

Solution

1.Let's divide the section into rolled profiles: I-beam and channel. Let's denote them by numbers 1 and 2.

3. We indicate the centers of gravity of each profile C 1 and C 2 using application tables.

4. Select a system of coordinate axes. The x axis is compatible with the axis of symmetry, and the y axis is drawn through the center of gravity of the I-beam.

5. Determine the coordinates of the center of gravity of the section. Coordinate y c = 0, since the axis X coincides with the axis of symmetry. We determine the x coordinate with the formula

According to the table 3 and 4 adj. I and the cross-sectional diagram we determine

Let's substitute the numerical values ​​into the formula and get

5. Let’s plot point C (the center of gravity of the section) using the found values ​​of x c and y c (see Fig. 9, a).

The solution must be checked independently with the axes positioned as shown in Fig. 9, b. As a result of the solution, we obtain x c = 11.86 cm. The difference between the values ​​of x c for the first and second solutions is 11.86 - 6.11 = 5.75 cm, which is equal to the distance between the y axes for the same solutions b dv /2 = 5.75 cm.

Answer: x c = 6.11 cm, if the y-axis passes through the center of gravity of the I-beam; x c = 11.86 cm if the y-axis passes through the left extreme points of the I-beam.

Example 6. The railway crane rests on rails, the distance between which is AB = 1.5 m (Fig. 1.102). The gravity force of the crane trolley is G r = 30 kN, the center of gravity of the trolley is at point C, lying on the line KL of the intersection of the plane of symmetry of the trolley with the plane of the drawing. The gravity force of the crane winch Q l = 10 kN is applied at the point D. The gravity force of the counterweight G„=20 kN is applied at point E. The gravity force of the boom G c = 5 kN is applied at point H. The outreach of the crane relative to the line KL is 2 m. Determine the stability coefficient of the crane in an unloaded state and what load F can be lifted with this crane, provided that the stability coefficient must be at least two.

Solution

1. When unloaded, the crane is in danger of tipping over when turning around the rail A. Therefore, relative to the point A moment of stability

2. Overturning moment relative to a point A is created by the force of gravity of the counterweight, i.e.

3. Hence the stability coefficient of the crane in an unloaded state

4. When loading the crane boom with cargo F there is a danger of the crane overturning when turning near rail B. Therefore, relative to the point IN moment of stability

5. Overturning moment relative to the rail IN

6. According to the conditions of the problem, operation of the crane is permitted with a stability coefficient k B ≥ 2, i.e.

Test questions and assignments

1. Why can the forces of attraction to the Earth acting on points of the body be taken as a system of parallel forces?

2. Write down formulas for determining the position of the center of gravity of inhomogeneous and homogeneous bodies, formulas for determining the position of the center of gravity of flat sections.

3. Repeat the formulas for determining the position of the center of gravity of simple geometric shapes: rectangle, triangle, trapezoid and half a circle.

4.
What is the static moment of area?

5. Calculate the static moment of this figure about the axis Ox. h= 30 cm; b= 120 cm; With= 10 cm (Fig. 8.6).

6. Determine the coordinates of the center of gravity of the shaded figure (Fig. 8.7). Dimensions are given in mm.

7. Determine the coordinate at figure 1 of the composite section (Fig. 8.8).

When deciding, use the reference data from the GOST tables “Hot-rolled steel” (see Appendix 1).

Rectangle. Since a rectangle has two axes of symmetry, its center of gravity is at the intersection of the axes of symmetry, i.e. at the point of intersection of the diagonals of the rectangle.

Triangle. The center of gravity lies at the point of intersection of its medians. From geometry it is known that the medians of a triangle intersect at one point and are divided in a ratio of 1:2 from the base.

Circle. Since a circle has two axes of symmetry, its center of gravity is at the intersection of the axes of symmetry.

Semicircle. A semicircle has one axis of symmetry, then the center of gravity lies on this axis. Another coordinate of the center of gravity is calculated by the formula: .

Many structural elements are made from standard rolled products - angles, I-beams, channels and others. All dimensions, as well as geometric characteristics of rolled profiles, are tabular data that can be found in reference literature in tables of normal assortment (GOST 8239-89, GOST 8240-89).

Example 1. Determine the position of the center of gravity of the figure shown in the figure.

Solution:

We select the coordinate axes so that the Ox axis runs along the bottommost overall dimension, and the Oy axis goes along the leftmost overall dimension.

We break a complex figure into a minimum number of simple figures:

rectangle 20x10;

triangle 15x10;

circle R=3 cm.

We calculate the area of ​​each simple figure and its coordinates of the center of gravity. The calculation results are entered into the table

Figure no.	Area of ​​figure A,	Center of gravity coordinates

Answer: C(14.5; 4.5)

Example 2 . Determine the coordinates of the center of gravity of a composite section consisting of a sheet and rolled sections.

Solution.

We select the coordinate axes as shown in the figure.

Let's designate the figures by numbers and write out the necessary data from the table:

Figure no.	Area of ​​figure A,	Center of gravity coordinates

We calculate the coordinates of the center of gravity of the figure using the formulas:

Answer: C(0; 10)

Laboratory work No. 1 “Determination of the center of gravity of composite flat figures”

Target: Determine the center of gravity of a given flat complex figure using experimental and analytical methods and compare their results.

Work order

Draw your flat figure in your notebooks in size, indicating the coordinate axes.

Determine the center of gravity analytically.

1. Divide the figure into the minimum number of figures whose centers of gravity we know how to determine.

   Indicate the area numbers and coordinates of the center of gravity of each figure.

   Calculate the coordinates of the center of gravity of each figure.

   Calculate the area of ​​each figure.

   Calculate the coordinates of the center of gravity of the entire figure using the formulas (the position of the center of gravity is plotted on the drawing of the figure):

The installation for experimentally determining the coordinates of the center of gravity using the hanging method consists of a vertical stand 1 (see figure) to which the needle is attached 2 . Flat figure 3 Made of cardboard, which is easy to punch holes in. Holes A And IN pierced at randomly located points (preferably at the furthest distance from each other). A flat figure is suspended on a needle, first at a point A , and then at the point IN . Using a plumb line 4 , attached to the same needle, draw a vertical line on the figure with a pencil corresponding to the thread of the plumb line. Center of gravity WITH the figure will be located at the intersection point of the vertical lines drawn when hanging the figure at the points A And IN .

In engineering practice, it happens that there is a need to calculate the coordinates of the center of gravity of a complex flat figure consisting of simple elements for which the location of the center of gravity is known. This task is part of the task of determining...

Geometric characteristics of composite cross sections of beams and rods. Often, design engineers of cutting dies have to face similar questions when determining the coordinates of the center of pressure, developers of loading schemes for various vehicles when placing cargo, designers of building metal structures when selecting cross-sections of elements and, of course, students when studying the disciplines “Theoretical Mechanics” and “Strength of Materials.” "

Library of elementary figures.

For symmetrical plane figures, the center of gravity coincides with the center of symmetry. The symmetrical group of elementary objects includes: circle, rectangle (including square), parallelogram (including rhombus), regular polygon.

Of the ten figures presented in the figure above, only two are basic. That is, using triangles and sectors of circles, you can combine almost any figure that has practical interest. Any arbitrary curves can be divided into sections and replaced with circular arcs.

The remaining eight figures are the most common, which is why they were included in this unique library. In our classification, these elements are not basic. A rectangle, parallelogram and trapezoid can be formed from two triangles. A hexagon is the sum of four triangles. A circle segment is the difference between a sector of a circle and a triangle. The annular sector of a circle is the difference between two sectors. A circle is a sector of a circle with an angle α=2*π=360˚. A semicircle is, accordingly, a sector of a circle with an angle α=π=180˚.

Calculation in Excel of the coordinates of the center of gravity of a composite figure.

It is always easier to convey and perceive information by considering an example than to study the issue using purely theoretical calculations. Let's consider the solution to the problem “How to find the center of gravity?” using the example of the composite figure shown in the figure below this text.

The composite section is a rectangle (with dimensions a1 =80 mm, b1 =40 mm), to which an isosceles triangle was added to the top left (with the size of the base a2 =24 mm and height h2 =42 mm) and from which a semicircle was cut out from the top right (with the center at the point with coordinates x03 =50 mm and y03 =40 mm, radius r3 =26 mm).

We will use a program to help you perform the calculations MS Excel or program OOo Calc . Any of them will easily cope with our task!

In cells with yellow we will fill it auxiliary preliminary calculations .

We calculate the results in cells with a light yellow fill.

Blue font is initial data .

Black font is intermediate calculation results .

Red font is final calculation results .

We begin solving the problem - we begin the search for the coordinates of the center of gravity of the section.

Initial data:

1. We will write the names of the elementary figures forming a composite section accordingly

to cell D3: Rectangle

to cell E3: Triangle

to cell F3: Semicircle

2. Using the “Library of Elementary Figures” presented in this article, we will determine the coordinates of the centers of gravity of the elements of the composite section xci And yci in mm relative to arbitrarily selected axes 0x and 0y and write

to cell D4: =80/2 = 40,000

xc 1 = a 1 /2

to cell D5: =40/2 =20,000

yc 1 = b 1 /2

to cell E4: =24/2 =12,000

xc 2 = a 2 /2

to cell E5: =40+42/3 =54,000

yc 2 = b 1 + h 2 /3

to cell F4: =50 =50,000

xc 3 = x03

to cell F5: =40-4*26/3/PI() =28,965

yc 3 = y 03 -4* r3 /3/ π

3. Let's calculate the areas of the elements F 1 , F 2 , F3 in mm2, again using the formulas from the section “Library of elementary figures”

in cell D6: =40*80 =3200

F1 = a 1 * b1

in cell E6: =24*42/2 =504

F2 = a2 *h2 /2

in cell F6: =-PI()/2*26^2 =-1062

F3 =-π/2*r3 ^2

The area of ​​the third element - the semicircle - is negative because it is a cutout - an empty space!

Calculation of center of gravity coordinates:

4. Let's determine the total area of ​​the final figure F0 in mm2

in merged cell D8E8F8: =D6+E6+F6 =2642

F0 = F 1 + F 2 + F3

5. Let's calculate the static moments of a composite figure Sx And Sy in mm3 relative to the selected axes 0x and 0y

in merged cell D9E9F9: =D5*D6+E5*E6+F5*F6 =60459

Sx = yc1 * F1 + yc2 *F2 + yc3 *F3

in the merged cell D10E10F10: =D4*D6+E4*E6+F4*F6 =80955

Sy = xc1 * F1 + xc2 *F2 + xc3 *F3

6. And finally, let’s calculate the coordinates of the center of gravity of the composite section Xc And Yc in mm in the selected coordinate system 0x - 0y

in merged cell D11E11F11: =D10/D8 =30,640

Xc = Sy / F0

in merged cell D12E12F12: =D9/D8 =22,883

Yc =Sx /F0

The problem has been solved, the calculation in Excel has been completed - the coordinates of the center of gravity of the section compiled using three simple elements have been found!

Conclusion.

The example in the article was chosen to be very simple in order to make it easier to understand the methodology for calculating the center of gravity of a complex section. The method is that any complex figure should be divided into simple elements with known locations of the centers of gravity and final calculations should be made for the entire section.

If the section is made up of rolled profiles - angles and channels, then there is no need to divide them into rectangles and squares with cut out circular “π/2” sectors. The coordinates of the centers of gravity of these profiles are given in the GOST tables, that is, both the angle and the channel will be the basic elementary elements in your calculations of composite sections (there is no point in talking about I-beams, pipes, rods and hexagons - these are centrally symmetrical sections).

The location of the coordinate axes, of course, does not affect the position of the figure’s center of gravity! Therefore, choose a coordinate system that simplifies your calculations. If, for example, I were to rotate the coordinate system 45˚ clockwise in our example, then calculating the coordinates of the centers of gravity of a rectangle, triangle and semicircle would turn into another separate and cumbersome stage of calculations that cannot be performed “in the head”.

The Excel calculation file presented below is not a program in this case. Rather, it is a sketch of a calculator, an algorithm, a template that follows in each specific case create your own sequence of formulas for cells with a bright yellow fill.

So, you now know how to find the center of gravity of any section! The complete calculation of all geometric characteristics of arbitrary complex composite sections will be considered in one of the upcoming articles in the “” section. Follow the news on the blog.

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A few words about the glass, coin and two forks, which are depicted in the “illustration icon” at the very beginning of the article. Many of you are certainly familiar with this “trick”, which evokes admiring glances from children and uninitiated adults. The topic of this article is the center of gravity. It is he and the fulcrum, playing with our consciousness and experience, that simply fool our minds!

The center of gravity of the “fork + coin” system is always located on fixed distance vertically down from the edge of the coin, which in turn is the fulcrum. This is a position of stable equilibrium! If you shake the forks, it immediately becomes obvious that the system is striving to take its previous stable position! Imagine a pendulum - a fixing point (= the point of support of a coin on the edge of a glass), a rod-axis of the pendulum (= in our case, the axis is virtual, since the mass of the two forks is spread out in different directions of space) and a load at the bottom of the axis (= the center of gravity of the entire “fork” system + coin"). If you begin to deflect the pendulum from the vertical in any direction (forward, backward, left, right), then it will inevitably return to its original position under the influence of gravity. steady state of equilibrium(the same thing happens with our forks and coin)!

If you don’t understand, but want to understand, figure it out yourself. It’s very interesting to “get there” yourself! I will add that the same principle of using stable equilibrium is also implemented in the toy Vanka-stand-up. Only the center of gravity of this toy is located above the fulcrum, but below the center of the hemisphere of the supporting surface.

I am always glad to see your comments, dear readers!!!

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