Regular quadrangular prism. How to take creative photos with a special effects crystal prism

This image is a "regular" street photo. Overpasses lead the eye to the image ... through a prism

A key element of any photography is how you use light. In this article, you will learn how to split it. The use of a prism in photography provides new opportunities and is another way to use the refraction of light.

What does a prism do to light?

Since the prism is a glass object, light refracts as it passes through it, creating several effects that you can use in photography.

There are two ways to use a prism.

• Rainbow projection - a prism, and in particular its triangular shape, acts by dividing the light and revealing waves of various lengths in the form of a rainbow. And you can take a picture of it.
• Light Redirection - Light can change direction abruptly when passing through a prism. This means that when you look through it, you will be able to see the painting at a 90 degree angle to yourself. This factor makes it possible to create a double exposure.

The picture clearly shows the rainbow light from the prism, as well as the remnants of light emitted at different angles.

Using a Crystal Prism to Create a Rainbow

A great way to use a prism is to create a rainbow. The larger the prism, the larger the resulting rainbow. Another way to increase its size is to increase the distance between the prism and the surface you are projecting the rainbow onto. The difference between these options is that as the aforementioned distance increases, the rainbow light becomes more diffuse and less intense.

Using a prism, you can create your own rainbow

Notice also how high the sun is in the sky. The angle at which sunlight strikes the prism affects the angle of the projected rainbow. It is easier to project a rainbow onto the ground at noon. To project a rainbow more horizontally, you need to photograph when the sun is lower in the sky, i.e. after sunrise or before sunset.

rainbow as photo detail

Rainbow light is very colorful and when projected onto a surface it can create an interesting effect. Look for a surface that has a neutral color (such as gray or white). Pay attention to surfaces with a pleasant texture.

Spin the prism until you can see the rainbow projected onto the surface you are photographing. You can, of course, take a picture by holding the prism and the camera. But it's good if you have a friend to help. Since this is a detailed photo, it is better to use a macro lens, but you can find equally interesting compositions using other lenses.

rainbow in portrait photography

Undoubtedly, one of the most popular forms of prism photography is projecting a rainbow onto the model's face. The rainbow won't end up being big, and it would be nice, again, to have another person hold the prism while you take the picture.

Three images in one frame

You can shoot through the glass those objects that appear inside the prism. Raise the prism and rotate it. You will see images inside. However, they will not be the same as those directly in front of you. Depending on how you rotate the glass prism, one or two images will be visible. These are the ones you can work with to create a one-click shutter.

Lens selection

For prism photography - wide angle and macro lenses.

• The wide-angle lens allows you to add a background image to your photo. However, the edge of the prism becomes more visible in the frame. It's not easy to blur an image with the aperture available on most wide-angle lenses.
• macro lens Most prism photography is done with it, as this lens allows you to focus close to the prism and avoid trapping your hand in the frame. The transition from the background to the prism image is also harder to detect.

The image was taken with a macro lens with a prism, and in the end it looks like an optical illusion.

Aperture for prism photography

Which one you use for these photos depends mostly on what you plan to do with the background and how sharp you want the image to be in the prism.

An open aperture of f/2.8 or more will certainly work to blur the background. Most photographs to achieve the feeling of multiple exposure. This means that an aperture of around f/8 is the right balance between background and detail, and avoids the prism line being too harsh when transitioning to the background.

background image

Due to the small width of the prism, even with a macro lens, the background takes up most of the frame. So what works as a backdrop for this type of photo?

• Leading lines - the background that draws attention to the images inside the prism - is used effectively. It could be a tunnel or a road leading to infinity.
• The texture background is more of a blank canvas for images in a prism. It can be a brick wall or leaves and flowers.
• Symmetry. Since the prism splits your image down the middle, using symmetry on both sides of that split is a pretty effective strategy.

Using background symmetry can work well in prism photography.

Image in glass

Now the hard part is getting a good image inside the prism. The images in it may be at 90 degrees to where you are looking, or perhaps at 60 degrees to the edge and in front of where the photographer is standing. Incorporating this into background composition is a tricky aspect of prism photography.

• Composition - You already have a good composition for your background. Now we need to save it while adding a point of interest that would look good through a prism. Just use trial and error. Change the angle of the prism or rotate it; You can also try stepping back and forth.
• Adding a model. An easier way to add interest to an image in a prism is to take it as a portrait photo. The advantage is that you can simply ask the model to stand in the desired position from which the refracted light passes through the prism.

Adding a model to the composition of this image made the sakura photo much more interesting.

Use fractals

Fractals are another element that uses refraction in photography. They produce prismatic effects, but are not triangular in themselves. You can shoot through them without worrying about the images being at a 90 degree angle to you. Fractals are often used to create creative soft-edged portrait photos or other abstract shots.

Time to go and share the light!

If you want to try something new in photography, you will definitely love . It's a bit difficult to photograph with her, but that's what makes the process really interesting. Right now it's time to take a crystal prism in your hands and go towards experiments!

It is necessary to build a development of the faceted bodies and draw on the development the line of intersection of the prism and the pyramid.

To solve this problem in descriptive geometry, you need to know:

- information about the development of surfaces, methods of their construction and, in particular, the construction of developments of faceted bodies;

- one-to-one properties between a surface and its unfolding and methods for transferring points belonging to the surface to the unfolding;

- methods for determining the natural values ​​of geometric images (lines, planes, etc.).

Procedure for solving the Problem

The scan is called a flat figure, which is obtained by cutting and unbending the surface until it is completely aligned with the plane. All surface unfolds ( blanks, patterns) are built only from natural values.

1. Since the scans are built from natural values, we proceed to determine them, for which a tracing paper (graph paper or other paper) of A3 format is transferred task No. z with all points and lines of intersection of polyhedra.

2. To determine the natural values ​​of the edges and the base of the pyramid, we use right triangle method. Of course, others are possible, but in my opinion, this method is more intelligible for students. Its essence lies in the fact that “on the constructed right angle, the projection value of the straight line segment is plotted on one leg, and on the other, the difference in the coordinates of the ends of this segment, taken from the conjugate projection plane. Then the hypotenuse of the resulting right angle gives the natural value of this line segment..

Fig.4.1

Fig.4.2

Fig.4.3

3. So, in the free space of the drawing (Fig.4.1.a) making a right angle.

On the horizontal line of this angle, we set aside the projection value of the edge of the pyramid DA taken from the horizontal projection plane - lDA. On the vertical line of the right angle, we plot the difference in the coordinates of the points D’ andA’ taken from the frontal projection plane (along the axis z down) - . Connecting the obtained points with a hypotenuse, we obtain the natural size of the edge of the pyramid | DA| .

Thus, we determine the natural values ​​\u200b\u200bof other edges of the pyramid D.B. and DC, as well as the base of the pyramid AB, BC, AC (fig.4.2), for which we construct the second right angle. Note that the definition of the natural size of the edge DC is made in those cases when it is given in projection on the original drawing. This is easily determined if we remember the rule: if a straight line on any projection plane is parallel to the coordinate axis, then on the conjugate plane it is projected in full size.

In particular, in the example of our problem, the frontal projection of the edge D’ C’ parallel to axis X, therefore, in the horizontal plane DC immediately expressed in natural size | DC| (fig.4.1).

Fig.4.4

4. Having determined the natural values ​​of the edges and the base of the pyramid, we proceed to the construction of a sweep ( fig.4.4). To do this, on a sheet of paper closer to the left side of the frame, we take an arbitrary point D considering that this is the top of the pyramid. Draw from a point D arbitrary straight line and set aside on it the natural size of the edge | DA| , getting a point A. Then from the point A, taking on the solution of the compass the full size of the base of the pyramid R=|AB| and placing the leg of the compass at the point A we make an arc. Next, we take on the solution of the compass the full size of the edge of the pyramid R=| D.B.| and placing the leg of the compass at the point D we make a second arc notch. At the intersection of arcs we get a point V, connecting it with points A and D get the edge of the pyramid DAB. Similarly, we attach to the edge D.B. facet DBC, and to the edge DC- edge DCA.

To one side of the base, for example VC, we attach the base of the pyramid also by the method of geometric serifs, taking the size of the sides on the compass solution ABandAWITH and making arc serifs from points BandC getting a point A(fig.4.4).

5. Building a sweep prism is simplified by the fact that in the original drawing in the horizontal plane of projections the base, and in the frontal plane - 85 mm high, it set at full size

To build a sweep, we mentally cut the prism along some edge, for example, along E, having fixed it on the plane, we will expand the other faces of the prism until it is completely aligned with the plane. It is quite obvious that we will get a rectangle whose length is the sum of the lengths of the sides of the base, and the height is the height of the prism - 85mm.

So, to build a sweep of the prism, we proceed:

- on the same format where the pyramid sweep is built, on the right side we draw a horizontal straight line and from an arbitrary point on it, for example E, successively lay off segments of the base of the prism EK, KG, GU, UE, taken from the horizontal projection plane;

- from points E, K, G, U, E we restore the perpendiculars, on which we set aside the height of the prism, taken from the frontal projection plane (85mm);

- connecting the obtained points with a straight line, we obtain a development of the side surface of the prism and to one of the sides of the base, for example, GU we attach the upper and lower bases using the method of geometric serifs, as was done when building the base of the pyramid.

Fig.4.5

6. To build a line of intersection on the development, we use the rule that "any point on the surface corresponds to a point on the development." Take, for example, the edge of a prism GU where the line of intersection with the points 1-2-3 ; . Set aside on the development of the base GU points 1,2,3 by distances taken from the horizontal projection plane. Restore the perpendiculars from these points and plot the heights of the points on them 1’ , 2’, 3’ , taken from the frontal projection plane - z 1 , z 2 andz 3 . Thus, we got points on the sweep 1, 2, 3, connecting which we get the first branch of the line of intersection.

All other points are transferred similarly. The constructed points are connected, getting the second branch of the line of intersection. Highlight in red - the desired line. Let us add that in case of incomplete intersection of faceted bodies, there will be one closed branch of the intersection line on the development of the prism.

7. The construction (transfer) of the intersection line on the development of the pyramid is carried out in the same way, but taking into account the following:

- since the sweeps are built from natural values, it is necessary to transfer the position of the points 1-8 lines of intersection of projections on the lines of edges of natural sizes of the pyramid. To do this, take, for example, the points 2 and 5 in the frontal projection of the rib DA we transfer them to the projection value of this right angle edge (fig.4.1) along communication lines parallel to the axis X, we get the required segments | D2| and |D5| ribs DA in natural values, which we set aside (transfer) to the development of the pyramid;

- all other points of the intersection line are transferred in the same way, including points 6 and 8 lying on the generators Dm and Dn why right angle (fig.4.3) the natural values ​​of these generators are determined, and then points are transferred to them 6 and 8;

- on the second right angle, where the natural values ​​\u200b\u200bof the base of the pyramid are determined, points are transferred mandn intersections of generators with the base, which are subsequently transferred to the development.

Thus, the points obtained on natural values 1-8 and transferred to the development, we connect in series with straight lines and finally we get the line of intersection of the pyramid on its development.

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A prism is a geometric body, a polyhedron, the bases of which are equal polygons, and the side faces are parallelograms. To the uninitiated, this may sound a bit intimidating. And, when your child needs to bring a prism made at home to a geometry lesson, you are at a loss, not knowing how to help your beloved child. In fact, everything is not so difficult and, using our tips on how to make a prism, you will adequately cope with this problem.

How to make a paper prism

We will immediately agree that we will do a straight prism, that is, a prism in which the side edges will be perpendicular to the bases. Making an inclined prism out of paper is very problematic (such layouts are usually made of wire).

We already know that two identical polygons lie at the bases of a prism. Therefore, our work will begin with them. The simplest of the polygons is the triangle. This means that we will first make a triangular prism.

How to make a triangular prism

We will need thick white paper for drawing, a pencil, a protractor, compasses, a ruler, scissors and glue.

We draw a triangle, any one is possible, but to make our prism especially beautiful, we will make the triangle equilateral. Such a prism in geometry is called "correct". We choose at our discretion the size of the side of the triangle, let's say 10 cm. With a ruler we put this segment on paper and with a protractor we measure an angle of 60 ∗ from one end of our segment.

We draw an inclined line. On it, using a ruler, set aside 10 cm from the end of the segment. Thus, we have found the third vertex of the triangle. We connect this point with the ends of the initial segment and the equilateral triangle is ready. It can be cut out. Similarly, we make the second triangle, or carefully trace the contours of the first on paper. Well, we already have two reasons.

We make the side edges. We decide what the height of the prism will be. Let's say 20 cm. We draw a rectangle in which the value of one side is the height of the prism (in our case, 20 cm), and the second side is equal to the value of the side of the base multiplied by the number of these sides (we have: 10 cm x 3 = 30 cm) .

On the long sides we make marks every 10 cm. We connect the opposite marks with straight lines. On them then it will be necessary to carefully bend the paper. These are the side edges of our prism. We outline narrow allowances for gluing along two long and one short sides of the rectangle (1 cm wide strips are enough). We cut out the rectangle along with the allowances, carefully bend them according to the markup. We bend the ribs.

We start assembly. We glue the rectangle along the side face into a tube of triangular section. Glue base triangles on top and bottom on the bent allowances. The prism is ready.

It is probably not worth going into the details of the question of how to make a prism out of cardboard. The entire assembly algorithm remains the same, only replace the paper with thin cardboard. By changing the number of sides of the base polygons, you can now independently make both a five- and a hexagonal prism.

In the school curriculum for the course of solid geometry, the study of three-dimensional figures usually begins with a simple geometric body - a prism polyhedron. The role of its bases is performed by 2 equal polygons lying in parallel planes. A special case is a regular quadrangular prism. Its bases are 2 identical regular quadrilaterals, to which the sides are perpendicular, having the shape of parallelograms (or rectangles if the prism is not inclined).

What does a prism look like

A regular quadrangular prism is a hexagon, at the bases of which there are 2 squares, and the side faces are represented by rectangles. Another name for this geometric figure is a straight parallelepiped.

The figure, which depicts a quadrangular prism, is shown below.

You can also see in the picture the most important elements that make up a geometric body. They are commonly referred to as:

Sometimes in problems in geometry you can find the concept of a section. The definition will sound like this: a section is all points of a volumetric body that belong to the cutting plane. The section is perpendicular (crosses the edges of the figure at an angle of 90 degrees). For a rectangular prism, a diagonal section is also considered (the maximum number of sections that can be built is 2), passing through 2 edges and the diagonals of the base.

If the section is drawn in such a way that the cutting plane is not parallel to either the bases or the side faces, the result is a truncated prism.

Various ratios and formulas are used to find the reduced prismatic elements. Some of them are known from the course of planimetry (for example, to find the area of ​​the base of a prism, it is enough to recall the formula for the area of ​​a square).

Surface area and volume

To determine the volume of a prism using the formula, you need to know the area of ​​\u200b\u200bits base and height:

V = Sprim h

Since the base of a regular tetrahedral prism is a square with side a, You can write the formula in a more detailed form:

V = a² h

If we are talking about a cube - a regular prism with equal length, width and height, the volume is calculated as follows:

To understand how to find the lateral surface area of ​​a prism, you need to imagine its sweep.

It can be seen from the drawing that the side surface is made up of 4 equal rectangles. Its area is calculated as the product of the perimeter of the base and the height of the figure:

Sside = Pos h

Since the perimeter of a square is P = 4a, the formula takes the form:

Sside = 4a h

For cube:

Sside = 4a²

To calculate the total surface area of ​​a prism, add 2 base areas to the side area:

Sfull = Sside + 2Sbase

As applied to a quadrangular regular prism, the formula has the form:

Sfull = 4a h + 2a²

For the surface area of ​​a cube:

Sfull = 6a²

Knowing the volume or surface area, you can calculate the individual elements of a geometric body.

Finding prism elements

Often there are problems in which the volume is given or the value of the lateral surface area is known, where it is necessary to determine the length of the side of the base or the height. In such cases, formulas can be derived:

• base side length: a = Sside / 4h = √(V / h);
• height or side rib length: h = Sside / 4a = V / a²;
• base area: Sprim = V / h;
• side face area: Side gr = Sside / 4.

To determine how much area a diagonal section has, you need to know the length of the diagonal and the height of the figure. For a square d = a√2. Therefore:

Sdiag = ah√2

To calculate the diagonal of the prism, the formula is used:

dprize = √(2a² + h²)

To understand how to apply the above ratios, you can practice and solve a few simple tasks.

Examples of problems with solutions

Here are some of the tasks that appear in the state final exams in mathematics.

Exercise 1.

Sand is poured into a box shaped like a regular quadrangular prism. The height of its level is 10 cm. What will the level of sand be if you move it into a container of the same shape, but with a base length 2 times longer?

It should be argued as follows. The amount of sand in the first and second containers did not change, i.e., its volume in them is the same. You can define the length of the base as a. In this case, for the first box, the volume of the substance will be:

V₁ = ha² = 10a²

For the second box, the length of the base is 2a, but the height of the sand level is unknown:

V₂ = h(2a)² = 4ha²

Insofar as V₁ = V₂, the expressions can be equated:

10a² = 4ha²

After reducing both sides of the equation by a², we get:

As a result, the new sand level will be h = 10 / 4 = 2.5 cm.

Task 2.

ABCDA₁B₁C₁D₁ is a regular prism. It is known that BD = AB₁ = 6√2. Find the total surface area of ​​the body.

To make it easier to understand which elements are known, you can draw a figure.

Since we are talking about a regular prism, we can conclude that the base is a square with a diagonal of 6√2. The diagonal of the side face has the same value, therefore, the side face also has the shape of a square equal to the base. It turns out that all three dimensions - length, width and height - are equal. We can conclude that ABCDA₁B₁C₁D₁ is a cube.

The length of any edge is determined through the known diagonal:

a = d / √2 = 6√2 / √2 = 6

The total surface area is found by the formula for the cube:

Sfull = 6a² = 6 6² = 216

Task 3.

The room is being renovated. It is known that its floor has the shape of a square with an area of ​​9 m². The height of the room is 2.5 m. What is the lowest cost of wallpapering a room if 1 m² costs 50 rubles?

Since the floor and ceiling are squares, that is, regular quadrilaterals, and its walls are perpendicular to horizontal surfaces, we can conclude that it is a regular prism. It is necessary to determine the area of ​​its lateral surface.

The length of the room is a = √9 = 3 m.

The square will be covered with wallpaper Sside = 4 3 2.5 = 30 m².

The lowest cost of wallpaper for this room will be 50 30 = 1500 rubles.

Thus, to solve problems for a rectangular prism, it is enough to be able to calculate the area and perimeter of a square and a rectangle, as well as to know the formulas for finding the volume and surface area.

How to find the area of ​​a cube