Dyson Sphere - what is it for? Part II: How to build a non-rigid Dyson Ring out of the elements?
So, the most practical and feasible version of the Dyson sphere in the form of a non-rigid Ring of individual elements (swarm) was selected in the first part of the article . Now it's time to describe in more detail the possible construction of such a Ring and calculate its parameters.
The most important moment in planning the future of the Dyson Ring is the right choice of the design of the standard Ring element. It should be standardized in size and shape for ease of production, unification of industry and tools, but its design and architecture should be flexible enough for further modernization and unpredictable improvements in advance.
The shape of the element should be simple and convenient to assemble. The element plane must have a concavity in order to focus the light on the pickup. The center of mass of the element should coincide with the geometric center for ease of towing and orientation.
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Taking the symmetrical and naturally occurring natural form as a basis, we can suggest the element of the Ring in the form of a giant hexagon. By the way, about the predecessors: similar elements form the mirror of the solar power station in this picture from the magazine “Technique of Youth” for March 1975 (note who painted the picture):
You can take another form - in one detailed article in Russian, elements-octagons were suggested (alas, I lost the link to it).
The size of a standard element should be optimal: not too small, so as not to cause problems when exposing and assembling a huge number of segments, not too large so that it can be towed to the place of assembly (of course not completely finished, but rather autonomous).
If we take for the width of the finished Ring a round value of 1 million km (this value is chosen, since it is small compared to the radius of the Ring in our system, but comparable to the diameter of the Sun, which is equal to 695700 * 2 = 1,391,400 km), it would be nice to cover at least one thousandth part of this width of the Ring with two or three elements. Therefore, let us take as the characteristic size of a fully assembled element of the Ring a value of 330 km from the edge to the edge of the hexagon, then along with the gaps three elements in a row will occupy about 1000 km of the width of the Ring (the horizontal axis in the picture below). The width of the Ring in 1 million km will accommodate 3,000 hexagonal elements - quite an observable number (but do not think that they can be made in 40-60 years, as in the video from the first part). Densely packed hexagonal elements with a size of 330 km occupy an average of (382 + 191 = 573) / 2 = 286.5 km in length along the Ring, as shown in the figure below (the vertical axis in the picture):
The parameters of such a regular hexagon are determined by two radii (in the picture below): a small radius of the inscribed circle r = 165 km and a large radius of the circumscribed circle R = 191 km, with r = SQR (3/4) * R. The area S = 3 * SQR (3) * R ^ 2/2 is approximately 94500 km2, we take it roughly over 95 thousand km2.
View of the element from the sun:
The element should be for the most part autonomous, self-orienting in space, self-correcting its orbit and self-repairing. At the beginning of the assembly, a separate tug can also drag it into place (into a stable orbit), but then the element needs to adjust the orbit, parry the deviations on its own.
As for the filling of this element of the Ring, it is necessary to have: mooring devices, warehouses for materials, repair capacities, receiving and energy transferring devices. The element must have all this in the geometric center - there is located the central module of metal (construction begins from it and all the main systems are in it). And around it are the outer zones of branching bent pipes (with not rigid, but elastic structure as on a wood leaf) plus openwork cobweb of thin, flexible tubes with a thin film stretched on them (with albedo about 100%), reflecting the light of the Sun.
Back in 1986, in this book (Griliches V.A. “Solar space power stations” - L .: Nauka, 1986.) in one of the sections the aspects of choosing the optimal film were considered .
But the easiest to manufacture, affordable and fairly light material from which to create a mirror of the external zones is in our home, in the kitchen. This is aluminum foil - the material is very visual.
Consider two alternatives:
a) thick foil with a thickness of 0.1 mm = 100 μm - a square meter weighs about 0.27 kg according to GOST (adding lithium to the foil can bring the weight up to 0.25 kg per 1m2 - approximately the same value of 245 grams per 1 m2 mentioned in Stewart Armstrong’s lecture on construction Roy Dyson from Mercury ).
b) thin foil with a thickness of 0.045 mm = 45 microns - a square meter weighs about 0.1 kg. Weight 100 m2 is only 10 kg.
Ideally, the entire hexagonal element of foil should have a weight equal to the weight of the film / foil multiplied by the number of square meters, which is:
a) 0.25 kg * 95000 * 10 ^ 6 m2 = 23.75 million tons (it will be a bit too much!).
Or
b) 0.1 kg * 95000 * 10 ^ 6 m2 = 9.5 million tons (not bad!).
The numbers are impressive, but not insanely large. So in the Russian Federation about 4 million tons of aluminum are produced annually and from it make 0.1-0.2 million tons of foil. In the world, in 2014, 54 million tons of aluminum were produced (which is noticeably more than 41 and 45 million tons of aluminum in 2010 and 2012). 16% of this volume is foil and other packaging materials.
The low strength of the film / foil is compensated by the ease of its processing and repair. And automated and permanent repair in any case is inevitable for a structure that is created not even for thousands, but for tens and hundreds of millions of years.
Alas, this is not all - the weight of the element should be much more than just the weight of the foil. Taking into account the thin openwork web of the tubes that support the shape of the foil itself, the weight of the element will approximately double. For every 100 m2 of foil, it is required to fix at least two tubes 10 meters long approximately centimeter in diameter, with walls about 1 mm thick (like ski poles). We can optimistically determine the weight of these aluminum tubes as 2 * 10 * 0.5 kg = 10 kg per 100 m2, in addition to 10 kg of the weight of 100 m2 of the foil itself. If for the manufacture of tubes instead of aluminum to take plastics, especially carbon plastics, the weight may be less.
A foil is deposited on this openwork web of tubes, attached from the sunny side.
This thin web should at least once at 0.1 - 1 km be added tubes / beams thicker of durable metals such as titanium - the supporting structure of the outer zone (with some mechanisms for changing the curvature and tension of the thin web and foil). Without going into detailed calculations, we are optimistic about the total weight of the supporting structure, a weight equal to the total weight of a thin openwork web. The total weight has tripled, for option b) the total weight of the element will be already 28.5 million tons.
Repair and assembly of all this design is performed by autonomous in motion (with its own engines) assembly robots collecting the hollow-tube frame and spider-robots crawling in vacuum between the frame tubes, collecting old foil scraps, pulling and securing the newly extruded foil directly onto these tubes.
In the center of the element, like an island in the sea, a rigid and partially hermetic central module rises high. Its size is from 2 to 10 km (can be both polygonal and round), thickness is up to 1 km. It has a rigid and durable three-dimensional frame made of a metal profile (aluminum, titanium, steel). Covered with steel, aluminum and glass / plastic sheets. For an area of 100 + 100 km2 square surfaces of the illuminated and shady side plus the area of the sides of such a module with a thickness of 0.5 km equal to 0.5 * 10 * 4 = 20 km2, the weight of the sheath made of steel sheets (with a specific weight of 6 kg per 1 m2) will be: 220 * 10 ^ 6 m2 * 6 kg = 1.32 million tons.
The central module of the island as a whole, with a filling, should weigh no more than 2.5 million tons (the weight of the element will then be packed at 30 million tons). The central module is the place of control, the location of the engines, ports and locks for the mooring of spacecraft, the place of energy storage and the place of sending energy. There are also cooling systems, processing of raw materials into fuel and materials, the production of body parts and parts of the external frame, the production of film-foil, as well as repair systems. On the illuminated side of the central module, you can make a separate mirror with fixed focus on the tower of the power receivers from the external metal panels of the module case so that the module has its own energy source from the very beginning.
The mooring ports and docking stations on the back, unlit side of the central module must be especially strengthened to transmit the impulse from the pusher ship to the entire element. Actually, at first, you can only bring the fully finished central module to the desired orbit, and then import to it billet pipes of the support structure, machines for the production of openwork tubes, tube collector robots, spider robots for attaching foils to finished tubes, materials for tubes and aluminum. foil, spare parts for cars and robots.
Power pipes or fencing profiles (steel, titanium) - for example, hexagonal in shape, are attached to the central module. From this enclosure to the corners of the element extend (but do not reach them) tapering tubes of supporting branching hardwood structure (titanium, aluminum). Flexible ropes (titanium, Kevlar, carbon fibers), which are needed to stretch and bend the entire structure (pipes, openwork structure and the entire surface of the foil) into a parabolic or spherical mirror, go to the ends and other anchor points of these pipes from the upper face of the sides of the central island. to concentrate the light of the sun into one focus of this gigantic structure. The focus of the mirror (which can be positioned 400-900 km from the central module, depending on the bending of the described mirror from the plane) is a suspended solar energy receiver connected to the central module with cables and cables. When maneuvering the whole element of the Ring, this energy receiver will either have to be dragged to the central island and fixed there, or supplied with its own engines for individual maneuvering. From the side, in profile, the element looks like this:
1. Energy payload: energy receivers, as well as energy emitters on the outer and inner side.
One energy pickup from the entire surface of the element is already described above in the previous paragraph. It is associated with the Ring element only with cables and cables. When it is dragged to the center of the element during maneuvers or for repair, it should attach to something in the center of the module. A second fixed motion receiver in the form of a high tower may well be the wharf for it.
What is the second power receiver for? To focus on it, the light reflected only from the central module itself (from its illuminated plane) is, so to speak, a permanent source of energy for the needs of the central module itself from the very beginning of work, even before assembling the external zones. The (upper) plane of the central module facing the Sun may also look like a parabolic or spherical mirror, only rigid, with constant geometry and focus. The light reflecting panels of this mirror direct it to the central fixed receiving receptacle in the form of a tower in the center. The tower is needed to move the focus point away from the central module to avoid overheating of the whole structure. Energy receivers consist of solar panels and heat exchangers for heating water / fuel / raw materials.
2. Production payload: raw material processing plants.
In any case, the necessary production capacity for the release of fuel (for the maneuvers of the element), materials and blanks (for the assembly of external zones) from raw materials / semi-finished products from asteroids. These products, in the form of fuel, materials, blanks, can then be transported to other elements.
3. Residential payload: a habitable colony with greenhouses, plants, living quarters for several hundred or thousands of people working on an element in laboratories, farms, factories and shipyards of the central module.
In this case, it makes sense to make the central module with the external toroidal part, which can be rotated (separately from the power receiving device, module center and external zones of the element), creating artificial gravity. A sort of space city donut is not only spinning separately in space, but strung on a central module rotating around a tower, and with external zones around it. In the residential version, the lighted upper surface of this bagel (torus) and its sides should have windows for greenhouses, farms, and living quarters.
Now we must determine the location and size of the Ring itself. If we build such a Ring for the Earth, in the Solar System, then there are two main alternatives:
1) The far option - Ring beyond the orbit of the Earth, but before the orbit of Mars - to intercept and use the light from the Sun, redirecting it to the Earth.
For a more elliptical orbit of Mars with a perihelion of 206 million km and a rounded orbit of the Earth with aphelion equal to 152 million km (which is not much more than the perihelion of the Earth at 147 million km), the circular orbit of the Ring elements with a radius will be optimally distant from both planets. about 190 million km.
The length of such a Ring will be 2 * Pi * R ~ 1200 million km. Densely packed hexagonal elements with a size of 330 km occupy an average of (382 + 191 = 573) / 2 = 286.5 km of length. About 1000 elements are needed per 1000 km. In total, the length of the Ring fits (1,200,000,000 km / 1,000 km) * 3.5 ~ 4.2 million elements. Multiplying this number by 3000, we get 12.6 billion elements of all in the Ring.
If the weight of each element is 30 million tons, then the total weight of the Ring will reach 378,000 trillion tons or 378 * 10 ^ 15 tons = 378 * 10 ^ 18 kg
And the area of such a Ring with a width of 1 million km will be about 1200 trillion km2.
This Ring will intercept 0.263% of the Sun’s light - only 1/380 of the area of the entire sphere - the area of the ring is = 2 * Pi * R * h, which is h / (2 * R) smaller than the area of the entire sphere (= 4 * Pi * R * R).
The earth will be slightly more lit and heated (due to the reflected light back), but if this overheating needs to be shielded with a mirror at L2, Lagrange or compensated by placing a shield of the right size of similar elements in front of the Earth, at Lagrange L1 (1.5 million km towards the Sun) - the day will be darker (because of this shield), and the night is lighter (because of the Ring).
2) The middle variant is the Ring inside the Earth's orbit, but outside of the orbit of Venus - in order to take the light from the Sun, preventing it from reaching the Earth.
At first glance, the idea of placing the Ring directly in front of the orbit of the Earth seems delusional - why intercept and not let the light of the Sun to Earth? But there are simple calculations concerning the fate of our Sun, which indicate that already after 1.5 billion years the intensity of the Sun’s radiation will begin to grow gradually (due to the accumulation of helium in the core of the Sun, the so-called triple helium reaction will be more active). Even an increase in the Sun’s energy reaching the Earth by 0.2–0.3% is catastrophic for the Earth’s climate - it is better prepared for this in advance, gradually obscuring most of the Earth’s orbit (with such a Ring or just a mirror at point L1 towards the Sun).
For a circular orbit of Venus with aphelion 109 million km and a circular orbit of the Earth with a perihelion equal to 147 million km optimally distant from both planets (taking into account the proximity of their masses) there will be a circular orbit of the Ring elements with a radius of about 125 million km.
The length of such a Ring will be 2 * Pi * R ~ 785 million km. The densely packed hexagonal elements in the amount of 3.5 cover 1000 km in length. In total, the length of the Ring fits (785 000 000 km / 1000 km) * 3.5 ~ 2.7475 million elements. Multiplying this number by 3000 (the number of elements in width) we get only 8242.5 million elements.
If the weight of each element is 30 million tons, then the total weight of the Ring = 237275 trillion tons or 237.275 * 10 ^ 15 tons = 237.3 * 10 ^ 18 kg.
Are there enough materials for the construction of such a ring in the solar system? Yes, it is: the total mass of the main belt asteroids between Mars and Jupiter is 3.0 - 3.6 * 10 ^ 21 kg. Of these asteroids, about 10% are M-type (metallic) asteroids. These metal asteroids weighing about 300 * 10 ^ 18 kg should be enough to build a ring 1 million km wide and weighing 237.3 * 10 ^ 18 kg (the closest variant is located behind Venus).
But for the long-range version with a mass of 378 * 10 ^ 18 kg (between the orbits of the Earth and Mars) iron in the asteroid belt (weighing 300 * 10 ^ 18 kg ) is not enough. We'll have to invent something with other materials.
For the Ring of a similar construction in the asteroid belt with an average belt radius of 3 AU. or 450 million km we get a circumference of 2 * Pi * R ~ 2830 million km, which is 2.5 times longer than the long-range version before Mars.
This means that the mass of the Ring is also 2.5 times larger, about 945 * 10 ^ 18 kg or about a third of the mass of all asteroids there, which is rather too small than enough.
The area of the Ring behind Venus, inside the Earth's orbit, with a width h of 1 million km will be equal to 785 trillion km2. Which is about 1.52 times smaller than the first version before Mars. Approximately 1.5 times less will be the weight of the Ring.
This Ring will intercept already 0.4% of the Sun’s light - 1/250 of the entire area of the sphere: the area of the ring is = 2 * Pi * R * h, which is h / (2 * R) less than the area of the entire sphere (= 4 * Pi * R * R ). The Earth will receive less light, but if it is necessary to compensate for this underheating, you can leave holes in the Ring or add re-reflectors of light to the Earth on the side of the Ring (or on the side of the Earth).
On the Earth's orbit, solar radiation with a capacity of 1.361 kW falls on a surface of 1 m2 normal to the direction, and on Venus orbit this power is about 2.7 kW.
Those. over the entire area of the Ring behind Venus will fall about 785 * 10 ^ 12 * 10 ^ 6 m2 * 2 kW = 1.570 * 10 ^ 21 kW = 1.6 * 10 ^ 15 GW. For the year it will be about 1.6 * 24 * 365 * 10 ^ 21 kW * hours = 1.4 * 10 ^ 25 kW * hours = 1.4 * 10 ^ 16 billion kW * hours . With an efficiency of generating electricity from this energy of about 10%, it is possible to get about 1.4 * 10 ^ 15 billion kW * hours, which (even taking into account losses during shipment) is about 10 ^ 10 times ( 10 billion times ) more than the entire production Electricity on Earth in 2015 ( 2.3 * 10 ^ 4 billion kW * hours ) and about 10 ^ 9 times more than the primary energy received on Earth (from all types of fuel in all processes), equal to (in 2012) 15.55 * 10 ^ 4 billion kWh
Is it not worth building such a Ring 10 times more for the sake of generated electricity?
Well, for the version of the Ring or Roy on the orbit of Mercury, look here - the already mentioned above (in the first part of the article ) idea of Stuart Armstrong disassembling Mercury weighing 3.3 * 10 ^ 33 kg. There the energy of the light falling on the Ring will be enormous.
In the third part of the article, the meaning and goals of constructing such a Ring, methods of its use and methods of non-standard use of individual autonomous elements of the Ring will finally be considered. The problem of finding such a giant structure from the outside will also be touched upon.
The most important moment in planning the future of the Dyson Ring is the right choice of the design of the standard Ring element. It should be standardized in size and shape for ease of production, unification of industry and tools, but its design and architecture should be flexible enough for further modernization and unpredictable improvements in advance.
The shape of the element should be simple and convenient to assemble. The element plane must have a concavity in order to focus the light on the pickup. The center of mass of the element should coincide with the geometric center for ease of towing and orientation.
')
Taking the symmetrical and naturally occurring natural form as a basis, we can suggest the element of the Ring in the form of a giant hexagon. By the way, about the predecessors: similar elements form the mirror of the solar power station in this picture from the magazine “Technique of Youth” for March 1975 (note who painted the picture):
You can take another form - in one detailed article in Russian, elements-octagons were suggested (alas, I lost the link to it).
The size of a standard element should be optimal: not too small, so as not to cause problems when exposing and assembling a huge number of segments, not too large so that it can be towed to the place of assembly (of course not completely finished, but rather autonomous).
If we take for the width of the finished Ring a round value of 1 million km (this value is chosen, since it is small compared to the radius of the Ring in our system, but comparable to the diameter of the Sun, which is equal to 695700 * 2 = 1,391,400 km), it would be nice to cover at least one thousandth part of this width of the Ring with two or three elements. Therefore, let us take as the characteristic size of a fully assembled element of the Ring a value of 330 km from the edge to the edge of the hexagon, then along with the gaps three elements in a row will occupy about 1000 km of the width of the Ring (the horizontal axis in the picture below). The width of the Ring in 1 million km will accommodate 3,000 hexagonal elements - quite an observable number (but do not think that they can be made in 40-60 years, as in the video from the first part). Densely packed hexagonal elements with a size of 330 km occupy an average of (382 + 191 = 573) / 2 = 286.5 km in length along the Ring, as shown in the figure below (the vertical axis in the picture):
The parameters of such a regular hexagon are determined by two radii (in the picture below): a small radius of the inscribed circle r = 165 km and a large radius of the circumscribed circle R = 191 km, with r = SQR (3/4) * R. The area S = 3 * SQR (3) * R ^ 2/2 is approximately 94500 km2, we take it roughly over 95 thousand km2.
The structure of the element of the Ring
View of the element from the sun:
The element should be for the most part autonomous, self-orienting in space, self-correcting its orbit and self-repairing. At the beginning of the assembly, a separate tug can also drag it into place (into a stable orbit), but then the element needs to adjust the orbit, parry the deviations on its own.
As for the filling of this element of the Ring, it is necessary to have: mooring devices, warehouses for materials, repair capacities, receiving and energy transferring devices. The element must have all this in the geometric center - there is located the central module of metal (construction begins from it and all the main systems are in it). And around it are the outer zones of branching bent pipes (with not rigid, but elastic structure as on a wood leaf) plus openwork cobweb of thin, flexible tubes with a thin film stretched on them (with albedo about 100%), reflecting the light of the Sun.
Back in 1986, in this book (Griliches V.A. “Solar space power stations” - L .: Nauka, 1986.) in one of the sections the aspects of choosing the optimal film were considered .
But the easiest to manufacture, affordable and fairly light material from which to create a mirror of the external zones is in our home, in the kitchen. This is aluminum foil - the material is very visual.
Consider two alternatives:
a) thick foil with a thickness of 0.1 mm = 100 μm - a square meter weighs about 0.27 kg according to GOST (adding lithium to the foil can bring the weight up to 0.25 kg per 1m2 - approximately the same value of 245 grams per 1 m2 mentioned in Stewart Armstrong’s lecture on construction Roy Dyson from Mercury ).
b) thin foil with a thickness of 0.045 mm = 45 microns - a square meter weighs about 0.1 kg. Weight 100 m2 is only 10 kg.
Ideally, the entire hexagonal element of foil should have a weight equal to the weight of the film / foil multiplied by the number of square meters, which is:
a) 0.25 kg * 95000 * 10 ^ 6 m2 = 23.75 million tons (it will be a bit too much!).
Or
b) 0.1 kg * 95000 * 10 ^ 6 m2 = 9.5 million tons (not bad!).
The numbers are impressive, but not insanely large. So in the Russian Federation about 4 million tons of aluminum are produced annually and from it make 0.1-0.2 million tons of foil. In the world, in 2014, 54 million tons of aluminum were produced (which is noticeably more than 41 and 45 million tons of aluminum in 2010 and 2012). 16% of this volume is foil and other packaging materials.
The low strength of the film / foil is compensated by the ease of its processing and repair. And automated and permanent repair in any case is inevitable for a structure that is created not even for thousands, but for tens and hundreds of millions of years.
Alas, this is not all - the weight of the element should be much more than just the weight of the foil. Taking into account the thin openwork web of the tubes that support the shape of the foil itself, the weight of the element will approximately double. For every 100 m2 of foil, it is required to fix at least two tubes 10 meters long approximately centimeter in diameter, with walls about 1 mm thick (like ski poles). We can optimistically determine the weight of these aluminum tubes as 2 * 10 * 0.5 kg = 10 kg per 100 m2, in addition to 10 kg of the weight of 100 m2 of the foil itself. If for the manufacture of tubes instead of aluminum to take plastics, especially carbon plastics, the weight may be less.
A foil is deposited on this openwork web of tubes, attached from the sunny side.
This thin web should at least once at 0.1 - 1 km be added tubes / beams thicker of durable metals such as titanium - the supporting structure of the outer zone (with some mechanisms for changing the curvature and tension of the thin web and foil). Without going into detailed calculations, we are optimistic about the total weight of the supporting structure, a weight equal to the total weight of a thin openwork web. The total weight has tripled, for option b) the total weight of the element will be already 28.5 million tons.
Repair and assembly of all this design is performed by autonomous in motion (with its own engines) assembly robots collecting the hollow-tube frame and spider-robots crawling in vacuum between the frame tubes, collecting old foil scraps, pulling and securing the newly extruded foil directly onto these tubes.
Element Central Module
In the center of the element, like an island in the sea, a rigid and partially hermetic central module rises high. Its size is from 2 to 10 km (can be both polygonal and round), thickness is up to 1 km. It has a rigid and durable three-dimensional frame made of a metal profile (aluminum, titanium, steel). Covered with steel, aluminum and glass / plastic sheets. For an area of 100 + 100 km2 square surfaces of the illuminated and shady side plus the area of the sides of such a module with a thickness of 0.5 km equal to 0.5 * 10 * 4 = 20 km2, the weight of the sheath made of steel sheets (with a specific weight of 6 kg per 1 m2) will be: 220 * 10 ^ 6 m2 * 6 kg = 1.32 million tons.
The central module of the island as a whole, with a filling, should weigh no more than 2.5 million tons (the weight of the element will then be packed at 30 million tons). The central module is the place of control, the location of the engines, ports and locks for the mooring of spacecraft, the place of energy storage and the place of sending energy. There are also cooling systems, processing of raw materials into fuel and materials, the production of body parts and parts of the external frame, the production of film-foil, as well as repair systems. On the illuminated side of the central module, you can make a separate mirror with fixed focus on the tower of the power receivers from the external metal panels of the module case so that the module has its own energy source from the very beginning.
The mooring ports and docking stations on the back, unlit side of the central module must be especially strengthened to transmit the impulse from the pusher ship to the entire element. Actually, at first, you can only bring the fully finished central module to the desired orbit, and then import to it billet pipes of the support structure, machines for the production of openwork tubes, tube collector robots, spider robots for attaching foils to finished tubes, materials for tubes and aluminum. foil, spare parts for cars and robots.
Power pipes or fencing profiles (steel, titanium) - for example, hexagonal in shape, are attached to the central module. From this enclosure to the corners of the element extend (but do not reach them) tapering tubes of supporting branching hardwood structure (titanium, aluminum). Flexible ropes (titanium, Kevlar, carbon fibers), which are needed to stretch and bend the entire structure (pipes, openwork structure and the entire surface of the foil) into a parabolic or spherical mirror, go to the ends and other anchor points of these pipes from the upper face of the sides of the central island. to concentrate the light of the sun into one focus of this gigantic structure. The focus of the mirror (which can be positioned 400-900 km from the central module, depending on the bending of the described mirror from the plane) is a suspended solar energy receiver connected to the central module with cables and cables. When maneuvering the whole element of the Ring, this energy receiver will either have to be dragged to the central island and fixed there, or supplied with its own engines for individual maneuvering. From the side, in profile, the element looks like this:
The payload of the central module of the element
1. Energy payload: energy receivers, as well as energy emitters on the outer and inner side.
One energy pickup from the entire surface of the element is already described above in the previous paragraph. It is associated with the Ring element only with cables and cables. When it is dragged to the center of the element during maneuvers or for repair, it should attach to something in the center of the module. A second fixed motion receiver in the form of a high tower may well be the wharf for it.
What is the second power receiver for? To focus on it, the light reflected only from the central module itself (from its illuminated plane) is, so to speak, a permanent source of energy for the needs of the central module itself from the very beginning of work, even before assembling the external zones. The (upper) plane of the central module facing the Sun may also look like a parabolic or spherical mirror, only rigid, with constant geometry and focus. The light reflecting panels of this mirror direct it to the central fixed receiving receptacle in the form of a tower in the center. The tower is needed to move the focus point away from the central module to avoid overheating of the whole structure. Energy receivers consist of solar panels and heat exchangers for heating water / fuel / raw materials.
2. Production payload: raw material processing plants.
In any case, the necessary production capacity for the release of fuel (for the maneuvers of the element), materials and blanks (for the assembly of external zones) from raw materials / semi-finished products from asteroids. These products, in the form of fuel, materials, blanks, can then be transported to other elements.
3. Residential payload: a habitable colony with greenhouses, plants, living quarters for several hundred or thousands of people working on an element in laboratories, farms, factories and shipyards of the central module.
In this case, it makes sense to make the central module with the external toroidal part, which can be rotated (separately from the power receiving device, module center and external zones of the element), creating artificial gravity. A sort of space city donut is not only spinning separately in space, but strung on a central module rotating around a tower, and with external zones around it. In the residential version, the lighted upper surface of this bagel (torus) and its sides should have windows for greenhouses, farms, and living quarters.
Ring Size and Weight
Now we must determine the location and size of the Ring itself. If we build such a Ring for the Earth, in the Solar System, then there are two main alternatives:
1) The far option - Ring beyond the orbit of the Earth, but before the orbit of Mars - to intercept and use the light from the Sun, redirecting it to the Earth.
For a more elliptical orbit of Mars with a perihelion of 206 million km and a rounded orbit of the Earth with aphelion equal to 152 million km (which is not much more than the perihelion of the Earth at 147 million km), the circular orbit of the Ring elements with a radius will be optimally distant from both planets. about 190 million km.
The length of such a Ring will be 2 * Pi * R ~ 1200 million km. Densely packed hexagonal elements with a size of 330 km occupy an average of (382 + 191 = 573) / 2 = 286.5 km of length. About 1000 elements are needed per 1000 km. In total, the length of the Ring fits (1,200,000,000 km / 1,000 km) * 3.5 ~ 4.2 million elements. Multiplying this number by 3000, we get 12.6 billion elements of all in the Ring.
If the weight of each element is 30 million tons, then the total weight of the Ring will reach 378,000 trillion tons or 378 * 10 ^ 15 tons = 378 * 10 ^ 18 kg
And the area of such a Ring with a width of 1 million km will be about 1200 trillion km2.
This Ring will intercept 0.263% of the Sun’s light - only 1/380 of the area of the entire sphere - the area of the ring is = 2 * Pi * R * h, which is h / (2 * R) smaller than the area of the entire sphere (= 4 * Pi * R * R).
The earth will be slightly more lit and heated (due to the reflected light back), but if this overheating needs to be shielded with a mirror at L2, Lagrange or compensated by placing a shield of the right size of similar elements in front of the Earth, at Lagrange L1 (1.5 million km towards the Sun) - the day will be darker (because of this shield), and the night is lighter (because of the Ring).
2) The middle variant is the Ring inside the Earth's orbit, but outside of the orbit of Venus - in order to take the light from the Sun, preventing it from reaching the Earth.
At first glance, the idea of placing the Ring directly in front of the orbit of the Earth seems delusional - why intercept and not let the light of the Sun to Earth? But there are simple calculations concerning the fate of our Sun, which indicate that already after 1.5 billion years the intensity of the Sun’s radiation will begin to grow gradually (due to the accumulation of helium in the core of the Sun, the so-called triple helium reaction will be more active). Even an increase in the Sun’s energy reaching the Earth by 0.2–0.3% is catastrophic for the Earth’s climate - it is better prepared for this in advance, gradually obscuring most of the Earth’s orbit (with such a Ring or just a mirror at point L1 towards the Sun).
For a circular orbit of Venus with aphelion 109 million km and a circular orbit of the Earth with a perihelion equal to 147 million km optimally distant from both planets (taking into account the proximity of their masses) there will be a circular orbit of the Ring elements with a radius of about 125 million km.
The length of such a Ring will be 2 * Pi * R ~ 785 million km. The densely packed hexagonal elements in the amount of 3.5 cover 1000 km in length. In total, the length of the Ring fits (785 000 000 km / 1000 km) * 3.5 ~ 2.7475 million elements. Multiplying this number by 3000 (the number of elements in width) we get only 8242.5 million elements.
If the weight of each element is 30 million tons, then the total weight of the Ring = 237275 trillion tons or 237.275 * 10 ^ 15 tons = 237.3 * 10 ^ 18 kg.
Are there enough materials for the construction of such a ring in the solar system? Yes, it is: the total mass of the main belt asteroids between Mars and Jupiter is 3.0 - 3.6 * 10 ^ 21 kg. Of these asteroids, about 10% are M-type (metallic) asteroids. These metal asteroids weighing about 300 * 10 ^ 18 kg should be enough to build a ring 1 million km wide and weighing 237.3 * 10 ^ 18 kg (the closest variant is located behind Venus).
But for the long-range version with a mass of 378 * 10 ^ 18 kg (between the orbits of the Earth and Mars) iron in the asteroid belt (weighing 300 * 10 ^ 18 kg ) is not enough. We'll have to invent something with other materials.
For the Ring of a similar construction in the asteroid belt with an average belt radius of 3 AU. or 450 million km we get a circumference of 2 * Pi * R ~ 2830 million km, which is 2.5 times longer than the long-range version before Mars.
This means that the mass of the Ring is also 2.5 times larger, about 945 * 10 ^ 18 kg or about a third of the mass of all asteroids there, which is rather too small than enough.
The area of the Ring behind Venus, inside the Earth's orbit, with a width h of 1 million km will be equal to 785 trillion km2. Which is about 1.52 times smaller than the first version before Mars. Approximately 1.5 times less will be the weight of the Ring.
This Ring will intercept already 0.4% of the Sun’s light - 1/250 of the entire area of the sphere: the area of the ring is = 2 * Pi * R * h, which is h / (2 * R) less than the area of the entire sphere (= 4 * Pi * R * R ). The Earth will receive less light, but if it is necessary to compensate for this underheating, you can leave holes in the Ring or add re-reflectors of light to the Earth on the side of the Ring (or on the side of the Earth).
Energy interception estimate
On the Earth's orbit, solar radiation with a capacity of 1.361 kW falls on a surface of 1 m2 normal to the direction, and on Venus orbit this power is about 2.7 kW.
Those. over the entire area of the Ring behind Venus will fall about 785 * 10 ^ 12 * 10 ^ 6 m2 * 2 kW = 1.570 * 10 ^ 21 kW = 1.6 * 10 ^ 15 GW. For the year it will be about 1.6 * 24 * 365 * 10 ^ 21 kW * hours = 1.4 * 10 ^ 25 kW * hours = 1.4 * 10 ^ 16 billion kW * hours . With an efficiency of generating electricity from this energy of about 10%, it is possible to get about 1.4 * 10 ^ 15 billion kW * hours, which (even taking into account losses during shipment) is about 10 ^ 10 times ( 10 billion times ) more than the entire production Electricity on Earth in 2015 ( 2.3 * 10 ^ 4 billion kW * hours ) and about 10 ^ 9 times more than the primary energy received on Earth (from all types of fuel in all processes), equal to (in 2012) 15.55 * 10 ^ 4 billion kWh
Is it not worth building such a Ring 10 times more for the sake of generated electricity?
Well, for the version of the Ring or Roy on the orbit of Mercury, look here - the already mentioned above (in the first part of the article ) idea of Stuart Armstrong disassembling Mercury weighing 3.3 * 10 ^ 33 kg. There the energy of the light falling on the Ring will be enormous.
In the third part of the article, the meaning and goals of constructing such a Ring, methods of its use and methods of non-standard use of individual autonomous elements of the Ring will finally be considered. The problem of finding such a giant structure from the outside will also be touched upon.
Source: https://habr.com/ru/post/369687/
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