CN101625432B - Fixed-point condensing reflector - Google Patents

Abstract

The invention discloses a fixed-point condensing reflector, and relates to the technical field of solar energy utilization, in particular to a solar fixed-point condensing reflector. The fixed-point condensing reflector is formed by taking a specific curve as a bus to rotate around a main axis and intercepting a certain area. When tracking the sun, the fixed-point condensing reflector rotates eta degrees around the main axis to make a main reflecting surface parallel to solar rays, and simultaneously, the fixed-point condensing reflector rotates 0.5[Delta]epsilon degrees around a vertex P and periodically tracks a declination difference value [Delta]sigma to make reflected rays of the reflector condensed at a fixed point F under any working condition. Light condensing points of the reflector are fixed and can be higher than, lower than or parallel to the reflector, the reflectors can be distributed into a reflector array, and the installation is not limited by the terrain. The fixed-point condensing reflector can be widely applied in the field of the solar energy utilization, and can be installed and distributed into a high-power point mode system such as a solar smelting furnace and the like.

Description

Fixed-point condensing reflector

Technical field

The application relates to technical field of solar utilization technique, particularly a kind of sun fixed-point condensing reflector.

Background technology

Existing solar thermal utilization equipment can be divided into slot type system, tower system and the large fundamental type of dish formula system 3, and the extinction body of slot type system is fixed, but its optically focused is smaller, the thermal efficiency is low; The optically focused of tower system compares greatly, the thermal efficiency is high, but its extinction body must be higher than catoptron, and its purposes is restricted; The optically focused of dish formula system is larger, the thermal efficiency is higher, but its extinction body must follow catoptron and rotate and follow the tracks of the sun, makes its heat transfer system complexity.Therefore develop that a kind of optically focused is larger, the thermal efficiency is higher, extinction body maintains static, and extinction body can higher than, lower than or the solar energy utilization system parallel with catoptron become the task of top priority, we are referred to as fixed-point condensing system, are called for short point type system; Well-known: the optically focused ratio of point focusing is greater than line focus, and the extinction open area of point focusing is less, under equal temperature, less the radiate heat of extinction open area is fewer, so the thermal efficiency of point focusing is higher than line focus; The core technology of point type system be to develop a kind of optically focused is larger, focal point is fixed and focal point can higher than, lower than or the catoptron parallel with catoptron, we are referred to as fixed-point condensing reflector.Application number is that 200610200777.8 Chinese patent discloses and a kind ofly rotates the solar furnace that aberration revised law is manufactured with spin-elevation tracking mode and ranks, its programme element is: the reflecting surface of heliostat is made up of multirow and the sub-mirror of multiple row, and extinction body is fixed, heliostat follows the tracks of when the sun that sub-mirror can be expert at and column direction carries out axial rotation with to making correction because angle of incidence of sunlight changes the potential difference that resembles causing; Is its inventive point high but do not understand in person: 1, how optically focused of its heliostat? 2, sub-mirror bending shaft and crisscross how rotation? 3, whether each viewpoint definition unclear? Because its sub-mirror is in row and column axial rotation, so its structure and tracker must be complicated.

Summary of the invention

The technical matters that the application will solve be to provide that a kind of optically focused is larger, focal point is fixed and focal point can higher than, lower than or parallel with catoptron and can be arranged to the catoptron of mirror battle array.

For solving the problems of the technologies described above, the application's fixed-point condensing reflector is to rotate around the spindle and intercept certain area to form take a specific curves as bus, and fixed-point condensing reflector is hereinafter to be referred as catoptron; Follow the tracks of the solar time, it rotates η degree around primary optical axis, corner η=arc cos{{sin γ s cosh cosh ₀sin (γ s-γ)+(cos ψ s sin h ₀-sin ψ s cos γ s cosh ₀) [cos ψ s sin h-sin ψ s cosh cos (γ-γ is s)] } × [cos ²h sin ²(γ-γ s)+[cos ψ s sin h-sin ψ s cosh cos (γ-γ s)] ²] ^-1/ ₂× [cos ²h ₀sin ²γ s+ (cos ψ s sin h ₀-sin ψ s cosh ₀cos γ s) ²] ^-1/ ₂, rotate 0.5 Δ ε degree, firing angle difference DELTA ε=ε-ε around summit P simultaneously ₀, firing angle ε=arc cos[sin ψ s sin h+cos ψ s cosh cos (γ-γ is s)], ε ₀for the position firing angle that begun, and regularly declination angle difference DELTA δ is followed the tracks of, in the time of primary optical axis position angle γ s=0 °, catoptron rotates 0.5 Δ δ, Δ δ=δ-δ around summit P ₀, δ=23.45sin[360 × (284+n)/365, declination angle], in formula, n is the number of days of starting at from annual January 1, δ ₀for the declination angle, position of having begun, in the time of γ s ≠ 0 °, catoptron rotates η degree around primary optical axis, make mainly to penetrate face and be parallel to sunray, rotate 0.5 Δ ε degree around summit P simultaneously.Said specific curves is Y ²=2af (Z-B) para-curve or radius-of-curvature ρ is greater than 0.5 ρ ₀be less than 2 ρ ₀optically focused camber line, particularly lower extreme point radius-of-curvature ρ > 0.5 ρ ₀, upper extreme point radius-of-curvature ρ < 2 ρ ₀round and smooth optically focused camber line, ρ ₀=af (1+tg ²Φ) ^3/2.

Said catoptron is to rotate around the spindle certain angle, the i.e. hyperboloidal mirror that intercepts certain width and highly form take a specific curves as bus, and its peak is that upper extreme point, minimum point are lower extreme point, and catoptron cross section can be oval or circular; Bus rotational trajectory can be circular arc line, also can be other optically focused camber lines.Para-curve Y ²in=2af (Z-B), Y, Z are coordinate figure, its O-XYZ coordinate is: fixed point F makes the Z axis sensing sun and is main shaft excessively, on Z axis, get FO=f, crossing true origin O point makes vertical line the energized north of Z axis and is Y-axis, be that YOZ coordinate surface is parallel to the earth's axis, cross O point and make the vertical line of Y, Z axis and point to west to be X-axis; A, B are constant, a=cos (ψ+θ z ₀)/tg Φ, constant B is the distance of para-curve summit to Y-axis, B=0.5fcos (ψ+θ z ₀) tg Φ, or B=f[1-sin (ψ+θ z ₀)-cos ²(ψ+θ z ₀)/2a], f is that focal length, said focal distance f are the distance of summit P to fixed point F; ψ is the angle of pitch, θ z ₀for the position zenith angle that begun, desirable θ z on the Northern Hemisphere ₀for the zenith angle in moment high noon in the Summer Solstice, in the Southern Hemisphere desirable θ z ₀for the zenith angle in moment high noon in Winter Solstice,

for local latitude, Φ is primary tangent inclination angle, Φ=0.5 (h ₀-ψ), i.e. the primary tangent inclination angle Φ position height angle h that equals to have begun ₀deduct the half of angle of pitch ψ, h ₀=90 °-θ z ₀; ρ ₀=af (1+tg ²Φ) ^3/2the title of middle a, f, Φ is equal to above-mentioned.Said main shaft is bus turning axle; Said optically focused camber line can be the round and smooth camber lines such as circular arc line, involute urve, ellipse, spiral of Archimedes, circular arc and parabolical complex line, saidly round and smoothly refers to that radius-of-curvature ρ gradually changes or constant.The said beginning is played the position of setting moment at the high noon spindle alignment sun when position is design and installation, can be the position of the Summer Solstice or moment spindle alignment at high noon in Winter Solstice sun; Said primary optical axis PF is through the light of summit P reflection, is also catoptron turning axle, the line of still fix a point F and summit P; A set point on the extinction mouth that said fixed point F is extinction body or light-guiding shade; Said extinction body is hot extinction equipment for thermal-arrest, can be flat plate collector or photovoltaic battery panel or pot and stove or boiler or smelting furnace or reacting furnace or processing stove etc.; Said light-guiding shade is to be installed on the condensing body towards catoptron on extinction body; Said extinction mouth is perpendicular to primary optical axis PF and cross the face of F of fixing a point on extinction body or light-guiding shade; Said catoptron summit P is the catoptron rotation centre of sphere, to be that the hinge axes of the hinged rotating shaft of catoptron is vertical mainly penetrate face and at the main point forming in face of penetrating, so summit P being treated as in the time that optically focused principle is described because its hinge is very little is to penetrate a point in specific curves in face master, catoptron summit P is called for short summit P in full; Said master penetrates the face that ray that face is sun incident summit P and primary optical axis form, i.e. the face that the ray of sun incident summit P and principal normal form, and the plane of incidence is the face that solar rays and incidence point normal form; Said principal normal is the normal of summit P, and primary tangent is the tangent line of summit P.

Some angle code names in the application and definition and the regulation on the Northern Hemisphere time: said firing angle ε is the angle of solar rays and primary optical axis; Said zenith angle θ z is the angle of solar rays and ground level normal; Sun altitude h is the angle of solar rays and its projection line on ground level,

solar azimuth γ be projection line on the ground of solar rays and Due South to angle,

regulation Due South is zero, westwards for just, eastwards for negative; Solar hour angle ω is the angle that represents the solar time, the ω of being equivalent to=15 ° per hour, and regulation high noon is zero, and the morning is for negative, and afternoon is for just; Declination angle δ is the angle of cut of ecliptic plan and equatorial plane, regulation: subsolar point under the line to the north of for just, under the line on the south for bearing; Primary optical axis position angle γ s is projection line and the due south angular separation of primary optical axis PF on ground level, and it is zero that regulation is pointed to Due South by P to F, westwards for just, eastwards for negative; Angle of pitch ψ is the angle of primary optical axis and the earth's axis, the angle of primary optical axis and ground level while being γ s=0 °, primary optical axis inclination angle ψ s is the angle of primary optical axis and ground level, regulation fixed point F is 0 ° of elevation angle ψ >, 0 ° of ψ s > during higher than summit P, and fixed point F is 0 ° of angle of depression ψ <, 0 ° of ψ s < during lower than summit P; Primary tangent inclination angle Φ is the angle of primary tangent and Y-axis; Tangent line inclination angle λ is the angle of tangent line and Y-axis; Cutting firing angle ξ is the angle of solar rays and xsect, ξ=ε+90 °-β; Effectively firing angle: efficient light according to time firing angle be called for short effective firing angle, selected while being design, be less than maximum elevation one specify firing angle; Main shaft and horizontal angle when main shaft angle β is ψ=0 °, position β has begun ₀main shaft and horizontal angle; Said corner η is that the begun master of position penetrates face and the instant main angle of penetrating between face, is the angle that catoptron rotate around primary optical axis, and regulation high noon is zero, and the morning is for bearing, and afternoon is for just.

The calculating formula of corner η is established by following inference: set up P-xyz coordinate system: take summit P as true origin, and x, the parallel ground of y axle, x axle points to due south, and y axle points to positive west, and z axle points to zenith; Leading if instant and penetrating face is I, and the primary optical axis surfaces of revolution that P is ordered is excessively II, and it is I that the master of the position of having begun penetrates face ₀, solar core is A, and | PA|=1: P (0,0,0), F (fcos ψ s cos γ s, fcos ψ s sin γ s, fsin ψ s), A (1cosh cos γ, 1cosh sin γ, 1sin h), vector $\stackrel{\&RightArrow;}{\mathrm{PF}}=f\cos \mathrm{\&psi;}s\cos \mathrm{\&gamma;si}+f\cos \mathrm{\&psi;}s\sin \mathrm{\&gamma;sj}+f\sin \mathrm{\&psi;sk},$ $\stackrel{\&RightArrow;}{\mathrm{PA}}=1\cosh \cos \mathrm{\&gamma;i}+1\cosh \sin \mathrm{\&gamma;j}+1\sinh k;$ Because penetrating face, instant master is parallel to solar rays PA, so the normal line vector of face I: $n_1=\stackrel{\&RightArrow;}{\mathrm{PF}}\&CenterDot;\stackrel{\&RightArrow;}{\mathrm{PA}}=n_{1x}i+n_{1y}j+n_{1z}k,$ N _1x=1f (cos ψ s sin γ s sin h-sin ψ s cosh sin γ), n _1y=1f (sin ψ s cosh cos γ-cos ψ s cos γ s sin h), n _1z=1fcos ψ s cosh (sin γ cos γ s-cos γ sin γ s)=(γ-γ s) for 1fcos ψ s cosh sin; The normal line vector of surfaces of revolution II: $n_2=\stackrel{\&RightArrow;}{\mathrm{PF}}=n_{2x}i+n_{2y}j+n_{2z}k,$ N _2x=fcos ψ s cos γ s, n _2y=fcos ψ s sin γ s, n _2z=fsin ψ s; Two sides intersection L ₁vector: s ₁=n ₁n ₂=s _1xi+s _1yj+s _1zk, s _1x=(n _1yn _2z-n _1zn _2y), s _1y=(n _1zn _2x-n _1xn _2z), s _1z=(n _1xn _2y-n _1yn _2x); Moment at high noon γ=0 °, A ₀(1cos h ₀, 0,1sin h ₀), ${\stackrel{\&RightArrow;}{\mathrm{PA}}}_0={1\cosh }_{0}i+1\sin h_{0}k;$ Face I ₀normal line vector: ${\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="S2008100293844E00061.png,S2008100293844E00071.png"\; state="\{\{state\}\}">n</figure-callout>}}_0=\stackrel{\&RightArrow;}{\mathrm{PF}}\&CenterDot;{\stackrel{\&RightArrow;}{\mathrm{PA}}}_0=n_{0x}i+n_{0y}j+n_{0z}k,$ N _0x=1fcos ψ s sin γ s sin h ₀, n _0y=1f (sin ψ s cos h ₀-cos ψ s cos γ s sin h ₀), n _0z=-1fcos ψ s sin γ s cosh ₀, face I ₀face II intersection L ₂vector: s ₂=n ₀n ₂=s _2xi+s _2yj+s _2zk, s _2x=(n _0yn _2z-n _0zn _2y), s _2y=(n _0zn _2x-n _0xn _2z), s _2z=(n _0xn _2y-n _0yn _2x); Vector s ₁with s ₂angle be both corner η, cos η=s ₁s ₂/ | s ₁|| s ₂|, by vectorial s ₁and s ₂coordinate figure substitution and simplify after: η=arc cos{{sin γ s cosh cosh ₀sin (γ s-γ)+(cos ψ s sin h ₀-sin ψ s cos γ s cosh ₀) [cos ψ s sin h-sin ψ s cosh cos (γ-γ is s)] } × [cos ²h sin ²(γ-γ s)+[cos ψ s sin h-sin ψ s cosh cos (γ-γ s)] ²] ^-1/ ₂× [cos ²h ₀sin ²γ s+ (cos ψ s sin h ₀-sin ψ s cosh ₀cos γ s) ²] ^-1/ ₂.Because firing angle ε is the angle of solar rays and primary optical axis, so $\cos \mathrm{\&epsiv;}=\stackrel{\&RightArrow;}{\mathrm{PF}}\&CenterDot;\stackrel{\&RightArrow;}{\mathrm{PA}}/\left|\mathrm{PA}\right|\&CenterDot;\left|\mathrm{PF}\right|,$ Will

and

coordinate figure substitution and simplify after: ε=arc cos[sin ψ s sin h+cos ψ s cosh cos (γ-γ is s)].

Said xsect is perpendicular to the face of main shaft; Cross sections is parallel to XOY coordinate surface, and carrying line at each cross sections internal reflector is the circular arc line take the intersection point of Z axis and cross sections as the center of circle, i.e. the projection of catoptron on XOY coordinate surface is the concentric arc line take true origin O as the center of circle; Because the application is condenser mirror, thus cross sections internal reflector can also be the F that fixes a point be other optically focused camber lines of focal point, the position catoptron that now begun is for some optically focused.If in addition with Y ²=2af (Z-B) para-curve be bus along X-axis move slot type catoptron.While having begun position because the spindle alignment sun, so ξ=90 °, solar rays in XOY coordinate surface for a bit; Learnt by reflection law: on minute surface, the reflected ray of arbitrfary point must be coplanar in the plane of incidence with the normal of this point and incident ray, and incident angle equals reflection angle; Because catoptron is to do circular arc rotation take specific curves as bus around Z axis to form minute surface, so in same xsect, the incident angle of each point is equal, reflection angle is equal, all planes of incidence penetrate that face intersects at Z axis and perpendicular to XOY coordinate surface, so the angle of each point equates with the main angle of penetrating the interior respective points of face in all planes of incidence with main.Specific curves is Y ²when=2af (Z-B) para-curve, fixed point F is the mid point of the normal of Z axis, arbitrfary point and the right-angle triangle of tangent line composition thereof, because Z axis is parallel to solar rays, so the reflection ray of the position incidence reflection mirror that begun focuses on fixed point F certainly; When following nothing special instruction, all with Y ²=2af (Z-B) para-curve is that the minute surface that bus is done circular arc line rotation around main shaft illustrates the optically focused principle of discussing catoptron.Due to fixed point, F is the fixed point on extinction body or light-guiding shade, and light-guiding shade is installed on extinction body, extinction body and catoptron are installed on ground, so fixed point F and summit P maintain static with respect to the earth, focal distance f is a constant, fixed point F can higher than, lower than or be parallel to summit P; The angle of primary optical axis and ground level while being γ s=0 ° due to ground normal perpendicular to local horizon, angle of pitch ψ, so the angle of primary optical axis and each xsect equals ψ+θ z in the time of ξ=90 °, in YOZ coordinate surface ₀, by formula Y ²=2af (Z-B) learns, para-curve is female curvature of a curve and slope is determined by constant a, B, and constant a=cos (ψ+θ z ₀)/tg Φ, B=0.5fcos (ψ+θ z ₀) tg Φ, and Φ=0.5 (h ₀-ψ), h ₀=90 °-θ z ₀, and when fixed point F higher than, lower than or while being parallel to summit P, according to angle of pitch ψ, focal distance f, local latitude and selected declination angle δ value, thereby determine female curvature of a curve and slope by calculative determination constant a, B value, make the each xsect of catoptron perpendicular to solar rays, thereby the position focusing mirror that makes to have begun is in fixed point F; While having begun position the focal point of catoptron fix and focal point can higher than, lower than or parallel with catoptron.

Because earth rotation causes that with revolving around the sun sunshine projects the firing angle variation of catoptron, so drive catoptron to follow the tracks of the sun with tracking means: the solar rays of establishing incident summit P has rotated Δ ε, make incident angle increase Δ ε, follow the tracks of the solar time, catoptron rotates η degree around primary optical axis, make master penetrate face and be parallel to sunray, simultaneously its principal normal rotates 0.5 Δ ε degree around summit P main penetrating in face, make incident angle reduce 0.5 Δ ε, be the actual 0.5 Δ ε that increased of incident angle, because principal normal has rotated 0.5 Δ ε, make reflection angle increase 0.5 Δ ε, because the incident angle of summit, the position P that begun equals reflection angle, reflection ray directive fixed point F, so the incident angle that after rotation is any operating mode summit P equals reflection angle, the reflection ray of ordering through P will be along primary optical axis PF directive fixed point F.When being followed the tracks of, firing angle ε comprises the tracking to declination angle δ, but every day, efficient light was according to approximately 10 hours, i.e. approximately 10 hours every days of tracking to firing angle ε, so also will carry out the tracking of other 14 hours to declination angle δ, regularly declination angle difference DELTA δ is followed the tracks of, make catoptron be concentrated on fixed point F; While realizing any operating mode thus the focal point of catoptron fix and focal point can higher than, lower than or parallel with catoptron.Catoptron is followed the tracks of sun process three kinds of operating modes: one is begun position, ε now ₀=h ₀=β ₀, ξ ₀=90 °, solar rays is parallel to main shaft, and reflection ray focuses on fixed point F; One is ε < ε ₀, 90 ° of ξ < now, one is ε > ε ₀, 90 ° of ξ > now, latter two operating mode catoptron is optically focused, reflection ray forms hot spot on extinction mouth; The size of its hot spot and incidence point to the limit distance and the firing angle difference of P are directly proportional, and the distance of incidence point and summit P is larger, hot spot distance is larger, the intersection point that said hot spot distance is reflection ray and extinction mouth and the distance of the F that fixes a point; Firing angle difference more large spot is just larger.And the larger extinction open area of hot spot is larger, and the heat loss through radiation of extinction mouth is directly proportional to extinction open area, and under extinction body equal temperature, larger the radiate heat of extinction open area is more, and thermal loss is larger, the extinction body thermal efficiency is lower; For dwindling hot spot in order to improve the thermal efficiency of extinction body, can control catoptron cross section and control by controlling the distance of incidence point and summit P the size of extinction open area.Because the hot spot distance of summit P is zero and the distance maximum of catoptron end points and summit P, thus the size in catoptron cross section be by selecting side apart from recently determining, said end distance is than being catoptron end points distance and its hot spot ratio of distances constant of P to the limit; Because the higher desired catoptron optically focused ratio of serviceability temperature of extinction body is larger, and optically focused than to smallest end apart from than square be directly proportional, end is apart from larger than larger optically focused ratio, the thermal efficiency of extinction body is higher, and catoptron cross section is less, relative tracking means cost is higher, so the size in catoptron cross section is determined by the serviceability temperature of extinction body.Can also realize by reducing firing angle difference for dwindling hot spot, because minimax firing angle is all the extreme value of firing angle difference, so can design with the mean value of minimum elevation with maximum; But during due to maximum elevation, be the maximal value that atmospheric envelope distance is passed in solar radiation, thus the application design with the mean value of effective firing angle and minimum elevation, take the elevation midpoint of effective firing angle and minimum elevation as the position firing angle ε that begun ₀; But effectively firing angle is a designated value but not the particular value of celestial bodies motion, for ease of calculate and explanation, generally take the firing angle in the Summer Solstice or moment high noon in Winter Solstice as the position firing angle ε that begun ₀, firing angle difference is minimized.

Can also one light-guiding shade be installed by the position towards catoptron on extinction body and realize in order to improve the thermal efficiency of extinction body, a light-guiding shade is installed between extinction body and catoptron and be realized; Light-guiding shade is a circle or oval cone, on its extinction body or be that bottom surface is less than the end face near catoptron near the end face of extinction body, end face near catoptron is extinction mouth, the normal three line unifications of light-guiding shade axis and primary optical axis and extinction mouth, slightly deviation; Its bottom surface and extinction mouth be perpendicular to primary optical axis PF, can mounting plane glass or collector lens on bottom surface and extinction mouth, and now light-guiding shade can be evacuated, to stop the heat loss through convection of light-guiding shade; The sticky plating of its conical surface inside surface reflectance coating, conical surface outside surface can add heat-insulation layer, outwards dispels the heat by light-guiding shade to stop extinction body.During due to any operating mode, the reflected ray of summit P is all along primary optical axis directive fixed point F, and normal three lines of light-guiding shade axis and primary optical axis and extinction mouth unify, so the reflected ray of summit P will inject extinction body along light-guiding shade axis; During due to any operating mode, catoptron is all concentrated on the fixed point F on extinction mouth, and the bottom surface that extinction mouth is coated with reflectance coating, light-guiding shade perpendicular to light-guiding shade axis, light-guiding shade conical surface inside surface is less than extinction mouth, so the reflected ray of catoptron will directly inject extinction body or inject extinction body after the refraction of light-guiding shade conical surface inside surface along light-guiding shade axis, make reflected ray optically focused again, carry high concentration ratio, thereby improve the thermal efficiency of extinction body.

Due to the focal point of catoptron can higher than, lower than or parallel with catoptron, so the installation of extinction body and catoptron is not limited by landform, multiple catoptrons can be mounted to along massif slope mirror battle array, its all around the reflected ray between catoptron stagger, do not block mutually, make multiple catoptrons be concentrated on a fixed point F, its extinction body is lower than catoptron and in the south of catoptron; Also extinction body can be arranged on to mountain top or tower top, make extinction body higher than catoptron, multiple catoptrons are arranged to mirror battle array on level land, make its all around the reflected ray between catoptron stagger, do not block mutually, make multiple catoptrons be concentrated on one fixed point F; Thereby expand daylighting area, improve the power input of single extinction body, widen solar energy heat utilization field, make the solar energy system of high-power single extinction body: as sun power smelting furnace etc. is achieved.While arranging on the Northern Hemisphere mirror array 1 system, extinction body is in the south of catoptron, in mirror battle array, only having a row catoptron and extinction body is that positive north and south is arranged, i.e. its primary optical axis position angle γ s=0 °, all the other each row catoptron γ s ≠ 0 °; Because the application's catoptron is to be parallel to set up a specific curves and intercept certain area take it as bus around the main shaft rotation of pointing to the sun in the YOZ coordinate surface of the earth's axis one to form, so the bus of each row catoptron of γ s ≠ 0 ° is identical with the specific curves in YOZ coordinate surface, the ξ of its any point of position that begun ₀=90 °, any row catoptron in the position of having begun all focuses on or is concentrated on fixed point F, has just begun position γ s=0 ° catoptron its lead and penetrate face perpendicular to ground level, and the catoptron of γ s ≠ 0 ° its lead and penetrate face tilt in ground level.Follow the tracks of solar time, all catoptron masters and penetrate face around primary optical axis rotation η degree separately, (catoptron its corner η of different γ s, ψ s and firing angle ε difference, be that each catoptron all has independent driving tracking means), be parallel to sunray so make master penetrate face, simultaneously its principal normal rotates 0.5 Δ ε degree around summit P separately each independently penetrating in face, while being any operating mode, all catoptron masters penetrate face and are parallel to sunray, the reflection ray of all summit P is directive fixed point F all, so all catoptrons are all concentrated on fixed point F.As can be seen here, the application's catoptron can be arranged to mirror battle array completely, forms thus the changeable solar thermal utilization point type system of power.

In sum, the application's catoptron is to rotate around the spindle and intercept the hyperboloidal mirror that certain area forms take a specific curves as bus, the size in its cross section is determined by the serviceability temperature of extinction body; Its specific curves is with respect to the height of summit P and distance, according to angle of pitch ψ, focal distance f, latitude according to fixed point F

and selected declination angle δ value establishes, the position focusing mirror or be concentrated on the F that fixes a point of making to have begun, its focus fixes a point that F fixes and on primary optical axis; And the position of having begun is to set with the mean value of the minimized effective firing angle of firing angle difference and minimum elevation, i.e. selected declination angle δ value; So the focal point of catoptron is fixed and focal point can higher than, lower than or parallel with catoptron and can be arranged to mirror battle array, make the installation of extinction body and catoptron not limited by landform.Tracking solar time, catoptron master penetrate face and rotate η degree around primary optical axis, while making master penetrate any operating mode of face, be all parallel to sunray, simultaneously its principal normal rotates 0.5 Δ ε degree around summit P main penetrating in face, and regularly declination angle difference DELTA δ is followed the tracks of, while making any operating mode, all directive fixed point F of the reflected ray of summit P, catoptron is all concentrated on fixed point F.Due to the optically focused of the application's catoptron than to smallest end apart from than square be directly proportional, selects higher end distance even to reach hundreds of times than making optically focused ratio reach tens times; Due to light-guiding shade optically focused again, the highest optically focused ratio that makes reaches thousands of times, thereby improves the thermal efficiency of extinction body; Because the application's catoptron can be arranged to mirror battle array, make multiple catoptrons be concentrated on a fixed point F, thereby improve the power input of single extinction body, widen solar energy heat utilization field.

Accompanying drawing explanation

Fig. 1 is split type solar furnace schematic diagram.Fig. 2 is that the position catoptron master of having begun penetrates face focused view.

Fig. 3 is optically focused figure after catoptron rotates 23.45 °.Fig. 4 be high noon in Winter Solstice moment master penetrate optically focused figure in face.

Fig. 5 is moment high noon in Winter Solstice left side optically focused figure.Fig. 6 is moment high noon in Winter Solstice left side optically focused vertical view.

Fig. 7 is the sun local horizon opticofacial winking reflex mirror optically focused figure that haunts in the Summer Solstice.Catoptron optically focused figure when Fig. 8 is effective firing angle.

Fig. 9 is split type solar cooker schematic diagram.Figure 10 is the elevation angle, the position focusing mirror figure that begun.

Figure 11 is moment high noon in Winter Solstice elevation angle catoptron optically focused figure.When Figure 12 is effective firing angle, elevation angle catoptron master penetrates face optically focused figure.

Embodiment

Below in conjunction with embodiment, the application is described further.

First embodiment is the embodiment of fixed-point condensing reflector composition solar furnace.Fig. 1 is split type solar furnace schematic diagram.Split type solar furnace is made up of extinction mouth 1, extinction body 2, catoptron 3, tracking means, support etc.; Extinction body 2 separates certain distance, is installed on ground by support 10 respectively with catoptron 3, extinction body 2 is a pot and stove, and extinction mouth 1 perpendicular to primary optical axis PF, has a fixed point F on extinction body 2 on extinction mouth 1; Catoptron 3 is hinged by hinge 11 and rotating shaft 7, hinge 11 axis are penetrated face and rotating shaft 7 perpendicular to catoptron 3 masters, be that catoptron 37 axis rotations around the shaft simultaneously can be around hinge 11 axis rotations, its rotation centre of sphere is summit P, on catoptron 3, there is a cross section 4, summit P is primary optical axis PF with the line of fixed point F, and primary optical axis PF is parallel to local horizon NS and points to due south, and primary optical axis PF and rotating shaft 7 axis are collinear; Rotating shaft 7 is enclosed within on support 10, one tracking means 6 is installed in rotating shaft 7, tracking means 6 is set a distance formula tracking means, support 10 is installed on the ground, one tracking means 9 is installed on support 10, tracking means 9 is to determine corner-type tracking means, and the details such as the following principle of tracking means can be applied for reference to me, application number is: the Chinese patent of 200610166899.X.Tracking means 9 rotates by gear 8 drive shafts 7, and rotating shaft 7 drives catoptron 3 and tracking means 6 together around primary optical axis PF rotation, and tracking means 6 is slidably connected by pull-push piece 5 and drives catoptron 3 around hinge 11 axis rotations.

Fig. 2 is that the position catoptron master of having begun penetrates face focused view.Cross fixed point F do Z axis aim at the sun be main shaft, its main shaft angle β ₀=90 °-θ z ₀; On Z axis, get FO=f, the vertical line energized north that mistake true origin O point is made Z axis is Y-axis, and YOZ coordinate surface is parallel to the earth's axis, and the vertical line sensing west that mistake O point is made Y, Z axis is X-axis, forms thus O-XYZ coordinate system.Make primary optical axis PF and be parallel to local horizon NS sensing due south, γ s=0 °, ψ=0 °, establishing specific curves is Y ²=2af (Z-B) para-curve is also made this para-curve on YOZ coordinate surface, constant a=cos θ z in formula ₀/ tg Φ, B=0.5fcos θ z ₀tg Φ or B=f (1-sin θ z ₀-cos ²θ z ₀/ 2a); Para-curve and primary optical axis PF joining P are catoptron summit, and mistake P point is made solar rays AP and is parallel to Z axis, crosses P point and makes principal normal PP ₁, hand over Z axis in P ₁, PP ₁with the angle of PF be reflection angle alpha ₀, with the angle of AP be incidence angle θ ₀, the angle of AP and local horizon NS is elevation angle h ₀, i.e. h ₀=β ₀; Cross P point and make Z axis vertical line PP ₂, P ₂for intersection point, because ground normal is perpendicular to local horizon NS, and AP ⊥ PP ₂so, ∠ FPP ₂=θ z ₀.Cross P point and make PP ₁vertical line be tangent line PP ₃, hand over Y-axis in P ₃, its coordinate is (0, Y _p3, 0), PP ₃with the angle of Y-axis be Φ; Because PP ₂parallel Y-axis, so ∠ P ₂pP ₃=Φ, the slope tg Φ=Y of ordering due to P _p/ af, Y _p=PP ₂=FPcos θ z ₀, a=cos θ z ₀/ tg Φ, so tg Φ=PP ₂tg Φ/fcos θ z ₀so, PP ₂=fcos θ z ₀, therefore FP=f=FO; Due to tangent line PP ₃equation be: Z _p-Z _p3=tg Φ (Y _p-Y _p3), and P ₃coordinate figure is (0, Y _p3, 0), so only have Y _p3=0 o'clock, just have a Z _p3=0, only have P ₃when point overlaps with true origin O, this equation is vertical, so tangent line PP ₃be primary tangent PO, the angle Φ of PO and Y-axis is primary tangent inclination angle Φ.Because PP ₁so ⊥ PO Δ PP ₂o ∽ Δ P ₁pO, so ∠ PP ₁f=Φ, because AP//P ₁so F ∠ APP ₁=∠ PP ₁f, i.e. Φ=θ ₀; Because so FP=FO ∠ FPO=∠ is FOP, ∴ θ ₀=α ₀=Φ; So FP ₁=FP=FO=f, F point is Z axis, principal normal PP ₁and the Δ P of primary tangent PO composition ₁the mid point of PO; Because β ₀=α ₀+ Φ=2 Φ, ∴ Φ=0.5 β ₀=0.5h ₀.By summit P coordinate figure (fcos θ z ₀, f-fsin θ z ₀) substitution equation Y ²=2af (Z-B) can try to achieve: B=f (1-sin θ z ₀-cos ²θ z ₀/ 2a), during due to Y=0, Zo=B, o is para-curve summit, so B=oO, B is the distance of para-curve summit to Y-axis; ∵ tg Φ=Y _p/ af=fcos θ z ₀/ af, ∴ a=cos θ z ₀/ tg Φ.Because 2f ²cos ²θ z ₀/ 2af=fcos θ z ₀tg Φ=P ₂pP ₂o/P ₂p=P ₂o=f-fsin θ z ₀so, f ²cos ²θ z ₀/ 2af=B, i.e. B=Y ² _p/ 2af=Z _p-b, ∴ 2B=Z _p, B=oO=P ₂o; Because tg is Φ=P ₂o/PP ₂so, Z _p=P ₂o=fcos θ z ₀tg Φ, B=0.5fcos θ z ₀tg Φ.As θ z ₀=30 ° time, a=1.5, its B=0.25f; As θ z ₀=0 ° time, a=1, its B=0.5f, summit o to the distance of Y-axis equal summit o to fixed point F distance, now para-curve Y ²=2af (Z-B) is nomianl para-curve y ²=2pz; Nomianl para-curve y ²the mathematics implication of=2pz is: have a moving some M to equate to the distance of fixed point F and boning out L, the track of moving some M is para-curve; It is take para-curve summit o as true origin, and its y axle is parallel with Y-axis, and its z axle and Z axis are collinear, its mathematical expression y ²in=2pz p be fixed point F to the distance of boning out L, y value equals Y value, i.e. y=Y, z=Z-B, p=f, Y-axis is equivalent to boning out L; As θ z ₀=0 ° time, a=1, formula Y ²=2af (Z-B) becomes Y ²=2f (Z-B), in o-yz coordinate system, its formula is y ²=2fz, so it is nomianl para-curve; Well-known: aim at the sun as long as nomianl para-curve main shaft is z axle, all reflected rays all focus on focal point F.

At para-curve Y ²on=2af (Z-B), get arbitrfary point M, mistake M point is made solar rays AM and is parallel to Z axis, connects MF, makes normal MM ₁hand over Z axis in M ₁, normal MM ₁divide ∠ AMF equally; Cross M point and make MM ₁vertical line be tangent line MM ₃, MM ₃hand over Z axis in M ₃, MM ₃with the angle of Y-axis be λ.Cross M point and make Z axis vertical line MM ₂, M ₂point is intersection point.Use numerical code expression for ease of understanding us: ray AM and tangent line MM ₃angle be ∠ 1, incident angle ∠ AMM ₁=∠ 2, reflection angle ∠ M ₁mF=∠ 3, ∠ FMM ₃=∠ 4; In addition ∠ MM ₃f=∠ 5, ∠ MM ₁f=∠ 6.Because ray AM is parallel to Z axis, thus ∠ 1=∠ 5, ∠ 2=∠ 6, again because MM ₁⊥ MM ₃so, 2=90 ° of ∠ 1+ ∠, ∠ 3+ ∠ 4=90 ° because ∠ 5+ λ=90 °, ∴ ∠ 1+ λ=90 °, ∠ 2=∠ 6=λ.Because MM ₃equation is: (Z _m-Z _m3)/Y _m=Y _m/ af, i.e. Z _m-Z _m3=2 (Y _m ²/ 2af+B)-2B, so Z _m=2B-Z _m3=P ₂o+OM ₃, because Z _m=M ₂o=M ₂p ₂+ P ₂o, ∴ M ₂p ₂=OM ₃; Because tg is λ=Y _m/ af=MM ₂tg Φ/fcos θ z ₀=MM ₂pP ₂/ P ₁p ₂pP ₂=MM ₂/ P ₁p ₂, tg λ=MM ₂/ M ₁m ₂so, M ₁m ₂=af, MM ₂/ P ₁p ₂=MM ₂/ M ₁m ₂so, P ₁p ₂=M ₁m ₂=af, therefore M ₁p ₁=M ₂p ₂=OM ₃, because FP ₁=FO=f, so M ₁f=FM ₃, F point is right-angle triangle Δ M ₁mM ₃the mid point of hypotenuse, so M ₁f=FM ₃=FM, F is Z axis, normal MM ₁and tangent line MM ₃the Δ M of composition ₁mM ₃mid point.So ∠ 6=∠ 3=∠ 2, incident angle ∠ 2 equals reflection angle ∠ 3, and the reflected ray of arbitrfary point M focuses on fixed point F, so fixed point F is para-curve Y ²the focus of=2af (Z-B).

Due to the present embodiment γ s=0 °, in the time of the δ changes delta δ of declination angle, drive catoptron 3 to rotate 0.5 Δ δ around summit P in face YOZ coordinate surface main penetrating with tracking means 6, thereby the sun is followed the tracks of.Fig. 3 is optically focused figure after catoptron rotates 23.45 °.The earth is by transmission moment high noon in Winter Solstice moment high noon in the Summer Solstice, and declination angular difference value is Δ δ=46.9 °, equals solar rays AP and has been rotated counterclockwise 46.9 °, now zenith angle θ z _d=θ z ₀+ Δ δ; Catoptron principal normal PP ₁rotate 0.5 Δ δ, equal to rotate 23.45 ° around summit P, Fig. 3 is the optically focused figure of moment catoptron at high noon in Winter Solstice.Because O-XYZ coordinate system rotates with catoptron, so the each point on para-curve and line do not become with respect to its value of O-XYZ coordinate system and angle, but 0.5 Δ δ, β are rotated with respect to P-xyz coordinate system _d=β ₀-0.5 Δ δ=90 °-θ z ₀-0.5 Δ δ.The solar rays AP of incident summit P has rotated Δ δ, has made incidence angle θ reduce Δ δ, but principal normal PP ₁rotated 0.5 Δ δ, made incidence angle θ increase 0.5 Δ δ, incident angle is by θ ₀become θ _d=θ ₀-Δ δ+0.5 Δ δ=θ ₀-0.5 Δ δ; Due to principal normal PP ₁rotated 0.5 Δ δ, made reflection angle alpha reduce 0.5 Δ δ, reflection angle is by α ₀become α _d=α ₀-0.5 Δ δ, because θ ₀=α ₀so, θ _d=α _d, incidence angle θ equals reflection angle alpha, i.e. and the reflection ray of ordering through P after catoptron rotation will be along primary optical axis directive fixed point F.In like manner, the solar rays AM of incident arbitrfary point M has rotated Δ δ, has made incident angle ∠ 2 _dreduce Δ δ, but normal MM ₁rotate 0.5 Δ δ, made incident angle ∠ 2 _dincrease 0.5 Δ δ, i.e. incident angle ∠ 2 _d=∠ 2-Δ δ+0.5 Δ δ=∠ 2-0.5 Δ δ.Catoptron is in the time that summit P rotates, and the above each point of para-curve P point is toward rotating near fixed point F direction, and the following each point of P point rotates toward leaving fixed point F direction, makes reflected ray form hot spot on extinction mouth 1.In order to reduce icon and interpreting blueprints convenience, following provisions arbitrfary point M is the above arbitrfary point of para-curve P point, and arbitrfary point M makes the parallel lines MM of primary optical axis PF excessively _phand over Z axis in M _ppoint, ∠ M ₁mM _p=β _d-λ; Due to incident angle ∠ 2 _dequal reflection angle ∠ 3 _d, and ∠ 2 _d=∠ 2-0.5 Δ δ, ∠ 2=λ, so ∠ 3 _d=∠ 2 _d=λ-0.5 Δ δ; If M ₄for the reflected ray MM through arbitrfary point M ₄with the intersection point of primary optical axis PF, there is ∠ M _pm M ₄=∠ 3 _d-∠ M ₁mM _p=λ-0.5 Δ δ-(β _d-λ)=2 λ+θ z ₀-90 °, ∵ ∠ M _pmM ₄=∠ M M ₄p, ∴ ∠ MM ₄p=2 λ+θ z ₀-90 °, because ∠ is M _pm M ₂=90 °-β _d, ∠ M ₄m M ₂=∠ M _pm M ₂-∠ M _pm M ₄so, ∠ M ₄m M ₂=90 ° of-2 λ+0.5 Δ δ, λ=arc tg Y _m/ af; Because ∠ is M ₄m P=∠ M ₄m M ₂+ ∠ PM M ₂, and ∠ PM M ₂=arc tg (Z _m-Z _p)/(Y _m-Y _p).Because PM/sin ∠ M is M ₄p=P M ₄/ sin ∠ M ₄m P, so P M ₄=PMsin ∠ M ₄m P/sin ∠ M M ₄p, PM ²=(Y _m-Y _p) ²+ (Z _m-Z _p) ².Following provisions arbitrfary point K is the following arbitrfary point of para-curve P point, crosses K point and makes solar rays AK, makes the normal KK that K is ordered ₁hand over Z axis in K ₁, cross K point and make the reflected ray KK that K is ordered ₄, K ₄for reflection ray KK ₄with the intersection point of primary optical axis PF, follow the example of line KK ₁with the intersection point of primary optical axis PF be K ₅.Due to ∠ PK ₅k=β _d-λ _k, ∠ AKK ₅=90 °-∠ PK ₅k-θ z _d=90 °-(90 °-θ z ₀-0.5 Δ δ-λ _k)-(θ z ₀+ Δ δ)=λ _k-0.5 Δ δ, λ _k=arc tgY _k/ af; ∠ K ₄kK ₅=∠ AKK ₅=λ _k-0.5 Δ δ, ∠ PK ₄k=∠ PK ₅k-∠ K ₄kK ₅=β _d-2 λ _k+ 0.5 Δ δ; Because PK ₄/ sin ∠ PKK ₄=PK/sin ∠ PK ₄k, so PK ₄=PKsin ∠ PKK ₄/ sin ∠ PK ₄k, PK in formula ²=(Y _k-Y _p) ²+ (Z _k-Z _p) ², ∠ PKK ₄=180 °-∠ FPK-∠ PK ₄k, ∠ FPK=∠ FPP ₂+ ∠ KPP ₂, ∠ FPP ₂=θ z ₀+ 0.5 Δ δ, ∠ KPP ₂=arc ctg (Y _k-Y _p)/(Z _k-Z _p).Therefore, β < β ₀time master penetrate in face the reflected ray of arbitrfary point and the intersection point of primary optical axis PF on incident para-curve and all can obtain; Get the reflected ray MM of m, k point for ordering through M, K ₄, KK ₄with the intersection point of extinction mouth 1: Fm=(PM ₄-f) tg ∠ MM ₄p, Fk=(PK ₄-f) tg ∠ PK ₄k, thus, β < β ₀time arbitrfary point hot spot distance all can obtain, above calculating formula is β < β ₀the calculating formula of hot spot distance while being 90 ° of ξ <.

Fig. 4 be high noon in Winter Solstice moment master penetrate optically focused figure in face.Penetrate main that in face, to get C on bus be that catoptron upper extreme point, E are lower extreme point; Crossing respectively C, E point makes solar rays AC, AE, makes normal CC ₁and EE ₁hand over Z axis in C ₁and E ₁, cross C point and make Z axis vertical line CC ₂, C ₂for intersection point, make respectively reflected ray CC ₄and EE ₄hand over primary optical axis PF in C ₄and E ₄, get c, e is respectively CC ₄and EE ₄intersection point with extinction mouth I.Below get concrete data and calculate the hot spot distance of upper and lower end points: in the present embodiment, all get latitude

the position of having begun:

the δ when Summer Solstice _x=23.45 °, β ₀=90 °-θ z ₀=70 °, Φ=0.5 β ₀=35 °; All get focal distance f=80m, all by figure ratio: draw at 1000: 1, f=80, a=cos θ z ₀/ tg Φ=1.342, B=0.5fcos θ z ₀tg Φ=26.32, P (0,75.175,52.638).Moment high noon in Winter Solstice: θ z _d=θ z ₀+ Δ δ=66.9 °, β _d=β ₀-0.5 Δ δ=46.55 °; Getting C point is C (0,90,64.042), obtains: λ according to the inference of M _c=arc tgY _c/ af=39.973 °, ∠ CC ₄p=2 λ _c+ θ z ₀-90 °=9.945 °; ∠ PCC ₂=arc tg (Z _c-Z _p)/(Y _c-Y _p)=37.569 °, ∠ C ₄cC ₂=90 ° of-2 λ _c+ 0.5 Δ δ=33.505 °, ∠ C ₄c P=∠ C ₄cC ₂+ ∠ PCC ₂=71.074 °; PC=18.704, PC ₄=PCsin ∠ C ₄c P/sin ∠ CC ₄p=102.447.Fc＝(PC ₄-f)·tg∠CC ₄P＝3.936。Getting E point coordinate is E (0,60,43.09), obtains: ∠ EPP according to the inference of K ₂=arc tg (Z _e-Z _p)/(Y _e-Y _p)=32.178 °, ∠ FPP ₂=43.45 °, ∠ FPE=∠ FPP ₂+ ∠ EPP ₂=75.628 °, λ _e=arc tg Y _e/ af=29.199 °, ∠ PE ₄e=β _d-2 λ _e+ 0.5 Δ δ=11.602 °, ∠ PEE ₄=180 °-∠ FPE-∠ PE ₄e=92.77 °; PE ²=(Y _e-Y _p) ²+ (Z _e-Z _p) ², PE=17.93; PE ₄=PEsin ∠ PEE ₄/ sin ∠ PE ₄e=89.05; Fe=(P E ₄-f) tg ∠ P E ₄e=1.86.Be that the hot spot distance of upper and lower end points is for Fc=3.936m, Fe=1.86m, by PC ₄and the calculating formula of Fc is learnt PC ₄be directly proportional to PC, Fc and PC ₄be directly proportional, so Fc is directly proportional to PC, in like manner Fe is directly proportional to PE, hot spot distance and its incidence point to the limit the distance of P be directly proportional.If FP and Z axis meet at D, cross F point and make Z axis vertical line FF ₁, DP=PP ₂/ sin β _d=Y _p/ sin β _d; DF=DP-f=Y _p/ sin β-f, so Y _f=FF ₁=DF sin β=Y _p-fsin β, Z _f=F ₁p ₂+ P ₂o=fcos β _d+ Z _p, the coordinate figure of moment high noon in Winter Solstice fixed point F in O-XYZ coordinate system becomes F (0, Y _p-fsin β _d, fcos β _d+ Z _p).

Be more than high noon in Winter Solstice moment master penetrate the optically focused situation analysis in face, for analyzing optically focused situation in other planes of incidence, get G point for upper surface left end point, J point be lower surface left end point, said go up or lower surface is by catoptron or lower extreme point and perpendicular to the face of main shaft; G, C ₂, E ₂, the J left side that is catoptron, left and right end face is penetrated face and is intersected at Z axis with main; If left and right end face with the main angle of penetrating face is

catoptron cross section 4 is rectangles that above bottom left right side intercepts, and catoptron cross section is generally oval or circular certainly.Due to the same xsect of G, C point, the same xsect of J, E point, because catoptron is with para-curve Y ²=2af (Z-B) does circular arc rotation for bus around main shaft and forms minute surface, so a year line for upper surface catoptron is with intersection point C ₂for the circular arc line in the center of circle, the normal that G is ordered and the intersection point of Z axis are C ₁point; A year line for lower surface catoptron is with intersection point E ₂for the circular arc line in the center of circle, the normal that J is ordered and the intersection point of Z axis are E ₁point.While having begun position because solar rays be parallel incident and be parallel to main shaft, so now solar rays perpendicular to all xsects, solar rays in XOY coordinate surface for a bit, learnt by reflection law: on minute surface, the reflected ray of arbitrfary point must be coplanar in the plane of incidence with the normal of this point and incident ray, and incident angle equals reflection angle, because catoptron is to do circular arc rotation take specific curves as bus around Z axis to form minute surface, so all planes of incidence are penetrated face with master and are intersected at Z axis, because the normal of arbitrfary point is crossing with Z axis and the vertical XOY coordinate surface of solar rays, so all planes of incidence are perpendicular to XOY coordinate surface, so the angle of each point equates with the main angle of penetrating respective points in face in all planes of incidence, in same xsect, the incident angle of each point equates, reflection angle equates, and incident angle equals reflection angle, be that on minute surface, the incident angle of arbitrfary point equals reflection angle, so begun, the reflection ray of position incidence reflection mirror focuses on fixed point F certainly.Specifically: the line that carries of left side upper reflector is the Y being penetrated on face by main ²=2af (Z-B) para-curve is done circular arc rotation around Z axis

degree forms, and within this year, line is still Y ²=2af (Z-B) para-curve, the G point on it is to lead the C point of penetrating on face around intersection point C ₂rotation

degree forms, so the intersection point of the normal that G is ordered and Z axis is C ₁point; Owing to having begun the parallel incident of position solar rays and be parallel to main shaft, so incident angle ∠ AGC ₁=∠ ACC ₁, reflection angle ∠ C ₁gF=∠ C ₁cF, and ∠ AGC ₁=∠ C ₁gF; On minute surface any point all therewith in like manner, so begun position focusing mirror in fixed point F.

Fig. 5 is moment high noon in Winter Solstice left side optically focused figure.Fig. 6 is moment high noon in Winter Solstice left side optically focused vertical view.Due to moment at high noon position angle r=0 °, Y-axis energized north, so solar rays is parallel to YOZ coordinate surface; Get A _gfor the intersection point (not shown) of solar rays AG and XOZ coordinate surface, moment at high noon ε=h, ξ=ε+90 °-β _d=h+90 °-β _d, get G point coordinate value and be: G (X _g, Y _g, Z _g), A _gpoint coordinate is: A _g(X _g, 0, Z _g+ Y _gtg ξ); Because C point coordinate is: C (0, Y _c, Z _c), C ₁point is: C ₁(0,0, af+Z _c), and G and the same end face of C point, so Z _g=Z _c; So ray vector $\stackrel{\&RightArrow;}{A_{G}G}=Y_{G}j-Y_G\mathrm{tg\&xi;k},$ Normal line vector $\stackrel{\&RightArrow;}{G_{1}G}=X_{G}i+Y_{G}j-\mathrm{afk}.$ If extinction mouth 1 is G with the intersection point of solar rays AG ₁, with normal C ₁the intersection point of G is G ₂; By vector

and learn GG ₁equation be: $\frac{Y-Y_G}{Y_G}=\frac{Z-Z_G}{-Y_G\mathrm{tg\&xi;}},$ GG ₂equation be: $\frac{X-X_G}{X_G}=\frac{Y-Y_G}{Y_G}=\frac{Z-Z_G}{-\mathrm{af}}.$ Extinction mouth 1 is crossed fixed point F and perpendicular to primary optical axis PF, and the normal line vector that PF is extinction mouth 1 is: $\stackrel{\&RightArrow;}{\mathrm{PF}}=(Y_F-Y_P)j+(Z_F-Z_P)k,$ So the equation of extinction mouth 1 is: (Y _f-Y _p) (Y-Y _f)+(Z _f-Z _p) (Z-Z _f)=0.Manage line C ₁the plane of incidence of G and incident ray AG composition is II, gets the intersection point of g point for reflected ray Gg and extinction mouth 1 on face II.For analyzing the size of hot spot, get with Fig. 4 identical data and calculate: C (0,90,64.042), E (0,60,43.09), C ₁(0,0,171.4), E ₁(0,0,150.45); Getting left and right end face with the main angle of penetrating face is

g (23.29,86.93,64.04), J (15.53,57.96,43.09), A _g(23.29,0,264.45), ξ=66.55 °, because the coordinate figure of fixed point F is: F (0,17.1,107.66), so the equation of extinction mouth 1 is: 58.08 (Y-17.1)-55.02 (Z-107.66)=0.By this equation and GG ₁equation composition linear equation in two unknowns group can solve G ₁the coordinate figure of point is G ₁(23.29,52.02,144.52), with GG ₂equation composition linear equation in two unknowns group can solve G ₂the coordinate figure of point is G ₂(9.60,35.84,127.14).Due to Δ GG in plane of incidence II ₁g ₂each length of side is respectively: GG ₁=87.73, GG ₂=82.34, G ₁g ₂=27.41, due to $\cos \&angle;\mathrm{AG}{C}_1=\frac{\stackrel{\&RightArrow;}{A_{G}G.}}{\stackrel{\&RightArrow;}{\left|A_{G}G\right|}}\frac{\stackrel{\&RightArrow;}{C_{1}G}}{\stackrel{\&RightArrow;}{\left|C_{1}G\right|}}=0.95,$ So ∠ AG C ₁=18.195 °; Because ∠ AG is C ₁=∠ C ₁gg, so ∠ AG g=2 ∠ AG C ₁=36.39 °, learnt by sine: sin ∠ G G ₁g=GG ₂sin ∠ AGC ₁/ G ₁g ₂=0.938, so ∠ GG ₁g=69.718 °; Learnt by interior and theorem: ∠ GgG ₁=180 °-∠ AG g-∠ GG ₁g=73.892 °, so G ₁g=GG ₁sin ∠ AG g/sin ∠ GgG ₁=54.17, so G ₂g=G ₁g-G ₁g ₂=26.76.If Ч _g=G ₂g/G ₁g ₂, Ч _g=0.97643, X _g-X _g2=Ч _g(X _g2-X _g1), Y _g-Y _g2=Ч _g(Y _g2-Y _g1), Z _g-Z _g2=Ч _g(Z _g2-Z _g1); So X _g=-3.77, Y _g=20.04, Z _g=110.17.In like manner: get j point for the reflected ray Jj of ordering through J and the intersection point of extinction mouth 1, can extrapolate: X _j=-4.07, Y _j=16.05, Z _j=106.56; So the distance of Fg, Fj is respectively: Fg=5.40, Fj=4.34; Because so the maximum radius of hot spot on left and right end surface symmetric moment high noon in Winter Solstice extinction mouth 1 is Fg=5.40m; Can obtain the hot spot distance of any point according to above reckoning.Catoptron is with C (0,90,64.042), E (0,59.324,42.709) is upper and lower end points, with U (15.337,74.662,53.376), V (15.337,74.662,53.376), while being the elliptic cross-section of left and right end points, hot spot distance in its upper and lower, left and right is: Fc=3.936, Fe=1.889, Fu=Fv=3.104 moment high noon in Winter Solstice.As can be seen here incidence point from summit P more away from its hot spot distance larger, so elliptic reflector is less than the hot spot of rectangular mirror.

The present embodiment γ s=0 °, so ψ=0 ° cos ε=cos hcos γ, the cosine of firing angle ε is directly proportional to the cosine of elevation angle h and position angle γ, and

be that the size of firing angle ε is by latitude

and declination angle δ and hour angle ω determine; Because catoptron is fixed with respect to the position of the earth, so local latitude

be a constant, therefore follow the tracks of the solar time and only need follow the tracks of hour angle ω and declination angle δ.Tracking means 9 drives the master of catoptron 3 to penetrate face around primary optical axis PF rotation η degree, corner η=arc cos sin h/ (cos ²h sin ²γ+sin ²h) ¹/ ₂, make master penetrate face and be parallel to sunray; Tracking means 6 drives catoptron 3 to rotate main penetrating in face around summit P around primary optical axis PF rotation, while tracking means 6 together with catoptron 3; In the time that firing angle ε changes delta ε spends, catoptron principal normal PP ₁rotate 0.5 Δ ε around summit P, thereby firing angle ε is followed the tracks of; And in the time that declination angle δ changes delta δ spends, because of γ s=0 °, catoptron principal normal PP ₁in face, rotate 0.5 Δ δ around summit P main penetrating, thereby declination angle δ is followed the tracks of.Comprised the tracking to declination angle δ when firing angle ε is followed the tracks of, but every day efficient light according to approximately 10 hours, i.e. approximately 10 hours every days of tracking to firing angle ε, so also will carry out the tracking of other 14 hours to declination angle δ; Because declination angle δ changes 93.8 in 1 year ⁰, 0.257 °/day of the Δ δ ≈ of every day, thus every day to firing angle ε follow the tracks of start front to declination angle δ carry out a secondary tracking, regularly declination angle difference DELTA δ is followed the tracks of just enough accurate.Being tracked as the sun while being γ s=0 °: catoptron 3, around primary optical axis PF rotation η degree, makes master penetrate face and is parallel to sunray, simultaneously its principal normal PP ₁in face, rotate 0.5 Δ ε degree around summit P main penetrating, and regularly rotate 0.5 Δ δ around summit P.Moment at high noon ω=0 °, γ=0 °, learnt by cos ε=cos hcos γ: ε=h, firing angle ε equals elevation angle h; Because the specific curves of the present embodiment is the zenith angle θ z with moment high noon in the Summer Solstice ₀design, and firing angle ε now ₀equal elevation angle h ₀, the spindle alignment sun, so the reflection ray of the position incidence reflection mirror that begun certainly focus on fixed point F; But not the optically focused situation in the position moment at high noon of having begun is discussed in declination angle δ is followed the tracks of.The sun haunt local horizon moment, elevation angle h=0 ⁰, corresponding firing angle ε equals position angle γ, and position angle now

formula is learnt thus: at Same Latitude, declination in Summer Solstice angle δ in the whole year _x=23 ⁰position angle maximum when 27 ' sun haunts local horizon, sunrise sunset in Summer Solstice moment firing angle ε maximum, declination in Winter Solstice angle δ _d=-23 ⁰position angle minimum when 27 ' sun haunts local horizon, i.e. sunrise sunset in Winter Solstice moment firing angle ε minimum, and also latitude is larger, position angle difference is larger the Summer Solstice in winter when sunrise sunset; Because firing angle now equals position angle, so the firing angle when Summer Solstice, the sun haunted local horizon in the whole year is maximum elevation ε _m.

Fig. 7 is the sun local horizon opticofacial winking reflex mirror optically focused figure that haunts in the Summer Solstice.The Summer Solstice, the sun haunted local horizon moment, the solar rays AP of incident summit P turned clockwise Δ ε, make incidence angle θ increase Δ ε, but principal normal PP ₁turn clockwise 0.5 Δ ε, make incidence angle θ reduce 0.5 Δ ε, incident angle is by θ ₀become θ _m=θ ₀+ Δ ε-0.5 Δ ε=θ ₀+ 0.5 Δ ε; Due to principal normal PP ₁turned clockwise 0.5 Δ ε, make reflection angle alpha _mincrease 0.5 Δ ε, reflection angle is by α ₀become α _m=α ₀+ 0.5 Δ ε, because θ ₀=α ₀so, θ _m=α _m, principal normal PP ₁the reflected ray of ordering through P after rotation will be along primary optical axis PF directive fixed point F.In like manner, solar rays AM, the AK of incident arbitrfary point M and K turned clockwise Δ ε, make incident angle increase Δ ε, but normal MM ₁, KK ₁turn clockwise 0.5 Δ ε, make its incident angle reduce 0.5 Δ ε, so its incident angle has increased 0.5 Δ ε; Main shaft angle β _mincrease 0.5 Δ ε: β _m=β ₀+ 0.5 Δ ε=90 °-θ z ₀+ 0.5 Δ ε.Because catoptron is in the time that summit P rotates, the above each point of para-curve P point rotates 0.5 Δ ε toward leaving fixed point F direction, and the following each point of P point, toward rotating 0.5 Δ ε near fixed point F direction, makes the reflected ray of catoptron form hot spot on extinction mouth.Cross arbitrfary point M and make Z axis vertical line MM ₂, M ₂for intersection point, M point is made the parallel lines MM of primary optical axis PF excessively _phand over Z axis in M _ppoint, ∠ M ₁mM _p=β _m-λ; Because incident angle ∠ 2 _mequal reflection angle ∠ 3 _m, and ∠ 2 _m=∠ 2+0.5 Δ ε, ∠ 2=λ, so ∠ 3 _m=∠ 2 _m=λ+0.5 Δ ε; If M ₄for reflected ray MM ₄with the intersection point of primary optical axis PF, ∠ M _pm M ₄=∠ 3 _m-∠ M ₁mM _p=λ+0.5 Δ ε-β _m+ λ=2 λ+θ z ₀-90 °, ∠ M _pmM ₂=β _m-90 °, ∠ M ₄mM ₂=∠ M _pm M ₂+ ∠ M _pm M ₄=β _m-90 °+2 λ+θ z ₀-90 °=2 Δ ε-90 °, λ+0.5, ∠ MM ₄p=∠ M _pm M ₄=2 λ+θ z ₀-90 °; λ=arc tg Y _m/ af; ∠ M ₄m P=∠ PMM ₂-∠ M ₄mM ₂, ∠ PMM ₂=arc tg (Z _m-Z _p)/(Y _m-Y _p), PM ²=(Y _m-Y _p) ²+ (Z _m-Z _p) ²; Because PM/sin ∠ is MM ₄p=PM ₄/ sin ∠ M ₄m P, so PM ₄=PMsin ∠ M ₄m P/sin ∠ M M ₄p.Cross arbitrfary point K and make solar rays AK, do normal KK ₁hand over Z axis in K ₁, make reflected ray KK ₄hand over primary optical axis PF in K ₄, normal KK ₁hand over PF in K ₅.Due to ∠ PK ₅k=β _m-λ _k, incident angle ∠ AKK ₁=ε _m-∠ PK ₅k=ε _m-(β _m-λ _k), wherein λ _k=arc tg Y _k/ af, firing angle

0.5 Δ ε=0.5 (ε _m-90 °+θ z ₀), so incident angle ∠ A K K ₁=λ _k+ 0.5 Δ ε, incident angle increases 0.5 Δ ε than the position of having begun; ∠ K ₄kK ₁=∠ AKK ₁=λ _k+ 0.5 Δ ε, ∠ PK ₄k=∠ PK ₅k-∠ K ₄kK ₁=β _m-λ _k-(λ _k+ 0.5 Δ ε)=β _m-2 λ _k-0.5 Δ ε; ∠ FPP ₂=0.5 Δ ε-θ z ₀, ∠ KPP ₂=arc tg (Z _k-Z _p)/(Y _k-Y _p), ∠ FPK=∠ KPP ₂-∠ FPP ₂, ∠ PKK ₄=180 °-∠ FPK-∠ PK ₄k.PK ²=(Y _k-Y _p) ²+ (Z _k-Z _p) ², by PK ₄/ sin ∠ PKK ₄=PK/sin ∠ PK ₄k, obtains PK ₄=PKsin ∠ PKK ₄/ sin ∠ PK ₄k.Fm=(f-P M ₄) tg ∠ MM ₄p, Fk=(f-P K ₄) tg ∠ P K ₄k, m, k point is respectively reflected ray MM ₄, KK ₄intersection point with extinction mouth 1.

Below with the size of concrete data analysis hot spot: maximum elevation

0.5 Δ ε=0.5 (ε _m-90 °+θ z ₀)=26.62 °, main shaft angle β _m=β ₀+ 0.5 Δ ε=96.62 °.In order to reduce, diagram and interpreting blueprints are convenient, to establish M (0,90,64.04) point be catoptron upper extreme point; : λ=39.973 °, ∠ M M ₄p=9.945 °, ∠ PMM ₂=37.564 °, ∠ M ₄m M ₂=16.566 °, ∠ M ₄m P=20.998 °, PM=18.703, PM ₄=38.806; Fm=7.223.If K (0,60,43.09) point is catoptron lower extreme point: λ _k=29.199 °, ∠ KPP ₂=32.178 °, ∠ FPP ₂=6.62 °, ∠ FPK=25.558 °, ∠ PK ₄k=11.602 °, ∠ PKK ₄=142.84 °; PK=17.93, PK ₄=53.85; Fk=5.37.Be maximum elevation ε _mtime upper and lower end points hot spot distance be Fm=7.223m, Fk=5.37m, this value is catoptron hot spot ultimate value; Hour angle now

the corresponding time is 4: 23 morning and 19: 37 afternoon; Due to maximum elevation ε _mtime be the maximal value that atmospheric envelope distance is passed in solar radiation, and solar radiation is longer through atmospheric distance, the energy of absorption, reflection and the scattering of atmosphere to solar radiation is more, is that the emittance of catoptron is fewer so arrive ground; Solar radiation is relevant with elevation angle h through atmospheric distance, and h=90 ° time, distance is the longest the shortest, the h=0 of distance ° time, and sinh is corresponding with cos ω, so every day, efficient light was according to approximately 10 hours, be the morning 7:00 to 17:00 in afternoon, corresponding hour angle is-75 ° to 75 °, the corresponding Summer Solstice

time effective firing angle ε=± 97.22 °.Winter Solstice the sun local horizon moment firing angle that haunts

corresponding hour angle ω=± 65.74 °, correspondence

hot spot distance when f=80m, M (0,90,64.04), K (0,60,43.09) is: Fm=1.38, Fk=0.82.

Catoptron optically focused figure when Fig. 8 is effective firing angle.Get effective firing angle ε=± 97.22 °, ω=± 75 °, 0.5 Δ ε=13.61 °, β=83.61 °; Get upper extreme point C (0,90,64.042), lower extreme point E (0,60,43.09), according to calculating of Fig. 7 M, K: λ _c=39.973 °, ∠ CC ₄p=9.945 °, ∠ PCC ₂=37.564 °, ∠ C ₄cC ₂=3.556 °, ∠ C ₄cP=34.008 °, PC=18.70, PC ₄=60.57; Fc=(f-PC ₄) tg ∠ CC ₄p=3.4.Lower extreme point λ _e=29.199 °, ∠ EPP ₂=32.18 °, ∠ FPP ₂=θ z ₀-0.5 Δ ε=6.39 °, ∠ FPE=∠ FPP ₂+ ∠ EPP ₂=38.568 °, ∠ PE ₄e=11.602 °, ∠ PEE ₄=129.83 °; PE=17.93, PE ₄=68.47; Fe=(f-P E ₄) tg ∠ PE ₄e=2.37.While being effective firing angle, the hot spot of upper and lower end points distance is Fc=3.4, Fe=2.37.With Fig. 5,6 in like manner, get G (23.29,86.93,64.04) point for upper surface left end point, J (15.53,57.96,43.09) point be lower surface left end point, G, C ₂, E ₂, the J left side that is catoptron, establish catoptron cross section is rectangle.Described in same Fig. 5,6: normal and the Z axis intersection point of upper left end point G and lower left end point J are respectively C ₁, E ₁point, gets A _gfor the intersection point of solar rays AG and XOZ coordinate surface, establishing extinction mouth 1 is G with the intersection point of solar rays AG ₁, with normal C ₁the intersection point of G is G ₂(A _g, G ₁, G ₂all not shown in figure); Cut firing angle ξ=ε+90 °-β=103.61 °, mainly penetrate face because solar rays is parallel to, so GG ₁, GG ₂same Fig. 5 of equation, 6 described in.Get the intersection point of g point for reflected ray Gg and extinction mouth I, due to Y _f=Y _p-fsin β=-4.33, Z _f=fcos β+Z _p=61.54, so fixed point F coordinate is: F (0 ,-4.33,61.54), the normal line vector of extinction mouth 1 is: $\stackrel{\&RightArrow;}{\mathrm{PF}}=-79.5j+8.9k,$ Its equation is: 79.5 (Y+4.33)-8.9 (Z-61.54)=0.By this equation and GG ₁equation composition solution of equations obtain: G ₁(23.29 ,-82.3 ,-634.94), with GG ₂equation composition solution of equations obtain: G ₂(1.88,7.01,162.75).Δ GG ₁g ₂each length of side is: GG ₁=719.17, GG ₂=129.48, G ₁g ₂=802.96, by $\cos \&angle;G_{1}G{G}_2=\frac{G{G_1}^2+G{G_2}^2-G_1{G_2}^2}{2G{G}_1\&CenterDot;G{G}_2}=-0.595,$ Obtain ∠ G ₁gG ₂=126.499 °, in like manner try to achieve ∠ GG ₂g ₁=46.053 °, ∠ G ₂gg=∠ AGG ₂=180 °-∠ G ₁gG ₂=53.501 °; ∠ GgG ₂=180 °-∠ G ₂gg-∠ GG ₂g ₁=80.446 °, G ₂g=GG ₂sin ∠ G ₂gg/sin ∠ GgG ₂=105.55, so G ₁g=G ₁g ₂-G ₂g=697.41.If Ч _g=G ₁g/g G ₂, Ч _g=6.60739, X _g-X _g1=Ч _g(X _g2-X _g), Y _g-Y _g1=Ч _g(Y _g2-Y _g), Z _g-Z _g1=Ч _g(Z _g2-Z _g); So X _g=4.69, Y _g=-4.73, Z _g=57.89.In like manner: get the intersection point of j point for reflected ray Jj and extinction mouth 1, can extrapolate: X _j=3.12, Y _j=-4.04, Z _j=64.14; So the distance of Fg, Fj is respectively: Fg=5.96, Fj=4.07; So on extinction mouth 1, the maximum radius of hot spot is Fg=5.96m during due to the effective firing angle of left and right end surface symmetric.Can obtain the hot spot distance of any point according to above reckoning; Catoptron is with C (0,90,64.042), E (0,59.324,42.709) is upper and lower end points, with U (15.337,74.662,53.376), V (15.337,74.662,53.376), while being the elliptic cross-section of left and right end points, effectively its upper and lower, left and right hot spot distance is when firing angle ε=± 97.22 °: Fc=3.4, Fe=2.443, Fu=Fv=3.0.As can be seen here incidence point from summit P more away from its hot spot distance larger, so elliptic reflector is less than the hot spot of rectangular mirror.

In sum: directive fixed point F of the reflected ray of summit P when any operating mode, the reflected ray of catoptron any point is all concentrated on fixed point F.Catoptron is followed the tracks of sun process three kinds of operating modes: one is begun position, ε now ₀=h ₀=β ₀, ξ ₀=90 °, solar rays is parallel to main shaft, and reflected ray focuses on fixed point F, and hot spot distance is zero; One is ε < ε ₀, β < β ₀, 90 ° of ξ <, hot spot distance is now obtained by the calculating formula of Fig. 3 to 6; One is ε > ε ₀, β > β ₀, 90 ° of ξ >, hot spot distance is now obtained by the calculating formula of Fig. 7 to 8; Therefore the hot spot of operating mode catoptron any point distance all can be obtained arbitrarily.And hot spot distance and incidence point to the limit distance and the firing angle difference of P are directly proportional, control incidence point to the limit the distance of P can control the size of hot spot, control end is apart from than the size that can control hot spot.As the present embodiment is low temperature solar furnace, to establish catoptron cross section be oval, get end apart from than=3; The upper extreme point of catoptron is C (0,128,102.622), lower extreme point is E (0,9,26.696), left end point is U (59.5,68.5,64.659), right endpoint is V (59.5,68.5,64.659), now the radius-of-curvature of upper and lower end points is ρ _c=af (1+tg ²λ _c) ^3/2=404.533, ρ _e=af (1+tg ²λ _e) ^3/2=108.495; Get the intersection point that c, e, u, v point are respectively reflected ray Cc, Ee, Uu, Vv and extinction mouth 1, its hot spot distance is respectively: Fc=24.09, Fe=23.376, Fu=Fv=19.103; The area of catoptron oval cross section is: A=π ab=13193.15m ², wherein major axis a=70.58, minor axis b=59.5; If corresponding extinction mouth 1 be fix a point F be the center of circle, take maximum hot spot apart from Fc=24.09m the circle (can certainly be other shapes that can absorb whole reflection rays) as radius, its area A=π Fc ²=1823.15m ², its optically focused ratio is: 13193.15m ²/ 1823.15m ²=7.236.Catoptron described in Fig. 5, Fig. 8 is with C (0,90,64.042), E (0,59.324,42.709) is upper and lower end points, with U (15.337,74.662,53.376), V (15.337,74.662,53.376), while being the elliptic cross-section of left and right end points, its maximum hot spot distance is: Fc=3.936; Its end distance is than being respectively: PC/Fc=4.752, PE/Fe=9.901, PU/Fu=PV/Fv=4.962, the area of its oval cross section is: A=π ab=900.15m ², corresponding extinction mouth 1 be fix a point F be the center of circle, take maximum hot spot apart from Fc=3.936m the circle as radius, its area A=π Fc ²=48.67m ², its optically focused ratio is: 18.5.While changing take summit P the elliptic reflector described in Fig. 5, Fig. 8 into as the center of circle, take PU as radius circular, its left and right end points coordinate is: U (15.337,74.662,53.376), V (15.337,74.662,53.376), upper and lower side point coordinate is: C (0,87.422,61.912), E (0,62.235,44.357), its maximum hot spot distance is: Fc=Fu=Fv=3.104; Its smallest end distance is than being: PU/Fu=PV/Fv=4.962, the area of its circular section is: Ac=π PU ²=741.482m ², corresponding extinction mouth 1 be fix a point F be the center of circle, take maximum hot spot apart from Fu=3.104m the circle as radius, its area A G=π Fu ²=30.269m ², its optically focused ratio is: 24.5.Because optically focused is than being that maximum daylighting area is the area in catoptron cross section and the ratio of maximum facula area, be circle and discuss and calculate with the extinction open area that is greater than maximum facula area with both, square being directly proportional of extinction open area and maximum hot spot distance, and maximum hot spot distance just smallest end apart from than denominator, catoptron area of section and smallest end apart from than square being directly proportional of molecule, so optically focused than equal smallest end apart from than square, i.e. Ac/AG=π PU ²/ π Fu ²=(PU/Fu) ²; And both when other shapes optically focused than to smallest end apart from than square be directly proportional.

The bus of catoptron can also be that other radius-of-curvature approach ρ ₀=af (1+tg ²Φ) ^3/2optically focused camber line, the present embodiment substitutes Y as the circular arc line with radius-of-curvature ρ=202.706m ²=2af (Z-B) para-curve, its summit P coordinate figure constant, get primary tangent inclination angle Φ _p=29 °, while getting its catoptron cross section and be major axis a=70.58, minor axis b=59.5 oval, its area is: A=13193.15m ², its maximum hot spot is apart from Fc=22.159m; If corresponding extinction mouth 1 be fix a point F be the center of circle, take maximum hot spot apart from Fc the circle as radius, its area A=1542.59m ², its optically focused ratio is: 13193.15/1542.59=8.55.Because the radius-of-curvature of the parabolic reflector summit P of the present embodiment is ρ 0=af (1+tg ²Φ) ^3/2=195.324m, the circular arc line of ρ=202.706m is to approach ρ thus ₀=af (1+tg ²Φ) ^3/2optically focused camber line; When identical by above-mentioned visible, catoptron area of section, the optically focused of circular arc line catoptron is than the optically focused ratio a little more than parabolic reflector, so can substitute parabolic reflector completely, other optically focused camber lines in like manner; Due to optically focused camber line is of a great variety and the calculating formula of its hot spot distance with above-mentioned, except parabolic reflector, the equal out-focus of other optically focused camber lines of position that begun, so do not enumerate at this.

Second embodiment is the embodiment of fixed-point condensing reflector composition solar cooker.Fig. 9 is split type solar cooker schematic diagram.Split type solar cooker is made up of extinction mouth 1, extinction body 2, catoptron 3, tracking means etc.; Extinction body 2 separates certain distance and extinction body 2 higher than catoptron 3 with catoptron 3, be that angle ψ > 0, the primary optical axis PF of primary optical axis PF and the local horizon NS projection line on ground level points to due south, be γ s=0 °, other structures are installed and are waited explanation identical with Fig. 1.Figure 10 is the catoptron elevation angle, the position focused view that begun.Point to due south at the main angle ψ > 0, the primary optical axis PF that make primary optical axis PF and local horizon NS in face of penetrating, be γ s=0 °, angle of pitch ψ > 0, the mapping at other points, line, angle is identical with Fig. 2 with proof: crossing F point is the main shaft sensing sun, i.e. β as Z axis ₀=h ₀=90 °-θ z ₀, get FO=f, set up O-XYZ coordinate system take O as true origin.If the specific curves of catoptron is Y ²=2af (Z-B) para-curve, in YOZ coordinate surface: ∠ FPP ₂=ψ+θ z ₀, ∠ P ₂pO=Φ, Y _p=PP ₂=fcos (ψ+θ z ₀), 2B=Z _p, B=oO=P ₂o, Z _p=P ₂o=fcos (ψ+θ z ₀) tg Φ, so B=0.5fcos (ψ+θ z ₀) tg Φ; ∵ θ ₀=α ₀=Φ, ∴ FP ₁=FP=FO=f, F point is Δ P ₁the mid point of PO; ∵ β ₀=ψ+α ₀+ Φ=h ₀, ∴ Φ=0.5 (h ₀-ψ).∵ tg Φ=Y _p/ af, ∴ a=cos (ψ+θ z ₀)/tg Φ; By Y _p, Z _pvalue substitution equation Y ²=2af (Z-B) can try to achieve: B=f[1-sin (ψ+θ z ₀)-cos ²(ψ+θ z ₀)/2a].At para-curve Y ²on=2af (Z-B), get arbitrfary point M: ∠ 2=∠ 6=λ, M ₁f=FM ₃=FM, F is Δ M ₁mM ₃mid point, ∠ 6=∠ 3=∠ 2, incident angle ∠ 2 equals reflection angle ∠ 3, the reflection ray of arbitrfary point M focus on fixed point F, the F that fixes a point is para-curve Y ²the focus of=2af (Z-B); When ψ < 0 demonstration with above-mentioned in like manner, therefore do not repeat.While having begun position because solar rays be parallel incident and be parallel to main shaft, so now solar rays perpendicular to all xsects, solar rays in XOY coordinate surface for a bit; Learnt by reflection law: on minute surface, the reflected ray of arbitrfary point must be coplanar in the plane of incidence with the normal of this point and incident ray; Because catoptron is with specific curves Y ²to be bus do circumference rotation around Z axis to=2af (Z-B) para-curve forms minute surface, so all planes of incidence penetrate that face intersects at Z axis and perpendicular to XOY coordinate surface with main, so the angle of each point equates with the main angle of penetrating respective points in face in all planes of incidence, in same xsect, the incident angle of each point equates, reflection angle equates and incident angle equals reflection angle, be that on minute surface, the incident angle of arbitrfary point equals reflection angle, so the reflection ray of the position incidence reflection mirror that begun focuses on fixed point F certainly.

Figure 11 is moment high noon in Winter Solstice elevation angle catoptron optically focused figure.At the main angle ψ > 0 that makes PF and NS in face that penetrates, the mapping of other points, line and Fig. 3,5 identical; When the earth by forwarding the positive period of the day from 11 a.m. to 1 p.m in Winter Solstice high noon in the Summer Solstice to, tracking means 6 drives catoptron 3 to forward noon position in Winter Solstice, catoptron principal normal PP by the position of having begun to around summit P ₁and O-XYZ coordinate system has rotated 0.5 Δ δ, now θ z main penetrating in face _d=θ z ₀+ Δ δ, β _d=90 °-θ z ₀-0.5 Δ δ.Summit P: incidence angle θ _d=θ ₀-0.5 Δ δ, reflection angle alpha _d=α ₀-0.5 Δ δ, because θ ₀=α ₀so, θ _d=α _d, the reflected ray of summit P will be along primary optical axis directive fixed point F.The arbitrfary point M that P point is above: ∠ 2 _d=∠ 3 _d=λ-0.5 Δ δ, ∠ M ₁mM _p=β _d-λ, λ=arc tg Y _m/ af; ∠ M _pmM ₄=∠ M ₁mM _p-∠ 3 _d=β _dΔ δ=90 ,-λ-λ+0.5 °-2 λ-θ z ₀, because ∠ is M _pmM ₄+ ∠ M M ₄p=ψ, so ∠ MM ₄p=ψ+2 λ+θ z ₀-90 °; Because ∠ is M _pm M ₂=90 °-β _d, ∠ M ₄mM ₂=∠ M _pmM ₂+ ∠ M _pmM ₄=90 ° of-2 λ+0.5 Δ δ, and ∠ PMM ₂=arc tg (Z _m-Z _p)/(Y _m-Y _p), so ∠ M ₄m P=∠ M ₄m M ₂+ ∠ PM M ₂.PM ²＝(Y _M-Y _P) ²+(Z _M-Z _P) ²，PM ₄＝PM·sin∠M ₄M?P/sin∠MM ₄P，Fm＝(PM ₄-f)·tg∠M?M ₄P。When ψ < 0: ∠ M ₄mP=∠ PMM ₂-∠ M ₄mM ₂, ∠ M ₄mM ₂=∠ M _pmM ₄-∠ M _pmM ₂=2 Δ δ-90 °, λ-0.5, all the other are identical during with ψ > 0.The following arbitrfary point K of P point: ∠ AKK ₁=∠ K ₄kK ₁=λ _k-0.5 Δ δ, λ _k=arc tgY _k/ af, ∠ PK ₅k=β _d-λ _k, ∠ PK ₄k=∠ PK ₅k+ ∠ K ₄kK ₁-ψ=β _d-0.5 Δ δ-ψ, ∠ FPP ₂=ψ+θ z ₀+ 0.5 Δ δ, ∠ KPP ₂=arc tg (Z _k-Z _p)/(Y _k-Y _p), ∠ FPK=∠ FPP ₂+ ∠ KPP ₂, ∠ PKK ₄=180 °-∠ FPK-∠ P K ₄k.PK ²=(Y _k-Y _p) ²+ (Z _k-Z _p) ², PK ₄=PKsin ∠ PKK ₄/ sin ∠ PK ₄k, Fk=(PK ₄-f) tg ∠ PK ₄k; When ψ < 0: ∠ PK ₄k=180 °-(2 λ _k-0.5 Δ δ)-(180 °-β _d-| ψ |)=β _d+ 0.5 Δ δ-ψ-2 λ _k, all the other are identical during with ψ > 0.Therefore, moment at high noon master penetrates in face the hot spot of arbitrfary point distance on para-curve and all can obtain γ s=0 ° time.The coordinate figure of fixed point F: γ s=0 ° time, F (0, Y _p-fsin (β-ψ), fcos (β-ψ)+Z _p).When the hot spot distance that when γ s=0 °, ψ ≠ 0 °, non-master penetrates any point in face is solved with γ s=0 °, ψ=0 ° in like manner, its ξ=ε+ψ+90 °-β just, refer to solving of Fg in Fig. 5, do not repeat at this.

When angle of pitch ψ ≠ 0 °, γ s=0 °, firing angle ε calculating formula is: (cos ψ cosh cos γ+sin ψ sin h) for ε=arc cos, follow the tracks of the solar time, catoptron 3 is around primary optical axis PF rotation η degree, make its master penetrate any operating mode of face and be all parallel to solar rays AP, η=arc cos{ (cos ψ sin h-sin ψ cosh cos γ)/[(cos ψ sin h-sin ψ cosh cos γ) ²+ cos ²h sin ²γ] ¹/ ₂, its principal normal PP simultaneously ₁in face, rotate 0.5 Δ ε around summit P main penetrating; O-XYZ coordinate system rotates with catoptron, so the each point on para-curve and line do not become with respect to its value of O-XYZ coordinate system and angle.When Figure 12 is effective firing angle, elevation angle catoptron master penetrates face optically focused figure.Its mapping is identical with Fig. 7; Ray AP has rotated Δ ε, has made the incident angle of summit P by θ ₀become θ=θ ₀+ 0.5 Δ ε; Principal normal PP ₁rotate 0.5 Δ ε, make reflection angle by α ₀become α=α ₀+ 0.5 Δ ε, because θ ₀=α ₀so, θ=α, i.e. principal normal PP ₁the reflection ray of ordering through P after rotation will be along primary optical axis PF directive fixed point F.In like manner, the incident angle of incident arbitrfary point M and K increased 0.5 Δ ε, so ∠ M ₁mM ₄=∠ AMM ₁=λ+0.5 Δ ε, M _pmM ₄=∠ M ₁mM ₄-∠ M ₁mM _p=ψ+2 λ+θ z ₀-90 °, wherein λ=arc tgY _m/ af, ∠ MM ₄p=∠ M _pmM ₄=ψ+2 λ+θ z ₀-90 °; Because ∠ is M _pmM ₂=ψ+θ z ₀-0.5 Δ ε, ∠ M ₄mM ₂=∠ M _pmM ₂-∠ M _pmM ₄=90 ° of-2 λ-0.5 Δ ε, ∠ M ₄mP=∠ PMM ₂+ ∠ M ₄m M ₂, ∠ PM M ₂=arc tg (Z _m-Z _p)/(Y _m-Y _p), PM ²=(Y _m-Y _p) ²+ (Z _m-Z _p) ²so, PM ₄=PMsin ∠ M ₄mP/sin ∠ MM ₄p, Fm=(f-PM ₄) tg ∠ MM ₄p.When ψ < 0: ∠ M _pmM ₂=0.5 Δ ε-ψ-θ z ₀, ∠ M ₄mM ₂=∠ M _pmM ₂+ ∠ M _pmM ₄=2 Δ ε-90 °, λ+0.5, ∠ M ₄m P=∠ PMM ₂-∠ M ₄m M ₂, all the other are identical during with ψ > 0.Due to ∠ K ₄kK ₁=∠ AKK ₁=λ _k+ 0.5 Δ ε, ∠ P K ₅k=90 °-θ z ₀+ 0.5 Δ ε-ψ-λ _kso, ∠ PK ₄k=∠ P K ₅k-∠ K ₄k K ₁=90 °-θ z ₀-ψ-2 λ _k; Because ∠ is FPP ₂=ψ+θ z ₀-0.5 Δ ε, ∠ KPP ₂=arc ctg (Y _k-Y _p)/(Z _k-Z _p), ∠ FPK=∠ KPP ₂+ ∠ FPP ₂, ∠ PKK ₄=180 °-∠ FPK-∠ PK ₄k, PK ₄/ sin ∠ PKK ₄=PK/sin ∠ PK ₄k, PK ²=(Y _k-Y _p) ²+ (Z _k-Z _p) ²so, PK ₄=PKsin ∠ PKK ₄/ sin ∠ PK ₄k, Fk=(f-PK ₄) tg ∠ PK ₄k.When ψ < 0: ∠ FPP ₂=0.5 Δ ε-ψ-θ z ₀, ∠ FPK=∠ KPP ₂-∠ FPP ₂, all the other are identical during with ψ > 0.Therefore, any time master penetrates in face the hot spot of arbitrfary point distance on para-curve and all can obtain γ s=0 ° time.

Claims (2)

1. a fixed-point condensing reflector, it is characterized in that rotating around the spindle and intercepting take a specific curves as bus certain area and form, follow the tracks of the solar time, it rotate η degree around primary optical axis, the while is rotated 0.5 △ ε degree around summit P, and regularly declination angular difference value △ δ is followed the tracks of, described specific curves is Y ²=2af (Z-B) para-curve or radius-of-curvature ρ is greater than 0.5 ρ ₀be less than 2 ρ ₀optically focused camber line, ρ ₀=af (1+tg ²φ) ^3/2, described Y ²=2af (Z-B) and ρ ₀=af (1+tg ²φ) ^3/2middle Y, Z are coordinate figure, and its O-XYZ coordinate is: fixed point F makes the Z axis sensing sun and is main shaft excessively, gets FO=f on Z axis, crossing true origin O point, to make vertical line the energized north of Z axis be Y-axis, cross O point and make the vertical line of Y, Z axis and point to west for X-axis, a, B are constant, a=cos (ψ+θ z ₀)/tg φ, B=0.5fcos (ψ+θ z ₀) tg φ, or B=f[1-sin (ψ+θ z ₀)-cos ²(ψ+θ z ₀)/2a], f is focal length, ψ is the angle of pitch, θ z ₀for the position zenith angle that begun, φ was primary tangent inclination angle, the φ=0.5 (h of summit P ₀-ψ), described primary optical axis PF is the line of fixed point F and summit P, is also catoptron turning axle, fixed point F and summit P are the set points fixing with respect to the earth, fixed point F is the set point on extinction mouth, summit P is the intersection point of primary optical axis PF and mirror surface, corner η=arc cos{{sin γ s cosh cosh0sin (γ s-γ)+(cos ψ s sin h ₀-sin ψ s cos γ s cosh ₀) [cos ψ s sin h-sin ψ s coshcos (γ-γ is s)] } × [cos ²h sin ²(γ-γ s)+[cos ψ s sin h-sin ψ s cosh cos (γ-γ s)] ²] ^-1/2 × [cos ²h ₀sin ²γ s+ (cos ψ s sin h ₀-sin ψ s cosh ₀cos γ s) ²] ^-1/2}, firing angle difference △ ε=ε-ε ₀, in formula, ε is firing angle, is the angle of solar rays and primary optical axis, ε ₀be the position firing angle that begun, γ s is primary optical axis position angle, is projection line and the due south angular separation of primary optical axis PF on ground level, and ψ s is primary optical axis inclination angle, is the angle of primary optical axis and ground level, and h is sun altitude, h ₀be the position sun altitude that begun, γ is solar azimuth.

2. fixed-point condensing reflector claimed in claim 1, is characterized in that described specific curves is Y ²=2af (Z-B) para-curve, and with Y ²to be bus make circular arc around main shaft to=2af (Z-B) para-curve rotates and intercept certain area and form minute surface.

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