CN101625432A - 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 dish formula system 3 big fundamental types, and the extinction body of slot type system is fixed, but its optically focused is smaller, the thermal efficiency is low; The optically focused ratio of tower system is big, 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 bigger, the thermal efficiency is higher, but its extinction body must be followed the catoptron rotation tracking sun, makes its heat transfer system complexity.Therefore develop that a kind of optically focused is bigger, the thermal efficiency is higher, extinction body maintains static, and extinction body can be higher than, be lower than or the solar energy utilization system parallel with catoptron becomes the task of top priority, and we are referred to as the fixed-point condensing system, abbreviation point type system; As everyone knows: the optically focused ratio of point focusing is greater than line focus, and promptly the extinction open area of point focusing is less, and the more little heat that radiate of extinction open area is few more under equal temperature, so the thermal efficiency of point focusing is higher than line focus; The core technology of point type system is to develop that a kind of optically focused is bigger, focal point is fixed and focal point can be higher than, be lower than or the catoptron parallel with catoptron, and 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 the aberration revised law is made 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, sub-mirror can be expert at and column direction carries out axial rotation with to making correction because of angle of incidence of sunlight changes the potential difference that resembles that causes when heliostat was followed the tracks of the sun; Its inventive point height but do not understand in person: 1, its heliostat optically focused how? 2, sub-mirror bending shaft and crisscross how the rotation? 3, whether each viewpoint definition unclear? Because its sub-mirror is in the row and column axial rotation, its structure and tracker must be complicated.

Summary of the invention

The technical matters that the application will solve provides that a kind of optically focused is bigger, focal point is fixed and focal point can be higher than, be lower than or is 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 be with a specific curves be bus around main axis rotation and intercept certain area and form, fixed-point condensing reflector is hereinafter to be referred as catoptron; Follow the tracks of the solar time, it rotates the 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 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 s)], ε ₀Be the position firing angle that begun, and regularly declination angle difference DELTA δ followed the tracks of, promptly 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, declination angle δ=23.45sin[360 * (284+n)/365], the fate of n for starting in the formula, δ from annual January 1 ₀Be the declination angle, position of having begun, when γ s ≠ 0 °, catoptron rotates the η degree around primary optical axis, make and mainly 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 are greater than 0.5 ρ ₀Less 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 ₀Slick and sly optically focused camber line, ρ ₀=af (1+tg ²Φ) ^3/2

Said catoptron is to be bus around main axis rotation certain angle, the hyperboloidal mirror that promptly intercepts certain width and highly form with a specific curves, and its peak is that upper extreme point, minimum point are lower extreme point, and the catoptron cross section can be oval or circular; The bus rotational trajectory can be circular arc line, also can be other optically focused camber lines.Para-curve Y ²Y, Z are coordinate figure among the=2af (Z-B), its O-XYZ coordinate is: fixed point F makes the Z axle sensing sun and is main shaft excessively, on the Z axle, get FO=f, crossing true origin O point makes the vertical line and the energized north of Z axle and is Y-axis, be that the YOZ coordinate surface is parallel to the earth's axis, vertical line and points west that mistake O point is made Y, Z axle are X-axis; A, B are constant, a=cos (ψ+θ z ₀)/tg Φ, constant B are 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 ₀Be the position zenith angle that begun, desirable on the Northern Hemisphere θ z ₀For high noon in the Summer Solstice constantly zenith angle, in the Southern Hemisphere desirable θ z ₀Be zenith angle constantly at high noon in Winter Solstice,

Be local latitude, Φ is the primary tangent inclination angle, Φ=0.5 (h ₀-ψ), be the primary tangent inclination angle Φ position height angle h that equals to have begun ₀Deduct half of angle of pitch ψ, h ₀=90 °-θ z ₀ρ ₀=af (1+tg ²Φ) ^3/2The title of middle a, f, Φ is equal to above-mentioned.Said main shaft is the bus turning axle; Said optically focused camber line can be slick and sly camber lines such as circular arc line, involute urve, ellipse, spiral of Archimedes, circular arc and parabolical complex line, and said slyness is meant that radius-of-curvature ρ gradually changes or is constant.The said beginning is played the position of setting spindle alignment constantly at the high noon sun when position is design and installation, can be the position of the Summer Solstice or high noon in Winter Solstice spindle alignment constantly sun; Said primary optical axis PF be through summit P reflection light, also be the catoptron turning axle, the line of still fix a point F and summit P; Set point on the extinction mouth that said fixed point F is extinction body or light-guiding shade; Said extinction body is a thermal-arrest with hot extinction equipment, can be flat plate collector or photovoltaic battery panel or pot and stove or boiler or smelting furnace or reacting furnace or handles stove etc.; Said light-guiding shade is the condensing body that is installed on the extinction body towards catoptron; Said extinction mouth is perpendicular to primary optical axis PF and cross the face of the F that fixes a point on extinction body or the 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 promptly mainly penetrate face and penetrate formed point in the face main, so summit P being treated as when the optically focused principle is described owing to its hinge is very little is to penetrate a point on the specific curves in the face main, 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 are formed, i.e. the face that the ray of sun incident summit P and principal normal are formed, the plane of incidence are the faces that solar rays and incidence point normal are formed; Said principal normal is the normal of summit P, and primary tangent is the tangent line of summit P.

Some angle code names among the application and definition and the regulation on the Northern Hemisphere the 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 solar rays on the ground projection line and Due South to angle,

The regulation Due South is zero, westwards for just, eastwards for negative; Solar hour angle ω is the angle of expression solar time, per hour is equivalent to ω=15 °, 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 the primary optical axis and the earth's axis, the angle of primary optical axis and ground level when promptly 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 elevation angle ψ＞0 °, ψ s＞0 ° when being higher than summit P, and fixed point F is angle of depression ψ＜0 °, ψ s＜0 ° when being 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; Effective firing angle: effectively the firing angle during illumination be called for short effective firing angle, selected when being design, specify firing angle less than one of maximum elevation; Main shaft and horizontal angle when main shaft angle β is ψ=0 °, position β has begun ₀Be main shaft and horizontal angle; Said corner η is that the begun master of position penetrates face and the instant main angle of penetrating between the face, promptly 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 the P-xyz coordinate system: with summit P is true origin, x, the parallel ground of y axle, and the x axle points to the due south, and the y axle points to positive west, and the z axle points to zenith; If leading immediately and penetrating face is I, the primary optical axis surfaces of revolution that mistake P is ordered is II, and it is I that the master of the position of having begun penetrates face ₀, sun center is A, and | PA|=1, then: P (0,0,0), F (fcos ψ s cos γ s, fcos ψ s sin γ s, fsin ψ s), A (1 cosh cos γ, 1 cosh sin γ, 1 sin h), vector $\stackrel{\&RightArrow;}{\mathrm{PF}}=f\cos \mathrm{\&psi;}s\cos \mathrm{\&gamma;si}+f\cos \mathrm{\&psi;}s\sin \mathrm{\&gamma;sj}+f\sin \mathrm{\&psi;sk},$ $\stackrel{\&RightArrow;}{\mathrm{PA}}=l\cosh \cos \mathrm{\&gamma;i}+l\cosh \sin \mathrm{\&gamma;j}+l\sinh k;$ Because instant master penetrates face and is parallel to solar rays PA, so the normal line vector of face I: ${\mathrm{<figure-callout\; id="1"\; label="n"\; filenames="A2008100293840017H1.png"\; state="\{\{state\}\}">n</figure-callout>}}_1=\stackrel{\&RightArrow;}{\mathrm{PF}}\&CenterDot;\stackrel{\&RightArrow;}{\mathrm{PA}}=n_{1x}i+n_{1y}j+n_{1z}k,$ 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)=1fcos ψ s cosh sin (γ-γ s); The normal line vector of surfaces of revolution II: ${\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="A2008100293840017H1.png,A2008100293840021H2.png"\; state="\{\{state\}\}">n</figure-callout>}}_2=\stackrel{\&RightArrow;}{\mathrm{PF}}=\begin{array}{cccc}{n}_{2x}& i+n_{2y}& j+n_{2z}& k\end{array},$ 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); High noon is γ=0 ° constantly, A ₀(1cos h ₀, 0,1sin h ₀), ${\stackrel{\&RightArrow;}{\mathrm{PA}}}_0=l\cos h_{0}i+l\sin h_{0}k;$ Face I ₀Normal line vector: ${\mathrm{<figure-callout\; id="0"\; label="n"\; filenames="A2008100293840023H1.png"\; state="\{\{state\}\}">n</figure-callout>}}_0=\stackrel{\&RightArrow;}{\mathrm{PF}}\&CenterDot;{\stackrel{\&RightArrow;}{\mathrm{PA}}}_0=n_{0x}i+n_{0y}j+n_{0z}k,$ 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 both be corner η, cos η=s ₁S ₂/ | s ₁|| s ₂|, with vectorial s ₁And s ₂The coordinate figure substitution and simplify after:

η＝arc?cos{{sinγs?cosh?cosh ₀?sin(γs-γ)+(cosψs?sin?h ₀-sinψs?cosγs?cosh ₀)[cosψs?sin?h-sinψs?cosh?cos(γ-γs)]}×[cos ²h?sin ²(γ-γs)+[cosψs?sin?h-sinψs?cosh?cos(γ-γs)] ²] ^-1/2×[cos ²h ₀?sin ²γs+(cosψs?sin?h ₀-sinψs?cosh ₀?cosγs) 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

The coordinate figure substitution and simplify after:

ε＝arc?cos[sinψs?sin?h+cosψs?cosh?cos(γ-γs)]。

Said xsect is perpendicular to the face of main shaft; Cross sections is parallel to the XOY coordinate surface, and carrying line at each cross sections internal reflector is that intersection point with Z axle and cross sections is the circular arc line in the center of circle, i.e. the projection of catoptron on the XOY coordinate surface is to be the concentric arc line in the center of circle with true origin O; Because the application is condenser mirror, so can also be to be other optically focused camber lines of focal point with fixed point F at the cross sections internal reflector, the position catoptron that begin this moment be an optically focused.In addition, if with Y ²=2af (Z-B) para-curve be bus along X-axis move the slot type catoptron.When having begun the position because the spindle alignment sun, so ξ=90 °, promptly solar rays in the XOY coordinate surface for a bit; Learn by reflection law: on the minute surface reflected ray of arbitrfary point must with the normal of this point and incident ray coplane in the plane of incidence, and incident angle equals reflection angle; To be bus with the specific curves do the circular arc rotation around the Z axle because catoptron is forms minute surface, so the incident angle of each point is equal in the same xsect, reflection angle is equal, all planes of incidence penetrate that face intersects at the Z axle and perpendicular to the 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 ²During=2af (Z-B) para-curve, fixed point F is the normal of Z axle, arbitrfary point and the mid point of the right-angle triangle that tangent line is formed thereof, because the Z axle is parallel to solar rays, so the reflection ray of the position incidence reflection mirror that begun focuses on the fixed point F certainly; When below not having special instruction, all with Y ²=2af (Z-B) para-curve is that the minute surface that bus is done the circular arc line rotation around main shaft illustrates the optically focused principle of discussing catoptron.Because fixed point F is the fixed point on extinction body or the light-guiding shade, and light-guiding shade is installed on the extinction body, extinction body and catoptron are installed on the ground, so fixed point F and summit P maintain static with respect to the earth, promptly focal distance f is a constant, and fixed point F can be higher than, is lower than or is parallel to summit P; Because the angle of ground normal primary optical axis and ground level when being γ s=0 °, so the angle of primary optical axis and each xsect equals ψ+θ z when ξ=90 °, in the YOZ coordinate surface perpendicular to local horizon, angle of pitch ψ ₀, by formula Y ²Curvature that=2af (Z-B) learns, para-curve is bus and slope are by constant a, B decision, 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 is higher than, is lower than or is parallel to summit P, according to angle of pitch ψ, focal distance f, local latitude

And selected declination angle δ value, thereby by calculating curvature and the slope of determining constant a, the definite bus of B value, make each xsect of catoptron perpendicular to solar rays, thus make the position focusing mirror that begun in fixed point F; The focal point of catoptron is fixed and focal point can be higher than, be lower than or parallel with catoptron when promptly having begun the position.

Because earth rotation causes that with revolving around the sun sunshine projects the firing angle variation of catoptron, so drive catoptron with tracking means follows the tracks of the sun: 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 the η degree around primary optical axis, make the master penetrate face and be parallel to sunray, its principal normal rotates 0.5 Δ ε degree around summit P main penetrating in the face simultaneously, 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, the incident angle of summit, position P equals reflection angle owing to begun, reflection ray directive fixed point F is so be that the incident angle of any operating mode summit P equals reflection angle after the rotation, will be through the reflection ray that P is ordered along primary optical axis PF directive fixed point F.Owing to comprised tracking when firing angle ε is followed the tracks of to declination angle δ, but about 10 hours of effective illumination every day, promptly to about 10 hours of every day of tracking of firing angle ε, so also will carry out other 14 hours tracking to declination angle δ, promptly regularly declination angle difference DELTA δ is followed the tracks of, make catoptron be concentrated on fixed point F; The focal point of catoptron is fixed and focal point can be higher than, be lower than or parallel with catoptron when realizing any operating mode thus.Catoptron is followed the tracks of sun process three kinds of operating modes: a kind of is begin position, ε at this moment ₀=h ₀=β ₀, ξ ₀=90 °, solar rays is parallel to main shaft, and reflection ray focuses on fixed point F; A kind of is ε＜ε ₀, this moment ξ＜90 °, a kind of is ε＞ε ₀, this moment ξ＞90 °, back two kinds of operating mode catoptrons are optically focused, promptly reflection ray forms hot spot on the extinction mouth; The size of its hot spot and incidence point distance and the firing angle difference of P to the limit are directly proportional, and promptly the distance of incidence point and summit P is big more, the hot spot distance is big more, 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 large spot more is just big more.And the big more extinction open area of hot spot is big more, and the heat loss through radiation of extinction mouth is directly proportional with the extinction open area, and promptly the big more heat that radiate of extinction open area is many more under the extinction body equal temperature, and promptly thermal loss is big more, the extinction body thermal efficiency is low more; For dwindling hot spot, can promptly control the size that the extinction open area is controlled in the catoptron cross section by the distance of control incidence point and summit P promptly in order to improve the thermal efficiency of extinction body.Because the hot spot of summit P distance is zero and the distance of catoptron end points and summit P is maximum, thus the size in catoptron cross section be by the selecting side apart from recently definite, said end distance is than being the catoptron end points ratio of distance and its hot spot distance of P to the limit; Because the high more desired catoptron optically focused of serviceability temperature of extinction body is bigger than more, and optically focused than with smallest end apart from than square be directly proportional, be that end distance is bigger than more than big more optically focused, the thermal efficiency of extinction body is high more, and the catoptron cross section is more little, relative tracking means cost is high more, so the size in catoptron cross section is determined by the serviceability temperature of extinction body.Can also realize by reducing the firing angle difference for dwindling hot spot, because the minimax firing angle all is the extreme value of firing angle difference, so can design with maximum mean value with minimum elevation; But owing to is the maximal value that the atmospheric envelope distance is passed in solar radiation during maximum elevation, so the application designs with the mean value of effective firing angle and minimum elevation, promptly the elevation midpoint with effective firing angle and minimum elevation is the position firing angle ε that begun ₀But effectively firing angle is a designated value but not the particular value of celestial bodies motion, for ease of calculate and explanation, generally with the Summer Solstice or high noon in Winter Solstice firing angle constantly serve as the position firing angle ε that begun ₀, the firing angle difference is minimized.

For the thermal efficiency that improves extinction body can also be realized by towards the position of catoptron one light-guiding shade is installed on extinction body, a light-guiding shade promptly is installed between extinction body and catoptron is realized; Light-guiding shade is a circle or oval cone, be on its extinction body or near the end face of extinction body to be that bottom surface is less than the end face near catoptron, end face near catoptron is the extinction mouth, the three-way unification of the normal 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 the extinction mouth, and this moment, light-guiding shade can be evacuated, to stop the heat loss through convection of light-guiding shade; The sticking 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.Because arbitrarily the reflected ray of summit P is all along primary optical axis directive fixed point F during operating mode, and the three-way unification of normal of light-guiding shade axis and primary optical axis and extinction mouth is so the reflected ray of summit P will inject extinction body along the light-guiding shade axis; Because arbitrarily catoptron all is concentrated on fixed point F on the extinction mouth during operating mode, and the extinction mouth is coated with the bottom surface of reflectance coating, light-guiding shade less than the extinction mouth perpendicular to light-guiding shade axis, light-guiding shade conical surface inside surface, 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 the light-guiding shade axis, make reflected ray optically focused once more, improve the optically focused ratio, thereby improve the thermal efficiency of extinction body.

Because the focal point of catoptron can be higher than, be lower than or be parallel with catoptron, so the installation of extinction body and catoptron is not limited by landform, a plurality of catoptrons can be mounted to along the massif slope mirror battle array, its all around the reflected ray between the catoptron stagger, do not block mutually, make a plurality of 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 installed in mountain top or cat head, make extinction body be higher than catoptron, a plurality of catoptrons are arranged to the mirror battle array on the level land, make its all around the reflected ray between the catoptron stagger, do not block mutually, make a plurality of catoptrons be concentrated on one the fixed point F; Thereby enlarge 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: be achieved as sun power smelting furnace etc.When arranging the mirror array 1 system on the Northern Hemisphere, extinction body is in the south of catoptron, having only a row catoptron and extinction body in the mirror battle array is that positive north and south arranges that promptly its primary optical axis position angle is γ s=0 °, all the other each row catoptron γ s ≠ 0 °; Because the application's catoptron is to be parallel to one to set up a specific curves in the YOZ coordinate surface of the earth's axis and be bus intercept certain area around the main axis rotation of the sensing sun forms with it, so the bus of each row catoptron of γ s ≠ 0 ° is identical with specific curves in the YOZ coordinate surface, its any any ξ of position begin ₀=90 °, any row catoptron in the position of promptly having begun all focuses on or is concentrated on fixed point F, just begin position γ s=0 ° catoptron its lead and to 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 the master penetrate face, simultaneously its principal normal is independently penetrated rotation 0.5 Δ ε degree in the face around summit P separately at each, when 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 all are concentrated on fixed point F.This shows, the application's catoptron can be arranged to the mirror battle array fully, forms the changeable solar thermal utilization point type system of power thus.

In sum, the application's catoptron be with a specific curves be bus around main axis rotation and intercept the hyperboloidal mirror that certain area forms, 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, promptly according to angle of pitch ψ, focal distance f, latitude according to fixed point F And selected declination angle δ value establishes, and makes and begun the position focusing mirror or be concentrated on the F that fixes a point, and its focus fixes a point promptly that F fixes and on primary optical axis; And the position of having begun is to set with the mean value of minimized effective firing angle of firing angle difference and minimum elevation, promptly selected declination angle δ value; So the focal point of catoptron is fixed and focal point can be higher than, be lower than or parallel with catoptron and can be arranged to the mirror battle array, make the installation of extinction body and catoptron not limited by landform.Tracking solar time, catoptron master penetrate face and rotate the η degree around primary optical axis, make the main sunray that all is parallel to when penetrating any operating mode of face, its principal normal rotates 0.5 Δ ε degree around summit P main penetrating in the face simultaneously, and regularly declination angle difference DELTA δ is followed the tracks of, when making any operating mode, the reflected ray of summit P directive fixed point F all, catoptron all is concentrated on fixed point F.Since the optically focused of the application's catoptron than with smallest end apart from than square be directly proportional, select higher end distance ratio can make the optically focused ratio reach tens times even reach hundreds of times; Because light-guiding shade is optically focused once more, 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 the mirror battle array, make a plurality of catoptrons be concentrated on a fixed point F, thereby improve the power input of single extinction body, widen solar energy heat utilization field.

Description of drawings

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

Fig. 3 is optically focused figure after catoptron rotates 23.45 °.Fig. 4 is the main constantly interior optically focused figure of face that penetrates at high noon in Winter Solstice.

Fig. 5 is moment high noon in Winter Solstice left side optically focused figure.Fig. 6 is moment high noon in a 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 synoptic 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.Elevation angle catoptron master penetrated face optically focused figure when Figure 12 was effective firing angle.

Embodiment

Below in conjunction with embodiment the application is described further.

First embodiment is the embodiment that fixed-point condensing reflector is formed solar furnace.Fig. 1 is split type solar furnace synoptic 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 the 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 the extinction mouth 1 on extinction body 2; Catoptron 3 is hinged by hinge 11 and rotating shaft 7, hinge 11 axis normal are penetrated face and rotating shaft 7 in catoptron 3 masters, be that catoptron 37 axis rotation around the shaft can be rotated around hinge 11 axis simultaneously, its rotation centre of sphere is summit P, one cross section 4 is arranged on the catoptron 3, 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 the due south, and primary optical axis PF and rotating shaft 7 axis are collinear; Rotating shaft 7 is enclosed within on the support 10, one tracking means 6 is installed in the rotating shaft 7, tracking means 6 is a set a distance formula tracking means, support 10 is installed on the ground, one tracking means 9 is installed on the support 10, tracking means 9 is to decide corner formula tracking means, and 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 drives rotating shaft 7 rotations by gear 8, and rotating shaft 7 drives catoptron 3 and tracking means 6 rotates around primary optical axis PF together, and tracking means 6 is slidingly connected by pull-push piece 5 and drives catoptron 3 and rotates around hinge 11 axis.

Fig. 2 is that the position catoptron master of having begun penetrates the face focused view.Crossing fixed point F makes the Z axle and aims at the sun and be main shaft, its main shaft angle β ₀=90 °-θ z ₀Get FO=f on the Z axle, the vertical line energized north that mistake true origin O point is made the Z axle is Y-axis, and promptly the YOZ coordinate surface is parallel to the earth's axis, and the vertical line points west that mistake O point is made Y, Z axle is X-axis, forms the O-XYZ coordinate system thus.Make primary optical axis PF and be parallel to local horizon NS and point to the due south, promptly γ s=0 °, ψ=0 °, establishing specific curves is Y ²=2af (Z-B) para-curve is also made this para-curve on the YOZ coordinate surface, constant a=cos θ z in the 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 the catoptron summit, and mistake P point is made solar rays AP and is parallel to the Z axle, crosses the P point and makes principal normal PP ₁, hand over the Z axle in P ₁, PP ₁With the angle of PF be reflection angle alpha ₀, with the angle of AP be incident angle θ ₀, the angle of AP and local horizon NS is elevation angle h ₀, i.e. h ₀=β ₀Cross the P point and make Z axle vertical line PP ₂, P ₂Be intersection point, because the ground normal is perpendicular to local horizon NS, and AP ⊥ PP ₂So, ∠ FPP ₂=θ z ₀Cross the P point and make PP ₁Vertical line be tangent line PP ₃, hand over Y-axis in P ₃, its coordinate be (0, Y _P3, 0), PP ₃With the angle of Y-axis be Φ; Because PP ₂So parallel Y-axis is ∠ P ₂PP ₃=Φ is because slope tg Φ=Y that P is ordered _P/ af, Y _P=PP ₂=FP cos θ z ₀, a=cos θ z ₀So/tg Φ is tg Φ=PP ₂Tg Φ/fcos θ z ₀So, PP ₂=fcos θ z ₀, so FP=f=FO; Because tangent line PP ₃Equation be: Z _P-Z _P3=tg Φ (Y _P-Y _P3), and P ₃Coordinate figure be (0, Y _P3, 0), so have only Y _P3=0 o'clock, just Z arranged _P3=0, promptly have only P ₃This equation was just set up when point overlapped with true origin O, 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 ₁So PO is ∠ PP ₁F=Φ is because AP//P ₁So F ∠ APP ₁=∠ PP ₁F, i.e. Φ=θ ₀Because so FP=FO ∠ FPO=∠ is FOP, ∴ θ ₀=α ₀=Φ; So FP ₁=FP=FO=f, promptly the F point is Z axle, 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 ₀With 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) because during Y=0, Zo=B, o is the para-curve summit, so B=oO, promptly 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 ₀In the time of=30 °, a=1.5, its B=0.25f; As θ z ₀In the time of=0 °, a=1, its B=0.5f, promptly summit o equals the distance of summit o to fixed point F to the distance of Y-axis, this moment para-curve Y ²=2af (Z-B) is standard para-curve y ²=2pz; Standard para-curve y ²The mathematics implication of=2pz is: have a moving some M to equate that to the distance of fixed point F and boning out L the track of then moving some M is a para-curve; It is a true origin with para-curve summit o, and its y axle is parallel with Y-axis, and its z axle and Z axle are collinear, its mathematical expression y ²P is the distance of fixed point F to boning out L among the=2pz, and promptly the y value equals the Y value, is y=Y, z=Z-B, and p=f, promptly Y-axis is equivalent to boning out L; As θ z ₀In the time of=0 °, a=1, formula Y ²=2af (Z-B) becomes Y ²=2f (Z-B), its formula is y in the o-yz coordinate system ²=2fz is so it is a standard para-curve; As everyone knows: aim at the sun as long as standard para-curve main shaft is the z axle, then all reflected rays all focus on focal point F.

At para-curve Y ²Get arbitrfary point M on the=2af (Z-B), mistake M point is made solar rays AM and is parallel to the Z axle, connects MF, makes normal MM ₁Hand over the Z axle in M ₁, normal MM ₁Divide ∠ AMF equally; Cross the M point and make MM ₁Vertical line be tangent line MM ₃, MM ₃Hand over the Z axle in M ₃, MM ₃With the angle of Y-axis be λ.Cross the M point and make Z axle 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; ∠ MM in addition ₃F=∠ 5, ∠ MM ₁F=∠ 6.Because ray AM is parallel to the Z axle, thus ∠ 1=∠ 5, ∠ 2=∠ 6, again because MM ₁⊥ MM ₃So, 2=90 ° of ∠ 1+ ∠, 4=90 ° of ∠ 3+ ∠ 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)-and 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 ₃Because 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, so M ₁P ₁=M ₂P ₂=OM ₃, because FP ₁So=FO=f is M ₁F=FM ₃, promptly the F point is right-angle triangle Δ M ₁MM ₃So the mid point of hypotenuse is M ₁F=FM ₃=FM, promptly F is Z axle, normal MM ₁And tangent line MM ₃The Δ M that forms ₁MM ₃Mid point.So ∠ 6=∠ 3=∠ 2, promptly 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).

Because s=0 ° of present embodiment γ penetrates the master around summit P when the δ changes delta δ of declination angle, with tracking means 6 driving catoptrons 3 and rotates 0.5 Δ δ in the face YOZ coordinate surface, thereby the sun is followed the tracks of.Fig. 3 is optically focused figure after catoptron rotates 23.45 °.The earth equals solar rays AP and has been rotated counterclockwise 46.9 ° by high noon in the Summer Solstice, constantly transmission high noon in Winter Solstice, declination angular difference value was Δ δ=46.9 ° constantly, this moment 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, promptly Fig. 3 is the optically focused figure of moment high noon in Winter Solstice catoptron.Because the O-XYZ coordinate system rotates with catoptron,, but 0.5 Δ δ, β have then been rotated with respect to the P-xyz coordinate system so each point on the para-curve and line do not have change with respect to its value of O-XYZ coordinate system and angle _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 incident angle θ reduce Δ δ, but principal normal PP ₁Rotated 0.5 Δ δ, made incident angle θ increase 0.5 Δ δ, promptly incident angle is by θ ₀Become θ _d=θ ₀-Δ δ+0.5 Δ δ=θ ₀-0.5 Δ δ; Because principal normal PP ₁Rotated 0.5 Δ δ, made reflection angle alpha reduce 0.5 Δ δ, promptly reflection angle is by α ₀Become α _d=α ₀-0.5 Δ δ is because θ ₀=α ₀So, θ _d=α _d, promptly incident angle θ equals reflection angle alpha, i.e. and catoptron rotation is after the reflection ray that P is ordered 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 ₁Rotated 0.5 Δ δ, made incident angle ∠ 2 _dIncrease 0.5 Δ δ, promptly incident angle ∠ 2 _d=∠ 2-Δ δ+0.5 Δ δ=∠ 2-0.5 Δ δ.When catoptron rotated around summit P, the above each point of para-curve P point was 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 conveniently, following provisions arbitrfary point M is the above arbitrfary point of para-curve P point, crosses the parallel lines MM that arbitrfary point M makes primary optical axis PF _PHand over the Z axle in M _PPoint, then ∠ M ₁MM _P=β _d-λ; Because incident angle ∠ 2 _dEqual reflection angle ∠ 3 _d, and ∠ 2 _d=∠ 2-0.5 Δ δ, so ∠ 2=λ is ∠ 3 _d=∠ 2 _d=λ-0.5 Δ δ; If M ₄Be reflected ray MM through arbitrfary point M ₄With the intersection point of primary optical axis PF, ∠ M is arranged then _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 °-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 ₄So M P is P M ₄=PMsin ∠ M ₄M P/sin ∠ M M ₄P, PM ²=(Y _M-Y _P) ²+ (Z _M-Z _P) ²Following provisions arbitrfary point K is the following arbitrfary point of para-curve P point, crosses the K point and makes solar rays AK, makes the normal KK that K is ordered ₁Hand over the Z axle in K ₁, cross the K point and make the reflected ray KK that K is ordered ₄, K ₄Be 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 ₅Because ∠ P K ₅K=β _d-λ _K, ∠ A K K ₅=90 °-∠ P K 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 ∠ PK K ₄=PK/sin ∠ P K ₄So K is P K ₄=PKsin ∠ PK K ₄/ sin ∠ P K ₄K, PK in the formula ²=(Y _K-Y _P) ²+ (Z _K-Z _P) ², ∠ PKK ₄=180 °-∠ FPK-∠ P K 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 ₀The time master penetrate in the face on the incident para-curve reflected ray of arbitrfary point and the intersection point of primary optical axis PF 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, then: Fm=(P M ₄-f) tg ∠ M M ₄P, Fk=(P K ₄-f) tg ∠ P K ₄K, thus, β＜β ₀The time arbitrfary point hot spot distance all can obtain, promptly above calculating formula is β＜β ₀The calculating formula of hot spot distance when being ξ＜90 °.

Fig. 4 is the main constantly interior optically focused figure of face that penetrates at high noon in Winter Solstice.Penetrate main that to get C in the face on the bus be that catoptron upper extreme point, E are lower extreme point; Crossing C, E point respectively makes solar rays AC, AE, makes normal CC ₁And EE ₁Hand over the Z axle in C ₁And E ₁, cross the C point and make Z axle vertical line CC ₂, C ₂Be intersection point, make reflected ray CC respectively 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 the hot spot distance of the upper and lower end points of concrete data computation: all get latitude in the present embodiment

The position of then having begun:

The δ during 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 f=80 then, a=cos θ z at 1000: 1 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 the C point is C (0,90,64.042), gets according to the inference of M: λ _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 °-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 the E point coordinate is E (0,60,43.09), gets according to the inference of K: ∠ EPP 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.The hot spot distance that is 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 Fc and PC with PC ₄Be directly proportional, so Fc is directly proportional with PC, in like manner Fe is directly proportional with PE, promptly hot spot distance and its incidence point to the limit the distance of P be directly proportional.If FP and Z axle meet at D, cross the F point and make Z axle vertical line FF ₁, DP=PP then ₂/ sin β _d=Y _P/ sin β _dDF=DP-f=Y _PSo/sin β-f is 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 F in the O-XYZ coordinate system of promptly fixing a point constantly high noon in Winter Solstice become F (0, Y _P-fsin β _d, fcos β _d+ Z _P).

It more than is main constantly optically focused situation analysis of penetrating in the face at high noon in Winter Solstice, is the lower surface left end point for analyzing optically focused situation in other planes of incidence, getting the G point for upper surface left end point, J point, said on or the lower surface be by on the catoptron or lower extreme point and perpendicular to the face of main shaft; Then G, C ₂, E ₂, J is the left side of catoptron, left and right end face is penetrated face and is intersected at the Z axle with main; If left and right sides end face with the main angle of penetrating face is

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

Degree forms, and promptly this year, line was still Y ²=2af (Z-B) para-curve, the G point on it are main to penetrate C point on the face around intersection point C ₂Rotation

Degree forms, so the intersection point of normal that G is ordered and Z axle is C ₁The point; So the parallel incident of position solar rays and be parallel to main shaft owing to begun is incident angle ∠ AGC ₁=∠ ACC ₁, reflection angle ∠ C ₁GF=∠ C ₁CF, and ∠ AGC ₁=∠ C ₁GF; On the minute surface arbitrarily a bit all therewith in like manner, so the position focusing mirror that begin is in the F that fixes a point.

Fig. 5 is moment high noon in Winter Solstice left side optically focused figure.Fig. 6 is moment high noon in a Winter Solstice left side optically focused vertical view.Because position angle r=0 ° constantly of high noon, the Y-axis energized north is so solar rays is parallel to the YOZ coordinate surface; Get A _GBe the intersection point (not shown) of solar rays AG and XOZ coordinate surface, high noon is ε=h constantly, ξ=ε+90 °-β _d=h+90 °-β _d, get G point coordinate value and be: G (X _G, Y _G, Z _G), A then _GPoint coordinate is: A _G(X _G, 0, Z _G+ Y _GTg ξ); Because the 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 _CSo ray vector $\stackrel{\&RightArrow;}{A_{G}G}=Y_{G}j-Y_G\mathrm{tg\&xi;k},$ Normal line vector $\stackrel{\&RightArrow;}{C_{1}G}=X_{G}i+Y_{G}j-\mathrm{af\; k}.$ 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 ₂By 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 PF to be the normal line vector of extinction mouth 1 be: $\stackrel{\&RightArrow;}{\mathrm{PF}}=(Y_F-Y_P)j+(Z_F-Z_P)k,$ 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 that G and incident ray AG form 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), then C ₁(0,0,171.4), E ₁(0,0,150.45); Getting left and right sides end face with the main angle of penetrating face is

Then 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.With this equation and GG ₁Equation form the linear equation in two unknowns group and can solve G ₁The coordinate figure of point is G ₁(23.29,52.02,144.52) are with GG ₂Equation form the linear equation in two unknowns group and can solve G ₂The coordinate figure of point is G ₂(9.60,35.84,127.14).Because Δ GG in the plane of incidence II ₁G ₂Each length of side is respectively: GG ₁=87.73, GG ₂=82.34, G ₁G ₂=27.41, because $\cos \&angle;\mathrm{<figure-callout\; id="1"\; label="AG\; C"\; filenames="A2008100293840017H1.png"\; state="\{\{state\}\}">AG</figure-callout>}{C}_{}{}_1=\frac{\stackrel{\&RightArrow;}{A_{G}G}.}{\left|\stackrel{\&RightArrow;}{A_{G}G}\right|}\frac{\stackrel{\&RightArrow;}{C_{1}G}}{\left|\stackrel{\&RightArrow;}{C_{1}G}\right|}=0.95,$ So ∠ AG C ₁=18.195 °; Because ∠ AG is C ₁=∠ C ₁So Gg is ∠ AG g=2 ∠ AG C ₁=36.39 °, learn: sin ∠ G G by sine ₁G=GG ₂Sin ∠ AGC ₁/ G ₁G ₂=0.938, so ∠ GG ₁G=69.718 °; Learn 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 then _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 the j point for the reflected ray Jj of ordering through J and the intersection point of extinction mouth 1, then 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 is Fg=5.40m on left and right sides end face symmetry moment high noon in Winter Solstice extinction mouth 1; Can obtain the hot spot distance of any arbitrarily according to above reckoning.Catoptron is with C (0,90,64.042), E (0,59.324,42.709) is end points up and down, with U (15.337,74.662,53.376), V (15.337,74.662,53.376) when being the elliptic cross-section of left and right sides end points, then high noon in Winter Solstice constantly its upper and lower, left and right hot spot distance be: Fc=3.936, Fe=1.889, Fu=Fv=3.104.This shows that incidence point is big more from summit P its hot spot distance far away more, so elliptic reflector is littler than the hot spot of rectangular mirror.

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

The size that is firing angle ε is by latitude

Reaching declination angle δ and hour angle ω determines; 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.The master that tracking means 9 drives catoptron 3 penetrates face around primary optical axis PF rotation η degree, corner η=arc cos sin h/ (cos ²H sin ²γ+sin ²H) 1/2, make the master penetrate face and be parallel to sunray; Tracking means 6 drives catoptrons 3 with catoptron 3 around primary optical axis PF rotation, while tracking means 6 and rotates main penetrating in the face around summit P; When 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 when declination angle δ changes delta δ spends, because of γ s=0 °, catoptron principal normal PP ₁Penetrate rotation 0.5 Δ δ in the face around summit P main, thereby declination angle δ is followed the tracks of.Owing to comprised tracking when firing angle ε followed the tracks of, but about 10 hours of effective illumination every day, promptly to about 10 hours of every day of tracking of firing angle ε, so also will carry out other 14 hours tracking to declination angle δ to declination angle δ; Because 93.8 ° of declination angle δ one annual variation, 0.257 °/day of the Δ δ ≈ of every day carries out a secondary tracking, promptly regularly declination angle difference DELTA δ is followed the tracks of that just foot is accurate declination angle δ so follow the tracks of before the beginning firing angle ε every day.Being tracked as the sun when being γ s=0 °: catoptron 3 makes the master penetrate face and is parallel to sunray, simultaneously its principal normal PP around primary optical axis PF rotation η degree ₁Penetrate rotation 0.5 Δ ε degree in the face around summit P main, and regularly rotate 0.5 Δ δ around summit P.High noon is ω=0 °, γ=0 ° constantly, learnt by cos ε=cos hcos γ: ε=h, and promptly firing angle ε equals elevation angle h; Because the specific curves of present embodiment is with zenith angle θ z constantly at high noon in the Summer Solstice ₀Design, and this moment firing angle ε ₀Equal elevation angle h ₀, the spindle alignment sun, so the reflection ray of the position incidence reflection mirror that begun focus on certainly the fixed point F; But not the position optically focused situation constantly at high noon that begun is discussed in declination angle δ is followed the tracks of.The sun haunts local horizon moment, elevation angle h=0 °, and corresponding firing angle ε equals position angle γ, and the position angle of this moment

Formula is learnt thus: the position angle maximum the when Summer Solstice, δ X=23 ° 27 ' sun in declination angle haunted the local horizon at same latitude, in the whole year, the Summer Solstice sunrise sunset moment firing angle ε maximum, Winter Solstice declination angle δ _dPosition angle minimum when=-23 ° of 27 ' sun haunt the local horizon, promptly Winter Solstice sunrise sunset moment firing angle ε minimum, and latitude is big more, the position angle difference is big more the Summer Solstice in winter during the sunrise sunset; Because this moment, firing angle equaled the position angle, the firing angle the when Summer Solstice, the sun haunted the 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 incident angle θ increase Δ ε, but principal normal PP ₁Turn clockwise 0.5 Δ ε, make incident angle θ reduce 0.5 Δ ε, promptly incident angle is by θ ₀Become θ _M=θ ₀+ Δ ε-0.5 Δ ε=θ ₀+ 0.5 Δ ε; Because principal normal PP ₁Turned clockwise 0.5 Δ ε, make reflection angle alpha _MIncrease 0.5 Δ ε, promptly reflection angle is by α ₀Become α _M=α ₀+ 0.5 Δ ε is because θ ₀=α ₀So, θ _M=α _M, principal normal PP ₁Rotation is after the reflected ray that P is ordered 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 when 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 makes the reflected ray of catoptron form hot spot on the extinction mouth toward rotating 0.5 Δ ε near fixed point F direction.Cross arbitrfary point M and make Z axle vertical line MM ₂, M ₂Be intersection point, cross the parallel lines MM that the M point is made primary optical axis PF _PHand over the Z axle in M _PPoint, then ∠ M ₁MM _P=β _M-λ; Because incident angle ∠ 2 _MEqual reflection angle ∠ 3 _M, and ∠ 2 _M=∠ 2+0.5 Δ ε, so ∠ 2=λ is ∠ 3 _M=∠ 2 _M=λ+0.5 Δ ε; If M ₄Be reflected ray MM ₄With the intersection point of primary optical axis PF, ∠ M then _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 λ+0.5 Δ ε-90 °, ∠ 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) ²Because PM/sin ∠ is MM ₄P=PM ₄/ sin ∠ M ₄So M P is PM ₄=PMsin ∠ M ₄M P/sin ∠ M M ₄P.Cross arbitrfary point K and make solar rays AK, do normal KK ₁Hand over the Z axle in K ₁, make reflected ray KK ₄Hand over primary optical axis PF in K ₄, normal KK ₁Hand over PF in K ₅Because ∠ PK ₅K=β _M-λ _K, incident angle ∠ AKK ₁=ε _M-∠ PK ₅K=ε _M-(β _M-λ _K), λ wherein _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, promptly 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 gets PK ₄=PKsin ∠ PKK ₄/ sin ∠ PK ₄K.Fm=(f-P M then 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 °.For reduce diagram and interpreting blueprints convenient, establish M (0,90,64.04) point and be the catoptron upper extreme point; Then: 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 the catoptron lower extreme point, then: 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 _MThe time upper and lower end points hot spot distance be Fm=7.223m, Fk=5.37m, this value is a catoptron hot spot ultimate value; The hour angle of this moment

The corresponding time is 4: 23 morning and 19: 37 afternoon; Because maximum elevation ε _MThe time be the maximal value that the atmospheric envelope distance is passed in solar radiation, and that atmospheric distance is passed in solar radiation is long more, then atmosphere is many more to the energy of absorption, reflection and the scattering of solar radiation, is that the emittance of catoptron is few more so arrive ground; It is relevant with elevation angle h that atmospheric distance is passed in solar radiation, in the time of h=90 ° the shortest, the h=0 of distance ° the time distance the longest, and sinh is corresponding with cos ω, so about 10 hours of effective illumination every day, be the morning 7:00 to 17:00 in afternoon, corresponding hour angle is-75 ° to 75 °, the corresponding Summer Solstice

The 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.The hot spot of upper and lower end points distance is Fc=3.4, Fe=2.37 when being effective firing angle.With Fig. 5,6 in like manner, get G (23.29,86.93,64.04) point and be lower surface left end point, then G, C for upper surface left end point, J (15.53,57.96,43.09) point ₂, E ₂, J is the left side of catoptron, establish , then the catoptron cross section is a rectangle.Same Fig. 5,6 described: go up left end point G and normal and the Z axle intersection point of left end point J are respectively C down ₁, E ₁Point is got A _GBe the intersection point of solar rays AG and XOZ coordinate surface, the intersection point of establishing extinction mouth 1 and solar rays AG is G ₁, with normal C ₁The intersection point of G is G ₂(A _G, G ₁, G ₂All not shown among the figure); Cut firing angle ξ=ε+90 °-β=103.61 °, because solar rays is parallel to and mainly penetrates face, so GG ₁, GG ₂Same Fig. 5 of equation, 6 described.Get the intersection point of g point, because Y for reflected ray Gg and extinction mouth I _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.With this equation and GG ₁Equation form solution of equations and get: G ₁(23.29 ,-82.3 ,-634.94) are with GG ₂Equation form solution of equations and get: 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 $\cos \&angle;{\mathrm{<figure-callout\; id="1"\; label="G"\; filenames="A2008100293840017H1.png"\; state="\{\{state\}\}">G</figure-callout>}}_1{\mathrm{<figure-callout\; id="2"\; label="GG"\; filenames="A2008100293840017H1.png,A2008100293840021H2.png"\; state="\{\{state\}\}">GG</figure-callout>}}_2=\frac{{{\mathrm{<figure-callout\; id="1"\; label="GG"\; filenames="A2008100293840017H1.png"\; state="\{\{state\}\}">GG</figure-callout>}}_1}^2+{{\mathrm{<figure-callout\; id="2"\; label="GG"\; filenames="A2008100293840017H1.png,A2008100293840021H2.png"\; state="\{\{state\}\}">GG</figure-callout>}}_2}^2-G_1{{\mathrm{<figure-callout\; id="2"\; label="G"\; filenames="A2008100293840017H1.png,A2008100293840021H2.png"\; state="\{\{state\}\}">G</figure-callout>}}_2}^2}{2{\mathrm{<figure-callout\; id="1"\; label="GG"\; filenames="A2008100293840017H1.png"\; state="\{\{state\}\}">GG</figure-callout>}}_1\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="GG"\; filenames="A2008100293840017H1.png,A2008100293840021H2.png"\; state="\{\{state\}\}">GG</figure-callout>}}_2}=-0.595,$ Get ∠ 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 ₂G g/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 then _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, then can extrapolate: X for reflected ray Jj and extinction mouth 1 _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 since left and right sides end face symmetry effectively during firing angle on the extinction mouth 1 maximum radius of hot spot be Fg=5.96m.Can obtain the hot spot distance of any arbitrarily according to above reckoning; Catoptron is with C (0,90,64.042), E (0,59.324,42.709) is end points up and down, with U (15.337,74.662,53.376), V (15.337,74.662,53.376) when being the elliptic cross-section of left and right sides end points, then effectively during firing angle ε=± 97.22 ° its upper and lower, left and right hot spot distance be: Fc=3.4, Fe=2.443, Fu=Fv=3.0.This shows that incidence point is big more from summit P its hot spot distance far away more, so elliptic reflector is littler than the hot spot of rectangular mirror.

In sum: any reflected ray of summit P directive fixed point F during operating mode, catoptron any reflected ray arbitrarily are concentrated on fixed point F.Catoptron is followed the tracks of sun process three kinds of operating modes: a kind of is begin position, ε at this moment ₀=h ₀=β ₀, ξ ₀=90 °, solar rays is parallel to main shaft, and reflected ray focuses on fixed point F, and the hot spot distance is zero; A kind of is ε＜ε ₀, β＜β ₀, ξ＜90 °, the hot spot distance of this moment is obtained by the calculating formula of Fig. 3 to 6; A kind of is ε＞ε ₀, β＞β ₀, ξ＞90 °, the hot spot distance of this moment is obtained by the calculating formula of Fig. 7 to 8; Therefore arbitrarily the operating mode catoptron arbitrarily any hot spot distance all can obtain.And hot spot distance and incidence point distance and the firing angle difference of P to the limit are directly proportional, and the control incidence point is the size apart from the may command hot spot of P to the limit, and promptly control end is apart from the size than may command hot spot.As present embodiment be the low temperature solar furnace, to establish the catoptron cross section be oval, gets end distance than=3; Then the upper extreme point of catoptron is that C (0,128,102.622), lower extreme point are that E (0,9,26.696), left end point are that U (59.5,68.5,64.659), right endpoint are V (59.5,68.5,64.659), and this moment, the radius-of-curvature of upper and lower end points was 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, then 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 is to be the center of circle, to be the circle (can certainly be other shapes that can absorb whole reflection rays) of radius, then its area A=π Fc with maximum hot spot apart from Fc=24.09m with fixed point F ²=1823.15m ², then its optically focused ratio is: 13193.15m ²/ 1823.15m ²=7.236.The described catoptron of Fig. 5, Fig. 8 is with C (0,90,64.042), E (0,59.324,42.709) be end points up and down, with U (15.337,74.662,53.376), V (15.337,74.662,53.376) and when being the elliptic cross-section of left and right sides end points, its maximum hot spot distance is: Fc=3.936; Its end distance ratio is 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 is to be the center of circle, to be the circle of radius with maximum hot spot apart from Fc=3.936m with fixed point F, its area A=π Fc ²=48.67m ², then its optically focused ratio is: 18.5.With the described elliptic reflector of Fig. 5, Fig. 8 change into summit P be the center of circle, when being radius circular with PU, its left and right sides end points coordinate is: U (15.337,74.662,53.376), V (15.337,74.662,53.376), then the 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=741.48m ², corresponding extinction mouth 1 is to be the center of circle, to be the circle of radius with maximum hot spot apart from Fu=3.104m with fixed point F, then its area A _G=π Fu ²=30.269m ², then 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 both with extinction open area greater than maximum facula area, 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/A _G=π PU ²/ π Fu ²=(PU/Fu) ²And both when other shapes optically focused than with smallest end apart from than square be directly proportional.

The bus of catoptron can also be that other radius-of-curvature are near ρ ₀=af (1+tg ²Φ) ^3/2The optically focused camber line, present embodiment substitutes Y as the circular arc line with radius-of-curvature ρ=202.706m ²=2af (Z-B) para-curve, be its summit P coordinate figure constant, get primary tangent inclination angle Φ _P=29 °, when 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 is to be the center of circle, to be the circle of radius with maximum hot spot apart from Fc with fixed point F, then 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 present embodiment is ρ ₀=af (1+tg ²Φ) ^3/2=195.324m, the circular arc line of ρ=202.706m is near ρ thus ₀=af (1+tg ²Φ) ^3/2The optically focused camber line; Than the optically focused ratio a little more than parabolic reflector, so can substitute parabolic reflector fully, other optically focused camber lines in like manner by above-mentioned optically focused visible, circular arc line catoptron when the catoptron area of section is identical; Because the calculating formula of of a great variety and its hot spot distance of optically focused camber line is with above-mentioned, except that parabolic reflector, the equal out-focus of other optically focused camber lines of position that begun, so do not enumerate one by one at this.

Second embodiment is the embodiment that fixed-point condensing reflector is formed solar cooker.Fig. 9 is split type solar cooker synoptic 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 with catoptron 3 and extinction body 2 is higher than catoptron 3, and promptly angle ψ＞0 of primary optical axis PF and local horizon NS, the primary optical axis PF projection line on ground level points to the due south, and promptly γ s=0 °, other structures are installed and waited explanation identical with Fig. 1.Figure 10 is the catoptron elevation angle, the position focused view that begun.Penetrate angle ψ＞0, the primary optical axis PF that make primary optical axis PF and local horizon NS in the face and points to the due south main, promptly γ s=0 °, angle of pitch ψ＞0, the mapping at other points, line, angle is identical with Fig. 2 with proof: promptly mistake F point is made the Z axle and is the main shaft sensing sun, is β ₀=h ₀=90 °-θ z ₀, get FO=f, be that true origin is set up the O-XYZ coordinate system with O.If the specific curves of catoptron is Y ²=2af (Z-B) para-curve is in the 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, promptly the 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; With 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 ²Get arbitrfary point M on the=2af (Z-B), then: ∠ 2=∠ 6=λ, M ₁F=FM ₃=FM, promptly F is Δ M ₁MM ₃Mid point, ∠ 6=∠ 3=∠ 2, promptly incident angle ∠ 2 equals reflection angle ∠ 3, the reflection ray of arbitrfary point M focus on the fixed point F, the F that promptly fixes a point is para-curve Y ²The focus of=2af (Z-B); In like manner, therefore ψ＜0 o'clock demonstration does not do repetition with above-mentioned.When having begun the position because solar rays be parallel incident and be parallel to main shaft, so this moment solar rays perpendicular to all xsects, promptly solar rays in the XOY coordinate surface for a bit; Learn by reflection law: on the minute surface reflected ray of arbitrfary point must with the normal of this point and incident ray coplane in the plane of incidence; Because catoptron is with specific curves Y ²To be bus do the circumference rotation around the Z axle to=2af (Z-B) para-curve forms minute surface, so all planes of incidence penetrate that face intersects at the Z axle and perpendicular to the XOY coordinate surface with main, so the angle of each point equates with the main angle of penetrating respective points in the face in all planes of incidence, the incident angle of each point equates in the same xsect, reflection angle equates and incident angle equals reflection angle, be that the incident angle of arbitrfary point equals reflection angle on the minute surface, so the reflection ray of the position incidence reflection mirror that begun focuses on the fixed point F certainly.

Figure 11 is moment high noon in Winter Solstice elevation angle catoptron optically focused figure.In main angle ψ＞0 of making PF and NS in the face of penetrating, the mapping of other points, line and Fig. 3,5 identical; When the earth drives catoptrons 3 and forwards noon position in Winter Solstice, catoptron principal normal PP by the position of having begun to around summit P by forwarding to high noon in the Summer Solstice the positive period of the day from 11 a.m. to 1 p.m in Winter Solstice, tracking means 6 ₁Reach the O-XYZ coordinate system and rotated 0.5 Δ δ in the face, at this moment θ z main penetrating _d=θ z ₀+ Δ δ, β _d=90 °-θ z ₀-0.5 Δ δ.Summit P: incident angle θ _d=θ ₀-0.5 Δ δ, reflection angle alpha _d=α ₀-θ .5 Δ δ is because θ ₀=α ₀So, θ _d=α _d, promptly the reflected ray of summit P will be along primary optical axis directive fixed point F.The arbitrfary point M that the 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 _d-λ-λ+0.5 Δ δ=90 °-2 λ-θ z ₀, because ∠ is M _PMM ₄+ ∠ M M ₄So P=ψ is ∠ MM ₄P=ψ+2 λ+θ z ₀-90 °; Because ∠ is M _PMM ₂=90 °-β _d, ∠ M ₄MM ₂=∠ M _PMM ₂+ ∠ M _PMM ₄=90 °-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 ₂PM 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。ψ＜0 o'clock: ∠ M ₄MP=∠ PMM ₂-∠ M ₄MM ₂, ∠ M ₄MM ₂=∠ M _PMM ₄-∠ M _PMM ₂δ-90 ° of=2 λ-0.5 Δ, all the other are o'clock identical with ψ＞0.The following arbitrfary point K of P point: ∠ A KK 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; ψ＜0 o'clock: ∠ 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 o'clock identical with ψ＞0.Therefore, the main constantly hot spot of the arbitrfary point distance of penetrating in the face on the para-curve all can be obtained high noon γ s=0 ° the time.The fixed point F coordinate figure: in the time of γ s=0 °, F (0, Y _P-fsin (β-ψ), fcos (β-ψ)+Z _P).When γ s=0 °, ψ ≠ 0 ° non-master penetrate in the face arbitrarily the hot spot of some distance when solving with γ s=0 °, ψ=0 ° in like manner, its ξ=ε+ψ+90 °-β just, see solving of Fg among Fig. 5 for details, do not do repetition at this.

Firing angle ε calculating formula is when angle of pitch ψ ≠ 0 °, γ s=0 °: ε=arc cos (cos ψ cosh cos γ+sin ψ sin h), 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 all be 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 ₁Rotate 0.5 Δ ε around summit P in the face main penetrating; The O-XYZ coordinate system rotates with catoptron, so each point on the para-curve and line do not become with respect to its value of O-XYZ coordinate system and angle.Elevation angle catoptron master penetrated face optically focused figure when Figure 12 was effective firing angle.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 is because θ ₀=α ₀So, θ=α, i.e. principal normal PP ₁Rotation is after the reflection ray that P is ordered 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 °, λ=arc tgY wherein _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 °-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.ψ＜0 o'clock: ∠ M _PMM ₂=0.5 Δ ε-ψ-θ z ₀, ∠ M ₄MM ₂=∠ M _PMM ₂+ ∠ M _PMM ₄=2 λ+0.5 Δ ε-90 °, ∠ M ₄M P=∠ PMM ₂-∠ M ₄M M ₂, all the other are o'clock identical with ψ＞0.Because ∠ 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 _KBecause ∠ 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.ψ＜0 o'clock: ∠ FPP ₂=0.5 Δ ε-ψ-θ z ₀, ∠ FPK=∠ KPP ₂-∠ FPP ₂, all the other are o'clock identical with ψ＞0.Therefore, any time master penetrates in the face on the para-curve hot spot of arbitrfary point distance and all can obtain γ s=0 ° the time.

Claims (3)

1, a kind of fixed-point condensing reflector, it is characterized in that with a specific curves being that bus is around main axis rotation and intercept certain area and form, follow the tracks of the solar time, it rotates the η degree around primary optical axis, rotate 0.5 Δ ε degree around summit P simultaneously, and regularly declination angle difference DELTA δ followed the tracks of.

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

3, the described fixed-point condensing reflector of claim 1 is characterized in that the normal and the three-way unification of light-guiding shade axis of described primary optical axis and extinction mouth, and the bottom surface of described light-guiding shade is less than the extinction mouth, and the inside surface of light-guiding shade glues the plating reflectance coating.

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