Quasi-optimum pseudo-lambertian reflecting concentrators: an analysis
|
|
- Amber Thomas
- 5 years ago
- Views:
Transcription
1 Quasi-optimum pseudo-lambertian reflecting concentrators: an analysis Antonio Luque n this work we analyze a two-stage concentrator in which the first stage is reflective and the second stage considers the first one as a Lambertian source in order to obtain the highest possible gain. We determine the profile and the position of this first stage, which happens to be a parabola, and we calculate the value of the gain as a function of the acceptance angle and the focal length of the first stage. 1. ntroduction The purpose of this paper is to analyze the first stage of a reflecting nature of a two-stage concentrator, whose second stage looks at the first one as if it were a Lambertian source.' t is assumed that the second stage is designed so that rays emitted by this source that penetrate through its entry aperture must reach the collector and that there exists conservation of etendue inside the second stage. The largest possible concentration of the system is obtained when the collector is illuminated by incident rays with leveling grade angle (900). n this case the etendue of the second stage is E = P, (1) where P is the collector's perimeter. n the case of monofacial and bifacial solar cells 3 submerged in a transparent medium of high index of refraction n the etendue is, respectively, E = nw, () E = 4nW, (3) where W is the width of the cell used. The use of bifacial solar cells submerged in a medium with a high index of refraction leads to the optimum photovoltaic concentration concept. 4 5 Figure 1 shows the relative arrangement of the first and second stages in schematic form. This type of ar- The author is with Universidad Politecnica de Madrid, nstituto de Energia Solar (ETST), Madrid-3, Spain. Received 5 January /80/ $00.50/ Optical Society of America. rangement, in which in the second stage solar cells were surrounded by air, was first suggested by Spectrolab and Arizona State University. 6 n Fig. 1 the second stage is suitable to obtain optimum photovoltaic concentration and therefore uses bifacial cells submerged in a medium with a high index of refraction. However, the nature of the second stage is irrelevant to this work. When the first stage is a reflecting surface that reflects a portion of the sky, considered in this case as a Lambertian source, the first stage is not itself a Lambertian, and therefore etendue E with which the second stage has been designed is higher than etendue E 1 produced by the sky on the first stage. As a consequence, the collector is bigger than this in an optimum system. n this work we try to analyze the gain of the proposed system and compare it with that of the optimum system to determine the shape and position of the first stage that gives a maximum gain. The technological interest of the configuration of Fig. 1 lies mainly in the reduction of the mirror area that can be obtained with respect to optimum concentrators of the compound parabolic concentrator (CPC) type 78 based on the fact that these concentrators intercept the rays at more oblique angles than in the concentrator under study. 11. Optimum Configuration n Fig. segment 00', which we assume for the moment to be of fixed length c, represents the secondstage entry aperture, while that of the first stage is represented by segments AA' with different subindices. The direction of an extreme ray is the vertical, while that of the other extreme ray is deflected an angle 0 clockwise. Etendue E of the second stage can be calculated using the method proposed by Winston and Welford 9 that results in 398 APPLED OPTCS / Vol. 19, No. 14 / 15 July 1980
2 ncident rays from sun where c is the length of 00'. from sun te o ncident rays G = D/c, (8) -j A. Maximum Concentration for a Given To obtain the maximum gain we must make the j! quotient D/E as large as possible, where D is the pro- '! j jection of segment AA' on the horizontal axis, and E is proportional to the cell's width (or to the perimeter of the collector). Let us assume that the extremes of primary mirror AA' are found in positions Al and A. Moving the points on their respective hyperbolae, the etendue of the second stage is not altered, arriving this way to points A and A of intersection with the outline of the allowed region and increasing D to its maximum value. Then, moving point A upward over vertical O'A the value Fig. 1. Quasi-optimum pseudo-lambertian reflecting concentrator of D is maintained, but the value of a is increased [we for bifacial solar cells. must keep in mind from Eq. (4) that the hyperbola's branches to the left of the MM' axis have negative values of semidiameter a], and the value of E is decreased. n this way we arrive at point A 3 of the intersection of said vertical with the vertical axis parabola and focus E = (A'O - A'') - (AO- AO') = a' - a. (4) at O' which crosses A. There is no continuous curve 1al locus f that could cross the vertical at a point above A This equation shows that the geometric 3 because points of equal etendue is constituted by] h locb o it would be necessary for the curve that represents the uiypterdas mirror to be at some point more horizontal than the from a set of hyperbolas that we will call e( with the foci at points 0 and 0', a' and c bein the parabola, and in this case it would reflect the vertical te ig he, ray that falls on that point into a direction that would largest semidiameters of the hyperbolas th at cross A' leave O to its left. The conclusion is that the vertical and A, respectively. )timum all axis parabola, with focus at O' that crosses A, is the On the other hand, if the system is to be 01 the sunrays comprised between the extrem( )timuo ll optimum outline for a given value of X (see Fig. ). intercept the entry aperture of the second stage. As a consequence, any mirror point X of the first stage must be the vertex of an oo' angle. This angle must be 0. This means that all points in the first-sstage mirror must be in the inside of the circle whose poihits subtend segment 00' with angle 0. AM For the same reason they must be to the left of the vertical that crosses O' so that the second sta ge does not shadow the first. The allowed space to extraames A and A' of the entry aperture of the first stage is indicated in Fig.. Let us suppose for the moment that, angle a (see Fig. ) of OO' with the vertical is given. Fo r given val- respect ues of 00' and 0 the circle and its position vvith to the vertical and consequently the allowedzone for the primary mirror are determined. f the extreme A' of the entry aperture is placed in a horizontal level above A, as is usually the case, the concentrator is not symmetric, and its may:imum gain occurs for the vertical beam of rays. The Nvalue GTH = DP = D/E, (5) where D is the projection of entry aperture AA' on the horizontal plane. According to Eqs. () and (3) the photovoltaic gain for monofacial and bifacial cells is respectively. is simply GPHM = ngth, (6) GPHB = ngth, (7) The gain of the first stage of this structure Fig.. Diagram of the position of the reflector of the first stage (points AA') with respect to the entry aperture 00' of the second stage. Only reflectors inside the shadowed contour can give high gain concentrators. The hyperbolas of equietendue are also represented by dotted lines. The outline of the reflector with the biggest gain is A3A July 1980 / Vol. 19, No. 14 / APPLED OPTCS 399
3 B. Maximum Concentration Outline We will try to analyze if it is better to move point A around the circumference in the direction of O' or in the opposite way. n the latter case a' decreases as A 3 goes down, which decreases a as well. t is impossible to predict if E increases or decreases without a more detailed analysis. Moreover, when A' moves around the circumference, projection D of mirror A A 3 on the horizontal plane changes, also changing the concentrator gain. An analysis of the variation of etendue E of the first-stage gain G 1 and of the thermodynamic gain GTH is accomplished in Appendix A as a function of angle o, which defines the position of point A over the circle that subtends 00' with angle 0. n Fig. 3 the GTH gain appears as a function of angle co of the A position for several values of acceptance angle 0 and of angle a of the second-stage position. t can be observed that the gain presents a maximum for small angles corresponding to parabolas in which point A 3 is located outside the circle subtending 00', and therefore they are unrealizable. When A 3 is outside the circle in Fig. 3, the gain appears as a dotted line. mmediately following there is a region of decreasing gain with co represented by a continuous line. Finally, there appears a minimum and last increasing portion that, in the case of high angular apertures, could give gains apparently higher than the optimum gain of the concentrator. n reality, these high gains cannot be obtained, since for very high angles X the mirror casts a shadow on itself hindering the rays' entrance through the 00' aperture. On the other hand, the concentrators in this region lack technological interest because they have their A point located above 0, which means they are deep concentrators that can be better achieved with the classical theory of the CPCs. The value of X at which A is placed over O' has been marked on each curve of Fig. 3. Beyond this point the concentrators are considered to be too deep. Besides, these concentrators present a very low first-stage gain G 1. As a consequence, the second stage must supply all the gain for the system; this would require a second stage of excessive dimensions. As a result, the only region of technological interest is the one with decreasing gain. t is clear that the value of co that gives the most gain is the smallest possible that locates the A 3 point inside the allowed region; this is in the A 4 position. The gain that can be achieved is the one that represents the transition from the dotted gain curve and the continuous curve, and it is somewhat. lower than the optimum gain. n this case the outline of the first stage is a vertical axis parabola with focus at O' that crosses point A 4. Point A3 constitutes the other extreme of the concentrator which is the best that can be obtained for a given orientation a of the second stage Quasi-Optimum Concentrations Orientation a of the second stage determines the position of the circle in Fig.. Since its radius is determined by c and 0, a also determines the focal distance O'A 4 of the parabolic concentrator of maximum gain that corresponds to angle 0 and the aperture horizontal projection D of said concentrator. The goal of this section is to determine the relationship between the maximum gain achievable and the relationship of focal distance to D (f/no.) of the concentrator. We also give the necessary parameters to build the concentrator. To obtain this relationship we will consider all the dimensions referred to the parabola's focal distance, which is taken as unity. The details of the calculations are given in Appendix B. The etendue of the second stage can be written as E = DS(D) sino - (D /4)(1 - coso), (9) where S(D) is a polynomial defined in Appendix B [Eq. (B)] in which S(O) = 1. The etendue of the first stage 9 is the difference of the projection of line A 4 A' 3 on a line inclined 0 with the vertical and on a vertical line. A simple calculation shows that this difference is E = D sino - (D /4)(1 - coso). () GTH E (D/4) tano 77 = -= (11) GTH E S(D)-(D/4) tano represents to which extent the two-stage concentrator differs from an ideal one (in which E 1 = E ). n the above expression the ideal thermodynamic gain is As we said previously, the gain of the concentrator is obtained from E by means of Eqs. (5)-(7). The quotient E 1 sino coso(1 - (D/4) tano) 1-1., - 11,,.. a hi U0 (1) Fig. 3. Gain of the concentrator as a function of position w of point A (see Fig. ). The parameters are the orientation a of the first stage and the acceptance angle APPLED OPTCS / Vol. 19, No. 14 / 15 July 1980
4 t can be observed that, when aperture D of the concentrator is very small, it approaches the ideal. The departure from ideality 1 is larger when acceptance semiangle 0 is high. For very high values of 0 it casts shadows over the rest of the concentrator so that Eqs. (9)-(1) become invalidated. The gain of the first stage is given by Eq. (8). Using expression c that appears in Eq. (B6) of Appendix B and also using the expression in Eq. (B3) in the same Appendix we have G, = D/c = -+-D + - D 4 1/ sino. (13) 3 56 / Curves with values of 1 and G 1 sino appear in Fig. 4 as a function of D. Even though 7j depends on acceptance angle 0, this dependence is very small for 0 <. The values of i7 shown in Fig. 3 have a margin of error smaller than ±1% for 0 <. Once these values are known, it is possible to design the first stage which will always be a vertical axis parabolic arc. ts horizontal projection D, normalized to the focal distance of the parabola, will be considered as an independent variable. The length of the aperture of the second stage will be obtained from G 1. The entry aperture of this second stage will contact its edges on the circle formed by the focus and the two edges of the parabolic arc, making sure that one of them coincides with the focus. Figure 5 shows values of the total gain of the concentrator as a function of D for several values of the semiangle of acceptance 0. V. Discussion The specific interest of this type of two-stage concentrator lies in the small dimension of the first stage, practically normal to the sunrays, as compared with the CPC in cases where high concentration and large acceptance angles are desired. The conditions for obtaining maximum gain with a second-stage concentrator have been discussed in detail. This gain increases when the F/No. of the first stage increases. The outline of the concentrator is, in this case, a vertical axis parabolic arc with focus at one of the second-stage entry aperture's edges. Though theoretically optimum this outline may not be the most desirable from a practical viewpoint. The reason for this is that any technological imperfection in the making of the mirror throws the extreme vertical rays that fall over the whole mirror, outside the second stage, resulting in a considerable reduction of the acceptance angle of the concentrator for rays near the extreme vertical ones. For the extreme rays falling with an inclination of 0 the problem is smaller since only those reflected near points A 4 and A 3 miss the second stage. The intermediate rays, even the ones outside the theoretical acceptance angle in fact, go into the second stage, producing a gradual fall of the effective concentration that extends outside the mirror's acceptance angle. n certain cases, it would be convenient to make the gain curve less asymmetric. This can be accomplished using curve arc A, A 4, located between points A in the circle in Fig., nearer O' than A3. Naturally, this system will have less gain and a larger acceptance angle 1.0. D ~ ~ * * Fig. 4. Approximate value to within +1% for semiacceptance angles smaller than of that of the structure's output GTH/GTHOPT and exact value of the gain of the first stage (multiplied by the sine of the semiacceptance angle) as a function of the first stage. 1- a 3 Go S ~ ~ ~ r 0 a 1 Fig. 5. Gain GTH as a function of entry aperture D (normalized to the focal length) with the semiangle of acceptance as the parameter. than the so-called optimum in this work. The etendue of the second stage of this concentrator increases not only because A' moves toward O' but because A 3 also moves toward A 4. The concentrator's outline is not fully defined in this case, giving additional freedom for the design that could take into account the possible errors in the fabrication of this stage. As the acceptance angle increases, the gain of the first stage decreases, which means that the second stage has a large entry aperture. This makes a very big second stage. The significance of the second-stage design is considerable, but the interest decreases as 0 takes higher values. Probably this interest is confined to acceptance semiangles lower than. want to thank Roland Winston for his helpful suggestions in planning this work, which was stimulated by the Ramon Areces Foundation. 15 July 1980 / Vol. 19, No. 14 / APPLED OPTCS 401
5 Appendix A The equation in polar coordinates with origin 0' and axis O'p (see Fig. ) of the equietendue hyperbolas is given by C- a a + c cosco (Al) When using this equation we adopted the formalism of not allowing negative values of p, considering outside the variability field of w all angles that do not meet this condition. This assignment to each value of a positive or negative sign, corresponds to only one branch of the hyperbola. The equations of the circle subtending 00' with an angle 0 and of the parabola of vertical axis and focus at O' are, respectively, c p = - sin(w + 0), (A) sin0 r 1 + cos(w-a) (A3) where r is the radial coordinate of point A 3 (see Fig. ). Given an arbitrary value of w corresponding to the A position, its radial coordinate p is given by Eq. (Al); r is obtained introducing the value p(w) in Eq. (A3): r = - sin(w + 0)[1 + cos(w - a)]. (A4) sin0 To calculate the value of a that corresponds to the equietendue hyperbola that contains point A 3 (r,a), we substitute r and a in Eq. (Al) and get (r + 4c - 4rc cosa) 1 - r (A5) Using an analogous procedure we can obtain the equietendue parabola corresponding to point A (P,w) getting (p + 4e - 4pc cos) 11 / - p (A6) This allows us to find the etendue of the second stage [see Eq. (4)]. The maximum gain of the concentrator appears when the rays fall vertically. n this case the entry aperture is D = p sin(w-a). (A7) The calculation of GTH is now performed by using Eq. (5) and represented vs w for several values of 0 and a in Fig. 3. n case the value of r, which is also calculated, is higher than its maximum allowable value rm = - sin(a + 0), (A8) sin0 curve GTH(w) is represented as a dotted line. Appendix B Setting the distance O'A 4 of Fig. to unity, the equation in rectangular coordinates of parabola A 4 A 3 and of the circle subtending 00' with angle 0, 00'- A3'A4 are given by, respectively, X + y - xxo-y = 0, x = 4y, (B1) (B) where x0 is the abscissa of the center of the circle. As both curves cross point A3 of abscissa D, it is possible to obtain the relation XO = -D + -D 3. (B3) 8 3 To calculate etendue E of the second stage it is necessary to calculate the length of segments A 4 0', A0', A 4 0 and A30. The first two are calculated using the expressions A 4 0' = 1 A 3 0' = 1 + (D /4). (B4) (B5) (This last equation establishes that the distance to the focus of a parabola's point is equal to the distance to the directrix.) For the calculation of A 4 0 and A30 it is necessary to solve triangles A 4 00' and A 3 00', which requires the previous calculation of 00'. This calculation is also needed for the gain of the first stage 00' = (1 + 4X) sin 0 = c. (B6) From the resolution of the first triangle it can be deduced that A 4 0' = cos0 - xo sin0. (B7) From the resolution of the second triangle it can be deduced that AO= (1 + D /4) cos0 + [(1 + D /4) + (1 + 4xo)11/ sin0. The etendue of the second stage, (B8) E = (OA3- 'A3) + ('A 4 -OAA (B9) using Eqs. (B3)-(B9), can be written in the form that is shown in Eq. (9). The function S(D) used there takes the value S(D) = 3 +D D + D 4 1/ \4 161 \ / (B) References 1. A. Rabl and R. Winston, Appl. Opt. 15, 880 (1976); R. Winston, U.S. Patent 3,957,031 (1976).. W. T. Welford and R. Winston, Optics of Nonimaging Concentrators (Academic, New York, 1978), pp A. Luque et al., in Proceedings, Photovoltaic Solar Energy Conference (D. Reidel, Dordrecht, Holland, 1978), pp A. Luque, Spanish Patent 453,575 (1976); U.S. Patent 4,169,738 (1979). 5. A. Luque et al., in Proceedings, COMPLES Conference, Milan, Cassa di Risparmio Calabria e Lucania, Comples talian Session, Roma, 1980; in press. 6. Ref., p R. Winston and H. Hinterberger, Sol. Energy 17, 55 (1975). 8. A. Luque et al., in Proceedings, SES Silver Jubilee Congress, K. W. B6er and B. H. Glenn, Eds. (Pergamon, New York, 1979), Vol. 3, pp , 9. R. Winston and W. T. Welford, J. Opt. Soc. Am. 68, 89 (1978). 40 APPLED OPTCS / Vol. 19, No. 14 / 15 July 1980
Finite Lambertian source analysis of concentrators: application to solar reflectors
Finite Lambertian source analysis of concentrators: application to solar reflectors A. Luque and J. M. Gmez The maximum power cast on a collector from a source of finite angular extension by a concentrator
More informationHigh Collection Nonimaging Optics
High Collection Nonimaging Optics W. T. WELFORD Optics Section Department of Physics Imperial College of Science, Technology and Medicine University of London London, England R. WINSTON Enrico Fermi Institute
More informationPARAMETRIC EQUATIONS AND POLAR COORDINATES
10 PARAMETRIC EQUATIONS AND POLAR COORDINATES PARAMETRIC EQUATIONS & POLAR COORDINATES 10.5 Conic Sections In this section, we will learn: How to derive standard equations for conic sections. CONIC SECTIONS
More information9.1 Circles and Parabolas. Copyright Cengage Learning. All rights reserved.
9.1 Circles and Parabolas Copyright Cengage Learning. All rights reserved. What You Should Learn Recognize a conic as the intersection of a plane and a double-napped cone. Write equations of circles in
More informationQUESTION BANK ON. CONIC SECTION (Parabola, Ellipse & Hyperbola)
QUESTION BANK ON CONIC SECTION (Parabola, Ellipse & Hyperbola) Question bank on Parabola, Ellipse & Hyperbola Select the correct alternative : (Only one is correct) Q. Two mutually perpendicular tangents
More informationPre-Calculus EOC Review 2016
Pre-Calculus EOC Review 2016 Name The Exam 50 questions, multiple choice, paper and pencil. I. Limits 8 questions a. (1) decide if a function is continuous at a point b. (1) understand continuity in terms
More informationThermal conversion of solar radiation. c =
Thermal conversion of solar radiation The conversion of solar radiation into thermal energy happens in nature by absorption in earth surface, planetary ocean and vegetation Solar collectors are utilized
More information(D) (A) Q.3 To which of the following circles, the line y x + 3 = 0 is normal at the point ? 2 (A) 2
CIRCLE [STRAIGHT OBJECTIVE TYPE] Q. The line x y + = 0 is tangent to the circle at the point (, 5) and the centre of the circles lies on x y = 4. The radius of the circle is (A) 3 5 (B) 5 3 (C) 5 (D) 5
More informationThe Distance Formula. The Midpoint Formula
Math 120 Intermediate Algebra Sec 9.1: Distance Midpoint Formulas The Distance Formula The distance between two points P 1 = (x 1, y 1 ) P 2 = (x 1, y 1 ), denoted by d(p 1, P 2 ), is d(p 1, P 2 ) = (x
More informationAngles and Applications
CHAPTER 1 Angles and Applications 1.1 Introduction Trigonometry is the branch of mathematics concerned with the measurement of the parts, sides, and angles of a triangle. Plane trigonometry, which is the
More informationTECHNICAL CONCEPT AND SOME POSSIBLE APPLICATIONS FOR HIGH-TEMPERATURE PARABOLIC BLIND-REFLECTING SOLAR CONCENTRATORS
EuroSun 98 III.2.52-1 V.Vasylyev TECHNICAL CONCEPT AND SOME POSSIBLE APPLICATIONS FOR HIGH-TEMPERATURE PARABOLIC BLIND-REFLECTING SOLAR CONCENTRATORS VIKTOR VASYLYEV Solar-Environmental Res. Center, Institute
More informationDesign Optimisation of Compound Parabolic Concentrator (CPC) for Improved Performance M. M. Isa, R. Abd-Rahman, H. H. Goh
Design Optimisation of Compound Parabolic Concentrator (CPC) for Improved Performance M. M. Isa, R. Abd-Rahman, H. H. Goh Abstract A compound parabolic concentrator (CPC) is a wellknown non-imaging concentrator
More informationJanuary 21, 2018 Math 9. Geometry. The method of coordinates (continued). Ellipse. Hyperbola. Parabola.
January 21, 2018 Math 9 Ellipse Geometry The method of coordinates (continued) Ellipse Hyperbola Parabola Definition An ellipse is a locus of points, such that the sum of the distances from point on the
More informationAdvanced Math. ABSOLUTE VALUE - The distance of a number from zero; the positive value of a number. < 2 indexes draw 2 lines down like the symbol>
Advanced Math ABSOLUTE VALUE - The distance of a number from zero; the positive value of a number. < 2 indexes draw 2 lines down like the symbol> ALGEBRA - A branch of mathematics in which symbols, usually
More informationEngineering Physics 1 Prof. G.D. Vermaa Department of Physics Indian Institute of Technology-Roorkee
Engineering Physics 1 Prof. G.D. Vermaa Department of Physics Indian Institute of Technology-Roorkee Module-04 Lecture-02 Diffraction Part - 02 In the previous lecture I discussed single slit and double
More information1. Graph each of the given equations, state the domain and range, and specify all intercepts and symmetry. a) y 3x
MATH 94 Final Exam Review. Graph each of the given equations, state the domain and range, and specify all intercepts and symmetry. a) y x b) y x 4 c) y x 4. Determine whether or not each of the following
More informationIntroduction to Computer Graphics (Lecture No 07) Ellipse and Other Curves
Introduction to Computer Graphics (Lecture No 07) Ellipse and Other Curves 7.1 Ellipse An ellipse is a curve that is the locus of all points in the plane the sum of whose distances r1 and r from two fixed
More informationQUESTION BANK ON STRAIGHT LINE AND CIRCLE
QUESTION BANK ON STRAIGHT LINE AND CIRCLE Select the correct alternative : (Only one is correct) Q. If the lines x + y + = 0 ; 4x + y + 4 = 0 and x + αy + β = 0, where α + β =, are concurrent then α =,
More informationChapter 10: Conic Sections; Polar Coordinates; Parametric Equations
Chapter 10: Conic Sections; Polar Coordinates; Parametric Equations Section 10.1 Geometry of Parabola, Ellipse, Hyperbola a. Geometric Definition b. Parabola c. Ellipse d. Hyperbola e. Translations f.
More informationSOLAR ENERGY CONVERSION AND PHOTOENERGY SYSTEMS Vol. I - High Temperature Solar Concentrators - Robert Pitz-Paal HIGH TEMPERATURE SOLAR CONCENTRATORS
HIGH TEMPERATURE SOLAR CONCENTRATORS Robert Institute of Technical Thermodynamics, German Aerospace Center (DLR), Germany Keywords: Optical concentration ratio, central receiver systems, dish/stirling,
More informationConcentration of fiber transmitted solar energy by CPC for solar thermal utilization
J Phys. IV France 9 (1999) Concentration of fiber transmitted solar energy by CPC for solar thermal utilization H. Yugami, M. Yano, H. Naito and H. ~ rashi~ Graduate School of Engineering, Tohoku University,
More informationPortable Assisted Study Sequence ALGEBRA IIB
SCOPE This course is divided into two semesters of study (A & B) comprised of five units each. Each unit teaches concepts and strategies recommended for intermediate algebra students. The second half of
More informationALGEBRA 2 X. Final Exam. Review Packet
ALGEBRA X Final Exam Review Packet Multiple Choice Match: 1) x + y = r a) equation of a line ) x = 5y 4y+ b) equation of a hyperbola ) 4) x y + = 1 64 9 c) equation of a parabola x y = 1 4 49 d) equation
More informationOrbit Characteristics
Orbit Characteristics We have shown that the in the two body problem, the orbit of the satellite about the primary (or vice-versa) is a conic section, with the primary located at the focus of the conic
More informationMCPS Algebra 2 and Precalculus Standards, Categories, and Indicators*
Content Standard 1.0 (HS) Patterns, Algebra and Functions Students will algebraically represent, model, analyze, and solve mathematical and real-world problems involving functional patterns and relationships.
More informationIntermediate Math Circles Wednesday, April 5, 2017 Problem Set 8
Intermediate Math Circles Wednesday, April 5, 2017 Problem Set 8 1. Determine the coordinates of the vertices and foci for each of the following ellipses. (a) + 9y 2 = 36 We want equation to be of the
More informationDAY 139 EQUATION OF A HYPERBOLA
DAY 139 EQUATION OF A HYPERBOLA INTRODUCTION In our prior conic sections lessons, we discussed in detail the two conic sections, the parabola, and the ellipse. The hyperbola is another conic section we
More informationSTEM-Prep Pathway SLOs
STEM-Prep Pathway SLOs Background: The STEM-Prep subgroup of the MMPT adopts a variation of the student learning outcomes for STEM from the courses Reasoning with Functions I and Reasoning with Functions
More informationTARGET : JEE 2013 SCORE. JEE (Advanced) Home Assignment # 03. Kota Chandigarh Ahmedabad
TARGT : J 01 SCOR J (Advanced) Home Assignment # 0 Kota Chandigarh Ahmedabad J-Mathematics HOM ASSIGNMNT # 0 STRAIGHT OBJCTIV TYP 1. If x + y = 0 is a tangent at the vertex of a parabola and x + y 7 =
More informationAlgebra and Trigonometry
Algebra and Trigonometry 978-1-63545-098-9 To learn more about all our offerings Visit Knewtonalta.com Source Author(s) (Text or Video) Title(s) Link (where applicable) OpenStax Jay Abramson, Arizona State
More informationMAT1035 Analytic Geometry
MAT1035 Analytic Geometry Lecture Notes R.A. Sabri Kaan Gürbüzer Dokuz Eylül University 2016 2 Contents 1 Review of Trigonometry 5 2 Polar Coordinates 7 3 Vectors in R n 9 3.1 Located Vectors..............................................
More informationConic Sections Session 2: Ellipse
Conic Sections Session 2: Ellipse Toh Pee Choon NIE Oct 2017 Toh Pee Choon (NIE) Session 2: Ellipse Oct 2017 1 / 24 Introduction Problem 2.1 Let A, F 1 and F 2 be three points that form a triangle F 2
More informationPreCalculus Honors Curriculum Pacing Guide First Half of Semester
Unit 1 Introduction to Trigonometry (9 days) First Half of PC.FT.1 PC.FT.2 PC.FT.2a PC.FT.2b PC.FT.3 PC.FT.4 PC.FT.8 PC.GCI.5 Understand that the radian measure of an angle is the length of the arc on
More informationExample 1: Finding angle measures: I ll do one: We ll do one together: You try one: ML and MN are tangent to circle O. Find the value of x
Ch 1: Circles 1 1 Tangent Lines 1 Chords and Arcs 1 3 Inscribed Angles 1 4 Angle Measures and Segment Lengths 1 5 Circles in the coordinate plane 1 1 Tangent Lines Focused Learning Target: I will be able
More informationTelescopes and Optics II. Observational Astronomy 2017 Part 4 Prof. S.C. Trager
Telescopes and Optics II Observational Astronomy 2017 Part 4 Prof. S.C. Trager Fermat s principle Optics using Fermat s principle Fermat s principle The path a (light) ray takes is such that the time of
More informationConic Sections Session 3: Hyperbola
Conic Sections Session 3: Hyperbola Toh Pee Choon NIE Oct 2017 Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 1 / 16 Problem 3.1 1 Recall that an ellipse is defined as the locus of points P such that
More informationCenterville High School Curriculum Mapping Algebra II 1 st Nine Weeks
Centerville High School Curriculum Mapping Algebra II 1 st Nine Weeks Chapter/ Lesson Common Core Standard(s) 1-1 SMP1 1. How do you use a number line to graph and order real numbers? 2. How do you identify
More informationPolarization properties of corner-cube retroreflectors: Theory and experiment
University of New Orleans ScholarWorks@UNO Electrical Engineering Faculty Publications Department of Electrical Engineering 3--997 Polarization properties of corner-cube retroreflectors: Theory and experiment
More informationApril 30, Name: Amy s Solutions. Discussion Section: N/A. Discussion TA: N/A
Math 1151, April 30, 010 Exam 3 (in-class) Name: Amy s Solutions Discussion Section: N/A Discussion TA: N/A This exam has 8 multiple-choice problems, each worth 5 points. When you have decided on a correct
More informationMATH Max-min Theory Fall 2016
MATH 20550 Max-min Theory Fall 2016 1. Definitions and main theorems Max-min theory starts with a function f of a vector variable x and a subset D of the domain of f. So far when we have worked with functions
More informationChetek-Weyerhaeuser High School
Chetek-Weyerhaeuser High School Advanced Math A Units and s Advanced Math A Unit 1 Functions and Math Models (7 days) 10% of grade s 1. I can make connections between the algebraic equation or description
More information3. A( 2,0) and B(6, -2), find M 4. A( 3, 7) and M(4,-3), find B. 5. M(4, -9) and B( -10, 11) find A 6. B(4, 8) and M(-2, 5), find A
Midpoint and Distance Formula Class Work M is the midpoint of A and B. Use the given information to find the missing point. 1. A(4, 2) and B(3, -8), find M 2. A(5, 7) and B( -2, -9), find M 3. A( 2,0)
More informationAn Evacuated PV/Thermal Hybrid Collector with the Tube/XCPC design
An Evacuated PV/Thermal Hybrid Collector with the Tube/XCPC design Lun Jiang Chuanjin Lan Yong Sin Kim Yanbao Ma Roland Winston University of California, Merced 4200 N.Lake Rd, Merced CA 95348 ljiang2@ucmerced.edu
More informationUnderstanding a halo is sometimes easier from the perspective of the halomaking
APPENDIX F Living on the (w)edge Understanding a halo is sometimes easier from the perspective of the halomaking wedge this is the insight that underlies the theory of halo poles. In this appendix we use
More informationFor a semi-circle with radius r, its circumfrence is πr, so the radian measure of a semi-circle (a straight line) is
Radian Measure Given any circle with radius r, if θ is a central angle of the circle and s is the length of the arc sustained by θ, we define the radian measure of θ by: θ = s r For a semi-circle with
More informationFresnel Number Concept and Revision of some Characteristics in the Linear Theory of Focused Acoustic Beams
resnel umber Concept and Revision of some Characteristics in the Linear Theory of ocused Acoustic Beams Yu.. Makov 1 and V.J. Sánchez-Morcillo 1 Department of Acoustics, aculty of Physics, Moscow State
More informationy 1 x 1 ) 2 + (y 2 ) 2 A circle is a set of points P in a plane that are equidistant from a fixed point, called the center.
Ch 12. Conic Sections Circles, Parabolas, Ellipses & Hyperbolas The formulas for the conic sections are derived by using the distance formula, which was derived from the Pythagorean Theorem. If you know
More informationAnalysis of reflected intensities of linearly polarized electromagnetic plane waves on parabolic boundary surfaces with different focal lengths
DOI 10.1007/s1596-013-0156-7 RESEARCH ARTICLE Analysis of reflected intensities of linearly polarized electromagnetic plane waves on parabolic boundary surfaces with different focal lengths Hossein Arbab
More informationDistance and Midpoint Formula 7.1
Distance and Midpoint Formula 7.1 Distance Formula d ( x - x ) ( y - y ) 1 1 Example 1 Find the distance between the points (4, 4) and (-6, -). Example Find the value of a to make the distance = 10 units
More information1. INTRODUCTION receiver concentrator Compound Parabolic Concentrator CPC
OPTIMIZATION OF HELIOSTAT FIELD LAYOUT FOR THE BEAM DOWN OPTICS by Akiba Segal Weizmann Institute of Science Chemical Research Support Department Solar Optics Design and Mathematical Modeling Unit Report
More informationUnit 1. Revisiting Parent Functions and Graphing
Unit 1 Revisiting Parent Functions and Graphing Precalculus Analysis Pacing Guide First Nine Weeks Understand how the algebraic properties of an equation transform the geometric properties of its graph.
More informationMATH-1420 Review Concepts (Haugen)
MATH-40 Review Concepts (Haugen) Unit : Equations, Inequalities, Functions, and Graphs Rational Expressions Determine the domain of a rational expression Simplify rational expressions -factor and then
More informationNumerical study of optical performance of a parabolic-trough concentrating solar power system
Numerical study of optical performance of a parabolic-trough concentrating solar power system *Raquel Miguez de Carvalho 1), Mavd R. Teles ) and Kamal A. R. Ismail 3) 1), ),3) Department of Energy, Faculty
More informationWritten test, 25 problems / 90 minutes
Sponsored by: UGA Math Department and UGA Math Club Written test, 5 problems / 90 minutes October, 06 WITH SOLUTIONS Problem. Let a represent a digit from to 9. Which a gives a! aa + a = 06? Here aa indicates
More informationLIGHT. A beam is made up of several rays. It maybe parallel, diverging (spreading out) or converging (getting narrower). Parallel Diverging Converging
LIGHT Light is a form of energy. It stimulates the retina of the eye and produces the sensation of sight. We see an object when light leaves it and enters the eye. Objects such as flames, the sum and stars
More informationSince x + we get x² + 2x = 4, or simplifying it, x² = 4. Therefore, x² + = 4 2 = 2. Ans. (C)
SAT II - Math Level 2 Test #01 Solution 1. x + = 2, then x² + = Since x + = 2, by squaring both side of the equation, (A) - (B) 0 (C) 2 (D) 4 (E) -2 we get x² + 2x 1 + 1 = 4, or simplifying it, x² + 2
More informationCONIC SECTIONS TEST FRIDAY, JANUARY 5 TH
CONIC SECTIONS TEST FRIDAY, JANUARY 5 TH DAY 1 - CLASSIFYING CONICS 4 Conics Parabola Circle Ellipse Hyperbola DAY 1 - CLASSIFYING CONICS GRAPHICALLY Parabola Ellipse Circle Hyperbola DAY 1 - CLASSIFYING
More informationChapter 1. Functions 1.3. Trigonometric Functions
1.3 Trigonometric Functions 1 Chapter 1. Functions 1.3. Trigonometric Functions Definition. The number of radians in the central angle A CB within a circle of radius r is defined as the number of radius
More informationMaths for Map Makers
SUB Gottingen 7 210 050 861 99 A 2003 Maths for Map Makers by Arthur Allan Whittles Publishing Contents /v Chapter 1 Numbers and Calculation 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14
More informationConic Sections. Geometry - Conics ~1~ NJCTL.org. Write the following equations in standard form.
Conic Sections Midpoint and Distance Formula M is the midpoint of A and B. Use the given information to find the missing point. 1. A(, 2) and B(3, -), find M 2. A(5, 7) and B( -2, -), find M 3. A( 2,0)
More informationPC.FT.3 Use special triangles to determine geometrically the values of sine, cosine, tangent for π 3, π 4, and π 6,
FIRST NINE WEEKS Text: Blitzer Pre-Calculus Chapters 4, 5, 6 Unit 1 Introduction to : Sections 4.1, 4.2, 4.3, 4.4 PC.FT.1 Understand that the radian measure of an angle is the length of the arc on the
More informationRadiation from planets
Chapter 4 Radiation from planets We consider first basic, mostly photometric radiation parameters for solar system planets which can be easily compared with existing or future observations of extra-solar
More informationConic section. Ans: c. Ans: a. Ans: c. Episode:43 Faculty: Prof. A. NAGARAJ. 1. A circle
Episode:43 Faculty: Prof. A. NAGARAJ Conic section 1. A circle gx fy c 0 is said to be imaginary circle if a) g + f = c b) g + f > c c) g + f < c d) g = f. If (1,-3) is the centre of the circle x y ax
More informationax 2 + bx + c = 0 where
Chapter P Prerequisites Section P.1 Real Numbers Real numbers The set of numbers formed by joining the set of rational numbers and the set of irrational numbers. Real number line A line used to graphically
More informationTrigonometric Functions and Triangles
Trigonometric Functions and Triangles Dr. Philippe B. Laval Kennesaw STate University Abstract This handout defines the trigonometric function of angles and discusses the relationship between trigonometric
More information1 Geometry of R Conic Sections Parametric Equations More Parametric Equations Polar Coordinates...
Contents 1 Geometry of R 1.1 Conic Sections............................................ 1. Parametric Equations........................................ 3 1.3 More Parametric Equations.....................................
More informationFundamentals of Mathematics (MATH 1510)
Fundamentals of Mathematics () Instructor: Email: shenlili@yorku.ca Department of Mathematics and Statistics York University March 14-18, 2016 Outline 1 2 s An angle AOB consists of two rays R 1 and R
More information2009 Math Olympics Level II Solutions
Saginaw Valley State University 009 Math Olympics Level II Solutions 1. f (x) is a degree three monic polynomial (leading coefficient is 1) such that f (0) 3, f (1) 5 and f () 11. What is f (5)? (a) 7
More informationSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Math 170 Final Exam Review Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Evaluate the function at the given value of the independent variable and
More informationSemester 2 Final Review
Name: Date: Per: Unit 6: Radical Functions [1-6] Simplify each real expression completely. 1. 27x 2 y 7 2. 80m n 5 3. 5x 2 8x 3 y 6 3. 2m 6 n 5 5. (6x 9 ) 1 3 6. 3x 1 2 8x 3 [7-10] Perform the operation
More informationVector diffraction theory of refraction of light by a spherical surface
S. Guha and G. D. Gillen Vol. 4, No. 1/January 007/J. Opt. Soc. Am. B 1 Vector diffraction theory of refraction of light by a spherical surface Shekhar Guha and Glen D. Gillen* Materials and Manufacturing
More informationAstro 500 A500/L-7 1
Astro 500 1 Telescopes & Optics Outline Defining the telescope & observatory Mounts Foci Optical designs Geometric optics Aberrations Conceptually separate Critical for understanding telescope and instrument
More informationPHY 205 Final Exam 6/24/2009 Second Semester2008 Part 1.
Part 1. Please read each question carefully. Each question worth s 1 point. or the following questions, please circle the correct answer. 1. Which one of the following statements concerning the index of
More informationPage 1
Pacing Chart Unit Week Day CCSS Standards Objective I Can Statements 121 CCSS.MATH.CONTENT.HSG.C.A.1 Prove that all circles are similar. Prove that all circles are similar. I can prove that all circles
More informationDesigning a Computer Generated Hologram for Testing an Aspheric Surface
Nasrin Ghanbari OPTI 521 Graduate Report 2 Designing a Computer Generated Hologram for Testing an Aspheric Surface 1. Introduction Aspheric surfaces offer numerous advantages in designing optical systems.
More informationEffects of Acceptance angle and Receiver radius on the Optical Performance of Compound Parabolic Collector
Nigerian Journal of Solar Energy, Vol. 28, 2017. Solar Energy Society of Nigeria (SESN) 2017. All rights reserved. Effects of Acceptance angle and Receiver radius on the Optical Performance of Compound
More information2. Find the side lengths of a square whose diagonal is length State the side ratios of the special right triangles, and
1. Starting at the same spot on a circular track that is 80 meters in diameter, Hayley and Kendall run in opposite directions, at 300 meters per minute and 240 meters per minute, respectively. They run
More informationTennessee s State Mathematics Standards Precalculus
Tennessee s State Mathematics Standards Precalculus Domain Cluster Standard Number Expressions (N-NE) Represent, interpret, compare, and simplify number expressions 1. Use the laws of exponents and logarithms
More informationMAT100 OVERVIEW OF CONTENTS AND SAMPLE PROBLEMS
MAT100 OVERVIEW OF CONTENTS AND SAMPLE PROBLEMS MAT100 is a fast-paced and thorough tour of precalculus mathematics, where the choice of topics is primarily motivated by the conceptual and technical knowledge
More information-,- 2..J. EXAMPLE 9 Discussing the Equation of a Parabola. Solution
670 CHAPTER 9 Analtic Geometr Polnomial equations define parabolas whenever the involve two variables that are quadratic in one variable and linear in the other. To discuss this tpe of equation, we first
More informationIUPUI Department of Mathematical Sciences Departmental Final Examination PRACTICE FINAL EXAM VERSION #1 MATH Trigonometry
IUPUI Department of Mathematical Sciences Departmental Final Examination PRACTICE FINAL EXAM VERSION #1 MATH 15400 Trigonometry Exam directions similar to those on the departmental final. 1. DO NOT OPEN
More informationOrthonormal vector polynomials in a unit circle, Part I: basis set derived from gradients of Zernike polynomials
Orthonormal vector polynomials in a unit circle, Part I: basis set derived from gradients of ernike polynomials Chunyu hao and James H. Burge College of Optical ciences, the University of Arizona 630 E.
More informationan off-summit observer
Mountain shadow phenomena. an off-summit observer 2: The spike seen by David K. Lynch The oblique spike or contrast edge seen by an off-summit observer on a mountain shadow when the sun is low is shown
More informationGeometrical optics and blackbody radiation Pablo BenÍTez ab ; Roland Winston a ;Juan C. Miñano b a
This article was downloaded by: [University of California, Merced] On: 6 May 2010 Access details: Access Details: [subscription number 918975015] ublisher Taylor & Francis Informa Ltd Registered in England
More informationFunctions and their Graphs
Chapter One Due Monday, December 12 Functions and their Graphs Functions Domain and Range Composition and Inverses Calculator Input and Output Transformations Quadratics Functions A function yields a specific
More informationMath 2414 Activity 1 (Due by end of class Jan. 26) Precalculus Problems: 3,0 and are tangent to the parabola axis. Find the other line.
Math Activity (Due by end of class Jan. 6) Precalculus Problems: 3, and are tangent to the parabola ais. Find the other line.. One of the two lines that pass through y is the - {Hint: For a line through
More information3. A( 2,0) and B(6, -2), find M 4. A( 3, 7) and M(4,-3), find B. 5. M(4, -9) and B( -10, 11) find A 6. B(4, 8) and M(-2, 5), find A
Midpoint and Distance Formula Class Work M is the midpoint of A and B. Use the given information to find the missing point. 1. A(, 2) and B(3, -8), find M 2. A(5, 7) and B( -2, -), find M (3. 5, 3) (1.
More informationChapter 8B - Trigonometric Functions (the first part)
Fry Texas A&M University! Spring 2016! Math 150 Notes! Section 8B-I! Page 79 Chapter 8B - Trigonometric Functions (the first part) Recall from geometry that if 2 corresponding triangles have 2 angles of
More informationDefinition (Polar Coordinates) Figure: The polar coordinates (r, )ofapointp
Polar Coordinates Acoordinatesystemusesapairofnumberstorepresentapointontheplane. We are familiar with the Cartesian or rectangular coordinate system, (x, y). It is not always the most convenient system
More information4.Let A be a matrix such that A. is a scalar matrix and Then equals :
1.Consider the following two binary relations on the set A={a, b, c} : R1={(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)} and R2={(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}. Then : both R1
More informationHomework Assignments Math /02 Fall 2014
Homework Assignments Math 119-01/02 Fall 2014 Assignment 1 Due date : Friday, September 5 6th Edition Problem Set Section 6.1, Page 178: #1, 2, 3, 4, 5, 6. Section 6.2, Page 185: #1, 2, 3, 5, 6, 8, 10-14,
More information1 Sets of real numbers
1 Sets of real numbers Outline Sets of numbers, operations, functions Sets of natural, integer, rational and real numbers Operations with real numbers and their properties Representations of real numbers
More informationMathematics Precalculus: Academic Unit 7: Conics
Understandings Questions Knowledge Vocabulary Skills Conics are models of real-life situations. Conics have many reflective properties that are used in every day situations Conics work can be simplified
More informationThermal Analysis of Solar Collectors
Thermal Analysis of Solar Collectors Soteris A. Kalogirou Cyprus University of Technology Limassol, Cyprus Contents Types of collectors Stationary Sun tracking Thermal analysis of collectors Flat plate
More informationCIRCLES: #1. What is an equation of the circle at the origin and radius 12?
1 Pre-AP Algebra II Chapter 10 Test Review Standards/Goals: E.3.a.: I can identify conic sections (parabola, circle, ellipse, hyperbola) from their equations in standard form. E.3.b.: I can graph circles
More informationMaterial and Design Requirements for Advanced Concentrators
Advances in Science and Technology Online: 2010-10-27 ISSN: 1662-0356, Vol. 74, pp 237-242 doi:10.4028/www.scientific.net/ast.74.237 2010 Trans Tech Publications, Switzerland Material and Design Requirements
More informationMath 5 Trigonometry Review Sheet for Chapter 5
Math 5 Trigonometry Review Sheet for Chapter 5 Key Ideas: Def: Radian measure of an angle is the ratio of arclength subtended s by that central angle to the radius of the circle: θ s= rθ r 180 = π radians.
More informationRAY TRAJECTORIES, BINOMIAL OF A NEW TYPE, AND THE BINARY SYSTEM
Alexander Yurkin A.M. Prokhorov General Physics Institute of the Russian Academy of Sciences e-mail: yurkin@fpl.gpi.ru RAY TRAJECTORIES, BINOMIAL OF A NEW TYPE, AND THE BINARY SYSTEM Abstract The paper
More informationMATH 32 FALL 2013 FINAL EXAM SOLUTIONS. 1 cos( 2. is in the first quadrant, so its sine is positive. Finally, csc( π 8 ) = 2 2.
MATH FALL 01 FINAL EXAM SOLUTIONS (1) (1 points) Evalute the following (a) tan(0) Solution: tan(0) = 0. (b) csc( π 8 ) Solution: csc( π 8 ) = 1 sin( π 8 ) To find sin( π 8 ), we ll use the half angle formula:
More information