



Answers to the quiz questions in the book "Candelas Lumens and Lux" Q1) Light of intensity 500 cd flows through a solid angle of 0.5, how many lumens is this? Flux = Intensity*SolidAngle = 500x0.5 = 250lm Q2) 6000 lm flow through a solid angle of 0.25 steradians, what is the luminous intensity of the light? You need to rearrange the formula in the book to get Intensity on the left: Intensity = Flux/SolidAngle =6000/0.25 = 24,000 cd Q3) A solid angle of 0.001 steradians has light of luminous intensity 1000 cd flowing through it, what is the flux? Flux = Intensity*SolidAngle = 1000x0.001 = 1 lm Q4) A ray of luminous intensity of 500 cd hits a surface 5 meters distant what is the illuminance? The ray is perpendicular to the surface. Illuminance = Intensity/(Distance*Distance) = 500/(5x5) = 20 lx Q5) A ray of luminous intensity of 1500 cd hits a surface 40 meters distant, what is the illuminance? The ray is perpendicular to the surface. Illuminance = Intensity/(Distance*Distance) = 1500/(40x40) = 0.936 lx Q6) A ray of luminous intensity of 750 cd hits a surface 2.5 meters distant, what is the illuminance? The ray is perpendicular to the surface. Illuminance = Intensity/(Distance*Distance) = 120 lx Q7) A luminaire emits a luminous intensity of 600 cd which hits a surface 2 meters distance at an incident angle of 30°. What is the illuminance at that point? In the book there are two ways of calculating the illuminance of a ray inclined to a surface, which is best depends on the data given. Here we have a distance, so we will use the formula with d in it: Illumination = Intensity*cos(alpha)/(dist*dist) = 600*cos(30)/4 = 129.9 lx Q8) A luminaire emits, at an incident angle of 70°, a luminous intensity of 1000 cd which hits a flat surface. The luminaire hangs 2.5 meters above the surface. What is the illuminance? Here we have a height and an angle (rather than a distance and an angle) so we can use the cosine cubed formula easier: Illuminance = Intensity*(cos(alpha)cubed)/(height*height) = 1000*0.040/(6.25) = 6.4 lx Q9) A LED emits a luminous intensity of 5 candela which hits a flat surface 3 centimeters distant at an incident angle of 15°. What is the illuminance at that point? Again, we get the distance from the surface rather than the height above the surface, so we can use this formula Illuminance = Intensity*cos(alpha)/(dist*dist) = 5xcos(15)/(0.03x0.03) = 5366 lx Q10) A LED emits, at a gamma angle of 10°, a luminous intensity of 3 cd and hits a flat surface. The LED is 1 cm alve the surface. What is the illuminance at the given point? Here we are given the height of the LED above the surface, rather than the distance from the LED to the point we need to calculated. So it will be easiest to use the cosine cubed formula: Illuminance = Intensity*(cos(alpha)cubed)/(height*height) = 3x0.955/(0.01x0.1) = 28650 lx Q11) A surface with reflectance 80% is illuminated to 450lx. What is its luminance? Luminance = Reflectance x Illuminance / Pi = 450 x 0.8/3.142 = 114.6 cd/m2rd Q12) A point on a surface is hit by a ray of light of 1000cd. The angle of incidence is 55°. The surface is 2m perpendicularly above the light and has a reflectance of 0.5. What is the illuminance and luminance? First you have to work out the illuminance, you have a height, so use the cosine cubed formula: Illuminance = Intensity*(cos(alpha)cubed)/(height*height) = 1000*0.189/(2*2) = 47.2 lx Now you can work out the luminance at that point: Luminance = Reflectance * Illuminance / Pi = 0.5 * 47.2 /3.142 = 7.508 cd/m2rd Q13) Imagine the luminaire which has a polar diagram shown on the previous page, has a version of 20,000 lm (or 20klm). What is the intensity, in candelas, at a gamma angle of 43° for that version of the luminaire? You'll need to open the book at page 62 to see the polar diagram. The upper diagram shows that at 43° there are 240 candelas per kilolumen. So if the luminaire has 20 klm you'll have 240 x 20 = 480 candelas at 43°. Q14) Imagine the same luminaire, has a version of 5klm. What is the intensity in candelas at a gamma angle of 45° for that version of the luminaire? Again, open the book at page 62, and you'll see in the cd/klm diagram that at 45° we have about 210 cd/klm, which means we have 210 * 5 = 1050 cd there. At the end of chapter 4, Internal Lighting, page 83, I ask you this: A 4m x 8m room's work surface is to be illuminated to 300lx. The room is 3.2m high, and the work surface is at 0.8m. The luminaires emitting areas hang at 0.2 below the ceiling. Assume that we are using the same luminaire as in the previous example in the book (so use the same flux, 6000lm, and the same UF table). The frieze has a reflectance of 0.7, the work surface has a reflectance of 0.1, the ceiling has a reflectance of 0.8 and the walls have a reflectance of 0.7. The maintenance factor is 0.8. As explained in the chapter you'll need to fnd the Room Index and from that the Utilization Factor. Room Index = (4 x 8)/(2.2 x (4 + 8)) = 1.21 2.2 is the mounting height, how far away the luminaire is from the worksurface. The column to use in the UF factor table is 8771 (80% for the ceiling, 70% for the frieze, 70% for the walls and 10% for the worksurface). There isn't a room index row of exactly 1.21, and we use the closest row of 1.25. So, looking up Room Index row of 1.25 and reflectancies column of 8871, we find a UF of 0.491. So the flux installed needs to be FluxInstalled = (300 x 4 x 8) / (0.8 x 0.491) = 24,440 lumens. Since the example luminaire in the book has 6000 lm we will need: Number of luminaires = 24,440/6000 = 4.07, round down to 4. US Road lightning examples, page 132. Remember to get the Section first (Short Medium or Long) by looking for the maximum intensity spot (in candleas), and then find the maximum extension, within that section of the 50% isocandela trace. You'll need to have the book open at page 132 to follow the reasoning below. Example A: The maximum isocandela point on the road is in the Long section, and within the Long section the 50% trace stays in the Type I area. So this is a Type I Long. Example B: The maximum isocandela point on the road is in the Medium section, and within the Medium section the 50% trace goes out into Type IV area. So this is a Type IV Medium. Example C: The maximum isocandela point on the road is in the Long section, and within the Long section the 50% trace goes as far as in the Type III area. So this is a Type III Long. Example D: The maximum isocandela point on the road is in the Medium section, and within the Medium section the 50% trace goes as far as in the Type II area. So this is a Type II Medium. 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