ARCH 315 School of Architecture University of Southern California Los Angeles, CA HOME SYLLABUS HOMEWORKS POP QUIZZES PRELIMS FINAL EXAMINATION BIBLIOGRAPHY SLIDES

DESIGN OF the Luminous and Sonic Environment spring 2004 Homework Assignment 2 Given: the illumination on a horizontal plane outside a room is 1200 lumens/ft^2 (fc), and the DF for the rear corner of the room is 2%: a.) is the DF enough (under the guidelines) for a drafting station? No. The guidelines indicate a DF of 6 for drafting. b.) what would be the actual footcandle level at that spot at that time? 1200 fc * .02 = 24 fc (or lumens/square foot) 2.) If you were designing a museum in Southern California and were working on a room used to display a specific, and large sculpture, which calculation method should be used to determine the illumination provided by natural sources? Why? Would that be different if you had one window to place anywhere in the wall, vs. a strip window the entire length of the wall? The Lumen Method is capable of calculating clear sky and direct sun levels. Southern California experiences a lot of clear sky. Use the Lumen Method. The Lumen Method assumes a strip window for the entire length of the window wall. The Daylight Factor method is capable of calculating asymmetrical windows. You might use the Daylight Factor method if you had an offset window, to find the minimum illumination on an overcast day. 3.) Now assume that the above room is approximately 20 ft. deep x 40 ft. long x 10 ft high. It is on the South facade of the building, and will have a continuous window strip, because that is what you wish to express in the facade (who knows why). Design and draw a good cross section (taken at the window) for that facade, to limit thermal overheating but provide natural illumination. (You may use any aesthetic you want, but add enough notes to the cross section to explain it. You do not need to do any calculations, yet.) The resultant design should incorporate a light shelf. This provides for reduced illuminance and heat gain at the front of the space and more even distribution of light from the front to the back. Even an overhang would help reduce the direct sun (solar gain) and still allow some diffuse sky illuminance. 4.) Assume that the above room had a simple facade with no overhangs, no setbacks, etc (unlike how you designed it, of course). The bottom of the window (the sill) is at 2.5 ft. and the top of the window (the head) is at 10 ft. You may assume whatever glass transmission, ground, wall and ceiling reflectances you like. What is the illumination on the work plane in the middle of the room at noon on a sunny September day? (The solar altitude angle is 56°, you must determine the azimuth.) Step I: altitude = 56° azimuth (always zero at noon) = 0° since the wall faces due South, the azimuth of the wall = 0° and the azimuth of the sun compared to the wall = 0°. Clear sky (sunny day) Exvk = 12 klux (from clear sky table, using 0° azimuth line, at altitude = 56°) ExHk = 15 klux (from clear sky table, using Horizontal line, or from table 7.8 p10) Exhk = 7.5 klux (half of ExHk) Exvu = 59 klux (from clear sun table, 0° azimuth, 56° altitude) ExHu = 82 klux (from clear sun table, using Horizontal line, or from table 7.8 p10) Exhu = 41 klux (half of ExHu) Let’s assume that the ground is covered with grass, so: Mg = ρ (Exhk + Exhu) = 0.06 * (7.5 klux + 41 klux) = 2.91 klux Step II: Find the net transmittance. The glass transmittance is assumed to be 85%. About 10% of the window area is assumed to be mullion, so the transmittance of the remainder is 90%. Assuming a clean area for this vertical surface, the Dirt Depreciation Factor is 0.9. Thus the net transmittance is: tnet = TR (Tc) LLF = .85 * .9 * .9 = 0.6885 = .69 Step III: Find the coefficients of utilization. CU is based on three factors: vertical illumince / half horizonal illuminance = (Exvk + Exvu) / (Exhk + Exhu) = (12 + 59) / (7.5 + 41) = 1.46 ~ 1.5 room depth / window height = 20 / 7.5 = 2.7 ~ 3.0 window width / window height = 40 / 7.5 = 5.3 ~ 6.0 CUk = .228 CUg = .157 Step IV: Substitute everything into the final equation. Exvg ~ Mg so substitute Ei = τ [(Exvk + Exvu) * CUk + Exvg* CUg)] = 0.69 [((12 klux + 59 klux) * .228) + (2.91 klux * .157))] = 0.69 [16.188 klux + .45 klux] = 11.43 klux (or about 1060 lumens/ft^2) That’s a lot of light for the middle of the room. It is primarily because of the direct sunlight coming into the space. We could afford to use the light shelf. 5.) Discuss the difference that it would make to your calculation if the room were on the 40th story of a downtown office building. You may assume whatever ground conditions you like. This depends on the surroundings, of course. The difference is the fact that at the 40^th floor, it’s not just grass or pavement outside the window, anymore. The ground reflectance would be the average of the surrounding roofs and a much broader area in front of the windows. It would be a judgement call. (This calculation is rarely done, since the worst case is usually the ground floor and all other cases have more light.) 6.) Using the rule of thumb for offices, what would be the maximum depth for that room if it were an office? The rule of thumb for offices says that the office shouldn’t be deeper than 1.5 times the height of the window head. In this case, that would have meant 1.5 x 10 feet or 15 feet. But we see that with a large strip window, we can have quite a lot of natural light coming into the space at 10 feet. We could calculate a value for 8:00 am at the back of the space, if we wanted to know exactly how deep we can go at weaker sun times. Again, if you use common sense, you can adjust the rule of thumb for a large clear window or a smaller, low transmissivity window.

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