qbs Posted January 18, 2012 Report Share Posted January 18, 2012 (edited) Dot no. 4: "For each height, calculate the ratio..." Should it be done manually or using technology? Because in the first case its hell lot of work.EDIT: Oh, nevermind. I'm dumb. Edited January 18, 2012 by qbs Reply Link to post Share on other sites More sharing options...
Kronos21 Posted January 19, 2012 Report Share Posted January 19, 2012 your function can be anything as long as the height and width are correct.Volume is just height*width*150Height= -kx^2 +36Area= 2x(-kx^2 + 36)So do I have to integrate the area or differentiate it? Reply Link to post Share on other sites More sharing options...
qbs Posted January 19, 2012 Report Share Posted January 19, 2012 I had a question that when you change the facade will the minimum and maximum height change to 75 and 112.5 respectively or remain 36 and 54. The reason for this is that the portfolio states that it is 50% and 75% of the width so is this width the width of the rectangular block (72 m constant) or the width of the facade? If it is the width of the rectangular block, it should remain the same at 36 and 54 right? Well, I expect they are 75 and 112.5. Aesthetic reasons matter, if height remained the same, the building would appear floppy, shapeless arc instead of this nice parabolic shape. So, it is the width of the facade that matters. Reply Link to post Share on other sites More sharing options...
timtamboy63 Posted January 28, 2012 Report Share Posted January 28, 2012 Hey guys!I'm working on the part where it says the office block isn't in the shape of a single cuboid, find max floor area. Now I can do it, but I can't seem to generalise it into a formula based off h? Does anyone have tips?I imagine it'd mean I'd have to find the maximum number of floors (h/2.5), then calculate the floor area of each floor independently, and add them together, but I have no idea how to generalise it? I was thinking about using sequences/series, but again, I don't know what I'd be doing?Cheers! Reply Link to post Share on other sites More sharing options...
banana.sh Posted February 8, 2012 Report Share Posted February 8, 2012 Hey!I have used the equation (-1/36)x^2+2x because my width is in the range 0...72. I have seen most of you used -36...+36 and got (-1/36)x^2+36. Is my way correct also?The maximum volume I found is 149649.1898 . However I have read from some of you that the volume should be 149616. Is difference because of the different equations used?Thanks xx Reply Link to post Share on other sites More sharing options...
rFumachi Posted February 8, 2012 Report Share Posted February 8, 2012 It's ok, I used this equation (-1/36)x^2+36 and the maximum volume I found was 149649.19 Reply Link to post Share on other sites More sharing options...
Antrikshy Posted February 12, 2012 Report Share Posted February 12, 2012 is there any formula to finding the maximum volume of the cuboid? and when i plot the graph,is it suppose to be a linear graph?Simply make an arbitrary rectangle on your parabola (that represents a facade). The parabola should have its vertex on x=0 (at least that's how I found it easy). Make the breadth of the rectangle 2x and let the height be the function of the parabola. Then find the maximum point of the area function. Find the x-coordinate of this point and double it. This is the width of the cuboid of maximum volume. Substitute 'x' in the parabola function to find the height of the cuboid. Then find the volume using V = l*b*h. Reply Link to post Share on other sites More sharing options...
Hexa Posted February 13, 2012 Report Share Posted February 13, 2012 I just finished this IA. I got lazy so I only have two graphics in mine. Also I think that I probably did stuff wrong, and at the minimum probably got a lot of the technical stuff wrong. But it's late and I'm too lazy to fix it. This is the least favorite IA I've ever had to do. Reply Link to post Share on other sites More sharing options...
Chaza65 Posted February 19, 2012 Report Share Posted February 19, 2012 #222 xuanmai022, it doesn't matter about that, because it's a question about floor area rather than volume, so you just find the maximum number of whole floors (9, for that height). Then, you simply multiply the width of your cuboid by the depth (150m) to find the floor area per floor, and multiply that result by your number of floors.Or, use the general equation:(Floor width * Floor depth) * Number of Floors = Office Floor Area per Floor * Number of Floors So essentially, it doesn't matter... Reply Link to post Share on other sites More sharing options...
Lazar Micovic Posted February 28, 2012 Report Share Posted February 28, 2012 Hey a question, i did everything up to the last point, where we change to the stair-like construction of the office. Did anyone manage to create a formula in excel which gives you total office volume through the total height of the building variable only? (so that the formula incoporates calculation of the width of every single floor, from h)? Reply Link to post Share on other sites More sharing options...
AndreaRuiz Posted March 9, 2012 Report Share Posted March 9, 2012 Unlike some people, I'm enjoying this IA quite a lot . I have a small question. In point 7, we have to make the stair-like model. However, I'm having trouble making the equation. I have an idea: When calculating the area, you know it's A=2xy In the y we need to put the function in it. If I take that function and make it equal to the, for example, first floor height (2,5), will I get the first floor area? I don't think it'll do, the x changes for that base, does it? And how do I take the second floor area, make it equal to 5? Help please? Reply Link to post Share on other sites More sharing options...
Muhasin Ibn Mohd Posted March 9, 2012 Report Share Posted March 9, 2012 Can any one help me out for the conclusion for the maths functional building. How do i conclude....and how do i apply the real life situation in my portfolio..The following table shows the floor area for a single cuboid and for many cuboids of 2.5 each, at a given height. The last column shows the percentage of the increased floor area of the cuboids with respect to the floor area of the single cuboid. As it can be seen, the area increases from 12 to almost 19 times.At present my state is till this much..."The following table shows the floor area for a single cuboid and for many cuboids of 2.5 each, at a given height. The last column shows the percentage of the increased floor area of the cuboids with respect to the floor area of the single cuboid. As it can be seen, the area increases from 12 to almost 19 times.Then, it can be said with reason, that the last model gives us the maximum approach of the volume and area of the building, being probably the best option. However, we still have to take into consideration the purpose of the building, as in some cases it is better to have higher rooms and the building may include a terrace.In conclusion, I liked the design of the building; however, it’s a very complex design to build and may also be a lot more expensive to build than a regular cuboid form building. It is important to know about the resistance of the materials used in the roof and count with a great amount of time, as it has to be work in parts. We also have to remember that in real life we need to take into consideration the space taken by the stairs and the halls. Also, if the building is composed of several cuboids the sunlight for each room will be very limited varying a lot during the day, so the building would need to use a lot of artificial light and energy, even during the day. Lastly, although an increase of height and doing the offices of 2.5m high cuboids increases the volume considerable, it is also good to remember that to be able to increase the number of offices and the height we will need more materials and more money."After this what to do??? Reply Link to post Share on other sites More sharing options...
Shahad Posted June 21, 2012 Report Share Posted June 21, 2012 Hey, I'm stuck on point three: "Use technology to investigate how changes to the height of the structure affect the dimensions of the largest possible cuboid." I am not so good with technology lol and i cant think of anything right now but doing it manually. Reply Link to post Share on other sites More sharing options...
The Economist Posted June 21, 2012 Report Share Posted June 21, 2012 Hey, I'm stuck on point three: "Use technology to investigate how changes to the height of the structure affect the dimensions of the largest possible cuboid." I am not so good with technology lol and i cant think of anything right now but doing it manually.I assume you know what variables you should use since you say you are doing it manually but the technology part should be covered by using excel. Reply Link to post Share on other sites More sharing options...
Shahad Posted August 6, 2012 Report Share Posted August 6, 2012 Thanks.I managed to do that but now i'm not too sure about the ratio of the volume of the wasted space to the volume of the office block.What i did was find a general equation for the curved roof and integrate to find the volume of wasted space and volume of the office block.So, volume of wasted space : volume of office blockThen i divided both by the volume of wasted space to get the ratio.I ended up with 1:1.37. I was pretty confident with my method.But my teacher said it might be incorrect.Can anyone help? Reply Link to post Share on other sites More sharing options...
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