Ben92 Posted December 27, 2010 Report Share Posted December 27, 2010 For the second to last part, I suggest you make the inner structure that looks like the following...In case you're wondering, you cannot use this image for your IA, since I took it straight out of mine. However, you can make one like that yourself, using Google SketchUp or any other 3D modeling software.why is it only 14 (by my count floors)? I mean is that your final model? I got mine to be 17 floors, maximizing floor space.Hey. What height are we supposed to use? The parameters I got were between 36 and 54. So I imagined that to maximize floor space one would use the maximum height, dividing it by the minimum height of a room. This causes me to have about 21 floors. Why is my building so much bigger than your guys' ones!? And I'm zetta proud of it. But really. How am I wrong? Edit : Oh and I got 14 when I used 36 as the height of the curved structureNo, you are right. However, I used 3m for the height of the room(taken into account the ceiling of the room and the floor of the room above). I just thought for the final model 14 was too little. Reply Link to post Share on other sites More sharing options...
jmmwng Posted December 28, 2010 Report Share Posted December 28, 2010 hey guyswhen calculating the equation of the parabola, are we required to use exact values for the constants? Reply Link to post Share on other sites More sharing options...
AAMomen Posted December 28, 2010 Report Share Posted December 28, 2010 hi guysI just started the task and I got stuck with the 3 point. What kind of technology should I use?Well I used graphing technology showing the changes of the parabola and the maximum height of the building. You could download the 30 day trial for autograph, a graphing software. The link is somewhere in this thread. Reply Link to post Share on other sites More sharing options...
foreverhunched Posted December 28, 2010 Report Share Posted December 28, 2010 hey guyswhen calculating the equation of the parabola, are we required to use exact values for the constants?I think it's better to leave it in exact value. I rounded off to 3 sig figs for the ratio though. Reply Link to post Share on other sites More sharing options...
Ajayalmighty Posted December 29, 2010 Report Share Posted December 29, 2010 hey guyswhen calculating the equation of the parabola, are we required to use exact values for the constants?I think it's better to leave it in exact value. I rounded off to 3 sig figs for the ratio though.Yeah I did that too, I just left the constant at 2 decimal points. Reply Link to post Share on other sites More sharing options...
Ajayalmighty Posted December 29, 2010 Report Share Posted December 29, 2010 How many examples do we have to give? For point 3 and point 5, do we do all the heights from 36-54? And de we have to deal with the same set of heights each time? Reply Link to post Share on other sites More sharing options...
jmmwng Posted December 29, 2010 Report Share Posted December 29, 2010 How many examples do we have to give? For point 3 and point 5, do we do all the heights from 36-54? And de we have to deal with the same set of heights each time?I did 6 examples for each model (2 ways of opening the roof). I got a question..are we required to do algebraic proof for the ratio? I never learned integration and it would be a pain to teach it to myself... Reply Link to post Share on other sites More sharing options...
tbnguyenphan Posted December 29, 2010 Report Share Posted December 29, 2010 How many examples do we have to give? For point 3 and point 5, do we do all the heights from 36-54? And de we have to deal with the same set of heights each time?i inserted the numbers into Excel then generated answers for all heights (36-54) It's super easy and quick, and you can use those numbers that you have calculated for further calculations later on. Reply Link to post Share on other sites More sharing options...
Nachoop Posted December 29, 2010 Report Share Posted December 29, 2010 For the second to last part, I suggest you make the inner structure that looks like the following...In case you're wondering, you cannot use this image for your IA, since I took it straight out of mine. However, you can make one like that yourself, using Google SketchUp or any other 3D modeling software.I have a similar picture in mine, but I don't know how to calculate the area of the space that is not used. Any idea? Reply Link to post Share on other sites More sharing options...
dessskris Posted December 30, 2010 Report Share Posted December 30, 2010 I have a similar picture in mine, but I don't know how to calculate the area of the space that is not used. Any idea?Isn't that half a cylinder? Then if you know how many blocks you have and their lengths, widths and heights; you can take the difference between the volume of the semi-cylinder and the volume of the blocks, I suppose? 1 Reply Link to post Share on other sites More sharing options...
dachensterr Posted December 31, 2010 Report Share Posted December 31, 2010 for point 6, where we need to investigate "what would happen" if the facade is placed on the longer side of the base, does this mean we have to repeat points 1,2,3,4,5 for the new facade? Reply Link to post Share on other sites More sharing options...
B4Z Posted December 31, 2010 Report Share Posted December 31, 2010 for point 6, where we need to investigate "what would happen" if the facade is placed on the longer side of the base, does this mean we have to repeat points 1,2,3,4,5 for the new facade?Pretty much...at least, thats what I interpreted to be. Try using Excel to aid with the calculations. Reply Link to post Share on other sites More sharing options...
szeherezada Posted December 31, 2010 Report Share Posted December 31, 2010 >>i inserted the numbers into Excel then generated answers for all heights (36-54) It's super easy and quick, and you can use those numbers that you have calculated for further calculations later on. <<Can you explain how exactly did you this calculations in Excel?Can someone tell how to use Excel for this portfolio, please? Reply Link to post Share on other sites More sharing options...
∫ Jorge δx Posted December 31, 2010 Report Share Posted December 31, 2010 I have a similar picture in mine, but I don't know how to calculate the area of the space that is not used. Any idea?Find the length of each cuboid by inputting the height at which it is into as the y value that your parabola is equal to (Assuming each floor is 3m, like in the picture I have, these heights would be 3, 6, 9, 12, etc...) and solving for x. Multiply this x you obtain by 2 (this is the length you were looking for), then by your floor height (3m in my case).Do this for all the floors you have, then add all the values.Then subtract the value you obtain from that last step from the definite integral of your parabola. Finding the ratio from this point should be pretty easy. Reply Link to post Share on other sites More sharing options...
Ajayalmighty Posted January 2, 2011 Report Share Posted January 2, 2011 (edited) >>i inserted the numbers into Excel then generated answers for all heights (36-54) It's super easy and quick, and you can use those numbers that you have calculated for further calculations later on. <<Can you explain how exactly did you this calculations in Excel?Can someone tell how to use Excel for this portfolio, please?Lol it's almost fun doing this in excel. THe tough part is just coming up with a formula, then once you do is just copy paste for the rest of the results.To do integration, you have to do integrate f(x) by hand and then input the equation with the upper limit (i.e. +36) into the spread sheet. Keep a column for your constant K and height H so it's easy to refer back. Then input the equation with the lower limit into a seperate column and finally subtract them both together and multiply by 150 to have the total volume of structure.Just making sure, the volume of the cuboid for bullet 2 is 149,616? Edited January 2, 2011 by Ajayalmighty Reply Link to post Share on other sites More sharing options...
kingsnqueens Posted January 2, 2011 Report Share Posted January 2, 2011 (edited) To find the total maximum floor area I was plugging in 2.5 m for y in the equation of the ellipse. This should give you the value for x (refers to the length of a rectangle inscribed in an ellipse). However, for some heights, the length of my cuboid was greater than 72 m ( this mean 75 or greater). What am I doing wrong? Need help asapdoes anyone have any idea what to do for floor area? Edited January 2, 2011 by Desy ♫ Reply Link to post Share on other sites More sharing options...
Ajayalmighty Posted January 4, 2011 Report Share Posted January 4, 2011 To find the total maximum floor area I was plugging in 2.5 m for y in the equation of the ellipse. This should give you the value for x (refers to the length of a rectangle inscribed in an ellipse). However, for some heights, the length of my cuboid was greater than 72 m ( this mean 75 or greater). What am I doing wrong? Need help asapdoes anyone have any idea what to do for floor area?How was the length of the cuboid greater than 72? the zeroes of the graph can only by -36 and +36 for all the heights.I just plugged in values for y to find x. i.e. y=2.5, 5, 7.5 etc. until the maximum height is reached. Then you just have to multiply 2x*150 for each floor where x is in terms of y.So ultimately I had to calculate the floor area for each floor for each height in excel, then add them up. Reply Link to post Share on other sites More sharing options...
iMesmerized Posted January 4, 2011 Report Share Posted January 4, 2011 >>i inserted the numbers into Excel then generated answers for all heights (36-54) It's super easy and quick, and you can use those numbers that you have calculated for further calculations later on. <<Can you explain how exactly did you this calculations in Excel?Can someone tell how to use Excel for this portfolio, please?Lol it's almost fun doing this in excel. THe tough part is just coming up with a formula, then once you do is just copy paste for the rest of the results.To do integration, you have to do integrate f(x) by hand and then input the equation with the upper limit (i.e. +36) into the spread sheet. Keep a column for your constant K and height H so it's easy to refer back. Then input the equation with the lower limit into a seperate column and finally subtract them both together and multiply by 150 to have the total volume of structure.Just making sure, the volume of the cuboid for bullet 2 is 149,616?i got 149550, i guess the contractors are going to prefer you used geograba Reply Link to post Share on other sites More sharing options...
prastha Posted January 4, 2011 Report Share Posted January 4, 2011 (edited) For the second to last part, I suggest you make the inner structure that looks like the following...In case you're wondering, you cannot use this image for your IA, since I took it straight out of mine. However, you can make one like that yourself, using Google SketchUp or any other 3D modeling software.I have a similar picture in mine, but I don't know how to calculate the area of the space that is not used. Any idea?do we really need such 3 d diagram with roomspecification from second part coz. part 7 only states about the maximising office space by not having single cubical block.and can you tell me how to draw those sort of figures in google sketch up coz. arcs are not perfectly parabolic and they seem more thansemicirclei.e.not correct figure.... Edited January 4, 2011 by Desy ♫ Reply Link to post Share on other sites More sharing options...
∫ Jorge δx Posted January 4, 2011 Report Share Posted January 4, 2011 Believe it or not, I actually used the arc tool to make that image.The key to it is making the cuboids first and then making an arc around them. Reply Link to post Share on other sites More sharing options...
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