tejjj3 Posted January 5, 2011 Report Share Posted January 5, 2011 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.dude!!! howd u do that?! Help me out here with this one give me like the first 2 steps or something plls wld help me alot. Reply Link to post Share on other sites More sharing options...
sid1729 Posted January 6, 2011 Report Share Posted January 6, 2011 (edited) Is the height of the largest cuboid always (0.66) / (2/3) of the height of the parabola??And also... to find the volume of the office block... how to calculate a general formula ? Edited January 6, 2011 by sid1729 Reply Link to post Share on other sites More sharing options...
I-Dont-Like-Boron Posted January 6, 2011 Report Share Posted January 6, 2011 What exactly are we supposed to do for part five??? Reply Link to post Share on other sites More sharing options...
∫ Jorge δx Posted January 6, 2011 Report Share Posted January 6, 2011 What exactly are we supposed to do for part five???Find the floor area of your cuboid. The floor area of one floor is the total length of the building multiplied by the width of your cuboid.And now of course, that number will need to multiplied by the number of floors you have. Reply Link to post Share on other sites More sharing options...
Depressed Posted January 7, 2011 Report Share Posted January 7, 2011 I managed to complete everything up to part 7.I'm stuck, any ideas? Reply Link to post Share on other sites More sharing options...
∫ Jorge δx Posted January 7, 2011 Report Share Posted January 7, 2011 I managed to complete everything up to part 7.I'm stuck, any ideas?I suggest you scroll up this page and you might see an image that could potentially give you an idea Reply Link to post Share on other sites More sharing options...
fan Posted January 10, 2011 Report Share Posted January 10, 2011 got mine done... pheww Think of everything that you doing. It all makes sense as you go step my step.. Reply Link to post Share on other sites More sharing options...
mbenassi1 Posted January 15, 2011 Report Share Posted January 15, 2011 I have a quick question... so I was given the portfolio, but out of laziness I'm just looking at this now.The first points sounds reasonable enough:general formula of a parabola in the form ax2 + bx + c. Since the height is 36 sub. that for c.then with the axis of symmetry formula (assuming that the range being used is -36 and 36),b turns out to be zero so the formula of the parabola is in the form y = ax2 + 36.Why is everybody saying it is in the form of y = -ax2 + 36? this is really bugging me...I am at a stage where I have to hand this in, in 5 days and cannot proceed without an answer.any help would be EXTREMELY appreciated. btw can you put in some advice as how to proceed? Reply Link to post Share on other sites More sharing options...
∫ Jorge δx Posted January 15, 2011 Report Share Posted January 15, 2011 Well, what I suggest is do the following...Since you already have y = ax2 + 36, why not enter the point (36,0) or (-36,0) (Either one is fine) to find the value of a? With that, you're set with a parabola. Reply Link to post Share on other sites More sharing options...
mbenassi1 Posted January 15, 2011 Report Share Posted January 15, 2011 Okay. so now that I have found the actual equation of the parabola i have to find the volume of the maximum cuboid that would fit into the building.this is how i would proceed... use technology to find the maximum area of the rectangle that would fit into the parabola/curve shape. then find the volume of that(just multiply by 150)??? right? O.o? then i experiment with different heights...if my previous way of procedure is correct, then the ratio is just the same process except that I find the integrate the parabola - the x-axis; with upper limit 36 and lower limit 0,multiply that answer by 150 and find the ratio of wasted space: office space??? right? all help and any TIPS I am EXTREMELY grateful for. Reply Link to post Share on other sites More sharing options...
mbenassi1 Posted January 16, 2011 Report Share Posted January 16, 2011 has anyone tried to approach the problem by using hyperbolic cosine functions? i.e. the Catenary Equations??? and use that instead of a quadratic?My dad glanced at the problem while I was working and mentioned that architects use them... too bad he didn't elaborate. does anybody know the implications of using catenary equations?and if the IB completely disregards them or if it is accepted? Reply Link to post Share on other sites More sharing options...
∫ Jorge δx Posted January 16, 2011 Report Share Posted January 16, 2011 You could, but I don't think you should. Hyperbolic cosine functions are (as far as I know) not covered in the HL Math Curriculum.If I were you, I'd stick to a parabola and do calculus Reply Link to post Share on other sites More sharing options...
lexie Posted January 16, 2011 Report Share Posted January 16, 2011 To calculate maximum office floor area, do we take the thickness of the ceiling of each level into account? or just go with 2.5m per level? :/Also, when we place the facade on a different side, does this mean the max height is now 50-75% of the longer side (150m)? Reply Link to post Share on other sites More sharing options...
Andoria Posted January 17, 2011 Report Share Posted January 17, 2011 Hi everybody!Just got the type II portfolio and I've come up with my first stupid question When saying "create a model" in the very beginning, do they just want us to find an appropriate function for the arc of the curved roof structure, or do they require some kind of planar equation describing the entire roof structure?P.S...if the second is the case, I'm already in trouble Reply Link to post Share on other sites More sharing options...
fan Posted January 17, 2011 Report Share Posted January 17, 2011 They want you to formulate an equation for the curved roof structure.. hope that answers your question Reply Link to post Share on other sites More sharing options...
eridanus Posted January 22, 2011 Report Share Posted January 22, 2011 it says that "investigate how changes to the height of the structure affect the dimensions of cuboid" well, i found same width each time, only the height of the cuboid changed. (increased when the height of the roof increased) is this correct? and in the meanwhile, i really have no idea how to do the steps about office area and the last two bullets... i really need some help pleeease? Reply Link to post Share on other sites More sharing options...
sid1729 Posted January 23, 2011 Report Share Posted January 23, 2011 Hey... im getting my number of floors from 9 ( when i use ht 36) to 14 (when i take ht as 54) ... this is becasue im considering that in point 5 we can only have floors inside the cuboid.. am i in the right track? Reply Link to post Share on other sites More sharing options...
ocfx Posted January 23, 2011 Report Share Posted January 23, 2011 Hey, I got this portfolio on friday and I think I know where I'm going with it, but just a general question with regards to the assessment criteria: what sort of variables, parameters, and constraints should I define. By the parameters do they mean height,width, and length? I suppose the variable is therefore the height which is then constrained by the given restricitons, whereas the width and length are constant?Thanks Reply Link to post Share on other sites More sharing options...
xbookgirlx Posted January 23, 2011 Report Share Posted January 23, 2011 In the third bullet what does it mean by "use of technology" does that mean a graphical representation of the change in the height and its relation to the area of the cuboid? Is there any other form of technology to use? If so then what?please help Reply Link to post Share on other sites More sharing options...
sid1729 Posted January 24, 2011 Report Share Posted January 24, 2011 YEah... youve got to form a general equation for the height of the cuboid ... when the structure of the roof changes ... take all heights between 36-54 and use excel or even your calculator.. But using excel is reallly helpful and fast and also covers use of technology!! Reply Link to post Share on other sites More sharing options...
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