maths internal assessment type 2 modeling a functional building

0BModelling a Functional Building

1BMathematics Higher Level

2BStudent: Joseph Crawley (000865-004)

3BWord Count: 2100 Words

4BTeacher: A. Harvey

5BSchool: The North London International School

6BSession: May 2012

7BInternal Assessment Type 2

Maths HL Type 2 IA Joseph Crawley (000865-004) of 14

This investigation is looking at a proposal for a design of an office block. The office is to have a width of 72 meters and a length of 150 meters, which have a parabolic roof arching over the office. In this design there is a maximum roof height of 75% of the width of the office and a minimum of 50%. The minimum room height is 2.5 meters. This investigation will be looking at how to maximise available office space given these constraints using mathematical modelling of the proposed office. This investigation will look at how to best place the roof (length ways or width ways), optimal height for the roof, total volume under this roof, and finally available office volume and floor space. The limitations of the model will be assessed and possible solutions will be presented.


The first situation this report will be looking at, is when the roof is placed width ways over the base, meaning that there is a parabola that is 72 meters across, which extends 150 meters, making the building a . The first model will use the lowest possible roof outlined in the design, which is 36 meters (50% of 72 meters). It can be modelled as the function; Y= -36-1+X2 +36 expanding 150 meter into the 3-d dimension. This is because, firstly the roof in the shape of an inverse parabola, so the minus symbol is added, secondly the vertex needs to be at This can be shown on a graph, created using the program Geogebra 4 and labelled with Photoshop CS 5, as:

8BGraph 1

The parabola extends 150 meters into 3-d space. The volume under this roof is given by product of the area under the parabola and the length of the roof. For this roof, the equation to find the area under the roof is:

This cross section of the graph, is 1728 m2, when multiplied by the length of 150 meters, gives the total area under the curve as 259200 m3.

This number however is not appropriate in estimating the total office space available, as the design requires that offices have a minimum height of 2.5 meters. The most basic method of having offices with a minimum height of 2.5 meters is a cuboid office space underneath the parabolic roof. However, when rendering this model on a 2-d graph, any cuboid used will appear as a rectangle. For this model a square will be used, as of all the rectangles, it has the greatest circumference to area ratio. To find the largest square that can fit under the curve, a line is drawn at 45o from the origin. Because when a line at 45o is drawn from the mid point of the base of a square, it eventually interests with the upper corner of the square, when considering its interaction with the parabola this will be the point where the line intersects with parabola. This means that the


algebraic solution is to find the point where a line which is a 45o from the normal (y=2x or y=-2x) intersects with parabola. Graphed it looks like:

9BGraph 2

This square is the largest that can fit under the current parabola. The area of the square is 29.822m2, or 889.2324m2. This means that the total volume of the cuboid is 133384.86m3. The wasted volume is equal to the total volume minus the useable volume, 259200m3-133384 m3= 125815.14 m3. This figure can’t be used to make comparisons between roofs of different heights, as roofs get taller, the volumes under them increase. So a ratio needs to be used, comparing the wasted volume to the total volume under the parabola. A ratio of total volume to wasted volume will be used, total volume: wasted volume, or in this case 259200m3:125815.14 m3. This means that 0.485 of the space is not being used for offices, or in more familiar terms, 48.5% of the volume under the parabola is being wasted. This is not the only way the roof can be laid out, the design allows roof heights from 50% of the width, all the way to 75 % of the width. This means that the heights of the roof can vary from, the already tested 36 meters, all the way to 54 meters high. This means that new curves are going to model these new higher roots, which means a general formula for out parabola is needed, in which height is a variable. This can be expressed as:

This expression allows for finding the parabolas that are greater in height than 36 meters. The same process is applied to them and the results are in tabular form below:

EquationHeight

ofArea

Under Volume Under

Side of Longest

Largest Square

Largest Cuboid

Wasted Space

Wasted Space

of RoofRoof (m)

Roof (m2)

Roof (m3)

Square (m)

Area (m2)

Volume (m3)

Volume (m3) Ratio

y=-36-1X2+36 36 1728 259200 29.82 889.23 133384.86 125815.14 0.485

y=-(33.2)-1X2+39 39 1872 280800 31.52 993.75 149062.96 131737.04 0.469


y=-(33.9)-1X2+42 42 2016 302400 33.12 1096.62 164493.60 137906.40 0.456

y=-(28.8)-1X2+45 45 2160 324000 34.05 1159.40 173909.25 150090.75 0.463

y=-27-1X2+48 48 2304 345600 36.00 1296.00 194400.00 151200.00 0.438

y=-(25.4)-1X2+51 51 2448 367200 37.31 1391.83 208774.43 158425.57 0.431

y=-24-1X2+54 54 2592 388800 38.53 1484.81 222721.48 166078.52 0.427


The equations on the table have been simplified so as to be presentable, but where derived from the above equation based on the desired height. Three of the lines from the table, y=- 36-1 X2+36 (in purple), y=- (28.8)-1 X2+45 (in green), and y=-24-1 X2+54 (in salmon), have been graphed below to show the increase in area that a higher, and therefore steeper, roof affords:

10BGraph 4

From the above table and graph is it clear that the taller the roof, the less wasted volume. The table also shows that as the roof gets taller, there is an increase in usable volume total. This means that the tallest roof is, in terms of total volume and % of usable volume, the best style of roof to have.


However this does not address the amount of office space each roof style provides. The amount of floor space provided is controlled by four factors, the width of the office space, the height of the office space, the length of the office space and finally the height of each floor. To find the floor area of an office, the number of floor that can fit inside the cube is needed. For this design the floor height is 2.5 meters, meaning that if the height of the square under the roof is taken, for example 38.53 meters from the previous graph, divided by the floor height (54/2.5= 15.412), and finally rounded down to the nearest whole number (15.412≈15), the number whole floors that can fit inside the given space are found. This is illustrated on the by this graph:

11BGraph 5

The office floor space is calculated by the product of the width of a floor (38.53), the length of the office, (150) and the number of floors (15). This gives a total office floor area of 86700 m2. The office floor area for the heights tested before is in the below:

Equation Height of Volume Under Side of Longest Number of Office of Roof Roof (m) Roof (m3) Square (m) whole floors Area (m2)

y=-36-1X2+36 36 259200 29.82 11 49203y=-(33.2)-1X2+39 39 280800 31.52 12 56743y=-(33.9)-1X2+42 42 302400 33.12 13 64575y=-(28.8)-1X2+45 45 324000 34.05 13 66397y=-27-1X2+48 48 345600 36.00 14 75600y=-(25.4)-1X2+51 51 367200 37.31 14 78345y=-24-1X2+54 54 388800 38.53 15 86700


This shows that as the height of the roof increases from 36 meters to 54 meters, the office space available nearly doubles. The model can be changed further to increase the office space, by rotating the office width wise, so that the parabolic roof now rests over the 150 meter side of the rectangle. The new style of roof can be shown by the equation:

This is because the only two factors that affect the cross section are the width and the height of the roof. Because the line formula has changes, the equation for area under the roof has changes, the new general statement for the area, which does not depend on the actual width of the build like the previous statement, now will have to have general terms for the limits of integration. The integration will now be from negative half the width and half the width, as the centre of the roof rests over x=0. This means that the new equation for the general area is:


This means that the total useable volume is now:

The previous method of finding the largest square under the roof will be used, using the line y=2x and y=-2x and finding its intersection with the roof. This is graphed as:

12BGraph 6

A range of graphs where tested for the new width, going from the shortest possible roof, 75 meters, to the tallest possible roof, 112.5 meters (which is shown on the above graph), the results are tabulated below.

EquationHeight

ofArea

Under Volume Under

Side of Longest

Largest Square Largest Cuboid

Wasted Space

Wasted Space

of RoofRoof (m)

Roof (m2) Roof (m3) Square (m)

Area (m2) Volume (m3)

Volume (m3) Coefficient

y= -(752/75)-1 X2+75 75 7500 540000 62.13 3860.39 277948.05 262051.95 0.485

y= -(752/81.25)-1 X2+81.25 81.25 8125 585000 65.67 4313.16 310547.83 274452.17 0.469

y= -(752/87.5)-1 X2+87.5 87.5 8750 630000 68.99 4759.65 342695.01 287304.99 0.456

y= -(752/93.75)-1 X2+93.75 93.75 9375 675000 72.09 5197.51 374220.39 300779.61 0.446

y= -(752/100)-1 X2+100 100 10000 720000 75.00 5625.00 405000.00 315000.00 0.438

y= -(752/106.25)-1 X2+106.25106.2

5 10625 765000 77.72 6040.93 434946.74 330053.26 0.431

y= -(752/112.5)-1 X2+112.5 112.5 11250 810000 80.28 6444.49 464003.08 345996.92 0.427


From this table, it becomes apparent that when using the square faced cuboid model, that the wasted space coefficient is the same for both tables.


The number of total floors that are available also increase, as the height is now, even at the lowest possible value, is much greater than the first roof width. If we tabulate the number of possible floors, using the same method as before:

Equation Height of Volume Under Side of Longest Number of Office

of Roof Roof (m) Roof (m3) Square (m) whole floors Area (m2)

Y= -(752/75)-1 X2+75 75 540000 62.13 24 107364

Y= -(752/81.25)-1 X2+81.25 81.25 585000 65.67 26 122943

Y= -(752/87.5)-1 X2+87.5 87.5 630000 68.99 27 134117

Y= -(752/93.75)-1 X2+93.75 93.75 675000 72.09 28 145341

Y= -(752/100)-1 X2+100 100 720000 75.00 30 162000

Y= -(752/106.25)-1 X2+106.25 106.25 765000 77.72 31 173479

Y= -(752/112,5)-1 X2+112.5 112.5 810000 80.28 32 184960

We can see that the building that offers the greatest area of office space, when using the square faced cuboid model the greatest office space is offered when the height is the greatest it can be, when the roof rests over longest side of the base. This can be graphed as:

13BGraph 7


However, even when the square models is optimised to provide the maximum office space, there is still about 40% of the volume not being used for office space, the following graph illustrated the amount of space not being used:

14BGraph 8


This is a large limitation of the model, as this space could be used. Instead of using a square model, a pyramidal model could be used, as on the lower floors of the building, more office space can fit, as there are larger areas in which the 2.5 meter height floors can fit. When you stack these floors, which progressively get smaller they resemble a pyramid. This can be graphed as:

15BGraph 9


This total office floor space offered by this model can’t be calculated using the previous formula, because the floor size is not regular, but instead decreases as the building gets taller. As each floor is on a 2.5 increment of y (or height), the line equation needs to be rearranged, so that with the information that is known, the height, it is possible to find the x value. If the current line equation (written simplified) of:

Is rearranged to be equal to x, then:

With this information and the knowledge that each floor is 2.5 meters high and that there are 44 floors, an equation for finding the total floor space can be given by:

This final equation, when resolved gives total office space of 318275.13m2. This is an increase of 133315.13m2 from the total office space offered by the previous model, 184960m2. The new volume used is the product of 318275.13m2 and the height of the floor of 2.5m, a total of 795687.825m3, which leaves 14312.175 m3 wasted of the total of 810000m3, meaning the new ratio of wasted to total space is 14312.175 m3 : 810000m3 . The percentage of space wasted is 1.77%. This is a great improvement over the square model.


maths internal assessment type 2 modeling a functional building

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