Modelling a Functional Building
0B
Mathematics Higher Level
1B
Student: Joseph Crawley (000865-004)
2B
Word Count: 2100 Words
3B
Teacher: A. Harvey
4B
School: The North London International School
5B
Session: May 2012
6B
Internal Assessment Type 2
7B
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Page 1 of 14
This investigation is looking at a proposal for a design of an office block. The office is to have a width of of 72 meters and a length of 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 room height is 2.5 meters. This investigation will be looking at how how to maximise 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 floor space. The limitations of of the model will will be assessed and possible solutions will be presented.
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Page 2 of 14
The first situation this report report will be looking at, is when the roof 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 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 created using the program Geogebra 4 and labelled with Photoshop Photoshop CS 5, as:
Graph 1
8B
The parabola extends 150 meters meters into 3-d space. The volume 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, 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 however is not appropriate in estimating 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, graph, any cuboid used 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.
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Page 3 of 14
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:
Graph 2
9B
This square is the largest largest that can fit under the current current parabola. 2 2 The area of the square is 29.82 29.82 m , 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, 259200m3133384 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 curves are going to model 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:
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Page 4 of 14
This expression allows for for finding the parabolas that are are greater in height than 36 meters. The same process is applied to them and the results are in tabular form below:
Equation of Roof
Area Volum Heig Und e ht of er Under Roof Roof Roof (m) (m2) (m3)
Side of Longe st Squar e (m)
Larges t Squar e Area (m2)
Largest Cuboid Volume (m3)
Wasted Space Volume (m3)
Waste d Space Ratio
y=-36-1X2+36 y=-(33.2) 1 2 X +39 y=-(33.9) 1 2 X +42 y=-(28.8) 1 2 X +45
36
1728
259200
29.82
889.23
133384.8 6
125815.1 4
0.485
39
1872
280800
31.52
993.75
149062.9 6
131737.0 4
0.469
42
2016
302400
33.12
1096.62
164493.6 0
137906.4 0
0.456
45
2160
324000
34.05
1159.40
y=-27-1X2+48 y=-(25.4) 1 2 X +51
48
2304
345600
36.00
1296.00
173909.2 5 194400.0 0
150090.7 5 151200.0 0
51
2448
367200
37.31
1391.83
y=-24-1X2+54
54
2592
388800
38.53
1484.81
208774.4 3 222721.4 8
158425.5 7 166078.5 2
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Page 5 of 14
0.463 0.438 0.431 0.427
The equations on the table have 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), purple), y=- (28.8)-1 X2+45 (in green), green), and y=-24-1 X2+54 (in salmon), salmon), have been graphed below to show the increase in area that a higher, and therefore steeper, roof affords:
Graph 4
10B
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.
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Page 6 of 14
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:
Graph 5
11B
The office floor space is calculated by the product 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 of Roof y=-36-1X2+36 y=-(33.2) 1 2 X +39 y=-(33.9) 1 2 X +42 y=-(28.8) 1 2 X +45
Maths HL Type 2 IA
Height of Roof ( m) 36
Volume Under der Roof (m3) 259200
39
Side Side of Long ongest Square (m) 29.82
Numbe umber r of whole floors 11
Off Office ice Area (m2) 49203
280800
31.52
12
56743
42
302400
33.12
13
64575
45
324000
34.05
13
66397
Joseph Crawley (000865-004)
Page 7 of 14
y=-27-1X2+48 y=-(25.4) 1 2 X +51
48
345600
36.00
14
75600
51
367200
37.31
14
78345
-1
54
388800
38.53
15
86700
2
y=-24 X +54
This shows that as the the height of the roof increases increases from 36 meters to 54 meters, the office space available nearly doubles. The model can be changed further further to increase the office space, 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 factors that affect the cross 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:
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Page 8 of 14
This means that the total total useable volume is now:
The previous method method of finding the largest square under 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:
Graph 6
12B
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.
Equation of Roof
Heig ht of Roof (m)
Area Unde r Roof (m2)
Roof (m3)
Volume Under
Side of Longest Square (m)
y= -(752/75)-1 X2+75 y= -(752/81.25)-1 X2+81.25
75 81.2 5
7500
540000
6 2 .1 3
8125
585000
6 5 .6 7
y= -(752/87.5)-1 X2+87.5 y= -(752/93.75)-1 X2+93.75
87.5 93.7 5
8750
630000
6 8 .9 9
9375
675000
7 2 .0 9
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Large st Squar e Area (m2) 3860.3 9 4313.1 6 4759.6 5 5197.5 1
Largest Cuboid Volume (m3) 277948.05 310547.83 342695.01 374220.39
Wasted Space Volum e (m3) 262051. 95 274452. 17 287304. 99 300779. 61
Page 9 of 14
Wasted Space Coeffici ent 0.485 0.469 0.456 0.446
y= -(752/100)-1 X2+100 y= -(752/106.25)-1 X2+106.25 y= -(752/112.5)-1 X2+112.5
100 106. 25 112. 5
1000 0 1062 5 1125 0
720000
7 5 .0 0
765000
7 7 .7 2
810000
8 0 .2 8
5625.0 0 6040.9 3 6444.4 9
405000.00 434946.74 464003.08
315000. 00 330053. 26 345996. 92
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.
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Page 10 of 14
0.438 0.431 0.427
The number of total floors floors that are available also increase, 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 of Roof 2
-1
2
Y= -(75 /75) X +75 Y= -(752/81.25)-1 X2+81.25 Y= -(752/87.5)-1 X2+87.5 Y= -(752/93.75)-1 X2+93.75 2
-1
2
Y= -(75 /100) X +100 Y= -(752/106.25)-1 X2+106.25 Y= -(752/112,5)-1 X2+112.5
Height of Roof (m)
Volume Under
Side of Longest
Roof (m3)
Square (m)
Number of whole floors
Office Area (m2)
75
540000
62.13
24
107364
81.25
585000
65.67
26
122943
87.5
630000
68.99
27
134117
93.75
675000
72.09
28
145341
100
720000
75.00
30
162000
106.25
765000
77.72
31
173479
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:
Graph 7
13B
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Page 11 of 14
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:
Graph 8
14B
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Page 12 of 14
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:
Graph 9
15B
Maths HL Type 2 IA
Joseph Crawley (000865-004)
Page 13 of 14
This total office floor space offered by 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 resolved gives total office space of 2 318275.13m . 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 HL Type 2 IA
Joseph Crawley (000865-004)
Page 14 of 14
