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The 30 St Mary Axe is a landmark skyscraper situated in London. Designed in 1998, completed in 2004, spending 138 million pounds to construct, its 180m height drove skyline boom in London years ago. The construction is highly recognized by the local, people call it the Gherkin, others say it is egg-shaped or looks like a bullet. Based on its shape, I think a conifer cone suits it as well. When I visited the UK two years ago, I was impressed by the Gherkin while making a visit to the Lloyd’s Building nearby.
The sun shone on the glass, making the construction looked pretty luminous on the street.
I rarely see a curved building in my city, and it still stood out in the megacity compared to architectures around it. Later I found that the building is so special because of its various floor area size. As long as I pile up all the different floors, I can obtain its volume. However, I can never do it until I met calculus in math class.
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I am still not satisfied with calculus exploration, it triggered me further to measure volume of the construction.
Basic rationale works as following: I come up with a curve equation (as illustrated by figure on the left below), making it rotate with x-axis for 360° (shown by figure on the right) and then volume of the construction can be found in a 3D solid. This is called the Solid of Revolution2. It works as the building can be considered as made up of infinite thin cyclical discs.
Suppose the height is 𝑏 − 𝑎 on the scale of x-axis 𝑏 is end point while 𝑎 is start point), for a regular solid, volume is 𝜋𝑟( 𝑎 − 𝑏 ), taking radius changes with height into account, general volume formula is 𝜋 𝑦(*+ 𝑑𝑥. fig 2. Basic solid of revolution application Returning back to the Gherkin, the outer surface can be represented by a curved equation. To find the equation, I need to rotate the vertical building clockwise for 90° so it fits to the coordinate axis.
To work out for the volume, the first thing I do is to be clear with the structure of the construction. There are 41 floor inside the building, including one single basement floor. Total 2 Coad, Mal et al. Mathematics For The International Student. Mathematical Studies SL.. 3rd ed., Haese Mathematics, 2012, p. 489 y=f(x) b a y=f(x) 3 height above ground is 179.8m with top floor height 167.1m3. However, some data source reports the figure is 163m4. So height of top floor (the dome) is 179.8-167.1=12.7m. Floor-floor is 4.15m5. However, according to data I found, 39 floors occupy 167.1m, so floor to floor distance is calculated by 167.1÷(40- 1)=4.28m. Inaccurate estimation is found here as height of ground floor is greater than this figure. fig 3. Structural elevation6 Measuring 49m at ground level, reaching the largest diameter 56.5m (some report the value is 56.15m7) on 17th floor, and then reduces to 26.5m on 40th floor.8 Therefore, 17th floor is point of inflection as it alters floor area trend significantly.
For floors from 17 to floor 40, I found the outer profile is in a parabola shape, which can be represented by a quadratic function. For floors below 17, the function is more like a natural exponential curve. The dome is similar to a cone. Based on this assumption, I tried to find each floor area and thus obtain floor radius, so the equation should be obtained. However, most floor area is not available. Therefore, I have to plot each floor on the graph and calculate its height to get points. Then I am going to find its scale (model versus real construction) and work out for the real volume.
I zoomed the graph to ensure precise plotted pointes. Plotted result is shown below with x-coordinates and corresponding y-coordinates. x-coordinate y-coordinate 0.033 1.500 0.513 1.540 0.780 1.567 1.047 1.580 1.553 1.633 1.820 1.643 2.313 1.660 2.580 1.687 3.087 1.699 4.113 1.700 4.340 1.713 Table 1. points on lower section According to the result I found on LoggerPro, the correlation of the function is 0.994571, which means the function is precise. The relationship between floor radius and floor height is 𝑦 = −0.2488× 𝑒67.897:; + 1.739, where y is floor radius and x is floor height. Equation is graphed below, it does seem like the exterior part of the lower section: fig 7: original scale of lower section From the table 1, it can be identified that range of x is 0 to 4.34, so volume for the lower section is π (−0.2488×𝑒67.897:; + 1.739)(8.:87 𝑑𝑥.
Since (−0.2488×𝑒67.897:; + 1.739)2= 0.06190144𝑒67.9E7F; − 0.8653264𝑒67.897:; + 3.024121 So the equation is transformed into: 𝜋 (0.06190144𝑒67.9E7F; − 0.8653264𝑒67.897:; + 3.024121)𝑑𝑥 8.:8 7 Let 𝑢 = −0.4903𝑥 y-intercept:1.4902 7 𝑑𝑢 = −0.4903 𝑑𝑥 ∴ 𝜋 (0.06190144×𝑒6(J − 0.8653264𝑒6J + 3.024121)𝑑𝑢 8.:8 7 = 𝜋 −0.03095072𝑒6(J + 0.8653264𝑒6J + 3.024121𝑢 78.:8 = 36.5436 cubic units Knowing that real ground floor radius is 49 ÷ 2 = 24.5m, ground floor radius in the function is equal to its y-intercept. (When x is 0, y is 1.4902m.) This means 1.4902 units in the picture represents 24.5m in reality. It gives 1 unit in the picture is equal to 16.441m in life.
So the scale based on floor radius is 1:(16.441)3, which is 1:4444. It gives the actual volume of lower section 162400m3. To examine the quantity, I assume the lower section is a cylinder, whose radius is identical in each floor. So maximum volume occurs when radius is in align with floor 17, height is 4.28 × 17 − 1 = 68.48 m. According to 𝜋𝑟(ℎ = 170488 m3. Minimum volume occurs when radius is equal to floor 1, which is 129135m3. So the result is accountable.
Again, by identifying points on the curve, coordinates can be generated below:
x-coordinate y-coordinate
4.338 1.724
4.587 1.720
8
4.844 1.716
5.102 1.707
5.360 1.698
5.609 1.692
5.867 1.680
6.133 1.653
6.373 1.627
6.631 1.600
6.907 1.564
7.138 1.538
7.413 1.502
7.662 1.440
7.964 1.396
8.204 1.351
8.453 1.289
8.711 1.218
8.960 1.158
9.244 1.067
9.500 1.000
9.760 0.889
10.009 0.773
Table 2. points on upper section
Based on calculation from LoggerPro, correlation is 0.994001, which makes the function precise. And the shape of the function can be matched to the exterior look of the upper section of the Gherkin. So the function is 𝑦 = −0.04594𝑥( + 0.4931𝑥 + 0.3891 . Since range of x-coordinate is 4.338 ≤ 𝑥 ≤ 10.009, so volume for upper section is 𝜋 (−0.04594𝑥( + 0.4931𝑥 + 0.3891)(𝑑𝑥N7.7798.::E . Since (−0.04594𝑥( + 0.4931𝑥 + 0.3891)( = 0.00211048𝑥8 − 0.04530602𝑥: + 0.2073971𝑥( + 0.38373042𝑥 + 0.15139881).
So the equation is converted into: 𝜋 ( N7.779 8.::E 0.00211048𝑥8 − 0.04530602𝑥: + 0.2073971𝑥( + 0.38373042𝑥 + 0.15139881)𝑑𝑥 = 𝜋 0.00042209𝑥O − 0.0113265𝑥8 + 0.06914323𝑥: + 0.19186521𝑥( + 0.15139881𝑥 8.::EN7.779 ≈ 38.43386 cubic units Real ground floor radius on 17th is 56.5 ÷ 2 = 28.25m while the radius in the function is 1.724. This means that 1.724 units in the picture represents 28.5m in reality.
It gives 1 unit in the picture is equal to 16.53m in life. So the scale based on floor radius is 1:(16.65)3, which is 1:4517. It gives the actual volume of lower section 173634m3. To examine the quantity, the volume of upper section should range between a cone and a cylinder. Height is 4.28× 39 − 17 = 94.16m. So maximum volume occurs when radius is in align with floor 17, According to 𝜋𝑟(ℎ = 236076m3. Minimum volume occurs when radius is equal to floor 40 (radius is 13.25m), which is 51933m3. So the result is reliable. I assume the dome is in a shape of cone, so according to formula N : 𝜋𝑟(ℎ = N : ×𝜋×13.35(×(179.8 − 167.1))) ≈ 2334m3.
Occupied volume of the gherkin is sun of lower section, upper section and the dome with formula 𝑉 = 𝜋 𝑦(*+ 𝑑𝑥 and 𝑉 = N : 𝜋𝑟(ℎ. 𝑦 = 𝜋 𝑓 −0.2488×𝑒 67.897:; + 1.739 −0.04594𝑥( + 0.4931𝑥 + 0.3891 ]0,4.34[ ]4.338,10.009[ (+ * 𝑑𝑥 𝑦 = 1 3 𝜋𝑟(ℎ (𝑟 = 13.25, ℎ = 12.7) lower section: V = π (−0.2488×𝑒67.897:; + 1.739)( 8.:8 7 𝑑𝑥 ≈ 162400𝑚: upper section: 𝑉 = 𝜋 (−0.04594𝑥( + 0.4931𝑥 + 0.3891)(𝑑𝑥 ≈ 173634𝑚: N7.779 8.::E the dome: 𝑦 = 1 3 ×𝜋×13.35(×(179.8 − 167.1) ≈ 2334𝑚: total volume: 𝑉XYX+Z = 𝑉ZY[\] ^\_X`Ya + 𝑉Jbb\] ^\_X`Ya + 𝑉cYd\ = 162400 + 173634 + 2334 = 338368𝑚.
From calculation above, total occupied volume of the Gherkin is 338368m3. However, there is no direct measure to this available, which makes calculate percentage error impossible. However, if I assume the construction is a regular cylinder, each floor is identical with 17 floor, based on 𝑉 = 𝜋𝑟(ℎ, it can be found that maximum volume of the building is 𝜋×28.25(×179.8 = 450792.2742m3. As there is much space empty in cylinder model, so the final result is convincible. However, there are some concerns need to discussed. Since all the data are obtained from website, there are different measurements towards floor area 11 and thus diameter. So there is risk that I didn’t select the most accurate data.
Moreover, my assumption towards two equations can be wrong, which means they cannot be totally represented by quadratic functions. In the following steps, I plotted points on the graph to obtain coordinates, there is possibility that axis of symmetry is not accurately drawn and I labeled inaccurate positions, leading to imprecise functions. Even if the initial function is perfectly true, scale that I calculated can be wrong. Noting that floor height of ground floor and peak floors is different, some adjustments towards scale can be made. In terms of processing the data, decimal digits that I preserved may not be enough, leading to imprecision.
1. Freiberger, Marianne. 'Perfect Buildings: The Maths Of Modern Architecture'. Plus.Maths.Org, 2007,
2. '30 St' Mary Axe (The Gherkin), London | Archinomy'. Archinomy.Com, 2011,
3. Carlson, Christopher. 'Twisted Architecture—Wolfram Blog'. Blog.Wolfram.Com, 2009.
Finding Occupied Volume of the 30 St. (2024, Feb 06). Retrieved from https://studymoose.com/document/finding-occupied-volume-of-the-30-st
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