One of the most frequent remarks that I get when I mention studying applied mathematics is "applied to what?.." It is not always easy to give convincing examples straight away, because for me the enormous role of this science in all aspects of life is just so obvious. This is why I like to discover the unusual cases where some elaborate techniques can be applied to in quite mundane areas.

This is one of the small problems I formulated for myself during my studies of maritime logistics, just for the sake of curiosity. Calculations are pretty simple and straightforward, but the results look nice, also with some unexpected turn in the middle.

In maritime transport one has to deal with the so-called bulk cargoes (iron ore, coal, fertilizers etc) which are often stored in the ports in large piles. I once considered determining some quantitative characteristics that are useful for designing the warehouse to store such type of cargo.

Let’s approximate the pile of cargo with the geometrical body of height \(H\) having a rectangle \(L_2\times B\) at its base and 4 facets inclined towards the centre at equal slope, thus forming the upper horizontal edge of length \(L_1\):

Its volume can be calculated as follows:

$$V=\int_0^HS(x)dx$$

Where

$$S(x)=a(x)b(x)=\left(L_{1}+\frac{L_{2}-L_{1}}{H}x\right)\frac{Bx}{H}=B\left(L_{1}\frac{x}{H}+\left(L_{2}-L_{1}\right)\left(\frac{x}{H}\right)^{2}\right)$$

So that

$$V=BH\int_0^1\left(L_1\frac{x}{H}+(L_2-L_1)\left(\frac{x}{H}\right)^2\right)d\left(\frac{x}{H}\right)=\frac{HB}{6}(2L_2+L_1)$$

Now let us determine the amount of energy required to form the pile (meaning mechanical work that has to be done against the forces of gravity). Assume that density of the cargo is \(\gamma\) kg/m3. To lift a small layer to height \(H-x\) requires:

$$dA=\gamma gdV(H-x)=\gamma gB\left(L_1\frac{x(H-x)}{H}+\frac{(L_2-L_1)x^2(H-x)}{H^2}\right)dx$$

Thus, total work equals:

$$A=\gamma gB\int_0^H(L_1\frac{x(H-x)}{H}+\frac{(L_2-L_1)x^2(H-x)}{H^2})dx=\frac{\gamma gBH^2}{12}(L_1+L_2)$$

No we can ask the following question: if volume is a given value, what should be length and breadth so that the pile occupies minimal warehouse area? To simplify calculations and cancel out some parameters let us introduce the value called “angle of natural slope”. This is the term from soil mechanics which means the angle that the substance makes against horizontal surface. It depends on density, viscosity and inter-particle friction and can be found in special tables.

Let:

$$\frac{2H}{B}=\tan \chi$$

$$H=\frac{1}{2}B\tan\chi;L_1=L_2-\frac{2H}{\tan\chi}=L-B$$

$$V=\frac{B^2\tan\chi}{12}(3L-B)$$

Assuming that V=const, obtain expression for L and insert it in the formula for the area S=LB:

$$L=\frac{4V}{B^2\tan\chi}+\frac{B}{3}$$

$$S=LB=\frac{4V}{B\tan\chi}+\frac{B^2}{3}$$

$$S'(B)=-\frac{4V}{B^2\tan\chi}+\frac{2B}{3}=0$$

$$B=\sqrt[3]{\frac{6V}{\tan\chi}}$$

$$L=\left(\frac{4}{\sqrt[3]{36}}+\frac{2}{3}\sqrt[3]{36}\right)\sqrt[3]{\frac{V}{\tan\chi}}$$

$$S=4\left(\frac{1}{\sqrt[3]{6}}+1\right)\left(\frac{V}{\tan\chi}\right)^{2/3}\approx 4.4\left(\frac{V}{\tan\chi}\right)^{2/3}$$

Now, assuming again that \(V=const\), let us determine the values of \(L\), \(B\) again which minimize the amount of energy required to form the pile. Again, substituting for \(L\), but now in the expression for \(A\) we obtain:

$$A=\frac{\gamma gBH^2}{12}(L_1+L_2)=\frac{\gamma g H^2}{48}\left(\frac{8VB}{\tan\chi}-\frac{B^4}{3}\right)$$

$$A'(B)=\frac{\gamma g \tan^2 \chi}{48}\left(\frac{8V}{\tan\chi}-\frac{4B^3}{3}\right)=0$$

$$B=\sqrt[3]{\frac{6V}{\tan\chi}}$$

Remarkably, we get the same values which were obtained to minimize the base area, although the approach is now quite different.

To complete the picture we can use the derived results to determine maximum allowed volume for a single pile if unitary pressure on the ground is limited by some given value \(q\) kg/m2:

$$\frac{V\gamma}{S}<q$$

$$V<\frac{qS}{\gamma}=\frac{q}{\gamma}4.4\left(\frac{V}{\tan\chi}\right)^{2/3}$$

$$V<\frac{(4.4q)^3}{\gamma^3\tan\chi}\approx 85.3\frac{q^3}{\gamma^3\tan\chi}$$

This is one of the small problems I formulated for myself during my studies of maritime logistics, just for the sake of curiosity. Calculations are pretty simple and straightforward, but the results look nice, also with some unexpected turn in the middle.

In maritime transport one has to deal with the so-called bulk cargoes (iron ore, coal, fertilizers etc) which are often stored in the ports in large piles. I once considered determining some quantitative characteristics that are useful for designing the warehouse to store such type of cargo.

Let’s approximate the pile of cargo with the geometrical body of height \(H\) having a rectangle \(L_2\times B\) at its base and 4 facets inclined towards the centre at equal slope, thus forming the upper horizontal edge of length \(L_1\):

Its volume can be calculated as follows:

$$V=\int_0^HS(x)dx$$

Where

$$S(x)=a(x)b(x)=\left(L_{1}+\frac{L_{2}-L_{1}}{H}x\right)\frac{Bx}{H}=B\left(L_{1}\frac{x}{H}+\left(L_{2}-L_{1}\right)\left(\frac{x}{H}\right)^{2}\right)$$

So that

$$V=BH\int_0^1\left(L_1\frac{x}{H}+(L_2-L_1)\left(\frac{x}{H}\right)^2\right)d\left(\frac{x}{H}\right)=\frac{HB}{6}(2L_2+L_1)$$

Now let us determine the amount of energy required to form the pile (meaning mechanical work that has to be done against the forces of gravity). Assume that density of the cargo is \(\gamma\) kg/m3. To lift a small layer to height \(H-x\) requires:

$$dA=\gamma gdV(H-x)=\gamma gB\left(L_1\frac{x(H-x)}{H}+\frac{(L_2-L_1)x^2(H-x)}{H^2}\right)dx$$

Thus, total work equals:

$$A=\gamma gB\int_0^H(L_1\frac{x(H-x)}{H}+\frac{(L_2-L_1)x^2(H-x)}{H^2})dx=\frac{\gamma gBH^2}{12}(L_1+L_2)$$

No we can ask the following question: if volume is a given value, what should be length and breadth so that the pile occupies minimal warehouse area? To simplify calculations and cancel out some parameters let us introduce the value called “angle of natural slope”. This is the term from soil mechanics which means the angle that the substance makes against horizontal surface. It depends on density, viscosity and inter-particle friction and can be found in special tables.

Let:

$$\frac{2H}{B}=\tan \chi$$

$$H=\frac{1}{2}B\tan\chi;L_1=L_2-\frac{2H}{\tan\chi}=L-B$$

$$V=\frac{B^2\tan\chi}{12}(3L-B)$$

Assuming that V=const, obtain expression for L and insert it in the formula for the area S=LB:

$$L=\frac{4V}{B^2\tan\chi}+\frac{B}{3}$$

$$S=LB=\frac{4V}{B\tan\chi}+\frac{B^2}{3}$$

$$S'(B)=-\frac{4V}{B^2\tan\chi}+\frac{2B}{3}=0$$

$$B=\sqrt[3]{\frac{6V}{\tan\chi}}$$

$$L=\left(\frac{4}{\sqrt[3]{36}}+\frac{2}{3}\sqrt[3]{36}\right)\sqrt[3]{\frac{V}{\tan\chi}}$$

$$S=4\left(\frac{1}{\sqrt[3]{6}}+1\right)\left(\frac{V}{\tan\chi}\right)^{2/3}\approx 4.4\left(\frac{V}{\tan\chi}\right)^{2/3}$$

Now, assuming again that \(V=const\), let us determine the values of \(L\), \(B\) again which minimize the amount of energy required to form the pile. Again, substituting for \(L\), but now in the expression for \(A\) we obtain:

$$A=\frac{\gamma gBH^2}{12}(L_1+L_2)=\frac{\gamma g H^2}{48}\left(\frac{8VB}{\tan\chi}-\frac{B^4}{3}\right)$$

$$A'(B)=\frac{\gamma g \tan^2 \chi}{48}\left(\frac{8V}{\tan\chi}-\frac{4B^3}{3}\right)=0$$

$$B=\sqrt[3]{\frac{6V}{\tan\chi}}$$

Remarkably, we get the same values which were obtained to minimize the base area, although the approach is now quite different.

To complete the picture we can use the derived results to determine maximum allowed volume for a single pile if unitary pressure on the ground is limited by some given value \(q\) kg/m2:

$$\frac{V\gamma}{S}<q$$

$$V<\frac{qS}{\gamma}=\frac{q}{\gamma}4.4\left(\frac{V}{\tan\chi}\right)^{2/3}$$

$$V<\frac{(4.4q)^3}{\gamma^3\tan\chi}\approx 85.3\frac{q^3}{\gamma^3\tan\chi}$$

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