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  1. (Convergence Institute of Biomedical Engineering and Biomaterials. Seoul National University of Science and Technology, KoreaChuncheon Campus of Korea PolytechnicⅢ, Dept. of Electrical Engineering, Korea)



Intersection angle, Optimization, Shortest connection, Specific point.

1. Introduction

Situation 1 : Shortest connection of two transmission lines with intersection angle $\theta$

In Fig. 1, we see an existing transmission line X (x axis) and another transmission line Y’ (y’ axis) that intersect at point O(0,0) with intersection angle $\theta$. A hydro power plant is under construction at point Q between transmission line X and Y’.

A new transmission line is planned to connect two lines X and Y’ via the power plant. If the coordinate of the power plant Q is (p,q), what is the shortest length and what are the coordinates of x and y’ intercept of the new transmission line? Assume that all transmission lines are straight.

Situation 2 : Minimal connection of two straight roads with intersection angle $\theta$

In Fig. 2, we see along the lakeside a road X (x axis) and another road Y’ (y’ axis) that meet at point O(0,0) with intersection angle $\theta$. A building lies at point Q between road X and Y’.

Fig. 1. New transmission line connecting existing line X and Y’ with intersection angle $\theta$ via Power plant Q(p,q)

../../Resources/kiee/KIEE.2019.68.11.1440/fig1.png

A military road is planned to connect two roads X and Y’ via this building. We want this new road to be as short as possible, but it must be straight without bends because it may be used for landing of airplanes in case of emergency. And then, the building is to be used as the control tower of the airport. If the coordinate of the building is (p,q), what is the shortest length and what are the coordinates of x and y‘ intercept of new road?

Fig. 2. New road connecting x and y’ axis with intersection angle $\theta$ via Building Q(p, q)

../../Resources/kiee/KIEE.2019.68.11.1440/fig2.png

This paper presents a derivation to determine the length of the straight line that minimally connects two axes with intersection angle $\theta$ via a specific point (p,q). A formula, represented by a cube equation w.r.t. the coordinate of specific point (p,q) and intersection angle $\theta$, is derived to obtain the minimal length and x-y’ intercepts of the connection line using optimization technique (1).

2. Shortest connection of perpendicular x-y axes via specific point (p, q)

Suppose that two axes x and y’ are perpendicular to each other for Situation 1 of Section 1. Let point O(0,0) be the origin where axis x and axis y(=y’) intersect with $\theta =90^{\circ}$ as shown in Fig. 3. Let the length of new transmission line be L, and x intercept of L be point A(a, 0) and y intercept be point H(0, h). Then, we have,

(1)
$L=\sqrt{a^{2}+h^{2}}$

Fig. 3. Straight line L connecting perpendicular axes x,y via specific point Q(p,q)

../../Resources/kiee/KIEE.2019.68.11.1440/fig3.png

If (p, q) is the coordinate of the power plant Q, L becomes the length of straight line $\overline{A H}$ that passes through specific point Q(p, q), forming the hypotenuse of a right triangle HOA. What is the shortest length of the new transmission line when $\theta =90^{\circ}$? This problem can be formulated as an optimization problem minimizing the length L that passes through specific point Q(p, q). Now we take the length L of new transmission line as the objective function to be minimized(1-4) such that:

(1)
Minimize $L=\sqrt{a^{2}+h^{2}}$

(2)
$$subject to q = -\dfrac{h}{a}p + h$$ $$w.r.t. a, h$$

Constraint (2) implies that $\overline{A H}$ is a straight line connecting points A(a,0) and H(0,h) via specific point Q(p, q). To solve above optimization problem in a simpler way, let us rewrite the objective function (1) and constraint (2) as follows:

(1-1)
Minimize $L^{2}=a^{2}+h^{2}$

(2-1)
subject to $\dfrac{p}{a}+\dfrac{q}{h}- 1=0$ $$w.r.t. a, h$$

Defining Lagrange dual function Λ with (1-1) and (2-1):

(3)
$\Lambda =a^{2}+h^{2}+\lambda\left(\dfrac{p}{a}+\dfrac{q}{h}-1\right)$

we obtain the optimality conditions:

(4)
$\dfrac{\partial\Lambda}{\partial a}=2a+\lambda\left(-\dfrac{p}{a^{2}}\right)=0$

(5)
$\dfrac{\partial\Lambda}{\partial h}=2h+\lambda\left(-\dfrac{q}{h^{2}}\right)=0$

where λ is the Lagrangian multiplier.

From (4),

(6)
$\lambda =\dfrac{2a^{3}}{p}$

From (5),

(7)
$\lambda =\dfrac{2h^{3}}{q}$

Thus, we have from (6) and (7):

(8)
$\dfrac{2a^{3}}{p}=\dfrac{2h^{3}}{q}$

from which we finally obtain:

(9)
$\dfrac{h^{3}}{a^{3}}=\dfrac{q}{p}$

From (9), we obtain h/a - the optimal slope of $\overline{A H}$ via specific point (p, q) - as follows:

(10)
$\dfrac{h}{a}=\sqrt[3]{\dfrac{q}{p}}$

Equation (10) implies L, the length of $\overline{A H}$, is minimized when the slope of the new transmission line is given by cube root of the coordinates of specific point Q(p, q).

Substituting (10) into (2) yields:

(11)
$q=-\dfrac{h}{a}p+h=-\sqrt[3]{\dfrac{q}{p}}p+h$

Thus, we obtain,

(12)
$h=\sqrt[3]{\dfrac{q}{p}}p+q$

and from (10) and (12),

(13)
$a =\dfrac{h}{\sqrt[3]{\dfrac{q}{p}}}=\dfrac{\sqrt[3]{\dfrac{q}{p}} p + q}{\sqrt[3]{\dfrac{q}{p}}}= p +\dfrac{1}{\sqrt[3]{\dfrac{q}{p}}} q$

Now we can obtain the following formula for minimal length of the straight line passing through specific point Q(p, q) (4).

(14)
$L^{2}=a^{2}+h^{2}$ $=\left(\sqrt[3]{\dfrac{q}{p}}p+q\right)^{2}\bullet\left\{1+\left(\dfrac{1}{\sqrt[3]{\dfrac{q}{p}}}\right)^{2}\right\}$

or

(15)
$\therefore L=\left(\sqrt[3]{\dfrac{q}{p}}p+q\right)\bullet\sqrt{1+\left(\dfrac{1}{\sqrt[3]{\dfrac{q}{p}}}\right)^{2}}$

2.1 Case study for (p, q) =(1,8)

Assume that (p, q)= (1, 8). Then, from (10) we obtain h/a - the optimal slope of $\overline{A H}$ that minimizes the length of new transmission line - as follows:

(16)
$\dfrac{h}{a}=\sqrt[3]{\dfrac{q}{p}}=\sqrt[3]{\dfrac{8}{1}}=2$

Using formula (15), we obtain the minimal length of new transmission line connecting x-axis and y-axis as follows:(4)

(17)
$L=\left(\sqrt[3]{\dfrac{q}{p}}p+q\right)\bullet\sqrt{1+\left(\dfrac{1}{\sqrt[3]{\dfrac{q}{p}}}\right)^{2}}$ $=(2*1+8)\sqrt{1+\left(\dfrac{1}{2}\right)^{2}}$ $=\sqrt{125}\approx 11.18$

Fig 4 and Fig 4-1 show that, for specific point Q(1,8), the length $L=\sqrt{125}\approx 11.18$ of new transmission line obtained by (17) is the shortest when h/a=2 as given by (16).

Fig. 4. Graph of slope h/a vs length L of new transmission line connecting axes x,y via specific point Q(1,8)

../../Resources/kiee/KIEE.2019.68.11.1440/fig4.png

Fig. 4-1. Graph of slope h/a vs length L focused on near the optimal point (2,$\sqrt{125}$)

../../Resources/kiee/KIEE.2019.68.11.1440/fig4-1.png

And we can get the coordinate of y intercept H(0, 10) and x intercept A(5, 0) from (12) and (13) as follows:

(18)
$h=\sqrt[3]{\dfrac{q}{p}}p+q=2\bullet 1+8=10$

and

(19)
$a=\dfrac{h}{\sqrt[3]{\dfrac{q}{p}}}=\dfrac{10}{2}=5$

3. Shortest connection of two axes with intersection angle $\theta$ via specific point (p,q)

In Section 2, we derived the minimal length of the straight line forming the hypotenuse of a right triangle via specific point Q(p, q). What if the intersection angle of the two axes is not 90° but acute or obtuse?

For Situation 2 of Section 1, let the road X be axis x and road Y’ be axis y’. Let point O(0,0) be the origin where axis x and axis y’ intersect with intersection angle $\theta$ as shown in Fig. 5.

Let the length of new road be L and the x-axis intercept of L be point A(a, 0). And let the y’ axis intercept of L be point H and the length $\overline{OH}$ be h. Then, the coordinate of point H becomes $(h\cos\theta ,\: h\sin\theta)$ as shown in Fig 5.

If the coordinate of the building is (p,q), how can we connect axis x and axis y’ via specific point Q(p,q) with minimal length? And what are the coordinates of x and y‘ intercept of new road?

Fig. 1. Straight line L connecting axes x and y’ with intersection angle $\theta$ via specific point Q(p,q)

../../Resources/kiee/KIEE.2019.68.11.1440/fig5.png

The problem of Situation 2 can be formulated as an optimization problem minimizing the length of the straight line $L=\overline{AH}$ that connects x-y‘ axes via specific point Q(p, q).

Length L can be represented by cosine rule as follows:

(20)
$L=\sqrt{a^{2}+h^{2}-2ah\cos\theta}$ $=a\sqrt{1+\left(\dfrac{h}{a}\right)^{2}-2\left(\dfrac{h}{a}\right)\cos\theta}$

Now in Fig 5 we take the length L as the objective function to be minimized(5-6), such that:

(21)
minimize $L=\sqrt{a^{2}+h^{2}-2ah\cos\theta}$

(22)
subject to $p = -\dfrac{a-h\cos\theta}{h\sin\theta}q + a$ $$w.r.t. a, h $$

Constraint (22) implies that $\overline{A H}$ is a straight line connecting points A(a,0) and H$(h\cos\theta ,\: h\sin\theta)$ via specific point Q(p, q).

To solve above optimization problem in a simpler way, let us rewrite the objective function (21) and constraint (22) as follows:

(21-1)
minimize $L^{2}=a^{2}+h^{2}-2 ah\cos\theta$

(22-1)
subject to $\dfrac{p\sin\theta -q\cos\theta}{a}+\dfrac{q}{h}-\sin\theta = 0$ $$w.r.t. a, h$$

Defining Lagrange dual function D with (21-1) and (22-1):

(23)
$D=a^{2}+h^{2}-2ah\cos\theta$ $+\mu\left(\dfrac{p\sin\theta -q\cos\theta}{a}+\dfrac{q}{h}-\sin\theta\right)$

we obtain the optimality conditions:

(24)
$\dfrac{\partial D}{\partial a}=2a-2h\bullet\cos\theta +\mu\left(-\dfrac{p\sin\theta -q\cos\theta}{a^{2}}\right)=0$

(25)
$\dfrac{\partial D}{\partial h}=2h-2a\bullet\cos\theta +\mu\left(-\dfrac{q}{h^{2}}\right)=0$

Rewriting (24) and (25) w.r.t. the Lagrangian multiplier μ, we have:

(26)
$\mu =\dfrac{a^{2}(2a-2h\bullet\cos\theta)}{p\sin\theta -q\cos\theta}$

(27)
$\mu =\dfrac{h^{2}(2h-2a\bullet\cos\theta)}{q}$

Equalizing (26) and (27) yields:

(28)
$\dfrac{a^{3}-a^{2}h\cos\theta}{p\sin\theta -q\cos\theta}=\dfrac{h^{3}-ah^{2}\cos\theta}{q}$

Let us rewrite Equation (28) so that variables a and h occur only on RHS such that

(29)
$\dfrac{p\sin\theta -q\cos\theta}{q}=\dfrac{a^{3}-a^{2}h\cos\theta}{h^{3}-ah^{2}\cos\theta}$

Dividing the numerator and denominator of RHS by $a^{3}$ and rearranging (29), we finally obtain the following formula: [See Appendix A]

(30)
$\left(\dfrac{h}{a}\right)^{3}-\cos\theta\left(\dfrac{h}{a}\right)^{2}+\dfrac{\cos\theta}{\dfrac{p}{q}\sin\theta -\cos\theta}\left(\dfrac{h}{a}\right)=\dfrac{1}{\dfrac{p}{q}\sin\theta -\cos\theta}$

or

(30-1)
$$\left(\dfrac{h}{a}\right)^{3}-\cos\theta\left(\dfrac{h}{a}\right)^{2}+\dfrac{\cos\theta}{k}\left(\dfrac{h}{a}\right)-\dfrac{1}{k}=0$$ $$where k=\dfrac{p}{q}\sin\theta -\cos\theta$$

Equation (30) is the formula that derives the optimal ratio h/a by which minimal connection of two axes with intersection angle $\theta$ via specific point (p,q) can be obtained. Note that we can obtain the optimal ratio h/a by solving only one Equation (30) which is composed of two variables a and h.

4. Case studies

4.1 Case of $\theta =90^{\circ}$

Let us investigate Equation (30) for $\theta =90^{\circ}$.

Substituting $\theta =90^{\circ}$ into formula (30),

(31)
$$\left(\dfrac{h}{a}\right)^{3}-\cos 90^{^{\circ}}\left(\dfrac{h}{a}\right)^{2}+\dfrac{\cos 90^{^{\circ}}}{\dfrac{p}{q}\sin\theta -\cos\theta}\left(\dfrac{h}{a}\right)$$ $$=\dfrac{1}{\dfrac{p}{q}\sin 90^{^{\circ}}-\cos 90^{^{\circ}}}$$

we obtain the following relation;

(32)
$\left(\dfrac{h}{a}\right)^{3}=\dfrac{q}{p}$

We see that (32) is exactly the same equation that appears in (9) of Section 2.

4.2 Case of $\theta =60^{\circ}$

Now, let us check a case of $\theta\ne 90^{\circ}$. Assume $\theta =60^{\circ}$ and $(p,\: q)=(\sqrt{3},\: 1)$ as shown in Fig. 6.

Fig. 6. Straight line L connecting axes x and y’ with intersection angle $60^{\circ}$via specific point $(\sqrt{3},\: 1)$

../../Resources/kiee/KIEE.2019.68.11.1440/fig6.png

We can get the minimum connection length L and x-y’ intercepts a, h by solving formula (20), (22) and (30) simultaneously. However, we can obtain the optimal ratio h/a by solving only one cubic Equation (30) without any other equation. Therefore, the following step-by-step solution is also possible.

Step 1

Substituting $\theta =60^{\circ}$ and $(p,\: q)=(\sqrt{3},\: 1)$ into (30), we obtain the following relation:

(33)
$\left(\dfrac{h}{a}\right)^{3}-\cos 60^{^{\circ}}\left(\dfrac{h}{a}\right)^{2}+\dfrac{\cos 60^{^{\circ}}}{\dfrac{\sqrt{3}}{1}\sin 60^{^{\circ}}-\cos 60^{^{\circ}}}\left(\dfrac{h}{a}\right)$ $=\dfrac{1}{\dfrac{\sqrt{3}}{1}\sin 60^{^{\circ}}-\cos 60^{^{\circ}}}$ $\Rightarrow\left(\dfrac{h}{a}\right)^{3}-\dfrac{1}{2}\left(\dfrac{h}{a}\right)^{2}+\dfrac{1}{2}\left(\dfrac{h}{a}\right)-1=0$ $\Rightarrow\dfrac{h}{a}=1$

Step 2

From (22) and (33), we have the x-intercept of L;

(34)
$a = p+\dfrac{a-h\cos\theta}{h\sin\theta}q = p+\dfrac{1-\left(\dfrac{h}{a}\right)\cos\theta}{\left(\dfrac{h}{a}\right)\sin\theta}q$ $=\sqrt{3}+\dfrac{1-1\bullet\dfrac{1}{2}}{1\bullet\dfrac{\sqrt{3}}{2}}\bullet 1 =\dfrac{4}{\sqrt{3}}$

We see in (34) that the coordinate of x axis intercept of L is given by $\left(\dfrac{4}{\sqrt{3}},\: 0\right)$.

Step 3

From (20), we have the minimal connection length of L;

(35)
$L=a\sqrt{1+\left(\dfrac{h}{a}\right)^{2}-2\left(\dfrac{h}{a}\right)\cos\theta}$ $=\dfrac{4}{\sqrt{3}}\bullet\sqrt{1+1^{2}-2\bullet 1\bullet\dfrac{1}{2}}$ $=\dfrac{4}{\sqrt{3}}$

Fig. 7 and Fig. 7-1 show that, for the specific point $(p,\: q)=(\sqrt{3},\: 1)$ and $\theta =60^{\circ}$, the length $L=4/\sqrt{3}\approx 2.31$ of new road obtained by (35) is the shortest when h/a=1 as given by (33).

Fig. 7. Graph of ratio h/a vs length L via specific point $(\sqrt{3},\: 1)$ with intersection angle $\theta =60^{\circ}$

../../Resources/kiee/KIEE.2019.68.11.1440/fig7.png

Fig. 7-1. Graph of ratio h/a vs length L focused on near the optimal point (1, $4/\sqrt{3}$)

../../Resources/kiee/KIEE.2019.68.11.1440/fig7-1.png

Step 4

We obtain;

(36)
$h=\left(\dfrac{h}{a}\right)\bullet a=1\bullet\dfrac{4}{\sqrt{3}}=\dfrac{4}{\sqrt{3}}$

Equations (34), (35) and (36) imply that a, h and L make an equilateral triangle. We see in (36) and in Fig 5 that the coordinate of y’ intercept of L is given by:

(37)
$(h\cos\theta ,\:h\sin\theta)=\left(\dfrac{4}{\sqrt{3}}\bullet\dfrac{1}{2},\:\dfrac{4}{\sqrt{3}}\bullet\dfrac{\sqrt{3}}{2}\right)=\left(\dfrac{2}{\sqrt{3}},\:2\right)$

4.3 Case of $\theta =30^{\circ}$

Let us investigate the case for $\theta =30^{\circ}$and $(p,\: q)=$ $(\sqrt{3},\: 1)$ as shown in Fig. 8.

Fig. 8. Straight line L connecting axes x and y’with intersection angle $30^{\circ}$via specific point $(\sqrt{3},\: 1)$

../../Resources/kiee/KIEE.2019.68.11.1440/fig8.png

Step 1

Substituting $\theta =30^{\circ}$into (30),

(38)
$$\left(\dfrac{h}{a}\right)^{3}-\cos 30^{^{\circ}}\left(\dfrac{h}{a}\right)^{2}+\dfrac{\cos 30^{^{\circ}}}{\dfrac{\sqrt{3}}{1}\sin 30^{^{\circ}}-\cos 30^{^{\circ}}}\left(\dfrac{h}{a}\right)$$ $$=\dfrac{1}{\dfrac{\sqrt{3}}{1}\sin 30^{^{\circ}}-\cos 30^{^{\circ}}}$$

we obtain the following relation:[See (A3) in Appendix A]

(39)
$\dfrac{h}{a}=\dfrac{2}{\sqrt{3}}$

Step 2

From (22) and (39), we have:;

(40)
$$a=p+\dfrac{1-\left(\dfrac{h}{a}\right)\cos\theta}{\left(\dfrac{h}{a}\right)\sin\theta}q$$ $$=\sqrt{3}+\dfrac{1-\dfrac{2}{\sqrt{3}}\bullet\dfrac{\sqrt{3}}{2}}{\dfrac{2}{\sqrt{3}}\bullet\dfrac{1}{2}}\bullet 1$$ $$=\sqrt{3}$$

We see in (40) that the coordinate of x axis intercept of L is given by $(\sqrt{3},\: 0)$.

Step 3

From (20), we have the minimal length of L;

(41)
$$L=a\sqrt{1+\left(\dfrac{h}{a}\right)^{2}-2\left(\dfrac{h}{a}\right)\cos\theta}$$ $$=\sqrt{3}\bullet\sqrt{1+\left(\dfrac{2}{\sqrt{3}}\right)^{2}-2\bullet\left(\dfrac{2}{\sqrt{3}}\right)\bullet\dfrac{\sqrt{3}}{2}}$$ $$=1$$

Step 4

We obtain;

(42)
$h=\left(\dfrac{h}{a}\right)\bullet a=\dfrac{2}{\sqrt{3}}\bullet\sqrt{3}=2$

Equations (40), (41) and (42) mean that a, h and L make a right triangle that has h as its hypotenuse. In fact, $\theta =30^{\circ}$and $(p,\: q)=(\sqrt{3},\: 1)$ imply that axis y' passes through the very specific point $Q(p,\: q)=(\sqrt{3},\: 1)$ as shown in Fig 8. In order for length L to become the shortest, L must be perpendicular to x axis, that is, $\angle QAO$ must be $90^{\circ}$. Since $Q(p,\: q)=(\sqrt{3},\: 1)$ and $\angle QAO=90^{\circ}$, we have:

$a=\sqrt{3}$,

$L=1$

and

(43)
$h=2$

We see that solutions in (43) are the same as (40)~(42). We confirm again that solutions derived by formula (30) are optimal. The case for $\theta =120^{\circ}$ (obtuse case) is described in Appendix B.

5. Conclusion

This paper has presented a derivation to determine the length of a straight line that minimally connects two axes x and y’ with intersection angle $\theta$ via a specific point.

A formula, represented by a cube expression w.r.t. the coordinate of the specific point and intersection angle, has been derived to obtain the minimal length and x-y’ intercepts of the connection line using optimization technique.

Case studies have been discussed to confirm if the solutions derived by the proposed formula are optimal.

It is expected that the proposed formula can be a reference for optimal routing of roads, optimal routing of power transmission or communication lines, optimal routing of gas pipes, optimal design of IC circuits and etc. for minimizing the construction, production and/or operation cost.

The authors also hope this paper be used as an optimization lecture note for students majoring in electrical engineering.

Appendix A

Dividing the numerator and denominator of RHS by $a^{3}$ and rearranging (29), we get:

(A1)
$\dfrac{p}{q}sin\theta -\cos\theta =\dfrac{1-\left(\dfrac{h}{a}\right)\cos\theta}{\left(\dfrac{h}{a}\right)^{3}-\left(\dfrac{h}{a}\right)^{2}\cos\theta}$

Equation (A1) also can be expressed in a general cubic expression as follows:

(A2)
$\left(\dfrac{h}{a}\right)^{3}-\cos\theta\left(\dfrac{h}{a}\right)^{2}+\dfrac{\cos\theta}{k}\left(\dfrac{h}{a}\right)-\dfrac{1}{k}=0$

or

(A3)
$k\left(\dfrac{h}{a}\right)^{3}-k\cos\theta\left(\dfrac{h}{a}\right)^{2}+\cos\theta\left(\dfrac{h}{a}\right)-1=0$

where

(A4)
$k=\dfrac{p}{q}sin\theta -\cos\theta$

Appendix B

Case study for $\theta =120^{\circ}$and $(p,\: q)=(\sqrt{3},\: 1)$.

Fig. 1. Straight line L connecting axes x and y’ with intersection angle $120^{\circ}$ via specific point $(\sqrt{3},\: 1)$

../../Resources/kiee/KIEE.2019.68.11.1440/fig1.png

Step 1

From (A4), we have:

(B1)
$k=\dfrac{p}{q}\\sin\theta -\cos\theta =\dfrac{\sqrt{3}}{1}\bullet\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}=2$

Substituting $\cos 120^{\circ}= -1/2$ into (A2), we obtain the following relation:

(B2)
$\left(\dfrac{h}{a}\right)^{3}-\cos\theta\left(\dfrac{h}{a}\right)^{2}+\dfrac{\cos\theta}{k}\left(\dfrac{h}{a}\right)-\dfrac{1}{k}=0$ $\Rightarrow\left(\dfrac{h}{a}\right)^{3}-\cos 120^{^{\circ}}\left(\dfrac{h}{a}\right)^{2}+\dfrac{\cos 120^{^{\circ}}}{2}\left(\dfrac{h}{a}\right)-\dfrac{1}{2}=0$ $\Rightarrow\left(\dfrac{h}{a}\right)^{3}+\dfrac{1}{2}\left(\dfrac{h}{a}\right)^{2}-\dfrac{1}{4}\left(\dfrac{h}{a}\right)-\dfrac{1}{2}=0$ $\Rightarrow\dfrac{h}{a}= 0.7427$

Step 2

From (34) and (B2), we have:

(B3)
$a=p+\dfrac{1-\left(\dfrac{h}{a}\right)\cos\theta}{\left(\dfrac{h}{a}\right)\sin\theta}q$ $=\sqrt{3}+\dfrac{1-0.7427\bullet\left(-\dfrac{1}{2}\right)}{0.7427\bullet\dfrac{\sqrt{3}}{2}}\bullet 1$

We see in (B3) that the coordinate of x axis intercept of line L is given(3.8639, 0).

Step 3

(B4)
$L=a\sqrt{1+\left(\dfrac{h}{a}\right)^{2}-2\left(\dfrac{h}{a}\right)\cos\theta}$ $=3.8639\sqrt{1+0.7427^{2}-2\bullet 0.7427\bullet\dfrac{1}{2}}$ $=5.8529$

Step 4

We have;

(B5)
$h=\left(\dfrac{h}{a}\right)\bullet a=0.7427*3.8639=2.8701$

and the coordinate of y’ intercept of line L:

(B6)
$(h\cos\theta ,\:h\sin\theta)=\left(2.8701\bullet -\dfrac{1}{2},\:2.8701\bullet\dfrac{\sqrt{3}}{2}\right)$ $=(-1.4350,\: 2.4855)$

Appendix C

Calculation of transmission line impedance for a nuclear power plant

In Fig C1, we see along the seashore a T/L(transmission line) X (x axis) and T/L Y’ (y’ axis) that meet at point O(0,0) with intersection angle $45^{\circ}$.

A nuclear power plant is to be constructed at point N between T/L X and T/L Y’. According to the code and standard, a nuclear power plant needs to be connected to at

Fig. C1. New T/L connecting line X and Y’ with intersection angle $45^{\circ}$ via power plant N(3km, 1km)

../../Resources/kiee/KIEE.2019.68.11.1440/figc1.png

least two independent substations of 154 kV or higher level against emergency situation of the nuclear plant. (E.g., substation A and substation H in Fig C1)

So, a new T/L is planned to minimally connect two lines X and Y’ via the power plant. If the coordinate of the power plant is N(3km, 1km), what is the location of substation A and what is the series impedance of the new T/L from site N to substation A? where the series impedance per circuit-km of 345 kV T/L constructed for the power plant is given 0.0015+j0.0255 %/C-km.(7) Assume that the new T/L is straight from substation A to substation H.

Solution

Step 1

Fig C1 is the case for $\theta =45^{\circ}$and $(p,\: q)=$$(3 km,\:$ $1km)$.

From (A4), we have:

(C1)
$k=\dfrac{p}{q}\\sin\theta -\cos\theta =\dfrac{3}{1}\bullet\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$

Substituting $\cos 45^{\circ}=\dfrac{1}{\sqrt{2}}$ into (A3), we obtain the following relation:

(C2)
\begin{align*} \sqrt{2}\left(\dfrac{h}{a}\right)^{3}-\left(\dfrac{h}{a}\right)^{2}+\dfrac{1}{\sqrt{2}}\left(\dfrac{h}{a}\right)-1=0\\ \\ \Rightarrow\dfrac{h}{a}= 0.9569 \end{align*}

Step 2

From (34) and (C2), we have:

(C3)
$a= 3+\dfrac{1-\left(\dfrac{h}{a}\right)\bullet\dfrac{1}{\sqrt{2}}}{\left(\dfrac{h}{a}\right)\bullet\dfrac{1}{\sqrt{2}}}\bullet 1 = 3.4780$

We see in (C3) that the location of substation A(Coordinate of x intercept of new T/L) is given (3.4780km, 0).

Distance L from site N and substation A is:

(C4)
$L=\sqrt{(p-a)^{2}+q^{2}}=\sqrt{(3-a)^{2}+1}= 1.1084 km$

Thus, the total series impedance per circuit of new T/L from site N and substation A is given:

(C5)
\begin{align*} z = 1.1084 km\bullet(0.0015+j0.0255%/C-km)\\ \\ = 0.00166+j0.0283% \end{align*}

Optimal routing of the superconducting cable - which requires very high operation cost for maintaining cryogenic temperature of the superconductor in addition to very high cost for construction - can be one example for path optimization in the power sector.

References

저자소개

김주철(Ju-Chul Kim)
../../Resources/kiee/KIEE.2019.68.11.1440/au1.png

He has worked for S-D E&GC Co., Ltd, for 12 years since 2002 and used to be the Chief Executive of R&D Center.

He has been a professor of Chuncheon Campus of Korea Polytechnic University since 2014.

His research interest includes Power system optimization, Quiescent power cut-off and Human electric shock.

He published many papers on ELCB(Earth Leakage Circuit-Breakers), Human body protection against electric shock, Improvement of SPD, Quiescent power cut-off, and etc.

이상중(Sang-Joong Lee)
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He proposed ‘Angle reference transposition in power flow computation’ on IEEE Power Engineering Review in 2002, which describes that the loss sensitivities for all generators including the slack bus can be derived by specific assignment of the angle reference on a bus where no generation exists, while the angle reference has been specified conventionally on the slack bus. He applied these loss sensitivities derived by ‘Angle reference transposition’ to ‘Penalty factor calculation in ELD computation’ [IEEE Power Engineering Review 2002], ‘Optimal MW generation for system loss minimization’ [IEEE Trans 2003, 2006] and etc.

He worked for Korea Electric Power Corporation(KEPCO) for 22 years since 1976, mostly at Power System Research Center.

He has been a professor of Seoul National University of Science and Technology since 1998.

His research interest includes power generation, large power system and engineering mathematics.

He received Ph.D. at Chungnam National University in 1995.