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)
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)
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,
Fig. 3. Straight line L connecting perpendicular axes x,y via specific point Q(p,q)
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:
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:
Defining Lagrange dual function Λ with (1-1) and (2-1):
we obtain the optimality conditions:
where λ is the Lagrangian multiplier.
From (4),
From (5),
Thus, we have from (6) and (7):
from which we finally obtain:
From (9), we obtain h/a - the optimal slope of $\overline{A H}$ via specific point (p, q)
- as follows:
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:
Thus, we obtain,
and from (10) and (12),
Now we can obtain the following formula for minimal length of the straight line passing
through specific point Q(p, q) (4).
or
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:
Using formula (15), we obtain the minimal length of new transmission line connecting x-axis and y-axis
as follows:(4)
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)
Fig. 4-1. Graph of slope h/a vs length L focused on near the optimal point (2,$\sqrt{125}$)
And we can get the coordinate of y intercept H(0, 10) and x intercept A(5, 0) from
(12) and (13) as follows:
and
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)
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:
Now in Fig 5 we take the length L as the objective function to be minimized(5-6), such that:
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:
Defining Lagrange dual function D with (21-1) and (22-1):
we obtain the optimality conditions:
Rewriting (24) and (25) w.r.t. the Lagrangian multiplier μ, we have:
Equalizing (26) and (27) yields:
Let us rewrite Equation (28) so that variables a and h occur only on RHS such that
Dividing the numerator and denominator of RHS by $a^{3}$ and rearranging (29), we finally obtain the following formula: [See Appendix A]
or
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),
we obtain the following relation;
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 $(\sqrt{3},\: 1)$
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:
Step 2
From (22) and (33), we have the x-intercept of L;
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;
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 $\theta =60^{\circ}$
Fig. 7-1. Graph of ratio h/a vs length L focused on near the optimal point (1, $4/\sqrt{3}$)
Step 4
We obtain;
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:
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 $(\sqrt{3},\: 1)$
Step 1
Substituting $\theta =30^{\circ}$into (30),
we obtain the following relation:[See (A3) in Appendix A]
Step 2
From (22) and (39), we have:;
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;
Step 4
We obtain;
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
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:
Equation (A1) also can be expressed in a general cubic expression as follows:
or
where
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 $(\sqrt{3},\: 1)$
Step 1
From (A4), we have:
Substituting $\cos 120^{\circ}= -1/2$ into (A2), we obtain the following relation:
Step 2
From (34) and (B2), we have:
We see in (B3) that the coordinate of x axis intercept of line L is given(3.8639, 0).
Step 3
Step 4
We have;
and the coordinate of y’ intercept of line L:
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)
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:
Substituting $\cos 45^{\circ}=\dfrac{1}{\sqrt{2}}$ into (A3), we obtain the following
relation:
Step 2
From (34) and (C2), we have:
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:
Thus, the total series impedance per circuit of new T/L from site N and substation
A is given:
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.
저자소개
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.
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.