2014년 4월 28일 월요일

현수선 곡선

셤기간도 끝났으니 포스팅을 재개해보지. 후훗

우선 최근 들어서 곡선에 관심이 많이 생겼어. 그래서 오늘은 현수선 곡선에 대해 알려주도록 하지.

정의를 살펴보자. 현수선 곡선은 무엇일까?????
일단, 사이언스올에서 찾은 결과는 다음과 같다.
"쌍곡선 코사인 함수 y=a cosh x/a=a(ex/a+e-x/a)/2의 그래프로 나타낼 수 있는 곡선. 포물선 y=x2/(4a)이 x축 위를 미끄러지지 않고 구를 때 그 포물선의 초점이 그리는 도형이며, 또한 A(0, a)에 있는 추를 길이 a의 실로써 x축위로 끌고 갈 때 그 추가 그리는 곡선(이것을 트랙트릭스〈tracktrix〉라 한다)의 곡률중심(曲率中心)이 그리는 도형이기도 하다."
흥미롭게도 포물선을 굴렸을때 생기는 도형이다.
밀도가 균일한 실의 양끝을 지탱하여 중력장에 매달 때의 실은 이 곡선의 형상이 된다.
현수선을 x축의 둘레로 회전시켰을 때 생기는 곡면을 현수면이라 한다.
회전면인 극소곡면(평균곡률이 모든 점에서 0인 곡면)은 현수면밖에 없다.

증명과정을 첨부하도록 할게.
Let the path followed by the chain be given parametrically by r = (x, y) = (x(s), y(s)) where s represents arc length and r is the position vector. This is the natural parameterizationand has the property that
\frac{d\mathbf{r}}{ds}=\mathbf{u}\,
where u is a unit tangent vector.

Diagram of forces acting on a segment of a catenary from c to r. The forces are the tension T0 at c, the tension T at r, and the weight of the chain (0, −λgs). Since the chain is at rest the sum of these forces must be zero.
A differential equation for the curve may be derived as follows.[41] Let c be the lowest point on the chain, called the vertex of the catenary,[42] and measure the parameter s from c. Assume r is to the right of c since the other case is implied by symmetry. The forces acting on the section of the chain from c to r are the tension of the chain at c, the tension of the chain at r, and the weight of the chain. The tension at c is tangent to the curve at c and is therefore horizontal, and it pulls the section to the left so it may be written (−T0, 0) where T0 is the magnitude of the force. The tension at r is parallel to the curve at r and pulls the section to the right, so it may be written Tu=(Tcos φ, Tsin φ), where T is the magnitude of the force and φ is the angle between the curve at r and the x-axis (see tangential angle). Finally, the weight of the chain is represented by (0, −λgs) where λ is the mass per unit length, g is the acceleration of gravity and s is the length of chain between c and r.
The chain is in equilibrium so the sum of three forces is 0, therefore
T \cos \varphi = T_0\,
and
T \sin \varphi = \lambda gs,\,
and dividing these gives
\frac{dy}{dx}=\tan \varphi = \frac{\lambda gs}{T_0}.\,
It is convenient to write
a = \frac{T_0}{\lambda g}\,
which is the length of chain whose weight is equal in magnitude to the tension at c.[43] Then
\frac{dy}{dx}=\frac{s}{a}\,
is an equation defining the curve.
The horizontal component of the tension, Tcos φ = T0 is constant and the vertical component of the tension, Tsin φ = λgs is proportional to the length of chain between the r and the vertex.[44]

Derivation of equations for the curve[edit]

The differential equation given above can be solved to produce equations for the curve.[45]
From
\frac{dy}{dx} = \frac{s}{a},\,
the formula for arc length gives
\frac{ds}{dx} = \sqrt{1+\left(\dfrac{dy}{dx}\right)^2} = \frac{\sqrt{a^2+s^2}}{a}.\,
Then
\frac{dx}{ds} = \frac{1}{\frac{ds}{dx}} = \frac{a}{\sqrt{a^2+s^2}}\,
and
\frac{dy}{ds} = \frac{\frac{dy}{dx}}{\frac{ds}{dx}} = \frac{s}{\sqrt{a^2+s^2}}.\,
The second of these equations can be integrated to give
y = \sqrt{a^2+s^2} + \beta\,
and by shifting the position of the x-axis, β can be taken to be 0. Then
y = \sqrt{a^2+s^2},\ y^2=a^2+s^2.\,
The x-axis thus chosen is called the directrix of the catenary.
It follows that the magnitude of the tension at a point T = λgy which is proportional to the distance between the point and the directrix.[44]
The integral of expression for dx/ds can be found using standard techniques giving[46]
x = a\ \operatorname{arcsinh}(s/a) + \alpha.\,
and, again, by shifting the position of the y-axis, α can be taken to be 0. Then
x = a\ \operatorname{arcsinh}(s/a),\ s=a \sinh{x \over a}.\,
The y-axis thus chosen passes though the vertex and is called the axis of the catenary.
These results can be used to eliminate s giving
y = a \cosh \frac{x}{a}.\,

Alternative derivation[edit]

The differential equation can be solved using a different approach.[47]
From
s = a \tan \varphi\,
it follows that
\frac{dx}{d\varphi} = \frac{dx}{ds}\frac{ds}{d\varphi}=\cos \varphi \cdot a \sec^2 \varphi= a \sec \varphi\,
and
\frac{dy}{d\varphi} = \frac{dy}{ds}\frac{ds}{d\varphi}=\sin \varphi \cdot a \sec^2 \varphi= a \tan \varphi \sec \varphi.\,
Integrating gives,
x = a \ln(\sec \varphi + \tan \varphi) + \alpha,\,
and
y = a \sec \varphi + \beta.\,
As before, the x and y-axes can be shifted so α and β can be taken to be 0. Then
\sec \varphi + \tan \varphi = e^{x/a},\,
and taking the reciprocal of both sides
\sec \varphi - \tan \varphi = e^{-x/a}.\,
Adding and subtracting the last two equations then gives the solution
y = a \sec \varphi = a \cosh \tfrac{x}{a},\,
and
s = a \tan \varphi = a \sinh \tfrac{x}{a}.\,

Determining parameters[edit]

In general the parameter a and the position of the axis and directrix are not given but must be determined from other information. Typically, the information given is that the catenary is suspended at given points P1 and P2 and with given length s. The equation can be determined in this case as follows:[48] Relabel if necessary so that P1 is to the left of P2 and let h be the horizontal and v be the vertical distance from P1 to P2. Translate the axes so that the vertex of the catenary lies on the y-axis and its height a is adjusted so the catenary satisfies the standard equation of the curve
y = a \cosh \tfrac{x}{a}\,
and let the coordinates of P1 and P2 be (x1, y1) and (x2, y2) respectively. The curve passes through these points, so the difference of height is
v = a \cosh \tfrac{x_2}{a} - a \cosh \tfrac{x_1}{a}.\,
and the length of the curve from P1 to P2 is
s = a \sinh \tfrac{x_2}{a} - a \sinh \tfrac{x_1}{a}.\,
When s2−v2 is expanded using these expressions the result is
s^2-v^2=a^2(-2+2\cosh \tfrac{x_2-x_1}{a})=4a^2\sinh^2 \tfrac{h}{2a},\,
so
\sqrt{s^2-v^2}=2a\sinh \tfrac{h}{2a}.\,
This is a transcendental equation in a and must be solved numerically. It can be shown with the methods of calculus[49] that there is at most one solution with a>0 and so there is at most one position of equilibrium.

Generalizations with vertical force[edit]

Nonuniform chains[edit]

If the density of the chain is variable then the analysis above can be adapted to produce equations for the curve given the density, or given the curve to find the density.[50]
Let w denote the weight per unit length of the chain, then the weight of the chain has magnitude
\int w\ ds,\,
where the limits of integration are c and r. Balancing forces as in the uniform chain produces
T \cos \varphi = T_0\,
and
T \sin \varphi = \int w\ ds,\,
and therefore
\frac{dy}{dx}=\tan \varphi = \frac{1}{T_0} \int w\ ds.\,
Differentiation then gives
w=T_0 \frac{d}{ds}\frac{dy}{dx} = \frac{T_0 \frac{d^2y}{dx^2}}{\sqrt{1+\left(\frac{dy}{dx}\right)^2}}.\,
In terms of φ and the radius of curvature ρ this becomes
w= \frac{T_0}{\rho \cos^2 \varphi}.\,

Suspension bridge curve[edit]


Golden Gate Bridge. Most suspension bridge cables follow a parabolic, not a catenary curve, due to the weight of the roadway being much greater than that of the cable.
A similar analysis can be done to find the curve followed by the cable supporting a suspension bridge with a horizontal roadway.[51] If the weight of the roadway per unit length is w and the weight of the cable and the wire supporting the bridge is negligible in comparison, then the weight on the cable from c to r is wx where x is the horizontal distance between c to r. Proceeding as before gives the differential equation
\frac{dy}{dx}=\tan \varphi = \frac{w}{T_0}x.\,
This is solved by simple integration to get
y=\frac{w}{2T_0}x^2 + \beta\,
and so the cable follows a parabola. If the weight of the cable and supporting wires are not negligible then the analysis is more complex.[52]

Catenary of equal strength[edit]

In a catenary of equal strength, cable is strengthened according to the magnitude of the tension at each point, so its resistance to breaking is constant along its length. Assuming that the strength of the cable is proportional to its density per unit length, the weight, w, per unit length of the chain can be written T/c, where c is constant, and the analysis for nonuniform chains can be applied.[53]
In this case the equations for tension are
T \cos \varphi = T_0,\,
T \sin \varphi = \frac{1}{c}\int T\ ds.\,
Combining gives
c \tan \varphi = \int \sec \varphi\ ds\,
and by differentiation
c = \rho \cos \varphi\,
where ρ is the radius of curvature.
The solution to this is
y = c \ln \sec \frac{x}{c}.\,
In this case, the curve has vertical asymptotes and this limits the span to πc. Other relations are
x = c\varphi,\ s = \ln \tan \tfrac{1}{4} (\pi+2\varphi).\,
The curve was studied 1826 by Davies Gilbert and, apparently independently, by Gaspard-Gustave Coriolis in 1836.

Elastic catenary[edit]

In an elastic catenary, the chain is replaced by a spring which can stretch in response to tension. The spring is assumed to stretch in accordance with Hooke's Law. Specifically, ifp is the natural length of a section of spring, then the length of the spring with tension T applied has length
s=\left(1+\frac{T}{E}\right)p,\,
where E is a constant.[54] In the catenary the value of T is variable, but ratio remains valid at a local level, so[55]
\frac{ds}{dp}=1+\frac{T}{E}.
The curve followed by an elastic spring can now be derived following a similar method as for the inelastic spring.[56]
The equations for tension of the spring are
T \cos \varphi = T_0,\,
and
T \sin \varphi = \lambda_0 gp,\,
from which
\frac{dy}{dx}=\tan \varphi = \frac{\lambda_0 gp}{T_0},\ T=\sqrt{T_0^2+\lambda_0^2 g^2p^2},\,
where p is the natural length of the segment from c to r and λ0 is the mass per unit length of the spring with no tension and g is the acceleration of gravity. Write
a = \frac{T_0}{\lambda_0 g}\,
so
\frac{dy}{dx}=\tan \varphi = \frac{p}{a},\ T=\frac{T_0}{a}\sqrt{a^2+p^2}.
Then
\frac{dx}{ds} = \cos \varphi = \frac{T_0}{T}
and
\frac{dy}{ds} = \sin \varphi = \frac{\lambda_0 gp}{T},
from which
\frac{dx}{dp} = \frac{T_0}{T}\frac{ds}{dp} = T_0(\frac{1}{T}+\frac{1}{E})=\frac{a}{\sqrt{a^2+p^2}}+\frac{T_0}{E}
and
\frac{dy}{dp} = \frac{\lambda_0 gp}{T}\frac{ds}{dp} = \frac{T_0p}{a}(\frac{1}{T}+\frac{1}{E})=\frac{p}{\sqrt{a^2+p^2}}+\frac{T_0p}{Ea}.
Integrating gives the parametric equations
x=a \operatorname{arcsinh}(p/a)+\frac{T_0}{E}p + \alpha,
y=\sqrt{a^2+p^2}+\frac{T_0}{2Ea}p^2+\beta.
Again, the x and y-axes can be shifted so α and β can be taken to be 0. So
x=a\ \operatorname{arcsinh}(p/a)+\frac{T_0}{E}p,\,
y=\sqrt{a^2+p^2}+\frac{T_0}{2Ea}p^2\,
are parametric equations for the curve.

Other generalizations[edit]

Chain under a general force[edit]

With no assumptions have been made regarding the force G acting on the chain, the following analysis can be made.[57]
First, let T=T(s) be the force of tension as a function of s. The chain is flexible so it can only exert a force parallel to itself. Since tension is defined as the force that the chain exerts on itself, T must be parallel to the chain. In other words,
\mathbf{T} = T \mathbf{u},\,
where T is the magnitude of T and u is the unit tangent vector.
Second, let G=G(s) be the external force per unit length acting on a small segment of a chain as a function of s. The forces acting on the segment of the chain between s ands+Δs are the force of tension T(s+Δs) at one end of the segment, the nearly opposite force −T(s) at the other end, and the external force acting on the segment which is approximately GΔs. These forces must balance so
\mathbf{T}(s+\Delta s)-\mathbf{T}(s)+\mathbf{G}\Delta s \approx \mathbf{0}.\,
Divide by Δs and take the limit as Δs → 0 to obtain
\frac{d\mathbf{T}}{ds} + \mathbf{G} = \mathbf{0}.\,
These equations can be used as the starting point in the analysis of a flexible chain acting under any external force. In the case of the standard catenary, G = (0, −λg) where the chain has mass λ per unit length and g is the acceleration of gravity.

위키에서 베꼈는데 워낙 내용이 좋아서 영어는 쉬우니까 읽어봐 ㅎㅎ.
작성자: 시간:

댓글 없음:

댓글 쓰기

피드 구독하기: 댓글 (Atom)