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Features: Discussing "Cosh"

Cosh (x) = ((exponent of x + exponent of -x) / 2.0) where a real (num), is raised to its natural antilogarithm.

y = cosh(x)

Catesian catenary formula: y = a cosh(x/a)

whereas a Cartesian parabola formula: y = ax2 + bx + c

Locus: "the set of all points whose location is determined by stated conditions"

The catenary (from the Latin word for chain) is the shape of a perfectly flexible chain suspended by its ends and acted on by gravity. Its equation was obtained by Leibniz, Huygens and Johann Bernoulli in 1691. They were responding to a challenge put out by Jacob Bernoulli to find the equation of the 'chain-curve'.

The catenary is the locus of the focus of a parabola rolling along a straight line.

The catenary is the evolute of the tractrix. It is the locus of the mid-point of the vertical line segment between the curves ex and e-x.

Euler showed in 1744 that a catenary revolved about its asymptote generates the only minimal surface of revolution.

Reference:
http://www-groups.dcs.st-and.ac.uk/~history/Curves/Catenary.html

Mathematical equations for St. Louis Arch:
http://www.nps.gov/jeff/equation.html


Dear Sir

Can you help me in calculating the length of a Catenary, which is generally y = a cosh(x/a).

Thanks

in the interval (u, v),
L = Integral sqrt(1 + y'^2)dx
= Integral sqrt(1 + sinh(x/a)^2)dx
= Integral sqrt(cosh(x/a)^2)dx (identity 1 + sinh(u)^2 = cosh(u)^2)
= Integral cosh(x/a)dx over (u, v)


Does anyone know the equation for the minimum tension of a catenary?
In other words, for a certain span, what length cable is required to minimize the tension at the ends of the cable. What is the depth of the catenary?

I think I have it worked out. Maybe I will have time to check
it over more closely tomorrow. I am assuming that the suspension
points are at the same elevation. We need to know the horizontal
distance between suspension points, and the linear weight density of
the cable.

Here are a few definitions:

a -- horizontal distance from the center to a suspension point
m -- linear weight density of the cable
W -- one half of the weight of the cable
T -- the tension at the middle (i.e., the horizontal component of the tension at any point on the cable)
F -- tension at a suspension point (We want to minimize F.)

The x-y coordinates are referenced to the origin at the lowest point on the cable. I am using this equation, taken from a web page I did last year: http://www.nas.com/~kunkel/hanging/hanging.htm

y = T/m[cosh(mx/T)-1]

At the right suspension point, the tension is F. Its horizontal component is T, and its vertical component is W. The cable is aligned in the direction of the net force, so its slope
is W/T.

W/T = y'(a) = sinh(ma/T)
W = T*sinh(ma/T)

F = sqrt[W^2 + T^2]
= sqrt[T^2*sinh^2(ma/T) + T^2]
= T*cosh(ma/T)

Minimize F by taking it derivative with respect to T and setting that equal to zero.

dF/dT = cosh(ma/T) - (ma/T)sinh(ma/T) = 0

==> tanh(ma/T) = T/(ma)

Remember, I am assuming that m and a are given conditions, so what I am looking for is T. I am not sure whether it is even possible to get an exact solution for this, but I do have this decimal approximation:

T = 0.8336ma approximately

To get answers to the questions, simply insert this value into the appropriate equations above.

tension at the suspension points:
F = T*cosh(ma/T)
= 0.8336ma*cosh(ma/0.8336ma)
= 0.8336ma*cosh(1.1997)
= 1.8509ma approximately

total length of cable:
S = 2W/m
= 2T/m*sinh(ma/T)
= 2*0.8336a*sinh(1.1997)
= 2.5156a approximately

depth of cable:
y = (0.8336ma/m)[cosh(ma/0.8336ma) - 1]
= 0.8336a[cosh(1.1997) - 1]
= 0.6753a approximately

As I said above, these calculations have not been checked closely, but you should have enough information here to check it yourself.

Kunkel


Two twenty foot tall poles are 50 feet apart. A 55 foot cable is suspended across them. What is the distance from the ground to the cable at the lowest point?

y=acosh(x/a)

Thank you for your help.
Sharon

If your cable shape is y = a*cosh(x/a) then:

You need to find an expression for the arc length of the cable between x = -25 an x = 25 (50 feet apart).

Set the arc length = 55 and solve for a,
The sag in the cable is a*cosh(25/a) - a*cosh(0) = a*cosh(25/a).

The height of the lowest point is 50 - a*cosh(25/a)

To find the arc length of y = f(x) from x=b to x=c,

ds = sqrt(1+f'(x)^2)*dx, where f'(x) = dy/dx.

Arc length = integral of ds from x=b to x=c


I have a question, my assignment is to determine the height of the lowest point of a 14ft cable hung between 2-5ft posts that are 10ft apart. Can you help? Must solve using Maple Language.

Equation of a catenary symmetrical with respect to the y-axis and with
the y-axis intercept B is

y = A[cosh(x/A) - 1] + B

Length of the catenary between x = -C and x = +C is

L = 2A sinh(C/A)

14 = 2A sinh(5/A)

Find the root of the function f(A) = 2A sinh(5/A) - 14 = 0 by
successive approximations:

A(k+1) = A(k) - f(A(k))/f'(A(k))

with an arbitrary initial approximation A(0), for example, A(0) = 1.

A = 3.40568... ft

5 = A[cosh(5/A) - 1] + B

B = 0.6212... ft


Two power poles are 25 feet tall and 27 feet apart. They are attached by a 30 foot long cable, that sags. If I wanted the sag center point to reach 10 feet above the ground, how close would I have to move the two poles towards each other to achieve this?

Notes
1) The solution is trivial. (Hint: One need not think about a catenary!)
2) This question (or one which was essentially the same) was used fairly
recently as "The Puzzler" on NPR's Car Talk.

Move poles one next to the other so that:
25ft-10ft=15ft , and
[15 ft]*2=30ft ,the length of the rope.
Panagiotis Stefanides
http://www.stefanides.gr

Presumably this is an exercise in a statics course. In the theory
of the catenary, does your text-book include the handy little equation
y^2 = c^2 + s^2 ? It looks like just what you need here.

Ken Pledger.



Please solve this equation:

b{sinh(10/b)-sinh(-10/b)}=21

I obtained it as the general solution of the length of arc of a catenary whose equation is y=bcosh(x/b)+c where b and c are constants with the limits of 10 and -10.length of arc was 21,we were to solve for b.
thank you

Hi,

With regards to the question on catenary on the 26th,i cant get an answer which satisfies the equation,the question says The eqn of a catenary is given as y=bcosh(x/b),and is hung btw 2 points of a, h and -a, h and the distance btw them is 20,the poles are 10metres above ground level(y=0).The catenary has a length of arc of 21. using this find the general eqn of the length of catenary, find b ,find c, and find the distance below the lowest point of the curve and the x axis from what i know,i thought b was the distance below the lowest point of the graph,so is this a trick question? secondly, I have obtained two eqns. 10=b cosh (10/b)(by saying x=0 as this makes c=o) 10.5=bsinh (10/b)(using the eqn for length of arc btw points x=0 and x=10.

I have tried solving both and im getting a negative value for b.Squaring both sides does not give me a value for b that when I substitute back into the length of arc eqn gives 10.5, pls help me out,i dont know what Iam doing wrong.

Thank you for your time, I appreciate it,

Are you related to Sharon? (Who posted this same problem on
math.undergrad on the 29th- although she gave the equation of the
catenary as y= b cosh(x/b)+ c, introducing a new unknown. You, on the
other hand, don't mention c except for that cryptic remark "10=b cosh
(10/b)(by saying x=0 as this makes c=o)" hmmm.)

You know that the catenary passes throught (-10,10) and (10,10) so
you know that 10= b cosh(10/b)- exactly what you have.

The derivative of y is sinh(x/b) so the differential of arc length
becomes sqrt(1+ y'^2)= sqrt(1+sinh^2(x/b))= cosh(x/b) and its integral
is bsinh(x/b) so the arclength between -10 and 10 is
2b sinh(10/b)= 21 (since sinh is an odd function) and therefore
b sinh(10/b)= 10.5- exactly as you have. My first thought was that
this is a transcendental equation that we would have to get a
numerical solution for but your idea works nicely: square both
equations to get b^2 cosh^2(10/b)= 100 and b^2 sinh^2(10/b)= 110.25.
OOOPS! b^2 cosh^2(10/b)- b^2 sinh^2(10/b))= b^2= -10.25. Yes, you
have a problem!

The problem is that IF y= b cosh(x/b) then the condition that when
x= 10, y= 10 alone would be enough to determine b- and so we could not
fit it to "arc-length 21".

May be that "c" is important after all! If y= b cosh(x/b)+ c
then the derivative is still the same and we still have, from the
arclength condition, that b sinh(10/b)= 10.5. But now, putting x= 10,
y= 10, we have b cosh(10/b)= 10- c. Squaring these two equations aand
subtracting gives b^2(cosh^2(10/b)- sinh^2(10/b))= b^2=
(10-c)^2- 10.5^2= c^2- 20c- 10.25. This gives a quadratic equation in
both b and c.

Looks to me like we have to solve the equation b sin(10/b)= 10.5
numerically after all. Use Newton's method for example. Then solve
the quadratic equation for c. The height of the cable at the center
is, of course, b cosh(0/b)+ c= b+ c.

I'm going to log off, solve for b and c and then get back to you.

Okay, I'm back. As I noted earlier (though it hasn't posted as I
type this), we really need y= b cosh(x/b)+ c to avoid get "b^2=
negative number". In that case, the height of the catenary at its
lowest point, x= 0, is given by b+ c.

We need to solve two equations: b sinh (10/b)= 10.5 (from the
arclength formula) and b cosh(10/b)+ c= 10 (height at x= 10). We can
rewrite the second equation as b cos(10/b)= 10- c. Now, proceed as
you did: square both equations and subtract:
b^2(cosh^2(10/b)- sinh^2(10/b)= b^2= 100- (10- c)^2= c^2-20c- 10.25.
(Without the c, this was b^2= -10.25, obviously impossible.)

I used Newton's method, with a programmable calculator (how very
nice that calculators have hyperbolic functions now!), to solve the
first equation, b sinh(10/b)= 10.5 to get b= 18.33 meters,
approximately. Putting that value for b into the second equation
gives the quadratic equation c^2- 20c- 384.54. Using the quadratic
formula, c= 31.18 or c= -11.18. It doesn't take much thought to see
that we need the negative value here. Since b= 18.33 meters and c=
-11.18 meters, the height of the catenary at x= 0 is b+c= 18.33-
11.18= 7.21 meters.


I need some help with the following problem please. there are two
poles that are 200m apart that a wire is spread across to make a
catenary. each post is 100m tall and the wire is 50m above the ground
at the lowest point. knowing the eq. y(x)=aCOSH(x/a)+b what values
must a and b be in order to fit the constraints? also, what is the
length of the cable. thanks.


A rope is suspended between two anchor points with a weight in the
middle. The angle of the rope at the weight is known as is the weight.
Is there a formula for calculating the load on the two anchor points?


Two twenty foot tall poles are 50 feet apart. A 55 foot cable is suspended across them. What is the distance from the ground to the cable at the lowest point?

y=acosh(x/a)

Thank you for your help.
Sharon

If your cable shape is y = a*cosh(x/a) then:

You need to find an expression for the arc length of the cable
between x = -25 an x = 25 (50 feet apart).

Set the arc length = 55 and solve for a,
The sag in the cable is a*cosh(25/a) - a*cosh(0) = a*cosh(25/a).

The height of the lowest point is 50 - a*cosh(25/a)

To find the arc length of y = f(x) from x=b to x=c,

ds = sqrt(1+f'(x)^2)*dx, where f'(x) = dy/dx.

Arc length = integral of ds from x=b to x=c


http://history.math.csusb.edu/Curves/Catenary.html

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