Algerbraic methods

Discussion in 'Math' started by Biggsy100, Jul 10, 2014.

Apr 7, 2014
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I 'BELIEVE' i am there with this question however just need a second opinion:

A power cable is strung between two pylons 40 metres apart at a height of 30.3 metres above level ground. At the midpoint between the two fixing points, the cable is 18.63 metres above the ground

• Determine the length of the cable given the following formulae:

L = 2a sinh (D/2)/a

which I understand
L= Length of cable
a = Lowest height of the cable
D = Distance between ends of the cable

So is this right?

L = (2x18.63)sinh (40/2)/18.63 = 8.158

2. studiot AAC Fanatic!

Nov 9, 2007
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Well I make it 48.14 m, if you read your formula correctly.

$L = 2a\sinh \left( {\frac{D}{{2a}}} \right)$

Which is not quite what you have calculated

See here for further examples and derivation.

http://euclid.trentu.ca/aejm/V4N1/Chatterjee.V4N1.pdf

Last edited: Jul 10, 2014

Apr 7, 2014
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Thank you for response

It clearly reads distance divided by 2 over a.

4. studiot AAC Fanatic!

Nov 9, 2007
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So are you clear on this one now?

Apr 7, 2014
88
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No because you have written it as

$L = 2a\sinh \left( {\frac{D}{{2a}}} \right)$

when the last part isclear written devided by 2?

Last edited by a moderator: Jul 13, 2014
6. studiot AAC Fanatic!

Nov 9, 2007
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The formula I have written is correct. It also corresponds to what you said in words in post#3

$\frac{{\frac{D}{2}}}{a} = \frac{D}{2}*\frac{1}{a} = \frac{D}{{2a}}$

Apr 7, 2014
88
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Ahhhr I see ok, do you know if there is a method of scanning and sending on here as then I can 'ping' over what I have so you are completely clear?

Apr 7, 2014
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9. studiot AAC Fanatic!

Nov 9, 2007
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Hello Biggsy,

You have two recent threads involving the soluution to differential equations, one involving sine functions and exponentials and this one involving hyperbolic sine function.

This is senior mathematics and I am not sure where you stand on that so I have prepared a simple explanation on the geometry of the curve taken up by the suspended cable. Please ask if you want more detail.

The solution to the catenary like yours is

y = a cosh(x/a), as shown in my diagram.

Now this is symmetrical about the y axis so if we measure from the origin where x=0 and double it we can calculate the curve length between two points on the curve.

The curve length, s is given by

s = a sinh(x/a)

So between two posts the horizontal distance is 2x and the curve length is 2s

My second diagram draws this out and puts the numbers in and shows how this leads to the formula I stated.

Please note that algebraic is not spelled as you have done with two r's.

Also if you wish to continue discussion on your other thread, feel free to post a question there and I will try for a similar explanation where it comes from.

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Biggsy100 likes this.

Apr 7, 2014
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Hi Thanks for clarifying that. However it is the formula at hand that I need the assistance with.

I have now completed the first one it is the second, regarding the instantaneous value.

I have scanned in my work sheet so everything is clear for you. Now, I am confident in the building process of putting the formula together. There are 2 parts that confuse me: -t/Tau, is the minus sign corresponding only to the time or both elements of the equation?

2nd and I need this slapped round my face with fish 2*pi*T? should this be inputted as seen?

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11. studiot AAC Fanatic!

Nov 9, 2007
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$y(t) = \left\{ {{e^{\left( {\frac{{ - t}}{\tau }} \right)}}} \right\}*\left\{ {\sin \left( {\frac{{2\pi t}}{T}} \right)} \right\}$

Does this help?

What subject / book is this paper from?

Do you understand what this equation is saying?

Apr 7, 2014
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Wow!Yes ,it is in an exponential so shows variables in change distribution

so I have t = 0.002 Tau=1.6 T=0.050

So if I understand it should be inputed as

e^-0.002/1.6 * sin(2pi/50) = 0.910

13. studiot AAC Fanatic!

Nov 9, 2007
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515

(and haven't you missed the t from the sin term?)

Apr 7, 2014
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The subject is from a HNC syllabus. When I enter the t as 2*pi*t it comes as a syntax error on my calculator?

15. MrAl Distinguished Member

Jun 17, 2014
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Hi,

When a question like the validity of a function comes up, sometimes you can use a straight line approximation to get a ballpark result that you can compare to several functions where you think only one is the right one.

For this example, we know the cable is 30.3m high and 40m long, and dips down to 18.63m. The shape of this cable can be approximated using two straight lines, one from each pole to the very center of the cable. That forms two right triangles, and then of course we can calculate the hypotenuse of the two triangles and add them to get an approximate cable length.
Doing this, the cable length approximation comes out to 46.3 meters. Since this is a straight line approximation we know the cable has to be longer than this, by a somewhat small amount.
This means that a length of around 48 meters would be reasonable while a length of about 8 or 9 meters would be completely impossible.
Note also anything under 46.3 meters would have to be incorrect because geometrically the cable must be longer than the straight line approximation. This means even 46 meters would be incorrect for example.

Biggsy100 likes this.

Apr 7, 2014
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Hi Studiot/all,

Here are my workings from our discussions, please can you tell me that I am on the right track or not?

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17. studiot AAC Fanatic!

Nov 9, 2007
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Remember that you should be working in radians, not degrees.

This emphasizes the need to always state your units