Math Is Fun Forum

dror

Guest

curve of string between two points

If you take a string of a fixed length, and stretch it between two points, then gradually take the edges of the string closer to each other, the string forms a curve. what would be the vertical distance from the original string line (when completely stretched) to the middle of the curve the string forms when the edges are getting closer?
I have done a little experiment at home, with a string of 100 cm, pulling the edges closer 10 cm at a time and measuring the distance from the middle of the curve to the original line, and got these results: (distance between edges --> vertical distance to middle of curve)
100 cm --> 0 cm
90 cm --> 16 cm
80 cm --> 24.3 cm
70 cm --> 30 cm
60 cm --> 34.3 cm
50 cm --> 38.4 cm
40 cm --> 41.2 cm
30 cm --> 43.6 cm
20 cm --> 45.2 cm
10 cm --> 46. 5 cm
I typed them into excel and got a very nice graph that looks exponential, but could not get a good function to relate to in (only high order polynomial functions, which turned out to be inconsistent with the results).
is there some geometric law that can help me figure out this relation theoretically? i would appreciate any help.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: curve of string between two points

Hi dror;

Can you provide a picture of the graph? The curve you are describing is a catenary. This is the curve a hanging chain will assume when acted upon by gravity.

Inconsistent? A long time ago I did that experiment. I found a useful fit using a parabola. That was when I did not know about anything but a polynomial.

Go here to get some background:

http://en.wikipedia.org/wiki/Catenary

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,649

Re: curve of string between two points

hi dror,

Using the Wiki equation

Put x = 0

So the lowest point is at (0,a)  see graph.

In your experiment you have measured down from the fixing points.  'a' is measured up from the x axis.

I have played about with your experimental values and found that a suitable position for the x axis is 60 cm below the fixing points

To obtain x values for the catenary equation I took your distances between the fixing points and halved them.  The table also shown below shows the resulting x values, the calculated catenary values (y), your experimental values and a last column which I shall now attempt to explain.

Comparing my y values and your experimental values, there is clearly a negative correlation.  I graphed the values on a scatter graph and found the line of best fit.  A formula of

gives the 4th column values.  It seems to be a good fit, so provides a mathematical model for your experiment.

Post back if you have any further questions on this.

Bob
999

Last edited by Bob (2011-07-19 01:05:55)

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