# Multi-spring system

1. Dec 4, 2008

### themiddle

Hi all,

I'm having some trouble understanding a multi-spring system.

Say I have 3 fixed nodes located at the following points in a 2D coordinate system:

node1: (2,4)
node2: (12,18)
node3: (30,6)

I have another node, node4, that is not fixed.

Then I connect node4 to all three fixed nodes using springs. The spring constants are as follows.

node4-node1: k1
node4-node2: k2
node4-node3: k3

My end goal is to determine the placement (coordinates) of node4, such that all the spring forces balance. i.e.

F_total = 0 = F1_4 + F2_4 + F3_4

How can I go about solving (or approximating) for the (x,y) coordinates of node4 such that the forces (or energy) of the system sums to zero?

2. Dec 4, 2008

### m.e.t.a.

Wouldn't you also need to know the rest lengths of the three springs and their respective maximum extensions? Or are we dealing with virtual springs which extend from 0 to infinity?

3. Dec 4, 2008

### themiddle

Yes, I'm hoping to use virtual springs that extend from zero to infinity.

I'm going to write an algorithm to approximate suitable (x,y) coordinates for node4 that result in equilibrium, but at this point I'm still trying to figure out how to set up the equation.

4. Dec 5, 2008

### m.e.t.a.

That is an interesting problem you have set yourself. I don't have much time to think about it at this moment but if the question is still unresolved by the time I get home then I'll take a proper look.

5. Dec 5, 2008

### themiddle

Thank you very much, any help would be greatly appreciated.

6. Dec 6, 2008

### m.e.t.a.

Just letting you know that I haven't forgotten about your question -- it is just that it is rather advanced for the likes of me. Having given up on solving algebraically, I have at the moment only two lines of thought, but both unsatisfactory.

My first instinct was to cheat by writing a simple algorithm which will rapidly home in on the coordinates of node4, for given values of k1, k2 and k3. The problem with this method is that it does not provide a general formula for the coordinates of node4, but instead must be re-run from scratch whenever you change any of the spring constants or node coordinates. Furthermore, while this method would provide a solution, it would not increase our understanding of how the solution was arrived at.

My second thought is that there might be a way of simplifying your 2-dimensional problem into a set of (easy) 1-dimensional problems, and then using these to somehow find the coordinates of node4. Clutching at straws, I wondered what would happen if you disconnected one of the springs (say, spring 3) from node4, allowing springs 1 & 2 to come to rest, then noted the centre-point between them; and then repeated the process for springs 1 and 2...as pictured below...would the centre-point of the new triangle (in red) be the location of node4? That is pure conjecture, though. I have no supporting proof other than that it "feels like node4 should be there". The following diagram shows the symmetrical case of an equilateral triangle with k1 = k2 = k3.

http://tinyurl.com/6qz3dl [Broken]

I'll work on the algorithm for the time being. Wish I could have been of help!

Last edited by a moderator: May 3, 2017
7. Dec 7, 2008

### m.e.t.a.

Well, I have a short program set up to find your node4 coordinates. If you have specific values of k1, k2 and k3 in mind then you're in luck; the coordinates of node4 can be found. For example, when k1 = k2 = k3 = 1, node4 lies very close to (12.81, 12.69). If you want node4's coordinates in terms of k1, k2 and k3, however, then I can't help you. Somebody else ought to be able to, though.

- m.e.t.a.

P.S. This disproves my guess that node4 might be found by drawing a kind of triangle. But then, it was quite a random guess.

8. Dec 7, 2008

Instead of messing with forces, you could try minimizing the potential energy. Specifically:

Given positions x1, x2, and x3 and spring constants k1, k2, and k3, you try and find x such that the quantity V = (k1|x - x1|2 + k2|x - x2|2 + k3|x - x3|2)/2 is minimized. You should be able to do this algebraically, since V is just a quadratic function of the coordinates of x.

9. Dec 11, 2008

### m.e.t.a.

Adriank: yes, that sounds a lot more sensible.

10. Dec 11, 2008

### hyperon

You don't have to do that actually, you can simply equate the sum of the forces to zero along the x and y axes.

$$F_1=k_1\left(\left(\begin{array}{c}2\\4\end{array}\right)-\left(\begin{array}{c}x\\y\end{array}\right)\right)=\left(\begin{array}{c}2k_1-k_1x\\4k_1-k_1y\end{array}\right)$$

$$F_2=k_2\left(\left(\begin{array}{c}12\\18\end{array}\right)-\left(\begin{array}{c}x\\y\end{array}\right)\right)=\left(\begin{array}{c}12k_2-k_2x\\18k_2-k_2y\end{array}\right)$$

$$F_3=k_3\left(\left(\begin{array}{c}30\\6\end{array}\right)-\left(\begin{array}{c}x\\y\end{array}\right)\right)=\left(\begin{array}{c}30k_3-k_3x\\6k_3-k_3y\end{array}\right)$$

$$F_{tot}=\left(\begin{array}{c}2k_1+12k_2+30k_3-(k_1+k_2+k_3)x\\4k_1+18k_2+6k_3-(k_1+k_2+k_3)y\end{array}\right)=0$$

$$\implies x=\frac{2k_1+12k_2+30k_3}{k_1+k_2+k_3}\mbox{, }y=\frac{4k_1+18k_2+6k_3}{k_1+k_2+k_3}$$