# Three Elastics Equilibrium Point

• Dec 29th 2011, 09:38 AM
fluxe1963
Three Elastics Equilibrium Point
Three fixed points in 3D space are each connected to a mobile fourth point with three elastic bands. The elastic bands are slack when shorter than their rest length, and they pull according to Hooke's law (simple linear, constant). The challenge is to find where the fourth point is when the pull vectors from all three elastics add to zero. In other words: where would the mobile fourth point come to rest? There will obviously be solutions involving iterative approximation, but is there a direct way to determine the location? It is easy if the elastics pull from length zero onwards, because then you can find the fourth point by a weighted average of the fixed three. Does the simple addition of the elastic slack-zone suddenly make this a difficult problem?
• Dec 29th 2011, 01:51 PM
corsica
Re: Three Elastics Equilibrium Point
Hmm, I'm certain that this can solved analytically. It will require some university-level mathematics though. The Virtual Work method might be a good way to tackle this problem (not fully certain about that though, it has been a long time since I've done this kind of stuff).
• Dec 29th 2011, 02:05 PM
ILikeSerena
Re: Three Elastics Equilibrium Point
Welcome to MHF, fluxe1963! :)

Let's see...

Let's define the 3 fixed points to be a, b, and c.
And let's define the mobile 4th point to be p.

If the rest length is zero and the elastic coefficient is k, then the force on p by a is:
$\displaystyle F_a = k(a-p)$

Then:

$\displaystyle \sum F = 0; k(a-p) + k(b-p) + k(c-p) = 0$

$\displaystyle p = (a + b + c)/3$

Yes! That is the average of the fixed 3 points!
Hurray! ;)

Now let's assume a rest length L, then the force on p by a (assuming all elastics are under tension) is:
$\displaystyle F_a = k(|a - p| - L) {a - p \over |a - p|}$

So:

$\displaystyle \sum F = 0;$
$\displaystyle k(|a - p| - L) {a - p \over |a - p|} + k(|b - p| - L) {b - p \over |b - p|} + k(|c - p| - L) {c - p \over |c - p|} = 0$

That does not look nice at all!!! :(

I'm pretty sure you can not solve this analytically.
This will require a numerical approximation.
• Dec 29th 2011, 02:24 PM
corsica
Re: Three Elastics Equilibrium Point
Quote:

Originally Posted by ILikeSerena
$\displaystyle \sum F = 0;$
$\displaystyle k(|a - p| - L) {a - p \over |a - p|} + k(|b - p| - L) {b - p \over |b - p|} + k(|c - p| - L) {c - p \over |c - p|} = 0$

That does not look nice at all!!! :(

I'm pretty sure you can not solve this analytically.
This will require a numerical approximation.

Yeah, I agree that the force equilibrium looks pretty nasty in this case. :p I'm still convinced that this problem can be solved analytically though. This problem has three unknown (mathematically independent) variables: x, y, and z position of the fourth point. But we have an equal amount of equations at our disposal: the three Hooke equations (one for every elastic band). That must mean that this system of equations is solvable, right?(Speechless)
• Dec 29th 2011, 02:37 PM
ILikeSerena
Re: Three Elastics Equilibrium Point
Quote:

Originally Posted by corsica
Yeah, I agree that the force equilibrium looks pretty nasty in this case. :p I'm still convinced that this problem can be solved analytically though. This problem has three unknown (mathematically independent) variables: x, y, and z position of the fourth point. But we have an equal amount of equations at our disposal: the three Hooke equations (one for every elastic band). That must mean that this system of equations is solvable, right?(Speechless)

We can simplify it a little bit to:

$\displaystyle a + b + c - 3p = L \left( {a - p \over |a - p|} + {b - p \over |b - p|} + {c - p \over |c - p|} \right)$

This equation is actually 3 equations with $\displaystyle p_x, p_y, p_z$ as variables.

And we have for instance:
$\displaystyle |a-p| = \sqrt{(a_x-p_x)^2 + (a_y-p_y)^2 + (a_z-p_z)^2}$

I'm afraid this really is the system of equations to solve and it will have a unique solution.
However the elastics are treated as springs, so we would not be done yet.

Since this is a square root with squares in it in a denominator and we have 3 of those in separate terms, this gets ugly in a hurry!
• Dec 29th 2011, 02:52 PM
corsica
Re: Three Elastics Equilibrium Point
Yeah, I don't think anyone would be looking forward to solving that equation. That's why for these kind of problems you usually abandon force equilibrium equations, and switch to more advanced mechanics methods (like the earlier mentioned Virtual Work).
• Dec 30th 2011, 08:27 AM
fluxe1963
Re: Three Elastics Equilibrium Point
Wow, exciting to get such well considered responses so quickly on a forum like this! Thanks. BTW, I'm curious how the math notation is created apparently so easily. Anyway, I had worked it out algebraically in the same way on paper, but then just sat staring at it. I find it so fascinating (surprising, frustrating) that the simple addition of the elastic slackness is enough to make the problem explode from trivial to something that doesn't even seem solvable analytically. I wonder what this actually means. Is it really a different class of problem all of a sudden?
• Dec 30th 2011, 08:55 AM
corsica
Re: Three Elastics Equilibrium Point
I think I probably wouldn't call it a different class of problem yet. Yes, the addition of elastic slackness makes the equations a lot more complicated, but I would bet my right arm that they're still solvable analytically. Perhaps most importantly: the number of degrees of freedom in the system hasn't changed. It's still three (the three position components of the 4th point).

If you want to see something that will really blow your mind, have a look at this movie:
Double Pendulum - YouTube
The movie starts with a single pendulum, which has a very stable repetitive motion. But then another degree of freedom is added, and the motion of the pendulum becomes total unpredictable chaos. This is an example of where the class of problem does totally change (it has now become a chaos theory problem).

Oh, you can use LaTeX code to create math notation in your posts:
http://www.mathhelpforum.com/math-he...orial-266.html

Glad you're enjoying yourself here. :)
• Dec 30th 2011, 11:58 PM
fluxe1963
Re: Three Elastics Equilibrium Point
In the sense you mention the problem has not changed with slack added, because it's got the same degrees of freedom. In another sense it's completely different because there was a trivial solution with the weighted average and now that is replace with something that seems like a showstopper. I was trying to think of numerical solutions where the slack zone starts zero and gradually grows to its full size, but nothing would beat a non-iterative solution.
• Jan 10th 2012, 07:47 PM
weddingcoo
Re: Three Elastics Equilibrium Point
(Rock)Yeah, I agree that the force equilibrium looks pretty nasty in this case.
• Jan 15th 2012, 08:22 AM
fluxe1963
Re: Three Elastics Equilibrium Point
So that's it then? I introduce elastics with slack and the math explodes? Failing a direction for solving it, I wonder if anybody has anything philosophically interesting to say about why a seemingly simple addition causes these difficulties?
• Jan 15th 2012, 08:27 AM
ILikeSerena
Re: Three Elastics Equilibrium Point
Real world problems typically explode in complexity when real parameters are introduced.
The challenge is to find a model that is simple enough to be solvable, but complex enough to yield realistic results.
In practice numerical methods are used to solve complex problems, since in real life one is usually not interested in perfect solutions, only in satisfactory results.
• Jan 15th 2012, 08:30 AM
fluxe1963
Re: Three Elastics Equilibrium Point
But the slackness doesn't even seem like a "real" parameter. It just extends the model of a spring ever so slightly. Can you imagine a numerical solution which is at least an algorithm of predictable extent?
• Jan 15th 2012, 08:39 AM
ILikeSerena
Re: Three Elastics Equilibrium Point
The slackness introduces a neutral length offset which is "in the way" for a neat solution.
I didn't even address the fact that its push/pull is zero when slack, although for a numerical solution this is easily introduced.

What do you mean by "predictable extent" in an algorithm?