# Is there a formula to calculate load on multiple ropes?

1. Aug 15, 2009

### bobpiet

I need to find a means to calculate the load on three or four ropes that are supporting a weight between them. These ropes radiate from the load at various angles, typically non-planer. The attached illustration shows a three rope example with the individual ropes, yellow, green, blue, projecting out at various angles, imagine they are attached to the inside of a large sphere for instance.

I would like to build a spreadsheet that would allow me to plug in the load weight, the angles of the 4 ropes and have the formulae calculate the load on each rope.

I am not a mathametician (often wish I were) but am comfortable building spreadsheets with complex formulas.

I'm not sure how to represent these angles in X, Y, Z space around the load either.

Anyone have a suggestion?

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2. Aug 16, 2009

### tiny-tim

Welcome to PF!

Hi bobpiet! Welcome to PF!
For four ropes, there's no unique solution …

for the same reason that you can't tell the load on the legs of a table with four legs.

For three ropes, just take components of force in the direction perpendicular to two of the ropes … that will leave only the weight and the load on the third rope.

Use the usual (x,y,z) components for each rope (with x2 + y2 + z2 = 1), and the cross-product to find the perpendicular.

3. Aug 17, 2009

### bobpiet

This is why I find math difficult.

I suspect that if I hang a weight from four ropes - emmenating from from various angles, and measure the load on each rope, each individual rope has a single load on it.

So in reality, the four ropes settle into a single unique solution, but in math it is unsolvable - interesting.

4. Aug 18, 2009

### tiny-tim

Yes, you need more conditions (for example, the deformability of the load) …

3 equations, 3 unknowns, is ok;

3 equations, 4 unknowns, gives you an infinite number of solutions, and if you want a unique solution, you need to find one more equation.

5. Aug 18, 2009

### Staff: Mentor

There is nothing wrong with the math. Just your description of the system is incomplete.

Your problem is a basic problem in statics. Simply specify the mechanical properties of your ropes and you will get a unique solution. The simplest specification that does the trick is the length of each rope and its Young's modulus.

Unfortunately, the Wikipedia entry on statics is useless and I didn't have time to find a good alternative. Sorry.

6. Aug 18, 2009

### bobpiet

I only need the loads on each rope as an estimate;

Thus the load can be idealized as a non-deformable sphere providing, therefore, a center of mass at the theoretical verticies of all four ropes within the load, cartisian coordinate 0,0,0.

The ropes as well can be idealized as non-deformable weightless entities, thus making thier length irrelevant and could be defined as passing from 0,0,0 (all of them would originate at 0,0,0) through some other coordinate, e.g. -10,10,10 if that was the best way to describe the angle in space.

Does this simplify or make possible a calculation with four ropes?

7. Aug 18, 2009

### Staff: Mentor

No, your ropes need to be deformable to have a sufficient number of equations. Let's say that you have three non-deformable ropes in specified positions. So the only unknowns are the tensions in the three ropes. That is three unknowns, so you need three equations. So you can write equilibrium equations for the sum of the forces in x, y, and z directions. Three equations in three unknowns.

Now, let's say that you add one more non-deformable rope in a specified position. Now, your number of unknowns has increased to 4 but there are still just 3 equilibrium equations. Three equations in four unknowns is not solvable.

So, instead of saying that the ropes are non-deformable let's use Hooke's law to relate the tension in each rope to the deformation of the rope. Now, we displace the load by an unknown amount, thereby deforming all of the ropes. That gives us 3 unknowns in the displacement and 4 unknowns in the tension. We can still make 3 equations from the equilibrium conditions, and now we can use Hooke's law 4 times (once for each rope) to generate four more equations. Then we have 7 equations in 7 unknowns and can solve.

8. Aug 19, 2009

### bobpiet

I find this very fascinating.

Any Idea what type of information regarding the ropes or cables I would need in order to plug them into the formula? e.g. would Young's modulus be in the specification or is that something that is derived from other data?

What would be the best spec to use in order to complete the four rope problem?

Thanks.

9. Aug 19, 2009

### kyiydnlm

Ropes have original length and are attached on certain points. you can assume a location of the load; then, find the equations of elongation and tension for equilibrium. You may need matrix if you do not want headaches.

10. Aug 20, 2009

### Staff: Mentor

The Young's modulus will vary based on the material used, but it is such an important material property that it is widely reported. Here is a page that gives a brief overview on the subject as well as some charts and tables with the Young's modulus of several kinds of fiber:
http://www-materials.eng.cam.ac.uk/mpsite/short/OCR/ropes/default.html

So you take Young's modulus, multiply by the cross-sectional area, and divide by the unstressed length, and you have a "spring constant" that you can use in your equations. If you want to simplify your problem then you can just have Young's modulus, the area, and the length all be the same for each rope. If you want to provide only the direction for each rope then you can specify that all of the ropes are attached to a point on a sphere centered on the origin of radius equal to the unstressed length of the ropes. Then, given some position of the load you can calculate the change in length of each rope, which when multiplied by the "spring constant" gives you the tension in each rope, which gives you the net force on the load. Solve for where that equals 0.