1. ## Minimum Distance

I have a probem which involves an object sliding along a rail supported by a number of hangers.
I have got a fair way through this problem to the point where I have the total force and the total couple required to keep all the hangers in contact with the rail.
Thus I end up with two equations:
Ft = SUM(Fh) : Total Force = sum of the individual hanger forces
Mt = SUM(Fh x Ph) : Total Moment = sum of the cross product of the forces and the hanger positions.
I Know Ft, Mt and all the values of Ph. I want to find sensible values for Fh

This can be solved as a simple simultaneous equation for two hangers.
For 3 hangers this can be viewed as two planes and the line of intersection gives the range of possible solutions.
Choosing the point on this line closest to the origin (all forces zero point) minimises the maximum force acting on any of the hangers.
I want to do an equivalent minimisation of the maximum force for any number of hangers.

Unfortunately I seem to have reached the edge of my maths ability, can you help?

Mark.

2. You know, if only you can show us a diagram of the positions of your hangers with the forces acting on them, and all weights/forcers involved in your system, maybe we could help you.

3. ## Am I in the correct forum?

Don't get too bogged down with the rail/hanger modelling, The problem boils down to the two simple equations:

Ft = SUM(Fh) : Total Force = sum of the individual hanger forces
Mt = SUM(Fh x Ph) : Total Moment = sum of the cross product of the forces and the hanger positions.
I Know Ft, Mt and all the values of Ph. I want to find sensible values for Fh.

This gives a single point solution for 2 hangers, a line of solutions for 3 hangers, a plane for 4 hangers...
What I want is to find the point within these solution spaces that minimises the maximum value of Fh.
Perhaps I should have posted this question in the inequalities and graphing forum?
Mark.

4. I don't know about you, but I cannot figure how you, or I, can get, or can understand your moments, without seeing the diagram or locations of the forces and the hangers.

5. ## Consider the 3 hanger problem...

For three hangers I was thinking along the lines of:
Assume the force on hanger 1 is zero, this reduces the problem to a 2 hanger problem, that can be solved (simultaneous equn.).
Then assume the force on hanger 2 is zero, solve for the hangers 1 and 3.
This gives two points on the line of possible solutions to:
Ft=F1+F2+F3
Mt=F1xH1+F2xH2+F3xH3
Find the perpendicular to this line that passes through the point F1=0, F2=0, F3=0.
I'm looking for a method for solving this for any values of Ft, Mt, H1, H2 and H3 (H1,H2,H3 are all different).
If I can get that working I would like to extend to any number of hangers.
Am I barking up the wrong tree? Am I in the correct forest?

Mark.

6. Diagram. Showing the locations of the hangers. Then distances. Then directions of the forces. As you know, moment is force times perpendicular distance.

7. If I was in a position to provide you with a diagram etc. then I wouldn't be asking for help, I could work it out for myself.

The whole point is that I want a fully general method where the locations and number of hangers will vary from case to case.

I basically want an algorithm that I can incorporate into a computer program to allow me to get a sensible estimate of the individual hanger loads given the overall object loads and the hanger locations.

I had hoped that there might be a nice elegent mathematical solution, if there isn't then I will adopt a dirty sledgehammer iterative approach.

Mark.