## 55 Reputation

12 years, 75 days

## Finding the Catenary...

Maple

Hi,

I posted a problem the other day to do with this specific problem and the issues got sorted out:)

The aim of the program is to find the catenary by iteration...

It's majorly diverging though and i'm not sure why.

Any ideas?

Thanks,

Rach

## Iterative scheme to get the catenary...

Maple

Hi everyone,

I recently posted about a problem I had with an iterative scheme in 2 dimensions.. someone answered that and it worked perfectly:)

I've now moved on to a slightly more complicated version. It's in 3d but only the r and z values will be used in the iteration (trying to get the catenary)

so it's not as hard as the main problem that I'm building towards.

Basically, I'm having trouble converting the points to polar coordinates and assigning them each a [j...

## HFloat undefined output: What does this ...

Maple

Hello,

This method works in 2d... but in 3d I get a HFloat undefined output when I try and substitute the actual values in to the function. If anyone could have look through the programming and see why this is, it would be really helpful!

Thanks,

Rach:)

The_Whole_Program_I.mw

## Repeating an iteration process...

Maple

Hey,

I know this might seem a bit trivial but I'm trying to get Maple to use iteration to get the shortest path between two points given initial conditions that aren't a straight line. (I'm learning how to do it in the simplest form before moving on to a 3D problem.)

So I have a initial set of points that are to be moved during the process and split them into their x and y components so there's a set of x[j]'s and y[j]s.Then I defined a couple of procedures to use in the iteration.

## Use known values for using alongside a p...

Hi,

I have a set of points defined for XX[j] and YY[j] where j=1..16,

I'm beginning to write a loop for these values and I was wondering how to get the actual numerical values in to say the following...

GX := proc (j)
local YY, XX, G;
G := (1/2)*(YY[j]-YY[j-1])^2/(XX[j]-XX[j-1])+(1/2)*(YY[j+1]-YY[j])^2/(XX[j+1]-XX[j]);
diff(G, XX[j]) end proc;

Then - for example - for GX(4), this gives an output in terms of XX[4], XX[5], etc but I'd...

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