issues w/new sdk

Replies to This Discussion

Permalink Reply by David Rutten on September 3, 2010 at 12:05pm

Looking though this code now. I see at least one bug so far:

crv.Domain.Set(0, 1);

this line of code does nothing.

Domain is of type Rhino.Geometry.Interval, which is a ValueType. I.e. when you call crv.Domain you get a copy of the domain. You then proceed to modify this copy. This is different from the old SDK where ON_Interval was a class (reference type), not a value type.

Write this instead:

crv.Domain = new Interval(0, 1);

--
David Rutten
david@mcneel.com
Poprad, Slovakia

Permalink Reply by David Rutten on September 3, 2010 at 12:21pm

Also, when you create your NurbsCurves, I don't think you want to create periodic curves:

Rhino.Geometry.NurbsCurve.Create(true, 2, cPts);

I think you want to use false as the first argument.

--
David Rutten
david@mcneel.com
Poprad, Slovakia

Permalink Reply by brendon_carlin on September 4, 2010 at 6:24am

yep, thanks david..

i debugged everything but even though the code has the same structure i am getting a different result, if you have a minute could you take a look, because i feel like its some fundamental logic difference between the old sdk and the new.

thanks

Attachments:

• shortest paths3.ghx, 94 KB

Permalink Reply by David Rutten on September 4, 2010 at 7:25am

I'm not entirely certain what the algorithm is supposed to do, but I rewrote the script as I would have written it. It seems to generate smooth curves, though I don't know if this is what you are looking for.

Do note that you switched SDKs and programming languages at the same time. I'm not at all certain some of the loops have been translated from VB to C# correctly.

Another thing to watch out for is the different division behaviour in VB and C#:

Dim A As Int32 = 50
Dim B As Int32 = 8
Print("A/B = {0}", A / B)

will yield "A/B = 6.25", whereas:

int A = 50;
int B = 8;
Print("A/B = {0}", A / B);

will yield "A/B = 6"


In C#, when you divide two integers, the result will be an integer. In VB, the result is a double.

--
David Rutten
david@mcneel.com
Poprad, Slovakia

Attachments:

• shortestpaths3.ghx, 41 KB

Permalink Reply by brendon_carlin on September 4, 2010 at 8:41am

perfect, thanks a lot david.

curious, is it possible to redraw in .net.

if not, might it be possible to add a progress bar..?

thanks again

Permalink Reply by David Rutten on September 4, 2010 at 9:02am

Problem is that if the script crashes, the feedback form is not properly destroyed. You should probably add a try...catch...finally clause around the script ensuring form destruction.

--
David Rutten
david@mcneel.com
Poprad, Slovakia

Attachments:

• shortestpaths4.ghx, 42 KB

Permalink Reply by Alexander on September 16, 2010 at 2:51am

But how to reparameterize a surface with VB?

The srf.domain(int) is a function not a Property of a surface,you can't write something like
srf.domain(0)=new Interval(0, 1);

best,

Permalink Reply by Jon Mirtschin on September 16, 2010 at 3:05am

Hi,

I asked David this yesterday, I haven't validated yet, but
srf.SetDomain(0,new Interval(0,1));
srf.SetDomain(1,new Interval(0,1));

should accomplish this.

Cheers,

Jon

Permalink Reply by David Rutten on September 16, 2010 at 3:07am

srf.SetDomain(1, New Interval(0, 1))

I would have preferred two properties on Surfaces (UDomain and VDomain respectively), but we've got a single get and set method now that takes an integer for the direction.

--
David Rutten
david@mcneel.com
Poprad, Slovakia

Permalink Reply by Kyle Chou on September 5, 2010 at 7:26am

Hi guys,
I been looking at this code as well and I seem to have hard time understanding why the following lines of codes makes a big difference on the final outcome.

If dist_local < dist Then
motion = pt - cv(i)
dist = dist_local
End If

I have print out dist and realize its a huge number that dist_local cannot be greater than...so can someone how this work logically?

Permalink Reply by David Rutten on September 16, 2010 at 3:16am

< means smaller than, not larger than.

dist is set to the biggest possible number before we start, so we know that every distance we calculate will be smaller than the initial value. This way we can guarantee that we indeed found the smallest distance.

Let's assume the distance calculation (stored in dist_local) in our loop would return the following results: 35.9, 40.5, 33.3, 80.2, 22.4, 3.7

When this loop starts dist will be equal to 1.7977e+308 which is the highest possible number in computing. Then, as the loop runs, we get the following:

iteration | dist_local |         logic                  | dist
    0          35.9      35.9 is less than 1.7977e+308    35.9
    1          40.5      40.5 is not less than 35.9       35.9
    2          33.3      33.3 is less than 35.9           33.3
    3          80.2      80.2 is not less than 33.3       33.3
    4          22.4      22.4 is less than 33.3           22.4
    5           3.7       3.7 is less than 22.4            3.7

--
David Rutten
david@mcneel.com
Poprad, Slovakia

Permalink Reply by Kyle Chou on September 16, 2010 at 3:55am

thanks david, this really helps