Math Bricking 2D

[Lingwisyer]

These are two lines within my 3D script dictated by a boolean:

louvre_seal_width = ( 2 * d - 2 * dp * (n - 1) - 2 * gs_louvre_thk * cos (gs_louvre_opening_angle) ) / ( 2 * sin (gs_louvre_opening_angle) )

louvre_seal_width = ( d - dp * (n - 1) + 3 * Louvre_over * sin (gs_louvre_opening_angle) - 2 * gs_louvre_thk * cos (gs_louvre_opening_angle) ) / ( 2 * sin (gs_louvre_opening_angle) )

Any idea why including the first line bricks my 2D projection while the second line does not? They both use the same variables in similar ways...



Ling.

[Kristian Bursell]

Use a process of elimination to determine the piece of code that is causing the issue.

Its difficult to spot the issue without deeper exploration; what are "d", "n", "dp", and "Louvre_over". Typically "n" is used for loops, so is this in a loop?

Send though more of the code and an explanation of the values and it should be easier to work out.

[Lingwisyer]

Kristian wrote:
Typically "n" is used for loops, so is this in a loop?

Woops. Completely overlooked that fact. Probably the cause of some of my other minor issues...

The issue is the 1 / sinθ and it turns out the issue isn't equation specific as if I swap them around, the one that fulfils that boolean still causes it to brick. If I set it to a static value not equal to 0, it works fine...



Ling.

[A_ Smith]

Lingwisyer wrote:

The issue is the 1 / sinθ and...
If I set it to a static value not equal to 0, it works fine...

Maybe occurs the situation when sin (x) = 0 and you can't divide by 0 (zero) => 1/0 => error ?

[Lingwisyer]

A. wrote:

Lingwisyer wrote:

The issue is the 1 / sinθ and...
If I set it to a static value not equal to 0, it works fine...

Maybe occurs the situation when sin (x) = 0 and you can't divide by 0 (zero) => 1/0 => error ?

That was a thought, but there is the fact that both equations have include that situation and it only has an issue with one of them... Then there is the fact that it equalling zero is impossible within my object since my script sets it to as close to 1 as possible without causing a geometry clash.



Ling.

[Lingwisyer · Accepted solution]

Fixed by solving indirectly. Used the same inputs to solve for a new horizontal value to allow the use of Pythagoras...

Side A cos θ to side A tan θ, Pythagoras



Ling.