Author Topic: Geometry problem: length of segments knowing the area, perimeter and width  (Read 8327 times)

0 Members and 1 Guest are viewing this topic.

Marc'Antonio Alessi

  • Swamp Rat
  • Posts: 1451
  • Marco
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #15 on: November 21, 2014, 12:33:04 PM »
Do you intend on making miter joints or butt joints?
I'm sorry but I did not understand the translation.

@owenwengerd:Thanks for answer.

 This should be the solution:

CAB

  • Global Moderator
  • Seagull
  • Posts: 10401
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #16 on: November 21, 2014, 01:20:59 PM »
See example
I've reached the age where the happy hour is a nap. (°¿°)
Windows 10 core i7 4790k 4Ghz 32GB GTX 970
Please support this web site.


Bethrine

  • Guest
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #18 on: November 21, 2014, 03:54:56 PM »
6550

You need the longer side.


CAB

  • Global Moderator
  • Seagull
  • Posts: 10401
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #20 on: November 21, 2014, 04:29:27 PM »
Well there is a better way. Need to add the blade widths
Green & magenta are flipped for cutting.
6550-6300=250 x250=62,500 in savings

You would need center line length from butt to butt.
« Last Edit: November 21, 2014, 04:42:13 PM by CAB »
I've reached the age where the happy hour is a nap. (°¿°)
Windows 10 core i7 4790k 4Ghz 32GB GTX 970
Please support this web site.

Marc'Antonio Alessi

  • Swamp Rat
  • Posts: 1451
  • Marco
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #21 on: November 21, 2014, 04:49:21 PM »
OK, this is valid. Now try with a sqared tube open on the top and closed with a cover so the magenta part can not be mirrored. Sorry for my english and the confusion.


CAB

  • Global Moderator
  • Seagull
  • Posts: 10401
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #22 on: November 21, 2014, 05:06:40 PM »
Not sure I understand.
I've reached the age where the happy hour is a nap. (°¿°)
Windows 10 core i7 4790k 4Ghz 32GB GTX 970
Please support this web site.

Marc'Antonio Alessi

  • Swamp Rat
  • Posts: 1451
  • Marco
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #23 on: November 22, 2014, 04:15:09 AM »
Sorry for my english:

1) a project may have from 20 to 200 of these conduits
2) the drawing of the conduits is 2D and with Polylines (closed or not > flag 70 = 0 or 1)
3) each conduits is made to measure and numbered
4) all joints are miter

Problem: find the total length of the conduits by considering each piece without any cut at 45 degrees.

I understand that with a program similar to the nesting is possible to calculate the optimized cutting considering all the conduits but, for simplicity, we consider as if the conduits are cut all at 90 degrees (before) and then 45 degrees where it serves to

I hope that my/google translation is understandable.

owenwengerd

  • Bull Frog
  • Posts: 451
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #24 on: November 22, 2014, 09:41:47 AM »
In that case the formula is (+ (/ enclosed-area conduit-width) (* conduit-width number-of-miters)).

Marc'Antonio Alessi

  • Swamp Rat
  • Posts: 1451
  • Marco
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #25 on: November 22, 2014, 09:58:55 AM »
In that case the formula is (+ (/ enclosed-area conduit-width) (* conduit-width number-of-miters)).
I had thought of such a thing, but how do I calculate the "number-of-miters"?

>>> 4) all joints are miter

In doubt, maybe I should specify: 4) all "L" joints are miter (not "T" joints)

CAB

  • Global Moderator
  • Seagull
  • Posts: 10401
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #26 on: November 22, 2014, 10:25:22 AM »
Wow, 25 post & we finally get to see the entire problem.

The conduit is a "Raceway" made of channel shape material
No interior obstructions allowed.
Connections will be miter joints.
Material is open one side & therefore can not be flipped upside down
Butt joints could be made on tee connections with a window cut in the "run" member.

Does that sum it up?  :)
I've reached the age where the happy hour is a nap. (°¿°)
Windows 10 core i7 4790k 4Ghz 32GB GTX 970
Please support this web site.

Marc'Antonio Alessi

  • Swamp Rat
  • Posts: 1451
  • Marco
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #27 on: November 22, 2014, 11:59:55 AM »
Wow, 25 post & we finally get to see the entire problem.

The conduit is a "Raceway" made of channel shape material
No interior obstructions allowed.
Connections will be miter joints.
Material is open one side & therefore can not be flipped upside down
Butt joints could be made on tee connections with a window cut in the "run" member.

Does that sum it up?  :)
In addition to the language problem is also that I do not deal with these constructions.
I humbly apologize...  :'( :oops:      :| now believe that nothing is missing. Thank you for your patience.

owenwengerd

  • Bull Frog
  • Posts: 451
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #28 on: November 22, 2014, 01:12:55 PM »
I had thought of such a thing, but how do I calculate the "number-of-miters"?

Each joint consists of two vertices. If the vertices are connected orthogonally, it is a butt joint; otherwise a miter joint. So the problem is reduced to grouping the vertices into orthogonal and non-orthogonal groups, then dividing the number of non-orthogonal vertices by two. I don't think this is a very difficult problem given that orthogonal vertices are always the conduit width apart.

Marc'Antonio Alessi

  • Swamp Rat
  • Posts: 1451
  • Marco
Re: Geometry problem: length of segments knowing the area, perimeter and width
« Reply #29 on: November 22, 2014, 02:57:03 PM »
I had thought of such a thing, but how do I calculate the "number-of-miters"?

Each joint consists of two vertices. If the vertices are connected orthogonally, it is a butt joint; otherwise a miter joint. So the problem is reduced to grouping the vertices into orthogonal and non-orthogonal groups, then dividing the number of non-orthogonal vertices by two. I don't think this is a very difficult problem given that orthogonal vertices are always the conduit width apart.
Thanks Owen, I think this is a good start. Good weekend!