Help with drawing. Is this over-defined and how do I draw it in 3D?

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robert123

Stainless

Joined

Apr 8, 2012

Location

AR, USA

• Sep 20, 2016

• #1

Please see the attached drawing. This is an old hand drawing that I am trying to decipher. It is a bit difficult to wrap your head around, but the 1.25" radius is looking down on the bent part. I think this means the curve of the flat pattern would not be a constant radius. I'm trying to figure out how to draw this in SolidWorks.

 

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sophtayl

Cast Iron

Joined

Apr 16, 2009

Location

Thirlmere, NSW, Australia

• Sep 20, 2016

• #2

Draw the centreline and use that as a path to follow and extrude a circle along path.
I expect they had a flat triangular ring of round material and bent it?

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sophtayl

Cast Iron

Joined

Apr 16, 2009

Location

Thirlmere, NSW, Australia

• Sep 20, 2016

• #3

I would guess it actually is a constant radius in the inclined plane. Its difficult to dimension it in the wrong plane tho.

It probably needs a separate view to be technically accurate but in the day people would use a bit of licence to get the message across as they were generally dealing with people used to seeing this type of thing.

Draw it flat. Kink it at the bend. Follow as per above and you should be fine

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robert123

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Apr 8, 2012

Location

AR, USA

• Sep 20, 2016

• #4

If I am to assume that they meant the 1.25" dimension is in the inclined plane, then what about the length of the inside in the bent condition? Is it 3.375" + 1.25" or 4.625"? Or do I assume that the 3.375" dimension also is the inflection point for the bend and use the 2.75" and 45 degree dimensions to determine the length? This yields about 4.8125".

 

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sophtayl

Cast Iron

Joined

Apr 16, 2009

Location

Thirlmere, NSW, Australia

• Sep 20, 2016

• #5

In drawing shapes like this you would typically draw the elevation and then project to find the plan intersection points. As its drawn the 3 3/8 projects throught the inflexion point. But the dimensions dont add up for the views shown. Is 2 3/4 the true length ot the projected elevation length. Reality is there arent enough dims to know 4 sure and you would need to scale off the drawing to clarify the discrepancies. It depends on what you are trying to achieve? Is this a real part you need to make or just an exercise in modelling?

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len_1962

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Joined

Dec 1, 2008

Location

Tempe

• Sep 20, 2016

• #6

Made this from one 2D sketch that I copied to the 45 deg plane.

then stared a 3D sketch and converted the lines and arcs of the center lines from the 2 planes.

then filleted the line intersections

then made a plane at the intersection of the fillet and line, drew a circle and the swept it.

 

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robert123

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Joined

Apr 8, 2012

Location

AR, USA

• Sep 21, 2016

• #7

This is a part we need to make. It appears the drawing is not correct. Looking at what len_1962 drew, there is significant distance between the 3.375" and 2.75" dimensions. This results in the side view looking different than the drawing (the legs are almost the same length). If drawn where the 3.375" and 2.75" dimensions meet, the side view looks right but the front view does not. I think I will have to make some measurements from the drawing and make some educated guesses.

 

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sophtayl

Cast Iron

Joined

Apr 16, 2009

Location

Thirlmere, NSW, Australia

• Sep 21, 2016

• #8

do you have a sample? maybe a few confirmation dimensions or clarification questions back to the client may clarify the issue if you can point to key data that is missing.

 

len_1962

Stainless

Joined

Dec 1, 2008

Location

Tempe

• Sep 21, 2016

• #9

I made a new one that is much closer.

So what I did different is I drew the side view using the 2.75 at 45 deg. then made the arcs tangent to the extents of the triangle sketch.

I extruded each sketch and deselected the merge results on the second extrude, then I used the combine tool, selected the common result circle, then applied .5 rads to the rest of the sharp edges, BAM! done.

here is a screen cast of the process.

Practical Machinist SW 1" dia bent triangle | Search | Autodesk Knowledge Network



even simpler, 2 sketches of the centerlines one on each plane, then use the curve\composite function, then make a circle sketch on the curve and sweep.

SW 1" bent triangle, 2 sketches, composit curve, sketched circle and a sweep | Search | Autodesk Knowledge Network

gotta love how one way leads into a simplistic faster more accurate model!

 

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R. Dan

Cast Iron

Joined

Sep 1, 2012

Location

NW Iowa

• Sep 24, 2016

• #10

No, this would be over-defined: