Old School Drawing Exercises and 2D CAD

[SpurryParticipant @spurry]

Ah, but that needs a set square, ruler and pencil in addition to the compasses. Is that allowed now?

Pete

[IanTParticipant @iant]

It's a rainy day outside (at least that's my excuse) so I've been amusing myself with Open SCAD.

There was a post recently about 3D printed T-Slot fillers. A nice simple thing to draw in 3D CAD (just sketch & extrude he says with new found confidence) but I thought I'd do it in SCAD (a much better use of coffee-time than Suduko!  )

Here is the script:

//
//
// T-Slot Filler – IanT
//
L = 75; // T-Slot Length
HR = 2; // Hole Radius
BW = 14; // Bottom (of Slot) Width
TW = 8; // Top (of Slot) Width
BH = 4; // Bottom (of Slot) Height
TH = 9; // Top (of Slot) Height
//
//
difference () {
union () {
cube ([L,BW,BH], center=false);
translate ([0,(BW/2-TW/2),0])
cube ([L,TW,TH]); }
translate ([L/2,BW/2,0])
cylinder (TH,HR,HR);
}

I've used arbitrary dimensions above, so just change the variables to what you need. I've not printed this myself (I think it might actually be quicker to just cut some wooden strips and glue them together?) but it should be fine.

You'll need to download Open SCAD and then just cut and paste the 'script' above into the Editor. For anyone not wanting to go "full-on" 3D CAD – Open SCAD can be learned in small steps and is pretty simple once you've got a few basics (it looks more complicated than it is) – and it also helps keep the little grey cells ticking over.

Regards,

IanT

Edited By IanT on 02/07/2020 17:18:16

[IanTParticipant @iant]

It also occurred to me that a printed T-Nut might be useful for holding things in place (where no or low loads were involved) – such as DTI mounts etc…

I just changed the length to 15mm and hole radius to 1.5mm (but would use tapping size in practice)

Regards,

IanT

[Michael GilliganParticipant @michaelgilligan61133]

Posted by SillyOldDuffer on 02/07/2020 12:33:00:

.

Steps:

1. Drop the vertical line AD
2. Draw an arc from the corner ABC that crosses AD (in Blue on Diagram)
3. Draw two circles from the each end of the blue arc where it meets AB and BC (Yellow on Diagram)
4. Draw a line from B to the intersection of the yellow circles. (Light Blue on diagram) This line bisects the angle ABC, and where it crosses AD is the centre of the biggest circle that will fit inside the triangle.

 

.

With respect, Dave … Your Step 1 is cheating, and disqualifies you.

But it’s simply a matter of adding the ‘mirror equivalent‘ of Steps 2&3 to locate the centre point.

… This also provides a legitimate way of placing that vertical, by construction.

MichaelG.

Edited By Michael Gilligan on 03/07/2020 00:15:36

[Nigel Graham 2Participant @nigelgraham2]

That SCAD code looks rather like something someone at work once gave me, called POV-Ray.

Now, POV-Ray makes no pretence to be a CAD programme. It is or was intended to be purely artistic, and you can make it produce the most wonderful renderings of very pretty but utterly unfeasible or useless objects!.

It too uses command-lines that are fairly simple to grasp and somewhat similar to that SCAD example; and although engineering-drawing is not its intended purpose I wonder if in fact it, or an up-to-date edition, could be used for making 3D-printer files.

Something POV-Ray , or at least the version I have, lacks, are a grid or ruler-set, but I suspect you could make it generate at least a grid for the individual project.

[SillyOldDufferModerator @sillyoldduffer]

Posted by Michael Gilligan on 03/07/2020 00:07:14:
Posted by SillyOldDuffer on 02/07/2020 12:33:00:
.
Steps:
1. Drop the vertical line AD
2. Draw an arc from the corner ABC that crosses AD (in Blue on Diagram)
3. Draw two circles from the each end of the blue arc where it meets AB and BC (Yellow on Diagram)
4. Draw a line from B to the intersection of the yellow circles. (Light Blue on diagram) This line bisects the angle ABC, and where it crosses AD is the centre of the biggest circle that will fit inside the triangle.

.
With respect, Dave … Your Step 1 is cheating, and disqualifies you.
But it’s simply a matter of adding the ‘mirror equivalent‘ of Steps 2&3 to locate the centre point.
… This also provides a legitimate way of placing that vertical, by construction.
MichaelG.

Posted by Spurry on 02/07/2020 16:30:38:
Ah, but that needs a set square, ruler and pencil in addition to the compasses. Is that allowed now?
Pete

To Pete, anything goes in my workshop. All's fair in Love, War and Model Engineering. Triple points for innovative cheating in my book, but don't try it in an Exam!

To Michael, no need for the 'with respect' – like Ian's comment on dividing, you are completely orthogonal…

Dave

[Michael GilliganParticipant @michaelgilligan61133]

Posted by SillyOldDuffer on 03/07/2020 08:56:29:

.

To Michael, no need for the 'with respect'

.

The respect is genuine, Dave

You have provided another thought-provoking exercise to keep the little grey cells active.

MichaelG.

[Michael GilliganParticipant @michaelgilligan61133]

Posted by SillyOldDuffer on 03/07/2020 08:56:29:

.

… you are completely orthogonal…

Dave

.

… upon which subject ^^^, allow me to throw this into the pot **LINK**

It’s a very nice proof of Thales’ Theorem

MichaelG.

[Michael GilliganParticipant @michaelgilligan61133]

What remains of my brain is at risk of frying … So can anyone help me, please ?

Taking Dave’s triangle as an example:

We know that

1. the centre of the inscribed circle is located at the intersection of the ‘bisectors’ of the three angles
2. the centre of the circumscribed circle is located at the intersection the ‘perpendicular bisectors’ of the three sides

So … There must be some elegant relationship between those two facts

But what is it ?

A geometric demonstration would be appreciated

MichaelG.

[Neil WyattModerator @neilwyatt]

Posted by Michael Gilligan on 03/07/2020 11:33:39:

What remains of my brain is at risk of frying … So can anyone help me, please ?

Taking Dave’s triangle as an example:

We know that

1. the centre of the inscribed circle is located at the intersection of the ‘bisectors’ of the three angles
2. the centre of the circumscribed circle is located at the intersection the ‘perpendicular bisectors’ of the three sides

So … There must be some elegant relationship between those two facts

But what is it ?

A geometric demonstration would be appreciated

MichaelG.

As a rider…

There are actually THREE points defined solely by the triangle itself that all lie in a straight line.

Michael has mentioned TWO, what is the THIRD?

And what are their correct names (I had to look these up)?

Neil

[Neil WyattModerator @neilwyatt]

Posted by Gary Wooding on 02/07/2020 12:22:41:

I made this knurling tool some years ago and kept it in a cupboard together with 4 pairs of different knurls in a plastic bag. I often toyed with the idea of making a fitted box for it, but was put off by the awkward shapes. Eventually I bit the bullet, designed the 3D model in Fusion and am about to 3D print it.

2D CAD would have been very difficult.

An excellent example, one great thing about 3D printing is the ability to incorporate clips and hinges.

Neil

[SillyOldDufferModerator @sillyoldduffer]

While I ponder Michael's problem and worry from his video if Thales is pronounced Thay Lees or Talez, here's another hard one. Well I think it's hard!

Replicate the curve of this moulding by whatever method. Clue – it's generated from a 1.5" radius semicircle, but the semi-circle's origin isn't on the 60° line. Whilst a CAD package can draw the spline curve given relatively few points to follow, producing a 'fair curve' manually needs several.

Nigel complained about hard-to-make drawings and this might be a good example! But shapes like this are common in the real-world. Streamlining and decorative work demand tricky curves, and made cheaply too! My stair bannister is an example, and dining room coving is another. No sympathy for the poor old tool-maker – the curve has to look right!

Dave

[GeorgineerParticipant @georgineer]

This thread is taking some very interesting directions. Keep it up, lads!

And Neil,your knurler looks useful. Would it make a suitable subject for MEW?

George B.

[GeorgineerParticipant @georgineer]

SOD, when my wife worked for the engineering firm Thales it was pronounced Tallis!

G.

[Neil WyattModerator @neilwyatt]

Posted by SillyOldDuffer on 03/07/2020 12:23:07:

While I ponder Michael's problem and worry from his video if Thales is pronounced Thay Lees or Talez, here's another hard one. Well I think it's hard!

Replicate the curve of this moulding by whatever method. Clue – it's generated from a 1.5" radius semicircle, but the semi-circle's origin isn't on the 60° line. Whilst a CAD package can draw the spline curve given relatively few points to follow, producing a 'fair curve' manually needs several.

Nigel complained about hard-to-make drawings and this might be a good example! But shapes like this are common in the real-world. Streamlining and decorative work demand tricky curves, and made cheaply too! My stair bannister is an example, and dining room coving is another. No sympathy for the poor old tool-maker – the curve has to look right!

Dave

With CAD the curve is best done as a simple bezier with a control point at each end. Trivially simple.

Manually just as easy using French curves, the hard bit being finding mine if they turn out not to be where I put them years ago…

Neil

[Neil WyattModerator @neilwyatt]

Posted by Georgineer on 03/07/2020 12:26:27:

This thread is taking some very interesting directions. Keep it up, lads!

And Neil,your knurler looks useful. Would it make a suitable subject for MEW?

George B.

It's not mine, it's Gary's!

Not 100% sure but I think the design looks familar.

Neil

[duncan webster 1Participant @duncanwebster1]

SODs latest:

assuming we are to use 2 arcs, unless both radii are specified there are an infinite number of solutions

[blowlampParticipant @blowlamp]

My best guess for the centre location of the 1.5" rad arc is at about 2.5" to the right & 1.5" down from the top/left corner. I'm suspecting a 0.625" rad arc at the top and a 3.125" rad arc at the bottom to join all three together tangentially.

Martin.

[Gary WoodingParticipant @garywooding25363]

How about this?

[Gary WoodingParticipant @garywooding25363]

Posted by Neil Wyatt on 03/07/2020 12:57:44:

Posted by Georgineer on 03/07/2020 12:26:27:

This thread is taking some very interesting directions. Keep it up, lads!

And Neil,your knurler looks useful. Would it make a suitable subject for MEW?

George B.

It's not mine, it's Gary's!

Not 100% sure but I think the design looks familar.

Neil

It was published in issue 72 of MEW, called Yet Another Knurling Tool

[SillyOldDufferModerator @sillyoldduffer]

Posted by Neil Wyatt on 03/07/2020 12:56:48:

Posted by SillyOldDuffer on 03/07/2020 12:23:07:

…

With CAD the curve is best done as a simple bezier with a control point at each end. Trivially simple.

Manually just as easy using French curves, the hard bit being finding mine if they turn out not to be where I put them years ago…

Neil

CAD may make drawing Bezier curves easy thank goodness, but are they trivially simple?

Doing this manually, I'd use a French curve at different angles to fair the curve, but it would be lucky to find an exact match. One of mine comes close, but not at both ends:

two or more

And someone designed that French curve on a drawing board in the first place! How was it done?

Re Duncan's comment, I'm afraid the curve isn't drawn with arcs. (Maybe it could be?)

Once the approach is explained, easy enough to do manually, with Qcad, or with any other competent 2D CAD package. Not sure about FreeCAD & Fusion360's sketch tools for this particular job. They might not have the 2D primitives needed to construct the curve. (I haven't checked.) They can both do spline curves, but in this example it might be easier to construct the outline in 2D CAD and import the answer rather than sketch it.

Dave

[Gary WoodingParticipant @garywooding25363]

Posted by SillyOldDuffer on 03/07/2020 13:55:13:

Once the approach is explained, easy enough to do manually, with Qcad, or with any other competent 2D CAD package. Not sure about FreeCAD & Fusion360's sketch tools for this particular job. They might not have the 2D primitives needed to construct the curve. (I haven't checked.) They can both do spline curves, but in this example it might be easier to construct the outline in 2D CAD and import the answer rather than sketch it.

Dave

My effort was done with Fusion sketch. What's wrong with it?

[SillyOldDufferModerator @sillyoldduffer]

Posted by Gary Wooding on 03/07/2020 13:45:41:

How about this?

 

Pretty close, your curve goes slightly deeper than mine:

I'm not complaining!   Fusion did better than I thought it would!

Dave

Edited By SillyOldDuffer on 03/07/2020 14:16:39

[Michael GilliganParticipant @michaelgilligan61133]

Posted by SillyOldDuffer on 03/07/2020 13:55:13:

And someone designed that French curve on a drawing board in the first place! How was it done?

…

.

Not sure … but I think it is literally a spline curve, so created with a suitably springy flexible strip and some pins.

MichaelG.

[Martin ConnellyParticipant @martinconnelly55370]

To me it looks like the curve produced by rotating a circle 45 degrees in two planes. At 1.5 radius this would produce an oval 3 high 2.12 wide. Rotate this 45 degrees anticlockwise to produce the example given.

Martin C

[Author]

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