Old School Drawing Exercises and 2D CAD

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

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.

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

G.

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

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

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

Neil

SODs latest:

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

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.

How about this?

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

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

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

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

MichaelG.

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

Michael and Neil have caused a complete meltdown here. Now I know I know nowt about triangles!

A guess; Neil means the centroid, incircle and orthocentre, all of which I had to look up and read four times. And there must be a million other triangle facts out there - my brain hurts.

Anyway, noticed this coincidence not mentioned by Wikipedia's Triangle entry. The inner circle is located by bisecting the angles (point I), whereas the outer circle (point O) is located by dropping verticals from the centre of each side. The coincidence is the white bisectors and green mid-verticals meet together on the perimeter of the outer circle when they're carried on outside the triangle (Points D, E and F).

More! The orthocentre is found by dropping the red verticals at a right angle into the triangle's opposite corner. They cross at the orthocentre (Point H). If the red and green lines are also carried across the circle, they too meet on the perimeter.

So all three centres I, H and O are all related to each other, and Point H is also the triangle's centre of gravity. There's deep logic in this I don't understand. Arggh!

Dave

Drawn with three arcs and it's not too far out, using SODs clue about the 1.5" arc.

Martin.

Edited By blowlamp on 03/07/2020 16:16:09

Hang on.. this is a moulding so to replicate it you need the negative. Release agent and plaster may do the job..(ducks and runs..)

pgk

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I’m sure you’re on the [long] road towards understanding, Dave

Sorry not to have participated ... I’ve been tinkering with microscopes, and leaving you to do all the work.

MichaelG.

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P.S. ___ Have you noticed that you are also, maybe, drawing a pentagram ?

https://en.wikipedia.org/wiki/Pentagram

Edited By Michael Gilligan on 03/07/2020 18:59:10

I'm still confused, the only constraints seem to be the 2 ends of the 60 degree construction line, and that fact that 1.5" radius comes into it somewhere. There are still an infinite number of solutions.

How did I do? one line produced from a 1.5R semicircle