Calculate curve to make two curved panel fit together

[CVH]

The panel is 16" wide because two are held together with an interlocking octagon (notched) to make 32".

I don't think that you can do that for all 4 equally:

Joining octagon supports.png (5.37 KiB) Viewed 683 times

Argh! New page
3 other replies are on the bottom of page 1.

Regards,
CVH

[franksbedford]

I'm having so much fun and this is helping so much.

After reviewing your drawing, I drew the cord to connect the two points (I will fix the radius of the panel and the size of the slot in decimal later). Then I drew an orthogonal line, and trimmed it to the height of the top of the slot (where the arc connects). I drew another cord from the top of the slot to the center of the arc, and took that measurement, copy and pasted it into the two point and radius tool and drew an arc that seems to fit.

Is this reasonable? It's a repeatable technique for all four size domes that I can do.

[franksbedford]

The panel is 16" wide because two are held together with an interlocking octagon (notched) to make 32".

I don't think that you can do that for all 4 equally:


Joining octagon supports.png


Argh! New page
3 other replies are on the bottom of page 1.

Regards,
CVH

Yes. I see. Cardboard also smushes together, so maybe I can get away with that one.

Your notes on cardboard not stretching on the outer edge and compressing on the inner edge is very true. I need to think about that.

I use double wall cardboard (currently) for one piece -- the very front, which needs to be rigid to hold the casting material in place. Trial casting is this week, so well see you well the design holds up.

[CVH]

After reviewing your drawing,
A) I drew the cord to connect the two points (...).
B) Then I drew an orthogonal line, and trimmed it to the height of the top of the slot (where the arc connects).
C) I drew another cord from the top of the slot to the center of the arc, and took that measurement, copy and pasted it into the two point and radius tool and drew an arc that seems to fit.

Agreed it fits, almost, part of C is bad practice, the values that you copy are rounded/truncated numbers.
That is if you want the arc to be tangent to vertical at the top of the slot of course ...

Wen you have the center position and two endings then simply draw an arc from Center, Point and Angles (AR).
- Select CW in the Option Toolbar for this case.
- Indicate the center position by snapping to an end or to the intersection in my case.
(Or even to a manual intersection of 2 lines that don't actually cross)
- Indicate the radius by indicating a point on the expected circle/arc (point S or E)
- Indicate a start-angle by indicating a point in the required direction (point S)
- Indicate an end-angle by indicating a point in the required direction (point E)

Now it is as exact as can be, QCAD snapped to reference positions in full floating point notation.

Note that arcs do not have a start or end point, they have a center, a radius and start/end angles.
The conversion from Polar to Cartesian or reversed can never be exact, only as exact as possible.

It's a repeatable technique for all four size domes that I can do.

Yes, the middle orthogonal of a chord always traverses the center.
An orthogonal to a tangent through the tangent point also traverses the center. (In this case the top horizontal of the slot)
The center is where both orthogonal lines intersect.

Basically the main technique exploited by several arc and circle tools.
But CT1 fails for the vertical under the slot and point S and E for a circle, there is no similar arc tool, there are but two exploiting a tangent.

Your notes on cardboard not stretching on the outer edge and compressing on the inner edge is very true. I need to think about that.

Not only 'think about', we need the specifics to estimate the required length of the side covers.
On the provided development marked as 17.8125276 + 0% for now.
How wide they should be to snugly fit at the interior is already defined, adapting is merely a linear transformation.
But you should transform the major Fit-Point splines, not the trimmed Control-Point splines nor those exploded to polylines.

BTW: Your last version has still the gap indicated by red remark circles in d32_dome_arch (3).dxf
Sorry, but I had to diversify the different versions with a suffix as it seems you don't define revisions.

Regards,
CVH

[franksbedford]

Agreed it fits, almost, part of C is bad practice, the values that you copy are rounded/truncated numbers.
That is if you want the arc to be tangent to vertical at the top of the slot of course ...

Wen you have the center position and two endings then simply draw an arc from Center, Point and Angles (AR).
- Select CW in the Option Toolbar for this case.
- Indicate the center position by snapping to an end or to the intersection in my case.
(Or even to a manual intersection of 2 lines that don't actually cross)
- Indicate the radius by indicating a point on the expected circle/arc (point S or E)
- Indicate a start-angle by indicating a point in the required direction (point S)
- Indicate an end-angle by indicating a point in the required direction (point E)

Now it is as exact as can be, QCAD snapped to reference positions in full floating point notation.

I knew that. But it was late and I couldn't find the AR tool. So that is now fixed.

I have started tracking versions with 0.1 (pre-production). I've always done that on design drawings, not sure why I wasn't doing it on production drawings.

I fixed the radius of the piece, and the size of the notches (they started life as an estimate at 3/16) in decimal. The 0.1 file is attached.

Time, I think, to move on to the cover design and fixed-point spline.

Thanks again.

[franksbedford]

I checked to see if the method is repeatable for other size arc's and learned something. I drew an orthogonal line from the cord that connects the two arc end point, and then drew two boxes to get vertical and horizontal lines from those end points -- which gives three intersections. The longest tangent gives the radius of the arc that fits.

I was able to draw the 36 and 40" pieces in just a couple of minutes building off the 32" model. Attached.

Fun, fun.

[CVH]

FB,

The arc in d36_dome_arch_0,1.dxf is CW and tangent to vertical at the right.
Similar as what I proposed for the 32" version.

- - - - - -

The arc in d40_dome_arch_0,1.dxf is CCW and tangent to the top plane.

The left side of the vertical box is orthogonal to the 5" line and goes trough the point I called E.
The top side of the horizontal box is orthogonal to vertical and goes trough the point I called S.

Bottom line: For the 40" version you have indicated the wrong intersection.
Making the arc not tangent at the left side but tangent to the top plate.

It can't be tangent to both ... = 90°... Because dx = 14.85 and dy = 13.925
An arc with two distinct radii is again an ellipse arc.
I have no intentions of finding the development of two elliptical cylinders cut at 22.5°
http://www.mathouriste.eu/Brunelleschi/ ... eschi.html

Regards,
CVH

[franksbedford]

Lived in Florence for 5 years and spent more time inside that dome than most. But it didn't seem to improve my math skills.

I think I picked the right center point for the arc, and drew it CW.

Time to dry to draw a correct fixed width spline for the cover pieces.

[CVH]

Time to try to draw a correct fixed width spline for the cover pieces.

Remind that each arc size has its associated spline by Math.

If you settle for a certain radius or radii I would gladly provide the fitpoints.
The quick draft Excel sheet is still open, merely a question of entering a radius and copy/paste the coordinates.

Have you measured up the corrugated nature of your cover boards?
Thickness and number of full wave period per foot is fine.
Then we can estimate the required elongation.

Regards,
CVH

[franksbedford]

I believe I have the bottom width, top width, and overall length of the piece defined by the sides of the top and bottom octagon and the length of the arc drawn in the dome arch panel.

On to math. I have to admit it left me a little in the dust while I was concentrating on getting the arch panels drawn properly.

The cardboard is 0.15". Standard single wall, and tested with a caliper. With any luck I am not touching the 32, 36 and 40" arch panels again. I have not drawn the 28" one yet.

[CVH]

... I think you messed up the title in that last post ...

An octagon that spans 35.7 has a face of 14.78742417671961107, not 14.79
An octagon that spans 10 has a face of 4.14213562373105137, not 4.14
The arc is presumed not to be 20.23 long.

That looks again as rounded/truncated values.
The splines by Math would fail connecting exactly.

Regards,
CVH

[franksbedford]

Yep. I have now moved the actual octagons to the correct locations, including the length of the arc to 6 decimals, and increased the dimension digits. Hopefully this makes me ready to understand the spline widths.

[CVH]

FB,
Used the arc of d36_dome_arch_0,1.dxf
R = 12.9252188715954
D = 25.8504377431908
Sweep = -89.66656284 (8 digits)
=> Length = -20.2276670292275

Then the 73 fitpoints are:

Code: Select all

5.35380095325549,0
5.33342812427165,1.12793812925123
5.27246468685015,2.25587625850246
5.17137460956051,3.38381438775369
5.03092724893074,4.51175251700492
4.85219149417969,5.63969064625615
4.63652763232459,6.76762877550738
4.38557699557558,7.89556690475861
4.10124946980645,9.02350503400984
3.78570895916996,10.1514431632611
3.44135691748061,11.2793812925123
3.07081407170112,12.4073194217635
2.67690047662774,13.5352575510148
2.26261405257053,14.663195680266
1.83110776936954,15.7911338095172
1.38566565039055,16.9190719387684
0.929677779124289,18.0470100680197
0.466614498604102,19.1749481972709
0.0,20.3028863265221
-0.466614498604103,21.4308244557734
-0.929677779124291,22.5587625850246
-1.38566565039055,23.6867007142758
-1.83110776936954,24.814638843527
-2.26261405257053,25.9425769727783
-2.67690047662774,27.0705151020295
-3.07081407170112,28.1984532312807
-3.44135691748061,29.326391360532
-3.78570895916996,30.4543294897832
-4.10124946980645,31.5822676190344
-4.38557699557559,32.7102057482857
-4.63652763232459,33.8381438775369
-4.85219149417969,34.9660820067881
-5.03092724893074,36.0940201360393
-5.17137460956051,37.2219582652906
-5.27246468685015,38.3498963945418
-5.33342812427165,39.477834523793
-5.35380095325549,40.6057726530443
-5.33342812427165,41.7337107822955
-5.27246468685015,42.8616489115467
-5.17137460956051,43.989587040798
-5.03092724893074,45.1175251700492
-4.85219149417969,46.2454632993004
-4.63652763232459,47.3734014285516
-4.38557699557558,48.5013395578029
-4.10124946980645,49.6292776870541
-3.78570895916996,50.7572158163053
-3.44135691748061,51.8851539455566
-3.07081407170111,53.0130920748078
-2.67690047662774,54.141030204059
-2.26261405257053,55.2689683333102
-1.83110776936954,56.3969064625615
-1.38566565039055,57.5248445918127
-0.92967777912429,58.6527827210639
-0.466614498604103,59.7807208503152
0.0,60.9086589795664
0.466614498604105,62.0365971088176
0.929677779124292,63.1645352380689
1.38566565039055,64.2924733673201
1.83110776936954,65.4204114965713
2.26261405257054,66.5483496258226
2.67690047662774,67.6762877550738
3.07081407170111,68.804225884325
3.44135691748061,69.9321640135762
3.78570895916996,71.0601021428275
4.10124946980645,72.1880402720787
4.38557699557559,73.3159784013299
4.63652763232459,74.4439165305811
4.85219149417969,75.5718546598324
5.03092724893074,76.6997927890836
5.17137460956051,77.8277309183348
5.27246468685015,78.9556690475861
5.33342812427165,80.0836071768373
5.35380095325549,81.2115453060885

If we put point #36 on the top left corner of the 35.7 wide octagon in d36_dome_cover_0,2.dxf ...
... Then we have a fairly fit at the left lower corner of the top octagon.
FAIRLY ... Within 0.0013 ... Acceptable but in my former D32 example that was within 0.0000000000001.

Triple checked it but I can't put my finger on it ...
It is what it is.

# EDIT # Found it ... Your top octagon center is positioned 0.00128791164034022 to the right in regard with your 35.7" one.

Still need how many full corrugated waves there are in a foot.
Standard can be A-F, I then suspect C with about 40 waves per foot.
Can you confirm this.

Regards,
CVH

[CVH]

About 40 flutes per foot = 12 inches would ideally segment the outside with edges 0.3 long.
The outer curve of R = 12.92522 + 0.15 is 20.46241354 long.
But segmented it is about 20.46196722 long.
We need to keep the width of the developed interior to make it fit laterally, but the length needs to be scaled up by a factor of ~1.0115.

It matters, it is 0.23262" or about 6mm longer.
We scale the Fit-Point splines before trimming, before converting them to polylines.

Some trial and error may be required.

Regards,
CVH

[franksbedford]

FB,
Used the arc of d36_dome_arch_0,1.dxf
R = 12.9252188715954
D = 25.8504377431908
Sweep = -89.66656284 (8 digits)
=> Length = -20.2276670292275

Thanks! Can you explain (1) how you calculate the fit points, so I can do it myself at home. Is there a formula or website that does that calculation and creates the 73 fit points? I need to do it for all four size models, and I need to better understand the process to be self-sufficient. There will be larger ovens in the future, for example.

(2). How do I import the 73 points into Qcad to create the fit point spline?

Basic stuff I don't know how to do yet. I understand putting point #36 on the top left corner of the wide octagon for each model. I've been experimenting with your previous spline on the new drawings, and see how it fits.

Thanks again for all of your help!