My very rough method (on page 1) also had a volume of 9868 - although the decimal places beyond that are different - could my method have almost averaged out the very complicated shape of the actual problem?
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durhamplumber
Vertices =corners,precisely.
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Hmmm did you produce the goods ? Certainly did not elaborate.
The attachment looks right, triple spike crown at each orifice.
I reckon the volume shown is correct, however or whoever produced it or at least we have made the same mistakes !!
So I recalculated using 5 mm radius for the cylindrical hole (I had previously used the original 10 mm rad)
Edit : Had written wrong figure for pyramid base area reposted edit. I then realised I had miscalced the volume of pyramid edit no.2
!
1) Pyramid at one corner = base area 32.476 mm² Height 3.535 mm. Volume 38.273 mm³ x 2 = 76.546 mm³ = vol of two pyramids.
2) Length between diametrically opposite corners of cube √(100² + 100² + 100²) = 173.205 mm Less two pyramid heights 173.205 - (2 x 3.535) = 166.135 mm
3) My 10 mm diam cylinder including fresh air pockets has volume of π5² x 166.135 = 13048.212 mm³ Add the two pyramid vols. 13048.212 mm³ + 76.546 mm³ = 13124.758 mm³
4) Calced one cylindrical wedge (the fresh air cut) Depth at wide end 2.5 mm Height of wedge 3.5355 mm angle 35.2644° (I used https://planetcalc.com/1442/) 22.2598 mm³. 6 x 22.2598 = 133.559 mm³
5) 13124.758 mm³ - 133.559 mm³ = 12991.2 mm³ total material removed, Being 12991.2 ÷ 1000 000 = 0.01299 Or approximately 1.3 %
So mm666 suggests in his result a Volume = 987009.09 mm³ I imagine mm666 means that is the material left in the processed cube.
So this suggests 12990.91 mm³ removed.
magicmushroom666 cad offering matched by my calculation, reckon 'tis correct !!
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EddieM
No, don't buy it, assumes straight edges, they're not. there are pyramidal edges with curved bases..
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Ok, my flat bottomed cutter removes the pyramid shape - then begins to plunge into the three edges as per the grey cyl shown in the sketch.
The 'crown points' form the corners of the pyramid CSA base, as the cut progresses the csa 'corners become growing radii, the straight line forming one base line of csa diminishes in length as the cut progresses.
The curve, or tongue of my cylindrical wedge is generated, all the time the cutter periphery is approaching the point where the radii originally generated at the corners of the pyramid csa become full circumference, at this point the cutter is entering the three cube faces and a full cylinder created.
See the junction of the grey cutter path above, and the pink and beige two cube corner faces in the sketch, there is one point of the three point crown.
The inside face of the cylindrical wedge shown in grey with closed end will be an extension of the adjacent pyramid / cube face with curves as shown.
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EddieM
Apologies I only skimmed it last night just before lights out and missed step 4, which is of course the nub of the problem. This is the bit that's always stumped me and now I see why, thats somewhat advanced mathematics.
So, Mr Pip, I would give you full marks, but I am afraid I'm gonna have to call foul on that!
So, Mr Pip, I would give you full marks, but I am afraid I'm gonna have to call foul on that!
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The cylindrical wedge faces above are in plane with the 'pyramid' faces and thus the corner faces of the cube - so yes, that is the side faces of the pyramid and the generated, never seen, faces with an elliptical curve as the cylindrical cutter enters the corner faces of the cube.
Look at the entry shape of a vertical cylindrical stack into an inclined roof - Ellipse !! - That is only one face we have three intersecting at each corner.
Thinking of what is removed when machining is more complex than first sight would suggest.
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