FreeCad

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This may be useful for someone thinking of getting into FreeCad...
A small test concerning accuracy of volumes, mass etc. compared to Solidworks.

Edit : - And how to ensure you have the capability within FreeCad.
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95.05 % of the original volume remains.
Started with 1000 000 mm³, removed 49515.008 mm³, leaving 950484.992 mm³.
950484.992 ÷ 1000 000 = 0.95048'ish hence 95.048% of original volume remains.

At one corner of the cube.
Need to visualise the triangular based 'pyramid'.
Then the three odd shaped 'cylindrical wedges' a little like a tilted cylindrical tank having an open end with fluid just dribbling out, the fluid is in the shape of a cylindrical wedge.
In this case the wedges represent the fresh air being cut as the tool works its way into the cube, the curves slowly disappearing as the tool fully enters the cube.
So calc the pyramid vols, then the distance between the pyramid bases use this to calc a solid cylinder vol from which we subtract the six cyl' wedges, this being the material removed between the pyramids.
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Sounds suspiciously like it may well be right, problem domain is well defined, and solution probably correct. Right then Mr. Pip looks like I owe you another pint at the Plough in Nempnett.

See bottom right side of pix vol = 950484.99856.mm³. Some 0.006 mm³ difference.
Proof at last !! Well done FreeCad !

Freecad Cube.png


Cube 100mm length side, hole 20mm diam. drilled / bored through opposite corners - Volume remaining ???
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