In my last post, I calculated the radius and diameter of a circle drawn within the external 'corner' of another circle - diagram below. The final ratio indicated that the diameter of the small circle is 10.8% of the larger circle. In this post, I'm going to extend this to look at circles in hexagons (something that I discussed a few years ago when I looked at square or cubic packing and hexagonal packing). I'll also then look to apply my findings to real life, by reviewing the sizes of atoms in alloys - how do they fit together?
Firstly, here's a reminder of the result from last time:
For a circle with radius DE = EF = 1 unit, then the radius of the smaller circle "in the corner" BC = CD = 0.108... units. The smaller circle has a radius of 11% of the larger one. The length AD is 40% of the radius of the larger circle.
In this blog, I'd like to study the relationship between a circle and the hexagon surrounding it (the circle is inscribed in the hexagon). The diagram below shows three such hexagons. The circles have radius r, and the distance from the centre of the circle (O) to the corner of the hexagon is h (the hypotenuse of the triangle formed by the radius and the tangent). The angle between the radius and the hypotenuse is 30 degrees.
Given that r = 1 unit, what's the length h?
Trigonometry tells us that cos 30 = r/h therefore h = r / cos 30 = r / 0.866
h = 1.154 r
This means that the additional distance ( h - r ) = 0.154 r or 15% of the radius of the larger circle. This additional distance is the maximum radius of a circle that would fit in the space between the three circles shown in the diagram. For example, if the larger circles have a radius of 1 metre, the radius of the smaller circles would be 0.15m or 15 cm. As I've calculated previously, there isn't much space between hexagonally-packed circles.
Atoms in metallic solids are typically hexagonally packed, and alloys are formed when other atoms are mixed with a pure metal. It occurred to me that it may be possible for these atoms to be small enough to fit into the gaps between the atoms - using the mechanism I've indicated above.
Here are the radii of some metal atoms, for reference, in picometres. 1 picometre, = 1 x 10 -12 metres
Fe (iron) 156
Sn (tin) 145
Al (aluminium) 118
Co (cobalt) 152
15% of these values gives us 24, 21, 17, and 22. The atomic radius of carbon (which is alloyed with iron to make steel) is 67 pm. So, there is no way that carbon will fit neatly into the gaps in the matrix of hexagonally-packed iron atoms. [Indeed, helium has the lowest atomic radius, and that's 31 pm, still too large].
Instead, in interstitial alloys, the smaller atoms in an alloy distort the close-packed arrangement of the metal and that's what affects its physical properties.
Further reading
Previous post on the radius of a circle in the corner of a circle
Space filling calculation 3D - sphere in cube, sphere in hexagonal prism
Hexagonal close packing - 2D-filling calculation
Interstitial alloys
Firstly, here's a reminder of the result from last time:
For a circle with radius DE = EF = 1 unit, then the radius of the smaller circle "in the corner" BC = CD = 0.108... units. The smaller circle has a radius of 11% of the larger one. The length AD is 40% of the radius of the larger circle.
In this blog, I'd like to study the relationship between a circle and the hexagon surrounding it (the circle is inscribed in the hexagon). The diagram below shows three such hexagons. The circles have radius r, and the distance from the centre of the circle (O) to the corner of the hexagon is h (the hypotenuse of the triangle formed by the radius and the tangent). The angle between the radius and the hypotenuse is 30 degrees.
Given that r = 1 unit, what's the length h?
Trigonometry tells us that cos 30 = r/h therefore h = r / cos 30 = r / 0.866
h = 1.154 r
This means that the additional distance ( h - r ) = 0.154 r or 15% of the radius of the larger circle. This additional distance is the maximum radius of a circle that would fit in the space between the three circles shown in the diagram. For example, if the larger circles have a radius of 1 metre, the radius of the smaller circles would be 0.15m or 15 cm. As I've calculated previously, there isn't much space between hexagonally-packed circles.
Atoms in metallic solids are typically hexagonally packed, and alloys are formed when other atoms are mixed with a pure metal. It occurred to me that it may be possible for these atoms to be small enough to fit into the gaps between the atoms - using the mechanism I've indicated above.
Here are the radii of some metal atoms, for reference, in picometres. 1 picometre, = 1 x 10 -12 metres
Fe (iron) 156
Sn (tin) 145
Al (aluminium) 118
Co (cobalt) 152
15% of these values gives us 24, 21, 17, and 22. The atomic radius of carbon (which is alloyed with iron to make steel) is 67 pm. So, there is no way that carbon will fit neatly into the gaps in the matrix of hexagonally-packed iron atoms. [Indeed, helium has the lowest atomic radius, and that's 31 pm, still too large].
Instead, in interstitial alloys, the smaller atoms in an alloy distort the close-packed arrangement of the metal and that's what affects its physical properties.
Further reading
Previous post on the radius of a circle in the corner of a circle
Space filling calculation 3D - sphere in cube, sphere in hexagonal prism
Hexagonal close packing - 2D-filling calculation
Interstitial alloys
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