Calculate and fill with circle

[danuthorincas]

Hello I nedd help .I dont how to start to calculate how many circles(30 mm)  and fill complite Circles.pdf (44.47 KB)(600x600 mm) >I atached one picture 

[plus]

Is that 30 a radius or diameter?

Regardless can you calculate the height of 1 circle stacked above 2 sitting side by side (all same radius)?
Hint: the centers form an equilateral triangle with side 2 radii and that sits 1 radius from bottom edge.

Come to think of it, I think it is same or very similar as hexagons.

[tsh73]

600/30 = 20
so

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will fill it nicely.

[tsh73]

Then again, hexagons could be more tightly packed
from www.euro-matic.co.uk/ball-calculator/

Circle Packing Surfaces – 2D Calculations
In geometry, circle packing is an arrangement of non-overlapping circles within a containing space and would apply to any balls sitting on a flat surface such as tanks, pools and reservoirs. The associated packing density is the proportion of the surface covered by the circles.

In two dimensional space, Joseph Louis Lagrange proved in 1773 that the highest-density lattice arrangement of circles is the hexagonal packing arrangement in which the centres of the circles are arranged in a hexagonal lattice (staggered rows like a honeycomb), and each circle is surrounded by 6 other circles. The density of this arrangement is

Density=(π√3)/6=90.690%
Using this density percentage of 90.7%, it is relatively easy to divide the area of a circle into the area of a surface you want to cover and apply the packing density to get a very accurate calculation of how many balls you will need. Note this does not allow for the part balls around the edges or disruption of the hexagonal packing but on larger areas contributes to far less than 1% error.

Have a look at Wikipedia, I think they have all the math
en.wikipedia.org/wiki/Close-packing_of_equal_spheres#Lattice_generation

[danuthorincas]

Apr 7, 2022 12:45:56 GMT plus said:

Is that 30 a radius or diameter?

Regardless can you calculate the height of 1 circle stacked above 2 sitting side by side (all same radius)?
Hint: the centers form an equilateral triangle with side 2 radii and that sits 1 radius from bottom edge.

Come to think of it, I think it is same or very similar as hexagons.

Diameter

[plus]

Of course if you can stack them all perfectly vertical you have 20 X 20 = 400 but they will fit in between 20, 19, 20, 19... per row and you might fit more, you lose 1 every other row but you gain maybe another row or 2 in height after a certain number of rows.

[tenochtitlanuk]

Takes me back to Crystallography at Uni in the Sixties!

Following code squeezes circles into 600 by 600.Up to you to decide what to do about the edge cases that are partly outside.




   nomainwin

   WindowWidth  =640
   WindowHeight =650

   graphicbox #w.g 4, 4, 600, 600

   open "Circle Packing" for window as #w

   #w "trapclose quit"

   diameter =40
   radius   =diameter /2

   for x =0 to 600 step diameter
       for y =600 to 0 step 0 -int( diameter *sin( 60 *3.14159265 /180))
           r =int( 256 *rnd( 1)): g =int( 256 *rnd( 1)): b =int( 256 *rnd( 1))
           #w.g "backcolor "; str$( r); " "; str$( g); " ";str$( b)
           if row =1 then
               #w.g "up ; goto "; x +radius; " "; y; " ; down ; circlefilled "; diameter /2
               row =0
           else
               #w.g "up ; goto "; x        ; " "; y; " ; down ; circlefilled "; diameter /2
               row =1
           end if
           scan
       next y
   next x

   #w.g "flush ; getbmp scr 0 0 600 600"
   bmpsave "scr", "packedSpheres.bmp"

   wait

   sub quit h$
       close #h$
       end
   end sub


[tsh73]

*working upon John's (tenochtitlanuk) code*

Run it.
Look that it does
READ what it printed to a mainwin - not obvious!

But still, hex packing fits more circles then rectangular one.
' nomainwin

   WindowWidth  =640
   WindowHeight =650

   graphicbox #w.g 4, 4, 600, 600

   open "Circle Packing - counting!" for window as #w

   #w "trapclose quit"

   pi=acs(-1)
   diameter = 30
   radius = diameter /2
   dRowY = diameter*sin(pi/3)
   evenRow=0   'starts from 1, 1 is odd


   'for y = 600-radius  to  dRowY  step 0-dRowY
   for y = 600-radius  to  0  step 0-dRowY
       dRowX=evenRow*radius
       for x =radius+dRowX to 600-radius step diameter
           r =int( 256 *rnd( 1)): g =int( 256 *rnd( 1)): b =int( 256 *rnd( 1))
           #w.g "backcolor "; str$( r); " "; str$( g); " ";str$( b)
           if row =1 then
               '#w.g "up ; goto "; x +radius; " "; y; " ; down ; circlefilled "; diameter /2
               row =0
           else
               '#w.g "up ; goto "; x        ; " "; y; " ; down ; circlefilled "; diameter /2
               row =1
           end if
           #w.g "up ; goto "; x+dx        ; " "; y; " ; down ; circlefilled "; diameter /2
           scan

           timer 10, [nxt]
           wait
[nxt]       timer 0
           k=k+1
           #w.g "place 10 30"
           #w.g "\ ";k;" "
       next
       evenRow=1-evenRow
   next

   print "So we *see* 23 rows of 20-19-20 circles"
   print k ;" circles"
   print "But, if we start *counting*"
   print "radius below + 22 spaces beween rows + radius above"
   print "gives us ";15+22*dRowY+15
   print "Since it is more the 600 we are given, upper row does not fit."
   print "That leaves us "; k-20 ;" circles."

   wait

   sub quit h$
       close #h$
       end
   end sub


[plus]

429 is what I got for air bubbles floating to top last row doesn't make cut inside 600 pixel square, almost but not quite.

Screenshot 600 x 600 window.

Now Easter Eggs is a different story!

Attachments:

[plus]

never mind... rethought my thoughts

[tsh73]

So, using constants from given program
After what number of lines we get extra line?
   for i = 1 to 20
       print i, i*diameter, radius+(i-1)*dRowY+radius, (i*diameter)-( radius+(i-1)*dRowY+radius)
   next

It happens that on 9th line for rectangular grid we get extra line at hex grid
And 2 lines on 16th line.

1             30            30.0          0.0
2             60            55.9807621    4.01923789
3             90            81.9615242    8.03847577
4             120           107.942286    12.0577137
5             150           133.923048    16.0769515
6             180           159.903811    20.0961894
7             210           185.884573    24.1154273
8             240           211.865335    28.1346652
9             270           237.846097    32.1539031
10            300           263.826859    36.173141
11            330           289.807621    40.1923789
12            360           315.788383    44.2116168
13            390           341.769145    48.2308546
14            420           367.749907    52.2500925
15            450           393.73067     56.2693304
16            480           419.711432    60.2885683
17            510           445.692194    64.3078062
18            540           471.672956    68.3270441
19            570           497.653718    72.346282
20            600           523.63448     76.3655198


[marshawn]

Apr 7, 2022 12:45:56 GMT plus said:

Is that 30 a radius or diameter?

Regardless can you calculate the height of 1 circle stacked above 2 sitting side by side (all same radius)?
Hint: the centers form an equilateral triangle with side 2 radii and that sits 1 radius from bottom edge.

Come to think of it, I think it is same or very similar as hexagons.

Is this why bees use hexagons for storage?

[plus]

I always thought it was the hexagons using the bees.

[marshawn]

wow, pretty cool variant of Boggle, B+

[danuthorincas]

Please switch diam to 50 .The result should by 150 because last row dosen fit complitly.I forgot to mention (haow many circles fit complitly)