David W. answered • 05/22/15
Tutor
4.7
(90)
Experienced Prof
Hi Ben,
Now, what kind of a silly question is this ?? Of course, there is a way to calculate it (weren't you assigned this problem?) … and there is a formula (or a procedure) for doing it. And, good mathematicians can produce such a formula or procedure with sometimes amazing skill. They might even take constraints like the number of tiles in a package or centering the tiles in a large surface (construction people think this is aesthetic, even though it costs more to cut extra tiles).
This is one reason that I love computer simulations (this one is quite impressive using Matlab) so we can visualize both the procedure and the result.
So, begin by picturing one regular hexagon – a 6-sided geometric figure with equal-length sides and equal angles (note: turtle graphics draws such a figure by first drawing x distance to the east, turning (sides/360) degrees left, drawing another x distance, etc., until the turtle arrives back at the starting point and pointing in the initial direction, thus turning 360 degrees in the process -- the sum of the external angles is 360).
Now, think about the maximum distance and the minimum distance across the hexagon. It seems reasonable that if we want to pack as many hexagons into a linear distance as possible, then putting them side-by-side using the minimum width does that (if that formation is possible). And, putting the hexagons side-by-side using the maximum width along a linear dimension produces the minimum number of hexagons.
Note, these two tiling arrangements require tiles to touch, but may leave large gaps in the pattern (if the tiles did not have to touch, it would look ‘creative.’)
O.K., put one row of hexagons side-by-side across the width. We must round down since we want to count whole hexagon tiles.
Now something wonderful happens ! The second row of regular hexagons doesn’t have to sit on top of the first row, but it can ‘slip’ down some (think of a soccer ball here) and not leave any empty space at all. That is the most dense pattern possible and all we have to do now is determine where the edges of the region ‘cut off’ our clever pattern (well, no we really didn’t invent the soccer ball, but we could use alternating black and white tiles, couldn’t we?).
Minimizing the gaps is the key. So, ask “How many hexagons widths can we fit into the width of the surface?” Then, carefully, ask how many shortened lengths (because we eliminated gaps) can we fit into the length of the surface?” O.K., now decide whether interchanging width with length of the surface would change the result. Well, since these are regular hexagons, they look exactly the same when they have turned 90 degrees, so I think that width and length can be interchanged and we would have the same result.
Again, an animated simulation is SOoo.. much more fun to watch than reading this.
However, it is the length (L) and the width (W) that are crucial here, not the area of the surface. This is true because we will use the MOD (remainder) function to determine the number of hexagons that will fit in any given distance.
Well, now the math.
For your hexagon, what is the minimum width? Given S=side width, you can calculate that.
And, given S=side, you can calculate the distance from the bottom of one hexagon to the top of a second hexagon that is sitting side-to-side with this hexagon and not allowing any gaps. Thus, the height of each hexagon after the first adds a little less to the total distance : D = D1 + (N-1)*D2 … again, you can easily calculate D1 and D2.
Oh, the number of hexagons would NOT be the number of hexagons in a row * number of hexagons in a column (seems silly to have to say this), but all the rows/columns don't have the same number (think of the stars in the American flag). So be careful to count the rows and columns correctly.
Now, for those aesthetic (real craftsman) decorators who CENTER the hexagons and leave the same amount of ‘extra’ on each end of the pattern, I have to ask, “Isn’t that the same number of tiles as if I had started at the edge of one tile and left all of the residual on the other edge?”
This is one reason that I love computer simulations (this one is quite impressive using Matlab) so we can visualize both the procedure and the result.
So, begin by picturing one regular hexagon – a 6-sided geometric figure with equal-length sides and equal angles (note: turtle graphics draws such a figure by first drawing x distance to the east, turning (sides/360) degrees left, drawing another x distance, etc., until the turtle arrives back at the starting point and pointing in the initial direction, thus turning 360 degrees in the process -- the sum of the external angles is 360).
Now, think about the maximum distance and the minimum distance across the hexagon. It seems reasonable that if we want to pack as many hexagons into a linear distance as possible, then putting them side-by-side using the minimum width does that (if that formation is possible). And, putting the hexagons side-by-side using the maximum width along a linear dimension produces the minimum number of hexagons.
Note, these two tiling arrangements require tiles to touch, but may leave large gaps in the pattern (if the tiles did not have to touch, it would look ‘creative.’)
O.K., put one row of hexagons side-by-side across the width. We must round down since we want to count whole hexagon tiles.
Now something wonderful happens ! The second row of regular hexagons doesn’t have to sit on top of the first row, but it can ‘slip’ down some (think of a soccer ball here) and not leave any empty space at all. That is the most dense pattern possible and all we have to do now is determine where the edges of the region ‘cut off’ our clever pattern (well, no we really didn’t invent the soccer ball, but we could use alternating black and white tiles, couldn’t we?).
Minimizing the gaps is the key. So, ask “How many hexagons widths can we fit into the width of the surface?” Then, carefully, ask how many shortened lengths (because we eliminated gaps) can we fit into the length of the surface?” O.K., now decide whether interchanging width with length of the surface would change the result. Well, since these are regular hexagons, they look exactly the same when they have turned 90 degrees, so I think that width and length can be interchanged and we would have the same result.
Again, an animated simulation is SOoo.. much more fun to watch than reading this.
However, it is the length (L) and the width (W) that are crucial here, not the area of the surface. This is true because we will use the MOD (remainder) function to determine the number of hexagons that will fit in any given distance.
Well, now the math.
For your hexagon, what is the minimum width? Given S=side width, you can calculate that.
And, given S=side, you can calculate the distance from the bottom of one hexagon to the top of a second hexagon that is sitting side-to-side with this hexagon and not allowing any gaps. Thus, the height of each hexagon after the first adds a little less to the total distance : D = D1 + (N-1)*D2 … again, you can easily calculate D1 and D2.
Oh, the number of hexagons would NOT be the number of hexagons in a row * number of hexagons in a column (seems silly to have to say this), but all the rows/columns don't have the same number (think of the stars in the American flag). So be careful to count the rows and columns correctly.
Now, for those aesthetic (real craftsman) decorators who CENTER the hexagons and leave the same amount of ‘extra’ on each end of the pattern, I have to ask, “Isn’t that the same number of tiles as if I had started at the edge of one tile and left all of the residual on the other edge?”
