Request to automate
That seems indeterminate to me. Surely only in a lucky combination of your input values could there be a solution with all the X's being equal. And in fact, yours are not -- the length of the horizontal dashed-yellow Lines is not the same as that of the diagonal ones, which means the X between Circles varies, and the Measurement values of the X Dimension objects also varies.] Something would usually need to be a slightly different resultant distance, for example if the X between Circles and between them and the top and bottom edges are the same, the distance between the end Circles and the right and left edges would need to be "sacrificed" and be different from the X value elsewhere.
There would also need to be some criterion for the approximate "target" ratio of X to D. Otherwise, in addition to your solution [if it really was one] with 8 Circles in the bottom row, there could be one with 6 or 7, and 3 rows, or one with 9 or 10, and maybe 5 rows -- even 11 Circles that size will fit in the L distance, with a small-enough X value.
I tried measuring distances between Circles [i.e. between intersections of Circles with yellow center-to-center Lines], comparing horizontal to diagonal where the yellow-Lines variance occurs, and the ones I checked were the same, so I wondered why the yellow Lines vary in length. In the process I found that your Circles are not all of the same radius. With two selected:
Some of them share a common radius, but there are a variety of slightly different sizes. That "allows" the variation in center-to-center Line lengths even if the spaces between are equal.
I really believe what you want is not going to be possible in most instances, if your intent is to specify L & W & D. For example, I set up a simple situation in which all the X dimensions and the Circle radius are 1 drawing unit, and I sized the rectangle around them to make that true, rather than starting from a specified rectangle size:
[The odd-ball decimal component of the vertical dimension, rounded to 3 decimal places, is because the sine of 60° is involved.]
Now suppose a routine is defined, and you ask it to do its thing in a rectangle 8.5 units wide as above, but 7 units tall. It could use the same arrangement of Circles to have their distances from each other [which must always be in 0°- and 60°-angle displacements] and to the sides of the rectangle remain at 1 unit. But it will simply not be possible for the distance from the Circles to the top and bottom of the rectangle to also be 1 unit [that is, not to both top and bottom -- it can be to one of them if you want]. Something will have to give -- you can't avoid having some variance between either the distances from the edges or the spacings/angles between the Circles. You can get the desired result only in rectangles whose dimensions happen to have that kind of magical relationship to the Circle radius and the sine of 60°.
I have a zigzag.lsp which does the circle rows in the desired pattern, but as you mention Kent there are to many variations, a minimum answer may be no circles at all or a single row, What is minimum edge clearance that is not mentioned. Its X in diagram, it may have to alter to make a pattern. Say how many rows.
As @Kent1Cooper pointed out, you can only have the same distance between circles, the horizontal rectangular boundary, and the vertical boundary if there is a specific relationship between the width and length of the rectangle.
Looking only at rectangle width for a moment and assuming that there are at least 2 rows of staggered circles we can determine the number of circles in a row from the following equation.
n = (W - 5/4 * D) / ( 3/2 * D) rounded up to the next integer.
The spacing x between circles and the rectangle can be found with:
x = (W - 1.5 * D) / (n + 1.5)
These equations were derived from the following illustration starting with the assumption that the maximum value for x was D/2. The yellow boxes highlight the spacing between circles and the rectangle.
For example, assuming D = 1, and W = 5.4 then n = 3 (2.76 rounded up) and x = 0.866.
A similar relationship can be used for determining the circle spacing based on the rectangle length L and circle diameter. This will probably yield a value for y that is different than x and you can choose to use either value for the final layout depending on your design requirements.
The following code with fill a rectangle as I think you wish given W, L and D. The side margins are calculated such that it will be the same as the distance between circles and less than or equal to the radius of the circles. The top and bottom margins are driven by the side margins and may be quite small. The top two solutions are a result of ensuring that the vertical margin is less than the circle radius. The bottom two do not make this assumption.
I've done a little testing and have found that it sometimes yields different vertical margins. I'll have to do some more debugging but I thought I would show the progress I have made.
The code could be modified to ensure a minimum margin if necessary.
Here's an improved edited version of the code 5/8/2022. It yields uniform margins for the top and bottom. The margins for the sides are uniform but probably different from the top/bottom margin.
(defun c:test (/ c_doc ms w L D pll pur n_per_row xh s xV i dx dy ptbase n_rows pt)
; Fills a rectangle of width W, length L, with circle of diameter D
; L. Minardi revised 5/8/2022
;
; xh = side margins
; xV = top and bottom margins
; s = space between circles
(setq c_doc (vla-get-activedocument (vlax-get-acad-object)))
(setq ms (vla-get-modelspace c_doc))
(setq w (getreal "\nEnter rectangle width: ")
L (getreal "nEnter rectangle length: ")
D (getreal "nEnter circle diamter: ")
)
(setq pll '(0.0 0.0)
pur (list L w)
)
(command "_rectangle" pll pur "")
(setq n_per_row (/ (- L (* 1.25 D)) (* 1.5 D)))
(setq n_per_row (+ (fix n_per_row) 1))
(setq xH (/
(- L (* n_per_row D) (/ D 2.))
(+ n_per_row 1.5)
)
)
(setq s xH
sin60 (sin (/ pi 3))
n_rows (+ (/ (- W (* 2 D)) (* (+ s D) sin60)) 1)
n_rows (+ (fix n_rows) 1)
xv (/ (- W (* (- n_rows 1) (+ D s) sin60) D) 2)
i 0
dx 0
dy 0
ptbase (list (+ xh (/ D 2.)) (+ xV (/ D 2.)))
)
(repeat n_rows
(repeat n_per_row
(if (> (rem i 2) 0) ; determine if odd or even row
(setq shift (/ (+ D s) 2.))
(setq shift 0.0)
)
(setq pt (mapcar '+ ptbase (list (+ dx shift) dy)))
(vla-addcircle ms (vlax-3d-point pt) (/ D 2.))
(setq dx (+ dx s D))
) ; repeat n_per_row
(setq dx 0
i (+ i 1)
dy (+ dy (* (+ s D) sin60))
)
) ; repeat n_rows
(princ)
)
@Sea-Haven wrote:
Looks good, would not the edge margin be same all round ?
....
[It can't be, if the Circles are also all the same distance from each other, except when the relationship is just right between Circle size and rectangle dimensions. See the second half of Message 5.]
@leeminardi and @Sea-Haven wrote:(setq sin60 (sin (/ (* 60.0 pi) 180.0)))
(setq sin60 (sin (/ pi 3)))
I have edited the code in my post #13 to fix a calculation bug for the number of rows and the vertical margins. To save several attoseconds of compute time I have defined a variable for the sine of 60°. Thanks for the suggestion @Kent1Cooper and @Sea-Haven.
Sample result.
@Sea-Haven, " ...would not the edge margin be same all round ?" In short, no. The aspect ratio of W / L is most likely different than what can be achieve by an array of equally sized circles with equal spacing. My code calculates a value for the side margins that will be the same as the circle spacing. This values drives the vertical margins.
