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# Paper size calculations

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Help:

My problem is based on calculating how many pieces of paper I get from a specific paper size. In printing one way to find how many pieces we get, is using the following math work. I write the size of the original paper, and then I write the size I need. By canceling the number I find how many pieces of paper come out of the original size. I work it out like an algebra problem. The original size on the top, on the bottom the size I want. By crossing (dividing o canceling) numbers I find out how many pieces I get. For example if the original paper size is 17â€ x 22â€, and I need pieces of 8.5â€ x 11â€, I will get four pieces from each original sheet size, depending the way I set the paper in the paper cutter. Iâ€™m using FM 7 & 8.

Is there any printer that has sold this problem the easy way?

Thanks

Humberto

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Hi Humberto

I'm not a printer, and I don't know about easy, but how about:

```Cuts=
Let
(
[
widthnum=
Case(AlignToggle=1;Div(PrintSheetWidth;OutputPageWidth);Div(PrintSheetWidth;OutputPageHeighth));
heighthnum=
Case(AlignToggle=1;Div(PrintSheetHeighth;OutputPageHeighth);Div(PrintSheetHeighth;OutputPageWidth ) )

];

widthnum*heighthnum

)
```

will give you the number of cuts you get from a print sheet. if the field AlignToggle=1, the width entered for the output page is aligned with the width of the sheet. if it is 0 (or null or any other value) it is turned 90Â°.

You can isolate widthnum and heighthnum in separate calcs to see how many pages to a side. call these calcs "Wide" and "Across"

you can then calculate the paper loss on the width and height by:

```widthloss=
PrintSheetWidth-(Wide*Case(AlignToggle=1;OutputPageWidth;OutputPageHeighth))

heighthloss=
PrintSheetHeighth-(Across*Case(AlignToggle=1;OutputPageHeighth;OutputPageWidth))
```

hope this is any use

kjoe

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Thanks kjoe

I will give it a try and let you know.

Humberto

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Humberto, your question caught my eye because I am indeed a printer. Kjoe's solution is about 50 feet above my head, so I came up with something that might be easier to relate to, if you're in the printing business. (Kjoe, I'll study your solution---it looks like it's got several components I could put to good use, but I need to get a little smarter first.)

If you take your 17" x 22" sheet, and lay out a letter size paper on it, you'll see that:

You can put TWO 11" letter lengths (A) on the 22" parent sheet length ©, and TWO 8.5" letter widths (B) on the 17" parent sheet width (D).

TWO times TWO = FOUR.

You can only put ONE 11" letter length (A) on the 17" parent sheet width (D), and only TWO 8.5" letter widths (B) on the 22" parent sheet length ©.

ONE times TWO = TWO

You can do this mathematically by:

(C / A) * (D / B) = X

and

(D / A) * (C / B) = Y

In this case, X is larger than Y and this is the result you want to get:

You can get a maximum of FOUR letter size pieces out of a 17" x 22" parent sheet.

So in Filemaker you can create a formula that determines which of the two formulas gives you the larger number, and this number is the result you are looking for.

You need to round down (using the "Floor" function) when you divide the parent sheet by the cut sheet.

I've attached a sample Filemaker 8.5 file that shows how to do it.

Thanks for asking this question---I've always drawn pictures to figure this out, and because of your question I now have an easier way!

By the way, this should work not only for printers, but for anyone who needs to calculate how to get the most rectangular pieces out of something. I can think of woodworkers and glass workers right off the bat.

Good Luck!

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If the goal is truely to find the most possible pieces, then some rectangles could be cut vertical and some horizontal. In the example from your demo, you'd be able to get 17 pieces (8.5 x 11) out the parent sheet (36 x 48).

To solve this, you'd need a recursive solution (either scripts or custom functions). We worked out the custom function to determine this count last fall here:

http://www.fmforums.com/forum/showtopic.php?tid/170643/

I've built CFs to actually work out the coordinates of the rectangles so they can be drawn with the xmChart plug-in. See attached.

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Those are some elegant examples! Looks like I've got some learning to do. Although, as mentioned in the posts from last Fall, the direction of the paper grain must often be dealt with. I think this might be important for Humberto. In my shop, we always try to keep the paper grain for any one job all going in the same direction. This is important whether we are printing on a press/duplicator or using high speed copiers. Any stock left over is marked with the grain direction, and stored for use later in a smaller job. The sample file I attached previously will return the maximum number of same-grain stock that can be pulled out of a parent sheet.

This works well for a smaller print shop, but where big print jobs are the norm, I would be going for the recursive calculations because the paper costs would more than make up for the additional time in cutting the stock.

On a related topic: Many years ago, I restored stained glass windows in churches and mounted sheets of polycarbonate to the outsides of the windows to protect them from weather and vandals. I spent untold hours on some jobs figuring out how to get the greatest number of little pieces out of 4x8 and 6x10 pieces. The recursive solutions would have been fantastic--but alas, that was w-a-a-a-a-y before FM (and desktop computers!).

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Yeah, I can see how the grain could be important.

It probably wouldn't be too hard to modify my CFs to accept a Grain parameter, so that a user could specify a grain in one direction or the other, or just ignore the grain and look for the max number of pieces. It could use your count calc for any grain-specific counts, or the recursive RectCount CF for those that ignore the grain. By doing this all within the custom functions, it would simplify the calc that calls it.

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• 5 years later...

been working on this problem for a little wile then I find this thread...

I have to do the slight modifications mentioned in the beginning about the grain also take a few other things into account like if the media comes in a roll or a sheet the print margins cutting margins we have already said the grain, but I was hoping to generate a result that would give no of pieces verticle no. Of pieces horizontal and no. Of pieces 'odd' (meaning those last couple that can be squeezed in but don't allign exactly with the rest)

Any help would be great....

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