# When making a box joint, how do I calculate the width of the board so that the teeth appear even?

I have made some box joints and thus far I've found it difficult to ensure that the teeth are evenly spaced; there's usually a pin that is cut down a bit. Is there an equation or method of determining what the board's width should be? Is it as simple as "always ensure board is a multiple of 2x the kerf of the dado set?

• The approach I've seen is to cut the joints, then trim the box to a tooth line.... – keshlam May 10 '15 at 22:55
• I cut mine with the band saw, so I just lay them out, usually 1/2" or 3/4" teeth – bowlturner May 10 '15 at 23:04
• Surely total board width is just a multiple of the tooth width or am I missing something obvious? If you're using stock that's fractionally wider than X number of teeth you then plane down to width after you've formed the joints. – Graphus May 10 '15 at 23:46

Start by taking a multiple of the pin width, then add a little extra and cut all the slots/kerfs on all sides of the box. Once you're done, trim each side flush with the edge of a pin or slot. If you prefer, you can trim it to size after assembling and gluing it.

To guarantee that the teeth are evenly-spaced, use a jig. Popular Woodworking has a great article on how to build a box joint jig, and Fine Woodworking has a nice interactive demo illustrating how a box joint jig works.

Note that FWW calls it a finger joint jig because many people use the terms finger joint and box joint interchangeably, but personally I like to differentiate the right-angle joining version as a box joint and the straight joining version (e.g., to "lengthen" a board) as a finger joint since the straight joining version also often has tapered fingers.

• I have an incra ibox jig, but my question was more about not having the width of the board be an exact multiple of the teeth, so that the last tooth on one side or the other might be larger than the others. I think @keshlam 's comment is probably the best answer to my question – Peter Grace May 13 '15 at 14:17

Is it as simple as "always ensure board is a multiple of 2x the kerf of the dado set?

If you want half fingers at top and bottom on the same side, then yes, using an even multiple of the finger width (or dado kerf) will do it.

If you want full fingers at top and bottom of the same side, use an odd multiple of the finger width.

Given any box or drawer finger joint height, W, you must custom size the width of the fingers for it. To do this you decide you want the fingers at the top and bottom configured. As said above, you will have an odd number of fingers on any side that has both both top and bottom fingers on the same side, so if "f" equals the width of both fingers and slots on either piece, where we want a whole number integer of fingers and "slots" then for n=3 then f=5; n=4-->f=7; n=5-->f=9; n=6-->f=11; etc. Note we can say that for a given integer n, that f=n+(n-1). That reduces to f=2n-1, and f(2n-1) must equal W. Chose any f depending on how big or small you want your equal fingers to be. This needs some thinking by the questioner, but just know that you will VIRTUALLY NEVER get a whole nice number of fingers to fit into a given width with a nice neat finger width size. For example, lets say that "W" is equal to the height of drawer we want to be 8" high on the sides. Logically thinking about it, we will pick a "nice" n for the joint, say 14. So for an f of fourteen we get f=27. W/f therefore is 0.296". If you make the fingers 0.296" wide, then you can get a perfect number of them, 14 fingers on the drawer side or end that has whole outside fingers at both top and bottom and 14 matching "slots" on the drawer side or end that has a "slot" at top and bottom and ALL the fingers and slots will be the exact same width! Note that the side that has the "slots" at top and bottom has one less projecting finger than does the side that will be joined to it! There's no way around it, ya'gotta do the math. Best way to understand this is to draw a drawing and study it if you can't accept it. If you look at it long enough, you can see that the side with the "outside fingers" has to have one more projecting finger than the joining side that has the "outside slots" at top and bottom, and THAT is what makes it so that the width of the fingers plus width of the slots, whose total MUST be equal to W, must be of a common width that will result in 2n-1 being a whole integer. Being an "integer" means a whole number of fingers and slots that are not fractional. Making tops and bottoms a half finger is attacking the problem from another direction. Enough, huh?

• Oh, and "picking a multiple of some width of fingers you will use and then multiplying it by 2n-1 is cheating. The drawer height always be demanded by the work, not the fingers in your joint! ... and cutting it down after you do the work is even worse! – user5624 Oct 7 '18 at 1:03
• Maybe a bit clearer in the example if I had said that there will be, on one of the two sides to be joined at the finger joint, fourteen fingers and thirteen slots and on the other side there will be fourteen slots and thirteen slots. Note that on each of the pieces being joined the fingers and slots total 27. Yeah, less muddier, perhaps. – user5624 Oct 7 '18 at 4:32
• Can't type today--Correct the second comment so that the there "fourteen slots and thirteen fingers on the second piece. – user5624 Oct 7 '18 at 4:39
• Well, OK, when I tried to correct it, I got a red box telling me I could not do it, but apparently it did it anyway. So, never mind ? – user5624 Oct 7 '18 at 4:41
• Hi, welcome to SE. Two quick points, first in the Answer paragraphs are your friend :-) It's very difficult to take in info from a big wall of text like this. Second, re. your first Comment, there is no cheating in woodwork. different people have different standards for what they find acceptable practice or a useful workaround — e.g. they choose to create a new drawer and they're fine with the final height being determined by finger height, using an existing jig, because they don't have the time to make a new one, can't be bothered or just don't have the space for another one. – Graphus Oct 7 '18 at 11:16

I was taught that the width of the box side must be an increment of the finger width.

Thus, if you decide to have 1/4" wide fingers, then the width of the workpiece must be an increment of 1/4" (e.g., a whole number, like 3",etc., or 3 1/4, 3 1/2, or 3 3/4...).

In summary, first you determine what width you want the fingers to be (for the sake of simplicity, I keep the finger width equal to the thickness of the boards I am joining), then determine what the height of the box wall will be, based on the increments of the finger width chosen. You then cut your stock to the width calculated.