Sooner or later, a homestead will need a gate. Or a door on something. Or some other swinging rectangle. Its inevitable. The part that swings is usually constructed by building a rectangular frame with a diagonal cross piece for stabilization and support, and then covered with boards or pickets, or whatever. The dimensions of the perpendicular pieces are fairly straightforward, but what about the diagonal brace? What size should that be and what angles should it have? Many folks just lay a board across the frame, mark where the edges should be, and cut along a line between the marks. But what if the frame is slightly out-of-square when the board is set on it? What if youre building with old warped wood (or twisty-turny, cool-looking logs) and the brace wont sit right on the frame? Wouldnt be nice if there were a way to calculate what the dimensions should be? Fortunately, there is! Time to pull out some triangle math and get measurin.
The first step is to decide on the dimensions of the gate. Once you know the height and width, and thickness of the wood (or other material) youre going to build it with, you can build the frame, minus the crosspiece. Then decide what type of diagonal brace you want.
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| Three common choices are the trapezoid (left), the parallelogram (middle), or the skinny hexagon (right). As a side note, the brace should be positioned with the bottom on the side with the hinges (i.e., in the figure, the hinges should be on the left side of each gate). That way, the weight of the gate compresses down on the brace, which gives more support than nails and screws can provide on their own, and helps mitigate sagging over time. |
Now its time to do math! Yay! Fortunately, we can get all the dimensions we need with just three equations: the Pythagorean theorem, SOH CAH TOA, and the law of sines, since we know the inner dimensions of the gate frame and the thickness of the wood were planning to use for the brace (e.g., 1.5" for the thin side of a 2 x 4).
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| Case 1: Trapezoid brace. Easy peasy. The length of the brace is the hypotenuse of the frame, z, which we calculate from the Pythagorean theorem, with L1 and L2 as the side lengths. The angle ?1 is the arctan of L2/L1. The angle ?2, which is the angle to cut the board at, is calculated by subtracting ?1 from 90°. The length, d, is the quotient of the brace board width, w, and tan ?2. The length d is often the most useful measure since we can measure down d from one corner, draw a line to cut along from there to the opposite corner (on the same end of the board), and the angles and other dimensions will take care of themselves. As a check that everythings kosher, x and y can be calculated as shown. Note: similar, but not the same, equations apply at the opposite end of the brace. There, tan ?1 = w/d. |
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| Case 2: Parallelogram brace. Probably the most common, but also the most math-intensive to figure out. The hypotenuse of the gate frame, z, is now the long diagonal of the parallelogram. The angle between z and L1 is ?1; the angle between y and L1 (the angle at which to cut the board), is ?2. z and ?1are calculated as in Case 1 above. Next, the SOH part of SOH CAH TOA means there are alternate equations for ?1 and ?2; namely using L2, w, x, and y, which will be helpful in the next step. For the triangle with x, y, and z as sides, we can use the law of sines, substituting for y and sin ?1, to find the difference between ?1 and ?2, and from there, ?2;. Once ?2 is known, substituting back into the equations from SOH makes for easy calculation of x and y, and since d is one side of a right triangle with w and x as the other sides, we can use the Pythagorean theorem to calculate d, which is what we really wanted in the first place. Phew! One advantage to this design is that its possible to use a slightly shorter board (by, like, fractions of an inch) since the hypotenuse of the frame is the diagonal of the board instead of the length. But if youve got a piece of scrap 2 x 4 thats a quarter of an inch too short for the other brace designs... |
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| Case 3: Skinny hexagon brace. This one is kind of like Case 1, with two trapezoids back to back. The hypotenuse of the frame lies along the centerline of the board, which means that to calculate d1 and d2 we can use the tangents of ?1 and ?2, which are the ratios of the sides and also of half the brace board width and one of the ds, as shown. As a check, x1, x2, and y can be calculated from the Pythagorean theorem and the SOH or CAH parts of the right triangle with (L2 - x2), (L1 - x1), and y as sides, respectively. |
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| With the equations above, its possible to make perfectly-fitting braces every time on gates, doors, and lots of other swinging rectangles around the homestead. (Within the experimental error of the craftsmans skills, of course). Clearly, the chickens appreciate the extra effort, since theyre always crowding into the door frame when the chicken tractor door opens. Theyre probably eager for a chance to admire the gate brace from a new angle, and not at all excited that they suddenly have more space around the feeder. |
Weve also compiled this information into a handy Excel-based calculator, free for download here. Let us know if you have any suggestions to improve it!
How do you size the diagonal braces for your gates, etc.? What was the last thing you attached a gate to? Let us know in the comments section below!
