Gate Brace Math

Sooner or later, a homestead will need a gate.  Or a door on something.  Or some other swinging rectangle.  It's 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 you're building with old warped wood (or twisty-turny, cool-looking logs) and the brace won't sit right on the frame?  Wouldn't 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) you're going to build it with, you can build the frame, minus the crosspiece.  Then decide what type of diagonal brace you want.

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 it's 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 we're planning to use for the brace (e.g., 1.5" for the thin side of a 2 x 4).

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 Θ₁ is the arctan of L₂/L₁.  The angle Θ₂, which is the angle to cut the board at, is calculated by subtracting Θ₁ from 90°.  The length, d, is the quotient of the brace board width, w, and tan Θ₂.  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 everything's 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 Θ₁ = w/d.

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 Θ₁; the angle between y and L₁ (the angle at which to cut the board), is Θ₂. z and Θ₁are calculated as in Case 1 above.  Next, the SOH part of SOH CAH TOA means there are alternate equations for Θ₁ and Θ₂; 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 Θ₁, to find the difference between Θ₁ and Θ₂, and from there, Θ₂;.  Once Θ₂ 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 it's 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 you've got a piece of scrap 2 x 4 that's a quarter of an inch too short for the other brace designs...

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 Θ₁ and Θ₂, which are the ratios of the sides and also of half the brace board width and one of the d's, 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.

With the equations above, it's 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 craftsman's skills, of course).  Clearly, the chickens appreciate the extra effort, since they're always crowding into the door frame when the chicken tractor door opens.  They're 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.

We've 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!