This post is a geometry exercise to work out the hinge offsets needed to make a gate rise from its closed to open position, but sit vertical at both closed and open positions. Between closed and open, it won’t be vertical.

You shouldn’t need to understand all this post just to hang a gate. I’ll be posting a calculator for this later. Click any of these diagrams for a bigger version.

Plan view of hinges, gate fully closed.

The diagram above shows a view looking down on the top of the gatepost when the gate is closed. Point A is the top hinge pin and point B is the bottom hinge pin. The gate is exactly vertical, although B is not directly below A, because the bottom hinge includes spacers to offset the hinge. The spacers offset the Hinge O_NS in the north-south direction (parallel to the gate) and O_WE in the east-west direction (perpendicular to the gate).

Plan view of hinges, gate half-open

This diagram shows the gate partially open, neither open nor closed, looking from above. The bottom bar of the gate, where the bottom hinge sits, is no longer directly underneath the top bar. The gate does not sit in a vertical plane. That’s OK. Only when the gate is fully open will it lie in a vertical plane again.

Plan view of hinges, gate fully open

The above diagram shows the gate fully open. This gate has been designed to be fully open at an angle β to its closed position. The diagram is confusing (sorry) because it’s a vertical plan view, and the gate has risen out of a 2D plane. Points B, K and F all lie in the horizontal plane through the lower hinge pin. Points M and H are in a horizontal plane raised above the plane BKF. It’s clear that M and H lie in a horizontal plane because

• The hinge offset MH is normal to the gate plane, and
• The gate plane is now vertical, because we’ve designed the gate plane to be vertical when the gate is open to an angle β

The line MH has a length O_WE when the gate is closed and MH is horizontal in the same plane as hinge pin B. MH is again horizontal with the gate open to angle β, so again the projection of line MH on the horizontal plane containing points BKF still has length O_WE.

The projection of the line MH in the plane BKF is at right angles to the intersection of the vertical gate plane and BKF. This is true however much the gate rotates in the vertical plane containing F and J, because MH is normal to the gate plane. So, line KB, parallel to MH but in the horizontal plane BKF, is at right angles to line FK, and also has length O_WE.

Triangles BFG and BFK are mirror images, because are both right triangles, they share a hypotenuse and side BG is the same length as BK. So, angle BFK is the same as angle BFG – let’s call this α.

Point G, F and L lie on a straight line:

β + α + α = 180°

α = 90° – β/2

or, in other words, the vertical plane containing the hinges must lie at a right angle to the vertical plane containing the bisector of the angle between the open and closed gate positions.

Or, in yet other words:

• Draw a line midway between the open and closed position of the gate
• Draw a line through the top hinge, at right angles to that line
• The bottom hinge must lie somewhere on that line
• and the further you move it out from under the top hinge, the more the gate will rise
  but the gate will always be in a vertical plane when full open.

And, equivalently,

O_WE / O_NS = tan α

O_WE = O_NS * tan (90° – β/2)   ♦ Equation 1