The previous post demonstrated that gate hinges should lie in a vertical plane normal to the bisector of the open and closed gate positions. Here’s the plan view of the hinges in the open position: (click for a bigger version)

Triangles BFG and BFK, lying in the horizontal plane, are mirror images. So, length FK must be O_NS

 

The points labelled here are exactly the same as the ones in the previous diagram. Length FK in the horizontal BFK plane is O_NS, because triangle BFK and BFG are mirror images. Length KM is also O_NS, because that’s how we made the spacer for the gate. Triangles AFK and AMK are therefore mirror images, because they are both right triangles, they share a hypotenuse and have side KM equal to side KF. So, angle KAF is equal to angle KAM. Let’s call this angle θ. And, call distance AF, the vertical distance between the hinges, O_UD.

tan θ = KF / AF = O_NS / O_UD

θ = tan^-1 (O_NS / O_UD)

Here’s how the gate looks when fully open.

 

The gate rises up at an angle 2θ from the horizontal. The effective length of the gate, JK, is O_GATE + O_NS.

JN / JK = sin (2θ)

Rise = JN = (O_GATE + O_NS) * sin (2θ)

Rise = (O_GATE + O_NS) * sin(2 * tan-1 (O_NS / O_UD))   ♦ Equation 2
Oddly, this doesn’t depend on β. Is that right?

So now I have some equations for the variables O_NS and O_WE. Here are the parameters fixed by the requirements of the gate design:

Symbol	Meaning
OUD	Vertical pitch between gate hinges
OGATE	Width of gate when closed, measured from centre of top hinge pin to outside edge of gate
Rise	Required Rise of gate when fully open
β	Angle between gate open and gate closed

Here are the parameters I’m trying to determine:

Symbol	Meaning
ONS	Offset of bottom hinge parallel to the gate
OWE	Outset of bottom hinge at right angles to the gate

And here are my equations:

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

Rise = (O_GATE + O_NS) * sin(2 * tan-1 (O_NS / O_UD))     ♦ Equation 2

I don’t actually know of a symbolic solution of (2) for O_NS. Fortunately, that doesn’t stop me solving it numerically, which will come in a post real soon now.

 

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.

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).

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.

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