“New and Improved” Mat Options

Since 2017, when we first created our webpage on Matting Embellishments^(Updated), we’ve gotten better:

• We’ve learned how to print on matboard,
• We bought a Dexter Manual Mat Cutter and know how to use it^Example.
• We found a better way to fit a picture to a standard-sized mat/frame. As discussed in Thoughts On Mat Layout, Choice D (our favorite), would be better if it were scalable (size independent). I’ll explain that in the next section.

Finding Proportional Moulding Widths

Instead of increasing the differences between the side moulding width and the top width, and then from the top width to the bottom width, by a fixed amount as shown in Choice D of the original article, there is a belief that increasing the sizes by the same percentage would appear more natural.

We will demonstrate this process using the same example as in “Thoughts On Mat Layout”. Suppose you want a 4″-high hole that’s 8″ wide in a standard 8″-high by 10″ mat. The horizontal difference between the mat size and the hole size is 10″ – 8″ = 2″. If you want the left and right borders to be the same, each will be 2″ ÷ 2 = 1”. The vertical difference between mat and hole size is 8″ – 4″ = 4″.

For Choice E, let’s call the ratio between the left (or right) border width and the top border width “r”, such that

\( \begin{array}{r c l} \mbox{(Side border width) }1” \times r & = & T \mbox{ (top border width)} \\ \mbox{Then the bottom border width, } B & = & T \times r \mbox{ or (substituting the last expression for T)} \\ (1” \times r) \times r & = & r = 1” \times r^2. \\ \mbox{Since } T + B & = & 4″, \mbox{then (substituting for T and B)} \\ (1 \times r) + (1 \times r^2) & = & 4, \mbox{ or} \\ r^2 + r – 4 & = & 0. \\ \mbox{Then, using the Quadratic Equation,} \\
r & = & \frac{-1 + \sqrt{1 – 4 \times 1 \times (-4)}}{2} \\ \mbox{meaning } r & \approx & 1.562 \\ \\ \mbox{Substituting back into our equations for T and B,} \\ \\ T & \approx & 1″ \times 1.562 \\ & \approx & 1.562″ \mbox{ or } 1 \frac{9}{16}” \\ \mbox{and } B & \approx & 1.562″ \times 1.562 \\ & \approx & 2.439″ \mbox{ or } 2 \frac{7}{16}” \\ \\ \mbox{note that } 1 \frac{9}{16}” + 2 \frac{7}{16}” & = & 4″. \end{array} . \)

Although a little more difficult than Choice D, this is still High School math, within the capability of most framers. It will put a little more weight in the bottom mat edge compared to Choice D. (In this case, that’s 2.439″ vs 2.333″, or less than ¹/₄“.) That difference will increase as the difference between the vertical edges and the horizontal edges increases, but whether that’s worth the effort is a decision for the framer (or their client).