## How We Digitally Stretch Our Gallery Wrap Edges Before Printing

As we discussed on the Services page of our website, we digitally “stretch” our image before wrapping it around the edge of our gallery-wrapped canvas images. Here’s how we do that:

Our gallery wraps are either 3/4” thick or 11/2“. On the thin ones, I usually take the 1/4” strip along the edges and stretch it to 1″, thus having an extra 1/4” to wrap around to the back side to cover for variations in the printing and stretching processes. On the larger ones, I take 1/2” and stretch it to 2″ (thus leaving 1/2” on the back). I wouldn’t stretch the image more than four times its original size, but you could go less. To do that, you would effectively be taking a wider margin to wrap around the side.

As an example, if I want a 12” x 18” image stretched around a 11/2” frame, I would crop the image to 13” x 19”. Then, after putting guides 1/2” in from each edge and another guide right on each edge, I would increase the canvas size 3” in both dimensions to get 16” x 22” with the image centered.

1. Click Image ⇨ Canvas Size…
2. Put a check in the Relative Box
3. Make Width and Height 3 Inches
4. Make sure Anchor dot is in center of the grid
5. Hit OK

I would then use a scale transform to digitally stretch the outermost 1/2” to 2” wide, filling the canvas.

1. Make sure Snap is checked in the View Menu
2. Use Rectangular Marquee tool to select the 1/2” strip between the guides along one of the edges
3. Click Edit ⇨ Transform ⇨ Scale
4. Place the mouse cursor over the little square in the middle of the outer edge of the selected area and drag to the edge of the canvas
5. Hit the check mark to finish the transform
6. Repeat Steps 2 through 5 with the 1/2” strips along the other three edges

(Actually, I first do the four corner squares separately, but since only a small bit along the edge of those squares has any chance of being seen, you could include them in either the horizontal or vertical strips (or even both)).

Then I add a blank (transparent) edge around the image representing the canvas I need for stretching the canvas around the frame by increasing the canvas size by double the required margins in both dimensions, the same way we did above. That margin would be at least the width of the moulding along the bottom (1″ for the 11/2” moulding we are using now) and enough extra to get a grip with the canvas pliers (for me that’s at least 3/4“). That would make the image’s final dimensions at least 191/2” x 251/2“. When I am finished, I add layers with cut lines, fold lines, staple lines, positioning marks for the hanging hardware, etcetera, but that is a personal matter beyond the scope of this article.

## A Solution To Second Mat(h) Problem

Sadly, we had no winners to this contest. Here is a solution to that math problem:

There is more than one way to solve this problem, but we will be exploiting three different relationships. First, in preserving the aspect ratio, the length of the image (we’ll call L) is 11/2 times the width (W). $L = 1.5W$. Then, adding up the components making up the overall width of the mat, the image width (less two overlaps of 1/8“) plus two mat widths (M) would equal 16 inches. $W - \frac{1}{4}" + 2M = 16"$ By the same token, the image length (less same overlaps) plus two mat widths would be 20 inches. $L - \frac{1}{4}" + 2M = 20"$

If you replace the L in the last equation with its W equivalent from the first equation, and then add 1/4” to both sides of both equations to combine constants, you are left with the following two equations to solve with two unknown variables:

$\begin{array}{r c l} 1.5W & + 2M = & 20.25 \\ W & + 2M = & 16.25 \end{array}$

From here you can use linear algebra (matrices) or algebraic manipulation to simplify until you are left with just one variable. For example, just subtracting the bottom equation from the top (subtracting the left sides separately from the right sides of each equation), you will wind up with

$0.5W = 4$

which means the image width is eight inches, which means its length is twelve inches, and the mat guide would be set to 41/8“.

### What’s Next

I’ve come up with one more printing-inspired math problem, which I will share as soon as I master a new plug-in for this blog.  After that, I’m not sure.  Response has been weak, but the former teacher in me feels a need to keep pointing out opportunities to use some of this stuff you learned in school (or is it just to torment those students who were the most difficult – I’m not telling).  This isn’t really costing anything, and I give enough warning for the math-averse to stay clear.  Stay tuned.

## Plans For Making A Cardboard Box For Framed Art

I just added an article to the website (Making A Cardboard Box To Ship Art) that has plans and instructions for making a cardboard box. To make any size box, all you have to do is plug your dimensions into the expressions for each measurement on the plans. You can even download and print the plans (but not the instructions – hmmm).

As per my last post, you can also reach this article from “Tips & Techniques” on the main menu. Click on “Printing & Beyond”, which will take you to the “Do-It-Yourself Ideas For Finishing Your Photographs“. The sitemap will also work.

Enjoy!

P.S.  I’ve promised to publish plans and instructions for making our booth display panels in a form that people can actually use, instead of just the notes and scribbles I made for myself at the time, but those won’t be ready for the beginning of the next Florida art festival season as I had hoped.  The cost of materials for one three-foot wide panel was about \$50 a few years back.  Stay tuned.

It occurred to me the other day that it might be good to be able to download the steps to adjust a Logan Sander, as I described in Another Method For Adjusting A Logan Precision Sander. While converting to the Acrobat .pdf format, I took the liberty of adding more information and, I hope, making it easier to follow. To download, click on the link below. Enjoy!

Get printable version(.pdf)

I posted our first Simple Mat(h) Problem on April 27, 2017, and Jim Farrington submitted a solution a couple of weeks later. Here are a few more comments on the problem that I published (but in the wrong place).

Although the mat cutter has no kerf, at the start of the cut the blade does swing down and could cut into the side of your finished piece around that square in the middle. For backing boards, this is not a problem, and because at least 1/8” will be hidden by the moulding, it is probably not a problem when cutting mats either. There is enough extra space that if you wanted to play it safe you could put a 1/2” between the four pieces as shown below.

## Could There Be More Than Four Pieces?

To be blunt, NO. There just aren’t enough scraps to possibly make another 16″ by 20″ piece. Five such pieces would have an area of 5 x 16″ x 20″ = 1,600 square inches. We started with a piece that was 39″ x 37″, or 1,443 square inches. It just can’t be done.

## Is There Another Way To Get The Correct Answer?

We’ve all managed to get a broom into a shorter closet by sliding it in at an angle. One question I had was “Would it be possible to squeeze the originally planned 40″ by 32″ rectangle needed for the four pieces into the 39″ high space by rotating slightly?” The short answer is NO. Because this is so much wider than a broom, as you rotate, the required height actually gets larger at first and doesn’t drop back below 39″ until you’ve rotated over 79°. By then the necessary width would be over 45″ (well over the 37″ available). To see the math, see the note below (Warning: this “solution” requires knowledge of trigonometry). Of course, this doesn’t guarantee that there is no other solution. If you find one, let me know.

To help with the math, I made a drawing.

The target 40″ by 32″ rectangular piece of foam board is shown by the dark blue rectangle. The red rectangle is the smallest “box” that can be placed around it. The length (L) of the outer box can be described as $L = l \cos \theta + w \sin \theta$.

This reminds me of a trigonometric identity for the sine or cosine of the sum or difference of two angles:

$\sin (x \pm y) = \sin x \cos y \pm \cos x \sin y,$ and $\cos (x \pm y) = \cos x \cos y \mp \sin x \sin y.$

Let d be the length of the diagonal of the inside rectangle. It can be found as $d = \sqrt{l^2 + w^2}$. Furthermore, $\frac{l}{d} = \sin \alpha$ and $\frac{w}{d} = \cos \alpha$, where α is the angle that the diagonal makes with the base of the rectangle.

If we divide both sides of our first equation for L by d, we get:

$\begin{array}{r c l} \frac{L}{d} & = & \frac{l}{d} \cos \theta + \frac{w}{d} \sin \theta \\ & = & \sin \alpha \cos \theta + \cos \alpha \sin \theta \\ & = & \sin (\alpha + \theta) \\ L & = & d \sin (\alpha + \theta) \end{array}$

By the same token, the width (W) of the outer box can be described as $W = l \sin \theta + w \cos \theta$. Following a similar path as above, we can show that $W = d \cos (\alpha - \theta)$.

Now you can plug 39″ in for L, 51.225″ (which is square root of 402 + 322) for d, and 51.34° for α (by taking the arcsine (which may be shown as sin-1 on your calculator) of 40″/51.225″) into the last equation for L, and use your calculator to find that α + θ = 49.583°. Unfortunately, that’s less than α and doesn’t meet our requirements. But around a circle, there are two angles with the same sine. Your calculator will find the one that is less than 90° (we’ll call it 90° – φ). The other would be 90° + φ, which makes θ = 79.192°. Plugging that, along with d and θ, into the last equation for W gives the result I mentioned above.

### A Shortcut?

That last equation for L could also have been found more directly by noticing the orange right triangle formed by the dashed orange line along the inside blue rectangle’s diagonal and through the lower right corner of the inside rectangle (as the hypotenuse), and the vertical dashed red line of length l (as the side opposite the angle), and the segment of the bottom edge of the outer rectangle that’s between those two other sides (as the adjacent side). The light blue arrows around that lower right corner show the angle of this right triangle is α + θ. Then by noticing the yellow right triangle formed by the other orange dashed diagonal line as hypotenuse and the other red dashed line of length w as the opposite side again we could have found that $W = d \sin(\theta + (90 - \alpha))$. Then, since $\sin x = \cos(90 - x)$,

$\begin{array}{rcl} W & = & d \cos(90 - (\theta + (90 - \alpha))) \\ & = & d \cos(\alpha - \theta) \end{array}$.