## 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. ## More About Adjusting A Logan Sander 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) ## Comments on Mat(h) Solution 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 see the Note click here.To hide the Note click here. 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 $\sin x = \cos(90 - x)$, $\begin{array}{rcl} W & = & d \cos(90 - (\theta + (90 - \alpha))) \\ & = & d \cos(\alpha - \theta) \end{array}$. Now, aren’t you glad you asked? ## Another Method For Adjusting A Logan Sander To download a printable version of this article (AdjustingSander.pdf), click here. We have a Logan Precision Sander Elite Model F200-2 disk sander for improving saw-cut miters for your picture frames to a “perfect 45°” after cutting the moulding to size on our miter saw. To maintain such perfection requires due diligence and occasional adjustment. ### How Do You Know When It’s Time To Adjust Your Sander? 1. You may notice that when you put your frames together, there is a small gap between the pieces of moulding either on the inside of all four corners or the outside of all four corners. If some corners have gaps on the inside and some have a gap on the outside, you have other problems. In the figures used in this article, the symptoms have been exaggerated for illustration purposes. If the condition of your sander gets this bad without you noticing, you may want to consider another profession or hobby. 1. When you are comparing the lengths of opposite pieces, and you have them side by side with the miters face up and their back sides touching, you may notice by running your finger over the miter that they are the same length on one end of the miter but not the other, or that one piece of moulding is higher at one end of the miter and the other piece is higher at the other end. In either of these cases, it’s time to adjust your sander. ### But What About The Miter Saw? It may be true that the miter saw also needs adjustment, but that would have minimal impact on your frames because even if the angle of the cut was wrong, the sander should correct that problem. Of course, it would take more sanding to correct, which besides taking more time and effort could, in the worst case, result in your frame being too small, so it should periodically be checked and corrected according to the manufacturer’s instructions (I currently have no improvements or suggestions for that process). An indication that the miter saw needed adjustment would be if as you are sanding the miter, sawdust builds up on top of one side of the moulding faster than it accumulates on the other. If it takes too many revolutions of the sander to perfect the edge, that could also be a clue, or it could be time to change the sandpaper. ### How To Adjust The Sander On the last page of the 4-page manual (available at www.logangraphic.com) are simple instructions for that adjustment that should work well if you are willing to follow Step 1 and remove the sandpaper. To see the Note click here.To hide the Note click here. The full instructions are as follows: Adjustment 45° 1. Remove sand paper 2. Place the 45˚ square flat against the wheel and up against bar (Fig. 7). Look for gaps against the bar. 3. Adjust the bar using adjustment wrench until gap disappears. When I don’t remove the sandpaper disk the technique doesn’t work as well, so I’ve come up with an alternate set of instructions: 1. Put miter cuts on both ends of two long scrap pieces of your widest moulding. 2. When you sand a piece of moulding, each end will use a different side of the sander. Call one side of the sander “A” and the other “B”. As you sand the two pieces of moulding, mark the back of each end of each piece with the side of the sander used (A or B). 3. Find a good right angle, either in a reliable carpenter’s square or using other methods. 4. Flip one of the pieces of scrap moulding upside down so you can join Corner “A” on both pieces to make a 90˚ (right) angle. Flipping is very important*. 5. Put one piece of moulding along one edge of the reference angle (carpenter’s square) and slide the reference toward the second moulding until it just touches at one end or the other (if it touches at both ends, you are finished with Side A – skip ahead to Step 7). Measure the error gap (I like millimeters only because they are so small) at the end of the moulding that’s away from the reference line. Then measure the length of that piece of scrap moulding (using the same units of measurement). 6. The adjustment screw on my sander had 32 threads per inch, and it was 109 millimeters from the pivot point. Based on that, your multiplier will be 25,000. To see the Note click here.To hide the Note click here. I’ll do the math just in case your sander has different measurements so that you can substitute the real numbers in for mine at the proper places. One complete turn of the adjustment screw is 360° or 1/32” and there are 25.4 millimeters per inch, so the constant multiplier would be 109 mm * 360° per turn of screw * 32 threads per inch ÷ 25.4 mm per inch ÷ 2 errors = 24,718.11. We’ll say 25,000. Divide the error you measured in the last step by the length of the moulding that you measured and multiply by 25,000. Your result will be the number of degrees you need to turn the adjustment screw. If the error gap was at the corner, then the angle is too large and you have to turn the screw counterclockwise to back it out. Conversely, if the error gap was at the end of the moulding, then the angle is too small and you have to turn the adjustment screw clockwise to push it out more. After making the adjustment, you may want to retest by repeating the process by starting at Step 2 and resanding the same two corners just used. Before sanding, I recommend drawing a line all the way across the end of the moulding with a pen or marker and then sanding until the line completely disappears. 7. Repeat this whole process (starting at Step 4) for the other two miters, labeled “B”. Remember, for these two you will be playing with the adjustment screw on the other side of the sander. That’s all there is to it. Congratulations. To see the Note click here.To hide the Note click here. #### Math Warning: a quick note about trigonometry (OPTIONAL)! This process was concerned with angles, not distances, but since angles are harder to measure with any precision we had to convert. When you take the ratio of the two perpendicular sides (the sides that are 90° apart) of a right triangle containing the angle you are interested in, that’s called the tangent of that angle, and you can have a good calculator app on your phone find it for you (for my Droid, I found RealCalc Plus by Quartic Software at the Play Store and was happy to pay$3.50. There are plenty of other options, though).

The error angle you measured (indirectly) was actually twice as large as the real error. One problem is that the tangent curve is not generally a straight line, which means that the tangent of twice some angle is not the same as twice the tangent of that angle. That’s why the normal procedure would be to convert to angles, do the adding, subtracting, or multiplying, and then convert back to distances we can measure again. We were able to use the small angle exception, however. It turns out that for angles less than say 10°, the tangent curve IS pretty straight and the error introduced by taking our shortcut isn’t worth worrying about. That’s what we did, and that’s why the problem was so easy. That’s your math lesson of the day week month. Let’s get back to work.

## Instructions For The Hanging, Exhibition, And Care Of Fine Art

There are three documents I made to be included in the Certificate of Authenticity package that, as Nancy has since pointed out, one would need before they received their certificate. Consequently, we now give them to our customers when they buy a limited-edition print of any of Nancy’s images. All of our customers who have previously bought an image will still receive this information in their certificate package, but I have also just made them available on our website for anybody to read and/or download. The documents are:

Enjoy!