## Thoughts On Mat Layout

The easiest and most common mat layout is one with the widths of all four borders equal. If you are forcing a picture into a standard-sized frame, however, that’s not always possible. And then there’s the matter of bottom-weighted mats.

### Bottom-Weighted Mats

Bottom-weighted mats, or mats with the bottom edge wider than the others, were introduced long, long ago. Some say that pictures centuries ago were hung very high on the wall and the bottom width of the mat was increased to compensate for the ‘distortion’ of that perspective. Unfortunately, that story makes no sense; top-weighting would be required to correct for the top being further from the viewer than the bottom. Another explanation involves the notion of a difference between the visual or optical center and the geometric center. Yet others claim it is to compensate for the drop of the mat in the frame due to tolerances necessary to account for expansion, etc. For whatever reason, bottom weighting could be seen as an attempt to fool your audience or overcome optical perceptions, whichever you prefer. As commonly practiced in “finer frame shops everywhere”, the bottom width is generally increased ¼” to 1″, depending on the size of the pictureref.

### Using Standard Mats

But how would one incorporate bottom weighting while fitting an image into a standard-sized mat? For example, if the vertical difference between the hole size and mat size is greater than the horizontal difference, and assuming the left and right borders will be the same width, is it better to:

 A Make the top and bottom borders equal, B Make the top the same size as the left and the right and put all of the extra width on the bottom, C Make the bottom larger than the top by some fixed amount, D Make the differences even more subtle by making the difference between the top border and the side borders the same as the difference between the top and bottom borders?

Let’s clarify your choices with an example. Suppose you want a 4″-high hole that’s 7″ wide in a standard 8″-high by 10″ mat. The horizontal difference between the mat size and the hole size is 10″ – 7″ = 3″, so if you want the left and right borders to be the same, each will be 3″ ÷ 2 = 1½”. The vertical difference between mat and hole size is 8″ – 4″ = 4″.

 Choice A Would make the top and bottom borders the same, making them each 4″ ÷ 2 = 2″. Choice B Would make the top 1½” like the left and right borders, leaving 4″ – 1½” = 2½” for the bottom border. Choice C Uses the customary bottom weighting, which the one reference I give above lists as ¼” for an 8″x10″ mat (personally, a ¼” bottom weight isn’t worth the trouble). That means the top border would be (4″ – ¼”) ÷ 2 = 1⅞” and the bottom would be ¼” more, or 2⅛” (notice as you check your work that 1⅞” + 2⅛” = 4″). Finally, Choice D Is a tad more complicated. Let’s call the difference between the left or right border width and the top border width “d”, such that 1½” + d = T (for top border width). Then the bottom border (B) would be T + d or (substituting the last expression for T) (1½” + d) + d = 1½” + 2⋅d. Since T + B = 4″, then (substituting for T and B) (1½” + d) + (1½” + 2⋅d) = 4″, meaning 3″ + 3⋅d = 4″ or 3⋅d = 1″, meaning d = ⅓”, so (substituting back into our equations for T and B) T = 1½” + ⅓” = 15/6” and B =15/6” + ⅓” = 21/6” (again noting that 15/6” + 21/6” = 4″) .

The choice you make would be an artistic decision, but I think A is the most common answer. Choice C could be used for traditional bottom-weighting, as in our example, or could be used for some other more artistic value. Technically, both Choices B and D are possible results of that equation. B would be exactly what you get when you want bottom-weighting and are not restricted to standard mats; it would work best if the resulting difference between the top and bottom borders is not too much greater than the customary bottom-weighting distances mentioned above. In our example, it yields 2½” for the bottom border, which is an inch larger than the other three borders and may just be too much.  In our example, C and D are very close, and remain close when we change the amount of weight in C from ¼” to ½” (as shown by the lighter blue opening).  D is more subtle than C, but may only be worth the effort when the difference between the left and top borders is small enough to fool somebody.  In other cases with different numbers, results may vary.

### With Larger Side Borders

If the horizontal difference between the hole size and the mat size is greater than the vertical difference, you could face up to the same number of choices as above, but you are working with less material for the top and bottom borders and I think it is usually better to keep things simple and make those borders equal.

### Differing Left And Right Borders?

Do the vertical borders always need be the same size? Although I can’t say I’ve ever seen or read about different-sized side borders, I’m not convinced that uniformity is strictly required. For example, in photography, as in older art forms, there a “rule” of spaceref that says, among other things, that there should be plenty of space on the side of the subject into which it is looking. If you have a “perfectly” centered and close-cropped picture of your mother looking to your left, could a mat with a wider border on the left side create the space that’s lacking in the image?  Maybe you could even choose a mat color that is a pastel version of the background to her right (your left)? Maybe a contrasting outer mat could be added with traditional (identical) vertical borders.

I present the above thoughts to give some background and (more importantly) stimulate your own creativity. If you think of other possibilities, I’d be thrilled to have you add them to the comments. Thank you!

## How To Find The Area Of An Object Using Photoshop

There are mathematical or drafting programs that may do a better job of finding areas of all sorts of seemingly random two-dimensional shapes, and I may have used one or two of these as a student, but I haven’t had any of them on my computer for many moons. So when I recently needed to compare the size of the visible sun at different times during a solar eclipse so I could compare exposure levels, I was out of luck. But then there was Photoshop. I just finished this article about how to find an object’s area, and put it on our website at www.beehappygraphics.com/find-area.html, mainly because I mentioned the technique in an earlier blog post, and was about to mention it again in an article I promised about the challenges of our newest eclipse image.  This probably isn’t the most common task you will be doing, but when you need it, this can be handy.  Enjoy!

## 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.

## Ideas For Shooting The Solar Eclipse In Miami With Phone Or Camera

I have some ideas for shooting the eclipse by either phone or SLR camera.  For those who haven’t heard, the next eclipse will be Monday, August 21st. In Miami, the eclipse will start around 1:30 pm, which is right after local apparent noon (when the sun crosses due south of us around 1:24 pm and is 77° above the horizon). The eclipse will last about three hours, by which time it will have reached an azimuth (compass bearing) of 261° and dropped to a height of 44°. At its peak just before 3 o’clock, it will be 64° above the horizon at a bearing of 243° (west-southwest). At that time, less than 1/5 of its diameter will be visible in South Florida, which means that about 22% of the sun’s area will still be showing, and the sun will still be a little less than 1/4 of its normal brightness (for lack of anything better at hand, I used Photoshop’s Count Tool to figure the sun’s brightnessHow).

In the news, they mentioned that you could use your smartphone to view the eclipse, but they warned that if your phone wasn’t eclipsing the sun (directly between you and the sun, obstructing your direct view) you could get seriously hurt, and since there are no nerves inside your eyeball, you wouldn’t immediately know the damage that was done. For that reason, you may want to use it in selfie mode.  You may also want to wait until the eclipse is close to its peak (although I have taken some test shots of the sun with no apparent damage to my phone).  There are a few problems with this approach, however. For one thing, the glare from your phone’s glass surface and/or the bright sunlight could make the image on the phone hard to see. On the other hand, if you actually wanted pictures, having yourself (or something else) in the foreground could improve the composition of the photograph.  But-

1. The resolution for the selfie camera may not be as great as on the regular camera. (I explain why bigger is better on the Bee Happy Graphics FAQ page).
2. My selfie camera doesn’t have controls for flash, exposure, white balance, and other things; these features being listed in the order of their importance.

You will need fill flash on your foreground subject, and the flash will probably need to be less than two feet away to be effective.  But that means the camera is in regular (non-selfie) mode and both aiming and pushing the shutter button could be a pain.  A short timer, if your app has one, could be helpful in pushing the button.

### Shooting With A Camera

First, you will need neutral density filters, not just for the proper exposure but unless you shoot in Live View mode it is more important that the filters can adequately protect you looking through the viewfinder.  For that, a 10-stop filter is not enough (but a 12-stop filter, if it existed, could be (at your leisure, you can check out the Bee Happy Graphics blog for another reason a 12-stop neutral-density filter would be better than a 10-stop). A 15 or 16-stop filter would be even better in this case. Focus on the horizon before attaching your filters and lock in your focus.

If using a zoom lens, begin as wide as possible; it is easier to find the sun before zooming and avoid the dangers of trying to peek around the camera.  You will need exactly the same focal length or amount of zoom that you needed when you took pictures of the moon. Most experts feel anything less than a 300mm lens is a waste of time. Remember that your shutter speed should be 1/(focal length x crop factor) or faster if you not using a tripod, but even with a tripod there may be no reason to go with less. The aperture (f-stop) setting is not critical since all the action is at infinity but should be small enough (large enough number) so that you can keep the ISO at its lowest value.

If you are planning to capture the whole eclipse in a sequential composite photograph, decide how many images you need, subtract one, and divide that number into 180 minutes (the duration of the eclipse). If you want a string of six suns in your picture, each picture will be 180/5 or 36 minutes apart. The camera will probably not be locked down to the tripod for the duration, but the focal length of the lens and other settings should be the same for the entire series.

The only way to get something in the foreground (for better composition), is to go for multiple exposures and combine them manually. At the designated time, take the sun shot and while the camera is strapped to the tripod, record your camera settings, remove the filters, change the settings as needed and shoot the foreground. For multiple exposure shots, they usually advise changing only the shutter speed, but I’m not sure it matters in this instance. If changing the shutter speed alone is not enough, I’d change the f-stop before changing the ISO. Now record the settings of the foreground shot so you can repeat as necessary. If you must change the focus for the foreground shot, be sure to refocus on the horizon before putting the filters back on. Return the camera settings to the sun shot values. You may now move the camera on the tripod to compose the next shot. I mentioned that the sun will be putting out only 1/4 of its normal light at the peak of the eclipse here in Miami. This means the exposure of your foreground shot will change by two f-stops. The exposure of your sun shots shouldn’t change.

### Final Words

Since this is such a rare event, you may not want to put all of your eggs in one basket. This means changing the settings of your camera (bracketing, if you will, checking the histogram, and perhaps rechecking the focus), which may mean taking several sequences simultaneously and taking good notes.

I’ve discussed some of your options, with some of the pros and cons of each one.  While I try to cover the technical aspects, you are the artist and the compositional issues are all yours.  It might be a good idea to get up early tomorrow and get some moon shots just for practice.  The moon will be just a waning (shrinking) crescent.  Moonrise here in Miami will be 4:37 am tomorrow and 5:40 Sunday (sunrise is 6:56 both days).

Well, that’s about it.  Have fun, don’t look directly at the sun, and let me know how it worked out for you.  I’d even be willing to post some of your pictures (with adequate credits of course).

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 $\sin x = \cos(90 - x)$,

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