A few weeks ago, I asked a few questions about a picture of me on the Turner Riverlink. I even offered a reward for the best answers. Here are my answers.
First Question
NO, the picture is not level. The photographers’ usual reference point for getting a picture level is the horizon. One of the rules of composition says that your horizon must be perfectly straight (unless it is so far off that the viewer will know that you did it on purpose).
To see the Note click here.To hide the Note click here.
But what if the horizon is not available, as in this picture? Nasim Mansurov, in his article The Importance of Straightening The Horizon and Aligning Lines, discusses (and shows examples of) several options for getting your picture straight. But he didn’t mention this situation.
Q2: How Do We Straighten It?
First, let me say what won’t work: allegedly horizontal elements on a canoe, like seats and thwarts, even in calm water, can be expected to deviate from horizontal as a matter of routine. This case, as it turns out, is no exception. So what can we use?
When you have a calm body of water, as we do here, and the horizon is not visible, you can still depend on the levelness of the water. As I discussed in Reflections – My Answer To “What’s Wrong With This Picture (Version 2)?”, the angle of incidence of a reflected light ray equals the angle of reflection, which means, since the surface of the water is a horizontal plane, that the reflected object will be directly under the object itself, or the line between the two will be vertical, or 90° from the horizon. In the picture above, I have identified four different reference lines. Once you are convinced that this works, you really only need one. Two things make this more challenging, however. There are floating obstacles obscuring good reflection candidates. Also, as I discussed in “Reflections…”, the reflection won’t look exactly like the reflected object due to the changed perspective. Since identifying the exact point reflected may be subject to slight errors in estimated position, the further they are apart (meaning the longer the line connecting them) the better because the error in the angle needed to rotate for the picture to be level is proportional to the positional error divided by the distance apart (for small error angles). In the above picture, the flower (labeled “A”) is an easy choice, but it and its reflection are close together. The hole in the canopy (“B”) and the more prominent branch (“C”), although less identifiable in the reflection, do have good separation distances. “D” shows that in a pinch, when no well-marked points are available, you could even use the point on a curved line where the slope of the curve and the slope of its reflection are the same (or parallel). Expect a higher positional error in cases like that.
Once you have a reference line, most editing software has a horizon-straightening feature, or at least the ability to rotate the image until your reference line is vertical. For what it’s worth, this image needs to be rotated about 8⅓° clockwise. To see the corrected version, go to the bottom of our Red Mangrove Maze image page, where you can also find the identity of the person who took this picture.
Bonus Question
For the last question of the article, which was a math problem to find my age, see the note below.
To see the Note click here.To hide the Note click here.
As you can see in the above illustration, there are three related timespans (years before 2008, years after 2008, and total years); if you know any two, you can find the third. This is true whether you are talking in years or percentages. We want the answer in years, but we only know one of the three. As a percentage, we know two of the three, so we can (and will) know all three. What we have to do, then, is find a relationship between years and percentages.
Twelve years is less than 19% of my current age,
or
Actually, I’m about 64½.
Your reward (including bonus) would be times the original award, meaning your bonus would be about 23%.
And The Winner Is…
There were four responses to the original post. All addressed the title question. Nobody addressed the follow-up or bonus questions. The judges have concluded that the first correct answer and winner of this contest, receiving ten dollars off of any Bee Happy Graphic product or service, and all bragging rights, is M. Alexander (former member of Kendall Camera Club). Congratulations!
I just recently looked a little deeper into some rainbow geometry and discovered that this challenge could be a lot harder than I first imagined. But for those of you who have no idea what I’m talking about, here’s a little history –
The Facebook Quiz
This all started with one of the little quizzes I put on my personal Facebook page from time to time. This one went like this:
– 11:59 pm on July 11, 2019
A friend, John Gilbert, responded:
Bogus question, Bruce. Answers D, E and F are unicorn answers. I’ll take B, high noon, for 500.
John Gilbert’s 4:56 pm Facebook comment on July 12, 2019 to above post
To see the Note click here.To hide the Note click here.
Although John was a photographer for the Navy, now he is an artist. You can check out his website www.johngilbertartist.com.
to which I responded,
You started well but still wound up with the fourth-best answer (just behind D)….
My 3:31 am response on July 22, 2019
Then:
– 8:50 am on July 22nd– 10:10 pm on July 22nd– 6:42 am on July 24th– 8:53 am on July 24th
To see the Note click here.To hide the Note click here.
By the way, the correct answer to the quiz could be either A or C. If you can’t handle more than one correct answer on a multiple-choice question, pick C (more large rainbows will be seen then because more people are awake around dusk than dawn). The animation below shows how a rainbow changes with the movement of the sun. (The black face on the ground represents the top of the shadow of the observer.)
The Contest
To speed things up, I then asked our local Kendall Camera Club for help by creating a contest for the best nighttime rainbow picture. I am now extending that contest to all of my readers. Here are the rules:
Contest Rules
I’ll give the equivalent of a six-pack of your favorite brew (within reason) for the best picture of a rainbow at night. I said six-pack equivalent because a corresponding discount for any of the products or services of Bee Happy Graphics (see our Products and/or Services pages) may be substituted at your request.
The contest will run at least another month but may continue until we get a chance to get our own rainbow pictures (which will not be part of this contest), or until I give up all hope of completing this quest. If there are not enough entries at the end of the contest, the prize need not be awarded.
Composition does count. Photos of real rainbows will be given precedence, but just in case I overstated the possibility of that occurring, “Photoshopped” rainbows will be accepted.
I will announce the close of the competition and the beginning of the voting process in a comment to this blog post. I will explain the voting process in that same comment.
At least three weeks after the announcement from Rule 5 above, a winner will be announced. If any entry has three or more votes, the one with the most votes will be the winner. If no entry has that many votes, then I will take an informal survey among my closest family and friends, and pick the winner. The decision of the judges (as defined above) is final. This prize may be combined with other promotions.
Since this is a photo contest, I should mention that the photographer retains all rights to his/her image. We get to use your image only to run and publish the results of this contest.
So far, there have been no entries.
The Physics
As I mentioned at the beginning of this article, the geometry doesn’t look good.
Figure 1: The average angle between the light source (usually the sun) and your eye from a single raindrop. Because of refraction, the angle for violet light is a couple of degrees less than the angle for red light.
Figure 1 shows the angle that a light ray bounces off of a single raindrop. In Figure 2 (which is a view looking down, with the blue-hashed region being a rain cloud), one can see that as the sun’s rays all come in as parallel lines from the same direction, the five representative raindrops, and all others from which the blue light rays bounce at the same angle into the observers eye are all along the same line of sight, and therefore all reinforce each other. A single raindrop doesn’t reflect a lot of light and it takes the light of a lot of raindrops to make the rainbow visible. Similarly, the raindrops reflecting red light into the eye from a slightly different angle form along a slightly different line of sight and also reinforce each other.
Figure 2: The geometry of light reflected/refracted from a distant source where the incoming light rays are all parallel
The Problem
Figure 3: The geometry of light reflected/refracted from a nearby light source
But using flash, the geometry is different; the light source and the eye can now be represented by single points. But unless you were the navigator of a Coast Guard buoy tender or something, you may have forgotten a lesson from high school geometry; that the set of all points for which the angle between those two objects is the same defines a circle, not a line. In Figure 3, the angle between the flash and the eye is the same for Representative Raindrops A and B, and all other dark blue points on the outer circle (there is no rain along the pale blue segments of that circle). Unfortunately, each point on that circle is along a different line of sight from the observer and the light rays do NOT accumulate in the eye. To make matters worse, the red light reflected from Representative Raindrop C (on the circle of red light) IS along the same line of sight as the blue light from Representative Raindrop B (and the other colors of the spectrum from raindrops directly between them), which tend to cancel out into white light. This means you will see no rainbow.
Where Does That Leave Us
The original purpose of this article was to throw in the towel and admit defeat, but as I was getting the materials together, I’ve already come up with a couple of new ideas that need to be explored. But the busiest part of our art festival season is now upon us and the rainy season doesn’t start in South Florida until around June, so this is not the rainbow of promise, but one of hope. Stay tuned. (And if you have a photograph of a nighttime rainbow, please send it.) Thanks.
Here’s another math problem inspired by real life! Each correct answer could earn you five dollars ($5) off any of our products or services (to redeem in person, just print and show your certificate). Good luck!
You are only allowed one try and have already submitted your quiz.
I recently posed the question “What’s Wrong With This Picture”blog about a modified landscape photograph of a foggy sunrise in Ten Thousand Islands National Wildlife Refuge in Goodland, Florida. It turns out Deborah Gray Mitchell, one of the commenters, was right; the image was upside down.
To see the Note click here.To hide the Note click here.
Ms. Mithcell has her own website (www.dgmfoto.com), but several other sites have information about her. Just Google “Deborah Gray Mitchell”.
To be more precise, I flipped the image vertically and took steps to remove ripples in the reflection and such so that the answer wouldn’t be so obvious. You can see the original picture at “Foggy Sunrise” on our website. Now I’d like to discuss reflections and the clues that should have given the answer away.
Reflections
Figure 1: Perspective showing differences between the direct and reflected image
. . . Of Your Subject
First of all, the reflected image should NOT look like a mirror copy of the unreflected image, because the photographer has a different perspective or viewing angle of the reflection. As your high school physics teacher may have told you, in reflections, the angle of incidence (e.g. α2 in Figure 1) equals the angle of reflection (α1), so the view you have of the reflected image would be the same as if the subject had been flipped below the reflecting surface, as shown in Figure 1 above. I know that may sound like I just contradicted myself, but it is the subject itself I just flipped, not the direct image of the subject. Notice in Figure 1 that in the reflection, the two trees appear the same height, as depicted with red sightline C, while in the direct image the far tree looks higher as shown by green sightlines B1 and B2. The further away the subject is, the less of a difference this makes.
Figure 2: Alternate perspective of reflected view that’s better for showing effects on the sun
. . . Of Celestial Bodies
Here’s another way to look at the effects of reflection; it is as if you had been flipped below the reflecting surface, as shown in Figure 2, instead of flipping the subject. Although possibly less intuitive, this interpretation yields the same results, as shown by lines B1, B2, & C, but makes the effects of the reflection of the sun more apparent. In the image under consideration, as in most cases, the sun would have been your biggest clue. The sun is 93 million miles from us, but even our closest celestial body, the moon, at under a quarter of a million miles (say 238,900 miles), is much further than what your lens considers to be infinity. All light rays from the sun are virtually parallel (or come in at the exact same angle), no matter where you are.
To see the Note click here.To hide the Note click here.
This detail helped Eratosthenes figure out how large the Earth was 2,260 years agoexplained and was crucial to celestial navigation. It is also important in the creation of rainbows. I might be addressing that aspect in an article about my quest for a midnight rainbow. Stay tuned! (See My Midnight Rainbow Quest – Tougher Than I Thought.)
This means that the sun will always be higher in the direct view than it appears in the reflection (compare the angle between sun ray A and line B1 to the difference between comparable sightlines D and C).
So There You Have It
I hope that clears things up. This information should make you better at spotting fake reflections, or as a photographer, help you create better forgeries by knowing what mistakes to avoid. Good luck!
Of course, you may share your reflections on this or any related material (or questions) in the comment section below. Thanks for stopping by.
This is our second finished test frame for this article. It uses 1¼”, 2″, and 3″ moulding widths. Incidentally, the image is a new one, Silverback.
In this article, the first of the “Weird Wood” seriesintro, we show how to build a picture frame using four strips of moulding that aren’t all the same width. Although Figure 1 uses a different width for each piece of moulding, we used three different sizes in our test frames (only because I couldn’t find four different sizes in the same moulding family).
Figure 1: Building a picture frame with different moulding widths (drawn to scale)
First The Math
Warning: This discussion includes a little trigonometry. Do Not Panic! It’s not as bad as it sounds.
Figure 2: Close-up of the lower right corner of Figure 1
Definition of “Tangent” (skip ahead To Next paragraph if you still remember this):
There are three sides to any right triangle (a triangle with a 90° corner). I will call the ones touching the right (90°) angle the height and the width. The last, longest side is the hypotenuse. It is opposite the right angle. You can use the ratio of the lengths of any two of these sides to find the size of the other two angles. Each possible ratio has a name, but we are only interested in one of them today. Probably the most common ratio and the one we will be using is called the tangent. The tangent is defined as the ratio between the height (the length of the side opposite the angle you are interested in) and the width (the length of the shorter of the two sides that create that corner that you are interested in). If you want to know the angle of corner α in the above drawing (Figure 2), for instance, you would calculate its tangent by dividing the height (3 inches in this case) by the width (1¼ inches), which is 2.4. Then you would use your calculator (or phone app – I use RealCalc Plus by Quartic Software (even though it cost $3.50)) to find the angle corresponding to that tangent. On your calculator, the tangent is abbreviated “tan”. If you enter 45 (degrees are assumed) and hit the “tan” button, you will get 1 because for a 45° angle the height is the same as the width, so their ratio is 1. To go the other way (to find the angle), like we are trying to do, we need the inverse of the tangent. Look for the “tan-1” button (it could be the same button, in which case you may need to hit a (yellow) shift or second-function key, and then hit the “tan” button). In this case, once we have the tangent of 2.4, we hit the inverse tangent button(s) to get 67.380135…. (the calculator is obligated to give you 8 or more digits – that doesn’t mean they mean anything. In Figure 2, I rounded that answer to 67.4 degrees and even that third digit should be suspicious.)
The Process
All you have to do is take the ratio between the widths of your two moulding pieces and take the inverse or arc-tangent to get the angle. Here are a few things you need to remember:
Which angle – the tangent gives you the angle that was touching the side whose length was used for the denominator (the width, which would be the second number in the division). The simplest way (but certainly not the only way) to get the other non-90° angle is to just subtract the first from 90° (since the two angles are complementary). Also remember that if the tangent was greater than one, the angle will be larger than 45°; if it was supposed to be a smaller angle (less than 45°), then you may have divided the two lengths in the ratio backward. Don’t worry, you just found the complementary angle and all you have to do to get the right answer is subtract what you got from 90.
It is up to you to keep track of whether that angle you are cutting should be to the left or the right. Making a drawing of your frame design might help. To be useful, the drawing doesn’t even need to be that good. This should also tell you if you calculated the complement (the other angle in that corner (for the other piece of moulding)).
Your saw may be measuring angle backward. My miter saw calls a cut perpendicular across the board 0°, not 90°. If that’s the case, just subtract the angle you calculated from 90.
As an exercise, go ahead and check the rest of my calculations in Figure 1. 😁
Make The Cuts
There is more than one way to make these cuts and more than one set of tools to help you. Which set of tools you should use will depend on such factors as how much of this work you intend to do, your skill set, what your budget is, and what tools you already have on hand.
To see the Note click here.To hide the Note click here.
Looking through the Framers’ Corner, the forum of the Professional Picture Framers Association, I found recommendations for the following tools for this application:
You would only need one of these (if any), not both.
(The Amazon.com descriptions are only used here as a reference. Although frequently competitive, Amazon isn’t always the only or the best place to buy something.)
Our workshop includes all of the tools listed in www.BeeHappyGraphics.com/about.html#BruceEquip, along with a number of other regular hand & power woodworking tools that Nancy has accumulated over the last several decades. For this project, I used our compound miter saw, but not without complications.
To see the Note click here.To hide the Note click here.
The precision on this saw looked fine; you should be able to get within ¼° of your target. The first picture (Image A) shows me trying for 22.6° (which would be one of the angles between a 3″ and a 1¼” moulding).
Image A: Peparing for a cut of 22.6°.
After cutting the 3″ piece, I ran into problems trying to cut the complementary angle (67.4°) on the 1¼” piece, as shown in Image B.
Image B: Trying to set up a cut of 67.4° exceeds the capabilities of the equipment.
I am not claiming that mine was the best path to reach our goal. In fact, I would love to see your ideas in the comment section about how to improve my techniques.
How I Did It
Working with one corner at a time, I cut both pieces of moulding square just a tad longer than their overall/outside measurement according to your diagram (you will see why in Step 4). If you don’t already have one, this is also when you would put a perfectly square cut on the alignment block you’ll see in Figure 5 to the left of the moulding. I grabbed a 2″ by 4″, but the wider the better.
I set the saw for the smaller of the two complementary angles, rechecking my diagram to confirm whether it should be to the left or right. In the setup shown below, the 3″ moulding would be clamped to the right of the blade.
I made the cut.
Without adjusting the angle of the saw, I set up the second cut. I positioned my (newly cut) alignment block to the left (opposite the side we placed the moulding for the cut (in Step 2)) so that I could also place the 2″ moulding to the left of the blade and perpendicular (at a right (90°) angle) to the miter saw fence. After clamping down the alignment block, I added a support block to the right of the moulding to keep it in place. I could still move the moulding in or out to position the cut. As you can see, I needed to precut this piece of moulding to keep it from extending too far into the aisle and getting in my way. Another reason is explained in Step 6.
I made the cut.
To see the Note click here.To hide the Note click here.
For those who noticed that the color of the moulding in Figure 4 was different than in Figure 5, I had to make two different frames while doing research for this article 1) to confirm and refine my techniques and 2) because I didn’t get enough pictures the first time.
Always check your work. Since you precut each piece of moulding a little longer than necessary, consider this first cut on each piece a test cut. See if the two pieces match up as expected. If, when you put the two pieces of moulding together, the miter edge on one piece is longer than the other, that is the angle that should have been larger. The angle on the other piece of moulding should have been smaller (by the same amount).
Figure 6: Example of cut with angle error ε.
Figure 7 shows the second setup from the right side. If you look close, you might notice that I didn’t cut enough to make a sharp corner and needed to recut.
Figure 7: Side view of the second setup
Moving to the next corner, I precut another piece of moulding and repeated Steps 2 through 6. Now that both angles are correct for the second piece of moulding, you can recut its last miter if necessary to get the length right.
I repeated Step 7. When both angles are correct on the third piece of moulding, recut the last miter on that piece as necessary to get the length right.
I repeated Step 7 one last time. When both angles are correct on the fourth piece of moulding, I used the second piece of moulding to mark the length of the fourth piece by matching the inside edges, as shown in Figure 8. Similarly, I used the third piece of moulding to mark the length of the first piece.
Figure 8: Before the cut: use the opposite piece of moulding to mark inside edge to be cut.
Figure 9: After the cut: verify that the inside corners and outside corners of opposite pieces of moulding match.
Finishing
As with my normal (45° miter) frames, I would next need to make sure the inner lengths on opposite pieces of moulding matched, and the outer lengths as well. Figure 9 shows a way to check to see if the outside and inside corners of the opposite sides match using two carpenter squares (or equivalent).
Some of the tools we normally use next to finish putting the frame together, namely our Logan Precision Sander and Logan Pro Joiner, are worthless for this application. After gluing (and clamping the pieces together until dry) we had to pound the V-nails in by hand (interestingly, the simpler Logan Studio Joiner can be adapted).
Figure 10: Nancy pounding V-nails.
The Back Side
For completeness, the left figure below shows what the backside of the lower left corner would look like. The gray section represents the rabbet, the equal-width (¼”) cut-out that holds the glass, mats, image, and backing of the picture inside the frame. Some of you might be surprised to see that there is a triangular notch in this rabbet in the corner along the miter cut. This notch has no effect on the functionality of the rabbet. To solve this “problem”, however, you could make a compound cut 45° in from the inner edge to the edge of the rabbet and 79.7° in from the outer edge to the same point, as shown in the right figure below (as an exercise, you can check my math on these angles also). But there is really no need to make these cuts. If the gray were to represent an equal-width feature on the front of the moulding, it might be worthwhile to take the extra trouble. Otherwise, don’t even think about it.
The End
Congratulations, you now have a fancy new picture frame. Of course, you still need to find a picture, cut mat(s) and backing, mount picture to same, cut glass, assemble the pieces without showing any annoying little specks, and apply a dust cover and hanging hardware, but all of that is beyond the scope of this article. Good luck!
As mentioned, this article is just the beginning of a series about “Weird Wood” that I announced months ago. Up next, we will look at handling moulding that is not of uniform width. You won’t find this moulding in any store; it is only an exercise to prepare you for our final project. But if it stimulates your creativity, that’s not always a bad thing. Stay tuned, and thanks for reading! Your comments are welcome and appreciated.
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″) .
Our Four Mat Choices (drawn to scale)
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!
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!
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). . 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. By the same token, the image length (less same overlaps) plus two mat widths would be 20 inches.
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:
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
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.
. 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. 
By the same token, the image length (less same overlaps) plus two mat widths would be 20 inches. 
