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

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

## Another Method For Adjusting A Logan Sander

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.

### 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.

The full instructions are as follows:

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.

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.
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.

#### 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.

## Neutral Density Filters And Another “Brilliant Idea”?

A little over a year ago I was shopping for neutral density filters.  While doing my research, I triggered the following e-mail from the owner of Breakthrough Photography:

Hello Bruce,

My name is Graham and I’m the founder of Breakthrough Photography. I wanted to take a second to say hello and welcome you.

Seriously, on behalf of myself and the entire Breakthrough team I want you to know that we’re truly excited and grateful that you decided to join us.

Here’s what you can expect from me…

We publish new and actionable photography content to our blog 1-2 times a month.  You can see that I’m sending these emails from my personal address, so when you respond I’ll get back to you.

Sound fair? GOOD!

Now for the long exposure guide: click here to download our Essential Guide to Long Exposure Photography ebook. I wrote it as a reference guide so you can also save it to your iPhone or iPad for easy reference when shooting. Just click the link above on those devices. . . .

Best regards,

Graham

email from the owner of Breakthrough Photography

I downloaded and read his guide, and sometime during this research process I got another of my “brilliant” ideas and sent the following response to tell him about it:

To: Graham Clark (founder of Breakthrough Photography) at grahamclarkphoto.com
Sep 5, 2015 at 11:51 pm

Thank you for your Long Exposure Guide. I just spent a good part of the day shopping for neutral density filters for my wife, Nancy (She is a nature & wildlife photographer . . . . After seeing that your X3 line follows tradition and has only a 3-stop, 6-stop, and 10-stop version, I thought I should make a suggestion to improve versatility. . . .

By the way, that base 2 thing reminds me of one . . . error I found at the top of that same section.  The first sentence of the second paragraph says “A 6-stop nearly doubles or triples the exposure time of what a 3-stop would have been . . .”  Since each f-stop lets in only half the light of the one before, technically, moving 3 stops would give one eighth (½ x ½ x ½) of the light and require a shutter speed eight times as long (instead of three).

my return email
Even though it has been over a year, the mistake still hasn’t been corrected.

He immediately replied

Hey Bruce,

Couldn’t agree more, 12 would be more logical given 3 and 6, however 10 is where the demand is.

“One should not create demand with a product, only channel existing demand onto a product.” Eugene Schwartz

Graham

his return email

While I was impressed with the quick response time, I would have preferred that he spend the extra time to think before he responded.  I sent him one more quick response,