Category Archives: Asked and Answered

Asked and Answered 4.0

I am one of those annoying persons who is kind and cooperative in everyday interactions but whose competitive bloodlust rises to the fore when Scrabble is involved.  Ask my wife.  She stopped playing Scrabble with me years ago, even after I began to spot her 100 points. That was little help, as far as she was concerned, because it only made me dig my heels in that much harder to overtake her.

I play online now, as I assume most players do, because who stops by your place to play Scrabble these days?  I am a good casual player but not tournament-level by any means — not that I care to be.  I have only recently begun to learn the two-letter-words and I am constantly surprised by the odd three-letter-words my opponents play.

Things were different in the analog days.  Back then, diligent players had to memorize the accepted words, because consulting a dictionary was verboten except to answer challenges. (VERBOTEN is an official Scrabble word, by the way.)  But there are no challenges in the online version — instead, players are allowed (and expected) to check whether words are valid before playing them.  And the two-letter-word list is now just a feature of the game.  So no one really needs to memorize much.

That said, there is still a place for vocabulary skills in online Scrabble, tempered as always by the luck of the draw of one’s letters.  And that, at long last, brings me to the point of this article.  I just completed a game with an anonymous online opponent named Micki.  Here is the board we played:

Scrabble Board - The 100 Billionth PersonYou can see that Micki defeated me, 339 to 308.  What you do not see is that Micki drew and played all five of the highest-scoring letters (Q, Z, J, X, K), along with seven of the ten next-highest-scoring letters (two each of F, H, V, W, Y) as well as the two blanks, each of which confer a 15 to 30 point advantage [Thomas, 2011].  Now perhaps I am biased, but this seemed to me to represent an especially good string of luck — for my opponent.

The competitive person that I am asked, what are the chances of such a lopsided draw?  The nerd that I am set about to answer it.

• • • •

At first I thought I would need to program a computer simulation of thousands of games with random word lengths, to see how likely it would be that a real game would have such a one-sided letter draw.  But then I made a few simplifying assumptions.  (That is what old engineers eat for breakfast, simplifying assumptions.  For us, they go down as easily as, say, eggs Benedict did for the last Pope.)

My first simplifying assumption was that each player will likely draw half of the stockpile of letters over the course of the game.  This assumption allows us to neglect the number of tiles drawn on a play-by-play basis.  Once we stipulate that each player will draw 50 of the 100 tiles over the course of the game, we can focus on the distribution of tiles between the two players, as if all the tiles were shuffled and dealt out to the players like a deck of cards.

To further simplify the problem, we can imagine that the special, high-scoring letters are dealt out first, followed by the low-scoring letters, using the following process.  Start with the Q.  To see which player gets the Q, the dealer flips a coin.  If the result is heads, the Q is dealt to my pile, otherwise it is dealt to my opponent.

The dealer repeats this process for the remaining high-scoring letters and the two blanks, seven tiles in all.  Since the chance of my being dealt a tile after a coin flip is 1/2, then the chance that I would be dealt all seven of the special tiles is 1/2 to the 7th power, or 1/128.  Needless to say, this is also the chance that my opponent would be dealt all seven tiles.

Though not common, landing all the best-scoring tiles is not as rare as I had first assumed. But wait — my opponent not only also played the top seven tiles but also seven of the ten four-point tiles.  So we need to do a bit more math.

In the second phase of the process, the dealer takes the ten four-point tiles and again flips a coin to determine which player is dealt each tile.  What is the chance that I will be dealt exactly three of the ten tiles?  Interestingly, I found that this specific problem was asked and answered on Mathematics Stack Exchange.  The general formula is

prob(k heads in n tosses) = C(n,k) * p k * (1-p) n-k

where C(n,k) is the number of combinations of n things taken k at a time (explained here) and p is the probability of a single event, which in our case is 1/2.  For n = 10 and k = 3, the answer is 15/128, or a little better than one chance in nine that I would be dealt three of the ten four-point tiles.*

So the overall likelihood of Micki’s lopsided draw is the product of the two probabilities we calculated, that is, 1/128 * 15/128 = 15/16384.  Expressed as odds, this would be 1091 to 1, or about the same likelihood as getting bumped from an airline flight due to overbooking.  If one looks at it this way, yes, I was a bit unlucky (or Micki lucked out, take your pick).

The big difference between one-sided Scrabble games and overbooked airline flights is that my bad luck did not result in my getting beaten up and hauled out of my recliner, kicking and screaming — instead, I wrote this post.  So there.  I’m vindicated.  But I still lost.

______________

* Some readers with penchant for details may be asking, wait, what about the rest of the tiles?  How does the dealer make the piles come out even, when one player has been dealt most of the special tiles?  Easy.  Take the remaining tiles, shuffle them and deal them out until one player has 50 tiles, then deal whatever is left to the other player.  For good measure, make sure the unlucky player gets most of the I’s and U’s.
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Asked and Answered 3.3

This is the third and final article in my series about hanging picture frames.  The first post, Why Frames Tilt Forward, discusses why frames tilt at the top and what you should and should not do about it.   My next post, The “Hang It with Two Hooks” Calculator, presents an online calculator to help you hang pictures with less forward tilt, using two wall hooks and 45° wire angles.  This post completes the picture, so to speak.  Here I try to illustrate, with intuitive examples, the role of physics in picture hanging.  Most of all, I want to help you understand why it is a bad idea to string a wire tightly across a frame to keep it from tilting forward.

2hookicon2I will also discuss the picture-hanging hardware I like and why.  Finally, I will provide another online calculator — this one evaluates your frame’s margin of safety by estimating the tension in the wire and the tendency of your frame to bend.  Click the icon at right to go directly to my safety-factor calculator.

Physics-minded readers, relax.  This article is for general readership.  So I am not going to distinguish between mass and weight, I am ignoring the gravitational constant, and I will use pounds, not newtons, as the unit of force, because that is how the people around here hang pictures.

The Graphical Physical Tour

FIGURE 1

FIGURE 1

Everyone else can relax too.  I’m going to walk us through some basic physics that my wife and I learned in high school, before we started dating.  I’m sure she remembers all of this.

Let’s start with something simple.  In Figure 1 (click to zoom) we see a 1o-pound weight hanging from the ceiling via a wire.  The weight is at rest, neither rising or falling — this means that the upward force (or tension) in the wire must be exactly equal to the downward force w of the weight.  Hanging a 10-pound weight on a single wire produces 10 pounds of tension in the wire.

FIGURE 2

Onto our next example.  In Figure 2 (at right), the weight is the same as before (1o pounds) but it is now hanging by two identical wires instead of one.  Once again, the downward force w of the weight is balanced by the upward pull of the wires.  Because there are two wires, each individual wire carries just half the load.  So the tension in each wire is now w/2 (or 5 pounds in this case).

Okay, time to use your intuition.  If we were to weld together the two wires in Figure 2 at the top, would this change the tension?  No — the tension in each section of the wire would still be w/2.  Ponder this until you’re comfortable with the idea.

Now that (in our minds) the two wires are connected at the top, let us take one more step:  slice the weight down the middle, so that each end of the wire supports half the original weight.  This action should also have no effect on the tension in the wire.  Agree?

FIGURE 3

FIGURE 3

We are now prepared to consider this model of a frame suspended from one hook (Figure 3).  The total weight w is the same, but it is divided into two equal weights supported on either end of a single continuous wire.

The wire passes over pulleys at the top and sides. The top pulley represents the wall hook; the side pulleys are the D-rings attached to the frame.

The weights create forces that pull downward at the top and inward at the sides.  We will take a closer look at this in the diagram below (Figure 4) which focuses on the left side of the setup (the right side is a mirror image).  It may be helpful if you click on the figure to view it full size.

FIGURE 4

Once again, we have a system at rest: the weights are not rising or falling and the pulleys are not moving.  This means that the downward forces are balanced by equal and opposite upward forces — and the same is true for the horizontal forces.

Let’s zoom in on the force on the left side.  The weight w/2 exerts a downward force at the pulley, which must be offset by an equal force upward.  But the wire does not extend upward — instead it heads away from the pulley on a diagonal.  How can a diagonal wire produce an upward force?

It helps to imagine that the pull of the wire is composed of vertical and horizontal parts that add up, so to speak, to a total force (tension) in the diagonal direction.  In the figure, I denote the vertical and horizontal parts as Ty and Tx, respectively.  Because there is no net motion in the vertical direction, we know that Ty (the upward force) must equal w/2 (the downward force).

How do we find Tx, the force in the horizontal direction, and T, the tension in the wire?  Here, we have to use some trigonometry.  The wire tension T equals Ty times the cosecant (csc) of the wire angle α, and the horizontal force Tx equals Ty times the cotangent (cot) of the wire angle α.  If you did not take trigonometry in school, please accept this on trust.

Sorry for the math, but I wanted to show how the wire angle has a multiplier effect on the tension T.  The smaller the wire angle (that is, the closer to horizontal the wire is strung), the greater the multiplier.

I have listed the multipliers for various wire angles in Figure 4.  The first column of the table is the wire angle, the second column is the tension multiplier, and the third column is the horizontal force multiplier.  These multipliers apply to Ty (which is w/2 in this case).

From the table, we see that if the wire is strung only 5° from horizontal, then the tension in the wire will be more than eleven times w/2.  For our 1o-pound frame, the wire tension would be 57.3 pounds and the inward pull on each side of the frame would be 57.1 pounds!

But if we were to string the wire 45° from horizontal, the wire tension would be 7.1 pounds and the force pulling in on the side would only be 5 pounds.  This shows why one should not string a wire tightly across a picture frame to reduce its forward tilt.

I like the idea of using two wall hooks and 45° wire angles, as discussed in my other posts, because it reduces both the wire tension and the inward pull on the frame.  But this does not mean that horizontal forces go away.  In any two-hook installation, there will be a net horizontal force on each hook, pulling them toward the center of the frame.

Figure 5: Forces on Wall Hook (Two-Hook Installation)

FIGURE 5

The diagram at left (Figure 5) depicts one wall hook in a two-hook setup.  The left end of the wire extends diagonally down to the frame, and the right end leads to the other hook.  The wire tension T is the same everywhere along the wire.

In the figure, the black arrows represent the forces that the wire exerts on the hook.  The net force on the hook Tz (red arrow) results from adding together the horizontal and vertical components of these forces.

Again, using some trigonometry, we find that the force Tz will always be somewhat higher than Ty (the exact formula is shown in Figure 5).  The direction of this force relative to vertical is one-half the wire angle.  In the case of our preferred 45° wire angle, the overall force on each hook would be 0.54 times the frame weight and the force would be directed 22.5° inward from vertical.   The horizontal component of this force would be 0.21 times the frame weight.  If one were to use a steeper wire angle — say, 60° from horizontal, as  some people suggest — it would increase the lateral force on each hook by almost 40%.

That’s it for the hard-core physics.  Let’s talk about what this means for picture framing.

Two-Hook Hardware

I do not intend to review all the various hardware available for hanging pictures.  Instead, I am going to focus on the parts and methods for a two-hook, low-forward-tilt installation. So the parts of interest here will be D-rings, wall hooks and wire.

FIGURE 6Let’s start with the hardware you use to attach the wire to the frame.  I much prefer D-rings (far left) because they lay flat and lead to less forward-tilt than eye-screws (right).  Also, D-rings are fastened to the frame with #6 or #8 screws which are larger and have deeper threads than the eye-screws that amateur framers often use.  This offers more resistance to sideways forces.

Next, the wall hook.  As I mentioned just a minute ago, each hook in a two-hook setup is subject to a lateral force.  When using 45° wire angles, the horizontal force on each hook will be about 20% of the weight of the frame.  But wall hooks are designed for vertical loads, not horizontal ones.  The wide base of the two-nail hook (Figure 5) offers extra stability in this situation.  I have not tested different brands but New England carpenter Doug Mahoney did, and Doug recommends the Floreat hangers sold by Ziabicki Imports.  I suggest you read his article on picture hangers – very thorough.

Finally, the wire.  I am always amazed by the types of wire I see on frames, old and new. Incredibly, I have seen framers re-use the wire from the customer’s old frame, even when the old wire was corroded and kinked.  I have also seen them use the thin consumer-grade wire that you find in drugstores and supermarkets.  Why do reputable people cut corners on a commodity like wire after so much money was put into the rest of the frame?

The strength of wire depends mostly on its thickness, somewhat on its construction.  It is difficult to find technical data (as opposed to marketing claims) on the breaking strength of picture-hanging wire.  I wrote to Wire & Cable Specialties, the Pennsylvania-based manufacturer of the Super Softstrand vinyl-coated stainless steel wire that I like to use — they replied that the breaking strength for this wire was about 2.5 times the so-called “maximum picture weight” that is printed on the spool.  The following chart summarizes the breaking strength vs. Super Softstrand wire size:

Wire Size “Maximum Weight” Breaking Strength
#2 15 lb 37 lb
#3 20 lb 50 lb
#4 25 lb 62 lb
#5 43 lb 107 lb
#6 60 lb 150 lb

“Wire size” for picture-hanging wire is a vague term that has less to do with its diameter than its weight rating.  I have one spool each of the #4 and #5 Super Softstrand.  I almost always use the #5 wire unless I’m hanging something very small and light.  The #5 is a seven-stranded wire that measures about 0.040 inches diameter (equivalent to 18 gauge) without the vinyl coating, and about 0.060 inches including the coating.  In my opinion, the #5 wire is as easy to thread and knot as any other size.

Unless you frame thousands of pictures, you will not save much money using thinner wire: this online framing supply store will sell you 500 feet of #5 or 1125 feet of #3 wire for only $15.95.  If the average frame needs 3 feet of wire and you frame 300 pictures a year, you would spend $29 a year on #5 wire, compared to $13 on #3 wire.  This works out to 5 cents a frame.  Is it really worth 5 cents to use a lower-rated wire?

The Picture Frame Safety Factor Calculator

At last, the calculator.  This calculator lets you estimate the tension in the wire and the inward deflection of the sides of the frame, based on your dimensions and wiring setup.  This necessarily involves a number of assumptions, which I will discuss after presenting the calculator.

To evaluate the safety factors in your frame, you will need to enter the dimensions shown in the figure below:

Picture Frame Safety Factor Dimensions

First, indicate whether you have one wall hook or two.  (Before doing this, you might want to consult The “Hang It with Two Hooks” Calculator for my two-hook recommendation.)  Next, select whether you will enter the weight of your frame or let the calculator estimate the weight from its construction.

Now enter the frame dimensions, starting with the overall width and height (W and H) and the total length of wire (L).  If you are using D-rings, enter the length (D) from the hole to the tip.  But if your wire is attached directly to the frame, enter zero for that value.

Next, enter the distance (B) between the D-ring fastening screws (or wire fastening points if there are no D-rings).   If you indicated you are using two wall hooks, you will be asked to enter the distance (X) between the hooks.

Finally, enter the dimensions of the frame molding and the breaking strength of the wire.  It is possible you may not know these values, so here is some guidance:

For the cross-section of the molding, enter the face width of the molding (F) and the average thickness of the molding (T).  Frame moldings can have complicated profiles, so do your best to estimate average thickness.  The more curves in the molding profile, the greater uncertainty there will be in the estimated deflection.

For the breaking strength of the wire, enter the value if you know it; otherwise enter 2.5 times the rated weight.  If you don’t know that, make a conservative guess such as 50 lbs. or less.  Corroded or kinked wire is likely to have a lower breaking strength than new wire — any wire is only as strong as the weakest point along its length.

When you are finished, click CALCULATE to validate your entries and show the results. The calculator will estimate the tension in the wire and tell you what percentage of the breaking strength this represents.  (With wire and cable, it is common to use a 5x safety factor, which implies the tension should be no more than 20% of its breaking strength.)  The calculator will also estimate the inward deflection of the sides of your frame.  I suggest that if the deflection is more than one-third the typical clearance (1/16th-inch all around) between the frame and its contents, then you are in danger of damaging the artwork and/or glass.  Do not count on the glass to reinforce a frame: it is the job of the frame to support the art and the glass.

As promised, here is a list of my assumptions:

  1. The estimated weight (if selected) assumes 2.5 mm soda-lime glass (if selected) with 2.5 specific gravity, wood frame molding with 0.4 specific gravity, and other materials at 0.002 lb / in².
  2. Elongation of the wire due to tension in the wire is ignored.
  3. The force pulling inward on the side of the frame is assumed to be concentrated at a point one-third of the way down from the top of the frame.  The corners of the frame are assumed to be stationary.
  4. The calculator does not evaluate the integrity of the miter joints or the fasteners.
  5. The side of the frame is assumed to bend as if it had a rectangular cross-section.
  6. The amount of bend in the frame is inversely proportional to the elastic modulus of the wood.  The elastic modulus is assumed to be 1,500,000 lbf / in², which is a mid-range value for typical framing woods (see reference).

If the calculator warns you about tension or excessive bending of your frame, I suggest you buy some #5 vinyl-coated wire and consult The “Hang It with Two Hooks” Calculator to find a more frame-friendly wiring method.  Also, be aware that the taller the frame or the narrower the molding, the more that its sides will bend inward for a given tension.

And now I must add my usual disclaimer.  This calculator makes it easy for you to estimate the safety factors in your framing situation — but because of the assumptions involved, the results should only be treated as estimates.   The calculator may indicate a problem where there is none, or it may fail to warn you that a problem exists.  I offer this calculator as a convenience but I assume no liability for damage of any kind, even if my suggestions are followed exactly.  You bear full responsibility for choosing to use this information.

That concludes my three-part series on framing with wall hooks and wires.  I believe this is one of the most exhaustive (hopefully not exhausting) treatments of this topic that you will find on the internet.  I have tried to make it as accessible as possible.  Please let me know if you find the calculator useful, or if you have problems or discover bugs while using it.

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Asked and Answered 3.2

Click to Use CalculatorTwo years ago, I posted an article here called Why Frames Tilt Forward, explaining why the top of a picture frame tilts away from the wall and what one should and should not do to address this.

The mistake that most people make (and a remedy that even some frame shops prescribe) is to fasten the wire tightly across the frame so that there is almost no slack in it.  But as I pointed out in my article, this can put considerable strain on the wire and the frame when the picture is hung.

If you don’t believe that a professional framer would make such a mistake, let me share a little story.  We recently took a favorite picture of ours to a local frame shop for reframing so that it would better suit our décor.  It was a large frame, about 46″ wide and 34″ high.  When we picked up our frame, I noticed how taut the wire was, and I mentioned this fact to the owner.  I also told her about my blog post on this topic.  She seemed disinterested.

Rather than argue with her, I decided I would rewire it when I got home.  As I was carrying the frame into our bedroom by its wire (the only practical way to carry such a large piece), the wire snapped and the corner of the frame hit the floor.  Luckily, I was carrying it only eight inches or so above the floor and, luckily, the floor was carpeted; otherwise, the frame or the floor or both would have been damaged.   Rest assured, I was not carrying the frame in such a way that it bounced around and tested the wire.

I rewired the frame with my own hardware, my own wire and according to my own rules, and I am confident that it will now not fall, bend or break.

On our next visit to the framer, I told this story to the owner, who remained unconcerned. Her response was that a frame should not be carried by its wire.  Apparently, she was not  familiar with the concept of a safety factor.  I thought her excuse was as flimsy as her wire.

Clearly, this frame shop is winging it.  They are content to string a wire across the frame and call it a day.  They are not mindful of physics — and they will not have a second chance to demonstrate this to me.

So… how do you know whether your frame shop is using their heads?

• • • •

But enough of cautionary tales.  In Why Frames Tilt Forward, I suggested that one way to achieve a low-tilt and low-tension installation is to use two wall hooks and 45° wire angles  (see figure below).  My instructions, however, were not so easy to follow — I know, because I tried to follow them myself.  The main difficulty was figuring out exactly how much wire to cut, something my instructions had not spelled out.Two Hook Frame Hanging Diagram

This is an ideal application for an online calculator — no fancy math, just basic geometry.  So, to help my readers, I have programmed my formulas into the wiring calculator below.  The user enters the outside dimensions of the frame (W and H), the size and position of the D-rings fastened to the frame (D and V), and the desired distance from the top of the frame to the bottom of the wall hook (Z).  The calculator returns the vertical position of the D-rings (Y), the spacing of the wall hooks (X), and the length of wire to cut, which includes three inches at each end for making knots.  To make things easier, some default values are suggested and results are rounded to the eighth-of-an-inch.

Notice that my illustration suggests the use of D-rings as well as double-nail wall hooks.  D-rings are preferable to screw-eyes because D-rings lay flat against the back of the frame, reducing the propensity for the frame to tilt forward.   And the wide base of double-nail wall hooks can help distribute the lateral forces associated with a two-hook installation.  More on this in the final post of this series, The Physics of Hanging Pictures.

In my low-tilt, low-tension scheme, the D-rings are placed one-fifth of the frame height below the wall hook.  The wall hook spacing and the wire length follow directly from this.  However, if the calculated wall hook spacing is less than one-third of the D-ring spacing, the calculator suggests values for a one-hook installation instead.  In the one-hook case, the wire angle varies with frame height, but the angle will be at least 33° above horizontal.

Click CALCULATE after editing your entries to view the installation instructions.

In a two-hook installation, there may be less slack in the wire than you expect.  To avoid frustration, try this: center the frame over the left hook and engage the wire into the hook, then shift the frame all the way to the right and engage the wire into the right hook.

I must end with a disclaimer.  This calculator makes it easy for a person to hang a picture with low forward tilt by using two hooks and 45° wire angles.  But whether this method is suitable in your situation is a judgment only you can make.  You bear full responsibility  for your installation.  I provide this calculator as a convenience but I assume no liability for damage of any kind, even if the suggestions offered in this post are followed exactly.

With that out of the way, happy hanging.  Returning readers may click the calculator icon at the top of this post to go directly to the calculator app.  Your suggestions and feedback are always welcome.

[Update 09-22-2017:  In the calculator, I increased the length of the wire to knot and twist from 2 inches per side to 3 inches per side.  There was no need to be so skimpy.]

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