Category Archives: Asked & Answered

Asked and Answered 5.0

I recently became a bitters-and-club-soda drinker.  As I was to discover, club soda has the annoying habit of going flat before one finishes the bottle.  To reduce waste, one can buy small bottles of national brands but they are much more expensive than the one-liter or two-liter bottles of store brands.  What’s a club soda drinker to do?  Glad you asked, and here’s your answer.

If club soda did not go flat, the choice would be clear.  A six-pack of 10-ounce glass bottles of White Rock Club Soda costs $3.99 here — 33.25 cents per 5-ounce serving — while four one-liter bottles (135.3 ounces) of our store-brand club soda cost only $3.00 — 11.09 cents per serving.  This means club soda in the 10-ounce bottles is three times more expensive… at least before any drinks are poured.

I tried a few one-liter bottles and was surprised to see how much club soda went to waste.  (Soda without sparkle belongs in the drain, not the drink.)  So I sat down to figure out whether the one-liter variety was a bargain or a boondoggle.  The answer would involve Henry’s Law, the Ideal Gas Law, bottling industry data and a few simplifying assumptions. (Full details are provided in the Appendix.)

But let’s start with how club soda goes flat (Figure 1).  The icon at far left represents the bottle as you buy it — filled to the neck with liquid and then topped off with high-pressure carbon dioxide (CO2).  Then, you open the bottle, pour a serving, and re-close the bottle.  The headspace is now filled with air from your room.  Over time, as the bottle rests in your refrigerator, CO2 escapes from the soda and collects in the headspace, until the pressure reaches an equilibrium (as determined by Henry’s solubility constant for carbon dioxide).  This equilibrium CO2 pressure will be lower than the original pressure — as suggested by the lighter shades of blue in the diagram — due to the CO2 that escaped when you opened the bottle along with the CO2 that you poured into your drink.

FIGURE 1: How Club Soda Goes Flat

The club soda gets flatter each time you open the bottle and pour a serving, until hardly any gas is left.  At some point, you may as well water your plants with what remains.

The question is, for a given serving size and bottle size, how many servings can one pour before the contents become undrinkable?  In my case, a serving is 5 oz (148 ml), I drink one serving a day, and I consider club soda too flat to drink if it has less than half of the original fizz (your mileage may vary).  This is the case I will address first, comparing the 10-ounce bottles to the one-liter bottles.

Let’s assume we refrigerate the bottle at 4C (39F).  Once a day, we take it out of the fridge, open it, pour the 5-oz serving, then close the bottle and return it to the fridge, all done as quickly as possible so that the contents do not get a chance to warm up.  We repeat these steps every day until the bottle is empty or the soda is flat.  The following chart (Figure 2) shows the resulting carbonation level for each serving, for both sizes of bottles:

FIGURE 2: Carbonation Level vs Serving Number (One 5-Oz Serving per Day)

We see that the 10-ounce bottle delivers two acceptable servings — the carbonation level of the second serving (57%) satisfies my 50% specification.  But the real surprise is that the one-liter bottle yields just one more drinkable (61%) serving — the fourth serving has only 36% of the original fizz and the fifth 17%.

So, by my criterion, almost 56% of the one-liter bottle is undrinkable.  This increases the effective cost-per-serving of a one-liter bottle from 11 cents to 25 cents.  Even so, it is still less expensive than the 33 cents-per-serving cost of a 10-ounce bottle.

Pouring two servings at a time (1o ounces) makes the one-liter bottle an even better value, but not by much.  The first two servings would be full strength of course, and the next two would have 72% of the original fizz.  The final two servings, however, would drop to 34%. The effective cost-per-serving in this scenario would be 19 cents, and 41% of the contents would still be undrinkable.

To completely eliminate waste from a one-liter bottle, you would have to either (a) lower your fizz-level standards, or (b) belt down half the bottle (3-plus servings) each time that you open it.  Only then will it cost you 11 cents per serving.

The bottom line is, a one-liter bottle of club soda at $0.75 is a better buy than a six-pack of glass bottles at $3.99, even if you have to share half of it with your houseplants.  But if the one-liter bottle costs $1.00, it is a break-even proposition.

Keep the Fizz Alive

Want to get the most fizz for your club-soda buck?  Research by Nestlé [1] suggests that soda in one-liter plastic (PET) bottles loses about 10% of its carbonation every 60 days when stored at room temperature.  This is because CO2 gas diffuses through the walls of the bottle.  (Bottlers compensate for the loss in shelf-life by adding extra carbonation to PET bottles.)  There are three takeaways from this:

(1) Buy your club soda from a busy supermarket.  This increases your chance of drinking  recently-bottled higher-carbonation product.

(2) Only buy as many bottles as you will use this week.  For the same reason.

(3) Refrigerate your club soda.  This reduces the internal pressure — and thus the rate of CO2 loss due to diffusion — by about one-third compared to room-temperature storage.  Not only that, cold liquid dissolves more CO2 so less gas is lost when you open the bottle.

Follow this advice and you will earn an A for effervescence.  And don’t forget to recycle.


To estimate the carbonation in a bottle of club soda at each step of its product life, I made a number of reasonable simplifying assumptions:

  • The bottle is completely rigid.
  • The gases and liquids follow Henry’s Law and the Ideal Gas Law.
  • When the bottle is filled, the headspace gas is assumed to be pure CO2 .
  • The carbonic acid formed by the reaction of CO2 and water is insignificant.
  • The diffusion of CO2 through the walls of the PET bottle is ignored.
  • Evaporation of the soda is insignificant.
  • The liquid and gas are at equilibrium before the bottle is opened.
  • Bubbles that escape from the soda just before it is poured are ignored.
  • The contents of the bottle remain at 4C (39F) while the soda is poured.
  • When the soda is poured, the gas in the headspace is replaced entirely by air.
  • The amount of CO2 and water vapor in the air can be ignored.

We begin by writing down the initial conditions in each bottle (Figure 3).  I will use liters as the unit of volume, atmospheres as the unit of pressure, and moles for the mass of CO2. (One atmosphere is essentially sea-level pressure, and one mole of CO2 is about 44 grams.)

FIGURE 3: Initial Conditions

A few notes on these facts and figures.  I measured the total volume of each bottle in my kitchen by filling them to the brim with water.  I assumed that the actual amount of soda in each bottle was exactly what the label claimed (bottling equipment is very accurate).  The initial headspace is the difference between those numbers.

According to Steen and Ashurst [2], bottle-filling is usually performed at 14C (about 57F) and the carbonation pressure for club soda is about 60 psi, or just over 4 atmospheres [3].  Assuming that all the gas in the headspace is carbon dioxide, we can use the Ideal Gas Law and Henry’s constant to calculate the amount of CO2 in the gas and liquid as bottled.

Initial moles CO2 in headspace (n0) from the Ideal Gas Law [4]:

\text{(1)}\hspace{5em}n_0 = p_0V_0 / RT_0

where p0 is the CO2 pressure (4.083 atm), V0 is the initial headspace volume (0.010 liters or 0.85 liters depending on bottle size), R is the gas constant (0.08206 liter-atm/mol-deg) and T0 is the filling temperature in degrees Kelvin (14C + 273.15 ≈ 287K).

Initial concentration of CO2 in liquid (c0) from Henry’s Law [5]:

\text{(2)}\hspace{5em}c_0 = p_0H_{14\text{C}}

where H14c is Henry’s constant at 14C (refer to above table). 

We now know the total moles (m0) of CO2 in the bottle:

\text{(3)}\hspace{5em} m_0 = n_0 + c_0L_0

where L0 is the initial liquid volume (0.296 or 1.000 liters depending on bottle size).

Chilling the bottle to 4C does not change the moles of CO2 in the bottle but it does affect the CO2 liquid-to-gas ratio.  We find the new liquid concentration (c1) at 4C by combining the Ideal Gas Law and Henry’s Law, then substituting for pressure p1 and rearranging:

\text{(4)}\hspace{5em} n_1 = p_1V_1/RT_1 = m_1 - c_1L_1\;,

\text{(5)}\hspace{5em} p_1 = c_1/H_{4\text{C}}\;,\hspace{.75em}\text{and so...}

\text{(6)}\hspace{5em} c_1 = m_1H_{4\text{C}}/(L_1H_{4\text{C}}+V_1/RT_1)

where m1m0 L1 = L0 and V1 = V0 (since nothing has been removed from the bottle).  Knowing ci lets us back-calculate n1 and p1 from Equations (4) and (5).

We now pour the first serving of soda, whose volume is LS .  The pour reduces the liquid in the bottle to L2 = L1Land the total moles of CO2 to m2 = m1n1c1 LS , as we assume all CO2 in the headspace is lost.  The headspace volume is now V2 = V1 + LS .  We quickly close the bottle and return it to the refrigerator, so that T2T1 .  While the bottle rests, the CO2 concentration of the liquid gradually decreases from c1 to cand the CO2 gas pressure in the headspace rises to p2 .  The values of c2 and p2 are unknown, but we can solve for them using Equations (5) and (6) with new subscripts:

\text{(5')}\hspace{5em} p_2 = c_2/H_{4\text{C}}\;,\hspace{.75em}\text{and...}

\text{(6')}\hspace{5em} c_2 = m_2H_{4\text{C}}/(L_2H_{4\text{C}}+V_2/RT_2)

This is the procedure I used to generate the CO2 concentration results in this article.


[1] Profaizer, Mauro. “Shelf life of PET bottles estimated via a finite elements method simulation of carbon dioxide and oxygen permeability.” Italian Food and Beverage Technology, vol. 48, 2007.

[2] Steen, David P., and Ashurst, Philip R. (editors).  Carbonated soft drinks: formulation and manufacture.  Oxford Ames, Iowa: Blackwell Publishers, 2006.

[3] Spangenberg, Craig. “Exploding Bottles.”  Ohio State Law Journal, vol. 24, 1963, p. 513.

[4] Khan Academy ( is one of thousands of sites with info on the Ideal Gas Law.  The Ideal Gas Law (pV=nRT) is to chemistry what Newton’s Second Law (F=ma) is to physics.

[5] Choose your own reference: Henry’s Law on Wikipedia or Henry’s Sparkling Water on ice.

More in  Asked & Answered | Read 5 comments | Subscribe

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.
More in  Asked & Answered | Read 2 comments | Subscribe

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



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.


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?



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.


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)


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.

More in  Art and Photo, Asked & Answered | Read 11 comments | Subscribe