Category Archives: Asked & Answered

Asked and Answered 3.4

Hello, and here we are again.  I thought I was done with this series on hanging pictures, but it seems physics never dies — it just gets more complicated.

Some commenters on my previous articles (Why Frames Tilt Forward, The “Hang It with Two Hooks” Calculator, and The Physics of Hanging Pictures) asked how they could hang items on wall studs, if the studs are off-center from the desired hanging spot.  This seemed to be a rather specialized topic, and beyond the scope of my series, so I deferred until now.  But a recent commenter rekindled my interest and finally inspired me to take a look.

Before I proceed, however, I must mention that I’m not the first to address this problem.  The number-one result (as of now) I found in searches for “hang item on off-center studs” is this article on instructables.com by an author named MolecularD.  The author describes the principles involved and offers a set of equations (minus the math) that are meant to show the reader where to place the wall hooks.  Unfortunately, some readers commented that they did not get the desired result when they followed the author’s instructions.

The solution provided in the inscrutables.com article is such a complicated equation that there is no way for me to verify it without essentially solving the problem myself.  Which is what I will do now, taking a somewhat simpler, more intuitive approach.

The four consecutive views in Figure 1 demonstrate the concept:

FIGURE 1: THE CONCEPT
Concept of Hanging a Frame on Two Studs

View (A) depicts a frame hanging on a wall, centered at our desired position (dotted line), using a wire on a single hook.  Because of the symmetry of the system, there is no tendency for the frame to rotate one way or the other.  Ignore for now the fact that the wire extends above the top of the frame.

View (B) shows the studs in the wall behind the frame (we use a stud-finder to spot them).  The two studs are different distances from the center of the frame.  We drive a nail into the center of each stud, just touching the underside of the wire.  This does not cause the frame to rotate.

In View (C), we attach a piece of wire (blue) to the original wire, from the point where the first nail touches the wire to where the second nail touches the wire, without any slack.  The load is now shared between the central hook and the nails in the studs.  But this still does not cause the frame to rotate.

In View (D), we snip away the original wire where it touched the nails, leaving our new wire in place.  The nails in the studs now assume all the load, with the higher nail bearing more than the lower.  Still the frame does not rotate, so we have found the solution.

Obviously, I don’t expect readers to repeat these steps to hang their pictures — this was just a demonstration of concept.  Instead I will offer a calculator, with instructions for taking measurements, placing the hooks and cutting the wire, to help the reader achieve the final result.

That is, if you really insist on using studs.  Personally, I think it would be easier in most cases to forget about the studs and use the Hang-It-With-Two-Hooks calculator that I presented in my earlier article.  You would fasten the hooks to the wall with toggle bolts, which can hold a significant amount of weight when paired with the appropriate hooks.  (This video shows how to install them.)  But in the end, it’s your call.

The Setup

Oh, you’re still here!  This must mean that you really, really want to use two studs to hang your item.  Okay then, onto the intricate details.  Please consult Figure 2 (below) to get a sense of the important lengths and measures:

FIGURE 2: USING TWO STUDS TO HANG A PICTURE
Diagram of Frame Hung on Two Studs

Start by measuring the height H and the weight of the item you want to hang.  Then mark the spot 0n the wall corresponding to the top-center of the item.  All other measurements will refer to this point.

Next, use your stud-finder to measure XA, the distance from top-center to the center of the closest stud, and XB, the distance from top-center to the center of the next-closest stud.

Now inspect your hanging hardware.  You want to (ideally) hide all your hardware behind the item you are hanging, which means the higher hook (A) should not show.  Therefore, you should choose a value for ZA, the distance from the top of the frame to the bottom of Hook A, that is slightly greater than the length of the hook.

While you are it, measure the length (D) of the D-rings attached to the item.  If you plan to attach the wire directly to the item, then this length is zero.

Your next measurement is WD, the distance between the D-ring attachment points.  If you have not yet attached the D-rings to your item, then mark the spots where you think they should be attached, and measure the distance between those marks.

Note that I have not asked you to specify Y, the distance from the top of the frame to the D-ring attachment point, or ZB, the distance from the top of the frame to the bottom of Hook B, or S, the length of wire to cut.  These values will be returned by the calculator.

There is one last thing you may have noticed on the diagram: to make the item hang true, you need to install a guide hook below Hook A to equalize the slack in the wire — and the forward tilt — on the left and right sides.  More on this later.

The Math and The Calculator

I provide geometric and algebraic solutions in this attachment.  The result we are most interested in is:

ZB = ZA+ (XBXA) tan θ

where θ is the wire angle, tan θ = (Y ZA)/(WC XA) and WC = ½ WD.

The formula for ZB assumes that Y, the D-ring attachment point, is a given.  But I don’t ask you to specify Y directly, as this involves a judgment call.  Ideally, the ratio Y/H would be about 1/5 (the “one-fifth rule”) to minimize forward tilt of the item.  But this might call for too small a wire angle and create too much tension in the wire.  On the other hand, if the  wire angle is too large, and Y/H is greater than 1/3, then the forward tilt could be excessive.

So what I did in the calculator is ask you to specify the wire angle, with 30° as the default. (The minimum entry is 20° and the maximum is the angle corresponding to Y/H = 1/3.) The wire angle is used to back-calculate Y as described in the attachment.

If the default angle seems to provide a reasonable value for Y/H, then go ahead with it, assuming the wire tension is not too high.

If the calculator flags one of your entries as out-of-bounds, don’t ignore it.  The calculator will not report any results if Y/H is greater than 1/3, and it will warn you if the estimated wire tension exceeds 25 lbs.  (You are responsible for selecting the appropriate hardware.)

Results are reported to the nearest one-eighth-inch.   The calculator provides guidance on positioning the guide hook and attaching the wire to the D-rings.  The suggested length of wire S includes 6 extra inches (3 inches per side) for tying the loose ends to the D-rings.

Final Notes

It may be a challenge to hang your item on three hooks.  I suggest you find a helper, if only for you to have someone to complain to while attempting it.  (Still, watch your language.)  You might start by feeding the slack of the wire through the guide hook and onto Hook A.  Then slide the item toward Hook B and feed the wire over Hook B.

You ask, do I have to use the guide hook?  If your item weighs much of anything, then yes.  The farther that Hook B is from the center, the more the item will tilt forward at Hook A,  since there is more slack in the wire on that side.  And the more front-heavy the item, the more uneven the forward tilt will be.  The guide hook helps keep the wire close to the wall on the Hook A side.

I end with my usual disclaimer.  My calculator makes it easier for a person to hang an item on two off-center studs using hooks and wire.  But whether this method is suitable in your situation is a judgment only you can make.  You assume full responsibility for your project. I offer this calculator as a convenience, but I accept no liability for damage of any kind, even if the suggestions offered in this post are followed exactly.

If you’re not confident how things are going to work out, you can always do a mock-up in your garage before you mark up your walls.

With that out of the way, good luck.  I would be interested to hear about any successes, failures or problems.  As always, your suggestions and feedback are welcome.

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Asked and Answered: 13.0

Last year, I told you about Spelling Bee, a word game that appears daily in the New York Times.  The point of the game is to form as many words as possible, using any of the seven letters provided, any number of times, as long as the center letter is used at least once.  It is a fairly undemanding diversion which, along with my cup of coffee, helps get me going in the morning.  Plus, it gives me something to kvetch about to my friend Eric, who also plays.

Eric and I often compare notes and complaints about the Bee of the Day.  My usual beef is about the exotic foodstuffs (e.g., BOBA, CALLALOO, GHEE) included in the answer list, and Eric (former chemistry professor) and I both gripe about the chemical words, such as NICOTINIC and PROPANOL, that of course should be accepted by the Bee but are not.

Nonetheless, we were pleasantly surprised by a recent Bee in which BORON (Element 5) and CARBON (Element 6) were among the answers.  This got me wondering: what is the greatest number of element names one can generate from a set of seven different letters?

My first step toward an answer was to create a spreadsheet to count the number of times each letter of the alphabet appears in the list of element names.  Note: I decided to limit the number of elements in my list to the first 100, i.e., from HYDROGEN to FERMIUM.  Elements 101 and beyond — all man-made — are unfamiliar even to Eric and me.  One of those is darmstadtium (Element 110) of which only a few atoms have ever been produced. So it’s not like I’m disrespecting Nature by excluding darmstadtium and its ilk.

Anyway, back to my spreadsheet.  I found that the consonants appearing most often in my 100-element list were M (50), N (36), R (33), L (22) and T (22).  And as you might guess, the most common vowels were I (56) and U (50).  After spending about 15 minutes playing around with frequently-appearing letters, I was able to find two different seven-letter sets which “contain” the names of four elements.  (Before I reveal, would you like to try?)

The first set I found was EIO/BMNR. (I’ll refer to these sets by their vowels/consonants). This set spells the elements BORON, BROMINE, IRON and NEON.  And my second set was AEIO/DNR, which spells IODINE, IRON, NEON and RADON.  Interesting, but…

I was, of course, not satisfied.  Humankind needed to know: are there other seven-letter sets that spell out four element names?  And more importantly, are there seven-letter sets that spell out five (or more) element names?  I did not yet have these important answers.

So, for humankind’s sake, I was obliged to resort to brute-force computation, employing the only modern programming language I know — PHP.  I am familiar with PHP because it is the language used by WordPress, the platform for this and millions of other blogs.  And though I have a PHP reference manual, most times when I want to write code for a new task, I just do an internet search — 99 percent of the time someone has already done the thing that I want to do and has provided functions and/or code for it.

And that is (mostly) how I wrote a PHP program to print out all the seven-letter sets that spell out four or more element names.  I’ll spare you the details, but suffice it to say that before now I had no idea there were PHP functions like count_chars (finds the number of unique letters in a word), array_intersect (lists the items that two sets have in common) and implode (combines a set of individual letters into a single word).  Those functions served me well here, but they (like many other PHP functions) are so special-purpose that I can’t imagine any programmer having good command of them all.

In any case, I ultimately wrote a program that evaluated all 213,333 of the seven-letter sets containing one or more vowels and one or more consonants found in the element list. Without further delay, here are the results.

SEVEN-LETTER SETS WHICH CONTAIN FOUR ELEMENT NAMES:

AEIO/DNR IODINE
NEON
IRON
RADON
AIU/BDMR BARIUM
RADIUM
IRIDIUM
RUBIDIUM
AIU/CDMR CADMIUM
IRIDIUM
CURIUM
RADIUM
AIU/DMNR INDIUM
RADIUM
IRIDIUM
URANIUM
AIU/LMNT ALUMINUM
TIN
TANTALUM
TITANIUM
EIO/BDNR BORON
IRON
IODINE
NEON
EIO/BMNR BORON
IRON
BROMINE
NEON
EIO/BNRT BORON
NEON
IRON
TIN
EIO/BNRX BORON
NEON
IRON
XENON
EIO/DNRT IODINE
NEON
IRON
TIN
EIO/DNRX IODINE
NEON
IRON
XENON
EIO/DNTX IODINE
TIN
NEON
XENON
EIO/GNRT IRON
NITROGEN
NEON
TIN
EIO/NRTX IRON
TIN
NEON
XENON
EIU/HLMT HELIUM
LUTETIUM
LITHIUM
THULIUM

SEVEN-LETTER SETS WHICH CONTAIN FIVE ELEMENT NAMES:

None.  Zero.  Not-a-single-one-ium.

So there you have it.  There are 15 different seven-letter sets which can be arranged to spell four element names, but there are no seven-letter sets that will spell five element names, at least not with respect to the first one hundred elements.

If some nerd ever uses this edition of Asked and Answered to win a bar bet, I will expect due credit, if not a beer.

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Map Challenge II

Asked and Answered 12

As a followup to last year’s Map Challenge post, I thought I would pose a few geo-nerd problems inspired by the Four Corners in the southwestern U.S.  As most of you know, there is only one spot in the U.S. — about 12 miles north of the Sinclair gas station in Beclabito, NM — where four states meet at a single point.*  But this begs the question: what about the near-misses?  Shouldn’t those places get a little bit of nerd-love too?

I will call these areas Four-State Circles — circles that include at least some part of four different states, other than the aforementioned Four Corners.  Here are my challenges:

  1.  What and where in the U.S. is the smallest Four-State Circle (and its runner-up)?
  2.  Similarly, what about the smallest Five-State Circle (and its runner-up)?
  3.  Finally, the smallest Six-State Circle (and its runner-up)?

I am sure you’ll all agree that we can safely stop at six states without feeling like we have shortchanged the topic.  So let’s go exploring — and here’s a map to help.

Four-State Circles

When scanning the map for potential four-state solutions, the key is to focus on regions where two nearby states don’t share a border.  This suggests a closer look at places like the Oklahoma panhandle and the contorted path of the Mississippi River south of Illinois.

But it turns out that both the winner and the runner-up for the smallest Four-State Circle lie east of the Appalachians.  Surprisingly, neither circle involves Rhode Island, and both include my home state of Pennsylvania.

THE RUNNER-UP

Zoom to 39.594ɴ, 78.350ᴡ, 100 feet from Pepper Lane in Doe Gully, West Virginia.  Next draw a circle of 8.83 miles radius around that point.  This circle touches four states: from north to south, Pennsylvania, Maryland, West Virginia and Virginia.  This may be the runner-up Four-State Circle, but I believe it is the shortest path that crosses four states, a subtly different problem.

THE WINNER

Now shift your focus 200 miles eastward to 39.676ɴ, 75.666 in Delaware, across the street from an industrial park in Christiana.  A 6.55-mile radius circle around this point is mostly Delaware, with slivers of New Jersey, Maryland and Pennsylvania.  This would be a conflicted place to be a sports fan: 7 miles from three different states with major league sports teams, but not one to call your own.

Five-State Circles

The Five-State Circles were more painstaking to sort through.  It’s not like one can just enlarge a Four-State Circle a few miles in order to pick up a fifth state.  And what’s worse, all of the obvious candidates were so similar in size that some very careful plotting was needed to find the winner.  Here are the results:

THE RUNNER-UP

Take U.S. 60 toward Grayridge, Missouri, then head a couple of miles south, and you will find — besides a lot of farmland — the center of the runner-up Five-State Circle.  This 28.8-mile-radius circle takes in parts of Kentucky, Tennessee and Missouri, plus razor-thin slices of Arkansas and Illinois.  S0, Grayridge, population 122, you are now on the map.  Welcome to internet fame.

THE WINNER

Here is where the Oklahoma panhandle gets its chance to shine.  A 27.1-mile radius circle centered on 36.892ɴ, 102.514, just north of Boise City, touches parts of five states: New Mexico, Colorado, Kansas, Oklahoma and Texas.  Although the center of the circle is now just a pile of rubble next to U.S. 385, just wait until some enterprising Okie gets wind of this and builds a McDonalds there.

Six-State Circles

There are two attractive candidates here — the area around Baltimore, MD, and the area around Springfield, MA.  Only one of these, the smallest Six-State Circle, can be crowned the winner — the other will be the runner-up who, a la Miss USA, will assume the title if for any reason the winner cannot fulfill its duties.  So, may I have the envelope please…

THE RUNNER-UP

Cold Bottom Road crosses Piney Creek and I-83 at 39.563ɴ, 76.665, just north of the National Lacrosse Hall of Fame in Sparks, Maryland.  The six states within 58.9 miles of this spot are: Maryland, Pennsylvania, Delaware, New Jersey, West Virginia and Virginia.  If Washington, D.C., were a state, this would be the winning Seven-State Circle but sorry — rules are rules.

THE WINNER

The center of the smallest Six-State Circle is American Legion Post 452 in Chicopee, MA, at 42.1456ɴ, 72.6140.  (The post meets on the fourth Thursday of each month at 17:30, so snap to it.)  This 43.2-mile radius circle covers parts of Massachusetts, Connecticut and Vermont plus thin slices of New York, New Hampshire and Rhode Island.  And yes, Rhode Island is a state, not a plantation.

So there you have it, asked and — exhaustively — answered.  But for those who need even more bar-bet material, here’s one last bit of U.S. state-border trivia.  The number of places where three (or four) states meet is (a) 55, (b) 68 or (c) 92.  Answer below.

_______________

* The four states which meet at one spot are Arizona, Utah, Colorado and New Mexico.  And by my count, there are 55 places in the U.S. where three or four states meet at a point.
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