## Introduction: Centering a Painting on a Wall With Off-Center Studs

An interesting challenge is posed to American families who want to decorate their new home by hanging a very nice but heavy painting on the wall. To our foreign readers: walls in modern American homes consist of a timber frame covered with sheets of so–called “drywall”. The latter is a ½-inch layer of gypsum wrapped in paper. As such, drywall is not strong enough to support heavy objects. Therefore, these objects must be anchored through drywall directly to the supporting vertical beams, made of wood, called “studs”. Studs can be detected with an electronic stud scanner.

For the reasons of visual aesthetics, it is often desirable to place a painting symmetrically; meaning the picture’s central axis should coincide with the central axis of the wall - see the above image.

Framed heavy paintings are sold with a strong hanging wire made of steel, whose ends are attached to the frame’s side beams. (If an upper part of the frame were used for hanging, then in time it would distort by picture's weight.) If a stud happens to be positioned exactly on the wall’s axis of symmetry, then one strong nail would solve our “picture problem”. Similarly, two studs positioned symmetrically on both sides of the central axis make the problem easy to solve. However, these ideal cases are rare since builders are constrained by the building code and most often studs are arranged non-symmetrically, as shown in the image above.

The non-symmetrical positioning of wall studs makes our picture problem difficult. Nevertheless, it is still solvable with just two cleverly positioned nails. If you are interested in the rationale of how it can be done, and maybe learn something, then proceed to Steps 1 and 2. If you are an impatient type and don't mind the "black box" answer, then proceed directly to Step 3.

## Step 1: How Is This Possible?

Consider the initial scenario illustrated by the first image. In there, the picture is hanging on a single nail N1. A1 and A2 in denote anchor points of the steel wire symbolized by thick black line. Nail N2 is only touching the wire and provides no support. Now, raise the nail N2 up as shown in the next image. This creates a moment of force that tries to rotate the picture counterclockwise. To compensate, a clockwise moment must be created. This is only possible when the N1 is to the left of picture's center of gravity. Which means the entire picture must shift to the right by {delta}x inches. In other words, shifting N2 up by {delta}y inches (within reasonable range) uniquely determines the corresponding horizontal displacement {delta}x. Therefore, an optimal arrangement of nails N1 and N2 may position the picture exactly in the center of the wall.

The obvious question is how to determine a proper {delta}y for a desired {delta}x? This is what this instructable is all about

## Step 2: The Underlying Principles

First, the principles of statics require that at a mechanical equilibrium the sum of all forces acting on our hanging picture must be zero (equation 1). The sum of all force moments must be zero, as well (equation 2). The latter requires *the same* center of rotation for the purpose of computing moments. However, at equilibrium one can choose any point as the center of rotation! To see that, suppose that one shifts the center of rotation by a vector **s** to another position. Consequently, all the rotation arm vectors, **r**_i, must be shifted by **s** - see equation 3. The equilibrium of moments still holds because the second sum disappears as a result of equation 1. (I must apologize to mechanical engineers at this point: I am sure this trivial result is taught as some named theorem; I just don’t happen to know it as such.)

In particular, the force vectors acting on our picture are quite simple. The gravity acts on the picture’s center of gravity with a down force m**g** (m=mass, **g**=Earth’s gravitational acceleration). It is countered by two wire tension forces, **T**1 and **T**2, acting on anchors A1 and A2, respectively - see the image above. The equilibrium of forces condition translates to a particularly simple equation 4. Choosing the center of rotation as a point exactly in between the anchors leads us to a simple expression for the equilibrium of moments - see equation 5. The first term there vanishes because **R**_g and m**g** are parallel.

You can easily verify that the above conditions are satisfied for a symmetric picture (in which **R**1 and **R**2 are equal in magnitude) only when the relations 6) hold where T_ix and T_iy {i=1,2} indicate vertical and horizontal components of tension vectors, respectively.

## Step 3: Final Result

The conditions derived in Step 2 lead to very nice geometrical constraints imposed on the hanging wire. In particular, the horizontal and vertical components of tension forces at both anchors must be equal in magnitude, respectively. This means that angles N1-A1-P1 and N2-A2-P2 in the first figure must be congruent! In turn, the two right triangles, A1N1P1 and A2N2P2, are similar. These congruence conditions are sufficient to solve our problem.

Instead of ambiguous “{delta}y” we will find a better defined {delta}h – the difference in vertical position of the nails, as shown in the first image. This is the sought unknown. The knowns are quantities easily measurable with a tape measure:

- W = wire length
- D = distance between anchors
- x = distance between studs
- {delta}x = desired horizontal shift

The unknowns are

- Partial wire lengths d1, d and d2
- Nail heights h1 and h2, as well as their difference {delta}h = h1 - h2,
- Partial lengths x1 and x2

The aforementioned congruence conditions, plus length constraints, provide the system of equations 7) sufficient to find {delta}h. But keep in mind that the {delta}x cannot exceed limit imposed by equation 8! Solution of

this system of equations is not difficult, but quite tedious. I will spare you the lengthy algebra by writing the final solution for {delta}h in equation 9, where all the partial quantities are explained in the collection 10.

I have attached Excel and OpenOffice Calc spreadsheets to ease the tedium of coding these formulas. In there, you will have to enter measured knowns, W, D, x as well as desired shift {delta}x.

For example, a painting in authors home, shown in the last image, has the anchor separation D=122.5 cm and wire W=126.5 cm long. Studs nearest the wall center are separated by x=70 cm. The picture had to be shifted horizontally by 4.8 cm for perfect centering. This was achieved by positioning one nail approximately 3.8 cm lower than the other nail.

Credits:

The unicorn image used in this tutorial, titled “Legend”, was created by a talented artist known only by his/her pseudonym “frenchromeo”. I have downloaded it many years ago from the www.deviantart.com website. The artist’s account is no longer there and relevant Google searches provide only dead links…

## 11 Discussions

7 months ago

Thanks for sharing. I pondered this a bit the last time I hung our 50+ lb. 36 in. tall x 46 in. wide mirror (five years ago) on annoyingly off-center-studs. The mirror had to be in the corner of the bedroom, centered over the matching dresser. Back then I started in on the math, but then just solved it empirically by marking the stud locations on the back of the mirror and then simulating wall anchor locations with two corners of a chunk of lumber that was slightly longer than the stud spacing.

Tonight I encountered this problem again with the same monstrous mirror in our new house. This time I Googled it and found your Instructable. Thanks for taking the time to document and share this. It finally got me motivated enough to do a bit more math.

I tried using your spreadsheet and got grossly incorrect answers. Not sure why that is. But since I was already in Excel I had enough momentum to try my own hand at it.

Anyway, after putting the kids to bed I realized that all the free body diagram equations are not necessary. All the forces acting on the picture frame must act though a common point or else there would be a net moment. Due to assumed symmetry of the picture frame and cable arrangement, both ends of the cable must make the same horizontal angle (theta).

I simply expressed the three cable segment lengths in terms of theta and the known geometry and then used Excel Goal Seek to find the angle that yielded the proper total length, along with h1 and h2 and delta h.

For verification purposes I then taped sheets of paper near the cable anchor locations and drew in reference angles with a protractor and straight edge. I also marked the stud locations on the back of the picture frame and proceeded to model cable behavior under the influence of the wall anchors until the cable theta angles were identical.

Turns out my calculated theta was off by a couple degrees, and my delta h was off by about 10%. Strangely, my h1 and h2 values were off by a lot. It’s such a simple calculation that it left me scratching my head.

I think part of the problem is that it’s difficult to measure the cable length accurately. And for cable lengths that are not much longer than the cable anchoring distance, the slightest change in length makes a big difference in the calculated h1 and h2 values.

So, I guess empirical is the way to go, at least for cables with small angles. Either that or span the off-center studs with inexpensive French cleat Z-hangers from Amazon.

9 months ago

I wish I had Goggled this yesterday. I spent more than an hour hanging a very heavy mirror at the end of the hall, where I had no flexibility to move the mirror to accommodate symmetrically spaced studs. I used trial and error, and I must have hung the mirror five or six times before I got it just right.

10 months ago

Thank you, exactly what I was looking for.

***UPDATE***

Sorry, I've tried it twice now and both times got answers from the sheet that were wrong. I commend you for putting all this work into it, but I think there is something that is being overlooked somewhere.

The first time I tried using the sheet, it was to hang a mirror in the hallway. After plugging in the data into the sheet, the sheet said that it wouldn't work. Yet I made it work by just hanging the mirror on the nail closest to the center line and then seeing where N2 would go by holding the string by hand. It's hanging perfectly now. I've attached a pic of it hanging as well as N1 (on the left) and N2 (on the right) marked with blue tape.

The second time I tried using the sheet, it was to hang a framed photo in the living room. The sheet told me to put N2 1.22 inches below N1. I've attached a picture of those results as well, along with where the nails are. In this pic, the blue dot is where the center line is.

I'm guessing it will be tempting to think that I may be a moron who can't use a measuring tape properly, but everything was measured three times by me and then measured again by my wife as a doublecheck. Believe me, the measurements were correct.

Again, thank you for putting the time into it. I wish I got better results.

1 year ago

Thanks for putting so much effort into sharing this.. much appreciated! Was looking for exactly this.

1 year ago

This instructable is actually hilarious

4 years ago

Fascinating but, why not put the main mountings in the regular way and use those adjustable, expanding toggle fastners in the areas without the studs?

Reply 4 years ago

Sure, there are many ways to skin a cat, according to one popular saying. I chose the simplest one. Fasteners - these I would have to buy form a

specialty store miles away. What could be simpler than just two nails?

Actually, my picture hangs on two

wood screws. But these are only marginally more complicated than nails :)

4 years ago

Finally a practical application for my math degree. Thanks for sharing.

Reply 4 years ago

I wonder what you do when your asked the time ?

Reply 4 years ago

I have been meaning to get a clock like this:

https://www.google.com/search?q=math+clock&rlz=1C1...

Reply 4 years ago

that's a great clock, I've wanted one myself. got to keep the mind working. ?