Wednesday, February 23, 2011

Low-cost Raised Beds

When we first bought our house, we were pretty strapped for cash, but I still wanted to put in a couple of raised garden beds and get some vegetables going right away.  I was building a fence at the time, and had some cedar fence pickets lying around that were too ugly to use on the fence.  So I tacked some together to make some raised beds.  Six years later, they're going strong, so I'm sharing this idea.  I wouldn't bother building anything more sturdy and expensive unless they needed to be much deeper for a really wet site.

For each bed you will need:

Three fence pickets.  You want 1x6 square top pickets, 6 ft long.  Don't mess with any nasty treated wood or try to do with plain pine or hemlock boards.  You want red cedar, and the ones with the darker red and brown colors are the most rot-resistant.  They should cost anywhere from $1.25 to $2.75 each, depending on the grade and where you get them.

Two 1 ft stakes, so get a couple of feet of cedar 1x2 material for each bed you plan to make.

Eight #8 x 2" and two #8 x 1-1/2" screws, flat head if you have a countersink, otherwise pan-head.  Get them in stainless if you like, but any decent galvanized or 'gold' coated will be fine.

Cut one of the boards into two equal lengths, about 3 ft.  These are the ends.  The uncut long boards are the sides.  Important:  Cedar tends to split, so predrill and countersink the sides, about 1" in from each end.  Screw the sides to the ends.

Put the screws an inch or so in from the end of the board to reduce splitting
Sit the frame on the ground where you want the bed to be, and drive a 1 ft. stake at the midpoint of each side.  If the soil is wet and soft, this should be easy and you probably won't even need to sharpen the stake.  Put a screw through each stake into the side of the bed frame, about an inch down from the top.  The stakes will keep the relatively flimsy boards from blowing out under the weight of the soil inside the bed.
Reinforcing stake at the middle of each long side

Now fill the bed with a nice mixture of compost and soil, and stir in any bone meal, lime, or other amendments you might need, and get gardening.

Two of my beds.  Herbs in the back, berries in the front (and a garlic escapee)

Thursday, February 3, 2011

Mission Finish

I looked at a bunch of recipes for a "Mission" finish for the TV stand project I've been working on.  I like to finish some parts before assembly, so I don't have to worry about glue squeeze out, and also because it's usually a lot easier to apply finish to the individual parts instead of trying to work it into all the corners of assembled furniture.

The first finish I tried was a gel stain.  General Finishes "Java" was recommended in some Popular Woodworking book or article.  It looked like a nice, easy, almost single-step finish, so I picked up a can at a local Woodcraft store, and tried it on a sample board.  It was too dark for my tastes, and more importantly, it colored the wood VERY differently depending on how the surface was finished.  Surfaces smoothed with a hand plane were darker and the stain absorbed differently than sanded surfaces. I'm sure I'll have some of each. Yikes.

Another Popular Woodworking finish recipe I found called for a layer of oil stain (Olympic "Special Walnut"), followed by a coat of Watco "Dark Walnut" followed by amber shellac.  As I am lazy, I skipped the first stain, and used Watco "Black Walnut."  I think it looks pretty good.  Watco is super easy to work with, and gives a little more protection than stain alone since it contains some varnish resin.  The color went on the same on sanded and planed surfaces, and it gives just the right amount of contrast to the ray flake in quartersawn oak for my tastes.  Here are some parts from the end panel assemblies.

I'm not sure what clear protective finish I'll apply on top of the Watco.  I've never used shellac before, so that's a little scary on such a big prominent piece of furniture.  Polyurethane varnish has always been my topcoat of choice, and it's tough, but also not very repairable.

Wednesday, February 2, 2011

Lutherie 101

Take a piece of string or wire and clamp it between a couple of points.  Pull it somewhat tight.  Now, put some energy into it, by hitting it or plucking it, or blowing on it.  It will naturally vibrate at a basic frequency determined by the length, tension, and diameter.  (that third factor is technically mass per-unit-length, but if the wires are the same material it comes down to diameter)  Shorter = higher frequency, tighter = higher frequency, and smaller diameter = higher frequency.  Then you make a way to clamp the wire at various points so you can change the length on the fly, and now each wire can make a series of frequencies depending on where it is clamped.  Get a few more pieces of wire, of different diameters, and tension them next to each other and you can make an arbitrarily wide range of frequencies. That's it!

Considering string instrument construction, there's a little terminology to start with. The two fixed ends of the wire are the nut (at the top) and the bridge (at the bottom). Strings are usually anchored at fixed points below the bridge, and wound around adjustable posts above the nut so that the tension on each string can be adjusted for tuning. The in-between clamping points are commonly little metal bars called frets, and the strings are pinched just above each fret. The open frequency of a string is defined by the nut-bridge length, string diameter, and tension, and the subdivision of each string's frequency is determined by the spacing and location of the frets.

Now for the relationship between frequencies, notes, and fret locations. Modern western music is dominated by a tuning system of 12 evenly-spaced notes per octave, based on a reference frequency of 440 Hz. Octaves are defined as a doubling of freqency, so an A note in the 4th octave (A4 = 440 Hz) is twice the frequency as an A in the 3rd octave (A3 = 220 Hz) which is twice the frequency of A in the 2nd octave (A2 = 110 Hz). The spacing between each individual note fits into the logarithmic pattern, so that the difference between any two notes is a factor of 2(1/12) Hz. - the frequency doubles every 12 notes. The first note in each octave is C, for whatever reason, and the twelve notes are:

C, C#, D, D#, E, F, F#, G, G#, A, A#, B

This is referred to as the chromatic scale. The # symbol means "sharp", but all the sharp notes are exactly the same frequency as the next regular note's "flat." Thus D-sharp is the same note as E-flat, except for the two notes that don't have a sharp: B and E. This convention makes little sense to me, but it's just the tip of that iceberg. Each note corresponds to a key on a standard piano keyboard. Here's one octave of keys, with a B and C on either end of the next octaves shown for reference.

Each note also corresponds to one fret on a guitar, so if you have a string tuned to a D, each fret position will play D#, E, F, F#, G, etc. down the fretboard. Say that the nut-bridge distance is 26 inches, and that D is D2. If you put a fret at 13 inches from the bridge, it will play D3, the same note an octave higher, because it is 1/2 the length.  If you put another fret at six and a half inches, it will now play D5, two octaves higher than the open string, because it's now 1/4 the length. Few stringed instruments actually have a full two-octave range, but many are close.

Another option is a diatonic scale, where you have only seven notes per octave.  This is mostly on some old timey instruments like mountain dulcimers, accordions, and harmonicas. They are eight of the same exact notes as a chromatic scale, just missing some to leave just what is called a major scale.  That is the familiar do, re, mi, fa, sol, la, ti, (do) thing. If the string is tuned to a C, the notes are the same as just the white keys on a piano, so you can't play any of the sharps.  That works ok for many simple tunes, and traditionalists like it.

The physical distance between fret positions is related in the same way the notes are, by a factor of 2(1/12) which I'll refer to as a to simplify the upcoming formula. Given the total nut-bridge length as L, the distance to a given n-th chromatic fret is: L - (L / an).