Calculating Radii for Armour Patterns, Hammer Faces etc.
by Scott Martin
Here is some math so you can calculate radii for curves (and don't have to draw approximate" curves by hand anymore).
I often find myself having to draw curved lines for armour patterns. Not a big deal right? But I usually find that these curved lines are because I'm trying to match two compound curves together (as in an articulation) or trying to figure out exactly how much metal to take off the edge of my hammer to give it an 8" radius curve! I realize that in period, the armourers often "fudged" these measurements, and as long as they are "close enough" it really doesn't matter. But if you are like me, and still trying to perfect your technique, it is often really frustrating to have two pieces of metal that just don't seem to want to go together... Is it my crappy technique, or is this pattern not "close enough" to work yet? I've wasted a lot of hours deciding that it must be the pattern, ground off 1/8" of an inch and then realized that I really needed to tap the metal a little over there. Now, since I don't have that extra metal, I either can't make it work (ever!) or I need to grind 1/8" off the entire remaining part of my curve, not a trivial chore in many cases.
So I reverted to my high school math: I knew I had to be able to calculate the raduis of a curve if I had enough other information on what I wanted the curve to look like. The first thing that I applied this formula to was articulating joint design, specifically laminations (or lames) and the joint itself. I knew how to calculate how big my joint should be from top to bottom (parallel to the bones of the limb in question) and I knew how wide it should be (parallel to the bend in the joint) but the angularity of the patterns I was producing was like nothing I had ever seen in pictures of medieval armour. To make a long story short(er) I wanted to have clean arcs, and they were never symmetrical when I drew them by hand.
So here's how to do it (using a circle, which is easy to draw ;) Figure out which two (of the following three) pieces you know the answer to.
1) Figure out how wide the "base" of your arc is: this will be the total width of a joint calculated a la mistress Hilary of Serendip, or the diameter of your hammer face. This measurement is (for the purposes of the calculation below) 2 y (y is 1/2 this value)
2) Figure out the "depth" that you need yur curve to be: this is the difference between the radius of the circle, and the distance from the center of the circle to the line that forms the "base" of your arc. (you know this if you are making a joint or similar pattern) This distance is "q" (why "q"? because it's a letter of the alphabet that doesn't get much air time ;)
3) Figure out the radius of the circle you need (you will know this if you are making a sinking [dishing] hammer). This value is "r" (that's what all the math texts call it: don't argue with success).
Now you can calculate the piece that you don't have!
Using the "x", "y", "r" and "q" in the diagram, we can make a general formula to figure out any curve!
The general equation for a circle is x^2 + y^2 = r^2
but you also know that (x+q)^2=r^2 (if you don't know this, take my word for it)
expanded out, (x+q)^2 is X^2 +2xq+q^2
but the r^2 is the same, so you can also say that
x^2+y^2 = r^2 = x^2 +2xq+q^2,
or x^2+y^2 = x^2 +2xq+q^2
Removing similar terms, you will get:
y^2 = 2xq + q^2 since you know the values for two of these, a calculator will give you the answer for the third: this is easy if you know "x" and "y" (that's why I derived this damn thing in the first place) but you can also calculate the "q" term if you know the radius because you know that r-x = q
A concrete example: say you have a hammer with a 2" diameter that you want to turn into a sinking hammer where the head has an 8" raduis. The "y" term is 1" (2"/2) the x term is r-q (yup, leave it in symbols) and the radius is 8" (so "x" is (8-q)"). For this type of calculation, it's easier to use the first equation to derive this stuff (if there are any math whizzes out there who can give an easier simplification, I'd really appreciate it!)
This will give an answer in the ballpark of q>1/4 (q is a little bigger than 1/4") sorry I can't remember the quadratic equation for an exact calculation.
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