How I made my sun-oven (or Death Ray, if you happen to be a small life-form at the focal-point of my parabolic reflector)
Nearly all of the DIY sun-ovens that I came across in a very brief survey of various blogs were shaped like a box with 4 reflective flaps. Well, that's great if your goal is to warm something up on a hot sunny cloudless day, but in my opinion, 4 reflectors is approximately equal to the magnitude of LAME! Come on! Anything worth doing yourself, is worth doing to a ridiculous extreme! So, here's how I made my sun-oven using 42 reflective surfaces made out of chip-board and tin-foil.
First, I plotted a simple parabola. I used the equation y=x2. This parabola has a focal point at (0, 0.25). (It is easy to identify this point because 0 is the center of the parabola, and 0.25 is the height at which the parabola forms a 45o angle. Note how the middle of the yellow-brown arrows reflects off of this parabola.)
Next, I picked two segments of the parabola (indicated with green and blue) that I wanted to cover with tin-foil reflectors. Of course, this is only a two-dimensional cross-section of my parabolic reflector. The final product, of course, will be 3-dimensional. Here is a top-down projection of my plans. (Note that the green and blue segments are also shown on this projection, with the labels "j" and "t".)
I also gave labels to several other sides, so I could figure out the precise size of the triangles that I needed to cut out. If you just want the instructions and don't care about the math, feel free to scroll down. (At times like these, I have to shake my head in pity of the poor souls who say that you don't really use math very often in real life. To me, the very purpose of breathing is so that I can live another day to create things--and math is the most useful tool for creating things of which I am aware. No, math may not be necessary to survive in life, but that is precisely why it is so wonderful. It is a power that those who merely struggle to survive will never understand. It is the magic staff that separates a wizard from a serf. But I digress...)
Okay, now let's do the math. I will distinguish between the edges on the 3-dimensional surface, and the corresponding edges on its 2-dimensional projection by putting an apostophe after the letters that represent edges on the 2-dimensional projection. I will compute the lengths of the edges on the 2-dimensional project first. If you carefully examine the two figures above, it should be pretty clear what I'm doing here.
j' = 0.2 t' = 0.3 f' = sqrt(0.32 - 0.152) = 0.259807621 g' = 0.5 - f' = 0.5 - 0.259807621 = 0.240192379 p' = sqrt((0.5 * cos(30o) - 0.5)2 + (0.5 * sin(30o) - 0)2) = 0.258819045 q' = p' / 2 = 0.129409523 u' = (0.3 + j' + t') * tan(15o) = 0.214359354Note that j' and t' were easily obtained just by looking at the figures above. To compute p', I used the distance formula to compute the distance between (0.5, 0) and (0.5 * cos(30o), 0.5 * sin(30o)), which are the vertices of that edge. To compute u', I observed that u' is one side of a right-triangle, such that u/(0.3 + j' + t') forms the tangent of 15o. I computed the other edges by using the Pythagorean theorem.
So far, we have only computed the edges on the flat projection. Don't forget, the actual triangles we want will be raised in three dimensions. So, next we compute the corresponding edges on the raised triangles. First, we identify the edges that are at a constant height. These have the same size as in the projection.
p = p' = 0.258819045 q = q' = 0.129409523 u = u' = 0.214359354Then we compute the edges that change in height. (All of these can be computed just using the Pythagorean theorem.)
g = sqrt(g'2 + 0.162) = sqrt(0.2401923792 + 0.162) = 0.288604191 j = sqrt(j'2 + 0.162) = sqrt(0.22 + 0.162) = 0.256124969 t = sqrt(t'2 + 0.392) = sqrt(0.32 + 0.392) = 0.492036584 h = sqrt(g2 + 0.152) = sqrt(0.2886041912 + 0.152) = 0.325257404 s = sqrt(t2 + u2) = sqrt(0.4920365842 + 0.2143593542) = 0.536702835 r = sqrt(s2 - q2) = sqrt(0.5367028352 - 0.1294095232) = 0.52086765
To compute m, n, and k, we set up some equations like this:
m + n = h m2 + k2 = j2 n2 + k2 = p2and then solve...
n = h - m m = sqrt(j2 - k2) n = h - sqrt(j2 - k2) (h - sqrt(j2 - k2))2 + k2 = p2 h2 - 2 * h * sqrt(j2 - k2) + (j2 - k2) + k2 = p2 h2 - 2 * h * sqrt(j2 - k2) + j2 = p2 sqrt(j2 - k2) = (p2 - j2 - h2) / (-2 * h) j2 - k2 = ((p2 - j2 - h2) / (-2 * h))2 k2 = j2 - ((p2 - j2 - h2) / (-2 * h))2 k = sqrt(j2 - ((p2 - j2 - h2) / (-2 * h))2) k = sqrt(0.2561249692 - ((0.2588190452 - 0.2561249692 - 0.3252574042) / (-2 * 0.325257404))2) k = 0.19960212 m = 0.160496085 n = 0.164761319
So, to sum up all that math, here's what you need to build a pseudo-parabolic reflector with a diameter of 1.6:
I cut my triangles out of chipboard.
I taped them together with boxing tape.
I covered it with aluminum foil. I put a layer of rubber-cement on the chip-board. That didn't really work very well, so then I used hot-melt glue around the edges of the foil. That worked very well. I recommend only using the hot-melt glue.
To give it some extra stability, I glued a nail at each corner, so I could tie a mesh of equilateral triangles over the top with fishing line. (The length of the string to make equilateral triangles is 1.434520755.) It is important to leave a little slack. If the lines are too tight, they will distort your parabola.
I also bent a hanger into a hexagon shape and taped it to the small opening in my parabola.
Then I tried it.
My only regret is that I didn't go hog-wild and build it ten times bigger. If you are going to use these plans to build one, I strongly recommend thinking big, or you might end up with a lame sun-powered water-simmerer like I did. Perhaps my next project will be to build a perfectly-shaped reflector at the focal point that will send the concentrated beam in a straight ray path down through the opening in the bottom, then bounce it off of a couple of mirrors to direct it into a weapon that I can use to TAKE OVER THE WORLD!!! (*Fades out in maniacal laughter...)