# How do I calculate the curve of a parabolic mirror to melt the ice from my street gutter?

I want to make a concave parabolic mirror that will have a focal point six inches below where I place it (on the edge of the sidewalk) so that it will melt the ice collecting in the gutter below and clear a path for the water to drain. I will probably be using shiny stainless steel for the reflector, so it won't be a fully reflective mirror. What would be the ideal size, and how do I calculate the curve needed to concentrate the heat to that focal point?

Edit: Here's a real life example of someone doing this unintentionally: Reflected light from London skyscraper melts car

• the focal point would be between the sun and the mirror so mirrors would be a poor choice for heating things on the ground. you should try lenses. checkout GREENPOWERSCIENCE on YouTube. you can make a huge lens with water pooling in a sagging stretched plastic plane – gregsan Dec 29 '13 at 7:53
• I edited my post with a link to a real world example. In higher latitudes the sun comes across the sky at a lower azimuth in the sky (from the south) so a refractive lens would probably not be ideal. This wouldn't be a simple parabola that points directly at the sun. – Nathan L Dec 29 '13 at 23:49
• @gregsan: I'm not sure if this is what the OP meant, but you don't need a complete parabola. – lionelbrits Dec 29 '13 at 23:56

A parabola with equation $x = \frac{y^2}{4a}$ (opens rightward) has a focus at $x=a$. So assuming the light comes in along the $x$ axis, the ground would some line passing through the point $(x=a,y=0)$ sloping downward. I.e., $y= \tan\theta\, (x-a)$. You only need the part of the parabola that is to the right of that line (i.e., above ground)