Hi Chris (& others) What a nice easy to use formula! I was wondering why you used a rectangular mirror instead of a circular or square one?
I've noticed that the eliptical sun spot, of course, has its major long axis oriented in a north south direction on the ceiling at apparent noon. Do you think that it would be possible to make the spot more circular by using a rectangular mirror who's long side is oriented in an east west direction? Shouldn't this compensate for some of the north/south stretch? (I must go purchase a larger mirror and experiment with masking tape to test this! I will also test your idea of two close mirrors. Also, I'm thinking the fuzzy region on the circumference might be sharpened a little if there was a thin frame around the mirror because the glass edge of the mirror might be throwing light rays off in several durections.) What do you think John John L. Carmichael Jr. Sundial Sculptures 925 E. Foothills Dr. Tucson Arizona 85718 USA Tel: 520-696-1709 Email: [EMAIL PROTECTED] Website: <http://www.sundialsculptures.com> ----- Original Message ----- From: "Chris Lusby Taylor" <[EMAIL PROTECTED]> To: "Dave Bell" <[EMAIL PROTECTED]> Cc: "John Carmichael" <[EMAIL PROTECTED]>; <sundial@rrz.uni-koeln.de> Sent: Tuesday, August 14, 2001 10:04 AM Subject: Re: diameter of reflected sun image > Dave Bell wrote: > > > On Mon, 13 Aug 2001, John Carmichael wrote: > > > > > Knowing the apparent diameter of the sun, the size of the mirror, and > > > the distance from the mirror to the ceiling, how could I calculate the > > > diameter of the sun spot? > > > > Take the limiting case, of a (nearly) zero-diameter mirror. This is > > exactly the same as a pinhole "lens", which casts an image no smaller than > > the angular diameter of the Sun, or 8.7 milliradians. (*Much* easier to > > work in than degrees!) If the spot on the ceiling is an average of 10 > > feet, or 120 inches from the mirror, the spot will be at least 1.04 inch > > diameter. Diffraction from the edges of the mirror will make the spot > > larger (and fuzzier), and a larger mirror will cast a larger image. > > Without modelling it (or thinking too hard), I'd also guess that the image > > spot would be the diameter of the mirror, PLUS the above figure derived > > from the 8.7 milliradian cone angle. So, a 1 inch mirror should make a 2 > > inch spot at 10 feet... > > <snip> > > Funnily enough, my brother was just asking why the spots produced by my mirror > dial are apparently round when the mirrors are 1cm x 5cm. As Dave Bell says, > the truth is that the spots are rounded-cornered rectangles with overall > dimensions of (1cm + 0.0087 x Distance) x (5cm + 0.0087 x Distance) and corner > radius of 0.0044 x Distance. When mirror-to-screen distance is 10M, this gives > 9.7 x 13.7 cm, which looks near enough circular. > > The optimum thickness of a string gnomon is such that the umbra has zero > width. If you use a mirror of equivalent width (0.0087 x Distance) only the > very centre of the spot is fully illuminated, but to the human eye this isn't > obvious, and in my experience a narrower mirror is better as the spot is quite > bright enough. Judging the centre of the spot is harder than judging the > edges. Maybe the ultimate would be two or more mirrors spaced so that their > spots just touch. Masking a large enough mirror to leave two small circles > would be worth trying. > > Chris Lusby Taylor > Newbury, Engalnd > 51.4N 1.3W > >
