I try to answer in part to the questions of Anselmo and to the problem suggested by John Bercovitz I apologize for the length of the message and for the great quantity of numbers :-) ------------------------ The questions of Anselmo
1) in the 74 camera obscura sundials that are in Italy, we find: - 28 in churches - 6 in convents - 8 in town hall - 17 in residences - 15 in universities , astronomic observatories and others then only the 38% are in churches. To my knowledge, there is no sundials of this type built in Mosques. 2) How the plane of the hole has to be placed, it is a matter of opinion and I try to summarize the most diffused or discussed cases, even if I think that the hole has often be placed how and where. it is possible :-) . In my opinion, if there are no limitations, the criterion to follow is to make brighter the image of the Sun in winter, when it is greater (illuminotecnic criterion). The illumination of the image must be enough greater than that of the surrounding zones of the floor, to have a good contrast and good possibilities to be well seen . The illumination is proportional to the ratio R between the projection of the hole's area on the plane normal to the rays and the area of the elliptic image on the floor. Dividing the ratio RS, on the day of Summer' Solstice , for the value RW, on the Winter' Solstice, we obtain a parameter C = RS/RW that gives an idea of the uniformity of the illumination in the two extreme cases. The values of RS and RW written below are proportional to the exact values. All the values are calculated for a Latitude = 44° Different methods: a) Axis of the hole in the direction of the Sun at noon on the the Equinoxes (axis inclined of [90-Lat] degrees on the horizontal plane): astronomical positioning. The rays from the Sun on the Solstices are equally inclined on the plane of the hole and the quantity of the incoming light is equal. RS=0.75; RW=0.051; C = 15 around b) Axis of the hole that reaches the mean point of the meridian line (equidistant from the Solstitial points) : geometric positioning. RS=0.68; RW=0,055; C = 12.5 c) Axis of the hole that reaches the point of the winter Solstice - illuminotecnic positioning. In my opinion this is the best, because in this way the quantity of the incoming light is greater on Winter Solstice, when the image on the floor is great, that in summer, when the image is small. RS=0.56; RW = 0.056; C = 10 d) Vertical Hole. Often the hole is made in a vertical metallic plate either to decrease the entrance of the rain, or because the room is separated from the outside only by a vertical wall. In this case the illumination of the image on Summer' Solstice decreases a lot (RS=0.29). RW = 0.051 C=5.56 . The illumination' uniformity increases at the expense of the summer image. e) Horizontal Hole. Necessary when the ceiling is horizontal. RS=0,77; RW = 0,021; C = 36. The intensity of the winter image decreases a lot. As we can see in the different cases the smaller illumination is that of the winter image, that is practically almost constant and improves slightly in the case c). If we want the mean illumination of the image of the Sun it is necessary to use the formula: J = R* 80000 * [D *107 / H]^2 lumen/mq or lux where 80000 lumen/mq is the mean value of the illumination from the Sun on a surfaces perpendicular to the rays, D the diameter of the hole in mm and H its height in mm. For example in a sundial with H=20m = 20000mm, D=20mm we obrtain J = R * 916 lux and therefore, for instance, in the case a) JS = 687, JW = 46 lux I remember that the illumination on a horizontal surface from the Sun, at sunset, is of around 380 lux; that required for a confortable reading is abaut 110 lux and that provided (by the architects) for an auditorium is about 50 lux . Obviously to make more visible the image of the Sun it is necessary to decrease the indirect illumination of the floor, reducing the dimension of the windows, painting the walls with dark colors,etc. ---------------------- The problem suggested by John Bercovitz The problem of the least dimension of the hole in a camera obscura, to have the most sharp and clean images as possible, was discussed in this List also 2 years ago. I copy at the end of this letter a part of a message that I sent on 4/14/2000 . In pin-hole photography the hole's minimum diameter is given by the formula D=SQR(L/k) where k = 745 (k can also be taken between 650 and 1000). Then D=0.0366*SQR(L) where L and D are in mm ( formula #1 ,given by John Bercovitz and Gordon Uber) The same formula with D and L in inches is D=SQR(L/18900) = SQR(L)/137.5 = 0.00727 SQR(L) If we decrease the diameter of the hole, the width of the diffraction disk increases, while if we increase it, the zone of fuzziness around the central bright disk increases. All this is true exactly only if we suppose the light source punctiform and his light monochromatic, things not true for the Sun! Otherwise the central disk (the Airy disk) disappears and there is the Sun 's image surrounded by an uncertain zone whose width is enough difficult to calculate. If we use the criterion of the pin-hole photography to calculate the hole's diameter in a camera obscuara sundial we obtain a hole too small that gives Sun's images very few illuminated. In a camera obscura sundial moreover the distance L, between the hole and the image, changes a lot from summer to winter: for ex. in a place with Lat=42d the ratio LMAX/LMIN = 2286 In this case, for the calculation of the hole's diameter, we can take an middle value as that on the days of the equinoxes L=H/cos(Lat) An example. Place with Latitude = 42d; hole's height H =21m (S.Maria degli Angeli in Rome). On the day of the Equinox L=28.2m and therefore ( formula #1) D=6.2 mm and L/D=4550. The illumination of the image of the Sun results [method a) ] JS = 60 lux in summer and only JW=4 lux in winter : values excessively low. In the existing sundial D=20 mm and therefore L/D = 1410 and the illumination is around 10.4 times greater : JS=623 lux and JW=42 lux . Another example. With Lat = 44d and H=3m we have L = 4.17m and then from the formula #1, D=2.37 mm , L/D = 1762, JS=427 lux and JW=30 lux. In conclusion, in my opinion, IT IS NOT POSSIBLE, except in very dark rooms, to use the values of the diameters used in the pinhole photography ( and calculated with formula #1), but it is necessary to take values notably greater. A formula that can be used is the following D=SQR(H/50) with D and H in mm or D=SQR(H/1250) with D and H in inches. [ Ferrari's Formula :-) :-) ] NOTE In the pin-hole photography we can use also hole's diameters very smaller because it is possible to increase the exposure time. Gianni Ferrari 44° 39' N 10° 55' E Mailto : [EMAIL PROTECTED] ------------------------------------ Message sent on 4/1/4/2000 .. Snip . An observation on the pin-hole photography Since the fuzzy zone has a width equal to the diameter D of the hole one can think to decrease the diameter D to obtain more and more sharp images. Apart the diminution of the brightness of the image, this it is not true. In fact, if the distance from the screen L is constant, and we decrease the diameter D of the pin-hole, the diffraction fringes begin to appear and extend the image. The pin-hole photographers use, as limit, the value of the diameter D that produces a diffraction disk (disk of Airy) with a diameter = D. Since the diameter of the disk of diffraction =2.44*L*w/d (formula also used for finding the separator power of a telescope) the better diameter results Dm=SQRT(2.44*w*L) where w= wave length , L = distance from the pin-hole and the film Using green-yellow light with w=550nm we have Dm * Dm = L/745 or Dm=SQR(L/745) Where Dm, the minimum diameter of the hole, and L are in mm. (this formula is also used with denominator = 650, 1000) This is the "optimal" diameter, that is the diameter which produces the sharpest possible image In pin-hole photography the distance L=745*D*D is called "focal length" of the pin-hole The ratio A = L/D = 745*D is called "Aperture" or "f-stop" of the pin-hole It is easy to show that the exposure time to have a photo is proportional to the square of this value; in fact the intensity of illumination of the image is J=Jo*(D*107/L)^2 = Jo*11500/(A*A) For ex. if we want a focal = 300mm (tele) we need a pin-hole with a diameter =0.63mm The f-stop of this objective = 472. The required time to make a photo is about 870 times that necessary with a normal image and a f-stop = 16 -
