Re: [pinhole-discussion] Angle of ligh
Hi Joao, Your U shaped film will result in good pictures. Film (b) and (c) will be well exposed than flat film's peripheral area. U shape bottom makes smooth transition of film plane. - Masaru
Re: [pinhole-discussion] Angle of ligh
- Original Message - From: "Joao Ribeiro" > The idea is producing a kind of a "triptico" (I don't know the name in > English). Triptych
Re: [pinhole-discussion] Angle of ligh
Hi Ishikawa, Thanks for the ilustration. The camera is ready, I made it yesterday, and is almost exactly as you pictured, I made it with cardboard. The main diference is that film placed at the position "a" is also bended, U shape. The idea is producing a kind of a "triptico" (I don't know the name in English). I am not sure if it will work, but I'll give it a try and post the results to the upload galery. The next one will probably be a pvc cilinder with the film placed against the wall, forming a cilinder. Thanks for the help with the matter. Joao
Re: [pinhole-discussion] Angle of ligh
- Original Message - From: "ISHIKAWA Masaru" > I don't know how the angle of image cone is detemined. But I think > it is equal to 2xArctan(d/t) where "d" is a diameter of pinhole, "t" is > the thickness of pinhole-material. Masaru, You are right, that is de formula that defines the geometric cone, unfortunately, the USABLE image cone is much more smaller. That is why I disregarded pinhole thickness in my previous post. Nice ASCII drawings, Masaru! Guillermo
Re: [pinhole-discussion] Angle of ligh
Ribeiro, I think "picture"s below help your thinking: (Sorry for lengthy text.) v--pinhole ___ _pinhole base / \ / \ / \ <--- image cone / \ / \ / \ / \ / \ / \ <- film (nominal) ~~~ < film (big enough) Light through pinhole forms image cone. On the film plane, image cone forms cricular edge (image circle). Usually film is a rectangular and utilize some part of image cone. If film is large enough, all the part of inside of image circle fits in the film. Image exposed film will be "disk" on rectanglar film. Once you make film on the perpendicular walls like next "picture," v--pinholea ___ _pinhole base / \ / \ / \ <--- image cone | / \ | film->| / \ |<-film (b) |/ \| (c) | | | | | | +-+ <- film(a) image cone exposes not only film(a) but also film (b) and (c). Image on (b) and (c) are distorted as you expected. Image on (a) is normal. I don't know how the angle of image cone is detemined. But I think it is equal to 2xArctan(d/t) where "d" is a diameter of pinhole, "t" is the thickness of pinhole-material. If t is very small, d/t becomes very large, and Arctan(d/t) becomes 90 degree (2x becomes 180). So, pinhome-material should be thin for wide angle image. - Masaru
Re: [pinhole-discussion] Angle of ligh
- Original Message - From: "Joao Ribeiro" > Now, what I intend to do is to place 2 sheets of film not at the end of > the camera, but at the walls that are perpendicular to the pinhole > plane. I imagine that the light absorbed by these walls will also form > images, certainly distorted (I believe someone in this list have already > done that). How far from the pinhole plane the film should be to be > completely covered by light? My imaginations tells me that if it is too > close I'll have a "V" shape image. > Is this formula you sent me able to give me that info? I'm asking because > it takes into account the film diagonal and I believe this cone is > independent of the film diagonal, I'm not sure. Joao, It was, I believe, a bit hard to formulate the question without help of drawings, the case is the same for when stating an answer. Let me see if a can make sense of what I am thinking, when I put it in writing: As you know, COS^4 law dictates the fall off at the edges of a FLAT film plane NORMAL (90 degrees) to the lens axis (pinhole, ZP or glass). There are 3 factor that make up the factor for "off lens axis" imaging, namely: 1) The round pinhole looks like an oval whose size decreases by a factor of COSine of the angle 2) The maze of light fall obliquely on the film plane, therefore covers and area 1/COSine larger and finally 3) the distance pinhole film increases by a factor of 1/COSine, which translates into a lose of light of 1/COS^2 (due to inverse square law). All the above adds up to the well known COS^4 LAW factor and for your particular application, is valid for calculations of the fall off on the "base" of the camera. For the fall off on the walls of the camera we have to make some analysis: 1) the round pinhole still looks like an oval as we "climb" the wall starting from the "base", that give us a factor of COSine of the angle in light lose, as above. 2) The maze of light stills falls obliquely, but this time the film plane being vertical rather than horizontal causes the lose of light to be proportional to 1/SINE of the angle, rather than 1/COSine as above. 3) and finally, the distance pinhole to (vertical) film plane increase changes at a rate of K / (2 * SIN of angle) , where "K" is the amount of times the focal length fit in the "base" of the camera, in other words, in the width of the film plane. All the above adds up to a factor of: Light fall off FACTOR = [ 4 * SIN^3 (angle) * COS (angle)] / K^2 (sorry I didn't simplify the above a bit more, but what I remember of my grade 8 trigonometry class does not help, probably it does not allow more simplification, but if someone out there thinks otherwise, I'd love to know how) You could use the above FACTOR result to find out the number of stops that angle will cause as light fall off on the camera walls. Number of fall of stops = [LOG ( 1/ FACTOR )] / 0.3 BTW, it's been said that for flat film planes the maximum image circle is 3.5 times the focal length of the camera. In such a camera, the fall off at the extreme edges would be 4 stops with respect to the center of the film plane. I mention this in case you want to find the angle that causes a similar fall off on the edges of the walls. Hope it helps Joao. (Answer is open to corrections) Guillermo
Re: [pinhole-discussion] Angle of ligh
Bill Erickson writes: > > Since you brought this up, there are two factors influencing the intensity > of light at the film plane, the distance from the pinhole and the angle off > axis. As you move off axis of a flat film plane, the distance from the > pinhole to the film grows, and the apparent shape of the pinhole changes > from round to narrower and narrower. The so called fourth power of the > cosine law governs. The intensity at any point on a flat film plane equals > the intensity at the axis point times the cosine, to the fourth power, of > the angle off axis. When you curve the film around the pinhole you > counteract half of it because the film is always the same distance, and the > only darkening you get at the edges is due to the change in the apparent > shape of the pinhole. That's true for half-cylinder cameras. For oatmeal box or paint can cameras, the film gets closer to the pinhole as it goes around the cylinder (at least in one dimension), so there is at least some correction for the pinhole shape factor also.
Re: [pinhole-discussion] Angle of ligh
Since you brought this up, there are two factors influencing the intensity of light at the film plane, the distance from the pinhole and the angle off axis. As you move off axis of a flat film plane, the distance from the pinhole to the film grows, and the apparent shape of the pinhole changes from round to narrower and narrower. The so called fourth power of the cosine law governs. The intensity at any point on a flat film plane equals the intensity at the axis point times the cosine, to the fourth power, of the angle off axis. When you curve the film around the pinhole you counteract half of it because the film is always the same distance, and the only darkening you get at the edges is due to the change in the apparent shape of the pinhole. - Original Message - From: "ragowaring" To: Sent: Tuesday, February 12, 2002 2:33 PM Subject: Re: [pinhole-discussion] Angle of ligh > Dear Joao > > I'm no mathematician but I think you will find that the parts of the film > nearest the pinhole will receive a greater amount of light for a given area > than parts of the film further away > > This is because of the inverse square law, which states simply that the > radiation falling on a surface from a point source will decrease inversely > proportinally with distance by 1/xsquared where x is the distance. > > This means that for every doubling of the distance from the source, the > amount of radiation reaching a given area is quartered (that is because the > same radiation has to cover four times the area covered at half the > distance). > > Imagine the area covered by a cone (of radiation if you like) - it is > actually easier to imagine this as a four sided pyramid, so I shall continue > with this visualisation. The square at the base of the pyramid is 1 square > unit. The point of the pyramid is the source. Radiation will reach the base > at a given rate, say one unit of radiation per second. > > If you double the height of the pyramid, which is equivalent to doubling the > distance from the source of radiation, you will find that the base of the > pyramid is now four times the area of the first pyramid - four squares of > one unit each or one big square four times the area. > > Now come the fun part. The radiation reaching this larger square in a given > time is the same as that reaching the 1 unit square at half the distance. > That is to say, one unit of radiation per second. But this time that one > unit has to cover four times the area as the radiation spreads out. > > Therefore each square unit at double the distance recieves a quarter of the > radiation per second. Therefore a doubling in the distance from a point > source of radiation results in one quarter of the radiation falling on one > unit area! > > This explains why on wide angle pinhole photographs, the sides of the > negative come out less dense - because they are further away from the > pinhole and therefore less light reaches them per given time. It is this > per given time that is all important when calculating exposures with focal > lengths etc. > > Now, when the film is parallel to the plane of the pinhole, i.e. at the back > of the camera, normally the inverse square law has a small effect, > particularly if the angle of acceptance or vision is small. > > However, if you put the film on the camera side walls, the effect becomes > very significant indeed. The parts nearer the pinhole will need a > considerably shorter exposure that those further away. > > This however, can be compensated for if the side wall of the camera are > short, that is to say, the camera has a short focal length. > > > Enough of theory, the thing is to EXPERIMEMT! > > It is so much easier with pictures > > By the way, the above explanation is an approximation because in real life > the base of the pyramid would be curved and not flat, but it is close > enought to get the picture - sorry no pun intended > > Alexis > > > > > > > > > > on 12/2/02 5:40 pm, Joao Ribeiro at jribe...@greco.com.br wrote: > > > Thanks Bill and Guillermo for your answers. > > > > But ... > > > >> Geometrically/mathematically speaking, the angle changes when the > >> pinhole diameter changes, the change is so small tho, that in practice you > >> can dismiss it. Since you want to calculate the "cone angle", otherwise > >> known as "angle of view", here is a formula I just derived that takes the > >> pinhole diameter into consideration: > >> > >> Cone angle = ArcTan [ (D+P) / (2 * B) ] > >> > >> Where" > >> D = Diagonal of your film format > >> P = Pinhole diameter > >> B = Bellows extensio
Re: [pinhole-discussion] Angle of ligh
Dear Joao I'm no mathematician but I think you will find that the parts of the film nearest the pinhole will receive a greater amount of light for a given area than parts of the film further away This is because of the inverse square law, which states simply that the radiation falling on a surface from a point source will decrease inversely proportinally with distance by 1/xsquared where x is the distance. This means that for every doubling of the distance from the source, the amount of radiation reaching a given area is quartered (that is because the same radiation has to cover four times the area covered at half the distance). Imagine the area covered by a cone (of radiation if you like) - it is actually easier to imagine this as a four sided pyramid, so I shall continue with this visualisation. The square at the base of the pyramid is 1 square unit. The point of the pyramid is the source. Radiation will reach the base at a given rate, say one unit of radiation per second. If you double the height of the pyramid, which is equivalent to doubling the distance from the source of radiation, you will find that the base of the pyramid is now four times the area of the first pyramid - four squares of one unit each or one big square four times the area. Now come the fun part. The radiation reaching this larger square in a given time is the same as that reaching the 1 unit square at half the distance. That is to say, one unit of radiation per second. But this time that one unit has to cover four times the area as the radiation spreads out. Therefore each square unit at double the distance recieves a quarter of the radiation per second. Therefore a doubling in the distance from a point source of radiation results in one quarter of the radiation falling on one unit area! This explains why on wide angle pinhole photographs, the sides of the negative come out less dense - because they are further away from the pinhole and therefore less light reaches them per given time. It is this per given time that is all important when calculating exposures with focal lengths etc. Now, when the film is parallel to the plane of the pinhole, i.e. at the back of the camera, normally the inverse square law has a small effect, particularly if the angle of acceptance or vision is small. However, if you put the film on the camera side walls, the effect becomes very significant indeed. The parts nearer the pinhole will need a considerably shorter exposure that those further away. This however, can be compensated for if the side wall of the camera are short, that is to say, the camera has a short focal length. Enough of theory, the thing is to EXPERIMEMT! It is so much easier with pictures By the way, the above explanation is an approximation because in real life the base of the pyramid would be curved and not flat, but it is close enought to get the picture - sorry no pun intended Alexis on 12/2/02 5:40 pm, Joao Ribeiro at jribe...@greco.com.br wrote: > Thanks Bill and Guillermo for your answers. > > But ... > >> Geometrically/mathematically speaking, the angle changes when the >> pinhole diameter changes, the change is so small tho, that in practice you >> can dismiss it. Since you want to calculate the "cone angle", otherwise >> known as "angle of view", here is a formula I just derived that takes the >> pinhole diameter into consideration: >> >> Cone angle = ArcTan [ (D+P) / (2 * B) ] >> >> Where" >> D = Diagonal of your film format >> P = Pinhole diameter >> B = Bellows extension (or focal length) >> >> As you can see, the effect of adding P to D is very small, i.e., for 8x10, >> "D" would be equal to about 325mm if you add to that a "P" of 0.5mm, you get >> 325.5mm, again, not a big change. The same happens if you change the >> pinhole diameter. > > I'm not sure this is the answer to my question. If I could send a drawing > attached to the list it would be easier, but this is what I want: > > I imagine a light entering the box/camera and forming a cone. This cone will > be > independent of film size. I believe it will vary with pinhole diameter but > maybe > not in a meaningful way. > Let's say I made a very long focal distance box, no matter the film size, and > at > the end of the box, parallel to the pinhole plane usually is put the film. But > I'll be using just a fraction of the image formed by this cone, the rest will > be > absorbed by the black walls of the camera box. > Now, what I intend to do is to place 2 sheets of film not at the end of the > camera, but at the walls that are perpendicular to the pinhole plane. I > imagine > that the light absorbed by these walls will also form images, certainly > distorted (I believe someone in this list have already done that). How far > from > the pinhole plane the film should be to be completely covered by light? My > imaginations tells me that if it is too close I'll have a "V" shape image. > Is this formula you sent me able to give me th
Re: [pinhole-discussion] Angle of ligh
Thanks Bill and Guillermo for your answers. But ... > Geometrically/mathematically speaking, the angle changes when the > pinhole diameter changes, the change is so small tho, that in practice you > can dismiss it. Since you want to calculate the "cone angle", otherwise > known as "angle of view", here is a formula I just derived that takes the > pinhole diameter into consideration: > > Cone angle = ArcTan [ (D+P) / (2 * B) ] > > Where" > D = Diagonal of your film format > P = Pinhole diameter > B = Bellows extension (or focal length) > > As you can see, the effect of adding P to D is very small, i.e., for 8x10, > "D" would be equal to about 325mm if you add to that a "P" of 0.5mm, you get > 325.5mm, again, not a big change. The same happens if you change the > pinhole diameter. I'm not sure this is the answer to my question. If I could send a drawing attached to the list it would be easier, but this is what I want: I imagine a light entering the box/camera and forming a cone. This cone will be independent of film size. I believe it will vary with pinhole diameter but maybe not in a meaningful way. Let's say I made a very long focal distance box, no matter the film size, and at the end of the box, parallel to the pinhole plane usually is put the film. But I'll be using just a fraction of the image formed by this cone, the rest will be absorbed by the black walls of the camera box. Now, what I intend to do is to place 2 sheets of film not at the end of the camera, but at the walls that are perpendicular to the pinhole plane. I imagine that the light absorbed by these walls will also form images, certainly distorted (I believe someone in this list have already done that). How far from the pinhole plane the film should be to be completely covered by light? My imaginations tells me that if it is too close I'll have a "V" shape image. Is this formula you sent me able to give me that info? I'm asking because it takes into account the film diagonal and I believe this cone is independent of the film diagonal, I'm not sure. Sorry for this long post Joao
Re: [pinhole-discussion] Angle of ligh
- Original Message - From: "Joao Ribeiro" > When the light enters the camera, it enters in an angle the depends on > the pinhole diameter or this angle is constant? > Imagine I have a pinhole of 0.5 mm. If I make a bellows camera and set > it to say 50 mm focal distance using an 8x10 film I'll have a very wide > angle image. But if I enlarge the bellows distance to 500 mm I'll then > have a telephoto image. Well, actually the image "cone" will be the > same, I'm just choosing a section of the cone farther away from the > pinhole or origin, and I am also selecting part of this cone to be > recorded. How can I calculate the cone angle? Will changes in the > pinhole diameter make any difference in this angle or it will always be > constant? Joao, Geometrically/mathematically speaking, the angle changes when the pinhole diameter changes, the change is so small tho, that in practice you can dismiss it. Since you want to calculate the "cone angle", otherwise known as "angle of view", here is a formula I just derived that takes the pinhole diameter into consideration: Cone angle = ArcTan [ (D+P) / (2 * B) ] Where" D = Diagonal of your film format P = Pinhole diameter B = Bellows extension (or focal length) As you can see, the effect of adding P to D is very small, i.e., for 8x10, "D" would be equal to about 325mm if you add to that a "P" of 0.5mm, you get 325.5mm, again, not a big change. The same happens if you change the pinhole diameter. Guillermo
Re: [pinhole-discussion] Angle of ligh
When you increase the focal length but leave the negative size the same all you do is decrease the "angle of acceptance" of the light beam. You sample a smaller portion of the potential image. I can't see how the size of the pinhole would make a difference, except that it casts an optimal potential image that is greater in radius. I figure out these things by drawing them out. I suspect Trigonometry would work but to me that's a dark science, one to which I have not been admitted. - Original Message - From: "Joao Ribeiro" To: Sent: Monday, February 11, 2002 6:51 PM Subject: [pinhole-discussion] Angle of ligh > Hi folks, > > I have a question, but I'm not sure I'll be able to put it properly. > Here it goes: > > When the light enters the camera, it enters in an angle the depends on > the pinhole diameter or this angle is constant? > Imagine I have a pinhole of 0.5 mm. If I make a bellows camera and set > it to say 50 mm focal distance using an 8x10 film I'll have a very wide > angle image. But if I enlarge the bellows distance to 500 mm I'll then > have a telephoto image. Well, actually the image "cone" will be the > same, I'm just choosing a section of the cone farther away from the > pinhole or origin, and I am also selecting part of this cone to be > recorded. How can I calculate the cone angle? Will changes in the > pinhole diameter make any difference in this angle or it will always be > constant? > > I hope I could make myself clear! > Thanks in advance for any info, > > Joao > > > ___ > Post to the list as PLAIN TEXT only - no HTML > Pinhole-Discussion mailing list > Pinhole-Discussion@p at ??? > unsubscribe or change your account at > http://www.???/discussion/ >
