If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below. |
|
|
Thread Tools | Display Modes |
|
#1
|
|||
|
|||
hyperfocal distance
When using the concept of hyperfocal distance, do we need to adjust for the
smaller sensor on 300D? |
#2
|
|||
|
|||
hyperfocal distance
No. It remains the same.
-- http://www.chapelhillnoir.com home of The Camera-ist's Manifesto The Improved Links Pages are at http://www.chapelhillnoir.com/links/mlinks00.html A sample chapter from my novel "Haight-Ashbury" is at http://www.chapelhillnoir.com/writ/hait/hatitl.html "leo" wrote in message k.net... When using the concept of hyperfocal distance, do we need to adjust for the smaller sensor on 300D? |
#3
|
|||
|
|||
hyperfocal distance
"Tony Spadaro" wrote in news:dJCEc.85007
: No. It remains the same. Yes - this is the intuitive answer. But it is not entirely correct. The formula for hyperfocal distans is (just as pointed out in another post): h = (f*f)/(N*c) f = focal length, N = f-ratio, c = "circle of confusion" diameter. The crucial factor here is c (circle of confusion). If you search further on the net you will find that it is 1/1740 of the diagonal of the sensor. Therefore, h will be bigger for a smaller sensors. So - the intuitive answer is wrong. The hyperfocal distance depends on how much you crop your image. Therefore, it is also only valid for the 35 mm film camera if you don't crop the image. /Roland |
#4
|
|||
|
|||
hyperfocal distance
Sorry - the intuitive answer is right.
Yes the more you enlarge the softer the image gets but it's just plain silly to assume any single size for the final image since as size of the image increases, viewing distance also increases and they cancel each other out. -- http://www.chapelhillnoir.com home of The Camera-ist's Manifesto The Improved Links Pages are at http://www.chapelhillnoir.com/links/mlinks00.html A sample chapter from my novel "Haight-Ashbury" is at http://www.chapelhillnoir.com/writ/hait/hatitl.html "Roland Karlsson" wrote in message ... "Tony Spadaro" wrote in news:dJCEc.85007 : No. It remains the same. Yes - this is the intuitive answer. But it is not entirely correct. The formula for hyperfocal distans is (just as pointed out in another post): h = (f*f)/(N*c) f = focal length, N = f-ratio, c = "circle of confusion" diameter. The crucial factor here is c (circle of confusion). If you search further on the net you will find that it is 1/1740 of the diagonal of the sensor. Therefore, h will be bigger for a smaller sensors. So - the intuitive answer is wrong. The hyperfocal distance depends on how much you crop your image. Therefore, it is also only valid for the 35 mm film camera if you don't crop the image. /Roland |
#5
|
|||
|
|||
hyperfocal distance
Tony Spadaro wrote:
Sorry - the intuitive answer is right. Yes the more you enlarge the softer the image gets but it's just plain silly to assume any single size for the final image since as size of the image increases, viewing distance also increases and they cancel each other out. What formula are you using to compute hyperfocal distance that uses increasing image sizes and increasing viewing distances? I don't see this anywhere. The formula that I've seen uses focal length, f ratio, and CoC. The first two are characteristics of the lens, so image size does not enter. If the CoC is taken to be the inverse of the resolution factor, and the rf is taken to be 1525/d, where d is the diagonal measurement of the sensor. In these computations, the image size is held constant at 25 cm, with an 8x10 image size. In all the discussions I read so far in the last couple of days, a family of CoCs can be generated for different sensor/image/viewing distance combinations, but to compare them across families is of no value. |
#6
|
|||
|
|||
hyperfocal distance
Don't get all bogged down in formulas. Take an 8x10 print and hold it
where you can see the entire print at one time -- this is proper viewing distance for an 6x10. Now take an 11x14 and do the same -- it is farther away, isn't it? A 16x20 will be farther away than the 11x14 and when you get up to 30x40 you should be halfway across the room. -- http://www.chapelhillnoir.com home of The Camera-ist's Manifesto The Improved Links Pages are at http://www.chapelhillnoir.com/links/mlinks00.html A sample chapter from my novel "Haight-Ashbury" is at http://www.chapelhillnoir.com/writ/hait/hatitl.html "M Barnes" wrote in message ... Tony Spadaro wrote: Sorry - the intuitive answer is right. Yes the more you enlarge the softer the image gets but it's just plain silly to assume any single size for the final image since as size of the image increases, viewing distance also increases and they cancel each other out. What formula are you using to compute hyperfocal distance that uses increasing image sizes and increasing viewing distances? I don't see this anywhere. The formula that I've seen uses focal length, f ratio, and CoC. The first two are characteristics of the lens, so image size does not enter. If the CoC is taken to be the inverse of the resolution factor, and the rf is taken to be 1525/d, where d is the diagonal measurement of the sensor. In these computations, the image size is held constant at 25 cm, with an 8x10 image size. In all the discussions I read so far in the last couple of days, a family of CoCs can be generated for different sensor/image/viewing distance combinations, but to compare them across families is of no value. |
#7
|
|||
|
|||
hyperfocal distance
On Thu, 1 Jul 2004 08:59:07 -0500, "M Barnes"
wrote: Tony Spadaro wrote: Sorry - the intuitive answer is right. Yes the more you enlarge the softer the image gets but it's just plain silly to assume any single size for the final image since as size of the image increases, viewing distance also increases and they cancel each other out. What formula are you using to compute hyperfocal distance that uses increasing image sizes and increasing viewing distances? I don't see this anywhere. The formula that I've seen uses focal length, f ratio, and CoC. The first two are characteristics of the lens, so image size does not enter. If the CoC is taken to be the inverse of the resolution factor, and the rf is taken to be 1525/d, where d is the diagonal measurement of the sensor. In these computations, the image size is held constant at 25 cm, with an 8x10 image size. Question: Using that formula, does it work for *any* sensor size, or the one the image size on the focal plane was designed for (in the case of the lenses in question, 35mm)? Or, to put it a different way, if you take a 35mm flm image at hyperfocal distance, does cropping that image alter the hyperfocal distance, or was the HD set when the pic was taken? I'm wondering, if the CoC formula includes 1525d (and I'm assuming it does), does d refer to the sensor, or the image on the focal plane, and the sensor size that image is designed for? I mean, in a DSLR, the lens uses a smaller part of the image on the focal plane than 35mm film does. In effect, it crops that image. As I ask above, does this really change the hyperfocal distance of that lens? In all the discussions I read so far in the last couple of days, a family of CoCs can be generated for different sensor/image/viewing distance combinations, but to compare them across families is of no value. Bill Funk Change "g" to "a" |
#8
|
|||
|
|||
hyperfocal distance
"Tony Spadaro" wrote in
. com: Sorry - the intuitive answer is right. Yes the more you enlarge the softer the image gets but it's just plain silly to assume any single size for the final image since as size of the image increases, viewing distance also increases and they cancel each other out. The formula I gave is the formula for hyper focal distance. It behaves exactly as I said. The hyper focal distance is inverse proportional to the size of of the sensor, everything else kept equal. You may not like the definition. You may want something else; but then it is not hyperfocal distance. You may call it Tony Spadero distance if you want. /Roland |
#9
|
|||
|
|||
hyperfocal distance
Roland Karlsson writes:
Yes - this is the intuitive answer. But it is not entirely correct. The formula for hyperfocal distans is (just as pointed out in another post): h = (f*f)/(N*c) f = focal length, N = f-ratio, c = "circle of confusion" diameter. The crucial factor here is c (circle of confusion). If you search further on the net you will find that it is 1/1740 of the diagonal of the sensor. Therefore, h will be bigger for a smaller sensors. Assuming that you keep the focal length the same. On the other hand, if you reduce the focal length in proportion to the sensor size change, to maintain the same field of view, then the hyperfocal distance becomes smaller not larger. That's because it depends on focal length squared. So - the intuitive answer is wrong. The hyperfocal distance depends on how much you crop your image. Therefore, it is also only valid for the 35 mm film camera if you don't crop the image. Yes - if you don't change the lens focal length. Dave |
#10
|
|||
|
|||
hyperfocal distance
|
Thread Tools | |
Display Modes | |
|
|