USB Cables 101 • A Guide to USB Connector Types - usb cable cable

 2024-09-26 22:37:07

Dots per inch is an measure of spatial resolution used in reference tomarking engines (printers) which lay down ink in discrete spots. Hue,Saturation and Value are comprised of multiple dots of inks laid downtogether. Digital cameras make images sized in pixel dimensions, and the physicalsizes of those pixels are related to the specific CCD or CMOS sensoryou have. Taking that pixel array and applying it to an output device,you size the pixels to a ppi standard. DPI is often confused with PPI because the people who wrote the specsheets and manuals were more familiar with printer terminology thanwith digital imaging terminology. Photoshop and other image processingsoftware perpetuates this confusion by presenting image size tools thatexpress PPI values as DPI. Example for illustration (and I ask for pardon for simple math errorsin advance ... ;-): My camera makes an image of 1920x2560 pixels (4.7 Mpixels) and I wantto make a highest quality print on an Epson 1270 printer, which iscapable of 1440x1440 dpi marking resolution. My experience shows thatthe Epson 1270 printer and its driver achieves the best quality imageit can make when it is rendering pixels which are at a 1:4 ratio ofpixel resolution to dots per inch marking capability, so my outputprint file should be set at 360 PPI and the printer set to 1440 ppi.That will achieve a print image size of 5.3x7.1 inches dimension. If the image was a macro photo of a grid at a 1:1 image:subjectmagnification, the finest grid that I could image accurately would beconstrained by the physical size of the CCD sensor in my camera and theactual pixel dimensions. I don't know those dimensions off the top ofmy head, but let's conjecture that the sensor is 12x16mm in size andthe effective pixel area represents the whole area of the sensor. Thatmakes each pixel .00625mm wide. Each pixel in the sensor can only takeon one value at a time, so the minimum width of a black-white line pairhas to be double that or .0125mm. So the sensor is limited to beingable image a grid with a line spacing of 80 LP@MM at most. Imaged as a5.3x7.1 inch print at maximum resolution, that 80LP@MM grid ismagnified approximately 15 times, resulting in a calculated 5.33 LP@MMgrid on paper. This is simplistic as it assumes a b&w grid and a sensor in which everypixel is used, not your typical case, as well as a lot of othersimplifications but I hope it illustrates the relationship of pixels,PPI, DPI and LP@MM. I hope that helps, and I truly hope I have not lost a decimal point ormade some other stupid error... ;-)GodfreyIn article <3E15142F...@integraonline.com>, Bill Wehrmacher

As for "absolutely beautiful" enlargements at any particular DPI value,that depends on how closely you look at the englargements. There is arule of thumb that you should have a pixel density of at least 200 DPIwhen printing, but that applies to prints that you hold about as closeas is reasonable to examine them - about 10 inches from your eye. Thus,for tiny prints up through 5x7 and maybe 8x10, you view the prints atthe same distance and you want about 200 DPI.But a 16x20 print is often viewed from further than 10 inches. If, infact, you view a 16x20 from twice the distance of an 8x10, it needs onlyhalf the DPI to have *the same* resolution at the eye. And if you areviewing your 36x44 inch print from 30 inches or more, you only needabout 67 DPI for an equal impression of sharpness.Looked at another way, if your viewing distance is proportional to printwidth, the same number of pixels will look equally sharp at any size,no matter what the DPI value.On the other hand, if you're looking at your large print from 10 inchesaway (people might do this if it's a panorama with a very wide field ofview), then you still need 200 DPI or so, and 72 DPI would look soft. Dave

The ocular lens provides additional magnification and is adjustable. Users can turn a knob or move the binocular lenses (on microscopes with two eyepieces), ...

Around here we usually forget something very basic: Every time you doublethe viewing distance, you double the *effective* resolution. Not directedat you, Jim or Jim, but I guess there is often little interest ineffectiveness when this issue comes up. ;-)

This is simplistic as it assumes a b&w grid and a sensor in which everypixel is used, not your typical case, as well as a lot of othersimplifications but I hope it illustrates the relationship of pixels,PPI, DPI and LP@MM. I hope that helps, and I truly hope I have not lost a decimal point ormade some other stupid error... ;-)GodfreyIn article <3E15142F...@integraonline.com>, Bill Wehrmacher

You dropped a power of 10. 8 x 10 x 700^2 is 39.2 megapixels, not 4.However, the 10 lp/mm figure is higher than what is recommended bythe sources I've seen. Recommendations for sharp prints range from 4-8lp/mm. At 4 lp/mm minimum, you need about 285 PPI, which means 6.5megapixels for an 8x10. Dave

A pixel is a picture element, each one comprised of Hue, Saturation andValue. "Pixels Per Inch" is a natural way to measure digital imagespatial resolution. Lines per Millimeter, or more accurately, Line Pairs per Millimeter(lp@mm) is a measure of resolution, most useful in analog imaging.Basically, it says some number of black/white lines fitted into amillimeter measure can be discerned, usually by eye or with some kindof optical sensor... In a digital world, tricks of discretemathematical entities make it less useful as a measure of resolutionalthough it's ultimately pretty simple if you don't consider the hardcases of "partial pairs per millimeter" .. If you can discern a whiteand a black space in the space of a millimeter, that's 2 lines permillimeter or 1 lp@mm. (The reason LP@MM is more accurate is that youcannot discern lines as distinct entities unless they alternate incolor, white/black, it is also easier to consider the alternating linesto be all the same width.) Dots per inch is an measure of spatial resolution used in reference tomarking engines (printers) which lay down ink in discrete spots. Hue,Saturation and Value are comprised of multiple dots of inks laid downtogether. Digital cameras make images sized in pixel dimensions, and the physicalsizes of those pixels are related to the specific CCD or CMOS sensoryou have. Taking that pixel array and applying it to an output device,you size the pixels to a ppi standard. DPI is often confused with PPI because the people who wrote the specsheets and manuals were more familiar with printer terminology thanwith digital imaging terminology. Photoshop and other image processingsoftware perpetuates this confusion by presenting image size tools thatexpress PPI values as DPI. Example for illustration (and I ask for pardon for simple math errorsin advance ... ;-): My camera makes an image of 1920x2560 pixels (4.7 Mpixels) and I wantto make a highest quality print on an Epson 1270 printer, which iscapable of 1440x1440 dpi marking resolution. My experience shows thatthe Epson 1270 printer and its driver achieves the best quality imageit can make when it is rendering pixels which are at a 1:4 ratio ofpixel resolution to dots per inch marking capability, so my outputprint file should be set at 360 PPI and the printer set to 1440 ppi.That will achieve a print image size of 5.3x7.1 inches dimension. If the image was a macro photo of a grid at a 1:1 image:subjectmagnification, the finest grid that I could image accurately would beconstrained by the physical size of the CCD sensor in my camera and theactual pixel dimensions. I don't know those dimensions off the top ofmy head, but let's conjecture that the sensor is 12x16mm in size andthe effective pixel area represents the whole area of the sensor. Thatmakes each pixel .00625mm wide. Each pixel in the sensor can only takeon one value at a time, so the minimum width of a black-white line pairhas to be double that or .0125mm. So the sensor is limited to beingable image a grid with a line spacing of 80 LP@MM at most. Imaged as a5.3x7.1 inch print at maximum resolution, that 80LP@MM grid ismagnified approximately 15 times, resulting in a calculated 5.33 LP@MMgrid on paper. This is simplistic as it assumes a b&w grid and a sensor in which everypixel is used, not your typical case, as well as a lot of othersimplifications but I hope it illustrates the relationship of pixels,PPI, DPI and LP@MM. I hope that helps, and I truly hope I have not lost a decimal point ormade some other stupid error... ;-)GodfreyIn article <3E15142F...@integraonline.com>, Bill Wehrmacher

The negative is actually 24x36 mm nominal. That changes the answer to4.2 megapixels - not a very big change.However, assuming 50 lp/mm instead of 25 means you need 16.9megapixels.

McNICHOLS carries one of the largest inventories of Steel and Aluminum Grating as well as Fiberglass Grating and Flooring. Browse all Grating materials and ...

>I understand what Pixels Per Inch means. I thought I understood what >lpmm implied, but now I am not sure. I assumed for a long time that 25 >lpmm would imply something like 25 PPI. Another reference I read >convinced me that in order to see a line, you need a space next to it, >therefore 25lpmm would translate to about 50PPI. Any truth here?>

But that math doesn't "say otherwise". The math says that what you seedepends on pixels per degree of angle, not pixels per inch. If you'reviewing something from 10 inches (a pretty standard assumption for 8x10and smaller size prints), then you need 200-400 PPI at the print for asharp-appearing image. But a 44 inch print *viewed from 44 inches*needs only 45-90 PPI for the same apparent sharpness.The math says you need the same number of pixels for the same visualangle. The print can be any size, as long as you maintain the sameratio of viewing distance to image width. On the other hand, if you areusing the larger print to cover a larger visual angle (e.g. the 6 footprint viewed from 15 inches), you need more pixels. Dave

A line pair is a spacial cycle so you could think of lp/mm likeelectrical cycles per second.Nyquist sampling theorem applies to both: two digital samples percycle are required. In actual practice, about 2.8 to 3.0 samples percycle are required for good quality. A pixel is a digital picturesample.The 35 mm film negative that you have processed at Walmart willprobably result in about 25 lp/mm resolution at best. 25 lp/mm X 2.8 samples/lp = 70 samples/mmor 70 pixels/mm which is 1,778 samples/inch.If the active area on the 35 mm negative is about 28 mm X 35 mm, howmany overall pixels are required to preserve the assumed 25 lp/mm?(28 mm X 25 lp/mm X 2.8 pixels/lp) X (35 mm X 25 lp/mm X 2.8pixels/lp) = 4,802,000 pixels.For prints, anything more than about 10 lp/mm is waisted on theunaided human eye. That is a little over 700 pixels per inch in bothdimensions. For an 8 X 10 print, this is about 4 megapixels overall.

I understand what Pixels Per Inch means. I thought I understood what lpmm implied, but now I am not sure. I assumed for a long time that 25 lpmm would imply something like 25 PPI. Another reference I read convinced me that in order to see a line, you need a space next to it, therefore 25lpmm would translate to about 50PPI. Any truth here?Finally, I wonder about DPI. It was explained to me at one time that Dots Per Inch is a printing term and that it takes about 4 dots to make one pixel. As a result, to print 300 PPI, one would need a printer that provides at least 1200 DPI. Any truth here?In some of these threads it seems as if the terms DPI and PPI are used interchangeably. What is the straight story here.I do appreciate all your efforts. I am getting an education that has long escaped me.Thanks,Bill

>I'm sorry, I don't understand any of this!>I have a Minolta D7i 5MP that puts out a file with 72DPI, Now I have>enlargements made at 36x44 inch that are absolutely beautiful. How can>that be?>Check my pbase site. I have also had a couple of panoramics made 1 13">x 6ft.

Line pairs per mmand pixel size

Nyquist sampling theorem applies to both: two digital samples percycle are required. In actual practice, about 2.8 to 3.0 samples percycle are required for good quality. A pixel is a digital picturesample.The 35 mm film negative that you have processed at Walmart willprobably result in about 25 lp/mm resolution at best. 25 lp/mm X 2.8 samples/lp = 70 samples/mmor 70 pixels/mm which is 1,778 samples/inch.If the active area on the 35 mm negative is about 28 mm X 35 mm, howmany overall pixels are required to preserve the assumed 25 lp/mm?(28 mm X 25 lp/mm X 2.8 pixels/lp) X (35 mm X 25 lp/mm X 2.8pixels/lp) = 4,802,000 pixels.For prints, anything more than about 10 lp/mm is waisted on theunaided human eye. That is a little over 700 pixels per inch in bothdimensions. For an 8 X 10 print, this is about 4 megapixels overall.

I *have* seen large prints (about 4x6 feet in size) made fromlarge-format negatives. You can stand back 6 feet and take in the wholescene, but you can also walk up so you're 15 inches away and examine allthe detailed things in the image - and I did do that. A large formatnegative has enough resolution to do that. It doesn't matter whetherthe print was made chemically, or whether the negative was scanned on adrum scanner and then printed on a large-format digital printer.A 6 megapixel image printed 4x6 feet large will look great when viewedfrom 6 feet, but not at 15 inches. It's the viewing distance thatmatters.

>For prints, anything more than about 10 lp/mm is waisted on the>unaided human eye. That is a little over 700 pixels per inch in both>dimensions. For an 8 X 10 print, this is about 4 megapixels overall.

The negative is actually 24x36 mm nominal. That changes the answer to4.2 megapixels - not a very big change.However, assuming 50 lp/mm instead of 25 means you need 16.9megapixels.

On the other hand, if you're looking at your large print from 10 inchesaway (people might do this if it's a panorama with a very wide field ofview), then you still need 200 DPI or so, and 72 DPI would look soft. Dave

If the image was a macro photo of a grid at a 1:1 image:subjectmagnification, the finest grid that I could image accurately would beconstrained by the physical size of the CCD sensor in my camera and theactual pixel dimensions. I don't know those dimensions off the top ofmy head, but let's conjecture that the sensor is 12x16mm in size andthe effective pixel area represents the whole area of the sensor. Thatmakes each pixel .00625mm wide. Each pixel in the sensor can only takeon one value at a time, so the minimum width of a black-white line pairhas to be double that or .0125mm. So the sensor is limited to beingable image a grid with a line spacing of 80 LP@MM at most. Imaged as a5.3x7.1 inch print at maximum resolution, that 80LP@MM grid ismagnified approximately 15 times, resulting in a calculated 5.33 LP@MMgrid on paper. This is simplistic as it assumes a b&w grid and a sensor in which everypixel is used, not your typical case, as well as a lot of othersimplifications but I hope it illustrates the relationship of pixels,PPI, DPI and LP@MM. I hope that helps, and I truly hope I have not lost a decimal point ormade some other stupid error... ;-)GodfreyIn article <3E15142F...@integraonline.com>, Bill Wehrmacher

Lines per Millimeter, or more accurately, Line Pairs per Millimeter(lp@mm) is a measure of resolution, most useful in analog imaging.Basically, it says some number of black/white lines fitted into amillimeter measure can be discerned, usually by eye or with some kindof optical sensor... In a digital world, tricks of discretemathematical entities make it less useful as a measure of resolutionalthough it's ultimately pretty simple if you don't consider the hardcases of "partial pairs per millimeter" .. If you can discern a whiteand a black space in the space of a millimeter, that's 2 lines permillimeter or 1 lp@mm. (The reason LP@MM is more accurate is that youcannot discern lines as distinct entities unless they alternate incolor, white/black, it is also easier to consider the alternating linesto be all the same width.) Dots per inch is an measure of spatial resolution used in reference tomarking engines (printers) which lay down ink in discrete spots. Hue,Saturation and Value are comprised of multiple dots of inks laid downtogether. Digital cameras make images sized in pixel dimensions, and the physicalsizes of those pixels are related to the specific CCD or CMOS sensoryou have. Taking that pixel array and applying it to an output device,you size the pixels to a ppi standard. DPI is often confused with PPI because the people who wrote the specsheets and manuals were more familiar with printer terminology thanwith digital imaging terminology. Photoshop and other image processingsoftware perpetuates this confusion by presenting image size tools thatexpress PPI values as DPI. Example for illustration (and I ask for pardon for simple math errorsin advance ... ;-): My camera makes an image of 1920x2560 pixels (4.7 Mpixels) and I wantto make a highest quality print on an Epson 1270 printer, which iscapable of 1440x1440 dpi marking resolution. My experience shows thatthe Epson 1270 printer and its driver achieves the best quality imageit can make when it is rendering pixels which are at a 1:4 ratio ofpixel resolution to dots per inch marking capability, so my outputprint file should be set at 360 PPI and the printer set to 1440 ppi.That will achieve a print image size of 5.3x7.1 inches dimension. If the image was a macro photo of a grid at a 1:1 image:subjectmagnification, the finest grid that I could image accurately would beconstrained by the physical size of the CCD sensor in my camera and theactual pixel dimensions. I don't know those dimensions off the top ofmy head, but let's conjecture that the sensor is 12x16mm in size andthe effective pixel area represents the whole area of the sensor. Thatmakes each pixel .00625mm wide. Each pixel in the sensor can only takeon one value at a time, so the minimum width of a black-white line pairhas to be double that or .0125mm. So the sensor is limited to beingable image a grid with a line spacing of 80 LP@MM at most. Imaged as a5.3x7.1 inch print at maximum resolution, that 80LP@MM grid ismagnified approximately 15 times, resulting in a calculated 5.33 LP@MMgrid on paper. This is simplistic as it assumes a b&w grid and a sensor in which everypixel is used, not your typical case, as well as a lot of othersimplifications but I hope it illustrates the relationship of pixels,PPI, DPI and LP@MM. I hope that helps, and I truly hope I have not lost a decimal point ormade some other stupid error... ;-)GodfreyIn article <3E15142F...@integraonline.com>, Bill Wehrmacher

I have just joined this NG and thought with all the conversations about National Geographic someone could help me with these concepts.I understand what Pixels Per Inch means. I thought I understood what lpmm implied, but now I am not sure. I assumed for a long time that 25 lpmm would imply something like 25 PPI. Another reference I read convinced me that in order to see a line, you need a space next to it, therefore 25lpmm would translate to about 50PPI. Any truth here?Finally, I wonder about DPI. It was explained to me at one time that Dots Per Inch is a printing term and that it takes about 4 dots to make one pixel. As a result, to print 300 PPI, one would need a printer that provides at least 1200 DPI. Any truth here?In some of these threads it seems as if the terms DPI and PPI are used interchangeably. What is the straight story here.I do appreciate all your efforts. I am getting an education that has long escaped me.Thanks,Bill

> All I am saying is you can get great enlargements with digital>cameras and the newer high resolution printers, even though the math>says otherwise.

Because you are not trying to pick out details from 2 inches away.a 4 mpixel picture scales quite well if you consider that theperson looking at it will usually be at a larger distance if theprint is huge.-- To get blacklisted please mail to lis...@listme.dsbl.orgWriting to the above address will blacklist your mailserver.Hvis du skriver til ovenstående e-mail bliver din mailserver blacklistet.

> All I am saying is you can get great enlargements with digital>cameras and the newer high resolution printers, even though the math>says otherwise.

>If the active area on the 35 mm negative is about 28 mm X 35 mm, how>many overall pixels are required to preserve the assumed 25 lp/mm?

The negative is actually 24x36 mm nominal. That changes the answer to4.2 megapixels - not a very big change.However, assuming 50 lp/mm instead of 25 means you need 16.9megapixels.

In some of these threads it seems as if the terms DPI and PPI are used interchangeably. What is the straight story here.I do appreciate all your efforts. I am getting an education that has long escaped me.Thanks,Bill

Line pairs per mmcalculator

Congratulations!Ian--Digital Photography NowUK-based digital photo Web magazinehttp://www.dp-now.com"Godfrey DiGiorgi" wrote in messagenews:020120032143518766%rama...@bayarea.net...

Line pairs per mmradiology

Choose from a credit, refund, or exchange. Order. Start Quick Order Curated Classroom Collections Electronic Catalog. Quote. How to ...

The negative is actually 24x36 mm nominal. That changes the answer to4.2 megapixels - not a very big change.However, assuming 50 lp/mm instead of 25 means you need 16.9megapixels.

Linepair resolution

Microscope calculations are a range of formulas used for digital microscopy applications to calculate the depth of field in microscope, field.

The negative is actually 24x36 mm nominal. That changes the answer to4.2 megapixels - not a very big change.However, assuming 50 lp/mm instead of 25 means you need 16.9megapixels.

lp/mm to pixel size

The math says you need the same number of pixels for the same visualangle. The print can be any size, as long as you maintain the sameratio of viewing distance to image width. On the other hand, if you areusing the larger print to cover a larger visual angle (e.g. the 6 footprint viewed from 15 inches), you need more pixels. Dave

>If the active area on the 35 mm negative is about 28 mm X 35 mm, how>many overall pixels are required to preserve the assumed 25 lp/mm?

Looked at another way, if your viewing distance is proportional to printwidth, the same number of pixels will look equally sharp at any size,no matter what the DPI value.On the other hand, if you're looking at your large print from 10 inchesaway (people might do this if it's a panorama with a very wide field ofview), then you still need 200 DPI or so, and 72 DPI would look soft. Dave

25 lp/mm X 2.8 samples/lp = 70 samples/mmor 70 pixels/mm which is 1,778 samples/inch.If the active area on the 35 mm negative is about 28 mm X 35 mm, howmany overall pixels are required to preserve the assumed 25 lp/mm?(28 mm X 25 lp/mm X 2.8 pixels/lp) X (35 mm X 25 lp/mm X 2.8pixels/lp) = 4,802,000 pixels.For prints, anything more than about 10 lp/mm is waisted on theunaided human eye. That is a little over 700 pixels per inch in bothdimensions. For an 8 X 10 print, this is about 4 megapixels overall.

300dpi from your D7i will give you an image around 9x7 inches, whichwill appear 'flawless' with close inspection, let's say a reasonable10 inch viewing distance. If you expand the image to your 44x36 inchsize, that's only about 60dpi, so you will need to use a viewingdistance of 50 inches to achieve the same perceived resolution interms of the 'circle of confusion' angle subtended at the eye. Now,you might consider that it's perfectly reasonable to judge the qualityof a 44x36 print from a distance of 4 feet, rather than sticking yournose on it, and I'm not going to argue with that at all.-- Stewart Pinkerton | Music is Art - Audio is Engineering

Dice Sizes Explained · 5mm Dice (3/16") · 8mm Dice (1/3") · 12mm Dice (1/2") · 16mm Dice (2/3") · 19mm Dice (3/4") · 25mm Dice (1") · 50mm Dice (2") · Footer ...

There's not much Wal-mart can do to reduce negative resolution. That isdetermined pretty much entirely by the film you chose, the quality ofthe lens, whether you used a tripod, the shutter speed, whether theautofocus worked, and so on. However, most any film and half-decentlens ought to be capable of 50 lp/mm at least. 25 lp/mm is a verypessimistic estimate of the film resolution. (On the other hand, theprint could be awful - that depends on Wal-mart's enlarger).

-- To get blacklisted please mail to lis...@listme.dsbl.orgWriting to the above address will blacklist your mailserver.Hvis du skriver til ovenstående e-mail bliver din mailserver blacklistet.

Finally, I wonder about DPI. It was explained to me at one time that Dots Per Inch is a printing term and that it takes about 4 dots to make one pixel. As a result, to print 300 PPI, one would need a printer that provides at least 1200 DPI. Any truth here?In some of these threads it seems as if the terms DPI and PPI are used interchangeably. What is the straight story here.I do appreciate all your efforts. I am getting an education that has long escaped me.Thanks,Bill

>If the active area on the 35 mm negative is about 28 mm X 35 mm, how>many overall pixels are required to preserve the assumed 25 lp/mm?

For prints, anything more than about 10 lp/mm is waisted on theunaided human eye. That is a little over 700 pixels per inch in bothdimensions. For an 8 X 10 print, this is about 4 megapixels overall.

However, the 10 lp/mm figure is higher than what is recommended bythe sources I've seen. Recommendations for sharp prints range from 4-8lp/mm. At 4 lp/mm minimum, you need about 285 PPI, which means 6.5megapixels for an 8x10. Dave

A lot of this is down to viewing distance. Pros insist on 300dpibecause their work must withstand very close inspection and stillappear flawless.300dpi from your D7i will give you an image around 9x7 inches, whichwill appear 'flawless' with close inspection, let's say a reasonable10 inch viewing distance. If you expand the image to your 44x36 inchsize, that's only about 60dpi, so you will need to use a viewingdistance of 50 inches to achieve the same perceived resolution interms of the 'circle of confusion' angle subtended at the eye. Now,you might consider that it's perfectly reasonable to judge the qualityof a 44x36 print from a distance of 4 feet, rather than sticking yournose on it, and I'm not going to argue with that at all.-- Stewart Pinkerton | Music is Art - Audio is Engineering

The negative is actually 24x36 mm nominal. That changes the answer to4.2 megapixels - not a very big change.However, assuming 50 lp/mm instead of 25 means you need 16.9megapixels.

Ian--Digital Photography NowUK-based digital photo Web magazinehttp://www.dp-now.com"Godfrey DiGiorgi" wrote in messagenews:020120032143518766%rama...@bayarea.net...

But a 16x20 print is often viewed from further than 10 inches. If, infact, you view a 16x20 from twice the distance of an 8x10, it needs onlyhalf the DPI to have *the same* resolution at the eye. And if you areviewing your 36x44 inch print from 30 inches or more, you only needabout 67 DPI for an equal impression of sharpness.Looked at another way, if your viewing distance is proportional to printwidth, the same number of pixels will look equally sharp at any size,no matter what the DPI value.On the other hand, if you're looking at your large print from 10 inchesaway (people might do this if it's a panorama with a very wide field ofview), then you still need 200 DPI or so, and 72 DPI would look soft. Dave

Glasses. $59.95. These matte black comfortable form-fitting glasses are perfect for any shooting conditions. Super light design only touches the face at 3 ...

Example for illustration (and I ask for pardon for simple math errorsin advance ... ;-): My camera makes an image of 1920x2560 pixels (4.7 Mpixels) and I wantto make a highest quality print on an Epson 1270 printer, which iscapable of 1440x1440 dpi marking resolution. My experience shows thatthe Epson 1270 printer and its driver achieves the best quality imageit can make when it is rendering pixels which are at a 1:4 ratio ofpixel resolution to dots per inch marking capability, so my outputprint file should be set at 360 PPI and the printer set to 1440 ppi.That will achieve a print image size of 5.3x7.1 inches dimension. If the image was a macro photo of a grid at a 1:1 image:subjectmagnification, the finest grid that I could image accurately would beconstrained by the physical size of the CCD sensor in my camera and theactual pixel dimensions. I don't know those dimensions off the top ofmy head, but let's conjecture that the sensor is 12x16mm in size andthe effective pixel area represents the whole area of the sensor. Thatmakes each pixel .00625mm wide. Each pixel in the sensor can only takeon one value at a time, so the minimum width of a black-white line pairhas to be double that or .0125mm. So the sensor is limited to beingable image a grid with a line spacing of 80 LP@MM at most. Imaged as a5.3x7.1 inch print at maximum resolution, that 80LP@MM grid ismagnified approximately 15 times, resulting in a calculated 5.33 LP@MMgrid on paper. This is simplistic as it assumes a b&w grid and a sensor in which everypixel is used, not your typical case, as well as a lot of othersimplifications but I hope it illustrates the relationship of pixels,PPI, DPI and LP@MM. I hope that helps, and I truly hope I have not lost a decimal point ormade some other stupid error... ;-)GodfreyIn article <3E15142F...@integraonline.com>, Bill Wehrmacher

First, your camera does not put out a "72 DPI" file, it just produces afile with a certain number of pixels. The "72 DPI" value is justPhotoshop's default when you read in a file that lacks a DPI figure inthe file header. You can change the "72" to any other value you want.This changes the printed image size without changing the number ofpixels (as long as the "resample image" box is unchecked).As for "absolutely beautiful" enlargements at any particular DPI value,that depends on how closely you look at the englargements. There is arule of thumb that you should have a pixel density of at least 200 DPIwhen printing, but that applies to prints that you hold about as closeas is reasonable to examine them - about 10 inches from your eye. Thus,for tiny prints up through 5x7 and maybe 8x10, you view the prints atthe same distance and you want about 200 DPI.But a 16x20 print is often viewed from further than 10 inches. If, infact, you view a 16x20 from twice the distance of an 8x10, it needs onlyhalf the DPI to have *the same* resolution at the eye. And if you areviewing your 36x44 inch print from 30 inches or more, you only needabout 67 DPI for an equal impression of sharpness.Looked at another way, if your viewing distance is proportional to printwidth, the same number of pixels will look equally sharp at any size,no matter what the DPI value.On the other hand, if you're looking at your large print from 10 inchesaway (people might do this if it's a panorama with a very wide field ofview), then you still need 200 DPI or so, and 72 DPI would look soft. Dave

>If the active area on the 35 mm negative is about 28 mm X 35 mm, how>many overall pixels are required to preserve the assumed 25 lp/mm?

I am assuming that you are talking about your run-of-the-mill compound light microscope. The iris diaphragm is found in the condenser, ...

Line pairs per mmconverter

>If the active area on the 35 mm negative is about 28 mm X 35 mm, how>many overall pixels are required to preserve the assumed 25 lp/mm?

--Digital Photography NowUK-based digital photo Web magazinehttp://www.dp-now.com"Godfrey DiGiorgi" wrote in messagenews:020120032143518766%rama...@bayarea.net...

It simplifies leveling work by easily adjusting your laser to virtually any height, making it ideal for hanging drywall, shelving or cabinets, laying tile, ...

The negative is actually 24x36 mm nominal. That changes the answer to4.2 megapixels - not a very big change.However, assuming 50 lp/mm instead of 25 means you need 16.9megapixels.

> All I am saying is you can get great enlargements with digital>cameras and the newer high resolution printers, even though the math>says otherwise.

a 4 mpixel picture scales quite well if you consider that theperson looking at it will usually be at a larger distance if theprint is huge.-- To get blacklisted please mail to lis...@listme.dsbl.orgWriting to the above address will blacklist your mailserver.Hvis du skriver til ovenstående e-mail bliver din mailserver blacklistet.

I hope that helps, and I truly hope I have not lost a decimal point ormade some other stupid error... ;-)GodfreyIn article <3E15142F...@integraonline.com>, Bill Wehrmacher

>If the active area on the 35 mm negative is about 28 mm X 35 mm, how>many overall pixels are required to preserve the assumed 25 lp/mm?

Length: 450cm (4.5m) Cotton Stretch Elastic ... Bermuda (USD $); Bhutan (AUD $); Bolivia (BOB Bs ... South Korea (KRW ₩); South Sudan (AUD $); Spain ...

Digital cameras make images sized in pixel dimensions, and the physicalsizes of those pixels are related to the specific CCD or CMOS sensoryou have. Taking that pixel array and applying it to an output device,you size the pixels to a ppi standard. DPI is often confused with PPI because the people who wrote the specsheets and manuals were more familiar with printer terminology thanwith digital imaging terminology. Photoshop and other image processingsoftware perpetuates this confusion by presenting image size tools thatexpress PPI values as DPI. Example for illustration (and I ask for pardon for simple math errorsin advance ... ;-): My camera makes an image of 1920x2560 pixels (4.7 Mpixels) and I wantto make a highest quality print on an Epson 1270 printer, which iscapable of 1440x1440 dpi marking resolution. My experience shows thatthe Epson 1270 printer and its driver achieves the best quality imageit can make when it is rendering pixels which are at a 1:4 ratio ofpixel resolution to dots per inch marking capability, so my outputprint file should be set at 360 PPI and the printer set to 1440 ppi.That will achieve a print image size of 5.3x7.1 inches dimension. If the image was a macro photo of a grid at a 1:1 image:subjectmagnification, the finest grid that I could image accurately would beconstrained by the physical size of the CCD sensor in my camera and theactual pixel dimensions. I don't know those dimensions off the top ofmy head, but let's conjecture that the sensor is 12x16mm in size andthe effective pixel area represents the whole area of the sensor. Thatmakes each pixel .00625mm wide. Each pixel in the sensor can only takeon one value at a time, so the minimum width of a black-white line pairhas to be double that or .0125mm. So the sensor is limited to beingable image a grid with a line spacing of 80 LP@MM at most. Imaged as a5.3x7.1 inch print at maximum resolution, that 80LP@MM grid ismagnified approximately 15 times, resulting in a calculated 5.33 LP@MMgrid on paper. This is simplistic as it assumes a b&w grid and a sensor in which everypixel is used, not your typical case, as well as a lot of othersimplifications but I hope it illustrates the relationship of pixels,PPI, DPI and LP@MM. I hope that helps, and I truly hope I have not lost a decimal point ormade some other stupid error... ;-)GodfreyIn article <3E15142F...@integraonline.com>, Bill Wehrmacher

lp/mm to resolution

>If the active area on the 35 mm negative is about 28 mm X 35 mm, how>many overall pixels are required to preserve the assumed 25 lp/mm?

>Nyquist sampling theorem applies to both: two digital samples per>cycle are required. In actual practice, about 2.8 to 3.0 samples per>cycle are required for good quality. A pixel is a digital picture>sample.

If the active area on the 35 mm negative is about 28 mm X 35 mm, howmany overall pixels are required to preserve the assumed 25 lp/mm?(28 mm X 25 lp/mm X 2.8 pixels/lp) X (35 mm X 25 lp/mm X 2.8pixels/lp) = 4,802,000 pixels.For prints, anything more than about 10 lp/mm is waisted on theunaided human eye. That is a little over 700 pixels per inch in bothdimensions. For an 8 X 10 print, this is about 4 megapixels overall.

(28 mm X 25 lp/mm X 2.8 pixels/lp) X (35 mm X 25 lp/mm X 2.8pixels/lp) = 4,802,000 pixels.For prints, anything more than about 10 lp/mm is waisted on theunaided human eye. That is a little over 700 pixels per inch in bothdimensions. For an 8 X 10 print, this is about 4 megapixels overall.

You don't look at the whole 36x44 from 10 inches away - you look atparts of it, in exactly the same way that when you drive over the top ofa hill and see a new chunk of landscape you look at it with a sequenceof glances at different parts of it, or the way you gradually look atthe contents of a room that you've never been in before.I *have* seen large prints (about 4x6 feet in size) made fromlarge-format negatives. You can stand back 6 feet and take in the wholescene, but you can also walk up so you're 15 inches away and examine allthe detailed things in the image - and I did do that. A large formatnegative has enough resolution to do that. It doesn't matter whetherthe print was made chemically, or whether the negative was scanned on adrum scanner and then printed on a large-format digital printer.A 6 megapixel image printed 4x6 feet large will look great when viewedfrom 6 feet, but not at 15 inches. It's the viewing distance thatmatters.

lp/mm calculator

DPI is often confused with PPI because the people who wrote the specsheets and manuals were more familiar with printer terminology thanwith digital imaging terminology. Photoshop and other image processingsoftware perpetuates this confusion by presenting image size tools thatexpress PPI values as DPI. Example for illustration (and I ask for pardon for simple math errorsin advance ... ;-): My camera makes an image of 1920x2560 pixels (4.7 Mpixels) and I wantto make a highest quality print on an Epson 1270 printer, which iscapable of 1440x1440 dpi marking resolution. My experience shows thatthe Epson 1270 printer and its driver achieves the best quality imageit can make when it is rendering pixels which are at a 1:4 ratio ofpixel resolution to dots per inch marking capability, so my outputprint file should be set at 360 PPI and the printer set to 1440 ppi.That will achieve a print image size of 5.3x7.1 inches dimension. If the image was a macro photo of a grid at a 1:1 image:subjectmagnification, the finest grid that I could image accurately would beconstrained by the physical size of the CCD sensor in my camera and theactual pixel dimensions. I don't know those dimensions off the top ofmy head, but let's conjecture that the sensor is 12x16mm in size andthe effective pixel area represents the whole area of the sensor. Thatmakes each pixel .00625mm wide. Each pixel in the sensor can only takeon one value at a time, so the minimum width of a black-white line pairhas to be double that or .0125mm. So the sensor is limited to beingable image a grid with a line spacing of 80 LP@MM at most. Imaged as a5.3x7.1 inch print at maximum resolution, that 80LP@MM grid ismagnified approximately 15 times, resulting in a calculated 5.33 LP@MMgrid on paper. This is simplistic as it assumes a b&w grid and a sensor in which everypixel is used, not your typical case, as well as a lot of othersimplifications but I hope it illustrates the relationship of pixels,PPI, DPI and LP@MM. I hope that helps, and I truly hope I have not lost a decimal point ormade some other stupid error... ;-)GodfreyIn article <3E15142F...@integraonline.com>, Bill Wehrmacher

The 35 mm film negative that you have processed at Walmart willprobably result in about 25 lp/mm resolution at best. 25 lp/mm X 2.8 samples/lp = 70 samples/mmor 70 pixels/mm which is 1,778 samples/inch.If the active area on the 35 mm negative is about 28 mm X 35 mm, howmany overall pixels are required to preserve the assumed 25 lp/mm?(28 mm X 25 lp/mm X 2.8 pixels/lp) X (35 mm X 25 lp/mm X 2.8pixels/lp) = 4,802,000 pixels.For prints, anything more than about 10 lp/mm is waisted on theunaided human eye. That is a little over 700 pixels per inch in bothdimensions. For an 8 X 10 print, this is about 4 megapixels overall.

A 6 megapixel image printed 4x6 feet large will look great when viewedfrom 6 feet, but not at 15 inches. It's the viewing distance thatmatters.

My camera makes an image of 1920x2560 pixels (4.7 Mpixels) and I wantto make a highest quality print on an Epson 1270 printer, which iscapable of 1440x1440 dpi marking resolution. My experience shows thatthe Epson 1270 printer and its driver achieves the best quality imageit can make when it is rendering pixels which are at a 1:4 ratio ofpixel resolution to dots per inch marking capability, so my outputprint file should be set at 360 PPI and the printer set to 1440 ppi.That will achieve a print image size of 5.3x7.1 inches dimension. If the image was a macro photo of a grid at a 1:1 image:subjectmagnification, the finest grid that I could image accurately would beconstrained by the physical size of the CCD sensor in my camera and theactual pixel dimensions. I don't know those dimensions off the top ofmy head, but let's conjecture that the sensor is 12x16mm in size andthe effective pixel area represents the whole area of the sensor. Thatmakes each pixel .00625mm wide. Each pixel in the sensor can only takeon one value at a time, so the minimum width of a black-white line pairhas to be double that or .0125mm. So the sensor is limited to beingable image a grid with a line spacing of 80 LP@MM at most. Imaged as a5.3x7.1 inch print at maximum resolution, that 80LP@MM grid ismagnified approximately 15 times, resulting in a calculated 5.33 LP@MMgrid on paper. This is simplistic as it assumes a b&w grid and a sensor in which everypixel is used, not your typical case, as well as a lot of othersimplifications but I hope it illustrates the relationship of pixels,PPI, DPI and LP@MM. I hope that helps, and I truly hope I have not lost a decimal point ormade some other stupid error... ;-)GodfreyIn article <3E15142F...@integraonline.com>, Bill Wehrmacher

> All I am saying is you can get great enlargements with digital>cameras and the newer high resolution printers, even though the math>says otherwise.

The negative is actually 24x36 mm nominal. That changes the answer to4.2 megapixels - not a very big change.However, assuming 50 lp/mm instead of 25 means you need 16.9megapixels.

... lens. It's just ... reading of what's on them with pancake lenses. With fresnel, your whole head often needs to move to read them clearly.