2 Pro One of my clients want a 36''x48''

[Mario McFly]

Too bad if I look newwbie wich I am anyway lolll I need to know if it is possible to print a 36'' x 48'' or larger or what is the maximum size to keep good definition at 6-8 feet distance. Drone use is the M2P 20 mega pixel camera . or maybe with the Air 2 and the 48 mega pixel(not a real one I know ) camera. I need to know how this maximum print size work. Thank you and Happy new year to all ???????

 

[Hummingbird.UAV]

You should shoot the image in PANO mode. The stitched image will have many more pixels that will result in a better enlargement.

 

[Mario McFly]

You should shoot the image in PANO mode. The stitched image will have many more pixels that will result in a better enlargement.

Thanks for the tip but will this method apply for house photos ?

 

[vindibona1]

A GOOD professional photo lab has incredible interpolation software that can keep decent detail even with files with 72dpi resolution. But as @Hummingbird.UAV said, take the image in the pano mode. You can crop it later if need be. I just pulled up a jpg Pano from my files and @300 dpi the uninterpolated size/resolution is 90x29 @ 300dpi.

 

[Hummingbird.UAV]

Thanks for the tip but will this method apply for house photos ?

I think so. It would require testing to position the drone to get the house in all of the pano shots.

 

[Mario McFly]

A GOOD professional photo lab has incredible interpolation software that can keep decent detail even with files with 72dpi resolution. But as @Hummingbird.UAV said, take the image in the pano mode. You can crop it later if need be. I just pulled up a jpg Pano from my files and @300 dpi the uninterpolated size/resolution is 90x29 @ 300dpi.

With covid all photo lab around here are close I will wait and check with them when they reopen . Thank You!

 

[vindibona1]

Thanks for the tip but will this method apply for house photos ?

In regard to what I said a moment ago, here is a sample of a pano and cropping from that pano. I could have been much closer to the house to begin with but I wanted a pano with the neighborhood. Yet still there is a lot of detail considering how far away I was. This was taken with a M2P.

[Note, the file sizes are reduced for the purposes of posting them here]

 

[DanMan32]

I never understood how dpi could be applied to a digital image file. There's no "inch" until you print or display it, and the print/display can vary in size for the same pixel resolution.

 

[vindibona1]

I never understood how dpi could be applied to a digital image file. There's no "inch" until you print or display it, and the print/display can vary in size for the same pixel resolution.

I’m not sure where to begin. The answer is simple and complex at the same time. A digital image file can be presented in a one dimension equation, a two dimensional equation or a three dimension equation.

The one dimension equation is represented in the file size, represented as bites, megabytes and sometimes gigabytes. When we break the file size (the total amount of bites) into a two dimension equation you will see an image file displayed as pixels. An example would be a the one dimensional 5.93 megabite file would be represented as 1920×1080 in pixels. It's in pixels because a resolution hasn't been assigned to it as of yet, so in 2 dimensions we haven't taken it to an "inch" representation. So typically, when you see that two dimensional equation representing the image file we assume the third dimension as 72 dpi. [We usually use 72dpi resolution as the base because it represents screen resolution.] And so given that the tacet 3rd part of the equation is 72dpi. To get inches you divide the pixels by 72 which gives you inches. 26.667" x 15" @72dpi. "DPI" stands for dots per inch sometimes notated as PPI (pixels per inch), nominally the same thing. So while inches aren't stated when you see only 1920 x 1080 pixels, inches are implied because the defacto and assumed ingredient that hasn't been stated yet is that 72dpi.

The requirements for good quality prints and electronic viewing are different. The standard resolution for print is usually 300 dpi but can often get by with lower resolution when we need a bigger print size but have that one file size to work with. I stated earlier 72dpi is often, if not usually used, but sometimes 96dpi is used for screen resolution. That doesn't matter. What matters is that the 3 dimensional equtions is fluid with the document size getting smaller or larger depending the designated resolution. Regardless of the resolution one and two dimensional representations of that file size stay the same. We can "resample" the resolution, which can interpolate the file to a larger or smaller file size and corresponding pixel depth. But we're still working with the same number of starting pixels. It's usually easier to interpolate down to a smaller file size than go up.

In the 1st illustration below it shows files size (one dimension), pixel width and height (2x) and document size (3x) @ 72dpi. In the second illustration we change the resolution from 72dpi to 150 dpi. You will note that the file size and pixel dimension stay the same. It's the same in the third illustration when we change the resolution to 300dpi. You can see that when you increase the resolution you proportionately decrease the "document" size.

To to answer your question directly, You divide the pixels by the resolution to get inches because that third dimension, the resolution is the key to converting pixels to inches.

Note below that in the sample illustrations below, as the resolution changes so does the document size, but file size and pixel representations stay the same.



I've tried to be as clear as possible a 2am. Just look at the examples and work through the math and it should be understandable.

 

[Mario McFly]

In regard to what I said a moment ago, here is a sample of a pano and cropping from that pano...

[Note, the file sizes are reduced for the purposes of posting them here]
View attachment 120504

View attachment 120505
View attachment 120506

Oh yeah that would work perfect thanks !

I’m not sure where to begin. The answer is simple and complex at the same time. A digital image file can be presented in a one dimension equation, a two dimensional equation or a three dimension equation.

The one dimension equation is represented in the file size, represented as bites, megabytes and sometimes gigabytes. When we break the file size (the total amount of bites) into a two dimension equation you will see an image file displayed as pixels. An example would be a the one dimensional 5.93 megabite file would be represented as 1920×1080 in pixels. It's in pixels because a resolution hasn't been assigned to it as of yet, so in 2 dimensions we haven't taken it to an "inch" representation. So typically, when you see that two dimensional equation representing the image file we assume the third dimension as 72 dpi. [We usually use 72dpi resolution as the base because it represents screen resolution.] And so given that the tacet 3rd part of the equation is 72dpi. To get inches you divide the pixels by 72 which gives you inches. 26.667" x 15" @72dpi. "DPI" stands for dots per inch sometimes notated as PPI (pixels per inch), nominally the same thing. So while inches aren't stated when you see only 1920 x 1080 pixels, inches are implied because the defacto and assumed ingredient that hasn't been stated yet is that 72dpi.

The requirements for good quality prints and electronic viewing are different. The standard resolution for print is usually 300 dpi but can often get by with lower resolution when we need a bigger print size but have that one file size to work with. I stated earlier 72dpi is often, if not usually used, but sometimes 96dpi is used for screen resolution. That doesn't matter. What matters is that the 3 dimensional equtions is fluid with the document size getting smaller or larger depending the designated resolution. Regardless of the resolution one and two dimensional representations of that file size stay the same. We can "resample" the resolution, which can interpolate the file to a larger or smaller file size and corresponding pixel depth. But we're still working with the same number of starting pixels. It's usually easier to interpolate down to a smaller file size than go up.

In the 1st illustration below it shows files size (one dimension), pixel width and height (2x) and document size (3x) @ 72dpi. In the second illustration we change the resolution from 72dpi to 150 dpi. You will note that the file size and pixel dimension stay the same. It's the same in the third illustration when we change the resolution to 300dpi. You can see that when you increase the resolution you proportionately decrease the "document" size.

To to answer your question directly, You divide the pixels by the resolution to get inches because that third dimension, the resolution is the key to converting pixels to inches.

Note below that in the sample illustrations below, as the resolution changes so does the document size, but file size and pixel representations stay the same.

View attachment 120552 View attachment 120553
View attachment 120554
I've tried to be as clear as possible a 2am. Just look at the examples and work through the math and it should be understandable.

Wowww thank you so much that will help me out for sure but still i need to sit down and figure that all out but thank for that great write up

 

[DVPOWERS]

There is a company called Topaz labs (www.topazlabs.com) and they offer software to upscale photos without losing detail. I have not used them but many photographers swear by them.

 

[Cymru]

GigaPixel AI upscales pretty well. Photoshop is pretty horrible at upscaling.

So shoot in pano up close for detail then scale up to whatever DPI you want for print in GigaPixel.

You need to work backwards - firstly know what print DPI you want for the size/distance image. From there you can work out the resolution needed.

 

[Mario McFly]

There is a company called Topaz labs (www.topazlabs.com) and they offer software to upscale photos without losing detail. I have not used them but many photographers swear by them.

Thanks for the info

GigaPixel AI upscales pretty well. Photoshop is pretty horrible at upscaling.

So shoot in pano up close for detail then scale up to whatever DPI you want for print in GigaPixel.

I will do that thanks for the tip

 

[Photo Beaty]

I know of two good labs that are still active. Bay Photo in San Francisco and McKenna in Waterloo, Iowa.

 

[Mario McFly]

I know of two good labs that are still active. Bay Photo in San Francisco and McKenna in Waterloo, Iowa.

Thank you but i'm in Canada/Quebec I would like to have one near me

 

[Bsnyder]

I’m not sure where to begin. The answer is simple and complex at the same time. A digital image file can be presented in a one dimension equation, a two dimensional equation or a three dimension equation.

The one dimension equation is represented in the file size, represented as bites, megabytes and sometimes gigabytes. When we break the file size (the total amount of bites) into a two dimension equation you will see an image file displayed as pixels. An example would be a the one dimensional 5.93 megabite file would be represented as 1920×1080 in pixels. It's in pixels because a resolution hasn't been assigned to it as of yet, so in 2 dimensions we haven't taken it to an "inch" representation. So typically, when you see that two dimensional equation representing the image file we assume the third dimension as 72 dpi. [We usually use 72dpi resolution as the base because it represents screen resolution.] And so given that the tacet 3rd part of the equation is 72dpi. To get inches you divide the pixels by 72 which gives you inches. 26.667" x 15" @72dpi. "DPI" stands for dots per inch sometimes notated as PPI (pixels per inch), nominally the same thing. So while inches aren't stated when you see only 1920 x 1080 pixels, inches are implied because the defacto and assumed ingredient that hasn't been stated yet is that 72dpi.

The requirements for good quality prints and electronic viewing are different. The standard resolution for print is usually 300 dpi but can often get by with lower resolution when we need a bigger print size but have that one file size to work with. I stated earlier 72dpi is often, if not usually used, but sometimes 96dpi is used for screen resolution. That doesn't matter. What matters is that the 3 dimensional equtions is fluid with the document size getting smaller or larger depending the designated resolution. Regardless of the resolution one and two dimensional representations of that file size stay the same. We can "resample" the resolution, which can interpolate the file to a larger or smaller file size and corresponding pixel depth. But we're still working with the same number of starting pixels. It's usually easier to interpolate down to a smaller file size than go up.

In the 1st illustration below it shows files size (one dimension), pixel width and height (2x) and document size (3x) @ 72dpi. In the second illustration we change the resolution from 72dpi to 150 dpi. You will note that the file size and pixel dimension stay the same. It's the same in the third illustration when we change the resolution to 300dpi. You can see that when you increase the resolution you proportionately decrease the "document" size.

To to answer your question directly, You divide the pixels by the resolution to get inches because that third dimension, the resolution is the key to converting pixels to inches.

Note below that in the sample illustrations below, as the resolution changes so does the document size, but file size and pixel representations stay the same.

View attachment 120552 View attachment 120553
View attachment 120554
I've tried to be as clear as possible a 2am. Just look at the examples and work through the math and it should be understandable.

Thanks for sharing this information, will be very helpful.

 

[DanMan32]

I’m not sure where to begin. The answer is simple and complex at the same time. A digital image file can be presented in a one dimension equation, a two dimensional equation or a three dimension equation.

The one dimension equation is represented in the file size, represented as bites, megabytes and sometimes gigabytes. When we break the file size (the total amount of bites) into a two dimension equation you will see an image file displayed as pixels. An example would be a the one dimensional 5.93 megabite file would be represented as 1920×1080 in pixels. It's in pixels because a resolution hasn't been assigned to it as of yet, so in 2 dimensions we haven't taken it to an "inch" representation. So typically, when you see that two dimensional equation representing the image file we assume the third dimension as 72 dpi. [We usually use 72dpi resolution as the base because it represents screen resolution.] And so given that the tacet 3rd part of the equation is 72dpi. To get inches you divide the pixels by 72 which gives you inches. 26.667" x 15" @72dpi. "DPI" stands for dots per inch sometimes notated as PPI (pixels per inch), nominally the same thing. So while inches aren't stated when you see only 1920 x 1080 pixels, inches are implied because the defacto and assumed ingredient that hasn't been stated yet is that 72dpi.

The requirements for good quality prints and electronic viewing are different. The standard resolution for print is usually 300 dpi but can often get by with lower resolution when we need a bigger print size but have that one file size to work with. I stated earlier 72dpi is often, if not usually used, but sometimes 96dpi is used for screen resolution. That doesn't matter. What matters is that the 3 dimensional equtions is fluid with the document size getting smaller or larger depending the designated resolution. Regardless of the resolution one and two dimensional representations of that file size stay the same. We can "resample" the resolution, which can interpolate the file to a larger or smaller file size and corresponding pixel depth. But we're still working with the same number of starting pixels. It's usually easier to interpolate down to a smaller file size than go up.

In the 1st illustration below it shows files size (one dimension), pixel width and height (2x) and document size (3x) @ 72dpi. In the second illustration we change the resolution from 72dpi to 150 dpi. You will note that the file size and pixel dimension stay the same. It's the same in the third illustration when we change the resolution to 300dpi. You can see that when you increase the resolution you proportionately decrease the "document" size.

To to answer your question directly, You divide the pixels by the resolution to get inches because that third dimension, the resolution is the key to converting pixels to inches.

Note below that in the sample illustrations below, as the resolution changes so does the document size, but file size and pixel representations stay the same.

View attachment 120552 View attachment 120553
View attachment 120554
I've tried to be as clear as possible a 2am. Just look at the examples and work through the math and it should be understandable.

I definitely agree in dpi when it is applied to physical media, but it is just arbitrary when it is applied to a virtual image.

One can have a 1080p monitor in 5" (smartphones), 13", 20", 27", 65" and so on. Since the resolution is fixed, thus the vertical pixel count, the dpi is variable at the media since the fixed number of pixels is spread out over different distances. So no, one can't assume dpi on a virtual image without specifying the size of the media.

I did calculate though that 75dpi for an image that is 1080p would be 14.4" long.

 

[vindibona1]

There is a company called Topaz labs (www.topazlabs.com) and they offer software to upscale photos without losing detail. I have not used them but many photographers swear by them.

I'm not familiar with Topaz labs or even any upscaling software, but I've not ever needed it. Where it might be useful is for SUPER sized prints where they will be viewed close up. The thing is that GOOD professional labs' printing software can do an outstanding job at interpolating files, even as low as 75dpi or sometimes lower and making a good print. A lot depends on the detail in the print as much as the file size itself. The Achilles Heel of digital imaging has always been DETAIL AT DISTANCE. So, the rule generally is get as close to the subject as possible to keep the maximum detail. This helps mitigate smaller than optimum file size. The Panorama approach really helps because it allows you to get closer than you normally would for the width needed in an image. As a side note, remember that professional labs might have millions of $$ invested in their equipment and software because they use it every day. In all likelihood, for the task of making prints what they have is generally better than what you and I can afford to purchase.

An example of the fine detail at close distance, below I am attaching a shot of a very detailed pocket watch that I photographed in 2001 (I found a web-size representation of the file in my archives below). This was taken with a 2.75 megapixel camera (Nikon D1x, if I remember correctly). I've taken a snapshot of the 20x24 inch print that hangs in my home (photographed with an iPhone that has a wonky camera) with every bit as much detail as you see here. I'm not sure what the actual file size of the print was, but I am certain that the lab had to interpolate it to get it to 20x24 size. Ergo, the closer you are the more detail you can capture regardless of file size.

Edit: Postscript.... In the photo of the framed watch you can see an image of a print of some pearls that I shot a few years later with a 6mp camera, printed as a 24x30 and framed.

 

[RockyMtns]

Really appreciate this discussion of resolution/dpi and image quality!
Thanks everyone..............

 

[vindibona1]

Thank you but i'm in Canada/Quebec I would like to have one near me

I cannot speak to anything up in Canada. I'm sure you can find a good lab somewhere suitable. I can only speak for us "south of the border", but I've used a lab, H&H Color Lab in Kansas City Missouri. I've used them for over 25 years and found them to have both excellent products, excellent quality control and outstanding customer service.