What's the conversion ratio between 8x10, 4x5, 645 and 645 50 MP CMOS Cropped?

[subrata1965]

While selecting the lens, what conversion ratio to apply?

For example: if an image was captured using 300mm lens in 8x10, what should be the equivalent focal length in 645 (60 - 100 MP) and 645 50 MP cropped sensor.

Thanks,

Subrata

 

[mjr]

This is pretty handy subatra

https://digitaltransitions.com/support/lens-visualizer-tools/

Covers all the formats, just add in the focal length you want.

Mat

 

[DougDolde]

I like this one

https://dl.dropboxusercontent.com/u/17811936/ALPA_CFL_Calc_V217B.xls

 

[torger]

The different ratios complicate things slightly, 5:4 for 8x10 and 4x5, 4:3 for 645 and 3:2 for 135.

So when I compare I first think about which aspect ratio I prefer and then crop all formats accordingly (as I would have to crop them in reality). I generally prefer 4:3, so the 135 format of 36x24mm becomes 32x24mm before comparing to 645.

Then you just need to calculate the diagonals using the Pythagorean theorem.

The film sizes has a bit border on the film which can vary a bit so 8x10 is a tad smaller than actual 8x10. Somewhat approximate:

8x10: 5:4 245x196mm (313mm diagonal), 4:3 245x184mm (306mm diagonal)
645 digital: 4:3 54x41mm (67mm diagonal), 5:4 51x41mm (65mm diagonal)

Then just divide diagonals of your prefered aspect ratio. Let's say we prefer 4:3, then the "crop factor" for 645 digital vs 8x10" film becomes 306/67 = 4.6, so a 300mm 8x10" lens correspons to a 300/4.6 = 65mm lens on 645 digital.

You can compare apertures in the same way. f/64 on 8x10 becomes 64/4.6 = f/14 on 645.

Note that in some cases the effective focal length differs quite much from the labeled focal length. That is wide angle lenses are usually not as wide as their labeling, and tele lenses are usually not as long. It's generally not a big issue though as you generally only need to know the approximate match. Likewise adjusting the formats like I did above is also often overkill as it doesn't make that much of a difference. It's more of a difference between 3:2 and 5:4 than 4:3 to 5:4 of course.

 

[Uaiomex]

I have trick that works for me.
If I want to know the coverage or the angle of view for wide views like when using wide angle lenses I only consider the longer sides to do the math. If I need to know the reach like when using telephotos, I only consider the shorter sides or just use the pyth theorem.
Try it, it may be useful to you too.
Eduardo

The different ratios complicate things slightly, 5:4 for 8x10 and 4x5, 4:3 for 645 and 3:2 for 135.

So when I compare I first think about which aspect ratio I prefer and then crop all formats accordingly (as I would have to crop them in reality). I generally prefer 4:3, so the 135 format of 36x24mm becomes 32x24mm before comparing to 645.

Then you just need to calculate the diagonals using the Pythagorean theorem.

The film sizes has a bit border on the film which can vary a bit so 8x10 is a tad smaller than actual 8x10. Somewhat approximate:

8x10: 5:4 245x196mm (313mm diagonal), 4:3 245x184mm (306mm diagonal)
645 digital: 4:3 54x41mm (67mm diagonal), 5:4 51x41mm (65mm diagonal)

Then just divide diagonals of your prefered aspect ratio. Let's say we prefer 4:3, then the "crop factor" for 645 digital vs 8x10" film becomes 306/67 = 4.6, so a 300mm 8x10" lens correspons to a 300/4.6 = 65mm lens on 645 digital.

You can compare apertures in the same way. f/64 on 8x10 becomes 64/4.6 = f/14 on 645.

Note that in some cases the effective focal length differs quite much from the labeled focal length. That is wide angle lenses are usually not as wide as their labeling, and tele lenses are usually not as long. It's generally not a big issue though as you generally only need to know the approximate match. Likewise adjusting the formats like I did above is also often overkill as it doesn't make that much of a difference. It's more of a difference between 3:2 and 5:4 than 4:3 to 5:4 of course.