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02-05-2014, 08:04 AM   #1
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Why do mm count more at the wide end?

I have often read that extra mm are more important on the wide end or are more substantial than on telephoto but I don't really see why.

I made a spreadsheet that calculates the angular field of view and the distance from subject required to have a 2 m field of view (i.e. taking a head to toe picture of an adult) and it is perfectly linear so I don't see why other than if you are backed up to a wall or pressed up to the edge of a cliff or guard rail.

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02-05-2014, 08:10 AM   #2
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A doubling (eg 8 to 16, 50 to 100) is linear, but each additional mm becomes less important.

Look at the progression 8-16-24. A difference of 8mm for each step, but the diagonal FOV decreases by ~40° in the first step, but only ~20° in the second step.
02-05-2014, 08:28 AM   #3
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QuoteOriginally posted by filoxophy Quote
A doubling (eg 8 to 16, 50 to 100) is linear, but each additional mm becomes less important.

Look at the progression 8-16-24. A difference of 8mm for each step, but the diagonal FOV decreases by ~40° in the first step, but only ~20° in the second step.
it is significant because of the visual area of coverage. Think of it this way, a 50 inch television has an area just over 1000 square inches, while a 60 inch television (a linear increase of only 20%) has over 1500 square inches of area (almost a 50% increase). The wider you get, adding that small amount of linear coverage nets an exponential amount more visual area (I'm sure there is probably an equation for it, though I don't know what it would be).
02-05-2014, 08:31 AM   #4
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The difference between 250mm and 300mm is hardly noticeable, the difference between 15 mm and 65 is huge, is the difference between UWA and telephoto on APS-c, both 50mm.

02-05-2014, 08:38 AM   #5
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Yeah, the field of view will have a more significant change. You will also need to come close to the subject, so the perspective will become more dramatic. There is a difference between 14mm and 15mm, but I doubt anyone would notice the difference between 200mm and 201mm. In fact, just loose QC can account for that difference at the tele end.
02-05-2014, 08:41 AM - 3 Likes   #6
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It's all about percentages. The difference between 20 and 22 is 10%; the difference between 200 and 202 is one percent.
02-05-2014, 08:54 AM   #7
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Ok, the percentage thing makes more sense but I think the products that get release take that into account. Was just thinking about this because of some of the stuff @heie wrote in the 55-300 review about magnification and the speculation/rumor about 135-380.
02-05-2014, 09:40 AM - 1 Like   #8
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I found through my own experimentation that it takes around a 40% change for a difference in focal length to be what I'd consider significant. It seems lens makers came to the same conclusion. Look at the limited lens series: 15*1.4 is 21. 21*1.4=29, but the 31 already exists, so close enough. 31*1.4=43. Magic. The classic 24/35/50 combo are at 1.4 ratios. It does seem, though, that once you get into the longer FL, the ratio gets even bigger. From 50 we go to 85 (1.7), from 43 to 77 (1.8), etc.

02-05-2014, 09:57 AM   #9
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QuoteOriginally posted by Kozlok Quote
I found through my own experimentation that it takes around a 40% change for a difference in focal length to be what I'd consider significant. It seems lens makers came to the same conclusion. Look at the limited lens series: 15*1.4 is 21. 21*1.4=29, but the 31 already exists, so close enough. 31*1.4=43. Magic. The classic 24/35/50 combo are at 1.4 ratios. It does seem, though, that once you get into the longer FL, the ratio gets even bigger. From 50 we go to 85 (1.7), from 43 to 77 (1.8), etc.
Excellent observation. I guess once you get into that middle range especially with primes it is where a couple steps forward or backwards is going to make up the difference between 85 and 50*1.4=70 and the situations you will be using it will be where you have the freedom to take those steps.
02-05-2014, 10:14 AM   #10
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QuoteOriginally posted by mikemike Quote
I have often read that extra mm are more important on the wide end or are more substantial than on telephoto but I don't really see why.

I made a spreadsheet that calculates the angular field of view and the distance from subject required to have a 2 m field of view (i.e. taking a head to toe picture of an adult) and it is perfectly linear so I don't see why other than if you are backed up to a wall or pressed up to the edge of a cliff or guard rail.

Attachment 205396
There are two aspects to this.

First of all, and I have found this over years of travelling in Europe , you simply can't back up the 5-10 meters needed to use a longer lens. Consider a narrow street in an old town. Likely it will be in the range of 3-6 meters side to side. Even the field of view of a 14-15mm lens is not enough to fit that magnificent church you want to photograph into the scene.

On the long end, you can always crop, and further reduce the FOV, but you can't ever increase it.
02-05-2014, 10:27 AM   #11
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The 18mm kit lens is 6/5th of the focal length of the DA15 ultrawide, but remember, the area is squared so the DA15mm ultrawide is capturing 144% of the image that the kit lens captures, (18/15)^2. The 18mm is is 1.5x the focal length of the 12mm ultrawide, but the area captured by the ultrawide is 225% of the kit lens, (3/2)^2.

For ultra ultra wides, the stretched corner effect gets more and more pronounced, faster than the angular FOV is increasing. So it is the square of the relative focal length that counts, not the relative angular FOV. It is true that sometimes you can't back up, but for ultrawides, it is not only about being able to be a few steps closer, and still capturing the image. It is also about the exaggerated perspective, as the subject gets much closer to the lens.

Last edited by sheld; 02-05-2014 at 11:10 AM.
02-05-2014, 10:41 AM   #12
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I think there is inversion between vertical and horizontal column at 2m
02-05-2014, 10:42 AM   #13
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It's like pxpaulx said. If my math isn't wrong, it's as follows:

The angle of view approximates a linear function of the focal length (AoV = 180 × d / π × f, with d = 31.7mm for APS-C), but the area of an image is a quadratic function (A = π × (d/2)²). Since the lenses are designed to project an image of roughly the same size (to cover the sensor), it follows that a linear change in the focal length has an exponential effect on how much/how little information you cram in the same photo. The side effect is that the resolution follows the inverse relation, because the pixels in the sensor are of constant size, so wider AoV -> less resolution.
02-05-2014, 04:57 PM   #14
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QuoteOriginally posted by hcarvalhoalves Quote
It's like pxpaulx said. If my math isn't wrong,
Isn't it FOV = 2 * arctan (x / (2 f)) where x is the film/sensor diagonal and f the focal length ?
02-05-2014, 08:13 PM   #15
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QuoteOriginally posted by kh1234567890 Quote
Isn't it FOV = 2 * arctan (x / (2 f)) where x is the film/sensor diagonal and f the focal length ?
Correct, that formula I gave is a linear approximation, just to make a point (linear -> quadratic). It's a good approximation for the normal focal lengths:

Angle of 50mm on APS-C: 35 deg

(2 * arctan (31.7 / (2 * 50))) * (180/pi) - Wolfram|Alpha

Approximation: 36 deg

(180 * 31.7) / (pi * 50) - Wolfram|Alpha
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