0

I've come from 3x Zoom magnification land and am having a hard time adapting to all these complex "mm" systems. I'm thinking about buying some lenses to enhance the range of zoom of my Sony A5000 camera.

I am already aware of the fact that you get Max magnification (times zoom) from dividing the big number by the little number. (55-210mm = 3.8181818xzoom). I'm aware that in essence, this isn't magnification from an unzoomed (1x) perspective, but is instead, just how many x's from the "unzoomed" lens position that lens can increase it.

From this I can see:

  • a 55-210mm lens has 3.8x zoom
  • a 650-1300 lens has 2x zoom.

The trouble is, from looking at those lenses on Amazon, I can't help but convince myself that the 650-1300 lens, from its shape, is made for zooming in on objects a lot farther away. Therefore, on top of the "adjustable zoom" difference above, there must be a "base zoom x" which is how much the x zoom already is, when the lens is unzoomed.

How do you work that out, so that you know how far in a lens will go with your camera? I'm assuming of course, that there is a way, and people don't just buy lenses without any idea of how far in it will zoom.

Please try to avoid jargon; I don't really understand big terms, but there must be a simple way of working all this out.

I don't think this is a duplicate of that other question, because that one only showed what i explained above - the "adjustable zoom range". I need a way to work it all out, not just the zoom range of a camera.

marked as duplicate by mattdm, Saaru Lindestøkke, James Snell, inkista, Michael C Jul 20 '15 at 4:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • Hi matt I already looked at this and it doesnt answer the question. – Crinkle Jul 19 '15 at 13:10
  • @Crinkle how does the linked answer lack in answering your question? Oh, and here's a Canon site simulating different focal lengths. Might be helpful. – Saaru Lindestøkke Jul 19 '15 at 13:10
  • So are you saying that from the same point, "full zoom" with both those lenses, the 650-1300 lens would be less zoomed in? – Crinkle Jul 19 '15 at 13:12
  • In a technical sense, yes, it would be less zoomed in. And the times-zoom numbers we kick around represent that. In the colloquial sense where zoomed in means "make it look bigger", the lens with the larger focal length would be more "zoomed in", but the number we use to represent that is simply the focal length itself. – mattdm Jul 19 '15 at 13:55
  • 1
    Really, the more I work at answering this, the more I am convinced that it really is covered by How do I convert lens focal length (mm) to x-times optical zoom?, as I really do think the answers there cover your concerns. – mattdm Jul 19 '15 at 19:18
1

I used the following formula that helped me better understand the focal magnifications when I went to DSLR’s from digital compacts.

First, I decided what I would consider to be a perfect focal range to mimic the human eye. this was a little short of 50mm, but I settled on 50mm. (please do note, that this is only a conceptual attempt to rationalise what could be considered the normal focal length for a human eye and in no way a proven fact)

I then took the crop factor into the equation (1.6x), which transformed the 50mm to 32mm for my camera, so I decided to settle on 35mm as my most normal focal length matching my vision.

From there on forth, I could work out that a 200mm lens was magnification of 5.70x, a 70mm was 2x magnification etc

The Sony a5000 has a crop factor of 1.52 which for me makes the ideal base focal to be 33mm. if I was in your shoes, I would use this or 35mm to be the base and work from there.

in your scenario, based on 35mm being the base focal length, a 55mm to 210mm would be 1.57x to 6x zoom lens.

  • +1 because this is actually the best answer for the question as posed by the OP. – Michael Borgwardt Jul 19 '15 at 21:11
  • 1
    I'm hesitant to upvote, because I don't want people to get the impression that this convention (using a normal focal length as the basis for "1×") is in any way typical or standard. It's not a bad idea, but it's just... not what people do, and because of that I'm afraid that it's going to spread more confusion than good in the long term. – mattdm Jul 19 '15 at 23:43
  • Maybe we could use a notation like "×normal" or "×N", as in: "a 1.6×N to 6×N zoom lens", or "a 10×N telephoto prime" — or "a 0.5×N wide angle". But it should be clear that this is inventing something new! – mattdm Jul 19 '15 at 23:45
1

The question is one of comparing apples and oranges.

Take this analogy: We can determine how strong an animal is by how many times an it can lift/pull it's own weight. That scale suggests we should have been using ants instead of horses to pull carts and work the land... but we know that's not right because we can ride a horse but would definitely crush an ant underfoot.

To do any meaningful farm work in that analogy we need an absolute figure of how much force an animal can provide so we need to use a different measure such as nm (newton-metres) and the same goes for photography. '3x' zoom is fine as a description of how versatile a lens is, but it does not give an absolute figure on how much magnification you can expect.

The amount of magnification you get will be different on an 18-105mm lens compared to a 70-200mm but the 'x' factor is not going to tell you which lens zooms farthest - it's not something we can use to compare real-world results. To compare different lenses we need something we can measure that is not relative to thing we're measuring and 'focal length' in mm is the measure we use.

1

In colloquial usage "zooming in" on something is to make it appear closer or larger. In technical terms, this is represented by the focal length. Your confusion comes from expecting the zoom ratio (the "times zoom") number to represent the amount of zoom in the colloquial sense, when in photography we use it simply in its technical sense: the amount any individual lens can change its focal length. This means you are focusing on the wrong number.

The thing you are looking for seems perfectly reasonable; it just happens that terminology evolved in a different direction.

In reading your comments, I think that you're also actually still misunderstanding what the times-zoom number for a zoom lens means. That is, it seems like you think that it represents magnification in some way. That's a logical assumption, but it's just wrong. That is, I know you understand the calculation outlined in How do I convert lens focal length (mm) to x-times optical zoom?, but I think you're misunderstanding what the result actually means.

It sounds to me like you assume that the number you get by dividing the higher focal length by the lower focal length tells you the magnification, or "how much zoomed in the lens will be". This is not the case. A zoom lens can vary the focal length, and the given numbers represent the range by which it can vary. But at any specific setting, the size of a distance object will be the same. That is, if one lens goes from 70-200mm and another goes from 100-500mm, if you set either one to 200mm, the resulting image will be the same (give or take other differences in the lens). In fact, if you have a non-zoom prime lens which is fixed at 200mm, it too will show the same image.

So:

Except when considering the very-close-up macro situation (which is covered at What does the magnification ratio number mean on a macro lens?), the only important number here is the focal length, which determines the angle of view. The smaller the angle of view, the larger far away things will appear — and, the larger the focal length, the smaller the angle of view. If you are concerned with how much "reach" you have, all you care about is the larger number given for a zoom lens.

There is no standard "base zoom" number — as you've figured out, and as explained in How do I convert lens focal length (mm) to x-times optical zoom?, it just the ratio of a lens's extreme focal lengths.

If you're looking to compare different lenses for the amount of reach they have, looking at the times-zoom number doesn't tell you much. However, comparing 210mm and 1300mm does tell you something. Don't worry about the amount that each lens's focal length can change (the zoom) and just look at the focal length number.

You can use a field of view calculator (there are many on the web, plus lots of apps) to figure out what this will mean in practical use, but you can also take advantage of the simple formula here: How to calculate the new field of view from an existent one?. 1300mm is about 6× 210mm, and so if you take a picture of the same object with both, it will appear about 6× larger (in each dimension) with the 1300mm lens.

Be aware that this is only relevant when the sensor size is the same. If your sensor (or film) is a different size, you have to take that into account too. See What is crop factor and how does it relate to focal length? for more. This is why the idea of "35mm equivalent" was invented — it normalizes the numbers to one standard so they can be compared.

  • From what you've said there, if the max length of one lens is comparable to the max length of another, then surely, there is just a certain ##mm amount which is "x2 zoom" and a further one which would be "x3 zoom"? – Crinkle Jul 19 '15 at 13:24
  • also, i looked at the formula link you gave, but its on about resizing images so i dont see how its relevant? – Crinkle Jul 19 '15 at 13:25
  • You say "surely there is a certain focal length which is x2 zoom". But, I'm telling you, no, there just isn't. If you want, you could pick an arbitrary number (50mm sounds good) as 1× and work from there, but since there's no standard like that, it wouldn't be useful for talking to other people. – mattdm Jul 19 '15 at 13:46
  • As for the link: crop and zoom are mathematically interchangeable. – mattdm Jul 19 '15 at 13:48
  • 1
    You are seriously overcomplicating things for yourself. If it's the ability to bring far things close that you are interested in, just compare the focal lengths. If you are interested in how flexible a given lens is, that's when you look at the ×. – mattdm Jul 19 '15 at 13:57
1

I've written one answer, but I'm going to try again with a different approach. Maybe you'll find this more helpful.

A prime lens — a lens with no zoom — has a certain fixed focal length. It always shows the same field of view. A zoom lens, on the other hand, can vary over a range of focal lengths, for wider and narrower fields of view.

When a zoom lens is at its longest focal length (the higher number in the specification), it has a narrow field of view, making far away subjects appear bigger. This is "zoomed in". At the other end of the range, the lens is "zoomed out", or as you say, "unzoomed".

The "times zoom" number (2×, 3×, 10×, whatever) is not useful for comparing lenses, because there is no standard base for what 1× means. It doesn't correspond to a fixed, standard value — it corresponds to the widest focal length of that lens.

So, for your 55-210mm lens, the base ("1×") is 55mm. For your 650-1300mm lens, the base is 650mm. 650mm is obviously much, much narrower than 55mm — for your camera, 650mm gives a horizontal angle of just about 2°, while 55m gives an angle of about 25°. The 650mm lens has a lot more "reach".

In fact, even when the 55-210mm lens is zoomed all the way in, it's still only at 210mm, much wider (6.5°) than the widest angle of the 650-1300mm lens. It doesn't really matter that it had to change by 3.8× to get there — it still doesn't equal 650mm.

The basic fact is that the zoom ratio just tells you something about that lens. You can't use it as a comparative measure of how big the lens will make distant subjects appear. For that, you only need to look at the focal length numbers, and specifically, the larger of the two. (On the other hand, to see how wide the lens will go for taking in a whole scene, look at the smaller number.)

To calculate the field of view for a given focal length, use 2 × arctan(11.6 ÷ focal length). You can even use Google's built-in calculator for this. For example, for a 50mm lens, you get (approximately) 26°.

  • OK, can anyone offer a good formula which can translate lens specs plus the camera sensor size (23.2 x 15.4 mm) into an actual angle of view? I tried looking for this, but all the answers seem to indicate that the camera + lens's distance for an object defines how much the camera's field of view can be, which is like saying the time it takes for a meal to be cooked depends on how close to the kitchen the eater will be. – Crinkle Jul 19 '15 at 19:32
  • 1
    @Crinkle Sure! We have that right here: How to calculate the FOV in degrees from focal length or distance? – mattdm Jul 19 '15 at 19:34
  • Or, you can use one of the dozens of web-based calculators, like this one. (Use 1.5× for "focal length multiplier" for your sensor size.) – mattdm Jul 19 '15 at 19:35
  • @Crinkle: The "zoom factor" (e.g. 20x) is a marketing term and never even existed in common use until a few years ago when digicam manufacturers wanted a new selling point to get potential customers like yourself to sweat over. As mattdm has been saying, please just ignore the zoom factor and instead ask yourself whether the lens angle(s) of view, size, weight, etc. is what you are looking for. – HamishKL Jul 25 '15 at 0:21
0

When the term zoom is used with lenses it simply means that you can change the magnification oft the lens. There are also other lenses which can't be zoomed, they are called prime lenses. The magnification numbers you see in point and shoot cameras are simply a quotient of the largest and smallest focal length oft the lens.

Lets assume you have a zoom lens with a focal length 15-150mm. That would be a 10x zoom. You can now buy 2 another zoom lenses instead, one 15-50mm (3.33x zoom) and one 50-150mm (3x zoom). What would be the difference between using the single superzoom and the other 2 zooms. Ignoring other aspects (quality, aperture, ...) you can do the same photos with the superzoom as well as with the other 2 lenses. The only difference is, when you use the 2 lens combo you need to change the lens when zooming from 15mm to 150mm.

You asked for a "base" focal length. When 35mm film cameras were used, 50mm was considered as the focal length which has the same field of view as the naked human eye. The same applies for so called full frame cameras today. Your camera has a smaller sensor with a crop factor oft 1.5. This means you just divide the above number: 50mm / 1.5 = 33.33mm. Lets round this up to 35mm. When you now use a lens with 350mm focal length you have about 10x magnification of that what you see without the camera.

However you can't compare this numbers with you 3x zoom point and shoot camera. It usually starts at a wide angle, and not at the 50mm equivalent mentioned above. In the spec of you old camera should be a focal range with the extension 35mm eqiv.. Divide that numbers by your crop factor 1.5 to compare them with your new camera.

Not the answer you're looking for? Browse other questions tagged or ask your own question.