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Thread started 27 Oct 2003 (Monday) 10:58
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Extension Tube Math?

 
Littlebike
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Oct 27, 2003 10:58 |  #1

Is there any formula for extension tubes?

i.e. a 12mm extension tube connected to a lens with a minimum focusing distance of 8' will reduce the focusing distance to x.

Is there any reason NOT to get a set of tubes from Vivitar or Kenko for example. There are no optics involved so i cannot see a reason to pay a premium.




  
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CyberDyneSystems
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Oct 27, 2003 11:21 |  #2

I don't know the math or formula,.

But in my opinion,. no there is no reason NOT to get a set. I have the Kenko set and even if I DID have a dedicated macro,. (I don't) it would allways be nice to have the option and flexibility to use the extension tubes.

Even with a Macro lens you can increase magnification.
And the adjustablility is nearly infinate with a set of three rings. I tend to use either the 12mm (smallest) or the largest (35mm?) rings most, on either my 70-200mm f/2.8 or on occasion the 17-40mm f/4. (I don't own a decent lens in the 28-70mm range yet)

But in some circumstances i have indeed stacked the rings to get even closer. (especially with the Sigma 70-200mm which has an obscenely long minimum focus distance to start with)

I have yet to try it,. but I have seen some amazing shots taken with a 500mm prime and extension tubes! I know it will be tricky shooting (tripod a given) but I do intend to try it! :)

Yes they are the poor mans macro,. but for about $100.00 you can't loose. The difficulties that the rings may impose are just that hurtles to the photographer but they in no way seem to disturb the lenses inherent optical quality.


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DaveG
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Oct 27, 2003 13:21 |  #3

Littlebike wrote:
Is there any formula for extension tubes?

i.e. a 12mm extension tube connected to a lens with a minimum focusing distance of 8' will reduce the focusing distance to x.

Is there any reason NOT to get a set of tubes from Vivitar or Kenko for example. There are no optics involved so i cannot see a reason to pay a premium.

There certainly is a formula but with a DSLR you don't need one. First you have a meter behind the lens and that will take into account the change in exposure. And then you can check the histogram to get another opinion.

You WILL lose your ability to focus these lenses to infinity as long as there is an extension tube attached and that's something to keep in mind.


"There's never time to do it right. But there's always time to do it over."
Canon 5D, 50D; 16-35 f2.8L, 24-105 f4L IS, 50 f1.4, 100 f2.8 Macro, 70-200 f2.8L, 300mm f2.8L IS.

  
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PrimoFelis
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Oct 27, 2003 15:05 |  #4

Littlebike wrote:
Is there any reason NOT to get a set of tubes from Vivitar or Kenko for example. There are no optics involved so i cannot see a reason to pay a premium.

Hi Littlebike,

I have had the Kenko extension tubes (a set of 12, 20, 36mm extension) for use with my Canon film camera bodies. I have been happy with them.

I'd point out, however, there are two potential issues that may or may not be a conern for you depending on which lenses and camera bodies you plan on using it, now and future:

1. EF-S mount (in)compatibility issue.

As most of us know by now, Canon introduced a new EF-S mount for the Digital Rebel "kit lens" (18-55mm f3.5-5.6).

If you plan on using your extension tube with this (or any other future) EF-S mount lenses, you will need to get one of the new Canon extension tubes that were also recently introduced. The older Kenko/Vivitar/Canon extension tubes won't work with the EF-S lens(es). (It won't fit -- physically.)

2. The (potential) vignetting issue

I don't know about Vivitar or Canon tubes (new or old), but my Kenko tubes have a circular opening of 35mm in diameter. (I measured mine.)

I know that, with some lenses (e.g., 600mm f/4) his turns out to be too small an opening to clear all the light going toward the image corners of a full-size 35mm format film/sensor. This results in a (mild) vignetting in the image corners.

This is NOT an issue with most smaller/slower lenses. Further, it should NOT be an issue with 10D/Digital Rebel users (probably with ANY lenses) thanks to the 1.6x "cropping factor".

But if you are a current/future user of 1Ds (and possibly 1D too?) you should be aware of this. (Film body users as well.)

(That said, I'm not entirely sure if the genuine Canon extension tubes do not have this same vignetting issue. A tiny picture of an allegedly Canon extension tube that I saw didn't look much different from my Kenko. Can anyone out there with the Canon or Vivitar extension tubes measure their tube opening?)




  
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PrimoFelis
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Oct 27, 2003 20:09 |  #5

Littlebike wrote:
Is there any formula for extension tubes?

i.e. a 12mm extension tube connected to a lens with a minimum focusing distance of 8' will reduce the focusing distance to x.

Hi Littlebike,

Okay, let me take a crack at this too, although it is going to get a bit long.

But first, let me say that I'm not sure if doing the math is always the best way to obtain an accurate or useful answer to your question, for a couple of reasons I am going to mention below. Might I suggest holding a lens of your interest a distance away from the camera's lens mount while looking through the view finder? (Then you can make measurments of your interest?)

But anyway, I aim to please :) and so I will give the math a try.

---

First, a short quote from John Shaw's excellent book "Closeups in Nature" (ISBN 0-8174-4052-6), that gives a basic formula on the use of extension tubes in macro photography:

John Shaw writes::
The total amount of extension you need in order to get to any given magnification depends on the focal length of the lens you are using. Notice that I say "total" extension; it does not matter how you gain the extension -- some could be added by using extension tubes, some could be in the focusing mount of the lens -- since the entire amount being used is what is important. A basic formula gives approximate magnification rates:

Total extension used / Focal length used = Magnification

If you want to photograph at 1/2x with a 50mm lens, you can see that you need 25mm of total extension. But on a 100mm lens, the same 25mm total extension yields only 1/4x, and on a 200mm lens it yields 1/8x. Any given amount of extension yields less magnification when used with a longer focal-length lens.

Note that, by "1/2x," etc., he is talking about the magnification of image at the film/sensor plane. So, as an example, at 1/2x magnification and using an EOS 10D, you would be filling the viewfinder with an object of a size 45.4mm x 30.2mm (double the size of the CMOS sensor -- 22.7mm x 15.1mm -- used in 10D). Hope this part is clear.

How to use the formula

In order to use the above basic formula to derive the answer you are seeking, you need to know the amount of (effective) extension built into the lens of your interest.

(I say "effective" because these days many lenses don't use a simple lens extension to achieve a close focus, instead rearranging position of lenses. This often produces a side effect of changing the actual focal length of the lens a bit -- even if it is a prime (i.e., non-zoom) lens -- introducing discrepancies between the theory and reality.)

Usually this "built-in extension" value is not part of a published lens spec.

However, you can derive this value using the lens' maximum image magnification value (or alternatively a minimum coverage area) and the lens' focal length, usually found in the lens spec. (How accurate these published values are remains a question. But let's not get too critical...)

Now, I will refer to such a chart from Canon...

Example: Step 1. Magnification

Let's pick EF50mm f/1.8 II. The table says the closest focus distance for this lens is 0.45 meters (= ~17.7 inches), and the maximam magnification is 0.15x. Using the above basic formula:

x / 50 = 0.15

Solving for x, the derived maximum built-in lens extension is: x = 50 * 0.15 = 7.5mm

If you add Canon's 12mm extension tube to this lens, you will achieve a maximum total extension of 7.5 + 12 = 19.5mm

Therefore, using a 12mm extension tube with EF50 1.8 II lens, you can obtain a maximum magnification of:

19.5 / 50 = 0.39

a magnification of 0.39x [Filling the frame with an object of actual size 58.2mm x 38.7mm]

Example: Step 2. Minimum focusing distance (a rough approximation)

[NOTE: The following was corrected and updated Nov. 7, '03]

To continue with the example and figure out new minimum focusing distance (MFD) from the new magnification, we need the help of a few additional formulas (found in any general physics book):

1) M = b/a
2) 1/a + 1/b = 1/f
3) MFD = a + b

where
M: Magnification
a: lens-to-subject distance
b: lens-to-film/sensor distance
f: focal length

[NOTE: for a and b above, the position of lens being referred to is that of a theoretical equivalent lens of single element, NOT of the actual camera lens which usually have multiple elements.]

So, given the new magnifcation of 0.39x and focal length of 50mm:

M = 0.39 = b/a
1/a + 1/b = 1/50

Solving for a and b, you get:

a=178.2mm
b=69.5mm
MFD = a + b = 247.7mm

That is to say, using EF50mm f/1.8 II lens with a 12mm extension tube, the minimum focus distance should be reduced from 0.45 meters to ~0.25 meters (17.7 inches to ~9.75 inches)

(Would anyone care to run a test to see if this calculated value matches the reality? I don't have an EF50mm 1.8II lens to test with.)

Anyway, I hope this kind of answers your original question...

--
PS. DISCLAIMER: Note that, as I originally stated at the very beginning of the post, there are a number of reasons why these simple formulas would not yield very accurate results. (Especailly the minimum focusing distance -- the formula for magnification should be more useful.) If what you are after is accuracy -- I recommend that you actually try it out and measure it. 8)




  
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Littlebike
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Oct 27, 2003 23:07 |  #6

Dear GOD man, you are insane, but thorough.

And yes, that does answer my question.

I knew there had a be a mathmatical way to figure that out.

I defenitely think I am going to order the Kenko extension tube set in the near future.




  
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PrimoFelis
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Oct 27, 2003 23:35 |  #7

Littlebike wrote:
Dear GOD man, you are insane, but thorough.

I'd say not as bad as Phil Askey at DPReview. HE is crazy. :)




  
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Littlebike
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Oct 31, 2003 16:18 |  #8

CyberDyneSystems wroteI have the Kenko set and even if I DID have a dedicated macro,. (I don't) it would allways be nice to have the option and flexibility to use the extension tubes.

Are you using a 10D?
Have you noticed any focusing issues with the Kenko Tubes?

I called Adorama today becuase I was thinking about buying the three tube set: http://adorama.com …67913190464&sku​=KNAETSEOS (external link)
And the sales person told me the set would not work with my camera, I would have to either buy the indevidual 12mm or 25mm, with a total cost being more than the three item set.

This sounds like hogwash to to me, just wondered what others have experienced.

I was planning adding a sigma 1.4x teleconverter as well.




  
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Daytripper
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Nov 05, 2003 19:14 |  #9

The three tube Kenko set works fine with the 10D. Only autofocus problems I have had are associated with the extremely shallow DOF when long extensions are attached.

Cheers,
Pete




  
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DonCoon
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Nov 06, 2003 14:37 |  #10

Hey PrimoFelis,

Got my tubes today and decided to check your math. I used a 75-300 to take the shots.

The math didn't work --- at least how I interpreted the math anyway.

The 75-300 has a published Max Mag of .26 at 300mm. The minimum focusing distance is 1.5 meter (59.1 inches.)

My initial test confirmed the starting point right to the tenth of an inch! 59 1/8 inches from the object to the focal plane of the sensor. Move it 1/4" closer and you can't focus. Great start.

My math:

Built-in lens extension is 78mm. 300*.26

Add 36mm tube gets us to 78+36 = 114mm.

114/300mm = .38 maximum magnification.

1.5 meters *.26 / .38 = 1.0263 meters = 40.4 inches.

Actual closest focusing distance is 51" : (

BTW, you can get the results quicker by multplying the intial distance by the before/after extension. 1.5 * 78/114 = 1.0263 meters = 40.4 inches.

I tried again with the full 68 mm of the 3 Keno tubes.
1.5 * 78/(78+68 ) = .8014 meters = 31.6 inches.

Actual closest focusing distance in this case is 44 inches.

BTW, with three tubes aboard, I had to use manual focus else the lens' USM focusing mechanism just chattered.

Is there an error in my calculations or something?




  
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DaveG
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Nov 06, 2003 15:23 |  #11

For what this is worth:

I do a fair bit of large format photography. There are no internal meters in these camera and bellows expansion factors are an everyday occurence.

Extension tubes are just fixed bellows extensions, so it's all the same thing. You are just moving the lens farther away from the film.

I have (somewhere) the formula for calculating the bellows extension factor. I'm sure that it works fine but I've never had to use it. What I do use is a device made by Calumet ($8 or so). It's two pieces of plastic, one is a target and the other is a dedicated ruler. I place the target in the shot and "measure" the size of that target on the camera's groundglass. The ruler is marked off in stops and it tells me, without math, the increase in exposure I need. And it's bang on, I should add.

Now my point. In photography we throw the term "telephoto" around, and telephoto pretty much means to us any lens with a longer than normal focal length. In all but large format photography that fine. But in LF the typical, longer than normal lenses are called "long focus" rather than telephoto.

In use Long Focus lenses need their camera to have a bellows as long as the len's focal length in order to focus at infinity. The same lens would require twice their focal length to focus at 1:1. So if I use a 210 mm lens on my camera I'd need 210 mm of bellows extension for infinity or 420 for 1:1 and somewhere around 375 mm for portrait distances.

The line in the new LF camera ads that everyone rushed to read, was the length of the bellows. The longer the bellows the longer the lens you could use and LF ad writers used it exactly the same way that car ads use horsepower.

Telephoto lenses, not long focus lenses, are in the LF context very specialized tools. They are designed to focus at infinity WITHOUT needing their focal length as bellows extension. These lenses were popular with nature photographers who would be dragging a camera with a smaller bellows into the field. There are some drawbacks of using telephoto lenses and a very important one is that that the bellows extension calculator DOESN'T WORK! The lens design has messed up the basic math of the calulation and it (and my trusty Calumet tool) can't be used.

So I wonder if the calculation formula bandied about here is moot when you use a true telepohoto lens.

But one more time: Why do you need to do this calculation? There's not one DSLR made without a TTL meter that will automatically compensate for the bellows extension. Even if you were using studio strobe flashes you could get close with a flash meter and then could fine tune with the histogram. I guess that I don't see any point in trying to PREDICT the exposure when you can REVIEW it.


"There's never time to do it right. But there's always time to do it over."
Canon 5D, 50D; 16-35 f2.8L, 24-105 f4L IS, 50 f1.4, 100 f2.8 Macro, 70-200 f2.8L, 300mm f2.8L IS.

  
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DonCoon
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Nov 06, 2003 18:42 |  #12

Well, there are at least two reasons.

1) Intellectual exercise -- which I think is the primary reason here : )
2) A way to determine, short of trial and error, the minimum focus distance and increase in magnification.

Why do people do crossword puzzles or have car computers to tell them their mpg even though it doesn't change their fuel economy.

Just my 2 cents




  
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Littlebike
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Nov 06, 2003 19:31 |  #13

DaveG wrote:But one more time: Why do you need to do this calculation?

When I wrote the initial question I was unaware of the cost effective Kenko extension tube set and thought I would have to buy each one (Canon) indevidually, quite expensive. So, if I could get an idea of what length extension tube would shorten my 70-200mm minimum focus down to about 2-3' I would buy only that tube.

Little did I know I was opening this can of worms.

As for metering, that never even crossed my mind. My camera knows how much light it is getting and I will trust to give me the coreect exposure. If not, I will fiddle with it until I get the picture I want.




  
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PrimoFelis
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Nov 07, 2003 09:59 |  #14

DonCoon wrote:
Hey PrimoFelis,

Got my tubes today and decided to check your math. I used a 75-300 to take the shots.

The math didn't work --- at least how I interpreted the math anyway.

...

Is there an error in my calculations or something?

Hi DonCoon,

Many thanks for actually carrying out the test and reporting back the results! There is nothing more valuable than actual measurements.

I think there are a number of possible reasons why the calculated estimates didn't agree with your measurments. I was half expecting this though -- especially for such a tele-zoom. (There were reasons why I chose for the example the simple 50mm lens that I didn't even have, and not one of the more complex zooms. :))

But in this particular case, the biggest reason for the discrepancies was my own error! :( I made a few mistakes in the final step of the example I gave. Sorry guys -- it did seem a little too simple at the time. I just corrected it in my original post above. Physics has never been my strength -- where are all the physics majors and PhDs on this forum when we need them?

Anyway, for what it's worth, here are the expected vs. actual (per DonCoon), one more time:

EF50 f/1.8 II + 12mm ext.: MFD of 9.75" (vs. actual ~10" -- is this right DonCoon?)
75-300@300 + 36mm ext.: MFD of 59.2" (vs. actual 51" ) [16% too long]
75-300@300 + 68mm ext.: MFD of 53.6" (vs. actual 44" ) [22% too long]

As you see, the result for the 50mm lens may be in the ballpark, though the errors in estimates for the zoom are large.

As for the reasons, I don't feel qualified to speculate with any credibility, but here are a couple of usual/possible suspects:

The physical sizes of these camera lenses.

The simple formula used for the math works by treating the camera lens as a theoretical equivalent lens of a single element. (I understand that the German mathematician Gauss -- as in the Gaussian lenses I suppose -- has something to do with this technique.) This is supposed to work pretty well for a lot of lens math, but I don't think subject-to-sensor distance (i.e., minimum forcusing distance) is one of them, since it would directly impact the total distance.

Another might be the effect of inaccuracies in data -- for example the published focal length and the maximum magnification. As I understand it, some tele-zooms -- though not just zooms and certainly not all zooms -- can "shrink" in actual forcal lengths as they focus closer. Just to see, I tried the math again with focal length of 270mm in place of 300mm:

@270mm + 36mm ext.: MFD of 52.4" [Still 2.7% too long]
@270mm + 68mm ext.: MFD of 47.5" [Still 8% too long.]

And I'm sure there are a number of other reasons.

(It might be interesting to run some more tests like these though, especially with simpler non-zoom lenses. Any takers?)

As I stated in the original post, these simple forumlas don't always give accurate or sufficiently useful results, and so I'd again recommend use of actual measurments for the focusing distances. (The magnification formula could be actually useful, though. It should be more accurate too.)

But I do think it was worth giving the math a try anyway, and for one I learned a few things in the process. And for me, like DonCoon said, it was mostly for intellectual exercise. I'd never claim any practical value in anything I'd post to this forum. :)

I have no doubt that there are many smart and far more qualified people on this forum who can talk about this with much more authority. If anyone is interested, please feel free to take this thread further or point out any errors

Best regards,




  
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DonCoon
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Nov 07, 2003 11:27 |  #15

PrimoFelis wrote:
That is to say, using EF50mm f/1.8 II lens with a 12mm extension tube, the minimum focus distance should be reduced from 0.45 meters to ~0.25 meters (17.7 inches to ~9.75 inches)

(Would anyone care to run a test to see if this calculated value matches the reality? I don't have an EF50mm 1.8II lens to test with.)

I think youv'e got it.... with one caveat.

I measured 9.375" -- close but not perfect. But... I recalled that my testing showed some of Canon's MFDs to be a little conservative. Therefore, I tested the 50 f/1.8 II without any extension tubes and got a MFD of 17" which is 96% of of the published 17.7".

AND.... 96% of your predicted 9.75" just happens to be 9.364" vs. my tested 9.375" --- a .12% difference! And since I'm using a flexible shop measuring tape, I'd call your prediction "spot on."

After I digest your new formulae, I'll crunch them into my spreadsheet and do some more tests.

Also two questions/confirmation​s needed:

1) If the base published MFD is incorrect, shouldn't the Maximun Magnification also be corrected? How? Proportionaly -- like .15 = .15/.96 = .156?? (Obviously some of the small errors result from Canon's rounding.)

2) Distances are measured from the subject back to the focal plane of the sensor? I'm pretty sure this is correct since it works : )

Nice job!




  
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Extension Tube Math?
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