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12-05-2011, 05:53 AM   #1
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How to calculate FL?

I saw this graph of actual focal length vs distance for a Sigma 150-500mm: Focal Length versus Subject Distance - Sigma 150-500mm

I don't think you can calculate the actual focal length in between min and max, but I expect you could calculate actual FL at the minimum focus distance because you have a magnification spec (assuming the manufacturer is being honest, maybe not a great assumption). Anyway is there a way to calculate FL if you know the magnification and MFD?

12-05-2011, 06:28 AM   #2
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you would need to look up the lens formulas and do a bit of math.

Strart on wikipedia for the formulas.

Magnification = focal length / (focal length - distance from the lens)


Magnification is quite simple to calculate, you know subject size, and you know the image size, because your sensor is 16 x 24 mm just look at the overall pixels in that distance and the height of your image in pixels, and scale the image size appropriately.
12-05-2011, 09:30 AM   #3
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Hi Lowell. I tried this calculation for my A*300mm f/4. The specs show magnification of .09 at a subject distance of 4m. Using 4000mm subject distance and .09 magnification, the result doesn't make sense.

I also tried calculating for the Sigma 150-500, S = 2200mm and m = 0.19, and again, the result was incorrect.

What am I doing wrong?
12-05-2011, 09:58 AM   #4
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QuoteOriginally posted by audiobomber Quote
Hi Lowell. I tried this calculation for my A*300mm f/4. The specs show magnification of .09 at a subject distance of 4m. Using 4000mm subject distance and .09 magnification, the result doesn't make sense.

I also tried calculating for the Sigma 150-500, S = 2200mm and m = 0.19, and again, the result was incorrect.

What am I doing wrong?
I will go back through my calculations, there may be a couple of points for errors, one is that the "lens extension" needs to be considered., i.e. the change in distance from infinity to min focus of the lens to sensor, the other is the sign of the magnification, do not forget that the image is inverted so magnification is negative.

12-05-2011, 11:52 AM   #5
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OK Here is what you need to do.

for any simple lens M =D2/D1 where
- M is magnification ratio
- D2 is lens to image and
- D1 is subject to lens.

But also know that 1/D1 + 1/D2 = 1/F where
- F is focal length.

the relationship between F and D2 is that D2 is the focal length plus lens extension (i.e. how muych ghe entire lens group for a non IF lens moves out

Using these formulas you can calculate for a non IF lens, the magnification as you move closer and closer to the subject.

To then consider an internally focused lens, and it's focal length, perhaps at minimum focus, simply measure the distance to the lens front element, and measure the magnification of the lens, based upon the image size on the sensor, and this will give you the effective focal length at MFD
12-05-2011, 12:32 PM   #6
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QuoteOriginally posted by audiobomber Quote
Hi Lowell. I tried this calculation for my A*300mm f/4. The specs show magnification of .09 at a subject distance of 4m. Using 4000mm subject distance and .09 magnification, the result doesn't make sense.

I also tried calculating for the Sigma 150-500, S = 2200mm and m = 0.19, and again, the result was incorrect.

What am I doing wrong?
I have a working spread sheet if you want it, the question I have is with the A300, is the distance the distance to film or distance to front element. when you are working with close focus, then the physical length of th elens is very important. Most times, mag ratios are quoted as the distance to film plane not the true working distance.

If the A300 is considered as about 250mm distance from fil,. that puts the working distance at 3.75 meters and my spreadsheet estimates the true focal length to be about 292mm.

for the sigma lens my sheet and the numbers you give suggest 325mm at min focus.
12-05-2011, 04:38 PM   #7
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Here's an experimental way to do it... I'll demonstrate it when I find my PK extension tubes...

Lens thickness must be considered for the measurements you seek. Here's a schematic of a object/lens/image setup.

|-------d1---------->| Lens |---------d2----->|
Object................Po.......Pi.....................Image

Where Po = Object Side "Principal Plane" & Pi = Image Side "Principal Plane".

The principal planes are where the lens appears to be when the thin lens formula is used to analyze the lens. The principal planes need not be located inside the physical lens. The lens' effective thickness, T, is the distance between Principal Planes.

IF
d1 = (object) to (object side principal plane) distance
d2 = (image) to (image side principal plane) distance
THEN the thin lens equation holds:
1/f = 1/d1 + 1/d2 where f is focal length
m = d1/d2 where m is magnification.

Using these definitions:
working.distance = f (1+1/m)
lens-image.distance = f (1+m) note these distances are measured from appropriate principal planes

Total.focusing.distance = working.distance + lens-image.distance + lens thickness = f(1+m)(1+1/m) +T

If focal length does not change when the lens' focusing ring is moved then measuring total focusing distance at two magnifications is sufficient to calculate both focal length (f) and lens thickness (T).

Unfortunately, many lenses change actual focal length when either the zoom ring or focusing ring is adjusted. This confusion can be unscrambled by using a short extension tube, X in length, as follows. Take a photo of an object at a particular lens setting (focal length and focus ring rotation), record total focus distance and magnification - then add the short extension tube and repeat the photo (move the object as needed to regain focus but do not touch the focus or zoom rings).

|-------d1'---------->| Lens |---------d2'----->| Total.focusing.distance' = f(1+m')(1+1/m') +T
Object..................Po......Pi.................... Image

|-------d1"---------->| Lens |[X]---------d2"----->| Total.focusing.distance" = f(1+m")(1+1/m") +T + X
Object...................Po......Pi+X..................... Image

These two measurements are sufficient to calculate focal length and lens thickness. The key trick is having a short extension tube to use (just the ends from a set of tubes).

Last edited by newarts; 12-06-2011 at 04:54 PM.
12-05-2011, 09:30 PM   #8
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QuoteOriginally posted by Lowell Goudge Quote
If the A300 is considered as about 250mm distance from fil,. that puts the working distance at 3.75 meters and my spreadsheet estimates the true focal length to be about 292mm.

for the sigma lens my sheet and the numbers you give suggest 325mm at min focus.
Thanks Lowell. That answers what I was trying to figure out; the A*300 stays pretty close to its nominal focal length, regardless of subject distance.

12-06-2011, 06:19 AM   #9
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QuoteOriginally posted by audiobomber Quote
Thanks Lowell. That answers what I was trying to figure out; the A*300 stays pretty close to its nominal focal length, regardless of subject distance.
As it should

the A300 focusing helix moves 100% of the elements in unison away from the flim/sensor plane. My guess would be that it has in total about 20mm of lens extension from infinity to min focus., The calculated focal length is actually within the reasonable tolorances in claiming focal length. If I had more accurate dimensions on the lens I could perhaps come closer, Of more interest is how the 150-500 stacks up with some accurate tests.

The spreadsheet I have i made to start looking at the issue of focal length loss as a function of internal focus.
12-06-2011, 07:25 AM   #10
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QuoteOriginally posted by Lowell Goudge Quote
Of more interest is how the 150-500 stacks up with some accurate tests.
These results were measured empirically using a D70, so should be accurate for Pentax dslrs:
Focal Length versus Subject Distance - Sigma 150-500mm
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