[quote name='genotypewriter' timestamp='1295119308' post='5545']
6000/(104.3/73.7) ~= 4240 pixels wide
[/quote]
I'll show you the error
[quote name='blende8' timestamp='1295119677' post='5546']
I don't know what you are calculating, Rainer.
Focal length is proportional to HFoV.
If you double the FL you get twice a much on your image.
[/quote]
No ... VoF (regardless of horizontal, diagonal or vertical) is only
approximatly linear ... and only if you stay quite near to the startingvalue ...
For everythig more global, FoV of a rectilinear lens follows the function:
FOV (rectilinear) = 2 * arctan (frame size/(focal length * 2))
with the arctan-function as a nonlinear element in it ... for small differences
of the term within the arctan-function, you can approximate this by a linear function ...
but going from 14 mm to 24mm is nearly a factor of 2 ... that is a very long way from
being a small change.
For a better and in depth explanation .. see here:
http://en.wikipedia.org/wiki/Angle_of_view
01-15-2011, 08:22 PM
(This post was last modified: 01-15-2011, 08:23 PM by Rainer.)
[quote name='Brightcolours' timestamp='1295120247' post='5549']
That is how we calculate equivalent focal lengths with different sized sensors, anyway.
[/quote]
Ok ... lets try it this way:
Asume you have the image of a 24mm on a fullframe camera.
Which focal length do you need for your crop-1.6 to have the same VoF?
24/1.6 = 15mm ... ok?
Now the TO had a 14mm lens ... nevertheless, lets do the rest here with 15mm...
If I have the 15mm on fullframe ... and crop out the portion that a crop-1.6 sensor
covers, I do get the VOF of a 24mm (on fullframe) ... now much megapixels will I get?
24mpix / (1.6 * 1.6) = 9.3mpix ....
and remember ... the TO had a 14mm lens .... so my calculation leading to 8mpix is
correct without any doubt.
01-15-2011, 08:40 PM
(This post was last modified: 01-15-2011, 08:40 PM by Rainer.)
[quote name='blende8' timestamp='1295114936' post='5537']
The dependence is linear, so it's 14 MP.
[/quote]
I now understand what you calulated ...
14mm -> 24mm is 14/24 or a factor 0.583 ...
24mpix * 0.5833 = 14mpix ...
and that is the error! ... you not only reduce the width ... but also the heigth ...
so you must calculate
24mpix * 0.5833 * 0.5833 = 8.16mpix
genotypewriter
Unregistered
[quote name='Rainer' timestamp='1295121930' post='5551']
I'll show you the error
[/quote]
Well my AOV numbers were pretty much the ones you gave, except I got mine from:
http://www.tawbaware.com/maxlyons/calc.htm
Unless I've overlooked something in the arithmetic, I can't see what the problem is in my calculation.
GTW
01-16-2011, 09:36 AM
(This post was last modified: 01-16-2011, 09:40 AM by Rainer.)
[quote name='genotypewriter' timestamp='1295119308' post='5545']
14mm = 104.3 degrees (horizontal)
24mm = 73.7 degrees (horizontal)
Assuming 24MP = 6000x4000 pixels,
6000/(104.3/73.7) ~= 4240 pixels wide
Therefore, at a 3:2 aspect ratio the image would be ~= 4240x2827 = 11986480 pixels = 11.99MP ~= 12MP
But as I said... 12MP from 864mm[sup]2[/sup] (FF area) is likely to be better than 12MP from 431mm[sup]2[/sup] (the cropped area on the 24MP FF sensor).
[/quote]
[quote name='genotypewriter' timestamp='1295158616' post='5563']
Unless I've overlooked something in the arithmetic, I can't see what the problem is in my calculation.
[/quote]
You are treating the image as if the 104 degrees of the 14mm lens would be distributed
evenly over the 6000 pixels ... would that be the case, your calculation would be correct ...
But now ask yourself ... when taking an image with a superwide ... why do you place persons
in the center ... and what happens if you place them at the sides?
I believe you already gathered it ... the outer 10 degrees on each side consume much more pixels
than the 10 degrees in the center (that comes from the arctan-function in the rectilinear
projection) ... so in order to go from 104 to 73 degrees ... you lose more pixels, since
you have to get rid ouf the outer 15 degrees on both sides.
--> 6000 / (104 / 73) ~= 4240 pixels wide
This is the invalid assumption in your calculation. If you do relatively small changes, you can
approximate the change by a calculation like yours ... but from 14 to 24mm is to much a step
for this.
genotypewriter
Unregistered
[quote name='Rainer' timestamp='1295170615' post='5567']
You are treating the image as if the 104 degrees of the 14mm lens would be distributed
evenly over the 6000 pixels ... would that be the case, your calculation would be correct ...
[/quote]
Good point... I was afraid this was the problem. Thanks for clarifying it.
Are all rectilinear lenses made the same? I once compared the standard 24 1.4L II against unshifted/center of the TS-E 24 3.5L II (which has a much much larger image circle) and was surprised to see that even the rectilinear/volume distortions were identical between the lenses. Wonder if the same holds with other lenses (ignoring barrel distortion), especially for lenses from different manufacturers?
GTW
01-16-2011, 11:06 AM
(This post was last modified: 01-16-2011, 11:06 AM by Rainer.)
[quote name='genotypewriter' timestamp='1295171709' post='5569']
Are all rectilinear lenses made the same?
[/quote]
Well, to be called "rectilinear corrected", the projection used in the lens
must follow the formula
FOV (rectilinear) = 2 * arctan (frame size/(focal length * 2))
as good as possible ... since it is this projection formula that
keeps the straight lines you see straight on the image you take.
So, yes, the better the correction of the lens, the more accurate it
will follow this formula.
genotypewriter
Unregistered
[quote name='Rainer' timestamp='1295175978' post='5571']
Well, to be called "rectilinear corrected", the projection used in the lens
must follow the formula
FOV (rectilinear) = 2 * arctan (frame size/(focal length * 2))
as good as possible ... since it is this projection formula that
keeps the straight lines you see straight on the image you take.
So, yes, the better the correction of the lens, the more accurate it
will follow this formula.
[/quote]
Thanks and that's good news. Because DXO has a correction for the volumetric distortions from wide rectilinears, so it's good to know that at least there's only one type of rectilinear projection for them to attempt to fix.
GTW
My calculation also results 8.16Mp. I like rainer's explanation of "the outer 10 degrees on each side consume much more pixels than the 10 degrees in the center"
' />