From:                 Self 
To:               eos@avocado.pc.Helsinki.FI
Subject:          Re: Question to users of 24TS L
Send reply to:    w.j.markerink@a1.nl
Date sent:        Sat, 19 Apr 1997 20:11:57

On 19 Apr 97 at 21:09, khtay@VNET.IBM.COM wrote:

> Roughly how tall (10-20 floors?) can a building be for the 24 TS L
> in shifted position to correct the converging verticals from say 30 metres?
> 
> Does the 24 TS L in shifted position gives an apparent (or real) wider
> angle of view than a 20mm?




1) First, Markerink's Shift Thesis: 

    "The Gain Is Larger Than The Loss."

(to be explained later)


2) Second, our core math:

    Angle of view = 2 x arctan ( L / 2 / F ) 

L = size of frame 
    [height (H) for vertical angle, width (W) for horizontal angle,
    and diagonal (D = SQR (H^2 + W^2)) for diagonal angle]

F = focal length 


3) In case of a static 24mm lens:

H = 24mm
W = 36mm
D = 43.27mm
F = 24mm

Vertical angle of view = 53.1°
Horizontal angle of view = 73.7° 
Diagonal angle of view = 84.1° 

4) In case of maximum shift in both directions, our 'working frame' is:
H = 46mm (24+11+11)
W = 58mm (36+11+11)
D = 65.27mm (43.27+11+11)
F = 24mm

Vertical angle of view = 87.6°
Horizontal angle of view = 100.8°
Diagonal angle of view = 107.3°


5) Total angular gain is:
Vertical gain = 87.6 - 53.1 = 34.5°
Horizontal gain = 100.8 - 73.7 = 27.1°
Diagonal gain = 107.3 - 84.1 = 23.2°


6) Obviously, you can only gain half of that, since you can only 
shift in either direction, never both (the image of the Magdeburger 
half spheres vacuum experiment pops up in my head for no reason). 
So our real world angle of gain is: 
Vertical gain = 17.3°
Horizontal gain = 13.6°
Diagonal gain = 11.6°


7) To translate this back in a gain of actual meters for a given 
distance, we need another line of math: 

   gain 
--------------   = tan (angle of gain)
distance

or

gain = distance x tan (angle of gain)

For your example of 30m distance, assuming shift in portrait mode:

gain = 30 x tan (13.6°) = 7.26m

Would you have shifted in landscape mode:

gain = 30 x tan (17.3°) = 9.32m

[Total height of the building goes similar, the only problem is that
for the life of me I can't calculate what the total angle of shifted
view is; thusfar we only calculated the gain on top, not the loss on
the bottom, and those are *not* equal (see 10). This means that a %
expression is also impossible, as that requires a total height/angle
of view as well.] 


9) Of course, if we combine the math in 3) and 7), we can simplify this 
multiplication factor to the following expression:

    tan [ arctan (Ls/2/F) - arctan (L/2/F) ]

[with Ls the double-shifted frame size, and L the original frame size]


10) For those who still didn't leave the room (I warned you!): there 
are three ways to explain Markerink's Shift Thesis ("The Gain Is Larger 
Than The Loss").

First, it has been stated before that the extreme tilt settings cause 
distortion similar to super wide angles, ie 'pulling' of objects in 
the edges (they look taller (vertical shift) or longer (horizontal 
shift) than in real life). This pulling only occurs on the side of 
shift, but the 'compression' on the other side of the frame is not 
equivalent to the pulling. Hence more pulling than compression, which 
can only be explained with an increase of the angle of view.

Second: if you look at the math in 3) carefully, you see that each
additional mm of shift results in a more than linear increase in
angle of view. This means that one mm shift 'loss' near the center
of the image (where you shift 'away from') results in less angular
loss as the same mm shift gains on the far edge.

Third: similar proof as above can be found in the difference between 
gain in meters in portrait mode vs landscape mode (7.26m vs 9.32m).
Tilt in mm's is the same, yet the gain on the shorter end is larger(?). 

Willem (you may go now) Jan

Date sent:        Mon, 21 Apr 1997 09:44:42 +0200 (MET DST)
From:             Janusz Martyniak 
To:               eos@avocado.pc.helsinki.fi
Subject:          Tilted math
Send reply to:    eos@avocado.pc.Helsinki.FI



Dear Willem-Jan,

 Youe have calculated the angle of view for  24mm lens and 24TS-E lens 
 to be

> Horizontal angle of view = 73.7° 
> Diagonal angle of view = 84.1° 

  and 

> Horizontal angle of view = 100.8°
> Diagonal angle of view = 107.3°

  respectively.

To calculate the gain in meters one should use  the formula :

> 
> gain = distance x tan (angle of gain)
> 
 But not for the gain but for the total angle of view, since tan is not a
linear function ...

 so for the distance of 30 meters,

  H_static = 30 * tan ( 73.7/2 ) = 22.5 [m]

  H_shift_up =30 * tan(100.8/2)  = 36.3 [m]

So one is able to photograph a building as tall as 22.5 m from the
distance of 30 m ( in portrait mode ) and a 36.3 m building with 
the maximum shift of 11mm. 
 The interesting thing is, that the gain is over 38% ( (36.3-22.4)/36.3 )
so even IF the loss were = gain, than one could not be able to eliminate
the ground level from the picture for this particular situation. As far as
the top part of the picture is concerned the shift operation is equivalent
to lifting the camera 13.9m (!) above the ground level and using the
static 24mm lens. One would need 50%  to have only a building 
(55m high, since we have total of 73.7 degrees in this case, look above )
in the frame ( still for loss=gain ).

 I have to perform some measurments with my 24TS-E to test this theory
however ..

              Janusz Martyniak

From:             Willem-Jan Markerink 
To:               eos@avocado.pc.Helsinki.FI
Date sent:        Mon, 21 Apr 1997 13:15:38 +0000
Subject:          Re: Tilted math
Send reply to:    eos@avocado.pc.Helsinki.FI

On 21 Apr 97 at 9:44, Janusz Martyniak wrote:

> 
> 
> Dear Willem-Jan,
> 
>  Youe have calculated the angle of view for  24mm lens and 24TS-E lens 
>  to be
> 
> > Horizontal angle of view = 73.7° 
> > Diagonal angle of view = 84.1° 
> 
>   and 
> 
> > Horizontal angle of view = 100.8°
> > Diagonal angle of view = 107.3°
> 
>   respectively.
> 
> To calculate the gain in meters one should use  the formula :
> 
> > 
> > gain = distance x tan (angle of gain)
> > 
>  But not for the gain but for the total angle of view, since tan is not a
> linear function ...

Yep, I realized that after my previous posting this morning.
I used the angle-difference to calculate gain in meters, whereas I 
should calculate the meter-difference straight away.
My method results in less gain than there actually is.
I somehow feel this non-linear tan function also represents the 
pulling effect in the corners quite effectively....might be a wild 
guess though.

>  so for the distance of 30 meters,
>  
>   H_static = 30 * tan ( 73.7/2 ) = 22.5 [m]
> 
>   H_shift_up =30 * tan(100.8/2)  = 36.3 [m]
> 
> So one is able to photograph a building as tall as 22.5 m from the
> distance of 30 m ( in portrait mode ) and a 36.3 m building with 
> the maximum shift of 11mm. 

Yep, correct. A gain on top of 13.8m.

>  The interesting thing is, that the gain is over 38% ( (36.3-22.4)/36.3 )

IMO that should be based on the initial 22.5m, ie a 13.8/22.5 = 61.3% 
increase over the non-shifted setting.
Or half of that given the total initial height (which is 2x22.5m).

> so even IF the loss were = gain, than one could not be able to eliminate
> the ground level from the picture for this particular situation. 

That can also be proved differently: the amount of shift never 
exceeds the center of the image, not even in landscape shift (11mm 
shift on a 24mm frame means that the horizont is still present on the 
bottom 1mm.

Btw, it keeps bugging me that none of the above really answers what the 
loss on the bottom is....

> As far as
> the top part of the picture is concerned the shift operation is equivalent
> to lifting the camera 13.9m (!) above the ground level and using the
> static 24mm lens. One would need 50%  to have only a building 
> (55m high, since we have total of 73.7 degrees in this case, look above )
> in the frame ( still for loss=gain ).
> 
>  I have to perform some measurments with my 24TS-E to test this theory
> however ..

Real world testing would take the suspense out of all this....;-)) 

-- 
Bye,

       _/      _/       _/_/_/_/_/       _/_/_/_/_/
     _/  _/  _/               _/       _/  _/  _/
     _/  _/ illem    _/     _/ an    _/  _/  _/ arkerink
                     _/_/_/  



      The desire to understand 
is sometimes far less intelligent than
     the inability to understand



[note: 'a-one' & 'en-el'!]

Date sent:        Mon, 21 Apr 97 02:16:32 UT
From:             "Jeff Conrad" 
To:               "EOS Mailing List" 
Subject:          Re: Question to users of 24TS L
Send reply to:    eos@avocado.pc.Helsinki.FI

Yet another approach ...
What's easy to explain with a picture isn't quite so simple with words
alone.

As a lens is shifted up, parallel to the film plane, the subject moves
down in the image by a corresponding amount (or if you prefer, the image
moves up).  The ratio of subject movement to lens shift is given by the
ratio of subject distance to lens focal length.

As Frank Sheeran mentioned, one possible benefit of lens shift is
minimizing convergence of vertical lines from a closer camera position
than would be possible without lens shift.  However, as he also pointed
out, choosing the shooting distance isn't always an option.  Lens shift
can still come to the rescue.

For a 24 mm TS-E lens, if we use a distance of 24 m, the subject will
move down 1 m for every mm that the lens is shifted up.  The maximum
subject height isn't affected by shifting; for a 24 mm lens at 24 m, the
subject heights are

    24 m for a horizontal composition
    36 m for a vertical composition

Of course, the above subject heights are possible only if the entire
subject is above ground level!  To do this without lens shift, the
camera must be 12 m above ground for the horizontal composition, and 18
m above ground for a vertical composition.  These positions might be
possible if there is a nearby building to which one can gain access, and
the pictures taken from an upper floor; some people swear by this
approach.  Such a building isn't always available, however, and at least
to my eyes, the resulting images usually look like vacation shots from a
hotel window.

If the 24 mm TS-E is shifted by the maximum amounts (11 mm for
horizontal, 8 mm for vertical), the required camera heights are reduced
to

     1 m for the horizontal composition
    10 m for the vertical composition

Alternatively, if the camera remains near ground level, the resulting
images will show more of the building and less unwanted foreground.

The 1 m camera position for the horizontal image should easily be within
the range of any reasonable tripod, so the problem is solved.

For the vertical composition, the nearby building might still make some
sense; to my eyes, at least, a building photographed from 10 m looks a
bit more natural than when done from 18 m.  But let's face it, such a
building isn't usually available.  In that case, the camera can be
pointed up slightly to cover the entire building.  The resulting
convergence is much less than without any lens shift.

Incidentally, although the 24 mm TS-E can be mechanically shifted 11 mm
in any direction, when this is done parallel to the long dimension of
the film, the image corners are outside the lens's 58.6 mm image circle,
and the vignetting is quite noticeable (don't even think of using a
filter).

Jeff Conrad
jeff_conrad@msn.com

From:             Willem-Jan Markerink 
To:               eos@avocado.pc.helsinki.fi
Date sent:        Mon, 21 Apr 1997 11:27:58 +0000
Subject:          Re: Question to users of 24TS L
Send reply to:    eos@avocado.pc.Helsinki.FI

On 21 Apr 97 at 2:16, Jeff Conrad wrote:

> For a 24 mm TS-E lens, if we use a distance of 24 m, the subject will
> move down 1 m for every mm that the lens is shifted up.  The maximum
> subject height isn't affected by shifting; for a 24 mm lens at 24 m, the
> subject heights are
> 
>     24 m for a horizontal composition
>     36 m for a vertical composition

The maximum subject height *is* affected by the amount of shift.
I think I even have to reverse my statement that the gain on top is 
larger than the loss on the bottom, as 'pulling' objects on top means that 
the magnification increases, with a relative magnification decrease 
on the bottom. This means that the gain on top is less than the gain 
on the bottom.
Main problem in all this is taking a linear approach, which IMO only 
applies if you take the entire frame, but not if you shift inside the 
image circle, using only part of the geometry.
The error might not be big, but my tan/arctan formula's showed a 
much lower gain (7.26m on 30m distance vs your 11m on 24m distance), 
*and* claims the same gain in both portrait and landscape shift (1m 
gain pro mm shift, whereas my result was 9.32m in landscape mode 
(7.26m in portrait mode)).


> If the 24 mm TS-E is shifted by the maximum amounts (11 mm for
> horizontal, 8 mm for vertical), the required camera heights are reduced

FWIW: one can shift 11mm in both directions....provided stopping down.


> Incidentally, although the 24 mm TS-E can be mechanically shifted 11 mm
> in any direction, when this is done parallel to the long dimension of
> the film, the image corners are outside the lens's 58.6 mm image circle,
> and the vignetting is quite noticeable (don't even think of using a
> filter).

I am pretty sure this 58.6mm is a wide open number, that is why you
must stop down (in contrast, the effect of vignetting as a result of
filters only increases when stopping down, as it makes the rim look
sharper and more defined).
I have no problems when shifting 11mm on the long side of the frame, 
nor using a polarizing filter. Of course, you are probably not done 
with buying a very slim polarizer, I use an extra-large B+W circular 
pol. I think I once tested this filter with tilt and shift combined 
in one direction (in contrast to the factory setting where tilt and 
shift are offset 90 degrees, it only takes 4 screws to do this 
modification yourself), and only then the filter problem gets nasty I 
recall. Going with 4x4" or larger foil seems the most likely solution 
in that case.

-- 
Bye,

       _/      _/       _/_/_/_/_/       _/_/_/_/_/
     _/  _/  _/               _/       _/  _/  _/
     _/  _/ illem    _/     _/ an    _/  _/  _/ arkerink
                     _/_/_/  



      The desire to understand 
is sometimes far less intelligent than
     the inability to understand



[note: 'a-one' & 'en-el'!]

If you have any question, remark, comment, want to share some philosophy or just want to express your opinion about these pages, feel free to send email to: w.j.markerink @ a1.nl