I want to change the lens_angle in a 3d scene, but with the objects
staying the same size, relative to the screen…
There has to be a simple formula for that, but I can’t seem to figure
out the relation between object size and lens_angle.
any clues?

brecht.

Using trigonometry, it should be simple to calculate the ratios.

sin(theta) = Oposite/adjacent

sin(LensAngle) = ( WidthOfObject/2.0 ) / DistanceFromCamera

sin(LensAngle) * DistanceFromCamera = ( WidthOfObject/2.0 )

sin(LensAngleOld) * DistanceFromCameraOld= ( WidthOfObjectOld/2.0 )

sin(LensAngleNew)*DistanceFromCameraNew = ( WidthOfObjectNew/2.0 )

since we want the object to stay the same size :
WidthOfObjectNew = WidthOfObjectOld

sin(LensAngleOld) * DistanceFromCameraOld
=
sin(LensAngleNew)*DistanceFromCameraNew

Now we solve for DistanceFromCameraNew or LensAngleNew. Since the lensangle and distance from camera are related. We must solve for one or the other.

LensAngleNew= arcsin(sin(LensAngleOld) * DistanceFromCameraOld/DistanceFromCameraNew )

DistanceFromCameraNew = DistanceFromCameraOld * sin
(LensAngleOld)/sin(LensAngleNew)

For example. If your object is 3 units away from the camera, and your old camera angle was 45.

Now lets say you want a new camera angle of 90.
you must move the object to

3*sin(45)/sin(90)
the actual new distance is 2.121

The easist way of thinking of this is

The relative distance is sin(oldLensAngle)/sin(NewLensAngle)

You should double check my math…

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On Aug 22, 2006, at 4:00 AM, Jason Schmitt wrote:
>
> If you want to change the lens angle but keep the relative size on
> the screen the same, you must move the object closer or farther away.
>
> Using trigonometry, it should be simple to calculate the ratios.
>
>
> sin(theta) = Oposite/adjacent
>
> sin(LensAngle) = ( WidthOfObject/2.0 ) / DistanceFromCamera
>
> sin(LensAngle) * DistanceFromCamera = ( WidthOfObject/2.0 )
>
>
> sin(LensAngleOld) * DistanceFromCameraOld= ( WidthOfObjectOld/2.0 )
>
> sin(LensAngleNew)*DistanceFromCameraNew = ( WidthOfObjectNew/2.0 )
>
>
> since we want the object to stay the same size :
> WidthOfObjectNew = WidthOfObjectOld
>
>
>
>
> sin(LensAngleOld) * DistanceFromCameraOld
> =
> sin(LensAngleNew)*DistanceFromCameraNew
>
>
> Now we solve for DistanceFromCameraNew or LensAngleNew. Since the
> lensangle and distance from camera are related. We must solve for
> one or the other.
>
>
>
> LensAngleNew= arcsin(sin(LensAngleOld) * DistanceFromCameraOld/
> DistanceFromCameraNew )
>
> DistanceFromCameraNew = DistanceFromCameraOld * sin
> (LensAngleOld)/sin(LensAngleNew)
>
>
> For example. If your object is 3 units away from the camera, and
> your old camera angle was 45.
>
> Now lets say you want a new camera angle of 90.
> you must move the object to
>
> 3*sin(45)/sin(90)
> the actual new distance is 2.121
>
>
> The easist way of thinking of this is
>
>
> The relative distance is sin(oldLensAngle)/sin(NewLensAngle)
>
>
> You should double check my math…
>
>

http://en.wikipedia.org/wiki/Dolly_zoom

best,
dan

On 8/23/06, Brecht Debackere wrote:
> Hm, that doesn’t seem to work… implemented the little example you
> gave at the bottom, and the results are different…
> The first numberbox is the distance, 2nd and 3rd the initial and
> changed lens angle.
>
> #P window setfont “Sans Serif” 9.;
> #P window linecount 1;
> #P newex 260 302 76 196617 unpack 0. 0. 0.;
> #P newex 260 282 61 196617 pak 0. 0. 0.;
> #P flonum 332 258 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
> #P flonum 296 258 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
> #P flonum 260 258 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
> #P flonum 260 347 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
> #P newex 260 323 140 196617 expr $f1* sin($f2)/sin($f3);
> #P fasten 6 2 0 2 331 321 395 321;
> #P fasten 6 1 0 1 298 321 330 321;
> #P connect 6 0 0 0;
> #P connect 0 0 1 0;
> #P fasten 4 0 5 2 337 277 315 277;
> #P fasten 3 0 5 1 301 277 290 277;
> #P connect 5 0 6 0;
> #P connect 2 0 5 0;
> #P window clipboard copycount 7;
>
> On Aug 22, 2006, at 4:00 AM, Jason Schmitt wrote:
> >
> > If you want to change the lens angle but keep the relative size on
> > the screen the same, you must move the object closer or farther away.
> >
> > Using trigonometry, it should be simple to calculate the ratios.
> >
> >
> > sin(theta) = Oposite/adjacent
> >
> > sin(LensAngle) = ( WidthOfObject/2.0 ) / DistanceFromCamera
> >
> > sin(LensAngle) * DistanceFromCamera = ( WidthOfObject/2.0 )
> >
> >
> > sin(LensAngleOld) * DistanceFromCameraOld= ( WidthOfObjectOld/2.0 )
> >
> > sin(LensAngleNew)*DistanceFromCameraNew = ( WidthOfObjectNew/2.0 )
> >
> >
> > since we want the object to stay the same size :
> > WidthOfObjectNew = WidthOfObjectOld
> >
> >
> >
> >
> > sin(LensAngleOld) * DistanceFromCameraOld
> > =
> > sin(LensAngleNew)*DistanceFromCameraNew
> >
> >
> > Now we solve for DistanceFromCameraNew or LensAngleNew. Since the
> > lensangle and distance from camera are related. We must solve for
> > one or the other.
> >
> >
> >
> > LensAngleNew= arcsin(sin(LensAngleOld) * DistanceFromCameraOld/
> > DistanceFromCameraNew )
> >
> > DistanceFromCameraNew = DistanceFromCameraOld * sin
> > (LensAngleOld)/sin(LensAngleNew)
> >
> >
> > For example. If your object is 3 units away from the camera, and
> > your old camera angle was 45.
> >
> > Now lets say you want a new camera angle of 90.
> > you must move the object to
> >
> > 3*sin(45)/sin(90)
> > the actual new distance is 2.121
> >
> >
> > The easist way of thinking of this is
> >
> >
> > The relative distance is sin(oldLensAngle)/sin(NewLensAngle)
> >
> >
> > You should double check my math…
> >
> >
>
>

–
***
http://danwinckler.com
http://share.dj

http://idmi.poly.edu

On Aug 24, 2006, at 7:27 AM, Dan Winckler wrote:

> are you going for the look a ‘dolly zoom’ or ‘trombone effect’? :)
>
> http://en.wikipedia.org/wiki/Dolly_zoom
>
> best,
> dan
>
>
> On 8/23/06, Brecht Debackere wrote:
>> Hm, that doesn’t seem to work… implemented the little example you
>> gave at the bottom, and the results are different…
>> The first numberbox is the distance, 2nd and 3rd the initial and
>> changed lens angle.
>>
>> #P window setfont “Sans Serif” 9.;
>> #P window linecount 1;
>> #P newex 260 302 76 196617 unpack 0. 0. 0.;
>> #P newex 260 282 61 196617 pak 0. 0. 0.;
>> #P flonum 332 258 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
>> #P flonum 296 258 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
>> #P flonum 260 258 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
>> #P flonum 260 347 35 9 0 0 0 3 0 0 0 221 221 221 222 222 222 0 0 0;
>> #P newex 260 323 140 196617 expr $f1* sin($f2)/sin($f3);
>> #P fasten 6 2 0 2 331 321 395 321;
>> #P fasten 6 1 0 1 298 321 330 321;
>> #P connect 6 0 0 0;
>> #P connect 0 0 1 0;
>> #P fasten 4 0 5 2 337 277 315 277;
>> #P fasten 3 0 5 1 301 277 290 277;
>> #P connect 5 0 6 0;
>> #P connect 2 0 5 0;
>> #P window clipboard copycount 7;
>>
>> On Aug 22, 2006, at 4:00 AM, Jason Schmitt wrote:
>> >
>> > If you want to change the lens angle but keep the relative size on
>> > the screen the same, you must move the object closer or farther
>> away.
>> >
>> > Using trigonometry, it should be simple to calculate the ratios.
>> >
>> >
>> > sin(theta) = Oposite/adjacent
>> >
>> > sin(LensAngle) = ( WidthOfObject/2.0 ) / DistanceFromCamera
>> >
>> > sin(LensAngle) * DistanceFromCamera = ( WidthOfObject/2.0 )
>> >
>> >
>> > sin(LensAngleOld) * DistanceFromCameraOld= ( WidthOfObjectOld/2.0 )
>> >
>> > sin(LensAngleNew)*DistanceFromCameraNew = ( WidthOfObjectNew/2.0 )
>> >
>> >
>> > since we want the object to stay the same size :
>> > WidthOfObjectNew = WidthOfObjectOld
>> >
>> >
>> >
>> >
>> > sin(LensAngleOld) * DistanceFromCameraOld
>> > =
>> > sin(LensAngleNew)*DistanceFromCameraNew
>> >
>> >
>> > Now we solve for DistanceFromCameraNew or LensAngleNew. Since the
>> > lensangle and distance from camera are related. We must solve for
>> > one or the other.
>> >
>> >
>> >
>> > LensAngleNew= arcsin(sin(LensAngleOld) * DistanceFromCameraOld/
>> > DistanceFromCameraNew )
>> >
>> > DistanceFromCameraNew = DistanceFromCameraOld * sin
>> > (LensAngleOld)/sin(LensAngleNew)
>> >
>> >
>> > For example. If your object is 3 units away from the camera, and
>> > your old camera angle was 45.
>> >
>> > Now lets say you want a new camera angle of 90.
>> > you must move the object to
>> >
>> > 3*sin(45)/sin(90)
>> > the actual new distance is 2.121
>> >
>> >
>> > The easist way of thinking of this is
>> >
>> >
>> > The relative distance is sin(oldLensAngle)/sin(NewLensAngle)
>> >
>> >
>> > You should double check my math…
>> >
>> >
>>
>>
>
>
> —
> ***
> http://danwinckler.com
> http://share.dj
> http://idmi.poly.edu
>
>

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