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• greetings, all! we all know the formulas to compute the number of images required at the horizon, given pano field of view and image overlap, but i would like
Message 1 of 6 , Jan 2, 2007
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greetings, all!

we all know the formulas to compute the number of images required at
the horizon, given pano field of view and image overlap, but i would
like to be able to calculate the numbers of images at non-zero
pitches. simple reasoning indicates they must be less than at the
horizon, but i can neither find nor deduce the formulas. can anyone help?

tia and cheers,
pedro
• ... Some trigonomtry should help - lets see: I assume rectilinear images since they have the same vertical and horizontal FoV throughout the whole image - for
Message 1 of 6 , Jan 2, 2007
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On Tuesday, January 02, 2007 at 11:28, pedro_silva58 wrote:

> we all know the formulas to compute the number of images required at
> the horizon, given pano field of view and image overlap, but i would
> like to be able to calculate the numbers of images at non-zero
> pitches. simple reasoning indicates they must be less than at the
> horizon, but i can neither find nor deduce the formulas. can anyone help?

Some trigonomtry should help - lets see:
I assume rectilinear images since they have the same vertical and
horizontal FoV throughout the whole image - for fisheyes it would be
too hard to calculate exactly since fullframe ones have a far larger
FoV along the edges than through the center and for cuircular ones
you probably won't need to calculate ;-)

Rectilinear images will have least overlap at the side which is
nearest to the horizon. Hence you need to know the pitch angle of
that side. You get this by subtracting half the vertical FoV from the
pitch angle of the image (center). If you get a negative value here
(this is the case if the image crosses the horizon) you are done.
Same formula as for horizontal shooting applies.

For a positive value the circumference which is to cover is by
cosinus of that angle smaller than the horizon. If you for example
pitch the camera 60° up and your image has 30° vertical F0V the lower
side will be at 45° and hence need to cover cos(45) = 0.7071 times
the horizon. Given a 20° horizontal FoV and 20% overlap you need
cos(45)*360/(20*0.8)=15.9 -> 16 images (instead of 23 for the
horizon). If you have the exact (floating point) number of images for
the horizon you can use the factor directly: 22.5 * 0.7071 = 15.9

best regards
--
Erik Krause
Resources, not only for panorama creation:
http://www.erik-krause.de/
• erik, many thanx for your help! ... anyone help? ... yes, of course... ;-) i too was assuming rectilinear, and should have been explicit. ... this is just
Message 1 of 6 , Jan 3, 2007
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erik, many thanx for your help!

--- In PanoToolsNG@yahoogroups.com, "Erik Krause" <erik.krause@...> wrote:
>
> On Tuesday, January 02, 2007 at 11:28, pedro_silva58 wrote:
>
> > we all know the formulas to compute the number of images required at
> > the horizon, given pano field of view and image overlap, but i would
> > like to be able to calculate the numbers of images at non-zero
> > pitches. simple reasoning indicates they must be less than at the
> > horizon, but i can neither find nor deduce the formulas. can
anyone help?
>
> Some trigonomtry should help - lets see:
> I assume rectilinear images since they have the same vertical and
> horizontal FoV throughout the whole image - for fisheyes it would be
> too hard to calculate exactly since fullframe ones have a far larger
> FoV along the edges than through the center and for cuircular ones
> you probably won't need to calculate ;-)

yes, of course... ;-) i too was assuming rectilinear, and should have
been explicit.

> Rectilinear images will have least overlap at the side which is
> nearest to the horizon. Hence you need to know the pitch angle of
> that side. You get this by subtracting half the vertical FoV from the
> pitch angle of the image (center). If you get a negative value here
> (this is the case if the image crosses the horizon) you are done.
> Same formula as for horizontal shooting applies.
>
> For a positive value the circumference which is to cover is by
> cosinus of that angle smaller than the horizon. If you for example
> pitch the camera 60° up and your image has 30° vertical F0V the lower
> side will be at 45° and hence need to cover cos(45) = 0.7071 times
> the horizon. Given a 20° horizontal FoV and 20% overlap you need
> cos(45)*360/(20*0.8)=15.9 -> 16 images (instead of 23 for the
> horizon). If you have the exact (floating point) number of images for
> the horizon you can use the factor directly: 22.5 * 0.7071 = 15.9

this is just what i needed! i guessed there should be a cosine
somewhere, but didn't quite know where. oh, and subtracting half the
vertical fov is a very thoughtful touch, but may not be strictly
necessary. with it, we get the minimum overlap, without, an average
overlap (i think). at any rate, a first rate reply, as usual!

thanx again,
pedro
• ... In my example with 20° horizontal FoV you would need only 12 images at 60° but 16 images at 45° (lower edge). 12 images cover 240° (without overlap!)
Message 1 of 6 , Jan 3, 2007
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On Wednesday, January 03, 2007 at 20:21, pedro_silva58 wrote:

> this is just what i needed! i guessed there should be a cosine
> somewhere, but didn't quite know where. oh, and subtracting half the
> vertical fov is a very thoughtful touch, but may not be strictly
> necessary. with it, we get the minimum overlap, without, an average
> overlap (i think).

In my example with 20° horizontal FoV you would need only 12 images
at 60° but 16 images at 45° (lower edge). 12 images cover 240°
(without overlap!) but at 45° you need 254° (=360*cos(45)). This
means you won't have any overlap at the bottom side.

best regards
--
Erik Krause
Resources, not only for panorama creation:
http://www.erik-krause.de/
• ... erik, i appreciate your efforts in getting this right. i agree that subtracting half the vertical fov, while not _always_ necessary, is the right way to
Message 1 of 6 , Jan 3, 2007
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--- In PanoToolsNG@yahoogroups.com, "Erik Krause" <erik.krause@...> wrote:
>
> On Wednesday, January 03, 2007 at 20:21, pedro_silva58 wrote:
>
> > this is just what i needed! i guessed there should be a cosine
> > somewhere, but didn't quite know where. oh, and subtracting half the
> > vertical fov is a very thoughtful touch, but may not be strictly
> > necessary. with it, we get the minimum overlap, without, an average
> > overlap (i think).
>
> In my example with 20° horizontal FoV you would need only 12 images
> at 60° but 16 images at 45° (lower edge). 12 images cover 240°
> (without overlap!) but at 45° you need 254° (=360*cos(45)). This
> means you won't have any overlap at the bottom side.

erik,

i appreciate your efforts in getting this right. i agree that
subtracting half the vertical fov, while not _always_ necessary, is
the right way to do it, and is safer.

on the other hand... how many images you need, depends on overlap
(yes, of course ;-) ). i understood your example to be an
illustrative example only. if you regularly shoot for only 20%
overlap you are a braver man than i am.

cheers,
pedro
• ... I regularly use 25% but I did shoot panoramas with as less as 5% overlap (well, not completely intentional I must admit ;-) I occasionally use my Zenitar
Message 1 of 6 , Jan 5, 2007
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On Wednesday, January 03, 2007 at 23:22, pedro_silva58 wrote:

> erik,
>
> i appreciate your efforts in getting this right. i agree that
> subtracting half the vertical fov, while not _always_ necessary, is
> the right way to do it, and is safer.
>
> on the other hand... how many images you need, depends on overlap
> (yes, of course ;-) ). i understood your example to be an
> illustrative example only. if you regularly shoot for only 20%
> overlap you are a braver man than i am.

I regularly use 25% but I did shoot panoramas with as less as 5%
overlap (well, not completely intentional I must admit ;-) I
occasionally use my Zenitar with a 5-around workflow which gives

However, even if you use 25% my example would result in 12 images if
you calculate with the image center tilted up to 60° and hence leave
holes. And the problem gets worse if you use a wider angle lens:
given 40°x60° FoV, 60° tilt up and 30% overlap you get 7 images if
you calculate with the image center. The lower edge will be 30° above
the horizon needing 312° total Fov but with 7 images you only get
280°.

Same for 60°x80° and 30% overlap: tilted up 60° you need 5 images for
30° overlap which results in 300° coverage where you need 339°.

best regards
--
Erik Krause
Resources, not only for panorama creation:
http://www.erik-krause.de/
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