What width & height to use for equirectangular jpeg?

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Huck

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Nov 12, 2010, 11:11:08 AM11/12/10
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When creating an equirectangular jpeg image, on the “Create Panorama”
tab, after stitching my panorama shows
Width = 17778
Height =8388.
What is the effect of changing Width and Height?
What is the best setting to have?

Erik Krause

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Nov 12, 2010, 2:16:42 PM11/12/10
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Am 12.11.2010 17:11, schrieb Huck:
> When creating an equirectangular jpeg image, on the �Create Panorama�

> tab, after stitching my panorama shows
> Width = 17778
> Height =8388.
> What is the effect of changing Width and Height?

You get a smaller or larger image. If PTGui recommends the above sizes
it means that in the result image you get approximately the same pixel
resolution like in your source images. However, depending on the
intended usage this might be too large. There is no point enlarging an
image beyond the maximum size recommendation since you won't get more
details.

> What is the best setting to have?

The best settings are those that fit your needs. PTGui offers two other
choices aside from "maximum" that might fit your needs better if you
click on "Set optimum size".

--
Erik Krause
http://www.erik-krause.de

JPS

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Nov 15, 2010, 5:48:13 AM11/15/10
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What camera+lens combo do you use to get a 17778x8388 dimensions
proposed as default by PTGui ?

I use a Nikon D700 (full format) with a fisheye Nikon 10.5mm. (DX
format), and PTGui offers an average ~6324x3162 or a "maximum size" of
~6740x3370 pixels !

Just curious...

;)
J-P.

Joergen Geerds

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Nov 15, 2010, 7:28:45 AM11/15/10
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On Nov 15, 5:48 am, JPS <jpscher...@infomaniak.ch> wrote:
> What camera+lens combo do you use to get a 17778x8388 dimensions
> proposed as default by PTGui ?
>
> I use a Nikon D700 (full format) with a fisheye Nikon 10.5mm. (DX
> format), and PTGui offers an average ~6324x3162 or a "maximum size" of
> ~6740x3370 pixels !

for cameras with a 6.4µm pixel pitch (i.e. the canon 5Dm2 or similar),
you can expect about 1000px per mm focal length (as a rule of thumb)
for a 360.

the D700 has of course much bigger pixel, you can use this:

image width in px (short side of the sensor) x number of images x (1 -
overlap) = size of the 360

(use your favorite pano calc to calculate the overlap)

joergen

JPS

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Nov 15, 2010, 11:58:48 AM11/15/10
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Thanks Joergen for the explanation !

Still, it's quite impressive to note that the difference in MP of the
2 cameras is less than the difference in size of the equirect proposed
by PTGui: 12.1 MP vs 21 MP !!! Also, the maximum resolution difference
isn't so big: 5616x3744 pixels (5Dm2) vs 4256x2832 (D700) !!!

The only thing that shows a HUGE difference is indeed the pixel
density: 2.4 MP/cm2 vs 1.4 MP/cm2 !!! Almost the double !

;)
J-P.

Erik Krause

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Nov 15, 2010, 1:59:35 PM11/15/10
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You shouldn't mix linear and square values in such comparisons. Half the
megapixel value means 0.7 times the linear value - which is almost the
case for the two cameras...

Huck

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Nov 16, 2010, 8:25:31 AM11/16/10
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The camera I’m using is a Sony DSC-HX5V. It has a 25mm lens and each
image is 3648X2746 pixels. It takes about 10 images in portrait
orientation to make a 360 degree circle (7 with no overlap). I did 3
circles, (1 horizontal, 1 tilted up, 1 tilted down). The 7 non-
overlapped images would give about 19383 pixels horizontally. So
stitching a lot of images would give a lot of pixels. Using a less
wide angle zoom would require more images to make the circle and would
produce more pixels for the 360 degree sweep.

The fisheye, having fewer images would give lower resolution.

Huck
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