panorama software,virtual tour software
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2004-11-18
#1

Allocating Memory in 3.01

Is it just me or is anyone else having problems with out-of-memory messages in Panoweaver 3.01?

I'm using a machine with 3GB of RAM, but Panoweaver keeps falling over with memory problems.

Is there a way of allocating more memory to the software?  I'm stitching 3 spheres from a Sigma 8mm/Canon 20D to successfully produce seamless images, but can't save the master files...

Any ideas out there...?

Cheers,

 

 


John M
Total VR Ltd
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2004-11-18
#2
John, are you on PC or Mac? If on Windows, do you have enough room available in your Windows temp file (at least one gigabyte)?
It's difficult to diagnose a problem without more info about your machine, operating system, etc.
Robert Jeantet
www.sav.org
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www.spherivue.com
www.webalpa.net/spheres
etc...
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2004-11-18
#3

Sorry Robert - fair point!

My machine is an Athlon 64 PC, 3Gb RAM, 5: 120 Gb hardrives - runing under XP SP2. Certainly not gasping for resources!

The system has several scratch disks, plus its standard system Temp. I don't think this is size limited is it? There is about 2GB clear headroom above the OS on its partition.

I normally avoiding writing temp files/scratch data to mty OS drives, as it's a recipe for messing up your OS in my experience. There seems to be no way to allocate memory/scratch disk to the EasyPano software - or am I missing something?

 

 


John M
Total VR Ltd
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2004-11-19
#4

John,  how large are your original single images ?  Stitched Output size?

I use Intel P4 2.8ghz  XPpro SP1 with 1gig ram and PW 3.01 using 2500 x 5000 output to 35mb file.  Each of my single images are 2048 x 1536 and TIF 9mb each.

Dave


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2004-11-19
#5

Each of the three single images is 4000 x 3000. Output size is as big as we can get the software to deliver - I think the max is 5000 px as you are using.

The software will stitch the images if they're knocked down to 1500 x 1000 ish, but we often find clients want to remap sections of our VR images for print, so we normally stitch 6 or 8 portrait images to give us big PSDs.

The current job - for which we've designed a new rig, based around the Sigma fisheye - is shooting the inside of a range of cars.  Sphere stitching is good for this as the close proximity of the vehicle body panels to the camera makes the multi-pane stitching more difficult, and the shooting sequence shorter.

There's probably an optimal pixel size the software won't fall over at, but the fact that it's falling over at all sugests that there's an upper limit being imposed by a registry line entry, for example. I'm hoping someone has come across this and knows a solution.


John M
Total VR Ltd
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2004-11-19
#6

Thank you for your question.  I don't have an answer .  However, I have forwarded it on to the Panoweaver folks.  Hopefully they will provide a quick response.

Dave


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2004-11-19
#7

The Canon 20D does not have a full size sensor! How is it you are using 3 x 20D Sigma 8mm FE files with Panoweaver?

Regards, Smooth


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2004-11-19
#8
For Panoweaver 3.01, the maximum limit on fisheye image size is
4500*3000, so John's fisheye images can be imported into Panoweaver to stitch panoramas.
The maximum limit on stitched panoramas is 5000*2500, Panoweaver 3.01 can meet John's need.

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2004-11-20
#9

Jane,

Doesn't the 20D sensor have a 1.6 multiplier like the 10D? I don't think you get "full circular" image from a Sigma 8mm FE on a 20D? You certainly don't on a 10D.

Regards, Smooth 


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2004-11-20
#10

I think the 20D has a 1.6 crop factor and for the forum archiving here is how that value is determined

The crop factor is the resultant value by comparing the sensors diagonal length with that of the know diagonal value of the 35mm film  form factor.

35mm form factor is:
24mm high x 36mm length with a diagonal length of 43.3mm.

Height = 24mm x 24mm = 576

Length = 36mm x 36mm = 1296

576 + 1296 = 1872

Square root of 1872 is 43.2666153  or 43.3mm diagonal length

--------

D20 sensor is:
15mm high x 22.5mm length

Hight 15 x 15 = 225

Length 22.5 x 22.5 = 506.25

225 + 506.25 =  731.25

Square root of 731.25 = 27.04163456 or 27.04 diagonal length

Now compare both 35mm and D20's diagonal lengths to determine the crop factor by  43.3mm / 27.04 = 1.600976114   or 1.6

------------ Extended Fun facts ---------

Lets say you are taking pictures with your new Canon D20 using a Canon EF-S 10 - 22mm  effectively that means the 35mm equivelant is 10 x 1.6 = 16mm    to  22 x 1.6 = 35.2.  So this lens is really a zoom lens that is 16 to 35 mm.

Since this is a zoom lens.. chances are that you probably zoomed in a bit or in between values.   Ok,  not to worry.  The images usually have an associated EXIF data set available when you preview the images in the camera "PLAY".  Thats where the image date and time, focal length, and other image specifications are located.  Look for the focal length value.  I think you have to use the 1.6 crop factor value to find the effective 35mm equivelant value.    So an image taken at  FL12mm really is 19.2mm (35mm).


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2004-11-20
#11

More on the D20

Image sizes in pixels

2366 x 3504 8.2mb

1696 x 2544 4.3mb

1152 x 1728 2.0mb

Also it takes RAW  with Fine/normal  but doesnt say what dimensions.  Probably it is the 2366 x 3504 .

Not sure where the 4000 x 3000 is coming from.


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