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I am using ArcGIS Pro 1.3.1 to share a Basemap, which has a single layer added from a georeferenced 17.2 Gb TIFF image (has world file), as a Tiled Web Layer on ArcGIS Online.

When I tried to share it I received an error from the Share Web Layer tool:

Staging failed

Upon investigation I realize that this will be due to the low amount of available local disk space that I currently have which is 4.9 Gb. I saw this being consumed as the staging was attempted.

I will start looking for what local disk space I can free up, and then re-test until it works, but is there a way to estimate how much local disk space I will need in advance?

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  • Are you trying to upload the 17,2 GB basemap to ArcGIS Online?
    – BERA
    Sep 17, 2016 at 7:59
  • @Bjorn I'm not sure what the staging process involves, but since I chose SHARE | Publish Web Layer and then Layer Type: Tiles on the Share Web Layer tool I think it is creating tiles on my local disk at various default scales and then uploading those tiles. I originally thought it might have been effectively copying the 17.2 Gb image (and not much more) up to ArcGIS Online and then creating the tiles there but I don't think that would consume local disk.
    – PolyGeo
    Sep 17, 2016 at 8:10

1 Answer 1

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The size of the pyramid will depend on the (geo)tiffs produced, in case the tiles will be the same as original image (in terms of tiff compression, number of bands, etc), and if the pyramid lods are in base 2 (e.g. 4 tile on higher level of detail are merged in one tile on lower level), the final pyramid will have (approximately) 4/3 * size of original image.

Approximately just because resolution of your original image might not be in power of two + metadata will be added to each tile as well.

EDIT: why 4/3...

If we have n pixels on level k, we have n/4 pixels on lod k-1 and so on. If we start on infinite level, we can write an geometric series, the sum of which boils down to 4n/3.

Since we never start on infinite level, but let's say on level k, 4/3 is just the leading part of the whole equation, which reads as

Σki=0(n/4i) = 4n/3 * (1 - 1/(4^(k+1)))

So in theory you end up with at most 4/3 of original number of pixels when you have pyramid with 2x2 tiles making up one on lower level of detail.

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  • It might help if you explain that the sum of the infinite series 1/n^2 is 4/3.
    – Vince
    Sep 17, 2016 at 13:52
  • @Vince I have edited post to include short calculation.
    – matej
    Sep 17, 2016 at 15:04
  • Compression varies per level, too, depending on the image characteristics, interpolation algorithm, and compression algorithm, but it generally washes out to 1/3 increase (unless the pyramid uses a different format/compression scheme, in which case it's an increase of 1/3 of what the base image would have been with those parameters)
    – Vince
    Sep 17, 2016 at 15:34
  • I'm trying to share this again now with 25.4 Gb which should give me a little margin over your estimate of about 22.9 Gb needed. I'll let you know the result either way.
    – PolyGeo
    Sep 18, 2016 at 5:41
  • It worked! It took about 5 hrs 40 mins to complete and reported that it had done so successfully. It's not yet appearing in a map viewer so I'll look into that tomorrow.
    – PolyGeo
    Sep 18, 2016 at 10:55

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