Is downloading raw Sentinel2-L2A data free with SentinelHub?

Apps / Services Python SDK
1

Hi!

I’m interested in downloading raw Sentinel2-L2A data (.jp2) from S3 using SentinelHub. Do I need to set up a SentinelHub account or purchase any license if I’m only using AwsTileRequest method in Python?

request = AwsTileRequest(tile=tile_name, time=time, aws_index=aws_index,bands=bands, metafiles=metafiles, data_folder=data_folder, data_source=DataSource.SENTINEL2_L2A)

I’m aware that the Sentinel S3 is requester-pays bucket but will be there other charges applied (apart from the AWS ones) by the SentinelHub if I only want to download the raw data?

2

If you wan tto download jp2, you do not need Sentinel Hub account. You will need AWS account due to requester pays policy.

1 Like
3

Some more info can be found in sentinelhub-py documentation.

TL;DR In this case only AWS will charge you for data transfer.

1 Like
4

Could you expand on this?
I would like to use the sentinelhub python package, but I was required to create a free trial account. Am i still able to query and download data using the sentinelhub package when this free trial is done?

I understand I can access the AWS server and download data (using awscli or boto3) on requester pays basis. If I were to use this route, would I go to file metadata.xml to get bounding geometries and cloud cover, then use this information to filter for scenes (e.g. at s3://sentinel-s2-l2a/tiles/10/S/DG/2018/12/31/0/)?

5

Hi Alex,

The Sentinel Hub python package is primarily developed to be used with the Sentinel Hub APIs. As I’m sure you know, there is some functionality for accessing the data straight from AWS. This is documented here. You should be able to achieve what you want to do with these tools.

However, it is important to note that many of the functionalities in this module have been depreciated and are no longer actively maintained by us, so couldn’t be guaranteed as a long term option.