# Request to use image

I recently retired from teaching Middle School and High School Science for 37 years in Texas. I am now writing a book series called "The Science Sleuths." Book #5 is called "Adventures in Astronomy." MAY I PLEASE USE YOUR DIAGRAM located at this post with your permission. I will give you full credit for the image.

Randal Lee Gritzner

• Google image search shows this in many places - perhaps one is the original. google.com/… – Keith McClary Nov 5 '20 at 19:46
• There are other images on wikimedia, which are free to use under their license. google.com/… – Keith McClary Nov 5 '20 at 19:52
• All the things here are free to use. – peterh - Reinstate Monica Nov 5 '20 at 19:58
• @peterh-ReinstateMonica They are supposed to be, but some people post copyright images. – Keith McClary Nov 5 '20 at 20:08
• @KeithMcClary Then it is problematic de jure, because the SE content is still free (maybe cc-ba-sa or so). De facto it is free. – peterh - Reinstate Monica Nov 5 '20 at 22:33
• @KeithMcClary peterh is correct. All images here are free to use. If one has been posted without permission of the copyright holder, it is the responsibility of the copyright holder to make a takedown request of anyone who uses it. It is not our responsibility as SE users to enforce unknown potential copyright restrictions on people referencing our site. We can enforce against the user who posted the image originally by requesting the source and/or flagging the answer. – called2voyage Nov 6 '20 at 16:15
• Randal, while my answer and comments here are still correct, you may want to take note in my comments below my answer that I have discovered the potential creator of the image has not been credited. If you need to create your content immediately, you may want to attribute them and us just to be safe. If you have time to wait and check back in a couple weeks, I should have the attribution situation resolved. – called2voyage Nov 6 '20 at 17:00
• That's not a very good image, Randal. It's quite low resolution, and it's not geometrically correct (eg, the focus is too close to the centre). Wouldn't you prefer something of higher resolution, and which is mathematically accurate? Is this for a printed book, or an online book? – PM 2Ring Nov 8 '20 at 10:47

Keep in mind that the content behind any external links that you follow away from this site is not covered by the license, so you have to respect whatever copyright/license restrictions are in place there.

* To the extent that it legally can be. If content has been shared without the copyright holder's permission, then the license is not legally valid.

• Unfortunately, userLTK didn't give any attribution for the image in question. Hopefully they can add that info to their answer... – PM 2Ring Nov 6 '20 at 15:34
• @PM2Ring That would be nice, but it is not technically required if they had the right to share the image without attribution. For example, if they were the image creator or if the image creator waived their rights to attribution. If you know the image creator, you are free to edit attribution in yourself or flag the answer for copyright violation (if you suspect license has not been given). – called2voyage Nov 6 '20 at 16:10
• Good point, but it is nice to be explicit about sources of images to avoid situations like this. ,) I did a Google image search, but it didn't find another copy of that image, only the one in userLTK's answer. – PM 2Ring Nov 6 '20 at 16:21
• @PM2Ring Never mind, I found evidence of prior use: web.archive.org/web/20160616121651/https://… – called2voyage Nov 6 '20 at 16:42
• @PM2Ring The site is still live and still using the image, so I'll reach out to the owner and see if they claim this image as their own. – called2voyage Nov 6 '20 at 16:42
• Nice detective work! – PM 2Ring Nov 6 '20 at 16:44
• @PM2Ring The site lists email reply times of up to a week or longer, so I'll give a couple weeks and if I don't hear back from them I'm going to assume it's theirs and follow their attribution and use guidelines. – called2voyage Nov 6 '20 at 16:45
• @PM2Ring Feel free to ping me if I forget to get back to this in a couple weeks. – called2voyage Nov 6 '20 at 16:51
• @PM2Ring My google image search (first comment under Question) found dozens. – Keith McClary Nov 6 '20 at 17:23
• @KeithMcClary You have to filter by time to exclude sites that used it after it was hosted here, but as you can see above there is one site that used this image prior. – called2voyage Nov 6 '20 at 17:24
• @KeithMcClary Interesting! Maybe Google (or my Samsung phone browser) is overzealous with its filtering... Or I don't know how to use Google Image search correctly. :) – PM 2Ring Nov 6 '20 at 17:28
• Even if content has been shared (on SE) with the copyright holder's permission, that probably does not constitute a CC-BY-SA license (for third parties to use it). Questions about this could go here: opensource.stackexchange.com/questions/tagged/cc-by-sa – Keith McClary Nov 8 '20 at 1:09
• @KeithMcClary If the copyright holder gave full knowing consent to share the content under the CC-BY-SA license, then it's usable by anyone who conforms with the terms of that license. OTOH, I Am Not A Lawyer, but I have read through a few versions of the CC-BY-SA license terms, and reviewed them again when the drama arose due to Stack Exchange changing the license version (and a decade or two ago I even managed to work my way through the Berne Convention :) ) – PM 2Ring Nov 8 '20 at 11:06

Here's another option. ;)

I wrote a small Python program to solve Kepler's equation and create mathematically accurate orbit diagrams. My code produces its output in the SVG vector graphics format, so there's no pixelization when you zoom in, or print it. Unfortunately, this site doesn't support SVG files (although they can be inserted into posts on Stack Overflow), so here's a medium resolution PNG preview.

The actual SVG file is fairly small, and you can see the SVG code & the image it produces, courtesy of SageMathCell, by clicking this link. (The SVG data is actually encoded into the URL). Or you can view the image & its source on GitHub. You can also click the PNG version to see the SVG on GitHub.

Any modern Web browser can display SVG files, just put that code into a plain text file and save it with a .svg or .xml extension, .htm & .html will probably work too. (And it's very easy to embed SVG into a HTML file).

This image & the SVG code is freely shareable & reusable under the conditions of the CC-BY-SA license linked at the foot of this page. That is, you can use it if you give me proper attribution, and where practical, link to this post.

• Why not edit this image into the answer to the linked question on Astronomy main? – gerrit Nov 12 '20 at 10:35
• @gerrit I was hoping to get some feedback first. ;) But I'll do it now, and if there are objections it can always be rolled back. – PM 2Ring Nov 12 '20 at 10:58
• +1 I really like this, wow! I once got worried about embedding SVGs running from arbitrary locations. Sagemath looks anything but arbitrary, in fact it looks really cool! – uhoh Nov 17 '20 at 13:34
• @uhoh Thanks! Sagemath is pretty impressive. And it's fun to be able to write & run Python on my phone. (I'm not a fan of ipython, but I guess it's tolerable). I've also been doing a bit of HTML + JavaScript with it, but SageMathCell won't create a permalink if the HTML has a script in it (including javascript: URIs), and I guess that'd apply to SVGs with JavaScript too. – PM 2Ring Nov 17 '20 at 15:08
• (cont) I was a bit hesitant about directly linking that PNG to the SVG on my GitHub. I guess I could change the link so it goes to the SVG source listing, so people can verify that it's safe before they run it. OTOH, that's only helpful for the people who can read SVG code, and it could be a little disorienting for people who expect to see a higher resolution / higher quality version when they click on an image. – PM 2Ring Nov 17 '20 at 15:08