Getting Image URL in Views with field_views_field I have an content type called Image that has two fields:  Title and Image

I have a Views block created so that it displays the images I want based on my criteria.

What I need is to have the raw image URL available.

Example:

/sites/default/files/background1.jpg


I found this article Rendering Drupal 7 fields (the right way) and quite frankly, I am at a loss.

From the Drupal API:

field_view_field($entity_type, $entity, $field_name, $display = array(), $langcode = NULL)


I know I have to code the views-view-fields--my_custom_view.tpl.php, but everything I do results in an error.  

What needs to go in this file so that I can simply get the URL from the image that I am looking for?

Also  point in the right direction to a tutorial would be great.  Everything I have found assumes you are an expert in Drupal and at this point, everything feels circular.
I was able to do this strictly using Views by creating a relationship to the File Entity.

I have detailed this in another post which can be found here Getting the URL from an Image (or Any File).

This eliminates the need for a URL Formatter module and is D8 compatible out of the box.