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Hank & Rosa
🕑 Added 2023-05-11 09:17:12 +0000 UTC(spoiler alert!)
Hey everyone!
This week I made some more art for the story of Hank & Rosa- you know, the parents of Mindy & Sandy :)
Now so far, they weren't getting along great. But that all changes in this scene...
And this is the sketch I made earlier.
Here Rosa finally realises that her daughters are grown up....and don't need her supervision anymore :)
So, Hank and Rosa talk it out- and decide to focus more on each other again.
And then they retreat to the Guestroom...to 'do their thing again'. Now I won't reveal what that is...but maybe you can already guess from how the room looks?
Or from the hint in my drawing process :)
Well, next time I'll let you know what Hank & Rosa are doing - and how Keen get's mixed up into it. ...I hope you like where this is going :)
All the best!
Robin
Comments
Robin
Haha, yeah ! Keen will always be friendly & thoughtfull - although he might still be a bit too shy for his own good :)
Robin
Yep, I really wanted to make this about Hank and Rosa as a couple & Also because it gives alternative options. However MMF would probably make keen blush so hard he'd pass out ;)
Dacer
Curious to see where this goes. I tend to think of Keen and the rest as characters in a story, rather than "porn puppets", so it will be interesting to see how Hank and Rosa mend their differences. if they do. One of the things I always liked about the writing for Keen is that he's basically a humble, friendly guy, not some sort of skeevy, arrogant pick-up-artist. The size of Keen's (*ahem*) spaceship aside, the story gives the sense that Keen gets so much female attention because he is a friendly, unassuming guy. This matches what I've seen in life too. It's just that the decent guys never brag, while the PUAs never do anything but :-) ... Cheers!
Kevin Taylor
I really appreciate this being about Hank and Rosa's reconciliation and not Keen being the partner for Rosa, as that would seem completely out of character for Keen. Now some MMF action would be an option π
