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Published: 2019-03-21 00:39:11 +0000 UTC; Views: 2578; Favourites: 57; Downloads: 3
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----IMPORTANT MESSAGE!! Requests are temporary closed! I'll finish those, which are already accepted, but i won't accept new requests for a while. I hope you understand!----
As requested by aaidnked .
After years of chasing Sonic to win his heart, Rose gave up her hunt and found a new blue boy, Mega Man. His sister Roll found her affection on Sonic instead.
I've not many practice in drawing Sonic or Mega Man characters, but i still hope you enjoy this very ful pic.
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Comments: 26
MetaKnightTheGreat [2024-12-30 05:27:49 +0000 UTC]
π: 1 β©: 1
Stefandorfer In reply to MetaKnightTheGreat [2024-12-30 19:26:55 +0000 UTC]
π: 0 β©: 1
MetaKnightTheGreat In reply to Stefandorfer [2024-12-31 03:39:26 +0000 UTC]
π: 1 β©: 0
MarkinoMoonlight [2024-07-22 11:24:14 +0000 UTC]
π: 0 β©: 1
Stefandorfer In reply to MarkinoMoonlight [2024-07-22 20:40:07 +0000 UTC]
π: 0 β©: 0
Zizum [2021-02-18 20:03:50 +0000 UTC]
π: 0 β©: 1
Stefandorfer In reply to Zizum [2021-02-18 20:11:41 +0000 UTC]
π: 0 β©: 1
Zizum In reply to Stefandorfer [2021-02-18 20:19:38 +0000 UTC]
π: 0 β©: 0
SuperdragonFeuragon [2019-03-27 09:58:28 +0000 UTC]
warte mal... ist roll nicht mega mans schwester?
π: 0 β©: 1
Stefandorfer In reply to SuperdragonFeuragon [2019-03-27 10:47:08 +0000 UTC]
Schon, aber mir ist kein besserer Titel fΓΌr eingefallen und dachte, es lΓ€sst sich dadurch rechtfertigen, dass sie ein Geschwister-"Paar" sind. ^^'
(AuΓerdem wollte ich die Request nicht hinterfragen)
π: 0 β©: 1
Stefandorfer In reply to MrNintMan [2019-03-23 12:46:38 +0000 UTC]
Danke. Wobei ich mit den Sonic Chars nicht ganz zufrieden bin, aber vielleicht bin ich da zu kleinlich. xD
π: 0 β©: 1
MrNintMan In reply to Stefandorfer [2019-03-23 23:50:11 +0000 UTC]
Keine Ursache^^.
Ist doch fΓΌr nen Zeichner normal, dass er nie 100%ig mit seinem Werk zufrieden ist...spornt ihn doch dazu an, sich zu verbessern^^.
π: 0 β©: 0
Stefandorfer In reply to aaidnked [2019-03-21 11:57:57 +0000 UTC]
I'm glad to know that.
π: 0 β©: 0