HOME | DD

#fanart #markers #mtmte #traditionalart #traditionalmedia #transformers #inkmarker #traditionalfanart #transformersfanart #markerstraditional #fanarttraditionalart #mtmtetransformers #transformersmorethanmeetstheeye #transformersmtmte #mtmte_art
Published: 2019-09-09 20:48:33 +0000 UTC; Views: 901; Favourites: 84; Downloads: 3
Redirect to original
Description
Rodimus is just thinkingRelated content
Comments: 41
fonaraww [2019-10-06 13:47:53 +0000 UTC]
ahaha this is very funny! and art is very cool XD
π: 0 β©: 1
Aklim04 In reply to N3ur0s [2019-09-17 21:27:55 +0000 UTC]
I think he tries to understand how to use his main processor right, probably Perceptor didn't explained it to him in a good wayΒ Β
π: 0 β©: 1
Aklim04 In reply to N3ur0s [2019-09-18 20:28:10 +0000 UTC]
XDΒ
And thank you for the kind words!Β Β
π: 0 β©: 0
Aklim04 In reply to Battledroidunit047 [2019-09-17 21:25:18 +0000 UTC]
Probably it's "To think or not to think?"Β Β Β
π: 0 β©: 0
Vampiric-Conure [2019-09-10 01:52:58 +0000 UTC]
LMAO! STOP RODDY! You'll kill the last remains of your processor....
π: 1 β©: 1
Aklim04 In reply to Vampiric-Conure [2019-09-10 20:34:33 +0000 UTC]
Too late for it, he started XDD
π: 0 β©: 0
Silvus-Prime [2019-09-10 01:42:54 +0000 UTC]
π: 1 β©: 1
Aklim04 In reply to Silvus-Prime [2019-09-10 20:33:36 +0000 UTC]
Hahahah THANKS!! XDDD I don't think it's perfect, I'd prefer to draw using sharper lines, but I still can't
π: 0 β©: 0
atram95 [2019-09-09 22:40:19 +0000 UTC]
Your draw is Magnificent!! Congratulation I love it!!
π: 0 β©: 1
Aklim04 In reply to atram95 [2019-09-10 20:23:43 +0000 UTC]
Aaww, thank you so much Β Β
π: 0 β©: 0
MysticTopaz04 [2019-09-09 22:08:14 +0000 UTC]
Such a wonderful group shot of people studying Rodimus using his two remaining Braincells.
π: 1 β©: 1
Aklim04 In reply to Ratchet301 [2019-09-09 21:56:04 +0000 UTC]
Hmm.. What a good question .. It seems that it took me about 10 days
π: 0 β©: 1
Aklim04 In reply to Ratchet301 [2019-09-10 20:20:29 +0000 UTC]
If wasn't lazy I would do it earlierΒ Β Β
π: 0 β©: 1