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Published: 2013-02-14 21:19:47 +0000 UTC; Views: 42654; Favourites: 1356; Downloads: 972
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Description
Dragonball Z Kai recently began broadcasting on British television. It seemed to rekindle the Android 18 fan girl in wifey, so I drew this picture for her! I had intended for it to be finished for her birthday in December, but that didn't happen - so she gets it for Valentine's day instead!I've made a high resolution version available for download. Enjoy!
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Dragonball Z and related characters are property of FUNimation, Toei Animation Co. and Akira Toriyama. No copyright infringement intended.
Brushes used: Skin and hair , dust particles (They're really great!)
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Comments: 63
Wasterbull [2015-03-17 17:04:39 +0000 UTC]
I like it, is a great work and downloading, thanks for share!
π: 0 β©: 0
ElsaIceNagaQueen [2014-10-16 22:38:54 +0000 UTC]
Awesome work! Β I really like this a lot!
π: 0 β©: 1
Shadowqueen17 In reply to Louistrations [2014-10-24 00:07:57 +0000 UTC]
You're welcomeΒ
π: 0 β©: 0
Speedforce1 [2014-03-02 05:39:23 +0000 UTC]
Best android 18 I've seen, I really mean it.
π: 0 β©: 1
Louistrations In reply to Speedforce1 [2014-03-13 13:49:12 +0000 UTC]
Aw, thanks so much~!!!
π: 0 β©: 0
Louistrations In reply to LivintheGoodLife13 [2013-12-31 15:32:51 +0000 UTC]
Thanks so much!
π: 0 β©: 1
LivintheGoodLife13 In reply to Louistrations [2014-01-02 22:30:19 +0000 UTC]
your welcome
π: 0 β©: 0
ColonelBSacquet [2013-08-02 21:36:34 +0000 UTC]
I say, that is a fine-looking android. Or rather, gynoid, if we are to stick with proper terms.
*readjusts eyeglass*
π: 0 β©: 0
hooksnfangs [2013-06-11 18:23:54 +0000 UTC]
Lovely pic of 18! I like how you did her hair!
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Nega5 [2013-06-11 07:53:27 +0000 UTC]
Best Android 18 I've EVER seen. She looks waaaaay better than the original.
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Louistrations In reply to Nega5 [2013-06-30 08:51:54 +0000 UTC]
Thank you, that means a lot to me!
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Katsumiyo [2013-06-03 02:53:21 +0000 UTC]
I love the beautiful lighting and colors here, and I really marvel at all the detail you put into the junk she's sitting on.
π: 0 β©: 1
Louistrations In reply to Katsumiyo [2013-07-08 12:48:42 +0000 UTC]
Thank you very much!
π: 0 β©: 0
Louistrations In reply to LeonsDarkDragon [2013-06-10 22:11:28 +0000 UTC]
Aw, thank you!!
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SammieWeiss [2013-05-19 17:23:11 +0000 UTC]
She looks so much better like this then the other pic
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Louistrations In reply to SuperSaiya-jinwolf [2013-03-28 07:55:10 +0000 UTC]
Thank you!
π: 0 β©: 1
SuperSaiya-jinwolf In reply to Louistrations [2013-03-31 19:30:49 +0000 UTC]
Your welcome :3
π: 0 β©: 0
Louistrations In reply to artBaptiste [2013-05-01 12:38:20 +0000 UTC]
Thank you!
(Had to look up the word "illin" XD)
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BritishMuffin [2013-02-24 23:30:08 +0000 UTC]
Really gorgeous lighting here, love it! And 18, hmmngh. :'D
The mirror looks especially striking, great addition.
Whoa it's showing on telly? I had no idea!
π: 0 β©: 1
Louistrations In reply to BritishMuffin [2013-05-01 12:42:28 +0000 UTC]
Yay, thanks so much!
Sorry for the slow reply! Yep, it's still showing on the Kix television channel, although they keep repeating the Frieza saga over and over and over!
π: 0 β©: 1
BritishMuffin In reply to Louistrations [2013-05-05 21:25:52 +0000 UTC]
Aw man I wish Virgin Media had Kix.. I'd absolutely settle for the Frieza saga. ;D
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Louistrations In reply to BritishMuffin [2013-05-08 19:03:23 +0000 UTC]
Aww, I'm sorry.
I'm not sure what to suggest. You might be able to get a similar experience by screaming at yourself in the mirror for 20 minutes, hehe!
π: 0 β©: 1
BritishMuffin In reply to Louistrations [2013-05-09 23:31:07 +0000 UTC]
Best thing I've read all day. 8)
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Louistrations In reply to Teacup-Princess [2013-02-16 09:22:02 +0000 UTC]
Teehee! Bear do well?
π: 0 β©: 1
Teacup-Princess In reply to Louistrations [2013-03-01 20:40:55 +0000 UTC]
Bear did very very well
π: 0 β©: 0
kissfull [2013-02-15 19:53:41 +0000 UTC]
Thank you for making such a nice piece of Android 18. <3 She looks beautiful.
π: 0 β©: 1
Louistrations In reply to kissfull [2013-02-16 09:21:31 +0000 UTC]
Thank you very much! And you're very welcome!
π: 0 β©: 0
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