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About austinthecity

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    Vizier - Molassed Harlequin
  1. QUOTE (Scoop @ Sep 21 2009, 04:16 PM) QUOTE (austinthecity @ Sep 21 2009, 12:00 PM) Brass, similar to this: http://www.hookahforum.com/?showtopic=16597 that one is pretty bad imo. threaded hookahs + common chamber = fail. may as well get a mya... Scoop, can you recommend a quality Syrian then? I have a KM and love it; I'm looking to get a Syrian. I would prefer the feel of a solid stem.
  2. Brass, similar to this: http://www.hookahforum.com/?showtopic=16597
  3. I saw the AF hookah - and I also saw that they improved a lot on them (downstem, removable 2nd tray, etc). However, it still doesn't have a solid brass downstem. Does anyone still sell these?
  4. I've been out of the forum loop for a while but I am back and needing some advice. I apologize for not using the search feature - I'm not sure the results would answer the specific question I have. I also apologize in advance for possibly mentioning a banned vendor; I just would like to be as specific as possible in describing the hookah I am looking for. I am looking for a solid brass stem Syrian like the ones sold by a banned vendor a year or so back. I am looking for a teardrop base if that design is still out there. I liked the stems that came apart in several pieces for easier storage/cleaning. Are there any good vendors out there that have something like this now? Thanks in advance! Austin
  5. I used to just "let it air dry"... but the water droplets won't go away inside the stem. To be effective, you need to get a coat hanger and push a bit of paper towel through the stem. However, air drying the outside is fine.
  6. Turbulent Indigo 30134 would take 2, 2nd being red bottom black top edit: shipping zip would be 30117
  7. in order to prove it, you have to use the squeeze theorem... you can't just say sqrt(x) goes to 0 and it's multiplied to the whole limit goes to 0. the limit may not exist... for squeeze thm: [i'm going to use "<" but it's really less than or equal to] we know -1 < sin(pi/x) < 1 so e^-1 < e^sin(pi/x) < e^1 so sqrt(x)*e^-1 < sqrt(x)*e^sin(pi/x) < sqrt(x)*e and since we know the limit of the outside functions -> 0, it "squeezes" our function to 0 as well.
  8. not quite... you can't divide by 0.
  9. yea.. use the squeeze theorem.. try the parameters: 1/e <= e^sin(pi/x) <= e
  10. find a shoe box or something and tape it to the wall over the smoke detector
  11. i'd be interested in a extra small funnel.. for short sessions and i could just use 2 coals..
  12. one of the things i like about coconaras over exoticas is that you don't have to break them...
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