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Thread: Pharmacokinetics equation ... anyone?

  1. 07-17-2006, 11:04 AM #1
    Two4the$$
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    Pharmacokinetics equation ... anyone?

    (sigma)[C*e^(ln(1/2)*x/h) - C*e^(ln(1/2)*(x+1)/h)]


    So far ... that is supposedly the basis of the roid calculator... which in my opinion, sucks as a website. Does anyone here disagree with that being the correct equation, or have access to one they know IS correct? I'm going to make a better site. Plus, roidcalc is down...
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  2. 07-17-2006, 12:20 PM #2
    Two4the$$
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    bump - my questions never get replies... lol
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  3. 07-17-2006, 12:22 PM #3
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    wish I could help, but I dont have a clue.. bump
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  4. 07-17-2006, 04:56 PM #4
    Two4the$$
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    bump
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  5. 07-17-2006, 06:19 PM #5
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    dude, what is this equation? maybe an explanation would help.
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  6. 07-17-2006, 08:41 PM #6
    Two4the$$
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    calculates the rate of metabolization of "gear" in your system.

    I.e., what the roidcalc did.

    I want something that calculates for the number of active mg in you after an injection, relative to the injected amount, ester, etc.
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  7. 07-18-2006, 11:17 AM #7
    Two4the$$
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    last bump attempt ... anyone?
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  8. 07-19-2006, 01:00 AM #8
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    I saw something like that online, it makes sense although I am pretty sure its a very rough and general estimate of levels left after X amount of days.

    And....bump

    GP
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  9. 08-12-2006, 11:05 PM #9
    Two4the$$
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    Thanks for the bump bro ... I'm trying again! :-)
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  10. 08-12-2006, 11:07 PM #10
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    Quote Originally Posted by Two4the$$
    (sigma)[C*e^(ln(1/2)*x/h) - C*e^(ln(1/2)*(x+1)/h)]


    So far ... that is supposedly the basis of the roid calculator... which in my opinion, sucks as a website. Does anyone here disagree with that being the correct equation, or have access to one they know IS correct? I'm going to make a better site. Plus, roidcalc is down...

    what the hell? summation of the speed of light times some random exponential equation?

    how is it possible that this has anything to do with steroids?
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  11. 08-12-2006, 11:09 PM #11
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    [QUOTE=Two4the$$](sigma)[C*e^(ln(1/2)*x/h) - C*e^(ln(1/2)*(x+1)/h)]
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  12. 08-12-2006, 11:13 PM #12
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    actually, now that i think about it, the way to calculate the amount of a certain chemical with a given half life is to plug it into an exponential eqn like that. kinda seems like more trouble than its worth for gear though.
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  13. 08-12-2006, 11:34 PM #13
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    [QUOTE=italianplayboy09]
    Quote Originally Posted by Two4the$$
    (sigma)[C*e^(ln(1/2)*x/h) - C*e^(ln(1/2)*(x+1)/h)]
    Sorry I had to. Not worth the time IMO...Sorry
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  14. 08-13-2006, 12:28 AM #14
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    wtf would you need to know that for??
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  15. 08-13-2006, 02:10 AM #15
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    to find out active mg after injection
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  16. 08-13-2006, 02:17 AM #16
    stupidhippo
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    do u have some link to where that formula is explained..
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  17. 08-13-2006, 07:28 AM #17
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    great Scott 1.21 Jiggawatts!!!!!!!!
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  18. 08-13-2006, 09:24 AM #18
    guns626
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    Lmfao!!!! 1.21.jiggawtts
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  19. 08-13-2006, 09:54 AM #19
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    Quote Originally Posted by rock75
    great Scott 1.21 Jiggawatts!!!!!!!!

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  20. 08-13-2006, 09:55 AM #20
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    Quote Originally Posted by Liftnainez
    to find out active mg after injection

    I get that. Why?
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  21. 08-13-2006, 12:05 PM #21
    Two4the$$
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    Don't be such meatheads!!! lol

    I don't need to justify the validity of my plight!
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  22. 08-13-2006, 11:35 PM #22
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    Quote Originally Posted by rock75
    great Scott 1.21 Jiggawatts!!!!!!!!

    man, ive been laughing about this post for the last 6 hours

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