Misc Half life question?

[keeping]

yeah ikr? my cousin gets clonazepam for his seizures?
perhaps theyre more preventative then to deal with an acute situation like actually fitting?

 

[Scrofula]

Yeah, epileptics are all different, and they'd want to match the drug kinetics to the illness.

I would bet most don't get benzos, though, they'd get stuff like your lamotragine instead, since benzos are always the best thing long-term.
And didn't mean to suggest Klonopin was bad or anything, just that you have many options.

 

[Mr Dobalina]

Yeah Im prescribed Diazepam but Klonopin I have to buy off someone & I dont wanna spend money because I'm tight ha. I guess I'll just settle for Diazepam.

Fuck I cant wait til this Lamictal is out of my system so I can take drugs.

 

[Scrofula]

Wait, the Valium isn't good enough?

I also can't find a good reason you'd need the lamotragine gone before taking anything else. There's a lot of woo about opiates out there, but mostly to see if it helped withdrawals (it doesn't).

 

[Mr Dobalina]

Wait, the Valium isn't good enough?

I also can't find a good reason you'd need the lamotragine gone before taking anything else. There's a lot of woo about opiates out there, but mostly to see if it helped withdrawals (it doesn't).

Valium is good to prevent me from having seizures. I just thought Klonopin was better. I need the lamotrigine gone because it blocks most drugs...i say this from personal experience and reports from people on the net saying they had the same problem.

 

[Scrofula]

I'll take your word for it (the drug interaction part). (And I didn't look very hard anyway)

 

[Lacquerman Steve]

But... back to half-lives...

The math equation for exponential growth or decay (growth might be interest on a bank account if compounded continuously, or populations with a birth:death ratio higher than 1:1... decay includes things like radioactive decay, and also metabolism) is:

N=No * e^(k*t)

where N is the number (amount) of something at time "t", No is the number (amount) at the start time, e is the exponentiation constant, t is the time, and k is the growth rate constant (a negative number applies to decay calculations) for the situation being considered.

Rearranging that, gives this: N / No = e^(k*t)

Given a half-life (t) of 29 hours, we need to find k. So, we need to separate the k from the t in that exponent of e. We can do that by taking the natural log of both sides:

ln(N / No) = ln(e^(k*t)), which simplifies (through the MAGIC of the "natural log" definition) to:
ln(N / No) = k * t

Because t (29 hours) is the "half life", the number (N) at that time IS half of whatever number we had originally. It doesn't matter if "No" is 2, 21, or 123,766, "N" will be half that much, and the fraction (N / No) is equal to 1/2. So, to find k, we can divide both sides of the equation by t:

ln(N / No) = k
t

or

ln(1/2) = k
29

Using a calculator for the "ln" function: ln(1/2) = -0.69315
-0.69315 / 29 = -0.0239 = k

NOW THAT WE KNOW "k", we can do some cool things:

To find the percentage of ONE full dose which is left after some time "t", we let "No" = 1, and it drops out of the original equation, so that N = e^(-0.0239*t). This will give a decimal like, say, 0.175, which is 17.5%.

To find the time at which the amount left will be some specified percentage (say, 1%, or 99% gone), write the "remaining" percentage (not the "eliminated" 99% ) as a decimal. In this example, it's 0.01. Then use:

ln(0.01) = t (this worked out to 192 hours and 40 minutes to be 99% eliminated, based on a 29 hour half-life)
-0.0239

FOR REPETITIVE DOSING, to calculate "steady state" amounts: First, I put "steady state" in quotes, because it isn't truly SS unless the dose is infused constantly over the entire period of time t. But, to calculate a "near SS" condition where the dose after each time interval "t" replaces the amount eliminated during that interval "t"... recognize that here, "N" (after time T) would equal "No" minus dose "D".

So: No-D = No * e^(k*t),

or: D = No - (No * e^(k*t))

or: D = No* (1-e^(k*t))

or: D / (1-e^(k*t)) = No
(where "No" here is the original starting dose, AND is the amount/concentration at the start of each new interval "t", AFTER administering each maintenance dose "D". The amount at the end of each interval (prior to each maintenance dose) is simply "No" minus "D", and the NSS amount is (roughly, slightly less than..) "No" minus "1/2 D".)

This is all pretty easy to build into an Excel spreadsheet - I'm tracking my MT2 dosing with one, and it's really informative...

 

[Scrofula]

Don't forget your enzyme kinetics! Ping Pong Poly Poly anyone? Only thing I remember.

Anyway, not all molecules are metabolized in an exponential rate (and of course, on top of all this, the reaction rate won't be constant anyway, as the cheeseburger you ate takes over).

A good example: alcohol to acetylaldehyde is zero-order kinetics. Just a line. A steady rate no matter how much you drank. Because you consumed such a vast excess of material, all the enzyme phone lines will be busy at maximum rate, and rate of clearance is dependent only on the gross number of alcohol dehydrogenases available (and their turnover, cofactor availabilty like Thiamin and zinc, genetics, lots of things).

 

[Slow_Mobius]

But... back to half-lives...

The math equation for exponential growth or decay (growth might be interest on a bank account if compounded continuously, or populations with a birth:death ratio higher than 1:1... decay includes things like radioactive decay, and also metabolism) is:

N=No * e^(k*t)

where N is the number (amount) of something at time "t", No is the number (amount) at the start time, e is the exponentiation constant, t is the time, and k is the growth rate constant (a negative number applies to decay calculations) for the situation being considered.

Rearranging that, gives this: N / No = e^(k*t)

Given a half-life (t) of 29 hours, we need to find k. So, we need to separate the k from the t in that exponent of e. We can do that by taking the natural log of both sides:

ln(N / No) = ln(e^(k*t)), which simplifies (through the MAGIC of the "natural log" definition) to:
ln(N / No) = k * t

Because t (29 hours) is the "half life", the number (N) at that time IS half of whatever number we had originally. It doesn't matter if "No" is 2, 21, or 123,766, "N" will be half that much, and the fraction (N / No) is equal to 1/2. So, to find k, we can divide both sides of the equation by t:

ln(N / No) = k
t

or

ln(1/2) = k
29

Using a calculator for the "ln" function: ln(1/2) = -0.69315
-0.69315 / 29 = -0.0239 = k

NOW THAT WE KNOW "k", we can do some cool things:

To find the percentage of ONE full dose which is left after some time "t", we let "No" = 1, and it drops out of the original equation, so that N = e^(-0.0239*t). This will give a decimal like, say, 0.175, which is 17.5%.

To find the time at which the amount left will be some specified percentage (say, 1%, or 99% gone), write the "remaining" percentage (not the "eliminated" 99% ) as a decimal. In this example, it's 0.01. Then use:

ln(0.01) = t (this worked out to 192 hours and 40 minutes to be 99% eliminated, based on a 29 hour half-life)
-0.0239

FOR REPETITIVE DOSING, to calculate "steady state" amounts: First, I put "steady state" in quotes, because it isn't truly SS unless the dose is infused constantly over the entire period of time t. But, to calculate a "near SS" condition where the dose after each time interval "t" replaces the amount eliminated during that interval "t"... recognize that here, "N" (after time T) would equal "No" minus dose "D".

So: No-D = No * e^(k*t),

or: D = No - (No * e^(k*t))

or: D = No* (1-e^(k*t))

or: D / (1-e^(k*t)) = No
(where "No" here is the original starting dose, AND is the amount/concentration at the start of each new interval "t", AFTER administering each maintenance dose "D". The amount at the end of each interval (prior to each maintenance dose) is simply "No" minus "D", and the NSS amount is (roughly, slightly less than..) "No" minus "1/2 D".)

This is all pretty easy to build into an Excel spreadsheet - I'm tracking my MT2 dosing with one, and it's really informative...

I would just take the half life and multiply it by five. That's how long it takes to eliminate 97% of the drug. At that point, virtually any interactions will be negligible.

 

[Bigdog379]

Not trying to threadjack but slow mobius could you check my thread out again?

 

[Mr Dobalina]

But... back to half-lives...

The math equation for exponential growth or decay (growth might be interest on a bank account if compounded continuously, or populations with a birth:death ratio higher than 1:1... decay includes things like radioactive decay, and also metabolism) is:

N=No * e^(k*t)

where N is the number (amount) of something at time "t", No is the number (amount) at the start time, e is the exponentiation constant, t is the time, and k is the growth rate constant (a negative number applies to decay calculations) for the situation being considered.

Rearranging that, gives this: N / No = e^(k*t)

Given a half-life (t) of 29 hours, we need to find k. So, we need to separate the k from the t in that exponent of e. We can do that by taking the natural log of both sides:

ln(N / No) = ln(e^(k*t)), which simplifies (through the MAGIC of the "natural log" definition) to:
ln(N / No) = k * t

Because t (29 hours) is the "half life", the number (N) at that time IS half of whatever number we had originally. It doesn't matter if "No" is 2, 21, or 123,766, "N" will be half that much, and the fraction (N / No) is equal to 1/2. So, to find k, we can divide both sides of the equation by t:

ln(N / No) = k
t

or

ln(1/2) = k
29

Using a calculator for the "ln" function: ln(1/2) = -0.69315
-0.69315 / 29 = -0.0239 = k

NOW THAT WE KNOW "k", we can do some cool things:

To find the percentage of ONE full dose which is left after some time "t", we let "No" = 1, and it drops out of the original equation, so that N = e^(-0.0239*t). This will give a decimal like, say, 0.175, which is 17.5%.

To find the time at which the amount left will be some specified percentage (say, 1%, or 99% gone), write the "remaining" percentage (not the "eliminated" 99% ) as a decimal. In this example, it's 0.01. Then use:

ln(0.01) = t (this worked out to 192 hours and 40 minutes to be 99% eliminated, based on a 29 hour half-life)
-0.0239

FOR REPETITIVE DOSING, to calculate "steady state" amounts: First, I put "steady state" in quotes, because it isn't truly SS unless the dose is infused constantly over the entire period of time t. But, to calculate a "near SS" condition where the dose after each time interval "t" replaces the amount eliminated during that interval "t"... recognize that here, "N" (after time T) would equal "No" minus dose "D".

So: No-D = No * e^(k*t),

or: D = No - (No * e^(k*t))

or: D = No* (1-e^(k*t))

or: D / (1-e^(k*t)) = No
(where "No" here is the original starting dose, AND is the amount/concentration at the start of each new interval "t", AFTER administering each maintenance dose "D". The amount at the end of each interval (prior to each maintenance dose) is simply "No" minus "D", and the NSS amount is (roughly, slightly less than..) "No" minus "1/2 D".)

This is all pretty easy to build into an Excel spreadsheet - I'm tracking my MT2 dosing with one, and it's really informative...

Ok you have just completely confused the hell out of me.
What on earth are you even talking about ?

 

[Scrofula]

Stick with Slo Mo's suggestion x5.

It's just your ol' "how to calculate interest" Pe^​rt equation. They used to teach that in the US, back when we had interest rates.

 

[Lacquerman Steve]

Ok you have just completely confused the hell out of me.
What on earth are you even talking about ?

That's an explanation of the half-life equation. If you know your dose of something, the time you took it, and its half life (if it has one - like a prior poster pointed out, some substances don't follow an exponential (half-life) metabolic curve...) - if you know those three things, and you apply the math I explained, you can figure out how much substance remains at a specified later time, and you can figure out how much time it will take for the amount remaining to equal some specified value.

You asked "with a 29 hour half life, how long until it has all cleared my system? Is 27 hours correct?"
No, 27 hours is not correct.
29 hours after you take it, HALF of it will have cleared your system (assuming your metabolism is "typical" - if yours isn't typical, then the half life FOR YOU will be something other than 29 hours, and you'll need carefully timed bloodwork to learn your specific half life for that substance.)

Anyway... lets say you take 1000mg of Tylenol. Here's data for Tylenol's half life:

The half-life of acetaminophen is approximately 2 to 3 hours after therapeutic doses, yet can be increased to more than 4 hours in patients with hepatic injury. Acetaminophen is extensively metabolized by the liver via three main hepatic pathways: glucuronidation, sulfation, and CYP450 2E1 oxidation.

Acetaminophen Toxicity: What Pharmacists Need to Know - Medscape

www.medscape.com/viewarticle/828260_2

So, let's assume Tylenol's half-life is 2.9 hours - that's in the published range, and its 1/10 of the substance you're asking about, so the times for your substance will be 10x what's given in this example. We're also assuming that Tylenol follows the half life curve no matter how much you take - that you can't take enough Tylenol to overwhelm your liver's ability to metabolize it (by using this assumption, we can set aside the "after therapeutic doses" in the quote above so we can focus on the half life behavior here - but this assumption is not true in real life)

So, you take 1000mg of Tylenol.
2.9 hours later, 500mg is cleared, and 500mg remains (because 500 is half of 1000, right?)
ANOTHER 2.9 hours later, from the 500mg that remained, 250mg have cleared, and 250mg still remain (because 250 is half of 500...)
Another 2.9 hours later, from the 250mg that remained, 125mg have cleared, and 125mg remain. (At this point, 8.7 hours after taking 1000mg, the TOTAL that has cleared is 500+250+125 = 875mg)
Still another 2.9 hours later (this is four half-lives later), from the 125mg that remained, 62.5mg have cleared and 62.5mg STILL remain. (62.5 remaining, is one sixteenth of the original 1000mg... 1/2 + 1/4 + 1/8 + 1/16 = 15/16 has cleared. That's 93.75% of the original dose which is now gone.)
Yet another 2.9 hours later (after 5 half lives have gone by, 14.5 hours since the 1000mg dose), from the 62.5mg that was remaining, another 31.25mg have cleared, but 31.25mg still remain. The original dose is 96.875% cleared after 5 TIMES the half-life have gone by.

So, how do you define "cleared from my system"? Is 96.9% good enough? 98.6%? 99.2%? 99.6%? Those are the percentages which have cleared out after 5, 6, 7, and 8 half lives, respectively.

But... all of that only allows you to estimate your levels at the points in time that are multiples of the half life. Those points are like mileposts on the highway. In the case of your substance, the mileposts are 29 hours apart - you can only estimate "where you are" in terms of remaining substance levels, once every 29 hours, when you're near the end of a half life period. Compared to that, being able to use the half life formulas in my earlier post, is like having a GPS with you on the highway - you can estimate "where you are" with pretty good accuracy any time you want to know. With the info in this post, all we can say about your "27 hours" guess is that 27 hours after taking your dose, you have "a bit more than half" still in you. With the formulas in my prior post, I can estimate that after 27 hours, you would still have about 52.4% of the original dose remaining in you.

 

[Mr Dobalina]

That's an explanation of the half-life equation. If you know your dose of something, the time you took it, and its half life (if it has one - like a prior poster pointed out, some substances don't follow an exponential (half-life) metabolic curve...) - if you know those three things, and you apply the math I explained, you can figure out how much substance remains at a specified later time, and you can figure out how much time it will take for the amount remaining to equal some specified value.

You asked "with a 29 hour half life, how long until it has all cleared my system? Is 27 hours correct?"
No, 27 hours is not correct.
29 hours after you take it, HALF of it will have cleared your system (assuming your metabolism is "typical" - if yours isn't typical, then the half life FOR YOU will be something other than 29 hours, and you'll need carefully timed bloodwork to learn your specific half life for that substance.)

Anyway... lets say you take 1000mg of Tylenol. Here's data for Tylenol's half life:


So, let's assume Tylenol's half-life is 2.9 hours - that's in the published range, and its 1/10 of the substance you're asking about, so the times for your substance will be 10x what's given in this example. We're also assuming that Tylenol follows the half life curve no matter how much you take - that you can't take enough Tylenol to overwhelm your liver's ability to metabolize it (by using this assumption, we can set aside the "after therapeutic doses" in the quote above so we can focus on the half life behavior here - but this assumption is not true in real life)

So, you take 1000mg of Tylenol.
2.9 hours later, 500mg is cleared, and 500mg remains (because 500 is half of 1000, right?)
ANOTHER 2.9 hours later, from the 500mg that remained, 250mg have cleared, and 250mg still remain (because 250 is half of 500...)
Another 2.9 hours later, from the 250mg that remained, 125mg have cleared, and 125mg remain. (At this point, 8.7 hours after taking 1000mg, the TOTAL that has cleared is 500+250+125 = 875mg)
Still another 2.9 hours later (this is four half-lives later), from the 125mg that remained, 62.5mg have cleared and 62.5mg STILL remain. (62.5 remaining, is one sixteenth of the original 1000mg... 1/2 + 1/4 + 1/8 + 1/16 = 15/16 has cleared. That's 93.75% of the original dose which is now gone.)
Yet another 2.9 hours later (after 5 half lives have gone by, 14.5 hours since the 1000mg dose), from the 62.5mg that was remaining, another 31.25mg have cleared, but 31.25mg still remain. The original dose is 96.875% cleared after 5 TIMES the half-life have gone by.

So, how do you define "cleared from my system"? Is 96.9% good enough? 98.6%? 99.2%? 99.6%? Those are the percentages which have cleared out after 5, 6, 7, and 8 half lives, respectively.

But... all of that only allows you to estimate your levels at the points in time that are multiples of the half life. Those points are like mileposts on the highway. In the case of your substance, the mileposts are 29 hours apart - you can only estimate "where you are" in terms of remaining substance levels, once every 29 hours, when you're near the end of a half life period. Compared to that, being able to use the half life formulas in my earlier post, is like having a GPS with you on the highway - you can estimate "where you are" with pretty good accuracy any time you want to know. With the info in this post, all we can say about your "27 hours" guess is that 27 hours after taking your dose, you have "a bit more than half" still in you. With the formulas in my prior post, I can estimate that after 27 hours, you would still have about 52.4% of the original dose remaining in you.

LMFAO quit messing with my brains receptors.

 

[Scrofula]

It brings up a point about long biological half-lifes too. Since it's squishy biology here, there are a lot of small adjustments involved. Just think about how many delays or sexy hitchhikers can happen on the way to work or on a cross-country drive. Small uncertainties build over time. And because it's a decay curve, those delays are more geometric. You're far more likely to unwittingly consume an inducer/inhibitor of a liver enzyme if the drug sticks around a while.

LMFAO quit messing with my brains receptors.

I think this is also a good time to explain the base of the natural log. I mean, what the hell does 2.7blah blah have to do with anything, right?

Well, you have to know a bit about calculus, and rates of change. You know how derivatives work, right? Not the banking things. Oh, and some things about infinite series. No, not really, because I don't even remember how it relates.

Yeah, fuck it use x5.

 

[Mr Dobalina]

It brings up a point about long biological half-lifes too. Since it's squishy biology here, there are a lot of small adjustments involved. Just think about how many delays or sexy hitchhikers can happen on the way to work or on a cross-country drive. Small uncertainties build over time. And because it's a decay curve, those delays are more geometric. You're far more likely to unwittingly consume an inducer/inhibitor of a liver enzyme if the drug sticks around a while.



I think this is also a good time to explain the base of the natural log. I mean, what the hell does 2.7blah blah have to do with anything, right?

Well, you have to know a bit about calculus, and rates of change. You know how derivatives work, right? Not the banking things. Oh, and some things about infinite series. No, not really, because I don't even remember how it relates.

Yeah, fuck it use x5.

Haha yeah I'll just use x5

 

[ConstanceG]

Not sure if it was stated but the "half life" for benzos is completely different. Especially depending on the way your digestion is, liver works, etc. Water intake and ouput.
I once had klonopin show up on my test almost 2 months later. Swear to it. Had to play stupid, because I was only scripted Xanax at the time. As someone said, when in doubt, just abstain...may not be fun but it has to be better than seizures.

 

[Scrofula]

It works the same for those too. And we've mentioned a lot of the extra variables.

And . . . see, I double-check before I holler at a n00b, and I learned some things just now, most of them from here:https://www.practicalpainmanagement...ying-benzodiazepine-urine-drug-screen-results

One thing is the half-life for your drugs: alprazolam at 12-15h is kinda medium, clonazepam has a whopping spread of 18-50h. (It looks like lorazepam never goes away). Wikipedia has a spread of 19-60.

So at the 97% guardband you get: 90 hours - 300 hours, or over 12 days.

I also had to see what metabolites they use, if they CAN differentiate between benzos. And they have limited ability to do so, which means more $/t for the tester. It looks like a pee test for those two is easy, although Klonopin seems to have some weirdness in immunoassays, at least as of 2014 you'd expect more negative results.

But yeah, n00b Ms.G, you had something else going on there. A leftover clam five days earlier you forgot or Someone swapped one out for a xanax for some reason.

 

[ConstanceG]

Thank you. I understand the science and reasoning behind it...however I have other health issues that I personally feel affect my absorption. I respect your post and yep...I am a noob. All gotta start somewhere, so Yell away.

 

[Jabberwocky]

Thank you. I understand the science and reasoning behind it...however I have other health issues that I personally feel affect my absorption. I respect your post and yep...I am a noob. All gotta start somewhere, so Yell away.

Just remember Constance, the only stupid questions are those you don't ask