# STEROIDS FORUM > ANABOLIC STEROIDS - QUESTIONS & ANSWERS > EDUCATIONAL THREADS >  Mathematical Formula For Steroid Blood Concentration

## exponent303

Ok guys i wondered why should we always calculate the amount of a steroid inside our body (after an initial amount injected) 
only in segments of halflifes?And if the steroid has a halflife of 
7 days and we inject every two days how do we calculate then?
Before we go on:By steroid halflife we mean the time required for 
a specific drug to be reduced at a quantity half the initial.
The math concept is simple:lets take a drug with a halflife of 7 days for example(nandrolone decanoate if you prefer)

The first day you inject Ymg:

DAY 1 ------ Ymg {every 7 days the drug amount gets half} 
DAY 2 ------ ? 
DAY 3 ------ ?
DAY 4 ------ ? 
...
DAY 8 -------Y/2mg
DAY 9 -------?
DAY 10 -----?
DAY 11 -----?
...
DAY 15 -----(Y/2)/2=Y/4mg
...
DAY 22 -----(Y/4)/2=Y/8mg
...
DAY 29 -----(Y/8)/2=Y/16mg
...
and so on.

NOW LETS PLAY WITH THE NUMBERS!!!

DAY1 = DAY{1+0x7} and Y:=Y/2*0 or Y

DAY8 = DAY{1+1x7} and Y:=Y/2*1 or Y/2

DAY15=DAY{1+2x7} and Y:=Y/2*2 or Y/4

DAY22=DAY{1+3x7} and Y:=Y/2*3 or Y/8

DAY29=DAY{1+4x7} and Y:=Y/2*4 or Y/16

DAY36=DAY{1+5x7} and Y:=Y/2*5 or Y/32
...

After a closer look at the numbers above we notice that a 
mathematical relation exists between the days and the subdivisions of the initial drug amount Y:

This relation can be expressed as :

DAY{1+Nx7} and Y:=Y/2*N which describes the above completely.(above: N=1,2,3,... )

So we have days 1,15,22,29,36,43,... and the corresponding amounts of drug but we dont have the days between those...
We'll do a mathematic trick:

lets say : 1+Nx7=M , then N=(M-1)/7 , so we can say instead:

DAY{M} and Y:=Y/2*[(M-1)/7]. M=1,2,3,...


Then: DAY{1} you'll have Y mg.

DAY{2} you'll have Y/2*(1/7) mg

DAY{3) you'll have Y/2*(2/7) mg

DAY{4} you'll have Y/2*(3/7) mg
...
DAY{8} you'll have Y/2*(7/7)=Y/2mg

and so on.

If you like mathematic equations here it is:
AMOUNT(day)=(AMOUNTinitial)/2*[(day-1)/7]

(day:1,2,3,4,5,...)

AMOUNT(day):the amount of drug at the beginning of day 1,2,3,4,...
AMOUNTinitial:first injection in mgs.

The same way we can show that for any drug halflife:

AMOUNT(day)=(AMOUNTinitial)/2*[(day-1)/HALFLIFE]
HALFLIFE:is a number for example 7 days.

What do you think guys?
I didn't find this anywhere i created the proof myself.
I want your opinions.The same proof can be done even for seconds or millisecs!!!!!!!!

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## CYCLEON

that sounds a lot like it was written by ANDY13 over at elite or maybe zygla.

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## exponent303

wrong - this proof is totally mine and i can prove it for every halflife and time unit-i would really like to find out what andy thinks about that.He has come up with some also very interesting stuff.

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## PaPaPumP

Ok, that just went in one eye, and out the other...I got an idea...How bout we just go to the site where it tells you half life and stop all this nonsense. :Big Grin:  

I stopped taking math in high school, and vow never to take it again, (unless held at gunpoint of course).

P

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## CYCLEON

I liked zyglas charts as well. was going to dothem myself but kinda hard ot find the time. 

If thats yours bro, my full apologies - and i already looked and nobody had it posted there  :Big Grin:  . some interesting insites- very nice, let me study on this some, give u my thoughts.

Cant wait till IG sees this tho and says "but it doesnt matter"  :Big Grin:

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## Nathan

You understand that that is all theory, I trust. The concentration isn't really halved once thedrug has been in your system for th duration of the half-life. Also, Y/2*(1/7) mg < Y/2 mg - - - > your day 2, 3, 4, 5, 6, 7 is less than day 8 above, which doesn't make sense as you're not injecting after day 1 until day 8 again. I think you need to edit that part. Care to go over it again?

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## The Iron Game

using enanthate , and a half life of 7 days, injecting 500mgs once a week for 10 weeks. lets see blood concentration levels  :EEK!:

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## The Iron Game

It doesnt matter  :Stick Out Tongue:

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## big_guy

exponent made it harder then it has to be with the damn formula. (no flame intended).. it's much easier to make a chart and add it all up. oh hey IG speakin of test e.,, i had made a chart of MY cycle. it shows the blood concentration at each injection time (once a week)

cycle looks like this.
weeks 1 and 2 - 600mgs test E.
weeks 3 through 9 - 400mgs test E.
start clomid on week 12, contemplating week 12.5 (at week 12 i would still have almost 100mgs in system, even though that is 3 weeks after the last injection.)


blood concentration levels at each injection.

week 1 - 600mgs
week 2 - 900mgs
week 3 - 850mgs
week 4 - 825mgs
week 5 - 812mgs
week 6 - 805mgs
week 7 - 802mgs
week 8 - 796mgs
week 9 - 793mgs
week 10 - 393mgs
week 11 - 193mgs
week 12 - 93mgs
week 13 - 43mgs

peace.

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## Tobey

Ok I read your post Big Guy and I still don't quite see your doseages adding up. I'm sure that it is just my idoitic way of comprehending but maybe you can clear things up.
Now at weeks 1-2 you are injecting 600 mgs of Test E right? Then in weeks 3-9 you have reduced the injections to 400 mgs per wk.
If you enject 600 mg your first week and you enject 600 mg during your second week and Test E's half life is 4 weeks how do you get having 900 mgs of Test in you at week 2? Should this not be 1200??? Dose the amount not begin to taper off at week 4 or are you some how calculating in possible conversion factors of Test arimatizing into Estrogen ect. I"m really interested in your post but I'm having a hard time understanding your numbers. Can you clear this up for me? Thanks
Tobey

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## exponent303

checking it out right now my friend-thank you for looking at it

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## exponent303

ok nathan i checked it out and i think its ok-look:
the equation says:
AMOUNT(day)=(AMOUNTinitial)/2*[(day-1)/7] 

the part (day-1)/7 is the EXPONENT of the number 2 not
a multiplication sign or something else.
Also AMOUNT(1) is defined as the amount of drug in your body
RIGHT AFTER you have injected.Thus AMOUNT(8) is the amount of substance in your body 7 days after day one and that means 
at the beginning of day 8.If you study this theory carefully you will notice that the time segments 1-8 , 8-15 , 15-22 , 22-29 ,...
are exactly 7 days. Considering the above lets see the concentration till day 8:(initial amount=Y , halflife=7)

DAY(1): Y/{2*(0/7)}=Y/{2*0)=Y/1=Y (we know that 2*0=1)

DAY(2): Y/{2*(1/7)}=Y/{2*0.14285}=Y/1.10409=0.9057 Y

DAY(3): Y/{2*(2/7)}=Y/{2*0.28571}=Y/1.21901=0.8203 Y

DAY(4): Y/{2*(3/7)}=Y/{2*0.42857}=Y/1.3459 =0.7429 Y

DAY(5): Y/{2*(4/7)}=Y/{2*0.57142}=Y/1.48599=0.6729 Y

DAY(6): Y/{2*(5/7)}=Y/{2*0.71428}=Y/1.64067=0.6095 Y

DAY(7): Y/{2*(6/7)}=Y/{2*0.85714}=Y/1.81144=0.5520 Y

DAY(8): Y/{2*(7/7)}=Y/{2*1}=Y/2=0.5 Y

...

I think this explanation solves many problems.This theory calculates the drug amount in your body after a single shot.Considering that you shoot in time periods of (lets say) 3 days you calculate what is left in your body at the end of the 3rd
day and then you add it to the next amount you shoot.You use this formula again and again until you shoot the last ampoule.
I know some guys will say its useless mathematic bullshit but with this theory you have flexibility of calculating drug amounts even when shooting at non-regular time periods.
I WANT FEEDBACK DUDES!!!!!

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## big_guy

TOBEY!... u don't understand it bcuz u think E has a half life of 4 weeks.. lol.. the half life of test E is 7 days. now u understand?

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## Tobey

Yeah, it makes perfect sense now. Somtimes when I'm typing with out referring back to certain notes that I have made I tend to assume shit. Like Deca 's half life=4 wks. Test half life=7days ect,ect. Thanks for the re-ply.
Tobey

Speaking of which can anybody list the different half lifes of different steriods or have I just not come across that post yet?

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## Iwan2bsolid2

That mathematical info went into my eye and then I farted it out(and it smells like dead cat)! :Big Grin:

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## exponent303

> _Originally posted by Iwan2bsolid2_ 
> *That mathematical info went into my eye and then I farted it out(and it smells like dead cat)!*


i love maths but bad smelling farts make me suddenly come up with such ideas.BWAHAHAHAHAHAHAHAHAHA.......

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## CYCLEON

not really - but if you look into the top article board ull see that it is a combo of half life and blood concentration that make up the optimal times to inject.

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## Nathan

What you have now looks better IN THEORY I suppose, but that's not what you had before. Oh well, it doesn't really matter especially. It isn't a complicated calculation especially and it is far easier to think of it graphically. Moreover, this theory can't be duplicated fully in oractice as I explained before.

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## exponent303

> _Originally posted by jetskidude_ 
> *no, really.
> 
> the formula assumes that you're in the middle of the *bell curve* for metabolism of the drug. we know that uptake and distribution of medications (including AS) is different from person to person for many reasons; basal metabolic rate, % body fat, liver / kidney function, --- the list goes on...
> 
> so basically you've done TONS of math and you still don't know for sure the best dose or when to give it.
> 
> that's why i called it "mental masturbation"*


The amount of drug being metabolised from person to person of course differs-but what's the point in saying that right now?Are you guys going to insult the statistic archives of all pharmaceutical companies in the world to figure out the magic dosage?Of course not-you are going to use a roid calculator.But roid calculators are program applications based on maths and thats exactly the math required TO CREATE ONE-I DID NOT CREATE THIS INFO FOR ANOTHER ROID CALC OR WHATSOEVER-I AM JUST TRYING TO SHOW THE EXACT RELATION BETWEEN DAYS AND DRUG AMOUNTS FOR EVERY DAY AND FOR EVERY DRUG AMOUNT FROM THE BEGINNING...didnt you guys ever wonder how could someone just wake up in the morning and come up with such a formula?No its not tons of maths its what this guy had in mind when CREATING this formula.

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## Tobey

Go get em exponent!
Good job in the math dep. 
IC

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## Iwan2bsolid2

I hate this post, It just makes me feel stupid and feel like my head is spinning, like I'm gonna throw up, but I don't, but I really want to, wait no I don't cause that would be a waste of good protein --YUM!

_SOLID

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## exponent303

> _Originally posted by Iwan2bsolid2_ 
> *I hate this post, It just makes me feel stupid and feel like my head is spinning, like I'm gonna throw up, but I don't, but I really want to, wait no I don't cause that would be a waste of good protein --YUM!
> 
> _SOLID*


DUDE i really am sorry for the for the abrupt disturbance of the mental tranquility that i have caused to you but i really didnt mean to...Come on bro its only numbers , they aint gonna bite you!!!  :Smilie:   :Smilie:   :Smilie:

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## exponent303

> _Originally posted by jetskidude_ 
> *settle down bro  i just don't share your passion for numbers 
> 
> didn't mean to offend -- and i understand and appreciate the importance of things such as half life of a drug. really. i use the concepts daily.*


Everything's ok bro...No offence taken honestly!!!! Basically i wanted to share this with everyone , but sometimes math just stinks and sucks for a lot of people and i can fully understand it!So for some people this post may appear very interesting , for others it may stink as dead cat.Everything's ok for me (except for that ugly dead cat smell...).I should have probably posted a warning or something before this post-LOL! 
 :Big Grin:

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## exponent303

> _Originally posted by jetskidude_ 
> *no worries bro. i'll send you the book "101 things to do with a dead cat" and the odor eaters from my gym shoes! 
> 
> glad to hear you weren't offended. hope you're not a cat lover -- i may have just done it again...
> jet*


BWAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA...

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## Andy13

There's a much easier way to calculate amount of esterifed AAS remaining.

k*e^(ln(1/2)x/h)

where 
k=amount injected
h=half life in days
x= day interested in

Andy

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## pureanger

I just take as much as I can fit in the pin and as often as I can

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## benji

JUST DO IT!

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## thepath

i dont think this formula can be accurately used to interpolate values between the peak blood concentration level and the half life of the drug because, as you should know, those values are parabolic, not linear.

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## thepath

> _Originally posted by thepath_ 
> *i dont think this formula can be accurately used to interpolate values between the peak blood concentration level and the half life of the drug because, as you should know, those values are parabolic, not linear.*


i.e. amount(day) = initial*2^(-day/halflife)

this is based off the half life decay formula. i'm surprised none of you caught this.

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## rexboy

kinda resembles the decay formula for a nuclear fallout  :Smilie:

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## bermich

Ouch. Whomever invented numbers should be shot along with that speed bump guy, because they both hurt when I see them in front of me. I just pay my smart girlfriend to figure out all that math shit.

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## Mudge

> TOBEY!... u don't understand it bcuz u think E has a half life of 4 weeks.. lol.. the half life of test E is 7 days. now u understand?


10.5 day half life for enanthate from the charts I've seen.

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## block

test E does not have a half life anywhere near 7days let alone 10.5even though the math is great and a fun read it also does not matter test E is in and out of your system in 7 days its peak blood levels happen at about 48 hours and drop below 200ng/dl buy day 8 plasma half life is 48hours so you need to inject every 3rd day to keep blood levels constant and to minimize the peaks and valleys. I know that some people still think that it last for 2 weeks well I am sorry you just happen to be wrong. also some factors to consider not all the drug is used some is lost in the needle and the body and unless its human grade gear your mg levels are going to be wrong because no one takes into account the ester weight. have fun useing your math now 

this is not a flame in any way it is just the truth.

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## Bulkingup

wow grade 9 math.... lemme guess you can change the variables too?




> Ok guys i wondered why should we always calculate the amount of a steroid inside our body (after an initial amount injected) 
> only in segments of halflifes?And if the steroid has a halflife of 
> 7 days and we inject every two days how do we calculate then?
> Before we go on:By steroid halflife we mean the time required for 
> a specific drug to be reduced at a quantity half the initial.
> The math concept is simple:lets take a drug with a halflife of 7 days for example(nandrolone decanoate if you prefer)
> 
> The first day you inject Ymg:
> 
> ...

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## TheSevnthWarrior

If you think you need another shot...do it.
In fact, do it until something bad happens, then back it off 1/2 a notch, wait until you're stronger, and your tolerance goes up, and start climbing again....simply do as much as you can...all the time.
More is Better....repeat after me, More...is Better.....
Haaaaaaaa! J/K  :Big Grin:

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## Billy_Bathgate

Half-life of enanthate is somewhere around 3-4.5 days.

The highest amount of the drug released is immediately. The rate decreases exponentially from then on.

These above formulas tell nothing. What would be useful is blood concentrations.

Blood concentrations = X ng/dl

X mg is not a concentration. Its a mass, which doesnt tell us anything about whats going on with the blood conentrations. All it can tell is how much is supposedly left in the depot. This has little relavance.

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## Andy13

> These above formulas tell nothing. What would be useful is blood concentrations.
> 
> Blood concentrations = X ng/dl
> 
> X mg is not a concentration. Its a mass, which doesnt tell us anything about whats going on with the blood conentrations. All it can tell is how much is supposedly left in the depot. This has little relavance.



I'm biased towards the theoretical plots.. probably because I was the first one to model the life of the ester in this way. Since then, I have seen all sorts of erroneous data coming from copy-cats implementing 'roid calc's and everything else. When used correctly (or, rather, when _carefully_ ripped off from the original) these plots do infact model the life of the ester quite well..


The points on the graph are: (y2-y1) over x2. y2-y1 = mg de-esterified per day. It is true that the values on the y axis are "arbitrary hormone density units" instead of "nanomolar" (as you would see experimental results). However I disagree with you that this y-axis data makes the model worthless.

I encourage you to explain what difference volume makes at all (mass/dL instead of mass). What if the total volume is 10L? 100L? how would that change the pharmacokinetics of the ester in the oil depot?

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## UrbanDawg

cant these fomulas be replaced with

X = 1/ (2^N)


where X is the amount in the depot and and N is number of days since injection

so making this a table it works as

day 0 i.e. 0 days from injection 
X = 1/(2^0) whcih is 1/1/ whioch is 100%
day 1 after injection
X = 1(2^1) == 1/2 = 50%
etc etc 
And like billyb said - we are modeling a dispersion of the depot which is not a perfect indicator of blodd plasma levels but its about the only indicator we have.

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## Cathex

Let's put an end to this nonsense once and for all. The equation for the quantity of a drug of a given half-life remaining after a given time after an initial quantity administered is:

y = a(0.5)^(x/h)

where:

y = amount remaining 
a = initial amount administered
x = particular time after first injection
h = half-life of drug

It should be noted that the day of the first injection is day 0

Example 1:

Let's say 200 mg of a drug with a half-life of 8 days is injected. How much is left on the day of the injection. If the maths is correct, this should be 200 mg, right. Okay let's see.

The equation tells us y = a(0.5)^(x/h)

In this instance we have:

Y = unknown (we do really, it should be 200mg)
a = 200mg
x = 0 (day 0 remember day of first shot)
h = 8 days (half-life)

Feeding the numbers into the equation we have

y = 200(0.5)^(0/8)

= 200(0.5)^0

= 200mg

Okay.

Example 2:

Same as above but how much is left on day 3?

Again we have y = a(0.5)^(x/h)

Feeding in the relavent numbers we have

y = 200(0.5)^(3/8)

= 200(0.771)

= 154.22mg (to 2 dp)

And just to show that as expected, after 1 half-life (8 days) has expired, we should have half the amount remaining, that is 100mg here goes again.

Y = 200(0.5)^(8/8)

= 200(0.5)^1

= 200(0.5)

= 100mg

So there you have it.

Also, if shots are give, say every 3 days the maximum blood concentration reached would be:

S = a/(1 - r)

Where 
S = max blood level
a = amount of injection in mg
r = (0.5)^(x/h) from above

So what would be the max blood level reached if 200mg of a drug with a half-life of 8 days was injected every 3 days?

We have

S = a/(1 - r)

= 200/(1 - 0.7711)

= 200/0.22889.....

= 874mg (to 3 sf)

The above is a sumation equation. Basically the regular injections form what is known as a geometric progression which can be summed as above. When the value r know as the common ratio is less that 1, a limit is reached after so many days, which is the max blood level.

There you have it. As regards to how the body actually absorbs the drug, that can, I as mentioned in a post above vary from one individual to another and so this does assume we are talking about 'Jo Average' but then so do a good many other matheamtical models. However, the maths is correct.

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