Frontloading!

[GeneralIx]

Though the original formula is, perhaps, not as skewed as I would have expected it to be, this still wouldn't validate its use over a more accurate, just as easy to use method*, especially given the option of using multiple front loads or even different injection sites with their different T1/2 if implemented correctly in an excelsheet. Given that even a monkey can be learned excel, this can't be an object.
* Moreover given so that rounding an inaccurate number might create a bigger inaccuracy, and unnecessarily so.

@ Sworder
Thanks.

 

[Bill Roberts]

Also, lastly, on the accuracy issue, in my experience in general it's very rare for those injecting steroids to choose any value other than in increments of 0.5 mL, and more commonly increments of 1 mL.

Just as we do cycles that are say 8 weeks exactly, or 10 weeks exactly (9 weeks apparently is unthinkable!) or say 12 weeks exactly, we like injecting exactly 100 mg, or 200, or 250, etc; or exactly 1 mL, 1.5 mL, 2 mL, 2.5 mL, or 3 mL.

And rarely will someone aim for say 740 mg per week: it will be 700 or 750 or a number like this. Generally the only way you'll see an odder number is if one of the preparations is say 75 mg/mL, which could then give a result that's in increments of 25 rather than 50 or 100.

It works fine.

It's not that there's a magic difference to these numbers, but that they are convenient and if suitable convenient numbers are chosen, results are excellent, without need to worry if instead they "should" have been different by 3% or whatever.

 

[GeneralIx]

Though you keep on hammering on the ever so slight difference (of which I feel that it is in potentia not) you neglect the other benefits of the formula's I've given.

I understand that you feel that as it's always been done is just fine, as perhaps were the results, but hanging on to old methods when newer are better is just idiocy. Also good results don't mean optimal results.

 

[Bill Roberts]

Well, the method I gave is exact, if one does the calculus (which this far out of graduate school I've forgotten how to do, but back in the day I did it), and if one neglects the release kinetics aspect, for which we have no data and your method does not account for either.

I disagree that experienced results of working perfectly well aren't important and sufficient.

A graph is not needed.

I'm glad that your method suits you, but other methods suit other people completely fine and are not wrong.

If you consider my method idiocy, it's your prerogative but many have found it entirely useful and, as mentioned, it is mathematically correct. Try it.

 

[GeneralIx]

I did not say your formula is that of an idiot, I did say that hanging on to it whilst it is of lesser quality and of the same usuability is.

Also, having good results is nice, but better is better. Though the difference in results in frontloads through our different methods is probably negligible on account for the use of longer esters for most people, side effects may come in to play especially when using shorter esters.

A graph is not necessary, yet it is insightful. Through my methods there is no direct need for a graph, it is a nice possibility though, again once more given the added possibility of split front loads or given different esters (or injection sites) and merge all that into an optimal scheme.

Why I defend this way is because it can be used in as basic a fashion as your older method, without being more complicated, with more accurate results, as well as it allows for rather complicated constructions for more advanced users and getting the scheme displayed in an easy, comprehensible way.

 

[Bill Roberts]

Well, as you don't grasp that yours is not more accurate, that what I posted is mathematically exactly correct (when disregarding release kinetics, which you also do) I guess that ends the discussion.

 

[GeneralIx]

You truly excell in ignoring any other point I make, as well as the fact that I clearly understand the mathematics better than you do.

If you want to hang on to your way, be my guest. But don't go pretending it's me that doesn't grasp the mathematics.

 

[Bill Roberts]

(Yawn)

I did the calculus on this problem, you did not. Try summing up how much drug is built up in the system at any given point when having reached steady-state. Go do it, compare with what I've posted, and then talk about how much you understand it -- if you actually do get to understanding it then you'll be retracting your above statements.

Easy and related math question for you, which requires calculus to solve exactly but which you should be able to guess correctly without it: What is the exact sum of the series 1/2 +1/4 + 1/8 + 1/16 + ..... ?

And how does it relate to this problem, and what then is the sum of the amount of drug built up in the system when at steady-state?

That ends my discussion with you.

 

[GeneralIx]

Let me tell you what ends the discussion. It's not you telling it's over. It's you failing to argue why your method is better than mine, or at least proving it has the same possibilities.

The only thing you're showing now is rather condescending an attitude to anyone who merely tries to improve on existing methods.

 

[Bill Roberts]

Well, okay, I can see where you can feel that so I take back that aspect of my choosing to not discuss further with you.

To explain why my above approach rather than providing proof: without spending considerable time I cannot now provide the actual proof, which takes calculus. It's a simple enough integration, about as simple as it gets, but I have not had to use calculus one time since my last course which required using it, which was I think in 1998 and I have completely forgotten how to do even the simplest integration and don't have the time today to go pick any of it back up.

So what I can do is provide you the fact of what calculus does prove to be the case, as well as the above illustration as to how you can see the principle without calculus.

You could also use your own spreadsheet method to iteratively find about what the total built up in the body is at steady state. But you didn't take the time to see how what I posted compares to spreadsheet, which while not as accurate as calculus (because of being step-wise rather than infinitesimal) would match closely enough for you to be able to see, had you tried.

I took the attitude with you that I did because I provided you the correct answer and -- ironically -- you continued to assert without proof that your method was more accurate when it is not, while haranguing me and being somewhat insulting. That's not a way to learn, and it was my impression that you were not out to learn. I provided the opporturnity, and it's still there. If you find it "condescending" that a person in some instances will view it as that they are providing something you can learn from, well, that is your attitude that you're entitled to have, but it may prove limiting for you.

Your method is not better: It is quite crude. I left that off before -- actually if you read my posts again you'll find I think that I treated you rather respectfully, but simply tired after a while of your insults and your intransigence and your repeated refusal to actually evaluate what I'd written instead of automatically dismissing it. What is the point in writing further when a person responds that way?

I do hope you have a good day. But really I have written enough in response on this and don't personally choose to write more, in any endless back-and-forth: there's no need or possible value, I think.

 

[GeneralIx]

Apology accepted. I don't feel that I've been insulting, rather insisting since no real argumentation came up from your side from my perspective.

I'll look into the way your formula was constructed, and research the formula I know exists for the question you posed earlier.

It is interesting to see that we feel the same about eachother in the discussion, so perhaps I owe you an apology as well.

I wish you a good day as well. As soon as I've got time, I'll check back in to give the discussion some new kicking.

 

[Bill Roberts]

 

[Bill Roberts]

Well, I went to see what could be determined with simple arithmetic, now that I can't remember how to do integrations.

Interesting result.

My original interest, way back when, had been calculating frontloads for trenbolone acetate and testosterone propionate, where injection frequency typically matches the half-life.

In the instance where half-life is estimated at one day and injection frequency is daily, then the amount built up in the body is:

From 1 day before: 1/2 the ongoing dose, because one-half life of time has passed.
From 2 days before: 1/4 that amount
From 3 days before: 1/8 that amount

Etc.

And the sum of 1/2 plus 1/4 plus 1/8 plus.... is exactly 1. At least I do remember that much!

So therefore the frontload is exactly the amount injected per half-life, plus the ongoing dose, in this instance where injection frequency matches the half-life.

With regard to your formula, actually I can't figure how you would use it to determine practical examples, or rather if I try what you seem to be saying and don't go and further do the main aspect of what I posted, I wind up with a wrong answer. Could you be more specific? For example, what would you calculate for the above trenbolone acetate example?

You had written.

Put in a formula you get this:
On Tx the blood level=0,5^x*T1
Tx= day x
x= amount of half lives
T1=starting amount
0.5= half life of course.

Half lives however, are hardly ever easy to use time units like a day or an hour.

Let me give an example: clenbuterol has a T1/2 of 35 hours. For us to use easily on 24-hour periods, we have to first convert the half life to the breakdown coefficient per hour, as follows:

x^35=0.5

Ergo: log(x)=35. This is easy to calculate in excel using target search. We then square the value 24 times (thus creating the breakdownfunction on daily basis), giving a value of: 0.98^24=0.62.

Thusly we have the bloodvalue of 0.62*intake of the day before+daily intake as your blood level at any given day.

So for trenbolone acetate we need to substitute 24 for the 35, and we come up with (correctly) 0.5 as the value, if we skip to your following part anyway and match that.

I don't know where the log(x)=35 comes from in your explanation though. For the clenbuterol I would do it as x = .5^(24/35), if by x we mean the fraction remaining after 24 hours, relative to 24 hours previously.

If I were going by your post I would figure that then for trenbolone acetate my blood level would be 0.5 times that of yesterday's dose and that would be what I should frontload with -- or for the clenbuterol, the 0.62 times is the only number appearing in the post for me to frontload with -- but that is not right in either case. With the trenbolone the blood level is fully as much as from the entirety of yesterday's dose: because of the day before yesterday, the day before that, etc. each making fractional contributions summing exactly to one dose.

However there is more to it than this -- for clarity though I wanted to break it into parts.

 

[viperbluelx]

If someone is already on trt at 200mg cyp a week but is planning a 10 week blast at 500mg a week does front loading still apply?

 

[Bill Roberts]

Yes, but less is needed because there's already an amount in the system corresponding to ongoing 200 mg/week use.

The extra amount, beyond the usual ongoing amount, would be based on the 300 mg difference, rather than the full 500 mg.

In other words, it will be sufficient to add only about 250 mg to what will be your ongoing dose. If adding 200 is more convenient, it also would be fine.

 

[GeneralIx]

I'll go into detail as soon as I get home, right now I don't have the means to explain it all. Hang in there.

 

[GeneralIx]

Well, I went to see what could be determined with simple arithmetic, now that I can't remember how to do integrations.

Interesting result.

My original interest, way back when, had been calculating frontloads for trenbolone acetate and testosterone propionate, where injection frequency typically matches the half-life.

In the instance where half-life is estimated at one day and injection frequency is daily, then the amount built up in the body is:

From 1 day before: 1/2 the ongoing dose, because one-half life of time has passed.
From 2 days before: 1/4 that amount
From 3 days before: 1/8 that amount

Etc.

And the sum of 1/2 plus 1/4 plus 1/8 plus.... is exactly 1. At least I do remember that much!

So therefore the frontload is exactly the amount injected per half-life, plus the ongoing dose, in this instance where injection frequency matches the half-life.

I see where this is headed indeed. It is indeed true that the sum of 1/x-->infinty will approach 1, so much so that 1 is the outcome with negligible deviation.

With regard to your formula, actually I can't figure how you would use it to determine practical examples, or rather if I try what you seem to be saying and don't go and further do the main aspect of what I posted, I wind up with a wrong answer. Could you be more specific? For example, what would you calculate for the above trenbolone acetate example?

You had written.

So for trenbolone acetate we need to substitute 24 for the 35, and we come up with (correctly) 0.5 as the value, if we skip to your following part anyway and match that.

I don't know where the log(x)=35 comes from in your explanation though. For the clenbuterol I would do it as x = .5^(24/35), if by x we mean the fraction remaining after 24 hours, relative to 24 hours previously.

Indeed. I see where you get confused, the log function is just another method of calculating the halflife per day or per hour given a half life in days or hours respectively (and then squaring it to your preferred amount of time of course). For instance, if you were to solve log(x)=35, you would get the degrading coefficient per hour, thus squaring this number 24 times, will get you the degradation per 24 hours. Mathematically it's the "cleanest" method, in practice you'll be easier off with the 0.5^(24/35) method.

If I were going by your post I would figure that then for trenbolone acetate my blood level would be 0.5 times that of yesterday's dose and that would be what I should frontload with -- or for the clenbuterol, the 0.62 times is the only number appearing in the post for me to frontload with -- but that is not right in either case. With the trenbolone the blood level is fully as much as from the entirety of yesterday's dose: because of the day before yesterday, the day before that, etc. each making fractional contributions summing exactly to one dose.

The 0.62 is the degradation coefficient per 24-hour timeframe. You then would get something like 0.62*x+daily dosage=the stable bloodvalue. Plotting this in either a table or graph, you can then experiment with any value to get the optimal frontload.

However there is more to it than this -- for clarity though I wanted to break it into parts.

I hope this clarifies a bit of what I said. Should you wish to see a more complex function of the way this formula can be used, I'll put some time in this. Show multiple frontload options and/or multiple half-lives.

 

[Bill Roberts]

Originally Posted by Bill Roberts
In the instance where half-life is estimated at one day and injection frequency is daily, then the amount built up in the body is:

From 1 day before: 1/2 the ongoing dose, because one-half life of time has passed.
From 2 days before: 1/4 that amount
From 3 days before: 1/8 that amount

Etc.

And the sum of 1/2 plus 1/4 plus 1/8 plus.... is exactly 1. At least I do remember that much!

So therefore the frontload is exactly the amount injected per half-life, plus the ongoing dose, in this instance where injection frequency matches the half-life.

Generallx: I see where this is headed indeed. It is indeed true that the sum of 1/x-->infinty will approach 1, so much so that 1 is the outcome with negligible deviation.

Not negligible, but zero error, if referring to actual steady state. The sum goes exactly to 1, not to some small amount off of 1.

Bill Roberts: With regard to your formula, actually I can't figure how you would use it to determine practical examples, or rather if I try what you seem to be saying and don't go and further do the main aspect of what I posted, I wind up with a wrong answer. Could you be more specific? For example, what would you calculate for the above trenbolone acetate example?

You had written.

So for trenbolone acetate we need to substitute 24 for the 35, and we come up with (correctly) 0.5 as the value, if we skip to your following part anyway and match that.

I don't know where the log(x)=35 comes from in your explanation though. For the clenbuterol I would do it as x = .5^(24/35), if by x we mean the fraction remaining after 24 hours, relative to 24 hours previously.

Generallx: Indeed. I see where you get confused, the log function is just another method of calculating the halflife per day or per hour given a half life in days or hours respectively (and then squaring it to your preferred amount of time of course). For instance, if you were to solve log(x)=35, you would get the degrading coefficient per hour, thus squaring this number 24 times, will get you the degradation per 24 hours.

Solving log(x)=35 gives x=100,000,000,000,000,000,000,000,000,000,000,000 so I don't see how this helps, let alone then "squaring it 24 times."

Generallx: Mathematically it's the "cleanest" method, in practice you'll be easier off with the 0.5^(24/35) method.

There is nothing mathematically wrong at all with what I wrote so I don't know why you say yours is "cleaner," especially as it would give perhaps the largest number I've ever encountered (after squaring the 10^35 24 times, which is higher than I'd want to calculate.) A googolplex is bigger, but I've never encountered it in an equation trying to solve a problem.

Bill Roberts: If I were going by your post I would figure that then for trenbolone acetate my blood level would be 0.5 times that of yesterday's dose and that would be what I should frontload with -- or for the clenbuterol, the 0.62 times is the only number appearing in the post for me to frontload with -- but that is not right in either case. With the trenbolone the blood level is fully as much as from the entirety of yesterday's dose: because of the day before yesterday, the day before that, etc. each making fractional contributions summing exactly to one dose.

Generallx: The 0.62 is the degradation coefficient per 24-hour timeframe. You then would get something like 0.62*x+daily dosage=the stable bloodvalue. Plotting this in either a table or graph, you can then experiment with any value to get the optimal frontload.

Well I asked the question which is at the heart of the matter: What is the frontload amount in this case? And rather than showing how your post gives even a method to go and calculate it as you instructed, let alone a better method, you say one "can experiment" with a table or a graph.

Even though asked you don't give the value, even though it would seem a blow to your cause to have left off of giving it when asked, particularly as you already did get yourself more time rather than it being an immediate exchange, and the entire POINT is calculating these things, as useful answers rather than math lessons or suggestion to go experiment with graphs.

So I don't see how you say your method is cleaner, better, simpler, or more accurate. You yourself are finding that you will have to put more time or experimentation to come up with a value!

Whereas with what I describe, a person can figure this example for himself in about 15 seconds or less in his head and be exactly correct.

Now there can be more but let's give you time to do as you wish and do more work in how to calculate at least this simple problem by your posted method.

 

[GeneralIx]

Not negligible, but zero, if referring to actual steady state.

Actually no. I do realise to the fullest that it is just being a dick to say so, but actually 1 is the asymptotic value that Y will near, but never actually reach. Though the value of Y will get to 0.99999999999999999999 and still closer, it will never be a full 100% 1. So the difference will get very negligible, not zero.

Log(35) is 1.5444 and "squaring that 24 times" yields a quite large number so I think you did not do what you are saying to do.

No, you're not doing what I said you should do. I see however that I didn't write it all out, which is plain stupid of me. This time in full: the 0.5th log of (x)=35. x^35=0.5, therefore this logarthmic function determines the degrading coefficient per unit of original time, in this case hours. Squaring that by 24, makes the degrade per day.

There is nothing mathematically wrong at all with what I wrote so I don't know why you say yours is "cleaner."

Talking math a log is "cleaner" than your way, since you're just coming close to the actual value, whereas a log will give you the exact value. There is nothing wrong with your method of calculating, it is however, mathematically speaking, not as clean. The difference will be small, but look at it like this:
Given f(x)=x^2
Dy/Dx (with an appropriately small value of x, like 0.01) approximately matches the actual f'(x)=2x, but is not the same. Like 0.667 is close to, but not the same as 2/3. Let me stress once more that this is just strictly mathematically speaking, there is no real extra value in this matter.

Well I asked the question which is at the heart of the matter: What is the frontload amount in this case? And rather than showing how your post provides it, you say one can experiment with a table or a graph. You don't give the value.

I read over your question, but here:
Given 500 mg testosterone enanthate, e5d, with an expected T1/2 of 6 days, your formula will put 1100mg as a frontload. My method however shows that 1139.5 is better. Even my method is just an approximation, but still more accurate than yours. That's why I said mine is more accurate.


So I don't see how you say your method is cleaner, better, simpler, or more accurate.

There you go.
Concluding:
It is more accurate, however it might not be evident for users because of the rounding of numbers, and rather slight a difference.
It is cleaner mathematically speaking, with no real life consequence.
It is just as simple (I never posed that it was simpler).
It has a broader useability for more advanced users.

 

[Bill Roberts]

1) I'm not going to argue that aspect of mathematics with you, but find someone who is a mathematician, or a suitable book, and you will find that mathematically the series in fact equals one rather than asymptotically approaches it. With reference to the mathematical series. It's the very basis of calculus actually.

With reference to biological application, very rapidly random factors exceed any difference resulting from the series in practice not being infinite. Pharmacologically, seven half-lives is considered equivalent to infinite (steady state) for this reason.

2) Yes, you didn't write your log function correctly: but in two different posts you didn't, so how were people supposed to follow your instructions?

3) Not so that the x = .5^(24/35) only comes close. It is exact. Try calculating these things before telling people they are wrong? You will find the result is the same as yours.

4) It is a fact that when the injection frequency differs from the half-life that a small difference results. Let's use your above example: your number, by a different method, is different by only under 4%.

You needed to experiment with a spreadsheet, you had to go through considerable procedures, it took you all this time to finally get a figure, you could never have done it in your head or explained in 30 seconds how do it as happens all the time with what I've posted, your earlier posts left no one able to get a figure from your instructions... and you call this better. Okay.

If it pleases you, good. But I suspect your method is not going to take off because it is not easier and your difference is in practice trivial. Less than 4% difference on the first injection, and that will rapidly close on following injections.

Two things for you to consider:

Having worked with well over a thousand people, I can tell you from experience that virtually no one chooses to measure out steroid amounts such as 1.73 mL. They are going to use 1.5 or 2.0, etc.

Second, any idea that the estimated half-life is anything like accurate to 4% would be quite mistaken. What you're doing is carrying calculations to 5 decimal places where the data is fuzzy in the first decimal place, and then arguing about how your third, fourth, and fifth places are different.

So perhaps it will make sense to you that people are unlikely to adjust their doses as you suggest, and when there's years of experience finding that -- whatever the true half-lives of the compounds may be -- the long-tested frontloads work just as desired, why change them?

You are free to do so of course. But I have a suspicion you're not going to convert many to your unwieldy method that even takes you a long time to produce an answer.