Should i raise my t score even more?

willnorthwest

New member
Hey guys, got my most recent bloodwork done. I inject every 5 days, and this test was done on day 5, immediately before my cyp injection. My test score was 696. First off, this score is alot better than the 218 that i had when i first started. 696 can be considered my "trough" score because it was done on the day of my next injection(but prior to, of course). I wanted to get some feedback as to what you think my t levels could be on days 1-4, could they be over 1000? Also, with a "trough" test level of 696, do you think it would be wise to try and raise it even higher? or do you think im at a good level?
Currently injecting 100mg cyp every 5 days, with 500iu hcG the day before cyp injection, and .50mg Adex the day after injection. Should i raise my cyp injection to 120mg every 5 days, or stay at 100mg? Thanks for the help!

Also, I would like to add that I am feeling pretty good on testosterone replacement therapy (TRT) so far. Just wondering if I should try and optimize my protocol even more so.
 
If you're feeling good, I'd stay at where you are. I'd bet your peak values are close to 1000ng/dl, but that's just a number to be honest and doesn't really matter in the grand scheme of things. If you feel like you could do a little more to see if you feel even better - by all means, talk to your doc and have at it!
 
Friend, I am no expert or even a terribly well-informed novice. But this is the way I have approached my own dosing. I went from 200mg @ 14 days to 200mg @10 days. I enjoyed a significant improvement with the increased dose. Because of a bit of end of cycle blahs, my doc has me on 150mg @ 7days. I think that will be perfect. The lower dose is less likely to be aromatized into estrogen and the blood level will be more stable. I feel real good on this amount and have no E issues, no man-boobs or other undesirable side effects. My suggestion is that you use as little as you can with the understanding that the more you use, the more you may have to contend with the physiologic consequences of more T than your body may want. Too much of a good thing can be bad for you.
Happy trails.
 
Appreciate all the input so far...thanks guys! So you think a "trough" score of 696 is healthy?

I've seen 5 days listed as the half-life of T Cyp, which would suggest a level of 1392 5 days earlier. That said, it took a day or two to reach peak from the time of injection, so the peak would be lower, probably within limits.
 
Appreciate all the input so far...thanks guys! So you think a "trough" score of 696 is healthy?
Yes, that is a healthy level of testosterone.

Edit: Removed my reply to mprtz since my value didn't seem right based on an exponential decay, so I'll leave it out. :p
 
Last edited:
Half-life doesn't quite work that way as it's exponential decay. I won't bore you with the math behind it, but assuming it's a straight A = A_0e^-kt equation, that comes out to ~800ng/dL at a day 1 peak.

I'm well familiar with exponential decay. And since the definition of half-life is the time in which the original amount is reduced by half, I fail to see what is wrong with what I wrote, i.e. if there is a known amount left at time t then one half life earlier there was twice as much of it, assuming that that time is still on the decay curve, which in this case it is not due to the time to ramp up to max.
 
I'm well familiar with exponential decay. And since the definition of half-life is the time in which the original amount is reduced by half, I fail to see what is wrong with what I wrote, i.e. if there is a known amount left at time t then one half life earlier there was twice as much of it, assuming that that time is still on the decay curve, which in this case it is not due to the time to ramp up to max.
Hah! You beat my edit while I was reading it and trying to figure out why it still didn't look right. Since you're familiar with exponential decay, maybe you can help me figure out where I went wrong with my calculation?

Sorry to other guys, I know math sucks - but I hate being wrong and not knowing where I messed up.

I started off with .5 = e^-5k, which gave me ln(.5) = -5k, k being ln(.5)/-5. I then plugged this back in with t = 1 for 696/x = e^-kt, and came to roughly 87% of the original value which seemed off. What step did I miss here? I've been doing stupid differential equations all day, so my brain refuses to work this out for some reason. :(
 
Hah! You beat my edit while I was reading it and trying to figure out why it still didn't look right. Since you're familiar with exponential decay, maybe you can help me figure out where I went wrong with my calculation?

Sorry to other guys, I know math sucks - but I hate being wrong and not knowing where I messed up.

I started off with .5 = e^-5k, which gave me ln(.5) = -5k, k being ln(.5)/-5. I then plugged this back in with t = 1 for 696/x = e^-kt, and came to roughly 87% of the original value which seemed off. What step did I miss here? I've been doing stupid differential equations all day, so my brain refuses to work this out for some reason. :(

I didn't really mean to imply that your estimate was wrong, though in hindsight it does seem low. Here is how I figure it: I get 0.139 for k, the same way that you do. Then I assume growth (rather than decay) at that rate for 4 days starting from the value of 696:
Asub4 = 696 e^(0.139(4)), which gives me 1213 for a level 4 days back. Just to verify, if I use t=5 instead, I get 1395, twice the value as expected. By using t=1, I think you got the value back one day from the min rather than one day away from the theoretical max.

The other way to do it would be Asub1 = 1395 e^(-.139(1)), which gives the same value of 1213.

Of course this is all only as accurate as the "5 day half life", assumption which is sure to be highly approximate and subject to all kinds of variation.
 
I didn't really mean to imply that your estimate was wrong, though in hindsight it does seem low. Here is how I figure it: I get 0.139 for k, the same way that you do. Then I assume growth (rather than decay) at that rate for 4 days starting from the value of 696:
Asub4 = 696 e^(0.139(4)), which gives me 1213 for a level 4 days back. Just to verify, if I use t=5 instead, I get 1395, twice the value as expected. By using t=1, I think you got the value back one day from the min rather than one day away from the theoretical max.

The other way to do it would be Asub1 = 1395 e^(-.139(1)), which gives the same value of 1213.

Of course this is all only as accurate as the "5 day half life", assumption which is sure to be highly approximate and subject to all kinds of variation.

Interesting, it must be taking a negative value of the natural log function that threw it off. I think the actual half-life is closer to 7-10 days, but that's irrelevant for the purposes of this discussion. Thanks for taking the time to help me figure this out, your explanation makes total sense. +rep :)
 
Inhave been experimenting and have lowered my dose from 140 mcg Test Cyp E7D to 100 mcg E7D. I'm 3 weeks in and can feel no decease in energy or libido. At 140/week my trough was about 550-650.

I'm of the opionion that a lot of guys running levels over 1200 are not really seeing any additional benefit to a more normal level.
 
Inhave been experimenting and have lowered my dose from 140 mcg Test Cyp E7D to 100 mcg E7D. I'm 3 weeks in and can feel no decease in energy or libido. At 140/week my trough was about 550-650.

I'm of the opionion that a lot of guys running levels over 1200 are not really seeing any additional benefit to a more normal level.

I think that depends partly on the goals... beyond energy and libido how much importance do you place on getting the best possible results in the gym.
 
Back
Top