@Captain Herb I can't dress up this script up and make a good presentation
but after
1) taking the curve in the OP
2) multiplying each value my 2000 ml we get the amount (in picograms) excreted in a day
3) We integrate we get the amount excreted (in picograms) over 250ish days
4) We convert that to mg we get about 0.01 mg so roughly 0.05% of the compound administered would be recovered as m3 in urine over 250 days. If we supposed that the amounts excreted in 500 days were identical (although models say it should be less) is double this. We still only recover 0.1% over 500 days.
I think it's pretty safe to say when t-bol is done secreting (after we've numerically integrated) we should get nowhere near the amount of m3 recovered in urine as amount of parent drug administered.
Here's the script in matlab/python(numpy)/octave/sci-lab if anyone wants to fuck with it. I don't have time to make illustrations right now.
t_singledose = linspace(0,252,22); %%time single dose
conc_singledose = [0,150,45,30,22,22,22,22,22,15,15,11.4,9.8,8.2,6.6,5,3.4,1.8,2.64,3.12,3.6,4.08]; %%concentration single dose
%%Time to interpolate
t_singledose_interp = linspace(0,252,253);
conc_interp = interp1(t_singledose,conc_singledose,t_singledose_interp,"linear");
% Plot
plot(t_singledose_interp,conc_interp)
xlabel('Time (days)')
ylabel('Concentration (pg/ml)')
% Supposing 2000 ml of urine a day the amount per day is concentration
% multiplied by 2000
amount_interp = 2000*conc_interp;
%Total amount excreted is the amount integrated (using trapezoidal
%numerical integration in this case)
I = trapz(t_singledose_interp,amount_interp);
% I (for Integral) is 10,142,400 picograms
%Convert to grams and then milligrams, divide by 10^12 to convert to
%grams,dvide by 10^(-3) to convert to milligrams
I_milligrams = I *(10^(-12))*(10^3);
%%I_milligrams = 0.01 milligrams