Talk:Science

Image versus table

Hi, my justification, why I remove that image here and only leave the table:

• Better readability: You have to click on the image to get a readable version
• Easier to edit
• “Loseless compressed”: Especially the small thumbnails look awful with the jpg artefacts.

— xZise [talk] 16:43, 18 October 2013 (CDT)

Formula for repeating experiments

According to the "Repeating experiments" section, when an experiment is repeated its future science potential is P - P * 0.8 where P is the current value. For ground samples on the launch pad, which give you 9.0 at first, you would expect this to be 1.8 the next time. However it is actually 2.3. Same for transmitting: P - P * 0.8 * E where E is the efficiency of the transmission seen on the blue "transmit" button. For this example that would be E = 0.5 so the next P should be 5.4, but it is 5.6. That formula is also missing a source (and google didn't yield any).

I noticed that the factor of reduction is defferent for some experiments: For crewreports it seems to be P - P * 0.627, for Evareports 0.8, for Mysterygoo 0.557 and for the Sc9001 0.715

Revised Formula

The way it appears to work is that the penalty is "science done" divided by the cap. For a crew report this is 5/8 so that's why it appears to be around 0.627 (rounding). It's actually 0.625. For EVA reports the base value is 8 and the cap is 10, hence 80%. For mystery goo the base is 10 and the cap is 18 so that results in 0.55 repeating (again, it may have appeared to be 0.557 due to the fact that KSP rounds in the GUI, though it doesn't round in the save files and you'll see that you get very tiny fractional science points). The material lab is 25/35 so that's 0.71428 etc.

The proper equation should be Y = T * S * (1 - P/C), where T is the transmission efficiency, S is the value of this experiment in this situation (base times the situation multiplier), P is the total number of points you've gained by doing this experiment in this situation (not just the last time, but the sum of all of them), C is the cap for this experiment in this situation (again this the cap from the cfg file times the situation multiplier), and Y is the yield of this experiment.

No source for this equation. But, it's in the save files more or less. Each experiment stores the science and the cap, and it also has a field called "scv" which is always equal to (1 - science / cap) and your next experiment always gives you whatever the first one did, times whatever scv was in your save file. Of course as I say it won't show up as exactly that value in the GUI due to rounding, but it will be added as exactly that amount in the save files--Akefay (talk) 23:52, 24 October 2013 (CDT)

Differential Equation

This looks like a differential equation to me. The solution should be P(t) = C - C * ( C / ( C - (S * T)))^(-t), where P(t) is the science obtained in total by doing it t times. Needs checking since I'm not confident about my maths, but at least this is correct for t = 0, 1 and infinity.

Calculate P(n+k) - P(n) for the science points you'll obtain by repeating it n times, when you've already done it k times. Or take log on both sides to calculate a particular t for some P(t). You'll only need the integer ts, though Tryourbreast (talk) 13:25, 26 October 2013 (CDT)

I don't know if you're right or wrong, but it's conceptually harder for the reader than the straight forward algebraic formula that is listed. I'd only update the page with this formula if it is more correct. --MCSquared (talk) 00:54, 29 October 2013 (CDT)

"Conceptually harder"? No, just no. Which part of it is conceptually hard at all? I didn't even include the steps on deriving it. You just plug in the value and get your result, period. Not difficult at all.

Besides, the current forumla isn't straightforward at all: each run depends on the previous run, and then the more previous run, and so on. It's convenient at calculating one step, but most of the times you want to calculate the total yield instead of each step, and this is the only way. I think it's needless to say that a equation of total yield is more useful when you want to calculate the science you'll earn, or the number of times you need to perform a experiment to get certain science.

If you still insist, though, just put both of them. It's simple.Tryourbreast (talk) 07:13, 29 October 2013 (CDT)

Now for t=0 the formula is obviously correct ($P(t=0)=C-C\cdot x^{0}=C-C=0$, $x$ is the base of the exponentation). For the first experiment the current formula in the article is $y=s\cdot m\cdot t$ while yours is $P(t=1)=C-C\cdot \left({\frac {C}{C-S\cdot T}}\right)^{1}=C-{\frac {C^{2}}{C-S\cdot T}}$ which doesn't look right. Although I'm not sure what this factor $T$ is. — xZise [talk] 11:25, 29 October 2013 (CDT)

The page used to denote the transmission efficiency $t$ as $T$, which I went with it because I used $t$ for the number of times you've did the experiment, and besides, constants should be in capital letters. Also you're wrong here; the power is $-t$, not $t$.

Let me restate the equation again: $P(t)=C\cdot (1-\left({\frac {C}{C-S\cdot T}}\right)^{-t})$. If C, S and T are all positive, then the equation reduces to $P(t)=C\cdot (1-\left({\frac {C-S\cdot T}{C}}\right)^{t})$ Tryourbreast (talk) 01:58, 30 October 2013 (CDT)

Ah darn, I didn't notice that the exponent is negative. So $P(t=1)=C-(C-S\cdot T)=S\cdot T$. But your formula doesn't take the celestial body modifier into account. Although I would guess you only need to multiply it to the transmission efficiency. $P(t)=C\cdot \left(1-\left({\frac {C}{C-S\cdot T\cdot M}}\right)^{-t}\right)$

Personally I'm also think we should show both formulas. As only your formula can give you the science yield without knowing how much you yielded before. — xZise [talk] 04:54, 30 October 2013 (CDT) Note, I fixed the formula for t=1 — xZise [talk] 05:51, 30 October 2013 (CDT)

Well, since the step equation has been changed again, let me fix the total yield equation: $P(t)=C\cdot M\cdot (1-({\frac {C-S\cdot T}{C}})^{t})$. Note that the asymptotic limit (doing the experiment infinite times) is $C\cdot M$.

Also, note that you do need to know how much you yielded before you calculate the total yield, because what you can get is the difference $P(n+t)-P(n)$, where n is the initial times of experiments you've done. When you've never done any science at all, it's $P(t)-P(0)$, and we have initial condition $P(0)=0$, so we can simply calculate $P(n)$ instead. When n is not 0, however, you need to subtract something.Tryourbreast (talk) 05:01, 30 October 2013 (CDT)

Okay your formula doesn't get you directly to science yielded with the n^th experiment, but you simply can take the difference of the n^th and (n-1)^th while you need to calculate all previous results (and not simply only the result before the n^th) with the formula currently given. Actually shouldn't $y=P(n+1)-P(n)$, where $y$ is the result from the formula in this article and where $p$ in that formula is equal to $P(n)$? Unfortunately I don't have enough time at the moment to verify this. — xZise [talk] 05:51, 30 October 2013 (CDT)

Yes.Tryourbreast (talk) 06:09, 30 October 2013 (CDT)

I'm pretty sure

$y=s\cdot m\cdot t\cdot (1-{\frac {st}{C}})^{i}$

is the incremental equation you're looking for. I've added it to the main article already -- if that was a no-no, feel free to remove it -- but my testing thus far shows it working well. It was derived using the difference equation for exponential growth/decay as presented on Wikipedia, where the equation $x_{t}=a\cdot x_{t-1}=x_{0}\cdot a^{t}$ is true. The incremental equation first put in the article is of the first (iterative) form, while the equation just added is of the second (non-iterative) form. NivvyDaSkrl (talk) 10:54, 30 October 2013 (CDT)

But we're talking about the total yield equation... What you've said was irrelevant. Also, the capital/small letter issue on that page still exists, so I fixed them. Tryourbreast (talk) 11:30, 30 October 2013 (CDT)

Ah. Sorry. I didn't intend to be irrelevant; I apologize. NivvyDaSkrl (talk) 19:08, 30 October 2013 (CDT)

Perhaps this will redeem me:

$\sum \limits _{i=0}^{n}smt(1-{\frac {st}{C}})^{i}=smt{\frac {C(1-(1-{\frac {st}{C}})^{1+n})}{st}}$ where $st\neq 0$. 10:52, 31 October 2013 (CDT)

And it simplified to $mC(1-({\frac {C-st}{C}})^{1+n})$, which is basically what I've derived above. Except in here it's $P(n+1)-P(0)$, which becomes $P(n+1)$ Tryourbreast (talk) 11:05, 31 October 2013 (CDT)

Eh, you're right. I should have read better. I'll think twice before attempting to contribute next time. NivvyDaSkrl (talk) 14:39, 31 October 2013 (CDT)

Data Scale?

How does the Data Scale value being listed for each experiment type tie into the formula? --MCSquared (talk) 00:54, 29 October 2013 (CDT)

Discussion on added equation.

I've added another form of the equation, based on the difference equation solution for exponential growth/decay equations on Wikipedia. It replaces "total science obtained" ($p$) with "number of experiments already run in the past" ($i$). I think it's worth adding; when finding the amount of science one will get from running an experiment for the nth time, it's no longer necessary to solve the equation for the other 1 through n-1 experiments previously run. It's also much more spreadsheet-friendly. NivvyDaSkrl (talk) 10:45, 30 October 2013 (CDT)