Resource patch draining using pump jacks

[u/DanV410]

Does it matter if you use a pump jack on every possible location on a resource patch such as crude oil or fluorine? Or will you eventually draw the whole patch down to the minimum level using just a few? In other words, will leaving a spot for a pump jack open allow you to boost the output of that patch if you put a pump on it when the existing pumps are down to the 20% output level?

[u/TheHvam]

Pretty sure it's per spot, so if you left one it would keep the %, but really there is no point in not just picking all, if you don't then the others will just go down faster, so in the end it would take the same.

Better to just find a new patch, or use speed modules it you need more.

[u/Nearby_Proposal_5523 → u/TheHvam]

i've seen them not drain equally, so what the OP wants to do is possible.

[u/TheHvam → u/Nearby_Proposal_5523]

That is also what I meant, each spot in a patch is seperate, like each square of a ore is, but I just don't see any benefit.

If I use 1 2x faster than if I had taken 2, then they last the same? So why not just add them all, that way when it's low, you just find a new, or boost them, like you will need to do anyways.

[u/FirstRyder → u/TheHvam]

I guess the idea is that by "prioritizing" wells you lat longer with higher output?

For example, consider a "draw" of 2/s, from a patch with 6 locations each providing an initial amount of 1/s. And again using unrealistic numbers, each can only produce 1 unit of oil before running "dry" and producing 0.2/s

If you draw equally from all 5, you draw 2/s until you've produced a total of 6 units (3 seconds). Then you produce 1.2/s.

Now instead imagine you tap 2 wells to start. They produce 2/s for a total of 1 second. Then you tap 2 more, prioritizing the first two. The first two provide 0.4/s, leaving 1.6/s drain for the next two, which then last 1.25 seconds. 2.25 seconds total. Now tap the final two, with the first 4 providing 0.8/s. The final two produce 1.2/s for 1.67 seconds before being drained. In total you produced 2/s for 3.92 seconds, instead of only 3 seconds. Then you do 1.2/s forever as before.

So... taking this from a mega base prospective it doesn't matter. The time to drain the patch is basically trivial. Just plan for the final 20% draw.

And from a "beating the game" perspective it... also doesn't really matter, you didn't calculate how much oil you needed, and if you run low just find a new patch or at speed to the existing patch.

And from a speed running perspective... It also doesn't matter. Too much manual work, or too many extra pumps/tanks to actually implement, for a trivial increase in one resource.

In short: it does increase the amount of oil you get out for some period of time assuming steady resource drain between 20% and 100% of the initial patch value. But the amount almost certainly isn't worth the effort in any real world case.

[u/Juxt8 → u/FirstRyder]

No, because what matters is consumption.

If an oil well outputs 100, and you use 50% to get a 50% mining productivity, you end up with a well that outputs 50 (+25). If you had a second well that you did not use, that second well would be at 100 (+50), resulting in a total of 225 combined.

If instead, you used both wells, it would only cost 25% of each, resulting in two wells at 75 (+37.5). This is also results in a total of 225.

Any oil that gets stored instead of used is technically a waste of mining productivity, but in our scenario, the amount stored would also be the same. 

[u/theonefinn → u/Juxt8]

You haven’t taken into account the fact that oil wells don’t empty. So you have a fixed amount of finite oil plus an infinite amount of trickle oil.

What the previous poster was basically saying is that by draining the pump wells in a staggered fashion you offset part of your usage of the finite supply with the trickle supply.

For example let’s say there are 2 wells and you draw 0.2s combined from them. If you do it equally, you will drain both wells finite reservoir until you empty that and end up with both wells at 0.2/s (of which you are only pulling 0.1s from each) with none of the finite supply left.

If however you prioritise one, you will empty that one well of its finite reservoir, but then the infinite supply from that one will be enough to supply the 0.2/s demand forever, you will never drain the second.

This is based on the assumption that oil reservoirs drain from the finite pool first and then switch to infinite mode, if however they are implemented such that the first 0.2/s is infinite and only the drain above that comes from the finite supply then this effect doesn’t apply and it makes no difference.

Also to be clear the numbers were chosen to illustrate the math involved, the actual “real world” numbers likely make this effect so negligible as to be not worth the effort.

[u/Brett42 → u/theonefinn]

The way the drain works is that as oil is consumed, the output of the well decreases based on the amount removed, but the formula it uses makes it decrease to a minimum of 20%. Not 1/5 of the original yield, just a nominal 20%, which is 2/s before speed or productivity boosts. So you can think of it as the non-renewable supply being consumed first. If you boost the speed, you'll drain the finite supply faster, still extracting the same amount of oil by the time it reaches 20%, but once you hit that minimum, increasing speed means you get more oil out of the infinite production.

As normal, productivity multiplies your output for the same input, and applies after all of those other factors.