Hybrid Solar Panels vs Hours to make

[zedekblue]

Say you want to make 100 Iridium plates, using nothing but solar energy. Disregarding the cost of making scrap, a Matter Fab, or anything besides the Iridium needed for Hybrid Solars, and pure UU matter for the final project, would it be faster to make a bunch of solar panels, or just wait it out with your current EU situation?
This has been a question that has plagued me since the beginning of FTB. I want to know at what point you should stop making solar panels and just wait for your UU matter to start piling up.
So I started doing some math.

This is what I have so far.

My real issue is column G. I would need a formula to decide how many new hybrids I need to make, and add them up one by one from column B until I have enough. (If that didn't make sense... Say I need to make 2 new solar panels, and I'm starting off with 8. So with 8 solar panels, I need to calculate how long it will take me to make a 9th, and then with 9, how long it will take me to make a 10th.)

With this information I can graph out and pretty much pinpoint exactly how many solar panels I should make to have it done the soonest, by just inputting how many solars I already have, and how many UU I need for the project.

Basically I need somebody to help me figure out how I should go about calculating this mess. I wanted to just create a Java or Computercraft program, but I don't really know how I would go about creating so many arrays of data and comparing it all.
Also:
I'm using Hybrid solar panels because they are the most efficient at making EU per Iridium used to make.
Also I'm finding hours needed by just dividing the game ticks needed by 20 (20 ticks per second), 60 (to find minutes), and then 60 again.

Thanks for your input!

 

[Adonis0]

I believe a simple sum formula should do.

Sum of column B up to the current row. That'll give the amount of time required to progress from 8 solars to the intended number. Because to get 11 hybrids starting at 8, it requires 200 hours to create the 9th, then another 100 hours on top of that to make the 10th, then 70 hours after that.

So, to make 11 hybrids starting at 8, you need 370 hours assuming you place down the solars as soon as you have the UU to do so.

 

[zedekblue]

Actually with 8 solars, it would take 25 hours to make a 9th. With 9, it would take 22. I'd have to start on different cells and end on different cells, depending on what D2 and the adjacent F column cell is. I don't think there would be a way to do that with just a sum function.

 

[Adonis0]

That is a good point, but, I think you can do it with a sum depending on your other formula.

How are you working out the time to make a new one?

 

[Bomb Bloke]

I'm using Hybrid solar panels because they are the most efficient at making EU per Iridium used to make.

Each hybrid solar requires two iridium plates (four iridium each) and one advanced solar (one iridium). Total is nine iridium, or 72 in total if you want to make eight.

You also need one bit of sunnarium at a cost of six UU, or 48 extra UU for eight panels.

Building an ultimate solar directly out of an advanced solar, to get the output of eight regular hybrids, requires 16 iridium plates (64 iridium) and one advanced solar (one iridium). Total is 65, or seven bits less.

You also need ten bits of sunnarium at a cost of 60 UU, which is 12 more then you'd need if going with hybrids, but that's not even worth two bits of iridium. And don't get me started on how much uranium you're saving by skipping regular hybrids and going straight to ultimates, let alone all the other components those extra advanced panels soak up...

(One of many edits: Forgot the extra two UU each advanced solar needs for the sunnarium part. That adds two UU to the cost of an ultimate, or sixteen to the cost of eight regular hybrids.)

A few months back, the advanced solars did not require iridium or UU and so hybrids were cheaper in regards to UU by a matter of those 12 units per eight.

Anyway, you want to use a loop to calculate this out. You can't really graph it 'cause your energy output doesn't scale up smoothly over time. I'll go with ComputerCraft's Lua, I guess, and ultimate solars.

local function timeWithXSolars(x)
  local timespent = 46666664800 / (512 * x)  -- Time to generate 100 plates with "x" solars.
  for i=1,x-1 do timespent = 8616666322 / (512 * i) + timespent end  -- Time to generate "x" solars, excluding the first.
  -- You may want to add in the amount of time it'll take your non-solar sources to generate your first solar here.
  return timespent
end
 
local lastTimeCost = 100000000  -- One ultimate should generate 100 plates in less then a hundred million ticks.
local curTimeCost = 0
local count = 1  -- A counter to track how many solars we're building.
 
while true do
  curTimeCost = timeWithXSolars(count)  -- How long would it take if I made "count" solars?
  if curTimeCost > lastTimeCost then break end  -- If it's more then it would've taken with "count"-1 solars, stop looping.
  lastTimeCost = curTimeCost  -- Otherwise, note that value down ready for the next check.
  count = count + 1
end
 
print("If I make "..(count-1).." solars I should have 100 plates in "..lastTimeCost.." ticks. Any other amount of solars will take longer.")

I think that's about right.

A few notes;

Ultimate solars take 8,616,666,322EU each to create, assuming the matter fabricator produces all 65 iridium ingots (455 UU), the 10 bits of sunnarium (60 UU), and the one sunnarium part (2 UU). This is not accounting for any EU spent of producing scrap or other materials.

This system does not assume that the advanced solars are being made and put onto the grid ASAP, then slowly upgraded to ultimates. Though you can probably see how to adapt it to do so, the code is a bit verbose and I can't be bothered typing it out right now. It also assumes the solars are producing EU full speed at all times; really they produce an average of (512 + 64) / 2 = 288EU per tick, assuming no rain or unexpected changes to the server time (eg, players using beds).

Because it assumes UU is the only "cost", this system will probably tell you regular hybrids would be faster, advanced ones faster still, and that plain ones could do it instantly. This is because the less UU it takes per solar, the less time it takes to get each solar onto the grid producing more solars. Ultimates are always going to be cheaper in the long run.

You need about as much uranium per solar as you do iridium. In fact, off the top of my head, I believe the cost is identical - ultimates again need less then eight regular hybrids do.

 

[zedekblue]

Hmm... That could work. Not exactly what I was looking for, but it might work.
Also
This is what my formulas are.
I haven't actually been calculating the UU matter for the sunnarium, although I probably should have been. I might have to change some of the formulas to account for them.

 

[Harvest88]

How much eu in total would it's cost to produce an Ultimate panel for every material needed including using recyclers without Overclockers for amplifying? As well as powering all the machines to do it most efficiently? (without any overclockers) Cause I only heard for the UUm and irriduim alone is 100 hours of max solar output.

 

[Bomb Bloke]

The short answer is "it can't be done"; it takes a random amount of effort to acquire resources, so the best anyone can do is an average estimate based on their personal collection methods. And no, you can't just generate the whole panel out of UU. Uranium is the biggest stumbling block there.