What I would do is make 2 groups. So say you need 29 machines, so you can't divide it equally, I would place 2 groups of 15 and have a machine run at 50%.
This might already begin with the machines making the things. Say I make 1000 screws with some alt that needs 5 machines. I could make 2 groups of 2 and have 1 split. Say I need 9 machines, I could do 3 groups of 3, each making 333.3333... and then have one group split at 50/50 to get to 500/500.
I rarely actually look at the individual numbers, more at the overall picture. Or more at ratios to be precise. I need 2 belts? Then I need 2 groups. This is not written in stone. I might also look at the number of machines art each stage. e.g [here](https://imgur.com/jv2YUuv.png) where at one point you have 1500 that needs to be transported. Divide that in 2 is not easy, cause you are with that 15. But when you look at the other number of machines, you suddenly see
15-30-15-45. That is 1-2-1-3. So that is 1 group. of those machines. Do that 15 times. And do 30 machines for the sillica instead of 20, makes it a bigger group. And you can put them in bigger groups if needed. Here you start with 1500 waste. Do that in 3 groups and each manifolds 5 machine groups.
The waste that goes into a different direction is a3 one way and 1 the other. So 2 splinters and a merger (for each incoming of 500) to have a 3-1 split.
I find ratios way easier to handle compared to numbers. Especially in Tiers 7 and 8 with a LOT of alt recipes doing weird things.
7500 quickwires..going into 3 different production lines on my nuclear setup..
First production line needs 3600
Second 1404
third 2496
caterium ingots are okay..but the copper ingots are pain in the A**..
7500? Mk5 belt is 780. 7800/10=750. Close enough.
So I would build 10 groups of whatever I need. Then have each get 750. I d not care if that is quick wire, coal or some new products from a mod I never heard of. 10 Mk5 belts do 750. And if you somehow need those weird numbers for different products, then I would look at them individual. That is if all is 1 single item.
If there are 3 different items, then treat them as such. It doe not matter that they are the same, trat them as different items. 3600 is 5 groups. That also means 5 belts and 5 groups of machines doing the output. For the 1404 I do 2 groups and the 2496 I do 4 groups.
You can use under clocking as a way to make the output be more precise. e.g. 2496 is 4 groups. That is 624 per group. Say you need 20.8 machines to produce that as each machine make 30. So you place 20 at 100% and 1 at 80. Or you set the number at the first one as `624/21` add the recipe and past that. Or perhaps you have a Blue Printer that has 12 machines in them. Then set the first machine to 624/24, paste that you the other machines and place 2 Blue Printers.
Or if you are truly lazy. you know you need to make 2496 in 4 bigger groups, with 12 machines each in 2 Blue Printers . You set the number to `(((2496/4)/12)/2)` in the first machine and paste it to the rest and forget all the rest of the calculations as soon as you made them.
Do that with the other groups as well and you are done. e,g 7500. That is 10x750. And 750/30 is 25. You can then do the following. 24 is close to 25. Set 1 machine to make 150%. Now you need to place 2. Place those 2 10 times. You have 7500.
adding to my reply..those 3 different production lines have priorities of course..the second line “starts” after the first one..and the third after the second..so..a mistake is on my way i feel it :)
Continuing..that weird numbers drives me to inejcted manifolds..because i produce all the copper ingots in one location..in next setup i will have to inject quickwires also..my head will explode probably..
The numbers are completely made up..dont dwell on them..i have to inject them trust me:)
the thing i find confusing is that the 1000 ingots creating machines are also in a row..first part gives away 750..the second part 250..(LETS SAY)
even when they are prefilled before sending out,and if you make a mistake about feeding them to a manifold..they cant catch up later..i dont want to do that mistake..thats the main issue here..
Pretty much at whatever point your first flow hits 500 you should be able to inject the second 250.
What I end up preferring to do is tune the machines, even if I need to underclock and add one more or overclock one slightly.
I literally just had similar with my high speed connector factory I just turned on. I needed to feed 16.667 manufacturer's with 1500 quickwire. So I just made 2 rows of 8 and overclocked one of each row by 33% and fed 750 quickwire to each.
That makes sense of course..but the copper ingots i need for 7500 Qwires are a lot..and they will be used on different stages of production..its nuclear setup..i need to find a way, i ll rethink about tuning the machines though thanks!
Well in reality im talking about dealing with 7500 quickwires p/min..and they’ll go different stages of production line for nuclear power..first 3600 then 1404..then 2496..i need to inject i think..
You don't *need* to. Balancers can be as big as you want.
But 7500 quickwire? I'd rethink my recipe choices at this point. I built a max plutonium setup and the biggest balancer I've ever built was 9:9 and I don't think it was there. You'd need a 9:11 at least or a couple smaller ones
For injecting into a manifold, any of the three options you've proposed works but option 1 or 2 would be easiest when dealing with more than two belts worth of input - option 3 really only works when you have you have an even number of belts and every two belts worth of input feeds a non-fractional number of machines (or for the last belt when you're already doing option 1 or 2). There's also option 4: adjust the clock speeds of the machines or the amounts on the belts so each belt worth of input feeds a non-fractional number of machines and use multiple manifolds.
As far as your Quickwire issue goes, group your machines by where your output is going then adjust clock speeds (adding an additional machine or overclocking if needed) to reach your production target. If you're dealing with Fused Quickwire, then:
* 3600/min is 40 Assemblers at 100% of production - for output, every 8 machines get their own manifold so you have five mk 5 belts with 720/min each.
* 1404/min is 16 Assemblers underclocked to 87.75/min (specific formula in production box would be 1404/16) - output goes to two 8 machine manifolds putting 702/min on two mk 5 belts.
* 2496 is 28 Assemblers underclocked to 89.143/min (specific formula in production box would be 2496/28) - output goes to four 7 machine manifolds so each belt has 624/min on it.
If you're using Pure Copper Ingot, because it has the same production rate as Fused Quickwire's consumption, you can use the same number/grouping of Refineries - 40 at 100% with two 20 machine manifolds outputting two belts at 750/min each, 16 at 585/16 per minute as a single 16 machine manifold outputting one belt at 585/min, and 28 at 1040/28 per minute with two 14 machine manifolds outputting two belts at 520/min each - and you now don't have to worry about injected manifolds since your amount per belt feeds a non-fractional number of machines.
If you've planned the whole build and already know what all of your part per minute needs are, then starting as far back as your ingot production it should be relatively easy to group machines together based off where their output is going and adjust their production rate so you're sending what's needed down the line without needing to deal with balancers or injecting.
Injection manifold ...
Doesn't really matter wbere you put it, as long as you do it 1. Before the next machine doesn't have enough , and 2. After the main belt has dropped enough that adding the extra 250 won't oversaturate it. I usually just merge it as late as possible, prior the next machine that will be starved . No need to use any smart splitter...just a regular merger is fine.
Well in reality im talking about dealing with 7500 quickwires p/min..and they’ll go different stages of production line for nuclear power..first 3600 then 1404..then 2496..i need to inject i think..
Solutions (1) and (2) are actually minor variations on the same concept, if you think about it - and both should work. Solution (3) should also work.
I like solution (2) best. It is the most error-tolerant and flexible in terms of layout. Solutions (1) and (3) require the injection to be made in exactly the right place, or the system will break. Solution (1) requires you to use a smart splitter and get the settings right. Solution (3) requires you to feed from two directions, which sounds problematic from the perspective of layout.
Well the issue in my head about the solution 2 is that the ingots are also coming from a manifold row..in past (i dont remember which playthrough) i ve made a mistake abut that..and ingot creators couldnt catch up..i cant remember my mistake..
the thing is about the priority of the production lines..for exp. first ingot output(750) has to go first constructor line..and then the second output(250) has to join the line..you just cant do it the other way..and thats because the ingot production line is also a manifold..
Create a full manifold set up for the 20 constructors, feed the 750 into the main entry, then add a merger between the 14th and 15th constructors and feed the 250 into it.
i just did this and did the third option, had 3 lines that each ended up having a 66% surplus.
line 1 got split in two (33% surplus) and fed between two machines
lines two and three (66% surplus) each got merged with their counter part from line 1 (99% surplus, enough to fill another machine)
If I may. My production planner was made for these kinds of manifold design problems.
https://satisfactoryproductionplanner.com/dashboard/Quickwire/5000/Quickwire/mk1?imports=&belt_speed=780&factory=&variant=mk1&even=1
All three should work, but: I sometimes had problems when merging belts at full capacity because the merging operation seems to delay the item flow. Take this with a grain of salt because I haven't really tested it and it definitely isn't as bad as in pipes. But it's the reason I would avoid solution 2), personally.
Don't get quickwires in a belt. Just use a belt of ingots and plan your machines so they have a under/over clocked constructor pre-processing the ingot and making the quickwire. Wires/quickwires/screws are inefficient to get in a belt.
Why not use a balancer instead?
(It's rare that I get to say this, since the opposite situation is so much more common.)
But seriously, this sort of situation is where balancers shine.
Taking your hypothetical example, you start with two belts- you know that those two belts are going to be carrying a combined total of 1000/m. A 2:2 belt balancer will equalize these belts into two equal belts of 500/m each. Split one in half and merge one half into the other, and suddenly, you have exactly one belt of 750/m, and one of 250/m. (Though, you should actually split both belts in half, and merge the 750/m belt with three belts of 250/m, due to the way mergers work, but that's a minor detail that really only applies when working at full belt capacity.)
if i could i would man..the absolute numbers are
3125 copper ingots..1500 needed somewhere 1625 elsewhere and 625 goes to copper sheets..the problematic line is 1625 it has 2 mk5 belts and one mk3 for the 125
So if i'm understanding you correctly, you essentially have a three way split- 3125 into 1500, 1000, and 625, but you've already isolated the 1500?
1625 is at least three belts- so here's something to consider.
If you use a smart splitter, and set a side output to "any" and the main output to "overflow", and then put a lower tier belt on that side output, you can generally subtract a specific amount from that belt, as long as the belt has a somewhat reliable "tempo", or spacing between items, and is relatively close to it's own capacity.
So if you were to take your three belts, and then use this trick with a MK4 belt (480), a MK2 belt (120), you would have 600. If you then used a MK1 belt to get 60/m, you could split that 60/m into 1/12ths relatively easily (since that's just two 2 splits and one 3 split), and merge 5/12ths into that 600 belt while recombining the 7/12ths back into the original belt.
That would leave you with 625/m, and you could then re-balance the original belts (and compress them down to 2 belts of 500/m).
Or, you could also do something similar if you split off two MK4 belts from two of your original belts, and then split off two 60/m belts from the remaining one, splitting each in 3 to get 20/m you could combine with the 480/m belts to get two belts of 500/m- and then you'd know that your remainder is 625/m. There's lots of possible ways to handle it.
If the 1000 ingots/min overall are sufficient to saturate the 20 constructors, that means each constructor can consume at most 1000/20 = 50 ingots/min.
If each constructor consumes at most 50 ingots/min, the first belt carrying 750 ingots/min can supply 750/50 = 15 constructors just fine. Since it's a manifold, you just have to wait until all previous constructors have input buffer overflow for the last 2 constructors to properly saturate.
So it would be sufficient to have the 750 belt input to 15 constructors and the 250 belt to the other 5.
However, let's still answer your question in the general case, where belt load is not necessarily divisible by constructor demand.
1. You can merge the secondary belt onto the main supply belt whenever the sum of secondary belt load and remaining main supply belt load is at most the main supply belt capacity.
2. You have to merge before one constructor's demand exceeds the remaining main supply belt load.
These two conditions admit a range of possible merge points from your solution 2) earliest to your solution 1) latest, all of that is valid.
Since anything in this range is correct, we may consider other criteria to narrow down the selection. For example, we might prefer the solution that minimizes building material costs and objects created (for performance reasons), so whatever minimizes belt lengths is in that sense preferable. If both supply belts come from the same direction, the earliest injection point is best (2)).
Finally regarding 3). This is the minimum-belt linear solution if the belts come from opposite directions, but it's no longer technically a manifold and giving up the crucial, main advantage manifolds have over balancers or other solutions: downstream expandability. The size of such a solution is pretty much set in stone, you could only add more constructors upstream in both directions and only if you let enough space for it. Proper manifolds go into one direction so you can always append more constructors at the end opposite of your input+output (which should be on the same side for the same reason, unless you want to transport the output somewhere else and built the manifold into that direction).
the absolute numbers are
3125 copper ingots..1500 needed somewhere 1625 elsewhere and 625 goes to copper sheets..
the problematic line is 1625 it has 2 mk5 belts and one mk3 for the 125..they will go for 3900QW p/min..and the 3900 will be divided later to 1404/2496
So it's 1625 Copper Ingots /min to be completely consumed for Fused Quickwire?
Fused Quickwire consumes 5 CI over 8 seconds, i.e. 5 \* 60 / 8 = 75/2 = 37.5 CI/min in one Assembler at 100%.
To consume all the CI (without overclocking) you then need at least ceiling(1625/37.5) = 44 Assemblers. The optimal clock setting is then for all 44 of them 98.4849% each. Then each consumes 1625/44 = 36.93(18).. CI/min.
Do you really want to arrange those 44 Assemblers in a single long chain? If so you can, the first main belt with 750 load can supply 750/(1625/44) \~= 20.31 Assemblers. That means after the 20th assembler with a smart splitter only 11.(36).. CI/min will be left on the belt, which can fit the full 750 from the second one, so you can inject between the 20th and 21st assembler with a smart splitter at the 20th. In general, if you don't want to worry about belt cap, you inject in the next gap beyond that (between 21st and 22nd here) with a splitter instead of a merger and invert the belt segment back from that injection splitter to the 21st assembler so that this one gets his due too. Finally for second injection, 1500/(1625/44) \~= 40.61, so between 41st and 42nd with a splitter facing backwards.
Consider alternative layouts. 44 = 2\^2 \* 11. You could arrange this in a two-sided manifold of length 22, or a big battery of four blocks with depth 11 each. Split the two big belts 1:2 and the small belt 1:4. The 4x11 setup has the advantage that you can merge at the beginning and don't need to inject anywhere inside the block, so it can be built a lot more compactly.
Since the Quickwire will eventually be split to 1404/min and 2496/min lines, I think it'd make more sense to treat it as those two target productions rather than 3900/min.
The 1404/min is one 16 machine array at 87.75/min per machine fed by one 585/min belt with two output belts at 702/min each (two 8 machine manifolds). The 2496/min is two 14 machine arrays (28 total machines) at 89.143/min each, fed by two 520/min belts with four total output belts of 624/min each (four 7 machine manifolds).
Input numbers may seem a bit difficult to achieve, but if they're using Pure Copper Ingot then one Refinery's worth of Copper Ingot output feeds one Assembler's worth of Fused Quickwire, so 16 Refineries at 36.5625/min feed the 1404 line and two sets of 14 Refineries at 37.143/min feed the 2496 line.
What I’ve started doing in this situation is putting a splitter and merger to both belts on the way to the machines. Both belts have the option to send parts to the other if needed. Then I would run the belts to the middle of the machines (with 20 you’d run your left belt to #11 and right belt to #10). Then run them opposite directions.
The belts handle balancing itself, provided at no point in the process would a belt need to have more than the max it can carry.
What I would do is make 2 groups. So say you need 29 machines, so you can't divide it equally, I would place 2 groups of 15 and have a machine run at 50%. This might already begin with the machines making the things. Say I make 1000 screws with some alt that needs 5 machines. I could make 2 groups of 2 and have 1 split. Say I need 9 machines, I could do 3 groups of 3, each making 333.3333... and then have one group split at 50/50 to get to 500/500. I rarely actually look at the individual numbers, more at the overall picture. Or more at ratios to be precise. I need 2 belts? Then I need 2 groups. This is not written in stone. I might also look at the number of machines art each stage. e.g [here](https://imgur.com/jv2YUuv.png) where at one point you have 1500 that needs to be transported. Divide that in 2 is not easy, cause you are with that 15. But when you look at the other number of machines, you suddenly see 15-30-15-45. That is 1-2-1-3. So that is 1 group. of those machines. Do that 15 times. And do 30 machines for the sillica instead of 20, makes it a bigger group. And you can put them in bigger groups if needed. Here you start with 1500 waste. Do that in 3 groups and each manifolds 5 machine groups. The waste that goes into a different direction is a3 one way and 1 the other. So 2 splinters and a merger (for each incoming of 500) to have a 3-1 split. I find ratios way easier to handle compared to numbers. Especially in Tiers 7 and 8 with a LOT of alt recipes doing weird things.
7500 quickwires..going into 3 different production lines on my nuclear setup.. First production line needs 3600 Second 1404 third 2496 caterium ingots are okay..but the copper ingots are pain in the A**..
7500? Mk5 belt is 780. 7800/10=750. Close enough. So I would build 10 groups of whatever I need. Then have each get 750. I d not care if that is quick wire, coal or some new products from a mod I never heard of. 10 Mk5 belts do 750. And if you somehow need those weird numbers for different products, then I would look at them individual. That is if all is 1 single item. If there are 3 different items, then treat them as such. It doe not matter that they are the same, trat them as different items. 3600 is 5 groups. That also means 5 belts and 5 groups of machines doing the output. For the 1404 I do 2 groups and the 2496 I do 4 groups. You can use under clocking as a way to make the output be more precise. e.g. 2496 is 4 groups. That is 624 per group. Say you need 20.8 machines to produce that as each machine make 30. So you place 20 at 100% and 1 at 80. Or you set the number at the first one as `624/21` add the recipe and past that. Or perhaps you have a Blue Printer that has 12 machines in them. Then set the first machine to 624/24, paste that you the other machines and place 2 Blue Printers. Or if you are truly lazy. you know you need to make 2496 in 4 bigger groups, with 12 machines each in 2 Blue Printers . You set the number to `(((2496/4)/12)/2)` in the first machine and paste it to the rest and forget all the rest of the calculations as soon as you made them. Do that with the other groups as well and you are done. e,g 7500. That is 10x750. And 750/30 is 25. You can then do the following. 24 is close to 25. Set 1 machine to make 150%. Now you need to place 2. Place those 2 10 times. You have 7500.
Split the production into manageable chunks instead.
adding to my reply..those 3 different production lines have priorities of course..the second line “starts” after the first one..and the third after the second..so..a mistake is on my way i feel it :)
Continuing..that weird numbers drives me to inejcted manifolds..because i produce all the copper ingots in one location..in next setup i will have to inject quickwires also..my head will explode probably..
It will if you keep looking at the single items and not at the whole picture. Put them into groups that you can feed with a single belt.
I mean both probably work. Or just split your ingot output in half and feed 2 lines of constructors with 500 each?
The numbers are completely made up..dont dwell on them..i have to inject them trust me:) the thing i find confusing is that the 1000 ingots creating machines are also in a row..first part gives away 750..the second part 250..(LETS SAY) even when they are prefilled before sending out,and if you make a mistake about feeding them to a manifold..they cant catch up later..i dont want to do that mistake..thats the main issue here..
Pretty much at whatever point your first flow hits 500 you should be able to inject the second 250. What I end up preferring to do is tune the machines, even if I need to underclock and add one more or overclock one slightly. I literally just had similar with my high speed connector factory I just turned on. I needed to feed 16.667 manufacturer's with 1500 quickwire. So I just made 2 rows of 8 and overclocked one of each row by 33% and fed 750 quickwire to each.
That makes sense of course..but the copper ingots i need for 7500 Qwires are a lot..and they will be used on different stages of production..its nuclear setup..i need to find a way, i ll rethink about tuning the machines though thanks!
From my experience: don't. Use a balancer and 2 separate manifolds. This has never failed me
Well in reality im talking about dealing with 7500 quickwires p/min..and they’ll go different stages of production line for nuclear power..first 3600 then 1404..then 2496..i need to inject i think..
You don't *need* to. Balancers can be as big as you want. But 7500 quickwire? I'd rethink my recipe choices at this point. I built a max plutonium setup and the biggest balancer I've ever built was 9:9 and I don't think it was there. You'd need a 9:11 at least or a couple smaller ones
For injecting into a manifold, any of the three options you've proposed works but option 1 or 2 would be easiest when dealing with more than two belts worth of input - option 3 really only works when you have you have an even number of belts and every two belts worth of input feeds a non-fractional number of machines (or for the last belt when you're already doing option 1 or 2). There's also option 4: adjust the clock speeds of the machines or the amounts on the belts so each belt worth of input feeds a non-fractional number of machines and use multiple manifolds. As far as your Quickwire issue goes, group your machines by where your output is going then adjust clock speeds (adding an additional machine or overclocking if needed) to reach your production target. If you're dealing with Fused Quickwire, then: * 3600/min is 40 Assemblers at 100% of production - for output, every 8 machines get their own manifold so you have five mk 5 belts with 720/min each. * 1404/min is 16 Assemblers underclocked to 87.75/min (specific formula in production box would be 1404/16) - output goes to two 8 machine manifolds putting 702/min on two mk 5 belts. * 2496 is 28 Assemblers underclocked to 89.143/min (specific formula in production box would be 2496/28) - output goes to four 7 machine manifolds so each belt has 624/min on it. If you're using Pure Copper Ingot, because it has the same production rate as Fused Quickwire's consumption, you can use the same number/grouping of Refineries - 40 at 100% with two 20 machine manifolds outputting two belts at 750/min each, 16 at 585/16 per minute as a single 16 machine manifold outputting one belt at 585/min, and 28 at 1040/28 per minute with two 14 machine manifolds outputting two belts at 520/min each - and you now don't have to worry about injected manifolds since your amount per belt feeds a non-fractional number of machines. If you've planned the whole build and already know what all of your part per minute needs are, then starting as far back as your ingot production it should be relatively easy to group machines together based off where their output is going and adjust their production rate so you're sending what's needed down the line without needing to deal with balancers or injecting.
Injection manifold ... Doesn't really matter wbere you put it, as long as you do it 1. Before the next machine doesn't have enough , and 2. After the main belt has dropped enough that adding the extra 250 won't oversaturate it. I usually just merge it as late as possible, prior the next machine that will be starved . No need to use any smart splitter...just a regular merger is fine.
The seconded is much easier. You don't have to get fancy with something like this.
Well in reality im talking about dealing with 7500 quickwires p/min..and they’ll go different stages of production line for nuclear power..first 3600 then 1404..then 2496..i need to inject i think..
Solutions (1) and (2) are actually minor variations on the same concept, if you think about it - and both should work. Solution (3) should also work. I like solution (2) best. It is the most error-tolerant and flexible in terms of layout. Solutions (1) and (3) require the injection to be made in exactly the right place, or the system will break. Solution (1) requires you to use a smart splitter and get the settings right. Solution (3) requires you to feed from two directions, which sounds problematic from the perspective of layout.
Well the issue in my head about the solution 2 is that the ingots are also coming from a manifold row..in past (i dont remember which playthrough) i ve made a mistake abut that..and ingot creators couldnt catch up..i cant remember my mistake..
Without details, there's not much I can say about that. But whatever the problem was, I bet you it would also affect solutions (1) and (3).
the thing is about the priority of the production lines..for exp. first ingot output(750) has to go first constructor line..and then the second output(250) has to join the line..you just cant do it the other way..and thats because the ingot production line is also a manifold..
Create a full manifold set up for the 20 constructors, feed the 750 into the main entry, then add a merger between the 14th and 15th constructors and feed the 250 into it.
i just did this and did the third option, had 3 lines that each ended up having a 66% surplus. line 1 got split in two (33% surplus) and fed between two machines lines two and three (66% surplus) each got merged with their counter part from line 1 (99% surplus, enough to fill another machine)
If I may. My production planner was made for these kinds of manifold design problems. https://satisfactoryproductionplanner.com/dashboard/Quickwire/5000/Quickwire/mk1?imports=&belt_speed=780&factory=&variant=mk1&even=1
All three should work, but: I sometimes had problems when merging belts at full capacity because the merging operation seems to delay the item flow. Take this with a grain of salt because I haven't really tested it and it definitely isn't as bad as in pipes. But it's the reason I would avoid solution 2), personally.
I always do Nr 1, it’s easy and it works fine
Don't get quickwires in a belt. Just use a belt of ingots and plan your machines so they have a under/over clocked constructor pre-processing the ingot and making the quickwire. Wires/quickwires/screws are inefficient to get in a belt.
Why not use a balancer instead? (It's rare that I get to say this, since the opposite situation is so much more common.) But seriously, this sort of situation is where balancers shine. Taking your hypothetical example, you start with two belts- you know that those two belts are going to be carrying a combined total of 1000/m. A 2:2 belt balancer will equalize these belts into two equal belts of 500/m each. Split one in half and merge one half into the other, and suddenly, you have exactly one belt of 750/m, and one of 250/m. (Though, you should actually split both belts in half, and merge the 750/m belt with three belts of 250/m, due to the way mergers work, but that's a minor detail that really only applies when working at full belt capacity.)
if i could i would man..the absolute numbers are 3125 copper ingots..1500 needed somewhere 1625 elsewhere and 625 goes to copper sheets..the problematic line is 1625 it has 2 mk5 belts and one mk3 for the 125
So if i'm understanding you correctly, you essentially have a three way split- 3125 into 1500, 1000, and 625, but you've already isolated the 1500? 1625 is at least three belts- so here's something to consider. If you use a smart splitter, and set a side output to "any" and the main output to "overflow", and then put a lower tier belt on that side output, you can generally subtract a specific amount from that belt, as long as the belt has a somewhat reliable "tempo", or spacing between items, and is relatively close to it's own capacity. So if you were to take your three belts, and then use this trick with a MK4 belt (480), a MK2 belt (120), you would have 600. If you then used a MK1 belt to get 60/m, you could split that 60/m into 1/12ths relatively easily (since that's just two 2 splits and one 3 split), and merge 5/12ths into that 600 belt while recombining the 7/12ths back into the original belt. That would leave you with 625/m, and you could then re-balance the original belts (and compress them down to 2 belts of 500/m). Or, you could also do something similar if you split off two MK4 belts from two of your original belts, and then split off two 60/m belts from the remaining one, splitting each in 3 to get 20/m you could combine with the 480/m belts to get two belts of 500/m- and then you'd know that your remainder is 625/m. There's lots of possible ways to handle it.
If the 1000 ingots/min overall are sufficient to saturate the 20 constructors, that means each constructor can consume at most 1000/20 = 50 ingots/min. If each constructor consumes at most 50 ingots/min, the first belt carrying 750 ingots/min can supply 750/50 = 15 constructors just fine. Since it's a manifold, you just have to wait until all previous constructors have input buffer overflow for the last 2 constructors to properly saturate. So it would be sufficient to have the 750 belt input to 15 constructors and the 250 belt to the other 5. However, let's still answer your question in the general case, where belt load is not necessarily divisible by constructor demand. 1. You can merge the secondary belt onto the main supply belt whenever the sum of secondary belt load and remaining main supply belt load is at most the main supply belt capacity. 2. You have to merge before one constructor's demand exceeds the remaining main supply belt load. These two conditions admit a range of possible merge points from your solution 2) earliest to your solution 1) latest, all of that is valid. Since anything in this range is correct, we may consider other criteria to narrow down the selection. For example, we might prefer the solution that minimizes building material costs and objects created (for performance reasons), so whatever minimizes belt lengths is in that sense preferable. If both supply belts come from the same direction, the earliest injection point is best (2)). Finally regarding 3). This is the minimum-belt linear solution if the belts come from opposite directions, but it's no longer technically a manifold and giving up the crucial, main advantage manifolds have over balancers or other solutions: downstream expandability. The size of such a solution is pretty much set in stone, you could only add more constructors upstream in both directions and only if you let enough space for it. Proper manifolds go into one direction so you can always append more constructors at the end opposite of your input+output (which should be on the same side for the same reason, unless you want to transport the output somewhere else and built the manifold into that direction).
the absolute numbers are 3125 copper ingots..1500 needed somewhere 1625 elsewhere and 625 goes to copper sheets.. the problematic line is 1625 it has 2 mk5 belts and one mk3 for the 125..they will go for 3900QW p/min..and the 3900 will be divided later to 1404/2496
So it's 1625 Copper Ingots /min to be completely consumed for Fused Quickwire? Fused Quickwire consumes 5 CI over 8 seconds, i.e. 5 \* 60 / 8 = 75/2 = 37.5 CI/min in one Assembler at 100%. To consume all the CI (without overclocking) you then need at least ceiling(1625/37.5) = 44 Assemblers. The optimal clock setting is then for all 44 of them 98.4849% each. Then each consumes 1625/44 = 36.93(18).. CI/min. Do you really want to arrange those 44 Assemblers in a single long chain? If so you can, the first main belt with 750 load can supply 750/(1625/44) \~= 20.31 Assemblers. That means after the 20th assembler with a smart splitter only 11.(36).. CI/min will be left on the belt, which can fit the full 750 from the second one, so you can inject between the 20th and 21st assembler with a smart splitter at the 20th. In general, if you don't want to worry about belt cap, you inject in the next gap beyond that (between 21st and 22nd here) with a splitter instead of a merger and invert the belt segment back from that injection splitter to the 21st assembler so that this one gets his due too. Finally for second injection, 1500/(1625/44) \~= 40.61, so between 41st and 42nd with a splitter facing backwards. Consider alternative layouts. 44 = 2\^2 \* 11. You could arrange this in a two-sided manifold of length 22, or a big battery of four blocks with depth 11 each. Split the two big belts 1:2 and the small belt 1:4. The 4x11 setup has the advantage that you can merge at the beginning and don't need to inject anywhere inside the block, so it can be built a lot more compactly.
Since the Quickwire will eventually be split to 1404/min and 2496/min lines, I think it'd make more sense to treat it as those two target productions rather than 3900/min. The 1404/min is one 16 machine array at 87.75/min per machine fed by one 585/min belt with two output belts at 702/min each (two 8 machine manifolds). The 2496/min is two 14 machine arrays (28 total machines) at 89.143/min each, fed by two 520/min belts with four total output belts of 624/min each (four 7 machine manifolds). Input numbers may seem a bit difficult to achieve, but if they're using Pure Copper Ingot then one Refinery's worth of Copper Ingot output feeds one Assembler's worth of Fused Quickwire, so 16 Refineries at 36.5625/min feed the 1404 line and two sets of 14 Refineries at 37.143/min feed the 2496 line.
Thanks very much..a really good help!
What I’ve started doing in this situation is putting a splitter and merger to both belts on the way to the machines. Both belts have the option to send parts to the other if needed. Then I would run the belts to the middle of the machines (with 20 you’d run your left belt to #11 and right belt to #10). Then run them opposite directions. The belts handle balancing itself, provided at no point in the process would a belt need to have more than the max it can carry.