a multiplayer game of parenting and civilization building
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So to run the plow it takes 1/24th a dose of kero. One dose of kero gives 5 iron, so one plow run is worth 0.2 iron.
With one iron a hoe can make 51 tilled rows, or 102 if you recycle the broken tool leftovers. So one plow run is equal to 0.2*102=22 (rounded up)
It's not enough to just maintain it's value though, it needs to pay off the cost of making the plow eventually too. So runs need to be at least 25 rows long to ever be able to pay off.
This isn't counting the soil saved. If someone can tell me the kerosene equivalent to get soil I'll factor that in. Otherwise, I'll do the maths later.
Last edited by NoTruePunk (2020-10-18 00:42:11)
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After much maths I figured out one bowl of soil costs aproximately 0.00569 kero doses. Most of that is from water, with a tiny amount of steel cost. It doesn't count when plowing for milkweed. I'll add this to the cost of the hoe:
102*0.00569+0.2=0.78038
The cost of the soil is significant. With each plowed furrow you save 0.00765 kero doses. That's compared to 0.00196 without factoring in the soil! So it starts paying off much earlier at run lengths of just 5.45, or 6 after rounding. If you want to use the plots for milkweed you'll have to lenthen them a bit to average out the cost, plus a little to pay off the engine eventually. I'd say runs of just 15 would very well be worth it.
It costs 4.4 kero doses to build a plow, assuming you only fire the newcomen 3 times. To pay that off you'd need to plow 568 tiles beyond the operating cost of 5.45 tiles per run. So at runs 15 tiles long it'll take 60 runs. That's not realistic. At runs 20 tiles long it'll take 39 runs to pay off the engine. Not in your lifetime, but maybe one day.
At runs 25 tiles long it will take 29 runs to pay off. Realistically it's going to take runs 30+, paying off after 23 runs. Lesson here is that engines are fucking expensive.
Last edited by NoTruePunk (2020-10-18 23:49:15)
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The problem here isn't strictly the lengths of the runs, but how much it's used past that initial 5.45 tiles. That's determined by your rate of consumption, which is determined by the size of your society. A population of 12 is still going to take days and days of IRL time to consume all the wheat and potatos from the plow whether your runs are 10 or 100 tiles long, which means the engine will likely never get paid off. By all means, if some dingus made 4 engines then go ahead and turn one into a plow. But never make one for this purpose. For some reason everybody likes making engines, so you can probably loot a couple from a dead town before you fire up that newcomen hammer.
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After much maths I figured out one bowl of soil costs aproximately 0.00569 kero doses. Most of that is from water, with a tiny amount of steel cost. It doesn't count when plowing for milkweed. I'll add this to the cost of the hoe:
102*0.00569+0.2=0.78038
The cost of the soil is significant. With each plowed furrow you save 0.00765 kero doses. That's compared to 0.00196 without factoring in the soil! So it starts paying off much earlier at run lengths of just 5.45, or 6 after rounding. If you want to use the plots for milkweed you'll have to lenthen them a bit to average out the cost, plus a little to pay off the engine eventually. I'd say runs of just would very well be worth it.
It costs 4.4 kero doses to build a plow, assuming you only fire the newcomen 3 times. To pay that off you'd need to plow 568 tiles beyond the operating cost of 5.45 tiles per run. So at runs 15 tiles long it'll take 60 runs. That's not realistic. At runs 20 tiles long it'll take 39 runs to pay off the engine. Not in your lifetime, but maybe one day.
At runs 25 tiles long it will take 29 runs to pay off. Realistically it's going to take runs 30+, paying off after 23 runs. Lesson here is that engines are fucking expensive.
Also theres no way I'd be able to make a engine in 3 newcomen rofl usually takes me 5 or 6.
Open gate now. Need truck to be more efficient!
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With one iron a hoe can make 51 tilled rows, or 102 if you recycle the broken tool leftovers.
So, you've assumed tilling of a shallow row or a 3 soil pile.
With one iron a hoe can make 51 tilled rows, or 102 if you recycle the broken tool leftovers. So one plow run is equal to 0.2*102=22 (rounded up)
(0.2 * 102) = 20.4. Rounding to the nearest natural number, gives us 20.
But that's treating a plow run like it tills only one crop plot.
So runs need to be at least 25 rows long to ever be able to pay off.
I don't know how you got 25 rows here.
Danish Clinch.
Longtime tutorial player.
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NoTruePunk wrote:With one iron a hoe can make 51 tilled rows, or 102 if you recycle the broken tool leftovers.
So, you've assumed tilling of a shallow row or a 3 soil pile.
Yes, most uses will be for reusing hardened rows. Occasionally you'll need to do milkweed, I don't count that though since that's the case for plows as well.
NoTruePunk wrote:With one iron a hoe can make 51 tilled rows, or 102 if you recycle the broken tool leftovers. So one plow run is equal to 0.2*102=22 (rounded up)
(0.2 * 102) = 20.4. Rounding to the nearest natural number, gives us 20.
But that's treating a plow run like it tills only one crop plot.
Can't ever round down since soil rows are integers. Have to round up to 21. Not sure how I got 22, I'll run through it again some time.
NoTruePunk wrote:So runs need to be at least 25 rows long to ever be able to pay off.
I don't know how you got 25 rows here.
That was just a quick estimate, I knew it didn't really count since it wasn't counting soil. Turns out it was a gross underestimate. See the followup post by me for a propper breakdown.
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Also theres no way I'd be able to make a engine in 3 newcomen rofl usually takes me 5 or 6.
Most of the cost comes from the 18 iron needed to make the plow, not counting tool use. The iron alone takes 3.6 doses to mine:
18*0.2=3.6
Water costs 0.25 kero per bucket, so with 6 buckets that's 5.1 kero. That's 666 hoe uses worth of kero needed to pay off beyond the operating cost of 5.45 per run.
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There's not much tool use needed to make a plow but if anybody wants I can try and work that in.
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Can't ever round down since soil rows are integers. Have to round up to 21.
Good point, you're right.
Danish Clinch.
Longtime tutorial player.
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Water costs 0.25 kero per bucket, so with 6 buckets that's 5.1 kero.
Did you mix up the position of the '5' and the '1'? Did you mean 1.5?
What number did you use for the hoe calculations?
Danish Clinch.
Longtime tutorial player.
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NoTruePunk wrote:Water costs 0.25 kero per bucket, so with 6 buckets that's 5.1 kero.
Did you mix up the position of the '5' and the '1'? Did you mean 1.5?
The digits being reversed is a coincidence
3.6 (steel) + 1.5 (water) = 5.1
What number did you use for the hoe calculations?
Each hoe use takes .00196 kero:
1/5/102=0.00196
Plus the cost of soil, which plow doesn't need. I can run through the cost of soil if you want, but it's fucking complicated and produces byproducts of mutton and wheat.
Last edited by NoTruePunk (2020-10-18 00:40:26)
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3.6 (steel) + 1.5 (water) = 5.1
1 steel and 1 water are both 1 kero? How did you reason that 1 steel is equivalent to 1 kero, and 1 water is equivalent to 1 kero?
1/5/102=0.00196
Division does NOT associate. ((1/5)/102) = (1/510) == .00196, where '==' means approximately equals. So, you used association to the left.
Danish Clinch.
Longtime tutorial player.
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NoTruePunk wrote:3.6 (steel) + 1.5 (water) = 5.1
1 steel and 1 water are both 1 kero? How did you reason that 1 steel is equivalent to 1 kero, and 1 water is equivalent to 1 kero?
A plow takes 18 steel, and each steel takes 0.2 kero to mine. A bucket of water takes 0.25 kero, and it takes approximately 6 buckets to fire the newcomen engine while making the diesel engine.
NoTruePunk wrote:1/5/102=0.00196
Division does NOT associate. ((1/5)/102) = (1/510) == .00196, where '==' means approximately equals. So, you used association to the left.
ACK, how embarrassing. Sorry I didn't format that correctly, but I think my conclusion is still accurate.
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Here's the breakdown for the kero cost of compost:
1 bowl of water = 0.025 kero
12 berry = 12/7 soil, 0.04286 kero (water)
2 carrot = 2/5 soil, 0.01392 kero (water and steel)
1 straw = 1 soil, 0.02696 kero (water and steel)
Total 0.10874 kero, 2.05714 soil
1 bowl of soil = 0.10874 / 21 == 0.00518
2.05714 * 0.00518 = 0.01065
New total 0.11939
Per bowl: 0.11939 / 21 = 0.00569
This requires an infinite series to find the limit, but I'm not going to do that.
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