Tutorial:Applied power math - Revision history

Bilka: typos

2025-02-07T17:29:48Z

typos

← Older revision		Revision as of 17:29, 7 February 2025
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	'''Expand to reveal answer'''		'''Expand to reveal answer'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	Since <math>E = P \cdot t</~~code~~>, a 20 MW factory will consume 20 MJ/s. Since coal contains 4 MJ, we'll need '''5 coal per second'''.		Since <math>E = P \cdot t</math>, a 20 MW factory will consume 20 MJ/s. Since coal contains 4 MJ, we'll need '''5 coal per second'''.

	<math>		<math>
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	'''Expand to reveal answer'''		'''Expand to reveal answer'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	Per the linked page, an electric mining drill ~~mine~~ 0.5 coal per second, so we'll need '''10 electric mining drills'''.		Per the linked page, an electric mining drill mines 0.5 coal per second, so we'll need '''10 electric mining drills'''.

	<math>\frac{5 \, \text{units}}{\text{s}}{0.5\, \frac{\text{units}}{\text{s}}} = 5</math>.		<math>\frac{5 \, \text{units}}{\text{s}}{0.5\, \frac{\text{units}}{\text{s}}} = 5</math>.

Orthotoma: /* Quiz */

2024-11-11T08:53:08Z

Quiz

← Older revision		Revision as of 08:53, 11 November 2024
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	'''Aside: why did the radar run for 20s and not 13.5s (50% of 27s)?'''		'''Aside: why did the radar run for 20s and not 13.5s (50% of 27s)?'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	Because it doesn't turn on and off instantly<!--. As you can see in the graph -->, the radar's demand of electrical power ramps up before reaching its operating power demand, therefore its electrical power usage is a function of time <math>p_\text{el}(t)</math>. When speaking of a machine using power over time, the common equation symbol is <math>W</math> for '''performed work''', which must be equal to the amount of '''energy supplied'''<math>E_{supply}</math>. To easier visualize the amount of energy or work, in the real world, instead of Joule, the unit in use is the kilowatt-hour (kWh).		Because it doesn't turn on and off instantly<!--. As you can see in the graph -->, the radar's demand of electrical power ramps up before reaching its operating power demand, therefore its electrical power usage is a function of time <math>p_\text{el}(t)</math>. When speaking of a machine using power over time, the common equation symbol is <math>W</math> for '''performed work''', which must be equal to the amount of '''energy supplied''' <math>E_{supply}</math>. To easier visualize the amount of energy or work, in the real world, instead of Joule, the unit in use is the kilowatt-hour (kWh).

	<math>E_{supply} = W = \int p_\text{el}(t) \mathop{}\!\mathrm{d} t</math>		<math>E_{supply} = W = \int p_\text{el}(t) \mathop{}\!\mathrm{d} t</math>

Untitled7 at 07:29, 11 November 2024

2024-11-11T07:29:45Z

Show changes

Dudasl at 21:22, 25 January 2020

2020-01-25T21:22:32Z

← Older revision		Revision as of 21:22, 25 January 2020
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	900,000 ÷ 1234.5679 = 729		900,000 ÷ 1234.5679 = 729

	800 pieces of coal per household, and a single steam engine could power ~~728~~ households ~~by approximate~~.		800 pieces of coal per household and a single steam engine could power 729 households exactly.
	</div>		</div>
	</div>		</div>

Dudasl: Change answer in second bonus to 729, which is exact value if we don't round off

2020-01-25T21:21:31Z

Change answer in second bonus to 729, which is exact value if we don't round off

← Older revision		Revision as of 21:21, 25 January 2020
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	3200 MJ ÷ 4 MJ = 800		3200 MJ ÷ 4 MJ = 800

	3200 MJ ÷ 30 days ÷ 24 hours ÷ 60 minutes ÷ 60 seconds = ~~1,235~~ W		3200 MJ ÷ 30 days ÷ 24 hours ÷ 60 minutes ÷ 60 seconds = 1234.5679 W
	900,000 ÷ ~~1235~~ = ~~728~~		900,000 ÷ 1234.5679 = 729

	800 pieces of coal per household, and a single steam engine could power 728 households by approximate.		800 pieces of coal per household, and a single steam engine could power 728 households by approximate.
	</div>		</div>
	</div>		</div>

Swan: Updated to 0.17. Removed entire section about 50% boiler efficiency, since that has been removed from the game. Updated coal fuel values. Updated pictures.

2019-05-05T00:01:22Z

Updated to 0.17. Removed entire section about 50% boiler efficiency, since that has been removed from the game. Updated coal fuel values. Updated pictures.

← Older revision		Revision as of 00:01, 5 May 2019
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	[[File:power-details.png\|800px]]		[[File:power-details.png\|800px]]

	Here we see one radar using 300 kW ~~(the 299 kW is a rounding error)~~. Factorio uses real science here. A Watt (W) is a measure of energy transfer. A kilowatt (kW) is 1000 W, a megawatt (MW) is 1000 kW, and if you're lucky enough to ever make a factory big enough, a gigawatt (GW) is 1000 MW.		Here we see one radar using 300 kW. Factorio uses real science here. A Watt (W) is a measure of energy transfer. A kilowatt (kW) is 1000 W, a megawatt (MW) is 1000 kW, and if you're lucky enough to ever make a factory big enough, a gigawatt (GW) is 1000 MW.

	So that answers the question of how much power our factory uses. To keep our factory running at full speed we need to maintain 300 kW. For ease of comparison, we'll convert that to 0.3 MW.		So that answers the question of how much power our factory uses. To keep our factory running at full speed we need to maintain 300 kW. For ease of comparison, we'll convert that to 0.3 MW.

	Next question! How much power is stored in coal? That's also easy, because it tells us when we hover over it: 8 MJ.		Next question! How much power is stored in coal? That's also easy, because it tells us when we hover over it: 4 MJ.

	[[File:coal-joules.png\|300px]]		[[File:coal-joules.png\|300px]]
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	== Quiz ==		== Quiz ==

	~~Assuming perfect efficiency, how~~ long could one piece of coal (8 MJ) run		How long could one piece of coal (4 MJ) run
	our single radar (0.3 MW)?		our single radar (0.3 MW)?

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	'''Expand to reveal answer'''		'''Expand to reveal answer'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	8 MJ = 0.3 MW × s		4 MJ = 0.3 MW × s
	s = 8 MJ ÷ 0.3 MW		s = 4 MJ ÷ 0.3 MW
	s = 26.67		s = 13.33

	~~Nearly 27~~ seconds.		A bit more than 13 seconds.
	</div>		</div>
	</div>		</div>

	~~Assuming perfect efficiency, what~~ is the maximum size factory (in watts)		What is the maximum size factory (in watts)
	a single piece of coal could run for 5 seconds?		a single piece of coal could run for 5 seconds?

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	'''Expand to reveal answer'''		'''Expand to reveal answer'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	8 MJ = MW × 5		4 MJ = MW × 5
	MW = 8 MJ ÷ 5		MW = 4 MJ ÷ 5
	MW = 1.6		MW = 0.8

	1.6 megawatts.		0.8 megawatts.
	</div>		</div>
	</div>		</div>
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	The energy ratios will match the weight ratios, so:		The energy ratios will match the weight ratios, so:

	8 MJ / 21000 MJ = weight ÷ 1000 kG		4 MJ / 21000 MJ = weight ÷ 1000 kG
	weight = (8 MJ ÷ 21000 MJ) × 1000 kG		weight = (4 MJ ÷ 21000 MJ) × 1000 kG
	weight = 0.38 kG		weight = 0.19 kG

	About ~~400g~~! Still doesn't explain how our character can carry so much of it...		About 190g! Still doesn't explain how our character can carry so much of it...
	</div>		</div>
	</div>		</div>

	~~== Efficiency ==~~

	Let's test it! We should be able to run a radar for 27 seconds, right? Using a setup with a single boiler, steam engine, and radar, we can drop a single piece of coal in the boiler and see how long the radar runs.

	~~[[File:radar-example.png\|800px]]~~

	~~We barely got 20s! What's wrong? We are generating less power than expected.~~
	~~Let's measure using an accumulator.~~

	~~[[File:Accumulator-example.png\|800px]]~~

	~~We started with 8 MJ in the coal, but only 4 MJ ended up in the accumulator.~~
	~~This is because, like in the real world, some energy is lost along the way as~~
	~~waste. In this case, only 50% is making it into our accumulator. In Factorio,~~
	~~this loss happens in the boiler, and it actually tells us this in the item details! You can think of the waste 50% as going into the pollution it generates...~~

	~~[[File:boiler-details.png]]~~

	Now we have everything we need to answer our initial question: '''how much coal do we need to power our factory?''' Well, the question actually needs to be a bit		Now we have everything we need to answer our initial question: '''how much coal do we need to power our factory?''' Well, the question actually needs to be a bit
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	'''Expand to reveal answer'''		'''Expand to reveal answer'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	Since <code>J = W × s</code>, a 20 MW factory will consume 20 MJ/s~~. We'll need 40 MJ/s input due to the 50% boiler efficiency loss~~. Since coal contains 8 MJ, we'll need '''5 coal per second''' (<code>5 × 8 MJ = 40 MJ</code>).		Since <code>J = W × s</code>, a 20 MW factory will consume 20 MJ/s. Since coal contains 4 MJ, we'll need '''5 coal per second''' (<code>5 × 4 MJ = 20 MJ</code>).
	</div>		</div>
	</div>		</div>
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	'''Expand to reveal answer'''		'''Expand to reveal answer'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	Per the linked page, an electric mining drill produces ~0.5/s, so we'll need '''10 electric mining drills''' (<code>10 × 0.5 = 5</code>).		Per the linked page, an electric mining drill produces 0.5/s, so we'll need '''10 electric mining drills''' (<code>10 × 0.5 = 5</code>).
	</div>		</div>
	</div>		</div>
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	3.2 GJ = 3,200 MJ		3.2 GJ = 3,200 MJ
	3200 MJ ÷ (8 MJ ~~× 50%)~~ = 800		3200 MJ ÷ 4 MJ = 800

	3200 MJ ÷ 30 days ÷ 24 hours ÷ 60 minutes ÷ 60 seconds = 1,235 W		3200 MJ ÷ 30 days ÷ 24 hours ÷ 60 minutes ÷ 60 seconds = 1,235 W

TomFryers: Spaces between units

2019-04-19T16:26:15Z

Spaces between units

← Older revision		Revision as of 16:26, 19 April 2019
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	[[File:power-details.png\|800px]]		[[File:power-details.png\|800px]]

	Here we see one radar using ~~300kW~~ (the ~~299kW~~ is a rounding error). Factorio uses real science here. A Watt (W) is a measure of energy transfer. A kilowatt (kW) is 1000 W, a megawatt (MW) is 1000 kW, and if you're lucky enough to ever make a factory big enough, a gigawatt (GW) is 1000 MW.		Here we see one radar using 300 kW (the 299 kW is a rounding error). Factorio uses real science here. A Watt (W) is a measure of energy transfer. A kilowatt (kW) is 1000 W, a megawatt (MW) is 1000 kW, and if you're lucky enough to ever make a factory big enough, a gigawatt (GW) is 1000 MW.

	So that answers the question of how much power our factory uses. To keep our factory running at full speed we need to maintain 300 kW. For ease of comparison, we'll convert that to 0.3 MW.		So that answers the question of how much power our factory uses. To keep our factory running at full speed we need to maintain 300 kW. For ease of comparison, we'll convert that to 0.3 MW.

SolinNull at 07:07, 6 October 2018

2018-10-06T07:07:25Z

← Older revision		Revision as of 07:07, 6 October 2018
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	{{Languages}}		{{Languages}}
	~~Bu sayfada "Bir Fabrikayı çalıştırmak için ne kadar kömür gerekiyor~~?~~" sorusunu cevaplayacağız.~~		In this tutorial we'll be answering the question: '''how much coal is needed to power a factory?'''

	~~Her şeyden önce~~, ~~fabrikamızın ne kadar enerji tükettiğini bilmeliyiz. Bunu görmek için elektrik teline tıklayıp~~ electricity ~~sekmesini incelemeniz kafidir~~.		First off, we need to know: how much power does our factory use? That's easy - you can check the electricity tab by clicking on a power pole.

	[[File:power-details.png\|800px]]		[[File:power-details.png\|800px]]

	~~Yukarıdaki görselde görülebileceği üzere fabrikamız~~ 300kW~~'lık güç tüketiyor.~~ (299kW ~~yazmasına aldırış etmeyin. Yuvarlama hatasından oluşmuş~~). Factorio ~~bilimi kullanır~~. Watt(W) ~~gücün birimidir~~. ~~1 Kilowatt~~(kW), ~~1000W'~~a ~~ve 1~~ megawatt(MW) ~~da 1000kW~~'a ~~eşittir. Eğer fabrikanızı yeterince büyütebilirseniz bu bilgi de işinize yarayacaktır: 1~~ gigawatt(GW) ~~da 1000MW'a eşittir~~.		Here we see one radar using 300kW (the 299kW is a rounding error). Factorio uses real science here. A Watt (W) is a measure of energy transfer. A kilowatt (kW) is 1000 W, a megawatt (MW) is 1000 kW, and if you're lucky enough to ever make a factory big enough, a gigawatt (GW) is 1000 MW.

	~~Fabrikamızın ne kadar güç harcadığını gördük~~. ~~Şimdi~~, ~~fabrikamızı tam gaz işletmek istiyorsak 300kW güç üretimini sürdürmeliyiz. Daha kolay işlem yapmak için - birim dönüştüreceğiz - 300kW~~'ı 0.~~3MW olarak alacağız~~.		So that answers the question of how much power our factory uses. To keep our factory running at full speed we need to maintain 300 kW. For ease of comparison, we'll convert that to 0.3 MW.

	~~Akabinde yeni bir soru~~! ~~Kömürden ne kadar güç elde edilir~~? ~~Bu soruyu yanıtlamak çok kolay~~, ~~yapmamız gereken imleci üzerine getirip beklemek akabinde 8MJ güç elde edilebildiğini görmüş olacağız~~.		Next question! How much power is stored in coal? That's also easy, because it tells us when we hover over it: 8 MJ.

	[[File:coal-joules.png\|300px]]		[[File:coal-joules.png\|300px]]

	Joule~~, enerjinin birimlerinden birisidir~~. 1 kilojoule(kJ), ~~1000J~~'~~e eşittir~~.		A Joule is another standard measure of stored energy. As with Watts a kilojoule (kJ) is 1000 J, and so on. There is a fixed relationship between Joules and Watts. '''1 Joule can provide 1 Watt of power for 1 second.''' As a formula:

	~~Enerji = Güç * Zaman~~		J = W × s
	J = W * s

	~~Eğer fark ettiyseniz; Zaman'ı her zaman 1 saniye olarak alırsak. Enerji ve Güç birbirlerinin aynısı oluyor. Yani fabrikamızı idame ettirmek için~~ 0.~~3MW güce ihtiyacımız var ise bir saniyede~~ 0.~~3MJ enerji tüketmemiz gerekecektir~~.		So to run our factory at 0.3 MW, we need to consume 0.3 MJ every second.

	== Quiz ==		== Quiz ==

	~~Mükemmel verimle çalıştığımızı düşünürsek~~, ~~bir parça kömür~~ (~~8MJ~~)~~; bir radarı~~ (0.~~3MW~~) ~~kaç saniye çalıştırır~~?		Assuming perfect efficiency, how long could one piece of coal (8 MJ) run
			our single radar (0.3 MW)?

	<div class="toccolours mw-collapsible mw-collapsed">		<div class="toccolours mw-collapsible mw-collapsed">
	'''~~Cevabı görmek için tıklayınız~~'''		'''Expand to reveal answer'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	8 MJ = 0.3 MW * s		8 MJ = 0.3 MW × s
	s = 8 MJ / 0.3 MW		s = 8 MJ ÷ 0.3 MW
	s = 26.67		s = 26.67

	~~Yaklaşık~~ 27 ~~saniye çalıştırır~~.		Nearly 27 seconds.
	</div>		</div>
	</div>		</div>

	~~Kusursuz verime sahip olduğumuzu varsayarsak, bir tane kömür~~, 5 ~~saniye içinde, azami güç tüketimde kaç watt güç üretir~~?		Assuming perfect efficiency, what is the maximum size factory (in watts)
			a single piece of coal could run for 5 seconds?

	<div class="toccolours mw-collapsible mw-collapsed">		<div class="toccolours mw-collapsible mw-collapsed">
	'''~~Cevabı görmek için tıklayınız~~'''		'''Expand to reveal answer'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	8 MJ = MW * 5		8 MJ = MW × 5
	MW = 8 MJ / 5		MW = 8 MJ ÷ 5
	MW = 1.6		MW = 1.6

	1.6 ~~megawatt üretir~~.		1.6 megawatts.
	</div>		</div>
	</div>		</div>

	'''~~İlave Soru~~!''' ~~Gerçek hayatta, bir~~ ton ~~kömür~~ [http://hypertextbook.com/facts/2006/LunChen.shtml ~~yaklaşık~~ 21 GJ] ~~enerji içerir~~. ~~Bu durumda bir tane~~ Factorio ~~kömürü kaç kg'dır~~?		'''BONUS!''' A ton of real world coal [http://hypertextbook.com/facts/2006/LunChen.shtml contains about 21 GJ]. How much does a
			piece of Factorio coal weigh?

	<div class="toccolours mw-collapsible mw-collapsed">		<div class="toccolours mw-collapsible mw-collapsed">
	'''~~Cevabı görmek için tıklayınız~~'''		'''Expand to reveal answer'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	~~Enerji oranları ile kütle oranları birbirlerini eşler~~, ~~yani~~:		The energy ratios will match the weight ratios, so:

	~~8MJ~~ / ~~21000MJ~~ = ~~kütle / 1000kG~~		8 MJ / 21000 MJ = weight ÷ 1000 kG
	~~kütle~~ = (~~8MJ / 21000MJ~~) * 1000kG		weight = (8 MJ ÷ 21000 MJ) × 1000 kG
	~~kütle~~ = 0.~~38kG~~		weight = 0.38 kG

	~~Yaklaşık~~ 400g! ~~Acaba bizim karakter bu kadar ağır kömürü nasıl taşıyor? Cevap yok~~...		About 400g! Still doesn't explain how our character can carry so much of it...
	</div>		</div>
	</div>		</div>

	== ~~Verim~~ ==		== Efficiency ==

	~~Hadi bu bilgileri~~ test ~~edelim~~! ~~Bir radarı~~ 27 ~~saniye boyunca çalıştırabilmeliyiz~~, ~~değil mi~~? ~~Bir kazan~~, ~~bir buhar makinesi ve bir~~ radar ~~ile sistemi kurduk. Şimdi~~, ~~kazanın içine bir parça kömür koyacağız ve radarın kaç saniye çalışacağını göreceğiz~~.		Let's test it! We should be able to run a radar for 27 seconds, right? Using a setup with a single boiler, steam engine, and radar, we can drop a single piece of coal in the boiler and see how long the radar runs.

	[[File:radar-example.png\|800px]]		[[File:radar-example.png\|800px]]

	~~Zar zor 20 saniye çalıştı~~! ~~Ne oldu~~? ~~Beklediğimizden daha az enerji ürettik~~. ~~Bir de akü kullanarak ölçelim:~~		We barely got 20s! What's wrong? We are generating less power than expected.
			Let's measure using an accumulator.

	[[File:Accumulator-example.png\|800px]]		[[File:Accumulator-example.png\|800px]]

SolinNull at 07:05, 6 October 2018

2018-10-06T07:05:36Z

← Older revision		Revision as of 07:05, 6 October 2018
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	[[File:power-details.png\|800px]]		[[File:power-details.png\|800px]]

	Yukarıdaki görselde görülebileceği üzere ~~radar~~ 300kW'lık güç ~~kullanıyor~~ (299kW yazmasına aldırış etmeyin. Yuvarlama hatasından ~~kaynaklı~~). Factorio ~~gerçek~~ bilimi kullanır. Watt(W) gücün birimidir. 1 Kilowatt(kW), 1000W ve 1 megawatt(MW) 1000kW'a eşittir. Eğer fabrikanızı yeterince büyütebilirseniz bu bilgi de işinize yarayacaktır: 1 gigawatt(GW) da 1000MW'a eşittir.		Yukarıdaki görselde görülebileceği üzere fabrikamız 300kW'lık güç tüketiyor. (299kW yazmasına aldırış etmeyin. Yuvarlama hatasından oluşmuş). Factorio bilimi kullanır. Watt(W) gücün birimidir. 1 Kilowatt(kW), 1000W'a ve 1 megawatt(MW) da 1000kW'a eşittir. Eğer fabrikanızı yeterince büyütebilirseniz bu bilgi de işinize yarayacaktır: 1 gigawatt(GW) da 1000MW'a eşittir.

	Fabrikamızın ne kadar güç harcadığını ~~anlamış olduk~~. ~~Eğer~~ fabrikamızı tam gaz işletmek istiyorsak 300kW güç üretimini sürdürmeliyiz. Daha kolay işlem yapmak için, 300kW'ı 0.3MW~~'a dönüştüreceğiz~~.		Fabrikamızın ne kadar güç harcadığını gördük. Şimdi, fabrikamızı tam gaz işletmek istiyorsak 300kW güç üretimini sürdürmeliyiz. Daha kolay işlem yapmak için - birim dönüştüreceğiz - 300kW'ı 0.3MW olarak alacağız.

	Akabinde yeni bir soru! Kömürden ne kadar güç elde edilir? Bu soruyu yanıtlamak çok kolay, yapmamız gereken imleci üzerine getirip beklemek akabinde 8MJ güç elde edilebildiğini görmüş olacağız.		Akabinde yeni bir soru! Kömürden ne kadar güç elde edilir? Bu soruyu yanıtlamak çok kolay, yapmamız gereken imleci üzerine getirip beklemek akabinde 8MJ güç elde edilebildiğini görmüş olacağız.
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	[[File:coal-joules.png\|300px]]		[[File:coal-joules.png\|300px]]

	A Joule ~~is another standard measure of stored energy~~. ~~As with Watts a~~ kilojoule (kJ) ~~is 1000 J~~, ~~and so on. There is a fixed relationship between Joules and Watts.~~ '~~''1 Joule can provide 1 Watt of power for 1 second~~.~~''' As a formula:~~		Joule, enerjinin birimlerinden birisidir. 1 kilojoule(kJ), 1000J'e eşittir.

	J = W × s		Enerji = Güç * Zaman
			J = W * s

	~~So to run our factory at~~ 0.~~3 MW, we need to consume~~ 0.~~3 MJ every second~~.		Eğer fark ettiyseniz; Zaman'ı her zaman 1 saniye olarak alırsak. Enerji ve Güç birbirlerinin aynısı oluyor. Yani fabrikamızı idame ettirmek için 0.3MW güce ihtiyacımız var ise bir saniyede 0.3MJ enerji tüketmemiz gerekecektir.

	== Quiz ==		== Quiz ==

	~~Assuming perfect efficiency~~, ~~how long could one piece of coal~~ (~~8 MJ~~) ~~run~~		Mükemmel verimle çalıştığımızı düşünürsek, bir parça kömür (8MJ); bir radarı (0.3MW) kaç saniye çalıştırır?
	~~our single radar~~ (0.~~3 MW~~)?

	<div class="toccolours mw-collapsible mw-collapsed">		<div class="toccolours mw-collapsible mw-collapsed">
	'''~~Expand to reveal answer~~'''		'''Cevabı görmek için tıklayınız'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	8 MJ = 0.3 MW × s		8 MJ = 0.3 MW * s
	s = 8 MJ ÷ 0.3 MW		s = 8 MJ / 0.3 MW
	s = 26.67		s = 26.67

	~~Nearly~~ 27 ~~seconds~~.		Yaklaşık 27 saniye çalıştırır.
	</div>		</div>
	</div>		</div>

	~~Assuming perfect efficiency~~, ~~what is the maximum size factory (in watts)~~		Kusursuz verime sahip olduğumuzu varsayarsak, bir tane kömür, 5 saniye içinde, azami güç tüketimde kaç watt güç üretir?
	~~a single piece of coal could run for~~ 5 ~~seconds~~?

	<div class="toccolours mw-collapsible mw-collapsed">		<div class="toccolours mw-collapsible mw-collapsed">
	'''~~Expand to reveal answer~~'''		'''Cevabı görmek için tıklayınız'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	8 MJ = MW × 5		8 MJ = MW * 5
	MW = 8 MJ ÷ 5		MW = 8 MJ / 5
	MW = 1.6		MW = 1.6

	1.6 ~~megawatts~~.		1.6 megawatt üretir.
	</div>		</div>
	</div>		</div>

	'''~~BONUS~~!''' A ton ~~of real world coal~~ [http://hypertextbook.com/facts/2006/LunChen.shtml ~~contains about~~ 21 GJ]. ~~How much does a~~		'''İlave Soru!''' Gerçek hayatta, bir ton kömür [http://hypertextbook.com/facts/2006/LunChen.shtml yaklaşık 21 GJ] enerji içerir. Bu durumda bir tane Factorio kömürü kaç kg'dır?
	~~piece of~~ Factorio ~~coal weigh~~?

	<div class="toccolours mw-collapsible mw-collapsed">		<div class="toccolours mw-collapsible mw-collapsed">
	'''~~Expand to reveal answer~~'''		'''Cevabı görmek için tıklayınız'''
	<div class="mw-collapsible-content">		<div class="mw-collapsible-content">
	~~The energy ratios will match the weight ratios~~, so:		Enerji oranları ile kütle oranları birbirlerini eşler, yani:

	~~8 MJ~~ / ~~21000 MJ~~ = ~~weight ÷ 1000 kG~~		8MJ / 21000MJ = kütle / 1000kG
	~~weight~~ = (~~8 MJ ÷ 21000 MJ~~) ~~× 1000 kG~~		kütle = (8MJ / 21000MJ) * 1000kG
	~~weight~~ = 0.~~38 kG~~		kütle = 0.38kG

	~~About~~ 400g! ~~Still doesn't explain how our character can carry so much of it~~...		Yaklaşık 400g! Acaba bizim karakter bu kadar ağır kömürü nasıl taşıyor? Cevap yok...
	</div>		</div>
	</div>		</div>

	== ~~Efficiency~~ ==		== Verim ==

	~~Let's~~ test it! ~~We should be able to run a radar for~~ 27 ~~seconds~~, ~~right~~? ~~Using a setup with a single boiler~~, ~~steam engine, and~~ radar, ~~we can drop a single piece of coal in the boiler and see how long the radar runs~~.		Hadi bu bilgileri test edelim! Bir radarı 27 saniye boyunca çalıştırabilmeliyiz, değil mi? Bir kazan, bir buhar makinesi ve bir radar ile sistemi kurduk. Şimdi, kazanın içine bir parça kömür koyacağız ve radarın kaç saniye çalışacağını göreceğiz.

	[[File:radar-example.png\|800px]]		[[File:radar-example.png\|800px]]

	~~We barely got 20s~~! ~~What's wrong~~? ~~We are generating less power than expected.~~		Zar zor 20 saniye çalıştı! Ne oldu? Beklediğimizden daha az enerji ürettik. Bir de akü kullanarak ölçelim:
	~~Let's measure using an accumulator~~.

	[[File:Accumulator-example.png\|800px]]		[[File:Accumulator-example.png\|800px]]

SolinNull at 06:27, 6 October 2018

2018-10-06T06:27:13Z

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	{{Languages}}		{{Languages}}
	~~In this tutorial we'll be answering the question: '''how much coal is needed to power a factory~~?~~'''~~		Bu sayfada "Bir Fabrikayı çalıştırmak için ne kadar kömür gerekiyor?" sorusunu cevaplayacağız.

	~~First off~~, ~~we need to know: how much power does our factory use? That's easy - you can check the~~ electricity ~~tab by clicking on a power pole~~.		Her şeyden önce, fabrikamızın ne kadar enerji tükettiğini bilmeliyiz. Bunu görmek için elektrik teline tıklayıp electricity sekmesini incelemeniz kafidir.

	[[File:power-details.png\|800px]]		[[File:power-details.png\|800px]]

	~~Here we see one~~ radar ~~using~~ 300kW (~~the~~ 299kW ~~is a rounding error~~). Factorio ~~uses real science here~~. A Watt (W) ~~is a measure of energy transfer~~. ~~A kilowatt~~ (kW) ~~is 1000 W~~, a megawatt (MW) ~~is 1000 kW, and if you~~'~~re lucky enough to ever make a factory big enough,~~ a gigawatt (GW) ~~is 1000 MW~~.		Yukarıdaki görselde görülebileceği üzere radar 300kW'lık güç kullanıyor (299kW yazmasına aldırış etmeyin. Yuvarlama hatasından kaynaklı). Factorio gerçek bilimi kullanır. Watt(W) gücün birimidir. 1 Kilowatt(kW), 1000W ve 1 megawatt(MW) 1000kW'a eşittir. Eğer fabrikanızı yeterince büyütebilirseniz bu bilgi de işinize yarayacaktır: 1 gigawatt(GW) da 1000MW'a eşittir.

	~~So that answers the question of how much power our factory uses~~. ~~To keep our factory running at full speed we need to maintain 300 kW~~. ~~For ease of comparison~~, we'~~ll convert that to~~ 0.~~3 MW~~.		Fabrikamızın ne kadar güç harcadığını anlamış olduk. Eğer fabrikamızı tam gaz işletmek istiyorsak 300kW güç üretimini sürdürmeliyiz. Daha kolay işlem yapmak için, 300kW'ı 0.3MW'a dönüştüreceğiz.

	~~Next question~~! ~~How much power is stored in coal~~? ~~That's also easy~~, ~~because it tells us when we hover over it: 8 MJ~~.		Akabinde yeni bir soru! Kömürden ne kadar güç elde edilir? Bu soruyu yanıtlamak çok kolay, yapmamız gereken imleci üzerine getirip beklemek akabinde 8MJ güç elde edilebildiğini görmüş olacağız.

	[[File:coal-joules.png\|300px]]		[[File:coal-joules.png\|300px]]