Fermi Estimate

注：本文为 “费米估算” 相关文章合辑。

英文引文机翻，未校。

如有内容异常，请看原文。

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Fermi Estimate 费米估算

July Thomas, Brilliant Physics, A Former Brilliant Member, and contributed

A Fermi estimate is one done using back-of-the-envelope calculations and rough generalizations to estimate values which would require extensive analysis or experimentation to determine exactly.
费米估计是使用粗略的计算和粗略的概括来完成的估计值，这需要广泛的分析或实验才能准确确定。

Physics is celebrated for its ability to make extremely accurate predictions about tough problems such as the magnetic moment of electrons, the deflection of light by the Sun’s gravity, or the orbit of the planets around the Sun. However, accuracy often comes at the cost of great difficulty in calculation.
物理学因其能够对棘手问题做出极其准确的预测而闻名，例如电子的磁矩、太阳引力引起的光偏转或行星围绕太阳的轨道。然而，准确性往往以计算的巨大难度为代价。

For example, calculating even a poor estimate for the temperature at which a Bose-Einstein condensate forms requires most people about 15 years of preparation, and an accurate calculation for the flight of a commercial jet is entirely intractable without the aid of sophisticated software systems that handle the difficult numerical calculations. If a nice, aesthetically pleasing analytical result were required, much of physics would not have happened. Indeed, the seduction of formal calculations can be a serious hindrance. Whenever the math gets out of hand, it is usually a good idea to relax demands and accept an approach that while imprecise, offers a prayer of moving forward.
例如，即使是对玻色-爱因斯坦凝聚态形成温度的糟糕估计，大多数人也需要大约 15 年的准备，而如果没有处理困难数值计算的复杂软件系统的帮助，商用喷气式飞机飞行的准确计算是完全困难的。如果需要一个漂亮、美观的分析结果，那么很多物理学就不会发生。事实上，正式计算的诱惑可能是一个严重的障碍。每当数学失控时，放宽要求并接受一种方法通常是一个好主意，这种方法虽然不精确，但提供了向前的祈祷。
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How many windmills are needed to power a country?
一个国家需要多少台风车？

Fermation of subproblems 子问题的费米化

If the cost of the wars in Afghanistan and Iraq were factored into the price of gasoline for a taxpayer in the United States, what would the effective war tax be per gallon? On the basis of energy per person, what is more effective, solar power or wind? How many cracked iPhone screen repairmen are there in the United States? Each of these questions is probably completely bewildering, at least if trying to guess the answer in one step. Who has an intuition for any of these numbers? They are outside normal experience for several reasons. In the context of the question of a war tax
如果阿富汗和伊拉克战争的成本被计入美国纳税人的汽油价格，那么每加仑的有效战争税是多少？根据人均能源，太阳能和风能哪个更有效？美国有多少破裂的 iPhone 屏幕维修工？这些问题中的每一个都可能完全令人困惑，至少如果试图一步猜出答案的话。谁对这些数字有直觉呢？他们超出正常体验有几个原因。在战争税问题的背景下

• Some of the numbers involved are incredibly big, and humans are poor judges of large numbers.
  其中一些涉及的数字非常大，而人类无法判断大数字。
• There are factors that are completely unknown, such as the number of miles driven by a typical person per year, the size of the war’s black budget (clandestine operations), or the stability of OPEC in the absence of United States foreign policy, that could be significant factors in finding an accurate answer.
  有一些因素是完全未知的，例如一个普通人每年驾驶的英里数、战争黑色预算的规模（秘密行动），或者欧佩克在没有美国外交政策的情况下的稳定性，这些因素可能是找到准确答案的重要因素。
• However, these concerns can be addressed by breaking the big problem into appropriate sub-problems.
  但是，可以通过将大问题分解为适当的子问题来解决这些问题。

How much revenue does a typical MLB team make per season? 一支典型的 MLB 球队每个赛季能赚多少收入？

 

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The main question is quite a tall order, but some reasonable assumptions can be made.
主要问题是一个相当高的要求，但可以做出一些合理的假设。
 
One of the major sources of team revenue must be ticket sales, or else they’d just play the games in one place and broadcast the proceedings to the respective cities. So, assume that ticket sales make up ∼$\frac{1}{3}$ the profits of a typical MLB team (this can always be revisited if it turns out to be wildly off).
球队收入的主要来源之一是门票销售，否则他们只会在一个地方进行比赛，并将比赛过程广播到相应的城市。因此，假设门票销售构成 $\sim \frac{1}{3}$ 典型 MLB 球队的利润（如果结果证明非常糟糕，这总是可以重新审视的）。
 

How much does a team make from ticket sales per year? Most MLB stadiums have room for $N_{seat}\approx 40,000$ fans to attend. Assuming that on most days the stadium is approximately$\gamma =80\%$ full, since they probably plan so that the stadium doesn’t look too empty on TV, but has enough room for extra fans should there be heightened interest.
一支球队每年从门票销售中赚多少钱？大多数 MLB 体育场都有空间容纳 $N_{seat}\approx 40,000$ 名粉丝参加。假设在大多数日子里，体育场大约为 $\gamma =80\%$ 满率，因为他们可能计划让体育场在电视上看起来不会太空，但如果兴趣更高，就有足够的空间容纳额外的球迷。
A fan can get a horrible seat for a team in a reasonably sized market for about $20 (personal estimate from going to Los Angeles Angels games). An extremely nice seat can cost as much as $2500. Let’s assume that the distribution is asymmetrical and that most seats are just okay, so that the average ticket costs slightly more than the ultra-bad seat $p_{ticket}\approx$ $30 .
一个球迷可以在一个合理规模的市场中以大约 20 美元的价格为一支球队获得一个可怕的座位（个人估计去看洛杉矶天使队的比赛）。一个非常好的座位可能要花费高达 2500 美元。让我们假设分布是不对称的，并且大多数座位都还可以，因此平均票价比超级糟糕的座位略高于 $p_{ticket}\approx$ $30 。
Finally, each team plays $N_{game}=162$ games per year, and on average, splits the profits of each game with the opposing team. Thus, the annual revenue of a typical MLB team is approximately $3\gamma N_{\text{seat}}{p}_{\text{ticket}}{N}_{\text{game}}\frac{1}{2}\approx$ $230 million.
最后，每支球队都进行比赛 $N_{game}=162$ 场比赛每年，平均而言，与对方球队平分每场比赛的利润。因此，典型 MLB 球队的年收入约为 $3\gamma N_{\text{seat}}{p}_{\text{ticket}}{N}_{\text{game}}\frac{1}{2}\approx$ $230 million。
The real answer is $226 million, which means the estimate is wonderfully accurate, certainly more so than it deserves to be.
真正的答案是 2.26 亿美元，这意味着这个估计非常准确，肯定比它应得的要准确。

Without knowing anything about the actual revenue of any given MLB team, we were able to break the big problem down into several more manageable sub-problems that we could reason about with our common sense:
在对任何给定 MLB 球队的实际收入一无所知的情况下，我们能够将大问题分解为几个更易于管理的子问题，我们可以用我们的常识来推理：

• stadium capacity 体育场容量
• typical attendance 典型出勤率
• ticket prices 票价
• the number of games in a season 一个赛季中的比赛数
• a reasonable guess for the contribution of ticket prices to revenue 对门票价格对收入贡献的合理猜测
   
  By multiplying the answers to all five sub-problems, we arrive at a remarkably good first approximation for team revenue.
  通过将所有 5 个子问题的答案相乘，我们得出了非常好的团队收入第一近似值。
   
  A Fermi problem is an approximate, back-of-the-envelope calculation of an arbitrary figure facilitated by identifying suitable factors of the figure, that are accessible to common experience. Depending on the difficulty of the problem, and the number of sub-problems required to get in touch with common sense, one can usually hope to be correct to within a factor of 2 or 3, and other times to within the correct order of magnitude.
  费米问题是任意图形的近似、粗略的计算，通过确定图形的合适因子来促进，这些因子是常识可以获得的。根据问题的难度，以及与常识取得联系所需的子问题的数量，人们通常希望正确到 2 或 3 倍以内，有时则在正确的数量级内。

0 250 2.5 25,000

You are an astronaut, it is a few hours before you are scheduled to launch for an intergalactic space mission, and you’re sitting on the shore of the Atlantic Ocean near Cape Canaveral, savoring the view one last time. This mission is especially trying; because of the relativistic speed of your spaceship, Earth will be one billion years into the future upon your return.
您是一名宇航员，距离您计划发射星际太空任务还有几个小时，您坐在卡纳维拉尔角附近的大西洋岸边，最后一次欣赏美景。这个任务特别令人费解; 由于你的宇宙飞船的相对论速度，当你回来时，地球将在未来 10 亿年。

Depressed beyond words, you decide, in one of your weaker moments, to pee into the Atlantic Ocean.
难以言喻的沮丧，你决定在你最虚弱的时刻，向大西洋撒尿。

Flash forward: 1 billion years. 快进：10 亿年。

You return to Earth, expecting a hero’s welcome, but instead you find that all of humanity has vanished. Instead, the Earth is run by a peaceful clan of telekinetic dolphins who made off with the lion’s share of Bitcoins that were abandoned by the last humans as they uploaded their souls to the singularity server. In a disillusioned haze, you bend down and fill your astronaut survival cup with refreshing lake water, hoping for some clarity. After drinking the glass, you realize that this far into the future, all the Earth’s water has been thoroughly well mixed since the time you took off on your mission.
你回到地球，期待英雄的欢迎，但你却发现全人类都消失了。相反，地球由一个和平的念力海豚氏族管理，他们偷走了最后一代人类在将灵魂上传到奇点服务器时丢弃的大部分比特币。在幻灭的雾霾中，你弯下腰，把你的宇航员救生杯装满清爽的湖水，希望能得到一些清澈。喝完杯子后，你意识到，在遥远的未来，自从你开始执行任务以来，地球上的所有水都已经彻底混合了。

Approximately how many molecules of your billion year old urine did you just consume?
您刚刚消耗了大约多少个十亿年前的尿液分子？

Assumptions and details 假设和详细信息

• You released $\text{1 L}$ of urine into the Atlantic Ocean.
  你释放了 $\text{1 L}$ 尿液进入大西洋。

• Approximate your urine as consisting of pure water.
  你的尿液近似由纯水组成。

How many iPhone screen repairmen are there in the United States?
美国有多少名 iPhone 屏幕维修工？

An important number for this problem is the number of iPhones in the United States. We can arrive at this number in two ways.
这个问题的一个重要数字是美国的 iPhone 数量。我们可以通过两种方式得出这个数字。

First, from common experience, it seems about 1 in 2 people has a smartphone, and those without them tend to be the very young or the very old. Furthermore, it is common to hear that Android phones dominate the market place, so let’s estimate the Android figure to be about $\frac{2}{3}$, and thus the iPhone share to be about $\frac{1}{3}$, approximating Windows, Blackberry, and other platforms to be minor players in the market. This yields $\sim 50,000,000$ iPhones.
首先，从一般经验来看，似乎大约每 2 个人中就有 1 人拥有智能手机，而没有智能手机的人往往是非常年轻或非常年长的人。此外，经常听到 Android 手机在市场上占据主导地位，因此让我们估计 Android 的数字约为 $\frac{2}{3}$，因此 iPhone 份额大约是 $\frac{1}{3}$，将 Windows、Blackberry 和其他平台近似为市场上的次要参与者。这样就会得到 $\sim 50,000,000$ 部 iPhone。

Another way is to think about the people you see on the street, and try to directly estimate how many of them have an iPhone. This number seems to me to be around 1 in every 5 people, which would make ∼60∼60 million iPhones in the country. Thus, an estimate in the range of fifty to sixty million seems to be approximately right.
另一种方法是想想你在街上看到的人，并尝试直接估计他们中有多少人拥有 iPhone。在我看来，这个数字大约是每 5 人中就有 1 人，这将使 ∼60 该国有 6000 万部 iPhone。因此，5000 万到 6000 万之间的估计似乎大致正确。

How many of these are cracked? Phone contracts usually last for two years, and most people take their upgrade, so let’s assume that the typical iPhone spends 2.5 years with the customer before being replaced. Now, how many of these screens will be cracked over the lifetime of the phone? I’d guess that this number sits somewhere around 20%.
其中有多少是破解的？电话合同通常持续两年，大多数人都会进行升级，所以让我们假设典型的 iPhone 在被更换之前与客户在一起会花 2.5 年。现在，在手机的整个生命周期内，这些屏幕中会有多少被破解？我猜这个数字在 20% 左右。

Every cracked screen doesn’t get replaced or else we wouldn’t see them around too often. Let’s assume that if a crack happens in the first $\frac{2}{3}$ of the time for which the customer owns the phone, they’ll get it fixed, but otherwise they’ll just wait for a new phone. This means that in a given year, $\frac{1\text{ iPhone}}{2.5\text{ years}}\times 20\%\times 66\%\approx 5\%$ of iPhones will need their screens replaced, or $5\%\times 55,000,000\approx 2,800,000$ iPhone screens.
每个破裂的屏幕都不会被更换，否则我们不会经常看到它们。我们假设，如果第一个 $\frac{2}{3}$ 客户拥有手机的时间，他们会修好它，但除此之外，他们只会等待新手机。这意味着，在给定的年份中， $\frac{1\text{ iPhone}}{2.5\text{ years}}\times 20\%\times 66\%\approx 5\%$ 的 iPhone 需要更换屏幕，或者 $5\%\times 55,000,000\approx 2,800,000$ 个 iPhone 屏幕。

How many repairmen does this support? Let us assume that each iPhone screen takes the average repairmen 1 hr1 hr to fix and that the average iPhone screen fixer spend about half their full time work week fixing iPhone screens (averaging over full and part time workers). Thus, we predict there to be enough broken iPhones to support the employment of approximately $\frac{2.8\times 10^6\text{ phones}\times (1\text{ hr/phone})}{52\text{ weeks}\times 20\text{ hrs/(iPhone fixer)/week}}\approx 2,700\text{ iPhone fixers}$ iPhone fixers in the entire country.
这支持多少维修工？让我们假设每个 iPhone 屏幕都需要普通的维修工 1 小时 1 小时来修复，并且 iPhone 屏幕修复者平均花费大约一半的全职工作时间来修复 iPhone 屏幕（全职和兼职工人的平均人数）。因此，我们预测将有足够多的破损 iPhone 来支持大约 $\frac{2.8\times 10^6\text{ phones}\times (1\text{ hr/phone})}{52\text{ weeks}\times 20\text{ hrs/(iPhone fixer)/week}}\approx 2,700\text{ iPhone fixers}$ iPhone 修复者 在全国。

This is roughly 10 times the number of Apple stores in the country, so it seems fairly reasonable.
这大约是该国 Apple 商店数量的 10 倍，因此似乎相当合理。

The accuracy of guess strings guess 字符串的准确性

As we saw with the MLB revenue problem, breaking down our big problems into small problems can be a big aid in tackling an estimate. However, there is potential for trouble. For instance, with the iPhone we are not necessarily getting more familiar with the numbers we have to estimate by breaking the problem down. Despite the number of cracked iPhone screens being a simpler problem than guessing the number of screen repairmen outright, few people are familiar with the frequency of iPhone screen breaks. We hope this number will be accurate to within a factor of two (we guessed 20% but it could easily be 10% or 30%). Similarly, we took a wild guess at the amount of time required to fix a given iPhone. As it turns out, the true frequency of screen breaks is close to one third of all iPhones and the average repair time is closer to half an hour. Thus in one of the numbers we were too low by a factor of $\frac{2}{3}$ and in another we were too high by a factor of two, which means that all told, those two factors place us about 33% too high in our estimate. With long strings of guesses, these errors can multiply, and drag us away from the right answer. However, we are unlikely to have a consistent bias (too high or too low) on a series of unique sub-problems, and thus, our mistakes will tend to balance each other out, as in a random walk. If we are too high in some numbers, we are likely to be too low in other numbers.
正如我们在 MLB 收入问题中看到的那样，将我们的大问题分解为小问题对解决估算有很大帮助。但是，可能会有麻烦。例如，对于 iPhone，我们不一定会更熟悉我们必须通过分解问题来估计的数字。尽管破裂的 iPhone 屏幕数量比直接猜测屏幕维修工的数量更简单，但很少有人熟悉 iPhone 屏幕破裂的频率。我们希望这个数字能精确到两倍以内（我们猜测是 20%，但也很容易是 10% 或 30%）。同样，我们对修复给定 iPhone 所需的时间进行了大胆的猜测。事实证明，屏幕破裂的真实频率接近所有 iPhone 的三分之一，平均维修时间接近半小时。因此，在其中一个数字中，我们太低了 $\frac{2}{3}$ 而在另一个 Factor 中，我们高出了 2 倍，这意味着总而言之，这两个因素使我们差不多 33% 在我们的估计中太高了。在一长串的猜测中，这些错误可能会成倍增加，并使我们远离正确的答案。然而，我们不太可能在一系列独特的子问题上有一致的偏见（太高或太低），因此，我们的错误往往会相互平衡，就像在随机游走中一样。如果我们在某些数字上太高，那么我们在其他数字上也可能太低。

Crudely, if we have a consistent uncertainty in estimating any numbers, we can model our guess as a random number drawn from a Gaussian distribution about the true answer with a characteristic variance ${\sigma }^2$. Breaking our guess into $n$ sub-problems means that our variance becomes $\approx n{\sigma }^2$ (the variance of random variables is additive), and thus the standard deviation (the square root of the variance) increases with the square root of $n$. This is the same behavior as a random walk (where the average displacement increases as $\sqrt{t}$ with time), which we might expect since we hope to guess slightly too high or slightly too low at each number we estimate. Thus, when possible, we should avoid making too many sub-guesses.
粗略地说，如果我们在估计任何数字时都有一致的不确定性，我们可以将我们的猜测建模为从高斯分布中抽取的关于真实答案的随机数，该随机数具有特征方差 ${\sigma }^2$ .将我们的猜测分解为 $n$ 个子问题意味着我们的方差变为 $\approx n{\sigma }^2$ （随机变量的方差是加法的），因此标准差（方差的平方根）随着 $n$.这与随机游走的行为相同（其中平均位移随着时间的推移增加 $\sqrt{t}$ ），这可能是我们预期的，因为我们希望在我们估计的每个数字上猜测稍微过高或稍微过低。因此，如果可能，我们应该避免进行过多的子猜测。

On the other hand, it is likely that our familiarity with the numbers in our sub-problems is significantly better than our knowledge of our big problem, so to a point it will always make sense to break up our big problems. It isn’t possible to precisely model this trade-off between random walks and accurate knowledge, so knowing when to stop finding sub-problems is a matter of intuition and building experience in making accurate predictions of this kind.
另一方面，我们对子问题中数字的熟悉程度可能明显优于我们对大问题的了解，因此在某种程度上，分解我们的大问题总是有意义的。不可能在随机游走和准确知识之间精确建模这种权衡，因此知道何时停止查找子问题是直觉问题，并需要积累做出此类准确预测的经验。

Dominant Part 主导部分

Another crucial technique in good estimates is to ignore everything but what you deem to be the most important factor(s) in the problem. In the baseball problem, for example, we assumed the major sources of revenue to be tickets, TV, and the merchandise that people buy at the stadium or elsewhere, and we took all other sources of income to be negligible. Likewise, if we estimate the daily energy budget of a person, we can assume that the energy used by their electric toothbrush makes a negligible contribution when compared with the energy that goes into making their food, driving their vehicle, heating their home, etc. If we consider the time required to get in a car and drive 100 miles, we can ignore the time required to open the door, et cetera.
良好估计的另一个关键技术是忽略所有内容，除了您认为是问题中最重要的因素。例如，在棒球问题上，我们假设主要收入来源是门票、电视以及人们在体育场或其他地方购买的商品，而我们认为所有其他收入来源都可以忽略不计。同样，如果我们估计一个人的日常能源预算，我们可以假设他们的电动牙刷使用的能源与制造食物、驾驶车辆、取暖等的能源相比，其贡献可以忽略不计。如果我们考虑上车行驶 100 英里所需的时间，我们可以忽略开门所需的时间，等等。

This takes some bravado to get started. Getting a sense for the dominant contribution(s) is not always obvious and may take some numerical comparisons to get a good sense for it. However, reducing the number of variables we need to keep track of always reduces the complexity of our mathematical problems, and can lead us toward more accurate solutions. For example, if we solve a problem first in one extreme case, then in another, we may be able to identify the short and long time behavior, or low and high energy behavior, and therefore know what we should be looking for when we