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# Incremental and full backup

Hi,

Is there a way I can use rclone to do incremental and full backup and upload backup to AWS S3 or GCP Cloud Storage?

Thanks in advance and I look forward to hearing from you.

Best Regards,

Kaushal

this is what i do, a simplified example of a full backup and forever forward incremental backups.

rclone sync \path\to\source remote:backup --backup-dir=remote:archive\20210121.083621

@asdffdsa Thanks for the reply and much appreciated. I am not sure if I completely understand the below command.

#rclone sync \path\to\source remote:backup --backup-dir=remote:archive\20210121.083621

I have the below source directory which I need to take a full and incremental backup. Once it is backed up (both Full and Incremental) it should be pushed to AWS S3 bucket.

GitLab SCM home directory.
/opt/gitlab/

Full Backup on Sunday late night
Incremental Backup on Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.

Please suggest. Thanks in advance and I look forward to hearing from you.

Best Regards,

Kaushal

hi,

that script will perform a forever forward incremental backup.

i have a script that generates the command, each time with a different time-stamp for the archive sub-folder
the latest set of files is always in remote:backup
each incremental is a sub-folder of remote:archive

each time that sync is run, rclone will compare local files to the files in remote:backup
let's say there is a local file named file.txt that is newer then the same file in remote:backup.
rclone will do the following

1. move file.txt from remote:backup to remote:archive\20210121.083621
2. copy the local file.txt to remote:backup

@asdffdsa I have a follow-up question.

that script will perform a forever forward incremental backup? Apologies I have not still understood. Are you referring to the below command?

#rclone sync /opt/gitlab/ remote:backup --backup-dir=remote:archive\20210121.083621

Does remote:backup means AWS S3 and remote:archive is local? I am not sure how this remote:archive\20210121.083621 is generated? I am finding it hard to understand. Apologies for bugging you. I will really appreciate it if you can explain me with some examples. Please suggest. Thanks in advance and I look forward to hearing from you.

Best Regards,

Kaushal

hi,
the remote: is a remote created by rclone, using aws s3.
so remote:archive\20210121.083621 points to aws a3.

i have a python script that generates this command, this case using wasabi, a s3 clone known for hot storage.

C:\data\rclone\scripts\rclone.exe sync "C:\data\u\keepass\database" "wasabi01:en07/kdbx/rclone/backup"  --stats=0 --fast-list --backup-dir=wasabi01:en07/kdbx/rclone/archive/20210201.120322   --log-level=DEBUG --log-file=C:\data\rclone\logs\kdbx_quick_wasabi\20210201.120322\rclone.log

on linux

dt=date +%Y%m%d.%I%M%S
rclone sync /path/to/local/folder remote:backup --backup-dir=remote:archive/$dt @asdffdsa is the right one for rclone. I too use this method since I really like that my backups are not reliant on another tool (except for if I use crypt). It is a good-enough strategy. But, also as @asdffdsa mentioned, it is forever forward incremental. What you end up with is your main backup that is a current view and then you have another directory of everything that would have been deleted or overwritten (and keep the logs around so you know which one). An alternative is something like restic which can use rclone as a backend. Restic works by breaking your files into chunks (blocks) that are decided based on the content and having a database of them. This means that if you make a small change in a big file, the whole file isn't copied again; just the block. The other benefit is that every backup is its own full snapshot and you can delete any snapshot you want and then recover (prune) the space. The (major) downside to this method is the complexity and requirement of using restic to get it all back. Whether you want that or not depends on you and your own use case. For now, I choose the inefficiencies of rclone for the benefit of simplicity. I just thought I'd put some other options out there. 1 Like Also, if it helps, below is my backup script: #!/usr/bin/env bash # # Simple rclone backup script. Note that since this is run in cron, # I've found I need full paths # cd to the right directory since this is where I have filters cd /home/<USER>/git_repos/backup_scripts/ NOW=$(date  +"%Y-%m-%dT%H%M%S")

echo "----------------- Starting Backup -----------------"
echo $NOW echo "---------------------------------------------------" SRC=/home/<USER>/ DST0=b2:<MYBUCKET> DST=$DST0/current
BCK=$DST0/backups/$NOW

# Log files
mkdir -p /home/<USER>/logs/backups/
LOGFILE=/home/<USER>/logs/backups/rclone_${NOW}.log /home/<USER>/bin/rclone \ --config /home/<USER>/.config/rclone/rclone.conf \ --fast-list \ --log-level INFO --log-file$LOGFILE \
--links # translate to .rclonelink \
# Exclude the log before the other filters \
--filter "- logs/backups/rclone_${NOW}.log" \ --filter-from filters.txt \ --exclude-if-present .ignore \ --track-renames \ # --track-renames-strategy modtime \ --transfers 25 \ --delete-excluded \ "$@" \
sync  \
--backup-dir $BCK \$SRC $DST /home/<USER>/bin/rclone \ --config /home/<USER>/.config/rclone/rclone.conf \ -v \ --no-check-dest # It is a new file. No need to check \ copyto \$SRC/logs/backups/rclone_${NOW}.log \$DST/logs/backups/rclone_${NOW}.log echo "----------------- Finished Backup -----------------" echo$NOW
echo "---------------------------------------------------"

well written

very well written

This topic was automatically closed 60 days after the last reply. New replies are no longer allowed.


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# Fourier Transform in wavefunction of Free particle

I was reading Introduction to Quantum Mechanics by David Griffiths and I am at Chapter 2, page 45. He says that

The general solution to the time-dependent Schrodinger equation is still a linear combination of separable solutions (only this time it's an integral over the continuous variable $$k$$, instead of a sum over the discrete index).

This means that we can write free particle with general wavefunction with $$\Psi(x,t)$$ as Fourier transform of eigenfunction of Schrodinger equation (I think). i.e., $$$$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \phi(k) e^{i(kx-\frac{\hbar k^2}{2m}t)}dk$$$$ where $$\phi(k)$$ is eigenfunction of Schrodinger Equation for free particle. (I think)
But then he writes $$$$\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \Psi(x,0)e^{-ikx}dx$$$$ from Plancherel's theorem. This clearly means that eigenfunction of the Schrodinger Equation will depend on the initial wave function of the particle.
Is my interpretation correct? I am not sure because this is not very intuitive to me.

Also, does this mean that Fourier Transform of any general free particle wavefunction $$\Psi(x,t)$$ is eigenfunction of Schrodinger equation?

$$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \phi(k) e^{i(kx-\frac{\hbar k^2}{2m}t)}dk$$ where $$\phi(k)$$ is eigenfunction of Schrodinger Equation for free particle.(I think)

No, you misunderstood your text book. $$\phi(k)$$ is a completely arbitrary function. For any function $$\phi(k)$$ this $$\Psi(x,t)$$ will be a solution of Schrödinger's equation $$i\hbar\frac{\partial}{\partial t} \Psi(x,t)= -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \Psi(x,t)$$ just because $$e^{i(kx-\frac{\hbar k^2}{2m}t)}$$ for any $$k$$ is a solution of Schrödinger's equation.

When you take the above solution $$\Psi(x,t)$$ at time $$t=0$$, then you have $$\Psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \phi(k) e^{ikx}dk$$

This means that $$\phi(k)$$ is the Fourier transform of $$\Psi(x,0)$$. You can invert this transformation and get $$\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \Psi(x,0) e^{-ikx}dx$$

Your problem lies in this sentence:

where $$\phi (k)$$ is eigenfunction of Schrodinger Equation for free particle. (I think)

Your interpretation is wrong. The eigen function is $$e^{ikx}$$ and $$\phi (k)$$ is just a number which tells you the coefficient of this eigen function in "constructing" the wave function. It might be a function of $$k$$ (each eigen function has a different amount in constructing the wave-function) but not a function of $$x$$, meaning it is not an eigen function. The way to calculate this number is by using the Fourier transform, which you wrote in your second equation (and that does depend on the initial wave function, but the eigen function themselves do not change.)

format: plain


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# The equations 3x - 4y = 5 and 12x - 16y = 20 have:

This question was previously asked in
UP TGT Mathematics 2016 Official Paper
View all UP TGT Papers >
1. no common solution
2. exactly one common solution
3. exactly two common solution
4. more than two common solutions

## Answer (Detailed Solution Below)

Option 4 : more than two common solutions
Free
CT 1: हिन्दी (आदिकाल)
5586
10 Questions 40 Marks 10 Mins

## Detailed Solution

Concept:

If two linear equations

x + b1y = c1 and x + b2y = c2. Then,

(a) If /a2 = b1/b2 = c1/c2 , the system is consistent and has infinitely many solutions.

(b) If /a2 = b1/b2 ≠ c1/c2 the system has no solution and is inconsistent

Calculation:

In equations 3x − 4y = 5 and 12x − 16y = 20

3/12 = -4/-16 = 5/20 ⇒ ¼ = ¼ = ¼

So the system is consistent and has infinitely many solutions

∴ They have more than two common solutions.


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 Trudy MIAN: Year: Volume: Issue: Page: Find

 Tr. Mat. Inst. Steklova, 2006, Volume 252, Pages 94–113 (Mi tm65)

A Cofinal Family of Equivalence Relations and Borel Ideals Generating Them

V. G. Kanovei, V. A. Lyubetskii

Institute for Information Transmission Problems, Russian Academy of Sciences

Abstract: An increasing $\omega _1$-sequence of Borel equivalence relations on a Polish space that is cofinal (in the sense of Borel reducibility) in the family of all Borel equivalence relations is defined as a development of Rosendal's construction. It is proved that equivalence relations from this sequence are generated by explicitly defined Borel ideals.

Full text: PDF file (345 kB)
References: PDF file   HTML file

English version:
Proceedings of the Steklov Institute of Mathematics, 2006, 252, 85–103

Bibliographic databases:

UDC: 510.225

Citation: V. G. Kanovei, V. A. Lyubetskii, “A Cofinal Family of Equivalence Relations and Borel Ideals Generating Them”, Geometric topology, discrete geometry, and set theory, Collected papers, Tr. Mat. Inst. Steklova, 252, Nauka, MAIK «Nauka/Inteperiodika», M., 2006, 94–113; Proc. Steklov Inst. Math., 252 (2006), 85–103

Citation in format AMSBIB
\Bibitem{KanLyu06} \by V.~G.~Kanovei, V.~A.~Lyubetskii \paper A~Cofinal Family of Equivalence Relations and Borel Ideals Generating Them \inbook Geometric topology, discrete geometry, and set theory \bookinfo Collected papers \serial Tr. Mat. Inst. Steklova \yr 2006 \vol 252 \pages 94--113 \publ Nauka, MAIK «Nauka/Inteperiodika» \publaddr M. \mathnet{http://mi.mathnet.ru/tm65} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=2255972} \transl \jour Proc. Steklov Inst. Math. \yr 2006 \vol 252 \pages 85--103 \crossref{https://doi.org/10.1134/S008154380601010X} \scopus{http://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-33746072961}


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# Have astronomers taken into account the fact that red-shifted light from far-away stars is also very old when studying the expansion of the universe?

I am aware that Hubble discovered that the further away a galaxy is the greater its redshift. What I don't understand is how astronomers can be sure that distant objects are still moving at greater velocities than nearer objects, as the redshifts light is also much older than the light from the nearer object. In other words, the galaxies billions of light years away may be moving fast away from us billions of years ago (when the light was emitted), but how do we know they haven't slowed down or even sped up more since?

For example, Galaxy A is 3 million light years away and has a slight redshift. Galaxy B is 3 billion light years away and has a much greater red shift and so it is assumed Galaxy B is moving away at a much greater velocity than Galaxy A even though Galaxy B's light is older.

I'm not an astronomer so please go easy on your explanation to me

I'm aware that this question may have been asked before, but I could not understand the answers given to them.

The graph below shows how the "Hubble law" departs from a straight line at redshift $$>0.2$$ (or light-travel distances greater than about 3 billion light years) (graphs from Goswami et al. 2014).


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# Modelling using the Binomial Distribution Formula

So I'm in my first semester at University studying Actuarial Science and one of my classes is Probability. Needless to say, I have fallen in love with the whole topic and given my passion, I constantly try to solve problems on my own. Except that this time, I really am puzzled.

I first started with a typical American Roulette game at the casino (zero and double zero are possible outcomes). I was interested in finding how many times would the casino need to spin the wheel before being almost sure of turning a profit; assuming there is only 1 bet per spin, the player always bets on red and the bet amount is always \$100. Well, this problem is not too difficult to solve. Intuitively we know that because the casino has an edge (expected value of 5,26 dollars for \$100 spin), it will beat the player over the long haul and turn a profit. But knowing the player could get lucky, and delay the inevitable (going bankrupt), how long could the "long haul" be? Would it be 10 spins? 50 spins? 100 spins?

As it turns out, I did the model for this projection and noticed that after 10 spins, the casino will only turn a profit 44.32% of the time. After 100 spins, it becomes 66.57% and after 1000 spins it is 94.89%. Conclusion, our intuition was confirmed. Over time, chances are the casino will crush the player.

I was able to do this using the Binomial Distribution Formula below:

Now, what if instead, my new game at the casino was to roll a fixed die. Possible outcomes would include (1,2,3,4,5,6) and their corresponding probabilities would be (10%,15%,20%,15%,10%,30%). Each number has the following profits for the casino (-\$200,-\$100,-\$50,-\$100,-\$200,\$500).

As you can see, the casino only makes money when a 6 is rolled (30% of the time), and although it doesn't win as often as with the Roulette game, its expected value is even higher (\$70 per roll). How would I find out how many events it would take for the casino to be profitable in this game? Intuitively I'm puzzled because the lower chance of winning tells me it would take much longer than the Roulette game but it's expected value being much higher tells me it would take fewer events. Can I use a Binomial Distribution once again? For instance, after 5 rolls, the results could have been (5,4,6,3,1) and in that case, the casino would have lost 50 dollars (-\$200,-\$100,\$500,-\$50,-\$200).

It's confusing because unlike the roulette game, the casino has 4 distinct outcomes (-\$200,-\$100,-\$50,\$500). Should I run my model with the binomial formula based on winning 150 dollars, 30 percent of the time and losing 80 dollars, 70 percent of the time? Those figures represent the sum of the expected values. I'm not sure how to solve this.

## 2 Answers

I will generalise your particular problem to show how to deal with this type of problem. Suppose you have a series of bets, each with $$m$$ possible outcomes. The probability vector for the outcomes, and the profit to the casino for each outcome, are given respectively by the vectors:

$$\mathbf{p} = (p_1,...,p_m) \quad \quad \quad \mathbf{\pi} = (\pi_1,...,\pi_m).$$

After $$n$$ rounds of play, the counts of outcomes follow a multinomial distribution. The profit to the casino after $$n$$ rounds of play is a random variable given by:

$$\Pi_n = \sum_{i=1}^m N_i \pi_i \quad \quad \quad \mathbf{N} = (N_1, ..., N_m) \sim \text{Multinomial}(n, \mathbf{p}).$$

The expected value and variance of the profit to the casino after $$n$$ rounds is:

\begin{aligned} \mu_n \equiv \mathbb{E}(\Pi_n) &= n \sum_{i=1}^m p_i \pi_i, \\[6pt] \sigma_n^2 \equiv \mathbb{V}(\Pi_n) &= n \sum_{i=1}^m p_i (1-p_i) \pi_i^2. \\[6pt] \end{aligned}

Probability of profit: The exact probability that the casino has profited after $$n$$ rounds is:

$$\mathbb{P}(\Pi_n > 0) = \sum_{\mathbf{n} \in \mathcal{S}_n(\mathbf{\pi})} \text{Multinomial}(\mathbf{n}|n, \mathbf{p}),$$

where $$\mathcal{S}_n(\mathbf{\pi}) \equiv \{ \mathbf{n} | \sum_i n_i \pi_i > 0, \sum_i n_i = n \}$$ is the set of count vectors that yield profit under the vector $$\mathbf{\pi}$$. For large $$n$$ it is computationally expensive to find this set, so it would be usual to approximate the multinomial distribution by the normal distribution to yield the approximation $$\Pi_n \sim \text{N}(\mu_n, \sigma_n^2)$$, which then gives an approximate probability of profit:

$$\mathbb{P}(\Pi_n > 0) \approx 1 - \Phi \Big( - \frac{\mu_n}{\sigma_n} \Big) = 1 - \Phi \Big( - \frac{\mu_1}{\sigma_1} \cdot \sqrt{n} \Big).$$

This can easily be calculated for any $$n$$ using the standard normal CDF $$\Phi$$. For large $$n$$ it will give a good approximation to the exact probability of profit.

Application to your example: In your case you have the vectors:

\begin{aligned} \mathbf{p} &= (0.10, 0.15, 0.20, 0.15, 0.10, 0.30), \\[6pt] \mathbf{\pi} &= (-200, -100, -50, -100, -200, 500). \\[6pt] \end{aligned}

This gives moments $$\mu_n = 70 n$$ and $$\sigma_n^2 = 62650 n$$. Hence, the approximate probability of profit after $$n$$ rounds is:

\begin{aligned} \mathbb{P}(\Pi_n > 0) &\approx 1 - \Phi \Big( - \frac{70n}{\sqrt{62650n}} \Big) \\[6pt] &= 1 - \Phi \Big( - \frac{14}{\sqrt{2506}} \cdot \sqrt{n} \Big) \\[6pt] &= 1 - \Phi ( - 0.2796646 \cdot \sqrt{n} ). \\[6pt] \end{aligned}

Coding this in R: You can code this in R to show how the probability of profit changes as we increase $$n$$. Note that the approximation is poor for small values of $$n$$ and so the early part of the curve is not an accurate representation of the true probability of profit.

#Generate function for approximate probability of profit
PROB <- function(p, pi, N) { R   <- sum(p*pi)/sqrt(sum(p*(1-p)*pi^2));
PPP <- rep(0,N);
for (n in 1:N) { PPP[n] <- pnorm(-R*sqrt(n),
lower.tail = FALSE); }
PPP; }

#Generate example
p  <- c(0.10, 0.15, 0.20, 0.15, 0.10, 0.30);
pi <- c(-200, -100, -50, -100, -200, 500);

#Plot probability of profit for n = 1,...,100
library(ggplot2);
DATA <- data.frame(n = 1:100, PROB = PROB(p, pi, 100));
FIGURE <- ggplot(data = DATA, aes(x = n , y = PROB)) +
geom_line(size = 1, colour = 'blue') +
expand_limits(y = c(0,1)) +
theme(plot.title    = element_text(hjust = 0.5, size = 14, face = 'bold'),
plot.subtitle = element_text(hjust = 0.5, size = 10, face = 'bold')) +
ggtitle('Probability of Profit') +
labs(subtitle = '(Using normal approximation to the multinomial distribution)') +
xlab('Number of bets') +
ylab('Probability of profit');
FIGURE;

N.B. My undergraduate degree was also in actuarial studies, and that is how I came to fall in love with probability and statistics. (So much so that I abandoned actuarial mathematics and became a statistician!) Good luck with your program.

If you're assuming that each outcome of the game is independent of the previous outcomes, you can just use the linearity property of the expectation (where $c$ is a constant): $$\mathbb E[c \cdot X] = c \cdot \mathbb E[X]$$

You just set up the equation as $x \cdot \mathbb E[X] = P$, where P is the profit you're interested in, and solving for $x$ will tell you how many games need to be played for casino to win $P$ in expectation.

• Hi @Martn thank you! I think that would tell you how many trials to meet the expectation you're shooting for, but doesn't address the question of how many trials are needed before being profitable. I'm looking to see how to do this in R. Maybe even C#. Otherwise, it gets pretty ugly. Nov 27, 2017 at 2:52


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# Does placing a lid on a pot affect the rate of boiling?

On the subject of cooking, someone once told me "If you have too much water in your pot after cooking a dish, just turn up the heat and let it boil without a lid for a few minutes".

Now I've thought about this, and as far as I can tell, it shouldn't matter if there is a lid or not. In both situations the pot will stay at the same temperature (around 100 degrees Celsius), and therefore the amount of heat energy entering the pot will be the same. The lid may cause some drops of water to condense and fall back in, but then it also traps heat and results in a higher rate of boiling inside the pot -- but the amount of water exiting the pot as vapour shouldn't change.

So by that reasoning it seems that it doesn't matter. Or does it?

EDIT: Please note, I'm not talking about the time it takes for the water to reach its boiling point; I'm talking about the rate of water vapor leaving the pot when it is boiling.

• The lid raises the partial pressure of water vapor, reducing the rate of evaporation. (Note that this has little to do with the "air" pressure, which is increased only slightly.) – Hot Licks Mar 9 '15 at 3:27
• @SteveJessop If the amount of heat energy entering the pot is X units/second, this value does not change considerably with the addition of the lid. Once equilibrium is reached the temperature of the pot is constant, so there must be X units/second heat energy leaving the pot. This can only happen via evaporation. So I think it's constant. – CaptainCodeman Mar 9 '15 at 11:07
• Heat also leaves the pot via radiation. Putting the lid on raises the temperature. – Hot Licks Mar 9 '15 at 12:44
• Came here looking for the most efficient way to reduce maple sap to syrup, thanks for asking! – adamdport Mar 19 '15 at 18:06

When cooking, keeping the lid on a pot does a few things if you think about it:

1. it decreases air circulation significantly so the air in the pot stays hotter, this cooks the food faster.
2. if it's snug, it should increase the air pressure.*
3. The pot lid captures condensation, so it will return water back into what you're cooking. That is, I think, the largest effect.

I can't prove that with science, it's just my observation. I cook a lot. Often when I lift a lid off a stew or broth, the lid is wet and dripping. Curiously, what this means is that water will get to a boil faster with the lid on, but it will boil away faster with the lid off.

* On Air pressure: I have a hard time believing that, unless it's a pressure cooker, this would be a significant increase, as air pressure is 14.4 lbs per square inch and what does the average lid weigh - a pound? But there might also be a small decrease in pressure (er, I think), similar to Bernouli's paper lift.

The boiling of water creates a wind direction of sorts off what you're cooing up through the pot and that wind direction should decrease air pressure, but that effect would likely also be small. Perhaps, the boiling point - no lid might be 98 degrees C and with lid, 101 degrees C - that's a guesstimate, but it shouldn't be a huge difference I would think.

• 14.4 psi air pressure is just the pressure of the atmosphere - it's not directed at lifting the lid, but the added pressure from heat would, and will lift a lid, but not from the 14.4 ambient pressure. – Dronz Mar 9 '15 at 3:52
• @Dronz I think userLTK is trying to say that the additional pressure on the gas in the pot would only be around 1 lb / (SA of the lid) before the lid flies off, which is marginal compared to the preexisting 14.4 psi atmospheric – k_g Mar 9 '15 at 4:46
• Just for reference here: 1 additional atmosphere of pressure elevates the boiling point of water by ~ 20 C. The pressure to lift a 1 lb 8" lid is 0.0013 atm, giving about 0.027 K of boiling point elevation. (And that's only relevant if the lid does't leak or lift at one edge, resulting in earlier pressure release). – user3823992 Mar 9 '15 at 5:26
• Yeah, that's what I meant. I was more going for ratio by bringing up 14.4 pounds per square inch, so obviously you can't increase the air pressure much under a lid before the lid is lifted and that's assuming the lid is even reasonably air tight, which most aren't. I think it increases the air pressure some, but logic suggests, only a tiny bit. – userLTK Mar 9 '15 at 5:27
• Any thoughts on the Bernoulli paper effect, or is that a stretch? The continuous creation of water vapor by the boiling and the expansion of it as it leaves the boiling water into the colder air creates a steady wind of sorts out of the pan which should (could?) create a low pressure system - but how low pressure, I'm not sure. Probably not much. – userLTK Mar 9 '15 at 5:35

This is the sort of question which is easily answered by actually doing the activity in question instead of looking at it and scratching your head. And when you actually do heat the water you will find that yes, the time to get from room temperature (for example) to boiling is shorter with the lid on the pot - and noticeably shorter - than without.

Also, when you do this with a lid on which is not a snug fit but rather the sort of lid which just rests on the rim of the pan, you will see how much easier it is to spin the pan lid when the water is boiling. During boiling you will see (and hear) the lid rattling as it rises very slightly and falls back. The steam does mechanical work to lift the lid. Now checkout James Watt.

• I agree that doing an experiment or two is pretty easy in this case and, strictly speaking, will answer the question at hand. It doesn't do very much to promote an understanding the heat transfer, thermodynamics and general "physics" of the problem. – user3823992 Mar 9 '15 at 5:41
• @user3823992 neither do most of the other answers to be honest, and many of the answerers seem equally unfamiliar with the common behavior of real-life pots. – jwg Mar 9 '15 at 12:58
• The OP acknowledges that time to boiling will be faster. The question is about how quickly the water is depleted, taking into account that with a lid some (most?) of the steam will condense on the lid and drip back in. – WinnieNicklaus Mar 9 '15 at 14:59
• Doing the experiment by yourself only gives a yes/no answer to the question. Does it boil faster or slower? It doesn't explain WHY that answer is true. There's no need to be so condescending. – user3932000 Nov 2 '15 at 1:07

The trick is here in the temperature of the pot.

As you know, evaporation does not require boiling at 100°C: else, towels would only dry in an oven, and as far as I know, nobody dries towels in their ovens. Instead, it depends on the partial pressure of water vapor near the water surface (as Hot Licks mentioned).

• If the lid is on the pot, the partial pressure of water vapor under the lid is about equal to atmospheric pressure (slightly higher).
• If the lid is removed, the partial pressure of water vapor above the water is by approximation equal to the partial pressure in your kitchen (especially if you have a steady stream of 'fresh' air).
• With a lower partial water vapor pressure, the liquid will start evaporating more readily. This will lower the temperature of the evaporated water significantly: the enthalpy of evaporation is orders of magnitude more than the heat capacity of water!
• With a lower temperature, both heat losses due to convection and evaporation at the walls of the pan will lessen, and the heat source will be able to transfer heat more efficiently.

So basically, with a lid, evaporation is a 'last resort' of keeping the water at 100°C, if 'normal' heat loss can't handle all of the incoming heat; without a lid, evaporation becomes the main way of losing incoming heat.

Although, of course, I challenge the OP to verify this with experiment, which is after all the core of science. Who knows this is just an urban legend, and your water will evaporate faster with a lid?

On the subject of cooking, someone once told me "If you have too much water in your pot after cooking a dish, just turn up the heat and let it boil without a lid for a few minutes".

That acquaintance was absolutely correct. There are a number of ways to cheat and quickly thicken a sauce, but (a) this is cheating, and (b) the sauce doesn't taste as good. Reducing the sauce is what the best cooks do.

Knowing when and when not to (and how and how not to) use a lid is Cooking 101. Presumably you can cook rice. You would never cook rice without a lid, would you? (Actually, you can cook rice without a lid, but this is beyond Cooking 101.)

Knowing how to properly reduce a sauce is just a bit beyond Cooking 101. Basically, you turn the heat up and you stir. Stir frequently.

So by that reasoning it seems that it doesn't matter. Or does it?

By way of analogy, suppose you go to a hotel or visit a friend in the far north that seems to think 60 Fahrenheit (15 Celsius) makes for a nice comfy sleeping temperature. Are you going to use a blanket, or say that it doesn't matter and sleep commando? Or does it? That's a rhetorical question. Of course it matters.

It matters even more with cooking.

Update

The reason using a lid (or not) matters more with cooking is twofold. I'll look at this in the context of the question raised in the title of the thread,

Does placing a lid on a pot affect the rate of boiling?

Note well: This is the wrong question with regard to the body of the question. I'll get to that later. The answer to the question raised in the title of the question is "yes".

Some of the answers have posited that the lid raises the boiling point. This effect is marginal. The Clausius-Clapeyron equation describes the relation between pressure, temperature, and the vaporization pressure of a liquid:

$$\frac{d\ln p}{dT} = \frac {\Delta H_\text{vap}}{RT^2}$$

At sea level, water boils at 373.15 K and has a heat of evaporation of 40.657 kJ/mole at that temperature. Linearizing the Clausius-Clapeyron equation yields a 0.2848 kelvin increase in the boiling point for each one percent increase in pressure. Unless one is using a lid that is held down with clamps, this is a small effect. A hefty lid might raise the pressure inside the pot by a few fractions of a percent. This is at best a small secondary effect of using a lid.

The primary effects of using a lid are to drastically reduce convective heat transfer from the steam/vapor to the atmosphere and to drastically reduce vaporization of the water. All other things being equal (same pot, same heat setting, same amount of water), a boiling pot of water covered with a lid will have a higher boiling rate than will an uncovered boiling pot of water. Some of the heat supplied to the uncovered pot goes toward countering the increased heat transfer rate due to vaporization and convection.

You can see this effect if you have pots with glass lids. Cover the pot, bring some water to a vigorous boil, and take the lid off. The boiling rate will decrease once the lid is removed. If you reduce the heat to a low simmer before taking the lid off the effect is even more dramatic: The water stops boiling. Yet another way to see these effects is to time how long it takes for a covered versus uncovered pot of water to come to a boil.

I mentioned above that this is the wrong question. The advice given to the questioner was not about the boiling rate. It was about increasing the evaporation rate for a dish that has a bit of excess liquid at the end of cooking. Presumably the advice was for a dish cooked at a low simmer. Very little water vapor escapes the pot or pan while covered. Most of the water vapor created by simmering instead condenses on the inside of the lid and soon rejoins the dish. The gas underneath the lid is mostly water vapor at close to the boiling point. There is very little evaporation from the food.

Removing the lid exposes the cooking food to air at a much lower absolute humidity. The entire surface evaporates, and the vapor is quickly carried away. Assuming no change in the temperature control, the rate at which water escapes increases significantly even though the boiling rate has decreased.

Note that the exact same advice given to the questioner was given here at the cooking stackexchange site.

• This seems like a tutorial on sauce making and a story about a camping trip rather than an answer to the question. – CaptainCodeman Mar 9 '15 at 9:56
• @CaptainCodeman - I updated the answer. This is the answer to the question. There is a problem with the question itself: The OP asked the wrong question. The reason for taking the lid off has little to do with the boiling rate. – David Hammen Mar 9 '15 at 12:21
• Still there is no explanation for why placing a lid increases total evaporation rate. I would argue that it does not matter: The same amount of heat energy enters the system, and the only way it can escape is via evaporation, so that must be constant as well. – CaptainCodeman Mar 10 '15 at 9:53
• @CaptainCodeman - Are you serious? I have to wonder whether you have cooked (other than microwave cooking). Do a physics experiment. Time how long it takes to bring a covered vs uncovered pot of water to a boil. Do another physics experiment. Simmer a dish to a point where it has a bit of excess water. If you keep the lid on, you will have to for a long time for that excess water to go away. Uncover the lid and the wait time is substantially less. – David Hammen Mar 10 '15 at 16:41
• I think you should keep your wonderings about my cooking experience to yourself as cooking has no relevance to this conversation. I'm trying to have an intellectual exchange, but you seem to be trying to assert your dominance in the kitchen. You don't seem to be able to focus on individual concepts and resort to condescendence and insults when you get confused. Furthermore you don't seem to be able to bring any scientific reasoning to support your claims. – CaptainCodeman Mar 10 '15 at 21:02

If you want to get technical, check out steam tables such as found on:http://www.engineeringtoolbox.com/saturated-steam-properties-d_273.html

Putting the lid on the pot will slightly raise the pressure inside the pot, therefore the boiling point will raised. The steam generated will be under somewhat higher pressure. When this happens the latent heat required to generate a pound of steam will go down. This means that you will actually generate more lbs of steam with the same amount of heat at the boiling point. But it will take more heat to raise the temperature of the water to the boiling point which offsets the lower latent heat requirement. So it would seem that you would actually generate less steam with the lid on, unless you raise the pressure and temperature extremely high, then things change drastically. But that is another subject.

With normal pots, I think any differences are negligible. Pressure cookers on the other hand cook foods faster because of the higher temperature (boiling point) inside the pot. Of course, when the temperature inside the pot is higher, less heat will enter the pot because of the smaller temperature difference between flame ant pot.

Also, I am told you should never watch a pot or it will not boil.


text:
## A surfboard is originally priced at $155. The online retailer gives a discount and the surfboard is now priced at$93. Enter the percentage

Question

A surfboard is originally priced at $155. The online retailer gives a discount and the surfboard is now priced at$93. Enter the percentage discount for
the cost of the surfboard.

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5 months 2021-08-28T09:02:34+00:00 1 Answers 0 views 0