I am looking for some help to understand what kind of battery could power a Dometic cfx50w 12/24/120V compressor fridge, while on-the-go. Here are the electrical specifications:     Input voltage (AC) 120-240 V   Input voltage (DC) 12/24 V   Rated input current (AC) 0.79 A   Rated input current (DC) 7.8 A   Input frequency 50/60 Hz     ()  Most people choose a dual battery DC/DC setup for their car. I am looking for a more compact and portable solution, to get a few hours (minimum 4) running in normal conditions. I am thinking of a battery pack like those used in the RC hobby. Would this work?  I am looking for best battery chemistry and recommended ah rating considering the fridge would run for 10 minutes every hour and battery would have to provide 4 hours of power without getting damaged. Fridge has a cutoff voltage of 11.1v. I am wondering if lipo or lifepo4 batteries could replace a flooded lead acid and what would be the usable capacity. Battery would be charged at home by small rc charger.
I have an interesting mathematical/electrical dilemma and I'm not sure where to start.  I am considering buying  (specs on page).  I will be hopefully buying a battery pack with it to support a long-road trip (a few months) wherein I'll keep medication in the fridge (powered by the battery pack when parked and by the car when driving). The medication needs to be keep between 2 and 8 degrees C at all times.  Here's the  I'm thinking of getting:  Can someone please help me understand what I need to do in order to calculate whether or not this fridge + battery combo will work, and if so, how much time will I be able to power the fridge for.  Ideally, I need a combination of fridge and battery which will allow me to have the fridge constantly powered, assuming I sleep in the car for approx. 8 hours a night and drive for the rest of the day (although, there will be times which I'm away from the car during the day for several hours + times where I hope to have access to a wall outlet).   Specs:   Fridge:   Input voltage (AC 220-240 V Input frequency 50/60 Hz Rated input power (AC) 64 W Rated input power (DC) 46W DC12V/50W DC24V W Input voltage (DC) 12/24 V   Battery:   Capacity: 453Wh Battery Type: Lithium-ion Input Voltage: AC 110-220 V, 50/60 Hz; Solar Power: 18 V; Car Input: 12 V Output Voltage: AC 110V 60 Hz; DC Output: 4 * 12V 4 A (Max.10 A); USB Output: 4 ports: 2* 5V 2.1 A and 2* 5V 1 A Power Output: Rate 500 W Peak/Surge 1000 W
Could anyone provide an example of a nonempty subset $U$ of $R^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $R^2$ So I was writing up a simple binary tree in Python and came across [...]  I don't believe this to be related to the Ellipsis object, more it seems to have something to do with an infinity loop (due to Python's shallow copy?). The source of this infinity loop and why it doesn't get expanded while expanding when accessed is something I'm completely lost to, however   &gt;&gt;&gt; a [[[[[], [], 8, 3], [[], [], 3, 2], 6, 3], [], 1, 4], [[], [], -4, 2], 0, 0] &gt;&gt;&gt; Keys(a)#With a+b [0, 1, 6, 8, 3, -4] &gt;&gt;&gt; Keys(a)#With [a,b] [8, [...], [...], 3, [...], [...], 6, [...], [...], 1, [...], [...], -4, [...], [...], 0, [...], [...]] &gt;&gt;&gt; Keys(a)[1]#?? [8, [...], [...], 3, [...], [...], 6, [...], [...], 1, [...], [...], -4, [...], [...], 0, [...], [...], 8, [...], [...], 3, [...], [...], 6, [...], [...], 1, [...], [...], -4, [...], [...], 0, [...], [...]]     Version using a+b  def Keys(x,y=[]):     if len(x):y+=[x[2]]+Keys(x[0],y)+Keys(x[1],y)#Though it seems I was using y=y[:]+, this actually outputs an ugly mess     return y   version using [a,b]  def Keys(x,y=[]):     if len(x):y+=[x[2],Keys(x[0],y),Keys(x[1],y)]     return y   So what exactly is [...]? How do I check if an object has a specific property in JavaScript? How do I check if an object has a specific property in JavaScript?  Consider:  x = {'key': 1}; if ( x.hasOwnProperty('key') ) {     //Do this }   Is that the best way to do it? I made a map with annotations in QGIS 3.0.  This is how it looks like in the layout view:    If I print the map as PDF, one would expect to get the PDF to be WYSIWYG, right? However, it looks like this (it's about half the size):    Is there a way to influence the the scaling of the Annoations? I already tried to play with the Scale and Magnifier in the map view. I also tried to change the Scale in the map item.   The strange thing is, I believe it worked a few days ago. But today I installed QGIS 3.0 on a new PC and it showed the same behaviour... Idiomatic expression for a difficult choice This has cropped up several times in the past couple of months, and I've been struggling to find  a fitting word to describe this phenomenon. I'll describe it:  You have two choices(no, it's not Hobson's choice). Both of them are desirable, yet both of them each have a tiny part which is most undesirable, such as a tough condition, or a demanding clause. You have difficulty deciding which one to pick.  Whenever something like that comes up, I think of the expression "Can't burn a candle at both ends", but I'm not looking for that.  I'm looking for an expression that describes a difficult choice. Any suggestions?  EDIT: It's idiomatic. I'm quite sure it is. It's an expression of some sort. And no, it isn't "dilemma" Functions Which are non differentiable on a Given Set. The title is probably very confusing, but here's the situation:  I'm animating a flat grid curve by shrink-wrapping it onto a moving plane in order to get a uniform wave animation, it looks like this:    Before I render it, i want a bevel. What I did then was simply adding a bevel to the curve, making the curve look like this:    Now, when activating the shrinkwrap modifier, I expexted that only the curve would be shrinkwrapped, but alas, the bevel did the same:    As seen by the blue spots, the curve has been completely flattened along with the bevel.  How can I avoid this? I tried using a wave modifier instead of shrinkwrap, but this modifier doesn't seem to make perfectly round waves like this:    So, any thoughs on how to solve this?  Thanks for reading! Shiny locked Legendaries in ORAS I would like to know what would be the best way to do unit testing of a servlet.   Testing internal methods is not a problem as long as they don't refer to the servlet context, but what about testing the doGet/doPost methods as well as the internal method that refer to the context or make use of session parameters?  Is there a way to do this simply using classical tools such as JUnit, or preferrably TestNG? Did I need to embed a tomcat server or something like that? I'm creating a function where I need to pass an object so that it can be modified by the function. What is the difference between:  public void myFunction(ref MyClass someClass)   and  public void myFunction(out MyClass someClass)   Which should I use and why? Let $\{s_n\}$ and $\{t_n\}$ be sequences.  I've noticed this inequality in a few analysis textbooks that I have come across, so I've started to think this can't be a typo:  $\limsup\limits_{n \rightarrow \infty} s_n + \liminf\limits_{n \rightarrow \infty} t_n \leq \limsup\limits_{n \rightarrow \infty} (s_n + t_n) \leq \limsup\limits_{n \rightarrow \infty} s_n + \limsup\limits_{n \rightarrow \infty} t_n$.  I understand the right two-thirds of the equation, and can prove it.  I'm stuck on the far-left third.  I don't understand how that is true. PCA and the train/test split I have a dataset for which I have multiple sets of binary labels. For each set of labels, I train a classifier, evaluating it by cross-validation. I want to reduce dimensionality using principal component analysis (PCA). My question is:  Is it possible to do the PCA once for the whole dataset and then use the new dataset of lower dimensionality for cross-validation as described above? Or do I need to do a separate PCA for every training set (which would mean doing a separate PCA for every classifier and for every cross-validation fold)?  On one hand, the PCA does not make any use of the labels. On the other hand, it does use the test data to do the transformation, so I am afraid it could bias the results.  I should mention that in addition to saving me some work, doing the PCA once on the whole dataset would allow me to visualize the dataset for all label sets at once. If I have a different PCA for each label set, I would need to visualize each label set separately. learn how to [verb] vs. learn to [verb] "learn to [verb]" "learn how to [verb]"   Is [1] merely a less formal version of [2]? If not, does [1] communicate something subtly different? Consider the following:  In [2], the object of learning is explicitly "how to [verb]" (for instance, "I will learn how to speak French").  Is the object of learning in [1] "to [verb]" or is it something implicit and abstract that allows the subject "to [verb]"? For instance, in "I will learn to speak French", the object of learning would be the prerequisite elements for speaking French: "French phonemes, grammar, syntax, etc.". Induction problem: "if the group has at least one player who is better than Messi, then all the members of the group are better than Messi" Power of a generator generates the group iff this power is coprime to the group order. Let $\langle g\rangle$ be a cyclic group of order $n$. Suppose $1\leq q \leq n-1$, I want to show that $g^q$ generates $\langle g\rangle$ if and only if $\gcd(n,q)=1$.  Suppose $g^q$ generates $\langle g\rangle$, then   $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are all distinct and $g^{nq}=1$, since $n$ is the order of the group. Note that this is the lowest multiple of $q$ for which this is the case. Therefore $\operatorname{lcm}(q,n)=qn$ and it follows that $q$ and $n$ are coprime.  On the other hand, if $\gcd(q,n)=1$, then $\operatorname{lcm}(q,n)=qn$, we have $g^{nq}=1$ and hence all of  $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are distinct again and $g^q$ is a generator.  Is my proof correct at all and is there maybe a more elegant argument? Gosper's unusual formula connecting $e$ and $\pi$ Here's my proof in my own words, does it stack up?   Showing $A$ is complete implies $A$ is closed. Let $(x_n)$ be a convergent sequence in $A$. $A$ is complete $\implies (x_n) \to p \in A$. Hence $A$ is closed. Showing $A$ is closed $\implies$ $A$ is complete. Let $(x_n)$ be a convergent sequence in $A$. $A$ is closed $\implies (x_n) \to p \in A$. As every convergent sequence is a Cauchy sequence, $(x_n)$ is a Cauchy sequence in $A$ that converges to $p \in A$ and hence $A$ is complete.   That sound ok?  As an aside, it seems to me that the definition of completeness and closed are basically identical, why are there two definitions for the same thing? Am I missing something here? What is the center of $D_{2n}/Z(D_{2n})$ What is the center of $D_{2n}/Z(D_{2n})$. I see that when $n=2^k$ then I have a $p$-group so the center is not trivial. But when $n$ is not power of $2$ how can i know what is the center of this group?. Surely when $n$ is odd then the center is trivial. How can I enable "clicklock"? Is there a cross-browser onload event when clicking the back button?