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July 8

Shared speakers ?

I have several sets of speakers (mostly "2.1" systems), for various TVs, computers, radios/CD players, etc. I was wondering if it would be possible to plug all the devices in one room into the same speakers. There would be the physical problem of splicing the lines together, then issues of the devices interfering with each other, especially if more than one was on at a time. They might also have different plugs, impedance/resistance, etc. So, is this possible ? Would it only work with a physical switch to connect only one at a time ? StuRat (talk) 01:04, 8 July 2016 (UTC)

Hang on you want to connect several devices into ONE set of speakers or into SEVERAL sets of speakers all at once? Vespine (talk) 04:46, 8 July 2016 (UTC)

Sorry after reading it multiple times I think I got it. You want to get rid of several sets of speakers and just use one set for everything. Yes this will work, but does depend on several things. Whether you get ground loops happening with all the equipment running straight into your speakers would be my biggest concern. The "real" solution is to get a mixer, which I don't think there's really any super cheap option, starting at about $50 for a 4 stereo channel mixer. Vespine (talk) 04:59, 8 July 2016 (UTC)

Thanks. Would I then need to manually select which channel(s) I want to hear, or would it be smart enough to only select the channel(s) with a real signal (as opposed to static) ? StuRat (talk) 16:08, 8 July 2016 (UTC)

You would need to do it manually, unfortunately, it's a very steep step into "pro audio" territory. You need something like a gated automixer, but i don't think there's such a thing as a consumer version of such a device so you probably won't find a cheap one. Ideally, you would have an AV receiver, if you keep an eye out, you might be able to find a cheap 2nd hand one, if you are lucky maybe you'll find one where the "video" part is broken and just use it for the "audio" part? I've had a quick look at a classified website and there's loads for $50. Mind you they're pretty huge so if you are short on space it's probably not the best solution either... Vespine (talk) 00:00, 11 July 2016 (UTC)

Thanks. I was just wondering if there was a way to simplify my life and combine all these speakers into one. These days flat screen TVs and computer monitors all have built-in sound that totally sucks, so now I have a plethora of these 2.1 speaker systems, one for each device. That gives me decent sound, but the clutter is bad. StuRat (talk) 03:17, 11 July 2016 (UTC)

How do I print the arithmetic mean symbol, "x bar", in Microsoft Word?

The symbol for the arithmetic mean is called "x bar", which is the letter "x", italicized (I believe), with a bar over the "x". You can see it in this article: arithmetic mean. My question is how do I get that symbol to print in Word? I looked through all of their symbols on the "Insert" tab. They have a million odd symbols, but I can't find this one anywhere. Thanks. Joseph A. Spadaro (talk) 03:22, 8 July 2016 (UTC)

This is the symbol: ${\bar {x}}$ (read $x$ bar). Joseph A. Spadaro (talk) 03:24, 8 July 2016 (UTC)

First, start the equation using Insert Equation. Then type the "x", select it, and in the Equation ribbon, select accents, and choose the bar. This instruction comes to you courtesy of Googling "type x-bar in Word".--Shantavira|^{feed me} 06:36, 8 July 2016 (UTC)

Thanks. But, I don't want an equation. I just want the symbol. Is it not available as a regular symbol, like the dozens of others? Joseph A. Spadaro (talk) 07:57, 8 July 2016 (UTC)

They have millions of symbols that I am sure no one ever uses. But they don't have a relatively commonplace symbol like the x-bar? Joseph A. Spadaro (talk) 08:01, 8 July 2016 (UTC)

@Joseph A. Spadaro: There are actually two parts in that question; one, how does Unicode provide a symbol for x̄, and two, how does one write it in MS Word.

Going by [1] (which includes a detailed procedure in Word at the end) there is no single-symbol "x bar" in standard Unicode, but some fonts accept diacritics so that you can effectively get what you want with "an x, with a bar over it". You will need to make sure the font supports it, and you will need to find the diacritic, but it works. Tigraan^{Click here to contact me} 09:11, 8 July 2016 (UTC)

@Joseph A. Spadaro and Tigraan: Possibly the Unicode U+0305, called a 'combining overline' is what you need; see Overline. --CiaPan (talk) 09:24, 8 July 2016 (UTC)

In MacOS with the Extended keyboard setting, option-shift-a (after the base letter) makes U+0304 COMBINING MACRON: x̄. —Tamfang (talk) 09:27, 8 July 2016 (UTC)

Doesn't Word have the option to add an overbar in character format? So, you type an italic x and add an overbar. Done. 209.149.113.4 (talk) 14:49, 8 July 2016 (UTC)

Thanks. Let me change my question. When I look at the symbols available in the Word "Inserts" tab. they have hundreds upon hundreds of extremely esoteric symbols. (Many of which are odd and bizarre. Many of which are useless, like "cute" little drawings.) Stuff that I am sure no one ever uses. Why would they not have a relatively "common" symbol, such as the x-bar? Thanks. Joseph A. Spadaro (talk) 17:57, 8 July 2016 (UTC)

You'd have to ask the Unicode Consortium about that. Presumably, x bar isn't included because it can be created by combining two pre-existing symbols as shown above. clpo13_(talk) 18:56, 8 July 2016 (UTC)

If you ask for ${\bar {x}}$, other people will ask for ${\bar {y}}$, ${\bar {M}}$, ${\bar {\theta }}$, etc... Mathematically, it is more appropriate to think of the bar as a modifier of whatever variable you have, whether it's called $x$ or anything else. --Wrongfilter (talk) 20:01, 8 July 2016 (UTC)

That could be said of many other symbols they already do include. Joseph A. Spadaro (talk) 20:13, 8 July 2016 (UTC)

You may want to read Precomposed character. Basically, combining characters are preferred since they are much more flexible (you can put a COMBINING OVERLINE character on almost any other character, while with precomposed characters you're limited to what's available). But some older software has trouble handling combining characters correctly, so they added some precomposed characters for what they considered to be common cases. x-bar apparently wasn't one of those cases. CodeTalker (talk) 00:06, 9 July 2016 (UTC)

Correct. And it wasn't just "what they considered to be common cases"; in general they included characters (that is, single-point or "precomposed" characters) only if they had already been included in ome other notable character set. (That is, they didn't trust their own judgement; they applied something similar to Wikipedia's Reliable Sources policy.) —Steve Summit (talk) 15:59, 9 July 2016 (UTC)

Thanks, all. Joseph A. Spadaro (talk) 04:28, 9 July 2016 (UTC)

Looking for a pointer

On Windows 7 (64-bit) every time I leave the PC for a while the mouse pointer disappears. Wiggling the mouse doesn't bring it back, but CTRL-ALT-DEL does, and it remains after I hit the Cancel button. What causes this and how do I stop it ? Note that it doesn't appear to have gone into sleep/hibernate mode, as the original screen is still displayed (not a screen saver). StuRat (talk) 18:03, 8 July 2016 (UTC)

Have you done virus checking firstly? -- Apostle (talk) 18:10, 8 July 2016 (UTC)

The physical interface is acting as if the screen locked. I have seen screensavers fail such that the screen locks, but the display doesn't change. So, you have to blindly unlock the screen - which can be difficult if you have to ctrl-alt-del and then type a password. 209.149.113.4 (talk) 18:21, 8 July 2016 (UTC)

UPS

1. What brand is highly praised?
2. Do they come with Lithum ION battery (or something better) like the Laptops? Does a UPS protect itself from ‘electric over powering’ the battery, like the Laptops which consist of circuts to mitigate over powering the battery…?

Apostle (talk) 18:08, 8 July 2016 (UTC)

It isn't a brand thing. I assume all brands make good and bad products. I have an APC 1500. It has been great. I also have an APC 650. It is crap. The output is so flaky that it sends my computer into power fault every few hours. So, I stopped using it. They do have circuits to stop charging the battery when it is fully charged. That should be obvious. Otherwise, a UPS would be pointless because you'd have to unplug it when charged and, somehow, plug it back in very quickly just before power goes out. 209.149.113.4 (talk) 18:28, 8 July 2016 (UTC)

They usually use lead batteries. Ruslik_Zero 20:34, 8 July 2016 (UTC)

Noted, and thank you both. Sorry for the delay, I was unwell. -- Apostle (talk) 18:07, 10 July 2016 (UTC)

Sectorless or blockless storage

Is it possible for memory storage, either magnetic or flash, without sector sizes or block sizes to exist? I am imagining an EEPROM that has no blocks. Would the construction of such be more difficult. — Melab±1 ☎

Read the flash memory article. NOR flash is byte-addressable. However NAND is cheaper and can achieve greater density, so it's typically used, unless the characteristics of NOR flash are needed. As for hard drives (which usually aren't referred to as "memory"), blocks are used in modern drives because the drive calculates and stores ECC for each block. This is a consequence of cramming data into such small areas. The signal is very weak, and read errors are a frequent occurrence. --71.110.8.102 (talk) 23:39, 8 July 2016 (UTC)

All NOR flash I've seen was divided into blocks just like SD cards are divided into sectors. The connection between "modern drives" using "blocks" instead of sectors due to ECC is dubious because other storage that uses "blocks" has no ECC (I haven't seen a hard drive that uses blocks, anyway). — Melab±1 ☎ 20:00, 9 July 2016 (UTC)

Maybe I'm missing something, but I don't see where it was claimed hard drives use blocks instead of sectors. The claim seem to be that hard drives are in blocks or sectors instead of being single byte addressable. Sector is arguably the better term to use for the physical chunks on hard drives since blocks often refers to the chunks used by the file system, Disk sector#Sectors versus blocks [2]. Still I don't know if you can say the usage is explicitly wrong. After all, consider Logical block addressing and Block (data storage). (Although as 512e shows, LBAs and presented sector sizes don't have to correspond to physical sector sizes. And nowadays we have the Shingled magnetic recording hard drives to complicate things even more when you consider writing.) Nil Einne (talk) 13:39, 11 July 2016 (UTC)

There are implementations that do not use addressable memory. I worked on a system that used track memory. A large rotating drum had a spiraling track that went from one end to the other. When a signal arrived, it was placed on the track at the current location of the head. If there wasn't an incoming signal, the head would read whatever track information was written and send it out as a repeating signal. It was, in effect, a repeating analog buffer - but it was memory and it wasn't broken into blocks or sectors or bits or addresses. It had a good use in that one particular case, but is rather useless for a modern computer. The argument to make is where the concept of "memory storage" ends. Is it memory storage if you are recording analog signals? Does it have to be binary? Is an old vinyl record memory storage? 209.149.113.4 (talk) 12:27, 11 July 2016 (UTC)

July 9

I²C over COM port

hello, does someone by chance know of a program or a library that can bitbang the I²C protocol over an RS-232 port (or, rather, an FTDI style breakout board with TTL levels), for communicating with I²C devices, preferably under Linux? I'm aware of the electrical differences between I²C and TTL Asmrulz (talk) 04:29, 9 July 2016 (UTC)

Our I2C article has some suitable source code. Tevildo (talk) 08:39, 9 July 2016 (UTC)

Wattage

Is it possible to measure an L.E.D.'s actual watt using a typical Multimeter (supposing it can measure Current, apart from Voltage, both in AC and DC) ? 210.56.110.175 (talk) 05:42, 9 July 2016 (UTC)

Not with one measurement. You need to measure the current, disconnect the multimeter, reconnect the supply, then measure the voltage. Tevildo (talk) 09:29, 9 July 2016 (UTC)

Incidentally, if you have a simple arrangement with a resistor in series with the LED, you can measure the current by measuring the voltage across the resistor, and using the equation $I={V \over R}$. You still need to make a second measurement of the voltage across the LED, though. Tevildo (talk) 09:41, 9 July 2016 (UTC)

Many thanks Tevildo. If it's not taking too much advantage of your generosity, I'd beg you to be a bit more elaborate. For instance in first instruction you say "measure the current", now shall I measure it along with LED attached or LED removed from the circuit. And is it true that current is never measured like voltage (in parallel) but in series ? As for putting resistor in series with LED that ain't difficult, why not simply jump to this step ? Please tell how voltage across resistor and across LED ultimately give us wattage. Unless I am very much wrong I think current and wattage are separate things. 124.253.145.142 (talk) 17:00, 9 July 2016 (UTC)

File:LED Annotated.PNG

Typical circuit

No problem, it's what we're here for. I'm assuming your circuit is basically the same as the one to the right (from our LED article, with annotations). To find the power, we need to find the voltage across the LED and the current through it, and multiply them together.

Measuring the voltage across the LED is easy:

• Set your multimeter to "DC Volts".
• Connect the positive lead to point B on the diagram, and the negative lead to point C.
• The reading on the multimeter is the voltage across the LED. Let's call this $V_{LED}$. For example, it might be 3 volts.

There are two ways to measure the current. The first is to measure it directly with the multimeter:

• Disconnect the wire between the power source and the LED (between points A and B on the diagram).
• Set your multimeter to "DC Amps".
• Connect the multimeter positive lead to point A on the diagram (the supply positive output), and the negative lead to point B (the LED anode).
• The LED should now be on. The reading on the multimeter is the current in the LED. Let's call this $I$, as in the diagram.

The second way is to measure the voltage across the resistor. This assumes you know what its resistance is.

• The circuit should be connected as in the diagram (with the LED on).
• Set the multimeter to "DC Volts".
• Connect the positive lead to point C, and the negative lead to point D.
• The reading on the multimeter is the voltage across the resistor. Let's call this $V_{R}$.
• If the resistance of the resistor is $R$, the current in the LED is $V_{R} \over R$. For example, if $R$ is 100 ohms, and $V_{R}$ is 2 volts, the current ($I$) is 20 mA (0.02 = 2 / 100).

Now we know $V_{LED}$ and $I$, the power of the LED (in watts) is $W=I\times V_{LED}$. Using the numbers from above, the power is 20 mA * 3 V = 60 mW. Hope this helps! Tevildo (talk) 18:08, 9 July 2016 (UTC)

I suggest you get a device like the Kill-A-Watt meter, which does the math for you and lists the wattage directly: [3]. StuRat (talk) 17:08, 9 July 2016 (UTC)

This is for mains equipment, and the OP just wants to measure the power for a (DC) LED. Tevildo (talk) 18:08, 9 July 2016 (UTC)

To be more specific, if you try to measure the watts of an alternating current circuit by measuring volts and amperes separately and calculating the result, you end up with volt-amperes, not watts. In direct current circuits watts and volt-amperes are identical, so the technique works. The math is easy: volts times amperes equals watts, as Tevildo explained in detail above. --Guy Macon (talk) 22:09, 9 July 2016 (UTC)

Note: High power LEDs, also called compact light sources, were operated in pulse mode due more efficient power supply. The pulses may not get captured correctly by the multimeter. An oscilloscope lets you know, but note, the oscilloscope is connected to ground. Aware of shorcuts. --Hans Haase (有问题吗) 20:53, 12 July 2016 (UTC)

Chinese (and other exotic) Characters

Whenever a server sends traditional or simplified Chinese (same about some other languages also) as text, instead of occuring as they should, they rather appear to be rectangles confused from within, as shown in the [4] picture here. Please tell me what should be done to make them occur naturally ? 124.253.145.142 (talk) 16:17, 9 July 2016 (UTC)

You will have to upgrade to an operating system with the Unicode fonts installed. You can install more fonts on your computer as well. Graeme Bartlett (talk) 22:35, 9 July 2016 (UTC)

Thanks, Graeme Bartlett...it works ! see → 活動写真 — Preceding unsigned comment added by 210.56.110.255 (talk) 10:04, 12 July 2016 (UTC)

Airlines use of continuous stationary

Why do airlines still use dot matrix printers and continuous stationery? I hear them at gates, and I assume they're printing passenger lists. The only advantage I can think of is that if you're clutching a wodge [spelling?] of continuous paper, you know you've got it all. Hayttom (talk) 20:55, 9 July 2016 (UTC)

One advantage I can think of is that if the ink runs out, you still have the indentations in the paper you can read, especially if you run a pencil lead over it. But while we are on the topic, why do most retail establishments in the US still use thermal printers for receipts ? StuRat (talk) 23:23, 9 July 2016 (UTC)

Great follow-up question. I hope we get some answers.Hayttom (talk) 14:21, 10 July 2016 (UTC)

One obvious factor is cost. The best technology to replace these old technologies, as far as readability, is likely a black-and-white laser printer (color seems like an unnecessary expense). I imagine they cost more initially (but maybe not by much, if it only prints the width of a receipt or airline ticket). However, for these high volume operations, the more important issue is what it costs per print. Can anyone provide data on this, for dot-matrix, thermal, and black-and-white laser printers ? StuRat (talk) 15:13, 10 July 2016 (UTC)

Also, (nearly) all laser printers are page printers, i.e. they print full pages. Dot matrix printers are line printers. If you need a permanent record immediately, a page printer will waste a full page on each transaction, even if you only need a single line. --Stephan Schulz (talk) 19:03, 10 July 2016 (UTC)

But is their any inherent reason why laser printers can't be designed to print smaller areas (hopefully at reduced cost) ? StuRat (talk) 22:25, 10 July 2016 (UTC)

StuRat You mean like a laser receipt printer? Probably far more economical to use a thermal printer. (See my answer below↓ too) If someone wanted to throw enough money at it, it is probably possible to shrink a laser printer down like that. I imagine you could write to thermal paper with a laser, thus avoiding the use of toner? - 220 of ^Borg 05:38, 11 July 2016 (UTC)

Those are most likely line printers, not dot-matrix. --Guy Macon (talk) 19:32, 10 July 2016 (UTC)

As to the use of thermal printers, they're very quick, but a key advantage may be that there's only one consumable (a thermal roll) rather than paper/ribbon or paper/ink or paper/toner - relatively speaking, this makes them a very much more easy to maintain than a printer which requires both. --Tagishsimon (talk) 22:34, 10 July 2016 (UTC)

re StuRats' query about Thermal Printers, I agree the big advantage of thermal printers may be speed. Where I live now they use a small thermal printer on a 'kiosk' at the library to print list of book etc borrowed. I have worked a lot (in the past) with inkjet log printers, and I was surprised that in this application were practically instant, i.e you tap the touchscreen asking for a 'receipt' and bang its there. Barely get the chance to lift my finger off the screen. Agree with Tagishsimon, in other words. I also think it's likely they can economically be made very small, likely more so than a laser. Thermals seem to be practically universal for cash register receipts.220 of ^Borg 05:38, 11 July 2016 (UTC)

Re the speed of thermal receipt printers, source here says up to 250 mm per second, so 10 inches/sec, 600 in/minute or ≈50 feet,(15 metres) per minute. That is pretty fast! 220 of ^Borg 05:49, 11 July 2016 (UTC)

Airlines were among the earliest earliest adopters of computer technology. It's entirely possible that some of what we see today just comes from the old adage of "if it ain't broke, don't fix it". If it works and fits their requirements, (including costs of consumables and break down frequency), sometimes it would cost a lot more to "re-engineer" something to work with a newer "peripheral" so you make do with the old one, until it becomes unviable. We have these old Dialogic Inc. fax cards called "brooktrout", they're old and obsolete but to replace the service will cost a fortune, so we just keep using them until they fail and then replace them with a different service. They might last 10 years or they might last 2 months but there's no point paying to change them over while they're still working. I'm not saying THAT is what's happening with line printers in airports, I'm just giving one example of why seemingly old obsolete computer hardware doesn't get replaced. Vespine (talk) 03:45, 11 July 2016 (UTC)

See If It Ain't Broke, Don't Fix It: Ancient Computers in Use Today. ---Guy Macon (talk) 04:32, 11 July 2016 (UTC)

July 11

Algorithm for Union of Sets

Let us denote $I=\{1,\dots ,n\}$, and denote $P_{k}(A)=\{B|B\subseteq A,|B|\leq k\}$.

I am looking for an efficient algorithm to transform $\{A_{i}\}\in P_{n}(P_{n}(I))$ into $\{\cup _{a\in A_{i}}{a}\}$.

Thanks in advance! 31.154.81.65 (talk) 10:25, 11 July 2016 (UTC)

Expressed with words, $P_{k}(A)$ is a set of k-element subsets of some $A$, right?

Then $P_{n}(I)$ is a set of n-element subsets of $I$. However, $I=\{1,\dots ,n\}$ is itself an n-element set, so it has just one n-element subset: $I\subseteq I$, hence $P_{n}(I)=\{I\}$.

Consequently $P_{n}(P_{n}(I))$ is a set of n-element subsets of a singleton $P_{n}(I)=\{I\}$, so it is $\{\{I\}\}$ if $n=1$, or an empty set otherwise.

Alas I can't get the $\{A_{i}\}\in \ldots$ part. Do you mean 'a singleton belonging to'...? If so, $\{A_{i}\}\in P_{n}(P_{n}(I))$ could only be $\{A_{i}\}=\{I\}=\{\{1\}\}$, so $A_{i}=\{1\}$ for $n=1$ (and does not exist at all for $n\neq 1$) and I can't see the reason for the $i$ index at $A_{i}$... --CiaPan (talk) 12:07, 11 July 2016 (UTC)

I'm sorry for the typo... I fixed my mistake (replacing equality sign with inequality sign). 31.154.81.65 (talk) —Preceding undated comment added 13:36, 11 July 2016 (UTC)

So, P_n(I) is simply the power set of I (where I = {1, 2, ... n}), and could be denoted P(I). Then P_n(P_n(I)), or more simply P_n(P(I)), is a truncated power set of P(I), containing only those sets of sets which have n or fewer elements from P(I). And your {A_i} is an element of that truncated power set of the power set?

In other words, you have a list of at most n lists of integers, the integers themselves ranging from 1 to n, the lists of integers presumably unsorted, with the additional constraints that the individual lists of integers have no duplication and that no two lists of integers from your list of lists would be the same even if sorted, and you want an efficient way to merge this list of lists into a single list containing, without duplication, exactly those integers which appeared in any list of your list of lists. Presumably you hope that these constraints will allow for a more efficient algorithm. Note that our article merge algorithm discusses the merging of sorted lists; I don't know what we have on the merging of unsorted lists without duplication. -- ToE 14:51, 11 July 2016 (UTC)

Okay, I'll read the article about merge algorithm carefully. Thank you for the idea to read this! 31.154.81.65 (talk) 05:55, 12 July 2016 (UTC)

If you can tell us your actual task, it would help us know if we are up against an XY problem. The multiple roles of n -- it is both the maximum value of the integers in your sets and is the maximum size of the elements which are in the truncated power set of the power set of I -- strike me as unlikely to arise naturally.

Your task, as I understand it, has the (perhaps slight) advantages over simply merging unsorted lists with possible duplication in that you know that the individual sets of integers themselves contain no duplication. Note that some languages offer set abstract data types which are often implemented "using more efficient data structures, particularly various flavors of trees, tries, or hash tables." Our article mentions the expected union and add operations, but I see that Python's built-in set type also offers an update operation, which "update the set, adding elements from all others." (In essence, a ∪= augmented assignment.) Even if you wish to roll your own solution, the library implementations may offer suggestions for optimization. -- ToE 20:31, 12 July 2016 (UTC)

• Assuming ToE did represent correctly the problem, a solution that is optimal in time (but disastrous in space) would be to read all the lists once to find the maximum value $n_{max}\leq n$, then create a boolean array of $n_{max}$ elements set to false, then read all the list and flip the bits in the array when the value is found. If you know $n$ you do not even need to do the first pass, and there is no way to do it more efficiently (you will eventually have to read all the elements of all the lists, and practically, the bit-flipping is easy to vectorize in many programming languages such as Python). Returning the sorted union is trivial then.

The downside of course is that if there is reason to suspect that $n_{max}$ is much larger than the size of the final list, the array will be much bigger than it would need to be by other methods. Tigraan^{Click here to contact me} 09:01, 12 July 2016 (UTC)

uninstall Yahoo! powered

Yesterday I installed an update of Freemake Video Downloader. It seemed to install a browser hijacker. It installed "Yahoo! Powered" and I can't uninstall it. Is that a browser hijacker? Bubba73 ^{You talkin' to me?} 16:44, 11 July 2016 (UTC)

Well, I ran Revo Uninstaller and that seemed to get it. Bubba73 ^{You talkin' to me?} 19:22, 11 July 2016 (UTC)

That would be Search.yahoo.com browser hijacker -- There are instructions for manual removal which are somewhat complicated (depending on OS & browser). This seems to be the most recent and comprehensive:[5]. --2606:A000:4C0C:E200:A073:98E5:BA6B:E905 (talk) 23:35, 12 July 2016 (UTC)

July 12

Would the graphics card R9 390X or gtx1070 fit on the motherboard GA-H81M-S1?

Would the graphics card R9 390X or gtx1070 fit on the motherboard GA-H81M-S1?201.79.70.197 (talk) 12:19, 12 July 2016 (UTC)

It should since those you mentioned are standard PCIe video cards anyway, but what matters more is if they fit inside the case you're using. Blake Gripling (talk) 12:57, 12 July 2016 (UTC)

And you want to make sure it works with the power supply. The GTX1070 is rated around 150W, so it should work easily with any power supply that is at least 500W. The R9 390X is rated around 250W. So, you will want 600+W. Because the price isn't much different between 500 and 600W power supplies, I usually get more than I need. If you are using a small power supply (like a 350W that comes free with many cases), you will have issues with using either card. 209.149.113.4 (talk) 14:10, 12 July 2016 (UTC)

Oh Yes, I know about the power supply thing, and I will buy one if needed. Anyway my current Graphics card is gt 730. Thanks for the help 201.79.78.164 (talk) 16:53, 12 July 2016 (UTC)

Movie maker sought

What is highly praised?

Requirement is, something that can make me select points (from and to) of a video clip, more than one time in one go… – An easy to use tool is sought. A portable software is desirable, however…

Apostle (talk) 18:11, 12 July 2016 (UTC)

Password Protector; Lockdown protection sought

Peeps, what’s highly praised? Btw, I possess Folderlock (currently installed; this provided Bitlocker – I believe bitlocker is unbreakable, and good, unless you guys re-correct me), Keepass (Portable acquired, and wired), AxCrypt (needs internet connection)... -- Apostle (talk) 18:17, 12 July 2016 (UTC)