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### Author Topic: Secret of Mana, Turbo - Beta 210915  (Read 619299 times)

#### hmsong

• Hero Member
• Posts: 678
##### Re: Secret of Mana, Turbo - Beta 200401
« Reply #1280 on: April 06, 2020, 06:44:29 am »
Thanks for letting me use the Thunder Saber and Poison Flower.

Oh, so 0x0C is the Mandala shop?  So the problem will be solved if I changed that?  Also, what's the value for other shops?  How do I find out?

Oh my.  I didn't know the existence of Mop's somedit.  Or maybe you already told me, but it didn't work for me in my pre-reformat computer.  In any case, I shall play around with it to see what it's capable of.  What's the most recent version of that?  I looked at the file, and it's dated 2014.  I don't know if there's a newer version, but I can't seem to find it.

#### Queue

• Hero Member
• Posts: 613
##### Re: Secret of Mana, Turbo - Beta 200401
« Reply #1281 on: April 06, 2020, 03:21:44 pm »
Shop data is in bank 0xD8, here's a quick copy and paste from someone else that probably shows which shop is which number:
Code: [Select]
[PointerTable_ShopData]D8/FC32: 52FC                             ;[00: ShopData_Potos]D8/FC34: 58FC                             ;[01: ShopData_Nekos]D8/FC36: 64FC                             ;[02: ShopData_Pandora_01]D8/FC38: 6BFC                             ;[03: ShopData_Kippo]D8/FC3A: 73FC                             ;[04: ShopData_DwarfVillage]D8/FC3C: 7FFC                             ;[05: ShopData_Pandora_02]D8/FC3E: 89FC                             ;[06: ShopData_MoogleNeko]D8/FC40: 95FC                             ;[07: ShopData_Matango]D8/FC42: 9FFC                             ;[08: ShopData_Kakkara]D8/FC44: A9FC                             ;[09: ShopData_TodoVillage]D8/FC46: B3FC                             ;[0A: ShopData_Southtown]D8/FC48: C1FC                             ;[0B: ShopData_Northtown]D8/FC4A: CEFC                             ;[0C: ShopData_Mandala]D8/FC4C: D7FC                             ;[0D: ShopData_GoldIsle]D8/FC4E: E3FC                             ;[0E: ShopData_Tasnica]D8/FC50: F0FC                             ;[0F: ShopData_TreePalaceNeko]
You found the most recent version, they stopped working on it. Mop's modern project has been the Secret of Mana Randomizer.

#### hmsong

• Hero Member
• Posts: 678
##### Re: Secret of Mana, Turbo - Beta 200401
« Reply #1282 on: April 06, 2020, 11:35:49 pm »
Hmm.

It worked (for mountain Neko, at least), but... it doesn't seem to work for Reloc.  Meaning, there's a separate code editing that I need to do for Reloc version.  I'm guessing it's different Event number (so I need to trace that event number, and change the shopping code there).

Do you have documents that compiles all the codes for Relocalized? (similar to "Disasm - Enker, Regrs, zhaDe.zip" thing)  That would help out immensely.

Oh man, RHDN is rather anal about the hacks.  I need to have at least one screenshot, and the initial document is picky about the words I use (ex: I can't use "I", "my", "me", etc).  What a pain.

#### Queue

• Hero Member
• Posts: 613
##### Re: Secret of Mana, Turbo - Beta 200401
« Reply #1283 on: April 07, 2020, 01:19:48 am »
Code: [Select]
@OFF $CA8962'RAW 06032D08\event \lock \gather \camera_center'RAW 50\event \text_open'RAW A8 9F A5 A9 C5AA 9592929292C0808994C29380838F8C84C87FTEXT ^N ^E ^K ^O : ^P u r r r r , _ i t  s _ c o l d ! \n'RAW 80AD 81968580998F95928087818D85CA80A9 92CATEXT _ ^S a v e _ y o u r _ g a m e ? _ ^O r ?'RAW 2800\event \wait_input'RAW 7FTEXT \n'RAW 5903\event \multispace= 03'RAW CBTEXT ('RAW 58\event \text_opt_start'RAW 5A05\event \text_opt_define= 05'RAW AD 819685TEXT ^S a v e'RAW 5A0C\event \text_opt_define= 0C'RAW 9C 9599TEXT ^B u y'RAW 5A12\event \text_opt_define= 12'RAW AD 858C8C80CCTEXT ^S e l l _ )'RAW 5B\event \text_opt_end=='RAW 1000\event \goto 000'RAW 1050\event \goto 050'RAW 1308\event \goto 308'RAW 1315\event \goto 315'RAW 00\event \endEvent 0x308 is where the shop number is set, but that event is also used by the Todo shopkeeper, so changing event 0x308 wouldn't be a good approach. The ideal probably would be to reorganize this event (0x65F) so that it sets event flag 0xFA, and calls event 0x312 instead of 0x308, so that this event has direct control over the shop number. There are lots of possible solutions, and no absolute "correct" way to do it. « Last Edit: April 10, 2020, 07:58:56 pm by Queue » #### hmsong • Hero Member • Posts: 678 ##### Re: Secret of Mana, Turbo - Beta 200408 « Reply #1288 on: April 08, 2020, 09:13:51 pm » Wow, modifying that Ice Country shop was complicated. Holy crap. Lots of trial and error (and half the time, I didn't even know what I was doing). But I got it. I think. I hope I didn't screw up anything. Could you check if I screwed up somewhere? Code: [Select] '[Event 318 - Mountain Neko]@OFF$C9CD58\event \flag== FA 01 ' Initial Neko Shop'[Event 65F - Ice Country Neko]@OFF $CA8962'\event \lock \gather \camera_center'\event \text_open'TEXT ^N ^E ^K ^O : ^P u r r r r , _ i t  s _ c o l d ! \n'TEXT _ ^S a v e _ y o u r _ g a m e ? _ ^O r ?'\event \wait_input'TEXT \n\event \flag== FA 01 ' Initial Neko Shop\event \lock \gather \camera_center\event \text_openTEXT ^N ^E ^K ^O : ^P u r r r r , _ i t  s _ c o l d ! \nTEXT _ ^S a v e _ y o u r _ g a m e ? _ ^O r \n@OFF$CA89B2 ' Ice Country Neko Shop Pointer\event \goto 312 ' Call Shop ListIF [Text]IF Relocalized'[Event 318 - Mountain Neko]@OFF $E20AB8\event \flag== FA 01 ' Initial Neko Shop'[Event 65F - Ice Country Neko]@OFF$E2E2A6'\event \lock \gather \camera_center'\event \text_open'TEXT ^N e k o : ^P u r r r r , _ i t  s _ c o l d ! \n'TEXT _ ^S a v e _ y o u r _ g a m e ? _ ^O r ?'\event \wait_input'TEXT \n\event \flag== FA 01 ' Initial Neko Shop\event \lock \gather \camera_center\event \text_openTEXT ^N e k o : ^P u r r r r , _ i t  s _ c o l d ! \nTEXT _ ^S a v e _ y o u r _ g a m e ? _ ^O r \n@OFF $E2E2FB ' Ice Country Neko Shop Pointer\event \goto 312 ' Call Shop ListENDIF x2 The problem was, I needed some space to call the shop, and I wasn't sure which codes I can get rid of (I feared I would have to change the Neko text). Also, I couldn't find anything for VWF, so I'm guessing it uses vanilla. I hope it doesn't screw around with [Ice_Country_Neko_Landing] or any other patches... If everything checks out, could you upload this into your package? I'll think of the description and post it soon. Call it... "Some Neko Shop Changes"? Maybe you can think of a better name, since I suck at naming things. EDIT: Do you think it would be a better idea for Mountain Neko to sell Tasnica items, instead of initial Neko items? That way, players would have early access of Tasnica shop (without having to clear all the element dungeons, although for double price). But of course, that also means Neko won't be selling the Barrel (only the initial and final Neko sells Barrel, as far as I know). And the only item unique to Tasnica is Gold Bracelet (for vanilla), which is weaker than the items sold in Gold Isle. EDIT2: Oh crap. I forgot to not capitalize the Neko name for Relocalized. Fixed. Also, must I put "\flag== FA 01" in the beginning, before dialog box opens, or can I put it anywhere before the "Buy" button? If I can put it anywhere, then I can just put it after the Neko talk (hence not messing with his dialog). « Last Edit: April 09, 2020, 03:31:12 am by hmsong » #### Queue • Hero Member • Posts: 613 ##### Re: Secret of Mana, Turbo - Beta 200408 « Reply #1289 on: April 09, 2020, 06:33:30 am » Once it appears that this patch is finalized (like, stuff works as intended, has necessary compatibility fixups, etc.), yes, I'd be happy to include it. Okay, so first off, VWF needs its own fixup: it doesn't use any vanilla events. Just like Relocalized, its events are in banks 0xE1 and 0xE2 (or rather, Relocalized is just like VWF). You can search for "10 00 10 50" to find event 65F in VWF's event data. VWF's text is encoded gibberish, but you'll be able to find 28 00 7F, so can just assume whatever byte comes before it is the question mark. As to your solution, probably that'll work, replacing: \event ? \wait_input \n with: \event \n \flag== FA 01 I'll need to go try it to see it in action; obviously the result is no more pause after Neko asks "Or?" and no more question mark. Nothing to worry about with Ice_Country_Neko_Landing; it doesn't modify events. I don't recall anything else touching event 65F for that matter. I've been pretty good about labeling events, so you can usually search the Turbo ZPS for "Event ###" and if there's no match, you're probably the first one modifying it. Kethinov's probably the only one who would, and I don't recall him having altered this one. I think a feature name that includes the word "copycat" would be a fun way to name this patch. Something like "Neko Not a Copycat" or "Neko No Longer Copycats". Hm, I can't find a phrase that I totally like. Re: EDIT: Tasnica sounds like a more interesting choice to me, but I imagine I'd have to playtest it to really get a feel for the result. Nobody uses Barrels, so no big loss there, and the Tasnica shop has high potential to be missed, so this gives it a second shot at being used. Just having Neko not repeat a shop that's 3 screens away seems like a good result though, so practically any shop would be an improvement. Re: EDIT2: You can put \event \flag== FA 01 anywhere before \event \text_opt_end== (because it has to come before \event \goto 312 (buy) and \event \goto 315 (sell), and to not break the selection menu, that means before \event \text_opt_end==). So yes, you could put it directly in place of \event ? \wait_input. #### hmsong • Hero Member • Posts: 678 ##### Re: Secret of Mana, Turbo - Beta 200408 « Reply #1290 on: April 09, 2020, 09:10:05 am » Okay, all tested and done. Works great. Thanks. Copycat is a great name. How is "Neko_is_not_a_Copycat"? I was also thinking, "Neko's Better Business-Sense". If you can think of a better name, feel free to use that instead. Description: Quote "(Patch)\n Author: hmsong\n Version: 0.0\n \n In the original game, some of the Neko shops sold items that really weren't benefitial, because the town shops (only few screens away) sold the exact same things for half the price. In short, those Neko shops were completely meaningless (except for the extremely rare situations).\n \n This hack changes those shop contents, so that they'd be more useful (maybe):\n \n - Ice Country Neko --> Initial Neko shop (sells all consumables, such as Royal Jam and Barrel)\n - Mountain Neko --> Tasnica shop (early access to Tasnica shop, although for double the price)\n Code: [Select] '[Event 318 - Mountain Neko]@OFF$C9CD58\event \flag== FA 0E ' Tasnica Shop'[Event 65F - Ice Country Neko]@OFF $CA8992'TEXT ?'\event \wait_input'TEXT \nTEXT \n\event \flag== FA 01 ' Initial Neko Shop@OFF$CA89B2 ' Ice Country Neko Shop Pointer\event \goto 312 ' Call Shop ListIF [Text]IF VWF_Edition'[Event 318 - Mountain Neko]@OFF $E20A4F\event \flag== FA 0E ' Tasnica Shop'[Event 65F - Ice Country Neko]@OFF$E2E205TEXT \n\event \flag== FA 01 ' Initial Neko Shop@OFF $E2E22A ' Ice Country Neko Shop Pointer\event \goto 312 ' Call Shop ListENDIF x2IF [Text]IF Relocalized'[Event 318 - Mountain Neko]@OFF$E20AB8\event \flag== FA 0E ' Tasnica Shop'[Event 65F - Ice Country Neko]@OFF $E2E2D6'TEXT ?'\event \wait_input'TEXT \nTEXT \n\event \flag== FA 01 ' Initial Neko Shop@OFF$E2E2FB ' Ice Country Neko Shop Pointer\event \goto 312 ' Call Shop ListENDIF x2
I'm probably gonna need those FILE <VWF> and FILE <Reloc> thing, since if both VWF and Reloc are enabled, VWF codes may overwrite some critical stuff from Reloc.  I'll leave that part to you, since I don't know which section this patch will go under.  Thanks.

And while I'm at it, do you think I should replace those temp Nekos shops in Gaia Cave and Haunted Forest with the initial Neko shop? (to not copy Dwarf shop that's nearby)  It shouldn't be a problem, probably.

Btw, I've been trying to use somedit (mostly map), but I don't know which map number is which.  Is there a way to find that out the map number by looking at the 0x08 bank document?
« Last Edit: April 09, 2020, 09:55:42 am by hmsong »

#### StarWyvern

• Full Member
• Posts: 113
• Family first, then gaming and music.
##### Re: Secret of Mana, Turbo - Beta 200408
« Reply #1291 on: April 09, 2020, 03:17:02 pm »
Version 2020-04-08:
https://ufile.io/o01kij4z

Changes:
- Updated hmsong's Enemy_Stat_Changes to v0.4
- Finished first complete version of Enhanced_World_Map; the flat world map is now dynamic
- Forgot to remove the (Preview) label from Enhanced_World_Map, whoops

Technical Changes:
- Added aliases for ADR.BANK, ADR.HIGH and ADR.LOW as ADR.BK, ADR.HI and ADR.LO, respectively, and replaced all prior uses of the longer versions

StarWyvwern, unfortunately, this Turbo project and the Randomizer are not compatible; both are huge projects that drastically change the ROM in incompatible ways.

There's probably a small subset of Turbo features that could be applied to a Randomizer patched ROM, but that would be an ill-advised pursuit.

hmsong, when you say "without changing the script" you mean without altering any of the text, I assume? Just wanted to verify that's what you meant.

$CA8962 is the correct address for event 0x65F. Unfortunately, that event is organized subtly differently from 0x318 making it harder to do what you want. The timing for gold display opening and closing, moving the players around, etc. is organized differently. I'll deconstruct it here: Code: [Select] @OFF$CA8962'RAW 06032D08\event \lock \gather \camera_center'RAW 50\event \text_open'RAW A8 9F A5 A9 C5AA 9592929292C0808994C29380838F8C84C87FTEXT ^N ^E ^K ^O : ^P u r r r r , _ i t  s _ c o l d ! \n'RAW 80AD 81968580998F95928087818D85CA80A9 92CATEXT _ ^S a v e _ y o u r _ g a m e ? _ ^O r ?'RAW 2800\event \wait_input'RAW 7FTEXT \n'RAW 5903\event \multispace= 03'RAW CBTEXT ('RAW 58\event \text_opt_start'RAW 5A05\event \text_opt_define= 05'RAW AD 819685TEXT ^S a v e'RAW 5A0C\event \text_opt_define= 0C'RAW 9C 9599TEXT ^B u y'RAW 5A12\event \text_opt_define= 12'RAW AD 858C8C80CCTEXT ^S e l l _ )'RAW 5B\event \text_opt_end=='RAW 1000\event \goto 000'RAW 1050\event \goto 050'RAW 1308\event \goto 308'RAW 1315\event \goto 315'RAW 00\event \endEvent 0x308 is where the shop number is set, but that event is also used by the Todo shopkeeper, so changing event 0x308 wouldn't be a good approach. The ideal probably would be to reorganize this event (0x65F) so that it sets event flag 0xFA, and calls event 0x312 instead of 0x308, so that this event has direct control over the shop number. There are lots of possible solutions, and no absolute "correct" way to do it.
Damn.  It would've been great to make a version of this hack that changes everytime I restart it. Maybe in the future...
Insert "Witty Text" here.

#### reconstructingmana

• Full Member
• Posts: 156
##### Re: Secret of Mana, Turbo - Beta 200408
« Reply #1292 on: April 10, 2020, 01:27:14 am »
Hi, Queue.  I received your message; thank you for getting in touch.

I'm happy to contribute.  Some things to know:

1)The retranslations were done on a spreadsheet on my hard drive that were then transferred to a different spreadsheet that zhaDe added tags to and somehow imported into the ROM.  I will have to locate the former; it has been some years.
2)Google Translate has changed their algorithm since the work was done.  Instead of translating word by word, they give a dynamic translation taking all the text into account.  Ironically, this makes finding deeper meaning in the text more difficult.  I can get around this by treading carefully and using other tools, but it will take a bit longer.  There are also some phrases that can not be translated, and hidden nuances that only a native speaker would know.  Here's a good example:

Quote
My translation: The Empire has been trying to spy here.  Just in case, we can't let anyone through.  Try back again later.

Fluent speaker's translation: We've obtained information that the empire is attempting [or "planning"...] to send a spy [or spies].

Just to be safe, we cannot allow anyone to pass through here for now. I'm sorry, but please leave.

As you can see, my translation is perfectly fine, but the fluent speaker's has a lot more detail.

3)I would likely want to rewrite what I did and take more time to make it better.  I have searched in vain for a script dump of the 2018 version to compare it to, but have not been successful.  We could really put together something nice if we can find one.

Let me know which step you'd like me to take next.
« Last Edit: November 18, 2020, 06:22:56 pm by ManaRedux »

#### budtzcbs@gmail.com

• Newbie
• Posts: 1
##### Re: Secret of Mana, Turbo - Beta 200408
« Reply #1293 on: April 10, 2020, 04:52:11 am »
Thx for this wonderful hack

Is there a way to upgrade to a new version without going through the entire thing and selecting and deselecting stuff all over again?

Could you provide some insigt into the backgound of the balance changes: monsters, spell overhaul ect. For instance the balance changes where monsters grant more exp in start of game. Hp changes. weaknes changes ect.

thx again

#### hmsong

• Hero Member
• Posts: 678
##### Re: Secret of Mana, Turbo - Beta 200408
« Reply #1294 on: April 10, 2020, 10:19:41 am »
@budtzcbs@gmail.com

Hmm.  The patch should already provide enough explanations (each with their own explanations).  If you select the options you want, it will have a brief description of what the patch does (and sometimes, why it was created).  I'm not sure what you mean by "balance changes", since that can mean many different things.

Assuming it's one of the Enemy Behavior/Stat/Element changes (my creations), I created them because I wanted enemies to use their attacks a bit more smartly.  For example, I wanted Turtlance (water type) to use Ice Saber on himself, so that he can get the extra DEF boost.  I also didn't want enemies to do something stupid, like Wolf Lord casting Cure Water on himself when he's at full health.  I also wanted enemies to have more varieties.  For example, the original game severely lacked Fire, Moon, or Light enemies, so I changed things to balance things out.

As for Spell Balance Overhaul, I wanted the spells to be as powerful as their MP cost (unless the spell had a special effect).  In vanilla (aka original game), some 4MP spells were only as powerful as 2MP spells of different element (ex: Thunder or Exploder were only as strong as Freeze or Dark Force).  Hell, Primm's Lucent Beam was 8MP, and it was only as powerful as Freeze (2MP).  It was ridiculous.

As for Enemy Reward Changes (this is a part of Enemy Stat Changes that I released on .NET), I changed things so that you would get extra experience in the beginning quarter of the game, based on how much exp is necessary to match the level of enemies that you're fighting (in vanilla, you'd have to grind for experience significantly more from Witch's castle and on).  However, that hack made it so that the extra exp you get from enemies becomes less and less as you progress the game, and by the time you get to Northtown Ruins, you'd get no extra exp from the enemies, and the grinding values will match the values of what was in vanilla.
« Last Edit: April 10, 2020, 10:29:57 am by hmsong »

#### LightBaneX

• Jr. Member
• Posts: 8
##### Re: Secret of Mana, Turbo - Beta 200408
« Reply #1295 on: April 10, 2020, 10:20:53 am »
Hello! Really like this mod. I have been playing around with it a bit today. I really love the removal of the stamina bar. And the inclusion of the blocking mechanic. It made the first boss pretty interesting! I also love the easily configurable zps patcher it uses.

One question, is it possible in this mod to change how long stuns last on enemies when you slap them around with your weapon? I noticed I can just walk up to an enemy and stunlock them by spamming attack and they can never hurt me. I see Quick Recovery under the Combat category in the zps patcher, but I am looking for something that is just for the enemies, quick recovery seems to be primarily for players. Maybe it's just these beginning enemies. But all the enemies so far are just pushovers you can stunlock. Or maybe add a few seconds of stun immunity after they recover from one so they can fight back a little bit.

Anyways fantastic work here! I am having a really fun time, thanks for doing this and sharing it with the community! I can't wait to get undine and try out the quickspell feature. I am also going to try and see if I can get my friend to play this with me

#### kethinov

• Full Member
• Posts: 146
##### Re: Secret of Mana, Turbo - Beta 200408
« Reply #1296 on: April 10, 2020, 10:28:30 am »
Nothing to worry about with Ice_Country_Neko_Landing; it doesn't modify events. I don't recall anything else touching event 65F for that matter. I've been pretty good about labeling events, so you can usually search the Turbo ZPS for "Event ###" and if there's no match, you're probably the first one modifying it. Kethinov's probably the only one who would, and I don't recall him having altered this one.

I have a hack in progress that modifies the text of that event, but does not change its length. It only modifies the "Purrrr, it's cold" line in a byte-neutral way and changes nothing else about the event.

#### StarWyvern

• Full Member
• Posts: 113
• Family first, then gaming and music.
##### Re: Secret of Mana, Turbo - Beta 200408
« Reply #1297 on: April 10, 2020, 01:35:32 pm »
Cannot download latest update. Page keeps giving me the same message. "Your IP address does not match the download token. Click download to generate a new one." Just keeps looping to the same message. Sites who pull shady crap like that should be burned. Advertising free file storage but not allowing access without a paid account.
Insert "Witty Text" here.

#### Queue

• Hero Member
• Posts: 613
##### Re: Secret of Mana, Turbo - Beta 200408
« Reply #1298 on: April 10, 2020, 07:14:17 pm »
hmsong, yes, I think you should consider changing any Neko that's even remotely near an identical shop. I don't know if anyone ever buys his stuff, just uses him as a save point, so trying to give his shop value seems worthwhile to me.

There's no exhaustive document naming each map; what's in the Bank08 disassembly text file is the best I'm aware of.

budtzcbs, when you use the patcher and apply the patch, it outputs a "SoM Turbo.######.ini" file. If you make the INI file match the filename of the ZPS file, it'll use the settings in the INI file.

So, if you had say, "SoM Turbo.200401.zps" and got settings how you wanted, and then applied the patch, you'd have a "SoM Turbo.200401.ini" file. Then if you downloaded "SoM Turbo.200408.zps", before opening the patcher, if you rename "SoM Turbo.200401.ini" to "SoM Turbo.200408.ini" and then open the patcher and load "SoM Turbo.200408.zps" it'll load the same settings you set for that previous patch (new features added will be at their defaults).

The other option is to rename the ZPS file to remove the date before you use the patch (for example, rename "SoM Turbo.200408.zps" to "SoM Turbo.zps"), so that your INI file is always just "SoM Turbo.ini".

hmsong made most, if not all, of the balance changes you happened to name. For anything I did / do, my changes are always in the name of fun: I consider this game incredibly easy, and there's no silver bullet to make it challenging and not tedious (and even when it is made tedious, I don't find it challenging), so I tend to focus on changes that reduce degenerate gameplay (simple examples of degenerate gameplay are spamming spells to lock down a boss or using the weapon overcharge glitch: both cause you to spend lots of time in menus when playing a game that is designed for you to be actively driving a little dude around swinging a weapon).

StarWyvwern, I suspect that file hosting site isn't doing that maliciously, but just having trouble, either on their end, or in their site functioning on your choice of browser. I found another file host I'll try for a bit. I'll update the download links soon after I post this.

reconstructingmana, great, whew, glad that message got through.

1) I can work with whatever format for the data you go with. Data processing isn't exactly my strong suit, but I'll manage. When I read about what you'd done, I just really didn't want the effort to go to waste if possible.

2) D'oh, well I'm in no hurry; I've been working on this Turbo project for over a year and haven't run out of steam yet.

3) Here's a dump of the 3D remake's English text:
https://filebin.net/7spaiz2xfys6b2i1/SoM18.textdump.zip
As far as I know, that's every bit of text that's not baked into an image.

As for what next step to take, it depends on what you feel up to: whether you want to use the second spreadsheet, or find the original, how much you want to rewrite, how much text from the remake you want to match up with the original (I have no idea how the text you were working off of was organized, and the remake textdump doesn't look especially organized to me)...

Questions:
- Do you want to stick with the vanilla fixed-width font? It allows for 28 characters per line (and 3 simultaneous lines), which is easy to plan for when formatting text for the game's textboxes.
- Do you want to use a variable-width font (VWF)? This requires either the creation of a specialized tool to handle previewing the text outside of the game for formatting purposes (of finding a pre-existing one that can be adapted for SoM), or in-game testing (which can vary from minorly to majorly intensive).

I bring that choice up because I would need / expect the text formatting to be done by you (or anyone working on the text), since it has an impact on how readable the result is and is best adjusted by the person writing it. I would prefer VWF, but that will require more work on your end formatting the text to fit in the game's textboxes (at least if we want it to look nice).

The next step for me is to move forward with "script insertion" prep, which I'm committing to since you got in touch. I need to essentially parse vanilla and Relocalized events into the format expected by this ZPS format (currently the latter is just a blob of raw bytes) and get that patching in byte-for-byte identical to how it is now.

Getting me an example formatted in your preference of data format (spreadsheet, text file, word document, fuzzy images of a cellphone screen, etc.) sooner rather than later would let me start figuring out how I'm going to process your new text for insertion.
« Last Edit: April 10, 2020, 07:58:25 pm by Queue »

#### hmsong

• Hero Member
• Posts: 678
##### Re: Secret of Mana, Turbo - Beta 200408
« Reply #1299 on: April 10, 2020, 07:28:03 pm »
@kethinov

My Neko shop modification actually changes one of the letters in the "Purrrr, it's cold" thing.  Specifically, it changes the last part -- "Or?" to "Or" (without the "?").  Would that affect things?

Also, for your Luka's Pet, as I emailed you earlier, I think you can change Luka's script.  I came up with this:

Code: [Select]
@OFF $C942A5 '(need 63 characters)TEXT ^L u k a : ^N o w , _ p l e a s e _ r e s t o r e _ t h e \nTEXT _ s e e d . _ ^Y o u _ c a n _ d o _ i t ! \nTEXT _ _ _ _ _ _ _ _ _ _ _ _IF [Text]IF Relocalized'sig 42 11EE 1 13F 50'[Event 136: Luka]@OFF$E149B0\event \if_flag== 11 EF@OFF $E14F6D '(need 63 characters)TEXT ^L u k a : ^N o w , _ p l e a s e _ r e s t o r e _ t h e \nTEXT _ s e e d . _ ^Y o u _ c a n _ d o _ i t ! \nTEXT _ _ _ _ _ _ _ _ _ _ _ _ENDIF x2 That way, it'll fit both for when you restore the seed after Jabberwocky, and during the Sunken Continent/Pure Land. @Queue Okay then. Here's some addition to the patch: Description: Quote "(Patch)\n Author: hmsong\n Version: 0.0\n \n In the original game, some of the Neko shops sold items that really weren't benefitial, because the town shops (only few screens away) sold the exact same things for half the price. In short, those Neko shops were completely meaningless (except for the extremely rare situations).\n \n This hack changes those shop contents, so that they'd be more useful (maybe):\n \n - Neko (Gaias Navel, Haunted Forest, Witch's Castle) --> Initial Neko shop (sells all consumables, such as Chocolate, Royal Jam, Faerie Walnut, and Barrel)\n - Ice Country Neko --> Initial Neko shop (sells all consumables, such as Royal Jam and Barrel)\n - Mountain Neko --> Tasnica shop (early access to Tasnica shop, although for double the price)\n Code: [Select] '[Event 1DF - Various Nekos in the Beginning]@OFF$C98144'TEXT ^N ^E ^K ^O : ^M e o w , _ b u s y _ t r a v e l e r s ! \n\event \flag== FA 01 ' Initial Neko ShopTEXT ^N ^E ^K ^O : ^I _ h a v e _ i t e m s - n y a a ! \n@OFF $C98195\event \goto 312 ' Call Shop ListIF [Text]IF VWF_Edition'[Event 1DF - Various Nekos in the Beginning]@OFF$E197AB'TEXT ^N e k o : ^M e o w , _ b u s y _ t r a v e l e r s ! \n\event \flag== FA 01 ' Initial Neko ShopTEXT ^N e k o : ^I _ h a v e _ i t e m s - n y a a ! \n ------------------ I'm pretty sure this doesn't work.@OFF $E19801\event \goto 312 ' Call Shop ListENDIF x2IF [Text]IF Relocalized'[Event 1DF - Various Nekos in the Beginning]@OFF$E19810'TEXT ^N e k o : ^M e o w , _ b u s y _ t r a v e l e r s ! \n\event \flag== FA 01 ' Initial Neko ShopTEXT ^N e k o : ^I _ h a v e _ i t e m s - n y a a ! \n@OFF \$E19866\event \goto 312 ' Call Shop ListENDIF x2
The problem is, I don't know what to do with VWF, because the letters are all weird.  Hmm.  Any idea?  Or can I still use TEXT thing, and VWF that you created will correct itself?  I know it the text starts at E197AB (and I can change the event to 312 at E19801).

Btw, do you know when[Event 1E6] happens?  I don't remember that script at all.  Or is it one of those unused scripts?
« Last Edit: April 10, 2020, 10:29:22 pm by hmsong »


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# Gravitational and electric fields

## Homework Statement

By considering the centripetal force which acts on a planet in a circular orbit, show that T^2 is proportional to R~^3 where T is the time taken for one orbit around the sun and R is the radius of the orbit

## Homework Equations

I'm thinking keplers third law?

## The Attempt at a Solution

C=2piR

P=2piR/ sqrt*GMm/r

rearranged to give P = 2pi sqrt*R^3/GM

Lost now..

P=2piR/ sqrt*GMm/r

Where did that expression come from?

Find an expression for the radial acceleration in terms of the orbital speed and R.

Then use Newton's second law.

format: plain


text:
I am multiplying radicals and I have a question like this:

square root of ((6x^7)(y^3)) multiplied by square root of ((7x^3)(y^-1))

In the end I got x^(5)y(square root of 42)

Does this look right at all?

2. Originally Posted by cjh824
I am multiplying radicals and I have a question like this:

square root of ((6x^7)(y^3)) multiplied by square root of ((7x^3)(y^-1))

In the end I got x^(5)y(square root of 42)

Does this look right at all?
yes, that is fine..

3. Hello, cjh824!

Simplify: . $\sqrt{6x^7y^3}\cdot\sqrt{7x^3y^{-1}}$

In the end I got: . $x^5y\sqrt{42}$ . . . . Right!

$\sqrt{6x^7y^3}\cdot\sqrt{7x^3y^{-1}} \;=\;\sqrt{(6x^7y^3)(7x^3y^{-1})}$

. . $=\;\sqrt{42x^{10}y^2} \;=\;\sqrt{42}\cdot\sqrt{x^{10}}\cdot\sqrt{y^2} \;=\;x^5y\sqrt{42}$

4. in addition, it should be true in the assumption that x and y are non negative or both negative..

simply stated, x and y should have the same sign..


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# FIN 419 Chapter 4 P4–48 Monthly loan payments Tim Smith is shopping for a used car. He has found

 FIN 419 Chapter 4 P4–48 Monthly loan payments Tim Smith is shopping for a used car. He has found FIN 419: Finance for Decision making FIN 419 C4 P4–48 Tim Smith is shopping for a used car. He has found one priced University of Phoenix Chapter 4 P4–48 Monthly loan payments Tim Smith is shopping for a used car. He has found one priced at $4,500. The dealer has told Tim that if he can come up with a down payment of$500, the dealer will finance the balance of the price at a 12% annual rate over 2 years (24 months). a. Assuming that Tim accepts the dealer’s offer, what will his monthly(end-of month) payment amount be? b. Use a financial calculator or Equation 4.15a (found in footnote 9) to help you figure out what Tim’s monthlypayment would be if the dealer were willing to finance the balance of the car price at a 9% annual rate. page 176 footnote 9

Price of car=$4,500 After downpayment of$500 4000 should be charged with interest of 12% a)...


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text:
## College Physics (4th Edition)

(a) The required torque is $~48.1~N \cdot m$ (b) Each child must push with a force of $19.2~N$
(a) Let $M_m$ be the mass of the merry-go-round and let $M_c$ be the mass of each child. We can find the rotational inertia of the system: $I = \frac{1}{2}M_mR^2+2M_cR^2$ $I = \frac{1}{2}(350.0~kg)(1.25~m)^2+(2)(30.0~kg)(1.25~m)^2$ $I = 367.2~kg~m^2$ We can express the final angular speed in units of rad/s: $25~rpm \times \frac{2\pi~rad}{1~rev} \times \frac{1~min}{60~s} = 2.618~rad/s$ We can find the angular acceleration: $\omega_f = \omega_0+\alpha~t$ $\alpha = \frac{\omega_f - \omega_0}{t}$ $\alpha = \frac{2.618~rad/s-0}{20.0~s}$ $\alpha = 0.1309~rad/s^2$ We can find the required torque: $\tau = I~\alpha$ $\tau = (367.2~kg~m^2)(0.1309~rad/s^2)$ $\tau = 48.1~N \cdot m$ The required torque is $~48.1~N \cdot m$ (b) We can find the force $F$ with which each child must push: $2~(R \times F) = \tau$ $F = \frac{\tau}{2R}$ $F = \frac{48.1~N \cdot m}{(2)(1.25~m)}$ $F = 19.2~N$ Each child must push with a force of $19.2~N$.


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## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 1 - Linear Functions - 1.4 Intercepts and Graphing - 1.4 Exercises - Page 63: 21

#### Answer

$x_{intercept} = (4,0)$ $y_{intercept} = (0,2)$

#### Work Step by Step

To find the intercepts, one simply equates the opposite variable to zero and solve the equation. In other words, to find the $x_{intercept}$: $$2x + 4(0) = 8$$ $$2x = 8$$ $$x = 4$$ And to find the $y_{intercept}$: $$2(0) + 4y = 8$$ $$4y = 8$$ $$y = 2$$. We plot the points on a graph and draw the line that connects them.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
Lemma 15.9.5. Let $A$ be a ring, let $I \subset A$ be an ideal. Let $f \in A[x]$ be a monic polynomial. Let $\overline{f} = \overline{g} \overline{h}$ be a factorization of $f$ in $A/I[x]$ such that $\overline{g}$ and $\overline{h}$ are monic and generate the unit ideal in $A/I[x]$. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a factorization $f = g' h'$ in $A'[x]$ with $g'$, $h'$ monic lifting the given factorization over $A/I$.

Proof. We will deduce this from results on the universal factorization proved earlier; however, we encourage the reader to find their own proof not using this trick. Say $\deg (\overline{g}) = n$ and $\deg (\overline{h}) = m$ so that $\deg (f) = n + m$. Write $f = x^{n + m} + \sum \alpha _ i x^{n + m - i}$ for some $\alpha _1, \ldots , \alpha _{n + m} \in A$. Consider the ring map

$R = \mathbf{Z}[a_1, \ldots , a_{n + m}] \longrightarrow S = \mathbf{Z}[b_1, \ldots , b_ n, c_1, \ldots , c_ m]$

of Algebra, Example 10.143.12. Let $R \to A$ be the ring map which sends $a_ i$ to $\alpha _ i$. Set

$B = A \otimes _ R S$

By construction the image $f_ B$ of $f$ in $B[x]$ factors, say $f_ B = g_ B h_ B$ with $g_ B = x^ n + \sum (1 \otimes b_ i) x^{n - i}$ and similarly for $h_ B$. Write $\overline{g} = x^ n + \sum \overline{\beta }_ i x^{n - i}$ and $\overline{h} = x^ m + \sum \overline{\gamma }_ i x^{m - i}$. The $A$-algebra map

$B \longrightarrow A/I, \quad 1 \otimes b_ i \mapsto \overline{\beta }_ i, \quad 1 \otimes c_ i \mapsto \overline{\gamma }_ i$

maps $g_ B$ and $h_ B$ to $\overline{g}$ and $\overline{h}$ in $A/I[x]$. The displayed map is surjective; denote $J \subset B$ its kernel. From the discussion in Algebra, Example 10.143.12 it is clear that $A \to B$ is etale at all points of $V(J) \subset \mathop{\mathrm{Spec}}(B)$. Choose $g \in B$ as in Lemma 15.9.4 and consider the $A$-algebra $B_ g$. Since $g$ maps to a unit in $B/J = A/I$ we obtain also a map $B_ g/I B_ g \to A/I$ of $A/I$-algebras. Since $A/I \to B_ g/I B_ g$ is étale, also $B_ g/IB_ g \to A/I$ is étale (Algebra, Lemma 10.143.8). Hence there exists an idempotent $e \in B_ g/I B_ g$ such that $A/I = (B_ g/I B_ g)_ e$ (Algebra, Lemma 10.143.9). Choose a lift $h \in B_ g$ of $e$. Then $A \to A' = (B_ g)_ h$ with factorization given by the image of the factorization $f_ B = g_ B h_ B$ in $A'$ is a solution to the problem posed by the lemma. $\square$

Comment #1334 by JuanPablo on

I do not understand why $A/I\cong A'/IA'$.

In the proof as currently written it does not seem to work.

Say $g=\Delta$ the resultant of the polynomials $x^n+\Sigma b_{i}x^{n-i}, x^m+\Sigma c_{j}x^{m-j}$ as in example 10.139.13 (tag 00UA) works to make $A\rightarrow B_g$ étale and maps to an invertible in $B\rightarrow A/I$ (that is, modulo $J$).

But looking at the universal property of $B_{\Delta}/IB_{\Delta}$ it seems to suggest that in any $A/I$ algebra with a second factorization $\bar{g}\bar{h}=\bar{g}'\bar{h}'$, $(\bar{g},\bar{h}),(\bar{g}',\bar{h}')$ generating the unit ideals, monic and of the same degree then $(\bar{g},\bar{h})=(\bar{g}',\bar{h}')$, which is in general false.

Comment #1354 by on

Thanks very much for pointing this out. I have fixed it here.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


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# Motivation for the Definition of an Affine Transformation

Given two groups $G$ and $H$, it is natural to declare that they have the same structure if there is a bijective map $f:G\to H$ such that $f(g_1g_2)=f(g_1)f(g_2)$ for all $g_1$ and $g_2\in G$. This says that the members of $H$ 'multiply the same way' as the corresponding members of $G$ under this mapping $f$.

This motivates the definition of a group homomorphism.

I recently read about affine spaces and the definition of an affine transformation. I cannot quite see how an affine isomorphism preserves the affine structure. Here is what I know.

Definiton. An affine space $P$ is a set along with a vector space $V$ acting on it transitively and freely.

Definition. Let $F$ be a field and $P$ be an affine space over the vector space $(V,F)$. Let $p_1,\ldots, p_n$ be points in $P$. Then for $a_1,\ldots, a_n\in F$ satisfying $a_1+\cdots +a_n=1$, we define $$p_1\cdot a_1+\cdots +p_n\cdot a_n = a_1(p_1-p)+\cdots +a_n(p_n-p)+p$$ where $p$ is a point in $P$.

It can be checked that this definition is unambiguous, i.e., it doesn't depend on the choice of the point $p$.

Definition. Let $F$ be a field. Let $P$ be an affine space over $(V,F)$ and $Q$ be an affine space over $(W,F)$. An affine transformation from $P$ to $Q$ is a mapping $f:P\to Q$ such that $$f(p_1\cdot a_1+\cdots p_n\cdot a_n)=f(p_1)\cdot a_1+\cdots +f(p_n)\cdot a_n$$

for all $p_1,\ldots, p_n\in P$ and $a_1,\ldots, a_n\in F$ satisfying $a_1+\cdots +a_n=1$.

Definition. An affine isomorphism is a bijective affine transformation.

I cannot see how an affine isomorphism, as defined above, captures the notion of structure preservation. The algebraic structure of an affine space is also making me a bit nauseous. It is a set acted upon by a vector space in a special way, where the latter itself has an algebraic structure.

Can somebody please enlighten me on how to understand an affine isomorphism as an affine structure preserving map?

Thank you so much for reading this long post.

• The affine stuff is roughly the linear stuff plus a displacement of the origin. I am not sure what the simpler case of linear transformations ought to preserve. Perhaps mapping linear spaces to linear spacese? – mvw May 9 '15 at 22:07
• In case of Linear isomorphisms, I think the definition is a natural one. The structure on a linear space is governed by the vector aaddition and scalar multiplication. If $T:V\to W$ is a linear isomorphism, then it's bijective and preserves additive structure: $T(u+v)=T(u)+t(v)$, and multiplicative structure $T(\alpha u)=\alpha T(u)$. – caffeinemachine May 9 '15 at 22:29


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# Isomorphism between groups and subgroups.

Say I know that $$G_1,G_2$$ are isomorphic groups, both with normal subgroups $$N_1,N_2$$ respectively. Can I take an isomorphism $$psi$$ between $$G_1,G_2$$ s.t $$psi(N_1)=N_2$$? Meaning that $$psi$$ also defines a isomorphism between $$N_1$$ and $$N_2$$?

The thm I’m trying to prove is if $$G_1 cong G2, N1cong N2,N1unlhd G1, N2 unlhd G2$$ then $$G1/N1 cong G2/N2$$, and proving the above statement I would be able to use the first isomorphism theorem and that’s it. Is it even true?

Any hint would be helpful on how to approach this problem.


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prove that $BA=\begin{bmatrix}9 & 0\\ 0 & 9\end{bmatrix}$ [duplicate]

I want to prove the following question but I have no idea to start!

let $A$ be a $3\times2$ matrix and $B$ be a $2\times3$ one. if $AB=\begin{bmatrix}8 & 2 &-2 \\2 & 5 & 4 \\-2 & 4 & 5 \end{bmatrix}$

then prove that $BA=\begin{bmatrix}9 & 0\\ 0 & 9\end{bmatrix}$. [ hint:if $AB=I$ then we have $BA=I$]

marked as duplicate by user26857, user147263, Paul Sinclair, Em., choco_addictedMay 17 '16 at 0:37

• I imagine that the spectral theorem has to play a role here as $AB$ is real symmetric... But I don't know how to play with it. – mathcounterexamples.net Feb 27 '16 at 21:26
• @mathcounterexamples.net It's true, for example eigenvalues of the matrix $3\times 3$ are 9;9;0. – echzhen Feb 27 '16 at 21:27
• I computed it directly. Not very elegant, but it works. Just solving polynomial equations with elimination method. – Dietrich Burde Feb 27 '16 at 21:30

Note that rank$(AB)=2$. But rank$(BA) \geq \text{rank}(AB)^2$ because $(AB)^2=(AB)(AB)=A(BA)B$. We compute $$(AB)^2=\begin{bmatrix}72 & 18 & -18\\18&45&36\\-18&36&45\end{bmatrix}=9(AB).$$ But rank$(AB)^2=2$(follows from above matrix equality), so rank$(BA) \geq 2$. Since $BA$ is $2 \times 2$, this means $BA$ is invertible. Now we have \begin{align*} (BA)^3 & = (BA)(BA)(BA)\\ &=B(AB)^2A\\ &=B(9AB)A\\ &=9(BA)(BA)\\ &=9(BA)^2\\ BA&=9I. \end{align*}

• you've mentioned that because $(AB)^2=A(BA)B$ so $rank(BA) \geq rank(AB)^2$. I can't understand why. Is there any special theorem which proves this claim? – F.K Feb 29 '16 at 8:18
• @C.B I am just using the fact that rank$(MN) \leq \min(\text{rank}(M), \text{rank}(N))$. In the proof above, $BA$ is one of the components in the product describing $(AB)^2$ so.... – Anurag A Feb 29 '16 at 9:59
• I got it. thank you so much – F.K Feb 29 '16 at 10:39

Probably there are shorter arguments, but maybe something like this:

1. $AB$ is symmetric and has an ON basis of eigenvectors $v_1$, $v_2$ and $v_3$ with corresponding eigenvalues $9$, $9$ and $0$.
2. $AB$ and $BA$ have the same non-zero eigenvalues (see for example here).
3. Multiplying by $B$ from the left, we find that $Bv_1$ and $Bv_2$ are eigenvectors corresponding to $BA$ with eigenvalues $9$. (Note that these vectors are non-zero.)
4. The vectors $Bv_1$ and $Bv_2$ are linearly independent, since, multiplying by $A$, we find that $Bv_1=kBv_2$ implies $9v_1=k9v_2$ which cannot hold since $v_1$ and $v_2$ are orthogonal.
5. This implies that $BA$ is a diagonalizable $2\times2$-matrix with double eigenvalue $9$. We conclude that $BA=9I$, since, we can write $BA=SDS^{-1}$ with $D$ diagonal with nines on the diagonal.
• To the down voter: Is something wrong with this solution? – mickep Feb 28 '16 at 8:50


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# MSSM

In the framework of the Minimal Supersymmetric Standard Model (MSSM), the Higgs sector has been extensively studied in the light of the recent Higgs discovery. In particular a Higgs at around 125 GeV with SM-like properties can be realized in the decoupling limit where the additional scalars and pseudoscalars are heavy (ma,H,H± >~300 GeV). In this regime, exotic decays of the type h→ A0 Z, h→ HH, h→ A0A0, h→ H± W are kinematically forbidden (here A0 denotes the CP-odd scalar).1 In general, the regime mA ≤ mh/2 is highly constrained. This is due to the fact that the masses of the H, A0, and H± scalars of the MSSM are closely tied to one another. In particular, at the tree level mH±2=mA2+mW2, leading to a charged Higgs boson already excluded by LEP searches, for mA <~60 GeV.
Additional Higgs exotic decays could be realized if some of the sparticles are lighter than the Higgs boson. In the MSSM with R-parity, this possibility is however very constrained by LEP and LHC searches. In particular, assuming a LEP bound at around 100 GeV for electrically charged sparticles, the only possible Higgs exotic decays, in the framework of the MSSM, are to sneutrinos or to neutralinos.2 However, in view of the LEP lower bound on the masses of the left handed sleptons, which are related through SU(2) symmetry to the sneutrino masses, the decay to sneutrinos are generically kinematically closed.

Figure 1: Branching ratios of the Higgs into neutralinos: Br(h→χ1χ1) and Br(h→χ1χ2) are shown in blue and red, respectively. The yellow region is the region excluded by the LEP bound on the Z invisible width. The region below the dashed green line is the region with a lightest chargino below the LEP bound of  ∼ 100 GeV. The input parameters are tanβ = 10 and M2=300 GeV (left), M2=150 GeV ( right).

The decay of the Higgs into neutralinos h→ χiχj is therefore typically the only accessible decay (here, as elsewhere, we suppress the superscript "0" on neutralinos to streamline notation). This decay mode is most easily realized in models with non-universal gaugino masses, for which the universality relation M1 ∼ [(M2)/2] ∼ [(M3)/7] at the electroweak scale is relaxed, allowing light LSPs while still satisfying the LEP and LHC bounds on chargino and gluino masses. As neutralinos which couple to the Higgs boson also typically couple to the Z, the main constraint on Higgs decays to neutralinos comes from the precise LEP measurements of the invisible and total widths of the Z boson, for mχi+mχj < mZ. However, as Fig. 1 shows, for mainly bino LSPs, it is possible to accommodate a sizable branching ratio for the decay h→ χ1χ1 while still maintaining compatibility with the LEP Z measurements (see also [7,8,9,10] for recent studies). The parameter space for which h→ χ1χ2 is open is presently strongly constrained by both LEP Z measurements (the yellow region in Fig. 1 is the region excluded by the LEP measurement of the Z invisible width) and chargino and neutralino searches at LEP and, especially, at the LHC (see e.g. [11]).
In summary, with the present constraints, the MSSM generally can only provide for Higgs decays into neutralinos. These neutralinos may either be detector-stable, in which case the Higgs decay is invisible (as discussed here), or, in models with gauge-mediated supersymmetry breaking, they may decay within the detector to photon-gravitino pairs [12] (as studied here). Higgs decays to other sparticles or to other (pseudo-)scalars in the extended MSSM Higgs sector are now strongly constrained by the LEP and LHC experiments.

## References

[1]K. Hagiwara, J. S. Lee, and J. Nakamura, Properties of 125 GeV Higgs boson in non-decoupling MSSM scenarios, JHEP 1210 (2012) 002, [arXiv:1207.0802].
[2]M. Drees, A Supersymmetric Explanation of the Excess of Higgs-Like Events at the LHC and at LEP, Phys.Rev. D86 (2012) 115018, [arXiv:1210.6507].
[3]P. Bechtle, S. Heinemeyer, O. Stal, T. Stefaniak, G. Weiglein, et. al., MSSM Interpretations of the LHC Discovery: Light or Heavy Higgs?, Eur.Phys.J. C73 (2013) 2354, [arXiv:1211.1955].
[4]T. Han, T. Li, S. Su, and L.-T. Wang, Non-Decoupling MSSM Higgs Sector and Light Superpartners, [ arXiv:1306.3229].
[5]G. Barenboim, C. Bosch, M.L.López-Ibañez, and O. Vives, Eviction of a 125 GeV "heavy"-Higgs from the MSSM, JHEP 1311 (2013) 051, [arXiv:1307.5973].
[6]B. Batell, C. E. M. Wagner, and L.-T. Wang, Constraints on a Very Light Sbottom, [arXiv:1312.2590].
[7]D. Albornoz Vasquez, G. Belanger, R. Godbole, and A. Pukhov, The Higgs boson in the MSSM in light of the LHC, Phys.Rev. D85 (2012) 115013, [arXiv:1112.2200].
[8]N. Desai, B. Mukhopadhyaya, and S. Niyogi, Constraints on Invisible Higgs Decay in MSSM in the Light of Diphoton Rates from the LHC, [arXiv:1202.5190].
[9]H. K. Dreiner, J. S. Kim, and O. Lebedev, First LHC Constraints on Neutralinos, Phys.Lett. B715 (2012) 199-202, [arXiv:1206.3096].
[10]B. Ananthanarayan, J. Lahiri, P. Pandita, and M. Patra, Invisible decays of the lightest Higgs boson in supersymmetric models, Physical Review D 87, 115021 (2013) [ arXiv:1306.1291].
[11]ATLAS collaboration, Search for direct production of charginos, neutralinos and sleptons in final states with two leptons and missing transverse momentum in pp collisions at sqrt s= 8 TeV with the ATLAS detector, JHEP 1405 (2014) 071, [arXiv:1403.5294].
[12]J. D. Mason, D. E. Morrissey, and D. Poland, Higgs Boson Decays to Neutralinos in Low-Scale Gauge Mediation, Phys.Rev. D80 (2009) 115015, [arXiv:0909.3523].

### Footnotes:

1SM-like Higgs bosons can also be achieved in a corner of parameter space where the additional scalar and pseudoscalars are lighter than mh (see for example [1,2,3]). Low energy flavor observables like b→ s γ, however, set important constraints on this region of parameter space [4,5]. Furthermore, the decays of the SM-like Higgs into lighter scalars are still not kinematically accessible.
2Light sbottoms are another possibility, but this is now almost entirely ruled out [6].
File translated from TEX by TTH, version 4.03. On 15 Dec 2013, 22:20.


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# Prismatoid

In geometry, a prismatoid is a polyhedron where all vertices lie in two parallel planes. (If both planes have the same number of vertices, and the lateral faces are either parallelograms or trapezoids, it is called a prismoid.)

If the areas of the two parallel faces are A1 and A3, the cross-sectional area of the intersection of the prismatoid with a plane midway between the two parallel faces is A2, and the height (the distance between the two parallel faces) is h, then the volume of the prismatoid is given by $V = \frac{h(A_1 + 4A_2 + A_3)}{6}$ (This formula follows immediately by integrating the area parallel to the two planes of vertices by Simpson's rule, since that rule is exact for integration of polynomials of degree up to 3, and in this case the area is at most a quadratic function in the height.)

## Prismatoid families

Families of prismatoids include:

## Higher dimensions

In general a polytope is prismatoidal if its vertices exist in two hyperplanes. For example in 4-dimension, two polyhedra can be placed in 2 parallel 3-spaces, and connected with polyhedral sides.

A tetrahedral-cuboctahedral cupola.


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text-2.0.2: An efficient packed Unicode text type.

Description

Functions used frequently when reading textual data.

Synopsis

# Documentation

type Reader a = Text -> Either String (a, Text) Source #

Read some text. If the read succeeds, return its value and the remaining text, otherwise an error message.

decimal :: Integral a => Reader a Source #

Read a decimal integer. The input must begin with at least one decimal digit, and is consumed until a non-digit or end of string is reached.

This function does not handle leading sign characters. If you need to handle signed input, use signed decimal.

Note: For fixed-width integer types, this function does not attempt to detect overflow, so a sufficiently long input may give incorrect results. If you are worried about overflow, use Integer for your result type.

Read a hexadecimal integer, consisting of an optional leading "0x" followed by at least one hexadecimal digit. Input is consumed until a non-hex-digit or end of string is reached. This function is case insensitive.

This function does not handle leading sign characters. If you need to handle signed input, use signed hexadecimal.

Note: For fixed-width integer types, this function does not attempt to detect overflow, so a sufficiently long input may give incorrect results. If you are worried about overflow, use Integer for your result type.

signed :: Num a => Reader a -> Reader a Source #

Read an optional leading sign character ('-' or '+') and apply it to the result of applying the given reader.

This function accepts an optional leading sign character, followed by at least one decimal digit. The syntax similar to that accepted by the read function, with the exception that a trailing '.' or 'e' not followed by a number is not consumed.

Examples:

rational "3"     == Right (3.0, "")
rational "3.1"   == Right (3.1, "")
rational "3e4"   == Right (30000.0, "")
rational "3.1e4" == Right (31000.0, "")
rational "e3"    == Left "input does not start with a digit"

Examples of differences from read:

rational "3.foo" == Right (3.0, ".foo")
rational "3e"    == Right (3.0, "e")

The syntax accepted by this function is the same as for rational.
Note: This function is almost ten times faster than rational, but is slightly less accurate.
The Double type supports about 16 decimal places of accuracy. For 94.2% of numbers, this function and rational give identical results, but for the remaining 5.8%, this function loses precision around the 15th decimal place. For 0.001% of numbers, this function will lose precision at the 13th or 14th decimal place.


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# Why are locally closed subschemes not open subschemes of closed subschemes?

Ravi Vakil gives the following argument for why open subschemes of closed subschemes are locally closed: "Clearly an open subscheme U of a closed subscheme V of X can be interpreted as a closed subscheme of an open subscheme: as the topology of V is induced from the topology of X, the underlying set of U is the intersection of some open subset U' on X with V. We can take V' = $V \bigcap U'$, and then $V' \rightarrow U'$ is a closed embedding, and $U' \rightarrow X$ is an open embedding.

What I don't understand is why this argument doesn't also give the converse. It seems to me that if the words "closed" and "open" are switched in the above argument, we will "prove" that locally closed subschemes are open subschemes of closed subschemes. But this is false; what is wrong with the "proof"?

-

Topologically this is correct: let $Z$ be a closed subscheme of an open subscheme $U$ of $X$. Then there exists a closed subset $F$ of $X$ such that $F\cap U=Z$ set-theoretically (take $F$ be the Zariski closure of $Z$ in $X$). But the scheme structure of $Z$ might not extend to $F$.
The natural attempt is to define the scheme-theoretical closure of $Z$ in $X$, and this actually exists when $X$ is locally noetherian or, more generally, when $Z\to X$ is quasi-compact. So under this assumption, $Z$ is an open subscheme of a closed subscheme of $Z$. See [EGA], I.4.1.3.
QiL has already gave a nice explanation. I just want to add one comment. Follows the notation of QiL, if $Z$ is reduced, the converse is also true. But in general the converse is not true. For a counter-example please see Example 24.3.4 in the Stacks Project by de Jong. All the above can be found in the chapter 24 of the Stacks Project.


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Showing that groups are isomorphic

gottfried
If one wants to show that two groups are isomorphic is simply finding a single isomorphism between them sufficient?

For example.

If G is an infinite cyclic group with generator g show that G is isomorphic to $Z$.

So suppose f(g)=ord(g)

then f is bijective and a homomorphism I believe?

Staff Emeritus
Homework Helper
If one wants to show that two groups are isomorphic is simply finding a single isomorphism between them sufficient?

Yes.

For example.

If G is an infinite cyclic group with generator g show that G is isomorphic to $Z$.

So suppose f(g)=ord(g)

then f is bijective and a homomorphism I believe?

Yes, this is true. However, you might want to give a bit of explanation on why it is bijective and a homomorphism.

gottfried
Cool thanks for clearing it up.


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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Interior Angles in Convex Polygons

## Angles inside a closed figure with straight sides.

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Practice Interior Angles in Convex Polygons
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Interior Angles in Convex Polygons

Below is a picture of Devils Postpile, near Mammoth Lakes, California. These posts are cooled lava (called columnar basalt) and as the lava pools and cools, it ideally would form regular hexagonal columns. However, variations in cooling caused some columns to either not be perfect or hexagonal.

First, define regular in your own words. Then, what is the sum of the angles in a regular hexagon? What would each angle be? After completing this Concept you'll be able to answer questions like these.

### Watch This

CK-12 Foundation: Chapter6InteriorAnglesinConvexPolygonsA

Watch the first half of this video.

James Sousa: Angles of Convex Polygons

### Guidance

Recall that interior angles are the angles inside a closed figure with straight sides. As you can see in the images below, a polygon has the same number of interior angles as it does sides.

A diagonal connects two non-adjacent vertices of a convex polygon. Also, recall that the sum of the angles in a triangle is \begin{align*}180^\circ\end{align*}. What about other polygons?

##### Investigation: Polygon Sum Formula

Tools Needed: paper, pencil, ruler, colored pencils (optional)

1. Draw a quadrilateral, pentagon, and hexagon.

2. Cut each polygon into triangles by drawing all the diagonals from one vertex. Count the number of triangles.

Make sure none of the triangles overlap.

3. Make a table with the information below.

Name of Polygon Number of Sides Number of \begin{align*}\triangle s\end{align*} from one vertex (Column 3) \begin{align*}\times\end{align*} (\begin{align*}^\circ\end{align*} in a \begin{align*}\triangle\end{align*}) Total Number of Degrees
Quadrilateral 4 2 \begin{align*}2 \times 180^\circ\end{align*} \begin{align*}360^\circ\end{align*}
Pentagon 5 3 \begin{align*}3 \times 180^\circ\end{align*} \begin{align*}540^\circ\end{align*}
Hexagon 6 4 \begin{align*}4 \times 180^\circ\end{align*} \begin{align*}720^\circ\end{align*}

4. Do you see a pattern? Notice that the total number of degrees goes up by \begin{align*}180^\circ\end{align*}. So, if the number sides is \begin{align*}n\end{align*}, then the number of triangles from one vertex is \begin{align*}n - 2\end{align*}. Therefore, the formula would be \begin{align*}(n - 2) \times 180^\circ\end{align*}.

Polygon Sum Formula: For any \begin{align*}n-\end{align*}gon, the sum of the interior angles is \begin{align*}(n - 2) \times 180^\circ\end{align*}.

A regular polygon is a polygon where all sides are congruent and all interior angles are congruent.

Regular Polygon Formula: For any equiangular \begin{align*}n-\end{align*}gon, the measure of each angle is \begin{align*}\frac{(n-2)\times 180^\circ}{n}\end{align*}.

#### Example A

Find the sum of the interior angles of an octagon.

Use the Polygon Sum Formula and set \begin{align*}n = 8\end{align*}.

\begin{align*}(8 - 2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ\end{align*}

#### Example B

The sum of the interior angles of a polygon is \begin{align*}1980^\circ\end{align*}. How many sides does this polygon have?

Use the Polygon Sum Formula and solve for \begin{align*}n\end{align*}.

\begin{align*}(n - 2) \times 180^\circ & = 1980^\circ\\ 180^\circ n - 360^\circ & = 1980^\circ\\ 180^\circ n & = 2340^\circ\\ n & = 13 \qquad \text{The polygon has} \ 13 \ \text{sides.}\end{align*}

#### Example C

How many degrees does each angle in an equiangular nonagon have?

First we need to find the sum of the interior angles in a nonagon, set \begin{align*}n = 9\end{align*}.

\begin{align*}(9 - 2) \times 180^\circ = 7 \times 180^\circ = 1260^\circ\end{align*}

Second, because the nonagon is equiangular, every angle is equal. Dividing \begin{align*}1260^\circ\end{align*} by 9 we get each angle is \begin{align*}140^\circ\end{align*}.

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter6InteriorAnglesinConvexPolygonsB

#### Concept Problem Revisited

A regular polygon has congruent sides and angles. A regular hexagon has \begin{align*}(6-2)180^\circ=4\cdot180^\circ=720^\circ\end{align*} total degrees. Each angle would be \begin{align*}720^\circ\end{align*} divided by 6 or \begin{align*}120^\circ\end{align*}.

### Guided Practice

1. Find the measure of \begin{align*}x\end{align*}.

2. The interior angles of a pentagon are \begin{align*}x^\circ, x^\circ, 2x^\circ, 2x^\circ,\end{align*} and \begin{align*}2x^\circ\end{align*}. What is \begin{align*}x\end{align*}?

3. What is the sum of the interior angles in a 100-gon?

1. From the Polygon Sum Formula we know that a quadrilateral has interior angles that sum to \begin{align*}(4-2) \times 180^\circ=360^\circ\end{align*}.

Write an equation and solve for \begin{align*}x\end{align*}.

\begin{align*}89^\circ + (5x - 8)^\circ + (3x + 4)^\circ + 51^\circ & = 360^\circ\\ 8x & = 224\\ x & = 28\end{align*}

2. From the Polygon Sum Formula we know that a pentagon has interior angles that sum to \begin{align*}(5-2) \times 180^\circ=540^\circ\end{align*}.

Write an equation and solve for \begin{align*}x\end{align*}.

\begin{align*} x^\circ + x^\circ + 2x^\circ + 2x^\circ + 2x^\circ&=540^\circ\\ 8x&=540\\x&=67.5\end{align*}

3. Use the Polygon Sum Formula. \begin{align*}(100-2) \times 180^\circ=17,640^\circ\end{align*}.

### Explore More

1. Fill in the table.
# of sides Sum of the Interior Angles Measure of Each Interior Angle in a Regular \begin{align*}n-\end{align*}gon
3 \begin{align*}60^\circ\end{align*}
4 \begin{align*}360^\circ\end{align*}
5 \begin{align*}540^\circ\end{align*} \begin{align*}108^\circ\end{align*}
6 \begin{align*}120^\circ\end{align*}
7
8
9
10
11
12
1. What is the sum of the angles in a 15-gon?
2. What is the sum of the angles in a 23-gon?
3. The sum of the interior angles of a polygon is \begin{align*}4320^\circ\end{align*}. How many sides does the polygon have?
4. The sum of the interior angles of a polygon is \begin{align*}3240^\circ\end{align*}. How many sides does the polygon have?
5. What is the measure of each angle in a regular 16-gon?
6. What is the measure of each angle in an equiangular 24-gon?
7. Each interior angle in a regular polygon is \begin{align*}156^\circ\end{align*}. How many sides does it have?
8. Each interior angle in an equiangular polygon is \begin{align*}90^\circ\end{align*}. How many sides does it have?

For questions 10-18, find the value of the missing variable(s).

1. The interior angles of a hexagon are \begin{align*}x^\circ, (x + 1)^\circ, (x + 2)^\circ, (x + 3)^\circ, (x + 4)^\circ,\end{align*} and \begin{align*}(x + 5)^\circ.\end{align*} What is \begin{align*}x\end{align*}?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 6.1.

### Vocabulary Language: English

Interior angles

Interior angles

Interior angles are the angles inside a figure.
Polygon Sum Formula

Polygon Sum Formula

The Polygon Sum Formula states that for any polygon with $n$ sides, the interior angles add up to $(n-2) \times 180$ degrees.


text:
silicagel 1 M NaOH aqueous standard, Oakland Schools Chemistry Resource Unit 1 answer. Bonding: Intermolecular Forces Then mole fraction of the dimer in the final mixture is, This site is using cookies under cookie policy. Add the volume of bromine solution to a second small, labeled beaker using a graduated cylinder. Brook R. Kirouac  David A. Consiglio, Jr.  Southfield‐Lathrup High School Southfield Public Schools    where the value of the rate constant, $$k$$, is dependent upon the temperature at which the reaction is run. Select OK to close. The invention of the transistor enabled the first radio telemetry capsules, which utilized simple circuits for in vivo telemetric studies of the gastro-intestinal tract. Solutions must be collected and placed in the hazardous waste container. The rate law for this reaction is of the form: $\text{Rate} = k [\ce{Br_{2}}]^{x}[\ce{Acetone}]^{y}[\ce{H_{3}O^{+}}]^{z} \label{2}$. Legal.

For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Construct a new table for the calculated concentrations as well as a column for the rate of the reaction calculated from the slope, $$-\Delta Abs/\Delta t$$. Free LibreFest conference on November 4-6! Note that this is actually the average rate of the reaction over the time interval $$\Delta t$$ since the concentration of $$\ce{Br_{2}}$$ and therefore the rate of the reaction is continuously decreasing during $$\Delta t$$. Under Operate, select Timed Acquisition to collect timed data. Make 50 mL of a 1.0 M HCl solution by measuring 12.5 mL of 4.0 M HCl with a graduated cylinder. carbon monoxide (CO), good night sweet dreams......... everyone, a. calcium and carbon aloxide.b. Using a timer, mix the two solutions by pouring the bromine into the first beaker containing acetone, water and HCl. (Hint: Use combinations of experiments to prove your point and isolate the effect of each solution.). An acid catalyst makes a reaction happen faster, but does not get consumed in the process. The values of the reaction orders, $$x$$, $$y$$, and $$z$$, are usually, though not always, small integers. Next In your lab notebook, write a statement that speculates on what caused the change you observed. Calculate the concentration of each reactant in each solution. Question options:

Not that it matters much... you may be having some reaction such as the slow chlorination of acetone.

Index

Place the acetone in a clean, 50 mL volumetric flask.

Index Repeat Steps 5 - 17 for remaining trails. At that point, the rate law can be used to find the value of $$k$$ for each trial. When the aspirated materials contain gastric acid, there is chemical injury to the lung parenchyma with infection as a secondary event usually 48 to 72 hours later. Set file name to Run 1. Next

For each of the four trials you collect, you will need to open this set up menu and rename each trial, i.e.

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# Applying the symmetric difference to multiple sets

I was working on set theory and I came across a rule: $$(A\cup B) - (A\cap B) = (A\cap B)'$$

I have 4 sets $A$, $B$, $C$, $D$. How can I apply the rule above to all set at same time? is the below mentioned result mathematically correct? $$(A\cup B\cup C\cup D) - (A\cap B\cap C\cap D) = (A\cap B\cap C\cap D)'$$

Regards,

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Your $(A ∪ B) - (A ∩ B)$ is called a symmetric difference. – draks ... Jul 24 '12 at 9:24
And your first identity is incorrect. – Tapu Jul 24 '12 at 9:26
@draks: thanks a ton. It helped me. Kindly put it as an answer so I can mark it :) :) – Priyank Thakkar Jul 24 '12 at 9:30
@Tapu why? ${ }$ – draks ... Jul 24 '12 at 9:32
@PriyankThakkar you're welcome (to Math.StackExchange.com ;-) – draks ... Jul 24 '12 at 9:37

If by $(\cdots)'$ you denote the complement of $(\cdots)$, $$(A ∪ B) - (A ∩ B) = (A\cap B)'$$ which is corresponds to the symmetric difference $A\,\triangle\,B\,$. As pointed by Martin your second equation is also just a way of writing the complement.

And, as already pointed out by Tapu, for multiple sets the $(\cap_k A_k )' \neq \triangle_k A_k$.

$\hskip2.7in$

Venn diagram of $~A \triangle B \triangle C$

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If you want a talk about complement of a set, you have first to state what is the universe $U$ you are working with. If you choose a universe then the complement of $A$ is $A'=U\setminus A$.

So if $U=A\cup B$ then $(A\cap B)'=U\setminus (A\cap B)$ and if $U=A\cup B\cup C\cup D$ then $(A\cap B\cap C\cap D)'=A\cup B\cup C\cup D\setminus (A\cap B\cap C\cap D)$.

So your formula is just the definition of complement, if the universe you're working with is $U=A\cup B$ in the first case and $U=A\cup B\cup C\cup D$ in the second one.

If you work with a different universe $U\supsetneqq A\cup B\cup C\cup D$, then the equality you wrote is not correct (and neither is the first one).

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@Marting Sleziak: I didn't use proper math terminology. What I exactly wanted to ask was how to apply Symmetric difference to multiple set. Thanks for the guidance :) – Priyank Thakkar Jul 24 '12 at 9:37

If we let $\triangle$ denote the symmetric difference of two sets, $$A\triangle B = (A\cup B) - (A\cap B) = (A-B)\cup(B-A),$$ then we have:

Theorem. If $A_1,\ldots,A_n$ are sets, then \begin{align*} x\in (A_1\triangle\cdots\triangle A_{n-1})\triangle A_n&\iff x\text{ is in exactly an odd number of }A_i\\ &\iff \Bigl|\{i\mid 1\leq i\leq n, x\in A_i\}\Bigr|\text{ is odd.} \end{align*}

Proof. We proceed by induction on $n$. The result is true if $n=1$ or $2$. Assume the result holds for $k$. Then $x\in (A_1\triangle \cdots\triangle A_n)\triangle A_{n+1}$ if and only if $x$ is in exactly one of $A_1\triangle\cdots\triangle A_n$ and $A_{n+1}$.

If $x$ lies in an even number of sets from among $A_1,\ldots,A_{n+1}$, then it either lies in an even number of sets from among $A_1,\ldots,A_n$ and not in $A_{n+1}$, in which case it lies in neither $A_1\triangle\cdots\triangle A_n$ (by the induction hypothesis) nor in $A_{n+1}$, hence not in the symmetric difference; or else it lies in an odd number of sets from among $A_1,\ldots,A_n$ (and hence lies in $A_1\triangle\cdots\triangle A_n$ by the induction hypothesis) and in $A_{n+1}$, and so it does not lie in the symmetric difference (since it lies in both operands).

If $x$ lies in an odd number of sets from among $A_1,\ldots,A_{n+1}$, then it either lies in an even number of sets from among $A_1,\ldots,A_n$ (and hence not in $A_1\triangle\cdots\triangle A_n$), and in $A_{n+1}$; or it lies in an odd number of sets from among $A_1,\ldots,A_n$ and not in $A_{n+1}$. Either way, it lies in exactly one of $A_1\triangle\cdots\triangle A_n$ and $A_{n+1}$, hence lies in $(A_1\triangle\cdots\triangle A_n)\triangle A_{n+1}$.

Thus, $x\in (A_1\triangle\cdots\triangle A_n)\triangle A_{n+1}$ if and only if it lies in exactly an odd number of sets from among $A_1,\ldots,A_{n+1}$. $\Box$

Moreover, since $(A\triangle B)\triangle C = A\triangle(B\triangle C)$ for all $A$, $B$, and $C$ (it follows easily from the theorem, or from truth tables), we can dispense with the parentheses.

Essentially, the characteristic function of the symmetric difference is the sum, modulo $2$, of the characteristic functions of the two sets: $$1_{A\triangle B} = 1_A\oplus 1_B,\qquad \oplus=\text{addition modulo }2.$$

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# Angular momentum objective

1. Dec 3, 2008

### sumitmanhas

1. The problem statement, all variables and given/known data
Which of the following is an accurate statement
1) The angular momentum of a moving particle depends on the specific origin with respect to which the angular momentum is calculated.
2) If the torque acting on a particle is zero about an arbitrary origin, then the angular momentum of the particle is also zero about that origin.
3) If the speed of a particle is constant, then the angular momentum of the particle about any specific origin must also be constant.
4) Consider a planet moving in a circular orbit about a star. Even if the planet is spinning it is not possible for its total angular momentum to be zero.
5) A particle moving in a straight line with constant speed necessarily has zero angular momentum.

2. Relevant equations
conservation of angular momentum and principles

3. The attempt at a solution
i am not sure whether which one is more accurate among 3,4 , could someone lead me on this.

2. Dec 3, 2008

### Carid

Perhaps if you could give us your reasoning for each answer we can understand your difficulty. I can see some fairly firm objections to both 3 and 4.

3. Dec 3, 2008

### sumitmanhas

so for the first 1 ) this is likely to be true , which i never thought of because evrytime you calculate your angular momentum you choose the point of rotation and from there using right hand rule determine the direction (in or out)
2) makes no sense because if there is no torque acting on the origin then the angular momentum is constant
3) if speed is constant then angular momentum = mrv is same since the mass remain the same throughout
4)if an object keeps rotating doesn't it have an initial momentum and final momentum which are same , so, i mean it is constant then.
5) particle in straight line can have angular momentum , as in the case of a person jumping of the merry go round with horizontal speed doesnot have zero angular momentum.

reply as soon as possible plz

Last edited: Dec 3, 2008
4. Dec 3, 2008

### sumitmanhas

i belive the first one is more accurate....


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Tag Info

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You can add \tracingmacros=1 to your code before the \draw command. Then you can look in the log-file and it is rather easy to see a few lines before the error (around line 84000) that pgf is just trying to calculate the reciprocal of a rather small number (0.00006). \pgfmath@reciprocaltemp ->0.00006 You could -- if you have lot time -- go ...

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You can define your own \dotMarkRightAngle as below. Four parameters need to be entered; size of the angle (square side) and three coordinates in parentheses (#2,#3,#4) taken clock-wise. \documentclass[tikz,border=2pt]{standalone} \usetikzlibrary{calc} \begin{document} \def\dotMarkRightAngle[size=#1](#2,#3,#4){% \draw ($(#3)!#1!(#2)$) -- ($($(#3)!...

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your angle radius is to small. enlarge it to 4mm. also observe small differences between yours and mine code: \documentclass[a4paper,12pt]{article} \usepackage{tikz} \usetikzlibrary{angles,decorations.markings} \tikzset{ doubledash/.style={decoration = { markings, % mark= at position 0.5 with{ \draw[thin] (-0.6pt,-2pt) -- (-0.6pt,2pt); ...

6

The easiest way to learn about arc construction is to use the verbose mode at first. So the command \draw [red, thick] (1.0,0) arc [start angle=0, end angle=30, radius=1cm] node [midway, right] {$\phi$}; says to start the arc at (1.0,0) with the provided parameters. A more concise way of saying that is to use \draw [red, thick] (1.0,0) arc (0:30:...

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angle expects names, not coordinates. The name argument is directly parsed to \pgfpointanchor, from tikzlibraryangles.code.tex: \def\tikz@lib@angle@parse#1--#2--#3\pgf@stop{% % ... \pgf@process{\pgfpointanchor{#2}{center}}% \pgf@xa=\pgf@x% \pgf@ya=\pgf@y% \pgf@process{\pgfpointanchor{#1}{center}}% \pgf@xb=\pgf@x% \pgf@yb=\pgf@y% \pgf@process{...

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Use "$\alpha$"{fill=white} instead of "$\alpha$" in \pic \documentclass{article} \usepackage[utf8]{inputenc} \usepackage{amsmath} \usepackage{tikz} \usetikzlibrary{calc,patterns,angles,quotes,arrows} \begin{document} \begin{figure} \begin{center} \begin{tikzpicture} %Define coordinates \coordinate (origin) at (0,0); ...

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The initial point of the angle needs to be located along the line, so using polar coordinates (20:0.7) is the easiest way to specify that: To add a caption you can use the caption package. Alternatively you can add a \node to place text below the axis. Notes: I added rotation to the blue node as that seemed the easiest way to display the text. If you don'...

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Chemfig documentation has an example, in French, about making a customized lewis macro accepting an arbitrary angle in degrees. I am reproducing it irresponsibly translated to English. (I am a native Brazilian Portuguese speaker, so I don't master neither English nor French). This example appeared on page 45 (46th page as there is a cover): \example*{Pairs ...

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You need to use named coordinates I think: \documentclass[12pt, a4paper]{standalone} \usepackage{tikz} \usetikzlibrary{angles} \begin{document} \begin{tikzpicture} \coordinate (O) at (0,0); \coordinate (A) at (0,1); \coordinate (B) at (1,0); \pic [draw, angle eccentricity=1.5] {angle=A--O--B}; \end{tikzpicture} \end{document}

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Maybe that? \documentclass{scrbook} %\input{configuration} \usepackage[utf8]{inputenc} \usepackage{amsfonts, amsthm, amsmath, braket} \usepackage{tikz} \usetikzlibrary{angles, quotes,calc,patterns} \begin{document} \begin{tikzpicture} \draw (0,0) node[anchor=north east] {$O$}; \draw[thick, ->] (-0.35,0) -- (5.5,0) node[anchor=...

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with use of the angles and quotes library and made a little bit different looks of diagram: \documentclass[tikz, margin=3mm]{standalone} \usetikzlibrary{angles, quotes} \usepackage{amssymb} \begin{document} \begin{tikzpicture}[ every edge quotes/.append style = {anchor=south, sloped} ] % axis \draw[thick, ->] (-0.35,0) -- (5....

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It seems to me that you must use coordinate names. My guess is that the internals use \pgfpointanchor or some such, which requires a named node. It's probably good form to name your coordinates anyway! To make the angle label (say) green, set the text property of the pic (or the entire path): \draw pic["$\alpha$", draw=red, text=green, ->, angle ...

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Try \documentclass[15pt,a4paper,headinclude, totoc]{scrartcl} \usepackage[usenames,dvipsnames]{xcolor} \usepackage{tikz} \usepackage{pgfplots} \pgfplotsset{compat=1.12} \usetikzlibrary{angles,calc,intersections,quotes,positioning,shapes} \begin{document} \begin{center} \begin{tikzpicture}[scale=1.5, my angle/.style ...

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"\ang{38}"scale=0.5. In general with the quotes syntax, "label text" {<options for label node>}, <other stuff>. If you only have one option, the braces aren't required. \documentclass[a4paper,12pt]{article} \usepackage{siunitx} \usepackage{tikz} \usetikzlibrary{angles, quotes} \begin{document} \begin{tikzpicture} % Declare Points \...

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Something like this: Here is the code: \documentclass[tikz, border=20mm]{standalone} \usetikzlibrary{angles,calc} \begin{document} \begin{tikzpicture} \newcounter{cntShader} \setcounter{cntShader}{60} \coordinate (o) at (0,0); \draw[->] (o) -- (+80:4) coordinate (b); \draw[->] (o) -- (-80:4) coordinate (c); \draw (o) -- (+170:3); \draw (...

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Perhaps you want something like \documentclass{article} \usepackage{tikz} \usetikzlibrary{quotes,angles} \tikzset{ ar/.style={angle radius=#1}, ar/.default=0.65cm, ae/.style={angle eccentricity=#1}, ae/.default=0.75, } \newcommand{\angs}[3][]{% \draw pic[ar,ae,#1,draw,"{\scriptsize $#3$}"] {angle = #2} } \begin{document} \begin{tikzpicture} \...

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It has nothing to do with the arrows. Which of the two possible angles is marked is determined by the order in which you specify the coordinates in \tkzMarkAngle. Hence, you just need to use \tkzMarkAngle[fill=orange,size=0.5,opacity=.4](D,E,C) instead of \tkzMarkAngle[fill=orange,size=0.5,opacity=.4](C,E,D). \documentclass{article} %\documentclass[tikz]{...

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As mentioned in comments, you can't. The manual says The three points <A>, <B>, and <C> must be the names of nodes or coordinates; you cannot use direct coordinates like (1,1) here. Hence, calculate and name the coordinates first. \documentclass{standalone} \usepackage{tikz} \usetikzlibrary{calc, angles} \begin{document} \begin{...

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pure tikz: \documentclass[tikz, border=3mm, ]{standalone} \usetikzlibrary{calc, positioning} \begin{document} \begin{tikzpicture}[ dotbox/.style = {draw, minimum size=3mm, inner sep=0pt, outer sep=0pt, anchor=#1, label=center:$\cdot$}, node distance = 0pt ...

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Partial answer: by removing the inter-word spacing in the angle's nodal definition and letting the tex engine know that it has to draw whatever we desire: \documentclass{article} \usepackage[utf8]{inputenc} \usepackage{tikz} \usetikzlibrary{angles} \begin{document} % a bit of a help from: https://tex.stackexchange.com/questions/267201/tikz-angle-between-two-...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# Problem: Carbon-12 contains 6 protons and 6 neutrons. The radius of the nucleus is approximately 2.7 fm (femtometers) and the radius of the atom is approximately 70 pm (picometers).Calculate the volume of the nucleus and the volume of the atom. (Assume two significant figures.)

###### FREE Expert Solution

The volume of a sphere is given by:

$\overline{){\mathbf{V}}{\mathbf{=}}\frac{\mathbf{4}}{\mathbf{3}}{{\mathbf{\pi r}}}^{{\mathbf{3}}}}$

Volume of nucleus:

###### Problem Details

Carbon-12 contains 6 protons and 6 neutrons. The radius of the nucleus is approximately 2.7 fm (femtometers) and the radius of the atom is approximately 70 pm (picometers).

Calculate the volume of the nucleus and the volume of the atom. (Assume two significant figures.)


text:
A Neat Random Walk Problem

As many of the people that read this blog probably know, I enjoy inventing math problems, especially problems that are probabilistic in nature, and then working out the math to solve that problem. In this blog post, I will introduce one problem that I solved that I think is particularly elegant:
You are randomly walking along the x-axis starting at $x=0$. When you are at $x = n$, you move to $x = n-1$ with probability $p = \frac{n}{n+1}$ and to $x = n+1$ with probability $p = \frac{1}{n+1}$. What is the expected number of steps until you return to $x = 0$ for the first time?
This problem differs from standard random walk problems because the probability of moving left and right along the x-axis depends on your current location. In this problem, there is a force that is pulling you towards $x = 0$, and the further away you get, the more of an effect this force has on you.


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# XCOPY files in batch to another folder but with structure of folders

I am trying to get the following done.

Let's say I have a directory stucture like this

C:\photos\all\abc\def\001.jpg
C:\photos\all\abc\def\002.jpg
C:\photos\all\abc\def\003.jpg
C:\photos\all\xyz\111.jpg
C:\photos\all\xyz\222.jpg
etc.

I want to use XCOPY to copy certain files to another base directory, but I want to keep the structure of the folders. So, for example I want to copy 3 files from the example above to another disk.

D:\newphotos\abc\def\001.jpg
D:\newphotos\abc\def\002.jpg
D:\newphotos\xyz\111.jpg

How can I achieve this?

I tried the following but none seems to do what I want (i ran those from C:\photos\all)

XCOPY "abc\def\001.jpg" "D:\newphotos\abc\def\001.jpg" (keeps asking if it is a file or folder)

XCOPY "abc\def\001.jpg" "D:\newphotos\" (doesn't create the target folder structure)

XCOPY /S "abc\def\001.jpg" "D:\newphotos\" (doesn't create the target folder structure)

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XCOPY "abc\def\001.jpg" "D:\newphotos\abc\def\001.jpg*"


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Speed - Stat

## DescriptionEdit

Speed determines two things:

• If the enemy has more speed than you, then they have the chance of doing an action more than once. This is the same as if you're faster than the enemy, then the enemy might not do an action.
• Note that if the enemy is much faster than you, then the number of turns they need to cast skills is also decreased.
• Speed affects the duration of Physical Skills, buffs, and debuffs. So the higher your speed, the longer these effects last relative to your turn.

## Speed and TimeEdit

Time, at the game's core, is measured in ticks. Each second is exactly six ticks. 1 turn = ((60/speed)-1 rounded up) ticks. Speed caps out at 60:

$\definecolor{litegray}{gray}{0.90}\definecolor{cqbg}{rgb}{.14,.08,.25} \pagecolor{cqbg}\color{litegray}1\text{ second} = 6\text{ ticks} = \left\lceil\frac{360}{speed}\right\rceil-6\text{ turns}$

## Base ValuesEdit

Base Speed
Normal Hard Suicide

Fighter

6 5 4
Thief 8 7 6
Wizard 6 5 4
Ranger 7 6 5


text:
John wrote:

> Maybe someone can expound a bit more on the notion of 'consistency' that I'm vaguely alluding to. All I mean is that a bunch of squares commute... but it's good to try to tease out the meaning of this, and express it in something resembling plain English.

Well, I'm not sure if this is exactly what you're referring to, but maybe it's close?

Generally speaking, homomorphisms identify a substructure of a target object that resembles (i.e. preserves the behavior of) the source object. This is why group homomorphisms identify subgroups of the codomain -- though you could also take the perspective that this is _why_ we define subgroups the way we do.

Functors are "just" homomorphisms between categories, so a functor identifies a subcategory of the codomain. Thus, given two functors from a common source category, we obtain two subcategories of the target category. It seems like a natural transformation ought to be a homomorphism between these two subcategories. Perhaps the strange part is that this homomorphism is "internal" to the containing category: it decomposes into morphisms between pairs of objects across the subcategories. I'm tempted to call this a "representation" of a functor between the two subcategories, but I know that word is heavily loaded in category theory.

On the other hand, there seems to be no guarantee that an object \$$F(X)\$$ must correspond with exactly one object \$$G(X)\$$. For instance, we can let \$$F\$$ project \$$C\$$ all the way down to the one-object one-morphism category, and let \$$G\$$ leave \$$C\$$ unchanged. If a natural transformation \$$\alpha\$$ exists between \$$F\$$ and \$$G\$$, it must necessarily associate a morphism \$$\alpha_X : \mathbf{1} \to X\$$ for every object in \$$C\$$. This is an extreme example, but if a situation even remotely like this is possible, then natural transformations _can't_ induce a functor, since it would need to associate multiple objects in \$$G(C)\$$ with a single object in \$$F(C)\$$.