Post/Author/DateTime	Post
Novacat 08-11-07, 02:03 AM	Sometimes, a player or DM goes to the trouble of producing an exact figure for some in-game value that either doesn't matter, or should just be estimated. Here's one of mine, based on an actual campaign encounter: 1. A human who spends 15 years in a timeless plane must consume 5475 days of rations before returning to a normal-time plane to prevent instant starvation. Given 3 casts per day of Create Food and Water at Caster Level 7, that would take 269 days of casting and resting (on a normal-time plane, of course) for a cleric to rescue him. This figure includes food necessary to cover the time spent casting, as well as the 15 years in question. So, please continue on with your own completely irrelevant calculations, figures, and whatnot. I'd like to see stuff along the lines of "how many pages would it take to add every spell in the PHB to a wizard's spellbook?" or "starting at level 1, how many kobold do you need to kill, one at a time, to achieve level 20?" or "exactly how many coins can one fit into each type of Bag of Holding, or the Portable Hole?" The more obscure and convoluted, the better, but remember to be accurate and thorough. Have fun with it.
Tedronai 08-11-07, 03:20 AM	That coins in the portable hole thing might actually be useful next time our party loots a Dragon's hoard When the DM describes several pile of coins waist high, then says we'll be unable to take them all with us even with the extra-dimensional spaces we also found in the hoard, it'd be nice to know how much we actually can get rather than just the 'fudge math' that either leaves us massively over-powered (the 10th level party with multiple +8 or better weapons) or pathetically under-funded (another 10th level party, but for the most part working with about 6th level wealth...except for those that died and had to bring in new PCs, they started with DMG lvl appropriate wealth)
Kensan_Oni 08-11-07, 03:33 AM	A Cure Light Potion cast at 5th level costs 250 gp's. A potion of Cure Moderate from the book costs 300. The average hit points for the mentioned Cure Light Hit Point Potion is 10. The average for the Cure Moderate is 12. The range for the cure light is 6-13. The range for the Cure Moderate is 6-19. Low end remains constant. The Cure Moderate Potion provides only the potential of 36% more healing, with only a 20% difference on average. Therefore, it's more economical to buy the fifth level Cure Light Potion.
shadzar 08-11-07, 04:44 AM	or "exactly how many coins can one fit into each type of Bag of Holding, or the Portable Hole?" i think in the last month i actually figured that up. no idea what thread it was in but i could try to do it again. let me check to see if i can find it again first.... found it. example: a recent game saw a player carrying 30+ thousand platinum or maybe gold pieces around. for any coin i will just use a simple US Roosevelt dime. the smallest i know of that could be found today. a roll consists of 50 dimes which is 2.75 inches long, 4 ounces, and .7 inches wide. so one pound worth of dimes would take roughly 4 cubic inches and be 200 dimes @ 16 ounces to a pound. now saying these dimes were the gold pieces carried in order to carry the 30+ thousand it would take 150 pounds worth of dimes taking up roughly 18.75 feet * 1.5 inches * 2.75 inches or 77.3 cubic feet. (hoping my math is correct at this late hour.) now exaclty how would someone carry around this mass of coins without geting hindered by its size if not its weight. and that is just for US dimes. not the standard larger coins you would expect a gold piece to be. i don't see how people can allow these things to go without taking them into account for a campaign. where does one put this mass of money, or say the 200 arros or bolts they carry? it doesn't seem possible even for a fantasy world they could carry it all all without some help. meaning bags of holding. a type II bag of holding by 3.5 standards wouldn't even hold it all for the dimensions. and would there be so many laying around that the players could easily get the higher "Ranked" bags of holding? it would take roughly 3 type I bags to carry this amount.
Tedronai 08-11-07, 04:46 AM	It's also incredibly more economical to just by 5 standard CLW potions than it is to buy that single CL5 version. The cure light at CL5 is actually an average of 9.5 HP, not 10. A small but significant difference when dealing with relatively small numbers as it bumps the average difference up to a little over 26% (of average CL5 CLW potion effect). The cost difference, on the other hand, is a flat 20% over the CL5 CLW. A CL5 CMW costs an average of 25 gp / HP A CL5 CLW costs an average of (slightly over) 26 gp / HP
Novacat 08-12-07, 02:15 PM	We're still short of 1001... anyone wanna expand on Kensan_Oni's foray into cure potion costs? Maybe include the higher level cures, along with gp per HP healed, and may even make out a graph showing the change in efficiency?
shadzar 08-12-07, 04:03 PM	math takes time. you have to find numbers before you can crunch them.
stargate525 08-12-07, 04:21 PM	I once figured out the amount of power that would be generated by a magnet falling through a pair of ring gates (through a wire coil) at the speed of light, as well as how long it would take to get the magnet to the speed of light. The power figures escape me, but I do remember that it would take roughly 5 hours and twenty minutes for it to go from dropped to lightspeed, assuming normal gravity conditions.
Wight_Chocolate 08-12-07, 04:23 PM	It's also incredibly more economical to just by 5 standard CLW potions than it is to buy that single CL5 version. Even more economical to buy a wand for the spell rather than potions. 50 CLW potions will set you back by 2500 GP. You can get a wand with 50 charges for 1/4 of that price, with the added bonus of not provoking attacks of opportunity. Of course, not everyone can use a wand, but if you happen to be of a divine spellcasting class and would rather not take up spell slots with healing, or if you have a high UMD skill, it's probably worth the investment.
Airos 08-12-07, 04:52 PM	"how many pages would it take to add every spell in the PHB to a wizard's spellbook?" Well, according to my calculations, (I can't remember if I updated these from 3.0 to 3.5 or not), it would be; lvl 0 19 pages lvl 1 39 pages lvl 2 100 pages lvl 3 126 pages lvl 4 164 pages lvl 5 215 pages lvl 6 258 pages lvl 7 245 pages lvl 8 280 pages lvl 9 216 pages I took a wizard from 3rd level to 35th, starting way back in 1st edition and through 3.5. When I converted him to 3.0 he had so much gold that I just scribed every spell in the PHB. Again, I can't remember if those numbers have been updated to the 3.5 PHB, but they're at least accurate for the 3.0 PHB.
Rustmonster 08-12-07, 06:24 PM	"starting at level 1, how many kobold do you need to kill, one at a time, to achieve level 20?" Well, this figure is very variable. Kobolds can be of any level. And, *sigh*, if we are talking the cannon fodder in the MM only, which I'm guessing is what most people think of when they say "kobold", then you can not get past level 9, according to standard RAW exp charts.
shadzar 08-12-07, 07:56 PM	Well, this figure is very variable. Kobolds can be of any level. And, *sigh*, if we are talking the cannon fodder in the MM only, which I'm guessing is what most people think of when they say "kobold", then you can not get past level 9, according to standard RAW exp charts. what those silly RAW charts probably don't account for is large/massive quantities of the critters. i would find it stretching credulity to say move form 19 to 20 by killing kobols because the number required would harldy be feasible for a single PC to take on. but you could theoretically calculate it.
Wight_Chocolate 08-12-07, 08:13 PM	what those silly RAW charts probably don't account for is large/massive quantities of the critters. "One at a time." Incidentally, it doesn't. However, it does have a note saying that the XP table can't cover every possible situation, and in such cases the DM should just use his judgement. But of course, you knew that anyway.
Adrez Nesnsid 08-12-07, 09:23 PM	Well, this figure is very variable. Kobolds can be of any level. And, *sigh*, if we are talking the cannon fodder in the MM only, which I'm guessing is what most people think of when they say "kobold", then you can not get past level 9, according to standard RAW exp charts. In that case, calculate it for Kobolds of the lowest character level which they can have while retaining experiance point value (or you could also calculate it for if you fought them all at once and then leveled up to level 20 in one go...)
Fallenangel359 08-12-07, 09:36 PM	remember, every time you double the number of kobolds, you increase the cr by 2. assuming all with only 1 npc class 2 kobold= cr 1, 4 kobold= cr 3, 8 kobold = cr 5, 16 kobold = cr 7, 32 kobold = cr 9, 64 kobold = cr 11, 128 kobold = cr 13, 256 kobold = cr 15, 512 kobold = cr 17, 1024 kobolds = cr 19, 2048 kobolds = cr 21. Whats sad is I put my 3rd level gestalt party that I was DMing for (CR 6 by my estimates) against a cr 17 kobold group. 1-3 class levels each, all heroic classes (rogue, fighter, and sorceror), 250 of them, quite a few of them being casters. I used different types of dice to represent different classes, and HD count. They didn't die... Using terrain to their advantage.
shadzar 08-12-07, 10:07 PM	"One at a time." Incidentally, it doesn't. However, it does have a note saying that the XP table can't cover every possible situation, and in such cases the DM should just use his judgement. But of course, you knew that anyway. i knew a DM should do what he sees fit when the rules don't cover something, but did NOT know that each instance of a creature was to be noted for threat purposes on a per case basis. makes it seem like no matter how many kobolds there is that they would only count as a single kobold for threat purposes to see if they are worth gaining XP from. :confused:
Wight_Chocolate 08-12-07, 10:34 PM	makes it seem like no matter how many kobolds there is that they would only count as a single kobold for threat purposes to see if they are worth gaining XP from. :confused: When I said "One at a time", I was referring to the original question. In other words, you cannot bump up the challenge rating of the encounter by using a group of kobolds. They must be fought and killed individually for the purposes of the calculation.
NthDegree256 08-12-07, 10:36 PM	Once upon a time, I was running a game in a campaign setting I had invented where the most notable feature was that all of the continents floated in the air, above a layer of clouds. Beneath the clouds was essentially a "hell-on-earth" scenario - the Evil-aligned planes had leaked onto the surface and the gods had lifted the continents into the sky to spare the mortal races. I hadn't really decided how high above the surface these continents were until it came time for the big epic fight scene in which the BBEG sacrifices two of his airships to create a portal to the surface and the players chase him down there. I picked a roughly-appropriate terminal velocity for the airships and, based on the number of rounds it would take for the PCs to navigate their way to the BBEG's portal to the surface, calculated just how far the airships would have to fall in order for them to come cinematically crashing down shortly after the PCs stepped out of the portal and onto the surface. And the whole time I knew this was absolutely ridiculous, but I just couldn't help myself.
Davith Bothain 08-12-07, 10:43 PM	Of course, we could just make the whole kobold issue easier and use the level-independent experience system as presented in UA. Unfortunately, I'ven't it with me, so I can't be the one to give a definite figure here...
multipleheads 08-12-07, 11:12 PM	Sorry to be completly of topic but Davith, your font just gave me a terrible headache. :weep:
Dragonlord20000 08-12-07, 11:54 PM	9. Using the level-independent experience awards from UA a normal MM version of the Kobold grants 75 exp upon death. Also a character needs 2.6 million exp to hit level 20. Therefore to gain 20th level a character needs to kill 34,667 kobolds one at a time. At higher levels that would make for a pretty quick level as you slaughter kobolds. I might edit this to do a non UA version, but I think this is sufficient.
shadzar 08-13-07, 12:41 AM	When I said "One at a time", I was referring to the original question. In other words, you cannot bump up the challenge rating of the encounter by using a group of kobolds. They must be fought and killed individually for the purposes of the calculation. exactly what i said. so one kobold is just as much a threat as 400?
Wight_Chocolate 08-13-07, 12:56 AM	...no. A group of 400 kobolds is a much greater threat than one kobold. That is why it would mess the calculation up, because the values are going to end up different. If a high-level character kills 400 kobolds one at a time, he won't get any experience for it at all. However, if he killed them all at once then he would get experience, because the CR for 400 kobolds would be high enough to warrant a challenging encounter for which XP can be awarded.
Bob Loblaw 08-13-07, 01:03 AM	Here are two things I worked on for a pet project of mine. The first is the how important the three types of saving throws are as you level as far as monsters are concerned. The second is similar, but for spells. The first table's Challenge Ratings are off by 1. For some reason Excel wants to add one to the table. I'll fix it when it becomes necessary. http://i83.photobucket.com/albums/j299/blzbob_photos/DnDSaves.jpg http://i83.photobucket.com/albums/j299/blzbob_photos/SpellSaves.jpg I am working on one for creatures and how much they need to worry about the different attacks that require opposed checks like Trip and Bull Rush.
shadzar 08-13-07, 01:07 AM	...no. A group of 400 kobolds is a much greater threat than one kobold. That is why it would mess the calculation up, because the values are going to end up different. If a high-level character kills 400 kobolds one at a time, he won't get any experience for it at all. However, if he killed them all at once then he would get experience, because the CR for 400 kobolds would be high enough to warrant a challenging encounter for which XP can be awarded. :ahem: sounds fishy to me. it would seem that no matter how many were killed at once the fact that the character was faced with 400 kobolds is of itself a threat. he can then take them out singularly or in groups. the initial threat is not lessened because he somehow got them to queue up and fight him one at a time. which i doubt any mob would do when faced with a single enemy. but if somehow he convinced them to fight him one on one and he beats them then he efectively overcame the challenge of the 400 kobolds since he defeated them with tactics.
Wight_Chocolate 08-13-07, 01:23 AM	:ahem: sounds fishy to me. it would seem that no matter how many were killed at once the fact that the character was faced with 400 kobolds is of itself a threat. he can then take them out singularly or in groups. the initial threat is not lessened because he somehow got them to queue up and fight him one at a time. which i doubt any mob would do when faced with a single enemy. but if somehow he convinced them to fight him one on one and he beats them then he efectively overcame the challenge of the 400 kobolds since he defeated them with tactics. You're missing the point entirely. The idea behind the calculation was to see how many fights with individual kobolds it would take for a character to reach level 20. It's a hypothetical situation. There isn't actually a big group of kobolds that the character has to work out how to deal with, and of course they aren't going to decide to fight him one at a time in a genuine combat situation. So let's say we have an empty room. Nothing in it but the character. One kobold appears, the character kills it, gets experience. He then rests, heals, buffs up, whatever. Then another kobold appears and is defeated. Rinse and repeat. The question was, how many times would that have to happen for the character to get enough XP to reach level 20? Having to fight a huge group of kobolds all at the same time is a completely different situation, and has nothing to do with the question that was being asked.
shadzar 08-13-07, 01:51 AM	i got that. but that is where i am talking about threat assessment. the one at a time ting means 0 kobolds are need to fight because the situation nullifies itself within the confines of the experiment. so in order to find out how many in general would be needed then you can just assign that number seeing it is possible that only a single kobold be killed at a time. mabe you can skewer two at a time with a long enough sword but who knows. so i read the question as per how many do you need to kill one at a time meaning killing one then the other, not just facing them one at a time before another appears. they can all be there at one time or come in waves so long as they are only killed one at a time.
Tedronai 08-13-07, 01:56 AM	The entire recent debate on the kobold issue is flawed. Increasing the number of opponents does NOT increase the CR of the encounter. It increases the EL. XP is determined individually per opponent. So, while those 400 kobolds might have an appropriate EL for a 10th level party, they provide no experience according to the chart (I would strongly argue for the DM to use the Ad hoc method in that instance, but otherwise they're SOL). The only method presented here for using 1st lvl npc class kobolds that MIGHT work is the all-at-once gamble and hoping that the DM doesn't either rule Ad hoc for extraordinary circumstances or simply that you cannot gain more than 1 level from experience gained in any one encounter (and I've seen DMs rule that by the session, which really hurts when a 3rd level party stubles unknowingly on that Deck of Many Things...bloody meddling demigods...:twitch: ).
Disciple_of_Salsa 08-13-07, 04:34 AM	Once I figured out how much damage a whale that fell from the upper atmosphere would do on average. I think it was 1206.
shadzar 08-13-07, 06:16 AM	Once I figured out how much damage a whale that fell from the upper atmosphere would do on average. I think it was 1206. are you sure it wasn't 42? ;)
Calsan 08-13-07, 07:27 AM	Isn't it actualy 70 the average of the 20d6 dice that is the max of falling damage acording to RAW? Something to do with friction ;) (and DMs lazyness)
Sajek 08-13-07, 09:33 AM	Yes, the average damage of dice is ([# of dice] x [dice value+1]/2. For example, the average of 20d6 is 20x7=140, divided by 2=70.
shadzar 08-13-07, 04:29 PM	Yes, the average damage of dice is ([# of dice] x [dice value+1]/2. For example, the average of 20d6 is 20x7=140, divided by 2=70. :confused: why do you assume that an average means dividing by two? figuring averages you divide by the number of things, not just some random number.
Novacat 08-13-07, 06:38 PM	:confused: why do you assume that an average means dividing by two? figuring averages you divide by the number of things, not just some random number. Dividing by the number of dice would result in the average individual die result, not the average total. His calculation is correct for determining the average total of a pool of dice.
lulzapalooza 08-13-07, 09:44 PM	my bbeg was knocked off the top of a pyramid, my party celebrated, and then, after using them fancy pythagoram formulas, I decided that he just slid down the side of the pyramid and didn't take much damage. The party quickly stopped celebrating when the wizard rolled his inteligence check and determined that he was still alive, and would be waking his way around the base of the pyramid and up the stairs leading to the top of the pyramid to continue the fight in sixty-six seconds, which is about enough time for each member to cast a buff spell eleven times.
DropsonExistance 08-13-07, 10:50 PM	When I explained the structure of my world to my friends, they had to do some calculations. I don't know if anyone's familiar with it, but my world is based off the Dyson sphere theory, which in science is basically making a giant sphere of solar panels that can completely encompass a star and collect energy from it. In science fiction, it has evolved to be an inside-out world, with life along the inside of the crust and the sun in the middle. Go wiki it if you're really interested :detect: Anyway, they immediately began pointing out problems with the design realistically. Like ventilation for the sun (solution: a pattern of cracks in the shell to allow ventilation), gravity (symmetrical spacing of the cracks to balance the shell, which rotates around the sun, hopefully kept balanced), and distance from the sun to the inside of the shell. This is where the real fun came in... because the sun would have to be really damned far away so as to avoid frying the people. So they started calculating the diameter of my planet. I think they came to 14.5 light minutes, or something like that.
shadzar 08-13-07, 11:03 PM	When I explained the structure of my world to my friends, they had to do some calculations. I don't know if anyone's familiar with it, but my world is based off the Dyson sphere theory, which in science is basically making a giant sphere of solar panels that can completely encompass a star and collect energy from it. In science fiction, it has evolved to be an inside-out world, with life along the inside of the crust and the sun in the middle. Go wiki it if you're really interested :detect: Anyway, they immediately began pointing out problems with the design realistically. Like ventilation for the sun (solution: a pattern of cracks in the shell to allow ventilation), gravity (symmetrical spacing of the cracks to balance the shell, which rotates around the sun, hopefully kept balanced), and distance from the sun to the inside of the shell. This is where the real fun came in... because the sun would have to be really damned far away so as to avoid frying the people. So they started calculating the diameter of my planet. I think they came to 14.5 light minutes, or something like that. your campaign was a Hollow World campaign?
Raymond_Luxury_Yacht 08-14-07, 07:24 AM	Isn't it actualy 70 the average of the 20d6 dice that is the max of falling damage acording to RAW? Something to do with friction ;) (and DMs lazyness) That's max from height. Damage from getting hit by a falling object includes damage from weight, and that does not cap. Hmm. Let's take a 60 ton whale, and have it fall from the upper atmosphere. That'd be 620d6 damage to anyone it hit. 2170 damage on average.
Uglysad 08-14-07, 09:24 AM	I don't remember where I saw it but someone figured out how many hit points (more or less) the earth would have. I don't remember exactly how it was calculated (it was a while ago) but I found it amusing. Perhaps someone else around here remembers it better.
Myers 08-14-07, 09:34 AM	are you sure it wasn't 42? ;) No I think that would be a bowl of petunia's that fell from the atmosphere.
Elodia7 08-14-07, 03:19 PM	So, here’s an interesting series of calculations, inspired by this thread. The results actually turned out to be more useful and revealing than I anticipated. For my own reference, I decided to determine once and for all how much space coins take up, how much they weigh and how many can be carried. As my players are getting up to higher levels, I’m beginning to question their ability to transport some of the treasures, particularly since only one party member has had the sense to invest in a bag of holding so far. First, I needed to choose a coin size. To give my players reference to something familiar, I decided for simplicity (not realism), that all copper, silver gold and platinum coins would be exactly the same size, thus having identical volumes. I decided to make them functionally identical in size to the Canadian 2$ coin, as the largest coin my players are familiar with. The coin is 1.10 inches in diameter and 0.07 inches thick. A quick calculation results in a fairly small volume, of 0.07 cubic inches. At this size it would take a whopping 25,976 coins to fill 1 cubic foot of space. A vast fortune, particularly if it is in gold coins. But coins also have mass. And each metal has a different density. A quick reference to a web page that lists metal densities reveals the following densities: Copper: 0.33 Pounds/cubic Inch Silver: 0.38 Pounds/cubic Inch Gold: 0.70 Pounds/cubic Inch Platinum: 0.77 Pounds/cubic Inch As such, we multiply density by volume to obtain the mass of each coin: Copper: 0.022 Pounds Silver: 0.025 Pounds Gold: 0.046 Pounds Platinum: 0.051 Pounds Now this may not seem like much, but it will add up quite quickly. Remember the number of coins that fit into one cubic foot? If we found 25,976 copper coinds, the total would weigh 571 pounds. This would be a challenging lift for any person. In fact, one pound can be represented by: Copper: 46 Coins Silver: 40 Coins Gold: 22 Coins Platinum: 19 Coins Not a huge number of coins. A quick check in the D20 SRD, we find out that a player would need a strength score of 23 to carry nothing but 1 cubic foot of copper coins at a heavy load. So, mass is the limiting factor, not volume. In fact, an average person (Strength 10) could only carry 4600 copper coins, and that would be a heavy load while they were wearing absolutly nothing else. Not terribly likely. Of course, we do have bags of holding, the adventurere’s best friend when it comes to carrying heavy loads. Another quick look at the D20 SRD gives us the following capacities for bags of holding TYPE Contents Weight (lb) Contents Volume (cubic feet) I 250 30 II 500 70 III 1000 150 IV 1500 250 Now, if we just look at volume, a bag of holding TYPE I can hold 779,276 coins of any type. Thus we can see, the weight limits will come into effect long before the volume limits. So, just looking at the weight limits of a bag of holding, an adventurer can carry the following numbers of standard coins: TYPE Copper Silver Gold Platinum I 11,558 9,907 5,390 4,861 II 23,116 19,814 10,780 9,722 III 46,233 39,628 21,559 19,444 IV 69,349 59,442 32,339 29,165 While a bag of holding can carry a significant fortune, it is by no means a silver bullet for recovering treasure. With these limitations, if high level characters find a dragon hoard it could be very challenging for a party to get it back to town. Of course, now that we have a defined mass for a coin, we can assign a value to putting swords and other items into the bag, as it will rapidly start counting against the coin capacity. A long sword at 4 lbs will cost 88 gold to place into a bag of holding which is currently filled with gold coins. A masterwork longsword can be sold for 157 gold, and so is more efficient to carry than gold coins, particularly considering that the volume limits for a bag of holding are extremely high. Naturally, if one was to adjust the coin size to make it larger or smaller, these numbers would be meaningless. I chose the 2$ coin because it’s relatively close to the size I tend to imagine these coins being. I hope you found my estimates and calculations interesting.
Chaze 08-14-07, 03:30 PM	When I explained the structure of my world to my friends, they had to do some calculations. I don't know if anyone's familiar with it, but my world is based off the Dyson sphere theory, which in science is basically making a giant sphere of solar panels that can completely encompass a star and collect energy from it. In science fiction, it has evolved to be an inside-out world, with life along the inside of the crust and the sun in the middle. Go wiki it if you're really interested :detect: Anyway, they immediately began pointing out problems with the design realistically. Like ventilation for the sun (solution: a pattern of cracks in the shell to allow ventilation), gravity (symmetrical spacing of the cracks to balance the shell, which rotates around the sun, hopefully kept balanced), and distance from the sun to the inside of the shell. This is where the real fun came in... because the sun would have to be really damned far away so as to avoid frying the people. So they started calculating the diameter of my planet. I think they came to 14.5 light minutes, or something like that. This is a bit off-topic, but wouldn't this world have sunlight all of the time, with no night... That is, if I am picturing this correctly.
Cavienn 08-14-07, 03:56 PM	Regarding the coin calculations above: A straight measure of volume isn't sufficient to accurately represent the space coins would take up. Even if they were neatly stacked, cylinders don't lock together perfectly, there is significant air space. And it is unlikely that a treasure hoard would be neatly stacked, which would create more air space. Thus, fewer coins could realistically fit into that cubic foot. I'm not about to try any crazy calculations, especially since it would be nearly impossible to judge how much wasted air space there was, but regardless, you can probably assume that weight is still the more limiting factor. The Draconomicon also has some calculations about the sizes of hoards, I believe, stating that an actual bed of coins for a dragon would represent vastly larger hoard than dragons actually have. Otherwise, very useful calculations--I may make use of those!
vegetalss4 08-14-07, 04:04 PM	No I think that would be a bowl of petunia's that fell from the atmosphere. you are both wrong, 42 is the meaning of life the universe and everything
Elodia7 08-14-07, 04:42 PM	Regarding the coin calculations above: A straight measure of volume isn't sufficient to accurately represent the space coins would take up. Even if they were neatly stacked, cylinders don't lock together perfectly, there is significant air space. And it is unlikely that a treasure hoard would be neatly stacked, which would create more air space. Thus, fewer coins could realistically fit into that cubic foot. I'm not about to try any crazy calculations, especially since it would be nearly impossible to judge how much wasted air space there was, but regardless, you can probably assume that weight is still the more limiting factor. The Draconomicon also has some calculations about the sizes of hoards, I believe, stating that an actual bed of coins for a dragon would represent vastly larger hoard than dragons actually have. Otherwise, very useful calculations--I may make use of those! Good point, I'd forgotten about that little detail about packing round objects. For real world spaces, such as a backpack, dealing with the packing issue would be necessary as players would encounter both volume and mass problems. However, I noticed when checking the bags of holding, there is far more space in the extra dimensional space than there is mass capacity. In other words we have lots of space left for our coins to pack when we hit the mass limit of a bag of holding. Just to confirm, here is a series of calculations to confirm the amount of free space when the mass limit is reached using coins. Copper coins are the lightest, and so we can carry the largest number of these. So, if we do not run out of space with copper coins, none of the other coin types will be an issue. So, starting with the basics from my earlier post: TYPE Copper I 11,558 II 23,116 III 46,233 IV 69,349 Now, ignoring the packing problem, we can calculate the total volume of those coins if they were melted into a solid block. It would take 25,976 copper coins melted down to fill one cubic foot. The resulting block would weight 571 pounds. Next we compare that 1 cubic foot mass to the volume of each bag of holding to determine how much free space will be left when we hit the mass limits of the bag, which will confirm if we need to worry about the packing issue: TYPE Coin Volume Bag Volume Space Left I 0.45 cubic ft 20 cubic ft 19.55 cubic ft II 0.88 cubic ft 70 cubic ft 69.12 cubic ft III 1.77 cubic ft 150 cubic ft 148.23 cubic ft IV 2.66 cubic ft 250 cubic ft 247.34 cubic ft We certainly have lots of space for those coins to rattle around. The large volume would help significantly with objects that do not pack neatly at all (such as swords, armor, etc). In fact, with the exception of very large, light or unusually shaped objects, mass is probably almost always the limiting factor when loading a bag of holding. If the coins weighed absolutely nothing, and packing wasn't an issue, a type I bag of holding could hold 779,276 coins of this type. For the sake of interest, I looked up the methods for determining packing of cylinders (assuming we take the coins and roll them conviniently). Let's just say the math is much uglier than I would like to take on for a hobby. :D However, if anyone does happen to be interested, this article (http://mathworld.wolfram.com/CirclePacking.html) shows the complexity of the problem just at two dimensions. While it's not an unsolvable equation, it does make things challenging. Thanks for the interesting feedback!
stargate525 08-14-07, 04:50 PM	This is a bit off-topic, but wouldn't this world have sunlight all of the time, with no night... That is, if I am picturing this correctly. A Dyson's Sphere that needs night would be accomplished by placing a second, incomplete, sphere within the first one, thereby making shadows on the second, inhabited sphere and simulating night. Better question. If your sphere is 100ft thick, how many earths-worth of material would it take to construct the entire assembly, assuming that the interior 'night-sphere' is 1/5th the mass of the outer one?
keelsc 08-14-07, 04:52 PM	Once I figured out how much damage a whale that fell from the upper atmosphere would do on average. I think it was 1206. now wat kinda whale did you do the calculations for, i did it for a blue whale at 200 tons (400,000lbs) falling from 6000mi (316,800,000ft). thats 2000d6 for weight and 31,680,000d6 for hieght. thats any where from 31,682,000 - 190,092,000 damage. feel free to correct my math
vaerdhdragonkin 08-14-07, 05:19 PM	you are both wrong, 42 is the meaning of life the universe and everything Forgive me, but you are also mistaken. 42 is the answer to The Ultimate Question Of Life, the Universe and Everything. Good luck finding the question itself. :P
stargate525 08-14-07, 11:58 PM	Forgive me, but you are also mistaken. 42 is the answer to The Ultimate Question Of Life, the Universe and Everything. Good luck finding the question itself. :P What do you get when you multiply nine by five?
shadzar 08-15-07, 12:25 AM	No I think that would be a bowl of petunia's that fell from the atmosphere. no. sorry, but the answer is always 42. ;)
shadzar 08-15-07, 12:27 AM	Naturally, if one was to adjust the coin size to make it larger or smaller, these numbers would be meaningless. I chose the 2$ coin because it’s relatively close to the size I tend to imagine these coins being. I hope you found my estimates and calculations interesting. $2 coins are about as queer as the $2 bill. :P and they don't help non-canadians very much. :mad:
B4_life 08-15-07, 01:36 AM	http://en.wikipedia.org/wiki/Toonie wiki article on the two dollar coin. images not to scale, but meh.
Elodia7 08-15-07, 11:22 AM	$2 coins are about as queer as the $2 bill. :P and they don't help non-canadians very much. :mad: Your objection is fair, I just chose to go with what I know the best. Here is how a 2$ canadian coin compares in size to other coins: USA Coins: It is almost exactly the same thickness as a USA Quarter. It's just a bit larger in diameter than the Quarter and a bit smaller in diameter than a USA 50 cent coin. Euro Coins: It is almost exactly the same thickness as a 0.10 Euro Coin. It is just a bit larger in diameter than the 2 Euro Coin. I could probably dredge up more but this should do the trick for now. Hope this helps. :)
Clawhound 08-15-07, 12:06 PM	Darkwood is worth 10 gp/pound. A mature oak weight 75 tons. Let's assume that a mature darkwood tree weights the same. A ton is 2,000 lbs. A ton of darkwood is 20,000 gp. Therefore, a darkwood tree is worth a potential 1.5 million gold pieces. Assuming 20 trees per acre (a modest estimate for a forest), you would have 30 million gold pieces per acre of forest. Assuming that your forest covered 1 square mile, that's x640 acres, or 19.2 billion gold pieces per square mile. Hmmm. Maybe this isn't a useless calculation.
Dazzer 08-15-07, 03:43 PM	This might be a little funny as well as un-needed. So in the Eberron campaign I am playing in Tuesday nights (this happened yesterday). Our warforged decides to take initiative and scout out our next adventure place, before we go there. The rest of the party has stuff to do in town for 2 days (sharn) and so we are delayed. He takes a lightning rail at 90 miles per hour, for 6 and a half hours (DM said this). So, later on, in Sharn, the party is asked for spot checks. The DM however had forgotten that our warforged has left to go scouting.... Well he picked up the dice, and roll a 12. Added his wisdom, and then proceeded to get a calculator and subtract 1 from his roll per 10ft he was away from the party.... He ended up scoring -93000 or so on the spot check... It was very funny at the time.
shadzar 08-15-07, 06:59 PM	Darkwood is worth 10 gp/pound. A mature oak weight 75 tons. Let's assume that a mature darkwood tree weights the same. A ton is 2,000 lbs. A ton of darkwood is 20,000 gp. Therefore, a darkwood tree is worth a potential 1.5 million gold pieces. Assuming 20 trees per acre (a modest estimate for a forest), you would have 30 million gold pieces per acre of forest. Assuming that your forest covered 1 square mile, that's x640 acres, or 19.2 billion gold pieces per square mile. Hmmm. Maybe this isn't a useless calculation. :OMG! any kingdom with a darkwood forest is rich!
stargate525 08-15-07, 08:55 PM	Darkwood is worth 10 gp/pound. A mature oak weight 75 tons. Let's assume that a mature darkwood tree weights the same. A ton is 2,000 lbs. A ton of darkwood is 20,000 gp. Therefore, a darkwood tree is worth a potential 1.5 million gold pieces. Assuming 20 trees per acre (a modest estimate for a forest), you would have 30 million gold pieces per acre of forest. Assuming that your forest covered 1 square mile, that's x640 acres, or 19.2 billion gold pieces per square mile. Hmmm. Maybe this isn't a useless calculation. Minus 50-60% for the weight of the unusable portions of the tree (leaves, twigs, bark), and a further 20-30% for constant replacement of axe heads, as well as payroll for your workers. So, you're actually making between 2 and 5 billion, and I haven't even begun to calculate the rarity of the tree, the length of time it takes to regrow, the decrease in density of trees by having other species, etc.
shadzar 08-15-07, 09:23 PM	Minus 50-60% for the weight of the unusable portions of the tree (leaves, twigs, bark), and a further 20-30% for constant replacement of axe heads, as well as payroll for your workers. So, you're actually making between 2 and 5 billion, and I haven't even begun to calculate the rarity of the tree, the length of time it takes to regrow, the decrease in density of trees by having other species, etc. well then give it to us with those calculations. i think the idea was for a forest entirely of this one tree. also the average weights of trees don't count the minute amount that leaves take up. and who says twigs and bark is useless? also who said that cuting it down require axes? bah im still working on them darn kobolds to find a way to get from 1 - 20 reasonably, but can't see, to find anyway a single PC could do it without a world covered in kobolds worth of space. :(
The Mighty Rex 08-16-07, 01:19 AM	Sometimes I'll roll all seven of my dice at once, keep the ones that roll their maximum, and reroll the rest. I'll repeat this until all the die have their maximum value, counting the total number of rolls. The best I've ever done is five rolls... I've never really bothered to figure out the average number of rolls, although it would be pretty straightforward, because that would ruin all the fun.
shadzar 08-16-07, 02:33 AM	Sometimes I'll roll all seven of my dice at once, keep the ones that roll their maximum, and reroll the rest. I'll repeat this until all the die have their maximum value, counting the total number of rolls. The best I've ever done is five rolls... I've never really bothered to figure out the average number of rolls, although it would be pretty straightforward, because that would ruin all the fun. what? :ahem: so you do or do not want to know the average number of rolls in order to get the maximum numebr on each dice in a standard polyhedral set? :thinks:
Sirkus 08-16-07, 04:16 AM	An epic spell capable of destroying the planet would require a spellcraft DC of about 153,012,284,979,068,099,263,689. Calculations (all values referenced from wikipedia): Assuming the Earth is more or less made of stone, it has 15 hit points per inch of thickness and hardness 8 (needless to say hardness is negligible). If we use the diameter of the Earth as thickness we have 6378200m(radius of earth) x 2 x 3.28084 ft/m x 12 in/ft x 15 hit points/in = 7533307327.68 HP Divide this by 3.5 to get the number of d6's required to do the job on average = 2152373522d6. In the rules for creating epic spells, the destroy seed gives you 20d6 to start (DC 29) with each extra dice bumping the DC up +2. So 2152373502 dice x 2 = +4304747004 for a total of 4304747033. But... the destroy seed only allows the target to have a total volume of 10 cubic feet. To increase the target volume, we must add +4 for every extra 10 cubic feet we add on (see table 2-2 Epic Spell Factors, EPH) The volume of the Earth is 1083206246123080894852 cubic metres, or 38253071244765948629164 cubic feet. This means we need to add 10 cubic feet 3825307124476594862917 times (rounded up), which, times 4, comes to an additional bonus of +153012284979063794516656. This makes the total Spellcraft DC up to 153,012,284,979,068,099,263,689. Happy smashing!
schpeelah 08-16-07, 04:57 AM	You are wrong. The surfacial layer of stone is only 40km thick, lower are just molten rocks, and the core is 90% iron. You have to include it in your calculations.
Sirkus 08-16-07, 11:39 AM	You are wrong. The surfacial layer of stone is only 40km thick, lower are just molten rocks, and the core is 90% iron. You have to include it in your calculations. You're right of course, but I didn't know how to deal with the molten part so I just made the assumption that the whole thing would act more or less like stone. If anyone can be more accurate that would be very interesting...
shadzar 08-16-07, 09:56 PM	just reposting this useless calculation. please number it whatever number it should be on the list. Something just occurred to me. It seems really stupid, but apparently, those who are complaining seemed to have overlooked it. Assuming that you consider OD&D, BD&D, D&D, AD&D, AD&D2, and AD&D2.5, then TSR created 6 editions in about 25 years, which comes out to something around 1 edition every 4-5 years. Now, lets compare this to Wizards. They released 3.0 in 2000, 3.5 in 2003, and assuming they release 4.0 in 2008/2009 that comes out to *gasp* a release about every 4-5 years! :eek: :eek: Wizards is *not* pumping out editions faster than TSR did. They just happen to space the editions slightly more evenly. :D and oddly you are talking about things that conflict each other. 3.5 was an upgrade to fix/replace 3.0 2e was an upgrade to fix/replace 1e 2.5 was an upgrade to augment 2e OD&D/BD&D were there own games. we have 3 games here. D&D, AD&D, and 3.x+ we will call them. D&D and AD&D were two very different games and by no means did they replace each other. Rules Cyclopedia for O/BD&D came out AFTER 1e did. which mean both game ran at the same time. now in your calculation is you seperate them into the two games they are rather than trying to say all these are the same game then what do you have? first we will remove 2.5 since it cannot stand alone. it requires 2.0 and is a supplement of it unlike the truely named 3.5 in comparison to 3.0. we have now taken your 6 down to 5 editions. rules cyclopedia was (i forget which) just a complied form of the OD&D or BD&D rules books into one place. so it can't really be seen as a standalone since it is just a compilation. so we have 5 edition for two games spanning over 26 years. (we will leave D&D as a standalone for reason i wont even go into) 3 editions for D&D over 27 years. (rules cyclopedia in 1991, chainmail in 1973/74) that comes out to one edition every 9 years. 2 editions of AD&D over 22 years (1e in 1978 until 3.0 in 2000) that comes out to one edition every 11 years. :thinks: 3.0 from 2000 until 2003. 3.5 from 2003 until 2008. 4.0 in 2008 3 editions in 8 years. that comes out to one edition every less than 3 years. the lifespan of the WotC editions are getting shorter by a LONG way! i mean the 3 WotC editions have all come out in the single span of one editions' lifespan for D&D by TSR. that is something that says a lot. if you also include the fact that WotC has the foundation of those other editions it would seem that they should be making a game with a lifespan logner than that of AD&D since it was the newer TSR edition and lasted longer. so why is it that WotC can't meet those standards? why does WotC need to keep fixing the game so often that it causes the consumer to keep repurchasing things and diluting the player base making it harder for people to find a game? by comparing all TSR editions you made a mistake since you failed to recognize they were two very different games. just like 3.0 and 3.5 are two very different games. now saying 4.0 is also vastly different it means that WotC has produced 3 games over the course of its control of D&D that keep replacing the previous game. doesn't look good from a reliability standpoint. if you had to constantly get your auto serviced because the mechanics put in fault parts wouldn't you look for a new mechanic, or new auto that doesn't break down so often? likewise players should be looking at why WotC can't make a game that lasts longer that just a few years without needing to replace it. ;) things to think about with your calculations.
Raymond_Luxury_Yacht 08-17-07, 02:35 PM	now wat kinda whale did you do the calculations for, i did it for a blue whale at 200 tons (400,000lbs) falling from 6000mi (316,800,000ft). thats 2000d6 for weight and 31,680,000d6 for hieght. thats any where from 31,682,000 - 190,092,000 damage. feel free to correct my math Damage from height caps at 20d6.
Sajek 08-17-07, 05:09 PM	Minus 50-60% for the weight of the unusable portions of the tree (leaves, twigs, bark), and a further 20-30% for constant replacement of axe heads, as well as payroll for your workers. So, you're actually making between 2 and 5 billion, and I haven't even begun to calculate the rarity of the tree, the length of time it takes to regrow, the decrease in density of trees by having other species, etc. Still, two billion is more than the combined wealth of most continents, if not worlds.
Rageheart 08-17-07, 05:30 PM	Still, two billion is more than the combined wealth of most continents, if not worlds. Don't forget that darkwood's density is far less than oak, so a darkwood tree would weigh a fraction of the weight of a similarly sized oak tree.
glorfon 08-19-07, 12:05 PM	9. Using the level-independent experience awards from UA a normal MM version of the Kobold grants 75 exp upon death. Also a character needs 2.6 million exp to hit level 20. Therefore to gain 20th level a character needs to kill 34,667 kobolds one at a time. At higher levels that would make for a pretty quick level as you slaughter kobolds. I might edit this to do a non UA version, but I think this is sufficient. My next plot hook: 34,667 kobolds are atempting to break into the city via a very narrow tunnel. Position yourself at the exit with your cross bow and kill them one at a time.
Simrobert2001 09-17-07, 07:16 PM	Actually, that won't work. At some point, mobs will no longer grant xp. so, therefore, you won't get level 20.
Bauglir 09-17-07, 07:52 PM	And, according to the PHB, you can't gain enough XP to level up twice from a single adventure, anyway. So either each kobold is a separate adventure, and so you stop at level 9, since they're no longer giving XP, or they're all the same encounter, and you thus stop at 1 XP short of level 3.
Disciple_of_Salsa 09-17-07, 08:29 PM	Why don't you just shout at the kobolds while you kill them and get to lvl 20 off RP XP. Something about this 33,667 kobold rampage seems a little Belkar-ish to me.