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Ravingdork
2010-04-08, 12:48 AM
My GM has allowed me to weaponize the stone shape spell, effectively shaping stone spikes that come out of the floor (instantaneously) to impale my foes.

My question is thus: If I am crafting spikes that are each 5 foot long cones with a 1 inch radius at the base, how many such cones can I fit into 17 cubic feet (the minimum volume for stone shape).

Please, if you will, explain to me how you came to the answer (all the while keeping in mind that I am a complete mathematical retard).

sonofzeal
2010-04-08, 12:53 AM

V = (1/3) * pi * r^2 * h

V = (1/3) * 3.14 * 1^2 * 60

V = 62.83 cubic inches

Your space is 17 cubic feet, or 29376 cubic inches. Therefor the upper bound on the number of spikes is 467, depending on how efficiently you can pack them.

Erom
2010-04-08, 12:55 AM
Volume of a cone: V = 1/3 * pi * r^2 * h

h = 5 feet = 60 inches
r = 1 inch

V = 62.8 cubic inches = 0.04 cubic feet per spike

17/0.04 = 467.7 spikes in 17 cubic feet

I'm not sure if you should truncate or round at this point, but there you go.

Note that depending on the type of stone, that's way too "pointy" - stone gets really brittle when it gets thin. Realistically, you should have to use "fatter" cones. But, you know, catgirls.

Edit, hehe my numbers were silly at first, me and google got in an argument about order of operations.

TheYoungKing
2010-04-08, 01:02 AM
Ha, I sighted the mistake in your original post, SoZ. Then I got ninjaed, as I figured I would.

sonofzeal
2010-04-08, 01:04 AM
I'm not sure if you should truncate or round at this point, but there you go.
Truncate. You can't have 0.7 spikes (though you could have a smaller one). Also, packing efficiency comes into play unless he makes them non-conical. Maximal packing efficiency for circles in a plane is roughly 0.9069, so my estimate then is 423 spikes.

2010-04-08, 01:07 AM
[professor's voice]Ahem,
I must concur with my esteemed colleague, sonofzeal. I find that the maximum number of spikes you could create out of 17 cu. ft. of stone with the dimensions given to be roughly 467 and a bit. Though I'm not sure, upon rereading your post, if you meant 17 cu. ft. to fit the spikes in, or 17 cu. ft. of material from which to make the spikes. My mathematical processes match sonofzeal's, and have been confirmed by Wolfram (http://www.wolframalpha.com/input/?i=((1/3)(pi)(5')((1%22)^2))) Alpha (http://www.wolframalpha.com/input/?i=17/.03636) (which is a very useful site for such things, if you can phrase your queries such that they are understood).[/professor's voice]

Oh, and I've been ninja'd. But still, Wolfram Alpha (http://www.wolframalpha.com/) is pretty nifty, so I guess I'll post my now mostly extraneous verbiage.

Ravingdork
2010-04-08, 01:13 AM
467! Goodbye opposing army! You've all been impaled!

I'm taking a (theoretically infinite) amount of stone (such as a cave floor), of which I can only effect 17 cubic feet, and shaping it into spikes. Since it's magic, I don't think packing efficiency is really a concern (especially since I'm not trying to fit them into any kind of "package").

I'm not too concerned about the thinness/brittleness either. If the spikes break off in my opponent's flesh, all the better!

EDIT: If each spike has a 2 inch diameter base, then I could potentially have a circle (5 ft. diameter) of 134 spikes pointed up from the floor and inwards from the circle impale a single medium target.

Ouch.

Eloel
2010-04-08, 02:02 AM
Theoratically, it's actually more than 467, since the stone you're 'digging' also helps you shape the spikes. So, half the spike is done because you removed stuff around it (leaving the spikey shape), and the other half is done because you created it with the stuff you removed. It's not a x2, but it's very well worth considering.

Ravingdork
2010-04-08, 02:17 AM
Theoratically, it's actually more than 467, since the stone you're 'digging' also helps you shape the spikes. So, half the spike is done because you removed stuff around it (leaving the spikey shape), and the other half is done because you created it with the stuff you removed. It's not a x2, but it's very well worth considering.

I thought volume calculations too that into account...?

Optimystik
2010-04-08, 05:43 AM
Hire a Geometer (CArc.)

...Wait, what were we talking about?

mint
2010-04-08, 06:44 AM
How many spikes with the radius, r and height, h can be made out of the Volume, V, is being solved.

Though my understanding is that what you are interested in is the solution to the problem: How many dangerous pointy edges, P, can be made by manipulating the volume, V.

Also, don't do it.

cheezewizz2000
2010-04-08, 06:59 AM
Probably wouldn't be too overpowering to allow the spell to create 17 spikes 5' apart that do (CL/2 d6, max 5d6) damage as a one off and create difficult terrain thereafter.

Personally, if you're weaponising stone shape, you haven't thought through its uses enough. I use it as a divine caster to create ad-hoc shackles, loosen roof tiles to force ballance checks and to make the key-stones of arches less supportive (good-bye stone bridges). Combine with "flesh to stone" and "stone to flesh" for horrific results. Arcane casters don't get the luxury of casting it on-the-fly, as they need to have a clay model of what they want to make before they can start.

Ossian
2010-04-08, 07:18 AM
I might be completely off track, but having them come out of both ceiling and floor, while the overall volume of the target area stays the same, could possibly double that amount (say 900 spikes). The would be half as long, roughly, but interlock nicely and turn targets into something about the consistency of toothpaste :smallcool:

Ravingdork
2010-04-08, 09:07 AM
I might be completely off track, but having them come out of both ceiling and floor, while the overall volume of the target area stays the same, could possibly double that amount (say 900 spikes). The would be half as long, roughly, but interlock nicely and turn targets into something about the consistency of toothpaste :smallcool:

I likely wouldn't have to reduce the length of the spikes anyway. The number that I can make with a single casting is pretty staggering already and I will most likely have more than enough to mess up a couple of people.

The problem is, in our games at least, stone dungeon ceilings are generally pretty high. 5-foot spikes just aren't going to reach far enough. Sadly, if I make them longer, they become more brittle and eat up a heckavalot of volume.

sonofzeal
2010-04-08, 09:55 AM
Note that, assuming you must fit strictly inside the given space, I've revised my estimate to 423 spikes due to projected packing inefficiency. That's just an estimate, but the actual number is no higher than that

DragoonWraith
2010-04-08, 10:12 AM
Arcane casters don't get the luxury of casting it on-the-fly, as they need to have a clay model of what they want to make before they can start.
No, they don't. Spell Component Pouches come with as many lumps of clay in any desirable shape as you want. The material component has no cost, and is therefore negligible, and is therefore in a Spell Component Pouch. Or removed by Eschew Materials.

IMO, you shouldn't ever try to make the components more than they are. They're a joke, and should be treated as such. Trying to wag your fingers at arcanists and saying "uh uh uh!" with regards to spell components is, again IMO, just annoying and makes the game unfun. Steal or sunder the pouch if you must, but don't start making them keep track of individual components. That's just a bookkeeping nightmare.

Autolykos
2010-04-08, 11:07 AM
I thought volume calculations too that into account...?
Let me illustrate this with an image:
Black is untouched stone, white is air, red is removed stone, green is added stone.
If you just calculate the volume of the spikes, you assume that you just pull the stone you need from anywhere and put it where you need your spikes. (like from under the spikes, as shown in the left image). But you can also get the needed material from the gaps between the spikes, making the spikes higher (or allowing more spikes if you stay with the same height). How many more is not trivial to calculate, especially if your spikes have a round base. I'll do it for a square base (the results are also correct for a hexagonal or trigonal base):

So each spike has the height h1 and the base square has the size a*a. The spikes are lowered h2 into the floor (later we will fit h2 so that we remove the same amount of stone as we add). The volume of stone to be removed is
(a*a*h2)-(a*a*h1/3-a*a*(h1-h2)*((1-h2/h1)^2)/3)
while the volume of stone to be added is
a*a*(h1-h2)*((1-h2/h1)^2)/3
If we set those two terms to be equal, we get:
(a*a*h2)-(a*a*h1/3-a*a*(h1-h2)*((1-h2/h1)^2)/3)=a*a*(h1-h2)*((1-h2/h1)^2)/3
(a*a*h2)-(a*a*h1/3)+(a*a*(h1-h2)*((1-h2/h1)^2)/3)=a*a*(h1-h2)*((1-h2/h1)^2)/3
(a*a*h2)-(a*a*h1/3)=0
(a*a*h2)=(a*a*h1/3)
h2=h1/3
We can now insert this in the second equation to get the volume of stone to be moved:
a*a*(2*h1/3)*((2/3)^2)/3=a*a*h1*8/81
So for h1=60" and a=2" one spike would need 23.7 cubic inches of stone to be moved. You can move 29376 cubic inches, which would make 1239.3 spikes, which would fill an area of 34.425 square feet, or a little more than a 5x5' tile.
Note that the area is unaffected by the actual shape and size of the spikes' bases and is inversely proportional to the spikes' height (as long as the bases allow to cover an area without gaps you can go completely nuts (http://en.wikipedia.org/wiki/Penrose_tiling) with their shape).

EDIT: Found and (hopefully) fixed the error...

ericgrau
2010-04-08, 11:14 AM
^ Just to ballpark the above the volume of the added mater is 1/3 * (2/2) * (2/2) * (60/2) = 1/3*1*1*30 = 10 cubic inches. Plus since the floor is 3D not 2D the volume removed around the spike would be greater than this, so in reality the added matter should be taller and the removed matter should be shorter, resulting in a greater volume. So, not sure what's up, but something doesn't seem right.

V = (1/3) * pi * r^2 * h

V = (1/3) * 3.14 * 1^2 * 60

V = 62.83 cubic inches

Your space is 17 cubic feet, or 29376 cubic inches. Therefor the upper bound on the number of spikes is 467, depending on how efficiently you can pack them.

In a square arrangement that's 21x21 spikes covering at least 21x21 inches. If you leave a 1.5 inch gap between them then you may cover a 5 foot square. Thing is if you leave gaps then Autolykos trick would only provide a small benefit. But gaps should still give more area. I think one of the two posters may have made a math error.

Ravingdork
2010-04-08, 03:13 PM
:confused:

sonofzeal
2010-04-08, 09:54 PM
^ Just to ballpark the above the volume of the added mater is 1/3 * (2/2) * (2/2) * (60/2) = 1/3*1*1*30 = 10 cubic inches. Plus since the floor is 3D not 2D the volume removed around the spike would be greater than this, so in reality the added matter should be taller and the removed matter should be shorter, resulting in a greater volume. So, not sure what's up, but something doesn't seem right.

In a square arrangement that's 21x21 spikes covering at least 21x21 inches. If you leave a 1.5 inch gap between them then you may cover a 5 foot square. Thing is if you leave gaps then Autolykos trick would only provide a small benefit. But gaps should still give more area. I think one of the two posters may have made a math error.
Er, no. He said 17 cubic feet, which is very very different from a 5 foot square. Your 5 foot square is 125 cubic feet. I answered the question he gave.

ericgrau
2010-04-08, 10:58 PM
I explained that he could get enough spikes to cover a 5 foot square with your spikes and the spacing I gave. That has nothing to do with a 5 foot cube; the spikes are only on the floor. I'm taking about the battlemat not the spell's volume.