The comparison to TV sizes was a really bad choice, because yes, in fact they are measured diagonally. This game on the other hand operates on a 5×5 grid system. A cube that is 10 feet across is 10x10x10.
So you would say that your TV is 50 inches across? Would you say that to someone? No. You would not. Because across would imply that you were measuring from one side to the other, instead of diagonally.
It's an inverse example. TVs are measured diagonally, thus, the use of the word across would be misleading. In this game, nothing is measured diagonally, so the meaning of across is very clear.
I think that most people probably wouldn't blink if you said 50 inches across, actually. And I don't think the definition of the word "across" precludes it from being on the diagnal. So technically and colloquially it seems like a bad example. Maybe if I was talking to a Best Buy employee or TV repairman, they'd correct such an oversight. But otherwise? Eh...
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Shape Stone says:Targets cube of stone 10 feet across or smallerHow big is this cube?
Someone says that 10 feet across means with 10 feets side, so the volume is 10x10x10, 1000 feet cubic.
But my interpretation is that 10 feet across means that 10 feets is the diagonal of two opposite corners of the cube and so the volume is 192.45 feet cubic as per below calculation:
https://encrypted-tbn0.gstatic.com/images?q=tbn%3AANd9GcSA-NdOPf42L_aya2RMx HpxL11XLMu6qoZQIKfVqqfAVPcfJ2d7
What do you think is the correct interpretation?
Mathematically the correct interpretation of the volume is indeed (10/sqrt(3))^3 or approx 192.45 square feets.
As a rules interpretation it is wrong. There is zero precedent of needing a calculator for calculations of square-roots or cubes. The limit of the Maths needed for Pathfinder is addition / subtraction and counting squares.
A 5 foot radius blast covers and area of 100 square feet (4 x 25) and not 78.5 which is mathematically correct. A 10 foot blast covers 350 square foot (14 x 25) and not 314.2 as mathematically correct.
On more than one occasion I have seen as GM players who needed their fingers to establish the AC they hit adding a d20 roll with a modifier <20.
I know it might feel wrong if you are mathematically inclined as the numbers in 3 dimensions start to differ a lot more then in dimensions. But that is a price to pay for simplicity.
Edit: Here is the actual Maths how it is done to get to a 10x10x10 foot cube even if we look at the diagonal of 10 feet. Diagonal of a cube with a 10 feet diagonal results in a side lengths of 5.774 feet. The center of the cube is at the origin at x=0, y=0, z=0. This means the eight corners are at x=+/- 2.887, y = =/- 2.887, z=+/- 2.887. Use all 8 permutations. As a last step we round to the closest 5 feet. This means x becomes +/- 5, y becomes +/- 5, z becomes +/- 5.
We now measure the side lengths of the resulting cube which is 10 feet. This means the volume = 10x10x10 = 1000 cubic feet.
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I agree that the game is about simplicity, but the difference between 1000 and 192 is way large that the difference between 100 and 78.5 and 350 and 314.2.
With the first interpretation I could create a cage around a large creature, while with the second I cannot exceed the measures of a medium creature.
See my edit.
And no - you can't create a cage around a large creature either way as the large creature is rounded to 10x10x10 as well and the cage needs to be > 10x10x10 for a large creature.
And no - you can't create a cage around a large creature either way as the large creature is rounded to 10x10x10 as well and the cage needs to be > 10x10x10 for a large creature.
Are you sure that the target requirements hold true for the resulting shape? Or can I make a half height 5x20x10 block out of the initial 10x10x10 block?
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I agree that the game is about simplicity, but the difference between 1000 and 192 is way large that the difference between 100 and 78.5 and 350 and 314.2.
With the first interpretation I could create a cage around a large creature, while with the second I cannot exceed the measures of a medium creature.
Some more Maths - and maybe you understand why Maths isn't being used.
Assumption 1 a volume of 192.45cubic feet.
Assumption 2 - I create a cage using this volume
Assumption 3 - the thickness of my cage = 2 inch.
I can generate a flat sheet of 192.45 * 6 = 1154.7 sqaure feet which is 2 inch thick.
A cube has 6 sides -> 1154.7 / 6 = 192.45 sqaure feet
sqrt(192.45) = 13.87 feet
Disregarding overlap of my 6 plates a volume of 192.45 cubic feet would create a cage of 13.87x13.87x13.87 feet.
Alas we round to the nearest 5 and my 2 inch get rounded to nothing and the nice cage just disappears into nothingness
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And no - you can't create a cage around a large creature either way as the large creature is rounded to 10x10x10 as well and the cage needs to be > 10x10x10 for a large creature.Are you sure that the target requirements hold true for the resulting shape? Or can I make a half height 5x20x10 block out of the initial 10x10x10 block?
I would guess you have 8 blocks of 5x5 - so the above should be perfectly fine.
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Some more Maths:
I worked out a formula for the overlap:
Overlap (assuming a cube) = 4/2 * Thickness / lengths of cube
So I should use approx. sqrt(42/41 * 192.45) for a mathematically more correct value of 14.04 feet and not 13.87 feet as in my first calculation.
Now it becomes mathematically truly interesting as we have a recursive formula. I need to the lengths to calculate the relative overlap but at the same time need the overlap to calculate the lengths.
I guess the next mathematical step would be to describe this as a mathematical series and to check the convergence.
Fun how much Maths a simple problem can yield if you want to do it properly.
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edit: I regard myself as pretty good in Maths - but I didn't see this one coming. My last calculation was building a cube of 14x14x14 feet and a thickness of 2 inch - assuming a volume of 192.45 cubic feet. Off course a cube is suboptimal. A sphere is more optimized. I assumed the result should be > 15 - but inquiring minds want to know.
What I did:
192.45 = 4/3 x pi x r^3 - 4/3 x pi x (r - 1/6)^3
V is the volume, I then calculate the volume for a sphere radius r and a spehere radius r-1/6 with 1/6 being 2 inch thickness.
Multiplying out the last part removes the cubic term and leads to a simple qudratic equation (using some rearrangements)
0 = r^2 - 1/6r - 3/2 x 192.45/pi - 1/216
Solving this leads to a radius of 9.67 feet or a diameter of 19.34 feet. This is nearly large enough to build a spherical cage around a gargantuan sized creature.
Guess by rearranging the formula I could calculate the thickness I get if I want a 20 foot diameter sphere. Must be just shy of 2 inch.
Guess if I carry on like that I get banned here - for trolling using Maths ...