help cheaking my math on Fire Ball. | Page 1 | The Dungeons & Dragons Guild | Guild Forums

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O.G. Gekko

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Posted: Thu Apr 02, 2020 8:51 am

ok first the question: how meany 5 foot entities can fit into a Fire ball. now hear is all the info needed.
1. the fireball is a sphere not a circle (witch is what makes me wonder if i have my math right.) that means it is 40ft from one side to the other(20 feet on all sides from the center. 20x2=40)

2. the Fire Ball should be broken down into 5ft cubes.(to make certain that no mater if the entity is lying down, siting, standing they will fit)

3. the fire ball will be treated as if it is being cast a point in space well above the ground to have max AOE.

after using both MS paint & 3d pant to make a model using cubes representing 5 square feet. i got 78 cubes/entities. it seems a lot but i think i have it right.

Posted: Thu Apr 02, 2020 1:52 pm

40 if measured from a cross point, 41 if measured from a center square

PupSage

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O.G. Gekko

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Posted: Fri Apr 03, 2020 3:37 pm

PupSage

40 if measured from a cross point, 41 if measured from a center square

well yes but, no. that is if it was a 2d object i am wanting to know how many if it was a 3D object. so a big ball the pic from the books are all 2d and a 2d object can be a cross section of a 3d object but it is not a 3d object. i used Ms 3d paint to replicated the 2d map then turned it 90 degrees upwards and did so again then 90 degrees to the left copied it again over lapped all thees cross section and filled in any missing parts keeping the width of the sphere at 40 cubes in all directions. and i came up with 78 but i am not center this is correct as i have bean told my my teachers that doing this is not the correct way & very inaccurate.

Posted: Fri Apr 03, 2020 7:10 pm

Assuming, I correctly remember algebra. sweatdrop 3d objects are multiplied by 3

Volume of a sphere = 4/3 π r^3

Saving you from the math.

33510.32 ft^3

And then divide that by 5 ft^3(the space you allocated for each victim) = 6,702.064

So 6,702 vic's and someone's right pinky toe will be affected by your fireball. Assuming you eliminated the ground from the equation and can shepherd everyone into a freefall ball.

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PupSage

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Posted: Sat Apr 04, 2020 9:21 am

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PupSage

40 if measured from a cross point, 41 if measured from a center square

well yes but, no. that is if it was a 2d object i am wanting to know how many if it was a 3D object. so a big ball the pic from the books are all 2d and a 2d object can be a cross section of a 3d object but it is not a 3d object. i used Ms 3d paint to replicated the 2d map then turned it 90 degrees upwards and did so again then 90 degrees to the left copied it again over lapped all thees cross section and filled in any missing parts keeping the width of the sphere at 40 cubes in all directions. and i came up with 78 but i am not center this is correct as i have bean told my my teachers that doing this is not the correct way & very inaccurate.

So extrapolate the equation dude

120 or 129

The traditional equation for volume of a sphere won't work here since we are working with a cubed sphere in technicality.

Posted: Sat Apr 04, 2020 10:32 am

Southern Cross Nemesis

Assuming, I correctly remember algebra. sweatdrop 3d objects are multiplied by 3

Volume of a sphere = 4/3 π r^3

Saving you from the math.

33510.32 ft^3

And then divide that by 5 ft^3(the space you allocated for each victim) = 6,702.064

So 6,702 vic's and someone's right pinky toe will be affected by your fireball. Assuming you eliminated the ground from the equation and can shepherd everyone into a freefall ball.

i am slitly sceptical as it seems a excedingly large number but you do seem to also know what your talking about and i like your wording.. did you by eny chance use a blender before you placed the "vic's" in to the area of the Fire Ball? if so i commend you for your creativity & dedication to maximum efficiency.

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Posted: Sat Apr 04, 2020 4:39 pm

Nah, I said there was a sale on toilet paper inside a building-sized dehydrator and locked the door behind them. There is a lot of water in elves... twisted

But, now that I had 8 hours rest and have been proven wrong. I thought of a simple way to find out.
This site has a printable area of effects circles. The 20-foot radius circle is in the upper left corner.

Count the number of 20's and times that by 4(number of cubicles upward)
Count the number of 15's and multiply that by 3
Follow the pattern with 10 and 5. Take those numbers and double 'em. Full cubic sphere count.

So,
2(48+60+24+8 )
2(140)
280

Posted: Sat Apr 04, 2020 5:41 pm

Figuring out your widest point

8 squares for crosspoint, 9 for square

square is easier to explain.

Layer 1 is 1 square. 1x2
Layer 2 is 5 squares 5x2
Layer 3 is 13 squares 13x2
Layer 4 is 25 squares 25x2
Layer 5 is your widest layer so 41x1

Add them all up

Or for crosspoint
4x2
12x2
24x2
40x1

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Southern Cross Nemesis
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Posted: Sat Apr 04, 2020 7:10 pm

Go with what Pupsage says, I stopped my education at an associate's degree in a s**t field from a s**t college(don't ask) and she is working on her masters.

That said, wouldn't you count your crosspoint layer 5 a few times, due to it being a sphere(equal on all sides) and you have four in layer 1? Not arguing just asking.

Posted: Sun Apr 05, 2020 8:55 am

I think it has to be measured from a crosspoint to be a "true" 5' radius (in cubic space!).

OO
OO

For example, in 2D battlefield, a 5' radius spell should only hit 4 targets.

In 3D space, 5' radius would hit 8 targets.

10' radius, it adds another 2x2 set of targets to each face of the above 2x2x2 cube. 8 + 24 = 32 targets.

15' radius, it extends out another 2x2 set of targets (+24), and the existing extension gets another "layer" of cubes around it. They overlap a bit though, so don't double-count. A single second layer gets 2 more cubes on each side, turning it into this thicc cross:

OOO
OOOO
OOOO
OOO

However, each of those second red O's will be shared with another layer extension. How I'm thinking of it at early-o-clock in the morning is +8 for the top layer, +8 for the bottom layer, and +8 for the middle four that don't double up on the top or bottom. Another +24. So we're at 8+24+24+24 = 80 targets.

20' radius, here we go! Extend out another 2x2 on the edge like before (+24), the previous extension will chonk up but be far enough removed from each other to not overlap so 6x8=+48, the previously extended layer chonks a second time:

OOOO
OOOOO
OOOOOO
OOOOOO
OOOOO
OOOO

The red O's will be shared like in the previous example, so you get the same situation of +8+8+8 (top, bottom, and middle that aren't shared). The blue is it diagonaling out, again getting shared but it only gets counted once for the 8 45-degree parts of this sphere so another +8. My final tally is 8+24+24+24+24+48+24+8 = 184 total targets.

That all said, even though I'm a numbers guy, I'm extremely bad at visualizing math graphically. Generally, spheres aren't measured in cubic intervals so that breaks my brain even more, and I'm not confident of my answer. I ran out of d6s to visualize past the 10' radius example. I know every cube face basically gets another cube stuck to it from the prior example as it expands outwards. If we weren't in a quarantine position, I'd run to the store and get a bunch of marshmallows and toothpicks and make a visual example to post here as a thinly-veiled excuse to get an extreme sugar high later. Alas, not today.

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