Math masters wanted. Will pay 2x going rate.

[Chemus]

I need to solve an equation for one value in the equation. The value is bracketed and has an exponent to the bracket.

I don't know how to get that value out of its brackets/exponential entanglement.

The most simplified form of the equation that I know of is:

(Volume of Spherical Shell Wall)/4.1888 = (Radius of the hollow portion of the spherical shell + Thickness of the shell wall)^3 - (Radius of the hollow portion of the shell)^3

With bolded characters above as the variables in the equation it reads thus:

V/41.888 = (R+t)^3-R^3

How do I get the cube root off? Is there a better way (other than integration of spherical shell slices, which I've completely forgotten how to do)?

Please help me. Thank you.

[sirpercival]

call V/thing -> W for simplicity

W = (R-T)^3 - R^3
W = -3R^2*T + 3R*T^2 - T^3
W = -T(3R^2 - 3RT + T^2)

This will give you complex answers...

[Garryl]

(R+t)^3-R^3
= (R^3 + 3*R^2*t + 3*R*t^2 + t^3) - R^3
= 3*R^2*t + 3*R*t^2 + t^3
= 3Rt(R+t) + t^3
... which may be an easier place to start, at least, depending on what you're looking for.

Oy! Sirp! You ninjaed me while I was trying to figure out why he's making domes out of Wall of Stone.
Also, it's R+t, not R-t.

[sirpercival]

Oopsh.

EDIT: The other thing to do is look at your limiting behaviors.  If R>>T then R + t ~ R, and you have W ~ t^3

[Chemus]

Oh dang! I forgot to say which variable I wanted to solve for! I need the radius R!

However with both your help, I might get there:

V/4.1888 = 3Rt(R+t) + t^3

V/t*4.1888 - t^2 = 3R(R+t)

(V/t*4.1888 - t^2)/3 = R^2 + Rt

aaand I think I'm stuck again. Still trying to solve for R.

(And yes, this is indeed for Wall of Stone.)

[sirpercival]

Clearly it can't be explicitly solved.  My suggestion is to graph it, or do it numerically.

[Chemus]

Dang! It sucks that it looks unsolvable. I'll have to make the table of spherical and hemispherical Walls of stone manually, rather than using a single formula. Not gonna leave an integral on a page, even if I can remember how.

Thanks a lot guys. I did not know, or perhaps I forgot (since it's vaguely familiar now that you two bring it up), that (x+y)^3 = (x^3 + 3x^2y + 3xy^2 + y^3)

Moving on to a cylindrical Wall of stone:

Solving for a Cylinder (with no top): Pi*r^2*h = cylinder vol. So subtract a smaller cylinder to get volume of a round wall of stone. t = thickness of wall/ differential between smaller and larger cylinder.

V = Pi*H*((r+t)^2 - r^2)

V = Pi*h*(r^2+2rt+t^2-r^2)

V/Pi*h = 2rt +t^2

(V/Pi*h*t - t)/2 = r

Did I miss anything?

[wotmaniac]

what exactly is it you're trying to do?  what is the whole context?
without integrating the function, you're gonna have to start replacing variables with knowns.

[nijineko]

he's trying to come up with a formula to calculate how big a sphere or cylinder or other geometric hollow shape he can construct from the wall of stone spell.

in the other thread we hijacked before he started this one, i was using a 12th level caster, for a flat wall of 3" thick and 12 5x5 squares.

assume the thickness is 1" is what he said before. chemus is looking for the final inner and outer radius. so "t" is 1". i believe my figure for the cubic volume of a standard wall was 10,800". but please double check my math. we should be able to plug the cubic volume in, as we are just changing the shape of the total volume. so 10800 should be "V". or "W" = 2578.30404889228.


(i'm assuming he, please forgive if i'm wrong.)

stone is assumed to be granite, as per material components, and i imagine that someone will ask for load bearing capacity of the geometric hollows, next. =P

[Chemus]

The spell wall of stone has an explicit amount of stone that it makes at any CL. That is 1/4CL inches by (60^2)CL inches. Or 900 cubic inches per CL. It's shapeable, and you can make it thicker by reducing the thickness area. I'm trying to make a formula for maximum Wall of Stone radii for Spheres and hemispheres, and for Cylinders.

I think I solved the Cylinder equation above for the Radius. Just plug in desired height (h), desired wall thickness (t) and volume of stone available at your CL (V), and you have your maximum radius. I already plugged the Cylinder equation above into a spreadsheet and got a 1-20 table that gives the radius for 5' tall cylinders from 1" to 12" thick.

I will have to fill in the thickness (t) in the above equation and see if I can then solve for R at different thicknesses. If so, then I can make another table.

edit: littha's proofreading is mean! j/k thanks

[nijineko]

may i request in advance a copy of your spreadsheet?

[littha]

you can make it thicker by reducing the thickness.

 

[wotmaniac]

I'm trying to make a formula for maximum Wall of Stone radii for Spheres and hemispheres, and for Cylinders.

Okay, that's what I was looking for.
Well, as I see it, you're just gonna have to decide on a minimum thickness, and go from there (bringing you down to only 1 variable, which will then solve itself)  ............. after you factor in the relevant caster level.

[Chemus]

I found an online source that had an automatic solver: Number empire

The other three i had tried before asking for help here did not solve it, and I assumed that it was too difficult.

Here's its solution:
r = (+/-) (sqrt(3)*sqrt(3*pi*t*V-pi^2*t^4)-3*pi*t^2)/(6*pi*t)

Of course I'm going to test the positive ones. The test for me is whether the original formula works backwards from the results...

Test success! Now I need to examine the equation again...see where I went wrong.

And my apologies for starting with 4.1888 (an approximation of 4/3*pi). That might be what sent folks on a wild goose chase. I'll return my findings here to spread the good word

[nijineko]

sweet. i was suspecting that taking of roots would be necessary. but i couldn't find anything that would convert an equation. i will be using that site, thanks!

i am looking forward to news!

[Chemus]

OK, so I haven't gotten my head around the the algebra needed to actually do the math myself, but I do have this nifty wall of stone spreadsheet.

Since I can't attach even a 30k file on the board, here's a 90-day link: Link to wiki send file. It's in Excel 2k7 format.

It will be slow d/l, but it's teeny tiny. It will die on or about 22 January 2013.

Ninjaedit: forgot to add this to the SS: (sqrt(pi*t*V+h*pi^2*t^3+h^2*pi^2*t^2)-pi*t^2-h*pi*t)/(pi*t)

It's the formula for a Cylinder with one closed end.

here it is formatted for that sheet: =SUM((SQRT(PI()*E5*F4+D4*PI()^2*E5^3+D4^2*PI()^2*E5^2)-PI()*E5^2-D4*PI()*E5)/(PI()*E5))

[nijineko]

what about googledocs? then you could upload edits and use the same link for as long as you need to...

[Agnostic Paladin]

If I understand what's going on here, the OP wants to determine the internal radius of a hollow hemisphere using a wall of stone.

Volume of a hemisphere is (2/3)πr^3
The volume of the hollow space inside is therefore (2/3)π(r-thickness)^3
The spell creates a fixed volume and fixed thickness; either 1" thick per four caster levels and 25 square feet per level, or 1/2" thick per four caster levels and 50 square feet per level. We'll call those Option A and Option B

In either case, we can solve for the inside radius as follows:
(V being the total volume of stone in cubic inches, r being the radius of the empty interior and t being the thickness in inches)
V = [(2/3)π(r+t)^3]-[(2/3)πr^3]
3V/2π=(r+t)^3-r^3
3V/2π=r^3 + 3(r^2t) + 3rt^2 + t^3 - r^3
3V/2π = 3tr^2 + 3t^2r + t^3
Okay, now we've got something that looks like a solvable quadratic once we enter in the values of V and t for a given caster level and option.

3tr^2 + 3t^2r + t^3 - 3V/2π = 0
Option A

Level 1: t=1" V = 5'x5'x1" = 3600 cubic inches
3r^2 + 3r + 1 - 3*3600/2π = 0
http://www.wolframalpha.com/input/?i=3r^2+%2B+3r+%2B+1+-+3*3600%2F2%CF%80+%3D+0 (click approximate form and ignore the negative result; obviously we're only interested in the positive value solution.)

So; now that I've shown how to get a solvable quadratic; I'll just give you the results for caster level 9-20, for options A and B.

Option A
Level 9: t=2" V = 5'x5'x2"x9 = 64,800 cubic inches r = 71"
Level 10: t=2" V = 5'x5'x2"x10 = 72,000 cubic inches r = 75"
Level 11: t=2" V = 5'x5'x2"x11 = 79,200 cubic inches r = 78"
Level 12: t=3" V = 5'x5'x3"x12 = 129,600 cubic inches r = 81"
Level 13: t=3" V = 5'x5'x3"x13 = 140,400 cubic inches r = 85"
Level 14: t=3" V = 5'x5'x3"x14 = 151,200 cubic inches r = 88"
Level 15: t=3" V = 5'x5'x3"x15 = 162,000 cubic inches r = 91"
Level 16: t=4" V = 5'x5'x4"x16 = 230,400 cubic inches r = 93"
Level 17: t=4" V = 5'x5'x4"x17 = 244,800 cubic inches r = 97"
Level 18: t=4" V = 5'x5'x4"x18 = 259,200 cubic inches r = 100"
Level 19: t=4" V = 5'x5'x4"x19 = 273,600 cubic inches r = 102"
Level 20: t=5" V = 5'x5'x5"x20 = 360,000 cubic inches r = 105"

Option B
Level 9: t=1" V = 5'x5'x1"x2x9 = 64,800 cubic inches r = 101"
Level 10: t=1" V = 5'x5'x1"x2x10 = 72,000 cubic inches r = 107"
Level 11: t=1" V = 5'x5'x1"x2x11 = 79,200 cubic inches r = 112"
Level 12: t=1.5" V = 5'x5'x1.5"x2x12 = 129,600 cubic inches r = 117"
Level 13: t=1.5" V = 5'x5'x1.5"x2x13 = 140,400 cubic inches r = 121"
Level 14: t=1.5" V = 5'x5'x1.5"x2x14 = 151,200 cubic inches r = 126"
Level 15: t=1.5" V = 5'x5'x1.5"x2x15 = 162,000 cubic inches r = 130"
Level 16: t=2" V = 5'x5'x2"x2x16 = 230,400 cubic inches r = 134"
Level 17: t=2" V = 5'x5'x2"x2x17 = 244,800 cubic inches r = 139"
Level 18: t=2" V = 5'x5'x2"x2x18 = 259,200 cubic inches r = 143"
Level 19: t=2" V = 5'x5'x2"x2x19 = 273,600 cubic inches r = 147"
Level 20: t=2.5" V = 5'x5'x2.5"x2x20 = 360,000 cubic inches r = 150"

So, the end result is that your thick dome can probably comfortably cover a single medium creature, or barely cover a single large create and the thin dome can either go a size larger or double the occupants.

[Chemus]

Thanks for the list. Earlier, I found an automated quadratic solver, got a general formula (above), and I made a spreadsheet. It's available for download above.

Perhaps I'm reading wall of stone too liberally; you read it that it only lets you halve the thickness once? Nor can you double the thickness for half the area?

This paragraph, which is not just flavor text, says that you can do almost anything you want, but that if it has to go through the air, or you want to make castle walls etc., you get 1/2 the area.

...Unlike a wall of iron, you can create a wall of stone in almost any shape you desire. The wall created need not be vertical, nor rest upon any firm foundation; however, it must merge with and be solidly supported by existing stone. It can be used to bridge a chasm, for instance, or as a ramp. For this use, if the span is more than 20 feet, the wall must be arched and buttressed. This requirement reduces the spell’s area by half. The wall can be crudely shaped to allow crenellations, battlements, and so forth by likewise reducing the area...

Edit: sorry, missed this.

what about googledocs? then you could upload edits and use the same link for as long as you need to...

OK. Here's the sheet.

[nijineko]

awesome! thank you, thank you. =D