Kastolite help?

[Jealdi]

So, my understanding is that kastolite is 90lbs/ft³. Trying to figure out how that equates to what I need to cast the inside of my forge. I'm guessing that's 90 lbs makes a solid cubic foot of hardened material.  So I was getting some weird calculations... like 26 lbs needed to cast the inside of my forge for 1/2" of kastolite.

I did find this below post from JHCC about how to figure it out.

 

So if that would be the case, because my diameter of my cylinder after Kaowool would be 8", that makes a 4" radius, which gives me a surface area of right at 425.

425 x 0.5 = 212.5

212.5/19.2 = 11.07 or 11 lbs.  Rounding that up, might as well buy 15 lbs since I can get it in 5 lb increments from Glenn? (This would also leave me almost 4 lbs to make baffles/doors)

Does that look right to people or am I just way off on my math skills today?

[Frosty]

Another way to calculate how much you'll need is to calculate the area of the outside diameter. Then subtract the diameter minus the thickness of your Kastolite. So, the outside radius would be 4" which has an area of 50.26" now subtract the area of the inside after applying 1/2" of Kastolite.  with a radius of 3.5" or 38.48 sq/in. for 11.78 cu/in per linear inch. Multiplied by the length gives you the volume of the hard refractory liner. I'd multiply by the length but I don't recall what you said it is and you didn't list it here.

Figuring the area of the inside of the forge and multiplying by the desired thickness only works if you calculate from the center of the hard refractory and then it can get a little iffy. 

I've found doing it like I described above works much better for me but I'm lousy at math. 

Frosty The Lucky.

[Jealdi]

The cylinder is 13 inches long. It might get shortened to 12 in my cleanup. Doing my calculations at 13 just to make sure I have enough though.

so 11.78 cu/in per linear inch would then be 11.78x13 = 153.14? I'm guessing that is then divided by the 19.2 from JHCC's post? Or not sure really where I go with it once I get that number. 

 

[Frosty]

I round numbers, this isn't precision work so 153/19.2= 8lbs. Rounding equals out over a few calcs. If you don't buy quite enough and you're a couple 100ths thinner than 1/2" its no big deal. Lots of guys are applying 3/8" hard liners with good results.

Frosty The Lucky.

 

[WilCo]

Have you considered using Wolfram Alpha for these kinds of computations? It's pretty good with formulas in as free-form text. In this case, for example, I used "cylindrical shell volume, height 13, inner diameter 7, outer diameter 8" and got the expected answer: 153.153, with a nice little graphic that shows me that it understood what I was asking, plus some other interesting bits of interpretation. It can also be used as a reference for formulas like this that you might not remember, like "cylindrical shell volume".

I have not been able to convince it to do more complicated expressions, like "(cylindrical shell volume, height 13, inner diameter 7, outer diameter 8)/19.2". Or, my earlier attempt, before remembering that it would be called a "shell", I tried calculating the volumes of the inner and outer cylinders and subtracting them, which it would not do directly either.

[Jealdi]

I ended up just ordering 15 lbs from Glenn. Figured it would give me extra to play around with for making door/baffles/another forge/etc.

[Frosty]

That's as good an example of why I don't try to calculate these things with one calc. Just subtract the volume of the 7" cylinder from the 8" cylinder. I'm a mathematical dumbbell and I can do that.

I have a high end scientific calculator that'll do it in a couple key strokes but it'd take me longer to figure out how to program the thing than make the forge.

Frosty The Lucky.