How can I control spread?

[Lisa Allen]

I have a project that I am working on that needs to be 3" wide. The top, unfired part is only 2 and 1/4" wide and 1/4" thick. So, how much base glass needs to go under it to get it to spread to at least 3" wide? It doesn't matter if it goes over 3" wide, it just needs to be at least that.

Thanks, Lisa

[charlie]

v1 = v2

v1 = 3" ** 2 * .25"
v2 = 2.25 ** 2 * .25 + (x ** 2 * .125)

solve for x

[Lisa Allen]

v1 = v2

v1 = 3" ** 2 * .25"
v2 = 2.25 ** 2 * .25 + (x ** 2 * .125)

solve for x

Hi Charlie-

I am not sure if I speak your language here. Are you using the double asterisk to denote, to the power of? And what are you solving for with "x", the number of layers of glass or the actual thickness needed to achieve the spread that I am wanting?

Or, you can just tell me how many layers of clear that I need to accomplish what I asked......

Lisa

[Carol B]

I have a project that I am working on that needs to be 3" wide. The top, unfired part is only 2 and 1/4" wide and 1/4" thick. So, how much base glass needs to go under it to get it to spread to at least 3" wide? It doesn't matter if it goes over 3" wide, it just needs to be at least that.

Thanks, Lisa

Hi Lisa,

I know there are mathmatical formulas you could use. But 2 and 1/4 inches is not that big. Why not run some tests with scrap? I would be interested to hear the results.

Carol

[Mira]

Re: controlling spread:

I've heard diet and excerise works, but I've never been successful . . .

[Lisa Allen]

I have a project that I am working on that needs to be 3" wide. The top, unfired part is only 2 and 1/4" wide and 1/4" thick. So, how much base glass needs to go under it to get it to spread to at least 3" wide? It doesn't matter if it goes over 3" wide, it just needs to be at least that.

Thanks, Lisa

Hi Lisa,

I know there are mathmatical formulas you could use. But 2 and 1/4 inches is not that big. Why not run some tests with scrap? I would be interested to hear the results.

Carol

Oh, Carol, that would be way too smart and patient for me......haha. So, not being able to decipher Charlie's equation and having recieved a call from Tony to help with the dilemma, I threw the piece in the kiln tonight and will just see what happens. If it doesn't work, I will consider it a test with scraps.....

[charlie]

v1 = v2

v1 = 3" ** 2 * .25"
v2 = 2.25 ** 2 * .25 + (x ** 2 * .125)

solve for x

Hi Charlie-

I am not sure if I speak your language here. Are you using the double asterisk to denote, to the power of? And what are you solving for with "x", the number of layers of glass or the actual thickness needed to achieve the spread that I am wanting?

Or, you can just tell me how many layers of clear that I need to accomplish what I asked......

Lisa

sorry, was running out the door to pack my truck full of boxes.

** means squared. solved, x would be the side (in inches) of the square of glass that would have to be added for 1/8" glass.