buickmarti
32 Cal.
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The coefficient of thermal expansion for lead is 16.3 X 10^-6/°F/inch. That would be 0.000016 per degree F per inch.Right now it measures.725”
3% shrinkage would mean I have to have it at .515” to make a .500” bullet. How much does lead shrink?
YesProjects like this are done for the fun of it and to see what we can learn. Why spend $120 on a beautiful mold from Accurate Molds when you can make your own in a week or two? I've made a bullet mold myself which is why I like to buy them.
Yes
Good idea.
So how does anyone make a ball mold from materials found around the house that is easy for a person to make in 2 weeks time? Let's say with tools available in 1840.
Most aluminum has a coefficient of thermal expansion of around 13 X 10^-6/degree F./inch or, thirteen millionths of an inch per degree F, per inch.
Written in so your calculator will know what your entering that would be: .000013
To use this figure, multiply it times the change in temperature in F., times the distance from one point to another.
Remembering the room temp is about 70°F if you want to know what the size of a 1" thick piece of aluminum at room temperature would be at 750°F, first subtract the 70°F from 750°F = 680° F temperature rise. Multiply this times .000013 = .00884 change in size so a 1 inch thick piece of aluminum would be 1.00884" thick. If the thickness is different than 1 inch, multiply the .00884 times the thickness. For instance a piece of aluminum that is 1/2" thick would expand .00884 X 0.5 = .00442".
Don't forget, the molten lead will also change size as it heats or cools.
Set the standards and materials allowed, I will try it, but I want soapstone included in the materials allowedI think the first things you'd have to do is to lower your standards. You, or at least me, are not going to be able to make a nice round cherry to ream out a ball mold. I'd be willing to bet that the vast majority of balls cast by the average guy in 1840 were nowhere near what we are getting out of modern molds. Beyond saying that, I'm not much help.
Yea. What you said. My original answer was going to be "just a hair" but I think you covered that.Most aluminum has a coefficient of thermal expansion of around 13 X 10^-6/degree F./inch or, thirteen millionths of an inch per degree F, per inch.
Written in so your calculator will know what your entering that would be: .000013
To use this figure, multiply it times the change in temperature in F., times the distance from one point to another.
Remembering the room temp is about 70°F if you want to know what the size of a 1" thick piece of aluminum at room temperature would be at 750°F, first subtract the 70°F from 750°F = 680° F temperature rise. Multiply this times .000013 = .00884 change in size so a 1 inch thick piece of aluminum would be 1.00884" thick. If the thickness is different than 1 inch, multiply the .00884 times the thickness. For instance a piece of aluminum that is 1/2" thick would expand .00884 X 0.5 = .00442".
Don't forget, the molten lead will also change size as it heats or cools.
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