Measuring dilemma

[Wayne Smith]

OK, machinists, I've got a problem. I have an old Bair powder measure and two drilled inserts and three blank inserts. If I want one of those blanks drilled to provide, say, 14.8gr 2400, how do I tell one of you how to drill it?? Or any local machinists, for that matter.

I can come up with three loads that I use frequently, but have no idea how to come up with the dimensions of the hole to accomplish this.

[oneokie]

Determine the volume of your desired weight of a specific powder. IIRC, the Lee manual has this information, or the formulas to determine.
Once you know how much volume it requires, do the math for a cylinder shape using the height of the blank bushings.

[Jon K]

Wayne,

I am not familiar with this powder measure, but if it is like a RCBS Little Dandy, then, it's just figuring the volume to weght ratio of the specific powder. So........Volume = dia*rad sq*height, ie how many grains of 2400 per cubic inch.

Jon

[94Doug]

I didn't even know that Dilemma's could be measured.


Doug

[deltaenterprizes]

Oneokie has it right, powder density x charge weight = volume = insert height x diameter.
The Lee manual will give density in gr/cc so it will have to be converted to cu/in unless you have metric drills in tenths of a milimeter sizes.

I can do that job and make some spare inserts.

[Bret4207]

Would it be possible to thread the insert and use a screw type plug to make it adjustable? The RCBS Little Dandy had an option of something like this and it worked great.

[Wayne Smith]

I didn't even know that Dilemma's could be measured.


Doug

Doug, if that was what I ment I hope I know enough English to use a plural! Thanks, guys, I have to spend more time in the Lee manual to find the information.

[oneokie]

Here is the information for determining the volume of various powders:

http://www.leeprecision.com/cgi-data/instruct/AP1704.pdf

The chart on page 6 gives the VMD of the popular powders.

Hope this helps. :drinks:

[deltaenterprizes]

What is the diameter of the cylinder?Do you have one that throws a charge that is close?

[Buckshot]

.............Right now I am currently overstocked on Dilemma's, as I seem to have been creating them wholesale for some time. If anyone is currently sailing a bit too smoothly, I can let some go in dozen lots without much arm twisting.

................Buckshot

[Linstrum]

Hi, Wayne, what is the height of the insert to within 0.001-inch? Once we know that we can then calculate the diameter of the hole for 14.8 grains of 2400.

Buckshot, you need to visit John Coffee, the giant in the movie "The Green Mile". He eats dilemmas, etc.


rl370

[Wayne Smith]

Hi, Wayne, what is the height of the insert to within 0.001-inch? Once we know that we can then calculate the diameter of the hole for 14.8 grains of 2400.

Buckshot, you need to visit John Coffee, the giant in the movie "The Green Mile". He eats dilemmas, etc.


rl370

Best I can do is an electronic caliper measure of .6795". I can reasonable vouch for the .679" measurement. It is a brass cylinder.

[Linstrum]

Hi, Wayne,

The basic formula that has already been given by others is
pi r^2 h = volume

the volume of one grain of Alliant 2400 is 0.0742 cubic centimeters, the height of the cylinder that will contain the 14.8 grains of 2400 powder is 0.679 inches. The volume of 14.8 grains of Alliant 2400 is 1.09816 cubic centimeters. Since the dimensions of the cylinder that will contain the powder in question is given in English units the cubic centimeters must be converted to the English unit of cubic inches. So 1.09816 cubic centimeters divided by 16.387 cubic centimeters per cubic inch = 0.067014 cubic inches. Plugging that into “pi r^2 h = volume” gives us

3.1415926 times r^2 times 0.679 = 0.067014 cubic inches.

We need to isolate the term we are solving for, so turning the math crank we get the equivalent formula of

r^2 = 0.067014 cubic inches divided by 3.1415926 times 0.679

simplifying that gives us

r^2 = 0.067014 / 3.1415926 (0.679)
Again turning the math crank we get r^2 = 0.0314157

Now we can solve for r^2 by taking the positive square root of 0.0314157, which is 0.1772. That is the radius, twice that is the diameter of the cylinder, 0.3544 inches. Since a ten thousandth of an inch is not required for close enough accuracy, the hole can simply be around 0.354 inch, or so. Going over to

http://bobmay.astronomy.net/misc/drillchart.htm

we see that either a 9mm or letter “T” drill will work for drilling the hole. Most machinist in the United States do not have metric drill and reamer sets so the hole will probably end up 0.358 inch and still be close enough to give satisfactory results, although I'd try a letter "S" drill at 0.348 inches first if you have that luxury since drills tend to cut a few thousandths oversize, although brass minimizes this.


Good luck!


rl375

[leftiye]

Drill it undersized first. Then open it up with an adjustable reamer to get exact weight. Aluminum will make this a lot easier.

[deltaenterprizes]

If it is like a little dandymeasure the hole does not go all the way through

[leftiye]

Then any size hole will work with varying length plugs for different charge weights. (Shades of RCBS et. al.)

[TAWILDCATT]

I am not sure but the size may be the same as pacific as they made pacific presses.I made a bunch of blanks and just made the hole the size of the original.It should be a hair smaller than the entrance.

[Doc Highwall]

Just remember when you have it perfect that when you change to a different lot of powder the weight will be different because of the difference in density from lot to lot, but the volume will be the same which is more important.