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Thread: LEE 356-125-2R to 115gr

  1. #1
    Boolit Master
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    LEE 356-125-2R to 115gr

    I was wonder if I could mill off the mold enough, to reduce the boolet, from 125gr to 115gr.????
    How do I figure out how much to take off????
    My Marlin Camp 9 likes 115gr boolets. 125gr not so much.

  2. #2
    AKA: GRMPS Conditor22's Avatar
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    Cast a boolit out of the alloy you plan on using, measure it then start filing off the base until you get your desired weight. Measure it again, do the math and that's how much you need to remove.


    My fear would be the boolit would no longer function properly. You're removing material from the drive band area only and none from the nose all this is calculated in the design of a boolit.

    I acquired a 9mm mold where someone had tried to do this and failed. I removed the lube grooves and just recently have started playing with it.
    Not sure what it was originally but it now cast 112 grn smooth-sided boolits.

    still, need to do a little more polishing

    Last edited by Conditor22; 08-28-2019 at 11:56 AM.

  3. #3
    Boolit Grand Master tazman's Avatar
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    Good answer ^^^^^^

  4. #4
    Boolit Master

    Tom Myers's Avatar
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    Quote Originally Posted by abunaitoo View Post
    I was wonder if I could mill off the mold enough, to reduce the boolet, from 125gr to 115gr.????
    How do I figure out how much to take off????
    My Marlin Camp 9 likes 115gr boolets. 125gr not so much.
    You can calculate the amount to remove like this:

    The Specific Gravity of wheel weight alloy is 11.0096 Grams per cubic centimeter.
    (If you are using a different alloy, substitute the Specific Gravity of the alloys shown at the bottom of the post in your calculation)

    There are 15.43236 grains in a gram.
    A cubic centimeter of ww alloy = 11.0096 x 15.43236 = 169.904 grains.

    We want to reduce the weight by 10 grains.

    So 10 / 169.904 = 0.058857 cubic centimeters to take off the base of the bullet.

    There are 2.54 centimeters in an inch.

    There are 2.54 x 2.54 x 2.54 = 16.387 cubic centimeters in one cubic inch.

    So we need to take off 0.58857 / 16.387 = 0.00359 cubic inches off the base of the bullet.

    Now we need to know how many square inches are in the base of the bullet.
    The base of the bullet measures 0.357" in diameter.
    So The radius of the bullet base is: 0.357" / 2 = 0.1785".
    Pi = 3.1416
    Pi x radius x radius = square inches of bullet base

    3.1416 x 0.1785 x 0.1785 = 0.1000090 sq inches in the bullet base.

    So we have 0.00359 cubic inches divided by 0.1000 square inches = 0.0359 inches to remove from the base of the bullet.

    Round that off, remove 0.036" from the base of the bullet and this is the result.


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    Elements in the Alloy Lead Tin Ant. Sp. Gr BHN
    Pure Lead 100.0% 0.0% 0.0% 11.3450 5
    WheelWeights 95.5% 0.5% 4.0% 11.0096 12
    Lyman #2 90.0% 5.0% 5.0% 10.6829 12
    Linotype 86.0% 3.0% 11.0% 10.3830 19
    Monotype 73.0% 9.5% 17.5% 9.6700 26
    Roto Metals HardBall
    (Terracorp)
    92.0% 2.0% 6.0% 10.7789 16
    Roto Metals Superhard 70.0% 0.0% 30.0% 9.3917 37
    50-50 Solder 50.0% 50.0% 0.0% 8.911 14
    10 : 1 Lead - Tin Alloy:
    10 Lead + 1 Tin = 11 units of Alloy
    9 .1 Lead + 0.9 Tin = 10 units of Alloy
    90.9% 9.1% 0.0% 10.8083 11.5
    16 : 1 Lead - Tin Alloy:
    5.88% Tin
    16 Lead + 1 Tin = 17 units of Alloy
    9.4 Lead + 0.6 Tin = 10 units of Alloy
    94.1% 5.9% 0.0 10.9918 10.53
    1/20 Tin/Lead Alloy:
    5.00% Tin
    19 Lead + 1 Tin = 20 units of Alloy
    9.5 Lead + 0.5 Tin = 10 units of Alloy
    95.0% 5.0% 0% 11.0434 10.15
    20 : 1 Lead Tin Alloy:
    4.76% Tin
    20 Lead + 1 Tin = 21 units of Alloy
    10 Lead + 0.5 Tin = 10.5 units of Alloy
    95.2% 4.8% 0% 11.0574 10.03
    30 : 1 Lead Tin Alloy:
    3.226% Tin
    30 Lead + 1 Tin = 31 units of Alloy
    9.7 Lead + 0.3 Tin = 10 units of Alloy
    96.8% 3.2% 0% 11.485 9.11
    Last edited by Tom Myers; 08-28-2019 at 03:00 PM.
    Respectfully,
    Tom Myers
    Precision Shooting Software


  5. #5
    Boolit Master

    gwpercle's Avatar
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    With the cost of a Lee mould being $22.00 wouldn't buying another mould be the wise thing to do .
    Ruing a good mould will result in buying another .
    Cutting down on the driving band area just might not the answer .
    Certified Cajun
    Proud Member of The Basket of Deplorables

  6. #6
    Boolit Master
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    Lee doesn't have a 115gr mold.
    They have a 102gr I may just have to try.

  7. #7
    Boolit Grand Master tazman's Avatar
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    I have used the 102 gr round nose in my 9mm with good results.
    I don't have one now since it wasn't any better than other options.
    You might even try the 95 grain RF mold from Lee. I use it in my 9mm as well. It makes a great plinking load and uses even less lead.

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Abbreviations used in Reloading

BP Bronze Point IMR Improved Military Rifle PTD Pointed
BR Bench Rest M Magnum RN Round Nose
BT Boat Tail PL Power-Lokt SP Soft Point
C Compressed Charge PR Primer SPCL Soft Point "Core-Lokt"
HP Hollow Point PSPCL Pointed Soft Point "Core Lokt" C.O.L. Cartridge Overall Length
PSP Pointed Soft Point Spz Spitzer Point SBT Spitzer Boat Tail
LRN Lead Round Nose LWC Lead Wad Cutter LSWC Lead Semi Wad Cutter
GC Gas Check