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Joined: Jan 2002
Posts: 623 Likes: 1
Sidelock
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OP
Sidelock
Joined: Jan 2002
Posts: 623 Likes: 1 |
Hey! I lost the formula for calculating the weigth reduction resulting from back boring barrels. Can anyone help me with the formula?
I know I found it on the web about 6-7 years ago!
Thanks-Mike
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Joined: Jan 2007
Posts: 638
Sidelock
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Sidelock
Joined: Jan 2007
Posts: 638 |
LC,
Wikipedia provides the formula which you should be able to find via Google or other seach engine. The formula is pretty complicated.
For most of us non-PhD's Bruce Buck provides the following from Shotgun Report:
"Does aftermarket (not factory) backboring have any benefit at all? You can bet your Junior Technoid magic slide rule ring that it does. While aftermarket backboring may not reduce recoil or improve patterns, it sure reduces weight. If the barrels on your gun feel too heavy and unresponsive, you may be able to put them on a diet. Most standard barrels have a wall thickness of around .040". This is a lot of meat and might be substantially reduced. On a standard 30" set of barrels, backboring .010" will reduce barrel weight by 2.77 ounces. A .020 backbore will take off a monumental 5.58 ounces. A change of 3 ounces is a lot, so go easy. Check first with your gunsmith. He will know what is safe. Be aware, however, that aftermarket backboring will void any factory warrantee. Briley charges about $150 per tube for backboring."
Regards, Bruce Buck Shotgun Report's Technoid
USMC Retired
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Joined: Jan 2005
Posts: 90
Sidelock
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Sidelock
Joined: Jan 2005
Posts: 90 |
Here is a simple formula.
weight change in ounces = 7.14 x IDB x (IDA-IDB) x L x N
IDB = barrel inner diameter before backboring, in inches IDA = barrel inner diameter after backboring, in inches L = length of backbored area in inches (e.g., barrel length minus chamber and choke) N = number of barrels
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Joined: Dec 2001
Posts: 12,743
Sidelock
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Sidelock
Joined: Dec 2001
Posts: 12,743 |
It doesn't really take a PHD to calculate this. You simply need to find the cubic inches of steel to be removed & multiply by the weight of a CuIn of steel. From a nominal bore dia of .730" & going to one of .740" for a pair of 30" bbls I come up with a wt reduction of about 3 1/8 oz.
Miller/TN I Didn't Say Everything I Said, Yogi Berra
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Joined: Jan 2007
Posts: 638
Sidelock
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Sidelock
Joined: Jan 2007
Posts: 638 |
Thanks Mike for a simple formula. The formula that I found on Wikipedia required a little more thought...
What does 7.14 represent?
USMC Retired
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Joined: May 2003
Posts: 264 Likes: 23
Sidelock
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Sidelock
Joined: May 2003
Posts: 264 Likes: 23 |
With all due respect to Bruce Buck and his Junior Technoid magic slide rule ring, I disagree. His answer overstates the results by a factor of 2. In a pair of 28" barrels the bore is length , less chamber and forcing cone, less choke and choke cone, @ 24.5 inches.Using metric units @ 25.4mm/"-2.54cm/" and a cylinder volume formulae of R2 x pi x L I come up with a.As recvd .729"dia = 18.52mm, r=9.26mm r sqd=85.74mm2 x 3.142= 269.4mm2 or 2.69 cm2 x length 62.23cm Volume #1=167.40 cm3 b After .010" overbore to .739"/18.77mm r=9.3853 sqd =88.08x pi 3,142 =276.76 mm2 or 2.77cm2 x l 62.23cm =172.37cm3 for volume #2 Vol 2 - Vol 1 = 4.97cm3. Steel has an sg of 7.35 gms/cm3 for 36.53 gms. Grams/ oz = 28.35 or 1.3 ounces per tube lighter. Aditional 2" for 30"bbls = a further reduction of 1/10oz/tube slightly less for choke and cone opening additional .010". Of further note I dont see many European Game guns with .030" thick walls let alone the .040" he quotes as common. Proceed with caution and do not expect major weight loss!! Regards to all from a Non PhD
Hugh Lomas, H.G.Lomas Gunmakers Inc. 920 876 3745
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Joined: Jan 2005
Posts: 90
Sidelock
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Sidelock
Joined: Jan 2005
Posts: 90 |
7.14 represents (density, in ounces per cubic inch) x pi / 2. My formula assumes that the change in bore diameter is small compared to the bore, a good assumption here. So the volume of metal removed is just 2 x pi x radius x (change in radius) x L.
Hugh, if I recall correctly Bruce Buck was referring to a pair of barrels, but the quoted text doesn't make that clear.
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Joined: Jan 2002
Posts: 623 Likes: 1
Sidelock
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OP
Sidelock
Joined: Jan 2002
Posts: 623 Likes: 1 |
Thanks guys for the formula and the discussion. I realize there is imprecision in estimating long forcing cones and length of choke. However, the formula worked well for gun I had back bored. It was a Day 500/Miroku. It had very heavy barrels around .050+ before work and was underbored .605 (a 20 gauge. After backboring about .021, my notes show it shed 4.3 ounces (28" bbls). The formula comes really close.I back bored it to rebalance the gun, after estimating the weight reduction with formula I lost. Now the gun weighs 6 pounds 3 ounces and is very well balanced! (I also drilled the stock a bit.)
I sent this information to my friend here in Anchorage who doubted the weight reduction.
Mike
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Joined: Jan 2002
Posts: 5,954 Likes: 12
Sidelock
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Sidelock
Joined: Jan 2002
Posts: 5,954 Likes: 12 |
I back bored it to rebalance the gun, after estimating the weight reduction with formula I lost. Now the gun weighs 6 pounds 3 ounces and is very well balanced! (I also drilled the stock a bit.)
Mike Mike, back-boring will reduce weight, move balance rearward, and decrease both unmounted and mounted swing efforts. Drilling the stock will do the same except balance will move forward. Stock and barrels can be weighted (weight added) for the opposite effects. Within reason, some combination of barrel and stock weighting or weight removal will give a combination of weight, balance, unmounted swing effort and mounted swing effort that suits the shooter. Note that each shooter's preference for a certain combination of handling dimensions is just as personal as is his/her preference for a certain set of stock dimensions.
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