doublegunshop.com - home Forums General Discussion DoubleGun BBS @ doublegunshop.com Two (2) Shotgun Formulas

Two formulas which should be of interest to all those having an interest in shotgun ballistics.
Formula #1:
F = MA or Force = Mass time Acceleration. now this formula tells us as was stated a while back that if for instance we accelerate an ounce of shot to say 1200 FPS the required force is the same, matters not whether it be from a 28 gauge or a 12 gauge. This is of course totally true.

Formula #2:
F = PA or in this case Force = Pressure times Area. To avoid confusion for further discussion I will use Ac for acceleration & Ar for area.

Since Force is on 1 side of each equation then the two can b combined & F eliminated & thus we have MAc = PAr or mass times acceleration = pressure times area.

Here of course is where it becomes Extremely obvious that as the gauge of the gun changes then Ar changes with it so P also has to change to maintain a balance.

Using nominal bore sizes of .550" for the 28 & .729" for the 12 then (.550"/.729")� shows the 28 gauge has only 57% of the area of the 12ga. Consequently pressure (P) has to be 175% that of the 12 to impart the same 1200 FPS to an ounce of shot as would the 12.

I fully realize that a 28 gauge does not normally produce 175% of the peak pressure of the 12 gauge. Peak or max pressure is however only a small part of the story. A shotgun does not work like a normal Hydraulic system where a pump is putting out a constant stream of fluid under a constant pressure.

Rather a shotgun burns a small amount of powder with a rapid rise to it's peak pressure, then a quick falling away with a reducing amount of pressure to the end of the barrel. What is needed in this formula is the entire average pressure under the curve for the length of the barrel. Thus if one gets the same velocity from the same weight of shot in different size guns it will be found this difference in "Total" pressure is an absolute essential.

If anyone has an interest in further discussion of this in a Civil manner feel free. Before one simply puts it down as Ignorant & Unscientific please make certain you know what you are talking about.

Note also that in order to move an ounce of shot the 28 has to move 4.2 oz per Sq Inch, the 12 only 2.4 oz per Sq inch. This fully explains why harder premium shot is more desirable to minimize shot deformation in the 28.

What is the point you are trying to make, is it that given same velocity and shot weight the smaller gauge recoils more because the shot or that the shot colomn is subject to greater pressure ?

I think the point is that the smaller the gauge the higher the chamber pressure is.
To me this would explain why a 12 gauge 6 1/4 lb. gun shooting a 3/4 oz. of shot has less felt recoil and less pressure as a 28 gauge weighing the same amount shooting the same load.

Colonel, Imagine your 20 something mistress walking around your vinyl laminate kitchen floor in 6" stiletto's. Now the navy pumps.
One will leave a mark you can't remove before your wife returns, and the other can leave unannounced.

He's saying chamber pressures have to be higher for the same ejecta in smaller bores.

Same ejecta (mistress), same breech face (floor) , different pressures (heel sizes) and different evidence of pressure (dents)

David- Not recoil. Pressures.

Neither of the two formulas relate to felt or actual recoil forces they simply explain why two shot charges of the same weight require higher chamber pressures to achieve the same velocity. Check out a few loading books and compare the difference in chamber pressure for loads fired in larger bores as compared to the same charge weight fired in smaller bores and it will become quite evident.

Clapper, God bless you

Interesting. Thank you.

I wonder if there's another equalizing formula that can be worked out for small vs big bores - I remember once reading a discussion between two chaps that argued whether a small-bore barrel is proportionally stronger or not to the big bore barrel of the same weight - something having to do with the area and the thickness increasing to different proportions with increase in barrel diameter. But for the life of me I can't remember any details

Recoil does not enter into these two formulas. Recoil is based on the weight of the total charge (including powder, wads & shot) & the velocity of the load against the weight of the gun. It does not always even mean a higher Chamber pressure, Chamber pressure is controlled to a large extent by the burning rate of the powder.
Remember, it is absolutely necessary to look at the total pressure curve. To push the same shot load to the same velocity from a smaller bore takes a "Total" pressure that is higher in proportion to the square of their bore diameters. This follows all well known Scientific Principals.

The main point of this post is that I have fumed for a few weeks over an attempt to make me look like an Ignorant Hill-Billy who did not know what I was talking about when I said all was not equal when firing the same shot load from two different sizes of bores, with the formula F = MA cited to prove it. Well F = MA is a correct formula BUT is not the whole story by any means. The F = PA formula is equally Correct & Scientific & shows how that force is obtained in the different bores.
The weight of the shot per square inch to be moved, coupled with higher pressure, shows why deformation is more likely the longer the shot charge is in the bore.

Humpty Dumpty, you are referring to hoop stress. You can look it up on Wikipedia. The formula takes into account the radius of the barrel and if I recall correctly the smaller radius has greater hoop strength. The engineers here will have to explain it.