Is there a ratio of velocity increase same weigh larger bore

[Ogre6br]

HI
If the same weight projectile is fired form a larger bore with equal amount of powder energy behind it- how come the velocity is usually higher with the larger bore than the control one??

same again but with the same weight slug etc etc fired from a smaller bore-- Is it slower?? when does the law of deminishing returns come in??
One bore size down, 2,

Is there an approximate ratio to work from so when you are doing "what if's" you have some sort of ballistics data to work from-- rough but plausable.

The different projectiles will have higher or lower BC's depending on whether you are stepping up a bore size or down- this will ultimately effect projectile performance but In the Q I;m just looking for the effect on velocity that bore size has.
same barrel length, bore friction unknown, everything that can be the same is for the sake of the argument aside from those things required to be different.


EG: 62 grain 224 cal projectile fired with a 100% density load providing X units or force

Now bring in a 62grain 20 calibre fired with the same X units or force- is it faster or slower than the 224 projectile

Now bring in a 62 grain 6mm projectile same x units of force applied-- faster or slower than the 224 projectile??

As this is a hypothetical Q I used 2 projectile weights I know exist in 224 and 6mm and made up the .20 cal projectile for the sake of the question.



Thanks

I haven't done any of this type of maths in 15 years and have no idea on how to do this.

later
P

 

[Asa_Yam]

Physics works agains the smaller diameter projectile.

- For an equal weight, the smaller diameter bullet has a longer bearing surface, and hence more friction;

- At the same maximum pressure,assuming an identical pressure curve), the smaller diameter projectile has less force propelling it. If pressure = force/area, force = pressure x area
For cartridges developing 60,000 PSI pressure:

Force = 60,000 x area

With a .224" diameter bullet:

area = pi x,,.224")^2)/4
= 0.0394 in^2

therefore force = 60,000 x 0.0394
= 2364 lbs.


With a .243" diameter bullet:

area = pi x,,.243")^2)/4
= 0.0463 in^2

therefore force = 60,000 x 0.0463
= 2778 lbs.

Since force = mass x acceleration, given projectiles of equal mass:
* The one with greater force pushing it will accelerate faster. Acceleration x time = velocity.

NOTE: When using "pounds" as a unit of force or weight, a conversion to "slugs" is required, as a pound is a unit of weight, not mass. However, the conversion is a constant value. Confusing? Yes. This is why most engineering calculations these days are done in SI units.

* Greater friction from a longer bearing surface results in even less force propelling the smaller diameter projectile.