# Calculating gas port location for a DI AR-15

Some hypothesizing on gas ports – size and location.

A typical loading for .223 showing pressure inside the barrel as a function of distance traveled

Whenever creating an AR-15 in a new caliber, one of the key challenges is determining the characteristics of the gas system.  Chiefly, there are two adjustments that are easily made: location of the gas port (distance from barrel shoulder) and diameter of gas port.

Because we are adapting an existing gas system, it makes sense to perform some analysis of what a known, working system ‘sees’.  The top figure shows a typical 55grain mid-velocity .223 load.  A rifle length AR-15 should work well with this load with a gas port at 12.375″ from the torque shoulder, that is 0.099″ in diameter.  Those are pretty typical values.

Examining the graph, we can generate some hypothesis about the impact of changing either of our two ‘knobs’ (distance and port size).  The effect of distance can be read almost directly off the graph – the farther along the graph we go, the lower the pressure at the gas port.  Also, the less time that pressure is present at the port.

With the described rifle, there is no pressure in the gas tube until the bullet passes the gas port at 12.375″.  There, according to the graph, there will be about 16000 psi inside the barrel.  The pressure inside the gas tube will be less – that’s a function of the gas port – but lets ignore that for the moment, to simplify the discussion of port location.  That pressure declines to about 10000 psi before the bullet leaves the bore, at which point the pressure will drop off steeply (not shown in the graph).  The time it takes the bullet to travel the last 6″ of this 18″ barrel is the time that the gas has to act on the bolt carrier.  Mathematicians and physicists would make some statement about the integral of this pressure function, or “the area under the curve” – that area represents the amount of energy available to do the work of moving the bolt.  (This is an over simplification, but our working hypothesis is that it will be sufficiently accurate to predict working gas port parameters).

The size of the gas port is the next adjustment we can make.  When pressure appears at the gas port (as the bullet passes the port) it moves through the port and into the gas tube.  Assuming that the port is not wide open (wide open would be equal to the internal diameter of the gas tube, or about 0.125″ diameter) then the port itself is a constriction, and results in a lower pressure on the opposite side of the port – that is if there’s 10000 psi in the barrel, the pressure in the gas tube will be less because the gas has to squeeze through the port to fill it.  This has the effect of scaling the curve we just examined.  The pressure in the barrel might be 16000 psi decaying to 10000 psi, but the pressure in the gas tube may only reach 8000 psi and decay to  5000 psi.  The area of the port is proportional to the square of its diameter, so small changes will have significant impact.

The area of a circle is Π (pi, or 3.14159) times the square of the radius.  A typical gas port, at 12.375″, will be about 0.099″ in diameter. That’s an area of 0.0077 sq. in.  A maximum sized gas port would be 0.0123 sq. in.   Note that the small change has nearly doubled the port size.  Without knowing exactly what the pressure in the gas tube reaches we can (again, oversimplifying) scale it in proportion to the port area.   So I’d expect the pressure at the 0.099″ hole to be (little area / big area) (0.0077 sq. in / 0.0123 sq. in.) 0.63 times the peak pressure that could be present.

Examining the pressure curve for a .45, we can see immediately that the port will have to be much closer to the chamber to get much work out of the gas at all.  To get the same peak pressure with the same port size, the port would have to be less than 1/2″ from the torque shoulder!  But, if we open the port to maximum dimension, we can get about 1.6 times the pressure of the smaller port, and that should let us move our port to a location with about 1/2″ of bullet travel!  (Note that 1/2″ of bullet travel is going to be about 1.5″ from the barrel torque shoulder.   That’s shorter than any standard gas tube, so we’ll be shortening one just for the occasion.

The foregoing has been a greatly simplified rule-of-thumb approach to the otherwise very complex field of Computational Fluid Dynamics (CFD). The foregoing ignores a lot of factors that are hopefully too small to greatly influence the results. Time will tell…

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### 6 Responses to Calculating gas port location for a DI AR-15

1. John says:

I am very pleased with finding this site. The explanation of location and diameter of the AR gas port was very well done. I look forward to the next months posts

2. Paul says:

Thanks for publishing this data. I realize that even getting it to this stage is a great effort. As for the .45 ACP estimate, it served as confirmation of my calcs on a .38 special design.

3. sirrrbosss says:

Adding to that, the .45 ACP empty casing is heavier and needs more energy to be extracted , but on the flipside its shorter than the M193 meaning less bolt traveling. In my opinion 0.45 caliber is not suited for modern assault rifle for many reasons among them short range which the bullet loses its velocity after 100 yards which also has great effect on accuracy and the recoil of the weapon in automatic firing. Other reasons the bulky magazines that soldier has to carry around. The 0.45 caliber machinegun of WWII was design to fire from the hip “spraying” now soldieries are trained to fire from the shoulder with short burst and accuracy.

4. cawpin says:

Did you ever finish this effort? I’m going to be building one myself shortly and the gas port location is the only thing I’m unsure of. I was actually going to do exactly what you did here with Quickload to determine my pressure needs.

• GsT says:

I have not. It’s still on the “to do” shelf, but it has a lot of company…