## Sunday, June 5, 2011

### 7.62x25 Blowback - Theoretical Bolt Mass and Spring Constant Calculations

Following the example given in George M. Chinn's The Machine Gun I will apply a similar treatment to the concept of a 7.62x25 blowback-operated semi-auto.

The Measurements.

Before going further, I will summarize the measurements that will be needed in later on in the discussion and calculations. These are presented as measured by me on my 7.62x25 AK conversion consisting of an AMD-65 kit and a bolt and carrier from a Bulgy 74.

Bolt and carrier weighth (without modifications) = 475 g
Unmodified gas piston weight = 82.24 g
Custom-made full diameter gas piston = 187.60 g
Stroke length (note that my barrel sits further into the receiver than a standard AKM barrel and that the bolt is fixed in the forward position in the carrier) = 11.2 cm
Spring constant (standard AMD return spring) = 94.75 N/m

The Bolt Mass.

George Chinn begins his discussion of blowback operation by stating that for a 20mm round (this is the caliber he uses throughout the discussion) the time during which the bolt travel must be minimized is 0.0015 seconds (without case lubrication). This is the time during which the pressure (rises and) falls from its peak value to a lower value -- one at which it is safe to begin extraction of the spent cartridge. This time period is determined experimentally and depends on the cartridge material, cartridge shape, amount of gunpowder, weight of the projectile and length of the barrel among other parameters. This is the graph that George Chinn presents for the 20mm round.

At 0.0015 seconds the pressure has fallen to approximately 30% of the maximum.We will use the same threshold in our analysis below. Lacking an experimentally generated time-pressure curve I used a program called QuickLOAD which is able to generate theoretical time-pressure curves based on input of cartridge, powder and bullet parameters. This is quite a powerful (and complicated) program, though I only have access to the demo version which comes with several pre-loaded cartriges, bullets and powders. I was very happy to find that one of the presets was the 7.62x25 cartridge. So, using QuickLOAD I was able to generate the following theoretical curve of the 7.62x25 time-pressure data:

I included the full screenshot so that you can see what values I used for the multitude of QuickLOAD's parameters. All of these are default, except:

Charge weight = 0.37 g (chosen so as to give a muzzle velocity of about 1600 fps)
Barrel length = 10.5 inches (this is the length of the barrel that I will be using)

What this graph shows us is that it will take approximately 0.4 ms for the pressure to drop to 30% of the maximum. Getting back to George Chinn's example, he mentions that the maximum distance that a 20mm brass case is allowed to travel during these first few milliseconds is 0.015 inch (without lubrication). This distance is once again an empirically established value which depends on the tensile strength of the case and the amount of reinforcement near the base of case. In other words, if you pull it too far either case separation will occur or the case will split down the side without the support of the chamber walls. By visual examination of a brass 7.62x25 case, I estimated the extent of allowable movement to be 1.5 mm. This makes the maximum initial velocity for the 7.62x25 bolt:

How does this translate to bolt mass? Recall the conservation of linear momentum:

Since for a small cartridge like 7.62x25 (unlike the 20mm cartridge) the amount of powder (and therefore gas) is rather small, to simplify the treatment it will be simply dropped.

Therefore, if the mass of the projectile is 6.026 g (93 gr), the muzzle velocity is 494 m/s (1622 fps) and the initial velocity of the bolt is 3.75 m/s, the mass of the bolt must be:

This is significantly higher than the stock weight of an AK-74 bolt-carrier combo (475 g). With a 475 g bolt (and carrier) the initial velocity of the bolt becomes 6.28 m/s which, needless to say, is far too fast.

So, how can we make the bolt heavier? Well, the most obvious modification is to turn a new, full-diameter gas piston as shown in the pictures below.

The new gas piston weighed in at 187.60 g compared to the old piston's weight of 82.24 g. This gives us 105.36 g of added mass, to bring the total bolt mass to 580 g. This is an improvement but we need to get closer to 795 g. Not sure what I'm going to do yet, but I will keep you posted.

The Spring Constant.

The initial resistance to premature opening of the bolt comes almost exclusively from the bolt's mass. Once the bolt reaches its full velocity however, it is up to the return spring to bring it to a stop and send it going the other way. Schematically, the situation can be drawn in the following manner:

Where, m = mass of bolt, k = spring constant, Vinit = initial velocity of bolt, l = maximum travel (bolt stroke), x = length coordinate (zero value is marked). There is no friction and the system evolves as a function of time, t. Note that in our idealistic Hooke's law treatment, it doesn't matter whether the spring is initially compressed or not - all that matters is the additional amount of compression during the operation of the system.

What we want to know is - given a certain mass and initial velocity of the bolt, as well as a maximum travel distance, what must the spring constant be in order for the bolt to come to come to a stop at the given distance (before going back)? Alternatively, we can ask - given a certain mass and initial velocity of the bolt, as well as a spring constant, how far will the bolt travel before coming to a stop (and going back)?

The following is the derivation of the equation of motion for the bolt. You don't necessarily have to follow exactly what's going on here, as I will highlight the final equation that is of interest to us at the end of the derivation.

So, the boxed equality above is what we will use to calculate the spring constant and bolt stroke. Let's answer the first question:

"Given a certain mass and initial velocity of the bolt, as well as a maximum travel distance, what must the spring constant be in order for the bolt to come to come to a stop at the given distance (before going back)?"

For the bolt mass, let's take our heavy bolt which weighs 0.795 kg. As discussed earlier its initial velocity will be 3.75 m/s and the measured bolt stroke on my 7.62x25 AK is 0.112 m. Therefore the spring constant must be:

Note that this is about 10x the measured value of the standard AMD spring (94.8 N/m). Realistically some resistance will be provided by the hammer spring as well as by friction (and don't forget real-life deviations from Hooke's law), so the true value is likely to be lower, however it seems that at least some stiffening of the spring is needed. Currently, there is not a whole lot of options to do that however, short of making your own spring. The only readily available solution is the extra-strength Wolff spring, but that only provides 15% extra stiffness (1.15x out of the theoretical 10x). Now for the sake of discussion, let's take a look at the second question:

"Given a certain mass and initial velocity of the bolt, as well as a spring constant, how far will the bolt travel before coming to a stop (and going back)?"

Again, the bolt mass will be 0.795 kg with an initial velocity of 3.75 m/s. The spring constant will that of the standard AMD spring = 94.8 N/m. Therefore, bolt stroke will be:

This is about 3x the length of the current bolt stroke and presents a danger of the bolt/carrier combo slamming into the rear trunnion with excessive force. As mentioned earlier, a stiffer spring (or a longer receiver -- what would that look like!?) would counteract this problem but may, at the same time, introduce a new one -- where the bolt/carrier slams into the front trunnion with too much umpf on the way back. A fine balance would need to be achieved, but alas, without options in spring strength this question may remain unanswered.

In conclusion.

The mathematical analysis presented here gives a value of 795g for the ideal bolt mass in a 7.62x25 plain blowback weapon firing from a closed bolt. This is a little less than double the mass of the standard AK-74 bolt/carrier. Furthermore, restricted to the length of an AK receiver the theoretical spring constant for the return spring works out to 891 N/m, which is about ten-times the strength of the standard AKM return spring.

We can see that making a full diameter gas piston gives us about 100g extra bolt weight, however the theoretical needed-value is higher still. The stock return spring appears to be too light and while a option for a slight strengthening exists, it is limited to 115% out of the theoretically needed 1000%.

#### 1 comment:

1. So how are you going to make your 7.62x25 AK work safely?