Buffer Spring Question?

[prepare]

Does anyone know if the buffer spring has a specified spring weight? Like a 15 lbs spring for example or does the TDP just go by length?

[lysander]

"Action Spring" . . . I was about to say the buffer has no spring.

Anyway, the spring particulars are:

At a compressed length of 6.90 inches, the spring should develop 5.96 +/-1.0 lbs

At a compressed length of 2.98 inches, the spring should develop 12.12 +/- 1.0 lbs

(You can calculate the spring rate from these two points.)

I never liked terms such as " a 15 pound spring", as it doesn't tell you much, but generally is the load developed at solid height, or the maximum force it can develop. Not very useful. (This would be a 12 pound spring) To confuse things, sometimes the term is used to denote the spring rate, (in which case the action spring would be around 1.5 pounds).

EDIT: That is for the 10.70 inch long, 38 coil, standard carbine spring.

[lysander]

deleted post (it was in error)

[prepare]

Thank you for the numbers and the explanation.

[joedirt199]

I was trying to do the same thing figuring out the compressed weight of the standard spring vs a flatwire spring. I wanted to get the same weight out of a cut down flatwire to reduce the closing force but still have the longer lasting of flat wire. I was using a rifle buffer extension and a fish scale to pull the spring compressed but the fish scale was not high enough and my bow scale started to high to measure.

[markm]

I was trying to do the same thing figuring out the compressed weight of the standard spring vs a flatwire spring. I wanted to get the same weight out of a cut down flatwire to reduce the closing force but still have the longer lasting of flat wire. I was using a rifle buffer extension and a fish scale to pull the spring compressed but the fish scale was not high enough and my bow scale started to high to measure.

Not for nuttin... but why not just buy some extra Colt springs?

[lysander]

You don't have one of these?

[lysander]

I was trying to do the same thing figuring out the compressed weight of the standard spring vs a flatwire spring. I wanted to get the same weight out of a cut down flatwire to reduce the closing force but still have the longer lasting of flat wire. I was using a rifle buffer extension and a fish scale to pull the spring compressed but the fish scale was not high enough and my bow scale started to high to measure.

Cutting coils off a spring increases the spring rate.

The standard carbine spring holds the bolt close with about 6 pounds force, when the bolt is fully to the rear the spring develops about 11.5 pounds.

If you clip off two coils, the force holding the bold closed drops to around 4 pounds, and the force with the bolt fully to the rear is about 10.2 pounds.

Your closing force drops a bit, but the force holding the bold closed drops substantially more. It is not something I would not recommend.

Flat wire springs have a lower spring rate than standard spring, they are just considerably longer, so the develop the same pre-load (~6 lbs). Chopping a flat wire spring off gives you the troubles of the reduced coil standard spring (low holding force), and the disadvantages of the standard length round wire spring (high spring rate). Why would you want to do that?

[tangolima]

I never liked terms such as " a 15 pound spring", as it doesn't tell you much, but generally is the load developed at solid height, or the maximum force it can develop. Not very useful. (This would be a 12 pound spring) To confuse things, sometimes the term is used to denote the spring rate, (in which case the action spring would be around 1.5 pounds).

Agreed. The confusion comes from that nobody cares to ask for a precise definition. I tried and got different answers from different people who were supposed to know (spring manufacturers). When pressed, they all ended saying "I'm not sure". Finally I got a good answer from an old timer. The spring's poundage is product of its Hook's constant (lbf/inch) and its natural length. Or it is 100x of the pound force produced for each percent change in length.

The carbine spring in your example

Lo = 10.7", K=1.53 lbf/inch

K*Lo=16.3lbf, or it is a 16 pound spring per this definition.

With that calculation actually becomes even simpler. Again for the same carbine spring, as per the forces given, the install length is 6.8", and the compressed length is 3.2".

1.53*(10.7-6.8)=6 lbf
1.53*(10.7-3.2)=11.5 lbf

Or
16.3*((10.7-6.8)/10.7)=6 lbf
16.3*((10.7-3.2)/10.7)=11.5 lbf

Note that the poundage is independent of the spring's natural length, unlike K, which is handy for calculating shortened spring.

2 coils off 38 coils. The new natural length is

36/38*10.7=10.1"

16.3*(1-6.8/10.1)=5.4 lbf
16.3*(1-3.2/10.1)=11.1 lbf

You can arrive at the same figures with the K method.

The force is weaker, but not really alarming. Actually in a quest to make a "gas assisted straight pull action" I cut the spring of an AR -10 pretty short. No problem, not that anyone should try though.

Does solid height mean compressing spring till it coil binds(cannot be compressed anymore)? According to the old timer who explained spring poundage to me, one should avoid that in use. It could pass the spring's limit and it could take a permanent set. There should be limiter in design that keeps this from happening.

-TL

Sent from my SM-N960U using Tapatalk

[lysander]

Does solid height mean compressing spring till it coil binds(cannot be compressed anymore)? According to the old timer who explained spring poundage to me, one should avoid that in use. It could pass the spring's limit and it could take a permanent set. There should be limiter in design that keeps this from happening.

-TL

Sent from my SM-N960U using Tapatalk

NOTE: When counting coils, only count active coils.

The action spring has closed ends, so the last coil on either end doesn't count - 36 active coils.

Solid height (or solid length) is simply, the number of coils (active and inactive) times the wire diameter, it is as short as the spring can physically be (action spring = 2.74"). If you want your spring to last a long time, you should stay at 130% of the solid length for maximum service defection. For the action spring, this would be 3.56", so its design is a little over-compressed. (Just about all gun recoil and magazine springs are. That's why they have relatively short lives, compared to, say car suspension springs.)