# HELP required calculating spring strength.

## Main Question or Discussion Point

I have recently redesigned an antique falling block single rifle action and need some help calculating the poundage of the firing pin spring.
A test that is often used to determine if a firing spring in a gun is still in serviceable condition is to place a coin, which weighs approx 10 grams on the breech face of a gun and fire it. If the firing pin has enough striking force to propell the coin approx 1.5m vertically into the air all is deemed well.

The prototype action currently uses a spring and firing pin to the following specifications
spring details; o.al. - 24mm, installed length - 20.5mm, compressed length 14mm, no of turns- 6.75, wire dia 2.00mm, external diameter 13.7mm, rated at 108"/lbs. (11.298482902 newton meters)
firing pin weight 28 grams, which is approx 1 oz (16 oz to the pound)
firing pin travel from cocked position = 6.5mm
height of 10 gram coin reached when placed on breech face and fired = 2.9m, which is obviously almost double required height.

If the firing pin remains the same weight, what reduction of the spring poundage will be required to propell the 10 gram coin only 1.5 meters vertically?

Related Other Physics Topics News on Phys.org
I'm assuming ideal conditions, such as an ideal spring (which is a reasonable approximation) and no air resistance (maybe not as much). If your spring constant is k (= 11.3 N m), the spring compression is x (= 6.5 mm), the mass of the coin is m (= 10 gram), and the maximum height of the coin is h (= 2.9 m), then kx2/2 = mgh, or h = kx2/mg. (I'm neglecting the firing pin weight, since it moves so little.) Thus, the height is proportional to the spring constant, and to the square of the spring compression.

Thus, if you want to get a height of 1.5m, adjust k to be about 52% (= 1.5 / 2.9) of what it was originally.

I have recently redesigned an antique falling block single rifle action and need some help calculating the poundage of the firing pin spring.
A test that is often used to determine if a firing spring in a gun is still in serviceable condition is to place a coin, which weighs approx 10 grams on the breech face of a gun and fire it. If the firing pin has enough striking force to propell the coin approx 1.5m vertically into the air all is deemed well.

The prototype action currently uses a spring and firing pin to the following specifications
spring details; o.al. - 24mm, installed length - 20.5mm, compressed length 14mm, no of turns- 6.75, wire dia 2.00mm, external diameter 13.7mm, rated at 108"/lbs. (11.298482902 newton meters)
firing pin weight 28 grams, which is approx 1 oz (16 oz to the pound)
firing pin travel from cocked position = 6.5mm
height of 10 gram coin reached when placed on breech face and fired = 2.9m, which is obviously almost double required height.

If the firing pin remains the same weight, what reduction of the spring poundage will be required to propell the 10 gram coin only 1.5 meters vertically?

Well, I just had a quick look at the specifications you gave, and I have to ask you how sure are you about the numbers? If the spring constant is really 11.3 newton meters, and the compression is 6.5 mm, I cannot see any way it is throwing a coin 2.9 meters in the air! It would take a much more powerful spring to do that. Can you please recheck those numbers? I also wonder why you are concerned that the firing pin is striking harder than it should? I would think that would help ensure that there are no mis-fires. Is there some other reason?

I just had another thought. You say it is an antique falling block fire? Then the firing pin spring may not be providing the power to the pin at all, but is actually working to return the pin to its original position, as well as limit the force with which the pin engages the primer. If that is the case, and the pin is striking too hard, you may need a stronger firing pin spring, not a weaker one. Can you verify the type of firing system?

Hi Schroder,
All the numbers regarding the spring I think are correct except possibly the number you are questioning, the 11.3 nm. I was informed by the guy who made it for me, has that the spring a rating of 108"/lbs, which I loaded into a conversion programme on the net and that is the answer it gave. He went on holiday on Friday. The coin is definitely shot 2.9 m vertically as I have measured that. The reason I need to reduce the spring pressure is that because it is an under-lever design, similar to a Ruger #1, Farquharson (if that means anything to you) and with the current spring there is too much force required to cock it. And rather than go through a re-design of the cocking mechanism I thought I would get the spring poundage right where it needs to be and see how it feels then. Hopefully this will alleviate the problem. There is also I believe the issue of potential ruptured primers if there is too much force/protrusion? I am an energy physics ignoramus,( hence posting this question), but given my dilemma was hoping that there would be some law of diminishing returns at play with this spring force. Like it takes 4x the horsepower to get a car to accelerate 1.5x faster.

Hi Schroder,
What I have done is put a newly designed firing mechanism in an old styled receiver. This mechanism uses an in line striker, similar to a bolt gun in concept. So the force exerted by the firing pin is in-line with the bore and the center of the primer. Most falling block single shot rifles use a heavyish hammer which strikes a very light firing pin. Given the results of the "coin test" I believe I can definitely reduce the power of the spring. Thanks

Hi Sorrocco, welcome to PF. I’m not a gun expert, but have had my share of experience with the M14 and M16 in the military and S&W .358 and a Ruger pistol. That’s about all except for the heavy stuff in a tank, but I don’t think that helps us much here. I still wonder if this is the firing pin spring which is driving the firing pin, or if it only helps to return the pin back after firing and also limits the force of the pin on the primer. Then the force for the pin would have to come from some other hammer and spring. Can you verify this for me? The reason I ask is because the spring constant of 11.3 nm is way to low by a factor of ten to drive that coin 2.9 meters in the air. But if it is true, I can calculate what the actual spring constant is now, and what it should be to get it down to 1.5 m.

Hi Schroder, Thanks for your help with this. The spring in question propells the firing pin, which weighs 28 grms forward in a straight line. The spring is quite large in diameter and short in length, compared to most other rifles. I just have to make sure there is enough force and preload to negate the force of the primer pushing the primer compound back through the firing pin hole, which would then lock the action solid, as it will prevent the block from dropping.

OK. I will drop off line now and crunch some numbers and see what I get. Just one last question: the spring constant you gave had a lot of decimal places, was that what the calculator spit out? I'm just wondering if the decimal point is in the right place.

Hi Schroder, I have cut and pasted the link to a conversion calculator below. I re-entered 108"/lbs and got a slightly different conversion of 12.202 nm. Maybe the figure I have been quoted by the spring maker was incorrect? Its still a bit early here in New Zealand to ring another spring maker but will do so shortly and see what numbers he provides as a comparison.

http://www.unitconversion.org/energy/inch-pounds-to-newton-meters-conversion.html

Hi Schroder, I just managed to get another spring manufacturer to run the spring specs through his computer and he came back with a figure of 20.8 newton mm. Does that make any sense?

Hi Schroder, I just managed to get another spring manufacturer to run the spring specs through his computer and he came back with a figure of 20.8 newton mm. Does that make any sense?
Well, that makes more sense than 12.2 as I can’t believe 12.2 can throw the coin 2.9 meters. Although there will be considerable momentum gain via the firing pin and the coin, and even more between the striker and the firing pin. And there may well be more between the spring and the striker, depending on how they are linked. I will stop calling that your firing pin spring and instead refer to it as your mainspring, to avoid any confusion in my own mind. I will stick to the original numbers you gave, until we can confirm something else. 108 in*lb is 9 ft*lb and is equal to 12.2 Joules or newton meters. To send that coin 2.9 meters high, the coin has to taking off at an initial velocity of 7.5 m/s. Now we can apply conservation of momentum to calculate the velocity of the firing pin as in mass*velocity of coin = mass*velocity of the firing pin: .01*7.5 = .028*X , X = 2.67 m/sec for the firing pin velocity. To give you an idea of how the calculation should continue, we need to reduce the coin launch velocity down to 5.4 meters/sec in order to limit the height to 1.5 meter. Then, the firing pin velocity will be 1.928 m/s. The reduction is about 25% It might be useful to also know the mass of the striker and any details about the mechanical advantage (leverage) between the spring and the striker. If it is a straight in-line configuration, then disregard the leverage. If it is a rotating hammer, or something similar, then the leverage is important. If the 25% reduction holds, then it is easy; just change the 12.2 to a 9 newton meter spring or 6.75 ft*lbs. Hopefully this will not cause any problems any where else. There are some other considerations, such as coefficient of restitution and lock time, but maybe we don’t need to consider all that. Let me know if can get any more data on the striker and spring.

Hi Schroder,
Thanks very much for your time and efforts with my problem. Its great there are people on the planet that enjoy sharing and utilising their knowledge for the pure enjoyment and challenge. It is appreciated.
The striker is the firing pin (28grams) in this instance and is in an in-line configuration, so there is no mechanical advantage. Without letting too much out of the bag because of patent application reasons, imagine the firing pin/striker is a cup, with a tit on the end, and a spring contained by it, pushing off a solid fixed point at the rear. After reading your response I have ordered 2 springs a 75% reduction of the original, as per your calcs and a 60% (in case I can get away with it and 'cos it was free). Obviously the lock time will be reduced however, compared to a hammer configuration, it will still be 2 or 3 times faster, so I have no concerns re this. All going well, I get the springs 2 morrow so will let you know how I get on. Cheers till then.

Hi Schroder,
Thanks very much for your time and efforts with my problem. Its great there are people on the planet that enjoy sharing and utilising their knowledge for the pure enjoyment and challenge. It is appreciated.
The striker is the firing pin (28grams) in this instance and is in an in-line configuration, so there is no mechanical advantage. Without letting too much out of the bag because of patent application reasons, imagine the firing pin/striker is a cup, with a tit on the end, and a spring contained by it, pushing off a solid fixed point at the rear. After reading your response I have ordered 2 springs a 75% reduction of the original, as per your calcs and a 60% (in case I can get away with it and 'cos it was free). Obviously the lock time will be reduced however, compared to a hammer configuration, it will still be 2 or 3 times faster, so I have no concerns re this. All going well, I get the springs 2 morrow so will let you know how I get on. Cheers till then.
It's comments like yours that make this place worthwhile. I hope the new spring works out. G'day mate!

Gidday Schroder et al,
Well I just finished testing both springs and am happy to report the following results. It seems that the 55 - 60% spring is right on the money in terms of force. It shoots that 10 grm coin between 1.5m & 1.65M (60" -67") everytime, I guess depending whether or not the coin is struck dead centre or just off. The 70-75% spring was still richocheting off the ceiling, ( I didn't bother going outside to see how much higher as the ceiling is approx 1.6m from my desk) So it would seem that the reduction in spring force in my particular set up has a corresponding linear reduction in height of vertical travel. This linear reduction, given my dilemma, has helped considerably with solving the problem. Thanks again for your assistance.

That's great news! I'm thinking maybe it was a 20 n-m to start out with, as that other fellow calculated, so now maybe you have about a 9 or so. Anyway, it all worked out, that's all that matters. Time to put on the barbee Q and have a tinnie I'd say.