josh smith

Good job Doug! Now we fully understand the science of the springs because we both wanted the truth. I agree with everything you just said. Maybe now curious people can calculate the force of a spring themselves since we exposed the secrets of the springs. lol later Doug JOSH

Sup Doug, Im not here for pissing contest aswell, I responded to this thread because I have few knowledge about springs since I majored in Physics. I know I'm not a spring engineer so I don't expect that I am right 100%. I'm a science person so I really want to know the truth. And I will accept that I am wrong if you give me logical proof that I made a mistake. I accept defeat is I deserve it. ok, your example here is not valid. You just picked random numbers. You assumed that if you cut 1 coil you reduce the length by .25". This is wrong. If the length of 1 coil is 0.25" then for 28 coils that would be 7 inches long. With that, I don't have to say more. The difference is NOT the spring constant. From Hooke's Law: F=kx, F is the force, k is the spring constant and x is the distance you compress the spring. So the difference is x. Results from internet calculators doesnt convince me. I studied computer programming myself, and I can make a calculator program that if you add 1 plus 1 you get 20. My point is I want the formula behind the calculator. Since Im not a spring engineer and I only majored in Physics, so my knowledge about spring is limited. So I researched again. and guess what? I found the HOLY GRAIL!!!, lol. So this is my conclusion: we were both partly right. spring rate increases if you reduce the number of active coil, which you are right about. But the increase is very little. In the end the Force required to compress a certain amout on a longer spring is still greater than the force to compress a shorter spring to the same amount. So if we cut the spring shorter it then it gets LIGHTER. Thats all I have to say. Im out of here...... heheh, jk. I know you're not happy with that, let me give you the HOLY GRAIL. Equation for spring constant/spring rate: k=(Gd^4)/(8nD^3), where k is spring constant/spring rate G is modulus of rigidity of spring material in lbs per square inches d is wire diameter n is number of active coils D is mean coil diameter. so k is inversely proporsional to n(number of active coils), which you are right about. But since G is very large number the effect of active coils on k is very small. And if we apply this in actual experiment, this will only have small effect and the lenght of compression counts more. example. To make it more valuable I'll try to use numbers close to a recoil spring. 8 inches recoil spring. lets use 0.25" per coil(length) for easy calculation. so theres 4 coils in 1 inches, hence 8 x 4 = 32 coils. 8" spring:32 coils if we cut 4 coils, that makes 7" spring:28 coils we need to find k for 8 inches and k for 7 inches. we can find k using the internet calculator you provided or by using the HOLY GRAIL of springs, which is k=(Gd^4)/(8nD^3). to compare our values lets use same parameters. d = Wire diameter .043" D = coil diameter .430" G = 11500000 sample from http://www.engineersedge.com/calculators/t...pring_k_pop.htm so if we plug in n =32 for 8 inches, we get spring rate, k = 1.9316 if we plug in n =28 for 7 inches, we get spring rate, k = 2.2076 Look at that, we if reduce the number of active coils it actually increases the spring rate. But we are not done yet. if we plug in the recoil spring in the gun. lets say the lenght of the spring in batterry is 3" for easy calculation. Using Hooke's Law, F=kx, where F is force, k is spring rate and x is distance if the compression. For 8" spring, k = 1.9316 and x = 8" - 3" = 5" So F=kx F=(1.9316)(5) = 9.658 lbs For 7 inches, k = 2.2076 and x = 7" - 3" = 4" So F=kx F=(2.2076)(4) = 8.8304 lbs hmmm, I think that means if we cut an 8" spring to 7" long, the Force require to compress them to batery went from 9.66 lbs to 8.83 lbs. So I was right all along that if we cut the spring shorter it gets lighter. But I give you credit for spring rate/spring constant variation. So Doug, since you already is a member in the engineering website. try plugging in the numbers I used, lets see what you get. I WANT TO KNOW THE TRUTH AND NOTHING BUT THE TRUTH, hehe. later Doug. peace

I want to see the website of spring engineers and manufactureres that you're talking about. I did search in the internet and I found informations that support what I learned from the university. I gave you examples thats proves that springs do not get stronger if you cut them. I showed you how I got my answers and what formula I used. But you didn't give me any formula and logical examples that proves you are right. Seriously though, I want to see the website that you're talking about. I already give the website where I found that Normal spring like recoil spring has a fixed spring rate. If noone else going to argue with me, them I'm out of here as well, lol.

yes spring rate is same as spring constant. and Normal springs has a fixed spring rate and recoil spring is a normal spring. I gave you the formula thats proves I was right. (F=kx). Now give me the formula that proves that you are right. I don't believe you actually tried cutting the screen door spring in half and measure the force. I think you just wrote that because thats what you believe. You can convince that Im wrong if you give the formula of what you believe. If you show me the formula of how you got twice the energy if you cut the door spring in half then I'll tell everything you want.

I just googled "spring rate". and I found this. "As the number of active coil decrease, the spring rate increases" TRUE if and only if the lenght of the spring stays the same. and I also found this. "Normal Springs has a fixed spring rate. Step Linear Springs are springs which have 2 different spring rates. Progressive Springs have a variable spring rate." I do believe recoil spring is under Normal Spring, so spring rate is constant. Spring rate is the same as Spring constant. For people thinks that springs get stronger if you cut them. Prove it by giving us the formula that shows that its true. If I'm wrong here I'm going to sue my university. lol reference: http://www.tuninglinx.com/html/springs.html

I'm confuse with the term spring rate. Define spring rate? I thought spring constant(k) depends on the material of the spring, diameter of the wire, diameter of the coil. But we are not changing those when we cut the spring, we just change the lenght, so k stays the same. Can you give me the formula that you use to get your calculation? I majored in physics and the formula I use is in the textbook and they never said that the spring constant changes when you cut the spring. And if the spring constant changes, Hooke Law(F=kx) will not work, will it? And I pay lots of money to go the university.

Hard to follow. Maybe because im not a spring person. But I know simple physics. Coil springs obey Hooke's law. And Hooke's Law is F=kx, where F is force, k is spring constant and x is the distance which the spring deformed example 1) Two springs, S1=10", S2=20", k = 3lbs/inch. Compress both springs to 5". F=kx F1=(3lbs/inch)(5inches) F1=15lbs F=kx F2=(3lbs/inch)(15inches) F2=45lbs This example tells us that it takes only 15 lbs of force to compress 10" to 5" and it takes 45 lbs to compress 20" to 5" example 2) (exaggeration) Two springs, S1=10", S2=20", k=3lbs/inch Compress both spring to 10 inches For S2 F=kx S2 =20" x2=20"- 10" = 10" k=3lbs/inch so, F2=(3)(10) = 30lbs. For S1 F=kx S1=10" x1= 0" since its only 10" long k=3lbs/inch so, F1=(3)(0) = 0 lbs. It make since to me. longer spring is stronger than shorter spring I can't tell you that you are wrong coz I can't follow your explaination, maybe bcoz im not a spring person. but can you tell me why am I wrong?

Scientific theories always right, just the matter of having the tools to do the experiment. You can't use the same formula for these two different things. On your example, if using the SAME rod for 1 mile long and 1 foot long with 1 foot long on one end for leverage, the force required to twist the the rod until breakage is the same. If you're talking about longer leverage for 1 mile rod and shorter leverage for 1 foot rod then its a different story. you can twist same materials the same amount no matter how long. if you change the length of the lever then its easy to twist with long lever.

You're are both right and wrong. Cutting a coil off changes the compression leght. but "k" stays the same. AND the "Force" lessen

I'm with dogmadog on this one. k for 20" spring = k for 10" spring spring constant will not change if you cut the spring in half. Did you actually do the experiment for your calculation? You're argument is like this, you're saying a 5 feet 10 year old boy will get to 10 feet tall by the time the boy turns 20. Don't make since does it? Try doing an experiment on your calculation and you'll find out that you are wrong.

thanks a lot. that was quick, JOSH

Hi guys, I've searched for over an hour but didnot find what I'm looking for. I want to change my RecoilMaster to a tungsten guide rod, I want to know if I need to change the reverse plug or not. Should I buy and different plug specipecally for Tungsten guide rod? or I can use my original STI severse plug with a tungsten guide rod. thanks, JOSH

WOW, thanks a lot for taking the time responding my questions, I really appreciate them. Lots of critical informations above and I will consider them all. I think what I'm gonna do is disables the grip safety, adjust the pretravel just enough so that if I pull the hammer to half cock and pull the trigger, the hammer wont fall. This is safe for me and also to people around me. I had lots of practise with the sear spring I know how to do this now. YOU RIGHT GUYS THIS IS THE RIGHT PLACE TO ASK QUESTIONS... Thanks again, JOSH

Just a theory of mine, I want to improve my double tap. I'm new so I want to experiment for improvement, but I want to do it safely. I thought maybe if I have almost zero pre-travel and almost zero over-travel I can have faster reset and therefore shorter split(time).

thanks for the responses guys. So, lets say I want almost zero pre-travel. I done this before but when I pull back the hammer to half-cock and if and only if I press the trigger, the hammer goes down and hit the firing pin. Is this still ok/safe? Is it okey if I do this and disable the grip safety too? Just left with the thumb safety. JOSH