Aespa

yes, N of [Newtons], Si units if anyone ever needs those units Btw I notcies some mistake in my calculations before, when I said "39 675 N*mm = k*(16^2) gives me a of 154 N/mm or aroun 880 punds/inch " it should have said 39 675 N*mm =0.5* k*(16^2) gives me a of 309 N/mm or aroun 1764 punds/inch  those "16" mm are the distance that I want to allow for deformation so I wasn't that far away.

So I ran the calculations and probably these are the rates of the spring for the LG LG25 : Spring Rate (or Spring constant), k : 1,726.708 Lbs/In roughly 300 N/mm and LG50: Spring Rate (or Spring constant), k : 2,424.196 Lbs/In roughly 500 N/mm

thanks, yeah I was kinda worried hahah

Thanks, but gotta ask is it really 0.0625'' the diameter in the 50 LG? cause the other one is 0.5'' which seems odd, anyway, thanks! at least I have an estimated rate now

Sorry for the double posting, but there's a simple experiment that someone who has a working hammer you can take the springg and measure his original Lenght then add weight on top of it (measure the weight) then measure how much of the spring is compressed with the weight on. with that info and the weight of the ram and the stroke of the hammer I can extraplate pretty much everything. and I can accuse empyrical evidence to benchmark the project

first, thanks for your time guys, and yeah the rate of the spring it's pretty much defined as that Force needed to compress 1 unit of distance. could you please give me acces to that data, dia, lenght, and coil count, I could work with that, but sadly I don't have access to that book, I live in Chile so...it's a big distance, and those books are kinda hard to get in here. I might have an idea of what it's needed for the spring, but I'm not completely sure and I've got nothing to compare at hand, so any info would be awesome as far as my simple reasoning goes: the spring needs to absorb the energy given the momentum of going from down to top the stroke, I've managed the calculate the speed of the ram when running at 3bps. at around 3.4 m/s. and the distance that I've allowed the arms of the dupont to move is pretty much 8 mm each so, 16 mm of compression, runnin math on that the kinetic energy of the Hammer would be roughly Ek=0.5*m*v^2 = 39 675 [N*mm] [My design is for a small hammer btw, 15 kg or 33 lbm] so I know that the energy of the spring is given by Es=k*x^2 where k is the rate and x it's the compression. so doing the math 39 675 N*mm = k*(16^2) gives me a of 154 N/mm or aroun 880 punds/inch the problem with that is that I've got nothing to compare and on youtube I've seen the stiffest springs to the softest working kinda fine, obviously the stiffer the spring the higher the rpm without resonance on the machine, so, kinda safe, but I don't really have something to compare atm as far as I've seen the Little giant has a xxxxxxx stiffer spring, given that it almost does not move anything BTW. sorry for using S.I. units, I use to work more with that than freedom units. so that would be everything I have, I'm literally throwing shots in the dark right now and that's pretty much the step I need to finish the calculation memory of the hammer

Hello, first of all, my Name is Alberto Esparza, I'm a mech engineering student working on a power hammer as a project for the course of "Mechanical Design" but I'm having some serious problems right now with the spring, so I don't know where to go more than you in the search for experience. What's the rate of the spring in the 50lb hammer? and those who have built their own power hammers with the Dupont linkage how did you selected the spring? any info would be awesome, thanks for your time in advance.