A few Questions about the Dupont Linkage

[Aespa]

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.

[Irondragon Forge ClayWorks]

Welcome to IFI Aespa. I can't help with the questions but this thread will help you get the bet out of the forum. READ THIS FIRST

[Frosty]

Welcome aboard Alberto, glad to have you. If you'll put your general location in the header it'll help considerably. Your translation software is going to take some getting used to some of your terms don't really make sense in American English. 

I'm having to guess what you mean by, "rate" of the spring. As an engineer you don't want opinions based on assumptions do you?

Frosty The Lucky.

[ThomasPowers]

spring rate = the weight necessary to compress a spring 1". It  can be linear exp: 100# => 1",  200# => 2"

or some function  exp: 100# => 1",   400# => 2"

[Frosty]

That makes sense, another glimmer of light shines on my ignorance. Thanks Thomas.

"The Little Giant Powerhammer", book only lists springs by size: Dia. Length, wire dia. and coil count. Nothing about rate. 

I'm afraid that's the best I can do for you.

Frosty The Lucky.

[swedefiddle]

Good Morning,

Racecar books have spring rate calculators. wire diameter, inside diameter of coil, number of working coils (the ends against the spring pad do not count, they are dead).

I will try to remember where I put my calculator.  Yes Thomas, springs are linear, rated per inch of compression.

Neil

[Aespa]

On 4/29/2019 at 4:40 PM, Frosty said:

That makes sense, another glimmer of light shines on my ignorance. Thanks Thomas.

"The Little Giant Powerhammer", book only lists springs by size: Dia. Length, wire dia. and coil count. Nothing about rate. 

I'm afraid that's the best I can do for you.

Frosty The Lucky.

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
 

 

[Aespa]

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

[Frosty]

This is from Appendix D, Pg. 208 in "The Little Giant Powerhammer."

The toggle spring in a 25 LG: OD 2.50", ID 1.5", Length 4.5", Wire Dia. 0.50", # of coils 6.5

The toggle spring in a 50 LG: OD, 3.0", ID 1.75", Length 5.50", Wire dia 0.0625", # of coils 6.75

I have a 50# LG but it's taken apart right now compressing the spring isn't practical. 

Frosty The Lucky.

[Aespa]

19 hours ago, Frosty said:

 

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

[Judson Yaggy]

Frosty just slipped a zero.  Read it as .625"

 

[Frosty]

Sneaky little decimals, that stinker sneaked over a place while I was typing!  Yes, 0.625" wire dia. My bad.

Frosty The Lucky.

[Aespa]

thanks,  yeah I was kinda worried hahah 

[Aespa]

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 

[Frosty]

 N/  Newtons? 

Frosty The Lucky.

[Aespa]

4 hours ago, Frosty said:

 N/  Newtons? 

Frosty The Lucky.

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.

[Frosty]

If you hadn't told us I never would've known you'd made a math mistake, that's way beyond me.

Frosty The Lucky.