How much force does my press exert?

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neck tension or bullet pull
This guy used a valve spring tester. .. PSI and Pounds , very different i think? 20170928_222357.png

I used a bathroom scale, to push bullets into the case.
The 5.56 needs 35 lbs minimum bullet pull. About .002" neck tension in my test.
The 45acp , different brands of brass , from 45 lbs to past 100 lbs. Neck tensions from .002" to .005"
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Uh OH, maybe I will need a bigger load cell? My first curiosity was measuring the uniformity of pulled bullets as in neck tension or bullet pull but the sizing force at the ram would be nice to measure too.

Ron

I am interested in bullet set back pressures .... for pistol calibers ... and maybe .223

I see I'm not alone ...!
 
I sent a email question asking the ft-lbs of force generated by the Forster CoAx today ... Who knows if they will answer it or not ...

I would think someone there would have the answer ...
 
I sent a email question asking the ft-lbs of force generated by the Forster CoAx today ... Who knows if they will answer it or not ...

I would think someone there would have the answer ...
Just about every time I have emailed questions like this to a manufacturer they have replied, so let's see what happens.

Ron
 
You can buy an arbor press that has a "force measurement ram & dial indicator"

http://www.lewilsondirect.com/kandmprecisionarborpress-1.aspx



Obviously (if you clicked on the links), this gauge remains on the press all the time so you can see how much force you are applying to the ram and how consistent the force required from piece of brass to piece of brass is. The gauge isn't to see just how much force is possible to produce from the press (in this press, it isn't much) but to see how much force is required to neck size a given piece of brass and whether the force is consistent for each piece of brass.

FWIW: I use this press to load a couple of my rifle cartridges (with LE Wilson hand dies) although I didn't get the dial indicator on mine.
 
A press does not have a top end force, it has a mechanical advantage.

If it has a mechanical advantage of 5, then a force of 10 lbs on the handle produces 50 lbs on the case (or whatever is between the ram and die).

The mechanical advantage can be determined by measuring the ratio of throw distances, but it is not usually constant for the whole handle motion. If if takes 5 inches of handle motion to produce the last 1 inch of ram motion, then the mechanical advantage is 5.

The top end of forces that are possible then depend on how much force the user can apply to the handle AND how much force the press can take before a part fails. Most adult men can apply a force to the handle that is close to their weight, or at least half. 100 lbs would be common. Most presses are strong enough that the possible forces do not result in failure. However, any time I've ever gotten anywhere near there, the shell holder shears the bottom of the brass cartridge off on the reverse stroke (handle up, pulling case down) so that I end up with a stuck cartridge. I've learned to STOP and try something different any time the required force on the handle exceeds a threshold that I recognize by feel (about 40 lbs).
 
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Am I guessing correct that Newtons is the term for force ?
Newton is something we stuff figs into resulting in the famous Fig Newton. :) I had to say that.

The Newton is a SI (System International) unit of force. It does have a definition of:
One newton is the force needed to accelerate one kilogram of mass at the rate of one metre per second squared in direction of the applied force. As taken from the WIKI.

Here in the US we like using pounds of force. The conversion would be 1.0 Newton = 0.224809 pound force. Just a matter of which engineering unit of measure we wish to use.

Ron
 
For you math guys ... here are some swing measurements of my old Bonanza CoAx press... to the best of my quick measuring...

Handle is 16" from tip to the center of the Fulcrum point ...

From the Center of Fulcrum to center of the pivot points:
0°= 0.0"
45° = 1.36"
90° = 1.90"
135° = 1.36"
180° = 0.0"

Co_Ax_Press_Arch.jpg

As Jmorris stated ... the last few degrees before the 180° mark it where most of the "work" is done when sizing ...
 
I have agree the simplest method is from calculations. Application of Physics and Statics from college.

However, since testing was suggested, I was hoping for testing with pressure sensitive powders. wile_burned.jpg
 
So Jesse can you calculate the force generated using the measurements above ... I was hoping some really smart math whiz would be able to .... I was dreaming about the girl in the next room during math class! There are no Holiday Inn Express around either ...

Say it you apply 50 lbs on the handle .... what would the force be generated.. . ft-lbs/Newtown ...
 
I read this, but couldn't figure out the why? I mean my co-ax will load anything thrown at it. Not sure I want to know the technical details regarding the force it's leverage gives me?
 
I read this, but couldn't figure out the why? I mean my co-ax will load anything thrown at it. Not sure I want to know the technical details regarding the force it's leverage gives me?

That is the whole point of this thread ... go back and start at the very first post ...

I ... along with others want to know ... and yes my Bonanza will handle most anything I use it for ...
 
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if you went 50 in, it would be a lot I am waiting on some seals I didn't have to get more precise results but it turns out on the co-ax at .100" off the top of the ram stroke,19.25-21.625 lbs in outputs 1000.

It also changes rapidly as it moves closer to the dead spot. Enough that I think I need to add a dial indicator to the mix so I can see in thousandths, the portion I am measuring and making sure it is they same spot in travel that is being measured subsequently from setup to setup.

I read this, but couldn't figure out the why?

I have to say that I agree, the fact that is hasn't been answered already is testament that few if any care.

Curiosity finally got the best of me and before I could only speak in qualitative terms on how presses "feel" but now I can say that a Lee breech lock takes very close to the same force as an RL550 to produce the same output and they are both about 4lbs more input needed than with the co-ax, again for the same output. Once I have everything the way I want it, I'll have a more extensive list (that many won't care about either ;) ).
 
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I read this, but couldn't figure out the why? I mean my co-ax will load anything thrown at it. Not sure I want to know the technical details regarding the force it's leverage gives me?
Among the possibilities of knowing force could be the following. Serious hand loaders for target shooting like things uniform. They trim their brass uniform, they sort head stamps for uniformity, they seat bullets uniform and everything is about being uniform. So one thought would be I could load 110 rounds and take note of how much force is applied to seat my bullets. Then with 110 rounds seated I could pull 10 bullets selected at random. If I charted each pull I could get a good feel for how my neck tension or bullet hold was. That just being an example of what could be done with the collected data.

Ron
 
I read the posts from top to bottom. No issues with it. Just wanted to know why the exercise? The reasons didn't seem to justify the effort. Continue on.
 
if you went 50 in, it would be a lot I am waiting on some seals I didn't have to get more precise results but it turns out on the co-ax at .100" off the top of the ram stroke,19.25-21.625 lbs in outputs 1000.

It also changes rapidly as it moves closer to the dead spot. Enough that I think I need to add a dial indicator to the mix so I can see in thousandths, the portion I am measuring and making sure it is they same spot in travel that is being measured subsequently from setup to setup.
That would be pretty cool as you could look at travel against force.

Ron
 
According to my simple brain ...If you applied 50 lbs force on the end of a 16" handle(like my CoAx) ... to the fulcrum ... with 1.9" past the fulcrum ...

The force generated is
421.05 ft-lbs @ 90 °
588.23ft-lbs @ 135°
1066.6 ft-lbs @ 157.5°

Most of the work happens between 157.5° & 180° for sizing
 
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I was just thinking I have stuck a couple cases before in a rock chucker and lifted one side of my 300 lb bench off the ground before the case rim pulled off. I wish I still had it so I could figure out the leverage ratio and figure out the approximate force.

Food for thought, if you have a press that cams over at the top the leverage ratio at the cam over point is almost infinite and the force applied basically depends on the stiffness of the press.
 
if you went 50 in, it

50 was just a number that my finger pushed in the calculator.... I don't remember ever putting that much force on any of the several brands/types presses that I have owned in my reloading history ...

If we were to use 25 lbs as a max input force ...then the numbers in post #46 could be cut in half ...

And the main work would still be in the last few degrees before TDC & I haven't tried to calculate from 157.5 ° to 180° ... yet.

What do you think would be a max input lbs ?
 
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I believe I had to retake college algebra at least once, but press>finger

HB
 
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