Using Longshot With 180XTP's?

[sqlbullet]

When I weighed my Para, I weighed the recoil spring plug and bushing with the slide.  Barrel is separate.

The spreadsheet calculates the velocity and displacement of the unsprung moving mass at the moment the bullet leaves the barrel.  I need to upgrade it to include information about the spring, but without spring constant values that will not be accurate.  Haven't had a chance to look at Wolff Springs site to see if he publishes spring constants.

[The_Shadow]

I don't think that there will be a constant for the recoil spring, it applies (X) amount of force at full battery and reaches it actual full weight forces at the point the slide hits the lock open spot, from what I have measured. 

The most important factor to the mass equation, would be the amount of pressure the spring applies (X) amount at full battery, where that value would add to the forces required to overcome the total mass, before any slide motion starts.  It is also at that point, where the cartridge has already peaked in its pressure in the curve with bullet movement and residuals are acting on the recoiling action.

If I recall the formula, work = force x distance, is where dwell time plays its role to hold the slide in full battery, before any travel distance.  There are forces are being applied to the spring rate and the mass as a whole, but within a framework of time, these forces are changing as the bullet moves and pressures are dropping.  Once the slide motion starts we would consider it out of battery, but the barrel remains locked to a particular amount of travel which can vary from one firearm to the next.

If the recoil spring's initial rate of force, to hold the side in battery is different, then wouldn't that also change the dynamics over time. 

This formula would reflect the slide in recoil motion
KE = (1/2)mv2, or kinetic energy = (1/2) times mass times the speed squared.

There may be certain frictional forces, during movement that would affect the slide's velocity.

[Desolo]

Bushing and spring plug with slide, then barrel.... got it,  will do. I may be able to get it this afternoon.

[sqlbullet]

A spring constant isn't.  In a linear spring, the constant k is measured in newtons / meter.

This is where you lost me:

where dwell time plays its role to hold the slide in full battery, before any travel distance.

This seems to imply that there is a period of time that is significant where the slide/barrel are not moving.  This is not true in my mind.  The acceleration is in the neighborhood of 1.5 million feet per second per second.

Also, taken in it's context:

If I recall the formula, work = force x distance, is where dwell time plays its role to hold the slide in full battery, before any travel distance.

W = force X displacement, not distance.  In either case, you are citing this formula as having relevance, but then state before any travel distance.  If no displacement has occured, then no work is done.

What is relevent here is:

F = M X A.

Functionally, as soon as the bullet starts moving forward, the slide starts moving to the rear, with the forces balanced:

M_bullet X A_bullet = M_slide X A_slide

Since the mass of the bullet is known (180 grains, .024670504 slugs) and the acceleration of the bullet is known by using the motion equation and solving for acceleration (1,642,680 feet per second^2), then all that is left to consider is the mass of the slide/barrel (reciprocating assembly) and the spring constant (k) over the distance traveled (F = -kX).

Incidentally, I don't think the spring value is going to have a significant impact.  The force of the bullet is 1,312.87 lbs.  The rated force of the springs is at most 28 lbs.

But I wanna do the math.

[The_Shadow]

How do we plug in these values?

Glock 20Sf Factory spring test
It takes 5lb 3oz to start the slide to move from the locked position

Glock 20SF spring test Wolff Gun Springs 22lb RSA installed
It takes 6lb 4oz to start the slide to move from the locked position (increased difference of 1lb 1oz to start the slide to move)

Glock 20SF spring test Wolff Gun Springs 24lb RSA installed
It takes 7lb 6oz to start the slide to move from the locked position (increased difference of 2lb 3oz to start the slide to move)

Here the spring is partially compressed, adding a known value of resistance to the forces, before the slide moves.

The formula you are working with, covers the movement of the slide and I understand it as the forces acting on the moving mass during the ejection cycle.  The recoil spring builds more resistance as it is further compressed to counteract those forces and controlling the mass velocity.

I only wish I had access to a high speed camera and a pressure rig, to measure the amount of pressure trace rise & drop or difference at the point, where the slide starts to move.  This is what I refer to as dwell time, as the bullet's movement and the opposite reaction or turned into the kinetic energy of recoil to cycle the slide.

When we observe a pressure curve like that below, where does the slide actually start to move? 


PMax is peak pressure.
If say the slide starts to move before PMax, then the casing will start to also move at the moment at which unlock occurs, increasing unsupported area.

The longer after PMax before the slide moves, the longer the till the unlock time and pressures are dropping more, before the casing starts to move.  This is where I look to control and eliminate "SMILES" and gain maximum performance while increasing safety.  I could be wrong in my assumptions!

As I observe some brass that have been badly bulged and "SMILED", it actually appears that the brass was being pulled backward before the pressures had dropped enough, this indicates to me the slide moved from battery too fast while pressures were still quite high.  How is that controlled, if all else remains the same.   More spring resistance at full battery or even more mass?  Reduced loads?   Both will change the out come to a point?

In this illustration it shows;
At PMax the bullet has moved estimated less than a 0.25"
100% powder burnt at 3.5" bullet travel, total was barrel is 5.0"(in a semi auto like the 10mm is more like 4.0" because the measure the entire chamber as well.  Total time bullet in barrel 0.435 ms

In rifles I have seen slower burning powders hold pressures too high, and the casing fully expanded to the chamber walls as ejection occurred and the case head was ripped off.  This is a timing event based on pressures being too high at the ejection cycle event.

I know I'm being a Pain!  But just thinking out the process and sharing these thoughts.

[Desolo]

This is as exact as i can get: 1# 1oz for the slide, bushing, and spring plug, barrel is about 5oz? Altogether it reads about 1# 5-6 oz? I hope this helps, im curious to see how this compares!

Unloaded the whole pistol is 2.5 lbs, loaded its 3lbs 2 oz

[sqlbullet]

Rifle systems aren't comparable since they are gas-operated and not recoil-operated.

In the example you give, P-Max occurs after bullet travel of about .15" as best I can estimate looking at the graph provided.  It would be great is quick load would time index this for us.  Maybe it does in another graph or table.  If so, I would love to have that.  I could use it to correlate my calculated time index against actual, among other things.

This translates to an average acceleration of 6,400,000.00 fps/s for a time period of 0.000062500 seconds, and a for a force on the slide of 5,683.37 lbs. Total travel distance of the slide/barrel during  this time is .003874" in my Para P16-40.


Even using "raw" advertised F values for the springs of 17, 22,24 lbs, the differences in slide movement are clear out there at the ten-thousandths of an inch

17lb = 0.12791740113265"
22lb = 0.12755887196960" (Diff = 0.000358529")
24lb = 0.12741546030438" (Diff = 0.000143412")

Of course, your measurements confirm what I suspected, which is the advertised spring values are near max compression.  Using your observed values for springs at "initial travel":

17lb = 0.12876424701576"
22lb = 0.12868823883319" (Diff = 0.000076008")
24lb = 0.12860721124235" (Diff = 0.000157035")


In both cases, a change of a few ten-thousandths of an inch on slide/barrel travel is unlikely to be relevant.

[sqlbullet]

This is as exact as i can get: 1# 1oz for the slide, bushing, and spring plug, barrel is about 5oz? Altogether it reads about 1# 5-6 oz? I hope this helps, im curious to see how this compares!

Unloaded the whole pistol is 2.5 lbs, loaded its 3lbs 2 oz

1911 barrels weigh in pretty close to 1oz/inch.  My Para barrel is 5 oz, so I put your longslide barrel in at 6oz.

So.

5" 10mm 1911 5 oz barrel, 12.7 oz slide/bushing/plug with a 180 grain load at 1220 fps has slide movement of .11484" during .00068 seconds of time

6" 10mm 1911 6 oz barrel,  17 oz slide/bushing/plug with a 180 grain loat at 1245 fps (+25 for extra inch of barrel) has a slide movement of .10542" during .00080 seconds of time.

[The_Shadow]

In some of these ultra slow motion playbacks you can see the slide is in motion before the bullet leaves the barrel.

And for your viewing pleasure and study
http://kuulapaa.com/home/highspeed/pistols.html

[The_Shadow]

In this picture of a Smith & Wesson SW 645 .45ACP single stack pistol model, you can see several things here...
The bullet has clearly left the barrel
The slide has retracted probably a 1/4" or so
And the barrel and slide appear to be still locked together or at the point of unlocking

[Desolo]

This is as exact as i can get: 1# 1oz for the slide, bushing, and spring plug, barrel is about 5oz? Altogether it reads about 1# 5-6 oz? I hope this helps, im curious to see how this compares!

Unloaded the whole pistol is 2.5 lbs, loaded its 3lbs 2 oz

1911 barrels weigh in pretty close to 1oz/inch.  My Para barrel is 5 oz, so I put your longslide barrel in at 6oz.

So.

5" 10mm 1911 5 oz barrel, 12.7 oz slide/bushing/plug with a 180 grain load at 1220 fps has slide movement of .11484" during .00068 seconds of time

6" 10mm 1911 6 oz barrel,  17 oz slide/bushing/plug with a 180 grain loat at 1245 fps (+25 for extra inch of barrel) has a slide movement of .10542" during .00080 seconds of time.

Now how do those numbers compare to a .45 cal 1911? If I understand this, the longslide took longer to  move a shorter distance, meaning slide velocity is slower?