In my Kimber STII if I go above 8.4 grains the groups start opening up. I would like to load a little higher for deer hunting. Any suggestions? Would magnum primer help? Also, using AA#9, if I go above 13.9 the groups open up.
How much do groups open up. The vitals target on a deer is pretty big, so you may have to live with some trade off. If the groups are opening from 2" to 3", no worries.
My Hodgdon manual says that 8.5gr is the starting charge for Longshot and 180gr jacketed hollow points. Any reason you're loading under the starting charge? My usual carry load is 9.8gr (over book max by .3gr) with the 180gr XTPs.
If he is looking for better accuracy, he may want to try Blue Dot, 10.6gr or 10.8gr might provide him with great accuracy and performance with the 180XTP's seated to 1.255". :)
I've shot 9.2gr of Longshot with 180gr Hornady XTPs, Nosler JHPs and Hornady FMJs in my Kimber STII and Glock 20's with KKM barrels with great results. Accuracy has been excellent. I seated them to 1.255" also.
The groups went from about 1 1/2" to three when I upped the powder charged. They were seated 1.260. I will accurate load any day over a high velocity load. You have to hit what you are shooting at.
K10, what comes to my mind is that the higher impulse ammo is causing your action to unlock early thus the groups are getting spread out. Have you tried other recoil spring weights, Flat Bottom Firing pin Stop or heavier main spring to help your use of the heavier ammo impulse? ???
A friend of mine has the Kimber Stainless Custom Target II with a 22# recoil spring. Not sure if he had other changes, but he was shooting some heavier loadings with his.
I put a Sprinco recoil unit in it. I did not change the factory recoil spring. Someone said they come with a 20LB, and I thought that would be enough with the sprinco unit. I put a Harrison design flat bottom firing pin stop in it. Should I order a new 20lb spring for it?
I would up the mainspring to 25 lbs before I upped the recoil spring.
I shot some Longshot loads through my stock Kimber STII last night. Recoil was sharp, fast and high impulse but accuracy was great. I load to 1.250". Love this pistol.
I was advised, not sure how true or not, that Longshot will not be available until March or April, which is when Hodgdon does that powder run. Fortunately, I was able to pick up a pound of 800x. (Helps to have Bruno's local ;-) )
If I up the weight of the main spring, will it make the trigger pull heavier? It has a 2 1/2" lb pull which is real good.
It will have an impact on the trigger pull. If the sear, disconnecter and hammer bow surfaces have been properly prepped, the impact will be negligible. If there are any imperfections in those surfaces it could introduce some gritty creep or change the pull weight noticeably.
Easy to change back if you don't like it. Be sure to measure with a trigger pull gauge, average of 10 pulls before the change, then again after. Otherwise you may get bitten by confirmation bias. By seeing that the pull weight did not change in a measurable fashion it will help your mind not manufacture the expected higher pull feeling.
I am not that familiar with the 1911 type guns, but changing the mainspring will not add to the dwell time to maintain lock up longer. Correct me if I'm wrong on this modification. However it may add to slow the slide's velocity during rearward travel.
The reason I suggest the stronger recoil spring is the added tension (at slide lock is still small and may only increase 1 pound say between a 20lb an 22lb) to hold the breech closed (added dwell time) just slightly longer would allow the bullet further travel and pressures to drop before the slide moves rearward and the breech unlocks and the barrel to drop down for ejection.
Nope, mainspring change won't affect unlock time in a 1911. Neither will changing the recoil spring.
As long as the bullet is in the barrel, the slide and barrel will remain locked together.
http://forum.m1911.org/archive/index.php/t-16298.html
I see two likely possibilities:
Higher impulse loads are causing more flinch. I know they do with me and I have been shooting handguns for many years. 44 Mag gets me every time. First shot, right down the pickle barrel. Nothing I do gets the second shot on paper. My autonomous nervous system is too full of NOPE. Third-Sixth shots I usually get it back together.
Or
You are walking your load out of a harmonic sweet spot in your particular barrel. Keep going up and you might see the groups tighten back up.
Oh...and if it is flinch, a flat bottom stop and stiffer main spring will soften the felt recoil, making it easier to overcome.
sqlbullet, I understood the swivel link and its use to pull the barrel downward out of slide engagement, I still have to believe the recoil spring's applied pressure pushing the slide forward, has to be overcome, even before the rearward travel of the slide, moves enough to start the link's action to pull the barrel out of the slide engagement. A recoil spring can also be so strong, that the forces of the fired ammo, can not overcome that spring pressure and too little slide travel or even no slide movement or ejection can take place, leaving the spent case inside the chamber or not fully ejected. Not trying to be argumentative just trying to show situational awareness. Alot of the physics involved has to do with the mass of the slide itself as well as the dynamics of the parts in action.
When talking about dwell timing, it is in microseconds, the higher impulse ammo seems to change these dynamics, so the further down the barrel the bullet is the better...as the slide does start to move, so too the recoil is being applied. It as that instant, that the shooter feels the recoil starting, but the duration is throughout the full rearward travel of the slide and as forward travel of the slide carries it back to battery.
I think of competition shooters who use usually faster burning powders, the pressure spikes more rapidly and also drops faster as the energy is dissipating, whereas when we use the slower powders to develop higher gas pressures for longer time periods (again microseconds) to give added push to our projectiles.
Me either (argument wise).
Argument here is whether spring rate influences dwell time, defined as the time the bullet spends traveling down the barrel.
You say yes. Spring rates influence dwell time. I say no, they don't. This is a long standing debate that can be found on lots of forums.
The simple answer to the question is so obvious it is obscure. The time the bullet is in the barrel before it leaves can be derived from it's muzzle velocity.
Does changing a recoil spring change muzzle velocity? No. Obviously not. The bullet spends the same time in the barrel whether you have a ten lb recoil spring or a million pound recoil spring. You might encounter some stoppages with that latter spring though ;D
Changing the spring does impact the slide velocity though. Heavier spring, lower velocity. That million lb spring would keep your action shut for sure. In fact, a 1500 lb spring would keep the slide/barrel still for these loads and probably any safe 10mm load (yep...did the math).
Here is the thing that surprised me (it shouldn't, but it did cause I wasn't thinking). I fact, I kept looking for an error that wasn't there.
Slide/barrel movement during dwell is static for a given bullet weight. That is right. The distance the slide moves during the dwell time is the same regardless of velocity for a given bullet weight.
Yes, more muzzle velocity means more slide velocity. But it also means proportionally less dwell time. The slide moves the same distance, faster. In fact, for a 180 grain bullet in a 5" barrel where the barrel/slide combo weighs 2lbs the distance moved is .064" regardless of velocity (no spring).
Might changing the spring impact accuracy. Almost certainly. In battery, the muzzle is locked against the bushing. The bushing is also in contact with the slide and the recoil spring plug which is acted on by the recoil spring. Changing the spring will therefore have an impact on the harmonics of the barrel, just like one of those harmonic tuners for a rifle barrel.
I think my second assessment is the most likely. You are walking your load up out of a harmonic sweet spot. Keep going up and you will walk back into one.
The way ive seen it explained as far as 1911s go is the small radius/flat firing pin stop just changes how the slide interacts with the hammer, and the recoil energy will always be the same, its just people feel it differently.
Anything else it does to slow the slide down is an added bonus in .45's but in a 10mm it helps keep the slide from rocketing back quite as violently after unlock. A long while ago I read about someone on The High Road forums ( May have been 1911Tuner) that had ran a 1911 WITHOUT a recoil spring at all ( had to line up the bushing and a flgr after every shot) just to prove that the pistol wouldnt destroy itself if the spring wasnt there. Its pretty interesting and wish I had a link!
It was 1911tuner that ran a gun without a recoil spring.
And you are correct. Recoil energy is determined by the weight and speed of the projectile leaving the gun. The energy is exclusively controlled by those values. Everything we do with springs and stops and geometry is to alter how that energy accelerates the slide. Ideally, the slide would loose all momentum .0001" short of full travel, and the recoil spring constant would be just strong enough to strip a round from the magazine and return the slide/barrel to battery.
Generally speaking, about a 10-12 lb spring is all you need to run the action forward. 14-16 lb springs are used for fudge factor in 45 acp duty guns.
It was Ned Christiansen (sp) of Michiguns who did empirical testing of flat bottom firing pin stops. His methodology was this:
He put a Wilson Shok-buff in a full size 1911 running 230 grain/830 fps ball ammo. He fired groups of rounds, I don't remember the counts, inspecting the shok-buff. Once the shok-buff failed, he stopped, noting the round count.
He then repeated the test with a Delta Elite. Round count diminished significantly.
Finally, he installed a flat bottom firing pin stop in the Delta Elite, and repeated. Round count was very similar to the 45 ACP gun. The implication was that the force the shok-buff experienced had been reduced to "near" 45 ACP ball levels.
Makes me wish I owned a really high speed camera and a Ransom rest. Should be pretty easy to just measure the slide velocity changes with and without the flat bottom stop, and from that calculate the difference.
It would indeed be a fascinating project, would also be neat to compare commander goverment and longslide variations aswell.
If someone has a longslide they can provide the weight of I will plug it into my spreadsheet and report the calculated difference.
sqlbullet, I still have to think that the recoil spring does indeed have an influence and that it is a measurable difference.
When I measured the different spring rates of my Glocks it requires more pulling pressure, just to start the slide moving rearward against the different RSA's.
Example:
Glock 20SF spring test Wolff Gun Springs 22lb RSA installed
It takes 6lb 4oz to start the slide to move from the locked position [The 22-Slightly over a pound difference 1lb 1oz to start the slide to move from the locked position over the factory RSA]
Full retraction at 22lb 9oz at lock open
Glock 20SF spring test Wolff Gun Springs 24lb RSA installed
It takes 7lb 6oz to start the slide to move from the locked position [The 24-Slightly over a two pound difference 2lb 3oz to start the slide to move from the locked position over the factory RSA]
Full retraction at 24lb 1oz at lock open
Glock 20Sf Factory spring test
It takes 5lb 3oz to start the slide to move from the locked position
Full retraction at 18lb 0oz at lock open
Not sure if it will make any difference in the 1911's like it does in the Glocks.
Can your mathematical model take that kind of difference into its equation? ???
I guess I wasn't clear.
Spring does make a difference. I was wrong about that.
We were both assuming the slide had moved MORE with the more energetic load. And were arguing about what could counter that.
But we were both wrong about the change in velocity making a difference in dwell distance.
Velocity of the fired round does NOT make a difference in the amount of slide travel when the bullet leaves the barrel. A faster bullet leaves the barrel faster, and the slide therefore moves that distance faster. But the barrel and slide in a given gun will always move the same amount when a 180 grain bullet leaves the barrel.
The slide doesn't move more with the faster load. It moves faster, but is in the same place when the bullet leaves.
Quote from: sqlbullet on September 19 2014 07:49:00 AM MDT
If someone has a longslide they can provide the weight of I will plug it into my spreadsheet and report the calculated difference.
Would a home postal scale reading work? I may be able to get a reading tommorrow, as a friend has a postal scale.
Yeah..Nearest oz or half oz would be ideal.
With barrel? Im assuming the barrel would be part of the reciprocating mass...
Yes, but ideally weigh them separately. My spreadsheet has different line items for them, though they are never used independently in the formula.
Plans changed so I wasnt able to meet up with my buddy with the scale yesterday :( I should be able to get to him next week. How many variables does the program cover, should I just weigh the barrel, then the rest of the slide components together? ( bushing and the assembled slide less barrel)
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.
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.
Bushing and spring plug with slide, then barrel.... got it, will do. I may be able to get it this afternoon.
A spring constant isn't. In a linear spring, the constant
k is measured in newtons / meter.
This is where you lost me:
Quote from: The_Shadow on September 22 2014 09:01:50 AM MDT
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:
Quote from: The_Shadow on September 22 2014 09:01:50 AM MDT
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
slideSince 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.
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?
(http://i1086.photobucket.com/albums/j441/_The_Shadow/HodgdonLongShot.jpg)
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. ;)
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
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.
Quote from: Desolo on September 22 2014 01:07:41 PM MDT
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.
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 (http://kuulapaa.com/home/highspeed/pistols.html)
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
(http://kuulapaa.com/Highspeed3D/FIN_SW_D_single_ORG.jpg)
Quote from: sqlbullet on September 22 2014 01:50:59 PM MDT
Quote from: Desolo on September 22 2014 01:07:41 PM MDT
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?