.22 Mag to 5.7X28 or Not?

[wachtelhund1]

I have a H&R model 700 .22 Mag, semi-auto which I’m considering converting to 5.7X28 mm or a version of it. I re-barreled this gun two years ago with a Green Mountain 22" fluted barrel and thought about converting it then, but decided not to based on the cost of ammo. Now once fired brass is available. I’m also concerned at what the long term .22 Mag ammo cost will be after this .22 LR shortage is over. Having a reloadable cartridge with slightly better .22 Mag ballistics would be nice. I shoot a lot in my back yard.

I wouldn’t consider this with any other .22 Mag semi-auto rifle, as they are all blow back actions. The 5.7x28 is a 50,000 PSI round vs. 25,000 PSI for the .22 Mag.

So why the H&R 700 .22 Mag?
The barrel is threaded into the receiver, 18 tpi; not pinned.
The new barrel is a Green Mountain 22”, 1:14 twist vs. the common 1:16 twist of rimfire barrels.
The H&R 700 has a double rail recoil system with recoil spring attached to the bolt to control the cycling, attached with two 5/8” nuts to the receiver.
An additional recoil spring can be slipped over the top of the original spring for the additional pressure of the 5.7X28 round. See picture.

The bolt is fairly simple and can be reproduced with the firing pin in the center.
This gun has 5 and 10 round magazines.

I don’t think there will be an issue with feeding, as I have worked the magazine to feed the 5.7 round, the bullet will enter the chamber before the case head clears the magazine lips.

Issues: 5.7x28 case is .082” longer than .22 mag, plus most .224 bullets are longer than those found in loaded .22 mag ammo.

COL for the 5.7x28 is 1.594" vs .22 mag @ 1.350”. So loading would be limited to 30, 35, 40 grain bullets.

The 5.7X28 shoulder needs to be set back about .070” to work in the magazine for this rifle. A collar could be used to short chamber the barrel and also make a sizing die.

This would help with COL and bullet seating. Case neck would be .182” plus .070” = .252, if left untrimmed. Loading a Speer .22 Hornet 33 Grain JHP, .438 length into the entire neck would leave .186 exposed for a COL of 1.323” just under the .22 mag COL. This bullet has the same shape as a .22 mag bullet.

Biggest Issue: Cycling. The 5.7X28 is a 50,000 psi pistol round. The .22 Mag is 25,000 psi. Barrel is CM .900" at the breech, receiver is 1.062 at the breech. PSI can be reduced some using slower rifle powder for the 22" barrel. Plus cycling can be controlled by adding additional recoil spring over the top of the existing one. Shoulder length increase will be an issue upon firing. Bottle neck case being fired in blow back action will lenghten.

Pros: Reloadable small round with slightly better .22 Mag ballistics,
.13 to 16 cents per round vs .20 to .25 cents for .22 Mag
Once fired brass now available.

Cons: Cost of conversion
Cost of dies

Thoughts Please!

 

[JD0x0]

IMO, you shouldn't convert it. It is a cool idea on paper but 5.7 ammo seems very limited and more expensive than .22wmr. And as far as ballistics, you don't gain a whole lot, for what you have to give up. If you really want something chambered in 5.7x28 you're probably better off buying something already designed for that round, but I think most people will tell you that it's more practical to trade up to .223rem. The 5.7x28 was mainly designed as an armor piercing SMG round. Since civilians can't use AP ammo the round becomes much less useful. Even more so, because a majority of civilians will not be using it in a full auto platform, which it was designed for. It's only advantages over .223rem is the low recoil and short case so it can be used in a very small weapon with a extremely high rate of fire, to make up for it's anemic firepower. So you're left with the 5.7 having less recoil than the .223 which has very little kick in a semi-auto.

 

[nanewt02]

Personally I think you've done your homework, I would try and find out what the bolt metal is, calculate bolt thrust and all that, I am rebarreling a mossberg 695 and am doing all kinds of stuff to make the cartridge fit. It can be done, is it worth it, that's up to you

 

[wachtelhund1]

True 5.7 ammo is limited, but at the moment there is a pretty good supply of once fired brass. This will only increase in time. Ballistics for the 5.7 is way better than the .22 Mag when compared in rifle barrels. Reloaders are getting 2700+ fps out of a 16" PS90 with 30 grain bullets. I've got a 22" barrel and with the right powder should equal or better that, even with a 7 percent case capacity reduction. The advantage is it is reloadable.

I set the should back .060" and the cases fit in my H&R 700 magazines and my H&K 300 magazines. This would result in a seven percent case capacity reduction. I also seated several bullets in 5.7 cases so the col would fit in the magazine. When the magazine is in the gun and I push the rounds forward out of the magazine the bullet enter the chamber before they leave the magazine lips.

A machineist, who I shoot F class with currently has the bolt and is doing a CAD drawing. I will see him tomorrow and discuss this further. Since he is making a new bolt, I'll retain the old one. If it does not work out, I can always set the barrel back and rechamber back to the .22 Mag.

 

[Scorch]

I wouldn't. I had a similar project a few years ago, I wanted to rebarrel a Rem 580 to 218 Bee. When I did the calculations, it was obvious I should not do it, bolt thrust was 5 times as high with the 218 Bee, and since the action was not heat treated its strength was unknown. I ended up building a 17 HMR single shot instead. I think I made the right choice, I like my eyes and fingers right where they are. I'll build a Low Wall in 218 Bee instead.

 

[natman]

I wouldn’t consider this with any other .22 Mag semi-auto rifle, as they are all blow back actions.

It's been a while since I owned an H&R 700, but IIRC it's a blowback also. All the pieces under the stock merely add another spring and some more reciprocating weight.

Converting it to a cartridge with twice as much pressure strikes me as being a spectacularly bad idea.

 

[Clark]

I found this BRUNO Lux 222 Remington for $125 at a gun show in 2003.
It looks to me like a factory manufactured rimfire design converted for mass production to centerfire .
Someone knew more about it than me:
http://www.thehighroad.org/showthread.php?t=35884&highlight=lux

It was such a bad idea, I had to buy it.
a) pinned barrel
b) tip off rings grooves
c) the only lock up is the bolt handle.

What does it all mean?
An H&R model 700 .22 Mag conversion to 5.7X28 mm seems like a good idea in comparison.

 

[LED]

I shot and compared 22mag and 5.7 The latter will give you a little more penetration and longer usable range, and a spectacular muzzle flash indoors. I don't think this proposition is worth your time and skill. The FiveSeven really stands out in a duty weapon, not in a sporting firearm.

 

[natman]

I did a bit of research and came up the following:

The 700 is indeed a blowback design.

The additional spring and carrier under the receiver is an add-on to allow what was originally a blowback 22LR to handle the longer pressure curve of the 22 magnum.

Now is it likely that this gun can be modified to successfully handle a bottlenecked cartridge at twice the pressure?

In a word, no. For starters you would need a LOT more reciprocating weight to have enough inertia for the new cartridge. I doubt if you could get enough even if you machined the entire bolt out of depleted uranium. Then you'd need a much stronger spring, like something off of a truck suspension.

As experience with the 17 HMR proves, it's much harder to design a blowback action for a bottle necked cartridge because it has a much lower expansion ratio, that is, each inch of barrel represents only a small percentage of the volume of the case. This keeps the pressure high until the bullet finally leaves the barrel. The combination of 50,000 psi pressures and the long pressure curve would make it very difficult to keep the action closed until then and if it opens too soon the case gets extracted while the pressure is still high and KABOOM, you get a faceful of hot brass shrapnel and flaming gas at ~40,000 psi.

In short, it is extremely unlikely that there is any way to safely perform this conversion. There's a reason Mini-14s use a locking bolt. Life is too short to make it shorter by fooling around with guns that explode.

 

[Clark]

I found this inside Google groups from 13 years ago.
If you can follow my process, you can fill in your own independent variables for your own project.

Clark Magnuson 5/9/01
I calculated this for someone on rec.guns last year who wanted a .223 blow back.
Clark
Subject: Re: blowback designs?

assume: Peak chamber pressure = 50kp/i/i
assume: average chamber pressure = 25kp/i/i
assume: Peak bullet velocity = 2500 f/s
assume: Barrel length = 16i = 1.33f
assume: brass case inside diameter = .35 i
calculate force from chamber = PA = [25kp/i][.35i/4][.35i/4][3.14]= 600 pounds
calculate time of chamber force = 2 1.33f/[2500f/s]=.001 s
assume: action 2.5 i long = .208 feet
assume: spring force = 20 pounds
calculate spring energy =fd=20 .208 =4.17 footpounds
calculate distance chamber pushes bolt = E/F =4.17fp/600 p = .0069 feet
This means the bolt will be accelerating back for .001 seconds until it has gone .0069 feet back and then it will be slowed down by the recoil spring for 2.42 inches where it just runs out of speed as it reaches the back of the action.
calculate bolt peak velocity = 2D/t = 2 .0069 f/.001s = 6.9 f/s
calculate mass of bolt = 2 E/VV = 2 4.17fp/ 6.9 f/s / 6.9 f/s = .175 pss/f
calculate weight of bolt = GM= [32.2 f/s/s] [.175pss/f] = 5.6 pounds

If you increase the recoil spring force, the bolt wieght requirement will go down.
Clark

 

[James K]

I say it won't work, and will be dangerous as all heck besides. I have what I think are good reasons, but I know expressing them will only result in an argument. So, I say, "Try it". Take videos and prove me wrong.

(But tie the gun to a tire, please.)

Jim

 

[wachtelhund1]

I found this inside Google groups from 13 years ago.
If you can follow my process, you can fill in your own independent variables for your own project.

Clark Magnuson 5/9/01
I calculated this for someone on rec.guns last year who wanted a .223 blow back.
Clark
Subject: Re: blowback designs?

assume: Peak chamber pressure = 50kp/i/i
assume: average chamber pressure = 25kp/i/i
assume: Peak bullet velocity = 2500 f/s
assume: Barrel length = 16i = 1.33f
assume: brass case inside diameter = .35i
calculate force from chamber = PA = [25kp/i][.35i/4][.35i/4][3.14]= 600 pounds
calculate time of chamber force = 2 1.33f/[2500f/s]=.001 s
assume: action 2.5 i long = .208 feet
assume: spring force = 20 pounds
calculate spring energy =fd=20 .208 =4.17 footpounds
calculate distance chamber pushes bolt = E/F =4.17fp/600 p = .0069 feet
This means the bolt will be accelerating back for .001 seconds until it has gone .0069 feet back and then it will be slowed down by the recoil spring for 2.42 inches where it just runs out of speed as it reaches the back of the action.
calculate bolt peak velocity = 2D/t = 2 .0069 f/.001s = 6.9 f/s
calculate mass of bolt = 2 E/VV = 2 4.17fp/ 6.9 f/s / 6.9 f/s = .175 pss/f
calculate weight of bolt = GM= [32.2 f/s/s] [.175pss/f] = 5.6 pounds

If you increase the recoil spring force, the bolt wieght requirement will go down.
Clark

Clark, Thanks for the info! This is exactly what I need to further evaluate this project. I'll recalulate the formula. The barrel is 22". Inside brass case is .297" The bolt moves moves 1.8". For the recoil spring, with a slight modification I can cut down the recoil springs from a Winchester Model 100, it uses two springs one inside the outer. I had one of mine apart today doing some measurements, they are larger and stiffer. This is a crude measurement, but it took 12 pounds to pull back the H&R 700 bolt. It takes 17.8 pounds to pull back a Winchester 100 bolt. Weight will also be added to the hammer and hammer spring can be increased. This system is a copy of the resiling submachine gun and is anything but conventional. I've added a schmatic of the bolt and hammer system. The hammer ( part number 29) is a round floating cylinder that is pushed forward upon trigger release, striking the firing pin and bolt. Upon firing the bolt pushs the hammer back 1.8" until it is caught by the trigger. Right now the hammer is mostly hollow and has a hole in the center making it lighter for the rimfire, a new solid one will be made so it will strike the firing pin. The bolt and hammer are round, 0.750". The bolt is 3.235" long and weighs 5.3 onces. The hammer is 1.550" long and weighs 1.7 onces.
A new hammer could approach 3 onces.

Your formula is interesting:

5.7 calulations:
assume: Peak chamber pressure = 50kp/i/i
assume: average chamber pressure = 25kp/i/i
assume: Peak bullet velocity = 2500 f/s
assume: Barrel length = 22i = 1.83f
assume: brass case inside diameter = .297 i
calculate force from chamber = PA = [25kp/i][.297i/4][.297i/4][3.14]= 430 pounds
calculate time of chamber force = 2 1.83f/[2500f/s]=.001 s
assume: action 2.5i long = .208 feet
assume: spring force = 18 pounds
calculate spring energy =fd=18 .208 = 3.74 footpounds
calculate distance chamber pushes bolt = E/F = 3.74 fp/430 p = .009 feet
This means the bolt will be accelerating back for .001 seconds until it has gone .009 feet back and then it will be slowed down by the recoil spring for 1.491 inches where it just runs out of speed as it reaches the back of the action.
calculate bolt peak velocity = 2D/t = 2 .009 f/.001s = 18 f/s
calculate mass of bolt = 2 E/VV = 2 3.74 fp/ 18 f/s / 18 f/s = .023 pss/f
calculate weight of bolt = GM= [32.2 f/s/s] [.023pss/f] = 0.676 pounds

Unless my math is wrong, I would need a 11 once bolt. Currently the bolt plus hammer is 7 onces and could go to 8.3 onces. Then there is the hammer spring resistance.

.22 Mag calulations:

assume: Peak chamber pressure = 25kp/i/i
assume: average chamber pressure = 12.5kp/i/i
assume: Peak bullet velocity = 1800 f/s
assume: Barrel length = 22i = 1.83f
assume: brass case inside diameter = .224 i
calculate force from chamber = PA = [12.5kp/i][.224i/4][.224i/4][3.14]= 123 pounds
calculate time of chamber force = 2 1.83f/[1800f/s]=.002 s
assume: action 2.5i long = .208 feet
assume: spring force = 12 pounds
calculate spring energy =fd=12 .208 = 2.5 footpounds
calculate distance chamber pushes bolt = E/F = 2.5 fp/123 p = .020 feet

calculate bolt peak velocity = 2D/t = 2 .020 f/.002s = 20 f/s
calculate mass of bolt = 2 E/VV = 2 2.5 fp/ 20 f/s / 20 f/s = .013 pss/f
calculate weight of bolt = GM= [32.2 f/s/s] [.013 pss/f] = 0.4025 pounds

This would require a 6.44 once bolt. The H&R 700 bolt weighs 5.3 onces. The hammer weighs 1.7 onces. Seven onces total, appears H&R included the hammer in the bolt weight.

 

[Scorch]

One reason high pressure cartridges are not often put into blowback actions is simple: you don't want them opening under pressure. One of the main purposes of locking mechanisms is to delay the opening until pressure has dropped in the chamber. According to your own calculations, your action will propel the bolt backwards at approximately 20 fps, With a case 1" long, this translates into .004 seconds, the same amount of time it will take the bullet to travel about 5". I would do more calculating.

 

[wachtelhund1]

Scorch, There are two or three pistols currently made for the 5.7X28 round, all are blow back guns. The 20 f/s was computed for the .22 Mag based on the 12 pounds to cock the bolt.

 

[Clark]

I did that calculation 13 years ago and have noticed it is does not match the experiments.

Some improvements to the calculation would be:
1) It takes less recoil spring if the gun frame has recoil movement.
2) It takes less recoil spring if there is an effort to cock the hammer.
3) It takes less recoil spring if there is action friction.

I have since fired a 45 Colt Commander from a 49 pound vise, because it will move less than my hand. I measured the wimpiest load that would cycle and compared that with the wimpiest round that would cycled while hand held. The difference is ~ 16% less forward momentum of gas and projectile needed.

I have measured the force with a force gauge.

Still, the amount of spring * mass product needed is half what I would predict.

 

[wachtelhund1]

Clark, I too have refined some calulations. Please explain the action lenght in your calulations. I assume you used 2.5" for a pistol action. With the H&R 700 wouldn't it be bolt + hammer + distance traveled to cock the hammer? Or 3.235 + 1.55 + 1.83 = 6.615". This is a long action. Also, if I were to do this, the shoulder on the 5.7 round would have to be set back .065", this would equate to a seven percent reduction in case cappacity and velocity. So using a velocity of 2300 fps. Does this make sense?

5.7 Short calulations:

assume: Peak chamber pressure = 50kp/i/i
assume: average chamber pressure = 25kp/i/i
assume: Peak bullet velocity = 2300 f/s
assume: Barrel length = 22i = 1.83f
assume: brass case inside diameter = .297 i
calculate force from chamber = PA = [25kp/i][.297i/4][.297i/4][3.14]= 430 pounds
calculate time of chamber force = 2 1.83f/[2300f/s]=.0016 s
assume: action 6.615i long = .551 feet
assume: spring force = 18 pounds
calculate spring energy =fd=18 .551 = 9.918 footpounds
calculate distance chamber pushes bolt = E/F = 9.918 fp/430 p = .023 feet
This means the bolt will be accelerating back for .001 seconds until it has gone .023 feet back and then it will be slowed down by the recoil spring for 6.592 inches where it just runs out of speed as it reaches the back of the action.
calculate bolt peak velocity = 2D/t = 2 .023 f/.001s = 46 f/s
calculate mass of bolt = 2 E/VV = 2 9.9 fp/ 46 f/s / 46 f/s = .009 pss/f
calculate weight of bolt = GM= [32.2 f/s/s] [.009 pss/f] = 0.2898 pounds, or 4.637 onces.

 

[Clark]

OK, I edited my post to put in a pic with the terms better defined.

Basically, if the energy stored in a blowback is force times distance.

We would like that converted to foot pounds, even if we start with inches.

The trouble is, not all the momentum of the bullet and gas goes into energy stored in the recoil spring.

Losses along the way getting energy to the recoil spring:
1) The case grips the chamber wall and makes friction during extraction.
2) The whole gun recoils with some velocity. That must be subtracted to find the bolt velocity relative to the frame velocity that compresses the spring.
3) The bolt has friction with respect to the frame.
4) The hammer has a spring that needs cocking and probably as much hammer friction as that again.


I can account, now, for 2), 3), and 4) in the Colt Commander experiment.
But I am still off by ~~ 50%.

..when you have excluded the impossible, whatever remains, however improbable, must be the truth. - Sherlock Holmes

It must be extraction friction, which is so dynamic, I do not know if I can measure it.

What does it all mean?
When you design the gun, you will have bolt mass and recoil spring force both sized per a calculation.
Then when trying to get the gun to work, you will have to reduce one or both of them.

 

[natman]

I say it won't work, and will be dangerous as all heck besides. I have what I think are good reasons, but I know expressing them will only result in an argument. So, I say, "Try it". Take videos and prove me wrong.

(But tie the gun to a tire, please.)

+1

It's like watching a train wreck in slow motion. You know it's going to be ugly, but you can't turn away.

The OP was right in the first post:

I wouldn’t consider this with any other .22 Mag semi-auto rifle, as they are all blow back actions.

Yet when it comes to light that the 700 is a blowback, no problem.

If you insist on this folly, PLEASE tie the gun to a tire, make a video remotely and have the honesty to post it.

Best of luck.

 

[natman]

There are two or three pistols currently made for the 5.7X28 round, all are blow back guns.

1) The FN 5.7 is a delayed blowback. Yes, the slide is powered by blowback rather than a gas piston or other mechanism. The CRUCIAL difference is there is a mechanical means of delaying opening until the bullet is gone. The 700 is a straight blowback. The only thing holding the action closed is the inertia of the bolt. The spring only serves to slow the bolt down once it's moving.

We have mentioned above that with light cartridges giving low
pressure it is quite possihle to use what is known as the straight blowback
breech mechanism, which depends entirely on the weight of
the breech block to keep it from opening too quickly.

Hatcher's notebook, page 38 (emphasis added)

2) A pistol slide is a lot larger and heavier than the bolt in a 700.

3) A pistol barrel is a LOT shorter than a rifle barrel, in this case about 17 inches shorter. This means that there is less barrel time to deal with.

I suggest you read this thread on Rimfire Central to get an idea of how difficult changing the cartridge can be on a blowback action. Keep in mind the conversion discussed is a 22LR to 17HM2, a much smaller difference than between a 22 Magnum and a bottlenecked cartridge at twice the pressure.

http://www.rimfirecentral.com/forums/showthread.php?t=68073

The way to deal with these things is to look at the evidence impartially and then decide if what you want to do is feasible, not decide what you want to do then cherry pick the evidence to support it.

"Experience is a dear teacher, but fools will learn at no other." - Benjamin Franklin

 

[LED]

What about the Excel? Saw a pistol and a rifle for sale, all black polymer and stainless steel, like $500 or so.