What is the difference in velocity of a .22LR from bolt vs semi-auto?

duckman!!

Does anyone know what the difference in muzzle velocity is of the same .22LR round shot from a bolt action or a semi-auto?

It seems like there has to be some loss of energy from a .22 bullet due to the semi-auto cycling the action?

Cass

Erm, no.

Would there even be a difference?

The issue of barrel length comes up all the time, but leaving aside all the crap about legality look at barrel length. A .22lr only uses 10-12 inches of barrel to fully expend the powder charge in the case. The average bullet (subsonic) travels at 1050fps. So in one second it travels 1050 feet. So 10-12 inches would be in micro-seconds.

In that minute, fraction of a second the bullet has left the case, traveled the 10-12 inches (or whatever length the barrel is), before a semi auto action begins to cycle. Meaning the energy taken from the "shot" should not have a dire effect on the round.

That's only guess work. You would need two rifles of equal barrel length, and using the same ammo, both on chronos to find out for sure.

clivej

Cass wrote: »

Erm, no.

Would there even be a difference?

The issue of barrel length comes up all the time, but leaving aside all the crap about legality look at barrel length. A .22lr only uses 10-12 inches of barrel to fully expend the powder charge in the case. The average bullet (subsonic) travels at 1050fps. So in one second it travels 1050 feet. So 10-12 inches would be in micro-seconds.

In that minute, fraction of a second the bullet has left the case, traveled the 10-12 inches (or whatever length the barrel is), before a semi auto action begins to cycle. Meaning the energy taken from the "shot" should not have a dire effect on the round.

That's only guess work. You would need two rifles of equal barrel length, and using the same ammo, both on chronos to find out for sure.

Newtons law 'n all that = for every action there is an equal and opposite reaction.

Google has the answer as

The bolt mass and return spring will keep the case in the chamber until after the bullet has left the muzzle

Cass

clivej wrote: »

Google has the answer as

The bolt mass and return spring will keep the case in the chamber until after the bullet has left the muzzle

There you have it.

Suppose it comes down to this. When the primer is struck, the propellant ignited, the case expanding, the bullet moving, etc. How far does the bullet travel before the action begins to cycle.

I'd guess (and i highlight guess) that the bullet is well on it's way, or as clivej said has left the barrel, by the time this happens. Therefore there should be none to no difference in muzzle velocity.


A chrono would be the only way. If there is a drop in speed it would stand to reason that the energy produced by identical ammo is not being fully used to propel the bullet.

hedzball

loads...

Out of a bolt they fly out... out of a semi you have to tip the crown down to let the bullet roll out

meathstevie

Staying with the Newtonian physics thingy; if action is equal to reaction the reaction should be split up in a bit for the bolt to cycle and the residual for your shoulder in a semi-auto. In a bolt action your shoulder gets all the good stuff.

clivej

If we read that '
Newtons law 'n all that = for every action there is an equal and opposite reaction.'

All would be true BUT for this to happen you would have to have nothing behind the case as it's ignited for it to be equal as you have nothing in front of the bullet at ignition time.

Things are not equal because you now have to factor in the bolt mass and return spring behind the case

Jonty

would it be right to say that the recoil forces of a bolt and semi auto are the same, its only the fact that the bolt travels back and compresses the spring means that the force is felt over a longer period, and therefore can perceived to be less of a kick than with the "fixed" bolt in a bolt action?

clivej

Jonty wrote: »

would it be right to say that the recoil forces of a bolt and semi auto are the same, its only the fact that the bolt travels back and compresses the spring means that the force is felt over a longer period, and therefore can perceived to be less of a kick than with the "fixed" bolt in a bolt action?

Yes and No

The mass of the bolt and the return spring will absorb some of the energy that won't be felt by the shooter. The same with a gas blow back system, part of the expanding gas of the discharge is fed back into the part that operates the reloading system.

Cass

clivej wrote: »

........... the return spring will absorb some of the energy ......

If i remember my secondary school physics a spring does not absorb energy, but stores it until is "springs" back.















I can do science.

clivej

Cass wrote: »

If i remember my secondary school physics a spring does not absorb energy, but stores it until is "springs" back

I can do science.

Correct, give that man a prize

deerhunter1

Cass wrote: »

If i remember my secondary school physics a spring does not absorb energy, but stores it until is "springs"









I can do science.

Thems were the days:D

extremetaz

clivej wrote: »

If we read that '
Newtons law 'n all that = for every action there is an equal and opposite reaction.'

All would be true BUT for this to happen you would have to have nothing behind the case as it's ignited for it to be equal as you have nothing in front of the bullet at ignition time.

Things are not equal because you now have to factor in the bolt mass and return spring behind the case

Sorry Clive but that has nothing to do with it.

The law quoted above is saying that in any exchange of energy the force acting in one direction is matched by a force acting in the other.

This has nothing to do with the nature of what the forces are acting on.

If the bullet is expelled by a force of say 100N (entirely arbitrary number), then the force experienced by the rear of the case is also 100N. The difference then comes down to the response of the mass that this force is acting upon.

The bullet is small and light, and hence it flies off in one direction.

The rifle is large and heavy and so it is pushed back in the other direction.

The recoiling action of a semi would most certainly absorb some of this opposition force from the firing of the bullet, and some more of the force would, as has been suggest earlier, be shifted out of phase with the initial impulse.

As to a difference in velocity between bolt and semi, then I would say that mathematically speaking there would certainly have to be *some* difference, however practically speaking there's every likelyhood that this difference is entirely negligable.

What really matters where the semi's are concerned though is not the magnitude of any such difference, but its consistancy, and that will come down to the nature and maintenance of the chambering mechanism.

Chiparus

would it also depend on the mass of the shooter, perhaps the shock absorbing qualities of the shoulder?

Cass

clivej wrote: »

Correct, give that man a prize

Gonna hold you to that.

deerhunter1 wrote: »

Thems were the days:D

Stop would ya. Was talking about it the other day and only realised it's been 20 years since i finished school. :eek:

FISMA

duckman!! wrote: »

It seems like there has to be some loss of energy from a .22 bullet due to the semi-auto cycling the action?

Did someone say Newton?

Duckman,
I would say that theoretically, the bolt action will have a greater muzzle velocity than the semi. All things considered equal, especially barrel length, the actual difference will not be much.

I think the easiest way to look at the scenario is a work-energy analysis.

In the case of the semi, there is work done on the bolt by the gasses in pushing the bolt back. This energy is unavailable to do work (speed up) on the bullet. Thus, the bullet fired from the semi will have a smaller muzzle velocity than that fired in the bolt action.

It is true that the spring will store much of that energy and cycle the bolt back. However, that energy will still not be able to speed up the bullet - increase the bullet's Kinetic Energy. The work done by friction, in cycling the bolt, is also unavailable to do useful work in speeding up the bullet.

You could make a similar argument using Momentum Conservation.

If you really want some fun, don't forget the rotational motion of the bullet! If the barrel forces the bullet to rotate, say clockwise as viewed by the shooter, then the bullet forces the barrel to rotate counterclockwise.