From: John Bercovitz <JHBercovitz@lbl.gov>

Ques.

Can someone clear up the difference between locked breech and blowback?

Ans.

Treatise alert!  Relinquish all hope:

There are so many questions in this thread that I would 
like to try to answer by narrowing the types of guns being 
considered to two, what one might call the "classical" locked 
breech pistol and the "classical" blowback pistol.  

In the "classical" locked breech pistol, the barrel is 
initially locked to the slide, but the slide is not locked 
to the gripframe (receiver).  Variations abound; some have 
tilting barrels with locking lugs on top of the barrel such 
as the 1911, High Power and Neuhausen, others have a separate 
locking block to connect barrel and slide such as the P-38, 
Lahti and Beretta, while still others such as the Luger (toggle 
action) use less common mechanisms.  All of these meet the 
requirements of the first sentence of this paragraph.

When the locked breech pistol is fired, the barrel and slide 
recoil backward, locked together, for a short distance.  
After this short distance has been traveled, the barrel/slide 
unit's position relative to the gripframe is by some means 
detected, the barrel unlocks from the slide, and then the 
barrel stops moving relative to the gripframe while the slide 
continues backward, extracting the spent cartridge's case 
and finally stopping against an abutment in the gripframe.  
(Whew!)  It is to be hoped that by the time the breech is
unlocking, the pressure in the barrel has fallen to a very low 
level.  Most usually, the bullet has left the barrel by the 
time the slide and barrel unlock from each other.  From a 
safety standpoint, the most important thing is that the pressure 
is low enough that the case wall won't rupture as the thin 
forward part of the case is pulled free of the chamber and 
hence becomes unsupported.  It is also important that the
pressure is low during the time the gun is unlocking because
as the gun unlocks, the amount of the locking surface which is 
actually engaged decreases to zero and so the force per unit area
of that remaining surface increases dramatically.  This is for 
most locked breech guns.  A toggle lock doesn't have this 
particular problem.

The recoil of the barrel/slide unit relative to the gripframe 
is essentially unimpeded except for the resisting force of 
the recoil spring.  The resisting force of the recoil spring 
on the barrel/slide unit is insignificant compared to the 
rearward thrust of the cartridge case on the barrel/slide 
unit.  A heavy recoil spring does decrease the cycle time of 
the action.  It also helps a given gun design handle a hotter 
cartridge through a secondary effect: When the barrel/slide 
moves rearward, it accelerates the frame of the gun and the 
hand of the shooter rearward by communicating through the 
recoil spring.  If the recoil spring is very stiff, the grip 
frame will move rearward a significant amount during the time 
the bullet is in the barrel.  The motion also compresses the 
web of the shooter's hand.  The effect is to make the 
barrel/slide unit have to travel farther (relative to the 
original inertial reference frame) before it reaches the 
point relative to the grip frame where the barrel and slide 
unlock from each other.  Note that the gun does not rotate 
appreciably before the bullet has left the barrel, witness 
the fact that the tops of the front and rear sights are very 
nearly the same distance from the centerline of the bore 
(unlike in the case of a revolver or a singleshot pistol).

Another way to allow a given design of locked breech pistol 
to handle a hotter cartridge is by increasing the mass of 
the barrel/slide unit.  This is because the bullet and the
barrel/slide unit have a momentum/impulse/momentum 
relationship with each other.  As a result, to the first 
order, the barrel/slide unit moves backward as far as the 
bullet moves forward times the ratio of the barrel/slide's 
mass to the bullet's mass.  By increasing the mass of the 
barrel/slide unit sufficiently, you can ensure that the 
bullet has left the barrel before the cartridge case wall 
becomes unsupported.

These are ways to increase how hot a cartridge can be 
tolerated by a given design; but be advised that one doesn't 
do these things and ignore other aspects: the barrel must be 
able to handle increased pressure and the locking lugs must 
be able to handle the increased force, for examples.

It's not necessary to ensure that the bullet has left the 
barrel by the time the thin case wall becomes unsupported.  
The pressure falls rapidly as the bullet progresses down the 
barrel.  In just a couple of inches of bullet travel, the 
pressure can fall to a few thousand psi if the cartridge is 
of small capacity and is loaded moderately.  The thin case 
wall can support on the order of two to ten thousand psi 
according to its diameter, wall thickness, and material 
strength.  

This brings us to the "classical" blowback pistol.  The 
blowback has the barrel fixed to the gripframe (receiver) and 
the slide is free to move.  The first thing the blowback 
depends on is that same momentum/impulse/momentum 
relationship which the locked breech pistol depends on.  The 
difference is that in this case the slide (rather than the 
slide _and_ barrel) is moving backward in response to the 
bullet's forward travel.  The consequence is that the 
cartridge case is coming out of the chamber while the bullet 
is proceeding down the barrel.  The blowback pistol also has 
the recoil spring's effect on the gripframe working for it.  
And lastly, the blowback action has the friction of the 
cartridge case on the chamber walls to slow the rearward 
motion of the case by decreasing the thrust on the slide.  
Surprisingly, this last can very significant.  If the case is 
long, there is a lot of case surface area and hence a lot of 
normal force pushing the case walls against the chamber's 
interior.  If the case is strong enough that its head won't 
tear off, a case can be fired in a gun which has _no_ 
breechblock*.  (Provided the primer is well enough supported 
that it doesn't pop out.) I think the most common example of 
this is the 30-30 Winchester.  I sure wouldn't recommend the 
practice of shooting a 30-30 without a breechblock, but 
under perfect conditions it can be done.  Pistol cases don't 
have this much surface area so this example is meant by way 
of illustrating the tendency.  When the pressure in the case 
falls low enough, the case shrinks in from the chamber walls 
and no longer drags on them.  *(Breechblock = the part which 
closes off the back end of the barrel, making sure the case 
doesn't get pushed out of the chamber.)

Since the slide moves in accordance with the ratio of its 
mass to the bullet's mass, as the slide of a blowback pistol 
is made lighter, the cartridge case has to be made stronger 
to handle the higher pressure it will see in the unsupported 
condition.  In the extreme case, the slide would have no 
significant amount of mass.  Then the case would completely 
exit the chamber while still at high pressure making quite a 
pop.  This configuration is called a zip gun.

The next thing to consider is the recoil which the shooter 
perceives.  First we ought to consider the difference between 
a single shot pistol and an auto.  In a singleshot, the 
momentum of the gun is equal to the momentum of the bullet.  
This is because of the law of conservation of momentum.  The 
total system starts out with zero momentum and must preserve 
it.  When we insert energy into the system, the  momentum the
bullet acquires must be equal to and opposite the momentum 
the gun acquires.  With this knowledge, you can calculate the 
rearward velocity of the gun and the kinetic energy it carries.  
OK, time for just one very simple but important derivation:

Let: 
      M = mass of gun
      V = velocity of gun
      E = kinetic energy of gun

      m = mass of bullet
      v = velocity of bullet
      e = kinetic energy of bullet

V = (m/M)*v            (1)
E = (1/2)*M*V^2        (2)

Substituting (1) into (2):

E = (1/2)*M*[(m/M)*v]^2 

Cancelling like terms and substituting in the definition 
of e (not shown) we get:

E = (m/M)*e

(Note: This sort of momentum problem is different from a 
colliding masses momentum problem so please don't try to 
use this equation in that application.)

We see the recoil energy of the gun goes up with bullet mass, 
down with gun mass and up with bullet energy.  The significant 
thing is the lighter the gun, the more the recoil energy.  
Digressing for a moment, there is no worldwide agreement 
on the measure of recoil.  In the US, recoil generally means 
recoil energy.  In other places, it is often recoil momentum.  
I'll stick with recoil energy for the purposes of this 
discussion.  Digressing further, most of the time what we 
calculate is technically the "free" recoil.  This is the 
recoil the gun would have if it were suspended from strings 
rather than being coupled to the mass of your hand in a hard-
to-characterize manner.

So what does this equation imply?  What it means is that an 
autopistol will have more recoil than a singleshot pistol.  
(When I say autopistol, I mean autoloading rather than fully 
automatic; fully automatic has _lots_ of recoil.)  The reason 
that the autopistol has more recoil is that it's the lighter 
mass of the slide or slide plus barrel which has momentum 
equal to that of the bullet, and not the heavier mass of an 
entire gun.  If the slide has half the weight of the total gun, 
then the recoil energy is doubled per the above equation, 
E = (m/M)*e.  If the slide weighs the same as the frame (half 
the total gun weight) and collides elastically with the abutment 
in the frame, then the slide transfers exactly its momentum 
(and recoil energy) to the frame when it hits the abutment.  
In other words, the slide stops dead and the frame continues 
at the slide's velocity much as billiard balls interact under 
certain conditions.  So the frame ends up getting a momentum 
equal to the momentum of the bullet.  Since the frame is light, 
that momentum makes for a lot of energy.  That's the difference 
between a singleshot (or revolver) and an autopistol when it comes 
to recoil.  Another difference is that in the autopistol, the recoil 
which you feel is separated into components, that is spread out 
over time.  The singleshot or revolver has a lesser but sharper 
recoil since its recoil comes all at once.  If you're attuned to 
it, you can really feel the difference between the two types of 
guns.  OK, I've ignored the fine points in this discussion, but 
let's keep this to treatise size; you haven't got time for a tome.
(Thought experiments: How should the feel of the recoil of an
autopistol change as the recoil spring is made progressively 
softer?  What effect will the revolver's higher boreline have?)

With the above in mind, what is the difference between a 
locked breech and a blowback?  Not a heck of lot in any major 
sense.  Most of it is in the fine points.  In tiny blowbacks, 
everything is too light and the recoil is vicious.  (And 
don't forget that noise is a lot of perceived recoil.  Hot 
popguns can be surprisingly noisy since pressures are high 
at bullet exit.)  In the big blowbacks like the Astra, I think 
what happens is that the strong recoil spring compresses or 
takes all the give out of your hand before the slide hits the 
stop.  When the stop is hit, the gripframe bangs into your bones.  
Another thing which precompresses your hand is the drag of the 
case in the chamber.  That drag travels through the gripframe to 
your hand.