Blowback formula needed

[Tod Glenn]

Greetings all,

I am hoping that someone can help me.

I am looking for a formula that will let me calculate the spring weight
required for a simple blowbacl operated gun where the cartridge
velocity, bolt weight and bolt movement are known. I am told that this
information can be found in volume 4 of Chinn's machinegun series.
Thanks,

Tod

--
----
Tod Glenn
mailto:webm...@cordite.com
http://www.cordite.com

------------------------------------------------------------------------
You can learn about rec.guns at http://doubletap.cs.umd.edu/rec.guns

[Clark Magnuson]

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
p
calculate time of chamber force = 2 1.33f/[2500f/s]=.001 s
assume: action 2.5 i long = .208 f
assume: spring force = 20 p
calculate spring energy =fd=20 .208 =4.17 fp
calculate distance chamber pushes bolt = E/F =4.17fp/600 p = .0069 f
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

[Aamund Breivik]

Clark Magnuson wrote in message <9dbicl$15n$1...@xring.cs.umd.edu>...
#
#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

Isn't there a variable missing? AFAIK when the case expands under pressure,
it "grips" the chamber wall and temporarily increases friction, thus
significantly reducing rearward force against the bolt face. As far as I can
see your calculation does not take this into account. But then, it might be
hard to calculate this friciton in any meaningful way. It varies, among
other things, with brass hardness, pressure, chamber wall
polish/"smoothness", accumulated crud etc.
Blowback or delayed-blowback weapons in high-pressure calibres typically
have fluted chambers to deal with this friction (The HK G3, for instance).

--
Aamund Breivik

[Ron Seiden]

And then there was the novel in which the protagonist designs a blowback
10mm pistol, his solution to the spring-rate problem being to load the first
round via a tip-up barrel like the little Berettas. (And his frame was made
of thin layers riveted together.) So just how unlikely *is* that idea?

[douglas Trabue]

Ron Seiden wrote:
#
# And then there was the novel in which the protagonist designs a blowback
# 10mm pistol, his solution to the spring-rate problem being to load the first
# round via a tip-up barrel like the little Berettas. (And his frame was made
# of thin layers riveted together.) So just how unlikely *is* that idea?
#
The thin layers riveted together sounds okay. I works for Sears robo
grip series of tools. It is how I made a wrench for the lug nut on the
wirewheels of my Austin Sprite. I didn't have tools to cut heavy metal,
made the wrench up of layers of aluminum, easier to cut and kept the
wrench from bending, and steel didn't round out.

Doug T

[JerryO]

"Tod Glenn" <webm...@cordite.com> wrote in message
news:9d9dul$q6v$1...@xring.cs.umd.edu...
# Greetings all,
#
# I am hoping that someone can help me.
#
# I am looking for a formula that will let me calculate the spring weight
# required for a simple blowbacl operated gun where the cartridge
# velocity, bolt weight and bolt movement are known. I am told that this
# information can be found in volume 4 of Chinn's machinegun series.
# Thanks,
#
# Tod

Since the forward momentum of the bullet (and gas) equals the rearward
momentum of the slide (and other rearward moveing parts) their velocities
are proportional to their mass.

Since energy is proportional to velocity times momentum, the ratio of
energy of the slide to bullet will be the same as the ratio of the mass of
the bullet to the slide.

ie. A .45 with a 1/2 oz bullet and a 20 oz slide will have 1/40 as much
energy in the slide as in the bullet. If said bullet has 400 foot-pounds,
the slide will have 10 foot-pounds.

This slide energy must be absorbed by the recoil spring (and hammer spring
and friction) which will store energy as average force (1/2 initial force
plus final force) times distance in feet.

This ignored the energy used to cock the hammer (or striker), but you can
subtract that from the available energy.

As to friction calculations, that is left for your enjoyment.

JerryO

[Aamund Breivik]

"Ron Seiden" <rse...@alphaaccess.net> skrev i melding
news:9dehei$8m6$1...@xring.cs.umd.edu...
# And then there was the novel in which the protagonist designs a blowback
# 10mm pistol, his solution to the spring-rate problem being to load the
first
# round via a tip-up barrel like the little Berettas. (And his frame was
made
# of thin layers riveted together.) So just how unlikely *is* that idea?
#

Dunno what happens with small riveted structures- but when I was in
demolitions training we were told to use twice as much explosive to cut a
given thickness of riveted steel, compared to solid steel. I think this is
because cracks cannot spread beyond the layer they are in; if a solid frame
cracks, it will eventually break clean off. If a riveted frame cracks, most
likely the crack will be limited to just one layer. So if you were to make a
gun out of junk metal, a riveted design might make sense in the safety
departement.

--
Aamund Breivik

[Robert Christman]

JerryO wrote:
#
# "Tod Glenn" <webm...@cordite.com> wrote in message
# news:9d9dul$q6v$1...@xring.cs.umd.edu...
# # Greetings all,

# #
# # I am hoping that someone can help me.
# #
# # I am looking for a formula that will let me calculate the spring weight

# # required for a simple blowbacl operated gun where the cartridge
# # velocity, bolt weight and bolt movement are known. I am told that this
# # information can be found in volume 4 of Chinn's machinegun series.
# # Thanks,
# #
# # Tod
#
# Since the forward momentum of the bullet (and gas) equals the rearward
# momentum of the slide (and other rearward moveing parts) their velocities
# are proportional to their mass.
#
# Since energy is proportional to velocity times momentum, the ratio of
# energy of the slide to bullet will be the same as the ratio of the mass of
# the bullet to the slide.
#
# ie. A .45 with a 1/2 oz bullet and a 20 oz slide will have 1/40 as much
# energy in the slide as in the bullet. If said bullet has 400 foot-pounds,
# the slide will have 10 foot-pounds.
#
# This slide energy must be absorbed by the recoil spring (and hammer spring
# and friction) which will store energy as average force (1/2 initial force
# plus final force) times distance in feet.
#
# This ignored the energy used to cock the hammer (or striker), but you can
# subtract that from the available energy.
#
# As to friction calculations, that is left for your enjoyment.

All correct, but misses the difficulty in making such a round as the
10mm into a blowback operated weapon. The spring resistance and inertia
of the slide must be sufficient to keep the action CLOSED until the
pressure drops to a safe level. With the pressures inherent in the 10mm
this would require either a truly monstrous spring, or a VERY heavy
slide (or both).

--
Bob C. NRA Patron USN (Ret)

[Ron Seiden]

"Michael Zimmet" <mzi...@voicenet.com> wrote in message
news:9dfof7$c13$1...@xring.cs.umd.edu...
# On 10 May 2001 13:00:02 -0400, "Ron Seiden"
# <rse...@alphaaccess.net>, wrote the following in rec.guns:
#
# # And then there was the novel in which the protagonist designs
# # a blowback 10mm pistol, his solution to the spring-rate problem
# # being to load the first round via a tip-up barrel like the little
# # Berettas. (And his frame was made of thin layers riveted together.)
#
# This is from "Pallas," by L. Neil Smith, right?

Yeah, it sort of made Robert Heinlein look like a mild mannered pacifist.
Pretty funny that the hero shoots a Grizzley .45 Magnum & the old guy he
hooks up with carries a .416 Rigby (and the scene where he wishes he had
solids instead of hollow points so he could get more than one attacking wolf
with each shot to save ammo). Interesting society where *the* social
activity is shooting contests.

[Ron Seiden]

I was wondering more about that tip-up barrel for loading the first 10mm
round (done so as to avoid having to rack the slide with the heavy recoil
spring).

[Aamund Breivik]

Robert Christman wrote in message <9djef0$i8n$1...@xring.cs.umd.edu>...
#
#
#All correct, but misses the difficulty in making such a round as the
#10mm into a blowback operated weapon. The spring resistance and inertia
#of the slide must be sufficient to keep the action CLOSED until the
#pressure drops to a safe level. With the pressures inherent in the 10mm
#this would require either a truly monstrous spring, or a VERY heavy
#slide (or both).
#

Or perhaps a variation of the weird blowback action used in the FN
"FiveseveN" pistol.
The breech is not locked so it's a true blowback, but the barrel is also
held forward by a spring and moves back a bit under recoil. Since both
barrel and breech move backwards for the first part of the recoil movement,
they stay close enough together to prevent case rupture.
FN calls this "advanced blowback", I'm not sure I buy that.

--
Aamund Breivik

[Trefor Thomas]

On 12 May 2001 10:47:29 -0400, "Ron Seiden" <rse...@alphaaccess.net> wrote:

#"Michael Zimmet" <mzi...@voicenet.com> wrote in message
#news:9dfof7$c13$1...@xring.cs.umd.edu...
## On 10 May 2001 13:00:02 -0400, "Ron Seiden"
## <rse...@alphaaccess.net>, wrote the following in rec.guns:

<SNIP>

## This is from "Pallas," by L. Neil Smith, right?

<SNIP>

#with each shot to save ammo). Interesting society where *the* social
#activity is shooting contests.

Sort of like Switzerland.

Trefor Thomas

To be civilized is to restrain the ability to commit mayhem.
To be incapable of committing mayhem is not the mark of the civilized,
merely the domesticated.

[Tloc...@aol.com]

Tod Glenn <webm...@cordite.com> wrote:

# I am looking for a formula that will let me calculate the spring

# weight required for a simple blowback operated gun where the
# cartridge velocity, bolt weight and bolt movement are known. I am
# told that this information can be found in volume 4 of Chinn's
# machinegun series.

That's interesting - didn't know that. Yet another reason to buy the
series if you can afford the tariff (I surely can't).

Regardless, you can do a pretty good job if you happen to have
retained your college or high school physics text.

To the first order, you can ignore friction (such as slide drag).
You provide extra travel at the rearward end of the bolt stroke to
cover variability of individual rounds and friction.

What are the constraints? Well first, if this is a simple fixed
firing-pin bolt, the bolt has to have enough velocity when it closes
the breech to set off the primer. Certainly the bolt has enough mass!
A minimum firing pin velocity to smartly set off a primer would be in
the 10 to 12 fps range. If you hit it more slowly, you'll still set
it off every time, but you might get into variable ignition.

Bolt weight depends on bullet weight because the heavier the bolt,
the less the case has backed out of the chamber while the case has
high internal pressure. (Of course with a fixed firing pin, one of
your concerns will be that the bolt might bounce off the breech
face, leaving the case unsupported at _just_ the wrong time.)
If it's a handgun, with a light frame, you can have a lighter bolt
if you increase the recoil spring stiffness because the spring can
then accelerate the frame backward, keeping the bolt more closed
during the pressure peak. But we'll keep it simple and say that's
not the case here.

bolt mass*bolt movement = bullet mass*bullet movement (1)

bolt velocity rearward = (bullet mass/bolt mass)*bullet velocity (2)

bolt energy = (1/2) m v^2 (3)

spring energy change = (1/2) k (X^2 - x^2) (4)
X is the greater change in length of the spring from its free length
x is the lesser change in length of the spring from its free length

We get bolt velocity forward by giving the spring's stored energy
to the bolt (potential energy -> kinetic energy) and subtracting a
little energy for stripping the cartridge from the magazine. Really
we should add about 1/3 of the spring's mass to the bolt's effective
mass, but in a blowback, the bolt weighs a lot more than the spring
so the spring's mass can be ignored.

The scheme is that kinetic energy of the bolt moving rearward is
absorbed by the spring, but the bolt still has some velocity left by
the time it gets back to the sear, so we're sure it will get past the
sear, reverse direction, and come back and hang on it.

Let's try an example to see how all this works. I don't know
specifically what sort of gun you're interested in, so let's just
say we're shooting a 9 mm round and its bullet weighs 125 gn
and exits at 1100 fps. Say the bullet travel in the barrel is 7".
Let's say we need 3.6" of travel from bolt-on-the-sear to bolt-
closed in order for the spring to give the bolt enough velocity
before it feeds a cartridge. Also, we want to make sure that after
a cartridge is fired the bolt hangs itself on the sear again, so
let's say that the cartridge imparts enough momentum to the bolt
to make the bolt go back an additional 30 or 40%: 135%*3.6 = 4.9".

I think if your chamber is pretty good - no large unsupported areas -
the cartridge can back out maybe .090" (about 3/32") without rupturing.
After all, the pressure has fallen substantially by the time the bullet
is about to exit, and we're going to make it .090 at bullet exit. What
bolt weight do we need to hold movement to .090"?
bolt mass*bolt movement = bullet mass*bullet movement (1)
rearranging and multiplying by g/g:
bolt weight = bullet weight*bullet movement/bolt movement
bolt weight = (125/7000)*7"/.090" = 1.4 lb

bolt's rearward velocity after bullet exits (if the bolt doesn't
bounce off the breechface) = (bullet wt/bolt wt)*bullet vel (2)
bolt velocity to rear = ((125/7000)/1.4)*1100 = 14 fps = 168 inches/sec
#From this we can see how much energy the spring must absorb when the
bolt goes back 4.9".
bolt energy = (1/2) m v^2 (3)
bolt energy = (1/2)*(1.4/386.1)*168^2 = 51.2 in-lb

How much energy must the spring give to the bolt to accelerate it
to 11 fps (132 inches/sec)? This is how fast the bolt must go when
accelerated toward the breech by the spring after being released
by the sear to travel a distance of 3.6".
bolt energy = (1/2) m v^2 (3)
bolt energy = (1/2)*(1.4/386.1)*132^2 = 31.6 in-lb
To this let's add 1.3 in-lb to cover the cost of stripping a
9mm cartridge from a full large-capacity magazine.
bolt energy = 31.6 + 1.3 = 32.9 in-lb

So now you see we have two distances (3.6 and 4.9 inches) and their
associated spring energies (32.9 and 51.2 in-lb). It's time to look
at the spring potential energy formula. The smaller x is the same
for both of our cases, but the larger X is different. Let's say the
smaller X is Xa and the larger X is Xb.
spring energy = (1/2) k (X^2 - x^2) (4)
Case 1:
51.2 in-lb = (1/2) k (Xb^2 - x^2)
where Xb - x = 4.9" -> rearranging: Xb = x + 4.9
substitute latter into former:
51.2 in-lb = (1/2) k ((x + 4.9)^2 - x^2) (5)
Case2:
32.9 in-lb = (1/2) k (Xa^2 - x^2)
where Xa - x = 3.6" -> rearranging: Xb = x + 3.6
substitute latter into former:
32.9 in-lb = (1/2) k ((x + 3.6)^2 - x^2) (6)
We have two equations and two unknowns, so solve (5) and (6)
simultaneously to get k and x, then find Xa and Xb from the
"where" formulas.
Solving for k, we get 2.0 lb/inch for a spring constant, and
plugging that back into either (5) or (6) we get x = 2.77
and then Xb must be 7.67 and Xa 6.37.

Any spring with a spring constant of 2.0 lb/in and a precompression
of 2 3/4" will satisfy what we have for constraints so far. However,
in addition, the spring has to mechanically fit in the space allotted.
The spring's collapsed height has to be less than the space left when
the bolt hits its rearward stop. (Never use a collapsed spring as a
stop - you'll beat the bejeebers out of it and it will fail shortly.)
Let's say there's another 0.4" beyond the 4.9" before the bolt
hits the stop and when the bolt hits the stop, there's 1 1/2 or
1 5/8 inches of space left for the spring. So now the spring's
free length is 2.77 + 4.9 + 0.4 + 1.56 = 9 5/8 inches. YMMV -
depends on your particular mechanical configuration.

What if this thing fires from a closed bolt instead of an open
bolt? You still have the rearward constraint, you still have
to strip a cartridge, but now you have to account for cocking
the hammer on the way back, and you need to keep the force
holding the bolt closed high enough that it doesn't fall open
as a nuisance, but low enough that you can manually operate it.

Are you going to buy one of those little hand-operated spring-
winder things and wind your own from music wire? If so, I can
probably look through my notes and find something on coil spring
design (haven't done it for years) and write you another snoozer.

John B, antique engineer - been there, done that, can't remember
JHBer...@lbl.gov tloC...@aol.com

[Guy Neill]

You wrote:

calculate force from chamber = PA = [25kp/i][.35i/4][.35i/4][3.14]= 600
p

shouldn't this be;

calculate force from chamber = PA = [25kp/i][.35i/2][.35i/2][3.14] =
2405

Guy

[John Bercovitz]

Tod Glenn <webm...@cordite.com> wrote:

------------------------------------------------------------------------

[John Bercovitz]

JerryO wrote:

# Since energy is proportional to velocity times momentum, the ratio of
# energy of the slide to bullet will be the same as the ratio of the mass

# of the bullet to the slide.

This is an important thing to keep in mind. Energy gives you a feel for
the physical volume the spring will have to occupy to do the job and not
be overstressed.

# This slide energy must be absorbed by the recoil spring (and hammer spring
# and friction) which will store energy as average force (1/2 initial force
# plus final force) times distance in feet.

As I indicated in my previous post, if you integrate f(x) dx, you get:

E = (1/2) k (X^2 - x^2)

Just to show JerryO's formula is the same, here's his formula:

E = [(k*X + k*x)/2 ]*[(X-x)]

which reduces to:

E = (1/2) k (X^2 - x^2)

It's just algebra. 8-) The choice of equation depends on what data you
have available.

John B

PS: However, one shouldn't infer averaging is the solution for everything!
One often sees the formula for striker fall time as the average velocity (initial vel = 0, final velocity = strike vel, ergo avg = strike vel/2)
times the distance the striker falls. But! Really the striker's velocity
curve is part of the velocity curve you see for natural resonance so
T = (m/k)^0.5 Arccos(x/X)
as laid out in Stuart Otteson's book, "The Bolt Action", vol 1, pg 247.
Depending on how much spring stroke is used, the two methods can give
significantly different results. Usually the average method is pretty
darned good, though.

[Aamund Breivik]

John Bercovitz wrote in message <9e0e1o$ho6$1...@xring.cs.umd.edu>...
#JerryO wrote:
#
## Since energy is proportional to velocity times momentum, the ratio of
## energy of the slide to bullet will be the same as the ratio of the mass
## of the bullet to the slide.
#

Exept the expanding powder gas also has velocity and mass. The mass of the
powder charge should be included in the formula, though it is probably
insignificant in mild target loads.

--
Aamund Breivik

[Mark]

That's why horn mounts on cars are made of a stack of plates and padlocks
are too.

--
mark

[ramy]

Clark Magnuson <cmag...@home.com> wrote in message news:<9dbicl$15n$1...@xring.cs.umd.edu>...
> ...

No u r incorrect it will most probobly go up.