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From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Need recommendation on good medium to big game hunting handgun
Organization: Lawrence Berkeley Laboratory
In article <37061@mimsy.umd.edu> ke4zv!gary@gatech.edu (Gary Coffman) writes:
#In any case, you want a gun with as long a barrel as you can comfortably
#carry, I like 7 inch. And you want a fast lock time and a strong hammer
#fall to insure reliable ignition. I really like the .454, but a .44
#Ruger is much cheaper and adequate for most any realistic handgun
#hunting situation. If you can accept single shot pistols, the Contender
#is far and away the favorite hunting handgun with a wide choice of
#interchangeable barrels ranging from the .22 LR all the way up to some
#heavy rifle calibers. I have a 14 inch .30 Herret barrel for mine that
#makes a fine hunter.
Bear in mind that the Ruger single-action revolvers have massive hammers
and long hammer falls. This doesn't really matter if you fire from a stable
position: in this case the hammer fall isn't slow enough that you could drift
off your target by the time the cartridge ignited. Similarly, the T/C
Contender has a heavy hammer but a somewhat faster hammer fall. Remember
that the Contender also has to be able to set off rifle cartridge primers
so its ignition must be heavy duty.
Fast lock times and light hammers become important when you stand up on
your hind legs and fire with one hand. The fast lock time is important
because you are adjusting the squeeze rate of the trigger such that you
cause the bullet's exit to coincide with the best alignment of sights with
target. If the lock time is long, you have to do longer-range "predicting"
of how things will be when the bullet exits.
Low-momentum hammer falls are also important in target shooting because every
time the hammer strikes, your grip's consistency is being tested. If you use
a tight grip one time followed by a looser grip the next time, the gun will
be jostled more the second time than the first. This will cause it to point
a different direction when the bullet leaves. A low-momentum hammer fall
minimizes this variation.
In conclusion, I would say that neither fast lock time nor low-momentum
hammer fall is important in a hunting situation if the gun is stably held
by any of several methods: two-handed hold, Creedmore, rested position, etc.
All you really need is reliable ignition whether from a fast, light hammer
or a slow, massive one. Incidentally, since primers respond well to higher
firing pin speeds, you can get by with lower energy and much lower momentum
ignition if your hammer is light and fast than if it is heavy and slow.
JHBercovitz@lbl.gov (John Bercovitz)
From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Coors Schuetzen Pistol Competition
Organization: Lawrence Berkeley Laboratory
In article <41796@mimsy.umd.edu> rhoten@master.ceat.okstate.edu writes:
#The Coors Beer folks have been sponsoring schuetzen rifle....
#They have now added a
#schuetzen PISTOL category I'd like to try. The rules, as I
#understand it, require a single shot action (no bolt guns),
#14" barrel length limit and lead bullets (no gas checks).
#I think a T/C Contender in 32-20 would be an inexpensive
#way to give this competition a try (I already have the
#Contender receiver).
Do the match rules required standing up on your hind legs and
firing one-handed as God intended? If so, the T/C may be a
poor choice. Set triggers were abandoned a long time ago by
benchresters and others because that little hammer in there
which hits the sear also knocks the gun out of line with the
target ever so slightly. Instead, match shooters use multi-
lever override triggers. Actually, this would make a heck of
an exciting project for the T/C: a three-lever trigger to replace
the set trigger. I've been considering it for a few months.
Wouldn't want to rush into something, ya know. 'Specially if
it entails work.
Some time back I got a T/C for the express purpose of having
a one-handed single shot pistol to shoot. I reworked the
trigger to take out the creep and changed springs so its weight
is less than a pound. I got a lot of vertical stringing
from it. So I asked a buddy who shoots in the Pan-American
games and the Olympics to try it out. Same problem - vertical
stringing. (Same _amount_ of vertical dispersion, too;
that made me feel good!) Next I clamped the barrel in my
Ransom Rest to test the barrel's accuracy. The barrel was just
fine, shooting into 3/8" at 25 yards.
The gun was also just fine with the two-handed hold or Creedmore
or any other way a T/C is usually fired. It's just not a one-
handed gun, near as I can tell.
John Bercovitz JHBercovitz@lbl.gov
From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Ti gun parts and physics
Organization: Lawrence Berkeley Laboratory, California
In article <48463@mimsy.umd.edu> rpalmer@Think.COM (Ralph Palmer) writes:
#I've seen a lot of gun catalogues pushing Titanium hammers, sears, and
#firing pins as the latest and greatest for your 1911 45. They all promise
#faster lockup but none attempt to quantify the improvement. I need some
#help with my basic physics to try to figure out what is going on. How much
#faster are we talking about? Is is worth the price?
That would be lock time rather than lockup. Lockup is the joining of
chamber and breechblock, in very simple terms. Picky, picky, picky -
Pat Paulsen.
#The two basic equations that come to mind are Force = Mass X Acceleration
#(F=MA) and Force = -Spring Constant X displacement (F=-kx). The simplest
#of the components to look at is the firing pin. The displacement and the
#spring constant do not change with the change in materials so the force
#remains constant. That results with changes in mass are proportional to
#the changes in acceleration.
#My material book lists the following densities in lbs/in3
#Carbon Steel 0.28
#Stainless Steel (300 series) 0.29
#Titanium Alloy ( with Al and V) 0.16
#Titanium parts are about 45% lighter than their steel counterparts. This
#means that their acceleration should be about %45 faster. This is where my
#physics hits the wall. I know that A=(dv/dt).
#So how about it mechanical engineers and physics buffs, can you quantify
#how much improvement Ti parts will make? Perhaps someone can provide the
#total travel distance of a 1911 45 firing pin to help with the
#calculations.
You're going to have a lot of trouble getting this calc out of us lazy bums,
I would predict. It's not that it's too hard to do, it just takes forever to
scribble that much.
One should note that a firing pin and a striker are two different animals.
A firing pin conveys the hammer's blow to the primer. In the case of the
1911, the firing pin is floating which means it is shorter than the distance
between hammer and primer when the hammer is down and so the firing pin
must carry enough whack to the primer by virtue of the momentum it gains
from the hammer. The firing pin spring in a 1911 serves only to retract the
firing pin after it's done its job.
A striker has the mainspring accelerating it towards the primer. There is
no hammer in this setup. I think this may be the configuration you were
referring to above. This is not found on the 1911 but it is found on the
P'08 (Luger), for example.
How about I give you an idea of how to proceed as a poor substitute?
You need to add a few equations to your arsenal. You'll want to follow
the energy trail and the momentum trail. A spring's potential energy
is (1/2)*k*(Xf^^2-Xi^^2) where Xf and Xi are the final and initial
compressions of the spring from its free length. The kinetic energy
of a striker is (1/2)*m*V^^2 where m and V are the usual. Just equate
these two equations, plug in the knowns and come out with V. You don't
have to use (1/2)*k*(Xf^^2-Xi^^2) for the spring's potential energy; an
equivalent is to measure the intial and final force of the spring, average
them, and multiply that average by the spring travel.
I just tried to write up how you would calculate the response of a 1911
to this change. It got to be three long paragraphs and I was just getting
warmed up. Let me put it in even more basic terms and you get a dynamics
text. 8-)
First you've got to figure out how fast the hammer is moving. You can find
this from the mainspring energy, which is about 6 in-lb, and the polar
moment of inertia of the hammer which you can get by hanging it as a
pendulum and measuring its period or by just integrating the mass times
radius squared. From these two values, you can find out how fast the part
of the hammer which contacts the firing pin is moving. Since the titanium
hammer looks like the steel hammer, it will be faster because it will
carry the same kinetic energy. See the equations in the paragraph
above. Equating the kinetic energy equations for the two hammers since
the spring energy into them is the same, you see that the titanium hammer
is (.28/.16)^.5 = 1.3 times faster. If you think about it, you may come up
with some experimental means of measuring your hammer's absolute speed.
You're going to need a dynamics book to see how the rotational momentum
of the hammer connects with the firing pin but basically, it's just
impulse times the radius the impulse is acting at. There are a few
subtleties like where this radius is relative to the radius of gyration
of the hammer but there just isn't enough space to explain all that trash.
For the momentum transfer from hammer to firing pin, you need the basic
momentum equation: Ma*Vai + Mb*Vbi = Ma*Vaf + Mb*Vbf where Ma is the
mass of the hammer converted to linear terms and Mb is the mass of the
firing pin. Vai is the initial velocity of the hammer, Vaf is the final
velocity of the hammer, etc. Also you need to know about the coefficient
of restitution equation: e = -(Vaf-Vbf)/(Vai-Vbi) where e is around
0.8 for hard steels. Solve these two equations simultaneously to get the
velocity of the firing pin.
And there you have it. With all parts' initial and final velocities,
you can calculate lock time. It's pretty simple, you've just got to
grind at it.
After you've done all this work, you can calculate the dynamic advantage
of a striker-fired gun over a hammer/floating firing pin-fired gun for us.
John Bercovitz (JHBercovitz@lbl.gov)
From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Ti gun parts and physics
Organization: Lawrence Berkeley Laboratory, California
In article <48738@mimsy.umd.edu> hays@SSD.intel.com (Kirk Hays) writes:
#In article <48701@mimsy.umd.edu>, JHBercovitz@lbl.gov (John Bercovitz) writes:
#>In article <48463@mimsy.umd.edu> rpalmer@Think.COM (Ralph Palmer) writes:
#> The firing pin spring in a 1911 serves only to retract the
#> firing pin after it's done its job.
#The firing pin spring also serves to keep the firing pin from
#impacting the primer too hard when the M1911 feeds and
#chambers a round.
Thanks, Kirk, I should have mentioned that and perhaps another nit,
which is that the spring can serve as the forward stop for the firing
pin's travel. If you dry fire, the firing pin stops with its tip
a loooong way out from the breech face. I'm not so sure this is very
good for the spring. Its coils must slam into each other pretty hard
if it fully collapses.
#Otherwise, an excellent article, although it seems to me like
#you assumed that the hammer's entire energy is transfered to
#the firing pin. I'm not a physicist, so I certainly could
#have misread that.
I'm sorry if I made misleading statements; I didn't try to.
(This time. heh, heh 8-)
#As far as the Ti parts, I can state from experience that they
#do shorten lock time, significantly, so much so that steel
#parts seem "slow and muddy" afterwards.
Now _that_ statement is worth more than all of the calculations.
Nevertheless........
After I wrote my reply to Rotten Ralph Palmer's article yesterday (Rotten
because he brought this subject up and now I gotta think about it.), I got
to thinking, well how much difference could it really make in firing pin
momentum if the hammer's equivalent mass drops according to the factor,
.16/.28? I mean, IF the hammer is still substantially more massive than the
pin, it shouldn't affect the firing pin's velocity much if the hammer's
mass is raised or lowered quite a bit, right? Since the hammer is now
going 1.3 times faster, the speed should more than compensate and we should
make out pretty well, I hope.
So I sat down with a pad of paper yesterday evening and scribbled up a
pretty little equation:
SPECIAL CASE OF MOMENTUM TRANSFER: u1 = 0
Variables:
m = "impacting" mass (linear equivalent of hammer's polar moment of inertia)
n = "impacted" mass (firing pin)
M = mass ratio = m/n
v1 = initial velocity of m
v2 = final velocity of m
u1 = initial velocity of n (in this special case, u1 = 0)
u2 = final velocity of n
e = coefficient of restitution (approx. 0.8 for hardened steel)
It is allowed to skip the following derivation and cut to the chase, below.
From yesterday's post, two equations:
momentum transfer: restitution:
m*v1 + n*u1 = m*v2 + n*u2 e = -(v2-u2)/(v1-u1)
Dividing the momentum transfer equation through by n and substituting M for m/n:
M*v1 + u1 = M*v2 + u2
Setting u1 = 0 :
M*v1 = M*v2 + u2 e*v1 = u2 - v2
v2 is the variable we're not particularly interested in at this time,
so solve both equations for v2:
v2 = (M*v1 - u2)/M = v1 - u2/M v2 = u2 - e*v1
Set the two equations for v2 equal to each other
and find relationship of u2 to v1:
v1 - u2/M = u2 - e*v1
v1 + e*v1 = u2 + u2/M
v1*(1+e) = [(M+1)/M]*u2
Chase starts here -> u2/v1 = [M/(M+1)]*(1+e)
8-)
Check the limits: If m and n are equal, then M = 1 and if e = 1 (maximum),
then u2/v2 = 1. That's correct: if a steel ball runs head on into a similar
steel ball which is not moving, the second ball acquires the velocity of the
first. The other tendencies as M and e vary also appear correct.
Implications: Since e is constant at 0.8 or more for collisions of hard
materials, about all we can mess with is M. In the case of the 1911, M is
surely? 3 or more and we can see that as M grows larger, there is very little
growth in u2 if v1 is held constant. So the velocity of the firing pin is
roughly one and one-half times the velocity of the hammer regardless. This
means increasing hammer speed by judicious density reduction may well be a
net gain: The firing pin speed will go up more or less as the hammer speed
as long as the hammer stays substantially more massive than the firing pin.
The conclusion is that the decreased lock time didn't cost us anything in
ignition reliability as long as the hammer's equivalent mass is still much
higher than the mass of the firing pin.
Let's see, what does this imply about changing the firing pin to titanium?
I haven't thought about this so I probably shouldn't write but that never
stopped me before. 8-) Well, again, changing the mass ratio, M, doesn't
have much effect on firing pin velocity, u2, with the above-mentioned caveat.
So firing pin velocity will stay the same but its mass will go down resulting
in lower ignition reliability. So it looks like it is valuable to change
the hammer to titanium but not the firing pin. As I said above, though,
all of this is predicated on the hammer being much more massive than the
firing pin. Someone needs to find out if this is true. It only appears
to me to be true; I haven't checked it. How are you coming on calculating
or experimentally determining the polar moment of inertia of the hammer,
Rotten Ralph? 8-)
Oh, wait a minute, the hammer's speed went up by 1.3 so the firing pin's
speed also went up by 1.3 although its mass went down by 1.3 so the
firing pin's momentum is a push. Besides, it may be energy which plays
the major part in setting off a primer since the primer cup has to be
deformed (Or does it? Maybe it's just the shock that sets the primer off
and the deformation occurs later?) Anyhow, it IS a push if impulse sets
of the primer and it's a net gain if kinetic energy sets off the primer
because energy goes up as m*v^^2 and our m is going down as the v goes
up proportionately. If you don't follow that, squawk at me and I'll do
better.
Well, we can certainly say for sure that you shouldn't change the firing pin
to titanium without first changing the hammer to titanium; that would give
you the worst of both worlds, so to speak.
John Bercovitz (JHBercovitz@lbl.gov)
From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Ti gun parts and physics
Organization: Lawrence Berkeley Laboratory, California
In article <48765@mimsy.umd.edu> bercov@bevsun.bev.lbl.gov
(John Bercovitz) writes:
#....... it may be energy which plays the major part in setting
#off a primer since the primer cup has to be deformed (Or does
#it? Maybe it's just the shock that sets the primer off and the
#deformation occurs later?) ...........
I've had a few more thoughts about this. I'm pretty sure the primer
doesn't have to be plastically deformed to be set off. Surely the
primer cup is elastically compliant enough that the pellet will be
receiving the shock and starting ignition before the primer cup is
permanently, plastically deformed. I think this deformation comes
later as the firing pin continues forward. Maybe the pellet isn't
well lit before the pin deforms the cup, but it's initiated anyhow.
So that makes it OK that the firing pin doesn't carry much energy
but has sufficient momentum to move the back of the primer cup at
sufficient speed into the pellet, right? Wrong. I think in many
cases it is necessary for the primer to form a hemisphere in the
back of the primer cup to hold in the pressure. It is also good
if the firing pin has some mass and some forward momentum to push
back against the developing chamber pressures.
You may remember that I tested primers a while back to see how fast
the 71 grain 1911 firing pin had to be moving relative to a primer
in order to reliably set off the primer. It looked like the answer
was maybe 40 feet per second (480 inches/sec). That's really fast
compared to a rifle striker but the striker has considerable mass
and a 1911 firing pin doesn't. Let's say a primer dent is .030"
deep. Average velocity will be roughly 240 ips so the time required
to form the dent will be 125 microseconds. The hemisphere appears
to indeed be formed before the chamber pressure rises very much.
The firing pin hole in the 1911 is about .09" in diameter. The peak
chamber pressure is maybe 16,000 psi. So the force on the unsupported
part of the cup is (pi/4)*(.09^2)*16000 = 102 pounds. The cup is
made out of gilding metal or similar with a yield strength in shear
of maybe 20,000 psi. The cup is about .01" thick so the shear area
resisting "blanking" failure is (pi)*(.09)*(.01) = .003 in^2. The
stress in this shear area is the force divided by the area or 102/.003
= 34,000 psi. This means the primer would blank or perforate if it
didn't have anything else going for it like a hemispherical shape and
the mass of the firing pin filling that hemisphere (if the firing pin
stays in place - I don't think it actually would bounce quickly off a
dead material like gilding metal).
As an aside, remember that problems have been had in this very area.
Remington down-sized its model 788 firing pin from the usual 078/080
to 058 because primers were blowing with its extra light (and extra
fast) striker. Colt down-sized the firing pin in the Delta Elite
from 090 to 060 because the 10mm has higher chamber pressure. [I bet
that really made them cry, they can hardly maintain enough machining
accuracy to hit the primer as it is, let alone with a smaller pin. 8-)
Maybe _that's_ why S&W put a small primer in the 40 S&W, so Colt couldn't
use the cartridge! 8-) 8-) ]
So where are you _going_ with this post, Johnboy? Where indeed. Well,
I'm trying to get around to saying that that low mass titanium firing
pin may allow primer blowouts if you hot up your 45 ACP loads. It
appears that there isn't all that much reserve strength left in the
primer cup as it is, so a well-formed hemisphere and some firing pin
mass would be a good idea. I think I'd stick with the steel firing
pin since the previous post showed it is unlikely that the titanium
firing pin is faster than the steel firing pin provided it's true that
the titanium hammer is still much more massive than either firing pin.
The kinetic energy of the steel firing pin is what allows it to form
a good hemisphere and its mass is what causes it to be able to help
hold the hemisphere's shape as the chamber pressure sets the primer
back.
John Bercovitz (JHBercovitz@lbl.gov)
From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Ti gun parts and physics
Organization: Lawrence Berkeley Laboratory, California
Summary: With the 6 in-lb spring energy input of a standard
mainspring, the titanium hammer takes 3.4 milliseconds to fall and
the steel hammer takes 4.5 milliseconds to fall. If your spring's
potential energy is not 6 in-lb, you can figure your hammer's time
of flight by using a correction factor of the square root of:
(6 divided by your spring's energy). As an example, if your spring's
stored energy is 3 in-lb, your hammer's time of flight would go
from 4.47 msec to 6.32 msec. The titanium hammer provides more
momentum to the steel firing pin than the steel hammer does. When
using the titanium hammer, the titanium firing pin saves an
additional 18 _micro_seconds and is probably not worth it since it
may also allow other problems to develop.
The first step was to determine the polar moments of inertia of two
spare Colt hammers I have. One is the standard spurred hammer and
the other the burr or rowel type Commander hammer. It seemed like
it would take longer to measure these moments by the experimental
pendulum method than to just calculate them. By drawing lines on
one side of a given hammer I was able to break it up into geometric
figures which could be calculated and summed to give the total
moment. I also weighed the hammers on a powder scale to get their
masses to serve as a check on the calculations. The agreement was
excellent. I like to work this sort of problem in units of force,
length, and time so where you see mass given, for example, the
units will be force-time^2/length.
As mentioned before, the polar moment of inertia is the integral of
all dmasses times (r of the dmass)^2 so it has the units of force-
time^2-length.
The center of percussion is the "sweet spot" on a bat. If you
don't hit the ball with the sweet spot of the bat, you get a
reaction at the pivot point, where your hands are, and your hands
sting. I calculated the radius to the center of percussion to make
sure it was somewhere near the point where the firing pin touches
the hammer, this to make sure there wasn't some large reaction at
the hammer pivot. The radius of the center of percussion is the
polar moment of inertia divided by the mass and by the radius to
the center of gravity.
Since a titanium hammer is perhaps similar to a burr hammer except
in mass (though I realize the Ti hammers are a bit more rakish
looking), it is fair to proportion the values for a burr hammer
according to the density ratio. I've also proportioned the mass of
the titanium firing pin although I think that additionally it is
cut down.
All of the figures given below are with respect to the pivot of the
hammer where applicable.
Standard Steel titanium
steel hammer burr-hammer burr-hammer
Polar moment
of inertia, 4.27E-5 3.85E-5 2.20E-5
lb-sec^2-inch
mass of hammer, 1.00E-4 1.01E-4 0.577E-5
lb-sec^2/inch
radius of center 0.47 0.45 0.45
of gravity, inch
radius of center 0.91 0.85 0.85
of percussion, inch
radius of centerline
of firing pin WRT the 0.77 0.77 0.77
hammer pivot hole, inch
mass of steel firing 2.63E-5 2.63E-5 2.63E-5
pin, lb-sec^2/inch
mass of Ti firing 1.50E-5 1.50E-5 1.50E-5
pin, lb-sec^2/inch
kinetic energy of 6 6 6
hammer, in-lb
As I mentioned in an earlier post, the standard amount of energy
the mainspring delivers to the hammer is around 6 in-lb. I
measured the stored energy of the mainspring of my Gold Cup and of
my Delta Elite. I believe the Gold Cup spring may be the one which
is stock for it, but I know I've cut a turn or so off the Delta
Elite mainspring. The stored energies came out 3.2 in-lb for the
GC and 4.3 in-lb for the DE. The stored energy of a spring is just
the spring travel times the average of the initial and final spring
forces.
If you know the energy put into the hammer and the polar moment of
inertia of the hammer, you can calculate the angular velocity of
the hammer. This is because the kinetic energy of a hammer is:
(1/2)(polar moment of inertia)(angular velocity, radians/sec)^2.
For 6 in-lb input:
Standard Steel titanium
steel hammer burr-hammer burr-hammer
angular velocity 530 558 738
of hammer, rad/sec
tangential velocity
of hammer at radius 408 430 568
of firing pin, in/sec
The effective mass of the hammer at the radius of the firing pin is
the polar moment of inertia of the hammer divided by the square of
the radius to the firing pin (.77^2). Dividing that mass by the
mass of the firing pin gives "M", the mass ratio alluded to in the
previous post. The travel of the firing pin is about .06 inches.
#From that previous post:
Vp = [M/(M+1)]*(1+e)*(Vh)
where:
Vp = velocity of firing pin
M = effective mass of hammer at .77" divided by mass of firing pin
e = coefficient of restitution, approximately = 0.8 for hard steels
Vh = velocity of hammer at radius of firing pin
Standard Steel titanium
steel hammer burr-hammer burr-hammer
velocity of steel 538 550 598
firing pin, Vp, in/sec
time of flight of 1.11E-4 1.09E-4 1.00E-4
steel pin, sec
momentum of steel 1.41E-2 1.45E-2 1.57E-2
firing pin, lb-sec
K.E. of steel 3.81 3.98 4.70
firing pin, in-lb
velocity of Ti 608 629 728
firing pin, in/sec
time of flight of 0.98E-4 0.95E-4 0.82E-4
Ti firing pin, sec
momentum of Ti 0.91E-2 0.94E-2 1.09E-2
firing pin, lb-sec
K.E. of Ti 2.77 2.97 3.97
firing pin, in-lb
As we can see, the maximum time saved by substituting titanium for
steel in the firing pin is 0.18E-4 seconds or 18 microseconds.
_Milliseconds_ are what you are trying to save in lock time, not
microseconds. So the titanium firing pin doesn't seem to have an
advantage in lock time. If you mistakenly use it with a steel
hammer (not recommended), kinetic energy, which forms the pressure-
resisting hemisphere, will fall. Also, after the titanium firing
pin dissipates its kinetic energy in deforming the primer, there
will be little mass available to help support the hemisphere as the
primer is being set back. All in all, not a very good deal.
I said in a previous post that the titanium hammer is 1.3 times
(SQRT density ratio) as fast as the steel hammer based on equal
kinetic energy input. That doesn't tell us the _absolute_ amount
of time saved by the titanium hammer. I made an attempt to measure
the hammer fall time but wasn't successful. I set up my ancient and
venerable and dim General Radio Strobotac for its maximum rate of
240 flashes per second, opened the shutter on my 4x5 and pulled the
trigger. I got good images of a cocked hammer and a fallen hammer
but I couldn't get any images in between. I guess the light was
too dim/the film was too slow. I could often see an extra image of
the hammer with my eye, but the film couldn't see it. Anyone in
these here parts got one o' them faster new-fangled bright
Strobotacs to loan for a worthy experiment? How about some
Polaroid Type 57 ISO 3000 speed film? Good, I need them both. 8-)
(Just kidding - I'm satisfied with the calculations below.)
The fall time of the hammer can be calculated, but it's not as much
fun as you'd think because the torque starts low and increases in a
funny skewed sinusoidal curve as the strut hole moves "out from
under" the hammer's pivot hole. It's analogous to the piston and
crank in your car: the pivot is somewhat offset from the line of
action (for different reasons, of course). Nevertheless, I
measured the force curve approximately with a spring scale so I
would have an idea of what its shape is. I set up Excel and
plugged in the values, did a piecewise integration, and got the
same kinetic energy and terminal velocity as I got above so I'm
confident that the measurements were accurate. For a 6 in-lb
input, the flight time of the steel burr hammer comes out to 4.5
milliseconds and the titanium hammer of the same configuration
comes out 3.4 msec. These _are_ in the ratio of the square root of
the density ratio (this was mentioned in a previous post) thereby
providing a further check on the calculations.
I was going to provide the derivations and calculations here at the
end of this post as a sort of appendix but this post is long enough
and I doubt that that would have general interest. Mainly this would
be the derivation of the concept of the effective mass of the
hammer and how I proportioned the torque curve of the hammer and
the equations associated with that process. If any of you are
really interested in this information, let me know and I'll write
it up.
John Bercovitz (JHBercovitz@lbl.gov)
From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Ti gun parts and physics
Organization: Lawrence Berkeley Laboratory, California
In article <49128@mimsy.umd.edu> dingbat@cix.compulink.co.uk
(Codesmiths) writes:
## I'm trying to get around to saying that that low mass titanium firing
## pin may allow primer blowouts if you hot up your 45 ACP loads.
#I've never seen a blown-out primer. What does it look like ? Does the
#primer cup move out of the pocket ? Does it split from the centre ?
#from the edge ?
There are lots of types of blown primers; a popular one is caused by
overpressure loads and results in smokey leaks around the primer.
The particular type of failure I was referring to evidences itself, in mild
cases, as a small hole in the middle of the dent. Higher pressure failures
may blow out most of the material which was the dent. These little holes,
without other pressure signs, are evidence of a firing pin which is too light
for its cross sectional area, at least at the given chamber pressure.
John Bercovitz (JHBercovitz@lbl.gov)
From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Ti gun parts and physics
Organization: Lawrence Berkeley Laboratory, California
In article <49339@mimsy.umd.edu> hays@SSD.intel.com (Kirk Hays) writes:
#|> For a 6 in-lb
#|> input, the flight time of the steel burr hammer comes out to 4.5
#|> milliseconds and the titanium hammer of the same configuration
#|> comes out 3.4 msec.
#I am pleased that I can still detect a difference in
#"fall-time" with my aging senses.
Actually, in the rifle world, designers fight hard for every millisecond
they can cut from the lock time. I guess the gun can wander a lot in that
time. As I mentioned before, my Gold Cup has a very weak mainspring - this
helps reduce the trigger pull along with the different placement of the
strut hole. It has an extremely slow lock time. Usually what I do is
yank the trigger when the sights are lined up and then realign them before
the gun goes off [NOT! 8-)]. There is another advantage to the titanium
hammer. If you multiply the polar moment of inertia of the hammer by its
angular velocity, you'll get its angular momentum. Angular momentum is a
lot of what disturbs the sights after the sear releases the hammer.
#Excellent post, John. Thank you.
My pleasure.
John Bercovitz (JHBercovitz@lbl.gov)
Hey - just had a weird thought. What if we decreased the hammer's inertia
by cutting it down and increased the firing pin's mass by making it out of
tungsten. Do this so that the effective mass of the hammer is equal to the
mass of the firing pin. Now when the hammer hits the firing pin, it comes
to a full stop and the firing pin takes off with 80% or more of the hammer's
velocity. Efficiency (on a KE basis) is maximized and you have less disturbing
momentum from the hammer. Lock time would be.... Hmmm... got to think about
that. Firing pin travel is short so it shouldn't hurt too much, might even
help.
Later,
John B
From: sfaber@ihlpb.att.com (Steven R Faber)
Subject: Re: Ti gun parts and physics
Organization: AT&T Bell Laboratories
#From article <49307@mimsy.umd.edu>, by bercov@bevsun.bev.lbl.gov (John Bercovitz):
# Summary: With the 6 in-lb spring energy input of a standard
# mainspring, the titanium hammer takes 3.4 milliseconds to fall and
# the steel hammer takes 4.5 milliseconds to fall. If your spring's
# potential energy is not 6 in-lb, you can figure your hammer's time
# of flight by using a correction factor of the square root of:
# (6 divided by your spring's energy).
I mentioned to the author that I had an interval timer/freq. counter
that could be used to measure the hammer fall time instead of a strobe,
and I could try it out on my Springfield Armory .45 (unmodified with
steel hammer).
John sent back some suggestions and warned that my SA would have
exceedingly long lock times compared to his Colts.
He knew I needed a little motivation there, so not to be undone,
I dragged out and dusted off my $5 hamfest special ancient vacuum
tube interval timer/freq counter and hooked up the apparatus:
A heavy wire was put in place of the firing pin so it extended
just past the pin block plate, and another wire rigged to just
contact the hammer when fully cocked. The hammer would ground
the first wire until released and then ground the firing pin wire.
I knew it would be difficult to avoid contact bounce, but managed
to get a couple measurements 4.43, 4.45 msec. Then the firing pin wire
got mashed and shorted. After repair, I had a hard time getting more
readings, so I hooked up a one-shot logic chip, set the time constant
long, and used the hammer release to trigger it, and the hammer fall
to clear it, resulting in a clean pulse. After this the measurements
were easy - 4.46, and 4.48, then the wire mashed in again and I got
readings of 5.01, 5.09. I believe the wire was shorting to ground
back by the chamber end of the hole.
In summary the measured time averaged about 4.45 msec to fall to
just touch the firing pin, and 5.0 msec to fall flush with the
pin block plate (and perhaps shove the wire/pin a little farther).
This is in excellent agreement with John's calculated 4.5 msec
value.
Steve Faber
From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Ti gun parts and physics
Organization: Lawrence Berkeley Laboratory, California
In article <49664@mimsy.umd.edu> sfaber@ihlpb.att.com (Steven R Faber) writes:
#I mentioned to the author that I had an interval timer/freq. counter
#that could be used to measure the hammer fall time instead of a strobe,
#and I could try it out on my Springfield Armory .45 (unmodified with
#steel hammer).
#John sent back some suggestions and warned that my SA would have
#exceedingly long lock times compared to his Colts.
Oops! When you wrote "SA 45" I thought you meant a single action revolver-
type 45 as in a Colt thumbuster. Seems like we have overlapping abbreviations
here. I apologize; I did not mean to insult Springfield Armory; I only
meant to say that Peacemaker-style SA revolvers have notoriously slow lock
times. I should have figured it out. Oh well....
Anyway, great post! Especially since your numbers agree with mine, saving
me from colossal humiliation. 8-)
John Bercovitz (JHBercovitz@lbl.gov)
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