.380 Magnum

[918v]

What has a higher slide velocity:

90gr. @ 1500 FPS

or

147gr. @ 1000 FPS

Is there a formula by which to calculate slide velocity?

[freakshow10mm]

The 147gr load has a higher PF (momentum) so I'd think the 147gr load would have more slide velocity. I'm terrible in physics but it makes sense to me.

[918v]

What about slide acceleration?

Would a faster burning powder accelerate the slide faster than a slower burning powder?

[MakeMineaP99]

Yes. How good is your calculus? I can do this calculus or algebra based.

[918v]

I suck at math. Please help.

[MakeMineaP99]

I suck at math. Please help.

Side note: Damn British units all the way to hell.

Alright, I doubt you'll understand any of this, if you haven't had calculus, but here goes.

Regarding slide velocity:

p=mv therefore (p=mv)1=(p=mv)2 (basic momentum equation).

90/7000 ibs *1500 fps= weight of slide (ibs)* velocity of slide (fps).

147/7000 ibs *1000 fps=weight of slide (ibs)* velocity of slide (fps).

I'll let you do the number crunching.

Regarding acceleration:

Acceleration is a function of change of velocity with respect to time, dV/dt. A rate of change of a rate of change measured in fps/s or fps^2.

Velocity is a function of change of position with respect to time, ds/dt. A rate of change measured in fps.

Therefore, would a faster burning powder change your position with respect to time at a great rate (velocity) than a slower burning powder?

Ex.

Faster powder changes position with respect to time (velocity) as represented by the function 3x. If V=3x, A=3 (and s=x^3).

Slower powder changes position with respect to time (velocity) as represented by the function 2x. If V=2x, A=2 (and s=x^2).

As you can see, the faster burning powder has a greater acceleration than the slower burning powder, modeled by the function.

If you want to discuss this at length, I'd be glad to do it via email. I can't write all the fancy math on here, but I can write this out and email a PDF to you (I'll also see if I can come up with a better way to explain the calculus). PM me if interested with your email.

[thorn]

Suddenly i'm having scary flashbacks of slopes and anti-derivatives...

thorn

[MakeMineaP99]

Suddenly i'm having scary flashbacks of slopes and anti-derivatives...

thorn

Ah, yes, sadly I use calculus quite frequently.

[918v]

So loads using faster powders will have a sharper recoil impulse because they accelerate the bullet faster.

Why is it then that 147gr loads using fast powders like Bullseye feel so light when compared to those using slower powders like Long Shot?

[Crazy4nitro]

*ADHD Kicking in*

Great Subject,Great Question......But DAMN!

IMO the same Bullet going roughly the same Fps has Less FELT recoil using Bullseye vs. Unique or Bullseye vs. 2400

IMO Faster powders result in less perceived FELT recoil due to Impulse/Time ratios......

Carry on......
'Nitro

[degoodman]

The powder burn rate really doesn't have all that much to do with the recoil. Follow along.

The only thing that the powder burn rate does is determine the rate of pressure build-up to peak pressure. From my understanding of internal ballistics, the point of peak pressure is usually the point at which the bullet starts to move. After that point, the volume available to the expanding gasses is expanding and expanding fast, so pressures are going down from then on.

With tapered cases the pressure build up does translate into some bolt thrust, and that can have an effect on recoil. Just not a big one because you have to consider the conservation of momentum on the case until it hits the slide and stops, and the case's small mass prevents it getting alot of momentum in a short acceleration.

The recoil contribution from the powder gasses is related to the mass and velocity of those gasses, just like the recoil contribution from the bullet. Until the gas moves, the increase in pressure is all potential energy. Since charges of faster powders are smaller than slower powders there is less momentum contribution from them, and you get less recoil.

Now the differences in acceleration on the bullet that you MAY get from powders of different burn rates (remembering that burn rate is a non-constant function of pressure too) might effect the preception of recoil, but it has no effect on the total recoil energy delivered. There is some doubt as to whether or not the human mind can preceive the difference between those accelerations, because they evidence themselves over such short time periods, of from .001 to .003 seconds, and after that the pressure curve is relatively the same across burn rates.

The "big physics" equations don't do a good job of capturing what happens on the (relatively) ultra short time scales we're deaing with here. It takes SUBSTANTIAL computing power to model the small parts of discharging a round, and the blunt tools we have for measuring pressure don't do a very good job of it.

[DssG19]

Thank god for smart people:thumbsup:

[Crazy4nitro]

I'll Buy That for a Dollar.

'Nitro

[jwp]

Suddenly i'm having scary flashbacks of slopes and anti-derivatives...

thorn


AMEN

:iagree::leaving:

[copdills]

2nd That

[Clark]

Only the sleazy consultants try to turn projects into something they have already done, so they can sell the same work again.

I am guilty as charged. And I had to call my father up to get the physics right.

My own calculation on "How far do cases fly?":
1) For a semi-automatic case fired 5 feet form the floor, the time to
reach the floor is
t= square root ((2)distance / acceleration )= root (5'/32'/sec/sec)= .54

seconds

2) In a Colt .45 the ejector contacts the case when the slide is back
1.3". The farthest the slide can travel is 1.8" where it hits a stop. If

the spring is perfectly sized for the gun and the round, then the slide
will just run out of energy at the stop. Assume Vslide = 0 at 1.8".

3) Energy Slide at 1.3 inches = (force) (distance)= (16lb
spring)(1.8-1.3=.5")=.66 foot pounds of kinetic energy left in slide
when it hits the case

4) Energy is also = 1/2 mass velocity squared = .5 (mass of slide=
weight
of slide/ grav accel=12 oz/32 ft/sec/sec)(V squared)

5) Combining equations 3) and 4): Vslide at 1.3" = square
root(E/(.5mass)) = root(.66 ft lb/((.5)(.023 lb sec sec /ft)) = 7.6
feet/sec

6)Center of gravity [this should be moment of inertia, but that would be

work] is .25" from extractor claw and ejector hits the
case at .35" from the extractor claw.
Velocity of case = (.25"/.35") velocity of slide at 1.3" = (.25/.35)7.6
feet/sec = 5.4 feet per sec = 3.6 miles per hour

7) Combining 1) and 6): Distance case travels=
(Velocity)(time)=(5.4ft/sec)(.54 sec)=2.9 feet horizontally from the
gun

And Wolff FAQ wants your empties to land from 3 to 6 feet. They must
want the slide to barely hit the stop.
http://www.gunsprings.com/1ndex.html
Clark

================================================== ===

Example:
Patriot 45acp
http://www.republicarmsinc.com/

15 pound to start and 42 pound spring at the rear [home made spring
assembly], is just right for 10.3 gr of AA#5. That is, it is just enough
powder to keep the Patriot from jamming with that spring.

The bullet leaves the gun at 1100 fps.
The bullet weighs 185gr = 185/7000 = .026 Lb.
The slide weighs .5 Lb.
The barrel weighs .1 Lb.
The spring is 15 Lb to start.
The spring is 42 lb at back.

The momentum of the bullet equals the momentum of the slide and barrel.
[1100][.026]=V[.5+.1}
V=47.7 feet per sec
Velocity of slide and barrel = velocity of slide
Energy of slide = .5mVV= one half mass velocity squared
Es=.5[mass of .5 pounds][47.7][47.7]
Mass = [wieght]/gravity= .5/32.2=.0155
Es=.5[.0155][47.7][47.7]=17.64 foot pounds of energy

The energy required to pull back the slide = [force][ distance]
Force = average force = [15+42]/2 = 28.5 pounds force
distance = 1.656"=1.656/12=.138 feet of slide stroke
Eslide = [28.5][.138]=3.93 foot pounds

BUT WAIT A MINUTE! 17.64 DOES NOT EQUAL 3.93!

Not all of the bullet momentum went into the slide and barrel.
Some of it accelerated the hand.
Only 3.933/17.64= 22% went into the slide and barrel.

What does this mean to someone who wants to calculate the spring needed
for a semi auto pistol?
Heavy firm hands need stiffer springs than soft light hands and there is
no good way to measure how light and soft one's hand is.
So, you cannot accurately calculate the spring you need.

However, a CZ52 9mm pistol takes 4.7 gr AA#5 to cycle with a 125 gr .357
bullet. Put in a Wolff +10% spring and it takes 5.3 gr AA#5 to cycle. So
you can calculate spring requirements over a small region of already
known spring requirements.

The 22% finagle factor [above] will go up for heavier slides with
lighter springs. The above example is an extreme case of light slide and
heavy spring. The numbers are usually around 30~33%.

Never say "limp wristing" again. Say, "If the gun jams, get a lighter
spring. If the slide slams the frame, get a heavier spring."
Clark

What does it all mean?
The slide velocity is the reaction of the slide and barrel mass to the acceleration on the projectile and gas.
But the math is fuzzy because the frame and hand are coupled through the recoil spring.
If the same slide, barrel, spring, and hand are used, then we can just compare the bullet momentum + gas momentum vs bullet momentum + gas momentum.

My father, the great gun designing chief engineer, always used 1.5X the projectile velocity to estimate the gas velocity.
The gas mass = powder mass
momentum = mass times velocity

[918v]

You're awesome!

[Tree Rat]

You dump that new fangled WST for good old Bullseye?



TR

[918v]

I didn't dump it. I got a new 9 and found it did not like the same loads my old 9's liked. So I started from scratch.

[soundwave]

Side note: Damn British units all the way to hell.

Alright, I doubt you'll understand any of this, if you haven't had calculus, but here goes.

Regarding slide velocity:

p=mv therefore (p=mv)1=(p=mv)2 (basic momentum equation).

90/7000 ibs *1500 fps= weight of slide (ibs)* velocity of slide (fps).

147/7000 ibs *1000 fps=weight of slide (ibs)* velocity of slide (fps).

I'll let you do the number crunching.

Regarding acceleration:

Acceleration is a function of change of velocity with respect to time, dV/dt. A rate of change of a rate of change measured in fps/s or fps^2.

Velocity is a function of change of position with respect to time, ds/dt. A rate of change measured in fps.

Therefore, would a faster burning powder change your position with respect to time at a great rate (velocity) than a slower burning powder?

Ex.

Faster powder changes position with respect to time (velocity) as represented by the function 3x. If V=3x, A=3 (and s=x^3).

Slower powder changes position with respect to time (velocity) as represented by the function 2x. If V=2x, A=2 (and s=x^2).



I just threw up a little bit.

Seriously though, nice work. I can't remember .001% of math from college.

[Crazy4nitro]

I just threw up a little bit.

Seriously though, nice work. I can't remember .001% of math from college.


I can...watch

1 Keg= 6.5 cases of Beer (roughly) or 15 Gallons of Beer:cheers2:

see it's simple, Momma is Proud!

'Nitro

[jawjaboy]

Keg = 15.5 gallons

Barrel = 31 gallons

[Clark]

There is a book, "How the mind works" that has examples of the same logic problem, but couched in two different stories. One is hard, the other is easy. The human mind is not averse to figuring out if it is being cheated.

What does it all mean?
The mind does better with incentive.

[Steve Koski]

God I love engineers. I actually understand them, unlike women. Are there any women engineers? That would blow my mind.

[Clark]

When I was at the UW EE in 76, there would be 600 students, all men but one woman. She would turn in her test first and get the highest score.

And she is still married to me, as a reward for getting me through school:)

[MakeMineaP99]

God I love engineers. I actually understand them, unlike women. Are there any women engineers? That would blow my mind.

Yep, I work with a few of them.

[Brass Nazi]

God I love engineers. I actually understand them, unlike women. Are there any women engineers? That would blow my mind.


There WAS at least one:

http://i57.photobucket.com/albums/g234/JRTRAPP1/train_wreck.jpg