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Old March 4, 2000, 08:02 PM   #1
Badger
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I recently bought a used 4 inch Python and using target velocity loads with a 158 grain bullet it's hitting several inches high at 50 feet. I can't adjust the sight to go any lower and from what I can see there are no aftermarket sights that are any different in height than the factory ones. If I drop to a 110 or 125 grain bullet will the point of impact drop significantly?
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Old March 4, 2000, 08:10 PM   #2
WESHOOT2
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Might.

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Old March 6, 2000, 01:38 AM   #3
Menos
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Yes it will make their point of impact lower, because of two reasons,1. the velocity is faster and the bullet is in the barrel less time during the recoil;2. the recoil effect on the gun is less because of the reduced enertia ratio or Gun to mass.
Plus , I think the Python was designed to shoot 125gr loads anyway.

What style of grips are you using, as slick wood grips that taper too much at the top accelerate the muzzle jump on recoil.
Are you shooting single or double action in these high shots?

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[This message has been edited by Menos (edited March 06, 2000).]

[This message has been edited by Menos (edited March 06, 2000).]
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Old March 6, 2000, 10:55 AM   #4
Ricciardelli
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It's truly amazing what you can wake-up and learn on a foggy Monday morning in northeastern Montana...especially in a reloading forum!

Why, I just found out that the Colt Python was developed around the 125 grain .357 Magnum cartridge! That's amazing! Considering the Colt Python was placed on the market in 1955, and the first commercial 125 grain, .357 load was not available until almost 15 years later...

Then I learned something about point of impact being lower for a high velocity bullet than the point of impact for a lower velocity bullet, because "...the velocity is faster and the bullet is in the barrel less time during the recoil...". This is earth shattering news, and I'm sure that Misters Einstein and Newton (as well as myself) are totally flabbergasted by this discovery! Here, I always thought that the recoil of a firearm was the result of the bullet and gasses venting from the barrel, and recoil wasn't a factor until these products of combustion exited...now I have "learned" that recoil starts the instant the hammer hits the primer...amazing!

Next I "learned" the "...the recoil effect on the gun is less because of the reduced enertia ratio or Gun to mass...".
And here I had been raised with the flawed concept that recoil was ((weight of the bullet in pounds x the velocity of the bullet in feet per second)+(4,700 x the weight of the powder in pounds))squared / (64.348 x the weight of the gun in pounds)! Oh, how have I managed to live such a fruitful life when my poor head was filled with such fallicies?

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Old March 7, 2000, 08:07 PM   #5
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Won't speculate on all the rest, but recoil does indeed start from the time the primer ignites. Technically, I guess, it starts from the time the firing pin starts to move forward. You cannot drive the bullet down the barrel without a counter reaction.
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Old March 7, 2000, 10:54 PM   #6
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Ricciardelli:

Which side of the bed did you get up on this morning? Perhaps the wrong side? Take two saucers of milk and call me in the morning...
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Old March 8, 2000, 02:09 AM   #7
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IN DEFENSE of sir Ricciardelli, I must chime in and say that recoil as a measurable effect upon barrel alignment typically does not begin until the bullet begins to move. Hammer fall and primer popping just don't move things much.

But Mr. R. is right about the Python's designers being amazingly prescient in giving it such a short front sight that most loads heavier than 125-gr strike noticeably above point of aim. Fooeey! Badger--sure a previous owner didn't mess with the front sight and shorten it? Or did someone put those nasty Millet adjustable sights on? You know, the ones that stick way up and deprive you of maybe six clicks from the bottom of the OEM adjustable sights' range of movement?

If you're shooting 158s at .38 Special velocities, consider this: the heavy weight and long barrel time make such loads strike about 3 inches higher at 75 feet, compared to Magnums.
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Old March 8, 2000, 01:01 PM   #8
Ricciardelli
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<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Herodotus:
Won't speculate on all the rest, but recoil does indeed start from the time the primer ignites. Technically, I guess, it starts from the time the firing pin starts to move forward. You cannot drive the bullet down the barrel without a counter reaction.[/quote]

What you are referring to is considered by some as "primary recoil". This is a theory shared by very few.

As for the momentum of the hammer/firing pin moving forward, the energy required to complete this process is "stored" in the compressed state of the spring.

As for the force required to drive the bullet down the barrel, it is different depending on whether you are firing a revolver or a closed breech gun.

The gasses generated by a modern load have a tendency to travel at 17,000 fps. As we know, there is no modern cartridge capable of delivering this velocity. However, the pressure is still there, and it is expanded and expended in all directions. The bullet is forced out of the barrel simply because it is a movable object, and as such allows the gasses to flow in that direction (as long as all other factors are not compromised). On a revolver the cylinder gap releases part of that pressure in a manner which vents equally in all directions axially to the bore of the cylinder. (Look at the front of the cylinder, the throat of the barrel, and especially the bottom of the topstrap (where severe gas-cutting can be seen). On semi-auto and closed breech guns this escape is not possible. Therefore all products of combustion must exit the muzzle. This is where "secondary recoil" comes into play.

If you were to take a barrel which had a bore large enough to contain all the gasses, and the projectile, without venting to the outside world, and in which the bullet would come to rest 1" proximal from the muzzle. There would be no recoil.

Now, chop 2 feet off the end of that barrel. Under the same conditions fire another round. As the projectile and the gasses exit the bore you will notice movement of the barrel. This movement is "recoil".

As I stated previously, "((weight of the bullet in pounds x the velocity of the bullet in feet per second)+(4,700 x the weight of the powder in pounds))squared / (64.348 x the weight of the gun in pounds)".

Until the bullet actually leaves the barrel it "technically" has no weight of its own. Nor does the weight of the powder. The weight of the bullet and powder, as long as it is in the barrel is "technically" part of the weight of the gun. Therefore, with a bullet/powder weight of "0", the formula shows zero recoil. When the bullet and gasses exit the barrel they become separate entities and their values can be inserted into the formula.

For a practical demo of this, get yourself a long piece of pipe, some fuse, and a few tennis balls. Lay the pipe on the ground, place a very small charge of powder in the pipe, and lodge the tennis ball agains the charge. (Make sure the charge is not strong enough to expell the ball from the pipe.) Light the fuse and watch the pipe. If the tennis ball does not exit the pipe, the pipe will not move. Next, increase the charge enough to just barely force the ball to exit the pipe. You may find that the pipe will show some movement. Next increast the charge enough to move the tennis ball at a sufficient velocity from the pipe, and you will definately see movement of the pipe.

As you increase the amount of powder, or increase the number of tennis balls placed over the powder, you will see increased movement of the pipe. If you use too much powder, or place too many balls in the pipe, you will see catastrophic results. The pipe will burst.

This is not intended to be a "flame", but rather as evidence of proof of my earlier post, which some have taken wrongly.

(And I prefer two cups of coffee in the morning, not milk; and since I have a round bed, I'm not really sure "which side" I got out on...grin)
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Old March 8, 2000, 05:16 PM   #9
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I wonder what the ATF would say about someone building a tennis ball cannon?

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Old March 10, 2000, 02:17 AM   #10
Ricciardelli
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The tennis ball canon was fun, but the spud-gun was a hell of a lot of fun!
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Old March 10, 2000, 09:25 AM   #11
WESHOOT2
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Hey RickyDicky,

a slower bullet of the same weight will impact higher. Go figure.......LOL

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Old March 10, 2000, 11:49 AM   #12
Ricciardelli
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<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by WESHOOT2:
Hey RickyDicky,

a slower bullet of the same weight will impact higher. Go figure.......LOL

[/quote]


Good grief, Charley Brown!

That is because your angle of elevation is higher to bring the point of impact with the slower bullet to the same point of impact as the faster bullet...

Therefore your MRT is higher with the slower bullet...


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Old March 10, 2000, 06:31 PM   #13
IK
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Ricciardelli,

Indeed it is truly amazing what you can wake-up and learn on a foggy morning in northeastern Montana...especially in a reloading forum!

I've fought this battle - and I lost! Truth is, the *CENTER OF MASS* won't move which comes to clarify that the gun itself will start moving as soon as the bullet started moving down the barrel - in order to keep the center of mass of (gun + moving bullet) in the same spot. No matter if it is revolver or semi-auto.
If you have some time, check this entertainig link (though extensive):
http://glocktalk.com/docs/gtubb/Foru...ML/001272.html

Got convinced?

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Old March 11, 2000, 05:28 PM   #14
tonyz
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All this for a simple question about shooting 2-3" high. Hmmmmmm.

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Old March 11, 2000, 07:44 PM   #15
WESHOOT2
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RickyDicky,

Read this sentence, "That is because your angle of elevation is higher to bring the point of impact with the slower bullet to the same point of impact as the faster bullet..." and explain yourself.

'Cause what I meant is if you change nothing else but velocity the faster load will impact lower.

You gotta lota esplaneing to do........

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