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If Elmer didn't say it, it probably ain't true.

Also mentioned is that fluting makes the barrel more rigid. I always thought of the barrel bending along the axis of the bore when it was shot. I wonder if the spiral flutes perform both tasks, keeping the barrel rigid along its axis and keeping the barrel from twisting?

I also wonder how much the barrel actually twists. Is there a measurable difference? Does anyone know if the bench-rest shooters use spiral fluting?

[This message has been edited by Ben_Wazzu (edited 04-18-2002).]

The only way the flutes provide a stiffer barrel is if you consider it on a pound for pound basis. In that scenario you will get a larger area moment of inertia with a fluted barrel than a same weight nonfluted barrel. Do you follow me.

For example. If I take two contour #4 barrels, one fluted the other is standard. The fluted barrel will be less stiff. If however the fluting is such that the fluted contour #4 barrel weighs the same as a standard contour #3 then the fluted barrel (contour #4) will be stiffer than the standard #3 contour.

If you utilize an helix angle (twist) in the fluting which is of an opposite hand to the rifling twist you will in fact increase the torsional stiffness of the barrel. The interesting part is that the majority of the bullet upsetting will occur within inches of the chamber. There want be much torquing going on near the muzzle. So the twisted fluting is most likely of little real advantage.

Todd E

[This message has been edited by Customstox (edited 04-18-2002).]

quote:

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Originally posted by Bill Leeper:
On the bright side your ER Shaw barrel wouldn't be any worse after fluting than it was before!

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It'll damn sure be less stiff!

Wally

If a "truely fluted" barrel is being cut into a piece, the cross section is much like a star. Most "fluted" barrels out there start out as a round barrel with only shallow grooves cut into it, Such "flutings" do not stand out as much as one that is "really fluted", nor is as effective, but it makes up a little bit of strength for the metal being cut.

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never to forgive. never to forget. never to give up.

BA

quote:

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Originally posted by Pyrotek:
Hey, remember an I-beam is more stiff than an equal weight, equal length round bar.

If a "truely fluted" barrel is being cut into a piece, the cross section is much like a star. Most "fluted" barrels out there start out as a round barrel with only shallow grooves cut into it, Such "flutings" do not stand out as much as one that is "really fluted", nor is as effective, but it makes up a little bit of strength for the metal being cut.

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Do vector analysis on an "I" beam then on a cylinder with several flutes cut right out of the meat.

Sure they look butch!

Wally

I didn't mean to offend you. I was only getting a little more specific. I agree with you 100% as I my last post says.

I don't know where the I beam discussion came from. Anyone ever see a barrel with a I beam section?

Todd E

Remember the inertia of the rifle must be overcome before you would begin to feel it. I have very lightly held on to the forearm of my .500 and it has laid over many times. I see this is why.

Since this force is relatively insignificant the spiral fluting is a maketing gimmick.

[This message has been edited by Roger Rothschild (edited 04-19-2002).]

This torsional excitation occurs so quickly that you cannot feel it. The barrel however does and reacts by vibrating (torsionally due to this input). The helical fluting, if designed correctly, can help to stiffen the barrel torsionally and reduce the effect of the vibrations.

Again the real benefit of fluting is to allow a greater section for the same weight. It is essentially a weight savings feature and does increase the available surface area for convective and radiate heat transfer. If you wanted a heavier contour but not the weight the fluting is the compromise which is struck.

Todd E

[This message has been edited by Todd E (edited 04-19-2002).]

I love you too Todd.

"The torsional load imparted to the barrel by the bullet getting upset"

What? "The bullet getting upset" from the following, don't you mean the bullet getting squashed?

"If you don't believe me try bushing that bullet through your bore!"

Do you mean "pushing"

"This torsional excitation"

Man you come up with the wildest concepts it's refreshing actually.

Ok: Sit down and let Dr. Rothschild show you how it's done.

Torque imparted to a barrel of a rifle.
I = mR^2 for a solid cylinder (the bullet)
Barrel length: 0.5207m
Muzzle velocity: 731.52m/s
Spin: 1:0.254m

Angular acceleration = (angular velocity final - angular velocity initial/ time)

Ang. accel.=(18095.57 Rads/s - 0)/6.872x10^-4sec.

Ang. accel. = 2.6332x10^7 Rads/sec

Sum of torque forces = Moment x ang. accel.
T= (1/2(mR^2))2.6332x10^7 Rads/sec
T= 21.52 Newton meters/s^2

1 Joule = 1 Newton x meter

1 Joule = 0.737 ft/lb

So: T = 21.52 Nxm/s^2(0.737) = 15.8 ft/lbs.

Thanks for bringing this to my attention Todd.

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[This message has been edited by Roger Rothschild (edited 04-19-2002).]

In addition, spiral flutes will decrease the axial stiffness. In fact it is mainly just the top and bottom "ribs" that account for the higher stiffness/weight ratio.

Todd. As to ever seeing a barrel with an I-Beam section. No, I haven't. BUT one that DID have such a section would have the highest stiffness/weight ratio. But it would also have the highest possible distortion due to heating as well. The higher the temperature gradient throughout the cross section the higher the probability that POI will "wander". Stiffness is great, but it's consistency that accuracy buffs are looking for. Any non symetrical machining done to a barrel runs the risk of creating or releasing internal stresses resulting in a "less than straight" barrel.

Long story short. I'll take mine round.

quote:

---

Originally posted by Roger Rothschild:
"You are in over your head."

I love you too Todd.

"The torsional load imparted to the barrel by the bullet getting upset"

What? "The bullet getting upset" from the following, don't you mean the bullet getting squashed?

"If you don't believe me try bushing that bullet through your bore!"

Do you mean "pushing"

"This torsional excitation"

Man you come up with the wildest concepts it's refreshing actually.

Ok: Sit down and let Dr. Rothschild show you how it's done.

Torque imparted to a barrel of a rifle.
I = mR^2 for a solid cylinder (the bullet)
Barrel length: 0.5207m
Muzzle velocity: 731.52m/s
Spin: 1:0.254m

Angular acceleration = (angular velocity final - angular velocity initial/ time)

Ang. accel.=(18095.57 Rads/s - 0)/6.872x10^-4sec.

Ang. accel. = 2.6332x10^7 Rads/sec

Sum of torque forces = Moment x ang. accel.
T= (1/2(mR^2))2.6332x10^7 Rads/sec
T= 21.52 Newton meters/s^2

1 Joule = 1 Newton x meter

1 Joule = 0.737 ft/lb

So: T = 21.52 Nxm/s^2(0.737) = 15.8 ft/lbs.

Thanks for bringing this to my attention Todd.

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ergo, rifling twist is a major factor in computing force.

We can carve, helically if you wish, to our hearts content. Very cute.

We could decide the characteristics we need in a barrel then design a trust barrel that would shift tension and compression where we want. Might not be pretty and might be difficult to chuck for threading and chambering.

Wally

Chic

Also a tube with only small hole will be about the same ( or stronger ? ) to bend than a steel rod of same outside diameter.

??

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As far as the other following posts, believe what you will. It is a fallacy and Shaw could not prove diddly in this issue and it is easily proved to be just the opposite.

In fact just consider this. I am trying to do it with common sense instead of formulae. Torgue will be induced by the bullet pressing, if you will, agains the edge of the rifling. That will take place on the side of the left side of the rifling in a right hand twist because that is the side that will be resisting movement.

That torque will be moving down the barrel with the bullet and the resistanc of the barrel will be more concentrated at the point of the bullet in time but the adjacent volume of barrel will be resisting it also but to a lesser degree as the distance from the bullet increases.

I think everyone agrees that a fluted barrel will have less resistance to this torque than one that is left unchanged in its round configuration. If you do not agree, then the nurses in the home need to put another bib on you to catch the drool and we will not worry about your confidence in this discussion.

The arguement is that helically fluted barrels will resist that torgue more than straight flutes. And I am assuming that all are in agreement with the concept that removing the steel at the outside of the barrel is detrimental to resisting torgue, hence the round barrels are stronger. Now consider how much material is removed in those flutes with straight flutes versus helical flutes. It is obvious that the helical flutes remove more material since the path is longer going around the barrel to reach the end versus straight. You have reduced the material even further that resists torgue. You are on a downward spiral trying to fight torque with this approach.

Hope that did not confuse the issue.

Also, assuming there is merit to this, wouldn't the flutes have to go in the SAME direction as the rifling to produce the greatest effect? - John

A barrel is a cantilevered beam and it is also a spring. The forces inputed to the barrel by the bullet and the expanding gases produce an excitation (a forcing function) upon the barrel.

So Roger you need to do the following to understand the solution to this problem:
1.) Determine the forces generated by the extruding of the bullet and resisted by the barrel.
2.) Determine the function which describes the dynamic force relationship from item 1 above.
3.) Determine the Eigenvalue of the barrel.
4.) Solve the differential equation. It should have at least 4 degrees of freedom.
5.) Then report out on what system is the best from a stiffness to weight ratio perspective.

Vibe,

An I beam is not stiff in torsion. It is an engineered section which provides maximum bending strength to weight ratio. We are not discussing bending, we are discussing torsional excitation and stiffness.

I am just a stupid LEO not a brilliant engineer like you Vibe or a Chemistry Major like Roger.

Todd E

[This message has been edited by Todd E (edited 04-20-2002).]

A "torque" is the sum of the rotational forces acting on something, whether it is your Crescent wrench adjusting your kids bicycle axle or spinning up a bullet, accelerating it rotationally (angular rotation) to a given angular velocity.

Now to find the "force" necessary to accelerate that bullet does two things. It finds us the "force" necessary to accelerate it and in so doing due to Newtons Second Law, we have the "force" on the barrel since it is the "thing" imparting the spin through the rifling. That "force" is the "torque". It is just giving a name to the force.

As for squashing the bullet I don't know what it takes to mash it into the bore, I really don't care. The lands have to engrave the bullet and that metal has to go somewhere. I was simply looking at just the average torque in an ideal system. Since I am taking the algebra based physics rather than the calculus based this is what I can get. I had asked my professor about integrating the acceleration over the time in the barrel but it doesn't tell me anything more than what I wanted to know. And that is 15.8 ft/lbs isn't going to concern me but does tell me why my .500 lays over when I do not grip it well.

Don't take things so personally Todd, it isn't like someone is calling into question your heritage though sometimes it seems as if you perceive that to be the case.
Have fun, I hear Cleveland has a wonderful symphony.

The torque applied to the barrel by the bullet after the bullet has been extruded is simply:

mass X acceleration X radius of the bullet.

The thrust imparted to the barrel by the bullets reaction against the riflings is simply:

mass X acceleration X cos helix angle. (longitudal)

mass X acceleration X sin helix angle.
(lateral)

This is all high school level physics.

To determine the force (thrust and torsional) due to the bullet extrusion you simply need to understand the total deformation of the bullet, the yield strengths of the bullet materials, and you need to make some assumptions about stress and strain at this point unless you have a stress strain plot for the particular bullet.

From all of this you can begin to derive a forcing function.

Anyway, I think this has gone on long enought to. I have not checked your work, but 15 lb-ft of torque is a significant quantity. You can only hand tighten to about 1.5 - 2 lb-ft.

I am not sure of the precise wave form generated in a barrel due to firing, but I would expect it to be manifest itself as a "circular whipping" of the muzzle.

Again the only advantage to be gain by fluting is to reduce weight and provide additional stiffness at that lower weight as opposed to a solid round barrel. The helical application of fluting my provide additional stiffness, but I cannot be certain without modelling it.

I don't get to the symphony much.

Todd E

[This message has been edited by Todd E (edited 04-20-2002).]

quote:

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Originally posted by Todd E:
I have not checked your work, but 15 lb-ft of torque is a significant quantity. You can only hand tighten to about 1.5 - 2 lb-ft.

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Girly-man!
I only weight lifting I do is 16 oz. curls (beer), but to test my hand tightening ability I grabbed a 1" diameter steel bar and a strap and a box of .40 cal bullets. The box weighed 21 lbs on my bathroom scale. Wrapping the strap around the box and the bar I can quite easily and rapidly turn the bar and raise the weight. 21 x 0.5 = 10.5 lb-ft. I would venture that I could double that, but time to go back to my 16 oz. curls.

[This message has been edited by pshooter (edited 04-20-2002).]

YOU are the girly man. You are only appling 0.875 foot pounds of torque. Even if you doubled it you would only be doing 1.75 foot pounds. Remember you are also using both hands to get to that measely 0.875 lb-ft of torque.

I suggest you take a good torque wrench grasp the socket with one hand and the handle of the wrench with the other and twist the socket has hard as you can. Watch your torque reading. If you are a real man you might get 3 foot pounds, but more likely you will achieve 1.5 to 2. I can get over 2 barely, but I am getting old and weak.

To put this in perspective. 15 foot pounds is approximately the torque on the cover plate bolts in the axles of your pick up trucks. Go out and see if you can tighten then up anymore by hand. Hell try loosening them by hand. Torque to loosen will most likely only be about 10-11 foot pounds.

Todd E

[This message has been edited by Todd E (edited 04-20-2002).]

quote:

---

Originally posted by Todd E:
dsverdrup,

YOU are the girly man. You are only appling 0.875 foot pounds of torque.

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Doh! You are right!
Note to self: never drink and calculate.
My math would have been inch-lbs. That is with one hand applying torque at a time so it is still a valid measuring. 'Just need to keep adding weight to see what my peak force is.

quote:

---

Originally posted by Roger Rothschild:
"You are in over your head."

I love you too Todd.

"The torsional load imparted to the barrel by the bullet getting upset"

What? "The bullet getting upset" from the following, don't you mean the bullet getting squashed?

"If you don't believe me try bushing that bullet through your bore!"

Do you mean "pushing"

"This torsional excitation"

Man you come up with the wildest concepts it's refreshing actually.

Ok: Sit down and let Dr. Rothschild show you how it's done.

Torque imparted to a barrel of a rifle.
I = mR^2 for a solid cylinder (the bullet)
Barrel length: 0.5207m
Muzzle velocity: 731.52m/s
Spin: 1:0.254m

Angular acceleration = (angular velocity final - angular velocity initial/ time)

Ang. accel.=(18095.57 Rads/s - 0)/6.872x10^-4sec.

Ang. accel. = 2.6332x10^7 Rads/sec

Sum of torque forces = Moment x ang. accel.
T= (1/2(mR^2))2.6332x10^7 Rads/sec
T= 21.52 Newton meters/s^2

1 Joule = 1 Newton x meter

1 Joule = 0.737 ft/lb

So: T = 21.52 Nxm/s^2(0.737) = 15.8 ft/lbs.

Thanks for bringing this to my attention Todd.

---

You win this argument, simply cause I don't have any good evidence to dispute it. Heck, I don't even understand it!

I was lost by about the second line!

Heck, just give me a good non-fluted barrel, a bunch of bullets, and I'm happy!

chic

The debate goes on for that those who disbelieve, simply disbelieve; those who believe, can only prove it in math and theory.

An experiment can be done to visually show the effectiveness of fluting. Barrels, made of homogeneous material, can be substituted with plastic. Here we got plastic barrels, rifled, fluted and un-fluted.

Now you jam a lead slug into the bore or simply twist these barrels from outside.
When these barrels showed their different ability to resist twisting, everything will become clear.

Until then, I stick to whichever barrel that is accurate.

That is my kinda torsional excitation! The term torsional excitation is used most often with engine crankshafts. There is a torsional damper on your crank hub. You crank pulley (sheaves) are bolted to it.

The flutes are there simply for mass savings. If the mass savings is significant enough the may afford one some stiffness improvement with no weight penalty. A better way to control vibration is with mass or better yet a tuneable damper.

Todd E

Chill dude. I was trying to highlight the point that helical fluting sacrifices logitudinal stiffness for the illusion of increased tortional stiffness. And I only claimed to be AN Engineer, not a briliant one. I am still human and thus quite capable of making mistakes (like calling people I don't know "stupid", but wait, I DIDN'T do that, you did in an attempt to put word in MY mouth) and overlooking other pertenant factors. I do not think that the torque generated during the bullets engraving is all that relevant to the discussion. I say this for several reasons.

One - because that section of barrel just ahead of the chamber is rarely included in the fluting. And imparts NO torque on the various versions of fluted barrels whose fluting would be in front of this section.

Two - the pressure/acceleration curve of the burning powder/bullet acceleration is still quite far from developing. Often it is just the pop of the primer that will push a bullet into and sometimes past this point.

Three - Due to #2 the rotational acceleration at this point is exceedingly low. The 15.8Ft-Lb figure is derived from accelerating the bullet to full RPM uniformly, and is assumed to be an average value. Since the acceleration of the bullet in not linear due to the pressure driving it also being non linear, then the rotational acceleration will also be non-linear. Thus allowing us to discount the torque generated this close to the chamber.

Particularly with larger bores and slow burning powders. Also since the 16 ftlb figure was arrived at using a 50 caliber, then yes I do see it as rather inconsequential in the larger scope of things. Particularly in the smaller calibers where this debate is the hottest.

"The flutes are there simply for mass savings. If the mass savings is significant enough the may afford one some stiffness improvement with no weight penalty. A better way to control vibration is with mass or better yet a tuneable damper."

I would agree with this completely.