kolbe twist formula

  • 781 Views
  • Last Post 23 December 2018
joeb33050 posted this 19 December 2018

I can find the calculator and some written matter, but can't find the formula. Nor can I find any reference to calculating Cg, the center of gravity.

Anybody know the formula?

Anybody know where Cg comes from or goes to?

Attached Files

Order By: Standard | Newest | Votes
Larry Gibson posted this 19 December 2018

From the NOE:

What is "C.G."  and what is it used for?

•   The "Center of Gravity", or Center of Mass, of a bullet is the point at which all the weight of the bullet seems to be concentrated. That is, there is as much mass, or  weight, behind the center of gravity as there is ahead of the center.

•   The relative positions along the bullet axis and distance between the Center of Gravity and Center of Pressure may be used, to evaluate the "Destabilizing Effect" of the Ballistic Wind upon bullet in flight..
The Center of Gravity may be physically determined by balancing a bullet on a knife edge.
The Center of Gravity and the Center of Pressure on a bullet such as a double ended wad-cutter would be at the same point, exactly at the center point of the axis of the bullet. However, if the bullet is composed of different profiles on the nose and base, then the CG and CP would be at different points along the axis of the bullet.




What is "C.P." and what is it used for?

•   The "Center of Pressure", or Profile Centroid, of a bullet is the point at which all the air pressure forces on the bullet seem to be concentrated. That is, there is as much air pressure force on the bullet behind the center of pressure as there are ahead of the center.

•   The relative positions along the bullet axis and distance between the Center of Pressure and Center of Gravity may be used, to evaluate the "Destabilizing Effect" of the Ballistic Wind upon bullet in flight.
The Center of Pressure of a bullet may be physically determined by balancing a a scaled up cardboard cutout of the Profile of the bullet on a knife edge.
The Center of Gravity and the Center of Pressure on a bullet such as a double ended wad-cutter would be at the same point, exactly at the center point of the axis of the bullet. However, if the bullet is composed of different profiles on the nose and base, then the C.G. and C.P. would be at different points along the axis of the bullet.



What is the "Destabilizing Effect"?

When a spin stabilized bullet travels along the trajectory path, the gyroscopic effect of the spin attempts to maintain a bullet in the initial nose-up position which was determined by the required sighting elevation.
This slight nose-up attitude presents the PROFILE of the bullet to the pressure of the Ballistic Wind of the trajectory. The base of an average bullet has a larger profile than the nose which applies more Ballistic Wind pressure to the bullet base and tends to lift the base of the bullet more into alignment with the path of the trajectory. However the downward angle of the trajectory is constantly being increased by gravity so the spin stabilized bullet never quite catches up and is always traveling in a slightly nose-up attitude in relation to the trajectory path.
Depending upon the design of the bullet, the Center of Pressure is usually closer to the nose of the bullet than the Center of Gravity.
A rotating bullet is seldom, if ever, perfectly balanced and therefore gyrates and precesses around the Center of Gravity of the bullet as it travels along the trajectory.
If the Center of Pressure is not at the same point as the Center of Gravity, the pressure of the Ballistic Wind acting upon the Center of Pressure will tend to increase the unbalanced condition of the bullet.
The distance from the Center of Pressure to the Center of Gravity may be considered as a lever to unbalance the bullet. The larger the distance between the two, the longer the lever, resulting in a larger destabilization force.
So, the generally accepted theory is that, the closer the Center of Pressure to the Center of Gravity, the more potential for a stable bullet.
 The "balance point" of a bullet would be the center of mass or center of gravity ( CG )
 The Center of Gravity may be calculated by finding the average of the distances from the bullet base of the Center of Gravity of each of the shanks, bands, grooves, bore rider, nose, etc. that make up the design of that particular bullet.
The Center of Gravity may be physically determined by balancing a bullet on a knife edge.
 The Center of Pressure may be calculated by the determining the balance point of the entire PROFILE of the bullet and can be defined as the average of the distances from the bullet base of the Area Center of each of the various section profiles of the bullet.
The Center of Pressure of a bullet may be physically determined by balancing a a scaled up cardboard cutout of the Bullet Profile of the bullet on a knife edge.
 A suggested procedure to move the center of pressure closer to the center of gravity on an existing design would be to produce that design in a hollow base version with the hollow base cavity volume calculated to move the Center of Gravity forward to coincide with the existing Center of Pressure.

Concealment is not cover.........

Attached Files

  • Liked by
  • M3 Mitch
John Alexander posted this 19 December 2018

Thanks Larry. Good contribution.

John

Attached Files

frnkeore posted this 19 December 2018

JoeB,

The program, calculates the CG, with the input for the nose profile. i.e. nose length, nose meplat and defaults to a tangent nose profile. A secant profile reduces the nose weight, compared to a tangent.

The nose profile, is based on a radius, with a tangent profile, only one radius will fit between the nose length and the meplat. A secant nose, uses a larger radius to join the two surfaces, and that's why you have to input the number of calibers the the radius consists of. A Hornady "Spire point" is a good example of a secant nose profile. Secant profiles are beneficial in super sonic flight.

You can't produce a truncated nose in this program but, you can use a larger secant radius, such a 10 cal and the meplat of the nose, to approximate that type nose.

Also note the history of the program a the 5% accuracy, it has.

Frank

Attached Files

joeb33050 posted this 19 December 2018

lOTS OF ANSWERS, BUT NOT TO THE QUESTION/S I ASKED.

what's the formula?

where does the formula use the Cg?

do the formula use the Cg?

(Frank says so.)

 

Attached Files

frnkeore posted this 19 December 2018

Joe,

Your the match wis, I can only tell you that when you change the weight of the nose, it changes the CG.

If you play with the program, keeping the nose, the same length and changing the profile, you may come up with a relationship, between CG change and twist required.

OR, you could contact them and ask for the formula BUT, they may not want people to know it.

All I can say, is that it has worked on every bullet profile/twist, that I have and is the only one I trust, for accurate results.

Frank

Attached Files

joeb33050 posted this 19 December 2018

Frank;

You claim that the Cg attects twist, while you don't know if the calculator calculates Cg, and you don't know WHAT the calculator uses in the formula, and you don't know if Cg has anything to do with twist. 

Cg slides backward as the nose gets smaller; and slides forward as the nose gets bigger. So what? Any HS geometer knows that. HOW Cg affects twist/stability, if it does  remains a mystery. You are the Cg exponent, explain or apologize.

Attached Files

Ken Campbell Iowa posted this 19 December 2018

i think the bullet cg doesn't have anything directly with twist rate to keep a perfect bullet spinning in "" stability "" ... but 

has to do with any upsetting forces acting on the bullet ....  such as defects ( unbalanced mass )  or unbalanced pressure-forces of any kind ...

the further the unbalanced force is from the center of gravity

the more torque ( force times distance ) it develops to "" unstabilize "" the bullet.

******************

the bullet doesn't know it has a center of gravity ... it is just a crutch for observing engineers to calculate probable effects of various forces.  reminds me that the average family has 2.3 kids ...

you can find cg by hanging objects from a string and making the continuation of the string through the object ... after a few different hangings, the cg is at the crosspoint of the extended string marks.   that is why god invented sharpies .

ken

oh, i have a question:: ... if a perfect bullet exits the muzzle and doesn't encounter any unbalanced forces ... does it need any spin at all ?? ... if so, why ?? ...  could that add to our confusion ??   maybe the spin formula guys are guessing at how much their bullet is out of balance ...  greenhill uses better bullets ?? ....

Attached Files

Larry Gibson posted this 20 December 2018

joe

I don't know the exact formula and I'm sure it's pretty involved.  I use a program which calculates the CG of cast bullets the same as I use several programs that calculate the necessary twist needed for stability.  The program is propriety and I purchased it ergo I don't want to put anything here that is propriety.  The program I use is;

Precision Cast Bullet Evaluation and Design Professional

You might google that as I don't know if they have free version for internet use.

Your question to Frank is answered in the information I posted from the NOE website.  "HOW Cg affects twist/stability, if it does  remains a mystery." It is no mystery but has been known for some time.  The answer to your question is also readily available in most ballistics works regard bullets in flight.  I'll once again recommend you purchase and read Robert A. Rinker's book Understanding Firearm Ballistics.  A more thorough and easily understood explanation is in there.

LMG

Concealment is not cover.........

Attached Files

  • Liked by
  • M3 Mitch
frnkeore posted this 20 December 2018

Frank;

You claim that the Cg attects twist, while you don't know if the calculator calculates Cg, and you don't know WHAT the calculator uses in the formula, and you don't know if Cg has anything to do with twist. 

Cg slides backward as the nose gets smaller; and slides forward as the nose gets bigger. So what? Any HS geometer knows that. HOW Cg affects twist/stability, if it does  remains a mystery. You are the Cg exponent, explain or apologize.

Joe,

First, I have nothing to apologize for. Certainly can't apologize for you not being able to figure out the math! Contact them, ask them what your doing wrong.

But, nothing changes BUT, the Cg, when you extend the nose of a spitzer bullet, leaving the meplat the same diameter.

The only other reason that I can think of, is that the extended forward Cp, helps balance the forces, that want the bullet turn over, if that's true, the Cg has to be closer to the base to work.

That would be like suspending a long object in two places, it makes it easier to control the yaw forces, than suspending it at the Cg.

THAT IS ONLY A GUESS, NOTHING ELSE! I don't even know if it's feasible.

Frank

 

Attached Files

Brodie posted this 20 December 2018

I am certain that the bullet "knows" nothing about it's center of gravity, but if you think that it doesn't affect bullet stability in flight try an extreme case:  Load up some hollow base wad cutters with the base forward in the case, and shoot them at 25 and 50 yards.  The few that hit the target will most likely hit it sideways due to the rearward CG. 

A bullet with a CG forward of its middle will stay point forward as it slows down in flight (see arrows for further explanation) .  A bullet with the CG behind the center will begin to tumble in flight because the ass end of the projectile wants to get ahead of the lighter front.  Brodie

B.E.Brickey

Attached Files

  • Liked by
  • M3 Mitch
joeb33050 posted this 20 December 2018

I am certain that the bullet "knows" nothing about it's center of gravity, but if you think that it doesn't affect bullet stability in flight try an extreme case:  Load up some hollow base wad cutters with the base forward in the case, and shoot them at 25 and 50 yards.  The few that hit the target will most likely hit it sideways due to the rearward CG. 

A bullet with a CG forward of its middle will stay point forward as it slows down in flight (see arrows for further explanation) .

HBWC is the rare example of the Bullet Cg >50% from the base, and I think I've frequently read that HBWC accuracy falls off > 50 yards.

 

  A bullet with the CG behind the center will begin to tumble in flight because the ass end of the projectile wants to get ahead of the lighter front. 

Wrong, Brodie. All pointed or sorta pointed bullet-looking bullets have Cg behind the center = < 50% of L from the base. Every spitzer. Little or no tumbling at normal V.

 

Brodie

Attached Files

  • Liked by
  • M3 Mitch
Larry Gibson posted this 21 December 2018

joe

"HOW Cg affects twist/stability, if it does  remains a mystery."

Apparently you don't want to read up on it so I'll give you a simplified answer.

The bullet tends to rotate/pivot around the CG either coaxially or rotationally.  The rotational spin provides the static and dynamic stability to the bullet which resist the forces applied to the CP.  The farther the CG is from the CP the more leverage the forces (precessions and nutants) have to act on the bullet.  The less rotationally stable the bullet is the greater the effect those forces can have on the bullet. 

Bottom line is; given identical bullets with equal distance from CP to CG and the same amount of pressure on the CP there will be greater effect on a marginally stabilized bullet than on a fully stabilized bullet. 

Given two different profiled bullets with equal stability with one having a longer distance between the CP and CG will be easier upset by the forces applying pressure.

LMG

Concealment is not cover.........

Attached Files

joeb33050 posted this 21 December 2018

Larry is repeating himself, over and over again.

Cg has only one value. A bullet has length, inches. Cg is at a distance from the base, or from the front. Db + Df = L, bullet length.  

So, Cg has only one value, Db. We can express Db as %. If bullet length = 1", and Db = .25", Cg location, Db, 25%.

Our pal Frank has repeatedly said that Kolbe uses Cg in twist calculations. 

"HOW Cg affects twist/stability, if it does  remains a mystery."

If you can tell us the relationship, do so. Tell us the numbers, the magnitude, the delta T/delta Cg location. DON'T copy some more nonsense and post it here; you ain't fooling us, Larry. (It appears that competitor Frank has retired from the field.) As Cg moves ? forward, what is the min twist change?

(This assumes that you know what stability is, and so far I don't think so. You too, Frank.)

 

Attached Files

M3 Mitch posted this 21 December 2018

I think it's a different ball game for a subsonic hollow base wadcutter flying backwards, and a supersonic spitzer bullet.

I have often read that hollow point bullets, when compared to similar bullets without the hollow point, tend to be more accurate.  This is contrary to the idea that having the Cg further behind the Cp (which would be the case here with a hollow point) tends to make the bullet less stable - or at least somehow less accurate.

Attached Files

Larry Gibson posted this 22 December 2018

 

Joe

 

You're correct; I did repeat myself because the answer remains the same.  I imagine Frank has bowed out because he understands everything in the world can not be entered into an excel spreadsheet and thus be defined. 

 

 

For all the formula you need  try;

 

 

 

 

 

http://site.iugaza.edu.ps/malqedra/files/lecture-9.11.pdf

 

 

 

You could have google’d that as easily as I did.  You also could get a book on ballistics and read it.  IApparently you just want to argue ,not learn, so I’m through attempting to explain it to you.  Get the answer yourself because the answer is there if you just bother to understand it.......or just continue as you have.

 

 

 

LMG

Concealment is not cover.........

Attached Files

Ed Harris posted this 22 December 2018

Then is the radius of gyration the distance between Cg and Cp?

73 de KE4SKY In Home Mix We Trust From the Home of Ed's Red in "Almost Heaven" West Virginia

Attached Files

frnkeore posted this 22 December 2018

 

Maybe this will help you figure something out, mathematically.

The top stabilization graph is for the 320366 bullet, as depicted in the drawing of it.

The next one down is the Ron Long bullet that I shorten the base, in the drawing, so that it would have the same Sg as the 320366 bullet both at 1500 fps.

The graph curve, is quite a bit different but, they both have similar low Sg's.

The Sg moves back 4.45% and the Cp moves back 3.39% on the 320366. This is why that I think that the Cg correlates to the twist required for stabilization.

Frank

Attached Files

joeb33050 posted this 22 December 2018

Then is the radius of gyration the distance between Cg and Cp?

 

No. https://www.bing.com/search?q=radius+of+gyration&form=EDGTCT&qs=PF&cvid=a9e78e646b4841af92ec0345546797e4&refig=ce75cfe25fab4349d275e47fdbe10eb4&cc=US&setlang=en-US&PC=LCTS

 

Attached Files

750k2 posted this 23 December 2018

Can you explain to me how to understand this graph?

This is one from one of my inputs.

Attached Files

joeb33050 posted this 23 December 2018

Sure. Start on the right. The minimum twist/smallest number is about 19" per revolution, a 19" twist, from 4000 fps to 1500 fps, speed is decreasing. As speed goes left, reduces, from 1500 fps to about 1100  fps, the minimum twist goes up to about 20", gets slower. Then a sudden drop at about 1100 fps, about the speed of sound, as speed falls a bit. Just under the speed of sound the minimum twist required falls to/gets faster about 16".

Start on the left. The minimum twist .required is 1 turn in 16" from 0 to speed of sound. At speed of sound the minimum twist leaps up/slows down to 1 turn in 19", and stays at 19" to 4000 fps.

The left hand 16" twist is faster than the right hand 19" twist; maybe slower pistol barrels at about 16", faster rifle barrels around 19" twist.

?

 

Attached Files

Close