Bearing Surface Area Of A Ball

[rodwha]

A discussion about the driving band of a ball in a standard Pietta's revolver chamber and the usefulness of using an oversized ball had me look into the cubic area that's created and how the differ. As the numbers were a bit over my head I went to a math forum and was given the formula, which is:

"Using Pythagoras, each flat spot's area is, for example, (.451 squared−.446 squared)pi/4 square inches."

My calculator didn't have enough characters for a complete pi number and so I went with what it allowed, which should have been rounded but being so far down the line I went with it.

This gives an area of 0.0035295" for a .451" ball,
an area of 0.0056548" for a .454" ball, and
an area of 0.0078013" for a .457" ball.

Interesting to note how much more surface area you get my increasing the ball's diameter such a small amount. That's more than a 50% increase with a .454" ball, and more than 100% with the .457" ball.

The greater friction caused by the increased area increases the pressures which in turn increase the velocity. This has been seen by a few people. I'm not sure that it actually increases accuracy though.

I'd certainly like to hear from those who have choreographed their comparisons as well as any who have claimed an increase in accuracy.

 

[galamb]

That may be the reason that Green Mountain has always suggested you shoot a "bore sized" ball for target shooting.

Never been too particular myself about "tiny" accuracy increases, but for those that are, might be something to have a look at...

 

[Zonie]

Not to be picky but I think your formula forgot to take the square root of something.

The way I calculate the width of the flattened surface on the ball uses a little different formula.

Flat width = (Ball diameter X Ball diameter)- (Bore diameter X Bore diameter) = a number I'll call J.

Now, I take the square root of J to give me the width of the flat that's formed when the ball is shoved thru the barrel.

For your .451 ball and a .446 bore (and the symbol ^2 meaning "squared") that would be:

.451^2 - .446^2 = J = .00449

The square root of .00449 = .06701 wide

Lets assume this is in a smoothbore so we don't have to mess with the effects of the rifling grooves. :grin:

OK. The bore diameter was .446 diameter so the circumference (distance around) the bore is .446 X Pi. Pi is 3.1416... so 3.1416 X .446 = 1.40115.

The width of the flat times the distance around the bore once gives us the area of the surface contacting the bore so I get 1.40115 X .06701 = .09389 square inches.

For your .457 diameter ball I get:

Sq root of (.457^2 - .446^2) = J = .09966

.09966 times the circumference of the bore = .09966 X 1.40115 = .13964

I hope you didn't mind that I noticed the missing square root and that got me going on all of this.

 

[jamesthomas]

You math guys :surrender: :bow: . Couldn't get past Trig. II.

 

[Zonie]

It's not that I like math. I just had to use it a lot in my previous job.

Oh, and I love to see the blank expressions on the faces of all you guys who thought adding and subtracting was more math than you ever wanted to deal with. :grin:

 

[necchi]

I've learned not to mention Math on this forum,, and have been content to just sit back and watch.

 

[mooman76]

It's not that I like math. I just had to use it a lot in my previous job.

Oh, and I love to see the blank expressions on the faces of all you guys who thought adding and subtracting was more math than you ever wanted to deal with. :grin:

It's enough for me. :surrender:

 

[Britsmoothy]

I love how maths can explain nearly everything but I too struggle with numbers.

Interesting post, but has anyone considered obduration? :haha:

B.

 

[Wes/Tex]

Yeah...once they got to dividing the cosine by the cube root of 13 leap year tangents, I was freakin' lost! :shocked2: :haha:

 

[Alden]

I was questioning the first calc when I chuckled about pi not fitting in his particular calculator -- it is an infinite number!

I suppose we are ignoring that balls twist and turn and swage and that the bearing surface may not be a constant (same for resistance).

I haven't studied it but Zonie seems to have found an issue which may or may not itself have to be reversed later in the calc as they often are in distilled actual results.

In any case, it was in fact those missing rifling grooves that most bothered me off the bat -- I think those are critical and the answers are probably significantly incorrect (although likely relative with a high level of confidence). Straight "rifling" is STILL used to reduce friction. Two examples just from Trap shooting include the a) lines on the outside of shotgun hulls and b) straight-rifled barrels (one Hastings of which I use on a Remington 870 Wingmaster Trap).

Friction, pressure, related velocity? When do we get to the bearing surface of the powder charge and magnitudes more important efficiency of the system with a larger surface of the bigger projectiles!?

 

[coloradoclyde]

I love how maths can explain nearly everything but I too struggle with numbers.

Interesting post, but has anyone considered obduration? :haha:

B.

Clever Brit,
The first time I read it I thought you meant obturation. :rotf: :rotf: :rotf:

 

[rodwha]

"I hope you didn't mind that I noticed the missing square root and that got me going on all of this."

Not at all. I quest after the truth of the matter, which is why I went to a math forum as that was a bit over my head. I assumed he knew what he was feeding me as he is a moderator there.

The numbers seemed a bit exaggerated to me, but I did figure that a sphere increases on both "sides" and that cutting it back a little would certainly increase the area to a fair extent, especially when the smaller of them (.451") would be shaving a fairly small amount. So I accepted it.

 

[Rifleman1776]

Who's on first???

 

[bpd303]

Excuse me while I go take an aspirin more commonly known as acetylsalicylic acid. :bow: :idunno: :surrender:

 

[necchi]

commonly known as acetylsalicylic acid.

:haha:
Now that, I understand.
But math being used to describe something? :td:

 

[Billnpatti]

Every good question disserves a good answer. If the question involves ballistics in any way, a good answer will, of necessity, involve math. When one gives an answer to a ballistics question and does not back it with the appropriate math, it is then just a :bull: opinion.

 

[Auldgoat]

Yeah...once they got to dividing the cosine by the cube root of 13 leap year tangents, I was freakin' lost! :shocked2: :haha:

Thas because ya forgots to multiply it by the Easter Bunny while whistling "Dixie". :hmm:

 

[rodwha]

This is the correction:


"His formula is correct, so I changed my previous answer.
As pi∗0.446=1.40115... and J = 0.004485, the resulting areas for ball diameters of 0.451", 0.454" and 0.457"
are respectively 0.0938 square inches, 0.1189 square inches and 0.1396 square inches (rounded to 4 decimal places).
Hence the percentage increases you wanted are 26.7% and 48.8% approximately."

 

[Archer 756]

:surrender: Before shooting that deer the deer must calculate how he is going to die, really who cares :rotf:

 

[Zonie]

I suppose someone who had an old original C&B revolver with a badly worn barrel might care.

That 26.7 or 48.8 increase in the area of the contact area would allow the ball to get a lot better grip on the worn rifling grooves.
That could turn a pistol that shoots shotgun sized groups to shoot small, very respectable groups.