estimating caliber of flattened ball?

[John Tice]

This is a different kind of question. I wasn’t sure which forum to post it in. Here it goes. If you have a spent round ball that has been flattened, can you estimate what the original caliber was based on its weight? I saw such a spent ball that was excavated by archaeologists. It dates to the 1700s. It is 19.46 grams (.686 ounces or 300 grains). What do you guys think?

 

[rice1817]

Absolutely: I have a book on the archeology done at Fort Necessity back in 1954. They dug down to the actual bottoms of the posts used in the fort to ascertain the true configuration. (Between 1932 and 1954 the reconstructed fort was diamond shaped...they assumed that the remnants of the trenches marked the outlines of the fort. After the dig they rebuilt the fort in its original round configuration.) They found a lot of melted led globs that were spent French Musket balls that imbedded into the palisades. When the French knocked down part of the walls, leaned them against the standing walls, and burnt them, the lead melted and stayed with the ashes for 200 years. Archeologists were able to determine the nominal size of the balls used by the French (.65 caliber) in their 69 caliber muskets. They also determined that there were probably no rifles present on either side.

 

This is a WAG, but it is the best I can do. I would contact the nearest (firearms friendly) university department of history and ask them if they know the answer or can map you to some place that can. Also - try emailng the Smithsonian Museum in DC. They have one heck of a firearms museum and their curators are top notch. Maybe they can help.

 

[JiminTexas]

If you go to the Member Resources forum and open up the caliber/weight article you'll see that a pure lead ball beween .575 and .595 woulde weigh in at 300 grains. That would put your balol at about .580 and with a .015 patch would make a musket of 59 caliber. That is assuming that A. all of the ball is there and none has been corroded away and or lost and B. that it is pure lead.

 

[bwhoffman]

I beleive a .58 rb will go 280gr so if its a rb, i'm thinking 58 to 62 cal, if its a minie or conical, I dont have a clue.

hope this helps some.

Brett

 

[paulvallandigham]

My .600" round balls for my 20 ga. fowler weigh in at 325 grains, more or less, using pure lead. So, I suspect that he is looking at a 58 or 59 cal. round ball. What gun it might have been made for will depend on other information from the site where it was found that can be used to date the events that surround its being fired and left.

 

[RedFeather]

Somewhere I dug up this formula for finding the weight of a roundball of pure lead:

K x DIAMETER = WEIGHT (in grains)

1502.6 x 0.584 = 299 grains

Yep, appears to be about a .58 caliber ball.

 

[Bob Riegl]

:hmm: Scientificaly speaking your assessment of the size of ball by relating the weight to calibre is IMHO---the only correct approach to use from the information provided. Huzzah!!!! :v

 

[mukluk]

I must be missing something here, but by my calculator and cypherin' I come up with 1502.6 X 0.584 = 877.518

 

[paulvallandigham]

What is the " K " in the formula, and where is it obtained? I also get a much larger number, suggesting that something got lost in the transcription, Red Feather. Can you help us out?

 

[AZ-Robert]

I must be missing something here, but by my calculator and cypherin' I come up with 1502.6 X 0.584 = 877.518

Yep, it should be the cube of the diameter times 1502.6 to get the approximate weight of a roundball made of (more or less) pure lead.

.584 x .584 x .584 = 0.1992
0.1992 x 1502.6 = 299

In this case, since we have the weight, we work backwards by dividing the weight by 1502.6 and then getting the cube root of the remainder.

300 / 1502.6 = 0.199654 --> cube root --> .584

...or a bit over 58 caliber. Bear in mind the math gets only an approximation and does NOT account for any lead that might have been lost in the target or lost to erosion over the years, or room for the patch (if used) and the practice of loose fit which (I believe) was in vogue at the time. So, as JimInTexas says, the bore of a gun that left behind a .584 ball was probably closer to .59 or .60 in diameter.

Numbers and formula from Bob Spencer's wonderfully usefull web site:
[url] http://home.insightbb.com/~bspen/math.html[/url]

 

[RedFeather]

I must be missing something here, but by my calculator and cypherin' I come up with 1502.6 X 0.584 = 877.518

Forgot to look at my Excel formula. Yep, the diameter should be cubed. :redface:

 

[Zonie]

Ain't that somethin??
Fer a round ball ye gots to cube it.

Ah wondars iffen ta fin tha wate of a squar blok ye got's ta round it ??
:rotf: :grin:

zonie

 

[Toomuch_36]

My .600" round balls for my 20 ga. fowler weigh in at 325 grains, more or less, using pure lead. So, I suspect that he is looking at a 58 or 59 cal. round ball. What gun it might have been made for will depend on other information from the site where it was found that can be used to date the events that surround its being fired and left.

As this is a spent ball I'm thinking that there would be some loss in weight retention upon impact, pluss corosion over time, kinda puts it at about right for a .60 cal. trade gun or a Fusile to my way of thinkin'.

Toomuch
...........
Shoot Flint

 

[paulvallandigham]

A loss of 5-10 grains of weight might be possible, but losing 25 grains is a bit much to accept. Of course, I would reserve final judgement until I actually saw and inspected the spent bullet, because if some part of the lead was lost, I would be able to see where it broke off. However, these are pure lead balls, and while some tin might be in them, no anitmony would have been in them. Antimony is what makes balls brittle, and has them crack off, rather than mushroom out.

The best we can guess without seeing it is that it is from a gun that would be nominally chambered for either the .58, or possibly a 62 cal. smootobore.

 

[N6398]

Ball weight and bore diameter are related.
See posts under the General Muzzleloading forum; subject, 'caliber as bore gauge'.