4 officers, and tons of iron.

[Cannoneer]

I saved this photo some time ago, and there was next to no information given, so I've been guessing at the location ever since. I assumed the time period was the Civil War era, but I wasn't even warm on the location. The pic had to be taken post war, because photographs weren't allowed during the war. I gleaned this info, and the identity of the building the other day when I ran across this PDF.



The Rock 

[subdjoe]

I looked at the photo and thought " that looks an awful lot like Mt. Tamalpias as seen from SF. Bay, between SF and Sausolito."  Given that, and the link to The Rock,  A History of Alcatraz, it must be taken on Alcatraz Island.  I would say towards the east end of the island.  Can't come close to the year though.  Well, between 1860 and I would guess 1875.

Here is another history of it http://books.google.com/books?id=1duAUPaHSgcC&pg=PA14&lpg=PA14&dq=alcatraz+island+civil+war+map&source=bl&ots=VogSf_lAco&sig=-ynNTY0Kf8tbj0wBQfMKnoZc8Rw&hl=en&ei=MIewSt7RNZSsMeLmzPIN&sa=X&oi=book_result&ct=result&resnum=1#v=onepage&q=alcatraz%20island%20civil%20war%20map&f=false

[cannonmn]

Thanks Boom, that pic is close to my view of what Heaven must be like, all the cannons and ammo you'd ever want, and a place where you can shoot them. 

I looked thru the full report, nearly 600 pages, and saw where a large number of Rodman guns proved defective-never had heard that berfore, so I learn something every day.  If I get time I may post a link to that report on the CMH board, there are probably two or three other people there interested in ordnance, and a few interested in forts. If your photo link is fairly permanent, with yr permission, I'll borrow it for that board, OK?

It was also interestiing how much thought and correspondence went into designing and building a custom-made fort in a somewhat atypical location.  A lot of it had to do with the unavailability of good bricks on the West Coast back then.

[cannonmn]

OK who knows how many shot are in the shot pyramid closest to the officers?  Any advanced artillery student should know the formulae, simple math.  Anyone?

I like they way they have a fender post next to that pyramid.  I guess they thought of what would happen if their shot-delivery wagon backed up a little too far and poked the pile...

[little seacoast]

Give the slant on that post, they probably were right to be a'feared.

[guardsgunner]

The photo just shows 4 men in standard heavy artillery uniform. Officers would not have cartridge boxes. Must be coming off or going to Dress parade as they have thier white gloves.

[cannonmn]

Here's one of those forbidden photos:

[BoomLover]

Does 2724 sound about right, cannon balls per pile? BoomLover

[Cannoneer]

If your photo link is fairly permanent, with yr permission, I'll borrow it for that board, OK?

Permission granted, Sir. If you post this on the CMH forum, would you ask about the soldiers uniforms? I can't think of a better place to get a definitive answer on that subject.

I also found out a lot of things about Alcatraz I hadn't known before; interesting piece.

[cannonmn]

Does 2724 sound about right, cannon balls per pile? BoomLover

I got 2434 by taking out one pyramid square at bottom (12 x 12) which contains 650 balls, then adding to it 23 diagonal layers of 78 balls each (another 1784 balls.)

Need a third party to see if either answer is right.??

[Ex 49'er]

I must be doing something wrong. My math only comes out to 1794 balls. Triple checked it!

[cannonmn]

My math only comes out to 1794 balls.

Like teacher always said "show your work."  Then we can see if you are right.

Or mebbe yr onna them danged jokesters who won't use the insipid yellow smilin' faces (I'm one too.)

[sseib]

24 x 1 = 24
25 x 2 = 50
26 x 3 = 78
27 x 4 = 108
28 x 5 = 140
29 x 6 = 174
30 x 7 = 210
31 x 8 = 248
32 x 9 = 288
33 x 10 = 330
34 x 11 = 374
35 x 12 = 420

the sum of the rows is 2,444 unless I miss counted the number in the top row, the number of rows or just messed up the math.

Steve Seib

[Double D]

Each row has 2 more than the one above. One on each end.

So one row of 24, two rows of 26, three rows of 28

[BoomLover]

Ok, I did it wrong, got some new figures...top row = 24, next row, add 2/row/layer = 52. Next layer, again add 2/row/layer = 84. Next layer, add 2/row/layer = 120. ext, till last row equals 45/row/layer = 540. When I added up all 12 layers, I came up with a new #, 3004. Next question, how much does each cannonball weigh? Is it enough to put the island in danger of sinking? BoomLover

[seacoastartillery]

     I get 2,444 by a different method.

Solve for 'n', the no. of cannon balls in the 4 equal-sided pyramid nearest the officers:  

Given:
k=no. of cannon balls on one side of pyramid base.

        k=12

Formula:     n=1/6k(1+k)(1+2k)
          
                 n=(13)(25)
                            6

                 n=650

Then count the remaining balls along the ridgeline of the rectangular pyramid which is 23.  Determine no. of balls in each of the 23 slanted layers which is 78 and multiply 78 by 23 which is 1,794 and finally add the no. of balls in the 4 equal-sided pyramid, 650 to 1,794 and you get the total number of cannon balls in the stack on the 'Rock' which is 2,444.

That method I learned at Waterveliet Arsenal across the Hudson from Troy, NY.

Regards,

Tracy
                  

[BoomLover]

Plus 12 to my last answer, = 3016... sorry if I lead anyone astray, I made the last layer 45/row instead of 46...thanks Double D, for making me check my math! BooLover

[BoomLover]

Tracy, afer doing a re-count, I have to say there is a great discrepancy between our two totals. Better not put me in charge of counting the chickens! Assuming Double D had the right idea, that was what I also came up with independantly. However, using higher math, such as your method which you were taught at Waterveliet, you would know when to call for more ammo at the right time! BoomLover

[Skunk]

Awesome pictures Boom J. & Cannonmn. I'm not really into cannons and mortars all that much, but I sure do appreciate good old pictures. Thanks for sharing them. If I enjoy these pictures, you big time cannon lovers must get goosebumps when seeing these old photos.

I like they way they have a fender post next to that pyramid. I guess they thought of what would happen if their shot-delivery wagon backed up a little too far and poked the pile...

Was thinking something similar when first viewing the photo. It appears that it would be silly easy to get that whole pyramid rolling.

[sseib]

BoomLover,

I think your adding to many per layer. In a pyramid the balls stack because of the offset. Top layer = 1, second layer is 2 x 2 = 4, third layer is 3 x 3 = 9, etc... if the row length is even, the next layer down must be odd or they will not stack, same for the width. For each layer down it gets one ball longer and one row wider.

Tracy,

I knew there was a formula but for a pyramid but couldn't remember from high-school.

Steve Seib

[Terry C.]

Each row has 2 more than the one above. One on each end.

So one row of 24, two rows of 26, three rows of 28

Each row only has one more than the one above it. The balls do not sit directly on top of each other, each ball is centered between the balls beneath it.

[Terry C.]

I think this will show you better than I can explain it.

This is an excerpt from a drawing I made when I was designing a ball rack to hold 100 balls. This view is the stack seen from directly overhead.

(8 x 5) + (7 x 4) + (6 x 3) + (5 x 2) + (4 x 1) = 100

[Double D]

now you know why I wasn't a math teacher or engineer...well they do have TV show that I would have starred in---Engineering disasters!

[Terry C.]

Don't feel bad, I usually figure things out by trial and error.

Many trials and lots'a errors.


Oh, and I came up with 2,444.

[Max Caliber]

I get 2444 if the top row is 24 balls long. Page 40 0f the Ordnance Manual of 1862 gives the formula for  calculating the number of balls in a pile and tables of the number of balls in a pile.

[GGaskill]

I looked thru the full report, nearly 600 pages, and saw where a large number of Rodman guns proved defective-never had heard that before, so I learn something every day.

I quit reading at 3 AM this morning so I didn't read the whole thing but it seemed that the guns that were defective weren't Rodmans but the earlier Columbiads made to Bomford's design from deficient material.  And apparently part of the "defect" was the guns would burst when fired with shot (versus the shell which they were designed for.)

[Artilleryman]

There are 2444 balls in the pile if there are 35 balls on the long side of the bottom row according to a table in the 1863 Ordnance Manual.  I would like to see what the man in the back of the pile is standing on.

[seacoastartillery]

Tracy, after doing a re-count, I have to say there is a great discrepancy between our two totals. Better not put me in charge of counting the chickens! Assuming Double D had the right idea, that was what I also came up with independantly. However, using higher math, such as your method which you were taught at Waterveliet, you would know when to call for more ammo at the right time! BoomLover
[/quote]

     After looking closely at your excellent mortar that you built, we would not hesitate to put you in charge of counting the chickens.  The execution of the machine, welding and hand-work proves you are a skilled craftsman, and besides, as a kid, I watched my grandfather and grandmother count the chickens on their farm in Connecticut, and not once did I see the chickens arrange themselves into a rectangular pyramid before the count.  They piled themselves up in that fashion in an attempt to confuse my grandfather, only if they saw him approaching the coup carrying his hatchet!!   

     Can anybody see those cannon well enough to know if they are Pattern 1844, 8" or 10" Columbiads, the first to be called Columbiads even though they had chambers and were designed to throw only shells?  The pic is labeled "1864 South Caponiere".

Regards,

Tracy and Mike

[Cannoneer]

Sseib was the first with the correct answer, he's the winner, and he got it solely through empirical observation. I wonder what kind of prize he's won? Oh now I remember: Great job Steve!
Welcome to the forum.

Another more convoluted way to get a total once you have the count of the top row, (24) and the tier beneath it, (50, as has been said add 1 to 24, and multiply by 2) is to keep adding 2 to the differance between the two figures, and add that number to the total of the preceding tier: 50 - 24 = 26, 26 + 2 = 28, 28 + 50 = 78, 30 + 78 = 108, 32 + 108 = 140, 34 + 140 = 174 etc.

[GGaskill]

Comparing them by eyeball with the following images from cwartillery.org, I would say they are 8".  The chase seems too small to be a 10".


8" M1844 Columbiad


10" M1844 Columbiad