American militarism pre-WW2 and the Ultimate Battleship

[TheMarcksPlan]

Having just run a very basic version of this into Springsharp I got the following returns for the basic hull with armour, no engines or guns at present;

Displacement: c2,000,000tons
Cost: $1,775,858 million.
It has a very high roll at 13.5 seconds, is an unsteady gunplatform at 14% (ave = 50%) and has the seaboat qualities of 0.08% where 1% is average.

Obviously these figures could be trimmed a fair amount, but on the positive side it will take 800k 6" shells or 1,800 torpedoes to sink!?

Wow those are some numbers.
My intuition about a free program like this is it's based on correlation with historical designs and has basically no underlying economic analysis of factors like economies of scale.
This is why I prefer a fundamentals-based approach from the main parameters of drag, propeller thrust, bending stress, etc.

That said, let me interrogate your inputs a bit:

It says a displacement of ~2mn tons. Is that empty displacement? Does the program break it into armor and hull (assuming no guns and armor as you said).
Basic arithmetic can tell us the weight of the armor I propose; this is nearly an order of magnitude above what that basic arithmetic indicates (~1250x300ft citadel, 2ft thick belt and deck armor, 490 lbs/ft3 steel density = ~300k tons citadel).

Hull weight would, I posit, be around 10% of gross displacement. This may be where the program is most off if it doesn't account for the fact that moment of inertia of a beam increases with the fourth power of its linear dimensions. It is for that reason that the structural efficiency of a ship (empty/full ratio of weight) improves dramatically with size. Do we know the basic structural/static engineering parameters of this program?

Probably to destroy all reserve bouyancy, the program is not clear on that sort of thing. I believe it is worked on the overall strength and size of the structure.

This is the kind of thing I'd want transparency on to credit a model...

A program that doesn't account for non-linear physical and economic factors that vary with size can easily create exponential errors when extrapolated beyond its sample set, even if it's fairly accurate within that sample.

[TheMarcksPlan]

A 2000 hp motor is huge.

Well so is this ship...

Note that because there's no armored barbette, turret rotation machinery can be placed outside the rotating structure instead of inside - on the electric deck - as with every other design. Between the armored deck and the magazines is ~20ft of vertical space in which to place as many large/medium motors as needed. That 2000hp motor would fit just fine in that space, along with dozens of others (if needed) around the giant circumference of the turret.

So definitely not a physical limitation on turret rotation power. Next up:

The 2000 hp motor will also require it's own ship's service electrical generator

Yep.
And as I already noted upthread, there's plenty of room in this hull for a 100MW generator (or larger if necessary) dedicated to the turret machinery. That's ~1/15th of the propulsion plant's output; the propulsion plant needs only ~8% of the hull volume.

Remember that, based on the fundamentals of drag (square/cube law and Froude number) this ship has the HP/ton ratio of a WW1 battleship for a top speed north of 35kn.

I'm also going to need more HP for rammers and for gun elevation than in other battleships, even on a per-gun basis.

Anyone have any idea on the cost of, say, a 2,000HP electric engine in 1940? I've been assuming (IMO reasonably) that the total bill for, say, Iowa's electric motors wouldn't exceed a few $million, given general impressions of cost drivers for BB's.

it takes about 3500 amps to run it at 450 VAC which is standard US Navy ship's electrical system voltage. I guess you could up the voltage to something non-standard

I don't think commonality between these ships' electrical systems and other ships would be an urgent concern. They're big enough that having crew/technicians specialized in them would be justified.

[TheMarcksPlan]

It has a very high roll at 13.5 seconds, is an unsteady gunplatform at 14% (ave = 50%) and has the seaboat qualities of 0.08% where 1% is average.

What parameters informed this output? Where did you set center of gravity? Is it based on the (IMO erroneous) apparent judgment that the armor would weigh over a million tons? Was there any ballast in the hull? (remember my design specifies that 35% of displacement is ballast during battle conditions).

[Takao]

(~1250x300ft citadel, 2ft thick belt and deck armor, 490 lbs/ft3 steel density = ~300k tons citadel).

That's only for 1 foot of armor...you said 2 feet thick armor. Also wasn't it 1,600 feet
The deck alone is going to be over 200k.

[Terry Duncan]

My intuition about a free program like this is it's based on correlation with historical designs and has basically no underlying economic analysis of factors like economies of scale.

Well, there are some problems with 'economy' involved here. Firstly, the US proved incapable of producing face hardened belt armour thicker than 12.2" without a loss in quality during the 1930s, which is why the same sort of belt was fitted to all three classes completed. You can go thicker, but the effectiveness returns are not as great as they first appear. Similarly, with the 18" guns, they are going to be required in numbers no single manufacturer is able to build without a plant about ten times the size as exists already.

That said, let me interrogate your inputs a bit:

It says a displacement of ~2mn tons. Is that empty displacement? Does the program break it into armor and hull (assuming no guns and armor as you said).
Basic arithmetic can tell us the weight of the armor I propose; this is nearly an order of magnitude above what that basic arithmetic indicates (~1250x300ft citadel, 2ft thick belt and deck armor, 490 lbs/ft3 steel density = ~300k tons citadel).

The program will break down weights, but it is also dependent on what details are put in. For example, what depth is your 24" belt? Given the variable draught hull idea, it is going to have to cover at least the waterline at deep load, but will it still cover it at light load? There is a lot of difference in a belt 12' deep to one 40' deep. The hull still has to cope with the belt weight for stability purposes at both load conditions.

Also, I had gone with rather more protection than you indicate here, using the 2/3 standard length.

This is the kind of thing I'd want transparency on to credit a model...

A program that doesn't account for non-linear physical and economic factors that vary with size can easily create exponential errors when extrapolated beyond its sample set, even if it's fairly accurate within that sample.

The program seems fairly well created, and is still used over a decade after it first appeared. As to how accurate, at least one naval architecht was invovled in checking it when it first appeared and they vouched for it being a good approximation of reality. More can be done with more exact details, such as how many guns, how many turrets, how high they are carried, what armour they have? Speed can be worked out easily enough with what is left, but without specifics it becomes hard to put someone elses ideas into practice.

[Terry Duncan]

A preliminary note on armor for MegaBB.

Assuming steel density of 490lbs/ft3 and:

• A citadel of length 1600ft, width 300ft, and deck armor of 2ft(!), that's 210k-t for deck armor
• For a citadel-belt 2ft thick, 50ft high [from 10ft below water to 40ft above given 140ft hull depth and 100ft battle draught] we have 70k-t belt armor
• Bulkheads of 200ft, 50ft high and 2ft-thick are ~9k-t
• Total citadel armor is ~300k-t

Our target displacement at 2400x360x40 and .7 block coefficient is ~700k-t so we're well within the limit with 100's of k-t for armament and turrets.

These were the basic figures I used, but now you say 1250ft main belt?

It also estimates the crew could be 35k!

[TheMarcksPlan]

These were the basic figures I used, but now you say 1250ft main belt?

Fair enough. Like everything I've posted about this ship so far, all figures are ballpark estimations of the ultimate ideal dimensions. I could see the ship being 2000x300ft instead of 2400x360ft, for example, or smaller. Or larger. Depends where the cost/benefit of guns intersects with mission capability and cost/ship. The larger the ship, the cheaper it is per ton of shells thrown. But at some point the marginal benefit of more shells is very low as there's no feasible configuration of ships or aircraft that would come within firing range of the ship (which means it's fulfilled its deterrence/strategic-dominance mission).

[TheMarcksPlan]

Similarly, with the 18" guns, they are going to be required in numbers no single manufacturer is able to build without a plant about ten times the size as exists already.

That's not a bug; it's a feature. You'd both have to and want to create an entirely new production infrastructure for producing the primary and secondary guns. Unlike OTL naval guns produced in small batches, these would be mass-produced with associated economies of scale. You'd bid out the contract for, say, 2,000 18in guns (depending on how many ships were ordered). Whoever bid the contract would integrate their own capital investment in plant into their bid price for the guns. Over a ~5-year production run it'd entirely possible for a private firm to recoup its investment in plant plus a healthy profit margin (as occurred in WW2 with massive production runs for weapons over ~5 years).

[OpanaPointer]

How much steel is needed for your supers?

[Takao]

Similarly, with the 18" guns, they are going to be required in numbers no single manufacturer is able to build without a plant about ten times the size as exists already.

That's not a bug; it's a feature. You'd both have to and want to create an entirely new production infrastructure for producing the primary and secondary guns. Unlike OTL naval guns produced in small batches, these would be mass-produced with associated economies of scale. You'd bid out the contract for, say, 2,000 18in guns (depending on how many ships were ordered). Whoever bid the contract would integrate their own capital investment in plant into their bid price for the guns. Over a ~5-year production run it'd entirely possible for a private firm to recoup its investment in plant plus a healthy profit margin (as occurred in WW2 with massive production runs for weapons over ~5 years).

The US is also paying a princely sum to do this. With production costs running about 33% higher in public shipyards vs. Government shipyards.

Public corporations also balked at such contracts, because of profits were limited by law(10% - which was done away with in WW2 and changed to cost+), handicapped by labor laws which had to be abided by while working the government contract, whatever improvements they made were only applicable to the one task(big guns and armor are not needed in the civilian world), nor was their any guarantee that the contract(especially a large one such as this) would be seen through to the finish(say, the next election for instance).

[Richard Anderson]

How much steel is needed for your supers?

According to his estimate, 300,000 tons for one. Given that prior to the full expansion in armor steel production facilitated by the requirements of the 1938 Naval Act, the US produced about 15,000 tons per year (at peak, essentially none was produced 1921-1926), the USS Behemoth should only take 20 years to build. Of course, assuming the US public and government collectively losses its minds and moves to production at levels achieved at peak in World War II in 1943, it might get completed in just six years...assuming of course that it is the sole item on the US Navy's plate for the entire period.

This thread is almost poetic in its utter absurdity. What makes it truly entertaining is reading only the replies since I have TMP on ignore. It makes the lack of answers to even the simplest of issues, such as the armor steel problem, glaringly obvious.

[TheMarcksPlan]

How much steel is needed for your supers?

For the 2400x360ft version, empty weight is ~700k so that's about the steel requirement.
For a 2000x300ft version about 2/3's of that for half the firepower.
For a 20% bigger version of the Super you'd get ~35% more firepower.

The key concept is that the ratio of firepower to empty weight increases with ship size.
The ideal Super size is probably smaller than 2400x360ft. Probably. But it may be larger too.

According to his estimate, 300,000 tons for one.

Once again, Richard needs to misrepresent a point to make his argument.
Any barely-literate person can see that 300k tons is for the citadel only.

Sad to see the quality of some responses here.

[OpanaPointer]

You know that the US steel industry was largely rubble in the '30s, right? That the plants were largely antiquated and/or inadequate? That the companies were wary of expanding their facilities after the bubble-busts in 1919?

[Richard Anderson]

You know that the US steel industry was largely rubble in the '30s, right? That the plants were largely antiquated and/or inadequate? That the companies were wary of expanding their facilities after the bubble-busts in 1919?

I do, you do, Takao does, and I think that any number of other sentient adults with a modicum of knowledge of the era knows that and also realize what an oxymoron "American militarism pre-WW2" is...but of course reality is irrelevant in a "what if", so 1.2-millon tons of armor steel to build the USS Behemoth class? No problem, just go with it.

[OpanaPointer]

Never mind where the money would come from, and the resistance of Sen. Gerald P. Nye, et al.