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Post by antonindvorak on Jan 12, 2020 17:49:10 GMT -6
Summary: Building a ship increases armor weight with growing speed. (Does not happen with rebuilds.) Game Version 1.14 Starting an all-defaults 1900 game, Great Britain. For demonstration, I shall build a 16kt CA. Yes, way oversized for January 1900, but I want to have space to increase the machinery. Here it is, with no engines (illegal design but a starting point): Here's the data: Max speed (knots)
| Weight of armor (tons)
| 0 (illegal, I know)
| 3390
| 1 to 16
| 3128
| 17
| 3195
| 18
| 3261
| 19
| 3293
| 20
| 3442
| 21
| 3503
| 22
| 3531
| 23
| 3559
| 24
| 3588
| 25
| 3617
| 26
| 3681 (100t overweight)
| 27
| 3744 (1400t overweight)
|
So basically the armor creeps from 3,128 by 490t (15%) to 3,618t just between 16 and 26 kts, with no other changes. Strange. It gets worse. Much worse.
Different game: still RTW 1.14 and 1900 start, but Germany, Very Large Fleets, Historical Resources, Varied technologies.
I have the auto-built 1899 Pre-Dreadnought of the "Wörth" class (overweight as so many autodesigned ones are): Opening it for rebuild, doing the following: - Change fire control to central firing
- Reduce secondary and tertiary guns to 12 each st save weight
- Remove all torpedo tubes to save weight
- Upgrade the main 9" to quality 0 and tertiary 3" to quality 1 (since we have them now)
This leaves me with 442t remaining: Since I got steam engines, let's rebuild the engine and see how fast we can get. 21kts(!), albeit we are now a (heavily armored) CA: Hmmm. That's expensive, how about cashing in on the better armor and shipbuilding I have now? So "Open Design" on it and just as above: - Change fire control to central firing
- Reduce secondary and tertiary guns to 12 each st save weight
- Remove all torpedo tubes to save weight
- Upgrade the main 9" to quality 0 and tertiary 3" to quality 1 (since we have them now)
And here we are: So, let's go to 21kts and get ourselves re-classified as a CA: So basically a heavier hull and heavier armor on the rebuild saves me over 1200t compared to a brand new design with everything else identical? How does that happen? That is more than just counter-intuitive, that breaks the whole weight/size modelling as it does not apply to rebuilds.
I was kind of hoping to use the significant savings to sneak in a torpedo defense 1 and increase the measly 80 rounds per barrel ...
Game2 is attached, just in case ...
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Post by dorn on Jan 13, 2020 0:30:59 GMT -6
I think it is WAD. With increased speed ship needs to have increased lenght to beam ration and larger machinery. It simulates that belt and deck armour needs to be longer adminship thus weights more.
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Post by wlbjork on Jan 13, 2020 2:46:29 GMT -6
I sort of agree, but I think that the formula may be a little bit high on additional armour required.
30 tons is a fair mass of steel - the best part of 4 m^3, and a few increases are double that (or more). Countering that is that we don't know the ship dimensions, the profile of the armour plate or even the dimensions of the belt and deck.
Not only that, there are limitations of the design engine to contend with - we can't just decide that we're increasing the number of boilers from 12 to 14 for example. So...yes, I think there should be some increase in armour weight for additional HP, but maybe not as much as we see here.
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Post by rimbecano on Jan 13, 2020 14:07:15 GMT -6
30 tons is a fair mass of steel - the best part of 4 m^3, and a few increases are double that (or more). Countering that is that we don't know the ship dimensions, the profile of the armour plate or even the dimensions of the belt and deck. The density of steel is on the order of 8000 kg / m^3. So, assuming a 15 ft belt height (not sure what it would tend to be historically), with the 5.5 inch thickness the OP lists there, we get about 1.5 tons per foot of citadel length. Assuming the width of the ship is approximately that of the historical Minotaur class, we can assume the beam to be 75 ft. With his thickness of 1.5 in for the deck, we get 2.1 tons per foot of citadel length. This adds up to 3.6 tons / ft, so 30 tons translates to an 8.5 ft extension of the citadel, 60 tons to a 17 ft extension. Between 20 and 26 knots (because below 20 knots is getting out of cruiser territory), his table averages about 40 tons per knot of speed increase, which, with the 3.6 tons / ft figure above, comes out to about 11 ft of citadel length per knot. This seems reasonable to me.
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Post by rimbecano on Jan 13, 2020 15:05:48 GMT -6
So basically a heavier hull and heavier armor on the rebuild saves me over 1200t compared to a brand new design with everything else identical? How does that happen? That is more than just counter-intuitive, that breaks the whole weight/size modelling as it does not apply to rebuilds.
Basically what's happening is that the game abstracts away a lot of the details of the actual dimensions of the ship so that you don't have to mess around with length, beam, draft, etc. Slower ships tend to be more efficient with a fuller hull form and lower length/beam ratio, faster ships with finer hull forms and higher length/beam ratios. So when you select CA, the game switches to a model that assumes a hull optimized for speed, which tends to have a narrower beam, which means more of the length of the ship is taken up by the same amount of machinery, which means the same thickness of armor takes more weight. Simply speaking, at this point in the game, it's not efficient to hang 9" of armor on a ship with a cruiser hull form. It's more efficient to use less armor and make the ship faster, so that it operates at a speed where the hull form is actually useful in keeping wave drag down. The rebuild turns out lighter because it was originally a battleship, so it has a battleship hull, which is wider and needs less length for machinery, so less armor is required for a given thickness over the machinery spaces, and with the machinery tech you have, wave drag at 21 knots doesn't require so much machinery as to make the ship infeasible.
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Post by aeson on Jan 13, 2020 15:15:21 GMT -6
30 tons is a fair mass of steel - the best part of 4 m^3, and a few increases are double that (or more). Countering that is that we don't know the ship dimensions, the profile of the armour plate or even the dimensions of the belt and deck. The density of steel is on the order of 8000 kg / m^3. So, assuming a 15 ft belt height (not sure what it would tend to be historically), with the 5.5 inch thickness the OP lists there, we get about 1.5 tons per foot of citadel length. Are we only armoring one side of the ship? A ship with a 5.5" armor belt 15' tall should require 165 inch-feet of steel per foot of citadel, which works out to about 3 tons per foot.
Beyond any lengthening of the citadel to cover more machinery or as a result of changing the length-beam ratio to support higher speeds, there is the possibility that a faster ship would require a deeper armor belt for similar protection, because where a ship sits in the water is affected by how fast it's moving.
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Post by rimbecano on Jan 13, 2020 16:26:55 GMT -6
Are we only armoring one side of the ship? Oops! EDIT: That gives us 5.2 long tons per foot, including the deck, not 3.6.
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Post by oldpop2000 on Jan 13, 2020 17:08:36 GMT -6
Are we only armoring one side of the ship? Oops! EDIT: That gives us 5.2 long tons per foot, including the deck, not 3.6. Oh thanks, because I came up with 5.5 tons per foot using Springsharp. Now, wouldn't the quality and type of armor make a difference? navweaps.com/index_nathan/metalprpsept2009.php
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Post by wlbjork on Jan 14, 2020 0:32:06 GMT -6
Not significantly. The mix is steel, albeit with slightly different mix of elements. The non-iron elements are less than 5% of the alloy, so they contribute little overall to the density.
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Post by rimbecano on Jan 14, 2020 6:46:37 GMT -6
Oops! EDIT: That gives us 5.2 long tons per foot, including the deck, not 3.6. Oh thanks, because I came up with 5.5 tons per foot using Springsharp. Now, wouldn't the quality and type of armor make a difference? navweaps.com/index_nathan/metalprpsept2009.phpClose enough. 8000 kg/m^3 is a figure from memory to one decimal place precision.
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Post by rimbecano on Jan 14, 2020 6:54:38 GMT -6
Not significantly. The mix is steel, albeit with slightly different mix of elements. The non-iron elements are less than 5% of the alloy, so they contribute little overall to the density. You have to be careful here. I don't know if it's ever the case with steel, but sometimes mixtures will have a different volume than the sum of their parts, so the density is different than you'd get by taking the total mass divided by the pre-mixing sum of volumes.
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