Custom Turret and turret sockets

[snrdg082102]

Howdy everyone,

I'm trying to figure out how to the standard 3 ton and 6 ton turrets and the associated turret sockets where constructed.

I understand that the volume and diameter are decided by the designer unfortunately I'm not sure how to figure out the cylinders length or surface area.

Can anyone point me to the correct pages in TNE FF&S please?

 

[aramis]

Howdy everyone,

I'm trying to figure out how to the standard 3 ton and 6 ton turrets and the associated turret sockets where constructed.

I understand that the volume and diameter are decided by the designer unfortunately I'm not sure how to figure out the cylinders length or surface area.

Can anyone point me to the correct pages in TNE FF&S please?

Given V=Volume (cubic meters) and D=diameter of the can...

L = V/(π(D/2)²)
A = 2π(d/2)² + πDL

L is length of the can
A is surface area of the can.

Surface Area taken is π(d/2)²

Page 12 & page 126 gives standard socket diameters:
3Td: D=3.6 L=4.2
6Td: d=4.5 L=5.25


Page 116 gives the standard formula for D and L given V:
D=1.4V^1/3
L=2D

 

[snrdg082102]

Hello aramis,

Given V=Volume (cubic meters) and D=diameter of the can...

L = V/(π(D/2)²)
A = 2π(d/2)² + πDL

L is length of the can
A is surface area of the can.

Surface Area taken is π(d/2)²

Page 12 & page 126 gives standard socket diameters:
3Td: D=3.6 L=4.2
6Td: d=4.5 L=5.25


Page 116 gives the standard formula for D and L given V:
D=1.4V^1/3
L=2D

Thank you for providing the formula and page numbers. I do know about the standard sockets which I couldn't match.

In the equations is the "n" actually PI?

 

[aramis]

Hello aramis,



Thank you for providing the formula and page numbers. I do know about the standard sockets which I couldn't match.

In the equations is the "n" actually PI?

It's actually the character for pi. Your computer is just not rendering it well.

 

[snrdg082102]

Morning aramis,

It's actually the character for pi. Your computer is just not rendering it well.

I had a feeling that what I was seeing as an "n" was the symbol for the spreadsheet PI. I've done an experiment and discovered that Verdana type does not render the pi symbol correctly.

Maybe this time when I try I'll be able to match the standard socket numbers.

Thank you again aramis.

 

[snrdg082102]

Update

Crud,

I'm obliviously on the highly disliked list of the math gods.

Using the standard 3 displacement ton socket and the formula D=1.4 x V^(1/3) my diameter did not work to to be 3.6.

D = 1.4 x 3^(1/3) = 1.4 x 1.4422 = 2.0191
D = 1.4 x 42^(1/3) = 1.4 x 3.4760 =4.8664

Which means that length is going to be incorrect using the formula L = 2 x D.

Next I tried the formula L = V ÷ (3.1416 x (D ÷ 2)^2) again using the 3 displacement ton socket with the result of

L = 3 ÷ (3.1416 x (3.6 ÷ 2)^2) =
L = 3 ÷ (3.1416 x 1.8^2) =
L = 3 ÷ (3.1416 x 3.24) =
L = 3 ÷ 0.9696 = 3.0940

L = 42 ÷ (3.1416 x (3.6 ÷ 2)^2) =
L = 42 ÷ (3.1416 x 1.8^2) =
L = 42 ÷ (3.1416 x 3.24) =
L = 42 ÷ 0.9696 = 43.3168

Another miss on my part. Going for area I used the 3 displacement ton socket's diameter of 3.6 and length 4.2

Area = (2 x 3.1416 x (D ÷ 2)^2) + (3.1416 x D x L)
Area = (2 x 3.1416 x (3.6 ÷ 2)^2) + (3.1416 x 3.6 x 4.2)
Area = (2 x 3.1416 x 1.8^2) + (3.1416 x 3.6 x 4.2)
Area = (2 x 3.1416 x 3.2400) + (3.1416 x 3.6 x 4.2)
Area = (6.2832 x 3.2400) + (3.1416 x 3.6 x 4.2)
Area = 20.3576 + (3.1416 x 3.6 x 4.2)
Area = 20.3576 + (11.0398 x 4.2)
Area = 20.3576 + 47.5010 = 67.8586

My last try using using the diameter of 3.6

Surface area = 3.1416 x (3.6 ÷ 2)^2
Surface area = 3.1416 x 1.8^2
Surface area = 3.1416 x 3.2400 = 10.1788 rounding to the nearest whole number equals 10 which matches the listed area for a 3 displacement ton socket on page 126.

What have I done wrong?

 

[Badbru]

Crud,

L = 42 ÷ (3.1416 x 3.24) =
L = 42 ÷ 0.9696 = 43.3168

What have I done wrong?

I'd say this bit.

3.1416 x 3.24 isn't 0.9696 it's 10.178.

L = 42÷ 10.178 = 4.126.

 

[snrdg082102]

Morning Badbru,

Originally Posted by snrdg082102
Crud,

L = 42 ÷ (3.1416 x 3.24) =
L = 42 ÷ 0.9696 = 43.3168

What have I done wrong?

I'd say this bit.

3.1416 x 3.24 isn't 0.9696 it's 10.178.
L = 42÷ 10.178 = 4.126.

Thank you for the cross check and finding my math error. My math error was the result of not noticing I pressed the division button located above the multiplication button calculator.

With the math error corrected I can now determine two specifications provided that the volume and diameter of the socket are provided.

Any idea of what I am doing wrong using the formula D=1.4V1/3?

Thank you and aramis for the help.

 

[snrdg082102]

Update 2

Morning all,

There appears to be a small problem with the standard turret and barbette socket heights as shown in the table on TNE FF&S Mk1 Mod1 (January 1994) page 12.

Using the information from TNE FF&S Mk1 Mod1 (January 1994) page 12 and the formulas worked out here height is off.

Turret: 3 displacement tons = 42 m^3 and diameter of 3.6 m

Height (m) = V (m^3) ÷ (3.1416 x (D ÷ 2)^2) =
Height (m) = 42 ÷ (3.1416 x (3.6 ÷ 2)^2) =
Height (m) = 42 ÷ (3.1416 x 1.8^2) =
Height (m) = 42 ÷ (3.1416 x 3.24) =
Height (m) = 42 ÷ 10.1788 =
Height (m) = 4.1262 or 4.1 m

Changing the diameter to 3.58 does match the listed height when rounding to the 0.1 position is used.

Barbette: 6 displacement tons = 84 m^3 and diameter of 4.5 m

Height (m) = V (m^3) ÷ (3.1416 x (D ÷ 2)^2) =
Height (m) = 84 ÷ (3.1416 x (4.5 ÷ 2)^2) =
Height (m) = 84 ÷ (3.1416 x 2.25^2) =
Height (m) = 84 ÷ (3.1416 x 5.0625) =
Height (m) = 84 ÷ 15.9044
Height (m) = 5.2816 or 5.2816 m

Changing the diameter to 4.515 does match the listed height when rounding to the 0.01 position is used.

Hopefully, my math is right this time.