29 replies to this topic

[cheapracer]

Interesting subject that information doesn't easily fall to hand for obvious National Security type reasons.

Anyone know much about them?

I believe some use a swash plate or "Dyna Cam" type engine for the required profile but what about air (is the torpedo like an industrial pressurized air bottle?) and exhaust?

Almost a shame all the mechanicals I imagine go into them for a short 1 minute or so life span ...!

Some good info here on an older (?) type http://www.pt-boat.c...do/torpedo.html

[malbear]

Interesting subject that information doesn't easily fall to hand for obvious National Security type reasons.

Anyone know much about them?

I believe some use a swash plate or "Dyna Cam" type engine for the required profile but what about air (is the torpedo like an industrial pressurized air bottle?) and exhaust?

Almost a shame all the mechanicals I imagine go into them for a short 1 minute or so life span ...!

Some good info here on an older (?) type http://www.pt-boat.c...do/torpedo.html

yes cheapy this is a favourite of mine. especially running on its own exhaust

Cross-section of a Junkers Jumo KM8 disc-valve engine, enclosed in its torpedo casing.
The disc valve is visible on the right side of the engine, just above the piston.

The engine had eight liquid-cooled cylinders of 90mm bore by 85mm, stroke arranged as a V-8 with a 90deg angle. The total swept volume was 4.34 litres and the compression ratio was 6.6 to 1. Output was 275 HP at 3650 rpm. It ran on a mixture of petrol, oxygen, and its own exhaust gas- the latter presumably to dilute the oxygen to a manageable content.

A production order for 100 engines was issued towards the end of WW2 but was never completed. A prototype was examined by British and American intelligence engineers, who concluded it was "a progressive trend in automotive development." It would appear they were wrong.


http://www.aqpl43.ds...taryValveIC.htm

http://www.aqpl43.ds.....torpedo a.gif[/img]

[Tony Matthews]

Aren't the Russians using some sort of rocket, or rocket/turbine cross, that was deemed too dangerous by the Americans? I beleive the Kursk was crippled by a torpedo motor exploding in the tube. Having written that, I'll now go a-Googling, and find I'm wrong. Torpedoes are very interesting.

[kikiturbo2]

Aren't the Russians using some sort of rocket, or rocket/turbine cross, that was deemed too dangerous by the Americans? I beleive the Kursk was crippled by a torpedo motor exploding in the tube. Having written that, I'll now go a-Googling, and find I'm wrong. Torpedoes are very interesting.

google. for supersonic torpedos... interesting stuff..

http://en.wikipedia....i/VA-111_Shkval

[Tony Matthews]

Many years ago a small group of us, like so many small groups, experimented with explosives. We made a lot of rockets as well as grenades and mines, but one day found ourselves in a local park with one rocket left. As we were standing by the kiddies paddling pool we decided to see if it would work as a torpedo - it was a proper rocket, with fins, etc., - so we lit the fuse and dropped it in. It fizzed and spluttered, then started moving, suddenly dipping below the surface and shooting across the pool. It was gloss black, about a foot long, and was about three inches below the surface of the greenish water, an angry orange glow at the back and a long trail of thick smoke and bubbles behind, which quickly surfaced and popped out small baloons of pale grey smoke. It must have hit 20 MPH, and was a fine end to a damp afternoon. No children or animals were hurt in any way, as there were non present, but it did impress some local thugs.

[Ross Stonefeld]

Hold on, a supersonic torpedo? That seems pretty difficult to achieve for drag reasons using the speed of sound at an altitude a plane would test; wouldn't the actual speed underwater be frighteningly high?

[J. Edlund]

Some pictures of a torpedo propulsion system can be found in this brochure:
http://www.gkllc.com...orpedo-2000.pdf

In this case a torpedo powered by high test peroxide (HTP) as oxidizer and paraffin as fuel. A catalyst is used to create steam and oxygen from the HTP and additional energy is provided by burning paraffin using the oxygen produced. This then powers a axial piston expander, which drives the pump jet.

Monofuels can also be used as a propellant, and the expander is commonly of the axial piston or turbine type.

The Kursk sinking was caused by bad welds in a dummy torpedo which led to HTP leaking into the torpedo, reacting with the metal forming steam and heat. This then caused an explosion of the kerosene fuel. Unfortunatly the internal door on the torpedo tube wasn't properly closed, so this blast took out the forward compartments (the tube should otherwise handle such a blast). Minutes later the warheads in the torpedos stored onboard exploded.
Since the torpedo was a dummy it had been considered unnecessary to inspect the welds.

[Tony Matthews]

Hold on, a supersonic torpedo? That seems pretty difficult to achieve for drag reasons using the speed of sound at an altitude a plane would test; wouldn't the actual speed underwater be frighteningly high?

Yes, but it's in a bubble of air that it produces as it moves. Still mind-boggling though!

[SteveCanyon]

Hold on, a supersonic torpedo? That seems pretty difficult to achieve for drag reasons using the speed of sound at an altitude a plane would test; wouldn't the actual speed underwater be frighteningly high?

AFAIK the Shkval does about 200 - 300 knots underwater.
The first versions couldn't be steered but later version can.

[Greg Locock]

Current US eels use Otto fuel, google it. The UK abandoned H2O2 as a propellant after one incident too many aboard HMS Exploder, an experimental German sub they took over after the war.

[BullHead]

HMS Exploder

[cheapracer]

The Kursk sinking was caused by bad welds in a dummy torpedo which led to HTP leaking into the torpedo, reacting with the metal forming steam and heat. This then caused an explosion of the kerosene fuel. Unfortunatly the internal door on the torpedo tube wasn't properly closed, so this blast took out the forward compartments (the tube should otherwise handle such a blast). Minutes later the warheads in the torpedos stored onboard exploded.

HMS Exploder,

I think I started a 3 Stooges thread!

[Greg Locock]

I think I started a 3 Stooges thread!

http://en.wikipedia....rer_(submarine)

Incidentally the MDI "Air" car now uses an engine very similar to the dry heater torpedo engines introduced in 1904, effectively turning their air "powered" car into one with a rather odd IC engine.

[Ross Stonefeld]

Ah, well that's different. 200knots is freaky fast but well short of supersonic.

I did like the theory that you could counter-attack and kill the other sub, hopefully knocking out the torpedo guidance, before the initial launch gets to you. It's like a pistol duel.

[ray b]

Hold on, a supersonic torpedo? That seems pretty difficult to achieve for drag reasons using the speed of sound at an altitude a plane would test; wouldn't the actual speed underwater be frighteningly high?

the torpedo projects a bubble and is rocket propelled inside the bubble
very low drag and high speed
guidance is a major problem or was when news broke 10 years or so ago

scientific american had an article about it

[mariner]

"Some pictures of a torpedo propulsion system can be found in this brochure:
[url="http://www.gkllc.com/history/other/Swed_torpedo-2000.pdf""]http://www.gkllc.com...to...-2000.pdf"[/url]

That was very interesting - strange to see the familar SAAB badge on a torpedo but I guess SAAB have long made planes ( as the adverts kept pointing out).

One question, does anybody know , or can calculate what sort of bhp these engines produce. Given the density of water versus air and speeds of 50 mph+ it must be quite large I think and all inside a 21" diameter.

[Greg Locock]

"can [anyone] calculate what sort of bhp these engines produce. Given the density of water versus air and speeds of 50 mph+ it must be quite large I think and all inside a 21" diameter.

Yes I could but I won't. You'd get very very close if you found the power and speed of an electric torpedo and scaled it by v^3 and dia^2.

[desmo]

I did a 15 minute google safari for torpedo tech and one of the pages gave 600bhp for a reasonably modern torpedo engine. Don't have the link but it came up pretty easily, it was a barrel engine running on Otto 2.

[cheapracer]

I noticed some of the costs of modern torpedos too, 2 to 3 million dollars - holy cow.

[desmo]

If one can take out a 200 or 300 million dollar warship I guess that's a real bargain.

I guess.

When was the last time a torpedo actually sank a warship?

[Greg Locock]

When was the last time a torpedo actually sank a warship?

1982 for sure, maybe the recent Korean frigate. And of course Kursk if you count own goals.

[OfficeLinebacker]

And of course Kursk if you count own goals.

You, sir, are twisted.

[cheapracer]

1982 for sure, maybe the recent Korean frigate. And of course Kursk if you count own goals.

Yeah the addition of "intentional or not" to the question was warranted.

and then theres subs sinking ships without any aids at all, I think American Subs have sunk more Japanese ships post war than during it .....

[Greg Locock]

and then theres subs sinking ships without any aids at all, I think American Subs have sunk more Japanese ships post war than during it .....

Fun facts

RN vs USN in WW2, subs only

which ally sunk the greatest tonnage of enemy shipping?
which ally sunk the greatest number of enemy naval vessels?
which ally sunk the greatest number of enemy aircraft carriers?

The last question is a big clue, and the answer is the same for all 3 questions.

However the Poms sunk far more enemy subs, basically the Japanese sub force was on the run from 1943 onwards, and the Pacific is a big place to hide in.

[Tony Matthews]

It has been suggested that a number of British trawlers were sunk by American and Russian subs during the Cold War.

[mariner]

"Yes I could but I won't. You'd get very very close if you found the power and speed of an electric torpedo and scaled it by v^3 and dia^2. "

Greg, I wasn't trying to be lazy, I am not a professional engineer so I was trying to use the following formula ( excuse the imperial units).

HP = D*V/375 where D=q*A*Cd

and q = dynamic pressure (lb/sq ft.As I understand it q (air) density and velocity and is is quoted as 0.00256(V^2).

My problems are

1) what is the Cd of a torpedo? - I would guess 0.20 or so as a perfect streamline shape is often quoted at 0.15.

2) Is the Cd above dimensionless with respect to the density of the flow i.e if it is 0.15 in air is it always 0.15 in water/oil etc. I seem to recall that vey high speed aircraft tests were sometimes done in water tanks to use the renolds number effect to reduce speeds to what can actually be produced- does this affect Cd? Or to put it another way is drag in an incompressible liquid fundementally diferent to compressable air?

3) I have looked upo the density of air versus water, both at 20 degrees C and they are 998.2071kg/m3 for water and 1.204 kg/m3 for air so that makes water 829 times denser than air. Does that mean I need 829 times as much power to push the same shape through water as air? - If a car needs about 20 bhp to do 50 mph then it would need 16,580 bhp in water (?).

If the car has 20 sq ft frontal area and say a Cd of 0.30 it has a CdA of 6.0; then a 21" torpedo has 2.4 sq ft area and ( say) 0 0.2 Cd so the torpedo would have a CdA of 0.48 versus 6.0 for the car. So is the torpedo bhp at 50 mph 16,850*0.48/6 = 1326 bhp.

Is this ( about ) right or is there a flaw/error in my logic

[J. Edlund]

"Some pictures of a torpedo propulsion system can be found in this brochure:
[url="http://www.gkllc.com/history/other/Swed_torpedo-2000.pdf""]http://www.gkllc.com...to...-2000.pdf"[/url]

That was very interesting - strange to see the familar SAAB badge on a torpedo but I guess SAAB have long made planes ( as the adverts kept pointing out).

One question, does anybody know , or can calculate what sort of bhp these engines produce. Given the density of water versus air and speeds of 50 mph+ it must be quite large I think and all inside a 21" diameter.

Saab is in this case the aerospace and defence group which no longer have any relations with the car manufacturer except in name.

[J. Edlund]

1982 for sure, maybe the recent Korean frigate. And of course Kursk if you count own goals.

If I recall correctly that 1982 sinking is the only time a nuclear powered submarine have sunken an enemy warship.

[Greg Locock]

"Yes I could but I won't. You'd get very very close if you found the power and speed of an electric torpedo and scaled it by v^3 and dia^2. "

Greg, I wasn't trying to be lazy, I am not a professional engineer so I was trying to use the following formula ( excuse the imperial units).

HP = D*V/375 where D=q*A*Cd

and q = dynamic pressure (lb/sq ft.As I understand it q (air) density and velocity and is is quoted as 0.00256(V^2).

My problems are

1) what is the Cd of a torpedo? - I would guess 0.20 or so as a perfect streamline shape is often quoted at 0.15.

2) Is the Cd above dimensionless with respect to the density of the flow i.e if it is 0.15 in air is it always 0.15 in water/oil etc. I seem to recall that vey high speed aircraft tests were sometimes done in water tanks to use the renolds number effect to reduce speeds to what can actually be produced- does this affect Cd? Or to put it another way is drag in an incompressible liquid fundementally diferent to compressable air?

3) I have looked upo the density of air versus water, both at 20 degrees C and they are 998.2071kg/m3 for water and 1.204 kg/m3 for air so that makes water 829 times denser than air. Does that mean I need 829 times as much power to push the same shape through water as air? - If a car needs about 20 bhp to do 50 mph then it would need 16,580 bhp in water (?).

If the car has 20 sq ft frontal area and say a Cd of 0.30 it has a CdA of 6.0; then a 21" torpedo has 2.4 sq ft area and ( say) 0 0.2 Cd so the torpedo would have a CdA of 0.48 versus 6.0 for the car. So is the torpedo bhp at 50 mph 16,850*0.48/6 = 1326 bhp.

Is this ( about ) right or is there a flaw/error in my logic

I think you are very much on the right track. I'd use power =v^3*1/2*Cd*A*rho in SI units. I think Cd=0.2 is good, so i'd get power=(50*1.6/3.6)^3*1/2*.2*pi*.267^2*1000 W

[Greg Locock]

"Yes I could but I won't. You'd get very very close if you found the power and speed of an electric torpedo and scaled it by v^3 and dia^2. "

Greg, I wasn't trying to be lazy, I am not a professional engineer so I was trying to use the following formula ( excuse the imperial units).

HP = D*V/375 where D=q*A*Cd

and q = dynamic pressure (lb/sq ft.As I understand it q (air) density and velocity and is is quoted as 0.00256(V^2).

My problems are

1) what is the Cd of a torpedo? - I would guess 0.20 or so as a perfect streamline shape is often quoted at 0.15.

2) Is the Cd above dimensionless with respect to the density of the flow i.e if it is 0.15 in air is it always 0.15 in water/oil etc. I seem to recall that vey high speed aircraft tests were sometimes done in water tanks to use the renolds number effect to reduce speeds to what can actually be produced- does this affect Cd? Or to put it another way is drag in an incompressible liquid fundementally diferent to compressable air?

3) I have looked upo the density of air versus water, both at 20 degrees C and they are 998.2071kg/m3 for water and 1.204 kg/m3 for air so that makes water 829 times denser than air. Does that mean I need 829 times as much power to push the same shape through water as air? - If a car needs about 20 bhp to do 50 mph then it would need 16,580 bhp in water (?).

If the car has 20 sq ft frontal area and say a Cd of 0.30 it has a CdA of 6.0; then a 21" torpedo has 2.4 sq ft area and ( say) 0 0.2 Cd so the torpedo would have a CdA of 0.48 versus 6.0 for the car. So is the torpedo bhp at 50 mph 16,850*0.48/6 = 1326 bhp.

Is this ( about ) right or is there a flaw/error in my logic

I think you are very much on the right track. I'd use power =v^3*1/2*Cd*A*rho in SI units. I think Cd=0.2 is good, so i'd get power=(50*1.6/3.6)^3*1/2*.2*pi*.267^2*1000 W