intercept target solutions ? geometrical method ?

target speed 10
torpedo speed 28

10 + 28 = 38

So how does 38 shooting angle come close to the 20 you used? The only way you can make it fit together is by using a different formula.


You are wrong ....

speed target of 10 and torpedo of 28 knots = angle of shoot of 20 not 38 lol .....
angle of shoot of 38 will be for a target of 22 knots.

on my video i shoot the torpedo at 348. the torpedo arrive to the boat at 0 angle so. angle of shoot was 12 degres. 348 + 12 you arrive to the 0 degres. front of your u-boat so.

12 degres that means a target of 6 knots around. 5 to 6.



again.....

Distance is not taking in consideration will using my method.

shooting at 1km or 10 or more ...... is the same.

Until you got the real speed of target and know your speed of torpedo = you have a angle of shoot + being at a correct angle = 90 degres = you shoot the boat same as shown in my video.



>>>> Distance does not change the amout of lead, or the torpedo shooting angle if you will, but it does require your aiming to be more accuracte.

dont need to aim more accurate or else......

speed target + speed of torpedo + being at 90 D = angle of shoot ^^

you can miss your target if you did do a wrong calcul of the speed target nothing else.



example in that video ( when am was not using magui as a gui ) good one btw ^^

angle of shooting here is 20. you see i shoot the torpedo a 20 the boat get sunk at 0.

target speed was 10 knots.
this video is just an example, cause i never destroy neutral boat in my carrier ^^



you can see when the first part of the boat cross the 20 angle am not shooting yet the torpedo, i wait a little bit, cause that way i can destroy the exactly part of the boat that i want to destroy..

i explain my self bad, in the second video angle of shoot still 90 degres. what just change here is speed target, torpedo speed still 28 knots one.

Originally Posted by Zosimus

I watched the video and didn't get it. My U-boat comes with a targeting computer, which has yet to let me down. Why not use it?

sorry man am french my english is not perfect, i dont get what you mean ^^

Originally Posted by MantiBrutalis

Completely baffled by sENoZ's procedure as well...

lol ^^ i take that as a compliment so

that the best method ever, dont need to complicate your life with useless calcul.

here what we only need is speed target and speed of torpedo = angle of shoot ^^ + being at a angle of 90 Degres = boom. :P

[Zosimus]

All right. Let's assume that we are firing a 28 knot torpedo at a ship that will be 2800 meters away from our original point of firing at the moment of impact. We also assume that the target will be at exactly 90º AOB at the moment of firing. Thus we have a right triangle with a hypotenuse of 2800.

Now let's say, for the sake of argument, that the lead angle is 20º – what will the target speed be?

Sin(20º) = 0.342 so the ship will have traveled 957.7 meters from the time of firing until the time of impact. Accordingly, the ship is traveling at 9.577 knots.

So your formula is:

Lead angle = torpedo speed - ship speed ?

28 - 9.6 = 18.4 not 20.

What if the ship is traveling more slowly? Let's try 2800 hypotenuse and 400 opposite (28 knots and 4 knots relatively). What's the angle? Well, if your formula is right, it should be 28 - 4 = 24? Not even close. The real angle is around 8.2º
------------------
Sorry. I don't get it.

Originally Posted by Zosimus

All right. Let's assume that we are firing a 28 knot torpedo at a ship that will be 2800 meters away from our original point of firing at the moment of impact. We also assume that the target will be at exactly 90º AOB at the moment of firing. Thus we have a right triangle with a hypotenuse of 2800.

Now let's say, for the sake of argument, that the lead angle is 20º – what will the target speed be?

Sin(20º) = 0.342 so the ship will have traveled 957.7 meters from the time of firing until the time of impact. Accordingly, the ship is traveling at 9.577 knots.

So your formula is:

Lead angle = torpedo speed - ship speed ?

28 - 9.6 = 18.4 not 20.

What if the ship is traveling more slowly? Let's try 2800 hypotenuse and 400 opposite (28 knots and 4 knots relatively). What's the angle? Well, if your formula is right, it should be 28 - 4 = 24? Not even close. The real angle is around 8.2º
------------------
Sorry. I don't get it.

______________

Now let's say, for the sake of argument, that the lead angle is 20º – what will the target speed be?


angle of 20 ? speed target at 10 with 28 knots torpedo.
as shown in my second video the wich without magui mod.


_______________


So your formula is:

Lead angle = torpedo speed - ship speed ?

no my formula still speed target + speed of torpedo + being at an 90 degres angles, nothing else.

the 90 degres will change if the target course and mine is not at 90 degres for sure. the rest will still the same so speed target and torpedo speed.. = you find shooting angles...


______________


What if the ship is traveling more slowly? Let's try 2800 hypotenuse and 400 opposite (28 knots and 4 knots relatively). What's the angle? Well, if your formula is right, it should be 28 - 4 = 24? Not even close. The real angle is around 8.2º



>>>> sorry my english isnt perfect enough to understand what you say here ^^

you ask me what about low speed target ?

always with an 90 degres angles. for example.

the minimum that i can have at speed target will be 5 knots for a 10 angle shooting always with an 90 degres position ( your uboat )

so i guess a 2,5 knots speed target ^^ will be like 5 angle of shoot...

a 6 knots speed target get a 12 angle of shoot always with 28 knots speed of torpedo.


but i think i explain bad my self.

when i say angle so 90 degres, its about positioning your uboat to about the course target.

and when i mean angle of shoot, i mean look trough obs periscop or attack periscop and wait till the boat came to the good number then shoot your torpedo.


for explain good so.



in that video my u boat is in a 90 degres position about the target course. and the shooting angle what i call shooting angle... anyway look in the periscop i just wait until the boat came at 20 wait a bit more cause i want to shoot a specific part of that boat, then shoot.

but i am at a 90 degres angles.

and so if shooting at 20 like in this video the speed of the target was at this moment 10 knots.

this work only for convoy mutlipe target with same speed, or single target that keep a constant speed.

sorry again my english isnt perfect ill do my best to explain ^^

hope you get it now lol.

[Pisces]

Originally Posted by sENoZ

target speed 10
torpedo speed 28

10 + 28 = 38

So how does 38 shooting angle come close to the 20 you used? The only way you can make it fit together is by using a different formula.


You are wrong ....

speed target of 10 and torpedo of 28 knots = angle of shoot of 20 not 38 lol .....
angle of shoot of 38 will be for a target of 22 knots.

Then show me your calculator (or abacus perhaps) with which you calculate that rule. I don't know how you make 10+28=20 How do you add?

That is what you do in this 1st video you posted:


You swing the periscope to 20 and wait. You do this on the knowledge or assumption that the target has speed of 10 knots. If you add the target speed 10 to the torpedo speed 28 then you get 38. Where do I see that in the video? Nowhere!

on my video i shoot the torpedo at 348. the torpedo arrive to the boat at 0 angle so. angle of shoot was 12 degres. 348 + 12 you arrive to the 0 degres. front of your u-boat so.

12 degres that means a target of 6 knots around. 5 to 6.

Well that is the 2nd video. This one:


And why is it 6 (or 5) knots? How do you calculate 6 knots from 12 degrees? Or the other way around?

>>>> Distance does not change the amout of lead, or the torpedo shooting angle if you will, but it does require your aiming to be more accuracte.

dont need to aim more accurate or else......

speed target + speed of torpedo + being at 90 D = angle of shoot ^^

you can miss your target if you did do a wrong calcul of the speed target nothing else.

That is exactly where the accuracy matters, in measuring the speed. If you got the wrong speed then you use the wrong amount of lead, and so you point the periscope wrong. So, distance does make it harder to hit accurately if you are not exactly sure of the speed. Because you need to hit something narrow.

the minimum that i can have at speed target will be 5 knots for a 10 angle shooting always with an 90 degres position ( your uboat )

so i guess a 2,5 knots speed target ^^ will be like 5 angle of shoot...

a 6 knots speed target get a 12 angle of shoot always with 28 knots speed of torpedo.

Ahhaaa!!

I am seeing a pattern with both videos, and your responses to Zosimus! In each video the lead angle is supposedly twice the target speed. Is that the solution to it? You double target speed to get the lead angle or shooting angle? If so, then you need to explain that with your rule. And don't say it as it is "this plus that", "target speed plus torpedo speed". Because in most cases that is simply wrong.

If you say that shooting angle is 2 times target speed then you are closer to the truth. A larger target speed makes the shooting angle larger by the same ratio.

ok..... i see you dont get it ^^ lol

i will try to make a video when possible.... And explain everything step by step.

remember ^^

i have to be at a angle of 90 degres that angle will cross the target course.

____________________________

You swing the periscope to 20 and wait. You do this on the knowledge or assumption that the target has speed of 10 knots. If you add the target speed 10 to the torpedo speed 28 then you get 38. Where do I see that in the video? Nowhere!

____________________________

i do this with knowledge ^^

Well me i dont see where you find 38
cause if target speed is 10 and my torpedo is 28 and i shoot at 38 angle ? angle will be way too much. the torpedo will for sure cross the course of the target, but way too late..


about the target speed i calcul his speed first then i know the angle of shooting ( when looking trough attack per or obs )

as i said i will make a video soon i think and show every detail and step one by one.

by the way you should be able to read that tablet ^^ it is for 30 knots torpedo but maybe you will understand..



>>> And don't say it as it is "this plus that", "target speed plus torpedo speed". Because in most cases that is simply wrong.



no no .... it is simply and again speed of target + speed of torpedo = you will know when to shoot your torpedo and destroy the boat.. ho yeah... and for sure being at an angle of 90 degres for this example for sure.


30kts.jpg

[Pisces]

Originally Posted by sENoZ

i do this with knowledge ^^

Well me i dont see where you find 38
cause if target speed is 10 and my torpedo is 28 and i shoot at 38 angle ? angle will be way too much. the torpedo will for sure cross the course of the target, but way too late..

I don't know what your language is that might cause misunderstandings, but I thought the language of mathematics and arithmetic was universal across most of the world:

1+2=3

38 is what I get when I apply your rule for shooting angle/lead with the known numbers:

target speed + torpedo speed = 10 + 28 =38

I know it is an incorrect number for shooting, resulting is a missed torpedo. But that is how you explained your rule. And I know you wrote that the uboat needs to be at 90 degrees to the course of the target when you fire. But that does not explain how your odd rule works and results to 20.

by the way you should be able to read that tablet ^^ it is for 30 knots torpedo but maybe you will understand..

Ok, that would be the missing link you left out in your explanation. That table calculates the exact same shooting angle as the TDC based on AOB and target speed. So, if that is what you use then you are using a paper TDC. It's the same mathematics! Not different. Not simpler.

Originally Posted by Pisces

I don't know what your language is that might cause misunderstandings, but I thought the language of mathematics and arithmetic was universal across most of the world:

1+2=3

38 is what I get when I apply your rule for shooting angle/lead with the known numbers:

target speed + torpedo speed = 10 + 28 =38

I know it is an incorrect number for shooting, resulting is a missed torpedo. But that is how you explained your rule. And I know you wrote that the uboat needs to be at 90 degrees to the course of the target when you fire. But that does not explain how your odd rule works and results to 20.


Ok, that would be the missing link you left out in your explanation. That table calculates the exact same shooting angle as the TDC based on AOB and target speed. So, if that is what you use then you are using a paper TDC. It's the same mathematics! Not different. Not simpler.

38 is what I get when I apply your rule for shooting angle/lead with the known numbers:

target speed + torpedo speed = 10 + 28 =38


>>>> and again me i dont understand why you add speed target and speed of torpedo ...... you are wrong, this calculation is just wrong, you will shoot at 38 angle for a target speed of 10 ? ok you will miss the target, easy as that




38 is what I get when I apply your rule for shooting angle/lead with the known numbers:


my rules ? no your rules, 38 is not what i get when i apply my rules ! this is the difference, you still dont get it lol !!!


you have the 30 knots tablet for example with speed target aand torpedo and different angles .......

i can not do more....

i did a video yesterday but the file was corrupt -___-

so you will have to wait until i make a new video of my method.

angle of 38 will be for a speed target of 22 knots .....just saying.


here is a screen.

bandicam 2015-05-11 01-09-31-373.jpg

i think am using the wrong word to explain you ^^


so i will try to find the good word to explain my method.

so i mean.

by knowing the torpedo speed this is easy to know...

in my case and for the moment 28 knots one.

+ knowing the speed target, you can use any method you want to know the speed target.

with these 2 things.

+ being on an angle of 90 degres see the screenshoot.

am able to find the angle of shooting of the torpedo ^^

ok...

you will understand i think ... with good english now ^^


>>>

The Dick O'Kane targeting method is a method submarine commanders can use to hit a target with a torpedo. What's great about it is that it only requires two things:

___________
knowing the angle by which you lead the target
having your sub at right angles to the target course
___________


so ????


But, how do you figure out how much you want to lead the target by? The posts I've seen usually give simple rules of thumb, like "10 degrees for a slow target, 20 degrees for a fast one", or "about 1/3 of the target's length", but I'm anal and like to know exactly.

The lead angle is a function of two things: the speed of the target (Vtarget), and the speed of the torpedo (Vtorp), and it's simply equal to atan(Vtarget, Vtorp). for example it will show you that if the target is going 10 knots, and your torpedoes go 36 knots, the optimal lead angle is 16 degrees:




now ... you should understand better then i tryed to explain you ^^

will upload a video soon.

i dont show how i do cause didnt have enough time to place my uboat and s**t else.

but you will pause the game when you will see the map, nothing difficult.

If i remember distance is around 3,2 km.

not my best record.

well the quality is s**t dont understand why, cause last video was available in 720p è__è

anyway the target was not exactly on my 90 degres, its because i am shooting the torpedo half of the boat and not before.

anyway resultat is the same >>> destroy the boat..


edit its good quality is here ^^ 720p .. better then 360p...


i will maybe do longer video in the futur..

[Pisces]

Originally Posted by sENoZ

...

But, how do you figure out how much you want to lead the target by? The posts I've seen usually give simple rules of thumb, like "10 degrees for a slow target, 20 degrees for a fast one", or "about 1/3 of the target's length", but I'm anal and like to know exactly.

The lead angle is a function of two things: the speed of the target (Vtarget), and the speed of the torpedo (Vtorp), and it's simply equal to atan(Vtarget, Vtorp). for example it will show you that if the target is going 10 knots, and your torpedoes go 36 knots, the optimal lead angle is 16 degrees:




now ... you should understand better then i tryed to explain you ^^

Ok, I understand now that your rule is not to be taken literally in a mathematical sense. With the + you mean both target speed and torpedo speed needs to be taken into account. The + does not mean adding in a numerical sense. But without further explanation in message # 7 this is how everyone interprets this. Be careful of this.

Ok, so then you have a method to calculate shooting angle from target speed (and torpedo speed). I understand that you are using this table attached in message #22. That is great. That was the unanswered question. Problem solved.

How I calculate shooting angle (lead angle)? Basically using the same formula. (mathematical Sine law: sin(lead angle)*torpedo speed=target speed* sin(AOB) ; or lead angle = asin (target speed*sin( AOB)/torpedo speed) )


Mostly I use the TDC as it is supposed to be used to make it steer the torpedo gyro angle by itself, but manually setting the dials. Not using the notepad method in any way. Sometimes I use a handheld calculator to calculate the above formula to check the leadangle, but if I am feeling nostalgic I calculate it with a handheld slideruler. The really old fashioned way. (And mess it up 9 times out of 10! )

The 'normal' german TDC procedure:
Set the periscope to 0,
Flip the TDC update switch, (I think it is that red button in between the TDC dials, but I am not sure with your GUI mod)
Set AOB to 90 (or whatever angle I am crossing the target path),
Set the target speed to the right value,
Set range as needed if I don't shoot straight ahead,
Flip the TDC update switch back,
point the periscope wherever I want,
the gyro angle follows correctly wherever I point the periscope.

^^

yes as i said i dont do target speed + speed of torpedo but by saying + it was like by knowing v target and speed target i get angle of shoot ^^

i dont use notepade i dont use also the book at all...

i just use so speed of torpedo and vtarget ^^ to get my angle of shoot.

yes its the green button with magui mod, very good gui btw. !