this morning i sat down, i had the mission editor opened so that i could see the map of the test mission, just staring at it,
not thinking of anything in particular. i thought, well we know course , aob, and speed in a 8010, the only thing missing was range,
if we had range, we would have all four parts of a torpedo solution, plugged into the tdc, miles before we see the target, so just staring at the triangle
in the test mission, it almost blew my mind, there in front of me was a way to determine target range, but it depends on the captains estimation
of how far away the collision point is for the target and submarine.
if your on an 8010, you can draw the right triangle on the map, or use radar to estimate target track, then range to the collision point.
but what ever the submarines speed is, and the disctance to the collision point is known or estimated, target speed and range can be known,
because in a 8010, the ratio of target distance to submarine disctance to the collision point is 5.4 to 1. this is because of the proprtions
of the 8010 right triangle, its a 5.4 to 1 ratio of range and speed to the collision point, this is the ratio of the test mission,
so the submarine is at 2 knots, times 5.4 = 10.8 knots, might as well call that 11, its more than, 10.5, target speed was 11 knots in test mission.
range of the submarine to the collision point is 5.4 km, or 2.9 nm, in this test mission,
range of the target is 5.4 times greater for the target, so range of the target would be 5.4 km times 5.4, or 29.16 km or 15.7 nm.
at least this is the ratio of this test mission, i dont know what other submarine speeds ratios are at this point, i think it will be the same ratio for all speeds of the sub, but alittle work with the calculator
will prove that.
5.4 is the ratio in km
2.9 is the ratio in nm

You can't develop AoB in an unseen target. You have no solution.

You can't develop target course in an unseen target. You still have no solution. If you're going to use radar you don't need the 8010 formulation. You'll never get it anyway without a high speed surface run.

If you do that you might as well put the target on a zero AoB so you don't have to move 10 miles submerged to attack. He's not going to run over top of you anyway because he'll change course before he gets there. But at least you are covered with an equal run to engage no matter which direction he zigs.

If you set up the 8010 and he zigs 10º away it's game over. You can't get there and he's gone. 8010 is a terrible and self-defeating strategy that eliminates the vast majority of contacts from ever being attacked. The Japanese love you.

Proper strategy is to be able to engage long range targets at the largest possible angle on the bow so that you engage more targets, scoring more tonnage per patrol. 8010 is the analog to the ostrich strategy of running submerged all day, cutting your contacts by 90% so you can feel "safe." You've found a way to eliminate 80% of the 10% they had left. Just think of the payoff from combining Ostrich with 8010! Eliminate 98% of potential targets! Enhance your career! 8010 Ostrich! Guaranteed or your money back.

Proper strategy is to develop targets while surfaced, maneuver at high speed while surfaced to achieve an attack position ahead that will minimize battery use during the attack, submerge on or very close to the track, waiting for the target and then attacking with a real targeting method. You then have full batteries at the time of attack and can evade with confidence that you can exercise those batteries with gusto to break contact, surface and repeat the end-around. In this way you can engage even a convoy detected at a 180º AoB. Instead of restricting yourself to a 20º wedge, you now can engage targets on any course, anywhere in the 360º search range. Why would your strategy be to cripple your ability to fight? Why would you then actually brag about that? On every level 8010 is a loser. There is NOTHING in it of value except in how to put yourself on a collision course by passive sonar.

Is there a single person who understands 8010, thinks it's a viable strategy, uses it and is willing to post about it? If so I'm asking them to come forward. Let's see who this amazing development has benefited.

you dont say?:wah:

your whine is getting old, i cant? i did!

Please show where you posted the method of determining the target's course without seeing it. Either I missed seeing it, or you've completely forgotten to post it (and missed several requests for same), or you don't have a method of doing it.

Isn't his very premise that the act of holding the target on an 80 degree bearing means the target AOB is 10?

That's the entirety of what I got from his arguments. Maybe it's just so convoluted I'm missing something, cause if his point was that IF you can get yourself into that geometry, it begs the question we're all asking: how do you know you are in the right geometry if you cannot tell the target what to do?

I have an office at Fort Meade, a parking pass for Langley, and have taken courses at Maxwell and in Carlisle. I have a junk drawer full of metal, and enough paper on the wall to train a sled team of dogs not to crap in my shoes. With that said, I have followed this thread from its inception and I can see no merit in putting the lives of sailors on the line based on the thin premise provided herein.

Ok I think we are over-killing it here.

Personally I like the collision course thingy and will continue to """work""" on it. In the end I think that it will turn out be a "good to know" thing, implementable with in-game (nav map) tools. But as is, the 8010 method is limited. I still maintain it is salvageable with "addons" (:DL) namely trying to establish the targets course (at least).

For an example of a potentially useful method of acquiring target info, based on collision course(s), have a look at this . More variants of the method proposed are on the way. Hope people will test it and comment on it. (Ok now I'll stop highjacking the thread :oops:).

... and going deep again ...:lurk:

Wow, after reading this whole thread i can only say that i am glad i use the auto-targeting option. Trying to figure this out made my head hurt and i'm still not sure whos right or wrong but it sure was fun to read! I have used realistic targeting in SH1, but these days i play for fun and let my crew (Auto) do the work. After all, I'm the Captain LOL!

reread razark, i havent got time to hold your hand

well, mister northbeach, i dont care about your paper, your shoes, or anything else you have, im not impressed by your military pencil pushing job in "snievel affairs".
i have a solution for you, you can use the dick okane method, you dont have to take risks with it, and it sounds like that would be the method for a timid personality like yours to use, so knock yourself out. rr vows its safe, he understands submarine warfare about as much as you, he says to fire at long range is what submarine warfare is all about, that way your "safe" , , ignoreing the fact that german u-boat were mixing it up with convoys, escorts right in the mist of them, on the surface, i would of loved to have him say that to those german commanders, they would laugh him right off the boat, like im laughing you right in this post.

you see, you can join the opposition if you like, for you people, you feel safety in numbers, but let it be said right here right now, i stand alone!, i do not ask nor do i want help from anyone here, i will accomplish the mission, with or without you, understand ?

Readable prose might be a good start on your part.

You seem to be saying that holding any contact @ 80 degrees makes his AOB 10, which is nonsense.

If you require a right triangle, you must know his AOB in advance.

Next thing we know here people will be dropping britches and comparing the sizes of their members....

Of course someone here may figure out how to make 3 inches 8...and think he's right because he's looking at it underwater and never cease at thinkings he's correct.

:up:

Three angles of a triangle. Add them up, get 180 degrees.

Therefore, if the bearing is 80 degrees, the AoB _must_ be 10. You know you are intercepting the target track at a 90 degree angle because... :06:

Because if you overshoot the bearing then you are going too fast and need to turn in and allow the target to catch up. If you can't match the bearing, then the target is too fast to be intercepted.

But this was posted as a method of determining target speed, based on your own speed. You vary your speed to hold the target on the 80 bearing, and use your speed in the formula to calculate the target's speed.

So, do you turn to keep him on the 80, or speed up/slow down to determine his speed?

I don't understand this statement greyrider.
Is the 8010 inherently a riskier attack method than the O'Kane?
How so?
Is the 8010 superior to the O'Kane because it is riskier?
Please explain.

The method that greyrider has put forward is not a hard rule type of method, it's more of a variable method that you do different things in different situations. You put the target on an 80 degree bearing (so at 080 or 280 relative bearing after determining which way the target is moving), then you try to keep the target on a constant bearing for a few minutes. If you can't keep the bearing constant, then the target is too fast and is one of the ones that you have to let go using this method. If you can match bearings, then you can use the chart he gives to work out a target's speed (although there is no gurantee that the AoB is 10, so the speed may not be accurate enough). If you find when trying to match bearing that your own speed is very low, then you need to wait to allow the target to 'catch up' (the details behind this would be that the target is either at very long range or has a very slight AoB and either situation would mean you would run accross his track and be on the wrong side).

theres been about 3000 views on this post, yet only a few have posted, so this post has certainly generated interest, but of those that did post, and not all, there seems to be a disconnect somewhere,
if it has been because of me, and the way i have explain it, then its my fault, guess im not a good explainer, but i did the best i could with it, in a number of ways.
then there are some that say there trying but there having trouble understanding, i understand, but im not!
those movies, sonar and radar, are campaign episodes, and cant be fudged, at least as far as i know, and i was able to do it, so the question is, if i can do it, how come you cant, if its easy for me, how come its hard for you.
most here have posted nothing but mud, so i consider them mudslingers, not gunslingers, and sometimes, i just couldnt resist throwing alittle myself, i had some fun to.
these people have posted the 8010 as junk, useless crap, ok, ill bite, it is useless, so why do you keep on coming back? if i thought something was useless, id be long gone, yet why do those people keep coming back?
because its intriguing isnt it?
i have one more thing to add to 8010,and then i will be done, and that is target position known. im working now to find the targets position along that right triangle, but if you use proportions, you wont need this way that im working on.
because proportion will put the target position right where its supposed to be.
you know what the 8010 is shaping up to be? example 25 of the maneuvering board manual, just in a different way.
below is example 25 from the mbm, notice the heading!
COURSE, SPEED, AND POSITION DERIVED FROM BEARINGS ONLY

and now, take a look at the range that was solved for this problem,
(2) Position of M at 1830: 274°.5 at 61 miles.
61 miles away,
but we cant do that i keep hearing here,
so im getting impatient with the bs, and if anyone wanted to try to work it out, they are going to be knee deep in negativity, thats not mine,
and so they will probably give up, nevertheless, ill use it, and im using it well with success, and to those that would be willing to try, i apoligize,
for the bs in this post, i tried to avoid alot of it, but it just keeps coming, and so the post is destroyed now. i would not want to try to read this now,
so i wont blame you for not reading any further either.
they threw every kind of diversion and cami they could at it, lies, bs, irrelevence, and ignorance, of something they knew little about didnt stop em.
a guy asks for help, then in his last sentence, spits, he then expects me to help him after that? sorry, not on his best days after that.
so if anyone is willing to learn, all i can say now, is good luck, and put your boots on, theres alot of poop on the floor.



EXAMPLE 25
COURSE, SPEED, AND POSITION DERIVED FROM BEARINGS ONLY
Situation:
Own ship is on course 090°, speed 15 knots. The true bearings of another ship
are observed as follows:
At 1600 own ship changes course to 050° and increases speed to 22 knots. The
following bearings of ship M are then observed:
Required:
(1) Course and speed of ship M.
(2) Distance of M at time of last bearing.
Solution:
(1) Draw own ship’s vector er1.
(2) Plot first three bearings and label in order observed, B1, B2, and B3.
(3) At any point on B1, construct perpendicular which intersects B2 and B3.
Label these points P1, P2, and P3.
(4) Measure the distance P1 to P2 and plot point X at the same distance from
P2 towards P3.
(5) From X draw a line parallel to B1 until it intersects B3. Label this intersection
Y.
(6) From Y draw a line through P2 until it intersects B1 at Z.
(7) From head of own ship’s vector er1, draw a line parallel to YZ. This establishes
the DRM on the original course and speed. The head of the em vector of
shipMlies on the line drawn parallel to YZ. It is now necessary to find the DRM
following a course and/or speed change by own ship. The intersection of the two
lines drawn in the direction of relative movement from the heads of own ship’s
vector establishes the head of vector em.
(8) Following course and speed change made to produce a good bearing drift,
three more bearings are plotted; the new direction of relative movement is obtained
following the procedure given in steps (3) through (7). The lines drawn
in the directions of relative movement from the heads of vector er1 and er2 intersect
at the head of the vector em. Ship M is on course 170° at 10 knots.
(9) From relative vector r2m, the SRM is found as 28.4 knots during the second
set of observations.
(10) Compute the relative distance traveled during the second set of observations
(MRM 56.8 mi.).
(11) On the line ZY for the second set of observations, lay off the relative distance
ZA. From A draw a line parallel to B4 until it intersects B6. Label this point
B. This is the position of M at the time of the last bearing.
Answer:
(1) Course 170°, speed 10 knots.
(2) Position of M at 1830: 274°.5 at 61 miles.
Note:
These procedures are based on bearings observed at equal intervals. For unequal
intervals, use the following proportion:
Time Bearing
1300 010°
1430 358°
1600 341°
Time Bearing
1630 330°
1730 302°
1830 274°.5
Time difference between B1 and B2
Distance from P1 to P2
--------------------------------------------------------------------------------------
Time difference between B2 and B3
Distance from P2 to X
= --------------------------------------------------------------------------------------.

no sergi, it is not, thats what the bs people would like you to think.

i will still help those willing to learn, im not turning my back on you,