Spiess TMA (aka Four Bearings)

[B_K]

I haven't spent much time for stunding the parabola but i think that this is true only when your own course is a linear one.In this case the parabola is tangent to all bearings and the axis of parabola is ,indead,parallel to DRM.
In the other case that our course is not linear ,as in this video, the parabola is tangent to all Spiess Lines produced by the three bearings and i haven't searched if the axis of this parabola is still parallel to DRM (99,9% it is,i just haven't study it).You can't 'imagine' if this axis is parallel in the video as it is impossible to draw this parabola (only one Spiess Line is showing).

I think you are right, Spiess Lines are equivalent to future bearings as if your course *was* linear. So axis of the parabola surely needs to be paralell to target's course. When own course is not linear, however, the only way to make the parabola tangent to bearing lines is to achieve such a position of your u-boat, that bearing line in time of observation will be drawn exactly on previously computed Spiess line (this is called singularity).

No, i haven't searched about it.Such case looks to me like having more a threoritical interest than a practical. I mean that ,even if you manage to make the three bearings intersect to one point (which practically will be extremelly hard to achieve), what would be the advantage ?
If i understand you right , you are asking to solve a problem with knowing only the first two bearings (neither target course or speed or range) and with (after the second bearing) proper own speed adjustment to have ,at the time of third observation, a third bearing which pass through the intersection point of the two previous two bearings. right ? if yes, i think that such a problem is not solvable.

In Dangerous Waters often all bearing lines crossed in one point, at least it looked like this on TMA screen. If we could discover speed component proportions (conditions of pure lag LOS) to achieve this, TMA could be based on two real bearings, one assumed bearing (in fact Spiess line beginning in assumed u-boat future position and crossing the common crossing point) and 4th bearing achieved by triangulation. With some aproximation and to minimize error - preferably for far distances - even not a single point but some surrounding of that point should be enough to assume all bearing lines intersect there and to conduct simplified calculations.

[makman94]

Originally Posted by B_K

I think you are right, Spiess Lines are equivalent to future bearings as if your course *was* linear. So axis of the parabola surely needs to be paralell to target's course. When own course is not linear, however, the only way to make the parabola tangent to bearing lines is to achieve such a position of your u-boat, that bearing line in time of observation will be drawn exactly on previously computed Spiess line (this is called singularity).

Exactly ! thats it !

Originally Posted by B_K

In Dangerous Waters often all bearing lines crossed in one point, at least it looked like this on TMA screen. If we could discover speed component proportions (conditions of pure lag LOS) to achieve this, TMA could be based on two real bearings, one assumed bearing (in fact Spiess line beginning in assumed u-boat future position and crossing the common crossing point) and 4th bearing achieved by triangulation. With some aproximation and to minimize error - preferably for far distances - even not a single point but some surrounding of that point should be enough to assume all bearing lines intersect there and to conduct simplified calculations.

Maybe this is happening becuase ,at DW, users usually 'lock' on TMA screen one (or more) of target's data.

About your question:
I gave a look at it and here is my conclusion:

The Problem you setted has infinite solutions if you let free target's course,speed and range.
It has infinite solutions if you 'lock' only the range
It has infinite solutions if you 'lock' only the speed

The problem has two solutions if you keep 'lock' both range and speed
The problem has one unique solution if you keep 'lock' only the course (direction)

All the aboves are proved (i can prepare the proves if you like to see them)

So,if you are looking for a solution you must have additionally known either target's range and speed or only its course (direction)

[B_K]

In my post I don't mean the final target solution which is achieved by locking any of those estimates. That's true ofc.

but

What I mean is that:
bearing lines alone intersect often in one point, which could help simplifying geometric calculations. When this is the case, only two first real bering lines are enough to use Spiess method. Instead of collecting third bearing for real, you can draw a Spiess line from the place where your uboat would be if you maintained current course and speed, and conduct the line through the common intersection point. Such a line is a Spiess line which is equivalent to assumed 3rd bearing if you didn't change course and speed, went there and collected it.
But in the meantime you turn the boat and by triangulation you collect real 4th bearing, intersecting freshly made Spiess line. You have target position and continue the procedure as in normal method.
However you just saved one time interval, which in turn can save your approach.
The point is - how do you know, having only two real first bearings, if all future bearings and Spiess lines go through one common intersection point.

It is defined by certain target-uboat geometry and Line of Sight parameters such as speed components and angles. Those conditions I would like to discover

EDIT:
Ok, I think I am closer to that.

page 42:

https://www.globalsecurity.org/milit...14308_ch10.pdf

describes Lagging geometry, where simplified Spiess could be utilized. I don't know yet if it's enough if speed vectors are just opposite, or should have specific component and angles proportions. But it is a good starting point.

[makman94]

Originally Posted by B_K

In my post I don't mean the final target solution which is achieved by locking any of those estimates. That's true ofc.

but

What I mean is that:
bearing lines alone intersect often in one point, which could help simplifying geometric calculations. When this is the case, only two first real bering lines are enough to use Spiess method. Instead of collecting third bearing for real, you can draw a Spiess line from the place where your uboat would be if you maintained current course and speed, and conduct the line through the common intersection point. Such a line is a Spiess line which is equivalent to assumed 3rd bearing if you didn't change course and speed, went there and collected it.
But in the meantime you turn the boat and by triangulation you collect real 4th bearing, intersecting freshly made Spiess line. You have target position and continue the procedure as in normal method.
However you just saved one time interval, which in turn can save your approach.
The point is - how do you know, having only two real first bearings, if all future bearings and Spiess lines go through one common intersection point.

It is defined by certain target-uboat geometry and Line of Sight parameters such as speed components and angles. Those conditions I would like to discover

EDIT:
Ok, I think I am closer to that.

page 42:

https://www.globalsecurity.org/milit...14308_ch10.pdf

describes Lagging geometry, where simplified Spiess could be utilized. I don't know yet if it's enough if speed vectors are just opposite, or should have specific component and angles proportions. But it is a good starting point.

i will have to disagree with you on this. Having three bearings intersecting to one point is an extremely rare situation like winning the lotto.

I understood what you are seeking for from your previous message.According to my point of view,without knowing anything else from target's data (speed ,course or range) don't expect for a solution.

The Spiess Line is the locus of all possible positions of target at the next time interval.At the case of having three bearings ,Spiess proved that ,at the fourth time, this locus is a straight line (ok,i know you know that) but (going to your theme now) , at the case of having only two bearings (and nothing more) the locus ,at the third time, is NOT a straight line (in fact there is no locus at all).By 'demanding' the third bearing ,at the third time, cross by the intersection point of two previous bearings,you are narrowing the theme to one specific situation.
Look at the following pic:



the blue bearing is the bearing from where your sub will be if you maintain course and speed.(as you can see is not pointing to the correct position of target)
the red bearing is from the point that your sub should have been in order to point to the correct position of target.

The only one specific situation that these two bearings matches and crossing from the intersection point of two previous bearings is when both courses (yours and target's) are parallel (look at pic below).
As long as both speeds are constant this is independent from speeds or ranges.

[B_K]

Now that's the good job! Thanks for explanations!
So my proposition turned out to be wrong, but at least we know that when you have three bearings intersecting in one point, the target course is exactly opposite to yours.
That's also a simplification anyway.

Wow, great thread !

Although I can't imagine to do myself all these drawings to attack a ship ... for now.

[3catcircus]

Originally Posted by Kendras

Wow, great thread !

Although I can't imagine to do myself all these drawings to attack a ship ... for now.

When you are stalking a ship or convoy in zero visibility, this definitely adds to your bag of tricks.