cacluating ship speed manually

[makman94]

Originally Posted by TorpX

.......
I used to use the following formula:

speed (kts.) = .59 target length (ft.) / time(sec.) + sub speed(kts.) * sin target bearing

if your sub and target are going in opposite directions then the last term sub speed(kts.) * sin target bearing is subtracted.

Hi TorpX,

.......i believe that this formula is wrong for getting target's speed,TorpX !

the problem is not so simple for been solved by this simple formula.
where did you find this formula written ? before i proceed and explain why this formula is not correct i would like you to explain it more detailed (for example write the formula more clearer or show with a little example how you are using it and getting speed)) becuase ,maybe, i am missing something at the way you have written this formula.


ps: if you can prove the above formula...would be even better.

sorry for being the 'Doupting Thomas' here but this formula is not working, TorpX. and there is no way to make it work becuase you are not considering the target's course (relative to your own course) factor which is a very important factor ! this factor has equal importance with your boat's speed factor (which you are correctly considering) and has ,also, equal importance with the bearing to target factor (which,also, you are correctly considering).

as i said ,is not so easy problem . a good tool that help in situations like this is the back side of attack disc (in case that we don't want to use digital-modern-calculators)

ps: @TorpX : don't feel offended ...i will be huppy if ,at the end, prooved to be me on the wrong side.

[Pisces]

Originally Posted by makman94

Hi TorpX,

.......i believe that this formula is wrong for getting target's speed,TorpX !

the problem is not so simple for been solved by this simple formula.
where did you find this formula written ? before i proceed and explain why this formula is not correct i would like you to explain it more detailed (for example write the formula more clearer or show with a little example how you are using it and getting speed)) becuase ,maybe, i am missing something at the way you have written this formula.

...

I agree, somehow the AOB should be worked into this too. I don't know how exactly though.

[TorpX]

@ makman94 and Pisces:

No offense taken.
You may be right. I dug the formula out of my old papers and have not used it in a long time. I will play around with it and check before I report back. I should have done this anyway.

[TorpX]

OK, you guys were right. Note to self- CHECK YOUR WORK!


I did some quick geometry to figure this out.
The formula I posted earlier is incorrect.

This is the correct formula:




Vt = .59 Tl / t + Vs * (sin Abg / sin Aob)

where:
Vt is target speed, in knots

Tl is target length, in feet

t is time, in seconds

Abg is bearing angle, in degrees

Aob is angle on the bow, degrees

Vs is sub speed, knots

Note that I define Abg as bearing angle, not relative target bearing. This is because the sine of angles between 180 and 360 deg. are negative and will give the wrong answer. If the relative target bearing is between 180 and 360, the "bearing angle" must be measured from the bow going counter-clockwise. Note also, that at relative target bearings of 0 or 180 deg., the last term becomes 0, and one need not know the Aob. The same goes for a sub that is not moving. Similerly, if the sub is on a parallel course, the ratio (sin Abg / sin Aob) becomes 1.

Also, as before, if viewing the port side of the target from port side of the sub, or starboard side of target form starboard side of sub, then subtract the sub speed. (This would mean you are heading in opposite direction of target.)

Well, I think I got it right this time. I did a numerical test case and it checked out.

One more thing; I would not try to use this for targets at a very sharp aspect, that is at Aob angles near 0 or 180 deg. This would make it very difficult the time the transit. I believe Don Reed mentioned this also. If anybody uses this, in game, please let me know how it works.

[makman94]

Originally Posted by TorpX

OK, you guys were right. Note to self- CHECK YOUR WORK!


I did some quick geometry to figure this out.
The formula I posted earlier is incorrect.

This is the correct formula:




Vt = .59 Tl / t + Vs * (sin Abg / sin Aob)

where:
Vt is target speed, in knots

Tl is target length, in feet

t is time, in seconds

Abg is bearing angle, in degrees

Aob is angle on the bow, degrees

Vs is sub speed, knots

Note that I define Abg as bearing angle, not relative target bearing. This is because the sine of angles between 180 and 360 deg. are negative and will give the wrong answer. If the relative target bearing is between 180 and 360, the "bearing angle" must be measured from the bow going counter-clockwise. Note also, that at relative target bearings of 0 or 180 deg., the last term becomes 0, and one need not know the Aob. The same goes for a sub that is not moving. Similerly, if the sub is on a parallel course, the ratio (sin Abg / sin Aob) becomes 1.

Also, as before, if viewing the port side of the target from port side of the sub, or starboard side of target form starboard side of sub, then subtract the sub speed. (This would mean you are heading in opposite direction of target.)

Well, I think I got it right this time. I did a numerical test case and it checked out.

One more thing; I would not try to use this for targets at a very sharp aspect, that is at Aob angles near 0 or 180 deg. This would make it very difficult the time the transit. I believe Don Reed mentioned this also. If anybody uses this, in game, please let me know how it works.

oh yes TorpX ...this formula is the correct one !

have in mind this: the angle that is named as 'AoB' in this formula is NOT trully the Aob but a 'compromised' angle that is very close to AoB.these angle as so close that 'allow' to this formula to give 'acceptable' results for target's speed

(for those that interested)...the 100% correct formula is :

a) length in meters and time in sec :

u(t) = 1,944 x [length/time] + u(b) x [sin(Abg)/sin(lb-Abg)]


b) length in feets and time in sec :

u(t) = 0,593 x [length/time] + u(b) x [sin(Abg)/sin(lb-Abg)]

where,

u(t)= target's speed
u(b)= u-boat's speed
Abg= bearing angle
lb= target's course angle relative to own course

ps:@TorpX:if there is interest in this theme i can give you the proof of the above formula

ps2:try the back side of attack disc ...it allows you with these data to get target's speed without having to use a digital calculator ! i believe that they did it that way back then.

bye

[TorpX]

@ makman94,

Unless I miss my guess, we are using different terms to arrive at the same result. I was puzzled by your "lb" variable, but looking at it, it seems like this is the same as the track angle. This is what the U.S. Navy calls the angle measured from the target's track to the sub's track, starting from ahead of the target. This means:

sin Aob = sin (Atr - Abg) [where Atr is the track angle]

I will provide an example to illustrate:

ship length.............320 ft.
time......................29 sec.
sub speed.............3.78 kts.
bearing angle.........100.5 deg.
angle on bow...........25 deg.

*target speed...............15.30 kts.

With the above example the track angle is 125.5 deg.
Tell me if you get a different answer.
(The numbers are odd because I drew this test case out on graph paper to check this.)

ps2:try the back side of attack disc ...it allows you with these data to get target's speed without having to use a digital calculator ! i believe that they did it that way back then.

To be honest, I've barely touched the TDC lately. I've been using manual fire control without the TDC. (S-boat style)

Bye

[makman94]

Originally Posted by TorpX

@ makman94,

Unless I miss my guess, we are using different terms to arrive at the same result. I was puzzled by your "lb" variable, but looking at it, it seems like this is the same as the track angle. This is what the U.S. Navy calls the angle measured from the target's track to the sub's track, starting from ahead of the target. This means:

sin Aob = sin (Atr - Abg) [where Atr is the track angle]

I will provide an example to illustrate:

ship length.............320 ft.
time......................29 sec.
sub speed.............3.78 kts.
bearing angle.........100.5 deg.
angle on bow...........25 deg.

*target speed...............15.30 kts.

With the above example the track angle is 125.5 deg.
Tell me if you get a different answer.
(The numbers are odd because I drew this test case out on graph paper to check this.)

To be honest, I've barely touched the TDC lately. I've been using manual fire control without the TDC. (S-boat style)

Bye

hi TorpX

yes , TorpX ...with these data the calculated speed is ,indeed, 15,3kts.

and yes ...angle lb is , at your example, 125.5 degrees(i named it that way...meaning the angle owncourselinetargetbowdirection....look at pic 1 to see what angle i name as lb....maybe this angle is called track angle ...don't know)

now, what i told you at the previous message is that the 25 degrees angle (at your example) is NOT trully the AoB but an angle that its value is very close to the AoB (the AoB at the time you start your stopwatch).look at pic 1 , when you say that bearing angle is 100,5 degrees you are talking for angle φ (this angle is not changing during the procedure) .when scope is at this angle(φ).... the 25 degrees angle is ,in fact, the angle θ and NOT the AoB (meaning the AoB at the momment of starting the stopwatch...look at pic 2...at AoB1).
the angle θ (also is not changing during the procedure) is the angle that you are using in your calculations(and leads to the above formula) and this angle can be found only through the angle lb (which ,also is not changing but this means that we have to know the target's course relative to our course).from pic 1 : θ=lb-φ ( θ = 125,5 - 100,5 = 25 degrees)

as you can see from pic 2 ,the AoB is changing during the procedure.The only situation that AoB is not changing is when you are on a collision course...but ,in this case, the target would never 'pass' your scope's vertical line

ps1: the back side of attack disc is(one of its uses) for helping you to get target's with the above data given.no tdc....etc
ps2: i have upload at my FF page (look at my sig) the proof for the spoken formula for you and whoever else is interest in.

bye

PIC1:

PIC2: