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Old 04-24-11, 05:51 PM   #1
Diopos
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3600 seconds in an hour
1852 meters in a nautical mile

.
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Old 04-25-11, 04:30 PM   #2
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The formula quoted is accurate as far as it goes, but does not take into account your subs movement. If you want to estimate speed while moving you must account for both your own speed and the relative bearing.


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.
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Old 04-25-11, 04:51 PM   #3
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I guess....
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Old 04-25-11, 05:11 PM   #4
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Quote:
Originally Posted by Armistead View Post
I guess....
I guess not everybody likes math.


Actually, I decided that the best way was to just plot it out like they did in real life.
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Old 04-25-11, 05:28 PM   #5
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in OM you can calculate speed using the Kiub interface, but you still have time the target as it passes the perpendicular line on the scope. Using donreed1's method if you are stopped skips the step with the interface. Also, in OM, the U-Boat's don't pick up the target map contact at great distance. If the target is close, saving time in the Kiub set up for manual targeting can be invaluable, especially if the target is moving at a high rate of speed.
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Old 04-25-11, 05:52 PM   #6
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Just keeping the periscope at 0 or 180 degrees and turning the sub with the bow or aft right in front of the target also corrects for own motion completely. Don't turn while you take the time though, or twist the periscope. Make sure the compass is steady. But other than that you could go at flank (forward or backwards) if you wish.
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Old 04-25-11, 06:24 PM   #7
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Quote:
Originally Posted by Pisces View Post
Just keeping the periscope at 0 or 180 degrees and turning the sub with the bow or aft right in front of the target also corrects for own motion completely. Don't turn while you take the time though, or twist the periscope. Make sure the compass is steady. But other than that you could go at flank (forward or backwards) if you wish.
This will work too.
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Old 04-25-11, 08:30 PM   #8
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Quote:
Originally Posted by Pisces View Post
Just keeping the periscope at 0 or 180 degrees and turning the sub with the bow or aft right in front of the target also corrects for own motion completely. Don't turn while you take the time though, or twist the periscope. Make sure the compass is steady. But other than that you could go at flank (forward or backwards) if you wish.
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Originally Posted by TorpX View Post
This will work too.
Huh? Huh? (one for each post.)

How do you keep periscope at 0 or 180 degrees and just in front of the bow of the target without turning your boat or "twisting" the periscope. I assume "twisting" does not mean bending the periscope, but refers the act of of rotating or swiviliing iit. Just for fun, let's assume the target is a big, fat Yamato BB, plodding along at a meager 24 kts. . This I have got see! (...er, read.)
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Old 04-25-11, 06:07 PM   #9
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Quote:
Originally Posted by TorpX View Post
The formula quoted is accurate as far as it goes, but does not take into account your subs movement. If you want to estimate speed while moving you must account for both your own speed and the relative bearing.


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.
Quote:
Originally Posted by Armistead View Post
I guess....
These math guys are taking over the world. Next thing you know, they will invent computers to use in every day life, connect them together, and start usikng computers to send messages, prepare business plans, solve problems, and heaven forbid, play chess and challenge Bobby Fischer. They may even try using them to send a man to the moon, can you believe that? It's totally nuts!
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Old 04-28-11, 12:09 PM   #10
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Quote:
Originally Posted by TorpX View Post
.......
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.
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Old 04-28-11, 02:44 PM   #11
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Quote:
Originally Posted by makman94 View Post
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.
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Old 04-28-11, 09:25 PM   #12
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@ 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.
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Old 04-29-11, 12:24 AM   #13
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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.
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Old 04-29-11, 02:12 PM   #14
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Quote:
Originally Posted by TorpX View Post
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
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