Zero gyro shooting

[Dorjun Driver]

The object of the exercise was to develop a solution with as little information about the target as possible, i.e., speed. All other data are up to the sub driver.

[Dignan]

Originally Posted by Pisces

Oh yeah, I forgot to add:

Dorjun Driver's formula can be significantly simplified if you know the AOB:

Deflection = arcsin ( Vship * sin AOB / Vtorpedo)

When you say AOB for this formula do you mean the AOB Of the target at the given moment you compute the solution or do you mean the intercept/track angle formed by the sub course and target course? I guess I dont understand the difference between aob and track angle in this situation.

EDIT: I continue to have a hell of a time getting this formula to work on my scientific calculator. It just doesn't want to recogize my A input as radians. I've tried everything. What would I have to multiply my angle input (A in this case) by to convert it to RADIANS within the formula? Maybe I'll have to do it that way. For instance:

Lead Angle = asin (V sin (conversion value for Radians)A / (V ^ 2 + T ^ 2 - 2VT cos (conversion value for Radians)A) ^ 1/2).

what would this value have to be? pi/180? A(pi)/180? A*(57.29577951 / pi), as mentioned before by pisces? I appreciate all the help thus far.

[TorpX]

Dignan,

I'm not sure whether you are using a scientific calculator or something else. Any dedicated calculator should have a degree mode. If not it would be computing in radians. Check this by trying this:

sin 90 = 1 and arcsin 1 = 90 [degree mode]

sin (3.1415927/2) = 1 and arcsin 1 = 1.570796... [radian mode]

[pi radians = 180 degrees]

Be careful about the track angle, intercept angle and the AoB. They are all different angles. IMO, the track angle is less confusing to use. This is a good way to calculate the Lead Angle:

La = arctan((Vt sin Ta)/(Vp + Vt cos Ta))

where
Vt is target speed,
Vp is torpedo speed
Ta is track angle

or, if you must use radians and convert:

Ta1 = Ta * (pi/180):
La1 = arctan((Vt sin Ta1)/(Vp + Vt cos Ta1)):
La = La1 * (180/pi)

There are diagrams which show the geometry of the set-up at the start of the Algebraic Firing Solution thread.

[I didn't get the right answer when I tried Pisces formula. Either he left something out or I'm fouling it up tonight.]

[Dorjun Driver]

Yeah, the intercept angle is simply 180-track angle (in degrees) or 3.1416-track angle (in radians). See complete diagram below—when I submitted the original post, I was still into the whole brevity thing.

[Dignan]

Originally Posted by TorpX

Dignan,

I'm not sure whether you are using a scientific calculator or something else. Any dedicated calculator should have a degree mode. If not it would be computing in radians. Check this by trying this:

sin 90 = 1 and arcsin 1 = 90 [degree mode]

sin (3.1415927/2) = 1 and arcsin 1 = 1.570796... [radian mode]

[pi radians = 180 degrees]

Be careful about the track angle, intercept angle and the AoB. They are all different angles. IMO, the track angle is less confusing to use. This is a good way to calculate the Lead Angle:

La = arctan((Vt sin Ta)/(Vp + Vt cos Ta))

where
Vt is target speed,
Vp is torpedo speed
Ta is track angle

or, if you must use radians and convert:

Ta1 = Ta * (pi/180):
La1 = arctan((Vt sin Ta1)/(Vp + Vt cos Ta1)):
La = La1 * (180/pi)

There are diagrams which show the geometry of the set-up at the start of the Algebraic Firing Solution thread.

[I didn't get the right answer when I tried Pisces formula. Either he left something out or I'm fouling it up tonight.]

Ah, I think I see my mis-step. I was failing to convert the final La result into radians too. I converted the input (Ta in your formula above) to radians but not my final La output. Regarding my calculator, its something I downloaded for free on my phone. It's possible the person who made it did not do it correctly or did not include a way to automatically convert to radians. Some of them are better than others.

Thanks a lot. I'll try this tonight. If it doesn't work I"ll probably just go back to drawing it out on the nav map. Not that big of a deal. I try to remind myself that somewhere under all this putzing around with mods and formulas and stuff there's a game to be played.

[Dignan]

Originally Posted by Dorjun Driver

when I submitted the original post, I was still into the whole brevity thing.

"The Dude abides"

Thanks for the new diagram. Just for my own edification, in your new diagram, the target AOB at the point of impact = the track angle, correct? And the Intercept angle is formed by the target course and your torpedo track course, correct?

[Dorjun Driver]

That's a big ten–fore, Dignan.

[BigWalleye]

Small quibble on nomenclature: In SLM-1, the Submarine Torpedo Fire Control Manual (http://www.hnsa.org/doc/attack/index.htm), what you are referring to as the "track angle", (theta sub track on your diagram) is designated the "torpedo track angle". This is distinct from the track angle, which SLM-1 defines as "The angle at the point of intercept target ships course and the submarine's course measured to port or starboard of the target ship's bow toward the submarine. Symbol: Ta." That is, track angle, as used in the WW2 USN and in many first-person accounts, is relative to the submarine's course, not the torpedo's course. Using the two interchangeably might cause confusion to someone reading first-person accounts or patrol reports.

[Dorjun Driver]

Correct. The first diagram represents a special case where they are the same. And indeed, the second diagram is mislabeled. Thanks for the heads up.

[BigWalleye]

The formula in your OP is correct (IMAO!), or at least agrees with my own analysis. It produces results which agree closely - but not exactly! - with the results shown on Plates XVII and XVIII in SLM-1. There are small differences, usually less than 1 degree, which I think are due to corrections for parallax and what SLM-1 calls "torpedo advance" - the distance the torpedo travels before it settles on its gyro-dictated course. It is interesting that the equation in your OP does not appear anywhere in SLM-1, and that the method used to calculate the curves of Plates XVII and XVIII is not specified either.

[Dorjun Driver]

No parallax, no reach, and no turning radius. KISS

[BigWalleye]

Since it would take far better eyesight than mine to set the firing bearing to within less than one degree, your equation is accurate enough for me. My point was that SLM-1 was the official USN publication on the subject, the one R/L sub officers were taught from, and that their plots, with Heaven only knows what unspecified second- and third-order refinements, agreed with your analysis to an accuracy better than we can use.

Although I prefer KISMIF.

[Dignan]

Originally Posted by Dorjun Driver

What Pisces said. It appears your calculator wants radians as input. Sooo:

asin (V sin RADIANS(A) / (V ^ 2 + T ^ 2 - 2VT cos RADIANS(A) ^ 1/2)

Or something to that effect. The output of asin(whatever), may be considered a scalar. But it ain't.

Your inputs from above should spit out 17.4 and 9.9 respectively.

Now. Who's working on the Q and D spread calculator?

Ok, I think I'm close to cracking this. I figured out the radian functions on my calculator are a different set of buttons with the "degree symbol" after it, ironically (see below)

La = asin˚ (V sin˚ A / (V ^ 2 + T ^ 2 - 2VT cos˚ A) ^ 1/2)

V=target speed
T=Torp speed
A=track angle

When I input the following like before
V=9
T=30
A=75
...I get 15.5 for a lead angle. Not the 17.4 Dorjun said I should get but closer. Anyone see any flaws with this formula now that the trig functions are set to radians?

[Dorjun Driver]

I don't know what to tell ya. Using your formula above I keep getting 17.4. I could be entering the same wrong numbers repeatedly, but...

[BigWalleye]

Maybe the point of confusion is that the intercept angle in Dorjun Driver's original diagram is NOT the track angle, but 180-the track angle. And what is marked as "theta sub track" in DD's second diagram is actually the intercept angle, again 180-the track angle. Track angle - whether ownship track angle or torpedo track angle - is equal to the AoB at the point where the tracks intersect, either ownship track and target track or torpedo track and target track. In either case, the intercept angle is 180-(AoB at intercept). I'm pretty sure Dignan will get the correct result if he uses 180-Track angle in his calculations. And DD's formula is correct for intercept angle not AoB and not track angle.

BTW, I believe that what is marked as "theta sub torpedo track" is in fact the torpedo track angle. It is only "theta sub track" which is on the wrong side of ownship track line.

Or I'm trying to figure this out too late at night and have it all wrong....

"I'm getting too old for this ****"! - Danny Glover, Lethal Weapon

Further clarification: Please see the diagram on Page 1-12, SLM-1.