Zero gyro shooting

[Dignan]

Originally Posted by Pisces

Asin, arcsin, sin with superscript -1 aka inv(erse) sine, all means the same function. It calculates an angle that corresponds with the slope of a vertical edge divided by the length of the sloping side (hypotenusa) in a right angled triangle. Though it can also be used with triangles that have no right angle (90 degrees) in them. The 'slope' must have a a value between -1 and 1, or else the angle cannot be calculated. No butts, no ifs, it can't!

In everyday household uses, the sine/cosine/tangent functions are calculated on degrees. As such the reverse kind of function: asin/acos/atan result in degrees. But most scientific calculators also have an option make them use angles in radians. A radian is the angle made when you wrap the radius of a circle around the circumference. Radians are a big thing in engineering and science. It's 57.29577951 degrees, or 180 degrees divided by Pi. 2 Pi radians make a full circle. On Casio calculators there should be a small 3 character abreviation in the top of the display: deg for degrees, rad for radians and gra for gradians. Gradians are a big thing in surveying.

To test your calculator for degree or radian mode: 0.707 asin should result in near 45 degrees, or 0.785398163 radians (=Pi/4), or 50 gradians (90 degrees angle is 100 gradians. The French trying to be funny there.)

So that could be the cause for unexpected results from the formula. Or the order of precedence in arithmetic: multiplication/division precedes adding/subtracting.

<-- is my face right now. I think I am beginning to understand what you are saying. I'm going to re-read this later and try to digest. Thanks for the explanation. So what I've done is plug this formula by Dorjun into an app called MagiCALc. It's a scientific calculator for the iphone. Here's what I've got.

Lead Angle = asin (V sin A / (V ^ 2 + T ^ 2 - 2VT cos A) ^ 1/2)
V= Target speed
T= Torpedo Speed
A= Track angle/Intercept angle.

The program automatically recognizes my variables and asks for input. When I input the following values for each as a test...

V(Target speed) = 9
T (Torp speed) = 30
A (Intercept angle) = 75

...it spits out (= -0.11)
I take this to mean an 11 degree lead angle. Is that how I should be doing this?

I tried it with a different lead angle variable again to be sure.
V(Target speed) = 9
T (Torp speed) = 30
A (Intercept angle) = 135

and it spit out (=0.02). or a 2 degree lead angle
That doesn't seem right to me. Can anyone see what I've done wrong here? Is my formula off? Do I need to make the calculator see my output "Lead Angle" as an angle and not a straight number value or something?

Pisces, thanks again for the trig refresher. While I love playing this game, this has reminded me why I was a "social science" person.

[Dorjun Driver]

Originally Posted by Dignan

<-- is my face right now. I think I am beginning to understand what you are saying. I'm going to re-read this later and try to digest. Thanks for the explanation. So what I've done is plug this formula by Dorjun into an app called MagiCALc. It's a scientific calculator for the iphone. Here's what I've got.

Lead Angle = asin (V sin A / (V ^ 2 + T ^ 2 - 2VT cos A) ^ 1/2)
V= Target speed
T= Torpedo Speed
A= Track angle/Intercept angle.

The program automatically recognizes my variables and asks for input. When I input the following values for each as a test...

V(Target speed) = 9
T (Torp speed) = 30
A (Intercept angle) = 75

...it spits out (= -0.11)
I take this to mean an 11 degree lead angle. Is that how I should be doing this?

I tried it with a different lead angle variable again to be sure.
V(Target speed) = 9
T (Torp speed) = 30
A (Intercept angle) = 135

and it spit out (=0.02). or a 2 degree lead angle
That doesn't seem right to me. Can anyone see what I've done wrong here? Is my formula off? Do I need to make the calculator see my output "Lead Angle" as an angle and not a straight number value or something?

Pisces, thanks again for the trig refresher. While I love playing this game, this has reminded me why I was a "social science" person.

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?

[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, thanks. I'll see if I can figure out how to switch it to Radian output.

Before I try, doesn't putting "sin" before my A make it read that input as radians?

[Dorjun Driver]

No no no! Just use the radians() function to convert your degree input to the format your calculators sin()/cos() functions want.

While hell is breaking loose we don't want to be dividing stuff by transcendental numbers! Next thing you know, you'll want to compute just how close you can shave that shot. Anywhere in the engine room will be fine. Honest.

[Dignan]

Originally Posted by Dorjun Driver

No no no! Just use the radians() function to convert your degree input to the format your calculators sin()/cos() functions want.

While hell is breaking loose we don't want to be dividing stuff by transcendental numbers! Next thing you know, you'll want to compute just how close you can shave that shot. Anywhere in the engine room will be fine. Honest.

Ha! 10-4 on that. The calculator I'm using removes the parentheses from the sin(A) and Cos(A)when I try to save it. That seems to be my problem. I'll putz with it some more. Thanks

[Pisces]

To switch between degrees and radian and gradians on a Casio you have to press the mode button and then 4, 5, or 6. See the legend under the display.

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

Agreed. If you get a number like 0.11 as a result for degrees, then it will infact be 0.11 degrees, NOT 11.

If it was a logarithmic sliderule then that decimal point fudging might make sense. But that method doesn't hold with a digital scientific calculator.

[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)

[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.

[TorpX]

BW, you are right about the angles not being marked off right. The track angle should be taken from where the target is going to where the sub is. The torpedo track angle, TTa is not always the same as the track angle, Ta, but as this thread is titled "Zero gyro shooting", the TTa should be the same as the Ta. Trying to make use of formulas derived for zero gyro shots in non-zero gyro shots will likely lead to disappointment and frustration.


BTW, for this:

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

When I input the following like before
V=9
T=30
A=75

I get a lead angle of 15.05 deg.

[Dorjun Driver]

Originally Posted by BigWalleye

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

Too late? Way too early my friend, way too early. Late or early, I think my second diagram is correct.

Whatever the correct nomenclature, if the target is steaming along the green line, and you fire your torpedo when the target arrives at the intersection of the green and blue lines, and the torpedo travels down the red line, the torpedo and target will meet at the intersection of the red and green lines. Much to the target's chagrin.

And for the record, using “the formula” as written, I achieve accuracy sufficient to put my torpedoes pretty much where I aim—under the stack, 5’ below indicated draft—every shot. Discounting duds and deep runners, I’ve yet to use more than one fish to sink ships displacing, up to and including, 8150 tons. So there.

TorpX, I'm still getting 17.4 degrees. Perhaps I transcribed my "working" Casio/Excel formula improperly. I'll check into it.

If I were Bill O'Reilly, I'ld be tempted to let this whole thing go with a simple "You can't explain that!" But I'm not, so I won't.

TMC/RSRDC/OTC

[Dorjun Driver]

I can find nothing amiss on either platform.

If anyone knows how to attach an Excel sheet, I'll be happy to share.

[TorpX]

Originally Posted by Dorjun Driver

TorpX, I'm still getting 17.4 degrees. Perhaps I transcribed my "working" Casio/Excel formula improperly. I'll check into it.

I think the difference is because you designate your angles differently. When someone says "track angle", I use the track angle as per the USN. I keep to using it, and it has made my SH life much simpler. When you first posted your formula, it gave me the right answer, but I had to see how you marked the angle, first.

Comparing the deflection angles generated from your equation with deflection angles picked off Plate XVIII from SLM-1 yields small discrepancies, which increase somewhat at higher target speeds. I had always assumed that Plates XVII and XVIII in SLM-1 were accurate. The text implies that they include various correction factors and that they are substantially the results generated by a WW2-vintage TDC. The text does not indicate directly how the curves were generated, or what factors were included in the calculations. I had also assumed that the curves were somewhat more accurate than a first-order trigonometric analysis, such as the one in your OP. Apparently, my assumptions were incorrect. I would appreciate if you could tell me why the SLM-1 data are not to be trusted. I do agree that a first-order analysis is perfectly adequate for our purposes.

I took another look at plates XVII and XVIII. I then broke out my trusty TI-85 and set it to graphing my formula for lead angle using their parameters. The results, as near as I could see, looked close to what the plates showed, but not exactly. This puzzled me at first, but on closer thought, I think I know why.

We are solving a simplified problem, where no allowance is made for the distance between the torpedo tube and the periscope, and none for any launch delay, torpedo acceleration and such. Any delay or sluggishness of the torps would have the effect of increasing the deflection angle and could account for the discrepancy.

"10 degrees. 15 if he's going fast."

Speaking of Q & D, in WAR IN THE BOATS, the author mentions using a rule of thumb, "speed plus three". It certainly has the advantage of being simple, but I haven't tried it.

[Dignan]

[QUOTE=BigWalleye;2022126] 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.

QUOTE]

That did it . When I subtract my angle input from 180 i get 17.4 also. To be clear, the angle I am using in my formula is what DD calls "track angle" in his second diagram, the angle formed by the intersection of my course and the target course. So to make this work we need the angle on "the other side" of that intersection, right? Hence subtracting from 180.

Dorjun, "prefered nomenclature" aside, that seems to be the way to go. "That" being using the angle formed by sub course and target course, subtracted from 180. I appreciate this guide. It's something I've been trying to find for a while.

Thanks BigWalleye and thanks DD, Pisces and TorpX for guiding my torps in the right direction throughout this. I've been "Mozarting" the heck out of this over the past few days (Mozarting is when you get obsessed with finishing a project or solving a problem and block out all other distractions and influences in your life...not good). Now hopefully I can actually play the dang game.