gotcha :up:

WTF 3.5 = distance the target travelled between the two sightings, which is really 3500 yds. if using imperial.

Heh well I can state as a matter of fact you don't suck at math as much as I do :}

Your instructions really cleared up a lot of what I wasn't getting in the first place but now I have a bit more confidence with my slide rule.

OK, everyone who has been using my instructions for targeting, little did you know that I am a double agent working for both the British and Japanese navies! If you followed the instructions I gave you, you were sure to miss!

Actually, I made a stupid mistake :damn:

But, I'm man enough to admit it when I do.

Thanks to don1reed for pointing out an egregious mistake I made. Go ahead and run through the instructions again. This time, I'm almost positive that they might be correct.

Aww no worries. Hell if it wasn't for this thread you started I wouldn't be the proud owner of two real slide rules :rock:

^^ I'm with jmr, this is the only thread I've been able to find where someone even suggests a method to calculate AOB with a standard slide rule.

Even if you did make a small mistake at least you gave the rest of us an idea of how to do it, and for that I am indebted to you:yep:

Ok, I went back and did this problem myself with my corrected instructions.

Here is what I got:


OBS A: 1700 345 ??? 0
OBS B: 1800 357 ??? 165

fiddling, I find that there are twelve degrees that fit between 1700 and 1800 starting at 68 degrees and ending on 80 degrees. Subtracting both from 180 (because target is moving *AWAY* from us) gives AOB's of 122 and 100, respectively:

OBS A: 1700 345 100 0
OBS B: 1800 357 122 165

This gives us a distance traveled of 380 meters, which gives a target speed of 2.3 meters per second, or 2.24 knots.


I actually plugged those numbers for OBS B into the TDC, and got a hit on the stern of that C2 in the training mission. That might be because I rounded my distance numbers to the nearest hundreds.

EDIT: I don't remember if I opened the tube before I fired! That could explain the hit that far astern.

After the torpedo hit, I had the weapons officer do a solution:

Range: 1869
AOB: 123
Speed: 0 (because I blew the prop off of him!)

Not too bad.

I hate to say this...Bill, you did it again. I think your shot hit aft because the C2 was doing 4 knots.

380m / 165s = 2.3 m/ps

2.3 / .51 = 4.5 knots

edit: If you're using the A & B scale on your sliderule...be aware those scales are mainly used for the squares & square root of the C & D scales. The left side for odd numbers of digits, i.e., 1, 123, 12345...and the right side for even numbers of digits 22, 4444....

All the simple math problems should be kept on the C & D...it's like mixing apples and pc's. (oranges). But, if you have a rule with a folded scale, i.e., CF, DF, CIF, then, those scales will integrate with the C, D, & CI scales.

I just re-did the problem from the beginning, and I got 3.1 knots, which actually squares pretty well: I was moving at about 1 knot towards the target.

By the way, that target has a known speed of 2 knots.

I have to use the A scale against my S scale, because my S scale goes from about 38 minutes to 90 degrees, instead of the standard S scale that goes from just under 6 degrees (the center index of my A scale falls right at 5 degrees, 45 minutes when the ends are aligned).

I used the C and D scales for everything else.

By the way, are my instructions at the beginning of this thread correct now?

A pic is worth 1000 words:

The center of the maneuvering board (mobo) is the sub/uboat. Let every ring = 200m so that the outter ring is 2000m from the sub.

http://img68.imageshack.us/img68/5664/amobocu1.th.jpg

Math:

1852m / 3600s = .51 (constant). 1/.51 = 1.96 (constant)

When the answer is in meter per second we can either multiply by 1.96 or divide by .51 to get knots.

therefore,

2.3 mps / .51 = 4.5 knots.

A line drawn from m2 (target) to the sub, subtending from DRM (direction of relative movement) = AoB 112°. 180° - 112° = 68° for sliderule use.

Mobo is handmade 1 cm rings total 10 rings.

Another way to test the speed of your known C2 is to clock the movement of the ship as it passes the center stationary reticule of the scope. End the timing when the stern passes.

Math: Known Length of ship in meters / time in seconds X 1.96 = ship speed in knots.

more math:

Imperial: 2026 yds. / 3600 s = .56 (constant). 1/.56 = 1.78 (constant). Either divide yds. per second by .56 or multiply by 1.78 to get speed in knots.

@Bill: Your initial instructions on p1 are in order.

Just curious, is there a chance that the statement:

"I move the slide until there is 20 degrees difference on the S between the 4 and the 7 on the S scale."

Should read:

"I move the slide until there is 20 degrees difference on the S between the 4 and the 7 on the A scale."

The reference to the 4 and 7 seem to be on the A scale in the picture, but I could be misunderstanding the procedure.

Thanks,
Ron

That statement is correct for the "type" of sliderule used by the author, which is, I believe, a Sterling, pocket, 6" sliderule. The std for the times was a 10" with different configurations and more scales. Although the major manufacturers of the times did build some elaborate 6" models; but, each had their own instructions.

You are correct. Fixed that in the original post.

Actually, it's a 10" Sterling 'Precision'.

This takes you to a decent picture of one:
http://www.sliderulemuseum.com/Sterl...txt_slider.jpg