Finding the range the hard way.

[Telemon]

Trying to do it the hard way I’m reading the method of determining range from The Submarine Torpedo Fire Control Manual found in Sub Skipper's Bag of Tricks-Techniques, tactics, tutorials, videos, in the forum and am somewhat puzzled.
The manual is reputed to be an official document and I am surprised that it seemingly contains errors. These errors are confusing me and I hope some one will clarify things for me.
From the Manual, page 5-2. The details of the periscopes states that the optical length of the Type II ‘scope is 40ft and that of the type IV 36ft. No other lengths are given.
The next paragraph states’ …examination of the tables reveals that …we have had to sacrifice about 6ft of periscope depth’. I have examined the tables but can only see a change of depth of 4ft! Where is the other 2ft?

Page 5-3. Reading the first paragraph on the page we are told in the figure (Plate II?) the target subtends 5 divisions (of the reticule) in high power and 1¼ divisions in low power. The Manual goes on to state ‘That it is known that at 1000yds 17 ½ yards or 52.5 ft subtends an angle of 1 degree’. It then says that ‘Using this information we can deduce the following formulas(sic):
The formulae given are. R(range) = (19.1h)/n
R(range) = (76.2h)/N

Where R = range in yards.
h = height in feet.
n = number scale divisions in low power.
N = number scale divisions in high power.

From the information given can some one explain to me how the figures 19.1 and 76.2 are derived and how they are related to 17.5 and 52.5?

Finally the figures referred to in Plate II.
The two worked examples.

(a) For the upper diagram; Range = 76.2 x 120 /5 = 1840 yards.

(b) For the lower diagram; Range = 120 x 19.1h/1.25 = 1840 yards.

Both answers are incorrect!

Upper dia. Range = 76.2 x 120 /5 = 1828.8 yards (rounded up would give 1830 yards)
Lower dia. Range = 120 x 19.1/1.25 = 1833.6 yards (rounded down would give 1830 yards)

Again can some one explain the discrepancy please or point out where I am misreading?

[WernherVonTrapp]

With my math, I certainly cannot provide the assistance that I think you need. Bearing that in mind, please bear with me as I try to understand why you are trying to figure this out, which if your math answers are correct, you already have.
The Fire Control Manual that you reference seems to have been distributed in the 1950s, after WWII. Is it possible that some postwar modifications had been made? Additionally, according to the official USS Cobia website (the sub from which that Fire Control Manual belongs), that particular Gato Class sub was a product of the Electric Boat Company of Groton, as opposed to the Manitowoc shipbuilders which also built Gato Class subs.
Might some of these variables account for the discrepancies that you point out? Again, my math is about as sharp as a soup spoon so I cannot offer assistance in that manner but, I am trying to aid you in an alternative manner.

[commandosolo2009]

why even complicate things? you got SONAR and now prefer doing it with a periscope? is this like hey Germans, we Americans dont need sonar anyways challenge? stop wasting time, head to the boat, order a fresh pair of cob pipes and use the sonar, the American way

[razark]

Originally Posted by commandosolo2009

why even complicate things?

Because sonar emits noise, and there's more than one way of doing things.

[commandosolo2009]

Originally Posted by razark

Because sonar emits noise, and there's more than one way of doing things.

Speaking of which, if I ping after the DD sonar, would they still discover me? If you seen ["Below" - 2002 ]?

[Rockin Robbins]

Okay, let me take the questions one at a time. In the first instance it isn't apparent in the table why the sub would have to give up six feet in periscope depth for a 4' difference in periscope length. It possibly could be that the radar had to be further out of the water to operate satisfactorily. Then they could figure that 4 is approximately equal to 6. They don't explain.

The second question is related to trigonometry. We're just taking the ajacent side of a very long right triangle and dividing it by the opposite side. That ratio is called the cosecant of the angle. To subtend one degree and object is 57.2986885... let's call it 57.3 times further away than it is tall. We can use that on a galaxy. If we think it is 100,000 light years across and it is one degree wide, then that galaxy has to be 5,793,000 light years away.

Notice that all our measurements can be in any unit at all, miles, feet, yards, light-years, it doesn't matter as long as the units are the same. Here they toss a curve in by measuring the masthead height in feet and the range in yards. So for our 1º angle, We have to take our computed distance and divide by three, or we could just divide our cosecant, 57.3 by 3 in our calculation. 57.3/3 equals 19.1. So now we know that for an object subtending 1º, the formula is Range=19.1 times height in feet. Looks suspiciously like part of the above formula!

So if each division is one degree, how much closer would an object be if it subtended two degrees? You can easily see that it would be the range at 1 degree divided by two. Now n is the number of divisions (degrees at low power). So take your range at one degree, 19.1h and divide it by the number of degrees and you have calculated the range: 19.1h/n=R.

Now if you have the range and want to compute the height, then you have to change the formula. Since we know the range, we have to multiply it by the ratio of the opposite side over the ajacent side of the triangle. That is called the sine and the sine of 1º is .01745.... okay, we'll call it .0175. So if R=1000 our height is .0175 x 1000 or 17.5 yards. And that is how 17.5 yards compares to 19.1 times the height in feet. They are two completely separate numbers used for two different purposes.


The calculations in plate 2 are just incorrect as you have shown. Fortunately the mistakes are still accurate enough to maintain the integerity of the firing solution as they are only a half a percent off the real range, much closer than the accuracy of their measuring device. It is very probable that they used a slide rule to calculate the formulae, and so just reported what they saw, which would be a close approximation of the real answer. That would explain the inaccuracy. Any error from the slide rule would be much less consequential than a garden variety arithmetic mistake, which could be of any magnitude, where the error from the slide rule is, as I showed, of such a small degree as to be inconsequential.