Finding the range the hard way.
 

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?

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.

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 :yeah:

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 ]?

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.

Is there a formula that works for periscope range finding in the game? I was under the impression that the magnification in the game did not match that of the real persicopes. At any rate, I have never been able to manually use the periscope range finder in the game. I have been able to do it with the split image device but not by observing the number of graduations and then doing the math. If it can be done I would like to try it. Thanks, Joe S

The short answer is no because the angle subtended by one division is dependent on your screen resolution. Last I knew Capn Scurvy was working on that and had a solution in the works.

Thanks! I like short answers! Joe S

I have been thinking about the short answer. Is the problem related to the mastheight? If so, we could use the height of the stack. It is much easier to see and less likely to be distorted by the resolution. Joe S

Thanks every one, for your comments particularly Rockin Robbins. This is a very interesting subject.

A point Rockin. You suggested, I think, that the formula is invalidated, or is dependant upon a particular screen resolutuion "...because the angle subtended by one division is dependent on your screen resolution". But surely screen resolution is irrelevant. If one knows the angular value of each division of the reticule then the angle subtended will always be the same.

My practice is to use the tangent rather than the cosec, this gives the range to the target at the waterline rather than to the masthead. My reasoning is that although, at longer ranges, the difference is small and can be accepted at shorter ranges i.e below 1000yds the error can be significant. I like to get in to c750 yds.

Joe S. I agree with you that trying to get a range using the stadimeter especially with 'no stabilised view' selected is a nightmare. Howver, why shouldn't you use the stack! If you look at the manual it is possible to estimate the height of the stack or any other part of the target, bridge, derrick et al. then do the trig calculation to get the range. Be aware though that because you are using a smaller height the error in range will be greater. On a pitch black night with rain though, it might be better than nothing.

Incidently I use an old Thorton PP221 slide rule fo my calculations.

IIRC there is a problem with one specific resolution, but in all others the image is correctly scaled.

The thread is here, for anyone interested in further reading: http://www.subsim.com/radioroom/showthread.php?t=175729

There's a natural tendency to want to pick an easy to see aspect of the target: stack, cabin height, flight deck, but you probably wonder why the real guys liked to use the masthead height. By the way, their recognition manuals tended to give heights for stacks, masthead, cabin top and deck of most targets. Real sub skippers could pick whatever they wanted for their stadimeter range measurements.

But the accuracy of your distance calculation depends on three factors. The closer the target is, the more accurate your determination will be of course.:D The more accurate your stadimeter angle measurement is the more accurate your distance will be.

But also the higher the measurement point is on the target, the more accurate your distance determination will be. Your error for a masthead height of 50' will be half that of a distance determination based on a stack height of 25' for instance. Your accuracy is directly proportional to the height of the aspect being measured. So whenever they could the real sub skippers used masthead heights, the tallest part of any target.

Stadimeters are nasty inaccurate enough without you handicapping yourself!:yep:

Thanks for the explanation about the potential for error. Would it not be true, however, that in a situation like ours, where there is a tendency for the image of the mastheight to be corrupted due to screen graphics issues, that the stack would be a more reliable measurement to use even though it might be theoretically less accurate? Joe S

I don't know about graphics corruption issues. I don't have any. I'd have to say the jury is out on that one. However, if the mast is twice as tall as the stack, then your error tolerance is doubled: you can have twice the error using the mast and still have the same distance error resulting. It's difficult for me to imagine that you could have twice the error sighting on the masthead. You might get a little more error, but it would still result in a more accurate distance measurement.

It can be difficult to resolve the masthead at extreme range. However, at that range, small errors make huge differences in your result, so range measurements that far out should be viewed as exercises in futility anyway. They just give you a starting number that you know will be very different before you shoot, no matter what aspect of the target you are measuring.